2018 AMC 12A Problem 7

Attempt Problem 7 of the 2018 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

$3$

$4$

$6$

$8$

$9$

Answer: E

Concepts:prime factorization exponent

Difficulty rating: 1380

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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