$36 \cdot 5$ Write down the natural numbers starting from 1, all the way to the 198th digit:
$$
\underbrace{123456789101112 \cdots}_{198\text{ digits}}
$$

Then the remainder of this number when divided by 9 is
(A) 4.
(B) 6.
(C) 7.
(D) None of the above.
(China Junior High School Mathematics League, 1987)
[Solution] It is easy to deduce that (the number of digits $1 \sim 9$ counts 9, two-digit numbers $10 \sim 99$ count 90, up to 198 digits) this number is
$$
1234 \cdots 9101112 \cdots 99100101102 \text{, }
$$

Since $1+2+\cdots+9$ is divisible by 9, that is,
$$
1+0+1+1+1+2+\cdots+9+9
$$

is also divisible by 9,
so the remainder when this number is divided by 9 is
$$
1+0+0+1+0+1+1+0+2=6 .
$$

Incidentally, $1+2+\cdots+9+1+0+1+1+\cdots+9+9$ has the same remainder when divided by 9 as $1+2+3+\cdots+9+10+11+\cdots+98+99$, and the latter sum is $\frac{99 \cdot 100}{2}=99 \cdot 50$, which is clearly divisible by 9.
Therefore, the answer is $(B)$.