Each letter represents a single digit number.

No two letters represent the same number. (Ex: if M=1 the no other letter in the problem can equal 1)

So far we've figured out that M=1, O=zero, and S is either 8 or 9

This is a very interesting problem, but we're a bit stuck and don't know where to go from here.

  • 1
    $\begingroup$ I can somehow agree that $M=1$ and $S \in \{ 8,9 \}$, but how did you conclude that $O=0$? $\endgroup$ – implicati0n Sep 12 '15 at 22:37
  • $\begingroup$ If M is 1 then S has to be 8 or 9 because 9 +1=10 or 8+1=9 and if you had a 1 carry from the hundreds place then (8+1)+1=10. It won't work with 7 or any number below that because 7+1 is 8 and even if it were to have a 1 carried from the hundred place it would still only equal 9 or a number below that. Now the reason O is 0 is that it can't be 1 because 1 is taken by M and it can't be any number larger than 1 because since M is 1, the largest number that you can have is 11; when you add (9+1)+1=11 and the only number that will carry from another place is 1. $\endgroup$ – John smith Sep 12 '15 at 22:50

Using your assumptions: $M=1, \ O=0, \ S=9$, by method of exhaustion of setting $E$ to be $2,3,4,\dots$ I got one (perhaps not the only) solution: $$ \textrm{SEND} = 9567, \textrm{MORE} = 1085, \textrm{MONEY} = 10652. $$

From your assumptions I deduced that $R=8$, because we would have to transfer $1$ in order to have $E\neq N$. Then you use the fact that when transferring the $1$ to the hundreds, you have $E=N+1$, and by exhausting digits, you find that for $E\in \{2,3,4\}$ there are no solutions which would yield that every letter is a unique digit. Then with $E=5$ you get this solution.


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