Determine all positive integers $n \geq 2$ such that, given any integers $a_1, a_2, ... , a_n$ , it is possible to find integers i and j ( $1 \leq i < j \leq n$ ) such that:

$a_i + a_{i+1}$ and $a_j + a_{j+1}$ leave the same remainder in the division by $n$

**Note:** Consider $a_{n+1} = a_1$.

Sorry, I can't really write in $\LaTeX$.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestRe Latex, you've got the code. You just needed to add in the Latex brackets of $\backslash ( \quad \backslash)$. I've edited your note so you can see my edits.

Log in to reply

@Calvin Lin Thank you very much! And do you have any idea of how I can proceed in this problem? I haven't made any progress.

Log in to reply

What have you tried? Where are you stuck?

What type of problem does this resemble? What approaches do you think should be pursued?

Log in to reply

@Calvin Lin

Log in to reply

I could do it if the integers didn't have to be onsecutive in the set of integers given, because there are n numbers with n possible remainders for them, so it would be easy. But like this, I really don't know how to start.

Log in to reply

That's a good start. You realized that if we set $b_i = a_i + a_{i+1}$, we would get $n$ numbers with $n$ possible remainders.

notbe different? Let's try small cases.Log in to reply

@Calvin Lin I think that if all sums had different remainders, the remainders of the numbers in odd positions of the set given would all be different and the ones in even positions would all be different.

Log in to reply

That's a thought. Continue developing it. Try small cases. Play around.

For $n = 2, 3, 4 , \ldots$, we have at most $n^2$ sets of remainders to test, so you should be able to make some initial guesses pretty quickly.

Log in to reply

@Calvin Lin I couldn't do it. I did what you said and tried a lot for $n=2 , 3, 4 ...$ but I couldn't make any progress. I should start with easier problems maybe. Have you solved this? How?

Log in to reply

I've not worked it out completely, but I believe the way I outlined below leads to a solution. I do believe that looking at small cases will help you get towards the approach, especially if you think of generalizing (instead of proving specific cases)

What conclusions did you get for small values? Is there a pattern?

This problem seems typical of pigeonhole principle, which is what you were suggesting when you said "n numbers, n possible remainders". We want to get some kind of conclusion / conclusion when all the remainders are distinct. For example, one thing that pops into mind, is the sum of all these remainders.

Log in to reply