Is there exist some positive integer $n$, such that the first decimal of $2^n$ (from left to the right) is $5$ while the first decimal of $5^n$ is $2$? [i](5 points)[/i]

The solution to the equation $\log_{3x} 4 = \log_{2x} 8$, where $x$ is a positive real number other than $\tfrac{1}{3}$ or $\tfrac{1}{2}$, can be written as $\tfrac {p}{q}$ where $p$ and $q$ are relatively prime positive integers. What is $p + q$? $\textbf{(A) } 5 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 31 \qquad \textbf{(E) } 35 $

Find the largest positive integer $k$ such that $\phi ( \sigma ( 2^k)) = 2^k$. ($\phi(n)$ denotes the number of positive integers that are smaller than $n$ and relatively prime to $n$, and $\sigma(n)$ denotes the sum of divisors of $n$). As a hint, you are given that $641|2^{32}+1$.

$n$ is a fixed natural number. Find the least $k$ such that for every set $A$ of $k$ natural numbers, there exists a subset of $A$ with an even number of elements which the sum of it's members is divisible by $n$.

Each positive integer $a$ undergoes the following procedure in order to obtain the number $d = d\left(a\right)$: (i) move the last digit of $a$ to the first position to obtain the numb er $b$; (ii) square $b$ to obtain the number $c$; (iii) move the first digit of $c$ to the end to obtain the number $d$. (All the numbers in the problem are considered to be represented in base $10$.) For example, for $a=2003$, we get $b=3200$, $c=10240000$, and $d = 02400001 = 2400001 = d(2003)$.) Find all numbers $a$ for which $d\left( a\right) =a^2$. [i]Proposed by Zoran Sunic, USA[/i]

Let $n$ be a positive integer, and let $1=d_1<d_2<\dots < d_k=n$ denote all positive divisors of $n$, If the following conditions are satisfied: $$ 2d_2+d_4+d_5=d_7$$ $$ d_3 d_6 d_7=n$$ $$ (d_6+d_7)^2=n+1$$ find all possible values of $n$.

(1) Let $a,b$ be coprime positive integers. Prove that there exists constants $\lambda$ and $\beta$ such that for all integers $m$, $$\left| \sum\limits_{k=1}^{m-1} \left\{ \frac{ak}{m} \right\}\left\{ \frac{bk}{m} \right\} - \lambda m \right| \le \beta$$ (2) Prove that there exists $N$ such that for all $p>N$ (where $p$ is a prime number), and any positive integers $a,b,c$ positive integers satisfying $p\nmid (a+b)(b+c)(c+a)$, there are at least $\lfloor \frac{p}{12} \rfloor$ solutions $k\in \{1,\cdots,p-1\}$ such that $$ \left\{\frac{ak}{p}\right\} + \left\{\frac{bk}{p}\right\} + \left\{\frac{ck}{p}\right\} \le 1 $$

Prove that there exists an integer $n \geq 1$, such that number of all pairs $(a, b)$ of positive integers, satisfying $$\frac{1}{a-b}-\frac{1}{a}+\frac{1}{b}=\frac{1}{n}$$ exceeds $2024.$

Determine the least $n\in\mathbb{N}$ such that $n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n$ has at least $2010$ positive factors.

The set A contains exactly $21$ integers. The sum of any $11$ numbers in $A$ is greater than the sum of the remaining numbers. It is known that the set $A$ contain thes number $101$, and the largest number in $A$ is $2014$. Find out the other $19$ numbers in $A$.

Let $(x_n)$ be a sequence of positive integers defined by $x_1=2$ and $x_{n+1}=2x_n^3+x_n$ for all integers $n\ge1$. Determine the largest power of $5$ that divides $x_{2014}^2+1$.

In base $10$ there exist two-digit natural numbers that can be factorized into two natural factors such that the two digits and the two factors form a sequence of four consecutive integers (for example, $12 = 3 \cdot 4$). Determine all such numbers in all bases.

Let $p$ be an odd prime number. Let $S$ be the sum of all primitive roots modulo $p$. Show that if $p-1$ isn't squarefree (i. e., if there exist integers $k$ and $m$ with $k>1$ and $p-1=k^2m$), then $S \equiv 0 \mod p$. If not, then what is $S$ congruent to $\mod p$ ?

Given that $n$ is a natural number such that the leftmost digits in the decimal representations of $2^n$ and $3^n$ are the same, find all possible values of the leftmost digit.

Find the sum of the four smallest prime divisors of $2016^{239} - 1$.

A paper cup has a base that is a circle with radius $r$, a top that is a circle with radius $2r$, and sides that connect the two circles with straight line segments as shown below. This cup has height $h$ and volume $V$. A second cup that is exactly the same shape as the first is held upright inside the first cup so that its base is a distance of $\tfrac{h}2$ from the base of the first cup. The volume of liquid that will t inside the first cup and outside the second cup can be written $\tfrac{m}{n}\cdot V$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] pair s = (10,1); draw(ellipse((0,0),4,1)^^ellipse((0,-6),2,.5)); fill((3,-6)--(-3,-6)--(0,-2.1)--cycle,white); draw((4,0)--(2,-6)^^(-4,0)--(-2,-6)); draw(shift(s)*ellipse((0,0),4,1)^^shift(s)*ellipse((0,-6),2,.5)); fill(shift(s)*(3,-6)--shift(s)*(-3,-6)--shift(s)*(0,-2.1)--cycle,white); draw(shift(s)*(4,0)--shift(s)*(2,-6)^^shift(s)*(-4,0)--shift(s)*(-2,-6)); pair s = (10,-2); draw(shift(s)*ellipse((0,0),4,1)^^shift(s)*ellipse((0,-6),2,.5)); fill(shift(s)*(3,-6)--shift(s)*(-3,-6)--shift(s)*(0,-4.1)--cycle,white); draw(shift(s)*(4,0)--shift(s)*(2,-6)^^shift(s)*(-4,0)--shift(s)*(-2,-6)); //darn :([/asy]

Let $r=0.d_0d_1d_2\ldots$ be a real number. Let $e_n$ denote the number formed by the digits $d_n, d_{n-1}, \ldots, d_0$ written from left to right (leading zeroes are permitted). Given that $d_0=6$ and for each $n \geq 0$, $e_n$ is equal to the number formed by the $n+1$ rightmost digits of $e_n^2$. Show that $r$ is irrational.

Let $n>1$ be an integer. Prove that there exists an integer $n-1 \ge m \ge \left \lfloor \frac{n}{2} \right \rfloor$ such that the following equation has integer solutions with $a_m>0:$ $$\frac{a_{m}}{m+1}+\frac{a_{m+1}}{m+2}+ \cdots + \frac{a_{n-1}}{n}=\frac{1}{\textrm{lcm}\left ( 1,2, \cdots , n \right )}$$ [i]Proposed by Navid Safaei[/i]

For each $n\in \mathbb{N}$, let $S(n)$ be the sum of all numbers in the set $\{ 1, 2, 3, \cdots , n \}$ which are relatively prime to $n$. $(a)$ Show that $2 \cdot S(n)$ is not a perfect square for any $n$. $(b)$ Given positive integers $m, n$, with odd $n$, show that the equation $2 \cdot S(x) = y^n$ has at least one solution $(x, y)$ among positive integers such that $m|x$.

You are in a strange land and you don’t know the language. You know that ”!” and ”?” stand for addition and subtraction, but you don’t know which is which. Each of these two symbols can be written between two arguments, but for subtraction you don’t know if the left argument is subtracted from the right or vice versa. So, for instance, a?b could mean any of a − b, b − a, and a + b. You don’t know how to write any numbers, but variables and parenthesis work as usual. Given two arguments a and b, how can you write an expression that equals 20a − 18b? (12 points) Nikolay Belukhov

The natural numbers $a_1,a_2,...,a_n, n \ge 12$, are smaller than $9n^2$ and pairwise coprime. Show that at least one of these numbers is prime.

Find all real number $x$ which could be represented as $x = \frac{a_0}{a_1a_2 . . . a_n} + \frac{a_1}{a_2a_3 . . . a_n} + \frac{a_2}{a_3a_4 . . . a_n} + . . . + \frac{a_{n-2}}{a_{n-1}a_n} + \frac{a_{n-1}}{a_n}$ , with $n, a_1, a_2, . . . . , a_n$ are positive integers and $1 = a_0 \leq a_1 < a_2 < . . . < a_n$

Consider the sequence of numbers $\left[n+\sqrt{2n}+\frac12\right]$, where $[x]$ denotes the greatest integer not exceeding $x$. If the missing integers in the sequence are $n_1<n_2<n_3<\ldots$ find $n_{12}$

Prove that for any positive integer $ m$ there exist an infinite number of pairs of integers $(x,y)$ such that $(\text{i})$ $x$ and $y$ are relatively prime; $(\text{ii})$ $x$ divides $y^2+m;$ $(\text{iii})$ $y$ divides $x^2+m.$

Find the largest real number $c$, such that for any integer $s>1$, and positive integers $m, n$ coprime to $s$, we have$$ \sum_{j=1}^{s-1} \{ \frac{jm}{s} \}(1 - \{ \frac{jm}{s} \})\{ \frac{jn}{s} \}(1 - \{ \frac{jn}{s} \}) \ge cs$$ where $\{ x \} = x - \lfloor x \rfloor $.