2007 AMC 12B 第 24 题

Clearing $b$ shows $\dfrac{14b^2}{9a}$ is an integer, and since $\gcd(a,b)=1,$ $14$ is divisible by $a$

Similarly $9$ is divisible by $b,$ so $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\};$ test the coprime pairs

Let $k$ be the integer value of the original expression. Multiplying by $b$ and subtracting $a$ gives $\dfrac{14b^2}{9a}=bk-a,$ an integer. Since $\gcd(a,b)=1,$ it follows that $14$ is divisible by $a.$ Multiplying instead by $9a$ and subtracting $14b$ gives $\dfrac{9a^2}{b}=9ak-14b,$ so $9$ is divisible by $b.$

Thus $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\}.$ Checking the coprime candidates, the expression is an integer only for $(a,b)=(1,3),$ $(2,3),$ $(7,3),$ $(14,3).$

So there are $4$ such pairs.

Thus, the correct answer is A.