Difference between revisions of "2021 AMC 12A Problems/Problem 12"
MRENTHUSIASM (talk | contribs) m |
Tecilis459 (talk | contribs) m (Fix i tag) |
||
(10 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
− | {{duplicate|[[2021 AMC 12A Problems | + | {{duplicate|[[2021 AMC 12A Problems/Problem 12|2021 AMC 12A #12]] and [[2021 AMC 10A Problems/Problem 14|2021 AMC 10A #14]]}} |
==Problem== | ==Problem== | ||
Line 7: | Line 7: | ||
==Solution 1== | ==Solution 1== | ||
− | By Vieta's formulas, the sum of the | + | 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 | ||
Line 13: | Line 13: | ||
==Solution 2== | ==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 | + | 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== | ==Video Solution by Hawk Math== | ||
Line 32: | Line 38: | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | ==Video Solution by CanadaMath== | ||
+ | https://www.youtube.com/watch?v=8D29aL7clFc (For AMC 10A) | ||
==See also== | ==See also== |
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
[hide]- 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.