Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "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? <math> \textbf{(A) }3...")
 
 
(44 intermediate revisions by 18 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2018 AMC 10A Problems/Problem 7|2018 AMC 10A #7]] and [[2018 AMC 12A Problems/Problem 7|2018 AMC 12A #7]]}}
 +
 +
==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?
  
Line 8: Line 11:
 
\textbf{(E) }9 \qquad
 
\textbf{(E) }9 \qquad
 
</math>
 
</math>
 +
 +
==Solution 1 (Algebra)==
 +
Note that <cmath>4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(2\cdot5^{-1}\right)^n=2^{5+n}\cdot5^{3-n}.</cmath>
 +
Since this expression is an integer, we need:
 +
<ol style="margin-left: 1.5em;">
 +
  <li><math>5+n\geq0,</math> from which <math>n\geq-5.</math></li><p>
 +
  <li><math>3-n\geq0,</math> from which <math>n\leq3.</math></li><p>
 +
</ol>
 +
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>
 +
 +
~MRENTHUSIASM
 +
 +
==Solution 2 (Observations)==
 +
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
 +
<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>
 +
 +
==Solution 3 (Brute Force)==
 +
The values for <math>n</math> are <math>-5, -4, -3, -2, -1, 0, 1, 2,</math> and <math>3.</math>
 +
 +
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.
 +
 +
In total, there are <math>\boxed{\textbf{(E) }9}</math> values for <math>n.</math>
 +
 +
~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 ==
 +
 +
{{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}}
 +
 +
[[Category:Introductory Number Theory Problems]]

Latest revision as of 04:26, 15 August 2024

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 \[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(2\cdot5^{-1}\right)^n=2^{5+n}\cdot5^{3-n}.\] 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_7&oldid=225044"