A pupil is writing on a board positive integers $x_0,x_1,x_2,x_3...$ after the following algorithm which implies arithmetic progression $3,5,7,9...$.Each term of rank $k\ge2$ is a difference between the product of the last number on the board and the term of arithmetic progression of rank $k$ and the last but one term on the bord with the sum of the terms of the arithemtic progression with ranks less than $k$.If $x_0=0 $ and $x_1=1$ find $x_n$ according to n.

Find the smallest positive integer $n$ that satisfies the following: We can color each positive integer with one of $n$ colors such that the equation $w + 6x = 2y + 3z$ has no solutions in positive integers with all of $w, x, y$ and $z$ having the same color. (Note that $w, x, y$ and $z$ need not be distinct.)

Prove that for all integers $ a > 1$ and $ b > 1$ there exists a function $ f$ from the positive integers to the positive integers such that $ f(a\cdot f(n)) \equal{} b\cdot n$ for all $ n$ positive integer.

Let $n\ge 2$ be an integer, and let $a_1, a_2, \cdots , a_n$ be positive integers. Show that there exist positive integers $b_1, b_2, \cdots, b_n$ satisfying the following three conditions: $\text{(A)} \ a_i\le b_i$ for $i=1, 2, \cdots , n;$ $\text{(B)} \ $ the remainders of $b_1, b_2, \cdots, b_n$ on division by $n$ are pairwise different; and $\text{(C)} \ $ $b_1+b_2+\cdots b_n \le n\left(\frac{n-1}{2}+\left\lfloor \frac{a_1+a_2+\cdots a_n}{n}\right \rfloor \right)$ (Here, $\lfloor x \rfloor$ denotes the integer part of real number $x$, that is, the largest integer that does not exceed $x$.)

On Christmas Eve, 1983, Dean Jixon, the famous seer who had made startling predictions of the events of the preceding year that the volcanic and seismic activities of $1980$ and $1981$ were connected with mathematics. The diminishing of this geological activity depended upon the existence of an elementary proof of the irreducibility of the polynomial \[P(x)=x^{1981}+x^{1980}+12x^{2}+24x+1983.\] Is there such a proof?

A magician intends to perform the following trick. She announces a positive integer $n$, along with $2n$ real numbers $x_1 < \dots < x_{2n}$, to the audience. A member of the audience then secretly chooses a polynomial $P(x)$ of degree $n$ with real coefficients, computes the $2n$ values $P(x_1), \dots , P(x_{2n})$, and writes down these $2n$ values on the blackboard in non-decreasing order. After that the magician announces the secret polynomial to the audience. Can the magician find a strategy to perform such a trick?

For any positive integer $b\ge2$, we write the base-$b$ numbers as follows: \[(d_kd_{k-1}\dots d_0)_b=d_kb^k+d_{k-1}b^{k-1}+\dots+d_1b^1+d_0b^0,\]where each digit $d_i$ is a member of the set $S=\{0,1,2,\dots,b-1\}$ and either $d_k\not=0$ or $k=0$. There is a unique way to write any nonnegative integer in the above form. If we select the digits from a different set $S$ instead, we may obtain new representations of all positive integers or, in some cases, all integers. For example, if $b=3$ and the digits are selected from $S=\{-1,0,1\}$, we obtain a way to uniquely represent all integers, known as a $\emph{balanced ternary}$ representation. As further examples, the balanced ternary representation of numbers $5$, $-3$, and $25$ are: \[5=(1\ {-1}\ {-1})_3,\qquad{-3}=({-1}\ 0)_3,\qquad25=(1\ 0\ {-1}\ 1)_3.\]However, not all digit sets can represent all integers. If $b=3$ and $S=\{-2,0,2\}$, then no odd number can be represented. Also, if $b=3$ and $S=\{0,1,2\}$ as in the usual base-$3$ representation, then no negative number can be represented. Given a set $S$ of four integers, one of which is $0$, call $S$ a $\emph{4-basis}$ if every integer $n$ has at least one representation in the form \[n=(d_kd_{k-1}\dots d_0)_4=d_k4^k+d_{k-1}4^{k-1}+\dots+d_14^1+d_04^0,\]where $d_k,d_{k-1},\dots,d_0$ are all elements of $S$ and either $d_k\not=0$ or $k=0$. [list=a] [*]Show that there are infinitely many integers $a$ such that $\{-1,0,1,4a+2\}$ is not a $4$-basis. [*]Show that there are infinitely many integers $a$ such that $\{-1,0,1,4a+2\}$ is a $4$-basis.[/list]

A rectangular array has 9 rows and 2004 columns. In the 9 * 2004 cells of the table we place the numbers from 1 to 2004, each 9 times. And we do this in such a way that two numbers, which stand in exactly the same column in and differ around at most by 3. Find the smallest possible sum of all numbers in the first row.

Freddy writes down numbers $1, 2,\ldots ,n$ in some order. Then he makes a list of all pairs $(i, j)$ such that $1\le i<j\le n$ and the $i$-th number is bigger than the $j$-th number in his permutation. After that, Freddy repeats the following action while possible: choose a pair $(i, j)$ from the current list, interchange the $i$-th and the $j$-th number in the current permutation, and delete $(i, j)$ from the list. Prove that Freddy can choose pairs in such an order that, after the process finishes, the numbers in the permutation are in ascending order.

In each cell, a strip of length $100$ is worth a chip. You can change any $2$ neighboring chips and pay $1$ rouble, and you can also swap any $2$ chips for free, between which there are exactly $4$ chips. For what is the smallest amount of rubles you can rearrange chips in reverse order?

We are given a bag of sugar, a two-pan balance, and a weight of $1$ gram. How do we obtain $1$ kilogram of sugar in the smallest possible number of weighings?

A finite number of coins are placed on an infinite row of squares. A sequence of moves is performed as follows: at each stage a square containing more than one coin is chosen. Two coins are taken from this square; one of them is placed on the square immediately to the left while the other is placed on the square immediately to the right of the chosen square. The sequence terminates if at some point there is at most one coin on each square. Given some initial configuration, show that any legal sequence of moves will terminate after the same number of steps and with the same final configuration.

Let $a,b,c>0$ such that $a+b+c=1$. Prove: \[\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2}) \]

To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y<0$, then the following operation is allowed: $x,y,z$ are replaced by $x+y,-y,z+y$ respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.

Prove that any positive integer can be represented as an aggregate of different powers of $3$, the terms in the aggregate being combined by the signs $+$ and $-$ appropriately chosen.