Monday Math 63

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 also use the rule that if n is odd.
The first tells us , and so is divisible by 11 (as 99=11*9) for even k.
The second tells us that
.
So is divisible by 11 for all non-negative integers k.
So we consider the alternating sum . The difference between n and a is then
, and as each term of the series is divisible by 11, the difference is a multiple of 11. Thus, n is divisible by 11 if and only if a is a multiple of 11.

An few examples of this test:
n=297: 7-9+2=0=0*11, so 297 is divisible by 11 (297=27*11)
n=12,529:
9-2+5-2+1=11 (12,529=1139*11)
n=4,596,836:
6-3+8-6+9-5+4=13, so 4,596,836 is not divisible by 11.
n=94,949,690:
0-9+6-9+4-9+4-9=-22=-2*11, so 94,949,690 is divisible by 11 (94,949,690=8,631,790*11).

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One Response to “Monday Math 63”

  1. Monday Math 64 « Twisted One 151’s Weblog Says:

    […] on, so that is divisible by 7 for all non-negative integers k. This creates a test by analogy to our divisibility test for 11. Thus, we split our large number into sets of three digits, and then take the alternating sum of […]

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