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

(Video Solution by pi_is_3.14 (Using Vieta's Formulas & Combinatorics))
(Solution)
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<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==
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==Solution 1:==
By Vieta's formulae, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, 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, <math>B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}</math>
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By Vieta's formulae, the sum of the 6 roots is 10 and the product of the 6 roots is 16. By inspection, we see the roots are 1, 1, 2, 2, 2, and 2, 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, <math>B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}</math>.
 
~JHawk0224
 
~JHawk0224
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==Solution 2:==
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Using the same method as solution 1, we find the roots are <math>2, 2, 2, 2, 1, 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}</math> products, we obtain <math>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} \cdot \binom {2}{2} \cdot 2 \right) = -(32+48+8) = \boxed{\textbf{(A)} -88}</math> ~ ike.chen.
  
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==

Revision as of 23:21, 11 February 2021

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 formulae, the sum of the 6 roots is 10 and the product of the 6 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, $B = -32 - 48 - 8 = \boxed{\textbf{(A)} -88}$. ~JHawk0224

Solution 2:

Using the same method as solution 1, we find the roots are $2, 2, 2, 2, 1, 1$. 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 $\binom {6}{3}$ products, we obtain $B= - \left(\binom {4}{3} \binom {2}{0} \cdot 2^{3} + \binom {4}{2} \binom{2}{1} \cdot 2^{2} \cdot 1 + \binom {4}{1} \cdot \binom {2}{2} \cdot 2 \right) = -(32+48+8) = \boxed{\textbf{(A)} -88}$ ~ ike.chen.

Video Solution by Hawk Math

https://www.youtube.com/watch?v=AjQARBvdZ20

Video Solution by OmegaLern (Using Vieta's Formulas & Combinatorics)

https://youtu.be/5U4MJTo3F5M

~ pi_is_3.14

See also

2021 AMC 12A (ProblemsAnswer KeyResources)
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

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