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Difference between revisions of "2018 AMC 10A Problems/Problem 7"

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The prime factorization of <math>4000</math> is <math>2^{5}\cdot5^{3}</math>. Therefore, the maximum number for <math>n</math> is <math>3</math>, and the minimum number for <math>n</math> is <math>-5</math>. Then we must find the range from <math>-5</math> to <math>3</math>, which is <math>3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}</math>.
 
The prime factorization of <math>4000</math> is <math>2^{5}\cdot5^{3}</math>. Therefore, the maximum number for <math>n</math> is <math>3</math>, and the minimum number for <math>n</math> is <math>-5</math>. Then we must find the range from <math>-5</math> to <math>3</math>, which is <math>3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}</math>.
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==Video Solution==
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https://youtu.be/ZiZVIMmo260
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 +
https://youtu.be/2vz_CnxsGMA
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 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==

Revision as of 14:02, 2 July 2020

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 (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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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