Show that 3100ⁿ - 650ⁿ - 558ⁿ + 117ⁿ is divisible by 2009 for every positive integer n.

(In reply to re: As years go by... by Ady TZIDON)

Because (a-b) is a factor of (a^n-b^n) in the specified cases.

Write X as (a-b) and Y as (c-d) choosing appropriate integers, so that we know that X divides (a^n-b^n) and Y divides (c^n-d^n)

All that then needs to be done is to multipy out (a^n-b^n) (c^n-d^n) and we have a ready-made Perplexus problem suitable for any year, XY, of our choice.

Edited on January 6, 2013, 10:33 am

Posted by broll on 2013-01-06 10:33:14