Difference between revisions of "2019 AMC 12A Problems/Problem 17"
Mathcat1234 (talk | contribs) 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 18:43, 9 February 2019
Contents
[hide]Problem
Let denote the sum of the th powers of the roots of the polynomial . In particular, , , and . Let , , and be real numbers such that for , , What is ?
Solution 1
Applying Newton Sums we get the answer as 10
Solution 2
Let , and be the roots of the polynomial. Then,
Adding these three equations, we get
can be written as .
We are given that is satisfied for , , , meaning it must be satisfied when , giving us .
Therefore, , and by matching coefficients.
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.