Solve each of the following Cryptarithms:

Two possible values of $\mathrm{A}$ are :

(i) If $7+\mathrm{B} \leq 9$

$\therefore 3+\mathrm{A}=9$

$\Rightarrow \mathrm{A}=6$

But if $\mathrm{A}=6,7+\mathrm{B}$ must be larger than 9 . Hence, it is impossible.

(ii) If $7+B \geq 9$

$\therefore 1+3+\mathrm{A}=9$

$\Rightarrow \mathrm{A}=5$

If $\mathrm{A}=5$ and $7+\mathrm{B}=5, \mathrm{~B}$ must be 8

$\therefore \mathrm{A}=5, \mathrm{~B}=8$