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Testing for Factors and Divisibility: Why Does the Rule of 9 Work?

Updated on August 16, 2013
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TR Smith is a product designer and former teacher who uses math in her work every day.

How can you tell when a number is divisible by 9? Why does this divisibility trick work?
How can you tell when a number is divisible by 9? Why does this divisibility trick work?

When elementary school students learn how to divide numbers to find factors, one of the first rules they learn is how to tell when a number is divisible by 9. The rule of 9 is simple: if the sum of the digits of the number is itself a multiple of 9, then the number you started with is a multiple of 9; if the sum is not a multiple of 9, then neither is the original number.

For example, the number 8703 is a multiple of 9 since 8 + 7 + 0 + 3 = 18 and 18 is a multiple of 9. But the number 5214 is not a multiple of 9 since 5 + 2 + 1 + 4 = 12 and 12 is not a multiple of 9. The rule is easy to apply, but the reason why it works involves much deeper mathematics than you learn in grade school.


Some Preliminary Arithmetic and Number Facts

Before proving that the rule of 9 works, there are some basic number properties you need to know. The first is that if you mutliply any integer by 9, the resulting product will be a multiple of 9.

The second fact is that if you have two multiples of 9, call them X and Y, then the sum X + Y and the difference X - Y are both multiples of 9.

Lastly, if X is a multiple of 9 but Z is not, then neither X + Z nor X - Z are multiples of 9. But by the first rule, the product X*Z is divisible by 9.


Proof of the Rule of 9

Let's take a four digit number |A|B|C|D| in which A represents the thousands (1000) digit, B represents the hundreds (100) digit, C represents the tens (10) digit, and D is the unit (1) digit. Each of A, B, C, and D is a digit between 0 and 9. Another way of writing the number |A|B|C|D| is as the sum

1000*A + 100*B + 10*C + D.

Now let's construct a multiple 9 using the digits of the number above. Since 999, 99, and 9 are all multiples of 9, then the products 999*A, 99*B, and 9*C are also divisible by 9, regardless of the values of A, B, and C. Using the second property, we can also see that the sum

999*A + 99*B + 9*C

is a multiple of 9. Now let's consider a different number formed from the digits of the original number, the sum A + B + C + D, the sum of the digits. We do not know whether the sum A + B + C + D is divisible by 9 or not, so there are two cases to consider.


Case I: Assume A + B + C + D is divisible by 9, i.e., a multiple of 9. Then the sum

(999*A + 99*B + 9*C) + (A + B + C + D)
= 1000*A + 100*B + 10*C + D
= |A|B|C|D|

is also a multiple of 9. In other words, if the sum of the digits is divisible by 9, then so is the original number.


Case II: Assume A + B + C + D is not a multiple of 9. Then the sum

(999*A + 99*B + 9*C) + (A + B + C + D)
= 1000*A + 100*B + 10*C + D
= |A|B|C|D|

is not a multiple of 9 either. In other words, if the sum of the digits is not a multiple of 9, then neither is the original number.


Thus, we have proved the rule of 9 for four-digit numbers. This argument can be applied to numbers of any size. The exact same argument can also be used to prove the rule of 3, that is, if the sum of an integer's digits is a multiple of 3, then so is the integer.

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      sigrid 3 years ago

      good explanation of something most people know in a superficial sense, but don't really know in a deeper sense. i, for example, didn't understand the underlying reason why it worked, but now i see why the rule works for 9 and 3.

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