Found problems: 85335
A ladder style tournament is held with $2016$ participants. The players begin seeded $1,2,\cdots 2016$. Each round, the lowest remaining seeded player plays the second lowest remaining seeded player, and the loser of the game gets eliminated from the tournament. After $2015$ rounds, one player remains who wins the tournament. If each player has probability of $\tfrac{1}{2}$ to win any game, then the probability that the winner of the tournament began with an even seed can be expressed has $\tfrac{p}{q}$ for coprime positive integers $p$ and $q$. Find the remainder when $p$ is divided by $1000$.
[i]Proposed by Nathan Ramesh
Prove that for every positive integer $t$ there is a unique permutation $a_0, a_1, \ldots , a_{t-1}$ of $0, 1, \ldots , t-1$ such that, for every $0 \leq i \leq t-1$, the binomial coefficient $\binom{t+i}{2a_i}$ is odd and $2a_i \neq t+i$.
Prove that $ 7^{2^{20}} + 7^{2^{19}} + 1 $ has at least $ 21 $ distinct prime divisors.
Find all functions $f : (0, +\infty) \to \mathbb{R}$ satisfying the following conditions:
$(i)$ $f(x) + f(\frac{1}{x}) = 1$ for all $x> 0$;
$(ii)$ $f(xy + x + y) = f(x)f(y)$ for all $x, y> 0$.
Find all nonzero polynomials $P(x)$ with integers coefficients that satisfy the following property: whenever $a$ and $b$ are relatively prime integers, then $P(a)$ and $P(b)$ are relatively prime as well. Prove that your answer is correct. (Two integers are [b]relatively prime[/b] if they have no common prime factors. For example, $-70$ and $99$ are relatively prime, while $-70$ and $15$ are not relatively prime.)
Let $K$ be a positive integer such that there exist a triple of positive integers $(x,y,z)$ such that
\[x^3+Ky , y^3 + Kz, \text{and } z^3 + Kx\]
are all perfect cubes.
(a) Prove that $K \ne 2$ and $K \ne 4$
(b) Find the minimum value of $K$ that satisfies.
[i]Proposed by Muhammad Afifurrahman[/i]
Lamija and Faris are playing the following game. Cards, which are numerated from $1$ to $100$, are placed one next to other, starting from $1$ to $100$. Now Faris picks every $7$th card, and after that every card which contains number $7$. After that Lamija picks from remaining cards ones divisible with $5$, and after that cards which contain number $5$. Who will have more cards and how many ? How would game end, if Lamija started with "$5$ rule" and Faris continues with "$7$ rule"?
suppose that $a=3^{100}$ and $b=5454$. how many $z$s in $[1,3^{99})$ exist such that for every $c$ that $gcd(c,3)=1$, two equations $x^z\equiv c$ and $x^b\equiv c$ (mod $a$) have the same number of answers?($\frac{100}{6}$ points)
Andrey and Sasha play the game, making moves alternate. On his turn, Andrey marks on the plane an arbitrary point that has not yet been marked. After that, Sasha colors this point in one of two colors: white and black. Sasha wins if after his move it is impossible to draw a line such that all white points lie in one half-plane, while all black points lie in another half-plane with respect to this line.
[b]a)[/b] Prove that Andrey can make moves in such a way that Sasha will never win.
[b]b)[/b] Suppose that Andrey can mark only integer points on the Cartesian plane. Can Sasha guarantee himself a win regardless of Andrey's moves?
[i](N. Naradzetski)[/i]
We say that a positive integer is super odd if all of its digits are odd. For example, 1737 is super odd and 3051 is not. Find an even positive integer that cannot be express as a sum of two super odd numbers and explain why it is not possible to express it thus.
Consider a $ 7\times 7$ numbers table $ a_{ij} \equal{} (i^2 \plus{} j)(i \plus{} j^2), 1\le i,j\le 7.$ When we add arbitrarily each term of an arithmetical progression consisting of $ 7$ integers to corresponding to term of certain row (or column) in turn, call it an operation. Determine whether such that each row of numbers table is an arithmetical progression, after a finite number of operations.
Simplify
i) $1+\frac{2a+\dfrac{2}{a}}{a+\dfrac{1}{a}}$
ii) $\frac{3b+\dfrac{3}{b}+\dfrac{3}{b^2}}{b+\dfrac{1}{b}+\dfrac{1}{b^2}}$
iii) $\frac{\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{ab}\right)a^6b^2-a^6-a^5b}{a^4b}$
Consider $\vartriangle A_0B_0C_0$ and points $C_1, A_1, B_1$ on its sides $A_0B_0, B_0C_0, C_0A_0$, points $C_2, A_2,B_2$ on the sides $A_1B_1, B_1C_1, C_1A_1$ of $\vartriangle A_1B_1C_1$, respectively, etc., so that
$$\frac{A_0B_1}{B_1C_0}= \frac{B_0C_1}{C_1A_0}= \frac{C_0A_1}{A_1B_0}= k, \frac{A_1B_2}{B_2C_1}= \frac{B_1C_2}{C_2A_1}= \frac{C_1A_2}{A_2B_1}= \frac{1}{k^2}$$
and, in general,
$$\frac{A_nB_{n+1}}{B_{n+1}C_n}= \frac{B_nC_{n+1}}{C_{n+1}A_n}= \frac{C_nA_{n+1}}{A_{n+1}B_n}
=k^{2n}$$ for $n$ even , $\frac{1}{k^{2n}}$ for $n$ odd. Prove that $\vartriangle ABC$ formed by lines $A_0A_1, B_0B_1, C_0C_1$ is contained in $\vartriangle A_nB_nC_n$ for any $n$.
Is it true that for any polynomial $P(x)$ with real coefficients of degree $2023$, there exists a natural number $n$ such that the equation $P(x) = n^{-100}$ has no rational root?
Defi ne an $n$-digit pair cycle to be a number with $n^2 + 1$ digits between $1$ and $n$ with every possible pair of consecutive digits. For instance, $11221$ is a 2-digit pair cycle since it contains the consecutive digits $11$, $12$, $22$, and $21$. How many $3$-digit pair cycles exist?
The number $2020$ has three different prime factors. What is their sum?
The polynomial \[P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}\] has 34 complex roots of the form $z_k=r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)], k=1, 2, 3,\ldots, 34$, with $0<a_1\le a_2\le a_3\le\cdots\le a_{34}<1$ and $r_k>0$. Given that $a_1+a_2+a_3+a_4+a_5=m/n$, where $m$ and $n$ are relatively prime positive integers, find $m+n$.
If $a$ and $b$ are positive integers such that
\[
\sqrt{8 + \sqrt{32 + \sqrt{768}}} = a \cos \frac{\pi}{b} \, ,
\]
compute the ordered pair $(a, b)$.
We have $ 2n\plus{}1$ elements in the commutative ring $ R$: \[ \alpha,\alpha_1,...,\alpha_n,\varrho_1,...,\varrho_n .\] Let us define the elements \[ \sigma_k\equal{}k\alpha \plus{} \sum_{i\equal{}1}^n \alpha_i\varrho_i^k .\] Prove that the ideal $ (\sigma_0,\sigma_1,...,\sigma_k,...)$ can be finitely generated.
[i]L. Redei[/i]
For a fixed natural number $m \geq 2$, prove that
[b]a.)[/b] There exists integers $x_1, x_2, \ldots, x_{2m}$ such that \[x_i x_{m + i} = x_{i + 1} x_{m + i - 1} + 1, i = 1, 2, \ldots, m \hspace{2cm}(*)\]
[b]b.)[/b] For any set of integers $\lbrace x_1, x_2, \ldots, x_{2m}$ which fulfils (*), an integral sequence $\ldots, y_{-k}, \ldots, y_{-1}, y_0, y_1, \ldots, y_k, \ldots$ can be constructed such that $y_k y_{m + k} = y_{k + 1} y_{m + k - 1} + 1, k = 0, \pm 1, \pm 2, \ldots$ such that $y_i = x_i, i = 1, 2, \ldots, 2m$.
The central square of the City of Mathematicians is an $n\times n$ rectangular shape, each paved with $1\times 1$ tiles. In order to illuminate the square, night lamps are placed at the corners of the tiles (including the edges of the rectangle) in such a way that each night lamp illuminates all the tiles in its corner.
Determine the minimum number of night lamps such that even if one of those night lamps does not work, it is possible to illuminate the entire central square with them.
Consider $n\geq 3$ points in the plane, no three of which are collinear. For every convex polygon with vertices among the $n$ points, place $k\cdot 2^k$ coins in every one of its vertices, where $k$ is the number of points strictly in the interior of the polygon. Show that in total, no matter the configuration of the $n$ points, there are at most $n(n+1)\cdot 2^{n-3}$ placed coins.
[i]Cristi Săvescu[/i]
What is the sum of the reciprocals of the roots of the equation
\[ \frac {2003}{2004}x \plus{} 1 \plus{} \frac {1}{x} \equal{} 0?
\]
$ \textbf{(A)}\ \minus{}\! \frac {2004}{2003} \qquad \textbf{(B)}\ \minus{} \!1 \qquad \textbf{(C)}\ \frac {2003}{2004} \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ \frac {2004}{2003}$
Positive sequence $(a_n)_{n=0}^{\infty}$ satisfies that $\sqrt{a_na_{n-2}}-\sqrt{a_{n-1}a_{n-2}}=2a_{n-1}(n\geq2),a_0=a_1=1$. Find $a_n$.
Let $\Gamma$ be the incircle of a non-isosceles triangle $ABC$, $I$ be it’s incenter. Let $A_1,
B_1, C_1$ be the tangency points of $\Gamma$ with the sides $BC, AC, AB$ respectively. Let $A_2 = \Gamma \cap AA_1$,
$M = C_1B_1 \cap AI$, $P$ and $Q$ be the other (different from $A_1$ and $A_2$) intersection points of $\Gamma$ and $A_1M$,
$A_2M$ respectively. Prove that $A$, $P$ and $Q$ are colinear.