Ya. V. Yashchenko

a) In Mexico City, to limit the traffic flow, each private car is assigned two days of the week when it cannot be driven on the city's streets. A family needs to have at least ten cars available every day. What is the smallest number of cars the family can manage with, if its members can choose the prohibited days for their cars themselves?

b) In Mexico City, each private car is assigned one day of the week when it cannot be driven on the city's streets. A wealthy family of ten people has bribed the police, and for each car, they name two days, one of which the police choose as the non-driving day. What is the minimum number of cars the family needs to buy so that each day every family member can drive independently, if the assignment of non-driving days for the cars goes sequentially?
a) 13 cars are not enough. Indeed, the total number of prohibitions is 26, and there are 7 days in a week, so on one of the days, at least four cars will be "resting," meaning only 9 can go out.

14 cars are sufficient: prohibit four cars from Monday and Tuesday, four from Wednesday and Thursday, two from Friday and Saturday, two from Saturday and Sunday, and two from Friday and Sunday.

b) 11 cars are not enough: on one of the days of the week, at least two cars will be "resting."

We will show that 12 cars are enough. When submitting the "application" for the next car, the family will indicate the pair of days when there is currently a prohibition on no more than one car. This can be continued until there are 6 days with two prohibitions, meaning this can be done for each of the 12 cars. As a result, on each day of the week, the family will have no more than two non-driving cars.

## Answer

a) 14; b) 12 cars.

Author: Senderov V.A.

Do there exist real numbers $a, b$, and $c$ such that for all real $x$ and $y$, the inequality

$$
|x+a|+|x+y+b|+|y+c|>|x|+|x+y|+|y| ?
$$

## Solution

Suppose such numbers $a, b$, and $c$ exist. Choose $x>0$ and $y>0$ such that $x+a \geq 0, x+y+b \geq 0, y+c \geq 0$. Then the difference between the left and right sides is $a+b+c$. If we take $x>0$ and $y>0$, on one side $a+b+c \geq 0$, and on the other $a+b+c<0$. Contradiction.

## Answer

No.