2018 AMC 10A Problems/Problem 7

Problem

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 $2^{5}\cdot5^{3}$. Therefore, the maximum number for $n$ is $3$, and the minimum number for $n$ is $-5$. Then we must find the range from $-5$ to $3$, which is $3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}$.

Video Solution

https://youtu.be/ZiZVIMmo260

https://youtu.be/2vz_CnxsGMA

~savannahsolver

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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