You may have wondered why certain numbers carry some special properties that allow a simplification of arithmetic processes, or, in other words, some arithmetic tricks. Let's consider some of these here. Both the number 11 and the number 9 are situated on either side of the number 10-which is the base of our number system-and therefore have very interesting properties. These properties give these numbers some unusual benefits in calculation. Let us begin by examining how we can determine divisibility by the numbers 9 and 11.
There are times in everyday-life situations when it is useful to know if a given number is divisible by 9 or by 3-especially if it can be done mentally! For example, suppose a restaurant bill of $71.22 needs to be split into three equal parts. Before actually doing the division, the thought about whether or not it is possible to split the bill equally into three parts may come into question. Wouldn't it be nice if there were some mental arithmetic shortcut for determining this? Well, here comes mathematics to the rescue. We are going to provide you with a rule to determine if a number is divisible by 3 or divisible by 9. The rule, simply stated, is:
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).
Perhaps an example would be best to introduce this technique. Consider the number 276,357. Let's test it for divisibility by 3 (or 9). The sum of the digits is 2 + 7 + 6 + 3 + 5 + 7 = 30, which is divisible by 3, but not by 9. Therefore, the original number (276,357) is divisible by 3, but not by 9. However, the number 14,688 is divisible by both 3 and 9, since the sum of its digits is 1 + 4 + 6 + 8 + 8 = 27, which is divisible by both 3 and 9.
Just to make sure you are comfortable with this procedure, we will consider another example. Is the number 457,875 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, 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 (perhaps because it might still be too large a sum), then take the sum of the digits of this just-found sum and continue the process until you can visually make a determination of divisibility by 3 or 9.
Now that you are an expert at determining whether a number is divisible by 3 or 9, we can go back to the original question about the divisibility of the restaurant bill of $71.22. Can it be divided into three equal parts? Because 7 + 1 + 2 + 2 = 12, and 12 is divisible by 3, then we can conclude that $71.22 is divisible by 3. (Notice that we need not be concerned with the decimal, since it is the number comprised of the digits with which we are concerned.) In case you are interested as to why this rule actually works, here is a brief explanation using very simple algebra. Consider the base-10 number N = ab,cde, where the letters a, b, c, d, and e represent the digits and, therefore, the value of the number can be expressed as follows:
N = 104a + 103b + 102c + 10d + e = (9 + 1)4a + (9 + 1)3b + (9 + 1)2c + (9 + 1)d + e. Gathering those multiples of 9, we get
N = [9M + (1)4]a + [9M + (1)3]b + [9M + (1)2]c + [9 + (1)]d + e (where 9M indicates a different multiple of 9 each time).
Factoring these multiples of 9, we get N = 9M[a + b + c + d] + a + b + c + d + e, which implies that the entire expression will be divisible by 9 when the sum of the digits a + b + c + d + e is divisible by 9.
Let us now consider if there is an analogous special property for divisibility by 11. If you have a calculator at hand, the problem is easily solved. But that is not always the case. Besides, there is such a clever “rule” for testing for divisibility by 11-one that it is worth knowing just for its charm.
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.
This sounds a bit complicated, but it really isn't. Let us take this rule a piece at a time. “The sums of the alternate digits” means you begin at one end of the number, taking the first, third, and fifth, digit (and so on), and add them. Then you add the remaining (even-placed) digits. Subtract the two sums, and then inspect this resulting difference for divisibility by 11. Perhaps it might be best to demonstrate this through an example. Suppose we test 768,614 for divisibility by 11. The sums of the alternate digits are: 7 + 8 + 1 = 16 and 6 + 6 + 4 = 16. The difference of these two sums is 16 – 16 = 0, which is divisible by 11. (Remember, ) Therefore, we can conclude that 768,614 is divisible by 11.
Another example might be helpful to firm up an understanding of this procedure. To determine whether 918,082 is divisible by 11, we once again find the sums of the alternate digits: 9 + 8 + 8 = 25 and 1 + 0 + 2 = 3. The difference of the two sums is 25 – 3 = 22, which is divisible by 11, and so the number 918,082 is divisible by 11. Here we have an example of a technique that not only can be helpful but also demonstrates the power and consistency of mathematics. We would be remiss if we did not provide a justification for this rather-unexpected technique for determining whether a number is divisible by 11. Here is a brief discussion about why this rule works as it does. Consider the base-10 number N = ab,cde, where the letters a, b, c, d, and e represent the digits and, therefore, the value of the number can be expressed as
N = 104a + 103b + 102c + 10d + e = (11 – 1)4a + (11 – 1)3b + (11 – 1)2c + (11 – 1)d + e.
If we let 11M represent a number which is a multiple of 11 (and it can be a different number each time, but still a multiple of 11), we can express the above equation as: N = [11M + (–1)4]a + [11M + (–1)3]b + [11M + (–1)2]c + [11 + (–1)]d + e or N = 11M[a + b + c + d] + a – b + c – d + e, which implies that divisibility by 11 of the number N is dependent on the divisibility of a – b + c – d + e, which written another way, is (a + c + e) – (b + d), which is actually the difference of the sums of the alternate digits.
Having now considered rules for divisibility by these two special numbers, 9 and 11, let's consider other properties that these numbers have, to simplify our arithmetic processes. Perhaps one of the simplest mathematical tricks is to multiply by 11, mentally! This trick often gets a rise out of the unsuspecting mathematics-phobic person, because it is so simple that it is even easier than doing it on a calculator.
The rule is very simple:
To multiply a two-digit number by 11, just add the two digits and place this sum between the two digits.
Let's try using this technique. Suppose you need to multiply 45 by 11. According to the rule, add 4 and 5 and place this sum between the 4 and 5 to get 495.
This can become a bit more difficult when the sum of the two digits you are adding results in a two-digit number. We no longer have a single digit to place between the two original digits. So, if the sum of the two digits is greater than 9, then we place the units digit between the two digits of the number being multiplied by 11 and “carry” the tens digit to be added to the hundreds digit of the multiplicand. (Recall: The multiplicand is the number that is multiplied by another number, the multiplier-in this case, 11.) Let's try this procedure by finding the product of 78 · 11. We first get the sum of the digits: 7 + 8 = 15. We place the 5 between the 7 and 8, and then we add the 1 to the 7, to get [7 + 1], or 858.
It is fair to ask whether this technique also holds true when a number consisting of more than two digits is multiplied by 11. Let's go straight for a larger number such as 12,345 and multiply it by 11. Here we retain the first and last digit, then we begin at the right-side digit and add every pair of digits, going to the left: 1[1 + 2][2 + 3][3 + 4][4 + 5]5 = 135,795.
As was the case earlier, if the sum of two digits is greater than 9, then we place the units digit appropriately and carry the tens digit. To better understand how this is done, consider the following multiplication, 456,789 · 11:
Follow along as we carry the process step-by-step:
4[4 + 5][5 + 6][6 + 7][7 + 8][8 + 9]9
4[4 + 5][5 + 6][6 + 7][7 + 8]9
4[4 + 5][5 + 6][6 + 7][7 + 8 + 1]9
4[4 + 5][5 + 6][6 + 7]9
4[4 + 5][5 + 6][6 + 7 + 1]9
4[4 + 5][5 + 6]9
4[4 + 5][5 + 6 + 1]9
4[4 + 5]9
4[4 + 5 + 1]9
[4 + 1]9
This technique for multiplying by 11 might well be shared with your friends. Not only will they be impressed with your cleverness, they may also appreciate knowing this shortcut.