What is the smallest positive integer $k$ such that the number $\textstyle\binom{2k}k$ ends in two zeros? $\textbf{(A) }3\hspace{14em}\textbf{(B) }4\hspace{14em}\textbf{(C) }5$ $\textbf{(D) }6\hspace{14em}\textbf{(E) }7\hspace{14em}\textbf{(F) }8$ $\textbf{(G) }9\hspace{14em}\textbf{(H) }10\hspace{13.3em}\textbf{(I) }11$ $\textbf{(J) }12\hspace{13.8em}\textbf{(K) }13\hspace{13.3em}\textbf{(L) }14$ $\textbf{(M) }2007$

We have a stick of length $2n{}$ and a machine which cuts sticks of length $k\in\mathbb{N}$ with $k>1$ into two sticks with arbitrary positive integer lengths. What is the smallest number of cuts after which we can always find some sticks whose lengths sum up to $n{}$?

A polynomial $p(x)$ has remainder three when divided by $x-1$ and remainder five when divided by $x-3$. The remainder when $p(x)$ is divided by $(x-1)(x-3)$ is $\textbf{(A) }x-2\qquad\textbf{(B) }x+2\qquad\textbf{(C) }2\qquad\textbf{(D) }8\qquad \textbf{(E) }15$

Let $n \ge 2$ be a positive integer, and let $\sigma(n)$ denote the sum of the positive divisors of $n$. Prove that the $n^{\text{th}}$ smallest positive integer relatively prime to $n$ is at least $\sigma(n)$, and determine for which $n$ equality holds. [i]Proposed by Ashwin Sah[/i]

Let $ABC$ be a triangle of perimeter $100$ and $I$ be the point of intersection of its bisectors. Let $M$ be the midpoint of side $BC$. The line parallel to $AB$ drawn by$ I$ cuts the median $AM$ at point $P$ so that $\frac{AP}{PM} =\frac73$. Find the length of side $AB$.

In the triangle $ABC$, $M$ is the midpoint of $AB$ and $B'$ is the foot of $B$-altitude. $CB'M$ intersects the line $BC$ for the second time at $D$. Circumcircles of $CB'M$ and $ABD$ intersect each other again at $K$. The parallel to $AB$ through $C$ intersects the $CB'M$ circle again at $L$. Prove that $KL$ cuts $CM$ in half.

Let $G$ be the point of intersection of the medians in the triangle $ABC$. Let us denote $A_1, B_1, C_1$ the second points of intersection of lines $AG, BG, CG$ with the circle circumscribed around the triangle. Prove that $AG + BG + CG \le A_1C + B_1C + C_1C$. (Yasinsky V.A.)

The number of distinct points common to the graphs of $x^2+y^2=9$ and $y^2=9$ is: $ \textbf{(A) }\text{infinitely many} \qquad\textbf{(B) } \text{four}\qquad\textbf{(C) }\text{two}\qquad\textbf{(D) }\text{one}\qquad\textbf{(E) }\text{none} $

Solve the equation $$[2 \sin x] =2\cos \left(3x+\frac{\pi}{4} \right)$$ ($[x]$ is the integer part of $x$, $[x]$ is equal to the largest integer not exceeding $x$. For example, $[3,33] = 3$, $[2] = 2$, $[- 3.01] = -4$).

Find all 3-digit positive integers $\overline{abc}$ such that \[ \overline{abc} = abc(a+b+c) , \] where $\overline{abc}$ is the decimal representation of the number.

For a sequence $x_1,x_2,\ldots,x_n$ of real numbers, we define its $\textit{price}$ as \[\max_{1\le i\le n}|x_1+\cdots +x_i|.\] Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price $D$. Greedy George, on the other hand, chooses $x_1$ such that $|x_1 |$ is as small as possible; among the remaining numbers, he chooses $x_2$ such that $|x_1 + x_2 |$ is as small as possible, and so on. Thus, in the $i$-th step he chooses $x_i$ among the remaining numbers so as to minimise the value of $|x_1 + x_2 + \cdots x_i |$. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price $G$. Find the least possible constant $c$ such that for every positive integer $n$, for every collection of $n$ real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality $G\le cD$. [i]Proposed by Georgia[/i]

Each subset of $97$ out of $1997$ given real numbers has positive sum. Show that the sum of all the $1997$ numbers is positive.

Let $n\ge 2$ be an integer. Ariane and Bérénice play a game on the number of the residue classes modulo $n$. At the beginning there is the residue class $1$ on each piece of paper. It is the turn of the player whose turn it is to replace the current residue class $x$ with either $x + 1$ or by $2x$. The two players take turns, with Ariane starting. Ariane wins if the residue class $0$ is reached during the game. Bérénice wins if she can prevent that permanently. Depending on $n$, determine which of the two has a winning strategy.

Does there exist such a power of ${2}$, that when written in the decimal system its digits are all different than zero and it is possible to reorder the other digits to form another power of ${2}$? Justify your answer.

Given that $n$ and $r$ are positive integers. Suppose that \[ 1 + 2 + \dots + (n - 1) = (n + 1) + (n + 2) + \dots + (n + r) \] Prove that $n$ is a composite number.

Consider an $ABCD$ parallelogram with $\overline{AD}$ $=$ $\overline{BD}$. Point E lies in segment $\overline{BD}$ in such a way that $\overline{AE}$ $=$ $\overline{DE}$. The extension of line $\overline{AE}$ cuts segment $\overline{BC}$ and $F$. if line $\overline{DF}$ is the bisector of the $\angle CED$. Find the value of the $\angle ABD$ $\textbf{4.1.}$ Point $E$ lies in segment $\overline{BD}$ means that exits a point $E$ in the segment $\overline{BD}$ in other words lies refers to the same thing found [i]Proposed by Alicia Smith, Francisco Morazan[/i]

Convex quadrilateral $ABCD$ is given. Lines $BC$ and $AD$ intersect at point $O$, with $B$ lying on the segment $OC$, and $A$ on the segment $OD$. $I$ is the center of the circle inscribed in the $OAB$ triangle, $J$ is the center of the circle exscribed in the triangle $OCD$ touching the side of $CD$ and the extensions of the other two sides. The perpendicular from the midpoint of the segment $IJ$ on the lines $BC$ and $AD$ intersect the corresponding sides of the quadrilateral (not the extension) at points $X$ and $Y$. Prove that the segment $XY$ divides the perimeter of the quadrilateral$ABCD$ in half, and from all segments with this property and ends on $BC$ and $AD$, segment $XY$ has the smallest length.

There is a pile of eggs. Joan counted the eggs, but her count was way off by $1$ in the $1$'s place. Tom counted in the eggs, but his count was off by $1$ in the $10$'s place. Raoul counted the eggs, but his count was off by $1$ in the $100$'s place. Sasha, Jose, Peter, and Morris all counted the eggs and got the correct count. When these seven people added their counts together, the sum was $3162$. How many eggs were in the pile?

Let $a,b,c$ be reals satisfying\\ $\hspace{2cm} 3ab+2=6b, \hspace{0.5cm} 3bc+2=5c, \hspace{0.5cm} 3ca+2=4a.$\\ \\ Let $\mathbb{Q}$ denote the set of all rational numbers. Given that the product $abc$ can take two values $\frac{r}{s}\in \mathbb{Q}$ and $\frac{t}{u}\in \mathbb{Q}$ , in lowest form, find $r+s+t+u$.

Find all pairs of solutions $(x,y)$: \[ x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1). \]

Denote $f(m) =\sum_{k=1}^m (-1)^k cos \frac{k\pi}{2 m + 1}$ For which positive integers $m$ is $f(m)$ rational?

Find a set of infinite positive integers $ S$ such that for every $ n\ge 1$ and whichever $ n$ distinct elements $ x_1,x_2,\cdots, x_n$ of S, the number $ x_1\plus{}x_2\plus{}\cdots \plus{}x_n$ is not a perfect square.

Let $ABCD$ be a cyclic quadrilateral. A point $P$ moves along the arc $AD$ which does not contain $B$ and $C$. A fixed line $l$, perpendicular to $BC$, meets the rays $BP$, $CP$ at points $B_0$, $C_0$ respectively. Prove that the tangent at $P$ to the circumcircle of triangle $PB_0C_0$ passes through some fixed point.

Let $n$ be a positive integer. An $n-\emph{simplex}$ in $\mathbb{R}^n$ is given by $n+1$ points $P_0, P_1,\cdots , P_n$, called its vertices, which do not all belong to the same hyperplane. For every $n$-simplex $\mathcal{S}$ we denote by $v(\mathcal{S})$ the volume of $\mathcal{S}$, and we write $C(\mathcal{S})$ for the center of the unique sphere containing all the vertices of $\mathcal{S}$. Suppose that $P$ is a point inside an $n$-simplex $\mathcal{S}$. Let $\mathcal{S}_i$ be the $n$-simplex obtained from $\mathcal{S}$ by replacing its $i^{\text{th}}$ vertex by $P$. Prove that : \[ \sum_{j=0}^{n}v(\mathcal{S}_j)C(\mathcal{S}_j)=v(\mathcal{S})C(\mathcal{S}) \]

Let $(a_{n})_{n\geq 1}$ be a sequence defined by $a_{n}=2^{n}+49$. Find all values of $n$ such that $a_{n}=pg, a_{n+1}=rs$, where $p,q,r,s$ are prime numbers with $p<q, r<s$ and $q-p=s-r$.