Show that from any five integers, one can always choose three of these integers such that their sum is divisible by 3.

Question

I wasn't sure how to solve this problem, can someone please help? Thanks!

FYI, this is a special case with $n = 3$ of a more general theorem of "Given any $2n−1$ integers, there is at least one subset of $n$ integers whose sum is divisible by $n$". The theorem statement & proof is given in Nine divisible by 9. For another specific case, there's Given any nine numbers, prove there exists a subset of five numbers such that its sum is divisible by $5$.. — John Omielan, May 27 '19 at 20:10

score 2 · Answer 1 · edited May 27 '19 at 16:47

First we assume it is possible for a set that does not satisfy the conditions to exist, and then we consider all numbers modulo $3$. Note that we cannot have three numbers the same mod 3 as adding them all together would yield a multiple of $3$. But this means that there is at least one number that is $0$, one number that is $1$ and one that is $2$ ($\text{mod }3$), and summing these gives a multiple of three, giving a contradiction.

Show that from any five integers, one can always choose three of these integers such that their sum is divisible by 3.

1 Answers1