Proof:
Let n be a positive integer . Consider the (n+1) integers 1 , 11 , 111 ,... ,11...1 n+1 ones (Where the last integer in this list with n+1 is in its decimal expansion).
Note that there are n possible remainders when an integer is divided by n .
Because there are n+1 integers in this list , by the pigeonhole principle there must be two with the same remainder when divided by n .
The larger of these integers minus .
The smaller one is a multiple of n , which has a decimal expansion consisting entirely of 0's and 1's .