Difference between revisions of "2018 AMC 10A Problems/Problem 7"
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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 | 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> | <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 Solutions== | ==Video Solutions== |
Revision as of 15:25, 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 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 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.