Mathematical Curiosities: A Treasure Trove of Unexpected Entertainments
382
Mathematical Curiosities: A Treasure Trove of Unexpected Entertainments
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Product Details
| ISBN-13: | 9781616149321 |
|---|---|
| Publisher: | Rowman & Littlefield Publishers, Inc. |
| Publication date: | 08/12/2014 |
| Sold by: | Barnes & Noble |
| Format: | eBook |
| Pages: | 382 |
| File size: | 11 MB |
| Note: | This product may take a few minutes to download. |
About the Author
INGMAR LEHMANN is retired from the mathematics faculty at Humboldt University in Berlin. For many years he led the Berlin Mathematics Student Society for gifted secondary-school students, with which he is still closely engaged today. He is the coauthor with Alfred S. Posamentier of Magnificent Mistakes in Mathematics and The Glorious Golden Ratio, and four other books.
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Mathematical Curiosities
A Treasure Trove of Unexpected Entertainments
By Alfred S. Posamentier, Ingmar Lehmann
Prometheus Books
Copyright © 2014 Alfred S. Posamentier and Ingmar LehmannAll rights reserved.
ISBN: 978-1-61614-932-1
CHAPTER 1
ARITHMETIC CURIOSITIES
The curiosities found in arithmetic and in numbers, in general, are probably boundless. They range from peculiarities of certain numbers to number relationships stemming from ordinary arithmetic processes. What makes these so entertaining are the unexpected results that are sometimes inexplicable. In this chapter we will be presenting to you some of these many arithmetic and numeric oddities in mathematics. Some are clearly errors that lead to correct results, while others are correct workings of mathematics that lead to wildly unexpected results. In either case, we hope that through the mathematics alone you will be entertained without having to apply it to other fields in either the sciences or the real world. Our intent here is to demonstrate a special beauty that can make mathematics fascinating and enjoyable.
HOWLERS
In the early years of schooling we learned to reduce fractions to make them more manageable. For this there were specific ways to do it correctly. Some wise guy seems to have come up with a shorter way to reduce some fractions. Is he right?
He was asked to reduce the fraction 26/65, and did it in the following way:
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That is, he just canceled out the 6's to get the right answer. Is this procedure correct? Can it be extended to other fractions? If this is so, then we were surely treated unfairly by our elementary-school teachers, who made us do much more work. Let's look at what was done here and see if it can be generalized.
In his book Fallacies in Mathematics, E. A. Maxwell refers to these cancellations as "howlers":
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Perhaps when someone did the fraction reductions this way, and still got the right answer, it could just make you howl. This simple procedure continues to give us the correct answers: As we look at this awkward—yet easy—procedure, we could begin by reducing the following fractions to lowest terms:
16/64, 19/95, 26/65, 49/98.
After you have reduced to lowest terms each of the fractions in the usual manner, one may ask why it couldn't have been done in the following way.
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At this point you may be somewhat amazed. Your first reaction is probably to ask if this can be done to any fraction composed of two-digit numbers of this sort. Can you find another fraction (comprised of two-digit numbers) where this type of cancellation will work? You might cite 55/55 = 5/5 = 1 as an illustration of this type of cancellation. This will, clearly, hold true for all two-digit multiples of eleven.
For those readers with a good working knowledge of elementary algebra, we can "explain" this awkward occurrence. That is, why are the four fractions above the only ones (comprised of different two-digit numbers) where this type of cancellation will hold true?
Consider the fraction 10x + a/10a + y, in which the second digit of the numerator and the first digit of the denominator match.
The above four cancellations were such that when canceling the a's the fraction was equal to x/y.
Let us explore this relationship: 10x+a/10a+y = x/y.
This yields y(10x+a) = x(10a + y), 10xy + ay = 10ax + xy, 9xy + ay = 10ax - y (9x + a) = 10ax, and so 10ax/9x + a.
At this point we shall inspect this equation. It is necessary that x, y, and a are integers since they were digits in the numerator and denominator of a fraction. It is now our task to find the values of a and x for whichy will also be integral. TO avoid a lot of algebraic manipulation, you will want to set up a chart that will generate values of y from y = 10ax/9x+a. Remember that x, y, and a must be single-digit integers. Below is a portion of the table you will be constructing. Notice that the cases where x = a are excluded, since x/a=1.
This small portion of the chart (figure 1.1) already generated two of the four integral values of y; that is, when x = 1, a = 6, then y = 4, and when x = 2, a = 6, and y = 5. These values yield the fractions 16/64 and 26/65, respectively. The remaining two integral values of y will be obtained when x = 1 and a = 9, yielding y = 5, and when x = 4 and a = 9, yielding y = 8. These yield the fractions 19/95 and 49/98, respectively. This should convince you that there are only four such fractions composed of two-digit numbers, excluding two-digit multiples of 11.
Let's extend this idea and investigate whether there are fractions composed of numerators and denominators of more than two digits where this strange type of cancellation holds true. Try this type of cancellation with 499/998. You should find that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].
A pattern is now emerging, and you may realize that
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Enthusiastic readers may wish to justify these extensions of the original howlers. Readers who, at this point, have a further desire to seek out additional fractions that permit this strange cancellation should consider the following fractions. They should verify the legitimacy of this strange cancellation, and then set out to discover more such fractions.
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Aside from requiring an algebraic solution, which can be used to introduce a number of important premises in a motivational way, this topic can also provide some recreational activities. Here are some more of these "howlers."
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This peculiarity shows how elementary algebra can be used to investigate an amusing number-theory situation. These are just some of the hidden treasures that mathematics continues to hold and that we will explore as we journey through this chapter.
A PAINTING TITLED A DIFFICULT ASSIGNMENT
At the end of the nineteenth century, the Russian artist Nikolai Petrovich Belsky (1868–1945) produced a painting with the title A Difficult Assignment. In the painting (figure 1.2), we see a group of students mulling around a chalkboard, apparently frustrated with an assignment of calculating an arithmetic challenge.
The problem is to find the value of 102 + 112 + 122 + 132 + 142/365.
Imagine trying to solve this problem without a calculator. It is certainly doable but somewhat time-consuming. However, through the amazing relationships that exist among numbers, we can see the following property that we can exploit. By partitioning the five numbers to be squared, we find that the sum of the first three squares has the same sum as the next two squares. In each case, the sum is 365, which then trivializes the original exercise.
(102 + 112 + 122) + (132 + 142)/365 = 365 + 365/365 =2.
Those who recognize this pattern might also be aware of the following pattern:
32 + 42 = 52 (= 25),
102 + 112 + 122 = 132 + 142 (= 365),
212 + 222 + 232 + 242 = 252 + 262 + 272 (= 2,030).
First, you will notice that on the left side of the equals sign in each case we have one more term than we have on the right side, and the numbers being squared are consecutive.
An ambitious reader might try to find the next equation, where five squared numbers would be on the left side of the equals sign and four on the right side.
Although it might be easier to do this with a calculator, it may be more fun to look for a pattern to make our calculation even easier.
THE MAGIC OF ALGEBRA
There are times when an overwhelming arithmetic problem can be nicely simplified with some basic algebra. Let us consider one such example now. In today's world, complicated calculations are easily disposed of using a calculator. However, it is entertaining to see how using algebraic manipulation can make a very complicated calculation practically trivial.
Consider the task of finding the value of √1999 · 2000 · 2001 · 2002+1. Surely, using a calculator, we can find that this cumbersome expression is equal to 4,001,999. However, it is interesting to see how we can generalize this expression to our advantage. Since the numbers being multiplied are consecutive, let's see if that gives us an advantage. We begin by letting n = 2,000 and express the other numbers under the radical sign in this way:
(n - 1) · n · (n + 1) · (n + 2) + 1.
Now for some algebraic gymnastics: by multiplying the terms of this long algebraic expression and then adding 1, we get:
(n - 1) · n · (n + 1) · (n + 2) + 1 = n4 + 2n3 - n2 - 2n + 1.
We shall now rearrange and dismantle these terms to suit our plan to get a workable expression:
n4 + 2n3 - n2 - 2n + 1 = n4 + n3—n2 + n3 + n2 - n - n2 - n + 1.
This allows us to form the following product of two trinomials:
n4 + n3 - n2 + n3 + n2 - n - n2 - n + 1 = (n2 + n - 1) · (n2 + n - 1) = (n2 + n - 1)2.
Replacing the original term under the radical sign with its equivalent established above, we are able to simplify the expression under the radical—a perfect square—which allows us to remove the radical sign. √(n-1) · n · (n+1) · (n+2)+1 = √(n2 + n -1)2 =| n2 + n - 1|.
Because we are working with natural numbers, we can conclude that √(n-1) · n · (n+1) · (n+2)+1 = √(n2 + n - 1)2 = n2 + n-1.
Therefore, when n = 2,000, we get √1999 · 2000 · 2001 · 2002+1 = 20002 + 2000 - 1 = 4,000,000 + 2,000 - 1 = 4,001,999, which is what we expected, since it conforms to the result obtained by using a calculator.
We have seen how algebra can help us understand and also simplify arithmetic processes, and we are now ready to explore some peculiarities embedded with particular numbers. Many numerals we take for granted we view only as the quantities they represent. Here are some rather-curious numeric insights and relationships that may make you consider and appreciate these numbers differently.
THE CURIOUS NUMBER 8
The number 8, which, in the Chinese culture, is the "lucky" number, has a unique arithmetic feature: it is the only cube number that is smaller than a square number by 1. That is, 8 = 23 = 9 - 1 = 32 - 1.
THE CURIOUS NUMBER 9
The number 9 is the only square number that is equal to the sum of the cubes of two consecutive natural numbers. That is to say, 9 = 13 + 23.
While we are considering the sum of the cubes, we can recall the finding by the famous Swiss mathematician Leonhard Euler (1707-1783), who stated that the smallest natural number that can be expressed as the sum of the cubes of natural numbers in two ways is 1,729.
That is, 13 + 123 = 1 + 1,728 = 1,729, and 93 + 103 = 729 + 1,000 = 1,729. Now returning to the number 9, we find that it can be expressed in fractional form using all ten digits exactly once, as we see in the following fractions, which are the only fractions that will give us this amazing result: 9 = 95742/10638, 9 = 95823/10647, and 9 = 97524/10836. However, if we allow the 0 to take a first position in any of the numbers, we get an additional three fractional equivalents to the number 9 also using all the digits exactly once.
9 = 57429/06381, 9 = 58239/06471, and 9 = 75249/08361.
THE CURIOUS NUMBER 11
The number 11 is truly a curious number. According to the British king George V, the armistice in 1918 occurred at the eleventh hour of the eleventh day of the eleventh month of the year.
In the American measuring system, the number 11 appears as a factor in linear measurements as follows: 1120 yards = 1 furlong, and 11160 yards = 1 mile.
We should also note that the number 11 is the only palindromic prime number that has an even number of digits. Some more curiosities with the number 11 are offered here to amuse the reader further.
First, we have 112 equal to the sum of five consecutive powers of 3 as follows:
112 = 121 = 30 + 31 + 32 + 33 + 34.
Then we have 113 as the sum of the squares of three consecutive odd numbers:
113 = 1,331 = 192 + 212 + 232.
We can also express the number 11 as the sum of a square and a prime in two different ways—and it is the smallest number that has this property!
11 = 22 + 7
11 = 32 + 2
Now here is one property of 11 that is really spectacular: if we reverse the digits of any number, which is divisible by 11, the resulting number will also be divisible by 11. To demonstrate this, let us take for an example the number 135,916, which is 11 times 12,356, and reverse the digits to get 619,531, which just happens to be 11 times 56,321, clearly a multiple of 11. You may wish to try this with other multiples of 11 and entertain your friends with this peculiarity.
Here is a nice little trick involving the number 11. Take any number where no two adjacent digits have a sum greater than 9. Then multiply that number by 11, and reverse the digits of this product. Then divide this result by 11. The resulting number will be the reverse of the original number. Let us consider as an example the number 235,412, which is a number where no two adjacent digits have the sum greater than 9. When we multiply it by 11 we get 235,412 · 11 = 2,589,532. Reversing the digits of this product, we get 2,359,852; and then dividing it by 11, we get 214,532, which is a number whose digits are in the reverse order of the original number.
Aside from being the fifth prime number, the number 11 is also the fifth Lucas number. You may recall the Lucas numbers are a sequence of numbers beginning with 1 and 3, where each succeeding number is the sum of the two previous numbers, as in the following sequence: 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, ... The sequence was popularized by the French mathematician Edouard Lucas (1842-1891), who also brought popularity to the Fibonacci numbers from which he got the idea of this sequence.
To the left of the Pascal triangle in figure 1.3, you will notice that the sum of the numbers in each row generates the powers of 2, while the oblique sums show the Fibonacci numbers, which are similar to the Lucas numbers in that they are also generated by the sum of consecutive numbers; however, this time beginning with 1 and 1 as the first two numbers. They are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
We can also find the powers of 11 on the first few rows of the famous Pascal triangle as shown in figure 1.3. Up to the fifth row, the powers of 11 appear directly: 114 = 14,641 = 1·104 + 4 · 103 + 6 · 102 + 4 · 101 + 1 · 100.
If we try to get the fifth power of 11, we notice that this row has two-digit numbers, so we will need to carry over the tens digit of each of these two-digit numbers to the next place (to the left) to get 115 = 161,051.
(Continues...)
Excerpted from Mathematical Curiosities by Alfred S. Posamentier, Ingmar Lehmann. Copyright © 2014 Alfred S. Posamentier and Ingmar Lehmann. Excerpted by permission of Prometheus Books.
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Table of Contents
Contents
Acknowledgments, 9,Introduction, 11,
Chapter 1: Arithmetic Curiosities, 15,
Chapter 2: Geometric Curiosities, 123,
Chapter 3: Curious Problems with Curious Solutions, 171,
Chapter 4: Mean Curiosities, 291,
Chapter 5: An Unusual World of Fractions, 315,
Conclusion, 343,
Notes, 345,
Index, 353,