2018 AMC 10A Problem 7

Question

For how many (not necessarily positive) integer values of

n

is the following value an integer?

4000 \cdot \left(\dfrac{2}{5}\right)^n

3

4

6

8

9

Write

4000

as

2^5\cdot5^3

The exponents of

2

and

5

must both be nonnegative

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.

Problem 7 in other years