$31 \cdot 47$ If $m$ is a positive integer, and the line $13x + 11y = 700$ and $y = mx - 1$ intersect at an integral point, then $m$ can only be
(A) 4.
(B) 5.
(C) 6.
(D) 7.
(E) one of 4, 5, 6, 7.
(17th American High School Mathematics Examination, 1966)
[Solution] Substituting $y = mx - 1$ into $13x + 11y = 700$, we get
$$
x = \frac{711}{13 + 11m} = \frac{3^2 \cdot 79}{13 + 11m}.
$$

Since $x$ is an integer, $13 + 11m$ must be a factor of the numerator, which means $13 + 11m$ can only be one of the following numbers:
$$
1, 3, 3^2, 79, 3 \cdot 79, 3^2 \cdot 79.
$$

Let $13 + 11m = d$, then $m = \frac{d - 13}{11}$. Since $m > 0$, we have $d > 13$, so we only need to consider the last three numbers of the above six.
(1) If $d = 79$, then $m = \frac{d - 13}{11} = \frac{66}{11} = 6$;
(2) If $d = 3 \cdot 79 = 237$, then $m = \frac{237 - 13}{11}$ is not an integer;
(3) If $d = 3^2 \cdot 79 = 711$, then $m = \frac{711 - 13}{11}$ is also not an integer.
In summary, the only solution is $m = 6$.
Therefore, the answer is $(C)$.