The Club of Six

Six friends, the men Tom, Dick and Harry and the women Anna, Cathy and Lucy, make up three married couples. Which man is married to which woman is exactly what we are asked to find.

One day they each went to market and bought some sheep. By chance, each of them bought as many sheep as the number of pounds they paid for a single one, so a person who bought $n$ sheep paid $n$ pounds apiece and spent $n^2$ pounds in all.

Comparing receipts afterwards, they noticed two things. Each husband had spent exactly £63 more than his own wife. And Tom had bought $23$ more sheep than Cathy, while Dick had bought $11$ more than Lucy.

Who is married to whom?

Solution

If a husband bought $h$ sheep and his wife $w$, he spent $h^2$ pounds and she spent $w^2$, and we are told $h^2 - w^2 = 63$. Written as $(h - w)(h + w) = 63$, and since $63$ is odd both factors must be odd. The three ways to split $63$ into two odd factors, $$(h - w,\ h + w) = (1, 63),\quad (3, 21),\quad (7, 9),$$ give $(h, w) = (32, 31)$, $(12, 9)$ and $(8, 1)$. So the husbands bought $32$, $12$ and $8$ sheep, and the wives bought $31$, $9$ and $1$.

Now use the clues. Tom bought $23$ more than Cathy. Among the husbands’ counts $32, 12, 8$ and the wives’ counts $31, 9, 1$, the only difference equal to $23$ is $32 - 9$, so Tom bought $32$ and Cathy bought $9$. Dick bought $11$ more than Lucy; of the men left, $12$ and $8$, and the women left, $31$ and $1$, the only difference equal to $11$ is $12 - 1$, so Dick bought $12$ and Lucy bought $1$. That leaves Harry with $8$ and Anna with $31$.

Finally, pair each husband with his wife using the counts found at the start: the man who bought $32$ is married to the wife who bought $31$, the one who bought $12$ to the wife who bought $9$, and the one who bought $8$ to the wife who bought $1$. Therefore $$\text{Tom is married to Anna,}\qquad \text{Dick to Cathy,}\qquad \text{Harry to Lucy.}$$ The pleasing twist is that neither woman named in the clues is the wife of the man she is compared with.