This may be the same as Broll's approach(?)
3^{2012} = (3^{4})^{503} = 3^{4} = 81 (mod 503) since 503 is prime (Fermat).
So 3^{2012} = 503n + 81 (1)
Since all even powers of 3 are congruent to 1 (mod 4), it follows from (1)
that 503n = 0 (mod 4) and therefore that n is divisible by 4.
Let n = 4m, then (1) gives: 3^{2012} = 503(4m) + 81 = 2012m + 81,
showing that the required remainder is 81.
Edited on April 15, 2012, 10:51 pm

Posted by Harry
on 20120415 19:38:18 