mathschallenge.net

Problem

Prove that every term in the infinite sequence 18, 108, 1008, 10008, 100008, ... , is divisible by 18.

Solution

The n th term, u_n , of the sequence 18, 108, 1008, 10008, ... is given by
u_n = 10ⁿ + 8.

un+1	= 10n+1 + 8
	= 10.10n + 8
	= (9 + 1).10n + 8
	= 9.10n + 10n + 8
	= 9.10n + 10n + 8

If u_n = 10ⁿ + 8 is divisible by 18, then so too will be u_n+1,
as 9 [even] will be divisble by 18.

We can see that u₁ = 18, hence 10ⁿ + 8 must be divisible by 18 for all n.