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 $63$ . $63/5$ has a remainder of $3$ . $\fbox{D}$

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1993 AHSME Problems/Problem 16

Problem

Solution

See also