A $ 16$-quart radiator is filled with water. Four quarts are removed and replaced with pure antifreeze liquid. Then four quarts of the mixture are removed and replaced with pure antifreeze. This is done a third and a fourth time. The fractional part of the final mixture that is water is: $ \textbf{(A)}\ \frac{1}{4}\qquad \textbf{(B)}\ \frac{81}{256}\qquad \textbf{(C)}\ \frac{27}{64}\qquad \textbf{(D)}\ \frac{37}{64}\qquad \textbf{(E)}\ \frac{175}{256}$

Let $a_n$ be the least positive integer the sum of whose digits is $n$. Find $a_1 + a_2 + a_3 + \dots + a_{20}$.

Prove that there exists a convex 1990-gon with the following two properties : [b]a.)[/b] All angles are equal. [b]b.)[/b] The lengths of the 1990 sides are the numbers $ 1^2$, $ 2^2$, $ 3^2$, $ \cdots$, $ 1990^2$ in some order.

Find the positive integer n such that $10^n$ cubic centimeters is the same as 1 cubic kilometer.

$f(x)=x^3-6x^2+17x$. If $f(a)=16, f(b)=20$, find $a+b$.

Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\]

There are $2n$ different numbers in a row. By one move we can interchange any two numbers or interchange any $3$ numbers cyclically (choose $a,b,c$ and place $a$ instead of $b$, $b$ instead of $c$, $c$ instead of $a$). What is the minimal number of moves that is always sufficient to arrange the numbers in increasing order ?

Consider a circle with center $O$ and radius $R,$ and let $A$ and $B$ be two points in the plane of this circle. [b]a.)[/b] Draw a chord $CD$ of the circle such that $CD$ is parallel to $AB,$ and the point of the intersection $P$ of the lines $AC$ and $BD$ lies on the circle. [b]b.)[/b] Show that generally, one gets two possible points $P$ ($P_{1}$ and $P_{2}$) satisfying the condition of the above problem, and compute the distance between these two points, if the lengths $OA=a,$ $OB=b$ and $AB=d$ are given.

Let $ABCDEF$ be a convex hexagon satisfying $AC=DF$, $CE=FB$ and $EA=BD$. Prove that the lines connecting the midpoints of opposite sides of the hexagon $ABCDEF$ intersect in one point.

Let $M$ be a finite set of lattice points and $n$ be a positive integer. A $\textit{mine-avoiding path}$ is a path of lattice points with length $n$, beginning at $(0,0)$ and ending at a point on the line $x+y=n,$ that does not contain any point in $M$. Prove that if there exists a mine-avoiding path, then there exist at least $2^{n-|M|}$ mine-avoiding paths. [hide=*]A lattice point is a point $(x,y)$ where $x$ and $y$ are integers. A path of lattice points with length $n$ is a sequence of lattice points $P_0,P_1,\ldots, P_n$ in which any two adjacent points in the sequence have distance 1 from each other.[/hide] [i]Ankit Bisain and Holden Mui[/i]

Let $ A$ be a set of positive integers such that no positive integer greater than 1 divides all the elements of $ A.$ Prove that any sufficiently large positive integer can be written as a sum of elements of $ A.$ (Elements may occur several times in the sum.)

Equiangular hexagon $ ABCDEF$ has side lengths $ AB \equal{} CD \equal{} EF \equal{} 1$ and $ BC \equal{} DE \equal{} FA \equal{} r$. The area of $ \triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $ r$? $ \textbf{(A)}\ \frac {4\sqrt {3}}{3} \qquad \textbf{(B)}\ \frac {10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac {17}{4} \qquad \textbf{(E)}\ 6$

Let the function $f:N^*\to N^*$ such that [b](1)[/b] $(f(m),f(n))\le (m,n)^{2014} , \forall m,n\in N^*$; [b](2)[/b] $n\le f(n)\le n+2014 , \forall n\in N^*$ Show that: there exists the positive integers $N$ such that $ f(n)=n $, for each integer $n \ge N$. (High School Affiliated to Nanjing Normal University )

[u]Round 1[/u] [b]p1.[/b] Goldilocks enters the home of the three bears – Papa Bear, Mama Bear, and Baby Bear. Each bear is wearing a different-colored shirt – red, green, or blue. All the bears look the same to Goldilocks, so she cannot otherwise tell them apart. The bears in the red and blue shirts each make one true statement and one false statement. The bear in the red shirt says: “I'm Blue's dad. I'm Green's daughter.” The bear in the blue shirt says: “Red and Green are of opposite gender. Red and Green are my parents.” Help Goldilocks find out which bear is wearing which shirt. [b]p2.[/b] The University of Washington is holding a talent competition. The competition has five contests: math, physics, chemistry, biology, and ballroom dancing. Any student can enter into any number of the contests but only once for each one. For example, a student may participate in math, biology, and ballroom. It turned out that each student participated in an odd number of contests. Also, each contest had an odd number of participants. Was the total number of contestants odd or even? [b]p3.[/b] The $99$ greatest scientists of Mars and Venus are seated evenly around a circular table. If any scientist sees two colleagues from her own planet sitting an equal number of seats to her left and right, she waves to them. For example, if you are from Mars and the scientists sitting two seats to your left and right are also from Mars, you will wave to them. Prove that at least one of the $99$ scientists will be waving, no matter how they are seated around the table. [b]p4.[/b] One hundred boys participated in a tennis tournament in which every player played each other player exactly once and there were no ties. Prove that after the tournament, it is possible for the boys to line up for pizza so that each boy defeated the boy standing right behind him in line. [b]p5.[/b] To celebrate space exploration, the Science Fiction Museum is going to read Star Wars and Star Trek stories for $24$ hours straight. A different story will be read each hour for a total of $12$ Star Wars stories and $12$ Star Trek stories. George and Gene want to listen to exactly $6$ Star Wars and $6$ Star Trek stories. Show that no matter how the readings are scheduled, the friends can find a block of $12$ consecutive hours to listen to the stories together. [u]Round 2[/u] [b]p6.[/b] $2013$ people attended Cinderella's ball. Some of the guests were friends with each other. At midnight, the guests started turning into mice. After the first minute, everyone who had no friends at the ball turned into a mouse. After the second minute, everyone who had exactly one friend among the remaining people turned into a mouse. After the third minute, everyone who had two human friends left in the room turned into a mouse, and so on. What is the maximal number of people that could have been left at the ball after $2013$ minutes? [b]p7.[/b] Bill and Charlie are playing a game on an infinite strip of graph paper. On Bill’s turn, he marks two empty squares of his choice (not necessarily adjacent) with crosses. Charlie, on his turn, can erase any number of crosses, as long as they are all adjacent to each other. Bill wants to create a line of $2013$ crosses in a row. Can Charlie stop him? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Let $S = \{1, 2, 3, \ldots , 2046, 2047, 2048\}$. Two subsets $A$ and $B$ of $S$ are said to be [i]friends[/i] if the following conditions are true: [list] [*] They do not share any elements. [*] They both have the same number of elements. [*] The product of all elements from $A$ equals the product of all elements from $B$. [/list] Prove that there are two subsets of $S$ that are [i]friends[/i] such that each one of them contains at least $738$ elements.

$ABCDEF$ is a cyclic hexagon with circumcenter $O$, and $AD, BE, CF$ are concurrent at $X$. $P$ is a point on the plane. The circumenter of $PAB$ is $O_{AB}$. Define $O_{BC}, O_{CD}$, $O_{DE}, O_{EF}, O_{FA}$ similarly. Prove that $O_{AB} O_{DE}, O_{BC}O_{EF}, O_{CD}O_{FA}$, $OX$ are concurrent.

Let \[N= \sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor).\] Find the remainder when N is divided by 1000. (Here $\lfloor x \rfloor$ denotes the greatest integer that is less than or equal to x, and $\lceil x \rceil$ denotes the least integer that is greater than or equal to x.)

Find all non-decreasing functions $f:\mathbb R^+\cup\{0\}\rightarrow\mathbb R^+\cup\{0\}$ such that for each $x,y\in \mathbb R^+\cup\{0\}$ \[f\left(\frac{x+f(x)}2+y\right)=2x-f(x)+f(f(y)).\]

Your answer to this problem will be an integer between $0$ and $100$, inclusive. From all the teams who submitted an answer to this problem, let the average answer be $A$. Estimate the value of $\left\lfloor \frac23 A \right\rfloor$. If your estimate is $E$ and the answer is $A$, your score for this problem will be \[\max\left(0,\lfloor15-2\cdot\left|A-E\right|\right \rfloor).\] [i]Proposed by Andrew Zhao[/i]

Let $ f: Z \to Z$ be such that $ f(1) \equal{} 1, f(2) \equal{} 20, f(\minus{}4) \equal{} \minus{}4$ and $ f(x\plus{}y) \equal{} f(x) \plus{}f(y)\plus{}axy(x\plus{}y)\plus{}bxy\plus{}c(x\plus{}y)\plus{}4 \forall x,y \in Z$, where $ a,b,c$ are constants. (a) Find a formula for $ f(x)$, where $ x$ is any integer. (b) If $ f(x) \geq mx^2\plus{}(5m\plus{}1)x\plus{}4m$ for all non-negative integers $ x$, find the greatest possible value of $ m$.

There are $64$ towns in a country and some pairs of towns are connected by roads but we do not know these pairs. We may choose any pair of towns and find out whether they are connected or not. Our aim is to determine whether it is possible to travel from any town to any other by a sequence of roads. Prove that there is no algorithm which enables us to do so in less than $2016$ questions. (Proposed by Konstantin Knop)

In this problem, we consider integers consisting of $5$ digits, of which the rst and last one are nonzero. We say that such an integer is a palindromic product if it satises the following two conditions: - the integer is a palindrome, (i.e. it doesn't matter if you read it from left to right, or the other way around); - the integer is a product of two positive integers, of which the first, when read from left to right, is equal to the second, when read from right to left, like $4831$ and $1384$. For example, $20502$ is a palindromic product, since $102 \cdot 201 = 20502$, and $20502$ itself is a palindrome. Determine all palindromic products of $5$ digits.

$PQR$ is an acute scalene triangle. The altitude $PL$ and the bisector $RK$ of $\angle QRP$ meet at $H$ ($L$ on $QR$ and $K$ on $PQ$). $KM$ is the altitude of triangle $PKR$; it meets $PL$ at $N$. The circumcircle of $\triangle NKR$ meets $QR$ at $S$ other than $Q$. Prove that $SHK$ is an isosceles triangle.

Let $n \ge 2018$ be an integer, and let $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ be pairwise distinct positive integers not exceeding $5n$. Suppose that the sequence \[ \frac{a_1}{b_1}, \frac{a_2}{b_2}, \dots, \frac{a_n}{b_n} \] forms an arithmetic progression. Prove that the terms of the sequence are equal.

For each $a$, $1<a<2$, the graphs of functions $y=1-|x-1|$ and $y=|2x-a|$ determine a figure. Prove that the area of this figure is less than $\frac13$.