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

m (Solution 2)
m (Solution 3)
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<math>mn = 126</math>.
 
<math>mn = 126</math>.
  
Factoring 126, we get <math>6</math> pairs of numbers: <math>(1,126), (2,63), (3,42), (6,21), (7,18) & (9,14)</math>.
+
Factoring 126, we get <math>6</math> pairs of numbers: <math>(1,126), (2,63), (3,42), (6,21), (7,18),</math> and <math>(9,14)</math>.
  
Looking back at our equations for <math>m</math> and <math>n</math>, we can solve for <math>k = \frac{2m + 2n}{18} = \frac{m+n}{9}</math>. Since <math>k</math> is an integer, there are only <math>2</math> pairs of <math>(m,n)</math> that work: <math>(3,42) & (6,21)</math>. This means that there are <math>2</math> values of <math>k</math> such that <math>u</math> is an integer. But looking back at <math>u</math> in terms of <math>k</math>, we have <math>\pm</math>, meaning that there are <math>2</math> values of <math>u</math> for every <math>k</math>. Thus, the answer is <math>2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}</math>.
+
Looking back at our equations for <math>m</math> and <math>n</math>, we can solve for <math>k = \frac{2m + 2n}{18} = \frac{m+n}{9}</math>. Since <math>k</math> is an integer, there are only <math>2</math> pairs of <math>(m,n)</math> that work: <math>(3,42)</math> and <math>(6,21)</math>. This means that there are <math>2</math> values of <math>k</math> such that <math>u</math> is an integer. But looking back at <math>u</math> in terms of <math>k</math>, we have <math>\pm</math>, meaning that there are <math>2</math> values of <math>u</math> for every <math>k</math>. Thus, the answer is <math>2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}
 
{{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:19, 2 May 2015

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 $9$, $9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b$

Since $b = 3n$ for some positive integer $n$, we can rewrite the fraction(divide by $9$ on both top and bottom) as $\frac{a^2 + 14n^2}{3an}$

Since the denominator now contains a factor of $n$, we get $n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2$.

But since $1=gcd(a,b)=gcd(a,3n)=gcd(a,n)$, we must have $n=1$, and thus $b=3$.

For $b=3$ the original fraction simplifies to $\frac{a^2 + 14}{3a}$.

For that to be an integer, $a$ must divide $14$, and therefore we must have $a\in\{1,2,7,14\}$. Each of these values does indeed yield an integer.

Thus there are four solutions: $(1,3)$, $(2,3)$, $(7,3)$, $(14,3)$ and the answer is $\mathrm {(A)}$

Solution 2

Let's assume that $\frac{a}{b} + \frac{14b}{9a} = m$ We get--

$9a^2 - 9mab + 14b^2 = 0$

Factoring this, we get $4$ equations-

$(3a-2b)(3a-7b) = 0$

$(3a-b)(3a-14b) = 0$

$(a-2b)(9a-7b) = 0$

$(a-b)(9a-14b) = 0$

(It's all negative, because if we had positive signs, $a$ would be the opposite sign of $b$)

Now we look at these, and see that-

$3a=2b$

$3a=b$

$3a=7b$

$3a=14b$

$a=2b$

$9a=7b$

$a=b$

$9a=14b$

This gives us $8$ solutions, but we note that the middle term needs to give you back $9m$.

For example, in the case

$(a-2b)(9a-7b)$, the middle term is $-25ab$, which is not equal by $-9m$ for whatever integar $m$.

Similar reason for the fourth equation. This elimnates the last four solutions out of the above eight listed, giving us 4 solutions total $\mathrm {(A)}$

Solution 3

Let $u = \frac{a}{b}$. Then the given equation becomes $u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}$.

Let's set this equal to some value, $k \Rightarrow \frac{9u^2 + 14}{9u} = k$.

Clearing the denominator and simplifying, we get a quadratic in terms of $u$:

$9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}$

Since $a$ and $b$ are integers, $u$ is a rational number. This means that $\sqrt{(9k)^2 - 504}$ is an integer.

Let $\sqrt{(9k)^2 - 504} = x$. Squaring and rearranging yields:

$(9k)^2 - x^2 = 504$

$(9k+x)(9k-x) = 504$.

In order for both $x$ and $a$ to be an integer, $9k + x$ and $9k - x$ must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let $9k + x = 2m$ and $9k - x = 2n$.

Then:

$2m \cdot 2n = 504$

$mn = 126$.

Factoring 126, we get $6$ pairs of numbers: $(1,126), (2,63), (3,42), (6,21), (7,18),$ and $(9,14)$.

Looking back at our equations for $m$ and $n$, we can solve for $k = \frac{2m + 2n}{18} = \frac{m+n}{9}$. Since $k$ is an integer, there are only $2$ pairs of $(m,n)$ that work: $(3,42)$ and $(6,21)$. This means that there are $2$ values of $k$ such that $u$ is an integer. But looking back at $u$ in terms of $k$, we have $\pm$, meaning that there are $2$ values of $u$ for every $k$. Thus, the answer is $2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}$.

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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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