Given sequence $\{a_n\}$ satisfying: $$ a_{n+1} = \frac{ lcm(a_n,a_{n-1})}{\gcd(a_n, a_{n-1})} $$ It is given that $a_{209} =209$ and $a_{361} = 361$. Find all possible values of $a_{2020}$.

Find all composite numbers $n$ such that for each decomposition of $n=xy$, $x+y$ be a power of $2$.

Let $a, b$ and $c$ be positive integers such that $a^2 + b^2 + 1 = c^2$ . Prove that $[a/2] + [c / 2]$ is even. Note: $[x]$ is the integer part of $x$.

Let $0 <a <b$ be two rational numbers. Let $M$ be a set of positive real numbers with the properties: (i) $a \in M$ and $b \in M$; (ii) if $x$ $\in M$ and $y \in M$, then $\sqrt{xy} \in M$. Let $M^*$denote the set of all irrational numbers in $M$. prove that every $c,d$ such that $a <c <d<b$, $M^*$ contains an element $m$ with property $c<m<d$

Is it true that every integer is a sum of finite number of cubes of distinct integers?

Determine all pairs $(x, y)$ of positive integers such that for $d = gcd(x, y)$ the equation $$xyd = x + y + d^2$$ holds. [i](Walther Janous)[/i]

Find all pairs of positive integers $(a, b)$, such that $S(a^{b+1})=a^b$, where $S(m)$ denotes the digit sum of $m$.

Find the minimal positive integer $m$, so that there exist positive integers $n>k>1$, which satisfy $11...1=11...1.m$, where the first number has $n$ digits $1$, and the second has $k$ digits $1$.

A Haiku is a Japanese poem of seventeen syllables, in three lines of five, seven, and five. In how many ways Can you add three integers Summing seventeen? Order matters here. For example, eight, three, six Is not eight, six, three. All nonnegative, Do not need to be distinct. What is your answer? [i]Proposed by Derek Gao[/i]

Let $F_n$ be the $n$th Fibonacci number ($F_1=F_2=1$ and $F_{n+1}=F_n+F_{n-1}$). Construct infinitely many positive integers $n$ such that $n$ divides $F_{F_n}$ but $n$ does not divide $F_n$.

Chandler the Octopus is making a concoction to create the perfect ink. He adds $1.2$ grams of melanin, $4.2$ grams of enzymes, and $6.6$ grams of polysaccharides. But Chandler accidentally added n grams of an extra ingredient to the concoction, Chemical $X$, to create glue. Given that Chemical $X$ contains none of the three aforementioned ingredients, and the percentages of melanin, enzymes, and polysaccharides in the final concoction are all integers, find the sum of all possible positive integer values of $n$. [i]Proposed by Taiki Aiba[/i]

Find some natural number $a$ such that $2a$ is a perfect square, $3a$ is a perfect cube, $5a$ is the fifth power of some natural number.

A finite set $S$ of positive integers has the property that, for each $s \in S,$ and each positive integer divisor $d$ of $s$, there exists a unique element $t \in S$ satisfying $\text{gcd}(s, t) = d$. (The elements $s$ and $t$ could be equal.) Given this information, find all possible values for the number of elements of $S$.

Do there exist natural numbers $a, b$ and $c$ such that $a^2+b^2+c^2$ is divisible by $2013(ab+bc+ca)$? [i]Proposed by Mahan Malihi[/i]

Find the number of positive integers $m$ for which there exist nonnegative integers $x_0,x_1,\ldots,x_{2011}$ such that \[ m^{x_0}=\sum_{k=1}^{2011}m^{x_k}. \]

Let $n$ be a positive integer with $n \ge 3$. Show that \[n^{n^{n^{n}}}-n^{n^{n}}\] is divisible by $1989$.

Let $S$ be the set of all positive integers less than $10,000$ whose last four digits in base $2$ are the same as its last four digits in base $5$. What remainder do we get if we divide the sum of all elements of $S$ by $10000$?

Let $n>4$ be a positive integer, which is divisible by $4$. We denote by $A_n$ the sum of the odd positive divisors of $n$. We also denote $B_n$ the sum of the even positive divisors of $n$, excluding the number $n$ itself. Find the least possible value of the expression $$f(n)=B_n-2A_n,$$ for all possible values of $n$, as well as for which positive integers $n$ this minimum value is attained.

Let $a,b,c,d$ be positive integers such that the number of pairs $(x,y) \in (0,1)^2$ such that both $ax+by$ and $cx+dy$ are integers is equal with 2004. If $\gcd (a,c)=6$ find $\gcd (b,d)$.

Petro and Vasyl play the following game. They take turns making moves and Petro goes first. In one turn, a player chooses one of the numbers from $1$ to $2023$ that wasn't selected before and writes it on the board. The first player after whose turn the product of the numbers on the board will be divisible by $2023$ loses. Who wins if every player wants to win? [i]Proposed by Mykhailo Shtandenko[/i]

For each positive integer $k$ we denote by $S(k)$ the sum of its digits, for example $S(132)=6$ and $S(1000)=1$. A positive integer $n$ is said to be $\textbf{fascinating}$ if it holds that $n = \frac{k}{S(k)}$ for some positive integer $k$. For example, the number $11$ is $\textbf{fascinating}$ since $11 = \frac{198}{S(198)} ($since $\frac{198}{S(198)}=\frac{198}{1+9+8}=\frac{198}{18} = 11)$. Prove that there exists a positive integer less than $2021$ and that it is not $\textbf{fascinating}$.

Let $\mathbb N$ denote set of all natural numbers and let $f:\mathbb{N}\to\mathbb{N}$ be a function such that $\text{(a)} f(mn)=f(m).f(n)$ for all $m,n \in\mathbb{N}$; $\text{(b)} m+n$ divides $f(m)+f(n)$ for all $m,n\in \mathbb N$. Prove that, there exists an odd natural number $k$ such that $f(n)= n^k$ for all $n$ in $\mathbb{N}$.

There are $n!$ empty baskets in a row, labelled $1, 2, . . . , n!$. Caesar first puts a stone in every basket. Caesar then puts 2 stones in every second basket. Caesar continues similarly until he has put $n$ stones into every nth basket. In other words, for each $i = 1, 2, . . . , n,$ Caesar puts $i$ stones into the baskets labelled $i, 2i, 3i, . . . , n!.$ Let $x_i$ be the number of stones in basket $i$ after all these steps. Show that $n! \cdot n^2 \leq \sum_{i=1}^{n!} x_i^2 \leq n! \cdot n^2 \cdot \sum_{i=1}^{n} \frac{1}{i} $

[u]Round 1[/u] [b]p1.[/b] Ash is running around town catching Pokémon. Each day, he may add $3, 4$, or $5$ Pokémon to his collection, but he can never add the same number of Pokémon on two consecutive days. What is the smallest number of days it could take for him to collect exactly $100$ Pokémon? [b]p2.[/b] Jack and Jill have ten buckets. One bucket can hold up to $1$ gallon of water, another can hold up to $2$ gallons, and so on, with the largest able to hold up to $10$ gallons. The ten buckets are arranged in a line as shown below. Jack and Jill can pour some amount of water into each bucket, but no bucket can have less water than the one to its left. Is it possible that together, the ten buckets can hold 36 gallons of water? [img]https://cdn.artofproblemsolving.com/attachments/f/8/0b6524bebe8fe859fe7b1bc887ac786106fc17.png[/img] [b]p3.[/b] There are $2023$ knights and liars standing in a row. Knights always tell the truth and liars always lie. Each of them says, “the number of liars to the left of me is greater than the number of knights to the right.” How many liars are there? [b]p4.[/b] Camila has a deck of $101$ cards numbered $1, 2, ..., 101$. She starts with $50$ random cards in her hand and the rest on a table with the numbers visible. In an exchange, she replaces all $50$ cards in her hand with her choice of $50$ of the $51$ cards from the table. Show that Camila can make at most 50 exchanges and end up with cards $1, 2, ..., 50$. [img]https://cdn.artofproblemsolving.com/attachments/0/6/c89e65118764f3b593da45264bfd0d89e95067.png[/img] [b]p5.[/b] There are $101$ pirates on a pirate ship: the captain and $100$ crew. Each pirate, including the captain, starts with $1$ gold coin. The captain makes proposals for redistributing the coins, and the crew vote on these proposals. The captain does not vote. For every proposal, each crew member greedily votes “yes” if he gains coins as a result of the proposal, “no” if he loses coins, and passes otherwise. If strictly more crew members vote “yes” than “no,” the proposal takes effect. The captain can make any number of proposals, one after the other. What is the largest number of coins the captain can accumulate? [u]Round 2[/u] [b]p6.[/b] The town of Lumenville has $100$ houses and is preparing for the math festival. The Tesla wiring company will lay lengths of power wire in straight lines between the houses so that power flows between any two houses, possibly by passing through other houses. The Edison lighting company will hang strings of lights in straight lines between pairs of houses so that each house is connected by a string to exactly one other. Show that however the houses are arranged, the Edison company can always hang their strings of lights so that the total length of the strings is no more than the total length of the power wires the Tesla company used. [img]https://cdn.artofproblemsolving.com/attachments/9/2/763de9f4138b4dc552247e9316175036c649b6.png[/img] [b]p7.[/b] You are given a sequence of $16$ digits. Is it always possible to select one or more digits in a row, so that multiplying them results in a square number? [img]https://cdn.artofproblemsolving.com/attachments/d/1/f4fcda2e1e6d4a1f3a56cd1a04029dffcd3529.png[/img] PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Let n be a positive integer. Prove the following statement: ”If $2 + 2\sqrt{1 + 28n^2}$ is an integer, then it is the square of an integer.”