SEND MORE MONEY PROBLEM

Solving cryptarithms[edit]

Solving a cryptarithm by hand usually involves a mix of deductions and exhaustive tests of possibilities. For instance the following sequence of deductions solves the example SEND+MORE = MONEY puzzle above (columns are numbered from right to left):

${\displaystyle {\begin{matrix}&&{\text{S}}&{\text{E}}&{\text{N}}&{\text{D}}\\+&&{\text{M}}&{\text{O}}&{\text{R}}&{\text{E}}\\\hline =&{\text{M}}&{\text{O}}&{\text{N}}&{\text{E}}&{\text{Y}}\\\end{matrix}}}$

1. From column 5, M = 1 since it is the only carry-over possible from the sum of two single digit numbers in column 4.
2. 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.
3. 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 3 and S = 9.
4. If there were no carry in column 3 then E = N, which is impossible. Therefore there is a carry and N = E + 1.
5. 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.
6. To produce a carry in column 2, we must have D + E = 10 + Y.
7. Y is at least 2 so D + E is at least 12.
8. 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.
9. Since N = E + 1, E can't be 7 because then N = 8 = R so D = 7.
10. E can't be 6 because then N = 7 = D so E = 5 and N = 6.
11. D + E = 12 so Y = 2.

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