OFFICIAL EXPLENATION


\(\text { 35. } \frac{g^2 x+5 g x}{x} \text { and } 6\left(\frac{g}{2}\right)-100\left(\frac{g}{2}\right)^2=\text { only. }\)


In the first choice, $x$ can be factored out and canceled:

$$
\(\frac{g^2+5 g x}{x}=\frac{x\left(g^2+5 g\right)}{x}=g^2+5 g\)
$$


Since $g$ is an integer, so too is $\(g^2+5 g\)$.

In the second choice, $\(g^2\)$ is certainly an integer, but $\(x^2\left(\frac{1}{3}\right)\)$ is only an integer if $x=3$ (since 3 is the only prime number divisible by 3 ), so the second choice is not necesarily an integer.

When the third choice is simplified, $\(6\left(\frac{g}{2}\right)-100\left(\frac{g}{2}\right)^2=\)$

$\(3 g-\frac{100 g^2}{4}=3 g-25 g^2\)$ results; since $g$ is an integer, $\(3 g-\)$ $\(25 g^2\)$ is also an integer.