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

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

@@ Line 1: / Line 1: @@
-How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and:
+==Problem==
-<math>\frac{a}{b} + \frac{14b}{9a}</math>
+How many pairs of positive integers <math>(a,b)</math> are there such that <math>a</math> and <math>b</math> have no common factors greater than <math>1</math> and:
+<cmath>\frac{a}{b} + \frac{14b}{9a}</cmath>
 is an integer?
@@ Line 7: / Line 9: @@
 <math> \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }9\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{infinitely many} </math>
-==Solution==
+== Solution 1 ==
-Let <math>x = \frac{a}{b}</math>. We can then write the given expression as <math>x+\frac{14}{9x} = k</math> where <math>k</math> is an integer. We can rewrite this as a quadratic, <math>9x^2 - 9kx + 14 = 0</math>. By the Quadratic Formula, <math>x = \frac{9k\pm\sqrt{81k^2-504}}{18} = \frac{k}{2}\pm\frac{\sqrt{9k^2-56}}{6}</math>. We know that <math>x</math> must be rational, so <math>9k^2-56</math> must be a perfect square. Let <math>9k^2-56 = n^2</math>. Then, <math>56 = 9k^2-n^2 = (3k - n)(3k + n)</math>. The factors pairs of <math>56</math> are <math>1</math> and <math>56</math>, <math>2</math> and <math>28</math>, <math>4</math> and <math>14</math>, and <math>7</math> and <math>8</math>. Only <math>2</math> and <math>28</math> and <math>4</math> and <math>14</math> give integer solutions, <math>k = 5</math> and <math>n = 13</math> and <math>k = 3</math> and <math>n = 5</math>, respectively. Plugging these back into the original equation, we get <math>\boxed{4 (A)}</math> possibilities for <math>x</math>, namely <math>\frac{1}{3}, \frac{14}{3}, \frac{2}{3},</math> and <math>\frac{7}{3}</math>.
+Combining the fraction, <math>\frac{9a^2 + 14b^2}{9ab}</math> must be an integer.
+Since the denominator contains a factor of <math>9</math>, <math>9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b</math>
+Since <math>b = 3n</math> for some positive integer <math>n</math>, we can rewrite the fraction (divide by <math>9</math> on both numerator and denominator) as <math>\frac{a^2 + 14n^2}{3an}</math>
+Since the denominator now contains a factor of <math>n</math>, we get <math>n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2</math>.
+But since <math>1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)</math>, we must have <math>n=1</math>, and thus <math>b=3</math>.
+For <math>b=3</math> the original fraction simplifies to <math>\frac{a^2 + 14}{3a}</math>.
+For that to be an integer, <math>a</math> must be a factor of <math>14</math>, and therefore we must have <math>a\in\{1,2,7,14\}</math>. Each of these values does indeed yield an integer.
+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>
+-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
+<math>9a^2 - 9mab + 14b^2 = 0</math>
+Factoring this, we get <math>4</math> equations:
+<math>(3a-2b)(3a-7b) = 0</math>
+<math>(3a-b)(3a-14b) = 0</math>
+<math>(a-2b)(9a-7b) = 0</math>
+<math>(a-b)(9a-14b) = 0</math>
+(It's all negative, because if we had positive signs, <math>a</math> would be the opposite sign of <math>b</math>)
+Now we look at these, and see that:
+<math>3a=2b</math>, <math>3a=7b</math>
+<math>3a=b</math>, <math>3a=14b</math>
+<math>a=2b</math>, <math>9a=7b</math>
+<math>a=b</math>, <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>.
+For example, in the case
+<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>.
+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>
+== 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's set this equal to some value, <math>k \Rightarrow \frac{9u^2 + 14}{9u} = k</math>.
+Clearing the denominator and simplifying, we get a quadratic in terms of <math>u</math>:
+<math>9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}</math>
+Since <math>a</math> and <math>b</math> are integers, <math>u</math> is a rational number. This means that <math>\sqrt{(9k)^2 - 504}</math> is an integer.
+Let <math>\sqrt{(9k)^2 - 504} = x</math>. Squaring and rearranging yields:
+<math>(9k)^2 - x^2 = 504</math>
+<math>(9k+x)(9k-x) = 504</math>.
+In order for both <math>x</math> and <math>a</math> to be an integer, <math>9k + x</math> and <math>9k - x</math> must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let <math>9k + x = 2m</math> and <math>9k - x = 2n</math>.
+Then:
+<math>2m \cdot 2n = 504</math>
+<math>mn = 126</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>.
+== 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>.
+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>.
+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>.
+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.
+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.
+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>.
+== 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>
+Thus, we have <math>9 \mid 9a^2 + 14b^2 \Rightarrow  3 \mid b.</math>
+Next, we have <math>ab \mid 9a^2+14b^2 \Rightarrow  ab \mid 14b^2 \Rightarrow a \mid 14b.</math>
+Thus, <math>a \in (1,2,7,14).</math>
+Next, we have <math>b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.</math>
+Thus, <math>b \in (1,3,9).</math>
+Now, we simply do casework on <math>b.</math>
+Plugging in <math>b = 1,3</math> and <math>9</math> gives that there are <math>4</math> total solutions for <math>(a,b).</math>
+~coolmath2017
+== 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.
+Case 1.
+<math>I+R=56/3</math>, <math>I-R=1/3</math>. This yields <math>I=19/2</math>, which is not an integer. This case produces 0 solutions.
+Case 2.
+<math>I+R=28/3</math>, <math>I-R=2/3</math>. This yields <math>I=5</math>. Substituting into our original equation yields: <math>r^2-5r+\frac{14}{9}=0</math>. Factoring gives: <math>r=\frac{1}{3}</math>, <math>r=\frac{14}{3}</math>. This case produces 2 solutions, namely (1,3) and (14,3).
+Case 3.
+<math>I+R=14/3</math>, <math>I-R=4/3</math>. This yields <math>I=3</math>. Substituting into our original equation yields: <math>r^2-3r+\frac{14}{9}=0</math>. Factoring gives: <math>r=\frac{2}{3}</math>, <math>r=\frac{7}{3}</math>. This case produces 2 solutions, namely (2,3) and (7,3).
+Case 4.
+<math>I+R=8/3</math>, <math>I-R=7/3</math>. This yields <math>I=5/2</math>, which is not an integer. This case produces 0 solutions.
+Altogether, we have 4 solutions, so our answer is <math>\boxed{(A)}</math>.
+~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==
 {{AMC10 box|year=2007|ab=B|num-b=24|after=Last question}}
+{{MAA Notice}}

2007 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last question
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 10 Problems and Solutions

Page

Toolbox

Search