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.

#Combinatorics #JOMO #JOMO4

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years 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

Finn Hulse - 7 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$