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Table of contents :
Copyright
TOC
Preface
Chapter 1: Number Box
Chapter 2: Number Cross
Chapter 3: Number Sequence
Chapter 4: Marbles in a Box
Chapter 5: Brick Wall
Chapter 6: Average Cell
Chapter 7: Mixed Arithmetic Cell
Chapter 8: Wisgo Number Tile
Chapter 9: Number Pyramid
Chapter 10: Average Number Pyramid
Chapter 11: I/O Arithmetic Box
Chapter 11: I/O Arithmetic Box
Chapter 13: Lock and Key
Appendix
Answers
Index
All rights reserved Copyright © 2019 Neelabh Kumar. All rights reserved. CCBYNCND 4.0
Publisher's note This book or any portions of this book, including the problems and solutions, may not be copied, reproduced, stored in a retrieval system, transmitted or distributed in any form by any means, electronically, mechanically, photocopying, microﬁlming, recording or otherwise, without prior written permission from the author and publisher The author uniquely designed every problem in the book while developing the concepts contained herein. We encourage you to use this book for selflearning and selfstudy. None of the problems, concepts or any other part of the book can be used for commercial purposes or coursework by any school, coaching institute, tutor and/or tutoring center, etc. without prior written permission from the author and publisher. For permission requests, write to the publisher, addressed “Attention: Permissions Logitica Book”. Refer to the site Logitica.com for contact details.
Logitica.com
About Author Neelabh Kumar is a thinker. Having memorized the first 1500 digits of Pi (π) using sequential recollection, he is ranked among the top 150 on the Pi World Ranking List. He is the creator behind Wisgo Logitica, which stimulates both sides of the brain. One of the Wisgo Logiticas Kumar created has a patent filing in Hong Kong. After earning a Masters Degree from one of the most prestigious universities in India (IIT), Kumar is now employed in Hong Kong at a large financial firm, while also creating and designing a new Logitica, with more to come.
Publisher’s CataloginginPublication data Wisgo Limited, Hong Kong Logitica: The Brain Behind the Brain Improve your critical thinking and problem solving skills First Edition
Dedication I lovingly dedicate this book to
my parents Mr. Tarkeshwar Prasad & Late. Mridulata Verma
Thanks for your love and support.
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Acknowledgment I express my thanks to Mr. Dilip Kumar Das for the invaluable comments and suggestions he has provided, from the book's initial concept to its completion. A very special thanks to Mrs. Lipica Das for encouraging me to write this unique book. I also thank my friends and family members for their suggestions and encouragement. Thank you Reetu, my wife and best friend.
You
inspire me with your ideas and suggestions. Without your continued support, this book would not have seen the light of day.
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Table of Contents Chapter 1  Number Box Overview Example1 Summary Exercise
...21
.........................21 .........................22 .........................31 .........................32
Keywords: Arithmetic Operations, Arithmetic Mean, Binary Operators.
Chapter 2  Number Cross Overview Example1 Summary Exercise
...37
.........................37 .........................38 .........................49 .........................50
Keywords: Arithmetic Operations, Arithmetic Mean, Binary Operators.
Chapter 3  Number Sequence Overview Example1 Advanced Sequences Summary Exercise
...60
.........................60 .........................61 .........................80 .........................113 .........................115
Keywords: Number Sequence, Prime Numbers, Fibonacci Numbers, Order of difference.
Chapter 4  Marbles in a Box Overview Example1 Example2 Summary Exercise
...117
.........................117 .........................117 .........................120 .........................125 .........................126
Keywords: Linear Equations.
Chapter 5  Brick Wall
...129
Overview Example1 Example2 Summary
.........................129 .........................130 .........................132 .........................138
Exercise
.........................139
Keywords: Linear Equations.
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Chapter 6  Average Cell Overview Example1 Example2 Summary
.........................143 .........................143 .........................148 .........................153
Exercise
.........................154
...143
Keywords: Linear Equations, Fractions, Arithmetic Mean.
Chapter 7  Mixed Arithmetic Cell Overview Example1 Example2 Summary
.........................157 .........................157 .........................161 .........................166
Exercise
.........................167
...157
Keywords: Linear Equations, Fractions, Arithmetic Mean.
Chapter 8  Wisgo Number Tile Overview
.........................169
Example1
.........................170
Exercise
.........................177
...169
Keywords: Wisgo Family, Stimulating the left and right side of the brain.
Chapter 9  Number Pyramid Overview Example1 Example2 Example3 Summary
.........................179 .........................179 .........................181 .........................183 .........................192
Exercise
.........................194
...179
Keywords: Linear Equations, Pascal's Triangle.
Chapter 10  Average Number Pyramid Overview Example1 Example2 Summary
.........................199 .........................200 .........................202 .........................211
Exercise
.........................214
...199
Keywords: Linear Equations, Pascal's Triangle, Arithmetic Mean.
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Chapter 11  I/O Arithmetic Box Overview Example1 Alternative Method of Solving Summary Exercise
...220
.........................220 .........................222 .........................232 .........................239 .........................240
Keywords: Arithmetic Operations, Reverse Step Process, Functions.
Chapter 12  Lost Ant
...241
Overview Example1 Example2 Summary
.........................241 .........................241 .........................247 .........................249
Exercise
.........................250
Keywords: Quadratic Equations: ax2 + bx + c = 0, Displacement, Distance, Vector.
Chapter 13  Lock and Key Overview
.........................251
Example1
.........................253
Example2
.........................255
...251
Optimal vs. Nonoptimal Strategy .........................258 Summary
.........................259
Exercise
.........................261
Keywords: Strategy, The WorstCase Scenario, The BestCase Scenario.
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Appendix (A) Sum of Natural Numbers
.........................263
(B) Sum of Odd Numbers
.........................264
(C) Sum of Even Numbers
.........................266
(D) Pascal's Triangle and Number Pyramid
.........................268
(E) Quadratic Equation
.........................272
(F) Binomial Expansion
.........................274
(G) Vector and Scalar
.........................277
(H) Linear Equations
.........................278
(I) Function and Set
.........................282
(J) More on Number Sequences
.........................286
(K) Fibonacci Sequence
.........................303
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Answers Chapter1 Number Box
.........................309
Chapter2 Number Cross
.........................309
Chapter3 Number sequence
.........................310
Chapter4 Marbles in a Box
.........................310
Chapter5 Brick Wall
.........................310
Chapter6 Average cell
.........................312
Chapter7 Mixed Arithmetic Cell
.........................316
Chapter8 Wisgo Number Tile
.........................318
Chapter9 Number pyramid
.........................319
Chapter10 Average Number Pyramid
.........................322
Chapter11 I/O arithmetic box
.........................326
Chapter12 Lost Ant
.........................327
Chapter13 Lock and Key
.........................327
Index
..............328
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Three important concepts in Logitica Logitica: The Brain Behind the Brain
I. Build true knowledge Continuous learning, along with building logic iteratively, is true knowledge. True knowledge is not acquired; it is rather built iteratively and continuously. Logitica is an excellent way to build true knowledge because you develop analytical and logical thinking iteratively by solving a variety of problems.
II. Don't memorize Build the logic, not the memory.
III. Do memorize Memorize the challenges posed by the problem, not its solution.
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Preface I have no special talent, I am only passionately curious. Albert Einstein
I. About this book and Logitica In this book, we are going to introduce a new concept called LOGITICA. The focus of Logitica is to improve critical thinking and problem solving skills by working through the challenges of solving a variety of problems. Each Logitica is carefully designed to stimulate iterative, analytical and logical thinking process. LOGITICA can be considered as, "The Brain Behind the Brain." Just like our brain tells us how to do things, Logitica is designed to train the brain how to think. The purpose of Logitica is described below in a few lines:
Logitica: The Brain Behind the Brain Improve your critical thinking and problem solving skills Purpose of Logitica: Logitica is a new concept, and new types of Logiticas are continually being created and designed by the author. However, this book is a good starting point for building logic in a unique way. Critical thinking is the cognitive ability to clearly analyze information and logically connect concepts and ideas. When you combine this with problem solving skills, you have the ability to rationally reason things out when presented with a problem and determine how to go forward in reaching the best possible solution. If you are reading this book, your neverending journey towards intelligence is well underway. This journey, taken many times by creators and inventors in the past while solving unknown problems, is now accessible to an inquisitive and passionate mind through the innovation of Logitica: The Brain Behind the Brain.
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II. Intended audience People of all ages can beneﬁt from the concepts and exercises covered in Logitica. Now that the developed world has entered the Digital Age, it is important that we all adapt to the advancements being made in technology, which often require logical and analytical thinking. An excellent way to develop these skills is by learning the concepts taught in Logitica and putting those concepts to use while solving the problems in the book. For students planning for college and wondering what they should major in, careers showing the most promise are in Science, Technology, Engineering and Mathematics (STEM). The world is currently experiencing a STEM revolution and the students of today will be on the forefront tomorrow. Those who choose STEM careers will be the leading innovators. Furthermore, more industries are relying on scientists, engineers and mathematicians to solve the world’s most serious problems, like cybersecurity and global warming. These challenges require people with advanced analytical and logical thinking, which you can start developing by mastering the concepts taught in Logitica. College graduates with a STEMfocused degree will have more job opportunities than most other graduates. These careers will inspiring, challenging, and very rewarding since these are the industries are the ones stimulating economic growth. Whether you plan on developing the next big app or video game, designing selfdriving vehicles, developing algorithms in machine learning, building career in robotics or working in artiﬁcial intelligence, you will need highly developed analytical and logical thinking skills. Why not start now with Logitica? In summary, since Logitica develops critical thinking and problem solving skills, this book is useful to anyone wanting to enhance their cognitive abilities. In general, this book would be a good starting point for secondary and high school students.
"Logitica: Improve your critical thinking and problem solving skills" is the perfect textbook for private tutors and tutoring centers to use in helping their students develop analytical and logical thinking. These are the precise skills they will need to come out on top academically and succeed later on in their professional careers.
III. What is the need for Logitica? The author takes the view that we can categorize knowledge in two ways: true knowledge and memorized knowledge. The idea of memorized knowledge can easily be visualized by a typical learning session as explained below: ✔ An instructor comes to a classroom full of students and writes a problem on the
blackboard. The instructor then solves the problem, writing the solution on the blackboard as well. 12
✔ The whole class happily copies both the problem and its solution, ﬁrst on their brains
and then in their notebooks or viceversa. ✔ The real questions are: • •
Where is the creativity? What are we learning here? The solution to the problem might have involved years of research that we are supposedly learning in a few minutes. Does it give us the same level of understanding that the individual who came up with the solution or formula has?
The above case may not be true or applicable to everyone. However, conceptually this situation might seem familiar to many students because it is very similar to what occurs in a lot of classrooms. This is what we call "learning by memorized knowledge." This method of teaching has the following traits: ✔ We only learn about a known problem. ✔ We only learn how to solve a problem in certain known ways. ✔ In general, we rarely think about alternative ways to solve the problem. That is why,
given an unknown problem, we either struggle to solve the problem or try only known ways to solve it. ✔ When we study new things, we tend to forget what we had learned previously. This
happens because when we are taught new concepts without understanding the underlying concepts, more often than not we simply try memorizing the new concepts. Even though our memory bank is large, it is ﬁnite. Therefore, we can only memorize a ﬁnite amount of information. Hence, as we try to take in a new concept, some of the old learned concepts get pushed out of our memory. However, if you initially learn the concept by understanding its logic, you can learn as much as you want, and even if you forgot it, you can always rebuild the concept iteratively. Memorized knowledge exists because in the classroom we were often given problems along with their solutions. At a time when we were supposed to be learning concepts using logic, the teaching method most often was based on memorization. We were never required to go through the thought processes that the individual who originally came up with the solution went through. So, what is true knowledge? True knowledge is developed through continuous and iterative learning. This is a journey that starts with understanding the fundamental concepts using analytical thinking, followed by building logic iteratively and expanding the concept to a higher level by learning alternative methods. The continuous and iterative learning in an exploratory way is how one acquires true knowledge. Read the following passage, it conveys a profound message:
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I do not know what I may appear to the world, but to myself I seem to have been only like a boy playing on the seashore, and diverting myself in now and then finding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me.
Sir Isaac Newton His message is that "Learning is continuous, incremental and iterative." If we have to summarize from the above, true knowledge is about building the logic in our thinking process, which can only be done stepbystep iteratively. This can be achieved by developing the following skills: • Analytical thinking: Analytical thinking is about analyzing and understanding the
structure of the problem and ﬁnding every possible solution to the problem. • Logical thinking: Logical thinking is about selecting the most logical and appropriate
solutions. Analytical thinking opens up the possibilities for diﬀerent solutions, whereas logical thinking helps you narrow down those possibilities to the most suitable solution. • Mathematical thinking: One of the three pillars of Logitica is the development of
mathematical thinking. Mathematical thinking improves numerical and problemsolving ability. In this book, we learn various mathematical concepts (e.g., quadratic equations, linear equations, and number sequences, etc.) in an interesting and innovative way.
IV. Concepts of Logitica Logitica is based on three important concepts as described below: Build true knowledge Continuous learning, along with building logic iteratively, is the right way to acquire true knowledge. If knowledge is built by understanding the fundamentals, we can learn to solve complicated problems, as each of the complicated problems can be deconstructed back to its simpler fundamental units. As you learn various types of problems in this book, you will begin to realize this. Don't Memorize the solution Build the logic so that a problem can be broken down to its fundamental units and solved iteratively. In general, people try to memorize solutions. However, this will only be temporary as memory is relatively ﬁnite. Hence, as we learn new concepts by memorization, previously learned concepts are erased from memory. This is why we keep forgetting the concepts we learned sometime back. Instead, if we take time to build concepts logically, the concepts are built from their basic units iteratively. This means we can always build up the previously learned concepts from their fundamentals. Logitica is 14
an excellent way to build logic using fundamental concepts, as each Logitica was created using simple mathematics and basic concepts. Do Memorize the structure of a problem If there is anything we need to memorize, then memorize the problem. Try to understand the problem, its structure, constraints, and components. If you can memorize the challenges the problem presents, you can think about it when you have time. This way you can analyze the problem, ﬁnd alternative solutions and even design a new problem.
Learning is iterative
V. True essence of Logitica The true essence of Logitica is summarized below in how best to approach any given problem: ✔ Step1: Deﬁne the problem. ✔ Step2: Since analysis is the key in Logitica, analyze each problem to gain an
understanding of the underlying principles. Then broaden your scope to learn as much as you can, and try to decode the pattern or logic behind the problem. ✔ Step3: Next, try solving the problem. If you still have issues, go back to step2 and
analyze further. You may also try solving a simpler version of the original problem ﬁrst. ✔ Step4: After successfully solving the problem, expand your scope of learning by
designing new problems that you can try solving. You will see that these steps are followed throughout the book. Here are a few examples: • In chapter3, we discussed Number Sequences. We began the chapter by deﬁning what
we mean by a Number Sequence. In the beginning, we presented a few problems to solve. Before trying to solve the problems we provided a detailed discussion of some wellknown sequences, and the various ways number sequences can be built, etc. Once there is suﬃcient understanding, we show how the original problem can be solved. Later in the chapter, we build the foundation of another more advanced concept, “Order of diﬀerence,” which helps us to solve a wider variety of Number Sequence problems. • In Chapter9, we discussed Number Pyramid. We started the chapter with the deﬁnition
of a Number Pyramid and presented a fairly simple problem to be solved. This helps us 15
build the foundation for understanding how Number Pyramids work. We followed this with a more advanced problem. To solve these types of problems we are required to learn concepts like Pascal’s triangle and the Coeﬃcient Rules, etc. Once we have learned those concepts, solving Number Pyramid problems became fairly straightforward.
VI. Structure of each chapter Each chapter starts with an Overview Section containing a general description of the problem related to that particular Logitica. Since learning by example is the most practical and eﬀective way to learn, each chapter contains one or more Example Sections, which includes the problems that we need to solve. Since a thorough analysis is the key to solving problems, we need to conduct a detailed analysis in order to understand the underlying principles required for solving any given problem. The Analysis Section provides an indepth explanation that will help you build logical and analytical thinking. Once we ﬁnd the answer, it is summarized in the Answer Section. Continuing on, sometimes it is possible to have more than one answer to a given problem. In some cases, only one answer may be provided. The reader is encouraged to ﬁnd alternative answers, when possible. Each chapter ends with an Exercise Section, which contains a list of problems. The Exercise Section includes the following three levels of problems: ✔ LevelI: These are the more straightforward problems for you to solve and are useful
in building a good basic understanding of the concepts discussed in the chapter. ✔ LevelII: The problems in this category are slightly more challenging than those in
LevelI. Readers will learn more about the relevant Logitica by solving problems on this level. ✔ For further practice: The problems in this category are designed to think beyond the
problems in LevelI and LevelII and are intended to challenge our understanding of the relevant chapter. The answers to the problems in this category are not provided in the book, as such problems need to be explored by the reader. At some point in the future, the solutions for this section will be revealed through a course and an app depending upon the level of awareness and popularity of Logitica. The Appendix contains the derivation of some of the mathematical concepts used in The various chapters. We will refer to the relevant section if and when necessary.
VII. Notion used in this book We used the following the symbols/notations in this book: (i) Multiplication (×): We will use this symbol for multiplication. (ii) Division: We will use the standard symbol like
2 to indicate division. In the Chapter3 16
11, we used
for division in order to express the concepts more clearly.
(iii) ♜Note: We will use this symbol to describe important notes on the relevant topic. (iv) ♛ Definition: We will use this symbol to deﬁne the important concepts in the relevant chapter. (v) ♞ Rule of thumb: We will use this symbol to describe some important rules or clues to help solve the relevant Logitica.
VIII. How to read this book The best way to read this book is to go through each chapter sequentially as arranged in the book. In general, you should spend two or more weeks on each Logitica. The reader is recommended to read the Overview and Example Sections to understand the problem given in the chapter. Subsequently, the reader should read the Analysis Section to learn to solve the corresponding problem in the chapter. Once a good level of understanding is achieved, readers should attempt the problem given in the Exercise Section, and verify the answer in the Answer Section (for LevelI and LevelII problems). Since learning is an iterative process and it takes time to build the logic. So, you should not speed read through all the chapters at once, rather you should take your time to study and analyze the material in each chapter. If you are not able to fully understand a chapter, it is okay if you want to skip the chapter and come back later to learn it. Referring to a Logitica: In simple terms, a speciﬁc Logitica generally refers to a category of problems that may span over multiple chapters. However, in this book, we have one chapter dedicated to one Logitica. Therefore, we will use the term "Logitica" or "problem" interchangeably whenever appropriate to mean the same the thing.
IX. Chapters in the book The contents of this book will start you on an amazing journey towards intelligence. Inside you will ﬁnd: • A discussion on how to solve Number Box problems using combinations of diﬀerent
arithmetic operations. • How to use various arithmetic operations to solve a variety of Number Cross problems
in Chapter2. • A detailed discussion on Number Sequences in Chapter3 and how to analyze and 17
solve various types of sequences. • A discussion on using linear equations for solving Marble in the Box and Brick Wall
problems. • Average and Mixed Cell problems that can be used in a rather unique way to derive a
set of linear equations. • Amazing Wisgo Number Tile problems that are designed to stimulate both the left
and right side of the brain. • Pascal's triangle and how it’s used in a Sum Number Pyramid and the innovative
Average Number Pyramid. • The Reverse Step Process of solving IO Arithmetic Box. • How quadratic equations are used to solve the Lost Ant problem, • A discussion on strategy related problem involving Key and Lock problems, • An enriched Appendix Section, which contains an introduction to various
mathematical concepts like linear equations, vectors, equations, quadratic equations, binomial expansions, the sum of natural numbers, etc. These concepts are used in solving problems in this book.
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SectionI
LOGITICA
™
The Brain Behind the Brain Improve your
Critical thinking AND Problem Solving skills
Offering a unique combination of
Analytical, Logical & Mathematical problems.
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If you are reading this book, your neverending journey towards intelligence is well underway. This journey, taken many times by creators and inventors in the past while solving unknown problems, is now accessible to an inquisitive and passionate mind through the innovation of LOGITICA: The Brain Behind the Brain.
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Chapter 1
Number Box Keywords: Arithmetic Operations, Arithmetic Mean, Binary Operators.
1.1 Overview In this chapter, we will discuss various arithmetic operations and learn how to use them to solve a problem involving three Number Boxes. Let us ﬁrst deﬁne what we mean by a Number Box. ♛ Definition  Number Box A Number Box is a box with three compartments: left, middle, and right. Each of these compartments contains a number as shown in ﬁgure1.1. The middle number of each box is calculated by using the numbers in the left and right compartments of the box. A problem in this chapter contains three boxes, and all the numbers in the ﬁrst two boxes have been provided. As a rule, the logic to determine the middle number must be the same for all of the boxes. Using this information, we need to ﬁnd the missing number in the third box. Refer to Example1 for further details on this. ♜Note: We will refer to the three numbers in the box as "number on the left," "middle number," and "number on the right" respectively.
Figure 1.1: A Number Box containing three number compartments 21
As stated before, a problem involving Number Boxes contains three boxes, and we need to ﬁnd the missing number in the third box. Since the logic to determine the middle number must be the same for all of the boxes, we need to ﬁgure out how the middle number is calculated for the ﬁrst two boxes and apply the same logic to determine the missing number in the third box. ♜Note: Logic or pattern in a Number Box As per the deﬁnition above, the middle number is obtained by applying a combination of arithmetic operations on the two numbers on the left and right sides of the box. So when we say "ﬁnd the missing number," it means that we are supposed to ﬁnd the appropriate combinations of arithmetic operations, which when applied to the numbers on the left and right, will yield the middle number. For example, as shown in ﬁgure1.1, the middle number can be obtained by multiplying the number on the left by the number on the right: 2 × 3 = 6. So, when we ask you to determine the missing number in this example, you need to ﬁnd the fact that the middle number is obtained by multiplying the number on the left by the number on the right. Once you discover the logic in one box, verify the logic in the other box, and then use the same logic to ﬁnd the missing number in the third box. Let us continue our discussion by analyzing the three questions laid out in Example1.
1.2 Example1 Number Box Logitica Find the missing numbers.
(a)
(b)
(c)
Figure 1.2: Example1 22
1.2.1 Analysis Since the number on the left and the number on the right are used to determine the middle number, we need to use a certain category of binary operators for solving a Number Box problem. Let us ﬁrst deﬁne what we mean by binary operators. ♛ Definition  Binary operators Binary operators are those operators, which require two operands. The wellknown binary operators are addition, subtraction, multiplication, and division. There are also some other types of binary operators, such as logical AND (&&), logical OR (), etc. However, in this chapter, we will only refer to the types of binary operators that have to do with arithmetic operations. Let us discuss some of the arithmetic operations that are useful for solving a Number Box problem: (i) Addition: Add the two numbers indicated on the left and right sides together to determine what the middle number is. For example, as shown in the ﬁgure on the right, the middle number is equal to 1 + 4 = 5. (ii) Subtraction: Subtract one number from another to determine the middle number. For example, in the ﬁgure on the right, the middle number is obtained by subtracting the number on the right from the number on the left, i.e., the middle number is 3  1= 2. Sometimes we can get the same result by applying a completely diﬀerent logic. For example, the middle number in this example
3 +1 4 = = 2. 2 2 Therefore, to select the correct logic to use, you must validate the logic with the other box provided in the problem. is also the average of the two numbers on the left and right sides of the box:
We must also remember that subtraction is noncommutative. Hence, when we change the order of numbers in a subtraction problem, the result also changes. In other words, each of the following operations is possible: • subtracting the number on the left from the number on the right. • subtracting the number on the right from the number on the left. • subtracting the smaller number from the larger number. • subtracting the larger number from the smaller number.
Please refer to section 1.2.1.1 Subtraction and division are noncommutative (p. 24) for further details on this topic. 23
(iii) Multiplication: Multiply the two numbers on the left and right sides together to calculate the middle number. For example, as shown in the ﬁgure on the right, the middle number is equal to 2 × 3 = 6. (iv) Division: Divide one number by another. For example, as shown in the ﬁgure on the right, we can calculate the middle number by dividing the number on the right by the number on the left, i.e., the middle number =
15 = 5. 3
We know that division is noncommutative, hence the result will change if we reverse the order of numbers in a division problem. In other words, each of the following operations is possible: • dividing the number on the left by the number on the right. • dividing the number on the right by the number on the left. • dividing the smaller number by the larger number. • dividing the larger number by the smaller number.
Please refer to section 1.2.1.1 Subtraction and division are noncommutative (p. 24) for further details on this.
1.2.1.1 Subtraction and division are noncommutative A binary operation is commutative if changing the order of operation does not change the result. Arithmetic operations like addition and multiplication are commutative. This is obvious as 3 + 4 = 4 + 3 or 5 × 7 = 7 × 5. On the other hand, division and subtraction are
5 7 ≠ . Understanding this concept crucial in 7 5 ﬁnding the correct answer, where sometimes a smaller (or larger) number is subtracted from a larger (or smaller) number to determine the middle number. Similarly, the middle number may be obtained as a result of dividing the smaller (or larger) number by the larger (or smaller) number. For example, as shown in the ﬁgure on the right:
noncommutative as 7  5 ≠ 5  7 or
✔ If we have to subtract the smaller number from the larger number to determine the
middle number, the value of the middle number will be 7  5 = 2. ✔ If we have to divide the smaller number by the larger number to determine the
middle number, the value of the middle number will be
5 . 7
In other cases, it could be positional, such as, subtract the number on the left from the number on the right or divide the number on the left by the number on the right to 24
determine the middle number or vice versa. For example, as shown in the ﬁgure on the right: ✔ If we have to subtract the number on the left from the
number on the right to determine the middle number, the value of the middle number will be 5  7 = 2. ✔ If we have to divide the number on the left by the number on the right to determine
the middle number, the value of the middle number will be
7 . 5
No matter what logic you choose, the same logic must apply to all boxes in the problem. Therefore, we need to validate the logic used in all of the boxes before we can ﬁnalize the answer.
1.2.2.2 Some more arithmetic operations There are also other kinds of arithmetic operations used in solving a Number Box problem. We are going to list some of them in this chapter. It is worth noting that the list of the arithmetic operations given here is not exhaustive and a given problem may use a diﬀerent combination of arithmetic operations. One can progressively learn to ﬁnd such combinations by solving and practicing these types of problems. (i) Sum of squares: The middle number could be the result of the number on the left squared plus the number on the right squared. As shown in 2
2
ﬁgure here, the middle number is equal to 3 + 4 = 9 +16 = 25. In addition to this, there may be many other possible combinations, such as, ﬁnding the diﬀerence between the squares or multiplying the squares together or calculating the average of squares, etc. (ii) Sum of cubes: The middle number could be the result of the number on the left cubed plus the number on the right cubed. As shown in the 3
3
ﬁgure here, the middle number is equal to 4 + 1 = 64 + 1 = 65. In addition to this, there may be many other combinations possible, e.g., ﬁnding the diﬀerence between cubes or multiplying the cubes together or calculating the average of cubes, etc. (iii) Arithmetic Mean: An arithmetic mean is the average of a list of numbers. It is deﬁned as the sum of the numbers divided by the count of these numbers. Here are the two examples on how to calculate the arithmetic mean: The arithmetic mean of 7 and 8 =
7 + 8 15 = 2 2
The arithmetic mean of 1, 2, and 3 =
1+ 2 + 3 6 = =2 3 3
In a box, if the middle number is to be determined by averaging, it will be the average 25
of the numbers on the left and right sides. As shown in the ﬁgure on the right, the middle number is
3+11 14 = =7 2 2 ♞ Rule of thumb: Start with simple arithmetic operations In general, you should try using the simplest logic ﬁrst, and if that does not work, try using another logic. The simplest logic that works is preferred. Here the simple logic is one that uses simple arithmetic operations.
1.2.3.3 Sequential dependency among the Number Boxes Some books or aptitude tests might have problems that require sequential dependency to solve them. This can be better explained with an example below:
Figure 1.3: An example of a sequential dependency In ﬁgure1.3, we have three boxes. We can use the steps described below to determine the middle number. In the steps below, we will use notations (a1, b1), (a2, b2), and (a3, b3) to represent the numbers on the left and the right sides of the three boxes: ✔ Add the number on the left to the number on the right for each box. The result of this
step will be (a1 + b1), (a2 + b2), and (a3 + b3) for the three boxes respectively. ✔ Add 1, 2, and 3 sequentially as per the position of boxes in the problem to the result of
the previous step to determine the missing numbers. In other words, the middle numbers in the ﬁrst, second and third boxes will be (a1 + b1 + 1), (a2 + b2 + 2), and (a3 + b3 + 3) as shown below: The first box: a1 = 4, b1 = 1 a1 + b1 = 4 + 1 = 5 a1 + b1 + 1 = 5 + 1 = 6. This equals the middle number 6. The second box: a2 = 5, b2 = 1 a2 + b2 = 5 + 1 = 6 a2 + b2 + 2 = 6 + 2 = 8. This equals the middle number 8. 26
The third box: a3 = 7, b3 = 2 a3 + b3 = 7 + 2 = 9 a3 + b3 + 3 = 9 + 3 = 12. This is the answer we get by assuming a sequential dependency in solving the problem. As you will see later, this is not the right way to design a Number Box problem. Notice how we have used numbers 1, 2, and 3 in each of the boxes sequentially to ﬁnd the answer. This is an example of a sequential dependency among the boxes. Here we have used the natural numbers (1, 2, 3, ...) sequentially to be added to the sum of the numbers on the left and right. In a diﬀerent problem, another variation of the sequential numbers (e.g., 0, 1, 3, or 1, 2, 3, etc.) can be added to the values that result from applying a diﬀerent combination of arithmetic operations on the two numbers in the box. While it is possible that some books and aptitude tests might have answers that need to be calculated in this manner, the author believes that all the logic required to ﬁnd the missing number should be encoded in the individual box, meaning that each box should contain its logic within. Hence, the problems in this book do not assume any form of sequential dependency among the boxes. However, readers are advised to follow the guidelines and norms of the books or aptitude tests to solve such problems in those books or tests and prepare accordingly. This brings us to the question: can the problem in ﬁgure1.3 be solved without using a sequential dependency? In cases where a problem is designed by assuming a sequential dependency, ﬁnding the solution using other ways might not even be possible. However, in this case, we can ﬁnd an answer as explained below: ✔ We will use notations (a, b) to represent numbers on the left and right sides of the box.
To be more speciﬁc, we will use notations (a1, b1), (a2, b2), and (a3, b3) to represent the numbers on the left and right sides for the three boxes respectively. ✔ Find the diﬀerence between two numbers by subtracting the number on the right
from the number on the left. This can be expressed as (a  b). ✔ Double the result from the previous step, which can be formulated as
2(a  b). This
formula can now be used to determine the middle number in each box as shown below: The first box:
a1 = 4, b1 = 1 a1  b1 = 4  1 = 3 27
2(a1  b1) = 2 × 3 = 6. This equals the middle number 6. The second box:
a2 = 5, b2 = 1 a2  b2 = 5  1 = 4 2(a2  b2) = 2 × 4 = 8. This equals the middle number 8. The third box:
a3 = 7, b3 = 2 a3  b3 = 7  2 = 5 2(a3  b3) = 2 × 5 = 10. This is the answer to the missing number in the third box. The answer diﬀers from what we found earlier where we assumed a sequential dependency among the boxes. Did you notice that we solved this problem without assuming any sequential dependency at all? These boxes were merely using the same logic to determine the middle numbers without using any sequential dependency, which is the right way to design such problems. Therefore, we will not assume any sequential dependency among the boxes when solving problems in this chapter. After discussing the basics of a Number Box problem, let us look at the ﬁrst question from Example1. Example1(a)
Figure 1.4: Example1(a) We should attempt these types of problems by trying diﬀerent arithmetic operations to see which one works: • Addition:
The ﬁrst box: 1 + 1 = 2. This result is not equal to the middle number 1. • Subtraction:
The ﬁrst box: 1  1= 0. This result does not equal the middle number 1. • Multiplication:
The ﬁrst box: 1 × 1 = 1. This result equals the middle number 1. The second box: 3 × 9 = 27. This result does not equal the middle number 6.
28
• Division:
The ﬁrst box: 1 = 1. This result equals the middle number 1. 1
The second box: 9 = 3 or 3 = 1 . Neither of these two numbers equals the middle 3
9
3
number 6. • Arithmetic Mean:
The ﬁrst box: 1+1 = 2 = 1. This result equals the middle number 1. 2
2
The second box: 3 + 9 = 12 = 6. This result equals the middle number 6. 2
2
Since this logic works for the ﬁrst two boxes, the missing number in the third box will be calculated by applying the same logic. Hence, the middle number is
7+ 3 10 = =5 2 2
Now that we have found the answer to question (a), let us discuss the next problem. Example1(b)
Figure 1.5: Example1(b) Let us ﬁrst try a few arithmetic operations to see which one solves this problem: • Addition:
The ﬁrst box: 2 + 3 = 5. This result equals the middle number 5. The second box: 4 + 6 = 10. This result does not equal the middle number 20. After trying a few combinations of arithmetic operations (similar to what we have seen in Example1(a)), we can ﬁnd the answer by calculating the diﬀerence between the square of the numbers as shown below. • The diﬀerence between the square of numbers:
We subtract the square of the smaller number from the square of the larger number: 2
2
The ﬁrst box: 3  2 = 9  4 = 5. This result equals the middle number 5. 2
2
The second box: 6  4 = 36  16 = 20. This result equals the middle number 20. 29
This logic determines both the values correctly, so the missing number for the third box will be calculated by applying the same logic. 2
2
Hence, the middle number is 3  1 = 9  1 = 8 2
2
♜Note: You cannot use 1  3 =1 9 = 8 in this case. Let us look at Example1(c) in the next section to discuss this further.
Example1(c)
Figure 1.6: Example1(c) If you notice carefully, there are similarities between the problems in ﬁgure1.6 and ﬁgure1.5, although the ordering is a bit diﬀerent. We can ﬁrst try the diﬀerence between squared numbers. This time, we subtract the square of the number on the left from the square of the number on the right: • The diﬀerence between the square of numbers: 2
2
The ﬁrst box: 2  3 = 4  9 = 5. This result equals the middle number, which is 5. 2
2
The second box: 6  4 = 36  16 = 20. This result equals the middle number 20. In the ﬁrst two boxes, the result is obtained by subtracting the square of the number on the left from the square of the number on the right. Since the logic works for the ﬁrst two boxes, we apply the same logic to determine the missing number in the third box as shown below: 2
2
The middle number is 1  3 = 1  9 = 8 Notice that sometimes we need to subtract the smaller number from the larger number or that sometimes we need to subtract the number on the left from the number on the right. Now that we have found the answers to all the questions, we can show them in the next section.
30
1.2.2 Answer The answers to the problems in Example1 are shown below: Number Box Logitica (a)
(b)
(c) Figure 1.7: Answers to the problems in Example1
1.3 Summary In a box, we use only two numbers to ﬁnd the missing number. Therefore, a Number Box problem is easy to understand. The only complexity lies in ﬁnding which logic to use to determine the missing number. We will discuss another type of problem called the "Number Cross" in Chapter2 (p. 37) where we use four numbers to ﬁnd the missing number. We started this chapter with three examples and discussed how to determine the missing number using some wellknown arithmetic operations. In each of the problems, the logic used to determine the middle number was diﬀerent. However, the steps followed to ﬁnd the logic were similar. We started with simple logic and continued applying diﬀerent types of logic to the problem until we found one that was valid for the ﬁrst two boxes. We then used that same logic to ﬁnd the middle number in the third box. However, you may ﬁnd there are more possible combinations of these operations when you are solving a Number Box problem. In other words, to determine the middle number, we might need to apply various arithmetic operations on the two numbers on either side of the
a2 + b2 , middle number like 2
a2 + b2 , (a + b +
ab), a − b , a + b , etc. 2
2
3
3
In some cases, there are multiple answers possible, but as mentioned previously in this chapter, the logic that uses the simplest arithmetic and is consistent with all the boxes is preferred. It is recommended that readers work on solving problems laid out in the Exercise Section to gain a better understanding of Number Box problems. 31
1.4 Exercise Find the missing numbers in the problems below:
LevelI (i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
32
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
(xvi)
(xvii)
(xviii)
(xix)
33
(xx)
LevelII (i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
34
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
(xvi)
(xvii)
(xviii)
(xix)
35
(xx)
For further practice (i)
(ii)
(iii)
(iv)
(v)
36
Chapter 2
Number Cross Keywords: Arithmetic Operations, Arithmetic Mean, Binary Operators.
2.1 Overview In this chapter, we will learn how to solve problems involving Number Cross. Let us ﬁrst deﬁne what we mean by a Number Cross. ♛ Definition  Number Cross A Number Cross is a structure consisting of ﬁve numbers that are arranged in a crosslike shape as shown in the ﬁgure on the right. The number in the center of a cross is determined by applying some combinations of arithmetic operations on the four surrounding numbers. A problem in this chapter consists of three crosses, and all the numbers in the ﬁrst two crosses have been provided. Using this information, we need to ﬁnd the missing number in the third cross. As a rule, the logic to determine the number in the center must be the same for all of the crosses. Refer to Example1 for further details on this. ♜Note: Logic or pattern in a Number Cross As deﬁned above, the number in the center is obtained by applying some combinations of arithmetic operations on the four surrounding numbers. So when we are asked to "ﬁnd the missing number," it means that we are supposed to ﬁnd the appropriate combinations of arithmetic operations, which when applied on the four surrounding numbers, will result in the same number that is in the center of the cross. For example, in the ﬁgure above, the number in the center is obtained by adding the four surrounding numbers: 2 + 3 + 4 + 7 = 16. In this case, "determining the missing numbers" requires you to ﬁnd the fact that the number in the center is the "sum of all four surrounding numbers," and use the same logic to determine the missing number in the problem. Since the logic used to determine the number in the center must be the same in all the three crosses, we need to validate the logic in the ﬁrst two crosses before we can use that same logic in the third cross to ﬁnd the answer. Let us continue our discussion by analyzing the two problems in Example1.
37
2.2 Example1 Number Cross Logitica Find the missing numbers:
(a)
(b)
Figure 2.1: Example1
2.2.1 Analysis To solve problems involving Number Crosses, we need to use some of the concepts that we have already discussed while solving Number Box problems. So, if you have not learned this yet, it is the time to go back and read about it in Chapter1 (p. 21). We have already discussed addition, subtraction, multiplication, division, and arithmetic mean, etc., in the context of solving Number Box problems. Now we are going to explore the same in the context of Number Cross problems. To ﬁnd the number in the center, we need to use some arithmetic operations. As we know, binary operators require two operands, but a cross has four surrounding numbers. So we need to ﬁnd a way to yield two numbers out of the four numbers seen in the cross. For this, we need to create two pairs of numbers out of the four surrounding numbers. To understand how numbers can be selected for pairings, we ﬁrst need to understand the structure of a cross. A cross consists of two numbers in a horizontal row intersecting with two numbers in a vertical column as shown in ﬁgure2.2. We can create pairs in three diﬀerent ways in a cross as listed below: (a) Horizontal pair: The two numbers in the horizontal row are paired up. As shown in ﬁgure2.2, the pair from the horizontal row is (1, 3). (b) Vertical pair: The two numbers in the vertical column are paired up. As shown in ﬁgure2.2, the pair from the vertical column is (2, 4). 38
(c) Sideways pair: Pairings of adjacent side numbers together are also possible. For example, in ﬁgure2.2, the pairs made from the sideways numbers are (1, 2), (2, 3), (3, 4), and (4, 1).
Figure 2.2: Structure of a Number Cross
2.2.1.1 An example of pairings in a cross Let us follow the steps below to see how pairing up numbers can be used to ﬁnd the number in the center of a cross: ✔ Step 1: Create two pairs of numbers using the surrounding numbers in the cross. ✔ Step 2: Apply various combinations of arithmetic operations to each pair, which will
give us two new numbers. ✔ Step 3: We can now calculate the number in the center by using the two newly
generated numbers from the previous step. We will use the cross shown on the right to explain the steps discussed above and to demonstrate how we obtained 24 at the center: ✔ Step 1: We create two pairs of numbers from the horizontal row
and vertical column: Horizontal row: (1, 3) Vertical column: (2, 4) ✔ Step 2: We add the numbers in each pair to generate two new numbers (4, 6) as
shown below: Horizontal row: 1 + 3 = 4 Vertical column: 2 + 4 = 6 ✔ Step 3: We can now multiply the two numbers to generate the number in the center,
i.e., the number in the center is 4 × 6 = 24. The above example demonstrates how a single number can be obtained by using the four surrounding numbers in the cross.
39
♞ Assumption: Ignore sideways pairings in a cross When solving Number Cross problems, we are going to ignore sideways pairings. This way we can focus on learning logic to solve the problem without worrying about the combinations of pairings possible in a cross. Aside from this, we can probably create a problem of similar complexity by just pairing the numbers from the horizontal row and vertical column. Readers should be aware that some books or certain tests might require the use of sideways pairings to ﬁnd the solution, so prepare accordingly as per the guidelines and norms of those books or tests. However, to ensure that we focus on learning the logic needed to solve the problem, we will exclude sideways pairings in this book.
2.2.2.2 Sequential dependency among the Number Cross Some books or aptitude tests may contain problems in which sequential dependency is required to solve them. This can be better explained with an example below:
Figure 2.3: An example of a sequential dependency In ﬁgure2.3, we have three crosses, and we need to ﬁnd the missing number in the third cross. We can use the following steps to determine the missing number: • Add up all four numbers in the cross. Let us call the totals for each cross c1, c2, and c3
respectively. • The number in the center is now determined by adding natural numbers (1, 2, and 3)
to c1, c2, and c3 as per their sequential position. In other words, the numbers in the center for each of the three crosses will be c1 + 1, c2 + 2, and c3 + 3 respectively as described below: The first cross: Add the four surrounding numbers: c1 = 5 + 2 + 1 + 3 = 11 Add 1: c1 + 1 = 11 + 1 = 12. This equals the number in the center. The second cross: Add the four surrounding numbers: c2 = 1 + 2 + 4 + 4 = 11 40
Add 2: c2 + 2 = 11 + 2 = 13. This equals the number in the center. The third cross: Add the four surrounding numbers: c3 = 3 + 4 + 2 + 5 = 14 Add 3: c3 + 3 = 14 + 3 = 17. This is the answer we found assuming a sequential dependency among the crosses. As you will see later, this is not the right way to design such problems. For each cross, we have added natural numbers (1, 2, and 3) to the sum of four surrounding numbers. In some other problems, an entirely diﬀerent set of sequential numbers (e.g., 0, 1, 3, or 1, 2, 3, or 2, 3, 4, etc.) could be added to the values that result from applying diﬀerent arithmetic operations on the numbers in the cross. This is what we mean by having a sequential dependency among the crosses. While it is possible that some books or aptitude tests might require problems to be solved this way, we believe that all the logic to determine the number should be encoded in the individual cross, i.e., each cross should contain the logic within itself. Hence, the answers in this book do not assume any form of sequential dependency among the crosses. However, readers are advised to follow the guidelines and norms of the books being read or the tests being taken to solve the problems in those books or tests and prepare accordingly. But, can the problem in ﬁgure2.3 be solved without using sequential dependency at all? In cases when a problem is designed by assuming sequential dependency, ﬁnding the solution in other ways might not even be considered. However, in this instance we can ﬁnd an answer without using any sequential dependency as explained below: ✔ Add the two numbers in the horizontal row and let us call the total A. ✔ Multiply the two numbers in the vertical column and let us call the result B. ✔ As shown below, the number in the center is A + B:
The first cross: A=5+1=6 B=2×3=6 A + B = 6 + 6 = 12. This equals the number in the center. The second cross: A=1+4=5 B=2×4=8 A + B = 5 + 8 = 13. This equals the number in the center. The third cross: A=3+2=5 B = 4 × 5 = 20 41
A + B = 5 + 20 = 25. This is the answer to the problem. This answer diﬀers from what we found earlier by assuming the sequential dependency among the crosses. However, did you notice how we solved this problem without assuming any sequential dependency at all? The crosses were merely using the same logic to determine the missing number, which is the right way to design such type of problems. Therefore, we will not consider any sequential dependency among the crosses when solving problems in this chapter.
2.2.3.3 Arithmetic operations Let us discuss some of the arithmetic operations that are useful in solving Number Cross problems: (i) Addition: We can calculate the number in the center by adding up all the four surrounding numbers. As shown in the ﬁgure, the number in the center is equal to 2 + 3 + 4 + 7 = 16. Besides this, other problems might use a variety of combinations, some of which are described below: ✔ Calculate the sum of two numbers from the horizontal row,
which we can call A. ✔ Calculate the sum of two numbers from the vertical column, which we can call B. ✔ Once we have A and B, we can choose the appropriate method described in section
2.2.4.4 Calculate the number in the center using A and B (p. 46) to ﬁnd the number in the center. For example, we can use the four surrounding numbers in the ﬁgure above, and: • add numbers from the horizontal row: A = 2 + 4 = 6 • add numbers from the vertical column: B = 3 + 7 = 10 • So, the number in the center can be obtained by using combinations like AB, A 2

2
B, B  A, A + B ,..., and so on. Here are a few examples for calculating the number in the center: • B + A = 10 + 6 = 16 • A  B = 6  10 = 4 • B  A = 10  6 = 4 2
2
2
2
• A + B = 6 + 10 = 36 + 100 = 136, and so on.
(ii) Multiplication: The number in the center can be calculated by multiplying all four of the surrounding numbers together. As shown in the ﬁgure here, the value of the number in the center is 1 × 4 × 5 × 3 = 60. Besides this, there are more possible combinations as described below: ✔ Multiply the two numbers from the horizontal row, which we can call A. 42
✔ Multiply the two numbers from the vertical column, which we can call B. ✔ Once we have A and B, we can choose the appropriate method described in section
2.2.4.4 Calculate the number in the center using A and B (p. 46) to ﬁnd the number in the center. For example, we can use the four surrounding numbers in the ﬁgure above, and: • multiply the two numbers from horizontal row: A = 1 × 5 = 5 • multiply the two numbers from vertical column: B = 4 × 3 = 12 • So, the number in the center can be obtained by using combinations like AB, A 2

2
B, B  A, A + B ,..., and so on. Here are a few examples: • A + B = 5 + 12 = 17 • A  B = 5  12 = 7 • B  A = 12  5 = 7 2
2
2
2
• A + B = 5 + 12 = 25 + 144 = 169, and so on.
(iii) Subtraction: This is explained below with an example: ✔ Let us call the diﬀerence between two numbers from the horizontal row as A. ✔ Let us call the diﬀerence between two numbers from the vertical column as B. ✔ The number in the center in the ﬁgure here is A
 B as shown
below: Horizontal row: A = 9  3 = 6 Vertical column: B = 6  1 = 5 The number in the center = A  B = 6  5 = 1. ✔ In addition to the steps in above example, we can choose the
appropriate method described in section 2.2.4.4 Calculate the number in the center using A and B (p. 46) to ﬁnd the number in the center. For example, we can use the four surrounding numbers in the ﬁgure above: • Find the diﬀerence between numbers from the horizontal row:
A = (9  3) or (3  9) = 6 or 6 So, A has two possible values, A = 6 or 6. • Find the diﬀerence between numbers from the vertical column:
B = (6  1) or (1  6) = 5 or 5 So, B also has two possible values, B = 5 or 5. • Thus, the number in the center can be obtained by using combinations like AB, A 2

2
B, B  A, A + B ,..., and so on. Since subtraction is noncommutative, we can arrive at more values of the number in the center. Here are a few examples: A + B : 6 + 5 = 11, 6 + 5 = 1, 6  5 = 1, 6  5 = 11. : ( 11, 1, 1, 11) 43
A  B : 6  5 = 1, 6  5 = 11, 6 + 5 = 11, 6 + 5 = 1. : ( 1, 11, 11, 1) AB : 6 × 5 = 30, (6) × 5 = 30, 6 × ( 5) = 30, (6) × ( 5) = 30. : ( 30, 30) (iv) Division: This is explained below with an example: ✔ Divide one of the numbers in the horizontal row number by the other and call the
result A. ✔ Divide one of the numbers in the vertical column number by the other and call the
result B. ✔ The number in the center in the ﬁgure here is
Horizontal row: A =
2 =2 1
Vertical column: B =
30 = 10 3
The number in the center =
B as shown below: A
B = 10 = 5 2 A
✔ In addition to the steps mentioned above, once we have A and B, we can choose the
appropriate method described in section 2.2.4.4 Calculate the number in the center using A and B (p. 46) to ﬁnd the number in the center. For example, using the four surrounding numbers in the ﬁgure above: Divide the numbers from the horizontal row with each other: A=
2 1 = 2 or A = 1 2
So, A has two values possible, A = (2,
1 ) 2
Divide the numbers from the vertical column with each other: B=
3 1 30 = 10 or B = = 30 10 3
So, B also has two values possible, B = (10,
1 ) 10 2
2
So the number in the center can be obtained by combinations like AB, A + B , A + B, A  B, ..., and so on. Since division is noncommutative, we can arrive at more values of the number in the center. Here are a few examples: 44
A + B: 2 + 10 = 12 2+
1 2 ×10 +1 21 = = 10 10 10
1 1+10 × 2 21 +10 = = 2 2 2 1 1 5 +1 6 3 + = = = 2 10 10 10 5 Hence, there are four possible values of A + B: (12,
21 21 3 , , ) 10 2 5
AB: 2 × 10 = 20 2×
1 1 = 10 5
1 × 10 = 5 2 1 1 1 × = 2 10 20 Hence, there are four possible values of AB: (20,
1 1 , 5, ) 20 5
Since division and subtraction are noncommutative, the results will vary if the order of these arithmetic operations is changed. Please refer to section 2.2.5.5 Subtraction and division are noncommutative (p. 46) for further discussions. (v) Sum of squares: The number in the center is the sum of squares of each of the surrounding numbers. As shown in 2
2
2
the ﬁgure here, the number in the center is: 2
1 + 2 + 3 + 2 = 1 + 4 + 9 + 4 = 18 (vi) Sum of cubes: The number in the center is the sum of the cubes of each of the surrounding numbers. As shown in the ﬁgure on the right, the number in the center is: 3
3
3
3
1 + 2 + 3 + 2 = 1 + 8 + 27 + 8 = 44 (vii) Arithmetic Mean: An arithmetic mean is the average of a list of numbers. It is calculated by dividing the sum of numbers by the count of the numbers in the list. In a cross, we have 4 surrounding numbers, so we can obtain the number in the center by 45
adding up the four numbers and dividing the total by 4. As shown in the ﬁgure on the right, the number in the center is
5+ 2+ 3+ 2 12 =3 = 4 4
2.2.4.4 Calculate the number in the center using A and B In the discussion for this section, we assume two numbers A and B as described below: ✔ Apply some arithmetic operations on the two numbers in the horizontal row and let
us call the result A. ✔ Apply some arithmetic operations on the two numbers in the vertical column and let
us call the result B. Depending on the problem, once we have A and B, the number in the center can be obtained by using one of the following patterns: 2
2
• AB, A B, AB
• A + B, A  B, B  A •
A, B B A A
B
A
B
2
2
2
 B , B  A , (AB) , etc.
• A +B ,A • A +B ,A
B 2
2
2
2
♞ Rule of thumb: Start with the simple arithmetic operations In general, you should try using the simplest logic ﬁrst, and if that does not work, then try another logic. The simplest logic is always preferred. Here the simplest logic is one that uses simple arithmetic operations.
2.2.5.5 Subtraction and division are noncommutative ✔ A binary operation is commutative if changing the order of operation does not change
the result. Arithmetic operations like addition and multiplication are commutative. This is obvious as 1 + 4 = 4 + 1 or 2 × 3 = 3 × 2. On the other hand, division and subtraction
1 3 ≠ . Understanding this concept crucial in 3 1 ﬁnding the right answer for a Number Cross problem, where sometimes the smaller (or larger) number is subtracted from the larger (or smaller) number to determine what the number in the center is. Similarly, the number in the center can be determined by dividing the smaller (or larger) number by the larger (or smaller) number. are noncommutative as 9  3 ≠ 3  9 or
✔ In a cross, we have four numbers, and we have seen before how to yield two numbers 46
out of these four numbers. For the discussion below, we will refer to these two newly generated numbers as “the number on the left” and “the number on the right.” ✔ Continuing on, the order in which subtraction or division is performed could be
positional, such as subtracting the number on the left from the number on the right and vice versa or dividing the number on the left by the number on the right and vice versa. it is obvious that the number in the center will vary depending on the order of subtraction or division. Now that we understand the basics of a Number Cross, let us try solving the two problems from Example1. Example1(a)
Figure 2.4: Example1(a) Let us try to solve this problem by attempting a few arithmetic operations to see which one works. • Addition:
The ﬁrst cross: 1 + 2 + 3 + 4 = 10. This equals the number 10 at the center. The second cross: 2 + 1 + 1 + 1 = 5. This equals the number 5 at the center. Since we have found the logic that works with the ﬁrst two crosses, we can apply that same logic for the third cross to determine the missing number. Hence, the number in the center is 2 + 2 + 2 + 3 = 9. Example1(b)
Figure 2.5: Example1 (b) Let us try a few arithmetic operations as listed below: • Addition:
The ﬁrst cross: 1 + 2 + 3 + 4 =10. It does not equal number 4 in the center.
47
• Subtraction:
The first cross: Diﬀerence between the larger and smaller number in the horizontal row (A1) = 3  1 = 2 Diﬀerence between the larger and smaller number in the vertical column (B1) = 4  2 = 2
The second cross: Diﬀerence between the larger and smaller number in the horizontal row (A2) = 2  1 = 1 Diﬀerence between the larger and smaller number in the vertical column (B2) = 3  1 = 2
Let us now try the following combinations to see which one works in ﬁnding the answer: A + B: The ﬁrst cross: A1 + B1 = 2 + 2 = 4. This result equals the number 4 in the center. The second cross: A2 + B2 = 1 + 2 = 3. This result does not equal the number 2 in the center.
A  B: The ﬁrst cross: A1  B1 = 2  2 = 0. This result does not equal the number 4 in the center. AB: The ﬁrst cross: A1B1 = 2 × 2 = 4. This result equals the number 4 in the center. The second cross: A2B2 = 1 × 2 = 2. This result equals the number 2 in the center. The third cross:
We found the logic that works for the ﬁrst two crosses, so we can apply that same logic to the third cross to determine the missing number in the center as shown below: Diﬀerence between the numbers in the horizontal row = A3 = 2  1 = 1 Diﬀerence between the numbers in the vertical column= B3 = 3  1 = 2 Hence, the answer is A3B3 = 1 × 2 = 2 Please note that in this case, we found the answer rather quickly, but in most cases, you will need to try more combinations to see which one works for a given problem. In the above section, we discussed the various types of logic and methods that would be useful in determining the correct answer. During the analysis, we have found the answers to both the problems, which we will show in the next section. 48
2.2.2 Answer The answers to the problems in Example1 are shown below: Number Cross Logitica
(a)
(b)
Figure 2.6: Answers to the problems in Example1
2.3 Summary As we have discussed in this chapter, we use the four surrounding numbers in the cross to determine the number in the center of the cross. Hence, the Number Cross problems are more challenging than the Number Box problems. Since both use arithmetic operations to determine the missing number, it is easy to observe similarities in the methods used in ﬁnding the missing number. We used two examples in this chapter to discuss the fundamentals behind a Number Cross. We followed a stepbystep approach to determine the missing number in the center. We discussed some common arithmetic operations that can be used to determine the number in the center. However, you may ﬁnd that there are even more combinations of these arithmetic operations possible that can be used in solving Number Cross problems. For example, the number in the center can be determined by using various combinations of arithmetic operations like (a1 + b1 + a2b2), 3
(a1 +
3
3
b1 )(a2 +
b23),
(a1 +
b1)(a1  b1),
(a
2 1
) (
+ b12 − a22 + b22 2
) , and etc. It is
recommended that readers practice with the problems provided in the Exercise Section to achieve a better understanding of solving Number Cross problems. 49
2.4 Exercise Find the missing numbers in the problems below:
LevelI
(i)
(ii)
(iii)
(iv)
50
(v)
(vi)
(vii)
(viii)
(ix)
51
(x)
(xi)
(xii)
(xiii)
(xiv)
52
(xv)
(xvi)
(xvii)
(xviii)
(xix)
53
(xx)
LevelII
(i)
(ii)
(iii)
54
(iv)
(v)
(vi)
(vii)
(viii)
55
(ix)
(x)
(xi)
(xii)
(xiii)
56
(xiv)
(xv)
(xvi)
(xvii)
(xviii)
57
(xix)
(xx)
For further practice
(i)
(ii)
(iii)
58
(iv)
(v)
(vi)
59
Chapter 3
Number Sequence Keywords: Number Sequence, Prime Numbers, Fibonacci Numbers, Order of difference.
3.1 Overview To start this chapter, we ﬁrst need to understand what we mean by a Number Sequence.
♛ Definition  Number Sequence A Number Sequence is a list of numbers arranged in a certain order using a particular pattern or logic. Usually, the logic to build the sequence is based on the wellknown arithmetic operations like addition, subtraction, multiplication, division, etc. Each number in the sequence is called a term. In this chapter, we may call a number sequence, for brevity, a sequence. The objective of a number sequence problem is to ﬁnd its missing term. Here is an example:
1, 1, 2, 3, 5, 8, 13, ? In the sequence above, the ﬁrst term is 1, the second term is 1, the third term is 2, the fourth term is 3, the ﬁfth term is 5, and so on. The objective here is to ﬁnd the missing term in the sequence. We will discuss and solve this problem in Example1. ♜Note: (i) We will use "term" and "number" interchangeably to mean the same thing. Hence, we could say "numbers in the sequence" or "terms in the sequence" to mean the same. (ii) We will use an ellipsis (...) whenever necessary to indicate the continuation of the sequence as per its logic. We can write a sequence by enclosing its terms within parentheses "()." For example, a sequence containing numbers 1, 1, 2, 3, 5, 8, 13, ... can be written as (1, 1, 2, 3, 5, 8, 13, ...). To solve a number sequence problem, we need to discover the logic or pattern behind the sequence to determine its missing term. As we will discover later, the terms in a sequence are usually generated by applying some of the wellknown arithmetic operations. So when we say "ﬁnd the missing term in a sequence," it means that we are supposed to ﬁnd a certain combination of arithmetic operations that are being used in generating the terms 60
of the sequence. Once we ﬁnd such a combination of arithmetic operations, we can use it to ﬁnd the missing term. This can better be explained with a few examples. Let us continue our discussion with three problems in Example1.
3.2 Example1 Number Sequence Logitica Find the missing numbers in the following sequences:
(a) 1, 1, 2, 3, 5, 8, 13, ? (b) 1, 2, 2, 4, 3, 6, ? (c) 2, 3, 10, ?, 55 Figure 3.1: Example1
3.2.1 Analysis Before we can solve the above problems, we need to understand the fundamentals behind the construction of the wellknown number sequences. Some of the more important ones are listed in the table below: Wellknown Number Sequences Positive integers
A sequence of positive integers is shown below:
(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...) We refer to the sequence as a sequence of positive integers because it excludes 0 and all the negative integers. ♜ Note: Positive integers are also referred to as natural numbers. Since these numbers are also used in counting things, they are also referred to as counting numbers.
Nonnegative integers
A sequence of nonnegative integers includes 0 and positive integers. Here is an example of a sequence of nonnegative integers:
(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...) ♜ Note: (i) Nonnegative integers are also referred to as whole numbers. (ii) Did you notice that 0 is the smallest whole number and 1 is the smallest natural number? 61
Negative integers
Negative integers can be obtained by negating all the positive integers. A sequence of negative integers is shown below:
(..., 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, ...) Integers
A sequence of integers contains nonnegative and negative integers. An example of a sequence of integers is shown below:
(..., 5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5, ...) Even numbers
An integer can be classiﬁed as an even number if it is evenly divisible by 2. This means that an even number yields no remainder when divided by 2. A sequence of even numbers is shown below:
(..., 10, 8, 6, 4, 2, 0, 2, 4, 6, 8, 10, ...) Odd numbers
An integer can be classiﬁed as an odd number if it is not evenly divisible by 2. A sequence of odd numbers is shown below:
(..., 11, 9, 7, 5, 3, 1, 1, 3, 5, 7, 9, 11, ...) Square numbers
A number can be classiﬁed as a square number or perfect square if it is the square of an integer. A sequence of square numbers is shown below:
(0, 1, 4, 9, 16, 25, 36, 49, 64, 81, ...) Explanation: 2
Square of 1 = 1 = 1 × 1 = 1 2
Square of 2 = 2 = 2 × 2 = 4 2
Square of 3 = 3 = 3 × 3 = 9 2
Square of 4 = 4 = 4 × 4 = 16 2
Square of 5 = 5 = 5 × 5 = 25 2
Square of 6 = 6 = 6 × 6 = 36, and so on.
♜ Note: A negative number becomes positive when squared as shown below: 2
Square of (1) = (1) = (1) × (1) = 1 2
Square of (2) = (2) = (2) × (2) = 4 2
Square of (3) = (3) = (3) × (3) = 9 2
Square of (4) = (4) = (4) × (4) = 16 2
Square of (5) = (5) = (5) × (5) = 25 2
Square of (6) = (6) = (6) × (6) = 36 62
Cube numbers A number can be classiﬁed as a cube number or perfect cube if it is the cube of an integer. A sequence of cube numbers is shown below:
(125, 64, 27, 8, 1, 0, 1, 8, 27, 64, 125, ...) Explanation: 3
Cube of 0 = 0 = 0 × 0 × 0 = 0 3
Cube of 1 = 1 = 1 × 1 × 1 = 1 3
Cube of 2 = 2 = 2 × 2 × 2 = 8 3
Cube of 3 = 3 = 3 × 3 × 3 = 27, and so on.
♜ Note: A negative number with an odd exponent is a negative number. Hence the cube of a negative number remains negative as shown below: 3
Cube of (3) = (3) = (3) × (3) × (3) = 27 3
Cube of (2) = (2) = (2) × (2) × (2) = 8 3
Cube of (1) = (1) = (1) × (1) × (1) = 1
Prime numbers A prime number is a natural number with exactly two factors, itself and 1. This means that a prime number can be divided evenly only by itself and and 1. A sequence of prime numbers is shown below: composite numbers
(2, 3, 5, 7, 11, 13, 17, ...) Composite numbers are positive integers (greater than 1) that are not prime. Hence, all natural numbers above 1 are either prime or composite. ♜Note: Prime and composite numbers: • 0 is not a prime number. • 1 is neither composite nor prime. • The ﬁrst prime number is 2, and it is the only even prime number.
All other even numbers (e.g., 4, 6, 8, 10, ...) are evenly divisible by 2, so they are not prime numbers. • Not all odd numbers are prime numbers. For example, 9 is a
composite number as it is evenly divisible by 1, 3, and 9. On the other hand, 5 is a prime number as it is only evenly divisible by 1 and 5. • No prime number greater than 5 ends with 5, as all such numbers
are also evenly divisible by 5 (i.e., 15, 25, 35, ..., etc.). • There are inﬁnite prime numbers. 63
Fibonacci sequence or
Fibonacci sequence starts with two numbers: 1 and 1. Each term in the sequence is the sum of two preceding terms together as shown below:
Fibonacci number sequence
Here are some of the initial numbers of Fibonacci sequence:
(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ...) ♜Note: As a side note, in some literature this sequence can also start with 0 and 1 as shown below:
(0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,...) Explanation: We start with two numbers 1 and 1. We then recursively add the previous two numbers in the sequence to get a new number as shown below: 1+1= 2 2+1= 3 3+2= 5 5 + 3 = 8, and so on. Interesting facts about Fibonacci sequence: rd
• Every 3 number is a multiple of 2 (e.g., 2, 8, 34, ...). th
• Every 4 number is a multiple of 3 (e.g., 3, 21, 144, ...). th
• Every 5 number is a multiple of 5 (e.g., 5, 55, 610, ...). th
• Every 6 number is a multiple of 8 (e.g., 8, 144, 2584, ...). • Following the same logic, Every
multiple of
nth term in this sequence is a
Fn. Refer to AppendixK (p. 303) for further details on
this. Figure 3.2: Wellknown Number Sequences 64
3.2.1.1 The n term and the index of a term th
When discussing sequences, we will frequently refer to two key terms: the n term and th
the index of a term. This will help us analyze the sequences that appear later in this chapter. Let us ﬁrst deﬁne what we mean by these two terms: ♛ Definition  The n term and the index of a term th
An index represents the position of a term in the sequence and the n term represents th
the term or number at the index value of n, which will be represented by the following notations:
tn = The nth term of a sequence. n = The index of the term in a sequence. Notice that the index of a term is, in fact, a natural number, i.e., n=1, 2, 3, ..... Let us try to understand the above notations using a few examples: (i) In an odd number sequence (1, 3, 5, 7, 9, ...), the terms corresponding to the diﬀerent values of n = (1, 2, 3, 4, 5, ...) are t1 = 1, t2 = 3, t3 = 5, t4 = 7, t5 = 9, and so on. We can can use the following formula to determine the n term of this sequence: th
The n term tn = 2n  1 th
For n = 1, t1 = 2 × 1  1 = 2  1 = 1 For n = 2, t2 = 2 × 2  1 = 4  1 = 3 For n = 3 ,t3 = 2 × 3  1 = 6  1 = 5 For n = 4, t4 = 2 × 4  1 = 8  1 = 7 For n = 5, t5 = 2 × 5  1 = 10  1 = 9
(ii) In a square number sequence (1, 4, 9, 16, 25, ...), the terms corresponding to the diﬀerent values of
n = (1, 2, 3, 4, 5, ...) are
t1 = 1, t2 = 4, t3 = 9, t4 = 16, t5 = 25, and so th on. The formula to determine the n term of this sequence is given below: The n term tn = n , where n = 1, 2, 3, ... th
2
For n = 1, t1 = 1 = 1 × 1 = 1 2
For n = 2, t2 = 2 = 2 × 2 = 4 2
For n = 3, t3 = 3 = 3 × 3 = 9, and so on. 2
65
(iii) In a Fibonacci sequence (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...), the terms corresponding to the diﬀerent values of n = (1, 2, 3, 4, 5, ...) are t1 = 1, t2 = 1, t3 = 2, t4 = 3, t5 = 5, and so on. We can write the n term of the sequence as: th
The n term tn = tn1 + tn2 , where t1 = 1, t2 = 1, and n > 2 th
For n = 3, t3 = t2 + t1 = 1 + 1 = 2 For n = 4, t4 = t3 + t2= 2 + 1 = 3 For n = 5, t5 = t4 + t3 = 3 + 2= 5 For n = 6, t6 = t5 + t4 = 5 + 3= 8, and so on.
Fn is also used to indicated the nth term of a Fibonacci sequence. An th explicit formula to determine the n term of a Fibonacci sequence exists, which is Alternatively,
derived in section AppendixK (p. 303). (iv) In a sequence of prime numbers (2, 3, 5, 7, 11, ...), the terms corresponding to the diﬀerent values of n = (1, 2, 3, 4, 5, ...) are t1 = 2, t2 = 3, t3 = 5, t4 = 7, t5 = 11, and so on.
3.2.2.2 Some more special sequences In this section, we will discuss a few more sequences that will help you to solve number sequence problems:
(i) Sum sequence: The nth term of this sequence is the sum of the ﬁrst n natural numbers as shown below: The 1st term of the sequence is 1. The 2nd term = 1 + 2 = 3 The 3rd term = 1 + 2 + 3 = 6, and so on.
To determine each term in this sequence, we can use the following formula: The formula1 for the sum of the ﬁrst n natural numbers =
n( n +1) 2
For example: The ﬁrst term in the sequence (n = 1) = t1= 1 The second term in the sequence (n = 2) = t2 =
(
2× 2+1 2
) = 2× 3 = 6 = 3 2
2
1. Refer to AppendixA (see p. 263) for the formula for the sum of natural numbers. 66
The third term in the sequence (n = 3) = t3 =
(
3× 3+1 2
) = 3× 4 = 12 2
2
=6
th Hence, the n term in the sequence = tn = n(n +1) 2
Some of the initial terms of the sum sequence are shown below:
(1, 3, 6, 10, 15, 21, 28, 36, ...) ♜Note: Triangular number and Average Number Pyramid Sum sequence discussed above also represents the number of objects arranged in an equilateral triangle as shown in the ﬁgure3.3. Hence, it is also referred to as a Triangular number or Triangle number sequence.
Figure 3.3: Triangular number and Allones pyramid However, an alternative and innovative way to represent the sum sequence is by adding all the numbers in an allones pyramid. This pyramid belongs to a new type of pyramid introduced in Chapter10 (p. 322) called the "Average Number Pyramid." Readers can refer to section 10.4 Summary (p.211) in Chapter10 Average Number Pyramid for further details.
(ii) Factorial sequence or multiplication sequence: The nth term of the sequence is
n natural numbers together. The multiplication of the ﬁrst n natural numbers represented by n! (n followed by an exclamation mark), is also referred to as factorial of n. As shown below, the multiplication of: obtained by multiplying the ﬁrst
• the ﬁrst 2 natural numbers = 2! = 2 × 1 = 2 • the ﬁrst 3 natural numbers = 3! = 3 × 2 × 1= 6 • the ﬁrst 4 natural numbers = 4! = 4 × 3 × 2 × 1= 24 • the ﬁrst 5 natural numbers = 5! = 5 × 4 × 3 × 2 × 1= 120, and so on. • the ﬁrst n natural numbers = n (n  1) (n  2) ..... × 3 × 2 × 1. 67
Here are some of the initial terms of the multiplication or factorial sequences: • If the sequence starts from 1!
⇒ (1, 2, 6, 24, 120, ...)
• If the sequence starts from 0!
⇒ (1, 1, 2, 6, 24, 120, ...)
A few variations of these sequences built by multiplying the consecutive natural numbers are shown below: • •
(1 × 2, 2 × 3, 3 × 4, 4 × 5, 5 × 6, ...) ⇒ ( 2, 6, 12, 20, 30, ...) (1 × 2, 3 × 4, 5 × 6, 7 × 8, ...) ⇒ ( 2, 12, 30, 56, ...)
Other variations of such sequences are also possible. Note that factorial is also used in a binomial expansion1.
♜Note: Some comments on factorial The factorials of 0 and 1 are both deﬁned to be 1. This is because of the deﬁnition and usage of factorial in set theory. In set theory2, factorial is used to determine the number of permutations in arranging unique elements of a set. Let us review a few examples to explain this concept: Empty set: An empty set is expressed as {}, and it can only be arranged in one way: {}. An empty set has no elements, i.e., the count of the elements in an empty set is 0. Hence, the factorial of 0 = 0! = 1. A set with one element: A set containing one element can be arranged in only one way. Therefore, the factorial of 1 is also one. For example, a set {4} can be arranged in only one way: (4), so, the number of arrangements is 1. Since the number of elements is 1 and the number of arrangements is 1, the factorial of 1 = 1! = 1. A set with two elements: A set containing two elements can be arranged in two ways. For example, a set {4, 5} can be arranged as (4, 5) and (5, 4). Hence, the number of arrangements = 2 × 1 = 2. Since the number of elements is 2 and the number of arrangements is 2, the factorial of 2 = 2! = 2. A set with three elements: A set with three elements can be arranged in six ways. For example, a set {3, 6, 7} can be arranged as (3, 6, 7), (3, 7, 6), (6, 3, 7), (6, 7, 3), (7, 3, 6), and (7, 6, 3), which means the number of arrangements is 3 × 2 × 1 = 6. Since the number of elements is 3 and the number of arrangement is 6, the factorial of 3 = 3! = 6.
n elements: Along the same lines, the number of arrangement can be calculated as n (n  1) (n  2) ..... × 3 × 2 × 1 = n!. A set with
We can summarize above discussion as follows: • The factorial of 0 = 0! = 1
1. Refer to AppendixF (p. 274) for a brief introduction to binomial expansion. 2. Refer to AppendixI (p. 282) for a brief introduction on set. 68
• The factorial of 1 = 1! =1 • The factorial of 2 = 2! =2 × 1 = 2 • The factorial of 3 = 3! =3 × 2 × 1 = 6 • The factorial of n = n! =n (n  1) (n  2) ..... × 3 × 2 × 1.
♜Note: Number of arrangements As discussed earlier, a set containing three elements can be arranged in 6 ways. For example, a set {3, 6, 7} can be arranged as (3, 6, 7), (3, 7, 6), (6, 3, 7), (6, 7, 3), (7, 3, 6), and (7, 6, 3). To ﬁgure out a systematic way to determine the number of arrangements, a more detailed explanation is needed here. Of the three numbers (3, 6, 7), there are three choices for the ﬁrst position: 3, 6, or 7. Let us assume that we select number 6 for the ﬁrst position. The number in the second position now will only be one of the two remaining numbers: 3 or 7. Hence, there are two choices for the second position. Let us assume we select number 3 for the second position. Since we have already chosen two numbers out of the three, we are left with only one number. Hence, there is only one choice for the number in the third position, which is 7 in this case. Therefore, the numbers in a set containing three elements can be arranged in six unique ways: 3 × 2 × 1 = 6. We can draw a similar conclusion that the numbers in a set containing four elements can be arranged in 24 unique ways: 4 × 3 × 2 × 1 = 24. Following the same logic, the numbers in a set containing n elements can be arranged in n! unique ways: n (n  1) (n  2) ..... × 3 × 2 × 1 = n!. The above discussion can also be alternatively explained by three colored marbles: 1 white, 1 black, 1 grey. As shown in ﬁgure3.4, there are 6 possible arrangements of these three marbles. We get the same number of arrangements by arranging the three unique numbers {3, 6, 7}.
Figure 3.4: Three diﬀerent colored marbles can be arranged in 6 diﬀerent ways.
(iii) Sequences using an exponent (or a power): In this sequence, each term is the result of a number raised to the power of another number. The base number or the exponent number can be a natural number, prime number, Fibonacci number, odd number, or even number, etc. Here are a few examples of how such type of sequences can be built: 69
(a) Consider a sequence built using a natural number as an exponent to the respective term of a natural number sequence as shown below: Natural numbers: 1, 2, 3, 4, 5, ... Sequence built using (Natural number)
Natural number
is shown below:
11, 22, 33, 44, ... ⇒ 1, 4, 27, 256, ... (b) A sequence using a natural number as an exponent to the respective term of a Fibonacci sequence is shown below: Natural numbers
: 1, 2, 3, 4, 5, ...
Fibonacci numbers : 1, 1, 2, 3, 5, 8, 13, ... Sequence built using (Fibonacci number)
Natural number
is shown below:
11, 12, 23, 34, 55, ... ⇒ 1, 1, 8, 81, 3125, ... (c) A sequence using a Fibonacci number as an exponent to the respective term of a prime number sequence is shown below: Prime numbers
: 2, 3, 5, 7, 11, ...
Fibonacci numbers : 1, 1, 2, 3, 5, 8, 13, ... Fibonacci number
Sequence built using (Prime number)
is shown below:
21 , 31 , 52 , 73 , ... ⇒ 2, 3, 25, 343, ... (iv) Combined number sequence: A combined number sequence is constructed by combining two diﬀerent sequences. Each alternate term in this type of sequence belongs to a diﬀerent sequence. To ﬁnd the missing number in a combined number sequence, we ﬁrst need to identify the sequences that make up the terms for the Combined Sequence. Here are two examples of this type of sequence: (a) In the example below, the combined number sequence is constructed by combining natural numbers and multiples of ﬁve as shown below. The respective terms of the two sequences are highlighted in italics and bold: Natural number sequence
:
1, 2, 3, 4, ....
The sequence of multiples of 5
:
5, 10, 15, 20, ...
The resulting combined number sequence :
1, 5, 2, 10, 3, 15, 4, 20 ...
(b) In the example below, the combined number sequence is comprised of natural numbers and odd numbers: Natural numbers
:
1, 2, 3, 4, ....
Odd numbers
:
1, 3, 5, 7, .... 70
1, 1, 2, 3, 3, 5, 4, 7, ...
The resulting combined number sequence :
(c) Consider another interesting example of an Combined number sequence: Adding 2 to the previous term
:
5, 7, 9, 11, ....
Subtracting 2 from the previous term
:
13, 11, 9, 7, ....
The resulting combined number sequence :
5, 13, 7, 11, 9, 9, 11, 7, ...
♜Note: The diﬀerence between the consecutive terms To analyze a certain category of sequences, we need to analyze the diﬀerence between each of the consecutive terms. Sometimes we are even required to examine the "diﬀerence of the diﬀerence" between the two consecutive terms. As shown in ﬁgure3.5, we have a sequence (1, 2, 7, 16, 29) represented as sequence (a). The numbers in sequence (b) are values of the diﬀerence between the consecutive terms in sequence (a). Continuing on, another sequence (c) is constructed from values of the diﬀerence between the consecutive numbers in sequence (b). One can notice that the numbers in sequences (b) and (c) could follow a pattern just like any other sequence. Hence, we will treat such type of sequences as number sequences in our discussion and analyze the patterns in their terms, whenever needed.
Figure 3.5: Diﬀerence between the consecutive terms To summarize the above discussion, we can list out the following sequences: • Sequence (a): This is the sequence in the problem: (1, 2, 7, 16, 29). • Sequence (b): As shown below, this sequence is constructed by calculating the
diﬀerence between the consecutive terms of sequence (a) and its terms are 1, 5, 9, and 13: 2  1 = 1, 7  2 = 5, 16  7 = 9, 29  16 = 13. • Sequence (c): As shown below, this sequence is constructed by calculating the
diﬀerence between the consecutive terms of sequence (b) and its terms are 4, 4, and 4: 5  1 = 4, 9  5 = 4, 13  9 = 4. Whenever necessary, we will use this approach to deﬁne sequences as illustrated in ﬁgure3.5 (e.g., sequences (b), (c), or even more if required). This will help us to analyze and solve such types of problems in a systematic way. 71
♜Note: All the terms in sequence (c) from ﬁgure3.5 are the same, which is a special property for a type of sequence called “quadratic sequence.” We will discuss this in detail later in the section "3.3.2 Order of diﬀerence (p. 88)."
3.2.3.3 Some more Number Sequences In this section, we will continue to explore the topic with a few more variations of the number sequences. Examplea: Find the missing term in the following sequence:
1, 8, 15, 22, 29, ?
Figure 3.6: Diﬀerence between the consecutive terms As shown in ﬁgure3.6, we have two sequences: • Sequence (a): This is the sequence in the problem: (1, 8, 15, 22, 29, ?). • Sequence (b): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (a) and the terms of this sequence are 7, 7, 7, and 7: 8  1 = 7, 15  8 = 7, 22  15 = 7, 29  22 = 7. While all the terms in sequence (a) are in increasing order, all the terms in the sequence (b) are the same. As shown in ﬁgure3.6, we use this constant value 7 to ﬁnd the next term in the sequence (a): 29 + 7 = 36. Hence, the completed sequence is:
1, 8, 15, 22, 29, 36 ♜Note: A sequence in which the diﬀerence between each of the consecutive terms is the same is called an arithmetic sequence. We will discuss this sequence in more detail later in the chapter. Exampleb: Find the missing term in the following sequence:
1, 3, 7, 13, 21, ? 72
As shown in ﬁgure3.7, we have three sequences: • Sequence (a): This is the sequence in the problem: (1, 3, 7, 13, 21, ?). • Sequence (b): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (a). As calculated below, the terms in this sequence are 2, 4, 6, and 8: 3  1 = 2, 7  3 = 4, 13  7 = 6, 21  13 = 8. • Sequence (c): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (b). As shown below, the terms in this sequence are 2, 2, and 2: 4  2 = 2, 6  4 = 2, 8  6 = 2
Figure 3.7: Diﬀerence between the consecutive terms As shown in ﬁgure3.7, the numbers in the sequence (a) are in increasing order, and so are the numbers in sequence (b). However, all the numbers in sequence (c) are the same as 2. Hence, we can follow the steps below to ﬁnd the missing term: ✔ Since all the numbers in sequence (c) are the same (i.e., 2), to determine the next
term in sequence (b), we add 2 to the last term of sequence (b): 8 + 2 = 10. ✔ Add the result obtained in the previous step to the last known term of sequence (a):
21 + 10 = 31, and this is the missing term we were looking for. Hence, the completed sequence is:
1, 3, 7, 13, 21, 31 It is easy to observe that each term in the sequence is calculated by adding 2, 4, 6, and 8 sequentially to the previous term. So, the next term in the sequence can be obtained by adding 10 to the last known term, which is 21 + 10 = 31. One might wonder whether it is necessary to analyze the values of the diﬀerence between the consecutive terms to ﬁnd the missing term. However, as you will see later, the analysis of the diﬀerence between the consecutive terms as outlined above is very useful in certain categories of sequences involving polynomials1 of diﬀerent degrees. We will discuss this further down in section "Order of diﬀerence (p. 88)." You will also learn in 1. Refer to AppendixJ for more details about sequence involving polynomials. 73
the next example how useful and eﬀective this method is when the pattern in the sequence is not so obvious. Examplec: Consider the following sequence:
9, 7, 6, 6, 7, 9, ? The numbers in this sequence start to increase after decreasing for the ﬁrst few terms. Since there is no obvious pattern visible, let us analyze the diﬀerence between the consecutive terms to see if we can ﬁnd any logic in this sequence.
Figure 3.8: Diﬀerence between the consecutive terms As shown in ﬁgure3.8, we have three sequences: • Sequence (a): This is the sequence in the problem: (9, 7, 6, 6, 7, 9, ?). • Sequence (b): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (a). As shown below, the terms in this sequence are 2,
1, 0, 1, and 2: 7  9 = 2, 6  7 = 1, 6  6 = 0, 7  6 = 1, 9  7 = 2. • Sequence (c): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (b). As calculated below, the terms in this sequence are 1, 1, 1, and 1:
(1)  (2) = 1, 0  (1) = 1, 1  0 = 1, 2  1 = 1. The numbers in the sequence (a) have no speciﬁc pattern, but the numbers in sequence (b) are in increasing order. Also, all the numbers in sequence (c) are the same (i.e., 1). To ﬁnd the missing term, we need to do the following: ✔ Since all the numbers in sequence (c) are the same, to ﬁnd the next number in the
sequence (b), we need to add this constant value 1 to the last term of sequence (b): 2 + 1=3 ✔ We now add the result obtained in the previous step to the last known term in
sequence (a): 9 + 3 = 12, and this is the missing term we were looking for. Therefore, the completed sequence is:
9, 7, 6, 6, 7, 9,12 Notice that when we initially saw the problem, there was no obvious pattern. However, once we calculated and analyzed the pattern in the numbers obtained by calculating the 74
diﬀerence, we found the logic behind the sequence. That is why the analysis of sequences constructed by calculating the diﬀerences between the terms (e.g., sequence (b) and (c)) is one of the most useful techniques for solving certain categories of sequences. We will discuss this in more detail in "3.3.2 Order of diﬀerence (p. 88)." ♜Note: You will see later that a sequence like this (9, 7, 6, 6, 7, 9) falls under a category of sequences called “quadratic sequences.” We will discuss this in detail later in the section "3.3.2 Order of diﬀerence (p. 88)." Exampled: Let us solve another interesting sequence:
19, 29, 40, 44, ?, 59
Figure 3.9: Diﬀerence between the consecutive terms As shown in the ﬁgure3.9, the numbers obtained by calculating the diﬀerence (10, 11, and 4) between the two consecutive terms of this sequence do not exhibit any visible pattern. Also, the alternate terms (19, 40, ?) and (29, 44, 59) do not indicate any obvious pattern. But the one thing we can say about this sequence is that the numbers are in increasing order as we move from left to right. Let us analyze this sequence: • The ﬁrst term is 19 and the ﬁrst term in sequence (b) is 10, which is also the sum the
two digits of 19: 1 + 9 = 10. • The second term is 29 and the second term in sequence (b) is 11, which is also the
sum the two digits of 29: 2 + 9 = 11. • The third term is 40 and the third term in sequence (b) is 4 and this is also the sum the
two digits of 40: 4 + 0 = 4. • So does it mean the fourth term can be calculated as shown below?
44 + (sum of digits of 44) = 44 + (4 + 4) = 44 + 8 = 52. Let us not jump to the conclusion that 52 is the missing term in sequence (a). We are yet to verify whether the above logic applies to the next term or not. • Following the same logic, the next term should be:
52 + (sum of digits of 52) = 52 + (5 + 2) = 52 + 7 = 59. And this number is indeed the last term in the sequence, which conﬁrms that this is the logic used for generating the terms in this sequence. Hence, the completed sequence is:
19, 29, 40, 44, 52, 59. 75
In this interesting sequence, each term is obtained by adding the sum of the digits in the previous term to the previous term itself. In this example, we have used the two digits, but a diﬀerent problem could involve more digits and apply diﬀerent combinations of arithmetic operations (e.g., multiplication, subtraction, division, etc.) to generate the next term in the sequence. Examplee: Let us review another sequence below:
3, 165, 219, 237, 243, ? We can observe that the numbers in the sequence are in decreasing order as we move from left to right. Also, as shown in the ﬁgure on the right, we have two sequences: • Sequence (a): This is the sequence being analyzed: (3, 165, 219, 237, 243). • Sequence (b): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (a). The terms in this sequence are 162, 54, 18, and 6. One can easily see that each number in sequence (b) is onethird (or
1 times) of the 3
previous number. The last number in sequence (b) is 6 and onethird of 6 is 2. Therefore, the missing number in sequence = 243 + 2 = 245. Hence, the completed sequence is:
3, 165, 219, 237, 243, 245 Examplef: Find the missing term in the following sequence:
5, 4, 4, 2, 4, 1, 3, ?
Figure 3.10: Diﬀerence between the consecutive terms This problem is slightly tricky, as the numbers in the sequence are neither increasing nor decreasing as we move from left to right. The alternate terms (5, 4, 4, 3) and (4, 2, 1) also do not exhibit any obvious pattern, so we can rule out a combined number sequence in this case. Let us examine the diﬀerence between the consecutive terms. As shown in ﬁgure3.10 here, we have two sequences: 76
• Sequence (a): This is the sequence from the example: (5, 4, 4, 2, 4, 1, 3, ?). • Sequence (b): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (a). The terms in this sequence are 1, 0, 2, 2, 3, and 2. There is no obvious pattern in either of the sequences (a) or (b). However, with further analysis, it is possible to discover the logic used in sequence (a), as described below: 51=4
[Subtracting 1 from the ﬁrst term]
4×1=4
[Multiplying 1 with the result of the pervious of step]
42=2
[Subtracting 2 from the result of the pervious of step]
2×2=4
[Multiplying 2 with the result of the pervious of step]
43=1
[Subtracting 3 from the result of the pervious of step]
1×3=3
[Multiplying 3 with the result of the pervious of step]
Hence, to ﬁnd the next two terms, we alternatively perform subtraction and multiplication as explained above. The last known term is 3, which was obtained by multiplying 3 and the previous term 1 together. Hence, the next term can be found by subtracting 4 from the last known term of 3. Therefore, the missing term = 3  4 = 1, and the completed sequence is:
5, 4, 4, 2, 4, 1, 3, 1 This completes our initial discussion on various types of sequences. You should be aware that this topic itself is very vast. There is a lot to learn and practice, so one chapter is not enough to discuss every trick required to solve these kinds of problems. However, we have given you some initial steps with speciﬁc examples of how to analyze such problems. This should give you a solid foundation on how to approach a number sequence problem. Having discussed the diﬀerent types of sequences, we are now prepared to solve the problems in Example1. Example1(a)
1, 1, 2, 3, 5, 8, 13, ? The terms in the above sequence are constantly increasing, but there is no obvious pattern to these increments. Let us examine the diﬀerence between each of the consecutive terms: 11=0 21=1 32=1 53=2 85=3 13  8 = 5 77
Figure 3.11: Diﬀerence between the consecutive terms As shown in ﬁgure3.11, we have two sequences: • Sequence (a): This is the sequence in the example: (1, 1, 2, 3, 5, 8, 13, ?). • Sequence (b): This sequence is constructed with the values obtained by calculating the
diﬀerence between the consecutive terms of sequence (a). The terms in this sequence are 0, 1, 1, 2, 3, and 5. It is easy to notice that except for the ﬁrst term, the numbers in sequence (b) are already present in sequence (a) as shown in ﬁgure3.12 here:
Figure 3.12: Diﬀerence between the consecutive terms Since sequence (b) is constructed with the values of the diﬀerence between the consecutive terms in sequence (a), this means that after the second term, each term can be obtained by adding the previous two terms together. As the last two terms are 8 and 13, the next term will be 8 + 13 = 21, which is the answer to this problem. If you notice, the sequence in the problem is a Fibonacci sequence. Hence the completed sequence is:
1, 1, 2, 3, 5, 8, 13, 21 Example1(b)
1, 2, 2, 4, 3, 6, ? In this problem, the numbers in the sequence are neither increasing nor decreasing. Let us see if we can ﬁnd a pattern in the diﬀerence between each of the consecutive terms: 21=1 22=0 78
42=2 3  4 = 1 63=3 There is no obvious pattern visible in the numbers obtained by calculating the diﬀerence between the consecutive terms. Let us check whether this sequence is a combined number sequence. For that, we will split the sequence into two parts by selecting alternate terms as shown below:
1, 2, 2, 4, 3, 6 , ? ⇒ Two sequences are: 1, 2, 3, ? and 2, 4, 6 Of the two constituting sequences above, the ﬁrst one is a natural number sequence and the second one is an even number sequence. Hence, the alternate terms of the original sequence are coming from two diﬀerent sequences: • Natural Number Sequence : (1, 2, 3, ...) • Even Number Sequence
: (2, 4, 6, ...)
The next number will come from the Natural Number Sequence (1, 2, 3, ...), and that number is 4. Hence, the completed sequence is:
1, 2, 2, 4, 3, 6, 4 Example1(c)
2, 3, 10, ?, 55 In this problem, the terms in the sequence are in increasing order. Let us ﬁnd the diﬀerence between each of the consecutive terms: 32=1 10  3 = 7 The third number is missing in the sequence, and there appears to be no consistent pattern in the increments. Could it be the result of the arithmetic on the two sequences? Let us try prime and Fibonacci numbers: Prime numbers: 2, 3, 5, 7, 11, ... Fibonacci numbers: 1, 1, 2, 3, 5, 8, 11, ... Adding the respective terms: Prime number + Fibonacci number: 2 + 1, 3 + 1, 5 + 2 , 7 + 3, ... this does not match the sequence given in the problem.
⇒ (3, 4, 7, 10, ...), and
Subtracting the respective terms: Prime number  Fibonacci number: 2  1, 3  1, 5  2 , 7  3, 11  5, ...
⇒ (1, 2, 3, 4,
6, ...), and this does not match the sequence given in the problem. Multiplying the respective terms together:
(Prime number) × (Fibonacci number): 2 × 1, 3 × 1, 5 × 2, 7 × 3, 11 × 5, ... ⇒ (2, 3, 10, 21, 79
55, ...), and this resulting sequence matches all the known terms of the sequence in the problem. Hence, this is the sequence we were looking for. Using this result, we can identify the missing term as 21, and the completed sequence will be:
2, 3,10, 21, 55 We have now found the answers to all the problems in Example1, which we will show in the next section.
3.2.2 Answer The answers to the problems given in Example1 are given below: Number Sequence Logitica (a) 1,
1, 2, 3, 5, 8, 13, 21
(b) 1,
2, 2, 4, 3, 6 , 4
(c) 2,
3, 10, 21, 55 Figure 3.13: Answers to the problems in Example1
Number Sequence is an interesting topic to learn. To get a better grasp of the topic, you need to learn the advanced concepts, which we will discuss in the next section.
3.3 Advanced Sequences The aim of a Number Sequence problem is to ﬁnd the missing term. There are no standard methods or set of rules that you can apply to solve these types of problems. Instead, we attempt diﬀerent types of logic and select the one that ﬁts exactly with all the known terms in a sequence. However, to solve these types of problems, some basic steps are recommended as described in the next section.
3.3.1 Steps for solving a Number Sequence problem ✔ Check if the sequence is of a wellknown type: We should ﬁrst check if the sequence
in the problem belongs to a wellknown category, e.g., odd numbers, even numbers, Fibonacci numbers, or prime numbers, etc. In that case, use the standard logic dictated by the sequence to determine its missing term. Consider the following sequence:
(0, 2, 4, 6, 8, ?) This is a sequence containing even numbers. We can ﬁnd the next term in the sequence by adding 2 to the last known term, i.e., the missing term is 8 +2 = 10, and 80
the completed sequence will be:
(0, 2, 4, 6, 8, 10) ✔ Analyze the diﬀerence between the consecutive terms: We should analyze whether
the numbers in the sequence are in increasing or decreasing order. In order to do that, we need to calculate the diﬀerence between each of the consecutive terms to check whether we notice any one of the following patterns: • Are the numbers obtained by calculating the diﬀerence between the consecutive terms constant, increasing or decreasing? • Do numbers obtained by calculating the diﬀerence between the consecutive terms exhibit any pattern? For example, as shown in the ﬁgure on the right, the terms in sequence (a) are 3, 165, 219, 237, and 243. Sequence(b) is built by the values of the diﬀerence between the consecutive terms, and its terms are 162, 54, 18, and 6. Notice how each number in this new sequence is onethird (or
1 times) the 3
previous number. This is the logic that will be applied to ﬁnd the next term in the sequence. In other words, the last known term of sequence (b) is 6, onethird of which is 2. Since sequence (b) is constructed with values of the diﬀerence of the consecutive terms of sequence(a), the missing term is 243 + 2 = 245. • In cases in which the numbers obtained by calculating the diﬀerence between the consecutive terms have no obvious pattern, we may need to analyze the sequence further by calculating the higherorder diﬀerence. We will discuss this in more detail in section Order of diﬀerence (p. 88). ✔ Observe any obvious pattern in the consecutive terms (tn1 and
tn) of the
sequence: We should also check if there is any obvious pattern in how the a term is related to its previous term. Let us review a few examples to explain this: (i) Consider the following sequence:
3, 6, 12, 24, 48, ? As shown below, each term in this sequence is twice the previous term: The n term = tn = 2tn1, , where th
n > 1 and t1 = 3
1st term = t1 = 3 2nd term = t2 = 2t1 = 2 × 3 = 6 3rd term = t3 = 2t2 = 2 × 6 = 12 81
4th term = t4 = 2t3 = 2 × 12 = 24 5th term = t5 = 2t4 = 2 × 24 = 48 Hence, the missing term = 6th term = t6 = 2t5 = 2 × 48 = 96
This is a geometric sequence1, and the formula for the n term in this sequence can th
be written as: tn =
rn1t1. Here r is the common ratio and t1 is the ﬁrst term in the
sequence. In the above example, t1 = 3, r = 2. n1
Hence, the n term = r th
t1 = 2n1× 3 = 3 × 2n1
We can use this formula to ﬁnd the terms of the sequence along with the missing term as shown below: 1st term = t1 = 3 × 2
11
0
= 3×2 =3×1=3
2nd term = t2 = 3 × 2
21
= 3×2 =3×2=6
3rd term = t3 = 3 × 2
31
= 3 × 2 = 3 × 4 = 12
1
2
4th term = t4 = 3 × 2
= 3 × 2 = 3 × 8 = 24
5th term = t5 = 3 × 2
= 3 × 2 = 3 × 16 = 48
41 51
3 4
The missing term = 6th term = t6 = 3 × 2
61
5
= 3 × 2 = 3 × 32 = 96
Hence the completed sequence is:
3, 6, 12, 24, 48, 96 (ii) Let us consider another sequence:
2, 5, 11, 23, 47, ? As shown below, to ﬁnd the current term in the sequence, we need to double the previous term and add 1 to it: The n term = tn = 2tn1 + 1, where th
n > 1 and t1 = 2
2nd term = t2 = 2t1 + 1 = 2 × 2 + 1 = 4 + 1 = 5 3rd term = t3 = 2t2 + 1 = 2 × 5 + 1 = 10 + 1 = 11 4th term = t4 = 2t3 + 1 = 2 × 11 + 1 = 22 + 1 = 23 5th term = t5 = 2t4 + 1 = 2 × 23 + 1 = 46 + 1 = 47 The missing term = 6th term = t6 = 2t5 + 1 = 2 × 47 + 1 = 94 + 1 = 95
1. For more details on geometric sequence refer to AppendixJ (p. 286). 82
Hence the completed sequence is:
2, 5, 11, 23, 47, 95 (iii) Let us consider the following sequence:
1, 4, 14, 45, ? We can ﬁnd the
nth term of this sequence by using the following formula:
The n term = tn = 3tn1 + n, where th
n > 1 and t1 = 1
2nd term = t2 = 3t1 + 1 = 3 × 1 + 1 = 3 + 1 = 4 3rd term = t3 = 3t2 + 2 = 3 × 4 + 2 = 12 + 2 = 14 4th term = t4 = 3t3 + 3 = 3 × 14 + 3 = 42 + 3 = 45 Hence, the missing term = 5th term = t5 = 3t4 + 4 = 3 × 45 + 4 = 135 + 4 = 139.
Hence the completed sequence is:
1, 4, 14, 45, 139 ✔ Verify whether it is a combined sequence: If the sequence is neither increasing nor
decreasing consistently, we should check whether or not it is a combined number sequence. There are exceptions to this rule, so you should also explore other possibilities in case the sequence turns out not to be a combined number sequence. A few examples of such sequences are given in section 3.2.2.2 Some more special sequences (p. 66). ✔ Verify whether the sequence is built by performing a combination of arithmetic
operations on two diﬀerent sequences: We should also analyze whether or not the sequence is the addition, multiplication, or division, etc., of two diﬀerent sequences. For example, a sequence can be built by adding the respective terms of a prime number and Fibonacci number sequences. Here are a few examples to explain this: (i) Let us consider the following sequence:
1, 3, 6, 9, 13, ? This sequence is constructed by adding the terms of an even number sequence (0, 2, 4, 6, 8, ...) to the respective terms of a Fibonacci sequence (1, 1, 2, 3, 5, ...). Since th
th
sequence in the problem is missing it 6 term, we need to add the 6 terms of even number and Fibonacci number sequences to ﬁnd the missing term: Even number sequence : 6th term = 10 Fibonacci sequence
: 6th term = 8
Therefore, the missing term in this sequence is 10 + 8 = 18.
83
Hence the completed sequence is:
1, 3, 6, 9, 13, 13. (ii) Let us consider another sequence:
2, 3, 5, 7, 10, ? The sequence above is built by adding the respective terms of natural number sequence and Fibonacci sequence as shown below. : 1, 2, 3, 4, 5, ...
Natural numbers
Fibonacci numbers : 1, 1, 2, 3, 5, ...
The n term = tn = n + Fn, th
where,
n = 1, 2, 3, ..., and Fn is the nth term in the Fibonacci sequence.
1st term = t1 = 1 + F1 = 1 + 1 = 2 2nd term = t2 = 2 + F2 = 2 + 1 = 3 3rd term = t3 = 3 + F3 = 3 + 2 = 5 4th term = t4 = 4 + F4 = 4 + 3 = 7 5th term = t5 = 5 + F5 = 5 + 5 = 10 Hence, the missing term = 6th term = t6 = 6 + F6 = 6 + 8 = 14.
Hence the completed sequence is:
2, 3, 5, 7, 10, 14 (iii) Let us consider the following sequence:
2, 10, 30, 68, 130, ? This sequence is built by using a combination of arithmetic operations applied to natural numbers as shown below: The n term = tn = th
n(n2+1)
1st term = t1 = 1(1 + 1) = 1(1 + 1) = 1 × 2 = 2 2
2nd term = t2 = 2(2 + 1) = 2(4 + 1) = 2 × 5 = 10 2
3rd term = t3 = 3 (3 + 1) = 3(9 + 1) = 3 × 10 = 30 2
4th term = t4 = 4 (4 + 1) = 4(16 + 1) = 4 × 17 = 68 2
5th term = t5 = 5(5 + 1) = 5(25 + 1) = 5 × 26 = 130. 2
84
Hence, the missing term = 6th term = t6 = 6(6 + 1) = 6(36 + 1) = 6 × 37 = 222. 2
Hence the completed sequence is:
2, 10, 30, 68, 130, 222. Similarly, one can build a completely diﬀerent sequence using another logic: The n term = tn = th
n(n2  1)
The sequence (2, 10, 30, 68, 130, ?) in this example is a cubic sequence, which will be discussed later in section Order of diﬀerence (p. 88). ✔ Check if the sequence is using the digits of its own terms: Sometimes a term in a
sequence can be generated by using the digits of the term prior to it. Let us review this with an example below:
12, 15, 21, 24, 30, ? In the sequence above, the current term is obtained by adding the previous term to the sum of digits the previous term has. This is illustrated below for this sequence: The n term = tn = tn1 + (sum of the digits in tn1), where th
n > 1 and t1 = 12
1st term = t1 = 12 2nd term = t2 = t1 + (sum of the digits in t1) = 12 + (1 + 2) = 12 + 3 = 15 3rd term = t3 = t2 + (sum of the digits in t2) = 15 + (1 + 5) = 15 + 6 = 21 4th term = t4 = t3 + (sum of the digits in t3) = 21 + (2 + 1) = 21 + 3 = 24 5th term = t5 = t4 + (sum of the digits in t4) = 24 + (2 + 4) = 24 + 6 = 30 6th term = t6 = t5 + (sum of the digits in t5) = 30 + (3 + 0) = 30 + 3 = 33.
Hence, the missing term in the sequence is 33, and the completed sequence is:
12, 15, 21, 24, 30, 33. ✔ Check if the terms in the sequence are grouped together in a certain way:
Sometimes a sequence is built by grouping three consecutive terms and applying a combination of arithmetic operations consistently on each group. This can be better explained with an example below:
1, 6, 5, 2, 10, 8, 3, 8, ? In this sequence, we form three groups by putting three consecutive numbers in each group as shown below:
(1, 6, 5), (2, 10, 8), (3, 8, ?) As shown above, we have the following 3 groups 1st group = (1, 6, 5) 85
2nd group = (2, 10, 8) 3rd group = (3, 8, ?)
And, we can observe the following pattern: • In each group, the middle number is the sum of two numbers on either side of it. • The middle number in the 1st group= 1 + 5 = 6 • The middle number in the 2nd group= 2 + 8 = 10 • Continuing on, the middle number in the 3rd group is 8, which is also the sum of 3 and the missing number. Hence the missing number is 8  3 = 5.
Therefore, the completed sequence is:
1, 6, 5, 2, 10, 8, 3, 8, 5 ✔ Check whether a sequence uses diﬀerent combinations of arithmetic operations:
in some cases, a sequence can be constructed by applying various combinations of arithmetic operations (addition, subtraction, division, and multiplication, etc.). Let us look at an interesting problem below:
1, 2, 3, 10, 39, ? In the sequence above, we will learn three diﬀerent ways of ﬁnding the missing term: (i) Let us use the following formula to ﬁnd the each term in this sequence:
tn = (n  1) tn1  1 , whenever n is odd. tn = (n  1) tn1 + 1, whenever n is even. where n>1, t1 = 1
t2 = (2  1) t1 + 1 = 1 × 1 + 1 = 1 + 1 = 2 t3
= (3  1) t2  1 = 2 × 2  1 = 4  1 = 3
t4
= (4  1) t3 + 1 = 3 × 3 + 1 = 9 + 1 = 10
t5
= (5  1) t4  1 = 4 × 10  1 = 40  1 = 39
t6 = (6  1) t5 + 1 = 5 × 39 + 1 = 195 + 1 = 196, and this is the missing term in the sequence.
t7 = (7  1) t6  1 = 6 × 196  1 = 1176  1 = 1175. We will explain later why we went ahead and calculated the 7th term.
(ii) An alternative answer exists for the above sequence as explained below:
tn =tn2 (tn2 + tn1) where n>2, t1 = 1, t2 = 2 t3 = t1(t1 + t2 ) =1(1 + 2) = 1 × 3 = 3 t4 = t2 (t2 + t3 ) = 2 (2 + 3) = 2 × 5 = 10 86
t5 = t3 (t3 + t4 ) = 3(3 + 10) = 3 × 13 = 39 t6 = t4 (t4 + t5 ) = 10(10 + 39)
= 10 × 49 = 490, and this is the missing term in the
sequence.
(iii) Another alternative answer can be obtained by using the formula below:
tn = (n  2)(tn2 + tn1), where n>2, t1 = 1, t2 = 2 t3 = (3  2) (t1 + t2 ) = 1(1 + 2) = 1 × 3
=3
t4 = (4  2) (t2 + t3 ) = 2(2 + 3) = 2 × 5 = 10 t5 = (5  2) (t3 + t4 ) = 3(3 + 10) = 3 × 13 = 39 t6 = (6  2) (t4 + t5 ) = 4(10 + 39) = 4 × 49 = 196, and this is the missing term in the sequence. Did you notice that the 6th term calculated using this method is the same as the one found in method (i)? As shown below the 7th term is diﬀerent, which conﬁrms that these two methods are completely diﬀerent from each other.
t7 = (7  2) (t5 + t6 )
= 5(39 + 196) = 5 × 235 =1175.
The reason we calculated the 7th term was to prove that the two methods ((i) and (iii)) are indeed diﬀerent and would generate two diﬀerent sequences, even though the ﬁrst six terms are exactly the same.
The sequence (1, 2, 3, 10, 39, ?) presented an interesting problem since it demonstrated the types of variations that number sequence problems may have. Notice that in this sequence, we got diﬀerent answers using diﬀerent combinations of arithmetic operations. All the answers were equally compelling as they were derived using simple arithmetic operations and were valid for all the known terms in the sequence. For this reason, there is no clear winner here. However, method (i) used only one term for ﬁnding the next term, while the other two methods needed two previous terms to do the same. Because of this, method (i) may have a slight edge over the others in this case. On the other hand, readers should also be aware that it is quite possible that a question in a particular aptitude test might prefer method (ii) or (iii). Hence, readers should evaluate and diligently select the appropriate method based on the options in the list of possible answers. The steps mentioned here are just basic guidelines. One can learn progressively by attempting to solve more problems from this chapter. Now it is time to learn about an advanced concept called "Order of diﬀerence." Order of diﬀerence is used for analyzing and solving certain types of sequences involving polynomials. We will explore these types of sequences later in the chapter. To keep the discussion on the topic of the "Order of diﬀerence" organized, we have divided our discussion into the following four sections: ✔ 3.3.2 Order of diﬀerence (p. 88): This section contains the deﬁnition of diﬀerent
types of order of diﬀerence. ✔ 3.3.3 Recursive sequence (p. 94): Before we embark on a detailed discussion of the 87
order of diﬀerence, we need to learn about a particular category of sequences called recursive sequence. This section provides an introduction to the diﬀerent types of the recursive sequence using a few examples. ✔ 3.3.4 Explanation of the order of diﬀerence (p. 98): This section explains how to use
the diﬀerent orders of diﬀerence in solving number sequence problems involving polynomials. ✔ 3.3.5 Summary of discussion on the order of diﬀerence (p. 109): This section
summarizes the overall discussion on the order of diﬀerence by outlining the key points we learned on this topic. ♜Note: Consecutive terms in a sequence In the chapter, whenever needed, we will analyze the pattern in the diﬀerence between the consecutive terms. Unless explicitly speciﬁed, when we mention "consecutive terms," we mean "two consecutive terms." Let us now discuss the topic "Order of diﬀerence" to understand what it really means and how it is used in ﬁnding the missing term of a sequence.
3.3.2 Order of diﬀerence As we have seen before, we have to analyze the pattern in the numbers obtained by calculating the diﬀerence between the consecutive terms in a sequence. Sometimes it was even required to analyze the sequence further by ﬁnding the "diﬀerence of the diﬀerence between the consecutive terms." Let us explore this concept bit more by revisiting the two examples: (i) Find the missing term in the following sequence:
1, 8, 15, 22, 29, ?
Figure 3.14: Diﬀerence between the consecutive terms The numbers in this sequence are in increasing order. Let us examine the pattern in the numbers by analyzing the diﬀerence between the consecutive terms. As shown in ﬁgure3.14 here, we have two sequences: • Sequence (a): This is the sequence we are trying to analyze: (1, 8, 15, 22, 29, ?). 88
• Sequence (b): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (a). The terms in this sequence are 7, 7, 7, and 7. We have already discussed this sequence earlier and deduced that all the terms in sequence (b) are 7. We used this information that "all the values in the diﬀerence between the consecutive terms are the same," to ﬁnd the next term in the sequence. As shown in ﬁgure3.14, the missing term = 29 + 7 = 36. Hence, the completed sequence is:
1, 8, 15, 22, 29, 36 (ii) Find the missing term in the following sequence:
1, 3, 7, 13, 21, ?
Figure 3.15: Diﬀerence between the consecutive terms As shown in c, we have three sequences: • Sequence (a): This is the sequence in the problem: (1, 3, 7, 13, 21, ?). • Sequence (b): This sequence is constructed with the values obtained by calculating
the diﬀerence between the consecutive terms of sequence (a). The terms in this sequence are 2, 4, 6, and 8. • Sequence (c): This sequence is constructed by calculating the diﬀerence between the
consecutive terms of sequence (b) and the terms in this sequence are 2, 2, and 2. The numbers in the sequence (a) are in increasing order, so are the numbers in sequence (b). However, all the numbers in sequence (c) are the same (i.e., 2); hence we will use this number to ﬁnd the missing term as explained below: ✔ Since all the terms in sequence (c) are the same (i.e., 2), we add this constant term to
the last known term of sequence (b) to ﬁnd the next term. The last known term of sequence (b) is 8. Therefore, the next term of the sequence (b) is 8 + 2 = 10. ✔ We now add the last term in sequence (b) to the last known term in the sequence (a).
As we have calculated in the previous step, 10 is the last term in sequence (b), and it is the diﬀerence between 21 and the missing term. So, we add 10 to 21 to ﬁnd the missing term, i.e., the missing term is 21 + 10 = 31. 89
As discussed in the previous two examples, it is apparent that there exists a certain type of sequence for which it is essential to study the diﬀerence between the consecutive terms in a much more systematic way. In order to do that we require an indepth understanding of a concept called "order of diﬀerence." Let us ﬁrst deﬁne what we mean by an "order of diﬀerence." ♜Note: Black/White colors used in figures "Order of diﬀerence" In ﬁgure3.14 and ﬁgure3.15, We have used a white box and ovals to indicate the numbers calculated using various orders of diﬀerence.
♛ Definition  Order of diﬀerence Depending on how the diﬀerence is calculated, we can deﬁne the diﬀerent types of the order of diﬀerence as listed below: i. The firstorder diﬀerence: It is the diﬀerence between each of the consecutive terms in a sequence. This is the lowest order of diﬀerence. ii. The secondorder diﬀerence: It is the diﬀerence between each of the consecutive terms in the ﬁrstorder diﬀerence of a sequence. This is the next level after the ﬁrstorder diﬀerence. iii. The thirdorder diﬀerence: It is the diﬀerence between each of the consecutive terms in the secondorder diﬀerence of a sequence. This is the next level after the secondorder diﬀerence. iv. The higherorder diﬀerence: We can ﬁnd the higherorder diﬀerence by calculating the diﬀerence between the consecutive terms in the previousorder diﬀerence of a sequence. ♜Note: (i) As you will see later that unless we ﬁnd the same value for all the terms of an appropriate order of diﬀerence (e.g., ﬁrstorder, secondorder etc.), we cannot use the techniques we will discuss in the "Order of diﬀerence" to solve such sequences. (ii) As we will see in the next example, the numbers obtained by calculating the ﬁrstorder, secondorder, or higherorder diﬀerences can have patterns like any other sequence. Hence we will treat these numbers in the order of diﬀerence as sequences. Let us review the above discussion with an example:
4, 12, 82, 286, 720, 1504, 2782, 4722, ... We have listed below the values of the diﬀerent orders of diﬀerence for this sequence: • the ﬁrstorder diﬀerence : (8, 70, 204, 434, 784, 1278, 1940). • the secondorder diﬀerence : (62, 134, 230, 350, 494, 662). • the thirdorder diﬀerence : (72, 96, 120, 144, 168). 90
• the fourthorder diﬀerence : (24, 24, 24, 24). Notice how all of the terms in this
sequence are the same. We will see later in the chapter that this is a unique property of a quartic sequence.
Figure 3.16: Calculation of various orders of diﬀerence
♛ Definition  Notation used in the order of diﬀerence We will discuss the individual order of diﬀerence with examples in the subsequent sections using the following notations: ✔ Let us assume a sequence of n terms: (t1, t2, t3, t4, t5, ..., tn), where n= 1, 2, 3, ... ✔ We will use dk(tn , tn1, tn2, ..., tnk) to indicate a k order diﬀerence using tn , tn1, tn2, th
..., tnk terms
of the sequence, where the following conditions must be met:
(n  k) > 0, k > 0, n > 1, i.e., n= 2, 3, 4, .... ✔ In other words: • d1(n, n1) is the ﬁrstorder diﬀerence of tn , tn1. • d2(n, n1, n2) is the secondorder diﬀerence of tn , tn1, tn2. • d3(n, n1,
n2, n3) is the thirdorder diﬀerence of tn , tn1, tn2, tn3.
• d4(n, n1,
n2, n3, n4) is the fourthorder diﬀerence of tn , tn1, tn2, tn3, tn4, and
so on. For example: • d1(t3, t2) represents the ﬁrstorder diﬀerence calculated using the terms t3 and t2. 91
• d2(t9, t8, t7) represents the secondorder diﬀerence calculated using the terms t9, t8,
and t7. • d3(t5, t4, t3, t2) represents the thirdorder diﬀerence calculated using the terms
t5, t4,
t3, and t2. Let us derive a formula for each of the orders of diﬀerence. In the discussion below we refer the sequence from ﬁgure3.16. (a) A few terms of the ﬁrstorder diﬀerence: d1(t2, t1) = t2  t1 = 12  4 = 8 d1(t3, t2) = t3  t2 = 82  12 = 70 d1(t4, t3) = t4  t3 = 286  82 = 204 d1(t5, t4) = t5  t4 = 720  286 = 434, and so on. (b) A few terms of the secondorder diﬀerence: d2(t3, t2, t1) = d1(t3, t2)  d1(t2, t1) = (t3  t2)  (t2  t1) = t3  t2  t2 + t1 = t3  t2  t2 + t1 = t3  2t2 + t1 As deﬁned earlier, d2(t3, t2, t1) is the value of the secondorder diﬀerence calculated using t3, t2, and t1. Following the same logic: d2(t4, t3, t2) = t4  2t3 + t2 d2(t5, t4, t3) = t5  2t4 + t3 d2(t6, t5, t4) = t6  2t5 + t4 Let us use the above formula in the sequence from ﬁgure3.16: d2(t3, t2, t1) = d1(t3, t2)  d1(t2, t1) = 70  8 = 62 Alternatively, d2(t3, t2, t1) = t3  2t2 + t1 = 82 2 × 12 + 4 = 86  24 = 62 d2(t4, t3, t2) = d1(t4, t3)  d1(t3, t2) = 204  70 = 134 Alternatively, d2(t4, t3, t2) = t4  2t3 + t2 = 286 2 × 82 + 12 = 298  164 = 134 d2(t5, t4, t3) = d1(t5, t4)  d1(t4, t3) = 434  204 = 230 Alternatively, d2(t4, t3, t2) = t5  2t4 + t3 = 720 2 × 286 + 82 92
= 802  572 = 230, and so on. We have discussed this sequence again in section 3.3.4 Explanation of the order of diﬀerence (p. 98) and explained how to ﬁnd the missing term for this sequence. (c) A few terms of the thirdorder diﬀerence: d3(t4, t3, t2, t1) = d2(t4, t3, t2)  d2(t3, t2, t1) = (t4  2t3 + t2)  (t3  2t2 + t1) = t4  2t3 + t2  t3 + 2t2  t1 = t4  3t3 + 3t2  t1 As deﬁned earlier, d3(t4, t3, t2, t1) is the value of the thirdorder diﬀerence calculated using t4, t3, t2, and t1. Following the same logic: d3(t5, t4, t3, t2) = t5  3t4 + 3t3  t2 d3(t6, t5, t4, t3) = t6  3t5 + 3t4  t3 Let us calculate the values of thirdorder diﬀerence of the sequence in ﬁgure3.16: d3(t4, t3, t2, t1) = d2(t4, t3, t2)  d2(t3, t2, t1) = 134  62 = 72 Alternatively, d3(t4, t3, t2, t1) = t4  3t3 + 3t2  t1 = 286  3 × 82 + 3 × 12  4 = 286  246 + 36  4 = 72 d3(t5, t4, t3, t2) = d2(t5, t4, t3)  d2(t4, t3, t2) = 230  134 = 96, and so on. Alternatively, d3(t5, t4, t3, t2) = t5  3t4 + 3t3  t2 = 720  3 × 286 + 3 × 82  12 = 720  858 + 246  12 = 96 In summary, to obtain a value of the higherorder diﬀerence, we start from the ﬁrstorder diﬀerence and calculate the next order of diﬀerence iteratively. This can be represented as follows: Steps to find the values of various orders of diﬀerence ➜Two consecutive terms of a sequence yield a value of the ﬁrstorder diﬀerence. ➜ Two consecutive values of a ﬁrstorder diﬀerence yield a value of the secondorder diﬀerence. ➜ Two consecutive values of a secondorder diﬀerence yield a value of the thirdorder diﬀerence. ➜ Two consecutive values of a thirdorder diﬀerence yield a value of the fourthorder diﬀerence. ➜ Continuing on, two consecutive values of a (k  1)th order diﬀerence yield a value of the kth order diﬀerence. 93
The question is, do we know in advance how many orders of diﬀerence (e.g., ﬁrstorder or secondorder diﬀerences, etc.) are to be calculated to solve a number sequence problem? We will address this question later in the chapter. The analysis of order of diﬀerence is a powerful technique in solving sequences involving polynomials1, which we will discuss in the next section.
3.3.3 Recursive sequence Before we can discuss in detail about the order of diﬀerence, we need to discuss a special category of sequence called recursive sequence.
♛ Definition  Recursive sequence A recursive sequence is constructed by a process called recursion, in which we apply a combination of arithmetic operations to one or more previous terms of the sequence to generate the next term. This can better be explained with an example:
1, 2, 4, 8, 16, 32, 64 In the sequence above, each term is twice the previous term as shown below: 2=2×1 4=2×2 8 = 2 × 4, and so on. We can use the following formula to express the relationship between the consecutive terms:
tn = 2tn1, where n>1 t1 = 2t2, t2 = 2t3, t4 = 2t3, and so on. ♜Note: The sequence (1, 2, 4, 8, 16, 32, 64) discussed above is an example of a geometric sequence2. It is easy to see that arithmetic, quadratic, cubic, and geometric sequences are of recursive types as explained below: (i) Arithmetic sequence: The
nth term of an arithmetic sequence3 can be expressed as:
The n term = tn = a0 + a1 n th
where
n= 1, 2, 3, ...
Here a0 and a1 are the coeﬃcients that we need to determine and n is the index of term in the sequence. We need two consecutive terms of an arithmetic sequence to determine
1. Refer to AppendixJ (see p. 286) for more details about polynomials of a higher degree. 2. For more details on geometric sequence refer to AppendixJ (p. 286). 3. Refer to AppendixJ (see p. 286) for more details about the arithmetic sequence. 94
these two coeﬃcients. Once we know these coeﬃcients, we can use them in the above formula to determine the missing term in the sequence. We have explained this approach with an example in AppendixJ (p. 286). An arithmetic sequence has the same value of the ﬁrstorder diﬀerence. Let us call this value as d1. We can then write the following formula relating the diﬀerent terms in an arithmetic sequence:
t2 = t1+ d1 t3 = t2 + d1 t4 = t3 + d1 Hence, tn = tn1+ d1 Below is an example of an arithmetic sequence. As shown in the ﬁgure on the right, all the values of the ﬁrstorder diﬀerence are the same:
(5, 9, 13, 17, 21, ...) (ii) Quadratic sequence: The n term of a quadratic sequence1 can be expressed as: th
The n term = tn = a0 + a1 n + a2 n where th
2
n= 1, 2, 3, ...
Here a0, a1 and a2 are the coeﬃcients that we need to determine, and n is the index of term in the sequence. We need three consecutive terms of a quadratic sequence to determine these three coeﬃcients. Once we know these coeﬃcients, we can use them in the above formula to determine the missing term in the sequence. We have explained this approach with an example in AppendixJ (p. 286). A quadratic sequence has the same value of the secondorder diﬀerence. Let us call this value as d2. We can then write the following formula relating the diﬀerent terms in a quadratic sequence:
t3 = d2 t1 + 2t2 t4 = d2  t2 + 2t3 Below is an example of a quadratic sequence:
(1, 6, 16, 31, 51, ...) As shown in the ﬁgure here, all the values of the secondorder diﬀerence are the same.
1. Refer to AppendixJ (see p. 286) for more details about the quadratic sequence. 95
(iii) Cubic sequence: The n term of a cubic sequence1 can be expressed as: th
The n term = tn = a0 + a1 n + a2 n + a3 n where th
2
3
n= 1, 2, 3, ...
Here a0, a1, a2, and a3 are the coeﬃcients that we need to determine, and n is the index of term in the sequence. We need four consecutive terms of a cubic sequence to determine these four coeﬃcients. Once we know these coeﬃcients, we can use them in the above formula to determine the missing term in the sequence. We have explained this approach with an example in AppendixJ (p. 286). A cubic sequence has the same value of the thirdorder diﬀerence. Let us call this value as d3. We can then write the following formula relating the diﬀerent terms in a cubic sequence: As discussed earlier, d3(t4, t3, t2, t1) = t4  3t3 + 3t2  t1 = d3 Hence, t4 = d3 + t1  3t2 + 3t3 Similarly, t5 = d3 + t2  3t3 + 3t4 Below is an example of a cubic sequence:
(2, 13, 48, 119, 238, ...) As shown in the ﬁgure on the right, in a cubic sequence, all the values of the thirdorder diﬀerence are the same. (iv) Quartic sequence: The n term of a th
quartic sequence2 can be expressed as: The n term = tn = a0 + a1 n + a2 n + a3 n + + a4 n where th
Here
2
3
4
n= 1, 2, 3, ...
a0, a1, a2, a3, and a4 are the coeﬃcients that we need to determine, and n is the
index of term in the sequence. We need ﬁve consecutive terms of a quartic sequence to determine these ﬁve coeﬃcients. Once we know these coeﬃcients, we can use them in the above formula to determine the missing term in the sequence. We have explained this approach with an example in AppendixJ (p. 286). A quartic sequence has the same value of the fourthorder diﬀerence. Below is an example of a quartic sequence:
(4, 12, 82, 286, 720, 1504, 2782, 4722, ...)
1. Refer to AppendixJ (see p. 286) for more details about the cubic sequence. 2. Refer to AppendixJ (see p. 286) for more details about the quartic sequence. 96
Figure 3.17: An example of a quartic sequence (v) Geometric sequence: The n term of a geometric sequence1 is shown below: th
The n term = tn = rtn1 = r th
n1
t1
where
n= 1, 2, 3, ...
Here r is the common ratio and t1 is the ﬁrst term of the sequence. We can use the above formula relating the diﬀerent terms in a geometric sequence as shown below:
t2 = rt1 t3 = rt2 = r(rt1) = r2t1 t4 = rt3 = r(r2t1) = r3t1 t5 = rt4 = r(r3t1) = r4t1 tn = rtn1 = r(rtn2) = r2tn2 = r2(rtn3) = r3tn3 = ... = rn1t1 Usually the notation "a" is used for the ﬁrst term in a geometric sequence, hence the formula for the n term is: th
The n term = th
arn1
where
n= 1, 2, 3, ...
Here r is the common ratio and a is the ﬁrst term of the sequence. Let us consider a example of geometric sequence starting from 3 with a common ratio of 2:
(3, 6, 12, 24, 48, 96, ...) (vi) Fibonacci sequence: The n term of a Fibonacci sequence can be written as: th
The n term = Fn = Fn1 + Fn2 where th
n > 2, Fn is the nth term of the sequence.
1. For more details on the geometric sequence, refer to AppendixJ (p. 286). 97
A formula exists to ﬁnd the
nth term of a Fibonacci sequence. Readers can refer to this
formula in AppendixK (p. 303). We will discuss later in this chapter how even the values from the higherorder diﬀerence in a Fibonacci sequence never converge to the same value. Therefore, we cannot use the techniques we discussed in the "order of diﬀerence" to ﬁnd the missing term in this sequence.
3.3.4 Explanation of the order of diﬀerence In this section, we will learn how to use the various orders of diﬀerence to ﬁnd the missing term. After discussing a few sequences involving polynomials, we will review each of the orders of diﬀerence and discuss two ways to ﬁnd the missing term as listed below: (a) Using the appropriate order of diﬀerence: To ﬁnd the missing term, we need to use the value from an appropriate order of diﬀerence, which depends on the type of sequence being analyzed. For example, as we will see later, a quadratic sequence has a property that all the values of secondorder diﬀerence are the same. We will use this property to ﬁnd the ﬁrstorder diﬀerence, which in turn will be used to determine the missing term thereafter. (b) Using a formula for the n term: We can also derive a formula for the n term of the th
th
sequence involving polynomials. To derive this formula, we need the required numbers of consecutive terms based on the type of sequence being analyzed. For example, as explained in AppendixJ (p. 286), to derive the n term: th
✔ two consecutive terms are needed in an arithmetic sequence ✔ three consecutive terms are needed in a quadratic sequence ✔ four consecutive terms are needed in a cubic sequence ✔ ﬁve consecutive terms are needed in a quartic sequence ✔ following the same logic, (k+1) consecutive terms are needed in a sequence involving
a polynomial of k degree. Once we have this formula, we can use the appropriate value of n to ﬁnd the missing term. Now we are ready to explore each order of diﬀerence individually:
i. Firstorder diﬀerence: As deﬁned earlier, the diﬀerence between each of the consecutive terms in a sequence is referred to as the firstorder diﬀerence. An arithmetic sequence has the same value of the ﬁrstorder diﬀerence as expressed below:
t2  t1= t3  t2 = t4  t3 = t5  t4 =.... = tn  tn1= d In order words, d1(2, 1) = d1(3, 2) = d1(4, 3) = .... = d1(n, n1) = d Let us take an example of an arithmetic sequence:
5, 9, 13, 17, 21, ? As shown in the ﬁgure ﬁgure3.18, an arithmetic sequence has the same value of ﬁrst98
order diﬀerence. We can calculate the value of ﬁrstorder diﬀerence as shown below: d1(2, 1) = d1(3, 2) = d1(4, 3) = .... = d1(n, n1) = d Using the above equation: 9  5 = 13  9 = 17  13 = 21  17 = 4 As expected, the values obtained by calculating the ﬁrstorder diﬀerence are the same. We can ﬁnd the missing term in the sequence using the two methods described below: (a) Using firstorder diﬀerence for an arithmetic sequence: To ﬁnd the next term in the arithmetic sequence, we just add the value of ﬁrstorder diﬀerence to the previous term as shown in ﬁgure3.18. So the missing term in the sequence is 21 + 4 = 25. While there is no need to ﬁnd the answer using the diﬀerence shown in ﬁgure3.18, but as you will see later that analyzing the values of the diﬀerence between the consecutive terms will assist greatly in solving sequences involving polynomials of a higher degree.
Figure 3.18: Using the firstorder diﬀerence to find the missing term
(b) Using a formula for the nth term: The formula to determine the
nth term of this arithmetic sequence is derived in
AppendixJ (p. 286). Readers can refer to it for further details. We are going to use this formula to ﬁnd all the terms (including the missing one) of this sequence as shown below:
tn = 4n + 1 Using n = 1: t1 = 4 × 1 + 1 = 4 + 1 = 5 Using n = 2: t2 = 4 × 2 + 1 = 8 + 1 = 9 Using n = 3: t3 = 4 × 3 + 1 = 12 + 1 = 13 Using n = 4: t4 = 4 × 4 + 1 = 16 + 1 = 17 Using n = 5: t5 = 4 × 5 + 1 = 20 + 1 = 21
We use the same formula to ﬁnd the missing term as shown below: Using n = 6: t1 = 4 × 6 + 1 = 24 + 1 = 25. 99
Note that the answer calculated using the formula matches with the answer we found using the value of the ﬁrstorder diﬀerence (refer to ﬁgure3.18). Hence, the completed sequence is:
5, 9, 13, 17, 21, 25 To recap: Referring to the example of the arithmetic sequence from ﬁgure3.18: ✔ The sequence being discussed is (5, 9, 13, 17, 21, 25). ✔ All the values in the ﬁrstorder diﬀerence terms are the same (i.e., 4), which was used
to ﬁnd the missing term. The constantness of the values in ﬁrstorder diﬀerence proves that this sequence is indeed an arithmetic sequence. ✔ Also, you need two consecutive terms of a sequence to determine the value of a ﬁrst
order diﬀerence. Hence, if a sequence has
n terms, you can calculate only (n  1)
values of the ﬁrstorder diﬀerence. In the sequence above, we have ﬁve known terms: 5, 9, 13, 17, and 21. Therefore, we have only four values of the ﬁrstorder diﬀerence, all of which is 4.
ii. Secondorder diﬀerence: The diﬀerence between the consecutive terms of a ﬁrstorder diﬀerence is referred to as secondorder diﬀerence. For a sequence (t1, t2, t3, t4,
t5,..., tn), here are a few terms of secondorder diﬀerence: d2(t3, t2, t1) = d1(t3, t2)  d1(t2, t1) = (t3  t2)  (t2  t1) = t3  2t2 + t1 d2(t4, t3, t2) = d1(t4, t3)  d1(t3, t2) = (t4  t3)  (t3  t2) = t4  2t3 + t2 d2(t5, t4, t3) = d1(t5, t4)  d1(t4, t3) = (t5  t4)  (t4  t3) = t5  2t4 + t3 Continuing on, we can write the secondorder diﬀerence using n , (n  1) , and (n th
th
th
2) as: d2(n, n1, n2) = tn 2 tn1+ tn2 , where n >2 Let us take an example of a quadratic sequence:
1, 6, 16, 31, 51, ? We can ﬁnd the missing term in the sequence using the two methods described below: (a) Using the secondorder diﬀerence for a quadratic sequence: As shown in ﬁgure3.19, we have calculated values of the ﬁrstorder and secondorder diﬀerences of this sequence as summarized below: • The values obtained by calculating the ﬁrstorder diﬀerence are 5, 10, 15, and 20. • The values obtained by calculating the secondorder diﬀerence are 5, 5, and 5, all of
which are the same (i.e, 5). This demonstrates the sequence is indeed a quadratic sequence. We can use the following steps to ﬁnd the missing term in this sequence: ✔ Since all the values of secondorder diﬀerence are the same, we add this value to the 100
last known term of the ﬁrstorder diﬀerence. In this sequence, the value of the secondorder diﬀerence is 5, and the last known value of the ﬁrstorder diﬀerence is 20. Hence, the next value of the ﬁrstorder diﬀerence is 20 + 5 = 25. ✔ As we have previously learned in the discussion of ﬁrstorder diﬀerence, we add the
appropriate value of the ﬁrstorder diﬀerence to the last known term in the sequence to determine its missing term. We have calculated in the previous step that 25 is the ﬁrstorder diﬀerence between 51 and the missing term, hence the missing term in the sequence is 51 + 25 = 76. ✔ In summary, the secondorder diﬀerence is used to determine the value of the ﬁrst
order diﬀerence, and an appropriate value of the ﬁrstorder diﬀerence is used in ﬁnding the missing term in a quadratic sequence.
Figure 3.19: Using the secondorder diﬀerence to find the missing term
(b) Using the formula for the nth term: The formula to ﬁnd the n term of this quadratic sequence is derived in AppendixJ (p. th
286). Readers can refer to it for further details. We are going to use this formula to ﬁnd all
the terms (including the missing one) of this sequence as shown below: Using formula: tn =
5 2 (n n) + 1 2
t1=
5 5 5 × (12  1) + 1 = × (1  1) + 1= × 0 + 1= 0 + 1 = 1 2 2 2
t2 =
5 5 5 × (22  2) + 1 = × (4  2) + 1= × 2 + 1= 5 + 1 = 6 2 2 2
t3 =
5 5 5 × (32  3) + 1 = × (9  3) + 1= × 6 + 1= 5 × 3 + 1= 15 + 1 = 16 2 2 2
t4 =
5 5 5 × (42  4) + 1 = × (16  4) + 1= × 12 + 1= 5 × 6 + 1= 30 + 1 = 31 2 2 2
t5 =
5 5 5 × (52  5) + 1 = × (25  5) + 1= × 20 + 1= 5 × 10 + 1= 50 + 1 = 51 2 2 2
We can use n = 6 in the above formula to ﬁnd the missing term as shown below: 101
t6 =
5 5 5 × (62  6) + 1 = × (36  6) + 1= × 30 + 1= 5 × 15 + 1= 75 + 1 = 76 2 2 2
As expected, the answer found using the formula and shown in ﬁgure3.19 matches. Hence, the completed sequence is:
1, 6, 16, 31, 51, 76 ♜ To recap: Referring to the example of the quadratic sequence from ﬁgure3.19: ✔ The sequence being discussed is (1, 6, 16, 31, 51, ?). ✔ The values obtained by calculating the ﬁrstorder diﬀerence are 5, 10, 15, 20, and 25. ✔ All the values in secondorder diﬀerence are the same number 5. The constantness of
all the values of the secondorder diﬀerence demonstrates that this sequence is indeed a quadratic sequence. ✔ Also, you need three consecutive terms of a sequence to determine one value of the
secondorder diﬀerence. For example, as shown in ﬁgure3.19, the ﬁrst value of the secondorder diﬀerence is obtained by using the ﬁrst three terms 1, 6, and 16. ✔ In other words, if a sequence has
n terms, you can have only (n  2) values of
secondorder diﬀerence. The sequence above has ﬁve known terms: 1, 6, 16, 31, and 51. Hence, we have only 3 values (5, 5, and 5) of secondorder diﬀerence. ✔ Did you notice that the sequence constructed by the values of the ﬁrstorder
diﬀerence in a quadratic sequence is, in fact, an arithmetic sequence?
iii. Thirdorder diﬀerence: The diﬀerence between the consecutive terms of a secondorder diﬀerence in a sequence is referred to as the thirdorder diﬀerence. For a sequence (t1, t2, t3, t4, t5,..., tn), here are the ﬁrst few terms of a thirdorder diﬀerence: d3(t4, t3, t2, t1) = d2(t4, t3, t2)  d2(t3, t2, t1) = (t4  2t3 + t2)  (t3  2t2 + t1) = t4  3t3 + 3t2  t1 Similarly, we can write the followings: d3(t5, t4, t3, t2) = t5  3t4 + 3t3  t2 d3(t6, t5, t4, t3) = t6  3t5 + 3t4  t3 Hence, we can write the thirdorder diﬀerence using tn, tn1, tn2, and tn3 terms as: d3(n, n1, n2, n3) = tn 3 tn1+ 3tn2  tn3 , where n >3 Consider an example of a cubic sequence:
2, 13, 48, 119, 238, ? 102
As shown in ﬁgure3.20, since all the values of the thirdorder diﬀerence are the same, this is a cubic sequence. We have calculated values of the ﬁrstorder, secondorder, and thirdorder diﬀerences of this sequence as summarized below: • The values obtained by calculating the ﬁrstorder diﬀerence are 11, 35, 71, and 119. • The values obtained by calculating the secondorder diﬀerence are 24, 36, and 48. • All the values in the thirdorder diﬀerence are the same (i.e., 12, 12).
Figure 3.20: Using the thirdorder diﬀerence to find the missing term We can ﬁnd the missing term in the sequence using the two methods described below: (a) Using the thirdorder diﬀerence for a cubic sequence: We can use the following steps to ﬁnd the missing term in this cubic sequence: ✔ Since all the values of the thirdorder diﬀerence are the same (i.e., 12), we add this
value to the last known value of the secondorder diﬀerence, which gives us the next value of the secondorder diﬀerence. Hence, the next value of the secondorder diﬀerence is 48 + 12 = 60 ✔ We add the value of the secondorder diﬀerence calculated in the previous step to the
last known value of the ﬁrstorder diﬀerence, which gives us the next value of the ﬁrstorder diﬀerence. Hence, the next value of the ﬁrstorder diﬀerence is 119 + 60 = 179. ✔ We now add the value of the ﬁrstorder diﬀerence calculated in the previous step to
the last known term in the sequence to ﬁnd the missing term. Hence, the missing term is 238 + 179 = 417.
(b) Using the formula for the nth term: We are going to use the formula for the
nth term for this cubic sequence (refer to
AppendixJ (p. 286) for the derivation of this formula): 3
The n term = tn = 2n  3n + 3 th
t1 = 2 × 13  3 × 1 + 3 = 2  3 + 3 = 2 103
t2 = 2 × 23  3 × 2 + 3 = 16  6 + 3 = 13 t3 = 2 × 33  3 × 3 + 3 = 54  9 + 3 = 48 t4 = 2 × 43  3 × 4 + 3 = 128  12 + 3 = 119 t5 = 2 × 53  3 × 5 + 3 = 250  15 + 3 = 238 t6 = 2 × 63  3 × 6 + 3 = 432  18 + 3 = 417. As expected, the answer found using the formula and shown in ﬁgure3.20 matches. Hence, the completed sequence is:
2, 13, 48, 119, 238, 417 ♜ To recap: Referring to the example of the cubic sequence in ﬁgure3.20: ✔ The sequence being discussed is (2, 13, 48, 119, 238, ?). ✔ The ﬁrstorder diﬀerence terms are 11, 35, 71, and 119 ✔ The secondorder diﬀerence terms are 24, 36, and 48. ✔ All the values in the thirdorder diﬀerence are the same (i.e., 12). The constantness of
the values in thirdorder diﬀerence demonstrates that this sequence is indeed a cubic sequence. ✔ Also, you need four consecutive terms of a sequence to determine one value of the
thirdorder diﬀerence. For example, as shown in ﬁgure3.20, the ﬁrst value of the thirdorder diﬀerence is obtained by using the ﬁrst four terms 2, 13, 48, and 119. Using the same logic, If a sequence has n terms, you can calculate only
(n  3) values of third
order diﬀerence. The sequence above has ﬁve known terms: 2, 13, 48, 119, and 238. Hence, we have only two values (12, 12) of thirdorder diﬀerence. ✔ Did you notice that the sequence constructed with the terms of the ﬁrstorder
diﬀerence of a cubic sequence is a quadratic sequence?
iv. Higherorder diﬀerence: Depending on the degree of the polynomial in a sequence, we need to use the higherorders of diﬀerence to ﬁnd the missing term. Consider the following quartic sequence1 (i.e., a polynomial of degree 4):
(4, 12, 82, 286, 720, 1504, 2782, 4722, ?) As shown in ﬁgure3.21, we have calculated ﬁrstorder, secondorder, thirdorder and fourthorder diﬀerences of this sequence as summarized below. The values obtained by calculating: • the ﬁrstorder diﬀerence are 8, 70, 204, 434, 784, 1278, and 1940. • the secondorder diﬀerence are 62, 134, 230, 350, 494, and 662.
1. Refer to AppendixJ for more details about the quartic sequence. 104
• thirdorder diﬀerence are 72, 96, 120, 144, and 168. • thirdfourth diﬀerence are 24, 24, 24, and 24.
Figure 3.21: Using the fourthorder diﬀerence to find the missing term We can ﬁnd the missing term in the sequence using the two methods described below: (a) Using the fourthorder diﬀerence for a quartic sequence: As shown in ﬁgure3.21, all the values of the fourthorder diﬀerence are the same (i.e., 24), which means that this sequence is a quartic sequence. The missing term is determined by using the values obtained from calculating the order diﬀerence iteratively as described below: ✔ Since all the values of the fourthorder diﬀerence are the same (i.e., 24), we can add
this value to the last known value of the thirdorder diﬀerence to obtain the next value for the thirdorder diﬀerence. Hence, the next value of the thirdorder diﬀerence is 168 + 24 = 192. ✔ We add the value of the thirdorder diﬀerence value calculated in the previous step to
the last known term of the secondorder diﬀerence. This will give us the next value of the secondorder diﬀerence. Hence, the next value of the secondorder diﬀerence is 662 + 192 = 854. ✔ We add the value of the secondorder diﬀerence calculated in the previous step to the
last known term of the ﬁrstorder diﬀerence, which gives us the next value of the ﬁrstorder diﬀerence. Hence, that the next value of the ﬁrstorder diﬀerence is 1940 + 854 = 2794. ✔ As we have discussed previously in the discussion of the ﬁrstorder diﬀerence, we add
an appropriate value of the ﬁrstorder diﬀerence to the last known term in the sequence to ﬁnd its missing term. As we have calculated in the previous step, 2794 is 105
the ﬁrstorder diﬀerence between the missing term and last known term of 4722. So, the missing term in the sequence is 4722 + 2794 =7516.
(b) Using the formula for the nth term: We are going to use the following formula for the n term for this quartic sequence (refer th
to AppendixJ (p. 286) for the derivation of this formula):
tn = n4+ 2n3  6n2  3n + 10 Let us verify the above formula for the ﬁrst few terms:
t1 = 14+ 2 × 13  6 × 12  3 × 1 + 10 = 1 + 2 × 1  6 × 1  3 + 10 = 1 + 2  6  3 + 10 = 4
t2 = 24+ 2 × 23  6 × 22  3 × 2 + 10 = 16 + 2 × 8  6 × 4  6 + 10 = 16 + 16  24  6 + 10 = 12
t3 = 34+ 2 × 33  6 × 32  3 × 3 + 10 = 81 + 2 × 27  6 × 9  9 + 10 = 81 + 54  54  9 + 10 = 82 Since the sequence is missing its 9 term, we can use n=9 in the above formula to ﬁnd th
the missing term as shown below:
t9 = 94+ 2 × 93  6 × 92  3 × 9 + 10 = 6561 + 2 × 729  6 × 81  27 + 10 = 6561 + 1458  486  27 + 10 = 7516 As expected, the answer found using formula and shown in ﬁgure3.21 matches. Hence, the completed sequence is:
4, 12, 82, 286, 720, 1504, 2782, 4722, 7516 If you have been paying attention, you will notice that the number of terms in the next order of diﬀerence is one less than the number of the terms in the current order of diﬀerence. Let us try to understand this with the example shown in ﬁgure3.21: • The ﬁrst two terms are 4 and 12, which yield the ﬁrstorder diﬀerence of 8. • The ﬁrst three terms are 4, 12, and 82, which give the secondorder diﬀerence of 62.
The next subsequent three terms (12, 82, and 286) yield the secondorder diﬀerence of 134, and so on. • The ﬁrst four terms are 4, 12, 82, and 286, which yield the thirdorder diﬀerence of 72.
The next four terms are 12, 82, 286, and 720, which yield the thirdorder diﬀerence of 96 and so on. • The ﬁrst ﬁve terms are 4, 12, 82, 286, and 720, which yield the fourthorder diﬀerence
of 24. Since we already know that this is a quartic sequence, all the values of the 106
fourthorder diﬀerence are the same. • In summary, this sequence has 8 known terms, using which we have calculated 7
values of the ﬁrstorder diﬀerence, 6 values of the secondorder diﬀerence, 5 values of the thirdorder diﬀerence, 4 values of the fourthorder diﬀerence. To generalize this discussion, we can say that: • A sequence of n terms has (n  1) values of the ﬁrstorder diﬀerence. • A sequence of n terms has (n  2) values of the secondorder diﬀerence. • A sequence of n terms has (n  3) values of the thirdorder diﬀerence. • Using the same logic, we can conclude that a sequence of
n terms has (n  k)
values of the k order of diﬀerence. th
To recap: Referring to the example of the quartic sequence in ﬁgure3.21: ✔ The sequence being discussed is (4, 12, 82, 286, 720, 1504, 2782, 4722, 7516). ✔ The ﬁrstorder diﬀerence terms are 8, 70, 204, 434, 784, 1278, 1940, and 2794. ✔ The secondorder diﬀerence terms are 62, 134, 230, 350, 494, 662, and 854. ✔ The thirdorder diﬀerence terms are 72, 96, 120, 144, 168, and 192. ✔ All the terms in the fourthorder diﬀerence are the same number 24. The constantness
of all the values in fourthorder diﬀerence demonstrates that this sequence is indeed a quartic sequence. ✔ Did you notice that the sequence constructed with the terms of the ﬁrstorder
diﬀerence of a quartic sequence is a cubic sequence? ✔ Also, a sequence of n terms has (n  k) values for the k order of diﬀerence. th
Fibonacci sequence and Order of diﬀerence Consider the following Fibonacci sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, ? As we can see in ﬁgure3.22, the values from the even higherorders of diﬀerence never converge to the same value. If you notice, the new terms are being introduced (highlighted in black) while the original terms in the sequence are getting pushed out (highlighted in gray). Even if you extend the sequence with more terms, you will ﬁnd out that the values in the higherorder diﬀerence still do not converge into a single value, something that we have observed with arithmetic, quadratic, cubic and quartic sequences. As pointed out earlier on, unless we all the values of an appropriate level of the order of diﬀerence are the same, we cannot use the value of the order of diﬀerence to ﬁnd the missing term. Due to this reason, the techniques we discussed in the "order of diﬀerence" 107
cannot be used to determine the missing term in Fibonacci sequence.
Figure 3.22: The values of higherorder diﬀerence in Fibonacci sequence This brings us to the question: do we know if there is a way to determine how many orders of diﬀerence (e.g., ﬁrstorder, secondorder, or thirdorder, etc.) are needed to analyze a number sequence problem? The answer to this question depends on the type of sequence we are analyzing. As we have discussed before, for certain types of sequences (e.g., Fibonacci sequence) even values of the higherorder diﬀerence do not converge to a constant value. On the other hand, for a sequence involving polynomials, all the values of an appropriate level of the order of diﬀerence are the same. Continuing on, unless we ﬁnd that all the values of an appropriate order of diﬀerence are the same, we cannot use the value of the order of diﬀerence to ﬁnd the missing term. On the other hand, once we ﬁnd that all the values of an appropriate order of diﬀerence are the same, the missing term in a sequence can be found by using this value. For example, the missing term in a cubic sequence can be found by using the appropriate values of the third, second and ﬁrstorder diﬀerences iteratively. However, it is also true that for a given problem we don't know in advance what type of sequence we are dealing with. In other words, just by looking at the sequence, we can't tell whether a sequence is quadratic, cubic, quartic, etc. Hence, there is no way to tell whether the techniques we discussed in the "order of diﬀerence" would be applicable in ﬁnding the missing term in a given problem. But there are some tips that you may ﬁnd useful when solving a number sequence problem: ✔ In practice, we are usually asked to solve arithmetic,
quadratic, cubic or quartic 108
sequences. ✔ Continuing on, it is easy to identify an arithmetic sequence. A few examples of the
arithmetic sequence are listed below: (i) 2, 5, 8, 11, 14, ...
[The diﬀerence between each of the consecutive terms is 3]
(ii) 1, 3, 5, 7, 9, ...
[The diﬀerence between each of the consecutive terms is 2]
(iii) 18, 14, 10, 6, ...
[The diﬀerence between each of the consecutive terms is 4]
✔ Also, we can make the following observations: • The ﬁrstorder diﬀerence of a quadratic sequence yields an arithmetic sequence. • The secondorder diﬀerence of a cubic sequence yields an arithmetic sequence. • The thirdorder diﬀerence of a quartic sequence yields an arithmetic sequence. ✔ Based on the above observations, we can use the following steps to ﬁnd the missing
terms: • Step1: We start from ﬁrstorder diﬀerence and try to ﬁnd a pattern in the values of
the subsequent orders of diﬀerence. • Step2: And, if we cannot see an arithmetic sequence on or before calculating the
thirdorder diﬀerence, either the sequence involves a higher degree of polynomials, which is less common, or it is a sequence that cannot be solved by using the techniques we discussed in section Order of diﬀerence (p. 88). • The Exception to this rule: There are exceptions to this rule, in which you may need
to consider whether or not it is a polynomial sequence of higher degree (5, 6 or 7, etc.), and hence may require further analysis using higherorder diﬀerence. However, these types of polynomials are rarely asked for in a number sequence problem. We now will summarize our discussion on the interesting of the topic "Order of diﬀerence" in the next section.
3.3.5 Summary of discussion on the order of diﬀerence We have summarized below all the key points on the order of diﬀerence discussed in this chapter: (a) Techniques: How to use the "Order of diﬀerence" to find the missing term Analyzing the order of diﬀerence helps us to identify the type of sequence (e.g., arithmetic, quadratic, or cubic, etc.) and provides a simple way of ﬁnding the missing term. As explained earlier, unless all the values of an appropriate order of diﬀerence are the same, we cannot use the techniques discussed in "order of diﬀerence" to ﬁnd the missing term in a sequence. In summary: • We noticed that for certain type of sequences (e.g., arithmetic, quadratic, cubic or
polynomials of a higher degree), all the values of an appropriate order of diﬀerence are the
same as we continued to calculate the higherorder diﬀerence. This means that the missing term of such type of sequences can be found using the value from an appropriate order of diﬀerence. 109
• For Fibonacci sequence, the values obtained by calculating the all the values of diﬀerence
were diﬀerent, even if we continued to calculate the higherorder diﬀerence. This means that this type of sequence cannot be solved using the techniques we discussed in the "order of diﬀerence".
(b)
kthorder diﬀerence
• All the terms in the ﬁrstorder diﬀerence of an arithmetic sequence are the same. • All the terms in the secondorder diﬀerence of the quadratic sequence are the same. • All the terms in the thirdorder diﬀerence of the cubic sequence are the same. • All the terms in the fourthorder diﬀerence of quartic sequence are the same. • Generalizing the above, we can conclude that all the terms in the kthorder diﬀerence of a
sequence involving polynomials of degree k are the same.
(c) Count of terms in diﬀerent orders of diﬀerence A sequence of n terms has: • (n  1) terms of ﬁrstorder diﬀerence. • (n  2) terms of secondorder diﬀerence. • (n  3) terms of thirdorder diﬀerence. • Generalizing the above, we can conclude that a sequence of n terms has (n  k) terms of
the kthorder of diﬀerence.
(d) Key observations The sequence constructed with the terms of the ﬁrstorder diﬀerence of: • a quadratic sequence is an arithmetic sequence. • a cubic sequence is a quadratic sequence. • a quartic sequence is a cubic sequence.
We have also derived a general formula for the nth term for arithmetic and quadratic sequence in AppendixJ (p. 286). Readers can refer to the appendix for a better understanding of how to derive these formulas. This completes our discussion on this interesting topic of "order of diﬀerence." Now we will review a few more sequences before we conclude this chapter.
3.3.6 Examples of a few more sequences In this section, we have provided a summarized list of diﬀerent types of numbers sequences as shown in the table in ﬁgure3.23. We have also added examples of a few more sequences to give a glimpse of how vast this subject is. Whenever needed, we have supplemented the example with hints in the square bracket on how the sequence in the example is constructed. In the discussion below, we will use the following notations: •
n for a natural number. 110
• •
F for a Fibonacci number. p for a prime number.
• We use notations t1, t2, t3, ..., tn to indicate ﬁrst, second, third, ... , n term of a sequence th
and n is the index of the term in a sequence. • Since the index of a sequence is also a natural number (e.g, 1, 2, 3, ...), we will use the
same notation for natural numbers also. Wellknown Number Sequences Integers
...,3, 2, 1, 0, 1, 2, 3, ...
Even numbers
0, 2, 4, 6, 8, 10, ...
Odd numbers
1, 3, 5, 7, 9, 11, ...
Factorial
1, 2, 6, 24, 120, ...
prime numbers
2, 3, 5, 7, 11, ...
Fibonacci numbers
1, 1, 2, 3, 5, 8, 13, 21, ... Square/Cubes
Square of natural numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, .. Cube of natural numbers
1, 8, 27, 64, ... Power or exponent
Using natural numbers
(i) 1, 4, 27, 256, ...
[ nn]
(ii) 1, 2, 9, 64, ...
[n ]
(iii) 1, 1, 3, 16, 125, ...
[n
n1
n2
]
n
[F ]
Using Fibonacci, prime (i) 1, 1, 8, 81, 3125, ... and natural numbers (ii) 2, 3, 25, 343, ...
[pF]
More arithmetics operations (i) 2, 6, 12, 20, 30, ... Arithmetics numbers
on
natural (ii) 0, 2, 6, 12, 20, ...
[n(n+1)] [n(n1)]
(iii) 0, 3, 8, 15, 24, ...
[(n+1)(n1)]
(iv) 6, 20, 42, 72, ...
[2n(2n+1)]
Using Fibonacci, prime (i) 2, 3, 10, 21, 55, ... and natural numbers (ii) 1, 2, 6, 12, 25, ... Some more arithmetic on (i) 4, 7, 10, 13, 16, ... natural numbers (ii) 3, 7, 11, 15, 19, ...
[pF] [nF] [(3n+1) or (tn = tn1 + 3, t1 =4)]
[4n  1]
Addition or subtraction Using Fibonacci, prime (i) 0, 1, 1, 1, 0, 2, 6, ... and natural numbers (ii) 2, 3, 5, 7, 10, 14, ...
[n  F] [F + n] 111
Polynomials (i) 8, 13, 18, 23, 28, ... [ tn = 5n + 3 or add 5 to the last term] Arithmetic sequence
(ii) 2, 1, 4, 7, 10, 13, ... [ tn = 3n  5 or add 3 to the last term] Note: We can use the ﬁrstorder diﬀerence to ﬁnd the missing term.
(i) 4, 11, 22, 37, 56, ... [ tn = 2n2 + n + 1 or add 7, 11, 15, 19, ...] Quadratic sequence
(ii) 1, 9, 27, 55, 93, ...
[tn = 5n2  7n + 3 or add 8, 18, 28, 38, ...]
Note: We can use the secondorder diﬀerence to ﬁnd the missing term.
(i) 1, 4, 25, 76, 169, ... Cubic sequence
[tn = 2n3  3n2 2n + 4]
(ii) 3, 24, 73, 162, 303, ... [tn = 2n3 + 2n2 + n  2] Note: We can use the thirdorder diﬀerence to ﬁnd the missing term.
(i) 4, 35, 176, 553, 1340, ... [tn = 2n4 + n3  n2  3n + 5] Quartic sequence
(ii) 5, 35, 159, 491, 1193, ... [tn = 2n4  n3 + 3n2  2n + 3] Note: We also can use the fourthorder diﬀerence to ﬁnd the missing term.
Grouping of terms (i) 1, 2, 3, 1, 4, 2, 2, 6, ... Hint: The next two numbers are obtained by adding and subtracting two previous numbers as shown below: 2+1=3
Group of numbers
21=1 1+3=4 1  3 = 2, and so on.
(ii) 1, 3, 5, 2, 6, 10, 4, 12, .... Hint: multiply each term in (1, 3, 6) by 2 to get the next 3 numbers
Recursive (i) 5, 14, 41, 122, 365, ... Hint: tn = 3tn1  1, n>1, t1 =5
(ii) 2, 3, 8, 31, 154, 923, ... Recursive sequence
Hint: tn = ntn1  1, n>1, t1 =1 (iii) 444, 147, 48, 15, ... Hint: tn =
tn−1  1 , n>1, t1 =444 3 112
Digits of the terms Arithmetic on digits and (i) 19, 29, 40, 44, ...
terms of the sequence
Hint: tn = tn1 + (sum of the digits in tn1)
where
n > 1 and t1 = 19
e.g., 19 + (1 + 9) = 29, 29 + (2 + 9) = 40 , 40 + (4 + 0) = 44, and so on.
(ii) 14, 18, 26, 38, ... Hint: tn = tn1 + (multiplication of the digits in tn1)
where
n > 1 and t1 = 14.
Figure 3.23: Some more examples of Number Sequence Readers should make a note that there are no limitations on how many diﬀerent types of Number Sequences can be constructed. The above list in the table can be extended further with the diﬀerent combinations of arithmetic operations, all of which cannot be fully listed or summarized in a single chapter. However, you can keep the above table for reference and continue practicing with diﬀerent types of number sequence problems.
3.4 Summary In this chapter, we have done a detailed discussion and analysis on how to solve number sequence problems. We have discussed some of the most important sequences like prime numbers, Fibonacci numbers, odd numbers, even numbers, square numbers, etc. We solved three examples in this chapter. In the ﬁrst problem, we learned how to analyze a consistently increasing sequence by analyzing the diﬀerence between each of the consecutive terms. As it turned out, the sequence given in the problem was a Fibonacci sequence. In the second problem, we had a combined number sequence to solve. We found the solution by identifying the constituents of the sequence. In the third problem, the numbers in the sequence were increasing. However, we could not ﬁnd any pattern that could lead us to the solution. In the end, we discovered that it was a sequence built by multiplying the respective terms of prime number and Fibonacci number sequences. During our discussion on certain categories of sequences, we discussed an interesting concept "Order of diﬀerence." Order of diﬀerence is a systematic way to analyze and solve number sequence problems involving polynomials. We have summarized the key points on this topic in section 3.3.5 Summary of discussion on the order of diﬀerence (p. 109). 113
We have discussed various ways of analyzing a number sequence in this chapter. However, a number sequence in a diﬀerent problem might use a diﬀerent logic other than what we have discussed in this chapter. Hence, to become proﬁcient in solving number sequence problems, you need to practice with a variety of problems. Before we conclude the chapter, you should learn about one more sequence called "Harmonic sequence." Readers can refer to AppendixH (p. 278) for a brief introduction on this. We have enriched this chapter with a variety of interesting problems in the Exercise Section. Readers should attempt to solve these problems to have a better understanding of how to solve number sequence problems.
114
3.5 Exercise Find the missing number in the following problems:
LevelI (i) (iii) (vi) (vii) (ix) (xi) (xiii) (xv) (xvii) (xix)
1, 3, 5, 7, 9, 11, ? 1, 2, 4, 8, 16, 32, ? 1, 4, 9, 16, 25, 36, ? 1, 1, 4, 8, 9, 27, 16, ? 1, 2, 4, 7, 11, 16, ? 7, 15, 31, 63, ?
(ii)
5, 9, 17, 33, ? 2, 10, 30, 68, ? 2, 3, 5, 7, 10, 14, ? 3, 4, 7, 10, 16, ?
(xiv)
(iv) (vi) (viii) (x) (xii)
(xvi) (xviii) (xx)
2, 3, 5, 7, 11, ? 1, 1, 2, 3, 5, 8, ? 1, 2, 3, 4, 5, 8, ? 3, 7, 16, 35, 74, ? 2, 12, 20, 42, ? 11, 13, 17, 25, ?
3, 5, 6, 10, 9, 15, ? 0, 6, 24, 60, ? 3, 5, 8, 11, 16, ? 2, 0, 3, 2, 5, 4, 7, ?
LevelII (ii)
(xxi)
2, 6, 12, 20, 30, ? 2, 10, 30, 68, 130, ? 2, 6, 30, 260, 3130, ? 1, 0, 2, 1, 3, 2, ? 0, 6, 24, 60, 120, ? 2, 3, 10, 21, 55, ? 2, 6, 12, 20, 30, ? 2, 4, 7, 11, 16, ? 2, 6, 21, 88, 445, 2676, ? 1, 0, 3, 2, 5, ? 5, 16, 50, 153, 463, ?
(xxiii)
3, 7, 13, 21, 31, ?
(xxiv)
(i) (iii) (v) (vii) (ix) (xi) (xiii) (xv) (xvii) (xix)
(iv) (vi) (viii) (x) (xii) (xiv) (xvi) (xviii) (xx) (xxii)
1, 4, 27, 256, ? 5, 15, 36, 79, 166, ? 0, 4, 18, 48, 100, ? 1, 2, 1, 3, 2, 5, 3, ? 3, 4, 7, 10, 16, ? 1, 3, 6, 8, 16, 18, ? 1, 1, 9, 256, ? 1, 2, 3, 4, 6, ? 1, 8, 22, 43, 71, ? 1, 1, 4, 27, 625, ?
0, 1, 2, 3, 6, 11, 20, ? 9, 17, 29, 45, 65, ? 115
(xxvi) 95, 86, 73, 56, 35, ? (xxvii) 0, 5, 20, 51, 104, ? (xxviii) 50, 130, 290, 560, (xxix) (xxx) 970, ? (xxv)
1, 4, 25, 76, 169, ? 2, 2, 16, 56, 134, ? 5, 31, 121, 341, 781, 1555, ?
For further practice Please ﬁnd the
(i)
nth term in the following sequence:
2, 12, 36, 80, 150, ....., n
0, 6, 20, 42, 72, 110, ....., n (iii) 0, 4, 18, 48, 100, ....., n (iv) 5, 9, 14, 20, 27, 35, ....., n (v) 3, 4, 5, 6, 8, 10, 12, 16, ....., n. Derive a formula in terms of n, not in (ii)
terms of (n1)th term.
(vi) 2, 4, 5, 7, 8, 9, 11, 12, 13, 14, ....., n. Derive a formula in terms of n.
116
Chapter 4
Marbles in a Box Keywords: Linear Equations.
4.1 Overview In this chapter, we are going to learn how to derive and solve a set of linear equations1. Let us start the chapter with the deﬁnition of "Marbles in a Box." ♛ Definition  Marbles in a Box A problem in this chapter consists of boxes containing marbles of diﬀerent colors (black, white, or gray). Each colored marble is assigned a numerical value. Also, each individual box is labeled with a number, which is the sum of the numerical values of the marbles in that box. The numerical value is only assigned to the marbles, whereas the box itself does not contribute to the number displayed on it. In this chapter, we aim to either ﬁnd the missing numerical value or ﬁnd the combination of marbles in a box based on the given information. Let us continue the discussion with a problem in Example1.
4.2 Example1 Marbles in a Box Logitica Find the missing number.
Figure 4.1: Example1
4.2.1 Analysis There are three boxes given in Example1. Based on the ﬁrst two boxes, we need to ﬁnd the numerical value for the third box. Each of the boxes provided in ﬁgure4.1 has a
1. Refer to AppendixH for more details about linear equations. 117
unique combination of black and white marble(s) as described below: • The ﬁrst box has 2 white marbles and 1 black marble. The sum of this combination of
marbles is 30. • The second box has 2 black marbles and 1 white marble. The sum of this combination of
marbles is 45. • The third box has 1 white and 1 black marble, and we need to determine the numerical
value for this combination of marbles. Based on the above observations, we can derive a set of linear equations by using the combination of marbles in each box. We assume the following symbols to derive the required equations:
w = Value of the white marble b = Value of the black marble From the ﬁrst box : 2w + b = 30 Equation (1) From the second box : w + 2b = 45 Equation (2) Now we have a pair of equations and two variables: w and b. We can solve these equations by following a stepbystep approach as shown below: ✔ Step 1: We multiply both sides of equation (1) by 2 to derive another equation (1a):
2(2w + b) = 30 × 2 4w + 2b = 60
[Multiply both sides by 2] [Equation (1a)]
✔ Step 2: To ﬁnd w, we need to subtract equation (2) from (1a) as shown below:
4w + 2b = 60
w + 2b = 45
[Equation (1a)] [Equation (2)]
Subtracting both sides of the equations: (4w + 2b)  (w + 2b) = 60  45 4w + 2b  w  2b = 15 3w = 15
w=
15 3
w=5 ✔ Step 3: We can substitute the value of w in either of the equations (1) or (2) to ﬁnd the
value of b. Let us take equation (1): 2w + b = 30 2 × 5 + b = 30
[Equation (1)] [Substituting w = 5]
10 + b = 30
b = 30  10 b = 20 118
The calculated values of marbles w and b are shown below:
w=5 b = 20 ✔ Step 4: This is the veriﬁcation step, where we substitute
w = 5, b = 20 in equations (1)
and (2) to verify that the calculated values are correct: Equation (1):
2w + b = 2 × 5 + 1 × 20 = 10 + 20 = 30 Equation (2):
w + 2b = 1 × 5 + 2 × 20 = 5 + 40 = 45 Since the abovecalculated values match with the righthand side of the respective equations, we conclude that the calculated values are correct. ✔ Step 5: We can now determine the value of the combination of marbles in the third
box as shown below:
w + b = 5 + 20 = 25
4.2.2 Answer Here is the answer to the problem given in Example1:
w = Value of the white marble = 5 b = Value of the black marble = 20 Value of the combination of marbles in the third box = 25 Marbles in a Box Logitica
Figure 4.2: Answer to the problem in Example1
In Example1 we had white and black marbles, which were represented by two variables w and b. Therefore, we needed two independent equations1 to ﬁnd the value of each variable. Once we were able to calculate the value of each variable, then determining the numerical value for the combination of marbles in the third box was fairly straightforward. Now we are ready to look at another problem in Example2.
1. Readers can refer to AppendixH (p. 278) for more details about independent equations. 119
4.3 Example2 Marbles in a Box Logitica Find the missing number.
Figure 4.3: Example2
4.3.1 Analysis In Example2 we have four boxes as shown in ﬁgure4.3. Each box has a unique combination of black, white and gray marble(s) as described below: • The ﬁrst box contains 1 gray marble, 2 white marbles, and 1 black marble. The sum of this
combination of marbles is 10. • The second box contains 1 black marble, 1 white marble, and 2 gray marbles. The sum of
this combination of marbles is 9. • The third box contains 1 white marble and 1 gray marble. The sum of this combination of
marbles is 3. • The fourth box contains 1 gray marble, 1 white marble, and 2 black marbles. We need to
ﬁnd the value of this combination of marbles. Based on the above information, we can derive a set of linear equations by using the combination of marbles in each box. We assume the following symbols to formulate the required equations:
w = Value of white marble b = Value of black marble g = Value of gray marble We can now derive the following equations: From the ﬁrst box: From the second box: From the third box:
g + 2w + b = 10 2g + w + b = 9 g+w=3
Equation (1) Equation (2) Equation (3)
Now we have three equations (1), (2), and (3) that can be used to determine the values of the three variables (w, g, and b) as shown below: ✔ Step 1: We derive a new equation (1a) by subtracting equation (2) from (1):
g + 2w + b = 10 2g + w + b = 9
[Equation (1)] [Equation (2)]
Subtracting both sides of the equations: 120
(g + 2w + b)  (2g + w + b) = 10  9
g + 2w + b  2g  w  b = 1 wg=1
Equation (1a)
✔ Step 2: Now we can add equation (1a) and (3) to determine the value of w:
wg=1
[Equation (1a)]
g+w=3
[Equation (3)]
By adding both sides of the equations (1a) and (3) together, we get:
wg +g+w=1+3 2w = 4
w=
4 2
w=2 ✔ Step 3: We can substitute the value of
w in equation (3) to calculate the value of g as
shown below:
g+w=3 g+2=3
[Equation (3)] [Substituting w = 2]
g=32 g=1 ✔ Step 4: We can use the values of w and g in either equation (1) or (2) to ﬁnd the value
of b. Let us use equation (1):
g + 2w + b = 10 1 + 2 × 2 + b = 10 1 + 4 + b = 10 5 + b = 10
[Equation (1)] [Using g = 1, w = 2]
b = 10  5 b=5 Now that we have found the values of all the marbles (black, white, and gray), we can summarize them as follows:
g=1 w=2 b=5 ✔ Step 5: Before we calculate the missing value in the fourth box, let us verify the value
of each marble in each of the equations: Equation (1):
g + 2w + b = 1 + 2 × 2 + 5 = 1 + 4 + 5 = 10 Equation (2):
2g + w + b = 2 × 1 + 2 + 5 = 2 + 2 + 5 = 9 121
Equation (3):
g+w=1+2=1+2 =3 Since the abovecalculated values match with the righthand side of the respective equations, we conclude that our calculated values are correct. ✔ Step 6: As we know the numerical value for each colored marble, we can now
determine the missing value of the fourth box. The fourth box contains 1 gray marble, 1 white marble, and 2 black marbles. Hence, the value of the fourth box is g + w + 2b = 1 + 2 + 2 × 5 = 1 + 2 + 10 = 13
4.3.2 Answer Here is the answer to the problem given in Example2: Value of gray marble = 1 Value of white marble = 2 Value of black marble = 5 Value of the combination of marbles in the fourth box = 13 Marbles in a Box Logitica
Figure 4.4: Answer to the problem Example2
In the section above, we have found the answer by using a stepbystep approach. However, sometimes it is possible to ﬁnd the answer with fewer calculations. Let us try to solve Example2 using two methods to demonstrate that sometimes such optimization is possible. ♜Note: Method1 : A quick way to solve the problem If you analyze the combinations of marbles carefully, the problem can be solved with fewer calculations as shown below: ✔ Step 1: Let us list the previously derived equations:
g + 2w + b = 10 [Equation (1)] 2g + w + b = 9 [Equation (2)] g + w = 3 [Equation (3)] 122
✔ Step 2: If we add equation (1) and (2), we can ﬁnd the value of b as shown below:
(g + 2w + b) + (2g + w + b) = 10 + 9 3g + 3w + 2b = 19 3(g + w) + 2b = 19 3 × 3 + 2b = 19
[Using equation (3) : g + w = 3]
2b = 19  9 = 10
b=
10 2
b=5 ✔ Step 3: We use the value of
b and equation (3) in (2) to ﬁnd the value of g as shown
below: 2g + w + b = 9
[Equation (2)]
g + g + w + 5 = 9 [Expanding 2g = g + g, b = 5] g + 3 + 5 = 9 [Using equation(3): g + w = 3] g+8=9 g=98 g=1 ✔ Step 4: To ﬁnd the value of w, we substitute the value of g in equation (3)
g + w = 3 [Equation (3)] 1 + w = 3 [Using g = 1] w=31 w=2 ✔ Step 5: We have found all the values :
g=1 w=2 b=5 This means value of the fourth box is :
g + w + 2b = 1 + 2 + 2 × 5 = 1 + 2 + 10 = 13 ♜Note: MethodII : Another quick way to solve the problem Occasionally it is possible to utilize combinations of marbles to ﬁnd the answer without even needing to ﬁnd the value of each type of marbles. If we analyze the problem in ﬁgure4.3 further, we can identify a more eﬃcient way to ﬁnd the value of the fourth box as described below: ✔ Step 1: Let us list all the equations:
For the ﬁrst box: For the second box:
g + 2w + b = 10 2g + w + b = 9
Equation (1) Equation (2) 123
g+w=3
For the third box:
Equation (3)
✔ Step 2: We have the following expression that represents the value of the fourth box
that we need to determine:
g + w + 2b = ? ✔ Step 3: We can multiply both sides of equation (3) by 2 to derive equation (3a):
g + w = 3 [Equation (3)] Multiply both sides by 2: 2(g + w) = 2 × 3 2g + 2w = 6
Equation (3a)
✔ Step 4: We can derive equation (4) by adding equation (1) and (2):
g + 2w + b = 10 [Equation (1)] 2g + w + b = 9 [Equation (2)] Adding both equations:
(g + 2w + b) + (2g + w + b) = 10 + 9 3g + 3w + 2b = 19 Equation (4) ✔ Step 5: We can derive equation (5) by subtracting equation (3a) from (4):
3g + 3w + 2b = 19 2g + 2w = 6
[Equation (4)] [Equation (3a)]
Subtracting both sides:
(3g + 3w + 2b)  (2g + 2w) = 19  6 g + w + 2b = 13
Equation (5)
The left side of equation (5) represents the combination of marbles in the fourth box (refer to step2). Hence, the right side of this equation is the answer we were looking for. Notice how we have cleverly used the combinations of marbles to ﬁnd answers with fewer calculations without even needing to ﬁnd the values of each type of marble. Note that such optimizations of solving problems in fewer steps depend on the combinations of marbles in the problem and sometimes such optimizations are not possible at all. In the above two methods, we utilized combinations of marbles to solve the problem with fewer calculations. Readers are advised ﬁrst to analyze the problem to spot any such opportunities, which can be helpful in solving problems with fewer calculations. However, we provided a detailed stepbystep approach earlier in this chapter to ensure that we learn the most common way of solving such problems, just in case a shorter approach is not possible in a given problem.
124
4.4 Summary In this chapter, we used two examples. Example1 contains two colored marbles (white and black), so we needed two independent equations to solve the problem. On the other hand, Example2 contains three colored marbles (gray, white, and black). Therefore, we required three independent equations1 to solve the problem. In general, to ﬁnd values of N variables, you need N independent equations. There are two ways of solving a set of linear equations: elimination and substitution2. In this chapter, whenever appropriate, we have preferred one over the other in order to solve the equations. We can modify a "Marbles in a Box" problem to make it more interesting, as we have done in the ﬁgure below in which the third box is missing its marbles. The objective here is to ﬁnd the combination of marbles using the number indicated on the box. In some cases, more than one answer is possible.
Figure 4.5: Marbles in a Box
Readers are advised to try and solve the problems provided in the Exercise Section of this chapter in order to improve their understanding of concepts in this chapter.
1. Refer to AppendixH (p. 278) for further details on independent equations. 2. Readers can refer to AppendixH (p. 278) for more details on the two methods of solving
linear equations: elimination and substitution. 125
4.5 Exercise Find the missing numbers in the following problems:
LevelI
(i)
(ii)
(iii)
(iv)
(v)
(vi)
126
LevelII (i)
(ii)
(iii)
(iv)
(v)
(vi)
For further practice
(i)
Find the combination(s) of marbles in the third box. (ii)
127
Find the combination(s) of marbles in the fourth box.
(iii)
128
Chapter 5
Brick Wall Keywords: Linear Equations.
5.1 Overview In this chapter, we need to ﬁnd the missing numerical values in a brick wall. Let us ﬁrst deﬁne what we mean by a Brick Wall problem. ♛ Definition  Brick Wall A Brick Wall problem contains a wall with seven bricks: top, bottom, left, right, and three middle bricks. Each of the bricks has been assigned with a numerical value. The numerical values of top, bottom, left, and right bricks depend on the inner middle bricks (A, B, and C) as listed below: Top brick
= A+B+C
Bottom brick = A  B + C Left brick
= A+B
Right brick
= B+C
♜Note: A variation of this problem may contain two walls, similar to what is shown in Example2. In a Brick Wall problem, the numerical values of some of the bricks are given while the values of the others are missing. We need to ﬁnd the missing values using the conditions deﬁned above and the numbers displayed on the bricks. As we will discuss later in the chapter, we need to derive a set of linear equations1 and solve them using a stepbystep approach. It is easy to notice that once we have the values of all the middle bricks (A, B, and C), the values of the top, bottom, left and right bricks can easily be determined using the composition of a wall as deﬁned above.
1. Refer to AppendixH (p. 278) for more details about linear equations. 129
Let us start the chapter with a problem in Example1.
5.2 Example1 Brick Wall Logitica Find the numerical values of the right and middle bricks (A, B, and C).
Figure 5.1: Example1
5.2.1 Analysis As shown in ﬁgure5.1, we need to ﬁnd the value of the right and middle bricks. Once we ﬁnd the values of middle bricks (A, B, and C), the value of the right brick can be determined easily by adding the values of B and C. Since we have three variables A, B, and C in this problem, we need three independent equations1 to determine the values of these three variables. We derive the following three equations using the conditions speciﬁed in Example1: Top brick
: A + B + C = 18
Equation (1)
Bottom brick : A  B + C = 8
Equation (2)
Left brick
Equation (3)
:
A+B=7
We now have three equations (1), (2) and (3), and three variables A, B, and C. We follow a stepbystep approach to solve these three equations to determine the values of variables A, B, and C: ✔ Step 1: We can use equation (1) and (3) to ﬁnd the value of C:
A + B + C = 18 7 + C = 18
[From Equation (1)] [A + B = 7 from equation (3)]
C = 18  7 C = 11 ✔ Step 2: We now can substitute the value of C in equation (2) to derive another
equation (2a) containing the variables A and B:
1. Refer to AppendixH (p. 278) for more details about independent equations. 130
AB+C=8
[Equation (2)]
A  B + 11 = 8
[Using C = 11]
A  B = 8  11 A  B = 3
Equation (2a)
✔ Step 3: We now have two equations (2a) and (3) that contain only two variables: A and
B. We can add these two equations together to determine the value of A as shown below: A+B=7
[Equation (3)]
A  B = 3
[Equation (2a)]
Adding both sides of equations (2a) and (3): (A + B) + (A  B) = 7  3 2A = 4 A=
4 2
A=2 ✔ Step 4: We can use either equation (3) or (2a) to determine the value of B. Let us
choose equation (3). As shown below, we can determine the value of B by substituting the value of A in equation (3): A+B=7
[Equation (3)]
2+B=7
[Using A = 2]
B=72 B=5 ✔ Step 5: We now have found values for variables A, B, and C. We should ﬁrst verify that
the calculated values are correct using the conditions of Example1: A = 2, B = 5, C = 11 Value of the top brick = A + B + C = 2 + 5 + 11 = 18 Value of the bottom brick = A  B + C = 2  5 + 11 = 8 Value of the left side brick = A + B = 2 + 5 = 7 All the values calculated above match with each of the bricks (top, bottom, and left) as shown in Example1. ✔ Step 6: We now can use the value of B and C to calculate the value of rightside brick.
The value of the rightside brick = B + C = 11 + 5 = 16 In the above steps, along with ﬁnding the values for the bricks A, B, C, and rightside brick, we have also veriﬁed that the calculated values are correct. Now we are ready to display the answer in the next section.
131
5.2.2 Answer The answer to the problem given in Example1 is displayed below: A = 2, B = 5, C = 11, and the value of the brick on the right = 16 Brick Wall Logitica
Figure 5.2: Answer to the problem given in Example1 The problem in Example1 was a simple problem with one wall. In the next example, we will make the problem more interesting by adding another wall.
5.3 Example2 Brick Wall Logitica In this problem, we have two walls: one on the leftside and another on the rightside. These two walls are joined by a common brick in the middle, which gives the same value to both sides of the wall: B + C = A + D. Find the values for the bricks A, B, C, D, and E and all the missing numbers.
Figure 5.3: Example2
132
5.3.1 Analysis In this problem, we have two walls: a leftside and rightside. The composition of the bricks on each wall is listed below: • The wall on the left contains the following bricks:
Middle bricks : A, B, C Top brick
:A+B+C
Bottom brick : A  B + C Left brick
:A+B
Right brick
:B+C
• The wall on the right contains the following bricks:
Middle bricks : A, D, E Top brick
:A+D+E
Bottom brick : A  D + E Left brick
:A+D
Right brick
:D+E
As stated in the problem, the following equation represents that fact that the common brick between the two walls has the same value: B+C=A+D It is easy to derive the following ﬁve equations based on the conditions given in the problem: A + B + C = 10
Equation (1)
A B+C=6
Equation (2)
A + D + E = 20
Equation (3)
A  D + E = 12
Equation (4)
B+C=A+D
Equation (5)
Given that we now have ﬁve independent equations and ﬁve variables (A, B, C, D and E), we can solve these equations to determine the value of these variables. A stepbystep approach to solve the above equations is provided below: ✔ Step 1: We can ﬁnd the value of B by subtracting equation (2) from (1) as shown
below: A + B + C = 10
[Equation (1)]
A B+C=6
[Equation (2)]
Subtracting equation (2) from equation (1) : (A + B + C)  (A  B + C) = 10  6 A+B+CA+ BC =4 2B = 4
133
B=
4 2
B=2 ✔ Step 2: We can ﬁnd the value of D by subtracting equation (4) from equation (3):
A + D + E = 20
[Equation (3)]
A  D + E = 12
[Equation (4)]
Subtracting equation (4) from (3): (A + D + E)  (A  D + E) = 20  12 A+D+EA+DE =8 2D = 8 D=
8 2
D=4 ✔ Step 3: We can use the value of B and D in equation (5) to derive equation (6)
containing variables A and C: B+C=A+D 2+C=A+4
[Equation (5)] [Using B = 2, D = 4]
CA=42 CA=2
Equation (6)
✔ Step 4: We can also derive another equation (7) containing A and C by using value of B
in equation (1): A + B + C = 10 A + 2 + C = 10
[Equation (1)] [Using B = 2]
A + C = 10  2 A+C=8
Equation (7)
✔ Step 5: Since the two equations (6) and (7) contain only two variables, A and C, we can
use these two equations to ﬁnd the values of A and C: A+C=8
[Equation (7)]
C A=2
[Equation (6)]
Adding both sides of the equations together: (A + C) + (C  A) = 8 + 2 2C = 10 C=
10 2
C=5
134
Substituting the value of C in equation (6) to determine the value of A: C A=2
[Equation (6)]
5A=2
[Using C = 5]
A=52 A=3 ✔ Step 6: We can calculate the value of E using the values of A and D in equation (3):
A + D + E = 20 3 + 4 + E = 20
[Equation (3)] [Using A = 3, D = 4]
E = 20  7 E = 13 ✔ Step 7: We summarize below the values of all variables used in the problem:
A = 3, B = 2, C = 5, D = 4, E = 13 ✔ Step 8: Let us verify whether the values of variables are correct using the conditions
of Example2: The leftside wall Top brick = A + B + C = 3 + 2 + 5 = 10 Bottom brick = A  B + C = 3  2 + 5 = 6 The rightside wall Top brick = A + D + E = 3 + 4 + 13 = 20 Bottom brick = A  D + E = 3  4 + 13 = 12 All the values calculated above match with top and bottom for both the walls shown in Example2. The common brick The common brick on leftside wall = B + C = 2 + 5 = 7 The common brick on rightside wall = A + D = 3 + 4 = 7 As expected, the common brick has the same value. ✔ Step 9: Using the values of A, B, C, D, and E, we can now determine the values of the
right and left bricks on each wall as shown below: The leftside wall: Left brick = A + B = 3 + 2 = 5 Right brick = B + C = 2 + 5 = 7 The rightside wall: Left brick = A + D = 3 + 4 = 7 Right brick = D + E = 4 + 13 = 17 In the above steps, along with ﬁnding the missing values, we have also veriﬁed that the calculated answer is correct, which we will display in the next section. 135
5.3.2 Answer Here is the answer to the problem in Example2: A = 3 , B = 2, C = 5 D = 4, E = 13 Brick Wall Logitica
Figure 5.4: Answer to the problem given in Example2 ♜Note: Another quick way to solve We are going to provide a quick way to solve the problem. let us ﬁrst start with the ﬁve equations we deﬁned for the problem in Example2: A + B + C = 10
Equation (1)
A B+C=6
Equation (2)
A + D + E = 20
Equation (3)
A  D + E = 12
Equation (4)
B+C=A+D
Equation (5)
The ﬁrst two steps are the same ones we used before in solving the problem. We have repeated them below for completeness: ✔ Step 1: We can ﬁnd the value of B by subtracting equation (2) from (1) as shown
below: A + B + C = 10
[Equation (1)]
A B+C=6
[Equation (2)]
Subtracting equation (2) from equation (1) : (A + B + C)  (A  B + C) = 10  6 136
A+B+CA+ BC =4 2B = 4 B=
4 2
B=2 ✔ Step 2: We can ﬁnd the value of D by subtracting equation (4) from equation (3):
A + D + E = 20
[Equation (3)]
A  D + E = 12
[Equation (4)]
Subtracting equation (4) from equation (3): (A + D + E)  (A  D + E) = 20  12 A+D+EA+DE =8 2D = 8 D=
8 2
D=4 ✔ Remaining Steps: We have calculated the values of B and D in the previous steps.
Now we are going to determine the values of the remaining variables: A, C, and E. For this let us analyze the equations again. We will ﬁrst substitute the value of D in equation (5) to derive another equation, (5a) as shown below: B+C=A+D
[Equation (5)]
Using D = 4 in equation (5): B+C=A+4
Equation (5a)
Now we use equation (5a) in (1) to determine the value of A as shown below: A + B + C = 10 A + (A + 4) = 10
[Equation (1)]
[Using B + C = A + 4 from equation (5a)]
2A + 4 = 10 2A = 10  4 = 6 A=
6 =3 2
Using the values of A and D in equation(3), we can calculate the value of E as shown below: A + D + E = 20 3 + 4 + E = 20
[Equation (3)] [Using A = 3, D = 4]
E = 20  7 E = 13
137
Substituting the values of A and B in equation (1), we can ﬁnd the value of C: A + B + C = 10
[Equation (1)]
3 + 2 + C = 10
[Using A = 3, B = 2]
C = 10  5 = 5 With these steps, we have found the values of all variables: A = 3, B = 2, C = 5, D = 4, E =13. Notice how we have utilized the positions and known values of the bricks to solve the problem with fewer calculations. Readers are advised to spot such opportunities and use them whenever possible. However, we have provided a detailed stepbystep approach earlier in the chapter to ensure that we learn the most common way of solving such problems, just in case no shorter approach is possible in a given problem.
5.4 Summary In this chapter, we used the composition of the wall and the conditions on individual bricks to derive the linear equations. These equations were solved systematically by using a stepbystep approach. We used two examples to provide an indepth understanding of this chapter. The problem in Example1 involved a single wall, whereas the problem in Example2 involved two walls. In both examples, we followed a stepbystep approach to determine the missing values. There are two ways of solving a set of linear equations: elimination and substitution1. In this chapter, whenever appropriate, we have preferred one over the other in order to solve the equations. The reader is advised that it would be best to practice the problems provided in the Exercise Section for a better understanding of solving problems in this chapter.
1. Readers can refer to AppendixH (p. 278) for more details about the elimination and
substitution methods used in solving linear equations. 138
5.5 Exercise Find the value of missing numbers in the following problems:
LevelI
(i)
(ii)
(iii)
(iv)
(v)
(vi)
139
LevelII Find the value of missing numbers in the following problems:
(i)
(ii)
140
(iii)
(iv)
141
For further practice You need to build the wall by selecting 7 numbers from the list given below in two questions. The number on each brick must be unique, and bricks must follow the relationship deﬁned in the deﬁnition section. Find all the possible answers using the numbers given below. (i) 1,
2, 3, 5, 2, 8, 6, 0, 7, 9, 1
♜Hint: For example, one of the possible combinations are: middle bricks=(2, 1, 6), top brick=7, bottom brick= 9, left brick=1, right brick= 5. Note that all the 7 numbers (2, 1, 6, 7, 9, 1, 5) are taken from the list of the numbers given above.
(ii) 4,
7, 3, 14, 0, 11, 10, 5, 16, 2, 18, 23.
142
Chapter 6
Average Cell Keywords: Linear Equations, Fractions, Arithmetic Mean.
6.1 Overview In this chapter, we will learn how to derive a set of linear equations1 in an interesting way, and use these equations to solve problems involving Average Cells. Let us ﬁrst deﬁne what we mean by an Average Cell. ♛ Definition  Average Cell In a cluster of cells, any cell that has more than one cell adjacent to it (above, below, left, or right), would be an Average Cell. The number indicated in the Average Cell is the average of all the numbers in the adjacent cells. We will use the number count of the adjacent cells as the divisor in calculating the average. Diagonal cells are not considered to be adjacent cells when solving problems in this chapter. In this chapter, we aim to ﬁnd the numerical value of each cell in a cluster of cells arranged in a particular way, similar to what is shown in ﬁgure6.1. Let us start the discussion with a problem in Example1.
6.2 Example1 Average Cell Logitica Find the numerical values of A, B, and C. ♜Note: Cell6C is not an Average Cell, as it has only one adjacent cell, which is cellC.
Figure 6.1: Example1
1. Refer to AppendixH for more details about linear equations. 143
6.2.1 Analysis To solve the problem, we ﬁrst need to identify which cells should be selected for calculating the average. According to the deﬁnition, an Average Cell contains the average of the numbers in the cells adjacent to it. As shown below in ﬁgure6.2, we can quickly identify three Average Cells in the problem. To indicate which cells we have selected, and the variable inside that will be used in calculating the average, we have marked them with circles in the images below. A black circle is used to highlight the variable inside the Average Cell in each of the images below:
B=
A+C 2
C=
B + 6C + A + B A + 2B + 6C = 4 4
A=
C−8 2
Figure 6.2: The cells selected for calculating the averages Based on ﬁgure6.2, we can derive three equations as described below: First image: In this image, B is the Average Cell, and it is the average of the values of A and C as expressed in the equation below:
A+C =B 2 A + C = 2B A  2B + C = 0
Equation (1)
Second image: In this image, C is the Average Cell, and it is the average of B, 6C, A, and B as shown in the equation below:
B + 6C + A + B =C 4 A + 2B + 6C =C 4 A + 2B + 6C = 4C A + 2B + 6C  4C = 0 A + 2B + 2C = 0
Equation (2)
144
Third image: In this image, A is the Average Cell, which is the average of C and 8 as shown in the equation below:
C−8 =A 2 C  8 = 2A 2A  C = 8
Equation (3)
We have listed down the derived equations below for legibility: A  2B + C = 0
Equation (1)
A + 2B + 2C = 0
Equation (2)
2A  C = 8
Equation (3)
We now have three independent equations that contain three variables: A, B, and C. We have laid down a stepbystep approach to ﬁnd the values of A, B, and C: ✔ Step 1: We can derive a new equation (4) containing variables A and C by adding
equations (1) to (2) as shown below: A  2B + C = 0
[Equation (1)]
A + 2B + 2C = 0
[Equation (2)]
Adding both sides of equations (1) and (2): (A  2B + C) + (A + 2B + 2C) = 0 + 0 2A + 3C = 0 3C = 2A C= −
2A 3
Equation (4)
Step 2: We use the relationship between A and C from equation (4) in (3) to calculate the value of A: 2A  C = 8 ⎛ 2A ⎞ 2A  ⎜ − ⎟ = 8 ⎝ 3 ⎠
[Equation (3)] [From Equation (4): C = −
2A ] 3
3(2A) + 2A = 8 3
6A + 2A = (8) × 3 8A = 24 A=
−24 8
A = 3 145
✔ Step 3: We can substitute the value of C in equation (4) to ﬁnd the value of A:
C= − =− =
2A 3
Equation (4)
2 × (−3) [Using A = 3 ] 3
6 3
=2 ✔ Step 4: By using the values of A and C in equation (1), we can ﬁnd the value of B:
A  2B + C = 0
[Equation (1)]
3  2B + 2 = 0 1  2B = 0 2B = 1 B= −
[Using A = 3, C = 2]
1 2
✔ Step 5: Now that we have calculated the values of all the variables, we can summarize
them below: A = 3 B= −
1 2
C=2 ✔ Step 6: This is the veriﬁcation step, where we will verify the calculated values of A, B,
and C using the deﬁnition of the Average Cell. We will use ﬁgure6.3 to verify our answers.
Figure 6.3: The cells selected for calculating the averages We will calculate the value of each of the Average Cells by averaging the numerical values of the adjacent cells. If we have derived and solved the equations correctly, the numerical value of each Average Cell will be equal to the average of the adjacent cells. We will use the values of the variables A, B and C in the table shown in ﬁgure6.4 to verify our 146
calculations.
1 2
A = 3, B = − , C = 2
First image
Second image
Third image
Value of the Average Cell = B
Value of the Average Cell = C
Value of the Average Cell = A
C=2
A = 3
B= −
B=
A+C 2
=
−3+ 2 2
1 2 C=
=
1 =− 2 This proves the answer.
=
=
A + 2B + 6C 4
⎛ 1⎞ −3+ 2× ⎜ − ⎟ + 6 × 2 ⎝ 2⎠ 4 −3−1+12 4
A=
C−8 2
=
2−8 2
=−
6 2
= 3 This proves the answer.
8 4
=2 This proves the answer. Figure 6.4: Verification steps Explanation of the table in figure6.4 In the table above, we have the following information: • The ﬁrst column refers to the ﬁrst image in ﬁgure6.3, in which B is the Average Cell. • The second column refers to the second image in ﬁgure6.3, in which C is the Average
Cell. • The third column refers to the third image in ﬁgure6.3, in which A is the Average Cell. • In each of the columns, we have shown the steps to verify the answer.
As shown in the table above in ﬁgure6.4, we have veriﬁed that the value of each Average Cell is indeed the average of its adjacent cells. We now can summarize the veriﬁed answer in the next section.
147
6.2.2 Answer The answer to the problem in Example1 is shown below:
1 2
A = 3, B = − , C = 2, and 6C = 6 × 2 = 12 Average Cell Logitica
Figure 6.5: Answer to the problem in Example1 We have now developed a basic understanding of how to solve problems involving Average Cells. Let us continue exploring the chapter with another problem in the next example.
6.3 Example2 Average Cell Logitica Find the numerical values of A, B, and C.
Figure 6.6: Example2
6.3.1 Analysis Similar to what we have already discussed in Example1, we ﬁrst need to identify the appropriate cells for calculating the average and derive the corresponding equations. As shown in ﬁgure6.7 below, we have three Average Cells in the problem. We have derived three equations based on the deﬁnition of an Average Cell. In the ﬁgure below, we have circled the variables that we have selected for calculating the average in each of the equations. We have also used the black circle to highlight the variable inside the Average 148
Cell in each of the images shown below:
B=
A + 20 2
20 =
B + B + C 2B + C = 3 3
C=
20 + 2B + B 3B + 20 = 3 3
Figure 6.7: The cells selected for calculating the averages As shown in ﬁgure6.7, we can write three equations for the variables A, B, and C using the formula for averaging. We derive these equations as shown below for each of the images in ﬁgure6.7: First image: In this image, B is the Average Cell, and it is the average of A and 20 as expressed in the equation below:
A + 20 =B 2 A + 20 = 2B 2B  A = 20
Equation (1)
Second image: In this image, the value of the Average Cell is 20, and it is the average of B, B, and C as expressed in the equation below:
B+B+ C = 20 3 2B + C = 20 × 3 2B + C = 60
Equation (2)
Third image: In this image, C is the Average Cell, and it is the average of 20, 2B and B as expressed in the equation below: 20 + 2B + B =C 3
20 + 3B = 3C 3C  3B = 20 3(C  B) = 20 149
C  B = 20 3
Equation (3)
We have listed down the derived equations below for legibility: 2B  A = 20
Equation (1)
2B + C = 60
Equation (2)
20 3
Equation (3)
CB=
We now have three variables and three independent equations as described above. We can now follow a stepbystep approach to ﬁnd the value of the variables A, B, and C: ✔ Step 1: We can transform equation (3) to (3a) so that the value of C can be expressed
as a function1 of B : CB=
20 3
C=B+
[Equation (3)]
20 3
Equation (3a)
✔ Step 2: We can now substitute the value of C from equation 3(a) in (2) to determine
the value of B as shown below: 2B + C = 60 2B + B +
[Equation (2)]
20 = 60 3 3B = 60 3B =
B=
[Using C = B +
20 from equation (3a)] 3
20 3
60 × 3 − 20 180 − 20 160 = = 3 3 3 160 160 = 9 (3× 3)
✔ Step 3: To ﬁnd the value of C, we can substitute the value of B in equation (3a):
C=B+
20 3
[Equation (3a)]
=
20 160 + 3 9
=
20× 3+160 60+160 220 = = 9 9 9
[Using B =
160 ] 9
1. Refer to AppendixI (p. 282) for more details about functions. 150
Hence, C =
220 9
✔ Step 4: We now substitute the value of B in equation (1) to ﬁnd the value of A as
shown below: 2B  A = 20 2×
160  A = 20 9 A=
=
A=
[Equation (1)] [Using B = 160 ] 9
320  20 9 320 − 20 × 9 320 −180 140 = = 9 9 9 140 9
The values of A, B, and C are summarized below: A=
220 140 160 ,B= ,C= 9 9 9
Now that we have found the values of all the variables, we need to verify that these calculated values are correct. We will use ﬁgure6.8 to verify the values we calculated for each cell.
Figure 6.8: The cells selected for calculating the averages We have already done a similar veriﬁcation for Example1. We will use the same process of veriﬁcation, where the value of each Average Cell is matched against the average of the numerical values on its adjacent cells. The value of each variable is shown below: A=
220 140 160 ,B= ,C= 9 9 9
We will use the values of the variables A, B and C in the table shown in ﬁgure6.9 to verify our calculations. 151
First image
Second image
Third image
Value of the Average Cell = B Value of the Average Cell = 20 Value of the Average Cell = C
B=
B=
160 9
A + 20 2
Average =
140 + 20 9 = 2 =
=
=
=
140+ 20× 9 2× 9
(
C=
)
140+180 18 320 18 160 9
2B + C 3 2×
=
C=
160 220 + 9 9 3
20+ 3B 3 20+ 3×
=
320 220 + 9 9 = 3
220 9
3
160 9
480 9 3
20+ =
=
320+ 220 (9× 3)
=
20× 9+ 480 (9× 3)
=
540 = 20 27
=
180+ 480 27
=
660 27
=
220 9
This proves the answer.
This proves the answer.
This proves the answer. Figure 6.9: Verification steps Explanation of the table in figure6.9: In the table above, we have shown the following information: • The ﬁrst column refers to the ﬁrst image in ﬁgure6.8, in which B is the Average Cell. • The second column refers to the second image in ﬁgure6.8. The value of the Average
Cell in this image is 20. • The third column refers to the third image in ﬁgure6.8, in which C is the Average Cell.
This completes our veriﬁcation step. Now that we have veriﬁed our calculations, we can summarize the answer in the next section.
152
6.3.2 Answer The answer to the problem in Example2 is shown below in ﬁgure6.10: The value of each variable: A =
Also, 2B = 2 ×
220 140 160 ,B= ,C= 9 9 9
160 320 = 9 9 Average Cell Logitica
Figure 6.10: Answer to the problem in Example2
6.4 Summary We started the chapter with a simple problem from Example1. While solving this problem, we calculated the average of the adjacent cells and followed a stepbystep approach to derive and solve a set of linear equations. In the end, we used the deﬁnition of the Average Cell to verify our calculations. In Example2, we discussed another problem with a diﬀerent cluster of cells. The approach taken to determine and verify the values in this problem was similar to that of Example1. The two examples discussed in the chapter should help to build a strong foundation for solving this type of problems. There are two ways of solving a set of linear equations: elimination and substitution1. In this chapter, whenever appropriate, we have preferred one over the other in order to solve the equations. Readers are advised to try and solve the problems provided in the Exercise Section of this chapter in order to improve their understanding of concepts in this chapter.
1. Readers can refer to AppendixH (p. 278) for more details about the elimination and substitution methods used in solving a set of linear equations. 153
6.5 Exercise LevelI Find the numerical values of A and B.
(i)
(iii)
(v)
(ii)
(iv)
(vi)
154
LevelII Find the numerical values of A, B, and C.
(i)
(iii)
(v)
(ii)
(iv)
(vi)
155
For further practice As shown in ﬁgure6.11 below, we have a list of cells sequentially arranged. Each of the interior cells is the average of the adjacent cells on its either sides. Can you ﬁnd the value th
of k term (ak) as a function of the a1, an and n? Where: n = 1, 2, 3, ... k = 2, 3, .. , n1 i.e., 1