Given that 5^j+6^k+7^d+11^m=2006 where j, k, d and m are different nonnegative integers, what is the value of j+k+d+m?
I made a table of powers of 5, 6, 7 and 11 and realized there wouldn't be many possibilities to check (72 in fact, although Dennis reduced it to 12)
I started by trying m=3, 11^3=1331, 2006  1331 = 675.
I couldn't help but notice this is 625+50 = 625+1+49.
So the solution is 4+0+2+3=9
A deceptive problem. It looks hard at first but is actually very easy.

Posted by Jer
on 20070226 11:54:34 