Solve each of the following cryptarithms.
$\begin{array}{cc}3& 7\\ +A& B\\ 9& A\end{array}$

$\begin{array}{cc}3& 7\\ +A& B\\ 9& A\end{array}$
Values oa A and B be from 0 to 9
In ten's digit 3+A=9
$\therefore$ A=6 or less.
$\therefore 7+B=A=6$ or less
$\therefore 7+9$ or $8=16$ or 15
$\therefore$ But it is two digit number
$\therefore$ B=8
Then A=5
$\begin{array}{cc}3& 7\\ +5& 8\\ 9& 5\end{array}$