3. The sum of $m$ consecutive natural numbers is equal to a prime number $p$. What can $m$ be?
Answer: 1 and 2.

Solution. The sum of $m$ consecutive numbers from $k$ to $k+m-1$ inclusive is $\frac{(2 k+m-1) m}{2}$. This sum must equal $p$, that is, the factors are 1 and $p$. If $m>2$ and $m$ is odd, then $m$ equals $p$. For $m=p: \frac{2 k+p-1}{2}=1, 2 k+p=3$, which is impossible. If $m>2$ and even, then $m / 2=p$. In this case $m=2 p, 2 k+2 p-1=1$, that is, $k+p=1$, which is impossible. Therefore, $m \leq 2$. For $m=1$, we need to take one number $p$. For $m=2$, we get $2 k+1=p$. Two consecutive numbers $k$ and $k+1$ are added, which are $\frac{p-1}{2}$ and $\frac{p+1}{2}$, respectively.

Comment. Correct solution - 20 points. Correct answer - 10 points, proof that there are no other answers - 10 points, points are cumulative. No example of implementation - 1 point is deducted for each case. Answer $m=1$ is omitted without explanation - 2 points are deducted. There is a gap in the proof - 15 points. The solution is started, there is some progress - 5 points. The solution is started, but the progress is insignificant - 1 point. Only the correct answer without a solution - 1 point.

4. In rectangle $A B C D$, side $B C=3$. Point $P$ is marked at the midpoint of side $A B$. A perpendicular $C Q$ is dropped from point $C$ to $D P$. Find the length of $B Q$.

Answer: 3.

Solution. Extend lines $C B$ and $D P$ until they intersect at point $M$. Triangles $M B P$ and $A P$ are equal, as they are right triangles with equal legs and equal vertical angles. Therefore, $M B=A D=B C$. Thus, $B Q$ is the median of the right triangle $M Q C$, and is equal to half the hypotenuse $M C$, which is 3.

Comment. Correct solution - 20 points. Potentially useful statements are proven - 5 points. The solution is started, but the progress is insignificant - 1 point.