2018 AMC 10A 第 7 题

Question

有多少个整数

n

（不一定是正数）使下式为整数？

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

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

把

4000

写成

2^5\cdot5^3

。

Write

4000

as

2^5\cdot5^3

2

和

5

的指数都必须是非负数。

The exponents of

2

and

5

must both be nonnegative

原式可改写为

(2^5 \cdot 5^3) \cdot \left(\dfrac{2}{5}\right)^n = 2^{5 + n} \cdot 5^{3 - n}\text{。}

要使它成为整数，两个指数都必须非负，即

\begin{aligned} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3 \end{aligned}\text{。}

所以

n

5 + 3 + 1 = 9

个取值。

因此正确答案是 E。

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.

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