AMC12 2007 B

AMC12 2007 B · Q24

AMC12 2007 B · Q24. It mainly tests Manipulating equations, GCD & LCM.

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?

有多少对正整数 $(a,b)$ 满足 $\text{gcd}(a,b)=1$ 且 $\frac{a}{b} + \frac{14b}{9a}$ 是整数？

(A) 4 4

(B) 6 6

(D) 12 12

(E) infinitely many 无穷多个

Answer

Correct choice: (A)

正确答案：（A）

Solution

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 $\mathrm{(A)}$

合并分式，$\frac{9a^2 + 14b^2}{9ab}$ 必须是整数。由于分母含有因子 $9$，有 $9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b$ 令 $b = 3n$（$n$ 为正整数），将分式上下同除以 $9$，得 $\frac{a^2 + 14n^2}{3an}$。由于分母含有因子 $n$，有 $n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2$。但因为 $1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)$，只能有 $n=1$，因此 $b=3$。当 $b=3$ 时，原式化为 $\frac{a^2 + 14}{3a}$。要使其为整数，$a$ 必须是 $14$ 的因子，因此 $a\in\{1,2,7,14\}$。这些取值确实都能得到整数。因此共有四组解：$(1,3)$、$(2,3)$、$(7,3)$、$(14,3)$，答案为 $\mathrm{(A)}$。

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