Here is problem 33 from "Putnam and Beyond" :
Given 50 distinct positive integers strictly less than 100, prove that some two of them sum to 99.
Now I answered the problem differently than the official solution in the book and I was hoping you could tell me if my answer is still correct, or if it is not, explain to me where I went wrong. So here is my answer:
As the book suggests I try and use the pigeon hole principle. In my answer the "pigeons" are all two element subsets of the given set and the "holes" are the possible values the sum of the two elements of any two element subset can take. There are 1225 two-element subsets of any such given set and the sum of the two members of these subsets can take any value from 1+2=3 to 99+98=197. Hence there are 1225 "pigeons" and only 195 "boxes". There must be more than two pigeons in every box and hence at least two subsets with sum of 99.
Thats my answer. Thanks for any help!