I4.2 If $x$ and $y$ be positive integers such that $x>y>1$ and $x y=x+y+A$. Let $B=\frac{x}{y}$, find the value of $B$.
$$
\begin{array}{l}
x y=x+y+22 \\
x y-x-y+1=23 \\
(x-1)(y-1)=23
\end{array}
$$
$\because 23$ is a prime number and $x>y>1$
$$
\begin{array}{l}
\therefore x-1=23, y-1=1 \\
x=24 \text { and } y=2 \\
B=\frac{24}{2}=12
\end{array}
$$