Let $\omega=\cos\frac{2\pi}{7}+i\cdot\sin\frac{2\pi}{7}$, where $i=\sqrt{-1}$. Find $$\prod_{k=0}^{6}(\omega^{3k}+\omega^k+1).$$

Is there a positive integer $N>10^{20}$ such that all its decimal digits are odd, the numbers of digits 1, 3, 5, 7, 9 in its decimal representation are equal, and it is divisible by each 20-digit number obtained from it by deleting digits? (Neither deleted nor remaining digits must be consecutive.)

Given a regular heptagon $A_1...A_7$. Prove that $$\frac{1}{|A_1A_5|} + \frac{1}{|A_1A_3| }= \frac{1}{|A_1A_7|}$$.

In triangle $ABC$ with $AB < AC$, let $H$ be the orthocenter and $O$ be the circumcenter. Given that the midpoint of $OH$ lies on $BC$, $BC = 1$, and the perimeter of $ABC$ is 6, find the area of $ABC$.

A [i]labyrinth[/i] is a system of $2024$ caves and $2023$ non-intersecting (bidirectional) corridors, each of which connects exactly two caves, where each pair of caves is connected through some sequence of corridors. Initially, Erik is standing in a corridor connecting some two caves. In a move, he can walk through one of the caves to another corridor that connects that cave to a third cave. However, when doing so, the corridor he was just in will magically disappear and get replaced by a new one connecting the end of his new corridor to the beginning of his old one (i.e., if Erik was in a corridor connecting caves $a$ and $b$ and he walked through cave $b$ into a corridor that connects caves $b$ and $c$, then the corridor between caves $a$ and $b$ will disappear and a new corridor between caves $a$ and $c$ will appear). Since Erik likes designing labyrinths and has a specific layout in mind for his next one, he is wondering whether he can transform the labyrinth into that layout using these moves. Prove that this is in fact possible, regardless of the original layout and his starting position there.

Let $ABCD$ be a trapezoid with $AB\parallel CD$. Points $E,F$ lie on segments $AB,CD$ respectively. Segments $CE,BF$ meet at $H$, and segments $ED,AF$ meet at $G$. Show that $S_{EHFG}\le \dfrac{1}{4}S_{ABCD}$. Determine, with proof, if the conclusion still holds when $ABCD$ is just any convex quadrilateral.

In \( \triangle ABC \) with \( AB > BC \), a tangent to the circumcircle of \( \triangle ABC \) at point \( B \) intersects the extension of \( AC \) at point \( D \). \( E \) is the midpoint of \( BD \), and \( AE \) intersects the circumcircle of \( \triangle ABC \) at \( F \). Prove that \( \angle CBF = \angle BDF \).

Given the integer $n > 1$ and the real number $a > 0$ determine the maximum of $\sum_{i=1}^{n-1} x_i x_{i+1}$ taken over all nonnegative numbers $x_i$ with sum $a.$

Let $n+1, n \geq 1$ positive integers be formed by taking the product of $n$ given prime numbers (a prime number can appear several times or also not appear at all in a product formed in this way.) Prove that among these $n+1$ one can find some numbers whose product is a perfect square.

Show that there exists a set of infinite positive integers such that the sum of an arbitrary finite subset of these is never a perfect square. What happens if we change the condition from not being a perfect square to not being a perfect power?

Say that a positive integer is MPR (Math Prize Resolvable) if it can be represented as the sum of a 4-digit number MATH and a 5-digit number PRIZE. (Different letters correspond to different digits. The leading digits M and P can't be zero.) Say that a positive integer is MPRUUD (Math Prize Resolvable with Unique Units Digits) if it is MPR and the set of units digits $\{ \mathrm{H}, \mathrm{E} \}$ in the definition of MPR can be uniquely identified. Find the smallest positive integer that is MPR but not MPRUUD.

There are $n\leq 99$ people around a circular table. At every moment everyone can either be truthful (always says the truth) or a liar (always lies). Initially some of people (possibly none) are truthful and the rest are liars. At every minute everyone answers at the same time the question "Is your left neighbour truthful or a liar?" and then becomes the same type of person as his answer. Determine the largest $n$ for which, no matter who are the truthful people in the beginning, at some point everyone will become truthful forever.

There are $n$ holes in a circle. The holes are numbered $1,2,3$ and so on to $n$. In the beginning, there is a peg in every hole except for hole $1$. A peg can jump in either direction over one adjacent peg to an empty hole immediately on the other side. After a peg moves, the peg it jumped over is removed. The puzzle will be solved if all pegs disappear except for one. For example, if $n=4$ the puzzle can be solved in two jumps: peg $3$ jumps peg $4$ to hole $1$, then peg $2$ jumps the peg in $1$ to hole $4$. (See illustration below, in which black circles indicate pegs and white circles are holes.) [center][img]http://i.imgur.com/4ggOa8m.png[/img][/center] [list=a] [*]Can the puzzle be solved for $n=5$? [*]Can the puzzle be solved for $n=2014$? [/list] In each part (a) and (b) either describe a sequence of moves to solve the puzzle or explain why it is impossible to solve the puzzle.

The set $ \{a_0, a_1, \ldots, a_n\}$ of real numbers satisfies the following conditions: [b](i)[/b] $ a_0 \equal{} a_n \equal{} 0,$ [b](ii)[/b] for $ 1 \leq k \leq n \minus{} 1,$ \[ a_k \equal{} c \plus{} \sum^{n\minus{}1}_{i\equal{}k} a_{i\minus{}k} \cdot \left(a_i \plus{} a_{i\plus{}1} \right)\] Prove that $ c \leq \frac{1}{4n}.$

For an integer $n$, define $f(n)$ to be the greatest integer $k$ such that $2^k$ divides $\binom{n}{m}$ for some $0 \le m \le n$. Compute $f(1) + f(2) + \cdots + f(2048)$.

In a group of $n$ people, each one knows exactly three others. They are seated around a table. We say that the seating is $perfect$ if everyone knows the two sitting by their sides. Show that, if there is a perfect seating $S$ for the group, then there is always another perfect seating which cannot be obtained from $S$ by rotation or reflection.

The diagonals of a convex quadrilateral $ABCD$ are perpendicular to each other and intersect at the point $O$. The sum of the inradii of triangles $AOB$ and $COD$ is equal to the sum of the inradii of triangles $BOC$ and $DOA$. $(a)$ Prove that $ABCD$ has an incircle. $(b)$ Prove that $ABCD$ is symmetric about one of its diagonals.

What is the least positive integer $n$ such that $\overbrace{f(f(\dots f}^{21 \text{ times}}(n)))=2013$ where $f(x)=x+1+\lfloor \sqrt x \rfloor$? ($\lfloor a \rfloor$ denotes the greatest integer not exceeding the real number $a$.) $ \textbf{(A)}\ 1214 \qquad\textbf{(B)}\ 1202 \qquad\textbf{(C)}\ 1186 \qquad\textbf{(D)}\ 1178 \qquad\textbf{(E)}\ \text{None of above} $

At Stanford in 1988, human calculator Shakuntala Devi was asked to compute $m = \sqrt[3]{61{,}629{,}875}$ and $n = \sqrt[7]{170{,}859{,}375}$. Given that $m$ and $n$ are both integers, compute $100m+n$. [i]Proposed by Evan Chen[/i]

Consider all sums that add up to $2015$. In each sum, the addends are consecutive positive integers, and all sums have less than $10$ addends. How many such sums are there?

In the triangle $ABC$, the medians $BB_1$ and $CC_1$, which intersect at the point $M$, are drawn. Prove that a circle can be inscribed in the quadrilateral $AC_1MB_1$ if and only if $AB = AC$.

Find all pairs of cubic equations $x^3 +ax^2 +bx +c =0$ and $x^3 +bx^2 + ax +c = 0$ where $a,b,c$ are integers, such that each equation has three integer roots and both the equations have exactly one common root.

Mary is $ 20\%$ older than Sally, and Sally is $ 40\%$ younger than Danielle. The sum of their ages is 23.2 years. How old will Mary be on her next birthday? $ \textbf{(A) } 7\qquad \textbf{(B) } 8\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$-by-$5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum? $\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

Let $ABC$ be an acute triangle. Find the locus of the points $M$, in the interior of $\bigtriangleup ABC$, such that $AB-FG= \frac{MF.AG+MG.BF}{CM}$, where $F$ and $G$ are the feet of the perpendiculars from $M$ to the lines $BC$ and $AC$, respectively.