The Gentle Art of Mathematics

Dan Pedoe

• Publisher: Dover Publications
• Year: 2012
• ISBN:
  9780486164069
• Topic: Technologie et ingénierie
• Subtopic: Sciences appliquées

This lighthearted work uses a variety of practical applications and puzzles to take a look at today's mathematical trends. In nine chapters, Professor Pedoe covers mathematical games, chance and choice, automatic thinking, and more.

CHAPTER I

MATHEMATICAL GAMES

MANKIND has always been fascinated by the ordinary integers, the natural numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, . . . . . At a very early age the normally endowed human being is made aware that he possesses 5 fingers on each hand and 5 toes on each foot. Sooner or later he believes that he himself is a unique individual; that he represents the number 1. The biological significance of the number 2 becomes only too clear. A mystic significance grows around the number 3. To bridge players 4 is a basic integer, and similarly 5, 6, 7, 8 and 9 all have their devotees. A favourite pastime in the not too distant past was to attach numbers to the letters of the alphabet so that, when you added up the numbers corresponding to the letters, the name of your enemy was shown to add up to the Number of the Beast, which is 6 6 6, according to Revelations. Much has been written on the association of integers with remarkable events. Good use was made of this knowledge by a Mr. Galloway, a Fellow of the Royal Society, when the Council of that august body first resolved to restrict the number of yearly admissions to the Society to fifteen men of science and noblemen ad libitum. Why fifteen, asked Mr. Galloway? He then continued, with true Victorian thoroughness:

Was it because fifteen is seven and eight, typifying the Old Testament Sabbath, and the New Testament day of the resurrection following? Was it because Paul strove fifteen days against Peter, proving that he was a doctor both of the Old and the New Testament? Was it because the prophet Hosea bought a lady for fifteen pieces of silver? Was it because, according to Micah, seven shepherds and eight chiefs should waste the Assyrians? Was it because Ecclesiastes commands equal reverence to be given to both Testaments — such was the interpretation — in the words “Give a portion to seven, and also to eight”? Was it because the waters of the deluge rose fifteen cubits above the mountains? — or because they lasted fifteen decades of days? Was it because Ezekiel’s temple had fifteen steps? Was it because Jacob’s ladder has been supposed to have had fifteen steps? Was it because fifteen years were added to the life of Hezekiah? Was it because the feast of unleavened bread was on the fifteenth day of the month? Was it because the scene of the Ascension was fifteen stadia from Jerusalem? Was it because the stone-masons and porters employed on Solomon’s temple amounted to fifteen myriads?

As this is not a book devoted only to mathematical curiosities we shall not spend too much time on these shallow numerists, which is what Cocker, the author of a famous 17th-century arithmetic book, called these numbermongers. In this chapter we consider the way in which ordinary integers can be represented, and describe a number of mathematical problems and games which arise from this representation.

The numbers we use in ordinary life are expressed in the scale of ten. Perhaps there is no need to stress this. Twenty means twice ten, thirty means three times ten, and so on. We use words like a hundred for ten times ten, a thousand for ten times a hundred, a million for a thousand thousand. The mathematician prefers to use unambiguous symbols instead of words, if he can, and writes

100 = (10). (10) = 10²,

where . is used as the symbol for multiplication, and the index 2 is used to show that 10 is multiplied by itself twice. With this notation

1000 = 10 . 10 . 10 = 10³,

and

1,000,000 = 10⁶.

The number 9824, say, stands for

9. 10³ + 8. 10²+ 2 . 10 + 4.

The position of the digits as we move from left to right in a number indicates which powers of ten are involved, and when we assess any number with a largish set of digits, like 54623108, we do a rapid mental calculation. We begin at the right, and mark off the digits in sets of three, so that we have

54, 623, 108,

and then we know that the number is fifty-four million, six hundred and twenty-three thousand, one hundred and eight. Or, since there are 8 digits involved, we also know that the number can be represented, using powers of 10, as

5 . 10⁷ + 4 . 10⁶ + 6 . 10⁵ + 2 . 10⁴
+ 3. 10³ + 1.10² + 0.10 + 8.

We note that the highest index involved is one less than the total number of digits.

How can we represent a general number with n digits? Here n represents any integer, such as 1, 2, 3, . . . . . We require n different symbols for the n digits. There are advantages in using a single symbol, with a digit suffix to distinguish one symbol from any other. We could use

a₀, a_1, a₂, a₃, . . . . . , a_n−1,

to represent n digits. We have brought in the cipher, or zero, 0, for use with the digits. We shall see that it is useful. The number with n digits will be written

a_n−1 a_n−2 a_n−3 . . . . . a₃ a₂ a₁ a₀

in the ordinary way, where the position of each digit is significant, as in ordinary numbers. We then know that the number represented is

10ⁿ⁻¹. a_n—1 + 10ⁿ⁻². a_n−2 + . . . + 10² . a₂ + 10¹ . a₁ + a₀.

The suffixes have indicated the power of 10 by which the corresponding digit is to be multiplied.

We make use of this discussion to solve a digital problem which runs as follows:

Find the smallest integer which is such that if the digit on the extreme left is transferred to the extreme right, the new number so formed is one and a half times the original number.¹

If the number were 5364, the digit on the extreme left is 5, and after transfer to the extreme right the number is 3645. But this is not the solution. The interest of this problem lies in the fact that the answer is so very large, and in the even more pertinent fact that a method of solution which is not systematic and logical will hardly obtain the result in a finite time! All the same, the reader is invited to have a go before he studies the solution. Since answers are usually given in examinations on the higher mathematics, we state the result: the smallest number satisfying the conditions is

1, 176, 470, 588, 235, 294.

When we transfer the digit from the extreme left to the extreme right we have

1, 764, 705, 882, 352, 941,

and a moment’s calculation will show that this second number is indeed one and a half times the first one.

The solution we now describe is fairly long, but all the steps in it are simple ones. A much shorter solution will follow, and the reader may turn to the shorter one first, if he so desires. We first write down the unknown number, and it is convenient to assume that it is a number of n + 1 digits, rather than a number of n digits. The second assumption is no more general than the first. The number will be

a_n a_n−1 a_n−2 . . . . . a₂ a₁ a₀,

and the digit on the extreme left being a_n, when we transfer it to the extreme right we obtain the number

a_n−1 a_n−2 . . . . . . . a₂ a₁ a₀ a_n.

Since we do not wish to be restricted by the positions of the various a’s in each number, we write the given number in the form

10ⁿ . a_n + 10ⁿ⁻¹ . a_n−1 + . . . . + 10 . a₁ + a₀.

We note that each of the various a’s is less than or equal to 9. This point will be significant later. When we have transferred the digit, the new number is

10ⁿ . a_n−1 + 10ⁿ⁻¹. a_n−2 + . . + 10² . a₁ + 10 . a₀ + a_n .

By the given conditions of the problem, this new number is 3/2 times the one we started with, and so we arrive at the equation:

3(10ⁿ . aⁿ + 10ⁿ⁻¹ . a_n−1 +. . . .+ a₀)
= 2(10ⁿ . a_n−1 + 10ⁿ⁻¹ . a_n−2 + . . + 10 . a₀ + a_n).

We notice that a_n occurs on both sides of this equation, and all that we do now is to collect together the terms containing a_n on to one side, using the elementary rules of algebra “taking over to the other ...

Table of contents

• Title Page
• Copyright Page
• PREFACE
• Table of Contents
• CHAPTER I - MATHEMATICAL GAMES
• CHAPTER II - CHANCE AND CHOICE
• CHAPTER III - WHERE DOES IT END?
• CHAPTER IV - AUTOMATIC THINKING
• CHAPTER V - TWO-WAY STRETCH
• CHAPTER VI - RULES OF PLAY
• CHAPTER VII - AN ACCOUNTANT’S NIGHTMARE
• CHAPTER VIII - DOUBLE TALK
• CHAPTER IX - WHAT IS MATHEMATICS?
• INDEX
• A CATALOG OF SELECTED DOVER BOOKS IN SCIENCE AND MATHEMATICS