Here is a problem that appeared in a Maths Battle in London a couple of weeks ago:
Donald, Jack, Peter, Richard and Steven have, in some order, the surnames Donaldson, Jackson, Peterson, Richardson and Stevenson. Donald is 1 year older than Donaldson, Jack is 2 years older than Jackson, Peter is 3 years older than Peterson, and Richard is 4 years older than Richardson. Who out of Steven and Stevenson is older, and by how much?
If you play around with this for long enough, you might get a feeling for what the answer must be; then pinning it down rigorously becomes the final hurdle. The following solution is stunning in its simplicity, and was the one presented by a 15-year-old student at the Maths Battle itself.
Let D, J, P, R and S be respectively the ages of Donald, Jack, Peter, Richard and Steven, and let D*, J*, P*, R* and S* be respectively the ages of Donaldson, Jackson, Peterson, Richardson and Stevenson. Since the forenames and surnames cover the same five people, we must have
D+J+P+R+S=D*+J*+P*+R*+S* .
But we are told that D=D*+1, J=J*+2, P=P*+3 and R=R*+4. Substituting all this into the equation above yields
(D*+1)+(J*+2)+(P*+3)+(R*+4)+S=D*+J*+P*+R*+S* ,
so that
S+10=S*.
Hence Stevenson is 10 years older than Steve.
I am a big fan of this problem – the statement is especially appealing, and the solution is wonderfully concise.