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Difference between revisions of "2007 AMC 10B Problems/Problem 25"

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<math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} </math>
 
<math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} </math>
  
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== Solutions ==
+
== Solution 1 ==
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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 ===
+
-Or we can see that the smallest choice is <math>4</math>, and just go with it after getting <math>a\in\{1,2,7,14\}.</math>
 +
 
 +
== 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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Now we look at these, and see that:
 
Now we look at these, and see that:
  
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<math>3a=2b</math>  
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<math>3a=2b</math>, <math>3a=7b</math>
−
 
 
−
<math>3a=b</math>
 
−
 
 
−
<math>3a=7b</math>
 
−
 
 
−
<math>3a=14b</math>
 
  
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<math>a=2b</math>  
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<math>3a=b</math>, <math>3a=14b</math>
  
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<math>9a=7b</math>
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<math>a=2b</math>, <math>9a=7b</math>
  
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<math>a=b</math>  
+
<math>a=b</math>, <math>9a=14b</math>
  
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<math>9a=14b</math>
+
This gives us <math>8</math> solutions, but we note that the middle term must be divisible by <math>-9</math>.
−
 
 
−
This gives us <math>8</math> solutions, but we note that the middle term needs to give you back <math>9m</math>.
 
  
 
For example, in the case  
 
For example, in the case  
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<math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not equal to <math>-9m</math> for any integer <math>m</math>.
 
<math>(a-2b)(9a-7b)</math>, the middle term is <math>-25ab</math>, which is not equal to <math>-9m</math> for any integer <math>m</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>
+
Apply similar reasoning to the fourth equation. This eliminates four solutions out of the above eight listed, giving us <math>4</math> solutions total <math>\mathrm {(A)}</math>
  
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=== Solution 3 ===
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== Solution 3 ==
 
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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<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),</math> and <math>(9,14)</math>.
+
Factoring <math>126</math>, 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)</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 ===
+
== Solution 4 ==
 
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, 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) ===
+
== Solution 5 (Similar to Solution 1) ==
 
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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=== Solution 6 (Similar to solution 3) ===
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== Solution 6 (Similar to solution 3) ==
 
Let <math>\frac{a}{b} = r.</math> So <math>r + \frac{14}{9r}=I</math>, where I is an integer. Algebraic manipulations yield: <math>r^2-Ir+\frac{14}{9}=0</math>. The discriminant of this must be the square of a rational number, call this R. So <math>I^2-\frac{56}{9}=R^2 \longrightarrow I^2-R^2=(I-R)(I+R)=\frac{56}{9}</math>. I is <math>\frac{1}{2}</math> the sum of <math>I-R</math> and <math>I+R</math>. To have an integer sum, <math>I-R</math> and <math>I+R</math> must have the same denominator, namely 3. We proceed with casework.
 
Let <math>\frac{a}{b} = r.</math> So <math>r + \frac{14}{9r}=I</math>, where I is an integer. Algebraic manipulations yield: <math>r^2-Ir+\frac{14}{9}=0</math>. The discriminant of this must be the square of a rational number, call this R. So <math>I^2-\frac{56}{9}=R^2 \longrightarrow I^2-R^2=(I-R)(I+R)=\frac{56}{9}</math>. I is <math>\frac{1}{2}</math> the sum of <math>I-R</math> and <math>I+R</math>. To have an integer sum, <math>I-R</math> and <math>I+R</math> must have the same denominator, namely 3. We proceed with casework.
  
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~Math4Life2020
 
~Math4Life2020
 +
 +
== Solution 7 ==
 +
Let <math>\frac{a}{b} = x</math> and <math>\frac{14b}{9a} = y.</math> Then for rational numbers <math>x,y</math> we have <math>xy = \frac{14}{9}</math> and <math>x+y \in \mathbb{Z}.</math> One can prove that <math>x</math> and <math>y</math> must both have the same denominator in order for their sum to be an integer. This denominator is <math>3</math> meaning <math>b=3</math> and <math>a</math> is any factor of <math>14.</math> This yields <math>\boxed{(A) 4}</math> solutions.
 +
~ab2024
  
 
==See Also==
 
==See Also==

Latest revision as of 13:14, 30 June 2024

Problem

How many pairs of positive integers $(a,b)$ are there such that $a$ and $b$ have no common factors greater than $1$ and:

\[\frac{a}{b} + \frac{14b}{9a}\]

is an integer?

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

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 numerator and denominator) 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 $\mathrm{(A)}$

-Or we can see that the smallest choice is $4$, and just go with it after getting $a\in\{1,2,7,14\}.$

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:

$3a=2b$, $3a=7b$

$3a=b$, $3a=14b$

$a=2b$, $9a=7b$

$a=b$, $9a=14b$

This gives us $8$ solutions, but we note that the middle term must be divisible by $-9$.

For example, in the case

$(a-2b)(9a-7b)$, the middle term is $-25ab$, which is not equal to $-9m$ for any integer $m$.

Apply similar reasoning to the fourth equation. This eliminates 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, $a \in (1,2,7,14).$

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

Solution 6 (Similar to solution 3)

Let $\frac{a}{b} = r.$ So $r + \frac{14}{9r}=I$, where I is an integer. Algebraic manipulations yield: $r^2-Ir+\frac{14}{9}=0$. The discriminant of this must be the square of a rational number, call this R. So $I^2-\frac{56}{9}=R^2 \longrightarrow I^2-R^2=(I-R)(I+R)=\frac{56}{9}$. I is $\frac{1}{2}$ the sum of $I-R$ and $I+R$. To have an integer sum, $I-R$ and $I+R$ must have the same denominator, namely 3. We proceed with casework.

Case 1. $I+R=56/3$, $I-R=1/3$. This yields $I=19/2$, which is not an integer. This case produces 0 solutions.

Case 2. $I+R=28/3$, $I-R=2/3$. This yields $I=5$. Substituting into our original equation yields: $r^2-5r+\frac{14}{9}=0$. Factoring gives: $r=\frac{1}{3}$, $r=\frac{14}{3}$. This case produces 2 solutions, namely (1,3) and (14,3).

Case 3. $I+R=14/3$, $I-R=4/3$. This yields $I=3$. Substituting into our original equation yields: $r^2-3r+\frac{14}{9}=0$. Factoring gives: $r=\frac{2}{3}$, $r=\frac{7}{3}$. This case produces 2 solutions, namely (2,3) and (7,3).

Case 4. $I+R=8/3$, $I-R=7/3$. This yields $I=5/2$, which is not an integer. This case produces 0 solutions.

Altogether, we have 4 solutions, so our answer is $\boxed{(A)}$.

~Math4Life2020

Solution 7

Let $\frac{a}{b} = x$ and $\frac{14b}{9a} = y.$ Then for rational numbers $x,y$ we have $xy = \frac{14}{9}$ and $x+y \in \mathbb{Z}.$ One can prove that $x$ and $y$ must both have the same denominator in order for their sum to be an integer. This denominator is $3$ meaning $b=3$ and $a$ is any factor of $14.$ This yields $\boxed{(A) 4}$ solutions. ~ab2024

See Also

2007 AMC 10B (Problems • Answer Key • Resources)
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