[b]p1.[/b] Find all real solutions of the system
$$x^5 + y^5 = 1$$
$$x^6 + y^6 = 1$$


[b]p2.[/b] The centers of three circles of the radius $R$ are located in the vertexes of equilateral triangle. The length of the sides of the triangle is $a$ and $\frac{a}{2}< R < a$. Find the distances between the intersection points of the
circles, which are outside of the triangle.


[b]p3.[/b] Prove that no positive integer power of $2$ ends with four equal digits.


[b]p4.[/b] A circle is divided in $10$ sectors. $90$ coins are located in these sectors, $9$ coins in each sector. At every move you can move a coin from a sector to one of two neighbor sectors. (Two sectors are called neighbor if they are adjoined along a segment.) Is it possible to move all coins into one sector in exactly$ 2004$ moves?


[b]p5.[/b] Inside a convex polygon several points are arbitrary chosen. Is it possible to divide the polygon into smaller convex polygons such that every one contains exactly one given point? Justify your answer.


[b]p6.[/b] A troll tried to spoil a white and red $8\times 8$ chessboard. The area of every square of the chessboard is one square foot. He randomly painted $1.5\%$ of the area of every square with black ink. A grasshopper jumped on the spoiled chessboard. The length of the jump of the grasshopper is exactly one foot and at every jump only one point of the chessboard is touched. Is it possible for the grasshopper to visit every square of the chessboard without touching any black point? Justify your answer.


PS. You should use hide for answers.
1. Given the system of equations:
   \[
   \begin{cases}
   x^5 + y^5 = 1 \\
   x^6 + y^6 = 1
   \end{cases}
   \]
2. From the second equation, we can derive that \( |x| \leq 1 \) and \( |y| \leq 1 \). This is because \( x^6 \) and \( y^6 \) are non-negative and their sum is 1.
3. Given \( x^5 + y^5 = 1 \), we can infer that neither \( x \) nor \( y \) can be negative. If both were negative, the left-hand side (LHS) would be negative, which contradicts the right-hand side (RHS) being positive.
4. If one of \( x \) or \( y \) were negative, the other would have to be greater than 1 to satisfy the equation \( (-\alpha)^5 + \beta^5 = 1 \Rightarrow \beta^5 = 1 + \alpha^5 \Rightarrow \beta > 1 \), which contradicts \( |y| \leq 1 \).
5. Therefore, the constraints \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \) must hold.
6. Combining both equations, we get:
   \[
   x^5 + y^5 = x^6 + y^6
   \]
7. Rearranging terms, we have:
   \[
   x^5 - x^6 = y^6 - y^5
   \]
   \[
   x^5(1 - x) = y^5(1 - y)
   \]
8. Let \( f(x) = x^5(1 - x) \). We need to find \( x \) and \( y \) such that \( f(x) = f(y) \).
9. In the interval \( 0 < x < 1 \), \( x^5 \) is positive and \( 1 - x \) is positive, so \( f(x) \) is positive. However, \( f(x) \) is decreasing in this interval because \( x^5 \) grows faster than \( 1 - x \) decreases.
10. The function \( f(x) \) is zero at the boundaries \( x = 0 \) and \( x = 1 \).
11. Therefore, the only solutions where \( f(x) = f(y) \) are at the boundaries \( x = 0 \) or \( x = 1 \).
12. Checking these boundary values:
    - If \( x = 0 \), then \( y^5 = 1 \) implies \( y = 1 \).
    - If \( x = 1 \), then \( y^5 = 0 \) implies \( y = 0 \).

Thus, the solutions to the system are:
\[
(x, y) = (1, 0) \text{ or } (0, 1)
\]

The final answer is \(\boxed{(1, 0) \text{ or } (0, 1)}\).