5. Point $M$ divides side $B C$ of parallelogram $A B C D$ in the ratio $B M: M C=2$. Line $A M$ intersects diagonal $B D$ at point $K$. Find the area of quadrilateral $C M K D$, if the area of parallelogram $A B C D$ is 1.

Task 1 Answer: 236 minutes or 173 minutes.
Solution. It was considered correct to count the moment 00:00 as both happy and not happy, with the solution practically unchanged.

Happy moments when the number of minutes is three times the number of hours: 01:03, $02:06, \ldots, 19:57$ and, perhaps, 00:00. The intervals between these happy moments are 1 hour 3 minutes from 01:03 (perhaps from 00:00) to 19:57 and 5 hours 6 minutes from 19:57 to 01:03 the next day (or 4 hours 3 minutes from 19:57 to 00:00 the next day).

Happy moments when the number of hours is three times the number of minutes: 03:01, $06:02, \ldots, 21:07$ and, perhaps, 00:00. The intervals between these happy moments are 3 hours 1 minute from 03:01 (perhaps from 00:00) to 21:07 and 5 hours 54 minutes from 21:07 to 03:01 the next day (or 2 hours 53 minutes from 21:07 to 00:00 the next day).

By overlaying these series of happy moments, we see that if the moment 00:00 is not considered happy, then from 01:03 to 19:57, the interval between happy moments does not exceed 1 hour 3 minutes, followed by an interval from 19:57 to 21:07 lasting 1 hour 10 minutes and an interval from 21:07 to 01:03 the next day lasting 3 hours 56 minutes. The maximum duration is the last interval, which is 236 minutes.

If the moment 00:00 is considered happy, then from 00:00 to 19:57, the interval between happy moments does not exceed 1 hour 3 minutes, followed by an interval from 19:57 to 21:07 lasting 1 hour 10 minutes and an interval from 21:07 to 00:00 the next day lasting 2 hours 53 minutes. The maximum duration is the last interval, which is 173 minutes.

Problem 2 Answer: $6,15,24,33,42$.

Solution. From the condition, it follows that the terms of the arithmetic sequence are natural numbers and the common difference, denoted by $d$, is positive. Let the letters A, B, C, D, E represent digits or numbers, and the corresponding two-digit numbers will be denoted by $\overline{\mathrm{A}}$ and so on.

Since $\overline{\Phi \mathrm{A}}-\overline{\mathrm{A}}=(10 \Phi+\mathrm{A})-(10 \mathrm{~A}+\Phi)=9(\Phi-\mathrm{A})=2 d$, the value of $d$ is a multiple of 9. Therefore, $d$ is either 9 or 18. The value 27 or more cannot be taken by $d$, because in this case $\overline{\text { ФА }}=$ D $+4 d \geq 4 \times 27>100$, which is impossible since the number $\overline{\text { ФА }}$ is a two-digit number.

First, assume that $d=9$. Then $\mathrm{E}=1, \mathrm{~A}=2, \mathrm{C}=3, \Phi=4$. Using the values of the found letters, we see that the record on the board becomes D, $1 \overline{\mathrm{E}}, 24,33,42$. Then $1 \overline{\mathrm{E}}=24-d=24-9=15$ and D $=1 \overline{\mathrm{E}}-d=15-9=6$. Thus, the numbers on the board are $6,15,24,33$ and 42.

Now assume that $d=18$. Since $\overline{\text { ФА }}=$ D $+4 d=Д+4 \times 18=Д+72$, then $81 \geq$ $\overline{\Phi А} \geq 72$, i.e., the letter Ф represents either 7 or 8. Consider both possibilities.

If the letter Ф corresponds to 7, then the record on the board has the form D, $\overline{\text { ЕЕ, }}, \overline{\mathrm{A}} 7, \overline{\mathrm{CC}}, 7 \overline{\mathrm{A}}$. Since $\overline{\mathrm{CC}}=$ $\overline{\mathrm{A}} 7+d=\overline{\mathrm{A}} 7+18$, the number $\overline{\mathrm{CC}}$ ends in 5 and the record on the board becomes D, $\overline{\text { ЕЕ }}, \overline{\mathrm{A}} 7,55,7 \overline{\mathrm{A}}$. Then $7 \overline{\mathrm{A}}=55+18=73, \overline{\mathrm{A}} 7=55-18=37, \overline{\mathrm{BE}}=55-2 d=55-36=19$ and D $=55-3 d=$ $55-54=1$. Thus, the letters D and B represent the digit 1, which contradicts the condition of the problem.

If the letter Ф represents 8, then the record on the board has the form D, $\overline{\text { ЕЕ }}, \overline{\mathrm{A}} 8, \overline{\mathrm{CC}}, 8 \overline{\mathrm{A}}$. Since $\overline{\mathrm{CC}}=\overline{\mathrm{A}} 8+$ $d=\overline{\mathrm{A}} 8+18$, the number $\overline{\mathrm{CC}}$ ends in 6 and the record on the board becomes D, $\overline{\text { БЕ }}, \overline{\mathrm{A}} 8,66,8 \overline{\mathrm{A}}$. Then $8 \overline{\mathrm{A}}=66+18=84$, which is impossible since $\overline{\Phi A}=8 \overline{\mathrm{A}} \leq$ 81 must hold.

Problem 3 Answer: 333.

Solution. Let $n$ be the smallest number written by Petya, and $n+q$ and $n+p, 0 \leq q<p \leq 7-$ be the numbers erased by Vasya. Then the sum of the remaining numbers is $\frac{(n+(n+7)) 8}{2}-(n+q)-(n+p)=6 n+$ $28-(q+p)=2022$. From this, $6 n=2022-28+(q+p)=1994+(q+p)=6 \times 332+(2+q+$ p). The value of $n$ is minimal when the right-hand side is minimal and must be a multiple of 6. Since the numbers $q$ and $p$ are non-negative, the right-hand side is minimal when $q+p=4$. Then $6 n=6 \times 333$, i.e., $n=333$.

Problem 4 Answer: proof.

Solution. There are infinitely many such triples. For example, for any natural number $n$ the triple $\left(n^{2023} ; 0 ; n^{2022}\right)$ is an integer solution. In variant 2 (and 3), where the equation is $x^{2023}+y^{2023}=z^{2024}$, possible solutions are $(n ;-n ; 0)$.

Problem 5 Answer: $\frac{11}{30}$.

Solution. According to the problem, the length of the segment $[B ; M]$ is $\frac{2}{3}$ of the length of the segment $[B ; C]$. Additionally, triangles $D A K$ and $B M K$ are similar with a similarity ratio of $\frac{2}{3}$. Therefore, the length of the segment $[B ; K]$ is $\frac{2}{5}$ of the length of the segment $[B ; D]$. Thus, for the areas of triangles $B C D$ and $B M K$, the relation $S_{B M K}=\frac{2}{3} \times \frac{2}{5} \times S_{B D C}=\frac{4}{15} \times \frac{1}{2}=\frac{4}{30}$ holds, and the area of the quadrilateral $C M K D$ is $S_{C M K D}=\frac{1}{2}-$ $\frac{4}{30}=\frac{11}{30}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7222f67593c2cda60ca3g-2.jpg?height=357&width=716&top_left_y=2346&top_left_x=747)