Prove that within any 39 consecutive natural numbers there must exist at least one number with the sum of its digits divisible by 11.
Bonus: Can the bound (39) be lowered?

(In reply to computer exploration -- bonus only is shown by Charlie)

Taking away the strict inequality and showing subsequent occurrences of the maximum gap shows:

38999980 to 39000019 is another such pair, as are 47999980 to 48000019, etc. where the first two digits add to 11 and 12 respectively.

In fact, come to think of it, doesn't this constitute proof? Whatever the sum of the digits in the 10 million and million positions is, or the congruence of any higher order digits mod 11, the cycle of 1 million in that range has only one occurrence that requires the 39 and none that require more. Quite a few require 29, which seems to be the second largest requirement--that is 28 is the second largest gap found in each set of 1 million numbers.

Posted by Charlie on 2012-02-19 12:49:08