**Divisibility Tests**

**Part 2: 3 and 9**

One may know that the tests for divisibility by 3 and 9 involve adding the digits of a number: a number is divisible by 3 if and only if the sum of its digits is divisible by 3, and divisible by 9 if and only if the sum of its digits is divisible by 9.

The key is to recall the factoring rule , used to prove the geometric series formulas. This means that . (This just means 9, 99, 999, 9999, etc. are all multiples of 9, which should be quite obvious).

Let us consider an integer *n*. Let *a*_{0} be the ones digit, *a*_{1} the tens digit, *a*_{2} the hundreds digit, and so on, with *a*_{m} the highest digit, so that *n* is an (*m*+1)-digit number. This means

. The sum of the digits, in turn, is . The difference between these two is

.

As this is a multiple of 9, we see that *n* is divisible by 9 if and only if *s* is divisible by 9, and as a multiple of 9 is also a multiple of 3, *n* is divisible by 3 if and only if *s* is divisible by 3.

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Tags: Divisibility, Divisibility Tests, Math, Monday Math, Number Theory

This entry was posted on March 9, 2009 at 12:03 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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March 27, 2009 at 12:10 am |

[…] Math 63 By twistedone151 Divisibility tests Part 3: 11 Last time, we proved the tests for divisibility by 3 and 9. Now, for 11, we again use the factoring rule . We […]