Difference between revisions of "2002 AMC 12P Problems/Problem 5"

Revision as of 23:23, 30 December 2023

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 $b \Longrightarrow \boxed{\mathrm{E}}$ .

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.

@@ Line 1: / Line 1: @@
 == Problem ==
 For how many positive integers <math>m</math> is
 <cmath>\frac{2002}{m^2 -2}</cmath>

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Difference between revisions of "2002 AMC 12P Problems/Problem 5"

Revision as of 23:23, 30 December 2023

Problem

Solution

See also