$10 \cdot 7$ Consider the non-decreasing sequence of positive integers
$$
1,2,2,3,3,3,4,4,4,4,5,5,5,5,5, \cdots
$$

where the positive integer $n$ appears $n$ times. The remainder when the 1993rd term is divided by 5 is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4
(44th American High School Mathematics Examination, 1993)
[Solution] From the given sequence, we can obtain: the last position where the positive integer $n$ appears in the sequence is
$$
1+2+\cdots+n=\frac{n(n+1)}{2} \text {. }
$$

Since $\frac{62 \cdot 63}{2}=1953$, and $\frac{63 \cdot 64}{2}=2016$, this indicates that the 1993rd term in the sequence is 63. The remainder when 63 is divided by 5 is 3. Alternatively, from the inequality
$$
\frac{n(n+1)}{2} \leqslant 19930$, solving for $n$ gives $n<64$.
$\therefore$ The 1993rd term is 63.
Therefore, the answer is $(D)$.