Difference between revisions of "2021 AMC 12A Problems/Problem 12"
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~ike.chen | ~ike.chen | ||
| + | ==Solution 3 (brute force)== | ||
| + | |||
| + | Same as solution 1, we find the roots are <math>1, 1, 2, 2, 2, 2</math> by Vieta's formula. This next part is brute force expansion. | ||
| + | <math>r_1r_2r_3+r_1r_2r_4+r_1r_2r_5+r_1r_2r_6+\\r_1r_3r_4+r_1r_3r_5+r_1r_3r_6+\\r_1r_4r_5+r_1r_4r_6+\\r_1r_5r_6+\\r_2r_3r_4+r_2r_3r_5+r_2r_3r_6+\\r_2r_4r_5+r_2r_4r_6+\\r_2r_5r_6+\\r_3r_4r_5+r_3r_4r_6+\\r_3r_5r_6+\\r_4r_5r_6</math>. | ||
| + | <math>=2 \times 4 + 4 \times 12 + 8 \times 4</math>\\ | ||
| + | <math>=8 + 48 + 32</math>\\ | ||
| + | <math>=\boxed{\textbf{(A) }{-}88}</math> | ||
==Video Solution (🚀 Just 2 min 🚀)== | ==Video Solution (🚀 Just 2 min 🚀)== | ||
https://youtu.be/s6MGGjPv1n0 | https://youtu.be/s6MGGjPv1n0 | ||
Revision as of 01:00, 30 August 2025
- 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 Solution 3 (brute force)
- 5 Video Solution (🚀 Just 2 min 🚀)
- 6 Video Solution by Hawk Math
- 7 Video Solution by OmegaLearn (Using Vieta's Formulas & Combinatorics)
- 8 Video Solution by Power Of Logic (Using Vieta's Formulas)
- 9 Video Solution by TheBeautyofMath
- 10 Video Solution by CanadaMath
- 11 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
Solution 3 (brute force)
Same as solution 1, we find the roots are 
by Vieta's formula. This next part is brute force expansion.
.
\\
\\
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
