Find the maximal $x$ such that the expression $4^{27} + 4^{1000} + 4^x$ is the exact square.

For which digits $a$ do exist integers $n \geq 4$ such that each digit of $\frac{n(n+1)}{2}$ equals $a \ ?$

Call a positive integer "useful but not optimized " (!), if it can be written as a sum of distinct powers of $3$ and powers of $5$. Prove that there exist infinitely many positive integers which they are not "useful but not optimized". (e.g. $37=(3^0+3^1+3^3)+(5^0+5^1)$ is a " useful but not optimized" number) [i]Proposed by Mohsen Jamali[/i]

Let $a, b, n$ be positive integers, $b < 10$ and $2^n = 10a + b$. Prove that if $n > 3$, then $6$ divides $ab$.

Find the five-hundredth-smallest positive integer that can be written using only the digits $1$, $3,$ and $5$ in base $7$? [i]Proposed by CaptainFlint[/i]

Let $n$ be a given positive integer. Prove that there is no positive divisor $d$ of $2n^2$ such that $d^2n^2+d^3$ is a square of an integer.

Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that \[ m^2 + f(n) \mid mf(m) +n \] for all positive integers $m$ and $n$.

[b]p1.[/b] Compute $45\times 45 - 6$. [b]p2.[/b] Consecutive integers have nice properties. For example, $3$, $4$, $5$ are three consecutive integers, and $8$, $9$, $10$ are three consecutive integers also. If the sum of three consecutive integers is $24$, what is the smallest of the three numbers? [b]p3.[/b] How many positive integers less than $25$ are either multiples of $2$ or multiples of $3$? [b]p4.[/b] Charlotte has $5$ positive integers. Charlotte tells you that the mean, median, and unique mode of his five numbers are all equal to $10$. What is the largest possible value of the one of Charlotte's numbers? [b]p5.[/b] Mr. Meeseeks starts with a single coin. Every day, Mr. Meeseeks goes to a magical coin converter where he can either exchange $1$ coin for $5$ coins or exchange $5$ coins for $3$ coins. What is the least number of days Mr. Meeseeks needs to end with $15$ coins? [b]p6.[/b] Twelve years ago, Violet's age was twice her sister Holo's age. In $7$ years, Holo's age will be $13$ more than a third of Violet's age. $3$ years ago, Violet and Holo's cousin Rindo's age was the sum of their ages. How old is Rindo? [b]p7.[/b] In a $2 \times 3$ rectangle composed of $6$ unit squares, let $S$ be the set of all points $P$ in the rectangle such that a unit circle centered at $P$ covers some point in exactly $3$ of the unit squares. Find the area of the region $S$. For example, the diagram below shows a valid unit circle in a $2 \times 3$ rectangle. [img]https://cdn.artofproblemsolving.com/attachments/d/9/b6e00306886249898c2bdb13f5206ced37d345.png[/img] [b]p8.[/b] What are the last four digits of $2^{1000}$? [b]p9.[/b] There is a point $X$ in the center of a $2 \times 2 \times 2$ box. Find the volume of the region of points that are closer to $X$ than to any of the vertices of the box. [b]p10.[/b] Evaluate $\sqrt{37 \cdot 41 \cdot 113 \cdot 290 - 4319^2}$ [b]p11.[/b] (Estimation) A number is abundant if the sum of all its divisors is greater than twice the number. One such number is $12$, because $1+2+3+4+6+12 = 28 > 24$: How many abundant positive integers less than $20190$ are there? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Prove that the set $\{2,2^2,\ldots,2^n,\ldots\}$ satisfies the following properties: [b](1)[/b] For every $a\in A, b\in\mathbb{N}$, if $b<2a-1$, then $b(b+1)$ isn't a multiple of $2a$; [b](2)[/b] For every positive integer $a\notin A,a\ne 1$, there exists a positive integer $b$, such that $b<2a-1$ and $b(b+1)$ is a multiple of $2a$.

Find the largest prime $p$ such that there exist positive integers $a,b$ satisfying $$p=\frac{b}{2}\sqrt{\frac{a-b}{a+b}}.$$

A few (at least $5$) integers are put on a circle, such that each of them is divisible by the sum of its neighbors. If the sum of all numbers is positive, what is its minimal value?

If the positive integers $a, b, c$ satisfy $a^2+b^2=c^2$, then $(a, b, c)$ is called a Pythagorean triple. Find all Pythagorean triples containing $30$.

Let $a,b,c\in\{0,1,2,\cdots,9\}$.The quadratic equation $ax^2+bx+c=0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number.

Find all primes $ p$ for that there is an integer $ n$ such that there are no integers $ x,y$ with $ x^3 \plus{} y^3 \equiv n \mod p$ (so not all residues are the sum of two cubes). E.g. for $ p \equal{} 7$, one could set $ n \equal{} \pm 3$ since $ x^3,y^3 \equiv 0 , \pm 1 \mod 7$, thus $ x^3 \plus{} y^3 \equiv 0 , \pm 1 , \pm 2 \mod 7$ only.

Determine all integers $n \geq 2$ for which the number $11111$ in base $n$ is a perfect square.

Let $f: \mathbb{N} \to \mathbb{N}$ be a function that satisfies: \[ f(1) = 2008, \] \[ f(4n^2) = 4f(n^2), \] \[ f(4n^2 + 2) = 4f(n^2) + 3, \] \[ f(4n(n+1)) = 4f(n(n+1)) + 1, \] \[ f(4n(n+1) + 3) = 4f(n(n+1)) + 4. \] Determine whether there exists a natural number $m$ such that: \[ 1^2 + 2^2 + \dots + m^2 + f(1^2) + \dots + f(m^2) = 2008m + 251. \]

Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \ldots, a_n$ such that \[ \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1. \] [i]Proposed by Dusan Djukic, Serbia[/i]

Let $n \geq 3$ be a positive integers and $p$ be a prime number such that $p > 6^{n-1} - 2^n + 1$. Let $S$ be the set of $n$ positive integers with different residues modulo $p$. Show that there exists a positive integer $c$ such that there are exactly two ordered triples $(x,y,z) \in S^3$ with distinct elements, such that $x-y+z-c$ is divisible by $p$.

Find all positive integer $n$ and nonnegative integer $a_1,a_2,\dots,a_n$ satisfying: $i$ divides exactly $a_i$ numbers among $a_1,a_2,\dots,a_n$, for each $i=1,2,\dots,n$. ($0$ is divisible by all integers.)

Let $\alpha\leq 22$ be non-negative integer. For which $\alpha$ does the equation $$8x^{23}-5^{\alpha}y^{23}=1$$ have the most integer solutions (x,y)? What can we say about $\alpha\geq 23$? [hide=Note]I believe the eqn has solutions only when $\alpha=0$. taking modulo 47, $\alpha\equiv 9,17\pmod{23}$ or ($23|\alpha$ and $47|x$). taking modulo 139 and 277 eliminates the $\alpha\equiv 9,17\pmod{23}$ cases. 139=23*6+1 , 277=23*12+1[/hide]

For an integer $n$ with $b$ digits, let a [i]subdivisor[/i] of $n$ be a positive number which divides a number obtained by removing the $r$ leftmost digits and the $l$ rightmost digits of $n$ for nonnegative integers $r,l$ with $r+l<b$ (For example, the subdivisors of $143$ are $1$, $2$, $3$, $4$, $7$, $11$, $13$, $14$, $43$, and $143$). For an integer $d$, let $A_d$ be the set of numbers that don't have $d$ as a subdivisor. Find all $d$, such that $A_d$ is finite.

Given $100$ integers, is it always possible to choose $15$ of them such that the difference of any two of the chosen numbers is divisible by $7$? What is the answer if $15$ is replaced by $16$?

Let $m$ and $n$ be positive integers and $p$ be a prime number. Find the greatest positive integer $s$ (as a function of $m,n$ and $p$) such that from a random set of $mnp$ positive integers we can choose $snp$ numbers, such that they can be partitioned into $s$ sets of $np$ numbers, such that the sum of the numbers in every group gives the same remainder when divided by $p$.

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. [i]Proposed by Amine Natik, Morocco[/i]

Let $f_n=\left(1+\frac{1}{n}\right)^n\left((2n-1)!F_n\right)^{\frac{1}{n}}$. Find $\lim \limits_{n \to \infty}(f_{n+1} - f_n)$ where $F_n$ denotes the $n$th Fibonacci number (given by $F_0 = 0$, $F_1 = 1$, and by $F_{n+1} = F_n + F_{n-1}$ for all $n \geq 1$