2018 AMC 12A Problem 7

Below is the professionally curated solution for Problem 7 of the 2018 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 12A solutions, or check the answer key.

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Concepts:prime factorization exponent

Difficulty rating: 1380

For how many (not necessarily positive) integer values of $n$ is the value of $4000 \cdot \left(\tfrac{2}{5}\right)^n$ an integer?

$3$

$4$

$6$

$8$

$9$

Solution:

Since $4000 = 2^5 \cdot 5^3,$ the expression equals $2^{5+n} \cdot 5^{3-n}.$ This is an integer exactly when both $5 + n \ge 0$ and $3 - n \ge 0,$ that is, $-5 \le n \le 3.$ There are $3 - (-5) + 1 = 9$ such integers.

Thus, the correct answer is E.

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