2. The digits 1 to 8 are placed into the cells of the grid on the right, making four three-digit numbers when read clockwise. For which values of $k$ from 2 to 6 is it possible to create an arrangement such that all four of the three-digit numbers are multiples of $k$ ?
SOLUTION
For $k=2$, any arrangement with the four even numbers in the four corners will work.
For $k=3$, there are plenty of possibilities. For example, the numbers $132,285,567$ and 741 reading clockwise will work.

For $k=4$, the corners would all have to be even to make the numbers even. In addition, the final two digits of each number would have to be divisible by 4 . But neither $14,34,54$ or 74 is divisible by 4 . Hence this is impossible.

For $k=5$, the last digit of each would need to be 5 . As there is only one 5 available, this is impossible.
For $k=6$, the corners would again have to be even. Moreover, the four digit sums would have to be multiples of 3 . This would mean that $1+3+5+7+2(2+4+6+8)$ is a multiple of 3 , but the sum is 56 , which is not. Therefore this is also impossible.