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?
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 cornersnwould again have to be even. Moreover,t he 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.
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