2020 AMC 8 Problems/Problem 17
Problem
How many positive integer factors of 
have more than 
factors? (As an example, 
has 
factors, namely 
and 
)
Solution 1
Since 
, we can simply list its factors: ![\[1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020.\]](http://latex.artofproblemsolving.com/f/2/6/f267298afe7942f84ed0d8e8483876d5e4d2278b.png)
There are of these; only 
(i.e. 
of them) have at most factors, so the remaining 
factors have more than factors.
Solution 2
As in Solution 1, we prime factorize as 
, and we recall the standard formula that the number of positive factors of an integer is found by adding 
to each exponent in its prime factorization, and then multiplying these. Thus has 
factors. The only number which has one factor is . For a number to have exactly two factors, it must be prime, and the only prime factors of are 
, , and 
. For a number to have three factors, it must be a square of a prime (this follows from the standard formula mentioned above), and from the prime factorization, the only square of a prime that is a factor of is 
. Thus, there are factors of which themselves have , , or factors (namely , , , , and ), so the number of factors of that have more than factors is 
.
Video Solution
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. 
