Consider a polynomial \(f(x)=x^{2014}+a_{2013}x^{2013}+\dots+a_1x+a_0\). Cody and Pi Han take turns choosing one of the coefficients \(a_0,a_1,\dots,a_{2013}\) 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 \(f(x)\) divisible by a fixed polynomial \(m(x)\), and Cody's goal is to prevent this.

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

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TopNewestBy 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 \(f(\omega)=0\), with \(\omega\) 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, \(\arg f(\omega)\) is not a multiple of \(\pi/3\). For the second point, if Cody plays in such a way that, just after his turns, \(\arg f(i)\) is not a multiple of \(\pi/4\), Pi Han's doomed again, since \(f(i)=0\) is a necessary condition in order to have \((x^4-1)\mid f(x)\). – Jack D'Aurizio · 3 years, 3 months ago

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– Cody Johnson · 3 years, 3 months ago

But Cody goes first.Log in to reply

Hardest question in the entire contest! :'( – Shaun Loong · 3 years, 3 months ago

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– Cody Johnson · 3 years, 3 months ago

Hint: look at the roots of \(m(x)\).Log in to reply

– Calvin Lin Staff · 3 years, 3 months ago

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