2. At one vertex of a cube, the number 2021 is written, and zeros are written at the other vertices. You can add one to the numbers at the ends of any edge. Is it possible to achieve that all numbers standing at the vertices of the cube are divisible by $2?$
Solution. Initially, the sum of the numbers at the vertices of the cube is 2021, which is odd. Each time we add one to two vertices, the sum remains odd. Since the sum of eight integers is odd (in this problem, the oddness of the sum is an invariant), the numbers at all vertices cannot become even. Answer: it is impossible.