# Wild About Math!Making Math fun and accessible

30Jul/083

## Winner for MMM #11

The prize for MMM #11 goes to Chao Xu. Here is Chao's solution.

A number divisible by 11 if it's alternating sum of it's digits is divisible by 11.
We only have 6 numbers, so we have a+b+c-d-e-f divisible by 11, where a to f are variables for each position. The largest possible number formed by a+b+c-d-e-f is 6+5+4-3-2-1 = 9, the smallest number is -9,
Then we have only one number that is divisible by 11 and between 9 and -9 is 0.
We have
a+b+c = d+e+f
a+b+c = (a+b+c+d+e+f)/2
and a+b+c+d+e+f = 21.
a+b+c = 10.5
It's not a whole number
It shows there is no number can be formed that's divisible by 11.

Jonathan, of the jd2718 blog made these observations (I've not checked his Math):

As a consolation, all 720 of them are multiples of 3.
Half of them (360) are even (multiples of 2)
One in 6 (120) are multiples of 5.
Eight in thirty (192) are multiples of 4.
None are multiples of 9.
Fourteen of 120 (84) are multiples of 8.
Multiples of 7?  New puzzle, good place to stop.

Check out Blinkdagger.com on Monday for another puzzle.