5. Jack Sparrow needed to distribute 150 piastres into 10 purses. After placing a certain number of piastres in the first purse, he put more in each subsequent purse than in the previous one. As a result, it turned out that the number of piastres in the first purse was not less than half the number of piastres in the last purse. How many piastres are in the 6th purse?
Answer: 16

Solution: Let there be $\mathrm{x}$ piastres in the first purse. Then in the second there are no less than $\mathrm{x}+1$, in the third - no less than $\mathrm{x}+2$... in the 10th - no less than $\mathrm{x}+9$. Thus, on the one hand, $x+$ $x+1+\cdots+x+9=10 x+45 \leq 150$, from which $x \leq 10$. On the other hand, $x \geq(x+$ 9 )/2, from which $x \geq 9$. Therefore, in the first purse, there are 9 or 10 piastres. But 9 cannot be, because then in the last one there will be no more than 18 and the sum will not reach 150.

So, in the 1st - 10, in the 2nd - no less than 11, in the 3rd - no less than 12, in the 4th - no less than 13, in the 5th - no less than 14, and in the 6th - no less than 15. But with 15, the sum is less than 150. Therefore, 16. And it cannot be more, because then in the last purse there will be 21.

For graders: A partial score can be given for a correct example of distribution (trial and error) without proof of uniqueness.

Find the smallest natural $N$ such that the decimal representation of the number $N \times 999$ consists entirely of sevens (the " $x$ " symbol denotes multiplication of numbers).

Answer 778556334111889667445223

Solution: $N \times 999=77 \ldots 7$, then $N$ is divisible by 7, denote $n=N / 7$. We get $999 n=$ $1000 n-n=11 \ldots 1$, so $1000 \mathrm{n}-111 \ldots 1=$ n. Write it as a subtraction in a column and repeat the found digits of $N$ with a shift of 3 to the left

********000
$* * * * * * 889$

$* * * * 889000$

1111111111

$* * * 777889$

777889000

111111111

6667778889

Notice that the digits repeat every 3, so we get $n=111222333444555666777889$.

Therefore, $N=7 n=778556334111889667445223$.

For graders: An alternative approach is to take the number 111... 1 and start dividing it by 999 in a column until it divides evenly. I think the score should not be reduced if at the very end the number $\mathrm{n}$ was incorrectly multiplied by 7.