# Problem 138 Solution

Posted June 25, 1999

The solution to this one is rather hard to explain so be patient with me.

At first we have SEND+MORE=MONEY.

The most MONEY can be is 9753+8642=18395, thus the M must be a 1.

SEND+1ORE=1ONEY.

What is the most SEND+1ORE can be? 9864+1753=11,617. The second digit in the sum (O) can't be a 1 because that is already taken by M. Since it can't be greater than 1 it must be 0.

SEND+10RE=10NEY.

S is obviously a large number. We have less than 1100 to add to SEND to get a five digit number. Lets consider that possibility that S=8, the most SEND+10RE could be is 8975+1064 = 10039. However N cannot be 0 since that was already taken by O, thus S can not be 8, leaving the only plausible number to be 9.

9END+10RE=10NEY.

Subtract 10000 from both sides (9000 from SEND, 1000 from 10RE, and 10000 from 10NEY).

Now we have END+RE=NEY.

It is obvious that N=E+1. ND+RE can not add not equal or exceed 200, thus N can not exceed E by more than 1, and also can not be equal to 1.

EED+RE+10=EEY+100.

EED+RE=EEY+90.

Next subtract 100E from both sides.

ED+RE=EY+90.

Lets rearrange that a little.

ED+RE-EY=90.

RE+D-Y=90.

R must be 8 for the sum to have hope to reach 90.

8E+D-Y=90.

E+D-Y=10.

Recall that N=E+1 so E can not be 7, since 8 is already taken. 0, 1, 8, and 9 have already been taken, leaving us with the middle numbers. For this to work E and D will need to be on the large end and Y on the small end. Without considering the N=E+1 constraint here are the possibilities given the remaining numbers:

1. E=7, D=6, Y=3
2. E=6, D=7, Y=3
3. E=5, D=7, Y=2
The first possiblility is not an option because E=7, which would make N=8, but N is already taken. The second possibility is not an option because N=E+1, or N=7, which is also taken by D. The third option is the one that works.

To put it all together:

``` 9567
+1085
-----
10652
```

Michael Shackleford, ASA