Difference between revisions of "2018 AMC 10A Problems/Problem 7"
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~Little ~MRENTHUSIASM | ~Little ~MRENTHUSIASM | ||
| − | ==Solution | + | ==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 | 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^ | + | <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> |
==Video Solutions== | ==Video Solutions== | ||
Revision as of 15:24, 21 July 2022
- The following problem is from both the 2018 AMC 10A #7 and 2018 AMC 12A #7, so both problems redirect to this page.
Contents
Problem
For how many (not necessarily positive) integer values of 
is the value of 
an integer?
Solution 1 (Algebra)
Note that ![\[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\]](http://latex.artofproblemsolving.com/d/1/b/d1bc2127024c06a05917905412d10cdb6c321596.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 (Brute Force)
The values for are 
and 
The corresponding values for 
are 
and 
respectively.
In total, there are 
values for
~Little ~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
Video Solutions
~IceMatrix
~savannahsolver
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
