2002 AMC 12P Problems/Problem 5

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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

If $\log_{b} 729 = n$, then $b^n = 729$. Since $729 = 3^6$, $b$ must be $3$ to some factor of 6. Thus, there are four (3, 9, 27, 729) possible values of $b \Longrightarrow \boxed{\mathrm{E}}$.

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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