$36 \cdot 4$ Let $x=0.1234567891011 \cdots 998999$, this decimal is formed by writing from 1 to 999 after the decimal point. Then the 1983rd digit to the right of the decimal point is
(A) 2 .
(B) 3 .
(C) 5 .
(D) 7 .
(E) 8 .
(34th American High School Mathematics Examination, 1983)
[Solution] Observe the first 1983 digits, let $z$ represent the digit at the 1983rd position. We can divide this sequence of numbers into three segments:
$$
0 . \underbrace{123456789}_{A} \underbrace{101112 \cdots 9899}_{B} \underbrace{100101 \cdots z}_{C}
$$

Segment $A$ contains 9 digits, and segment $B$ contains $2 \cdot 90=180$ digits. Thus, segment $C$ contains $1983-189=1794$ digits.
Dividing 1794 by 3, the quotient is 598, so segment $C$ contains the first 598 three-digit numbers.
Since the first three-digit number is 100, the 598th three-digit number is 697, so $z=7$.
Therefore, the answer is $(D)$.