Show that among any 52 integers, there are two whose sum or difference is divisible by 100.
Assume all the integers are placed into boxes based on their remainders, then there must be one box y that contains two integers.
If, (a - b) = (100x + y) - (100z + y), where x,z are integers.
= 100(x - z), which means 100*integer = (a - b).
Is this valid? I did some other examples, but the number of given integers was odd, which meant that there must be two integers in one box...so when given 52 that changes things up and I am wondering how that affects the problem and how I solve it? Any further explanation would be helpful. Thanks!