$31 \cdot 41$ Let real numbers $a, b, c$ satisfy $a+b+c=0, abc=1$. Then the number of positive numbers among $a, b, c$ is
(A) 3.
(B) 2.
(C) 1.
(D) 0.
(China Guangzhou, Wuhan, Fuzhou, etc. five cities junior high school mathematics league, 1990)
[Solution] Given $a+b+c=0$, if there is a positive number among them, then there must be a negative number (since $abc=1$, hence $a, b, c$ do not include 0).
Also, $abc=1>0$, so the number of negative numbers is 0 or 2. But from the above, the number of negative numbers cannot be 0.
Therefore, the number of positive numbers is 1.
Hence, the answer is (C).