In a triangle $ABC ~(\overline{AB} < \overline{AC})$, points $D (\neq A, B)$ and $E (\neq A, C)$ lies on side $AB$ and $AC$ respectively. Point $P$ satisfies $\overline{PB}=\overline{PD}, \overline{PC}=\overline{PE}$. $X (\neq A, C)$ is on the arc $AC$ of the circumcircle of triangle $ABC$ not including $B$. Let $Y (\neq A)$ be the intersection of circumcircle of triangle $ADE$ and line $XA$. Prove that $\overline{PX} = \overline{PY}$.

Function $f : \mathbb{R^+} \rightarrow \mathbb{R^+}$ satisfies the following condition. (Condition) For each positive real number $x$, there exists a positive real number $y$ such that $(x + f(y))(y + f(x)) \leq 4$, and the number of $y$ is finite. Prove $f(x) > f(y)$ for any positive real numbers $x < y$. ($\mathbb{R^+}$ is a set for all positive real numbers.)

Let $p$ be an odd prime. Let $A(n)$ be the number of subsets of $\{1,2,...,n\}$ such that the sum of elements of the subset is a multiple of $p$. Prove that if $2^{p-1}-1$ is not a multiple of $p^2$, there exists infinitely many positive integer $m$ for any integer $k$ that satisfies the following. (The sum of elements of the empty set is 0.) $$\frac{A(m)-k}{p}\in\mathbb{Z}$$

Find all positive integers $n$ satisfying the following. $$2^n-1 \text{ doesn't have a prime factor larger than } 7$$

Given a positive integer $n$, there are $n$ boxes $B_1,...,B_n$. The following procedure can be used to add balls. $$\text{(Procedure) Chosen two positive integers }n\geq i\geq j\geq 1\text{, we add one ball each to the boxes }B_k\text{ that }i\geq k\geq j.$$For positive integers $x_1,...,x_n$ let $f(x_1,...,x_n)$ be the minimum amount of procedures to get all boxes have its amount of balls to be a multiple of 3, starting with $x_i$ balls for $B_i(i=1,...,n)$. Find the largest possible value of $f(x_1,...,x_n)$. (If $x_1,...,x_n$ are all multiples of 3, $f(x_1,...,x_n)=0$.)

For positive integer $n\geq 3$ and real numbers $a_1,...,a_n,b_1,...,b_n$, prove the following. $$\sum_{i=1}^n a_i(b_i-b_{i+3})\leq\frac{3n}{8}\sum_{i=1}^n((a_i-a_{i+1})^2+(b_i-b_{i+1})^2)$$($a_{n+1}=a_1$, and for $i=1,2,3$ $b_{n+i}=b_i$.)