2021 AMC 12A Problems/Problem 12
- The following problem is from both the 2021 AMC 12A #12 and 2021 AMC 10A #14, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution (🚀 Just 2 min 🚀)
- 5 Video Solution by Hawk Math
- 6 Video Solution by OmegaLearn (Using Vieta's Formulas & Combinatorics)
- 7 Video Solution by Power Of Logic (Using Vieta's Formulas)
- 8 Video Solution by TheBeautyofMath
- 9 Video Solution by CanadaMath
- 10 See also
Problem
All the roots of the polynomial are positive integers, possibly repeated. What is the value of ?
Solution 1
By Vieta's formulas, the sum of the six roots is and the product of the six roots is . By inspection, we see the roots are and , so the function is . Therefore, calculating just the terms, we get .
~JHawk0224
Solution 2
Using the same method as Solution 1, we find that the roots are and . Note that is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the products we obtain ~ike.chen
Video Solution (🚀 Just 2 min 🚀)
~Education, the Study of Everything
Video Solution by Hawk Math
https://www.youtube.com/watch?v=AjQARBvdZ20
Video Solution by OmegaLearn (Using Vieta's Formulas & Combinatorics)
~ pi_is_3.14
Video Solution by Power Of Logic (Using Vieta's Formulas)
Video Solution by TheBeautyofMath
https://youtu.be/t-EEP2V4nAE?t=1080 (for AMC 10A)
https://youtu.be/ySWSHyY9TwI?t=271 (for AMC 12A)
~IceMatrix
Video Solution by CanadaMath
https://www.youtube.com/watch?v=8D29aL7clFc (For AMC 10A)
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 |
2021 AMC 12A (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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.