A probability problem by Paola Ramírez

Make a partition of the given set as follows:

$\left\{ 1,8,15,22,29,36,43,50 \right\}$ , $\left\{ 2,9,...,44 \right\}$ ,..., $\left\{ 7,14,...,49 \right\}$

So we have $7$ subsets; the ${ i }^{ th }$ , with $1\le i\le 7$ , contains the elements $k\in N$ such that $k\equiv i(mod\quad 7)$

Thus, if we choose an element from the ${ 6}^{ th }$ subset and one that belong to the ${ 1}^{ th }$ subset, their sum will be divisible by $7$ . But, if we choose any number of elements from the same subset (that is not the ${ 7 }^{ th }$ ), the sum of any pair won't be divisible by $7$ .

As the first subset has $8$ elements and in is the biggest, we pick the whole subset in order to optimize the quantity of elements in the final subset. Then, if we choose the ${ 2}^{ th }$ , we cannot choose any element that belong to the ${ 5}^{ th }$ . Similarly, if we pick the ${ 3 }^{ th }$ , we're not able to pick the ${4}^{ th }$ . Finally, we can pick at most one element from the ${7 }^{ th }$ subset, because if we have two or more, their sum will be divisible by $7$ .

Hence, we have picked $8+7+7+1=23$ elements from the initial subset. Note that it does not matter that we pick the ${ 2 }^{ th }$ , we could have picked the ${ 5 }^{ th }$ instead, and it would not have affected the result.

P.S.: This problem has been posted several times, I posted it time ago and deleted it because I realized that someone posted before me.