Found problems: 5802
Given a set $S$ of positive real numbers, let $$\Sigma (S) = \Bigg\{ \sum_{x \in A} x : \emptyset \neq A \subset S \Bigg\}.$$
be the set of all the sums of elements of non-empty subsets of $S$. Find the least constant $L> 0$ with the following property: for every integer greater than $1$ and every set $S$ of $n$ positive real numbers, it is possible partition $\Sigma(S)$ into $n$ subsets $\Sigma_1,\ldots, \Sigma_n$ so that the ratio between the largest and smallest element of each $\Sigma_i$ is at most $L$.
Let $n$ be a positive integer. A sequence of $n$ positive integers (not necessarily distinct) is called [b]full[/b] if it satisfies the following condition: for each positive integer $k\geq2$, if the number $k$ appears in the sequence then so does the number $k-1$, and moreover the first occurrence of $k-1$ comes before the last occurrence of $k$. For each $n$, how many full sequences are there ?
A positive integer is written on the board. Every minute Maxim adds to the number on the board one of its positive divisors, writes the result on the board and erases the previous number. However, it is forbidden for him to add the same number twice in a row. Prove that he can proceed in such a way that eventually a perfect square will appear on the board.
Let $k\geq 2$,$n_1,n_2,\cdots ,n_k\in \mathbb{N}_+$,satisfied $n_2|2^{n_1}-1,n_3|2^{n_2}-1,\cdots ,n_k|2^{n_{k-1}}-1,n_1|2^{n_k}-1$.
Prove:$n_ 1=n_ 2=\cdots=n_k=1$.
Let $n$ be a positive integer. Tasty and Stacy are given a circular necklace with $3n$ sapphire beads and $3n$ turquoise beads, such that no three consecutive beads have the same color. They play a cooperative game where they alternate turns removing three consecutive beads, subject to the following conditions:
[list]
[*]Tasty must remove three consecutive beads which are turquoise, sapphire, and turquoise, in that order, on each of his turns.
[*]Stacy must remove three consecutive beads which are sapphire, turquoise, and sapphire, in that order, on each of her turns.
[/list]
They win if all the beads are removed in $2n$ turns. Prove that if they can win with Tasty going first, they can also win with Stacy going first.
[i]Yannick Yao[/i]
A square is divided into congruent rectangles with sides of integer lengths. A rectangle is important if it has at least one point in common with a given diagonal of the square. Prove that this diagonal bisects the total area of the important rectangles
Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$, $y$ the identity \[ f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2\]
[i]Proposed by Japan[/i]
(a) Find all positive integers $ n$ for which $ 2^n\minus{}1$ is divisible by $ 7$.
(b) Prove that there is no positive integer $ n$ for which $ 2^n\plus{}1$ is divisible by $ 7$.
A diabolical combination lock has $n$ dials (each with $c$ possible states), where $n,c>1$. The dials are initially set to states $d_1, d_2, \ldots, d_n$, where $0\le d_i\le c-1$ for each $1\le i\le n$. Unfortunately, the actual states of the dials (the $d_i$'s) are concealed, and the initial settings of the dials are also unknown. On a given turn, one may advance each dial by an integer amount $c_i$ ($0\le c_i\le c-1$), so that every dial is now in a state $d_i '\equiv d_i+c_i \pmod{c}$ with $0\le d_i ' \le c-1$. After each turn, the lock opens if and only if all of the dials are set to the zero state; otherwise, the lock selects a random integer $k$ and cyclically shifts the $d_i$'s by $k$ (so that for every $i$, $d_i$ is replaced by $d_{i-k}$, where indices are taken modulo $n$).
Show that the lock can always be opened, regardless of the choices of the initial configuration and the choices of $k$ (which may vary from turn to turn), if and only if $n$ and $c$ are powers of the same prime.
[i]Bobby Shen.[/i]
Given a finite sequence $ S \equal{} (a_1,a_2,\ldots,a_n)$ of $ n$ real numbers, let $ A(S)$ be the sequence
\[ \left(\frac {a_1 \plus{} a_2}2,\frac {a_2 \plus{} a_3}2,\ldots,\frac {a_{n \minus{} 1} \plus{} a_n}2\right)
\]of $ n \minus{} 1$ real numbers. Define $ A^1(S) \equal{} A(S)$ and, for each integer $ m$, $ 2\le m\le n \minus{} 1$, define $ A^m(S) \equal{} A(A^{m \minus{} 1}(S)).$ Suppose $ x > 0$, and let $ S \equal{} (1,x,x^2,\ldots,x^{100})$. If $ A^{100}(S) \equal{} (1/2^{50})$, then what is $ x$?
$ \textbf{(A) } 1 \minus{} \frac {\sqrt {2}}2\qquad \textbf{(B) } \sqrt {2} \minus{} 1\qquad \textbf{(C) } \frac 12\qquad \textbf{(D) } 2 \minus{} \sqrt {2}\qquad \textbf{(E) } \frac {\sqrt {2}}2$
Let $S$ be a set of integers such that [list][*] there exist $a, b \in S$ with $\gcd(a, b)=\gcd(a-2,b-2)=1$, [*] if $x,y\in S$, then $x^2 -y\in S$.[/list] Prove that $S=\mathbb{Z}$.
An empty $2020 \times 2020 \times 2020$ cube is given, and a $2020 \times 2020$ grid of square unit cells is drawn on each of its six faces. A [i]beam[/i] is a $1 \times 1 \times 2020$ rectangular prism. Several beams are placed inside the cube subject to the following conditions:
[list=]
[*]The two $1 \times 1$ faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are $3 \cdot {2020}^2$ possible positions for a beam.)
[*]No two beams have intersecting interiors.
[*]The interiors of each of the four $1 \times 2020$ faces of each beam touch either a face of the cube or the interior of the face of another beam.
[/list]
What is the smallest positive number of beams that can be placed to satisfy these conditions?
[i]Proposed by Alex Zhai[/i]
Let $ n\ge 3$ be an integer. Let $ f(x)$ and $ g(x)$ be polynomials with real coefficients such that the points $ (f(1),g(1)),(f(2),g(2)),\dots,(f(n),g(n))$ in $ \mathbb{R}^2$ are the vertices of a regular $ n$-gon in counterclockwise order. Prove that at least one of $ f(x)$ and $ g(x)$ has degree greater than or equal to $ n\minus{}1.$
For positive integers $n$ and $k \geq 2$, define $E_k(n)$ as the greatest exponent $r$ such that $k^r$ divides $n!$. Prove that there are infinitely many $n$ such that $E_{10}(n) > E_9(n)$ and infinitely many $m$ such that $E_{10}(m) < E_9(m)$.
Let $A_{1}, ..., A_{n}$ be different subsets of an $n$-element set $X$. Show that there exists $x\in X$ such that the sets
$A_{1}-\{x\}, A_{2}-\{x\}, ..., A_{n}-\{x\}$ are all different.
Consider an $n$-by-$n$ board of unit squares for some odd positive integer $n$. We say that a collection $C$ of identical dominoes is a [i]maximal grid-aligned configuration[/i] on the board if $C$ consists of $(n^2-1)/2$ dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: $C$ then covers all but one square on the board. We are allowed to slide (but not rotate) a domino on the board to cover the uncovered square, resulting in a new maximal grid-aligned configuration with another square uncovered. Let $k(C)$ be the number of distinct maximal grid-aligned configurations obtainable from $C$ by repeatedly sliding dominoes. Find the maximum value of $k(C)$ as a function of $n$.
[i]Proposed by Holden Mui[/i]
Let $P_1P_2\dotsb P_{100}$ be a cyclic $100$-gon and let $P_i = P_{i+100}$ for all $i$. Define $Q_i$ as the intersection of diagonals $\overline{P_{i-2}P_{i+1}}$ and $\overline{P_{i-1}P_{i+2}}$ for all integers $i$.
Suppose there exists a point $P$ satisfying $\overline{PP_i}\perp\overline{P_{i-1}P_{i+1}}$ for all integers $i$. Prove that the points $Q_1,Q_2,\dots, Q_{100}$ are concyclic.
[i]Michael Ren[/i]
Let $(a_n)_{n\geq 1}$ be a sequence of positive real numbers with the property that
$$(a_{n+1})^2 + a_na_{n+2} \leq a_n + a_{n+2}$$
for all positive integers $n$. Show that $a_{2022}\leq 1$.
Determine all positive integers $M$ such that the sequence $a_0, a_1, a_2, \cdots$ defined by \[ a_0 = M + \frac{1}{2} \qquad \textrm{and} \qquad a_{k+1} = a_k\lfloor a_k \rfloor \quad \textrm{for} \, k = 0, 1, 2, \cdots \] contains at least one integer term.
Given a natural number $n\ge 3$, determine all strictly increasing sequences $a_1<a_2<\cdots<a_n$ such that $\text{gcd}(a_1,a_2)=1$ and for any pair of natural numbers $(k,m)$ satisfy $n\ge m\ge 3$, $m\ge k$, $$\frac{a_1+a_2+\cdots +a_m}{a_k}$$ is a positive integer.
For a sequence $x_1,x_2,\ldots,x_n$ of real numbers, we define its $\textit{price}$ as \[\max_{1\le i\le n}|x_1+\cdots +x_i|.\] Given $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price $D$. Greedy George, on the other hand, chooses $x_1$ such that $|x_1 |$ is as small as possible; among the remaining numbers, he chooses $x_2$ such that $|x_1 + x_2 |$ is as small as possible, and so on. Thus, in the $i$-th step he chooses $x_i$ among the remaining numbers so as to minimise the value of $|x_1 + x_2 + \cdots x_i |$. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price $G$.
Find the least possible constant $c$ such that for every positive integer $n$, for every collection of $n$ real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality $G\le cD$.
[i]Proposed by Georgia[/i]
For $ a_i \in \mathbb{Z}^ \plus{}$, $ i \equal{} 1, \ldots, k$, and $ n \equal{} \sum^k_{i \equal{} 1} a_i$, let $ d \equal{} \gcd(a_1, \ldots, a_k)$ denote the greatest common divisor of $ a_1, \ldots, a_k$.
Prove that $ \frac {d} {n} \cdot \frac {n!}{\prod\limits^k_{i \equal{} 1} (a_i!)}$ is an integer.
[i]Dan Schwarz, Romania[/i]
Find the triplets of natural numbers $(p,q,r)$ that satisfy the equality $$\frac{1}{p}+\frac{q}{q^r -1}=1.$$
Let $r$ and $s$ be positive integers. Define $a_0 = 0$, $a_1 = 1$, and $a_n = ra_{n-1} + sa_{n-2}$ for $n \geq 2$. Let $f_n = a_1a_2\cdots a_n$. Prove that $\displaystyle\frac{f_n}{f_kf_{n-k}}$ is an integer for all integers $n$ and $k$ such that $0 < k < n$.
[i]Evan O' Dorney.[/i]
Let us consider a simle graph with vertex set $ V$. All ordered pair $ (a,b)$ of integers with $ gcd(a,b) \equal{} 1$, are elements of V.
$ (a,b)$ is connected to $ (a,b \plus{} kab)$ by an edge and to $ (a \plus{} kab,b)$ by another edge for all integer k.
Prove that for all $ (a,b)\in V$, there exists a path fromm $ (1,1)$ to $ (a,b)$.