2018 AMC 10A Problem 7

The problem says $n$ does not have to be positive, so check positive, zero and negative values.

1. Break $4000$ into primes. $4000 = 32 \cdot 125 = 2^5 \cdot 5^3$. So $4000$ has five $2$s and three $5$s.
2. Positive $n$: each step divides by $5$. Multiplying by $\tfrac25$ multiplies by $2$ and divides by $5$.
- $n = 1$: $4000 \cdot \tfrac25 = 1600$.
- $n = 2$: $1600 \cdot \tfrac25 = 640$.
- $n = 3$: $640 \cdot \tfrac25 = 256$.
- $n = 4$: $256 \cdot \tfrac25 = \tfrac{512}{5}$, not a whole number.

Once the three $5$s are used up, the result stops being a whole number. So $n = 1, 2, 3$ work.
3. $n = 0$. Anything to the power $0$ is $1$, so the value is $4000$. It works.
4. Negative $n$: each step divides by $2$. For example $\left(\tfrac25\right)^{-1} = \tfrac52$, so each step multiplies by $5$ and divides by $2$. There are five $2$s in $4000$, so we can do this $5$ times but not $6$:
- $n = -1, -2, -3, -4, -5$ work ($n = -5$ gives $4000 \cdot \tfrac{5^5}{2^5} = 125 \cdot 3125$, a whole number).
- $n = -6$ would need a sixth $2$, which is not there.
5. Count. $3$ positive values, $1$ zero, and $5$ negative values: $3 + 1 + 5 = 9$.

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

Multiplying by $\tfrac25$ swaps a $5$ for a $2$; multiplying by $\tfrac52$ swaps a $2$ for a $5$. The result stays a whole number only while there is still a $5$ (or a $2$) left to cancel. So the prime factorization of $4000$ tells you exactly how far you can go in each direction. The words "not necessarily positive" are a hint to check zero and the negatives.

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 $\boxed{\textbf{(E)}\ 9}$.