OK, here's how I solve
this type of problem
. It also shows you that there's just this one solution.
We have to find values for S,E,N,D,M,O,R,Y (8 digit
s out of 10
). Now, we're adding 2
s. Since 9999
+9999 < 20000, M cannot be >=2
. And by the "usual rules" for this kind of question, it can't be 0. So M=1
Now, looking at the fourth (left-most) column, we have either S+1>=10 (if there's no carry) or 1+S+1>=10 (if there's carry). So S=8 or 9, and O=0 or 1. Since 1 is already taken, O=0 (just as well that, or the typography would get confusing).
In the third column, we can't have E+0=N (no carry), so E+1=N and there's carry from the second column. So in the second column either N+R=10+E=9+N, and R=9, or there's carry and 1+N+R=10+E=9+N, and R=8. So R=8 or 9, just like S.
IF S=8 and R=9,
we're looking at
But this cannot possibly work: we need to get either E+0=10+N or 1+E+0=10+N in the third column, to get carry in the fourth column. Neither is possible (we've already used up both 9 and 0). So...
S=9 and R=8.
We're looking at
We've already used up the digits 0,1,8,9, and N=E+1
, so the only choices for E are 6,5,4,3,2.
We know we must have carry from the first column into the second, so D+E>=10. D is at most 7, so we immediately rule out E=2 (7+2<10). Also E=3 is impossible (because then either D=7 and Y=0=O, or D<7 and E+D<10, both of which are impossible).
If E=4, then D=7 or D=6 don't work (because then Y=1 or Y=0, and both are already taken), and D<6 doesn't work because then E+D<10.
If E=6 then N=7, so D<=5. But D=5 yields Y=1 and D=4 yields Y=0, both taken, and D<=3 gives E+D<10.
So E=5 and N=6. D=7 (the alternative, D<=4, is again too small), so Y=2 and the solution is
We also see there are no other solutions to this cryptarithm