Let $\varphi(n)$ denote the number of positive integers less than $n$ that are relatively prime to $n$. Prove that there exists a positive integer $m$ for which the equation $\varphi(n)=m$ has at least $2015$ solutions in $n$. [i]Proposed by Iurie Boreico[/i]

Let $M_n$ be the $n\times n$ matrix with entries as follows: for $1\leq i \leq n$, $m_{i,i}=10$; for $1\leq i \leq n-1, m_{i+1,i}=m_{i,i+1}=3$; all other entries in $M_n$ are zero. Let $D_n$ be the determinant of matrix $M_n$. Then $\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{8D_n+1}$ can be represented as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. Note: The determinant of the $1\times 1$ matrix $[a]$ is $a$, and the determinant of the $2\times 2$ matrix $\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]=ad-bc$; for $n\geq 2$, the determinant of an $n\times n$ matrix with first row or first column $a_1\ a_2\ a_3 \dots\ a_n$ is equal to $a_1C_1 - a_2C_2 + a_3C_3 - \dots + (-1)^{n+1} a_nC_n$, where $C_i$ is the determinant of the $(n-1)\times (n-1)$ matrix found by eliminating the row and column containing $a_i$.

A circle in the first quadrant with center on the curve $y=2x^2-27$ is tangent to the $y$-axis and the line $4x=3y$. The radius of the circle is $\frac{m}{n}$ where $M$ and $n$ are relatively prime positive integers. Find $m+n$.

Find all pairs of positive integers $(m,n)$ for which there exist relatively prime integers $a$ and $b$ greater than $1$ such that $$\frac{a^m+b^m}{a^n+b^n}$$ is an integer.

For a positive integer $n$, define the set $S_n$ as $S_n =\{(a, b)|a, b \in N, lcm[a, b] = n\}$ . Let $f(n)$ be the sum of $\phi (a)\phi (b)$ for all $(a, b) \in S_n$. If a prime $p$ relatively prime to $n$ is a divisor of $f(n)$, prove that there exists a prime $q|n$ such that $p|q^2 - 1$.

A function $f : \mathbb{Z}_+ \to \mathbb{Z}_+$, where $\mathbb{Z}_+$ is the set of positive integers, is non-decreasing and satisfies $f(mn) = f(m)f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8)f(13) \ge (f(10))^2$.

Find all pairs $(x, y)$ of positive rational numbers such that $x^{y}=y^{x}$.

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 $m + n$.

For each integer $n$ ($n \ge 2$), let $f(n)$ denote the sum of all positive integers that are at most $n$ and not relatively prime to $n$. Prove that $f(n+p) \neq f(n)$ for each such $n$ and every prime $p$.

Let n be an integer which is greater than 1, not divisible by 1997. Let $ a_m\equal{}m\plus{}\frac{mn}{1997}$ for all m=1,2,..,1996 $ b_m\equal{}m\plus{}\frac{1997m}{n}$ for all m=1,2,..,n-1 We arrange the terms of two sequence $ (a_i), (b_j)$ in the ascending order to form a new sequence $ c_1\le c_2\le ...\le c_{1995\plus{}n}$ Prove that $ c_{k\plus{}1}\minus{}c_k<2$ for all k=1,2,...,1994+n

Bob has invented the Very Normal Coin (VNC). When the VNC is flipped, it shows heads $\textstyle\frac{1}{2}$ of the time and tails $\textstyle\frac{1}{2}$ of the time - unless it has yielded the same result five times in a row, in which case it is guaranteed to yield the opposite result. For example, if Bob flips five heads in a row, then the next flip is guaranteed to be tails. Bob flips the VNC an infinite number of times. On the $n$th flip, Bob bets $2^{-n}$ dollars that the VNC will show heads (so if the second flip shows heads, Bob wins $\$0.25$, and if the third flip shows tails, Bob loses $\$0.125$). Assume that dollars are infinitely divisible. Given that the first flip is heads, the expected number of dollars Bob is expected to win can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a + b$. [i]Proposed by Lewis Chen[/i]

For integer $n\ge 4$, find the minimal integer $f(n)$, such that for any positive integer $m$, in any subset with $f(n)$ elements of the set ${m, m+1, \ldots, m+n+1}$ there are at least $3$ relatively prime elements.

Let $k$ be a fixed integer greater than 1, and let ${m=4k^2-5}$. Show that there exist positive integers $a$ and $b$ such that the sequence $(x_n)$ defined by \[x_0=a,\quad x_1=b,\quad x_{n+2}=x_{n+1}+x_n\quad\text{for}\quad n=0,1,2,\dots,\] has all of its terms relatively prime to $m$. [i]Proposed by Jaroslaw Wroblewski, Poland[/i]

Let $x$ satisfy $(6x + 7) + (8x + 9) = (10 + 11x) + (12 + 13x).$ There are relatively prime positive integers so that $x = -\tfrac{m}{n}$. Find $m + n.$

Find all polynomials $P$ with integer coefficients which satisfy the property that, for any relatively prime integers $a$ and $b$, the sequence $\{P (an + b) \}_{n \ge 1}$ contains an infinite number of terms, any two of which are relatively prime.

Two points $A$ and $B$ are selected independently and uniformly at random along the perimeter of a unit square with vertices at $(0,0)$, $(1,0)$, $(0,1)$, and $(1,1)$. The probability that the $y$-coordinate of $A$ is strictly greater than the $y$-coordinate of $B$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $100m+n$. [i]Proposed by Rajiv Movva[/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$.

Six children stand in a line outside their classroom. When they enter the classroom, they sit in a circle in random order. There are relatively prime positive integers $m$ and $n$ so that $\tfrac{m}{n}$ is the probability that no two children who stood next to each other in the line end up sitting next to each other in the circle. Find $m + n$.

Let $a$, $b$, $c$ be positive real numbers for which \[ \frac{5}{a} = b+c, \quad \frac{10}{b} = c+a, \quad \text{and} \quad \frac{13}{c} = a+b. \] If $a+b+c = \frac mn$ for relatively prime positive integers $m$ and $n$, compute $m+n$. [i]Proposed by Evan Chen[/i]

Find all the pairs of positive numbers such that the last digit of their sum is 3, their difference is a primer number and their product is a perfect square.

For $\pi\leq\theta<2\pi$, let \[ P=\dfrac12\cos\theta-\dfrac14\sin2\theta-\dfrac18\cos3\theta+\dfrac1{16}\sin4\theta+\dfrac1{32}\cos5\theta-\dfrac1{64}\sin6\theta-\dfrac1{128}\cos7\theta+\ldots \] and \[ Q=1-\dfrac12\sin\theta-\dfrac14\cos2\theta+\dfrac1{8}\sin3\theta+\dfrac1{16}\cos4\theta-\dfrac1{32}\sin5\theta-\dfrac1{64}\cos6\theta+\dfrac1{128}\sin7\theta +\ldots \] so that $\tfrac PQ = \tfrac{2\sqrt2}7$. Then $\sin\theta = -\tfrac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

For how many ordered pairs of positive integers $ (m,n)$, $ m \cdot n$ divides $ 2008 \cdot 2009 \cdot 2010$ ? $\textbf{(A)}\ 2\cdot3^7\cdot 5 \qquad\textbf{(B)}\ 2^5\cdot3\cdot 5 \qquad\textbf{(C)}\ 2^5\cdot3^7\cdot 5 \qquad\textbf{(D)}\ 2^3\cdot3^5\cdot 5^2 \qquad\textbf{(E)}\ \text{None}$

Let $m$ be a positive integer and $x_0, y_0$ integers such that $x_0, y_0$ are relatively prime, $y_0$ divides $x_0^2+m$, and $x_0$ divides $y_0^2+m$. Prove that there exist positive integers $x$ and $y$ such that $x$ and $y$ are relatively prime, $y$ divides $x^2 + m$, $x$ divides $y^2 + m$, and $x + y \leq m+ 1.$

The repeating decimal $0.328181818181...$ can equivalently be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Let $x_1$, $x_2$, and $x_3$ be the roots of the polynomial $x^3+3x+1$. There are relatively prime positive integers $m$ and $n$ such that $\tfrac{m}{n}=\tfrac{x_1^2}{(5x_2+1)(5x_3+1)}+\tfrac{x_2^2}{(5x_1+1)(5x_3+1)}+\tfrac{x_3^2}{(5x_1+1)(5x_2+1)}$. Find $m+n$.