[b]p1.[/b] There are $16$ students in a class. Each month the teacher divides the class into two groups. What is the minimum number of months that must pass for any two students to be in different groups in at least one of the months?


[b]p2.[/b] Find all functions $f(x)$ defined for all real $x$ that satisfy the equation $2f(x) + f(1 - x) = x^2$.


[b]p3.[/b] Arrange the digits from $1$ to $9$ in a row (each digit only once) so that every two consecutive digits form a two-digit number that is divisible by $7$ or $13$.


[b]p4.[/b] Prove that $\cos 1^o$  is irrational.


[b]p5.[/b] Consider $2n$ distinct positive Integers $a_1,a_2,...,a_{2n}$ not exceeding $n^2$  ($n>2$).  Prove that some three of the differences $a_i- a_j$ are equal .


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
To solve the functional equation \(2f(x) + f(1 - x) = x^2\), we will follow these steps:

1. **Define the functions:**
   Let \( A = f(x) \) and \( B = f(1 - x) \). We are given:
   \[
   2A + B = x^2
   \]

2. **Substitute \(1 - x\) into the given equation:**
   Substituting \(1 - x\) into the given equation, we get:
   \[
   2f(1 - x) + f(x) = (1 - x)^2
   \]
   This simplifies to:
   \[
   2B + A = (1 - x)^2 = 1 - 2x + x^2
   \]

3. **Form a system of linear equations:**
   We now have the following system of equations:
   \[
   \begin{cases}
   2A + B = x^2 \\
   2B + A = 1 - 2x + x^2
   \end{cases}
   \]

4. **Solve the system of equations:**
   To solve for \(A\) and \(B\), we can add the two equations:
   \[
   (2A + B) + (2B + A) = x^2 + (1 - 2x + x^2)
   \]
   Simplifying, we get:
   \[
   3A + 3B = 2x^2 - 2x + 1
   \]
   Dividing by 3:
   \[
   A + B = \frac{2}{3}x^2 - \frac{2}{3}x + \frac{1}{3}
   \]

5. **Express \(B\) in terms of \(A\):**
   From the first equation \(2A + B = x^2\), we can solve for \(B\):
   \[
   B = x^2 - 2A
   \]

6. **Substitute \(B\) into the sum equation:**
   Substitute \(B = x^2 - 2A\) into \(A + B = \frac{2}{3}x^2 - \frac{2}{3}x + \frac{1}{3}\):
   \[
   A + (x^2 - 2A) = \frac{2}{3}x^2 - \frac{2}{3}x + \frac{1}{3}
   \]
   Simplifying, we get:
   \[
   x^2 - A = \frac{2}{3}x^2 - \frac{2}{3}x + \frac{1}{3}
   \]
   Solving for \(A\):
   \[
   A = x^2 - \left(\frac{2}{3}x^2 - \frac{2}{3}x + \frac{1}{3}\right)
   \]
   \[
   A = x^2 - \frac{2}{3}x^2 + \frac{2}{3}x - \frac{1}{3}
   \]
   \[
   A = \frac{1}{3}x^2 + \frac{2}{3}x - \frac{1}{3}
   \]

7. **Express \(f(x)\):**
   Since \(A = f(x)\), we have:
   \[
   f(x) = \frac{1}{3}x^2 + \frac{2}{3}x - \frac{1}{3}
   \]

The final answer is \(\boxed{f(x) = \frac{1}{3}x^2 + \frac{2}{3}x - \frac{1}{3}}\).