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

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== Solution ==
 
== Solution ==
 
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</math>, and <math>1001</math>.  
 
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</math>, and <math>1001</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>
 
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>
  

Revision as of 02:09, 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 $1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286$, and $1001$.

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

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