A wedding is going to be held in a city with $25$ types of meals, to which some of the $2017$ citizens will be invited. All of the citizens like some meals and each meal is liked by at least one person. A "$suitable$ $list$" is a set of citizens, such that each meal is liked by at least one person in the set. A "$kamber$ $group$" is a set that contains at least one person from each "$suitable$ $list$". Given a "$kamber$ $group$", which has no subset (other than itself) that is also a "$kamber$ $group$", prove that there exists a meal, which is liked by everyone in the group.

Let $ABCD$ be a quadrilateral such that line $AB$ intersects $CD$ at $X$. Denote circles with inradius $r_1$ and centers $A, B$ as $w_a$ and $w_b$ with inradius $r_2$ and centers $C, D$ as $w_c$ and $w_d$. $w_a$ intersects $w_d$ at $P, Q$. $w_b$ intersects $w_c$ at $R, S$. Prove that if $XA.XB+r_2^2=XC.XD+r_1^2$, then $P,Q,R,S$ are cyclic.

Denote the sequence $a_{i,j}$ in positive reals such that $a_{i,j}$.$a_{j,i}=1$. Let $c_i=\sum_{k=1}^{n}a_{k,i}$. Prove that $1\ge$$\sum_{i=1}^{n}\dfrac {1}{c_i}$

Let $d(n)$ be number of prime divisors of $n$. Prove that one can find $k,m$ positive integers for any positive integer $n$ such that $k-m=n$ and $d(k)-d(m)=1$

Let $x_0,\dots,x_{2017}$ are positive integers and $x_{2017}\geq\dots\geq x_0=1$ such that $A=\{x_1,\dots,x_{2017}\}$ consists of exactly $25$ different numbers. Prove that $\sum_{i=2}^{2017}(x_i-x_{i-2})x_i\geq 623$, and find the number of sequences that holds the case of equality.

Finite number of $2017$ units long sticks are fixed on a plate. Each stick has a bead that can slide up and down on it. Beads can only stand on integer heights $( 1, 2, 3,..., 2017 )$. Some of the bead pairs are connected with elastic bands. $The$ $young$ $ant$ can go to every bead, starting from any bead by using the elastic bands. $The$ $old$ $ant$ can use an elastic band if the difference in height of the beads which are connected by the band, is smaller than or equal to $1$. If the heights of the beads which are connected to each other are different, we call it $valid$ $situation$. If there exists at least one $valid$ $situation$, prove that we can create a $valid$ $situation$, by arranging the heights of the beads, in which $the$ $old$ $ant$ can go to every bead, starting from any bead.