by rcoker7 » Mon Apr 15, 2013 7:14 pm
Here's another possible solution, though it's still not exactly elegant. It involves the binomial expansion.
13^2006 = (10 + 3)^2006 = 10^2006 + 2006 * 3 * 10^2005 + ... + 2006 * 3^2005 * 10 + 3^2006.
Only the last 2 terms are relevant. The only problem is dealing with those massive powers of 3. This is why the solution still isn't elegant. The last two digits of the powers of 3 repeat every 20 powers(figuring this out take a long time, and it isn't fun writing 20 powers of 3), so 3^2005 is equivalent to 3^5 and 3^2006 is equivalent to 3^6. From here, the calculations are simple, giving you 9.
What test was this on, by the way? This seems like a ridiculously long problem.
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