Use numbers from 0 to 9

SEND + MORE = MONEY

Answer fast before 4th of july

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Solving a cryptarithm by hand usually involves a mix of deductions and exhaustive tests of possibilities. 

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 3. But if there were a carry in column 3, 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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