p1. How many $10$-digit strings are there, such that all its digits are only zeros or ones and the sum of its even-place digits equals the sum of the odd-place digits.


p2. Find all pairs $(x, y)$ of nonnegative integers, such that $x! + 24 = y^2$.


p3. Consider a function $f: Z \to Q$ such that $$f(1) = 2015 \,\,\, \,\, f (1) + f (2) + ...+ f (n) = n^2 f (n).$$
Determine the value of $f(2015)$.


p4. Let $ABC$ be an isosceles triangle with $AB = AC$, and let $D$ be the midpoint of $BC$, $E$ the foot of the perpendicular on $AB$ from $D$, and F the midpoint of $DE$. Show that $AF$ is perpendicular to $CE$.


p5. On an island there are only two tribes: the Mienteretes who always lie and the Veritas who always tell the truth. On a certain day there is an assembly attended by $2015$ inhabitants of the island. They sit at random in a circle and each one declares: "the two people next to me are Mienteretes." The next day the assembly continues but one of them became ill, for which $2014$ inhabitants attend, again they sit at random in a circle and each one declares: “The two people next to me are from the same tribe, which is not mine ”.
Deduct the number of inhabitants of each tribe and the type of inhabitant to which the patient belongs.
1. We need to find the number of 10-digit strings consisting only of 0s and 1s such that the sum of the digits in the even positions equals the sum of the digits in the odd positions.
2. Let's denote the positions of the digits in the string as follows:
   - Odd positions: 1, 3, 5, 7, 9
   - Even positions: 2, 4, 6, 8, 10
3. Let \( S \) be the sum of the digits in the odd positions and the sum of the digits in the even positions. We need to find the number of strings for each possible value of \( S \) from 0 to 5.
4. For each value of \( S \), we need to count the number of ways to distribute \( S \) ones among the 5 odd positions and \( S \) ones among the 5 even positions.

Case \( S = 0 \):
- There is 1 way to place 0 ones in the odd positions.
- There is 1 way to place 0 ones in the even positions.
- Total: \( 1 \times 1 = 1 \)

Case \( S = 1 \):
- There are \( \binom{5}{1} = 5 \) ways to place 1 one in the odd positions.
- There are \( \binom{5}{1} = 5 \) ways to place 1 one in the even positions.
- Total: \( 5 \times 5 = 25 \)

Case \( S = 2 \):
- There are \( \binom{5}{2} = 10 \) ways to place 2 ones in the odd positions.
- There are \( \binom{5}{2} = 10 \) ways to place 2 ones in the even positions.
- Total: \( 10 \times 10 = 100 \)

Case \( S = 3 \):
- There are \( \binom{5}{3} = 10 \) ways to place 3 ones in the odd positions.
- There are \( \binom{5}{3} = 10 \) ways to place 3 ones in the even positions.
- Total: \( 10 \times 10 = 100 \)

Case \( S = 4 \):
- There are \( \binom{5}{4} = 5 \) ways to place 4 ones in the odd positions.
- There are \( \binom{5}{4} = 5 \) ways to place 4 ones in the even positions.
- Total: \( 5 \times 5 = 25 \)

Case \( S = 5 \):
- There is 1 way to place 5 ones in the odd positions.
- There is 1 way to place 5 ones in the even positions.
- Total: \( 1 \times 1 = 1 \)

5. Summing up all the cases, we get:
\[ 1 + 25 + 100 + 100 + 25 + 1 = 252 \]

The final answer is \( \boxed{252} \)