Difference between revisions of "2018 AMC 10A Problems/Problem 7"
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− | {{duplicate|[[2018 AMC | + | {{duplicate|[[2018 AMC 10A Problems/Problem 7|2018 AMC 10A #7]] and [[2018 AMC 12A Problems/Problem 7|2018 AMC 12A #7]]}} |
==Problem== | ==Problem== | ||
Line 12: | Line 12: | ||
</math> | </math> | ||
− | ==Solution== | + | ==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 | |
− | ==Video | + | ==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 | https://youtu.be/ZiZVIMmo260 | ||
+ | ==Video Solution== | ||
https://youtu.be/2vz_CnxsGMA | https://youtu.be/2vz_CnxsGMA | ||
~savannahsolver | ~savannahsolver | ||
− | https://youtu.be/ | + | ==Video Solution by OmegaLearn== |
+ | https://youtu.be/ZhAZ1oPe5Ds?t=1763 | ||
− | + | ~ pi_is_3.14 | |
== See Also == | == See Also == |
Latest revision as of 05:26, 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.
Contents
Problem
For how many (not necessarily positive) integer values of is the value of
an integer?
Solution 1 (Algebra)
Note that
Since this expression is an integer, we need:
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.