42

Consider the thousands column. The letters represent different digits so, as $S$ is 3, $M$ is 2 and there is a carry of 1 from the hundreds column. Therefore $A$ is 9, $U$ is 0 and there is also a carry of 1 from the tens column. In the units column, $O+Y$ produces a units digit of 3, so $O+Y=$3 or $O+Y=$13. However, $O+Y=$3 requires one of $O$, $Y$ to equal zero (impossible as $U=$0) or 2 (also impossible as $M=$2). So $O+Y=$13. We can also deduce that $N$ is 8, since, in the tens column, 1 $+$ 3 $+N=$12. The pairs of digits that produce a sum of 13 are 4 and 9, 5 and 8, 6 and 7. As $A$ is 9 and $N$ is 8, the only possible values for $O$ and $Y$ are 6 and 7. These are interchangeable, but in both cases $Y\times O=$42.

*This problem is taken from the UKMT Mathematical Challenges.*