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

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Problem 24

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?

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

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,$ (Error compiling LaTeX. Unknown error_msg)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

@@ Line 122: / Line 122: @@
 Next, we have <math>ab \mid 9a^2+14b^2 \Rightarrow  ab \mid 14b^2 \Rightarrow a \mid 14b.</math>
-Thus,
+Thus, <math>a \in (1,2,7,4).</math>
+Next, we have <math>b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.
+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).$
+~coolmath2017
 ==See Also==
 {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}

2007 AMC 12B (Problems • Answer Key • Resources)
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