2018 AMC 12A Problem #7

$2$ and $5$ are coprime, so the numerator will not affect whether the expression is an integer. For example, no matter how many times a number is doubled, it will not change whether the product is divisible by $5$ . Therefore, we will only consider the denominator.

If $n$ is positive, the denominator will be a power of $5$ . If $n$ is negative, the denominator will be a power of $2$ . $4000 = 2^5 \cdot 5^3$ , so $n$ may not exceed these exponents. Therefore, $-5 \le n \le 3$ (including $0$ ), so there are $3-(-5)+1 = \boxed{9}$ possible values of $n$ .

Note for $4000(\frac{2}{5})^{n}$ the denominator of $(\frac{2}{5})^{n}$ must cancel with 4000.

Trivially we get $n=0 \Rightarrow 4000(\frac{2}{5})^{n}=4000$

Since $4000=2^{5}\cdot 5^{3}$ then if n is positive, then $5^n$ must cancel with $5^3$ , it follows $n\leq 3 \enspace [[n\in\mathbb{N}]]$ .

If n is negative, then $2^n$ must cancel with $2^5$ , it follows $n\geq -5 \enspace [[n\in\mathbb{N}]]$ .

Note if n is not an integer, $(\frac{2}{5})^{n}$ Is irrational.

Thus $-5\leq n\leq 3 \enspace [[n\in\mathbb{N}]]$ .

With $3-(-5)+1=9$ solutions.