The side lengths of a trapezoid are $\sqrt[4]{3}, \sqrt[4]{3}, \sqrt[4]{3}$, and $2 \cdot \sqrt[4]{3}$. Its area is the ratio of two relatively prime positive integers, $m$ and $n$. Find $m + n$.

Let $a,b$ be positive integers such that $a+b=10$. Let $\tfrac{p}{q}$ be the difference between the maximum and minimum possible values of $\tfrac{1}{a}+\tfrac{1}{b}$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$.

In chess, there are two types of minor pieces, the bishop and the knight. A bishop may move along a diagonal, as long as there are no pieces obstructing its path. A knight may jump to any lattice square $\sqrt{5}$ away as long as it isn't occupied. One day, a bishop and a knight were on squares in the same row of an infinite chessboard, when a huge meteor storm occurred, placing a meteor in each square on the chessboard independently and randomly with probability $p$. Neither the bishop nor the knight were hit, but their movement may have been obstructed by the meteors. The value of $p$ that would make the expected number of valid squares that the bishop can move to and the number of squares that the knight can move to equal can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a + b$. [i]Proposed by Lewis Chen[/i]

Let the sequence $a_1,a_2,\ldots,a_n,\ldots$ is defined by the conditions: $a_1=2$ and $a_{n+1}=a_n^2-a_n+1$ $(n=1,2,\ldots)$. Prove that: (a) $a_m$ and $a_n$ are relatively prime numbers when $m\ne n$. (b) $\lim_{n\to\infty}\sum_{k=1}^n\frac1{a_k}=1$ [i]I. Tonov[/i]

One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane. The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color. The ratio of the total area of the green regions to the area of the circle of radius 100 can be expressed as $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Four consecutive three-digit numbers are divided respectively by four consecutive two-digit numbers. What minimum number of different remainders can be obtained? [i](A. Golovanov)[/i]

Heather and Kyle need to mow a lawn and paint a room. If Heather does both jobs by herself, it will take her a total of nine hours. If Heather mows the lawn and, after she finishes, Kyle paints the room, it will take them a total of eight hours. If Kyle mows the lawn and, after he finishes, Heather paints the room, it will take them a total of seven hours. If Kyle does both jobs by himself, it will take him a total of six hours. It takes Kyle twice as long to paint the room as it does for him to mow the lawn. The number of hours it would take the two of them to complete the two tasks if they worked together to mow the lawn and then worked together to paint the room is a fraction $\tfrac{m}{n}$where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that \[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \frac{h_n}{k_n}.\] Let $L_n$ denote the least common multiple of the numbers $1, 2, 3,\cdots, n$. For how many integers $n$ with $1 \le n \le 22$ is $k_n<L_n$? $\textbf{(A)} ~0 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~7 \qquad\textbf{(D)} ~8 \qquad\textbf{(E)} ~10 $

Suppose that $A=1,2,$ or $3$. Let $a$ and $b$ be relatively prime integers such that $a^{2}+Ab^2 =s^3$ for some integer $s$. Then, there are integers $u$ and $v$ such that $s=u^2 +Av^2$, $a =u^3 - 3Avu^2$, and $b=3u^{2}v -Av^3$.

Find all pairs $(a, b)$ of positive integers for which $a + b = \phi (a) + \phi (b) + gcd (a, b)$. Here $ \phi (n)$ is the number of numbers $k$ from $\{1, 2,. . . , n\}$ with $gcd (n, k) = 1$.

Prove that between every $27$ different positive integers , less than $100$, there exist some two which are[color=red] NOT [/color]relative prime. [u]babis[/u]

Show that if $x, y, z$ are positive integers, then $(xy+1)(yz+1)(zx+1)$ is a perfect square if and only if $xy+1$, $yz+1$, $zx+1$ are all perfect squares.

Determine all positive rational numbers $r \neq 1$ such that $\sqrt[r-1]{r}$ is rational.

Alice and Bob play the following game. They alternate selecting distinct nonzero digits (from $1$ to $9$) until they have chosen seven such digits, and then consider the resulting seven-digit number by concatenating the digits in the order selected, with the seventh digit appearing last (i.e. $\overline{A_1B_2A_3B_4A_6B_6A_7}$). Alice wins if and only if the resulting number is the last seven decimal digits of some perfect seventh power. Please determine which player has the winning strategy.

A [i]triangulation[/i] of a polygon is a subdivision of the polygon into triangles meeting edge to edge, with the property that the set of triangle vertices coincides with the set of vertices of the polygon. Adam randomly selects a triangulation of a regular $180$-gon. Then, Bob selects one of the $178$ triangles in this triangulation. The expected number of $1^\circ$ angles in this triangle can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$. [i]Proposed by Lewis Chen[/i]

For every positive integer $ n$, let $ \phi(n)$ denotes the number of positive integers less than $ n$ that is relatively prime to $ n$ and $ \tau(n)$ denote the sum of all positive divisors of $ n$. Let $ n$ be a positive integer such that $ \phi(n)|n\minus{}1$ and that $ n$ is not a prime number. Prove that $ \tau(n)>2009$.

For integers $n \ge k \ge 0$ we define the [i]bibinomial coefficient[/i] $\left( \binom{n}{k} \right)$ by \[ \left( \binom{n}{k} \right) = \frac{n!!}{k!!(n-k)!!} .\] Determine all pairs $(n,k)$ of integers with $n \ge k \ge 0$ such that the corresponding bibinomial coefficient is an integer. [i]Remark: The double factorial $n!!$ is defined to be the product of all even positive integers up to $n$ if $n$ is even and the product of all odd positive integers up to $n$ if $n$ is odd. So e.g. $0!! = 1$, $4!! = 2 \cdot 4 = 8$, and $7!! = 1 \cdot 3 \cdot 5 \cdot 7 = 105$.[/i]

In the diagram below, fill the $12$ circles with numbers from the following bank so that each number is used once. Two circles connected by a single line must contain relatively prime numbers. Two circles connected by a double line must contain numbers that are not relatively prime. $$\text{Bank: } 20, 21, 22, 23, 24, 25, 27, 28, 30 ,32, 33 ,35$$ [asy] real HRT3 = sqrt(3) / 2; void drawCircle(real x, real y, real r) { path p = circle((x,y), r); draw(p); fill(p, white); } void drawCell(int gx, int gy) { real x = 0.5 * gx; real y = HRT3 * gy; drawCircle(x, y, 0.35); } void drawEdge(int gx1, int gy1, int gx2, int gy2, bool doubled) { real x1 = 0.5 * gx1; real y1 = HRT3 * gy1; real x2 = 0.5 * gx2; real y2 = HRT3 * gy2; if (doubled) { real dx = x2 - x1; real dy = y2 - y1; real ox = -0.035 * dy / sqrt(dx * dx + dy * dy); real oy = 0.035 * dx / sqrt(dx * dx + dy * dy); draw((x1+ox,y1+oy)--(x2+ox,y2+oy)); draw((x1-ox,y1-oy)--(x2-ox,y2-oy)); } else { draw((x1,y1)--(x2,y2)); } } drawEdge(2, 0, 4, 0, true); drawEdge(2, 0, 1, 1, true); drawEdge(2, 0, 3, 1, true); drawEdge(4, 0, 3, 1, false); drawEdge(4, 0, 5, 1, false); drawEdge(1, 1, 0, 2, false); drawEdge(1, 1, 2, 2, false); drawEdge(1, 1, 3, 1, false); drawEdge(3, 1, 2, 2, true); drawEdge(3, 1, 4, 2, true); drawEdge(3, 1, 5, 1, false); drawEdge(5, 1, 4, 2, true); drawEdge(5, 1, 6, 2, false); drawEdge(0, 2, 1, 3, false); drawEdge(0, 2, 2, 2, false); drawEdge(2, 2, 1, 3, false); drawEdge(2, 2, 3, 3, true); drawEdge(2, 2, 4, 2, false); drawEdge(4, 2, 3, 3, false); drawEdge(4, 2, 5, 3, false); drawEdge(4, 2, 6, 2, false); drawEdge(6, 2, 5, 3, true); drawEdge(1, 3, 3, 3, true); drawEdge(3, 3, 5, 3, false); drawCell(2, 0); drawCell(4, 0); drawCell(1, 1); drawCell(3, 1); drawCell(5, 1); drawCell(0, 2); drawCell(2, 2); drawCell(4, 2); drawCell(6, 2); drawCell(1, 3); drawCell(3, 3); drawCell(5, 3); [/asy]

Let $a$ and $b$ be positive integers with $\gcd(a, b)=1$. Show that every integer greater than $ab-a-b$ can be expressed in the form $ax+by$, where $x, y \in \mathbb{N}_{0}$.

In the diagram $ABCDEFG$ is a regular heptagon (a $7$ sided polygon). Shown is the star $AEBFCGD$. The degree measure of the obtuse angle formed by $AE$ and $CG$ is $\dfrac{m}{n}$ where m and n are relatively prime positive integers. Find $m + n$. [asy] size(150); defaultpen(linewidth(1)); string lab[]={"A","B","C","D","E","F","G"}; real r = 360/7; pair A=dir(90-r),B=dir(90),C=dir(90+r),D=dir(90+2*r),E=dir(90+3*r),F=dir(90+4*r),G=dir(90+5*r); draw(A--E--B--F--C--G--D--cycle); for(int k = -1;k <= 5;++k) { label("$"+lab[k+1]+"$",dir(90+k*r),dir(90+k*r)); } [/asy]

Let $ n$ be a positive integer, let $ A$ be a subset of $ \{1, 2, \cdots, n\}$, satisfying for any two numbers $ x, y\in A$, the least common multiple of $ x$, $ y$ not more than $ n$. Show that $ |A|\leq 1.9\sqrt {n} \plus{} 5$.

[i]The Game.[/i] Eric and Greg are watching their new favorite TV show, [i]The Price is Right[/i]. Bob Barker recently raised the intellectual level of his program, and he begins the latest installment with bidding on following question: How many Carmichael numbers are there less than $100,000$? Each team is to list one nonnegative integer not greater than $100,000$. Let $X$ denote the answer to Bob’s question. The teams listing $N$, a maximal bid (of those submitted) not greater than $X$, will receive $N$ points, and all other teams will neither receive nor lose points. (A Carmichael number is an odd composite integer $n$ such that $n$ divides $a^{n-1}-1$ for all integers $a$ relatively prime to $n$ with $1<a<n$.)

Ten positive integers are arranged around a circle. Each number is one more than the greatest common divisor of its two neighbors. What is the sum of the ten numbers?

If you roll six fair dice, let $\mathsf{ p}$ be the probability that exactly five different numbers appear on the upper faces of the six dice. If $\mathsf{p} = \frac{m}{n}$ where $ m $ and $n$ are relatively prime positive integers, find $m+n.$

An integer-sided triangle has angles $ p\theta$ and $ q\theta$, where $ p$ and $ q$ are relatively prime integers. Prove that $ \cos\theta$ is irrational.