Solution to Problem 11: "C Hunt"

My personal strategy with this type of problem is to write down 25-x for every value and then attempt to find four of the new numbers whose sum is 0.

We see that -2 + -16 = -18 and 4 + 14 = 18, so their corresponding values in the original table (9 23 29 39) total 100.

Comments: In the contest, I had done what I described but didn't see the sum after five minutes of unorganized looking, so I passed it up.  That evening, I saw it the moment I turned to this page. 

In retrospect, I think that I would make two lists with all the positive and negative numbers separated and then make sums of pairs of each list and find the common values.   If that didn't work, then you'd need one value in one list to be equal to the sum of three numbers in the other list, which is similarly easy to find once you start looking for it.

Nick Baxter suggests that a better strategy is to examine the twelve values modulo 3:

Since the sum of the four numbers has to be 1 modulo 3, it is evident that the two 0 entries are in the solution, so the problem is reduced to finding two values from the remaining ten that total 52.