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CHAPTER 1

ARITHMETIC NOVELTIES

When you think of arithmetic, you typically consider the four basic arithmetic operations. With a little more thought, you tend to tag on the square root operation as well. Unfortunately, most of our school curriculum focuses on ensuring that we have a good mechanical command of the arithmetic operations and know the number facts as best we can to service us efficiently in our everyday life. As a result, most adults are not aware of the many amazing relationships that can be exhibited arithmetically with numbers. Some of these can be extremely useful in our everyday life as well. For example, just by looking at a number and determining if it is divisible by 3, 9, or 11 can be very useful, especially if it can be done at a glance. When it involves determining divisibility by 2, we do this without much thought, by simply inspecting the last digit. We shall extend this discussion to considering divisibility by a prime number, something that clearly is not presented in the school curriculum, with which we hope to motivate the reader to investigate further primes beyond those shown here. We truly expect that the wonders that our number system holds, many of which we will present in this book, will motivate you to search for more of these curiosities along with their justifications. Some of the units in this chapter will also provide you with a deeper understanding for our number system beyond merely arithmetic manipulations. Our introduction to a variety of special numbers will generate a greater appreciation of arithmetic than the typical school courses provide. Let us begin our journey through numbers and their operations.

WHEN IS A NUMBER DIVISIBLE BY 3 OR 9?

Teachers at various grade levels often neglect to mention to students that in order to determine whether a number is divisible by 3 or 9, you just have to apply a simple rule: If the sum of the digits of a number is divisible by 3 (or 9), then the original number is divisible by 3 (or 9).

An example will best firm up your understanding of this rule. Consider the number 296,357. Let's test it for divisibility by 3 (or 9). The sum of the digits is 2 + 9 + 6 + 3 + 5 + 7 = 32, which is not divisible by 3 or 9. Therefore, the original number, 296, 357, is not divisible by 3 or 9.

Now suppose the number we consider is 457,875. Is it divisible by 3 or 9? The sum of the digits is 4 + 5 + 7 + 8 + 7 + 5 = 36, which is divisible by 9 (and then, of course, divisible by 3 as well), so the number 457,875 is divisible by 3 and by 9. If by some remote chance it is not immediately clear to you whether the sum of the digits is divisible by 3 or 9, then continue with this process; take the sum of the digits of your original sum and continue adding the digits until you can visually make an immediate determination of divisibility by 3 or 9.

Let's consider another example. Is the number 27,987 divisible by 3 or 9? The sum of the digits is 2 + 7 + 9 + 8 + 7 = 33, which is divisible by 3 but not by 9; therefore, the number 27,987 is divisible by 3 and not by 9.

If this divisibility rule is mentioned in school settings, what is typically missing from the instruction of this rule is why it works. Here is a brief discussion about why this rule works as it does. Consider the decimal number abcde, whose value can be expressed in the following way:

N = 104a + 103b + 102c + 10d + e = (9 + 1)4a + (9 + 1)3b + (9 + 1)2c + (9 + 1)d + e.

After expanding each of the binomials, we can now represent all of the multiples of 9 as 9M to simplify this as

N = [9M + (1)2]a + [9M + (1)3]b + [9M + (1)2]c + [9 + (1)]d + e.

Then, factoring out 9M, we get N = 9M[a + b + c + d] + a + b + c + d + e, which implies that the divisibility of N by 3 or 9 depends on the divisibility of a + b + c + d + e by 3 or 9, which is the sum of the digits.

As you can see, things become so much better understood and appreciated when the reason for a "rule" is presented.

WHEN IS A NUMBER DIVISIBLE BY 11?

When a teacher shows the class something that is not directly specified in the school curriculum, it often generates some enjoyment and can be motivating. Take, for example, a method of determining whether a number is divisible by 11, without actually carrying out the division process. The problem is easily solved if you have a calculator at hand, but that is not always the case. Besides, there is such a clever "rule" for testing for divisibility by 11 that it is worth knowing just for its cleverness.

The rule is quite simple: If the difference of the sums of the alternate digits is divisible by 11, then the original number is also divisible by 11. That sounds a bit complicated, but it really isn't. Finding the sums of the alternate digits means that you begin at one end of the number, and you take the first, third, fifth, etc., digits and add them together. Then you add the remaining (even-placed) digits. Subtract the two sums and inspect for divisibility by 11.

This rule is probably best demonstrated through an example. Suppose we test 918,082 for divisibility by 11. We begin by finding the sums of the alternate digits: 9 + 8 + 8 = 25 and 1 + 0 + 2 = 3. Their difference is 25 - 3 = 22, which is divisible by 11, and so the number 918,082 is divisible by 11. We should point out that if the difference of the sums is equal to zero, then we can conclude that the original number is divisible by 11, since zero is divisible by all numbers. We see this in the following example: testing the number 768,614 for divisibility by 11, we find that the difference of the sums of the alternate digits (7 + 8 + 1 = 16 and 6 + 6 + 4 = 16) is 16 - 16 = 0, which is divisible by 11. Therefore, we can conclude that 768,614 is divisible by 11.

In case you may be wondering why this technique works, we offer the following. Consider the decimal number N = abcde, which then can be expressed as

N = 104a + 103b + 102c + 10d + e = (11 - 1)4a + (11 - 1)3b + (11 - 1)2c + (11 - 1)d + e.

This can be written as

N =[11M + (-1)4]a + [11M + (-1)3]b + [11M + (-1)2]c + [11 + (-1)]d + e,

where, after expanding each of the binomials, 11M represents the terms which are multiples of 11 written together. Factoring out the 11M terms, we get N = 11M[a + b + c + d] + a - b + c - d + e, which leaves us with an expression that would be divisible by 11, but only if this last part of the previous expression is divisible by 11, namely, a - b + c - d + e = (a + c + e) - (b + d), which just happens to be the difference of the sums of the alternate digits. This is a handy little "trick" that can also enhance your understanding of arithmetic. By the way, another way of looking at this trick is to say that the number 24,847,291 is divisible by 11 if and only if we obtain a number that is divisible by 11; let's see what we get: 2 - 4 + 8 - 4 + 7 - 2 + 9 - 1 = 15. Therefore, since the difference of the sums was 15, which is not divisible by 11, we know that 24,847,291 is not divisible by 11.

DIVISIBILITY BY PRIME NUMBERS

In today's technological world, arithmetic skills and competencies seem to be relegated to a back burner, since a calculator is so easily available. We can assume that most adults can determine when a number is divisible by 2 or by 5, simply by looking at the last digit (i.e., the units digit) of the number. That is, if the last digit is even (such as 2, 4, 6, 8, 0), then the number itself will be divisible by 2. Furthermore, if the number formed by the last two digits is divisible by 4, then the original number itself is divisible by 4. Also, if the number formed by the last three digits is divisible by 8, then the original number itself is divisible by 8. This rule can be extended to divisibility by higher powers of 2 as well.

Similarly, for the number 5: If the last digit of the number being inspected for divisibility by 5 is either a 0 or 5, then the number itself will be divisible by 5. If the number formed by the last two digits is divisible by 25, then the original number itself is divisible by 25. This is analogous to the rule for powers of 2. Have you guessed what the relationship here is between powers of 2 and 5? Yes, they are the factors of 10, the basis of our decimal number system.

Having completed in the previous discussions, the nifty techniques for determining whether a number is divisible by the primes 3, 9, and 11, the question then is: Are there also rules for divisibility by other prime numbers? Let's consider divisibility rules by prime numbers.

Aside from the potential usefulness of being able to determine whether a number is divisible by a prime number, the investigation of such rules will provide for a better appreciation of mathematics, that is, divisibility rules provide an interesting "window" into the nature of numbers and their properties. Although this is a topic that is typically neglected from the school curriculum, it can prove useful in everyday life.

The smallest prime number that we have not yet discussed in our quest for divisibility rules is the number 7. As you will soon see, some of the divisibility rules for prime numbers are almost as cumbersome as an actual division algorithm, yet they are fun, and, believe it or not, can come in handy. As we begin our quest for divisibility rules for the early prime numbers, we will begin with the following rule for divisibility by 7.

The rule for divisibility by 7: Delete the last digit from the given number, and then subtract twice this deleted digit from the remaining number. If the result is divisible by 7, then the original number is divisible by 7. This process may be repeated until we reach a number that we can visually inspect as one that is divisible by 7.

Let's consider an example to see how this rule works. Suppose we want to test the number 876,547 for divisibility by 7. Begin with 876.547 and delete its units digit, 7, and subtract its double, 14, from the remaining number: 87,654 - 14 = 87,640. Since we cannot yet visually inspect the resulting number for divisibility by 7, we continue the process. We delete the units digit, 0, from the previously resulting number 87,640, and subtract its double (which is still 0) from the remaining number to get 8,764 - 0 = 8,764. It is unlikely that we can visually determine whether this number, 8,764, is divisible by 7, so we continue the process. Again, we delete the last digit, 4, and subtract its double, 8, from the remaining number to get 876 - 8 = 868. Since we still cannot visually inspect the resulting number, 868, for divisibility by 7, we again continue the process.

Continuing with the resulting number, 868, we once again delete its units digit, 8, and subtract its double, 16, from the remaining number to get 86 - 16 = 70, which is divisible by 7. Therefore, the number 876.547 is divisible by 7.

Before continuing with our discussion of divisibility of prime numbers, you might want to practice this rule with a few randomly selected numbers, and then check your results with a calculator.

Now for the beauty of mathematics! Why does this rather strange procedure actually work? To see why things work is the wonderful aspect of mathematics — it enlightens us!

To justify the technique of determining divisibility by 7, consider the various possible terminal digits (that we are "dropping") and the corresponding subtraction that is actually being done after dropping the last digit. In the chart below you will see how in dropping the terminal digit and doubling it, we are essentially subtracting a multiple of 7. That is, we have taken "bundles of 7" away from the original number. Therefore, if the remaining number is divisible by 7, then so is the original number, because you have separated the original number into two parts, each of which is divisible by 7, and therefore, the entire number must be divisible by 7.

There is another way to argue why this method always works, and you may also want to give this some thought: Removing the final digit and then subtracting twice this digit from the remaining number is equivalent to subtracting 21 times the final digit from the number and then dividing the resulting number by 10. (The latter is certainly possible, since the number resulting from the first step must terminate in the digit 0.) Since 21 is divisible by 7, and 10 is not, the resulting number is divisible by 7 if and only if the original number was divisible by 7.

The next prime number that we have not yet considered for divisibility is the number 13.

The rule for divisibility by 13: The procedure here is similar to that used for testing divisibility by 7, except that instead of subtracting twice the deleted digit, we subtract nine times the deleted digit each time.

Perhaps it is best for us to do an example applying this rule. Let us check for divisibility by 13 for the number 5,616. We begin with our starting number, 5,616, and delete its units digit, 6, and subtract nine times 6, or 54, from the remaining number to get 561 - 54 = 507.

Since we still cannot visually inspect the resulting number for divisibility by 13, we continue the process. With this last resulting number, 507, we delete its units digit, 7, and subtract nine times this digit, 63, from the remaining number, which gives us 50 - 63 = -13, which is divisible by 13; therefore, the original number, 5,616, is divisible by 13.

In this rule for divisibility by 13, you might wonder how we determined the "multiplier" to be 9. We sought the smallest multiple of 13 that ends in a 1. That was 91, where the tens digit is 9 times the units digit. Once again consider the various possible terminal digits and the corresponding subtractions in the following table.

In each case, a multiple of 13 is being subtracted one or more times from the original number. Hence, if the remaining number is divisible by 13, then the original number is divisible by 13.

Divisibility by 17: Delete the units digit and subtract five times the deleted digit from the remaining number until you reach a number small enough to determine its divisibility by 17.

We justify the rule for divisibility by 17 as we did for the rules for 7 and 13. Each step of the procedure subtracts a "bundle of 17s" from the original number until we reduce the number to a manageable size and can make a visual inspection for divisibility by 17.

The patterns developed in the preceding three divisibility rules (for 7, 13, and 17) should lead you to develop similar rules for testing divisibility by larger primes. The following chart presents the "multipliers" of the deleted terminal digits for various primes.

You may want to extend this chart. It's fun, and it will increase your perception of mathematics. You may also want to extend your knowledge of divisibility rules to include composite (i.e., nonprime) numbers.

Divisibility by composite numbers: A given number is divisible by a composite number if it is divisible by each of its relatively prime factors.

The chart below offers illustrations of this rule. You might want to complete the chart to include composite numbers up to the number 48.

You now have a rather comprehensive list of rules for testing divisibility, as well as an interesting insight into elementary number theory. An interested reader may want to test these rules (to instill even greater familiarity with numbers) and try to develop rules to test divisibility by other numbers in base ten and to generalize these rules to other bases.

SQUARING NUMBERS QUICKLY

We all learned in school how to multiply two multidigit numbers using pencil and paper. However, if we want to multiply a number by itself (that is, to square the number), there exist shortcuts to get the answer. Moreover, the multiplication of any two numbers can be written as a combination of squares of sums and differences of these numbers. Hence, knowing how to add, subtract, and square numbers is actually enough to compute the product of any two numbers.

Squaring Numbers with a Last Digit of 5

Here is a quick way to square any number with a last digit of 5: We delete the last digit and we are left with some number N. Multiplying N by N + 1 and appending the digits 2 and 5 at the end yields the correct result.

For example, to compute 85, we delete 5, multiply the remaining digit, 8, by 9, giving 72, and append 25 at the end. The result is 7,225, which is 85.

Why does this rule work? If we let N denote the number that remains after we have dropped the last digit, then we can write the square of the number as (10 · N + 5)2 = 100 · N2 + 100 · N+25 = 100 · N · (N+ 1) + 25. The product N · (N + 1) represents the amount of hundreds in the result. But by writing its numerical value in front of the digits 2 and 5, we assign the place value of a hundred to this number and, according to our little calculation, will end up with the square of the original number.

(Continues…)

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Excerpted from "The Joy of Mathematics"
by .
Copyright © 2017 Alfred S. Posamentier, Robert Geretschläger, Charles Li, and Christian Spreitzer.
Excerpted by permission of Prometheus Books.
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