It is immediately obvious that a carry will requre that $M$=1.

   S E N D
 + 1 O R E
-----------
 1 O N E Y

From this $S$ must be either 8 or 9. If $S$=9 and there is a carry from the 100's position, $O$=1, but this is not allowed since we have already determined that $M$=1. If there is no carry, then $O$=0. If $S$=8, there must have been a carry from the 100's which would again mean that $O$=0. We have then established that $O$=0. If there is a carry from the 100's position, it must be the case that $E$=9. But if this is so, then $N$=0, which is not possible since we have already determined that $O$=0. Since there is no carry we must have $S$=9:

   9 E N D
 + 1 0 R E
-----------
 1 0 N E Y

Since we cannot have $E = N$, it must be the case that there is a carry from the 10's position and that $N=E+1$. In general, the following equation must be also true (we have to add 10 to account for the carry).

\[ c + N + R = E + 10. \]

Where $c$ is the carry from the one's position. If we subtract $N=E+1$ from this equation, we have

\[ c + R = 9. \]

If there is no carry (i.e. $c=0$) then $R=9$, but $9$ is already spoken for, so $c=1$ and $R$ must be $8$:

   9 E N D
 + 1 0 8 E
-----------
 1 0 N E Y

Since there is a carry from the one's position, it must be the case that

\[ D+E = Y + 10 \]

Since $Y\ne 0, 1$, $D+E \gt 11$. Since $8$ and $9$ have already been used, we must have $ D,E \leq 7$. Together these facts mean that $13 \ge D+E \ge 12$. To satisfy this we must select $D$ and $E$ from the set ${7,6,5}$. Just plugging in the possible combinations, we find that only $D=7$ and $E=5$ works, so we have $Y=2$, $N=6$ and . . .

   9 5 6 7
 + 1 0 8 5
-----------
 1 0 6 5 2