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.

QED

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