2007 AMC 10B Problem 25

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.