Found problems: 800
A sequence $x_1, x_2, \ldots$ is defined by $x_1 = 1$ and $x_{2k}=-x_k, x_{2k-1} = (-1)^{k+1}x_k$ for all $k \geq 1.$ Prove that $\forall n \geq 1$ $x_1 + x_2 + \ldots + x_n \geq 0.$
[i]Proposed by Gerhard Wöginger, Austria[/i]
Assign to each side $b$ of a convex polygon $P$ the maximum area of a triangle that has $b$ as a side and is contained in $P$. Show that the sum of the areas assigned to the sides of $P$ is at least twice the area of $P$.
Let $G$ be a simple graph with $3n^2$ vertices ($n\geq 2$). It is known that the degree of each vertex of $G$ is not greater than $4n$, there exists at least a vertex of degree one, and between any two vertices, there is a path of length $\leq 3$. Prove that the minimum number of edges that $G$ might have is equal to $\frac{(7n^2- 3n)}{2}$.
Let $f : \{ 1, 2, 3, \dots \} \to \{ 2, 3, \dots \}$ be a function such that $f(m + n) | f(m) + f(n) $ for all pairs $m,n$ of positive integers. Prove that there exists a positive integer $c > 1$ which divides all values of $f$.
Determine all integers $ n\geq 2$ having the following property: for any integers $a_1,a_2,\ldots, a_n$ whose sum is not divisible by $n$, there exists an index $1 \leq i \leq n$ such that none of the numbers $$a_i,a_i+a_{i+1},\ldots,a_i+a_{i+1}+\ldots+a_{i+n-1}$$ is divisible by $n$. Here, we let $a_i=a_{i-n}$ when $i >n$.
[i]Proposed by Warut Suksompong, Thailand[/i]
There are 60 empty boxes $B_1,\ldots,B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game.
In the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps:
(a) Bob chooses an integer $k$ with $1\leq k\leq 59$ and splits the boxes into the two groups $B_1,\ldots,B_k$ and $B_{k+1},\ldots,B_{60}$.
(b) Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group.
Bob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning.
[i]Czech Republic[/i]
Let $P$ be a point inside triangle $ABC$. Let $AP$ meet $BC$ at $A_1$, let $BP$ meet $CA$ at $B_1$, and let $CP$ meet $AB$ at $C_1$. Let $A_2$ be the point such that $A_1$ is the midpoint of $PA_2$, let $B_2$ be the point such that $B_1$ is the midpoint of $PB_2$, and let $C_2$ be the point such that $C_1$ is the midpoint of $PC_2$. Prove that points $A_2, B_2$, and $C_2$ cannot all lie strictly inside the circumcircle of triangle $ABC$.
(Australia)
Prove that if $a,b,c,d$ are nonnegative integers satisfying $(a+b)^2+2a+b= (c+d)^2+2c+d$, then $a = c $ and $b = d$.
Show that the same is true if $a,b,c,d$ satisfy $(a+b)^2+3a+b=(c+d)^2+3c+d$, but show that there exist $a,b,c,d $ with $a \ne c$ and $b \ne d$ satisfying $(a+b)^2+4a+b = (c+d)^2+4c+d$.
$n>1$ and distinct positive integers $a_1,a_2,\ldots,a_{n+1}$ are given. Does there exist a polynomial $p(x)\in\Bbb{Z}[x]$ of degree $\le n$ that satisfies the following conditions?
a. $\forall_{1\le i < j\le n+1}: \gcd(p(a_i),p(a_j))>1 $
b. $\forall_{1\le i < j < k\le n+1}: \gcd(p(a_i),p(a_j),p(a_k))=1 $
[i]Proposed by Mojtaba Zare[/i]
Given a $n\times n$ table, where $n$ is odd. There is either $1$ or $-1$ in its every field. A product of the numbers in the column is written under every column. A product of the numbers in the row is written to the right of every row. Prove that the sum of $2n$ products doesn't equal to $0$.
Suppose that $1000$ students are standing in a circle. Prove that there exists an integer $k$ with $100 \leq k \leq 300$ such that in this circle there exists a contiguous group of $2k$ students, for which the first half contains the same number of girls as the second half.
[i]Proposed by Gerhard Wöginger, Austria[/i]
A sequence of real numbers $a_1,a_2,\ldots$ satisfies the relation
$$a_n=-\max_{i+j=n}(a_i+a_j)\qquad\text{for all}\quad n>2017.$$
Prove that the sequence is bounded, i.e., there is a constant $M$ such that $|a_n|\leq M$ for all positive integers $n$.
Let the set $S = \{P_1, P_2, \cdots, P_{12}\}$ consist of the twelve vertices of a regular $12$-gon. A subset $Q$ of $S$ is called communal if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.)
Let $a, n \in \mathbb{Z}^{*}_{+}$. $a$ is defined inductively in the base $n$-[i]recursive[/i]. We first write $a$ in the base $n$, e.g., as a sum of terms of the form $k_tn^t$, with $0 \le k_t < n$. For each exponent $t$, we write $t$ in the base $n$-[i]recursive[/i], until all the numbers in the representation are less than $n$. For instance,
$1309 = 3^6 + 2.3^5 + 1.3^4 + 1.3^2 + 1.3 + 1$
$ = 3^{2.3} + 2.3^{3+2} + 1.3^{3+1} + 1.3^2 + 1$
Let $x_1 \in \mathbb{Z}$ arbitrary. We define $x_n$ recursively, as following: if $x_{n-1} > 0$, we write $x_{n-1}$ in the base $n$-[i]recursive[/i] and we replace all the numbers $n$ for $n+1$ (even the exponents!), so we obtain the successor of $x_n$. If $x_{n-1} = 0$, then $x_n = 0$.
Example:
$x_1 = 2^{2^{2} + 2 + 1} + 2^{2+1} + 2 + 1$
$\Rightarrow x_2 = 3^{3^{3} + 3 + 1} + 3^{3+1} + 3$
$\Rightarrow x_3 = 4^{4^{4} + 4 + 1} + 4^{4+1} + 3$
$\Rightarrow x_4 = 5^{5^{5} + 5 + 1} + 5^{5+1} + 2$
$\Rightarrow x_5 = 6^{6^{6} + 6 + 1} + 6^{6+1} + 1$
$\Rightarrow x_6 = 7^{7^{7} + 7 + 1} + 7^{7+1}$
$\Rightarrow x_7 = 8^{8^{8} + 8 + 1} + 7.8^8 + 7.8^7 + 7.8^6 + ... + 7$
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Prove that $\exists N : x_N = 0$.
Given positive integer $k$, prove that there exists a positive integer $N$ depending only on $k$ such that for any integer $n\geq N$, $\binom{n}{k}$ has at least $k$ different prime divisors.
Suppose that $a_0, a_1, \cdots $ and $b_0, b_1, \cdots$ are two sequences of positive integers such that $a_0, b_0 \ge 2$ and \[ a_{n+1} = \gcd{(a_n, b_n)} + 1, \qquad b_{n+1} = \operatorname{lcm}{(a_n, b_n)} - 1. \] Show that the sequence $a_n$ is eventually periodic; in other words, there exist integers $N \ge 0$ and $t > 0$ such that $a_{n+t} = a_n$ for all $n \ge N$.
Determine whether there exists an infinite sequence of nonzero digits $a_1 , a_2 , a_3 , \cdots $ and a positive integer $N$ such that for every integer $k > N$, the number $\overline{a_k a_{k-1}\cdots a_1 }$ is a perfect square.
Determine all functions $f$ defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions:
[list]
[*] $(i)$ $f(n) \neq 0$ for at least one $n$;
[*] $(ii)$ $f(x y)=f(x)+f(y)$ for every positive integers $x$ and $y$;
[*] $(iii)$ there are infinitely many positive integers $n$ such that $f(k)=f(n-k)$ for all $k<n$.
[/list]
Find all positive integers $n$ with the following property: the $k$ positive divisors of $n$ have a permutation $(d_1,d_2,\ldots,d_k)$ such that for $i=1,2,\ldots,k$, the number $d_1+d_2+\cdots+d_i$ is a perfect square.
$\{a_{n}\}_{n\geq 0}$ and $\{b_{n}\}_{n\geq 0}$ are two sequences of positive integers that $a_{i},b_{i}\in \{0,1,2,\cdots,9\}$. There is an integer number $M$ such that $a_{n},b_{n}\neq 0$ for all $n\geq M$ and for each $n\geq 0$
$$(\overline{a_{n}\cdots a_{1}a_{0}})^{2}+999 \mid(\overline{b_{n}\cdots b_{1}b_{0}})^{2}+999 $$
prove that $a_{n}=b_{n}$ for $n\geq 0$.\\
(Note that $(\overline{x_nx_{n-1}\dots x_0}) = 10^n\times x_n + \dots + 10\times x_1 + x_0$.)
[i]Proposed by Yahya Motevassel[/i]
Let $n \ge 5$ be an integer. Prove that $n$ is prime if and only if for any representation of $n$ as a sum of four positive integers $n = a + b + c + d$, it is true that $ab \ne cd$.
Suppose $a_1, a_2, \ldots, a_n$ and $b_1, b_2, \ldots, b_n$ are real numbers such that \[ (a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 -1)(b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 - 1) > (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n - 1)^2. \] Prove that $a_1 ^ 2 + a_2 ^ 2 + \cdots + a_n ^ 2 > 1$ and $b_1 ^ 2 + b_2 ^ 2 + \cdots + b_n ^ 2 > 1$.
Let \[ f(x) = \sum_{i=0}^{i=n} a_i x^{n - i}\] be a polynomial of degree $n$ with integral coefficients. If $a_0, a_n,$ and $f(1)$ are odd, prove that $f(x) = 0$ has no rational roots.
In the simple and connected graph $G$ let $x_i$ be the number of vertices with degree $i$. Let $d>3$ be the biggest degree in the graph $G$. Prove that if :
$$x_d \ge x_{d-1} + 2x_{d-2}+... +(d-1)x_1$$
Then there exists a vertex with degree $d$ such that after removing that vertex the graph $G$ is still connected.
Proposed by [i]Ali Mirzaie[/i]
Let $n$ be a positive integer. Given is a subset $A$ of $\{0,1,...,5^n\}$ with $4n+2$ elements. Prove that there exist three elements $a<b<c$ from $A$ such that $c+2a>3b$.
[i]Proposed by Dominik Burek and Tomasz Ciesla, Poland[/i]