$28 \cdot 10$ Let $P=3659893456789325678$ and 342973489379256's product, then the number of digits in $P$ is 
(A) 36.
(B) 35.
(C) 34.
(D) 33.
(E) 32.
(20th American High School Mathematics Examination, 1969)
[Solution] Let the multiplier and multiplicand be $x, y$, respectively, then
$$
\begin{array}{l}
3.6 \cdot 10^{18}<x<3.7 \cdot 10^{18}, \\
3.4 \cdot 10^{14}<y<3.5 \cdot 10^{14} .
\end{array}
$$

Therefore,
$$
3.4 \cdot 3.6 \cdot 10^{32}<x y<3.5 \cdot 3.7 \cdot 10^{32} \text {, }
$$

which means
$$
1 \cdot 224 \cdot 10^{33}<x y<1 \cdot 295 \cdot 10^{33} \text {. }
$$

We know that $P=x y$ is a 34-digit number.
Thus, the answer is $(C)$.