The teacher calculates and writes on the board all the numbers $a^b$ that satisfy the condition $n = a\times b$ for the natural number $n.$ Here $a$ and $b$ are natural numbers. Is there a natural number $n$ such that each of the numbers $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ is the last digit of one of the numbers written by the teacher on the board? Justify your opinion.

Find all the integer solutions of the equation: $$\sqrt{x} + \sqrt{y} = \sqrt{x+2023}$$

Let $m$ be a positive integer. Find polynomials $P(x)$ with real coefficients such that $$(x-m)P(x+2023) = xP(x)$$is satisfied for all real numbers $x.$

To open the magic chest, one needs to say a magic code of length $n$ consisting of digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9.$ Each time Griphook tells the chest a code it thinks up, the chest's talkative guardian responds by saying the number of digits in that code that match the magic code. (For example, if the magic code is $0423$ and Griphook says $3442,$ the chest's talkative guard will say $1$). Prove that there exists a number $k$ such that for any natural number $n \geq k,$ Griphook can find the magic code by checking at most $4n-2023$ times, regardless of what the magic code of the box is.

The incircle of the acute-angled triangle $ABC$ is tangent to the sides $AB, BC, CA$ at points $C_1, A_1, B_1,$ respectively, and $I$ is the incenter. Let the midpoint of side $BC$ be $M.$ Let $J$ be the foot of the altitude drawn from $M$ to $C_1B_1.$ The tangent drawn from $B$ to the circumcircle of $\triangle BIC$ intersects $IJ$ at $X.$ If the circumcircle of $\triangle AXI$ intersects $AB$ at $Y,$ prove that $BY = BM.$