Authors: Bogdanov I.i., Sukho K.

Given the polynomial $P(x)=a_{2 n} x^{2 n}+a_{2 n-1} x^{2 n-1}+\ldots+a_{1} x+a_{0}$, where each coefficient $a_{i}$ belongs to the interval $[100,101]$.

For what minimal natural $n$ can such a polynomial have a real root?
Example. Let's call a polynomial that satisfies the condition of the problem beautiful. The polynomial $P(x)=100\left(x^{200}+x^{198}+\right.$ $\left.\ldots+x^{2}+1\right)+101\left(x^{199}+x^{197}+\ldots+x\right)$ is beautiful and has a root -1. Therefore, for $n=100$ the required is possible.

Estimate. We will prove that for $n101\left(t^{2 n-1}+t^{2 n-3}+\ldots+t\right)$ for all $t>0$. Multiplying it by $t+1$, we get the equivalent inequality $100\left(t^{2 n+1}+t^{2 n}+\ldots+1\right)>101\left(t^{2 n}+t^{2 n-1}+\ldots+t\right)$, or $100\left(t^{2 n+1}+1\right)>t^{2 n}+t^{2 n-1}+\ldots+t . \quad(*)$

Notice that $\left(t^{k}-1\right)\left(t^{2 n+1-k}-1\right) \geq 0$ for all $k=1, \ldots, n$ (both brackets have the same sign for $t>0$). Expanding the brackets, we get

$t^{2 n+1}+1>t^{2 n+1-k}+t^{k}$

Adding all such inequalities and considering that $n<100$, we get $t^{2 n}+t^{2 n-1}+\ldots t \leq n\left(t^{2 n+1}+1\right)<100\left(t^{2 n+1}\right.$ $+1)$, which proves ( $($ )

## Answer

$n=100$.