6. Find the largest possible number of elements in the set $\mathrm{M}$ of integers, which has the following property: for every triple of distinct numbers $z \mathrm{M}$, it is possible to select some two of them whose sum is a power of 2 with an integer exponent.
SOLUTION. Let us note at the outset that the mentioned power as an integer sum must have a non-negative exponent. We will only consider such powers of the number 2 further on.

We will prove that the set M can have at most 6 elements, as does, for example, the suitable (as we will verify immediately) set

$$
\{-1,3,5,-2,6,10\} \text {. }
$$

The sum of any two numbers from the triplet $-1,3,5$ is indeed a power of two, and the same is true for the triplet $-2,6,10$. Whenever we select three numbers from the mentioned set, some two of them will belong to one of the two mentioned triplets, and these two numbers will thus have a sum equal to a power of two.

Now, let us assume that there exists a suitable set with more than six elements. If the set M contained three non-positive numbers, the sum of any two of them would be negative, while every power of two is positive. Therefore, the set M contains at most two non-positive numbers, and thus at least five positive numbers. Let $x$ be the largest of them and $a, b, c, d$ some four other positive numbers from M.

Consider the four sums $x+a, x+b, x+c, x+d$. All of them are greater than $x$ and less than $2 x$. However, in the interval $(x, 2 x)$, there can be at most one power of two, so at most one of these four sums is equal to a power of two, and the remaining three (without loss of generality, $x+a, x+b, x+c$) are not powers of two. By applying the condition of the problem to the triples $(a, b, x)$, $(a, c, x)$, and $(b, c, x)$, we find that all three sums $a+b, a+c$, and $b+c$ must be powers of two. Without loss of generality, let $a=\max \{a, b, c\}$. Similarly as above, let us focus on the open interval $(a, 2 a)$. In it, there can be at most one power of two, but at the same time, it contains both (different) numbers $a+b$ and $a+c$, which we have already concluded must be powers of two. Thus, we have reached the desired contradiction.

Answer. The largest possible number of elements in the set $\mathrm{M}$ is 6.