by RonPurewal Fri Sep 14, 2007 5:57 am
There's not much I could do to improve on the solution given here, but here's an item to take away from this problem:
* Consider PRIME FACTORIZATIONS of numbers with exponents. In this case, a 'factor tree' of 4^11 would consist of eleven 4's, which would then break down into twenty-two 2's (giving the correct representation even if you've forgotten the 'common base' concept). Additionally, 10^n would break down into n 5's and n 2's, which would turn the whole right side (including the standalone 2) into n 5's and (n + 1) 2's.