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

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</math>
 
</math>
  
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==Solution==
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==Solution 1==
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The prime factorization of <math>4000</math> is <math>2^{5}\cdot5^{3}</math>. Therefore, the maximum integer value for <math>n</math> is <math>3</math>, and the minimum integer value 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 = \boxed{\textbf{(E) }9}</math>.
  
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The prime factorization of <math>4000</math> is <math>2^{5}\cdot5^{3}</math>. Therefore, the maximum integer value for <math>n</math> is <math>3</math>, and the minimum integer value 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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==Solution 2==
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Note that <cmath>4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.</cmath>
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Since this expression is an integer, we need
 +
<ol style="margin-left: 1.5em;">
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  <li><math>5+n\geq0,</math> from which <math>n\geq-5.</math></li><p>
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  <li><math>3-n\geq0,</math> from which <math>n\leq3.</math></li><p>
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</ol>
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Taking the intersection gives <math>-5\leq n\leq3.</math> So, there are <math>\boxed{\textbf{(E) }9}</math> integer values of <math>n.</math>
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 +
~MRENTHUSIASM
  
 
==Video Solutions==
 
==Video Solutions==

Revision as of 07:15, 13 August 2021

The following problem is from both the 2018 AMC 12A #7 and 2018 AMC 10A #7, so both problems redirect to this page.

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 1

The prime factorization of $4000$ is $2^{5}\cdot5^{3}$. Therefore, the maximum integer value for $n$ is $3$, and the minimum integer value for $n$ is $-5$. Then we must find the range from $-5$ to $3$, which is $3-(-5) + 1 = 8 + 1 = \boxed{\textbf{(E) }9}$.

Solution 2

Note that \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\] Since this expression is an integer, we need

  1. $5+n\geq0,$ from which $n\geq-5.$

  2. $3-n\geq0,$ from which $n\leq3.$

Taking the intersection gives $-5\leq n\leq3.$ So, there are $\boxed{\textbf{(E) }9}$ integer values of $n.$

~MRENTHUSIASM

Video Solutions

https://youtu.be/ZiZVIMmo260

https://youtu.be/2vz_CnxsGMA

~savannahsolver

https://youtu.be/vzyRAnpnJes

Education, the Study of Everything

https://youtu.be/ZhAZ1oPe5Ds?t=1763

~ pi_is_3.14

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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