Here is a nice high school olympiad math problem:

Can you choose 4 distinct positive integers so that the sum of each 3 of them is prime? How about 5?

It looks that just by looking at reminders mod 2,3,6 is not enough.


1 Answer 1


One important issue to consider in choosing the $4$ integers is the parity. For the sum of each of them to be prime means that you must have $2$ or less being even as the sum of $3$ even numbers is even. If you have $2$ or $1$ even (and $2$ or $3$ odd) numbers, then choosing $1$ even and $2$ odd numbers makes the sum even. Thus, you need to have all $4$ numbers being odd.

Considering remainders modulo $3$, note if $3$ or $4$ of the numbers have the same remainder, the sum of $3$ of them would be a multiple of $3$ and, thus, not prime. As such, you need to have $2$ groups of $2$ different remainders modulo $3$.

One set which meets these requirements is $\{1,3,7,9\}$ giving sums, when taken $3$ at a time, of $11,13,17,19$. In this case, there are $2$ numbers each with a remainder of $0$ and $1$ modulo $3$.

For $5$ (or more) integers, once again, you can't have $3$ or more with the same remainder modulo $3$. However, with at most $2$ having any the same remainders modulo $3$, there must be at least one integer each with a remainder of $0, 1$ and $2$, so the sum of those $3$ would, once again, be a multiple of $3$, so it's not a prime. Thus, you can't have $5$ or more such integers where the sum of each subset of $3$ is a prime.

FYI, this property that a set of $5$ integers always has a subset of $3$ that sum to a multiple of $3$ is an example of the more general case of, for all $n \in \mathbb{N}$, that a set of $2n - 1$ integers always has at least one of subset of $n$ integers which sum to a multiple of $n$. For larger specific cases, there's Given 7 arbitrary integers,sum of 4 of them is divisible by 4 and Among any $11$ integers, sum of $6$ of them is divisible by $6$. At a math seminar I attended just under $40$ years ago, we were told the general case was not yet proven. I just did a brief online search but could not find whether or not this is still the case.


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