# Is the sum divisible by 7?

Prove that given any 37 positive integers it is possible to choose 7 whose sum is divisible by 7.

Note by Yan Yau Cheng
4 years, 1 month ago

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Here's what I submitted. FYI this is the awesomest problem in the whole world!

We use mathematical induction. WLOG, because we are only considering terms $$\mod{7}$$, we can assume that all terms are $$1, 2, 3, 4, 5, 6,$$ or $$7$$. There are two super-general cases for some arbitrary $$7$$-term subset.

Case 1: They can all be the same, such as $$1, 1, 1, 1, 1, 1, 1$$.

Case 2: They can all be different, such as $$1, 2, 3, 4, 5, 6, 7$$.

Obviously, for either one of these conditions, their sum is divisible by $$7$$. Can we create a $$37$$-term set so that neither of these cases is fulfilled? Let’s try having $$6$$ $$1$$’s, $$6$$ $$2$$’s, $$6$$ $$3$$’s, etc. As soon as we’ve reached $$6$$ $$6$$’s, we are at a dilemma. If we put $$7$$, then we'll have a "Case 2" subset of $$7$$ different numbers. If we put anything else, we’ll already have $$6$$ of them and we'll have created a "Case 1". So we can never get to $$37$$ without fulfilling one of these general cases. In fact, we haven't even proven that there isn't a way to have a $$7$$-term subset divisible by $$7$$ in this $$36$$-term set. But that doesn't matter! We’re finished! Great problem! :D

- 4 years, 1 month ago

@Yan Yau Cheng In my submission, I see you marked a point off of one of my proof problems. Which one was it and why? Thanks. :D

- 4 years, 1 month ago