Determine all positive integers n≥2 such that, given any integers a1,a2,...,an , it is possible to find integers i and j ( 1≤i<j≤n ) such that: ai+ai+1 and aj+aj+1 leave the same remainder in the division by n Note: Consider an+1=a1.
This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
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.
@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.
@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?
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.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
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
Re Latex, you've got the code. You just needed to add in the Latex brackets of \(\). I've edited your note so you can see my edits.
@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?
@Calvin Lin
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 bi=ai+ai+1, we would get n numbers with n possible remainders.
@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,…, we have at most n2 sets of remainders to test, so you should be able to make some initial guesses pretty quickly.
@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.