Forceful Subtraction

Relevant wiki: Cryptogram

We can rewrite the subtraction as an addition (because $A- B=C$ is equivalent to $C+B = A$ ):

$\Large{\begin{array}{ccccccc} && & & & \square & 0 &1 \\ -&& & & & & \square &2\\ \hline & & & & & & &\square \\ \hline \end{array}} \qquad \Leftrightarrow \qquad \Large{\begin{array}{ccccccc} & & & & & & &\square \\ +&& & & & & \square &2\\ \hline && & & & \square & 0 &1 \\ \hline \end{array}}$

By looking at the rightmost column, we see that the last digit of $\square + 2$ is 1, so the rightmost square is equal to 9 only. Thus, the cryptogram simplifies to

$\Large{\begin{array}{ccccccc} & & & & & & &\boxed9 \\ +&& & & & & \square &2\\ \hline && & & & \square & 0 &1 \\ \hline \end{array}}$

If the middle square is any positive single digit less than 9, then the sum $9 + \overline{\square \; 2 }$ cannot be a 3-digit number, so the middle square is forced to be a 9,

$\Large{\begin{array}{ccccccc} & & & & & & &\boxed9 \\ +&& & & & & \boxed9 &2\\ \hline && & & & \square & 0 &1 \\ \hline \end{array}}$

The last square can simply be evaulated by simple addition, $9 + 92 = \boxed101$ . Converting the cryptogram to the original form,

$\Large{\begin{array}{ccccccc} && & & & \boxed1 & 0 &1 \\ -&& & & & & \boxed9&2\\ \hline & & & & & & &\boxed9 \\ \hline \end{array}}$

and the answer is $1 + 9 + 9 = \boxed{19}$ .

The numbers must be very close to each other to give a single digit answer. The only case possible is when the two digit number is in 10th decade and 3 digit number is in 11th decade, or 101-92=9. So answer = 9+9+1=19