Consider a polynomial f(x)=x2014+a2013x2013+⋯+a1x+a0. Cody and Pi Han take turns choosing one of the coefficients a0,a1,…,a2013 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) divisible by a fixed polynomial m(x), and Cody's goal is to prevent this.
- Which of the players has a strategy if m(x)=x2+x+1 and Cody goes first?
- Which of the players has a strategy if m(x)=x4−1 and Cody goes first?
#Proofathon
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:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Hardest question in the entire contest! :'(
Log in to reply
Hint: look at the roots of m(x).
Log in to reply
I felt that remainder-factor theorem was a recurring theme in the questions about polynomials.
By considering unassigned coefficients as they were zero, the polynomial f(x) changes move-by-move. For the first point, Pi Han's goal is to have f(ω)=0, with ω 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) is not a multiple of π/3. For the second point, if Cody plays in such a way that, just after his turns, argf(i) is not a multiple of π/4, Pi Han's doomed again, since f(i)=0 is a necessary condition in order to have (x4−1)∣f(x).
Log in to reply
But Cody goes first.