Monday Math 62

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 a0 be the ones digit, a1 the tens digit, a2 the hundreds digit, and so on, with am 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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One Response to “Monday Math 62”

  1. Monday Math 63 « Twisted One 151’s Weblog Says:

    […] 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 […]

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