11.7. Given a 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 $n$ can such a polynomial have a real root? (I. Bogdanov, K. Sukhov)
Answer. $n=100$.

Solution. Let's call a polynomial that satisfies the condition of the problem beautiful. The polynomial $P(x)=100\left(x^{200}+x^{198}+\ldots+x^{2}+\right.$ $+1)+101\left(x^{199}+x^{197}+\ldots+x\right)$ is beautiful and has a root -1. Therefore, when $n=100$, the required is possible.

It remains to show that for $n<100$, the required is impossible. Indeed, for any beautiful polynomial $P(x)$, the inequality $P(x)>0$ must hold for all $x$. This inequality is obviously satisfied for $x \geqslant 0$; for negative $x=-t$, it follows from the inequality

$$
100\left(t^{2 n}+t^{2 n-2}+\ldots+t^{2}+1\right)>101\left(t^{2 n-1}+t^{2 n-3}+\ldots+t\right)
$$

Thus, it is sufficient to prove this inequality for all $t>0$. Multiplying (*) by $t+1$, we obtain 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

$$
100\left(t^{2 n+1}+1\right)>t^{2 n}+t^{2 n-1}+\ldots+t
$$

Note that for all $k=1, \ldots, n$, the inequality $\left(t^{k}-1\right)\left(t^{2 n+1-k}-1\right) \geqslant 0$ holds, since both parentheses have the same sign for $t>0$. Expanding the parentheses, we get

$$
t^{2 n+1}+1 \geqslant t^{2 n+1-k}+t^{k}
$$

Adding all such inequalities and considering that $n<100$, we obtain

$$
t^{2 n}+t^{2 n-1}+\ldots+t \leqslant n\left(t^{2 n+1}+1\right)<100\left(t^{2 n+1}+1\right)
$$

which proves $(* *)$.

Comment. Only the answer - 0 points.

An example of a polynomial satisfying the condition of the problem for $n=100$ - 1 point.

It is proven that $n \geqslant 100$ - 4 points.

If this inequality is not proven but reduced to the inequality $(* *)$ - 2 points.