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 $2002$.

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 $\boxed{\textbf{(C) } \text{three}}.$

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
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

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