PK 2019 Q22

It is clear that none of the integers can be 1, since then the diagonally opposite integer will certainly be a multiple of it.

Let the smallest integer be \(a\), and the diagonally opposite integer be \(b\). Since \(a\) is the smallest integer, the other two numbers must both be multiples of \(a\), say \(ma\) and \(na\), for some integers \(m\), \(n\). Now \(b\) cannot be a multiple of \(a\) since \(a\) and \(b\) are diagonally opposite, so \(b\) cannot be a multiple of its neighbours \(ma\) and \(na\); hence both \(ma\) and \(na\) are multiples of \(b\).

Suppose \(a\) and \(b\) have a factor \(k\) in common (with \(k > 1\)). Then the other two integers also have this factor since they are multiples of \(a\) and of \(b\). But then each of the four integers could be divided by this factor to produce four smaller integers with the desired properties. But we are looking for the smallest possible sum, hence \(a\) and \(b\) must not have any factors in common. The smallest pair of integers with no common factors to consider would be \(a = 2\) and \(b = 3\).

The other two integers must be multiples of both 2 and 3, but not of each other. We cannot use 6 because this would be a factor of any other multiple of both 2 and 3. The smallest possible pair is 12 and 18.

This gives a total of 2 + 3 + 12 + 18 which is 35.