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

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== Problem ==
 
== Problem ==
 
For how many positive integers <math>m</math> is  
 
For how many positive integers <math>m</math> is  
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<cmath>\frac{2002}{m^2 -2}</cmath>
 
<cmath>\frac{2002}{m^2 -2}</cmath>
 +
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a positive integer?
  
 
<math>
 
<math>
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== Solution ==
 
== Solution ==
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If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
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For <math>\frac{2002}{m^2 -2}</math> to be an integer, <math>2002</math> must be divisible by <math>m^2-2.</math> Again, memorizing the prime factorization of <math>2002</math> is helpful. <math>2002 = 2 \cdot 7 \cdot 11 \cdot 13</math>, so its factors are <math>1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001</math>, and <math>2002</math>.
 +
 
 +
Since <math>m^2-2</math> equals all of these, adding <math>2</math> to our list and checking if they are perfect squares will suffice. <math>4, 9,</math> and <math>16</math> end up being our perfect squares, giving us an answer of <math>\boxed{\textbf{(C) } \text{three}}.</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=4|num-a=6}}
 
{{AMC12 box|year=2002|ab=P|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:10, 2 July 2024

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 (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. AMC logo.png

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