Gigantic Polynomial Game

Algebra Level 4

f ( x ) = □ x 2016 + □ x 2015 + … + □ x 2 + □ x + □ \large f(x) = \Box x^{2016} + \Box x^{2015} + \ldots + \Box x^2 + \Box x + \Box

The above shows a polynomial of degree 2016, but with all of its coefficients left blank. Two players, Euler and Fermat, take turns to fill in the gaps with any real number . Euler makes the first turn.

When the gaps are all filled, if the equation f ( x ) = 0 f(x) = 0 has at least 1 real solution, Euler wins; otherwise, Fermat wins.

Assume that Euler and Fermat both play optimally. What's the probability that Fermat wins?

Pick the closest answer.

0 % 0\% 20 % 20\% 40 % 40\% 80 % 80\% 100 % 100\%

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Tran Quoc Dat by Tran Quoc Dat , 19, Vietnam

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4 solutions

Relevant wiki: Factoring Polynomials

Euler will always get the chance to fill the gap in the constant term. If the constant term is 0 0 , f ( x ) f(x) will have at least one real root i.e. 0 0 .So Euler will always win.

Hence probability that Fermat will win is 0 0 .

9 Helpful 0 Interesting 0 Brilliant 0 Confused

I miscounted and therefore put 100%. I hate myself sometimes lol.

Sam Maltia - 5 years ago
Gaurav Chahar
May 19, 2016

Dis is an idiotic problem as the sample space can't be determined by any means so how can one calculate the exact probability.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I agree with it.

Nivedit Jain - 5 years ago

Like all people saying that he can fill zero.. How do you know he is not an idiot. Maybe he fills some iodically.

Nivedit Jain - 5 years ago
Praful Jain
May 18, 2016

Since euler plays first ,he will get the chance to fill the constant term and if constant is filled as 0 then the product of 2016 roots should include 0 and hence we get 0 as one of root. So euler has 100% chance of winning and the other one has 0% chance of winning.

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Manuel Kahayon
May 19, 2016

Euler can actually just fill all his boxes with zero and sleep through the entire game. Since this ensures that the polynomial will have an odd leading coefficient then the function will range through all real numbers, so f ( x ) = 0 f(x) = 0 will have at least one real zero. If Pascal fills all his boxes with 0 0 then we get f ( x ) = 0 f(x) = 0 , which will totally crush Pascal as the polynomial will have infinitely many zeroes.

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