Hamilton 2019 Q1

Suppose that there were \(c\) couples. Then each couple interacted with \(c - 1\) other couples (every couple except themselves).

Then \(c \times (c - 1)\) gives double the total number of interactions between couples (since we have counted each interaction exactly twice – once for each couple involved in it). So the total number of interactions is \(\tfrac{1}{2}c(c - 1)\).

Each interaction between couples results in four handshakes, so the total number of handshakes was \(4 \times \tfrac{1}{2}c(c - 1)\), that is, \(2c(c - 1)\).

Therefore \(2c(c - 1) = 31\,000\), so that \(c^2 - c - 15\,500 = 0\). Hence \((c - 125)(c + 124) = 0\), so that \(c = 125\) (ignoring the negative answer, which is not practicable here).

Thus there were 125 couples.

If you have not yet seen how to solve a quadratic equation, you could also notice that you are looking for two consecutive numbers which multiply to 15 500. Since one of these needs to be a multiple of 25 (can you see why?), completing a methodical search is not too onerous and, if explained thoroughly, is just as valid a solution.