8.13 If $a^{x}=b^{y}=1994^{z}$ (where $a, b$ are natural numbers), and $\frac{1}{x}+\frac{1}{y}=$ $\frac{1}{z}$, then all possible values of $2 a+b$ are
(A) 1001.
(B) 1001 or 3989.
(C) 1001 or 1996.
(D) 1001 or 1996 or 3989.
(China Junior High School Mathematics League, 1994)
[Solution] From $a^{x}=1994^{z}$ and $b^{y}=1994^{z}$, we get
$$
a^{\frac{1}{z}}=1994^{\frac{1}{z}} \text { and } b^{\frac{1}{z}}=1994^{\frac{1}{y}} \text {. }
$$

Thus, $(a b) \frac{1}{z}=1994^{\frac{1}{x}}+\frac{1}{z}=1994^{\frac{1}{z}}$.
Therefore, $a b=1994=1 \times 1994=2 \times 997$.
However, $a, b$ are not equal to 1, so we have $a=2, b=997$ or $a=997, b=2$.
Hence, $2 a+b=1001$ or 1996.
Therefore, the answer is $(C)$.