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Sequence Divisibility

Problem

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


Solution

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

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

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

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

Problem ID: 75 (Apr 2002)     Difficulty: 3 Star

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