1. A number of couples met and each person shook hands with everyone else present, but not with themselves or their partners.
There were 31000 handshakes altogether.
How many couples were there?
Solution
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 $\frac{1}{2} c(c-1)$.
Each interaction between couples results in four handshakes, so the total number of handshakes was $4 \times \frac{1}{2} c(c-1)$, that is, $2 c(c-1)$.
Therefore $2 c(c-1)=31000$, so that $c^{2}-c-15500=0$. Hence $(c-125)(c+124)=0$, so that $c=125$ (ignoring the negative answer, which is not practicable here).
Thus there were 125 couples.
Note
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 15500 . 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.