2007 AMC 10B 第 25 题

Combine into the single fraction $\dfrac{9a^2+14b^2}{9ab}$ and use $\gcd(a,b)=1$

Deduce that $14$ is divisible by $a$ and $9$ is divisible by $b,$ then test the finitely many pairs

Combining, the expression is $\dfrac{9a^2+14b^2}{9ab}.$ For this to be an integer, $a$ must divide $9a^2+14b^2,$ hence $14b^2$ is divisible by $a.$ Since $\gcd(a,b)=1,$ we get that $14$ is divisible by $a.$ Similarly, $9a^2$ is divisible by $b,$ so $9$ is divisible by $b.$

So $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\}.$ Checking these, only $b=3$ makes the expression an integer for each allowed $a.$

The valid pairs are $(1,3),$ $(2,3),$ $(7,3),$ and $(14,3),$ for a total of $4.$

Thus, the correct answer is A.