Send More Money!

From column 5, M = 1 since it is the only carry-over possible from the sum of two single digit numbers in column 4. Since there is a carry in column 5, O must be less than or equal to M (from column 4). But O cannot be equal to M, so O is less than M. Therefore O = 0. Since O is 1 less than M, S is either 8 or 9 depending on whether there is a carry in column 4. But if there were a carry in column 4, N would be less than or equal to O (from column 3). This is impossible since O = 0. Therefore there is no carry in column 4 and S = 9. If there were no carry in column 3 then E = N, which is impossible. Therefore there is a carry and N = E + 1. If there were no carry in column 2, then ( N + R ) mod 10 = E, and N = E + 1, so ( E + 1 + R ) mod 10 = E which means ( 1 + R ) mod 10 = 0, so R = 9. But S = 9, so there must be a carry in column 2 so R = 8. To produce a carry in column 2, we must have D + E = 10 + Y. Y is at least 2 so D + E is at least 12. The only two pairs of available numbers that sum to at least 12 are (5,7) and (6,7) so either E = 7 or D = 7. Since N = E + 1, E can't be 7 because then N = 8 = R so D = 7. E can't be 6 because then N = 7 = D so E = 5 and N = 6. D + E = 12 so Y = 2

Remember cryptarithmetic conventions

Each letter or symbol represents only one digit throughout the problem; When letters are replaced by their digits, the resultant arithmetical operation must be correct; The numerical base, unless specifically stated, is 10; Numbers must not begin with a zero; There must be only one solution to the problem.

The puzzle SEND + MORE = MONEY, after solving, will appear like this:

        S E N D
        9 5 6 7
      + M O R E
        1 0 8 5
      ---------
      M O N E Y
      1 0 6 5 2

Thus, S+E+N+D+M+O+R+Y = 38 Hope its clear enough.

Let the numbers "to carry" be t, u, v, and w that I've written above the addition:

tuvw

SEND

+MoRE

MoNEY

The numbers "to carry" are either 0 or 1. Since the letter O looks like the digit 0 (zero), I will use Q instead to avoid confusion.

Obviously t = M = 1, so we have

1uvw SEND

+1oRE

1oNEY

Case 1: u=1

11vw SEND

+1oRE

1oNEY

Then S = 8, Q = 9

11vw 8END

+10RE

10NEY

This makes v=1 to prevent E=N But in order to cause a carry of 1, E could only be 9, causing N=0 but o is already 0. That rules out case u=1.

So u=0

10vw SEND

+1oRE

1oNEY

so S=9, o=0

10vw 9END

+10RE

10NEY

v=1 to prevent E=N

101w 9END

+10RE

10NEY

E + 1 = N w + N + R = 10 + E D + E = Y + 10w

Try w=0

E + 1 = N N + R = 10 + E D + E = Y

Substitute E + 1 for N in the second equation

E + 1 + R = 10 + E

That simplifies to

R = 9

But S = 9,

so w = 1

1011

9END

+10RE

10NEY

N + R = E + 9

1 + E = N

D + E = Y + 10

Substitute 1 + E for N in the 1st equation

1 + E + R = E + 9

    R = 8

1011

9END

+108E

---

10NEY

E + 1 = N D + E = Y + 10

0 and 1 are used, so Y > 1, so

D + E > 11

So the only possibility is

{D,E} = {6,7} or {D,E} = {5,7}

Try {D,E} = {6,7}

So Y = 3

1011

9END

+108E

---

10NE3

But E has to be 1 more than E

E can't be 7 because that would make N=8=R.

E can't be 6 because that would make N and D

both be 7.

So

{D,E} = {5,7} which makes Y = 2

1011 9END

+108E

10NE2

Since N is 1 more than E, E can't be 7,

for that would make N=8=R, so E=5, and D=7

1011

95N7

+1085

---

10N52

So N=6

1011

9567

+1085

---

10652

so answer is 38

The cryptarithm is:

$9567 + 1085 = 10652$

& the value of $\mathsf{S} + \mathsf{E} + \mathsf{N} + \mathsf{D} + \mathsf{M} + \mathsf{O} + \mathsf{R} + \mathsf{Y} = 9 + 5 + 6 + 7 + 1 + 0 + 8 + 2 = \boxed{38}$ .