There are $n$ students in a circle, one behind the other, all facing clockwise. The students have heights $h_1 <h_2 < h_3 < \cdots < h_n$. If a student with height $h_k$ is standing directly behind a student with height $h_{k-2}$ or lesss, the two students are permitted to switch places Prove that it is not possible to make more than $\binom{n}{3}$ such switches before reaching a position in which no further switches are possible.

Let $n$ be a positive integer. Denote by $S_n$ the set of points $(x, y)$ with integer coordinates such that \[ \left\lvert x\right\rvert + \left\lvert y + \frac{1}{2} \right\rvert < n. \] A path is a sequence of distinct points $(x_1 , y_1), (x_2, y_2), \ldots, (x_\ell, y_\ell)$ in $S_n$ such that, for $i = 2, \ldots, \ell$, the distance between $(x_i , y_i)$ and $(x_{i-1} , y_{i-1} )$ is $1$ (in other words, the points $(x_i, y_i)$ and $(x_{i-1} , y_{i-1} )$ are neighbors in the lattice of points with integer coordinates). Prove that the points in $S_n$ cannot be partitioned into fewer than $n$ paths (a partition of $S_n$ into $m$ paths is a set $\mathcal{P}$ of $m$ nonempty paths such that each point in $S_n$ appears in exactly one of the $m$ paths in $\mathcal{P}$).

On a bookcase there are $ n \geq 3$ books side by side by different authors. A librarian considers the first and second book from left and exchanges them iff they are not alphabetically sorted. Then he is doing the same operation with the second and third book from left etc. Using this procedure he iterates through the bookcase three times from left to right. Considering all possible initial book configurations how many of them will then be alphabetically sorted?

In a blackboard, it's written the following expression $ 1-2-2^2-2^3-2^4-2^5-2^6-2^7-2^8-2^9-2^{10}$ We put parenthesis by different ways and then we calculate the result. For example: $ 1-2-\left(2^2-2^3\right)-2^4-\left(2^5-2^6-2^7\right)-2^8-\left( 2^9-2^{10}\right)= 403$ and $ 1-\left(2-2^2 \left(-2^3-2^4 \right)-\left(2^5-2^6-2^7\right)\right)- \left(2^8- 2^9 \right)-2^{10}= -933$ How many different results can we obtain?

In a simple graph $G$, we call $t$ pairwise adjacent vertices a $t$[i]-clique[/i]. If a vertex is connected with all other vertices in the graph, we call it a [i]central[/i] vertex. Given are two integers $n,k$ such that $\dfrac {3}{2} \leq \dfrac{1}{2} n < k < n$. Let $G$ be a graph on $n$ vertices such that [b](1)[/b] $G$ does not contain a $(k+1)$-[i]clique[/i]; [b](2)[/b] if we add an arbitrary edge to $G$, that creates a $(k+1)$-[i]clique[/i]. Find the least possible number of [i]central[/i] vertices in $G$.

Let $ a,b,c,x,y$ be five real numbers such that $ a^3 \plus{} ax \plus{} y \equal{} 0$, $ b^3 \plus{} bx \plus{} y \equal{} 0$ and $ c^3 \plus{} cx \plus{} y \equal{} 0$. If $ a,b,c$ are all distinct numbers prove that their sum is zero. [i]Ciprus[/i]

The $100$ vertices of a prism, whose base is a $50$-gon, are labeled with numbers $1, 2, 3, \ldots, 100$ in any order. Prove that there are two vertices, which are connected by an edge of the prism, with labels differing by not more than $48$. Note: In all the triangles the three vertices do not lie on a straight line.

Let $n$ be an integer greater than 2. A positive integer is said to be [i]attainable [/i]if it is 1 or can be obtained from 1 by a sequence of operations with the following properties: 1.) The first operation is either addition or multiplication. 2.) Thereafter, additions and multiplications are used alternately. 3.) In each addition, one can choose independently whether to add 2 or $n$ 4.) In each multiplication, one can choose independently whether to multiply by 2 or by $n$. A positive integer which cannot be so obtained is said to be [i]unattainable[/i]. [b]a.)[/b] Prove that if $n\geq 9$, there are infinitely many unattainable positive integers. [b]b.)[/b] Prove that if $n=3$, all positive integers except 7 are attainable.

Find the least positive integer $ n$ for which $ \frac{n\minus{}13}{5n\plus{}6}$ is non-zero reducible fraction. $ \textbf{(A)}\ 45 \qquad \textbf{(B)}\ 68 \qquad \textbf{(C)}\ 155 \qquad \textbf{(D)}\ 226 \qquad \textbf{(E)}\ \text{none of these}$

Prove that for all natural $n$ there exists $a,b,c$ such that $n=\gcd (a,b)(c^2-ab)+\gcd (b,c)(a^2-bc)+\gcd (c,a)(b^2-ca)$.

Determine whether it's possible to cover a $K_{2012}$ with a) 1000 $K_{1006}$'s; b) 1000 $K_{1006,1006}$'s. [i]David Yang.[/i]

A group of people is lined up in [i]almost-order[/i] if, whenever person $A$ is to the left of person $B$ in the line, $A$ is not more than $8$ centimeters taller than $B$. For example, five people with heights $160, 165, 170, 175$, and $180$ centimeters could line up in [i]almost-order[/i] with heights (from left-to-right) of $160, 170, 165, 180, 175$ centimeters. (b) How many different ways are there to line up $20$ people in [i]almost-order[/i] if their heights are $120, 125, 130,$ $135,$ $140,$ $145,$ $150,$ $155,$ $160,$ $164, 165, 170, 175, 180, 185, 190, 195, 200, 205$, and $210$ centimeters? (Note that there is someone of height $164$ centimeters.)

In the following triangular lattice distance of two vertices is length of the shortest path between them. Let $ A_{1},A_{2},\dots,A_{n}$ be constant vertices of the lattice. We want to find a vertex in the lattice whose sum of distances from vertices is minimum. We start from an arbitrary vertex. At each step we check all six neighbors and if sum of distances from vertices of one of the neighbors is less than sum of distances from vertices at the moment we go to that neighbor. If we have more than one choice we choose arbitrarily, as seen in the attached picture. Obviusly the algorithm finishes a) Prove that when we can not make any move we have reached to the problem's answer. b) Does this algorithm reach to answer for each connected graph?

Find all prime numbers $ p,q,r$, such that $ \frac{p}{q}\minus{}\frac{4}{r\plus{}1}\equal{}1$

For positive integers $m$ and $n$, let $r(m, n)$ be the remainder when $m$ is divided by $n$. Find the smallest positive integer $m$ such that \[r(m, 1) + r(m, 2) + r(m, 3) +\cdots+ r(m, 10) = 4.\]