2002 AMC 12P Problems/Problem 5

Problem

For how many positive integers $m$ is

$\frac{2002}{m^2 -2}$

a positive integer?

$\text{(A) one} \qquad \text{(B) two} \qquad \text{(C) three} \qquad \text{(D) four} \qquad \text{(E) more than four}$

Solution

For $\frac{2002}{m^2 -2}$ to be an integer, $2002$ must be divisible by $m^2-2.$ Again, memorizing the prime factorization of $2002$ is helpful. $2002 = 2 \cdot 7 \cdot 11 \cdot 13$ , so its factors are $1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001,$ and $1002.$ Since $m^2-2$ equals all of these, adding $2$ to our list and checking if they are perfect squares will suffice. $4, 9,$ and $16$ end up being our perfect squares, giving us an answer of $\boxed{\textbf{(C) three}}.$

2002 AMC 12P (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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
All AMC 12 Problems and Solutions

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

Page

Toolbox

Search

2002 AMC 12P Problems/Problem 5

Problem

Solution

See also