1
$\begingroup$

As stated on the title,

Is it possible to find $4024$ positive integers such that the sum of any $2013$ of them is not divisible by $2013$?

I used to assumed they have to be distinct. Being inspired by the answer below, I am thinking maybe we have to select those $4024$ numbers by two types: the first group of $2012$ integers being those that are divisible by $2013$, and the rest $2012$ integers are those not divisible by $2013$. So that if you pick any $2013$ of them you will end up having to pick at least $1$ of those $2012$ not-divisible-by-$2013$ integers. Does that work out?

If the numbers don't have to be distinct then the following suggestion seems to work well, but I am not sure whether they have to be distinct. I will update the question if there are any changes. Thank you guys.

$\endgroup$
3
  • $\begingroup$ If you just pick them so that none of them are divisible by 2013, then certainly no subset of them of size 2013 will have any numbers divisible by 2013. $\endgroup$ – Brendan W. Sullivan Oct 1 '12 at 5:49
  • $\begingroup$ Did you mean "the sum of any 2013 of them is not divisible by 2013" ? $\endgroup$ – Brendan W. Sullivan Oct 1 '12 at 5:50
  • $\begingroup$ oh yeah sorry that's what i meant, now the typo is corrected $\endgroup$ – fmat Oct 2 '12 at 5:38
5
$\begingroup$

Half of them 1, half of them 2013.

$\endgroup$
2
  • $\begingroup$ I definitely think this is working well when we are allowed to choose an integer more than 1 time. But just wondering if the condition that all the 4024 numbers have to be distinct, is my way of thinking above still correct? thank you. $\endgroup$ – fmat Oct 2 '12 at 6:34
  • $\begingroup$ If you want them to be distinct, you can take the numbers $1+2013k$ and $2013k$ for $k=1,2,\dots,2012$. $\endgroup$ – Gerry Myerson Oct 2 '12 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.