Consider 10^a - 10^b where a, b, member of Natural numbers and a > b
Let k = 10^a - 10^b
k = 10^b.(10^(a-b) - 1)
but we know from before the any power of 10 minus 1 is divisible by 9
Assume that (10^(a-b) - 1) = 9c for some c member of natural numbers
k = 10^b.9c
Thus k is a multiple of 9.