$32 \cdot 7$ in the sequence $1,2,3, \cdots, 10000$, the sum of the digits of all numbers is
(A) 180001.
(B) 154756.
(C) 45001.
(D) 154755.
(E) 270001.
(21st American High School Mathematics Examination, 1970)
[Solution] Consider $0,1,2,3, \cdots, 9999$ each number multiplied by $10^{-4}$, which does not change their digit sum. At this point, these numbers become
$$
0.0000,0.0001,0.0002, \cdots, 0.9999 \text {. }
$$

Clearly, the ten digits $0,1,2, \cdots, 9$ appear the same number of times in different decimal places, i.e., 1000 times, so each digit appears 4000 times in the four decimal places.
Thus, the sum of the digits of the above decimals is
$$
(1+2+3+\cdots+9) \cdot 4000=180000 .
$$

In the sequence of numbers considered, since 0 is added and 10000 is removed, the original digit sum should be $180000+1=180001$.
Therefore, the answer is $(A)$.