2018 AMC 10A Problem 7

We can rewrite the expression as $$(2^5 \cdot 5^3) \cdot \left(\dfrac{2}{5}\right)^n = 2^{5 + n} \cdot 5^{3 - n}.$$ For this to be an integer, both exponents must be nonnegative. This means that $$\begin{aligned} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3. \end{aligned}$$ This gives us $5 + 3 + 1 = 9$ values for $n.$ Thus, E is the correct answer.