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

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 $3$ .

$\fbox{D}$

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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