Difference between revisions of "2020 AMC 8 Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | We list out the factors of <math>2020</math>: <cmath>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.</cmath> Of these, only <math>1, 2, 4, 5, 101</math> (<math>5</math> of them) do not have more than <math>3</math> factors. Therefore the answer is <math>\tau{2020}-5=\boxed{\textbf{(B) }7}</math>. | + | We list out the factors of <math>2020</math>: <cmath>1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.</cmath> Of these, only <math>1, 2, 4, 5, 101</math> (<math>5</math> of them) do not have more than <math>3</math> factors. Therefore the answer is <math>\tau\left({2020}\right)-5=\boxed{\textbf{(B) }7}</math>. |
==See also== | ==See also== | ||
Revision as of 01:08, 18 November 2020
How many positive integer factors of 
have more than 
factors?
Solution
We list out the factors of : ![\[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\]](http://latex.artofproblemsolving.com/f/2/6/f267298afe7942f84ed0d8e8483876d5e4d2278b.png)
Of these, only 
(
of them) do not have more than factors. Therefore the answer is 
.
See also
| 2020 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
