Difference between revisions of "2021 AMC 12A Problems/Problem 12"
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All the roots of the polynomial <math>z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16</math> are positive integers, possibly repeated. What is the value of <math>B</math>? | All the roots of the polynomial <math>z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16</math> are positive integers, possibly repeated. What is the value of <math>B</math>? | ||
− | <math>\textbf{(A) }-88 \qquad \textbf{(B) }-80 \qquad \textbf{(C) }-64 \qquad \textbf{(D) }-41\qquad \textbf{(E) }-40</math> | + | <math>\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40</math> |
==Solution 1== | ==Solution 1== |
Revision as of 16:26, 4 December 2021
- The following problem is from both the 2021 AMC 12A #12 and 2021 AMC 10A #14, so both problems redirect to this page.
Contents
[hide]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 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, 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 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
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.