2007 AMC 10B Problems/Problem 25

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?

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

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 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 $\mathrm{(A)}$

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 to $-9m$ for any integer $m$ .

Apply similar reasoning to 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, $a \in (1,2,7,14).$

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

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