Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2019 AMC 12A Problems/Problem 17"

m (→‎Solution)
Line 5: Line 5:
 
<math>\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26</math>
 
<math>\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26</math>
  
−
==Solution==
+
==Solution 1==
 
Applying Newton Sums we get the answer as 10
 
Applying Newton Sums we get the answer as 10
 +
 +
==Solution 2==
 +
 +
Let <math>p, q</math>, and <math>r</math> be the roots of the polynomial. Then,
 +
 +
<math>p^3 - 5p^2 + 8p - 13 = 0</math>
 +
 +
<math>q^3 - 5q^2 + 8q - 13 = 0</math>
 +
 +
<math>r^3 - 5r^2 + 8r - 13 = 0</math>
 +
 +
Adding these three equations, we get
 +
 +
<math>(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0</math>
 +
 +
<math>s_3 - 5s_2 + 8s_1 = 39</math>
 +
 +
<math>39</math> can be written as <math>13s_0</math>.
 +
 +
<math>s_3 = 5s_2 - 8s_1 + 13s_0</math>
 +
 +
We are given that <math>s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}</math> is satisfied for <math>k = 2</math>, <math>3</math>, <math>....</math>, meaning it must be satisfied when <math>k = 2</math>, giving us <math>s_3 = a \, s_2 + b \, s_1 + c \, s_0</math>.
 +
 +
Therefore, <math>a = 5, b = -8</math>, and <math>c = 13</math> by matching coefficients.
 +
 +
<math>5 - 8 + 13 = \boxed{\textbf{(D)}10}</math>
  
 
==See Also==
 
==See Also==

Revision as of 19:43, 9 February 2019

Problem

Let $s_k$ denote the sum of the $\textit{k}$th powers of the roots of the polynomial $x^3-5x^2+8x-13$. In particular, $s_0=3$, $s_1=5$, and $s_2=9$. Let $a$, $b$, and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$, $3$, $....$ What is $a+b+c$?

$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$

Solution 1

Applying Newton Sums we get the answer as 10

Solution 2

Let $p, q$, and $r$ be the roots of the polynomial. Then,

$p^3 - 5p^2 + 8p - 13 = 0$

$q^3 - 5q^2 + 8q - 13 = 0$

$r^3 - 5r^2 + 8r - 13 = 0$

Adding these three equations, we get

$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$

$s_3 - 5s_2 + 8s_1 = 39$

$39$ can be written as $13s_0$.

$s_3 = 5s_2 - 8s_1 + 13s_0$

We are given that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ is satisfied for $k = 2$, $3$, $....$, meaning it must be satisfied when $k = 2$, giving us $s_3 = a \, s_2 + b \, s_1 + c \, s_0$.

Therefore, $a = 5, b = -8$, and $c = 13$ by matching coefficients.

$5 - 8 + 13 = \boxed{\textbf{(D)}10}$

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_17&oldid=101612"