$29 \cdot 5$ is formed by the consecutive two-digit integers from 19 to 92 arranged in sequence to form the integer $N=$ $19202122 \cdots 909192$. If $3^{k}$ is the highest power of 3 that is a factor of $N$, then $k$ equals
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) Greater than 3.
(43rd American High School Mathematics Examination, 1992)
[Solution] A natural number $n$ is divisible by 3 or 9 $\Leftrightarrow$ the sum of its digits is divisible by 3 or 9.
Consider
$$
\begin{aligned}
& 1+9+2+0+2+1+\cdots+9+1+9+2 \\
= & 1+9+(2+3+4+5+6+7+8) \cdot 10+(1+2 \\
& +3+\cdots+7+8+9) \cdot 7+9+0+9+1+9+2 \\
= & 3 \cdot 235,
\end{aligned}
$$

Since 3 divides $3 \cdot 235$, and $9 \times 3 \cdot 235$, we have $k=1$. Therefore, the answer is $(B)$.