2018 AMC 10A Problems/Problem 7: Difference between revisions

Revision as of 17:25, 8 February 2018

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

$\textbf{(A) }3 \qquad \textbf{(B) }4 \qquad \textbf{(C) }6 \qquad \textbf{(D) }8 \qquad \textbf{(E) }9 \qquad$

Solution

The prime factorization of $4000$ is $2^{5}5^{3}$ . Therefore, the maximum number for $n$ is $3$ , and the minimum number for $n$ is $-5$ . The range from $-5$ to $3$ , which is $3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}$ .

  ~Nosysnow

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 {{AMC10 box|year=2018|ab=A|num-b=6|num-a=8}}
+{{AMC12 box|year=2018|ab=A|num-b=6|num-a=8}}
 {{MAA Notice}}

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2018 AMC 10A Problems/Problem 7: Difference between revisions

Revision as of 17:25, 8 February 2018

Solution

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