2018 AMC 10A Problems/Problem 7: Difference between revisions
Ishankhare (talk | contribs) 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..." |
|||
| (46 intermediate revisions by 19 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 00:59, 3 November 2025
- 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 
is the value of 
an integer?
Solution 1 (Algebra)
Note that ![\[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(2\cdot5^{-1}\right)^n=2^{5+n}\cdot5^{3-n}.\]](http://latex.artofproblemsolving.com/d/a/1/da18869f564298c403c1de14e39521bd6a19ab29.png)
Since this expression is an integer, we need:
from which 
from which 
from which 
from which 
Taking the intersection gives 
So, there are 
integer values of 
~MRENTHUSIASM
Solution 2 (Observations)
Note that 
will be an integer if the denominator is a factor of 
. We also know that the denominator will always be a power of 
for positive values and a power of 
for all negative values. So we can proceed to divide
by 
for each increasing positive value of until we get a non-factor of and also divide by 
for each decreasing negative value of . For positive values we get 
and for negative values we get 
. Also keep in mind that the expression will be an integer for 
, which gives us a total of 
for
Solution 3 (Brute Force)
The values for are 
and 
The corresponding values for are 
and 
respectively.
In total, there are values for
~Little ~MRENTHUSIASM
Video Solution (HOW TO THINK CREATIVELY!)
Education, the Study of Everything
Video Solution
Video Solution
~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 | |
These problems are copyrighted © by the Mathematical Association of America. 
