Difference between revisions of "2021 AMC 12A Problems/Problem 12"
(→Solution) |
Tecilis459 (talk | contribs) m (Fix i tag) |
||
| (35 intermediate revisions by 14 users not shown) | |||
| Line 1: | Line 1: | ||
| + | {{duplicate|[[2021 AMC 12A Problems/Problem 12|2021 AMC 12A #12]] and [[2021 AMC 10A Problems/Problem 14|2021 AMC 10A #14]]}} | ||
| + | |||
==Problem== | ==Problem== | ||
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== | ||
| + | By Vieta's formulas, the sum of the six roots is <math>10</math> and the product of the six roots is <math>16</math>. By inspection, we see the roots are <math>1, 1, 2, 2, 2,</math> and <math>2</math>, so the function is <math>(z-1)^2(z-2)^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)</math>. Therefore, calculating just the <math>z^3</math> terms, we get <math>B = -32 - 48 - 8 = \boxed{\textbf{(A) }{-}88}</math>. | ||
| − | |||
| − | |||
~JHawk0224 | ~JHawk0224 | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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 | ||
| + | |||
| + | ==Video Solution (🚀 Just 2 min 🚀)== | ||
| + | https://youtu.be/s6MGGjPv1n0 | ||
| + | |||
| + | <i>~Education, the Study of Everything</i> | ||
| + | |||
| + | ==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== | ==See also== | ||
| + | {{AMC10 box|year=2021|ab=A|num-b=13|num-a=15}} | ||
{{AMC12 box|year=2021|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2021|ab=A|num-b=11|num-a=13}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:15, 16 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.
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 ![\[B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \binom {2}{2} \cdot 2 \right) = -\left(32+48+8 \right) = \boxed{\textbf{(A) }{-}88}.\]](http://latex.artofproblemsolving.com/f/2/4/f24718ee669124e49dff8e9563ca68a3e55125aa.png)
~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. 
