Waste less time on Facebook — follow Brilliant.

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
2 years, 5 months ago

No vote yet
1 vote


Sort by:

Top Newest

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 Finn Hulse · 2 years, 5 months ago

Log in to reply

@Finn Hulse @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 Finn Hulse · 2 years, 5 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...