2021 AMC 12A Problems/Problem 12: Difference between revisions

Revision as of 00:06, 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.

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

Solution 3 (brute force)

Same as solution 1, we find the roots are $1, 1, 2, 2, 2, 2$ by Vieta's formula. This next part is brute force expansion. $r_1r_2r_3+r_1r_2r_4+r_1r_2r_5+r_1r_2r_6+\\r_1r_3r_4+r_1r_3r_5+r_1r_3r_6+\\r_1r_4r_5+r_1r_4r_6+\\r_1r_5r_6+\\r_2r_3r_4+r_2r_3r_5+r_2r_3r_6+\\r_2r_4r_5+r_2r_4r_6+\\r_2r_5r_6+\\r_3r_4r_5+r_3r_4r_6+\\r_3r_5r_6+\\r_4r_5r_6\\=2 \times 4 + 4 \times 12 + 8 \times 4\\ =8 + 48 + 32\\ =\boxed{\textbf{(A) }{-}88}$

~yhbettysun

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)

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. Error creating thumbnail: File missing

@@ Line 19: / Line 19: @@
 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>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\\=2 \times 4 + 4 \times 12 + 8 \times 4\\ =8 + 48 + 32\\ =\boxed{\textbf{(A) }{-}88}</math>
-<math>=2 \times 4 + 4 \times 12 + 8 \times 4</math>
-<math>=8 + 48 + 32</math>
+~yhbettysun
-<math>=\boxed{\textbf{(A) }{-}88}</math>
-yhbettysun
 ==Video Solution (🚀 Just 2 min 🚀)==
 https://youtu.be/s6MGGjPv1n0

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