$31 \cdot 49$ satisfies the system of equations $\left\{\begin{array}{l}x y+x z=255 \\ x y+y z=31\end{array}\right.$, the number of positive integer solutions $(x, y, z)$ is
(A) 3.
(B) 2.
(C) 1.
(D) 0.
(Chinese Hubei Province Huanggang Junior High School Mathematics Competition, 1993)
［Solution］The given system of equations can be transformed into
$$
\left\{\begin{array}{l}
x(y+z)=5 \cdot 3 \cdot 17, \\
y(x+z)=1 \cdot 31 .
\end{array}\right.
$$

Assuming the solutions to the equations are positive integers, then $x+z>1$. From (2), we know $y=1$, and $x+z=31$. The system of equations can be simplified to
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = 3 , } \\
{ y + z = 8 5 , } \\
{ y = 1 , } \\
{ x + z = 3 1 ; }
\end{array} \quad \left\{\begin{array}{l}
x=5, \\
y+z=51, \\
y=1, \\
x+z=31 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x = 1 5 , } \\
{ y + z = 1 7 , } \\
{ y = 1 , } \\
{ x + z = 3 1 ; }
\end{array} \quad \left\{\begin{array}{l}
x=17, \\
y+z=15, \\
y=1, \\
x+z=31 .
\end{array}\right.\right.
\end{array}
$$

From the last two systems of equations, we can find the solutions $(15,1,16),(17,1,14)$. Therefore, the answer is $(B)$.