Difference between revisions of "2018 AMC 10A Problems/Problem 7"
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+ | ==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? | ||
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==Solution== | ==Solution== | ||
− | The prime factorization of <math>4000</math> is <math>2^{5} | + | The prime factorization of <math>4000</math> is <math>2^{5}\cdot5^{3}</math>. Therefore, the maximum number for <math>n</math> is <math>3</math>, and the minimum number for <math>n</math> is <math>-5</math>. Then we must find the range from <math>-5</math> to <math>3</math>, which is <math>3-(-5) + 1 = 8 + 1 = \fbox{\textbf{(E) }9}</math>. |
− | + | ==Video Solution== | |
+ | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
+ | https://youtu.be/2vz_CnxsGMA | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2018|ab=A|num-b=6|num-a=8}} | {{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}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] |
Revision as of 14:02, 2 July 2020
Contents
Problem
For how many (not necessarily positive) integer values of is the value of an integer?
Solution
The prime factorization of is . Therefore, the maximum number for is , and the minimum number for is . Then we must find the range from to , which is .
Video Solution
~savannahsolver
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.