Find all functions $ f:\mathbb{N} \rightarrow \mathbb{ N }_0 $ satisfy the following conditions: i) $ f(ab)=f(a)+f(b)-f(\gcd(a,b)), \forall a,b \in \mathbb{N} $ ii) For all primes $ p $ and natural numbers $ a $, $ f(a)\ge f(ap) \Rightarrow f(a)+f(p) \ge f(a)f(p)+1 $

Let $I\subseteq \mathbb{R}$ be an open interval and $f:I\to\mathbb{R}$ a strictly monotonous function. Prove that for all $c\in I$ there exist $a,b\in I$ such that $c\in (a,b)$ and \[\int_a^bf(x) \ dx=f(c)\cdot (b-a).\]

Let \[ f(x_1,x_2,x_3,x_4,x_5,x_6,x_7)=x_1x_2x_4+x_2x_3x_5+x_3x_4x_6+x_4x_5x_7+x_5x_6x_1+x_6x_7x_2+x_7x_1x_3 \] be defined for non-negative real numbers $x_1,x_2,\dots,x_7$ with sum $1$. Prove that $f(x_1,x_2,\dots,x_7)$ has a maximum value and find that value.

Find all functions $f : \mathbb Z \to \mathbb Z$ such that \[f (n |m|) + f (n(|m| +2)) = 2f (n(|m| +1)) \qquad \forall m,n \in \mathbb Z.\] [b]Note.[/b] $|x|$ denotes the absolute value of the integer $x.$

Prove that there are no distinct positive integers $x$ and $y$ such that $x^{2007} + y! = y^{2007} + x! $

let $m,n\in \mathbb{N}$ and $p(x),q(x),h(x)$ are polynomials with real Coefficients such that $p(x)$ is Descending. and for all $x\in \mathbb{R}$ $p(q(nx+m)+h(x))=n(q(p(x))+h(x))+m$ . prove that dont exist function $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all $x\in \mathbb{R}$ $f(q(p(x))+h(x))=f(x)^{2}+1$

$S$ is the set of functions $f:\mathbb{N} \to \mathbb{R}$ that satisfy the following conditions: [b]I.[/b] $f(1) = 2$ [b]II.[/b] $f(n+1) \geq f(n) \geq \frac{n}{n + 1} f(2n)$ for $n = 1, 2, \ldots$ Find the smallest $M \in \mathbb{N}$ such that for any $f \in S$ and any $n \in \mathbb{N}, f(n) < M$.

Let $ f:[0,1]\longrightarrow\mathbb{R} $ a continuous function. Prove that $$ \int_0^1 f^2\left( x^2 \right) dx\ge \frac{3}{4}\left( \int_0^1 f(x)dx \right)^2 , $$ and find the circumstances under which equality happens.

Find all surjective functions $f:\mathbb{N}\to\mathbb{N}$ such that for all positive integers $a$ and $b$, exactly one of the following equations is true: \begin{align*} f(a)&=f(b), <br /> \\ f(a+b)&=\min\{f(a),f(b)\}. \end{align*} [i]Remarks:[/i] $\mathbb{N}$ denotes the set of all positive integers. A function $f:X\to Y$ is said to be surjective if for every $y\in Y$ there exists $x\in X$ such that $f(x)=y$.

A calculating ruler is a ruler for doing algebric calculations. This ruler has three arms, two of them are sationary and one can move freely right and left. Each of arms is gradient. Gradation of each arm depends on the algebric operation ruler does. For eaxample the ruler below is designed for multiplying two numbers. Gradations are logarithmic. [img]http://aycu05.webshots.com/image/5604/2000468517162383885_rs.jpg[/img] For working with ruler, (e.g for calculating $x.y$) we must move the middle arm that the arrow at the beginning of its gradation locate above the $x$ in the lower arm. We find $y$ in the middle arm, and we will read the number on the upper arm. The number written on the ruler is the answer. 1) Design a ruler for calculating $x^{y}$. Grade first arm ($x$) and ($y$) from 1 to 10. 2) Find all rulers that do the multiplication in the interval $[1,10]$. 3) Prove that there is not a ruler for calculating $x^{2}+xy+y^{2}$, that its first and second arm are grade from 0 to 10.

The product of the positive real numbers $x, y, z$ is 1. Show that if \[ \frac{1}{x}+\frac{1}{y} + \frac{1}{z} \geq x+y+z \]then \[ \frac{1}{x^{k}}+\frac{1}{y^{k}} + \frac{1}{z^{k}} \geq x^{k}+y^{k}+z^{k} \] for all positive integers $k$.

The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from $ 1$ to $ 15$ in clockwise order. Committee rules state that a Martian must occupy chair $ 1$ and an Earthling must occupy chair $ 15$. Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is $ N\cdot (5!)^3$. Find $ N$.

Find all functions $f:\mathbb{Q}^{+} \to \mathbb{Q}^{+}$ such that for all $x\in \mathbb{Q}^+$: [list] [*] $f(x+1)=f(x)+1$, [*] $f(x^2)=f(x)^2$. [/list]

Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that the following conditions are true for every pair of positive integers $(x, y)$: $(i)$: $x$ and $f(x)$ have the same number of positive divisors. $(ii)$: If $x \nmid y$ and $y \nmid x$, then: $$\gcd(f(x), f(y)) > f(\gcd(x, y))$$

Let $\mathbb{Z}^+$ be the set of positive integers. Determine all functions $f : \mathbb{Z}^+\to\mathbb{Z}^+$ such that $a^2+f(a)f(b)$ is divisible by $f(a)+b$ for all positive integers $a,b$.

Find all function $f,g: \mathbb{Q} \to \mathbb{Q}$ such that \[\begin{array}{l} f\left( {g\left( x \right) - g\left( y \right)} \right) = f\left( {g\left( x \right)} \right) - y \\ g\left( {f\left( x \right) - f\left( y \right)} \right) = g\left( {f\left( x \right)} \right) - y \\ \end{array}\] for all $x,y \in \mathbb{Q}$.

The function $g$ is defined about the natural numbers and satisfies the following conditions: $g(2) = 1$ $g(2n) = g(n)$ $g(2n+1) = g(2n) +1.$ Where $n$ is a natural number such that $1 \leq n \leq 2002$. Find the maximum value $M$ of $g(n).$ Also, calculate how many values of $n$ satisfy the condition of $g(n) = M.$

An integer number $m\geq 1$ is [i]mexica[/i] if it's of the form $n^{d(n)}$, where $n$ is a positive integer and $d(n)$ is the number of positive integers which divide $n$. Find all mexica numbers less than $2019$. Note. The divisors of $n$ include $1$ and $n$; for example, $d(12)=6$, since $1, 2, 3, 4, 6, 12$ are all the positive divisors of $12$. [i]Proposed by Cuauhtémoc Gómez[/i]

Find a non-zero polynomial $f(x, y)$ such that $f(\lfloor 3t \rfloor, \lfloor 5t \rfloor) = 0$ for all real numbers $t$.

Let $G$ be a non-empty set of non-constant functions $f$ such that $f(x)=ax + b$ (where $a$ and $b$ are two reals) and satisfying the following conditions: 1) if $f \in G$ and $g \in G$ then $gof \in G$, 2) if $f \in G$ then $f^ {-1} \in G$, 3) for all $f \in G$ there exists $x_f \in \mathbb{R}$ such that $f(x_f)=x_f$. Prove that there is a real $k$ such that for all $f \in G$ we have $f(k)=k$

Let $ f(n)$ be a function defined on the set of all positive integers and having its values in the same set. Suppose that $ f(f(n) \plus{} f(m)) \equal{} m \plus{} n$ for all positive integers $ n,m.$ Find the possible value for $ f(1988).$

Let $a,b$ be two constants within the open interval $(0,\frac{1}{2})$. Find all continous functions $f(x)$ such that \[f(f(x))=af(x)+bx\] holds for all real $x$. This is much harder than the problems we had in the 1st TST...

Show that there are no integers $a,b,c$ for which $a^2+b^2-8c=6$.

Fin the functions $ f:\mathbb{N}\longrightarrow\mathbb{N} $ such that: $$ \frac{f(x+y)+f(x)}{2x+f(y)} =\frac{2y+f(x)}{f(x+y)+f(y)} ,\quad\forall x,y\in\mathbb{N} . $$

Prove that for all real numbers $x,y,z \in [1,2]$ the following inequality always holds: \[ (x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq 6(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}). \] When does the equality occur?