Difference between revisions of "2018 AMC 10A Problems/Problem 7"
m (Added Problem heading) |
m |
||
| Line 13: | Line 13: | ||
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>. | 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 | ||
== See Also == | == See Also == | ||
Revision as of 20:46, 23 January 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
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. 
