Bogonov I.I.

Given an infinite supply of white, blue, and red cubes. Any \$N\$ of them are arranged in a circle. A robot, starting at any point on the circle, moves clockwise and, until only one cube remains, repeatedly performs the following operation: it destroys the two nearest cubes in front of it and places a new cube behind it of the same color if the destroyed ones are the same, and of the third color if the destroyed ones are of different colors. We call a placement of cubes good if the color of the remaining cube at the end does not depend on the starting point of the robot. We call \$N\$ lucky if, for any choice of \$N\$ cubes, all their placements are good. Find all lucky \$N\$.
Let's assign the remainders $0,1,2$ from division by 3 to colors arbitrarily. All operations with them will also be performed modulo 3. Then the operation performed by the robot is as follows: if cubes of colors $ are destroyed,

If $\$ N=2 \wedge k \$$, then after each full round, the number of cubes is halved, and their sum changes sign. Therefore, at the end, a cube of color $\$(-1)^{\wedge} \mathrm{k}\left(\mathrm{a} \_1+\ldots+\mathrm{a}-\mathrm{N}\right) \$$ will be obtained, regardless of the starting point. Thus, powers of two are successful.

If $\$ \mathrm{~N}=2 \wedge \mathrm{k}+\mathrm{d} \$$, where $1 \leq \$ \mathrm{~d} \$ \leq \$ 2 \wedge k-1 \$$, then consider an arrangement with one red cube and $\$ \mathrm{~N}-1 \$$ white cubes. If the robot starts before the red cube, then after $\$ \mathrm{~d} \$$ moves, one blue cube and $\$ 2 \wedge \mathrm{k} \$-$ 1 white cubes will remain. If the robot starts immediately after the red cube, then after \$d\$ moves, one red cube and $\$ 2 \wedge k \$ 1$ white cubes will remain. The arguments provided for powers of two show that in these two situations, the final colors will be different, so \$N\$ is unsuccessful.

## [ quadratic trinomial (miscellaneous) $\quad]$ Problem 67002 Topics: [ [raffles and GMT on the coordinate plane] [ Monotonicity, boundedness $]$  Difficulty: 4 Classes: 10,11  Author: Lopatnikov A.A.  Can the plane be tiled with parabolas, none of which are equal? (It is required that each point of the plane belongs to exactly one parabola and that no parabola can be translated into any other parabola by a movement.)

## Solution

We introduce Cartesian coordinates \$(x; y)\$ on the plane and consider all possible parabolas with equations of the form $\$ \mathrm{y}=\mathrm{ax}{ }^{\wedge} 2+$ \operatorname $\{\ln \} \mathrm{a} \$$, where $\$ \mathrm{a} \$$ is an arbitrary positive number. Among these parabolas, there are no identical ones, since the coefficients at \$x^2\$ are different. (If the vertices of two parabolas and their axes, taking into account the direction, are aligned, then the parabola with the larger \$a\$ lies between the "horns" of the other.)

Fix an arbitrary \$x\$ and let \$a\$ run through the positive half-axis. Then \$y=ax^2+ \operatorname $\{\ln \}$ a $\$$ is a continuous function of \$a\$. Since \$x^2 \geq 0\$, the value \$ax^2\$ is non-decreasing

![](https://cdn.mathpix.com/cropped/2024_05_06_d9b26d382f9c77aca61cg-06.jpg?height=51&width=2035&top_left_y=998&top_left_x=0)
non-decreasing and strictly increasing functions. As \$a \to +\infty\$, we have \$y \to +\infty\$, and as \$a \to 0\$, we have \$y \to -\infty\$. Therefore, each value of \$y\$ for a given \$x\$ will appear exactly once. Hence, every point on the plane belongs to exactly one parabola.

## Answer

Yes.