2018 AMC 10A Problem 7

Below is the professionally curated solution for Problem 7 of the 2018 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2018 AMC 10A solutions, or check the answer key.

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Concepts:exponentprime factorizationcounting integers in a range

Difficulty rating: 1070

7.

For how many (not necessarily positive) integer values of nn is the following value an integer? 4000(25)n4000 \cdot \left(\dfrac{2}{5}\right)^n

33

44

66

88

99

Solution:

We can rewrite the expression as (2553)(25)n=25+n53n. (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, the exponents must be positive. This means that 5+n0n53n0n3. \begin{align*} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3. \end{align*}

This gives us 5+3+1=95 + 3 + 1 = 9 values for n.n.

Thus, E is the correct answer.

Problem 7 in Other Years