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

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==Problem 24==
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== Problem ==
 
How many pairs of positive integers <math>(a,b)</math> are there such that <math>\text{gcd}(a,b)=1</math> and <math>\frac{a}{b} + \frac{14b}{9a}</math> is an integer?
 
How many pairs of positive integers <math>(a,b)</math> are there such that <math>\text{gcd}(a,b)=1</math> and <math>\frac{a}{b} + \frac{14b}{9a}</math> is an integer?
  
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<math>\mathrm {(A)} 4\quad\mathrm {(B)} 6\quad\mathrm {(C)} 9\quad\mathrm {(D)} 12\quad\mathrm {(E)} \text{infinitely many}</math>
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<math>\mathrm {(A)}\ 4\quad\mathrm {(B)}\ 6\quad\mathrm {(C)}\ 9\quad\mathrm {(D)}\ 12\quad\mathrm {(E)}\ \text{infinitely many}</math>
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==Solution 1==
 
  
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== Solutions ==
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=== Solution 1 ===
 
Combining the fraction, <math>\frac{9a^2 + 14b^2}{9ab}</math> must be an integer.
 
Combining the fraction, <math>\frac{9a^2 + 14b^2}{9ab}</math> must be an integer.
  
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Thus there are four solutions: <math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math> and the answer is <math>\mathrm{(A)}</math>
 
Thus there are four solutions: <math>(1,3)</math>, <math>(2,3)</math>, <math>(7,3)</math>, <math>(14,3)</math> and the answer is <math>\mathrm{(A)}</math>
  
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==Solution 2==
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=== Solution 2 ===
 
Let's assume that <math>\frac{a}{b} + \frac{14b}{9a} = m</math> We get
 
Let's assume that <math>\frac{a}{b} + \frac{14b}{9a} = m</math> We get
  
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Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total <math>\mathrm {(A)}</math>
 
Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total <math>\mathrm {(A)}</math>
  
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==Solution 3==
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=== Solution 3 ===
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Let <math>u = \frac{a}{b}</math>. Then the given equation becomes <math>u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}</math>.
 
Let <math>u = \frac{a}{b}</math>. Then the given equation becomes <math>u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}</math>.
  
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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>.
 
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>.
  
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==Solution 4==
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=== Solution 4 ===
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Rewriting the expression over a common denominator yields <math>\frac{9a^2 + 14b^2}{9ab}</math>. This expression must be equal to some integer <math>m</math>.
 
Rewriting the expression over a common denominator yields <math>\frac{9a^2 + 14b^2}{9ab}</math>. This expression must be equal to some integer <math>m</math>.
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Thus, <math>\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm</math>. Taking this <math>\pmod{a}</math> yields <math>14b^2 \equiv 0\pmod{a}</math>. Since <math>\gcd(a,b)=1</math>, <math>14 \equiv 0\pmod{a}</math>. This implies that <math>a|14</math> so <math>a = 1, 2, 7, 14</math>.
 
Thus, <math>\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm</math>. Taking this <math>\pmod{a}</math> yields <math>14b^2 \equiv 0\pmod{a}</math>. Since <math>\gcd(a,b)=1</math>, <math>14 \equiv 0\pmod{a}</math>. This implies that <math>a|14</math> so <math>a = 1, 2, 7, 14</math>.
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We can then take <math>9a^2 + 14b^2 = 9abm \pmod{b}</math> to get that <math>9 \equiv 0 \pmod{b}</math>. Thus <math>b = 1, 3, 9</math>.
 
We can then take <math>9a^2 + 14b^2 = 9abm \pmod{b}</math> to get that <math>9 \equiv 0 \pmod{b}</math>. Thus <math>b = 1, 3, 9</math>.
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However, taking <math>9a^2 + 14b^2 = 9abm \pmod{3}</math>, <math>b^2 \equiv 0\pmod{3}</math> so <math>b</math> cannot equal 1.
 
However, taking <math>9a^2 + 14b^2 = 9abm \pmod{3}</math>, <math>b^2 \equiv 0\pmod{3}</math> so <math>b</math> cannot equal 1.
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Also, note that if <math>b = 9</math>,  <math>\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}</math>. Since <math>a|14</math>, <math>\frac{14}{a}</math> will be an integer, but <math>\frac{a}{9}</math> will not be an integer since none of the possible values of <math>a</math> are multiples of 9. Thus, <math>b</math> cannot equal 9.
 
Also, note that if <math>b = 9</math>,  <math>\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}</math>. Since <math>a|14</math>, <math>\frac{14}{a}</math> will be an integer, but <math>\frac{a}{9}</math> will not be an integer since none of the possible values of <math>a</math> are multiples of 9. Thus, <math>b</math> cannot equal 9.
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Thus, the only possible values of <math>b</math> is 3, and <math>a</math> can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is <math>\mathrm{(A)}</math>.
 
Thus, the only possible values of <math>b</math> is 3, and <math>a</math> can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is <math>\mathrm{(A)}</math>.
  
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==Solution 5 (Similar to Solution 1)==
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=== Solution 5 (Similar to Solution 1) ===
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Rewriting <math>\frac{a}{b} + \frac{14b}{9a}</math> over a common denominator gives <math>\frac{9a^2 + 14b^2}{9ab}.</math>
 
Rewriting <math>\frac{a}{b} + \frac{14b}{9a}</math> over a common denominator gives <math>\frac{9a^2 + 14b^2}{9ab}.</math>
  
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~coolmath2017
 
~coolmath2017
  
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==See Also==
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== 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 00:05, 19 October 2020

Problem

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

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

Solutions

Solution 1

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 be a factor of $14$, 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 any integer $m$.

Similar reason for the fourth equation. This eliminates 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 $2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}$.

Solution 4

Rewriting the expression over a common denominator yields $\frac{9a^2 + 14b^2}{9ab}$. This expression must be equal to some integer $m$.

Thus, $\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm$. Taking this $\pmod{a}$ yields $14b^2 \equiv 0\pmod{a}$. Since $\gcd(a,b)=1$, $14 \equiv 0\pmod{a}$. This implies that $a|14$ so $a = 1, 2, 7, 14$.

We can then take $9a^2 + 14b^2 = 9abm \pmod{b}$ to get that $9 \equiv 0 \pmod{b}$. Thus $b = 1, 3, 9$.

However, taking $9a^2 + 14b^2 = 9abm \pmod{3}$, $b^2 \equiv 0\pmod{3}$ so $b$ cannot equal 1.

Also, note that if $b = 9$, $\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}$. Since $a|14$, $\frac{14}{a}$ will be an integer, but $\frac{a}{9}$ will not be an integer since none of the possible values of $a$ are multiples of 9. Thus, $b$ cannot equal 9.

Thus, the only possible values of $b$ is 3, and $a$ can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is $\mathrm{(A)}$.

Solution 5 (Similar to Solution 1)

Rewriting $\frac{a}{b} + \frac{14b}{9a}$ over a common denominator gives $\frac{9a^2 + 14b^2}{9ab}.$

Thus, we have $9 \mid 9a^2 + 14b^2 \Rightarrow 3 \mid b.$

Next, we have $ab \mid 9a^2+14b^2 \Rightarrow ab \mid 14b^2 \Rightarrow a \mid 14b.$

Thus, $a \in (1,2,7,4).$

Next, we have $b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.$

Thus, $b \in (1,3,9).$

Now, we simply do casework on $b.$

Plugging in $b = 1,3$ and $9$ gives that there are $4$ total solutions for $(a,b).$

~coolmath2017

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AMC 12 Problems and Solutions

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