# Problem 9.

Given a polynomial $P(x)$ of degree 10 with the leading coefficient 1. The graph of $y=P(x)$ lies entirely above the $O x$ axis. The polynomial $-P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be expressed as the product of two non-constant polynomials). It is known that at $x=2020$, all the obtained irreducible polynomials take the value -3. Find $P(2020)$.
Answer: 243

Solution. Note that since the graph of $y=P(x)$ is entirely above the $O x$ axis, the polynomial $P(x)$ has no real roots. This means that all irreducible polynomials in the factorization have degree 2, and their number is 5. Since the polynomial $-P(x)$ will have the same number of irreducible polynomials in its factorization, and all of them take the value -3 at the point $x=2020$, then $-P(2020)=(-3)^{5}=-243$. Hence, $P(2020)=243$.

## B-2

Given a polynomial $P(x)$ of degree 12 with the leading coefficient 1. The graph of $y=P(x)$ is entirely above the $O x$ axis. The polynomial $P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be represented as a product of two non-constant polynomials). It is known that at $x=2020$ all the obtained irreducible polynomials take the value -3. Find $P(2020)$.

Answer: 729

## B-3

Given a polynomial $P(x)$ of degree 20 with the leading coefficient 1. The graph of $y=P(x)$ is entirely above the $O x$ axis. This polynomial is factored into irreducible factors (i.e., polynomials that cannot be represented as a product of two non-constant polynomials). It is known that at $x=2020$ all the obtained irreducible polynomials take the value 2. Find $P(2020)$.

Answer: 1024

## B-4

Given a polynomial $P(x)$ of degree 18 with the leading coefficient -1. The graph of $y=P(x)$ is entirely below the $O x$ axis. The polynomial $P(x)$ is factored into irreducible factors (i.e., polynomials that cannot be represented as a product of two non-constant polynomials). It is known that at $x=2020$ all the obtained irreducible polynomials take the value -2. Find $P(2020)$.

Answer: -512

## B-5

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-210 x+11000}{\sqrt[2021]{x^{2021}-420 x^{2020}}}>1
$$

Answer: 421

Solution. We will solve the problem in general, specifying the constraints on the parameters $a, b, c$: Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-(a+b) x+a b}{\sqrt[2021]{x^{2021}-c x^{2020}}}>1
$$

We assume that $a, b, c$ are natural numbers, $1 \leqslant a < b$, and $c \geqslant a+b$. For $x \leqslant a$, the numerator is non-positive, and the denominator is positive, so the left side of the inequality is non-positive. For $x \geqslant b$, the numerator is non-negative, and the denominator is positive, so the left side of the inequality is non-negative. For $a < x < b$, the numerator is positive, and the denominator is positive, so the left side of the inequality is positive. We need to find the smallest $x$ such that the left side of the inequality is greater than 1.

For $x = a$, the numerator is 0, and the denominator is positive, so the left side of the inequality is 0. For $x = b$, the numerator is 0, and the denominator is positive, so the left side of the inequality is 0. For $x = c$, the numerator is positive, and the denominator is positive, so the left side of the inequality is positive. We need to find the smallest $x$ such that the left side of the inequality is greater than 1.

For $x = c+1$, the numerator is positive, and the denominator is positive, so the left side of the inequality is positive. We need to check if the left side of the inequality is greater than 1.

The numerator is

$$
x^{2}-(a+b) x+a b = (x-a)(x-b)
$$

The denominator is

$$
\sqrt[2021]{x^{2021}-c x^{2020}} = x \sqrt[2021]{1 - \frac{c}{x}}
$$

For $x = c+1$, the denominator is

$$
(c+1) \sqrt[2021]{1 - \frac{c}{c+1}} = (c+1) \sqrt[2021]{\frac{1}{c+1}} = (c+1) \frac{1}{(c+1)^{1/2021}} = (c+1)^{1 - 1/2021}
$$

The left side of the inequality is

$$
\frac{(c+1-a)(c+1-b)}{(c+1)^{1 - 1/2021}}
$$

Since $c \geqslant a+b$, we have $c+1 > a+b$, so $(c+1-a)(c+1-b) > 1$. Since $(c+1)^{1 - 1/2021} < c+1$, we have

$$
\frac{(c+1-a)(c+1-b)}{(c+1)^{1 - 1/2021}} > 1
$$

Therefore, the smallest integer satisfying the inequality is $c+1$.

Answer: $c+1$

## B-6

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-171 x+7290}{\sqrt[2021]{x^{2021}-215 x^{2020}}}>1
$$

Answer: 216

## B-7

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-251 x+15730}{\sqrt[2021]{x^{2021}-374 x^{2020}}}>1
$$

Answer: 375

## B-8

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-294 x+21600}{\sqrt[2021]{x^{2021}-517 x^{2020}}}>1
$$

Answer: 518

## B-9

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-210 x+11000}{\sqrt[2021]{x^{2021}-425 x^{2020}}}>1
$$

Answer: 426

## B-10

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-171 x+2290}{\sqrt[2021]{x^{2021}-219 x^{2020}}}>1
$$

Answer: 220

## B-11

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-251 x+15730}{\sqrt[2021]{x^{2021}-379 x^{2020}}}>1
$$

Answer: 380

## B-12

Find the smallest integer satisfying the inequality

$$
\frac{x^{2}-294 x+21600}{\sqrt[2021]{x^{2021}-519 x^{2020}}}>1
$$

Answer: 520

## Lomonosov Olympiad for Schoolchildren

Preliminary Round 2020/21 academic year for $10-11$ grades

## B-1

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