Editing Verbal arithmetic (section)

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

<math>\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}</math>

#From column 5, '''M = 1''' since it is the only carry-over possible from the sum of two single digit numbers in column 4.
#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'''.
#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 (generated by the addition of column 3), N would be less than or equal to O. This is impossible since O = 0. Therefore there is no carry in column 4 and '''S = 9'''.
#If there were no carry in column 3 then E = N, which is impossible. Therefore there is a carry and N = E + 1.
#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'''.
#To produce a carry in column 2, we must have D + E = 10 + Y.
#Y is at least 2 so D + E is at least 12.
#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.
#Since N = E + 1, E can't be 7 because then N = 8 = R so '''D = 7'''.
#E can't be 6 because then N = 7 = D so '''E = 5''' and '''N = 6'''.
#D + E = 12 so '''Y = 2'''.
Another example of TO+GO=OUT (source is unknown):

<math>\begin{matrix}
     &   & \text{T} & \text{O}  \\
   + &   & \text{G} & \text{O} \\
 \hline
   = & \text{O} & \text{U} & \text {T}\\
\end{matrix}</math>

# The sum of two biggest two-digit-numbers is 99+99=198. So '''O=1''' and there is a carry in column 3.
# Since column 1 is on the right of all other columns, it is impossible for it to have a carry. Therefore 1+1=T, and '''T=2'''.
# As column 1 had been calculated in the last step, it is known that there isn't a carry in column 2. But, it is also known that there is a carry in column 3 in the first step. Therefore, 2+G≥10. If G is equal to 9, U would equal 1, but this is impossible as O also equals 1. So only '''G=8''' is possible and with 2+8=10+U, '''U=0'''.

The use of [[modular arithmetic]] often helps.  For example, use of mod-10 arithmetic allows the columns of an addition problem to be treated as [[simultaneous equations]], while the use of mod-2 arithmetic allows inferences based on the [[parity (mathematics)|parity]] of the variables.

In [[computer science]], cryptarithms provide good examples to illustrate the [[brute force search|brute force]] method, and algorithms that  generate all [[permutation]]s of ''m'' choices from ''n'' possibilities. For example, the Dudeney puzzle above can be solved by testing all assignments of eight values among the digits 0 to 9 to the eight letters S,E,N,D,M,O,R,Y, giving 1,814,400 possibilities. They also provide good examples for [[backtracking]] paradigm of [[algorithm]] design.