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

This entry was posted on March 16, 2009 at 12:20 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 22, 2009 at 11:20 pm |

[…] 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 […]