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Difference between revisions of "2018 AMC 10A Problems/Problem 7"
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{{duplicate|[[2018 AMC 10A Problems/Problem 7|2018 AMC 10A #7]] and [[2018 AMC 12A Problems/Problem 7|2018 AMC 12A #7]]}}
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==Problem==
 
For how many (not necessarily positive) integer values of <math>n</math> is the value of <math>4000\cdot \left(\tfrac{2}{5}\right)^n</math> an integer?
 
For how many (not necessarily positive) integer values of <math>n</math> is the value of <math>4000\cdot \left(\tfrac{2}{5}\right)^n</math> an integer?
  
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</math>
 
</math>
  
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==Solution==
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==Solution 1 (Algebra)==
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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:
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<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>3-(-5)+1=\boxed{\textbf{(E) }9}</math> integer values of <math>n.</math>
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~MRENTHUSIASM
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==Solution 2 (Observations)==
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Note that <math>4000\cdot \left(\frac{2}{5}\right)^n</math> will be an integer if the denominator is a factor of <math>4000</math>. We also know that the denominator will always be a power of <math>5</math> for positive values and a power of <math>2</math> for all negative values. So we can proceed to divide
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<math>4000</math> by <math>5^n</math> for each increasing positive value of <math>n</math> until we get a non-factor of <math>4000</math> and also divide <math>4000</math> by <math>2^{-n}</math> for each decreasing negative value of <math>n</math>. For positive values we get <math>n= 1, 2, 3</math> and for negative values we get <math>n= -1, -2, -3, -4, -5</math>. Also keep in mind that the expression will be an integer for <math>n=0</math>, which gives us a total of <math>\boxed{\textbf{(E) }9}</math> for <math>n.</math>
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==Solution 3 (Brute Force)==
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The values for <math>n</math> are <math>-5, -4, -3, -2, -1, 0, 1, 2,</math> and <math>3.</math>
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The corresponding values for <math>4000\cdot \left(\frac{2}{5}\right)^n</math> are <math>390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,</math> and <math>256,</math> respectively.
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In total, there are <math>\boxed{\textbf{(E) }9}</math> values for <math>n.</math>
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~Little ~MRENTHUSIASM
  
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The prime factorization of <math>4000</math> is <math>2^{5}5^{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>. 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 (HOW TO THINK CREATIVELY!)==
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https://youtu.be/vzyRAnpnJes
  
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  ~Nosysnow
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Education, the Study of Everything
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 +
 
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==Video Solution==
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https://youtu.be/ZiZVIMmo260
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 +
==Video Solution==
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https://youtu.be/2vz_CnxsGMA
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 +
~savannahsolver
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==Video Solution by OmegaLearn==
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https://youtu.be/ZhAZ1oPe5Ds?t=1763
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 +
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==
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{{AMC12 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}}
 
{{MAA Notice}}
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[[Category:Introductory Number Theory Problems]]

Revision as of 13:39, 3 July 2023

The following problem is from both the 2018 AMC 10A #7 and 2018 AMC 12A #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 (Algebra)

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 $3-(-5)+1=\boxed{\textbf{(E) }9}$ integer values of $n.$

~MRENTHUSIASM

Solution 2 (Observations)

Note that $4000\cdot \left(\frac{2}{5}\right)^n$ will be an integer if the denominator is a factor of $4000$. We also know that the denominator will always be a power of $5$ for positive values and a power of $2$ for all negative values. So we can proceed to divide $4000$ by $5^n$ for each increasing positive value of $n$ until we get a non-factor of $4000$ and also divide $4000$ by $2^{-n}$ for each decreasing negative value of $n$. For positive values we get $n= 1, 2, 3$ and for negative values we get $n= -1, -2, -3, -4, -5$. Also keep in mind that the expression will be an integer for $n=0$, which gives us a total of $\boxed{\textbf{(E) }9}$ for $n.$

Solution 3 (Brute Force)

The values for $n$ are $-5, -4, -3, -2, -1, 0, 1, 2,$ and $3.$

The corresponding values for $4000\cdot \left(\frac{2}{5}\right)^n$ are $390625, 156250, 62500, 25000, 10000, 4000, 1600, 640,$ and $256,$ respectively.

In total, there are $\boxed{\textbf{(E) }9}$ values for $n.$

~Little ~MRENTHUSIASM

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/vzyRAnpnJes

Education, the Study of Everything


Video Solution

https://youtu.be/ZiZVIMmo260

Video Solution

https://youtu.be/2vz_CnxsGMA

~savannahsolver

Video Solution by OmegaLearn

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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