, 2024

The three different integers selected from $1$ to $6$ and whose sum is $7$ must be the integers $1$, $2$, $4$. Thus, the vertical column contains the integers $1$, $2$, $4$ in some order. (Can you see why no other combination of three of the given integers has a sum of $7$?) The three different integers selected from $1$ to $6$ and whose sum is $11$ must be $1$, $4$, $6$ or $2$, $4$, $5$ or $2$, $3$, $6$. If the integers in the horizontal row are $1$, $4$, $6$, then there are two integers in common with those in the vertical column, namely $1$ and $4$. Since there have to be five different integers used in the squares, then there cannot be two integers in common between the two lists, and so $1$, $4$, $6$ cannot appear in the horizontal row. Similarly, $2$, $4$, $5$ cannot appear in the horizontal row. Thus, the horizontal row must contain the integers $2$, $3$, $6$ with $2$ appearing in the centre square since it is the integer in common between the two lists. The integer not appearing in any square is $5$.

The figure shows a possible arrangement of the integers.