For a positive integer $n$, a cubic polynomial $p(x)$ is said to be [i]$n$-good[/i] if there exist $n$ distinct integers $a_1, a_2, \ldots, a_n$ such that all the roots of the polynomial $p(x) + a_i = 0$ are integers for $1 \le i \le n$. Given a positive integer $n$ prove that there exists an $n$-good cubic polynomial.

Let $P(x)$ be a polynomial of degree $n$ with real coefficients and let $a\geq 3$. Prove that \[\max_{0\leq j \leq n+1}\left | a^j-P(j) \right |\geq 1\]

Let $n$ be a positive integer and $M = \{1, 2, 3, 4, 5, 6\}$. Two persons $A$ and $B$ play in the following Way: $A$ writes down a digit from $M$, $B$ appends a digit from $M$, and so it becomes alternately one digit from $M$ is appended until the $2n$-digit decimal representation of a number has been created. If this number is divisible by $9$, $B$ wins, otherwise $A$ wins. For which $n$ can $A$ and for which $n$ can $B$ force the win?

Greta Grasshopper sits on a long line of lily pads in a pond. From any lily pad, Greta can jump $5$ pads to the right or $3$ pads to the left. What is the fewest number of jumps Greta must make to reach the lily pad located $2023$ pads to the right of her starting point? $\textbf{(A)}~405\qquad\textbf{(B)}~407\qquad\textbf{(C)}~409\qquad\textbf{(D)}~411\qquad\textbf{(E)}~413$

Two players by turns draw diagonals in a regular $(2n+1)$-gon ($n>1$). It is forbidden to draw a diagonal, which was already drawn, or intersects an odd number of already drawn diagonals. The player, who has no legal move, loses. Who has a winning strategy? [i]K. Sukhov[/i]

A real number $f(X)\neq 0$ is assigned to each point $X$ in the space. It is known that for any tetrahedron $ABCD$ with $O$ the center of the inscribed sphere, we have : \[ f(O)=f(A)f(B)f(C)f(D). \] Prove that $f(X)=1$ for all points $X$. [i]Proposed by Aleksandar Ivanov[/i]

The number of integral points satisfy $\begin{cases} y\leq 3x\\ y\geq \frac{x}{3}\\ x+y\geq100 \end{cases}$ on the coordinate plane is________.

Consider the two square matrices \[A=\begin{bmatrix} +1 & +1 &+1& +1 &+1 \\+1 &+1 &+1&-1 &-1 \\ +1 &-1&-1 &+1& +1 \\ +1 & -1 & -1 & -1 & +1 \\ +1 &+1&-1 &+1&-1 \end{bmatrix} \quad \text{ and } \quad B=\begin{bmatrix} +1 & +1 &+1& +1 &+1 \\+1 &+1 &+1&-1 &-1 \\ +1 &+1&-1& +1&-1 \\ +1 &-1& -1& +1& +1 \\ +1 & -1& +1&-1 &+1 \end{bmatrix}\] with entries $+1$ and $-1$. The following operations will be called elementary: (1) Changing signs of all numbers in one row; (2) Changing signs of all numbers in one column; (3) Interchanging two rows (two rows exchange their positions); (4) Interchanging two columns. Prove that the matrix $B$ cannot be obtained from the matrix $A$ using these operations.

Each of two angles of a triangle is $60^{\circ}$ and the included side is $4$ inches. The area of the triangle, in square inches, is: $ \textbf{(A) }8\sqrt{3}\qquad\textbf{(B) }8\qquad\textbf{(C) }4\sqrt{3}\qquad\textbf{(D) }4\qquad\textbf{(E) }2\sqrt{3} $

Let $S$ be the set of all points $t$ in the closed interval $[-1, 1]$ such that for the sequence $x_0, x_1, x_2, ...$ defined by the equations $x_0 = t, x_{n+1} = 2x_n^2-1$, there exists a positive integer $N$ such that $x_n = 1$ for all $n \ge N$. Show that the set $S$ has infinitely many elements.

a) Prove that one cannot assign to each vertex of a cube $ 8$ distinct numbers from the set $\{0, 1, 2, 3, . . . , 11, 12\}$ such that, for every edge, the sum of the two numbers assigned to its vertices is even. b) Prove that one can assign to each vertex of a cube $8$ distinct numbers from the set $\{0, 1, 2, 3, . . . , 11, 12\}$ such that, for every edge, the sum of the two numbers assigned to its vertices is divisible by $3$.

Consider an $n \times n$ grid consisting of $n^2$ until cells, where $n \geq 3$ is a given odd positive integer. First, Dionysus colours each cell either red or blue. It is known that a frog can hop from one cell to another if and only if these cells have the same colour and share at least one vertex. Then, Xanthias views the colouring and next places $k$ frogs on the cells so that each of the $n^2$ cells can be reached by a frog in a finite number (possible zero) of hops. Find the least value of $k$ for which this is always possible regardless of the colouring chosen by Dionysus. [i]Proposed by Tommy Walker Mackay, United Kingdom[/i]

The game of Nelipe is played on a $16\times16$-grid as follows: The two players write in turn numbers $1, 2,..., 16$ in different squares. The numbers on each row, column, and in every one of the 16 smaller squares have to be different. The loser is the one who is not able to write a number. Which one of the players wins, if both play with an optimal strategy?

Each of the $36$ squares of a $6 \times 6$ table is to be coloured either Red, Yellow or Blue. (a) No row or column is contain more than two squares of the same colour. (b) In any four squares obtained by intersecting two rows with two columns, no colour is to occur exactly three times. In how many dierent ways can the table be coloured if both of these rules are to be respected?

Start with a finite sequence $ a_1,a_2,\dots,a_n$ of positive integers. If possible, choose two indices $ j < k$ such that $ a_j$ does not divide $ a_k$ and replace $ a_j$ and $ a_k$ by $ \gcd(a_j,a_k)$ and $ \text{lcm}\,(a_j,a_k),$ respectively. Prove that if this process is repeated, it must eventually stop and the final sequence does not depend on the choices made. (Note: $ \gcd$ means greatest common divisor and lcm means least common multiple.)

Suppose that $n$ people $A_1$, $A_2$, $\ldots$, $A_n$, ($n \geq 3$) are seated in a circle and that $A_i$ has $a_i$ objects such that \[ a_1 + a_2 + \cdots + a_n = nN \] where $N$ is a positive integer. In order that each person has the same number of objects, each person $A_i$ is to give or to receive a certain number of objects to or from its two neighbours $A_{i-1}$ and $A_{i+1}$. (Here $A_{n+1}$ means $A_1$ and $A_n$ means $A_0$.) How should this redistribution be performed so that the total number of objects transferred is minimum?