Math CentralQuandaries & Queries


Question from JYOTI, a student:

the letters in the alphanumeric addition are all different. find the numbers indicated in codes. all letters are digits from 0 to 9. SEND+MORE=MONEY.

Hi Jyoti.

This will be a bit tricky to follow but I'll try to explain my reasoning as we go along.

When you add two single digit numbers, the largest sum is 19 and the smallest sum is 00. So the "carry" would be a 1 or nothing at all. If we presume that there are no leading zeros in this puzzle, then that means S + M = MO and the M is not zero. Therefore the M is a 1.


So S + 1 + (the carry from the 100s column) = 1O. Since the carry from the 100s cannot exceed 1, S must be at least 8. But if S is 8, then the carry must be 1 (otherwise we don't get a total more than 9). S might be 9, too. But if so, then we have 9 + 1 = 10 or 9 + 1 + 1 = 11. But this last scenario is wrong, because then we'd have MM instead of MO. So there are two possibilities: S = 8 and the 100s carry is a 1 or S = 9 and the 100s carry is a 0. Either way, the total is 10. So the O is a zero.

SEND + 10RE = 10NEY.
S = 8 or 9

But we know that E + 0 + carry from the 10s column = N perhaps with a carry. But if there is a carry going to the 1000s column, E + 0 + 1 = 10, in which case the sum digit is a zero, which is an M. Since we have MONEY, not MOMEY, this can't be the case, so there is no carry of the 100s column to the 1000s column. As well, this means that S must
be a 9.


Since E is not equal to N, there must be a carry from the 10s to the 100s, so ND + RE = 1EY, or in other words, N = E + 1. As well, at least one of N or R is higher than 4.

Now we are getting enough information just to try solutions and check them.

Try N = 8. Then E = 7, Then 78D + R7 = 87Y But then R couldn't be more than 6 and that's not enough to make the sum 87Y. So N is not 8.

Try N = 7. Then E = 6. Then 67D + R6 = 76Y. Here R must be an 8. Thus 67D + 86 = 76Y, with a carry obvious from the ones to tens column. To generate such a carry, D > 3. But D=4 makes Y=O and D=5 makes Y = M and 6,7,8,9 are taken. So N is not 7.

Try N = 6. Then E = 5. Then 56D + R5 = 65Y. Again, it looks like R must be 8 and a carry exists from the ones to the tens, so D >4. For similar reasons, D can only be 7. 567 + 85 = 65Y. So Y = 2.

9567 + 1085 = 10652.

We have a solution. Note that my "guess and check" method towards the end doesn't exclude other possible solutions, though.

Stephen La Rocque

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