1993 AHSME Problems/Problem 16

Problem

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

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4$

Solution

You want to find the largest integer x that satisfies $\frac{x*(x+1)}{2}<1993$. You then can find that value of x, which is $62$. Therefore, the next value of the sequence is $63$. $\frac{63}{5}$ has a remainder of $3$. $\fbox{D}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 16
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