Agakhanov N.K.

Two cars, an Audi and a BMW, are moving along two intersecting roads at equal constant speeds. It turned out that both at 17:00 and at 18:00, the BMW was twice as far from the intersection as the Audi. At what time could the Audi have passed the intersection?
Let at 17:00 "Audi" (A) be at a distance $x$, and BMW (B) - at a distance $2x$ from the intersection (P), $v$ - the speed of the cars. If at 17:00 A and B had already passed P, then at 18:00 they would be at distances $x+v$ and $2x+v$ from P, but the ratio of these distances is not equal to 2. Similarly, it is impossible for the cars to be moving towards P at 18:00. Therefore, one of them must have crossed P between 17:00 and 18:00. This could not have been B, because if A crosses P later, then at 17:00 B is closer to P than A; and if A crosses P earlier, then at 18:00 A is farther from P than B. Therefore, between 17:00 and 18:00 A crossed P, but B did not.

Then at 18:00 A was at a distance $z-x$ from P, and B - at a distance $2x-v$ (if it was moving towards the intersection at 18:00), or $2x+v$ from P. In the first case, $2x-v=2(v-x)$, that is, $3v=4x$. In the second case, $-2x+v=2(z-x)$, that is, $v=4x$. Therefore, at 18:00 A was either at a distance $x/3$ or at a distance $3x$ from P and thus he crossed P either at 17:45 or at 17:15.

## Answer

At 17:15 or 17:45.

![](https://cdn.mathpix.com/cropped/2024_05_06_388b3f8bbd0a17840215g-02.jpg?height=231&width=1121&top_left_y=-1&top_left_x=14)

Author: Sadikov P.

In the cells of a $11 \times 11 \times 11$ cube, the numbers $1, 2, \ldots, 1331$ are placed. Two worms start from one corner cube and head to the opposite corner cube. Each worm can crawl to an adjacent cube by a face, with the first worm moving if the number in the adjacent cube differs by 8, and the second worm moving if it differs by 9. Is there a way to arrange the numbers so that both worms can reach the opposite corner cube?

## Solution

Assume such a number arrangement exists.

Let the numbers in the initial and final corner cubes be $a$ and $b$ respectively. We can assume that $a < b$. Also, we can assume that each worm does not crawl into any cube more than once.

Then the first worm must sequentially crawl through cubes with numbers $a, a+8, a+16, a+24, \ldots$, $a+72, \ldots, b$ (obviously it makes more than 10 "moves"), and the second worm must sequentially crawl through cubes with numbers $a, a+9, a+18, a+27, \ldots, a+72, \ldots, b$.

Now consider a chessboard coloring of our cube. We can assume that the cube with number $a$ is black.

Notice that adjacent cubes by a face must have different colors. Therefore, cubes with numbers $a, a+18, a+36, \ldots, a+72$ are black, and cubes with numbers $a+8, a+24, a+40, \ldots, a+72$ are white.

Thus, the cube with number $a+72$ must be both black and white. Contradiction.

## Answer

No.