Difference between revisions of "2018 AMC 10A Problems/Problem 7"
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==Solution 3 (Brute Force)== | ==Solution 3 (Brute Force)== | ||
The possible numbers for <math>n</math> are: <math>-5, -4, -3, -2, -1, 0, 1, 2, 3</math>. In total there are <math>9</math> possible values for <math>n</math>. Hence answer is <math>\boxed{\textbf{(E) }9}</math>. | The possible numbers for <math>n</math> are: <math>-5, -4, -3, -2, -1, 0, 1, 2, 3</math>. In total there are <math>9</math> possible values for <math>n</math>. Hence answer is <math>\boxed{\textbf{(E) }9}</math>. | ||
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~ Little | ~ Little | ||
Revision as of 10:24, 29 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 ![\[4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\frac{2}{5}\right)^n=2^{5+n}\cdot5^{3-n}.\]](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 (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 possible numbers for are: 
. In total there are 
possible values for . Hence answer is 
.
~ Little
Video Solutions
~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. 
