Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 12A Problems/Problem 12"

(→‎Problem)
(→‎Solution 2)
Line 15: Line 15:
 
Using the same method as Solution 1, we find that the roots are <math>2, 2, 2, 2, 1,</math> and <math>1</math>. Note that <math>B</math> is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the <math>\binom {6}{3} = 20</math> products <math>r_a \cdot r_b \cdot r_c,</math> we obtain <cmath>B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A) }{-}88}.</cmath>
 
Using the same method as Solution 1, we find that the roots are <math>2, 2, 2, 2, 1,</math> and <math>1</math>. Note that <math>B</math> is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the <math>\binom {6}{3} = 20</math> products <math>r_a \cdot r_b \cdot r_c,</math> we obtain <cmath>B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A) }{-}88}.</cmath>
 
~ike.chen
 
~ike.chen
−
 
  
 
==Video Solution (🚀 Just 2 min 🚀)==
 
==Video Solution (🚀 Just 2 min 🚀)==

Revision as of 12:23, 10 July 2024

The following problem is from both the 2021 AMC 12A #12 and 2021 AMC 10A #14, so both problems redirect to this page.

Problem

All the roots of the polynomial $z^6-10z^5+Az^4+Bz^3+Cz^2+Dz+16$ are positive integers, possibly repeated. What is the value of $B$?

$\textbf{(A) }{-}88 \qquad \textbf{(B) }{-}80 \qquad \textbf{(C) }{-}64 \qquad \textbf{(D) }{-}41\qquad \textbf{(E) }{-}40$

Solution 1

By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$. By inspection, we see the roots are $1, 1, 2, 2, 2,$ and $2$, so the function is $(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$. Therefore, calculating just the $z^3$ terms, we get $B = -32 - 48 - 8 = \boxed{\textbf{(A) }{-}88}$.

~JHawk0224

Solution 2

Using the same method as Solution 1, we find that the roots are $2, 2, 2, 2, 1,$ and $1$. Note that $B$ is the negation of the 3rd symmetric sum of the roots. Using casework on the number of 1's in each of the $\binom {6}{3} = 20$ products $r_a \cdot r_b \cdot r_c,$ we obtain \[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A) }{-}88}.\] ~ike.chen

Video Solution (🚀 Just 2 min 🚀)

https://youtu.be/s6MGGjPv1n0

~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)

https://youtu.be/5U4MJTo3F5M

~ pi_is_3.14

Video Solution by Power Of Logic (Using Vieta's Formulas)

https://youtu.be/rl6QtVnIbdU

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_12&oldid=222014"