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 $3$ . $\fbox{D}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 16
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1993 AHSME Problems/Problem 16

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