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Answer: 42

S is 3

The letters are all different so M is not 3, so 1 has been carried and M is 2

A and U are different so 1 has been carried


  
A + 1 makes 1 be carried into the last column,
so A + 1 = 10 and A = 9

So U = 0



 3 + N = 12 so N = 9 without 1 being carried,
or N = 8 with 1 being carried

N cannot be 9 because 9 is taken

N = 8 and 1 was carried


 
O + Y = 13

4 + 9 = 13 but 9 is taken
5 + 8 = 13 but 8 is taken
6 + 7 = 13


O and Y are 6 and 7 so O$\times$Y = 42



This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.