OK, here's how I

solve this type of

problem. It also shows you that there's just this one solution.

SEND
+ MORE
======
MONEY

We have to find values for S,E,N,D,M,O,R,Y (

8 digits out of

10). Now, we're adding

2 4-digits

numbers. 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

8END
+ 109E
======
10NEY

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

9END
+ 108E
======
10NEY

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

9567
+ 1085
======
10652

We also see there are no other solutions to this

cryptarithm.