AMC 10/12/A/B 2013 Discussion Thread

with_one_voice_5
Fri, 15 Mar 2013 11:42:59 +0000

I got 200 as well. Multiple of five (end in 0 or 5) and and first and last digit are the same (can’t have a leading 0), so it is 5ABC5. So for the sum to be a multiple of five, A+B+C=0 (mod 5). A+B can be any number between 0 and 18 inclusive. For each of those possible sums, there are 2 numbers for C that satisfy the condition. Therefore, for every possible A+C, there are 2 solutions. A can be 0 to 9, B can be 0 to 9, so 1010=100=>1002=200.

For the triangle fold one, I got 459. From left to right, let the points created by folding be D and E. Then I set AD=12-BD and AE=12-CE. Angle C and B are both 60, so you can find AD and AE in terms of BD and CE. Then set the two equations for AD equal to each other and the same for AE. Then I used law of cosines again on angle A (60) and AD and AE. Sorry, I don’t remember the exact numbers, but they were kinda ugly. Anyway, they added to 459. I might have made an arithmetic error though, so correct me if I’m wrong.