Given that n is a positive integer, determine the remainder (in terms of n) whenever 3^(2^n) – 1 is divided by 2^(n+3)
Note: (a)^b implies 'a' raised to the power of 'b', ((a)^b)^c implies 'a' raised to the power 'bc', but a^(b^c) implies 'a' raised to the power 'b' raised to the power 'c'
Very nice problem.
We are going to start by doubling 3^(2^n) – 1. This gives these numbers in base 3: 121,12221,122222221,1+15 2's+1, 1+31 2's+1, 1+63 2's+1 etc.
The first is evenly divisible by 16, the second by 32, the third by 64 etc. (the other divisor rapidy becomes enormous). So the remainder is always zero.
But since in the problem as set we are taking exactly half that sum, the remainder is invariably 2^(n+2).
Edited on April 19, 2011, 1:28 pm

Posted by broll
on 20110419 13:24:34 