p1. In the rectangle $ABCD$ there exists a point $P$ on the side $AB$ such that $\angle PDA = \angle BDP = \angle CDB$ and $DA = 2$. Find the perimeter of the triangle $PBD$.


p2. Write in each of the empty boxes of the following pyramid a number natural greater than $ 1$, so that  the number written in each box is equal to the product of the numbers written in the two boxes on which it is supported.
[img]https://cdn.artofproblemsolving.com/attachments/3/f/395e1c09fd955ed7ba6f9fea23ef68a00a4880.png[/img]


p3. In how many ways can seven numbers be chosen from $ 1$ to $9$ such that their sum is a multiple of $3$?


p4. Simplify the product $$p = (10 + 1) (10^2 + 1) ((10^2)^2 + 1) ((10^2)^4 + 1) ((10^2)^8 + 1)...((10^2)^{1024} +1)$$


p5. A [i]lattice [/i] point is a point on the Cartesian plane with integer coordinates. We randomly select five lattice points. Show that there are two of them that form a line segment whose midpoint is also a lattice point.
To solve this problem, we need to find the number of ways to choose seven numbers from the set \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) such that their sum is a multiple of 3.

1. **Sum of the Set**:
   The sum of all numbers from 1 to 9 is:
   \[
   1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
   \]
   Since 45 is a multiple of 3, the sum of any 7 numbers chosen from this set will be a multiple of 3 if and only if the sum of the 2 numbers that are not chosen is also a multiple of 3.

2. **Finding Pairs**:
   We need to find pairs of numbers from the set \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) whose sum is a multiple of 3. We can categorize the numbers based on their remainders when divided by 3:
   \[
   \{1, 4, 7\} \quad \text{(remainder 1)}
   \]
   \[
   \{2, 5, 8\} \quad \text{(remainder 2)}
   \]
   \[
   \{3, 6, 9\} \quad \text{(remainder 0)}
   \]

3. **Valid Pairs**:
   - Pairs from \(\{3, 6, 9\}\): Any two numbers from this set will sum to a multiple of 3. There are \(\binom{3}{2} = 3\) ways to choose 2 numbers from this set.
   - Pairs from \(\{1, 4, 7\}\) and \(\{2, 5, 8\}\): We need one number from each set to sum to a multiple of 3. There are \(3 \times 3 = 9\) ways to choose one number from each set.

4. **Total Pairs**:
   Adding the number of valid pairs from both cases, we get:
   \[
   3 + 9 = 12
   \]

Thus, there are 12 ways to choose 7 numbers from the set \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) such that their sum is a multiple of 3.

The final answer is \(\boxed{12}\)