p1. Katia and Mariela play the following game: In each of their three turns, Katia replaces one of the stars in the expression $\star \star\star\star\star\star\star\star\star$ for some a digit of between $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$ that has not been used before in the game. In shifts of Mariela, she replaces two of the stars with two different digits that have not been used. Katia starts playing and plays alternately. Mariela wins if the resulting number at the end of the game is divisible by $27$. Does Mariela have any to ensure the triumph?


p2. Four line segments divide a rectangle into $ 8$ regions, such as illustrated in the figure. For the three marked regions, the area is as follows: $3$, $5$ and $ 8$. How much is the area of the gray quadrilateral?
[img]https://cdn.artofproblemsolving.com/attachments/b/5/60d70b8f12fcc57e63a1d221fbe8e5a324d090.png[/img]


p3. In the interior of a square with side $ 1$, $2016$ points are marked. Prove that it is possible, regardless of the position of the points, to join them all using a path continuous whose length is not greater than $146$.


p4. Find all pairs of prime numbers $(p, q)$ for which the following equality be true $$7pq^2 + p = q^3 + 43p^3 + 1$$

PS. Senior p1 was also [url=https://artofproblemsolving.com/community/c4h2690898p23356626]Juniors p2[/url].
To solve the given equation \(7pq^2 + p = q^3 + 43p^3 + 1\) for prime numbers \(p\) and \(q\), we will analyze the parity of the equation and consider different cases for \(p\) and \(q\).

1. **Parity Analysis**:
   - If both \(p\) and \(q\) are odd, then the left-hand side (LHS) \(7pq^2 + p\) is odd, and the right-hand side (RHS) \(q^3 + 43p^3 + 1\) is even (since \(q^3\) and \(43p^3\) are odd, and adding 1 makes it even). This is a contradiction.
   - Therefore, at least one of \(p\) or \(q\) must be 2 (the only even prime).

2. **Case 1: \(p = 2\)**:
   - Substitute \(p = 2\) into the equation:
     \[
     7 \cdot 2 \cdot q^2 + 2 = q^3 + 43 \cdot 2^3 + 1
     \]
     Simplify the equation:
     \[
     14q^2 + 2 = q^3 + 344 + 1
     \]
     \[
     14q^2 + 2 = q^3 + 345
     \]
     Rearrange the equation:
     \[
     q^3 - 14q^2 = -343
     \]
     Factorize the equation:
     \[
     q^2(q - 14) = 343
     \]
     Since \(343 = 7^3\), we can write:
     \[
     q^2(14 - q) = 7^3
     \]
     The only integer solution for \(q\) is \(q = 7\).

3. **Case 2: \(q = 2\)**:
   - Substitute \(q = 2\) into the equation:
     \[
     7p \cdot 2^2 + p = 2^3 + 43p^3 + 1
     \]
     Simplify the equation:
     \[
     28p + p = 8 + 43p^3 + 1
     \]
     \[
     29p = 9 + 43p^3
     \]
     Rearrange the equation:
     \[
     29p - 43p^3 = 9
     \]
     Factorize the equation:
     \[
     p(29 - 43p^2) = 9
     \]
     Since \(p\) is a prime number, \(p\) must divide 9. The only prime factors of 9 are 3, but substituting \(p = 3\) does not satisfy the equation. Therefore, there are no solutions in this case.

Thus, the only solution is when \(p = 2\) and \(q = 7\).

The final answer is \( \boxed{ (p, q) = (2, 7) } \).