$32 \cdot 5 N$ is a 1992-digit natural number, where the digits $1,2,3, \cdots, 8$ appear in $N$ a number of times that are all multiples of 9. If the sum of the digits of $N$ is $N_{1}$, the sum of the digits of $N_{1}$ is $N_{2}$, and the sum of the digits of $N_{2}$ is $N_{3}$, then the value of $N_{3}$ is
(A) 3 .
(B) 6 .
(C) 9 .
(D) 12 .
(China Zhejiang Province Junior High School Mathematics Competition, 1993)
[Solution] From the given, we know that the number of times the digits $1, 2, \cdots, 8$ appear are all multiples of 9, and since 9 is a multiple of itself, it follows that the sum of the digits of $N$ is a multiple of 9.

Moreover, the sum of the digits of $N$, $N_{1}$, does not exceed $1992 \cdot 9=17928$, which is a five-digit number divisible by 9.
The sum of the digits of $N_{1}$, $N_{2}$, is less than $5 \cdot 9=45$, which is a two-digit number divisible by 9.

Therefore, the sum of the digits of $N_{2}$, $N_{3}$, is less than $2 \cdot 9=18$, and it is also divisible by 9, so $N_{3}=9$.
Thus, the answer is $(C)$.