2020 AMC 8 Problems/Problem 12
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (variant of Solution 1)
- 4 Solution 3 (using answer choices and elimination)
- 5 Solution 4
- 6 Video Solution by NiuniuMaths (Easy to understand!)
- 7 Video Solution by Math-X (First understand the problem!!!)
- 8 Video Solution (🚀 Fast 🚀)
- 9 Video Solution by North America Math Contest Go Go Go
- 10 Video Solution by WhyMath
- 11 Video Solution
- 12 Video Solution by Interstigation
- 13 See also
Problem
For a positive integer , the factorial notation represents the product of the integers from to . What value of satisfies the following equation?
Solution 1
We have , and . Therefore, the equation becomes , and so . Cancelling the s, it is clear that .
Solution 2 (variant of Solution 1)
Since , we obtain , which becomes and thus . We therefore deduce .
Solution 3 (using answer choices and elimination)
We can see that the answers to contain a factor of , but there is no such factor of in . The factor 11 is in every answer choice after , so four of the possible answers are eliminated. Therefore, the answer must be . ~edited by HW73
Solution 4
We notice that
We know that so we have
Isolating we have
~mathboy282
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=bHNrBwwUCMI
~NiuniuMaths
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/UnVo6jZ3Wnk?si=aiVOk6HkZErYYi6P&t=1568
~Math-X
Video Solution (🚀 Fast 🚀)
~Education, the Study of Everything
Video Solution by North America Math Contest Go Go Go
https://www.youtube.com/watch?v=mYs1-Nbr0Ec
~North America Math Contest Go Go Go
Video Solution by WhyMath
~savannahsolver
Video Solution
Video Solution by Interstigation
https://youtu.be/YnwkBZTv5Fw?t=504
~Interstigation
See also
2020 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.