Difference between revisions of "2018 AMC 10A Problems/Problem 7"
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==Solution 3 (Brute Force)== | ==Solution 3 (Brute Force)== | ||
− | The | + | 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. | 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> | + | In total, there are <math>\boxed{\textbf{(E) }9}</math> values for <math>n.</math> |
− | ~Little | + | ~Little ~MRENTHUSIASM |
==Video Solutions== | ==Video Solutions== |
Revision as of 01:47, 30 October 2021
- The following problem is from both the 2018 AMC 12A #7 and 2018 AMC 10A #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 (Arithmetic)
The prime factorization of is Therefore, the maximum integer value for is and the minimum integer value for is Then we must find the range from to which is
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.