Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AMC 12B Problems/Problem 24"

(→‎Problem 24)
(→‎Solution)
Line 35: Line 35:
  
 
<math>m \in (1,2,7,14)</math>
 
<math>m \in (1,2,7,14)</math>
 +
 +
Checking back to the fraction, all of these <math>m</math> do indeed yield integers.
  
 
Now, returning to <math>a</math> and <math>b</math>
 
Now, returning to <math>a</math> and <math>b</math>

Revision as of 14:01, 20 February 2008

Problem 24

How many pairs of positive integers $(a,b)$ are there such that $gcd(a,b)=1$ and \[\frac{a}{b}+\frac{14b}{9a}\] is an integer?

$\mathrm {(A)} 4$ $\mathrm {(B)} 6$ $\mathrm {(C)} 9$ $\mathrm {(D)} 12$ $\mathrm {(E)} \text{infinitely many}$

Solution

Combining the fraction, $\frac{9a^2 + 14b^2}{9ab}$ must be an integer.

Since the denominator contains a factor of $9$,

$9 | 9a^2 + 14b^2$

$9 | b^2$

$3 | b$

Rewriting $b$ as $b = 3n$ for some positive integer $n$, we can rewrite the fraction $\frac{a^2 + 14n^2}{3an}$

Since the denominator now contains a factor of $n$,

$n | a^2 + 14n^2$

$n | a^2$

$n | a$

Rewriting $a$ as $a = mn$ for some positive integer $m$, we can rewrite the fraction again as $\frac{m^2 + 14}{3m}$

Since the denominator contains $m$,

$m | m^2 + 14$

$m | 14$

$m \in (1,2,7,14)$

Checking back to the fraction, all of these $m$ do indeed yield integers.

Now, returning to $a$ and $b$

$a \in (n,2n,7n,14n)$ and $b = 3n$

Since $gcd(a,b) = 1$, $n$ must be $1$. This yields four possible pairs $(1,3)$, $(2,3)$, $(7,3)$, $(14,3)$

$4 \Rightarrow \mathrm {(A)}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&oldid=23522"