[b]p1.[/b] In triangle $ABC$, the median $BM$ is drawn. The length $|BM| = |AB|/2$. The angle $\angle ABM = 50^o$. Find the angle $\angle ABC$.


[b]p2.[/b] Is there a positive integer $n$ which is divisible by each of $1, 2,3,..., 2018$ except for two numbers whose difference is$ 7$?


[b]p3.[/b] Twenty numbers are placed around the circle in such a way that any number is the average of its two neighbors. Prove that all of the numbers are equal.


[b]p4.[/b] A finite number of frogs occupy distinct integer points on the real line. At each turn, a single frog jumps by $1$ to the right so that all frogs again occupy distinct points. For some initial configuration, the frogs can make $n$ moves in $m$ ways. Prove that if they jump by $1$ to the left (instead of right) then the number of ways to make $n$ moves is also $m$.


[b]p5.[/b] A square box of chocolates is divided into $49$ equal square cells, each containing either dark or white chocolate. At each move Alex eats two chocolates of the same kind if they are in adjacent cells (sharing a side or a vertex). What is the maximal number of chocolates Alex can eat regardless of distribution of chocolates in the box?


PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Let \( a_1 = x \) and \( a_2 = y \). According to the problem, each number is the average of its two neighbors. Therefore, we have:
   \[
   a_3 = \frac{a_2 + a_4}{2}
   \]
   Since \( a_2 = y \) and \( a_4 = 3y - 2x \), we can write:
   \[
   a_3 = 2y - x
   \]
   Similarly, we can find:
   \[
   a_4 = 3y - 2x
   \]
2. We observe a pattern and hypothesize that \( a_n = (n-1)y - (n-2)x \) for all \( n \leq 19 \). We will prove this by induction.
   
   **Base Case:** For \( n = 1 \) and \( n = 2 \):
   \[
   a_1 = x \quad \text{and} \quad a_2 = y
   \]
   These satisfy the hypothesis since:
   \[
   a_1 = (1-1)y - (1-2)x = x \quad \text{and} \quad a_2 = (2-1)y - (2-2)x = y
   \]
   
   **Inductive Step:** Assume \( a_k = (k-1)y - (k-2)x \) holds for some \( k \). We need to show it holds for \( k+1 \):
   \[
   a_{k+1} = \frac{a_k + a_{k+2}}{2}
   \]
   Using the inductive hypothesis:
   \[
   a_{k+1} = \frac{(k-1)y - (k-2)x + (k+1)y - kx}{2} = \frac{2ky - 2kx}{2} = ky - kx
   \]
   This matches our hypothesis for \( k+1 \), thus completing the induction.

3. Now, we use the pattern to find \( a_{20} \):
   \[
   a_{20} = 19y - 18x
   \]
   We also know:
   \[
   2a_{20} = a_{19} + a_1 \implies 2(19y - 18x) = 18y - 17x + x \implies 38y - 36x = 18y - 16x \implies 20y = 20x \implies y = x
   \]
4. Since \( y = x \), we have:
   \[
   a_1 = a_2 = \cdots = a_{20} = x
   \]
   Therefore, all the numbers must be equal.

The final answer is \( \boxed{ x = y } \).