2016 is absolutely awesome 16

Algebra Level 5

$x^{2016}+cx^{2015}+\binom{c}{2}x^{2014}+\binom{c}{3}x^{2013}+\cdots+\binom{c}{2016}$

How many real numbers $c$ are there such that the above polynomial has 2016 real roots (counting multiplicities)?

Details and Assumptions

We define the generalized binomial coefficient as $\displaystyle\binom{c}{k}=\dfrac{c(c-1)(c-2)\cdots(c-k+1)}{k!}$ .
If there are infinite real numbers $c$ satisfied, submit $-1$ as the answer.

This is a part of the Set .

The answer is 2017.

1 solution

Arghyadeep Chatterjee
Jun 20, 2019

For $c = 0$ to $c = 2016$ The equation could be written as $(x^{2016-c})(x+1)^{c}$ . 0 has multiplicity $2016 - c$ and $-1$ has multiplicity $c$ . SO there are $2017$ values of $c$ Now if c be a integer greater than 2016 . Then the equation would be $(x+1)^{2016} + \binom{c}{n}$ which is bound to have some complex roots. Now if c be a negative integer then $\binom{c}{n}$ is undefined . Also this is the case for $c$ being a real number other than integers. Now I do not have proofs for the last two arguments that I made. But I can only assume that the factorial notation was not extended to reals other than natural numbers .

2016 is absolutely awesome 16

This is a part of the Set .

The answer is 2017.

1 solution

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