Does there exist a set of exactly five positive integers such that the sum of any three distinct elements is prime? Does there exist $A \subseteq \mathbb N $ such that $|A|=5$ and sum of any $3$ distinct elements of $A$ is a prime number ?
 A: If $A$ contains at least three even elements, the sum of three even elements is an even number $> 2$, hence composite. If $A$ contains at least one even element, and at least two odd elements, the sum of an even element and two odd elements is an even number $> 2$, hence composite.
Thus such a set would necessarily have only odd elements, and the sum of three elements must then be at least $1 + 3 + 5 = 9 > 3$.
Let $B$ be a set of five odd natural numbers. If $B$ contains three elements that are congruent modulo $3$, their sum is a multiple of $3$, and hence composite. If in each residue class modulo $3$ there are at most two elements of $B$, then by the pigeon hole principle, for each $k \in \{0,1,2\}$ $B$ must contain an element $b_k \equiv k \pmod{3}$. Then $b_0 + b_1 + b_2 \equiv 0 \pmod{3}$.
So every set of five natural numbers contains a subset of three elements whose sum is composite.
A: There is exactly one prime which is $\equiv2\pmod6$, namely, $2$.
There is exactly one prime which is $\equiv3\pmod6$, namely, $3$.
All other primes are $\equiv1,5\pmod6$.

If the element $2$ is in the set, then the sum of every triplet containing it is divisible by $2$.
If $3$ elements are all $\equiv1\pmod6$, then their sum is $\equiv3\pmod6$ hence divisible by $3$.
If $3$ elements are all $\equiv5\pmod6$, then their sum is $\equiv3\pmod6$ hence divisible by $3$.

The only remaining option is a set with:


*

*$2$ elements which are $\equiv1\pmod6$

*$2$ elements which are $\equiv5\pmod6$

*The element $3$


Take an element $\equiv1\pmod6$, an element $\equiv5\pmod6$, and the element $3$.
Then their sum is $\equiv3\pmod6$ hence divisible by $3$.
