SEND MORE MONEY..

Relevant wiki: Algebra Warmups - Cryptograms

In the above addition problem, each letter represents a different digit. Break the code so that the resulting numbers add up correctly.

$M$ is obviously 1, since there is no way there can be carryover than one when only two digits are added. Similarly, $O$ must be zero and $S$ either 8 or 9. A quick check shows that if $S=8, E \text{ must be equal to } 9$ , and $N$ will equal 0. But this is impossible because $O=0.$ So $S=9$ , making the result so far:

$9 \ E \ N \ D + 1\ 0 \ R \ E = 1 \ 0\ N \ E \ Y$ .

Since $E$ plus zero plus carry over equals $N$ , it follows that $N= E+1$ and that $N+R >9$ , there must be a carryover, or $N+R=E+10$ , if there is no carryover from the first column; on the other hand, if there were a carryover, we would have $N+R=E+9.$ Substituting for $N$ its equivalent value $E+1$ in these last two equations gives $E+1+R=10 \text { and } E+1+R=E+9 \text{ or } R=9 \text{ or } 8.$

Since $S=9$ , $R$ must equal $8 \text { and } E+D=12$ or more (there must be a carryover and the digits 0 and 1 have already been assigned). Therefore, the only possible values for $E, D, \text { and } N$ are $5, 6, \text { and } 7.$

$N= E+1 \text { and } E+D=12$ or more $\implies E=5, N=6, D=7 \text { and } Y=2\implies SEND+ MORE= MONEY \iff 9567+1085=10652$

Therefore, sum of digits is: $9+5+6+7+1+0+8+2= \boxed{38}$