An integer $n \geq 3$ is called [i]special[/i] if it does not divide $\left ( n-1 \right )!\left ( 1+\frac{1}{2}+\cdot \cdot \cdot +\frac{1}{n-1} \right )$. Find all special numbers $n$ such that $10 \leq n \leq 100$.

A game of solitaire strats of with $25$ cards. Some are facing up and sum are facing down. In each move a card that's facing up should me choosen, taken away, and turning over the cards next to it (if there are cards next to it). The game is won when you have accomplished to take all the $25$ cards from the table. If you initially start with $n$ cards facing up, find all the values of $n$ such that the game can be won. Explain how to win the game, independently from the initial placement of the cards facing up, justify your answer for why it is impossible to win with other values of $n$. Two cards are neighboring when one is immediately next to the other, to the left or right. Example: The card marked $A$ has two neighboring cards and the one marked with only a $B$ has only one neighboring card. After taking a card there is a hole left, such that the card marked $C$ has only one neighboring card, and the one marked $D$ does'nt have any.

Let $ n$ be the number of integer values of $ x$ such that $ P \equal{} x^4 \plus{} 6x^3 \plus{} 11x^2 \plus{} 3x \plus{} 31$ is the square of an integer. Then $ n$ is: $ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 0$

Let $f$ be a primitive polynomial with integral coefficients (their highest common factor is $1$ ) such that $f$ is irreducible in $\mathbb{Q}[X]$ , and $f(X^2)$ is reducible in $\mathbb{Q}[X]$ . Show that $f= \pm(u^2-Xv^2)$ for some polynomials $u$ and $v$ with integral coefficients.

In this problem, we explore a variant of the Monty Hall Problem. There are $n$ doors numbered $1, \dots, n$. A single gold coin is placed randomly behind one of the doors, with the probability it is placed behind door $i$ equalling $p_i.$ There are $r$ "rounds" in which we may make at most $k$ "turns" each, defined as follows: During a "turn", you pick two doors and send them to the game host. Then, the host picks one of the two doors in the following manner: [list] [*]If neither door contains the coin, the host randomly picks one with equal probability. [*] If one of the doors contains the coin, the host picks the door which does not have the coin. [/list] The host reveals that the picked door does not contain the coin, and opens it. A "round" consists of Alice performing at least 1 and at most $k$ turns. After all the turns in round $j$ are complete, suppose there are $d_j$ doors remaining. Bob will compute the probability of each individual door containing the coin. Let $m_j$ be the minimum of these probabilities computed during the $j$th round. Then, Bob awards Alice $d_jm_j$ points. Note that Bob will pay Alice $r$ times, and Alice's total payout is the sum of the $r$ individual payouts. Note that opened doors remain revealed between rounds. Suppose $S$ is some strategy that determines which doors Alice sends to the host. Let $F(S, n,k,r,(p_1, \dots, p_n))$ be the minimum possible amount Alice could earn with strategy $S$ for parameters $(n,k,r,(p_1, \dots, p_n))$, and let \[G(n,k,r,(p_1, \dots, p_n))= \max\limits _S F(S, n,k,r,(p_1, \dots, p_n)).\] Find all tuples $(n,k,r,(p_1, \dots, p_n))$ for which $G(n,k,r,(p_1, \dots, p_n))=r.$ You may find it useful to consider lists where each element of a list is at most double the prior element. [i]Proposed by Hari Desikan[/i]

Find all the pairs of natural numbers $(a, b),$ such that \[a!+1=(a+1)^{(2^b)}\]

Find the smallest positive integer $n$ such that $\underbrace{2^{2^{...^{2}}}}_{n}> 3^{3^{3^3}}$. (The notation $\underbrace{2^{2^{...^{2}}}}_{n}$ is used to denote a power tower with $n$ $2$’s. For example, $\underbrace{2^{2^{...^{2}}}}_{n}$ with $n = 4$ would equal $2^{2^{2^2}}$.)

For any finite sets $X$ and $Y$ of positive integers, denote by $f_X(k)$ the $k^{\text{th}}$ smallest positive integer not in $X$, and let $$X*Y=X\cup \{ f_X(y):y\in Y\}.$$Let $A$ be a set of $a>0$ positive integers and let $B$ be a set of $b>0$ positive integers. Prove that if $A*B=B*A$, then $$\underbrace{A*(A*\cdots (A*(A*A))\cdots )}_{\text{ A appears $b$ times}}=\underbrace{B*(B*\cdots (B*(B*B))\cdots )}_{\text{ B appears $a$ times}}.$$ [i]Proposed by Alex Zhai, United States[/i]

Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABO$ and $ACO$, respectively. The circumcircle of $\triangle AO_1O_2$ intersects segment $BC$ at two distinct points $P$ and $Q$, with the four points $B, P, Q, C$ appearing in this order along $BC$. Let $O_3$ be the circumcenter of $\triangle OPQ$. Prove that points $A, O, O_3$ are collinear.

A polygon is $\textit{monochromatic}$ if all its vertices are coloured by the same colour. Suppose now every point of the plane is coloured red or blue. Show that there exists either a monochromatic equilateral triangle of side length $2$, or a monochromatic equilateral triangle of side length $\sqrt{3}$, or a monochromatic rhombus of side length $1$.

Let $m, n$ be positive integers. Consider a sequence of positive integers $a_1, a_2, ... , a_n$ that satisfies $m = a_1 \ge a_2\ge ... \ge a_n \ge 1$. Then define, for $1\le  i\le  m$, $b_i =$ # $\{ j \in \{1, 2, ... , n\}: a_j \ge i\}$, so $b_i$ is the number of terms $a_j $ of the given sequence for which $a_j  \ge i$. Similarly, we define, for $1\le   j \le  n$, $c_j=$ # $\{ i \in \{1, 2, ... , m\}: b_i \ge j\}$ , thus $c_j$ is the number of terms bi in the given sequence for which $b_i \ge j$. E.g.: If $a$ is the sequence $5, 3, 3, 2, 1, 1$ then $b$ is the sequence $6, 4, 3, 1, 1$. (a) Prove that $a_j = c_j $ for $1  \le j  \le n$. (b) Prove that for $1\le  k \le m$: $\sum_{i=1}^{k} b_i = k \cdot b_k + \sum_{j=b_{k+1}}^{n} a_j$.

Let $a$ be a positive integer. We say that a positive integer $b$ is [i]$a$-good[/i] if $\tbinom{an}{b}-1$ is divisible by $an+1$ for all positive integers $n$ with $an \geq b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.

Evaluate the expression \[ \frac{121 \left( \frac{1}{13} - \frac{1}{17} \right) + 169 \left( \frac{1}{17} - \frac{1}{11} \right) + 289 \left( \frac{1}{11} - \frac{1}{13} \right)}{ 11 \left( \frac{1}{13} - \frac{1}{17} \right) + 13 \left( \frac{1}{17} - \frac{1}{11} \right) + 17 \left( \frac{1}{11} - \frac{1}{13} \right)} \, . \]

Find all positive integer pairs $(a,b)$ such that, $$\frac{10^{a!} - 3^b +1}{2^a}$$ is a perfect square.

Let $I,G,O$ be the incenter, centroid and the circumcenter of triangle $ABC$, respectively. Let $X,Y,Z$ be on the rays $BC, CA, AB$ respectively so that $BX=CY=AZ$. Let $F$ be the centroid of $XYZ$. Show that $FG$ is perpendicular to $IO$.

Let $ ABC$ be an isosceles triangle with $ AB \equal{} AC$. $ M$ is the midpoint of $ BC$ and $ O$ is the point on the line $ AM$ such that $ OB$ is perpendicular to $ AB$. $ Q$ is an arbitrary point on $ BC$ different from $ B$ and $ C$. $ E$ lies on the line $ AB$ and $ F$ lies on the line $ AC$ such that $ E, Q, F$ are distinct and collinear. Prove that $ OQ$ is perpendicular to $ EF$ if and only if $ QE \equal{} QF$.

Compare numbers $A=5+2\sqrt{5}$ and $B=\sqrt{45+20\sqrt{5}}$

For integers $m,n \ge 3$ we consider a $m \times n$ rectangular frame, consisting of the $2m+2n-4$ boundary squares of a $m \times n$ rectangle. Renate and Erhard play the following game on this frame, with Renate to start the game. In a move, a player colours a rectangular area consisting of a single or several white squares. If there are any more white squares, they have to form a connected region. The player who moves last wins the game. Determine all pairs $(m,n)$ for which Renate has a winning strategy.

Suppose $n$ is a natural number such that $4^n + 2^n + 1$ is a prime. Prove that $n = 3^k$ for some nonnegative integer $k$.

We have $k$ switches arranged in a row, and each switch points up, down, left, or right. Whenever three successive switches all point in different directions, all three may be simultaneously turned so as to point in the fourth direction. Prove that this operation cannot be repeated infinitely many times.

Let $ABCD$ be a cyclic quadrilateral with opposite sides not parallel. Let $X$ and $Y$ be the intersections of $AB,CD$ and $AD,BC$ respectively. Let the angle bisector of $\angle AXD$ intersect $AD,BC$ at $E,F$ respectively, and let the angle bisectors of $\angle AYB$ intersect $AB,CD$ at $G,H$ respectively. Prove that $EFGH$ is a parallelogram.

Shade in some of the regions in the grid to the right so that the shaded area is equal for each of the 11 rows and columns. Regions must be fully shaded or fully unshaded, at least one region must be shaded, and the area of shaded regions must be at most half of the whole grid. [asy] size(200); defaultpen(linewidth(0.45)); real[][] arr = { {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0},}; for (int i=0; i<11; ++i){ for (int j=0; j<11; ++j){ if(arr[10-j][i] == 1){ fill((i,j)--(i+1,j)--(i+1, j+1)--(i,j+1)--cycle, grey); } } } draw((0,0)--(4,0)--(4,3)--(3,3)--(3,1)--(1,1)--(1,2)--(2,2)--(2,1)--(2,3)--(3,3)--(1,3)--(1,2)--(1,3)--(0,3)--(0,0)--(0,5)--(1,5)--(1,4)--(4,4)--(4,3)--(6,3)--(6,2)--(5,2)--(5,1)--(4,1)--(4,0)--(6,0)--(6,3)--(9,3)--(9,0)--(8,0)--(8,2)--(7,2)--(7,1)--(8,1)--(7,1)--(7,0)--(6,0)--(11,0)--(11,2)--(10,2)--(10,1)--(9,1)--(9,3)--(11,3)--(11,2)--(11,4)--(8,4)--(8,3)--(8,4)--(6,4)--(6,3)--(6,4)--(3,4)--(3,5)--(2,5)--(2,6)--(1,6)--(1,5)--(1,7)--(0,7)--(0,5)--(0,8)--(2,8)--(2,6)--(3,6)--(3,7)--(4,7)--(4,6)--(3,6)--(4,6)--(4,5)--(3,5)--(5,5)--(5,4)--(5,5)--(7,5)--(7,4)--(7,5)--(9,5)--(9,4)--(9,5)--(11,5)--(11,4)--(11,6)--(10,6)--(10,7)--(9,7)--(9,5)--(9,7)--(8,7)--(8,5)--(8,6)--(7,6)--(7,7)--(8,7)--(6,7)--(6,6)--(7,6)--(6,6)--(6,5)--(5,5)--(5,6)--(4,6)--(4,8)--(2,8)--(2,9)--(0,9)--(0,8)--(0,10)--(1,10)--(1,9)--(1,10)--(2,10)--(2,9)--(3,9)--(3,8)--(3,11)--(0,11)--(0,10)--(0,11)--(5,11)--(5,10)--(3,10)--(4,10)--(4,8)--(5,8)--(5,6)--(5,7)--(6,7)--(6,8)--(5,8)--(5,9)--(4,9)--(6,9)--(6,10)--(5,10)--(5,11)--(8,11)--(8,10)--(6,10)--(7,10)--(7,9)--(6,9)--(7,9)--(7,8)--(6,8)--(7,8)--(7,7)--(8,7)--(8,8)--(7,8)--(8,8)--(8,10)--(9,10)--(9,9)--(8,9)--(10,9)--(10,8)--(9,8)--(9,7)--(10,7)--(10,6)--(11,6)--(11,8)--(10,8)--(11,8)--(11,10)--(9,10)--(11,10)--(11,11)--(8,11)); [/asy] You do not need to prove that your answer is the only one possible; you merely need to find an answer that satisfies the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.)

For a finite graph $G$, let $f(G)$ be the number of triangles and $g(G)$ the number of tetrahedra formed by edges of $G$. Find the least constant $c$ such that \[g(G)^3\le c\cdot f(G)^4\] for every graph $G$. [i]Proposed by Marcin Kuczma, Poland [/i]

Let $(a_n)_{n=1}^{\infty}$ be a real sequence such that $a_1=1, a_3=4$ and for every $n\geq 2$, $a_{n+1}+a_{n-1}=2a_n+1$. What is $a_{2011}$? $\textbf{(A)}\ 2^{2010} \qquad\textbf{(B)}\ 2021056 \qquad\textbf{(C)}\ 1010528 \qquad\textbf{(D)}\ 3016 \qquad\textbf{(E)}\ 2011$

Let $ABCD$ be a square, point $E$ be the midpoint of the side $BC$. The point $F$ belongs to the side $AB$, and $DE \perp EF$. The point $G$ lies inside the square, and $GF = FE$ and $GF \perp FE$. Prove that: a) $DE$ is the bisector of the $\angle FDC$ b) $FG$ is the bisector of the $\angle AFD$ c) the point $G$ is the center of the circle inscribed in the triangle $ADF$. (Ercole Suppa, Italy)