Consider the non-decreasing sequence of positive integers
\[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots\]
in which the $n^{th}$ positive integer appears $n$ times. The remainder when the $1993^{rd}$ term is divided by $5$ is
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4$
The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of $n$'s ends at position $1+2+\dots+n$.
Therefore we want to find the smallest integer $n$ that satisfies $\frac{n(n+1)}{2}\geq 1993$.
By trial and error, the value of $n$ is $63$, and $63 \div 5$ has a remainder of $3$.
$\fbox{D}$