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.


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.


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.


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.



Next subtract 100E from both sides.


Lets rearrange that a little.



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



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:


Michael Shackleford, ASA