2018 AMC 10A Problem 7

Factor $4000 = 2^5 \cdot 5^3$. The sign of $n$ decides which prime is being divided out.

Case $n \ge 0$. The expression is $\dfrac{2^5 \cdot 5^3 \cdot 2^n}{5^n}$. The power of $2$ is harmless; we need $5^n \mid 5^3$, so $n \le 3$. This gives $n = 0, 1, 2, 3$, four values.

Case $n < 0$. Put $n = -k$ with $k > 0$. Then $\left(\tfrac25\right)^{-k} = \left(\tfrac52\right)^k$ and the expression is $\dfrac{2^5 \cdot 5^3 \cdot 5^k}{2^k}$. We need $2^k \mid 2^5$, so $k \le 5$. This gives $k = 1, 2, 3, 4, 5$, five values.

Altogether $n$ ranges over $-5, -4, \dots, 3$, which is $4 + 5 = 9$ integers.

The answer is $\boxed{\textbf{(E)}\ 9}$.

Multiplying by $(2/5)^n$ trades powers of $2$ for powers of $5$ (or the reverse when $n$ is negative). The expression stays an integer exactly as long as the prime being divided out has not run out, so the two exponents in $4000 = 2^5 \cdot 5^3$ are the two limits. The phrase "not necessarily positive" is the signal that both directions matter.

• Considering only nonnegative n and answering 4, or forgetting n = 0 and answering 8.
• Mixing up which prime limits which direction, e.g. allowing $n$ up to $5$ because $4000$ has five factors of $2$; positive $n$ divides by $5$, not by $2$.