2020 AMC 12A Problems/Problem 21
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution by Richard Rusczyk
- 6 Video Solution by OmegaLearn
We set up the following equation as the problem states:
![\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}.\]](https://latex.artofproblemsolving.com/e/4/1/e4182bee46f971119822c1847de1c6dd5e26a37f.png "art of problem solving amc 12a")
Breaking each number into its prime factorization, we see that the equation becomes
![\[\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}.\]](https://latex.artofproblemsolving.com/3/5/6/356f8d9a66c1cdd0ad995e7193a7bed51528f404.png "art of problem solving amc 12a")
Like the Solution 1, we start from the equation:
As in the previous solutions, we start with
![\[\text{lcm}(5!,n) = 5\text{gcd}(10!,n)\]](https://latex.artofproblemsolving.com/6/0/6/6068059d0e947abfe440e5b9b1ce3449076c1c0a.png "art of problem solving amc 12a")
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=8S85536jpYw&list=PLyhPcpM8aMvJvwA2kypmfdtlxH90ShZCc&index=1&t=25s - AMBRIGGS
Video Solution by OmegaLearn
https://youtu.be/CWZkTCNu42o?t=846
~ pi_is_3.14
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