Determine the minimum value of a positive integer N > 2 such that:
the number 199...991, containing precisely N nines, is divisible by 1991.

(2*10^n-9)/11; cyclic, {0,4}, so no problems there, and n is odd.

1/181 has a period of 180, so I'm guessing the next n is big, and if n=3 is a solution, then so is n=183.

So there will be 182 nines.

Edited on January 29, 2016, 11:14 am

Posted by broll on 2016-01-29 09:30:19