5. (20 points) Alexei came up with the following game. First, he chooses a number $x$ such that $2017 \leqslant x \leqslant 2117$. Then he checks if $x$ is divisible by 3, 5, 7, 9, and 11 without a remainder. If $x$ is divisible by 3, Alexei awards the number 3 points, if by 5 - then 5 points, ..., if by 11 - then 11 points. The points earned for the number are summed. Which number should be chosen in this game to score the maximum number of points?
Answer: $2079=11 \cdot 9 \cdot 7 \cdot 3$.

Solution: Note that the divisibility of $x$ by 9 immediately gives 12 points, as it also implies divisibility by 3. If the number $x$ is not divisible by 11, it will score no more than $9+7+5+3=24$ points, and if it is not divisible by 9, it will score no more than $11+7+5+3=26$ points. If the number is divisible by both 11 and 9, then it is a multiple of 99. In the given range, the only such number is 2079. Since 2079 is also divisible by 7, it gives $11+9+7+3=30$ points.

It remains to note that more than 30 points cannot be scored. Indeed, such a result can only be achieved by a number that is simultaneously divisible by $3, 5, 7, 9$, and 11. The smallest such number is $11 \cdot 9 \cdot 7 \cdot 5=3465$, but it is greater than 2117.