Polynomial Game - Problem 4

Consider a polynomial f(x)=x2014+a2013x2013++a1x+a0f(x)=x^{2014}+a_{2013}x^{2013}+\dots+a_1x+a_0. Cody and Pi Han take turns choosing one of the coefficients a0,a1,,a2013a_0,a_1,\dots,a_{2013} and assigning a real value to it. Once a value is assigned to a coefficient, it cannot be changed anymore. The game ends after all of the coefficients have been assigned values. Pi Han's goal is to make f(x)f(x) divisible by a fixed polynomial m(x)m(x), and Cody's goal is to prevent this.

  • Which of the players has a strategy if m(x)=x2+x+1m(x)=x^2+x+1 and Cody goes first?
  • Which of the players has a strategy if m(x)=x41m(x)=x^4-1 and Cody goes first?
#Proofathon

Note by Cody Johnson
7 years, 2 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Hardest question in the entire contest! :'(

Shaun Loong - 7 years, 2 months ago

Log in to reply

Hint: look at the roots of m(x)m(x).

Cody Johnson - 7 years, 2 months ago

Log in to reply

I felt that remainder-factor theorem was a recurring theme in the questions about polynomials.

Calvin Lin Staff - 7 years, 2 months ago

By considering unassigned coefficients as they were zero, the polynomial f(x)f(x) changes move-by-move. For the first point, Pi Han's goal is to have f(ω)=0f(\omega)=0, with ω\omega being a third root of unity, with its last move. Cody can prevent this by playing in such a way that, just after his turns, argf(ω)\arg f(\omega) is not a multiple of π/3\pi/3. For the second point, if Cody plays in such a way that, just after his turns, argf(i)\arg f(i) is not a multiple of π/4\pi/4, Pi Han's doomed again, since f(i)=0f(i)=0 is a necessary condition in order to have (x41)f(x)(x^4-1)\mid f(x).

Jack D'Aurizio - 7 years, 2 months ago

Log in to reply

But Cody goes first.

Cody Johnson - 7 years, 2 months ago
×

Problem Loading...

Note Loading...

Set Loading...