$31 \cdot 50$ satisfies the system of equations $\left\{\begin{array}{l}a b+b c=44 \\ a c+b c=23\end{array}\right.$ the number of positive integer triples $(a, b, c)$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(35th American High School Mathematics Examination, 1984)
[Solution] From the given conditions and the requirement to find the positive integer solutions of the system of equations, from the second equation $a c + b c = 23$, we have $c(a + b) = 23$.

Noting that 23 is a prime number, we can conclude that one of the two factors on the left side must be 1, and the other must be 23.

Since $a$ and $b$ are required to be positive integers, $a + b > 1$, so it must be that $c = 1$ and $a + b = 23$.

Substituting $c = 1$ and $b = 23 - a$ into the first equation, and after rearranging, we obtain a quadratic equation:
$$
a^{2} - 22a + 21 = 0.
$$

The two roots of this equation are $a = 1$ and $a = 21$. The corresponding values of $b$ are 22 and 2, respectively. Therefore, the positive integer solutions $(a, b, c)$ of the original system of equations can only be $(1, 22, 1)$ and $(21, 2, 1)$. Substituting them back into the original system of equations, both satisfy the equations.
Hence, the answer is (C).