Find the smallest two-digit positive integer that is a divisor of 201020112012.

To every natural number $k, k \geq 2$, there corresponds a sequence $a_n(k)$ according to the following rule: \[a_0 = k, \qquad a_n = \tau(a_{n-1}) \quad \forall n \geq 1,\] in which $\tau(a)$ is the number of different divisors of $a$. Find all $k$ for which the sequence $a_n(k)$ does not contain the square of an integer.

Prove: For each positive integer is the number of divisors whose decimal representations ends with a 1 or 9 not less than the number of divisors whose decimal representations ends with 3 or 7.

Let \( a_1 \) be an integer greater than or equal to 2. Consider the sequence such that its first term is \( a_1 \), and for \( a_n \), the \( n \)-th term of the sequence, we have \[ a_{n+1} = \frac{a_n}{p_k^{e_k - 1}} + 1, \] where \( p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \) is the prime factorization of \( a_n \), with \( 1 < p_1 < p_2 < \cdots < p_k \), and \( e_1, e_2, \dots, e_k \) positive integers. For example, if \( a_1 = 2024 = 2^3 \cdot 11 \cdot 23 \), the next two terms of the sequence are \[ a_2 = \frac{a_1}{23^{1-1}} + 1 = \frac{2024}{1} + 1 = 2025 = 3^4 \cdot 5^2; \] \[ a_3 = \frac{a_2}{5^{2-1}} + 1 = \frac{2025}{5} + 1 = 406. \] Determine for which values of \( a_1 \) the sequence is eventually periodic and what all the possible periods are. [b]Note:[/b] Let \( p \) be a positive integer. A sequence \( x_1, x_2, \dots \) is eventually periodic with period \( p \) if \( p \) is the smallest positive integer such that there exists an \( N \geq 0 \) satisfying \( x_{n+p} = x_n \) for all \( n > N \).

For each positive integer $n$ larger than $1$ with prime factorization $p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$, its [i]signature[/i] is defined as the sum $\alpha_1+\alpha_2+\cdots+\alpha_k$. Does there exist $2020$ consecutive positive integers such that among them, there are exactly $1812$ integers whose signatures are strictly smaller than $11$?

Let $n > 6$ be a perfect number. Let $p_1^{a_1} \cdot p_2^{a_2} \cdot ... \cdot p_k^{a_k}$ be the prime factorisation of $n$, where we assume that $p_1 < p_2 <...< p_k$ and $a_i > 0$ for all $ i = 1,...,k$. Prove that $a_1$ is even. Remark: An integer $n \ge 2$ is called a perfect number if the sum of its positive divisors, excluding $ n$ itself, is equal to $n$. For example, $6$ is perfect, as its positive divisors are $\{1, 2, 3, 6\}$ and $1+2+3=6$.

Let $ m_1$, $ m_2$, $ \cdots$, $ m_r$ (may not distinct) and $ n_1$, $ n_2$ $ \cdots$, $ n_s$ (may not distinct) be two groups of positive integers such that for any positive integer $ d$ larger than $ 1$, the numbers of which can be divided by $ d$ in group $ m_1$, $ m_2$, $ \cdots$, $ m_r$ (including repeated numbers) are no less than that in group $ n_1$, $ n_2$ $ \cdots$, $ n_s$ (including repeated numbers). Prove that $ \displaystyle \frac{m_1 \cdot m_2 \cdots m_r}{n_1 \cdot n_2 \cdots n_s}$ is integer.

Let us call the number of factors in the prime decomposition of an integer $n > 1$ the complexity of $n$. For example, [i]complexity [/i] of numbers $4$ and $6$ is equal to $2$. Find all $n$ such that all integers between $n$ and $2n$ have complexity a) not greater than the complexity of $n$. b) less than the complexity of $n$. (Boris Frenkin)

Two players $A$ and $B$ play the following game. Before the game starts, $A$ chooses 1000 not necessarily different odd primes, and then $B$ chooses half of them and writes them on a blackboard. In each turn a player chooses a positive integer $n$, erases some primes $p_1$, $p_2$, $\dots$, $p_n$ from the blackboard and writes all the prime factors of $p_1 p_2 \dotsm p_n - 2$ instead (if a prime occurs several times in the prime factorization of $p_1 p_2 \dotsm p_n - 2$, it is written as many times as it occurs). Player $A$ starts, and the player whose move leaves the blackboard empty loses the game. Prove that one of the two players has a winning strategy and determine who. Remark: Since 1 has no prime factors, erasing a single 3 is a legal move.

Let $A$ be the greatest possible value of a product of positive integers that sums to $2014$. Compute the sum of all bases and exponents in the prime factorization of $A$. For example, if $A=7\cdot 11^5$, the answer would be $7+11+5=23$.

Find all the positive integers $p$ with the property that the sum of the first $p$ positive integers is a four-digit positive integer whose decomposition into prime factors is of the form $2^m3^n(m + n)$, where $m, n \in N^*$.

Find all surjective functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $m,n\in \mathbb{N}$: \[m \vert n \Longleftrightarrow f(m) \vert f(n).\]

For any positive integer $n$, let $\tau (n)$ denote the number of its positive divisors (including 1 and itself). Determine all positive integers $m$ for which there exists a positive integer $n$ such that $\frac{\tau (n^{2})}{\tau (n)}=m$.

If you multiply all positive integer factors of $24$, you get $24^x$. Find $x$.

Find four positive integers each not exceeding $70000$ and each having more than $100$ divisors.

Four dice are thrown. What is the probability that the product of the number equals $ 36?$

Let $n$ a positive integer, $n > 1$. The number $n$ is wonderful if the number is divisible by sum of the your prime factors. For example; $90$ is wondeful, because $90 = 2 \times 3^2\times 5$ and $2 + 3 + 5 = 10, 10$ divides $90$. Show that, exist a number "wonderful" with at least $10^{2002}$ distinct prime numbers.

Let $n$ be a given number greater than 2. We consider the set $V_n$ of all the integers of the form $1 + kn$ with $k = 1, 2, \ldots$ A number $m$ from $V_n$ is called indecomposable in $V_n$ if there are not two numbers $p$ and $q$ from $V_n$ so that $m = pq.$ Prove that there exist a number $r \in V_n$ that can be expressed as the product of elements indecomposable in $V_n$ in more than one way. (Expressions which differ only in order of the elements of $V_n$ will be considered the same.)

Let $Q$ be a set of prime numbers, not necessarily finite. For a positive integer $n$ consider its prime factorization: define $p(n)$ to be the sum of all the exponents and $q(n)$ to be the sum of the exponents corresponding only to primes in $Q$. A positive integer $n$ is called [i]special[/i] if $p(n)+p(n+1)$ and $q(n)+q(n+1)$ are both even integers. Prove that there is a constant $c>0$ independent of the set $Q$ such that for any positive integer $N>100$, the number of special integers in $[1,N]$ is at least $cN$. (For example, if $Q=\{3,7\}$, then $p(42)=3$, $q(42)=2$, $p(63)=3$, $q(63)=3$, $p(2022)=3$, $q(2022)=1$.)

Find all positive integers $n$ such that the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ has exactly $2011$ positive integer solutions $(x,y)$ where $x \leq y$.

Let $Q$ be a set of prime numbers, not necessarily finite. For a positive integer $n$ consider its prime factorization: define $p(n)$ to be the sum of all the exponents and $q(n)$ to be the sum of the exponents corresponding only to primes in $Q$. A positive integer $n$ is called [i]special[/i] if $p(n)+p(n+1)$ and $q(n)+q(n+1)$ are both even integers. Prove that there is a constant $c>0$ independent of the set $Q$ such that for any positive integer $N>100$, the number of special integers in $[1,N]$ is at least $cN$. (For example, if $Q=\{3,7\}$, then $p(42)=3$, $q(42)=2$, $p(63)=3$, $q(63)=3$, $p(2022)=3$, $q(2022)=1$.)

Denote by $d(n)$ the number of divisors of the positive integer $n$. A positive integer $n$ is called highly divisible if $d(n) > d(m)$ for all positive integers $m < n$. Two highly divisible integers $m$ and $n$ with $m < n$ are called consecutive if there exists no highly divisible integer $s$ satisfying $m < s < n$. (a) Show that there are only finitely many pairs of consecutive highly divisible integers of the form $(a, b)$ with $a\mid b$. (b) Show that for every prime number $p$ there exist infinitely many positive highly divisible integers $r$ such that $pr$ is also highly divisible.

Let $t(n)$ be the sum of the digits in the binary representation of a positive integer $n,$ and let $k \geq 2$ be an integer. [b]a.[/b] Show that there exists a sequence $(a_i)_{i=1}^{\infty}$ of integers such that $a_m \geq 3$ is an odd integer and $t(a_1a_2 \cdots a_m)=k$ for all $m \geq 1.$ [b]b.[/b] Show that there is an integer $N$ such that $t(3 \cdot 5 \cdots (2m+1))>k$ for all integers $m \geq N.$

For any integer $k\ge1$, let $p(k)$ be the smallest prime which does not divide $k$. Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$, and $X(k)=1$ if $p(k)=2$. Let $\{x_n\}$ be the sequence defined by $x_0=1$, and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\ge0$. Find the smallest positive integer, $t$ such that $x_t=2090$.

The ${\textit{distinct prime factors}}$ of an integer are its prime factors listed without repetition. For example, the distinct prime factors of $40$ are $2$ and $5$. Let $A=2^k - 2$ and $B= 2^k \cdot A$, where $k$ is an integer ($k \ge 2$). Show that, for every integer $k$ greater than or equal to $2$, [list=i] [*] $A$ and $B$ have the same set of distinct prime factors. [*] $A+1$ and $B+1$ have the same set of distinct prime factors. [/list]