Find all prime numbers $p$ for which number $p^2 + 11$ has exactly six different divisors (counting $1$ and itself).

Is it true that every positive integer greater than 30 is a sum of 4 positive integers such that each two of them have a common divisor greater than 1?

Find the least natural number $n$, which has at least 6 different divisors $1=d_1<d_2<d_3<d_4<d_5<d_6<...$, for which $d_3+d_4=d_5+6$ and $d_4+d_5=d_6+7$.

For a natural number $n$, let $D (n)$ be the set of (positive integers) divisors of $n$. Furthermore let $d (n)$ be the number of divisors of $n,$ that is, the cardinality of $D (n)$. For each such $n$, prove the equality $$\sum_{k\in D(n)} d(k)^3=\left( \sum_{k\in D(n)} d(k)\right) ^2.$$

On a piece of paper, we write down all positive integers $n$ such that all proper divisors of $n$ are less than $30$. We know that the sum of all numbers on the paper having exactly one proper divisor is $2397$. What is the sum of all numbers on the paper having exactly two proper divisors? We say that $k$ is a proper divisor of the positive integer $n$ if $k | n$ and $1 < k < n$.

Determine all pairs $(a, b)$ of integers having the following property: there is an integer $d \ge 2$ such that $a^n + b^n + 1$ is divisible by $d$ for all positive integers $n$.

The number $4$ has an odd number of odd positive divisors, namely $1$, and an even number of even positive divisors, namely $2$ and $4$. Is there a number with an odd number of even positive divisors and an even number of odd positive divisors?

Determine the smallest prime that does not divide any five-digit number whose digits are in a strictly increasing order.

Is there such a natural number that the product of all its natural divisors (including $1$ and the number itself) ends exactly in $2001$ zeros?

Prove that for any integers $ n> m> 0 $ the number $ 2 ^n-1 $ has a prime divisor not dividing $ 2 ^m-1 $.

Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the number of divisors of $sn$ and of $sk$ are equal.

Every positive integer is either [i]nice [/i] or [i]naughty[/i], and the Oracle of Numbers knows which are which. However, the Oracle will not directly tell you whether a number is [i]nice [/i] or [i]naughty[/i]. The only questions the Oracle will answer are questions of the form “What is the sum of all nice divisors of $n$?,” where $n$ is a number of the questioner’s choice. For instance, suppose ([i]just [/i] for this example) that $2$ and $3$ are nice, while $1$ and $6$ are [i]naughty[/i]. In that case, if you asked the Oracle, “What is the sum of all nice divisors of $6$?,” the Oracle’s answer would be $5$. Show that for any given positive integer $n$ less than $1$ million, you can determine whether $n$ is [i]nice [/i] or [i]naughty [/i] by asking the Oracle at most four questions.

Let a pair of positive integers $(n, m)$ that are relatively prime be called [i]intertwined[/i] if among any two divisors of $n$ greater than $1$, there exists a divisor of $m$ and among any two divisors of $m$ greater than $1$, there exists a divisor of $n$. For example, pair $(63, 64)$ is intertwined. [b]a)[/b] Find the largest integer $k$ for which there exists an intertwined pair $(n, m)$ such that the product $nm$ is equal to the product of the first $k$ prime numbers. [b]b)[/b] Prove that there does [b]not[/b] exist an intertwined pair $(n, m)$ such that the product $nm$ is the product of $2025$ distinct prime numbers. [b]c)[/b] Prove that there exists an intertwined pair $(n, m)$ such that the number of divisors of $n$ is greater than $2025$. [i]Proposed by Stijn Cambie, Belgium[/i]

Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the number of divisors of $sn$ and of $sk$ are equal.

Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the number of divisors of $sn$ and of $sk$ are equal.

For a positive integer $n$, we dene $D_n$ as the largest integer that is a divisor of $a^n + (a + 1)^n + (a + 2)^n$ for all positive integers $a$. 1. Show that for all positive integers $n$, the number $D_n$ is of the form $3^k$ with $k \ge 0$ an integer. 2. Show that for all integers $k \ge 0$ there exists a positive integer n such that $D_n = 3^k$.

Let $n$ be a positive integer. Let $\sigma(n)$ be the sum of the natural divisors $d$ of $n$ (including $1$ and $n$). We say that an integer $m \geq 1$ is [i]superabundant[/i] (P.Erdos, $1944$) if $\forall k \in \{1, 2, \dots , m - 1 \}$, $\frac{\sigma(m)}{m} >\frac{\sigma(k)}{k}.$ Prove that there exists an infinity of [i]superabundant[/i] numbers.

Find all positive integers $a, b,c$ greater than $1$, such that $ab + 1$ is divisible by $c, bc + 1$ is divisible by $a$ and $ca + 1$ is divisible by $b$.

Let $n$ be a given positive integer. Determine all positive divisors $d$ of $3n^2$ such that $n^2 + d$ is the square of an integer.

For a positive integer $n$, let $d(n)$ be the number of all positive divisors of $n$. Find all positive integers $n$ such that $d(n)^3=4n$.

A positive integer $n{}$ is called [i]discontinuous[/i] if for all its natural divisors $1 = d_0 < d_1 <\cdots<d_k$, written out in ascending order, there exists $1 \leqslant i \leqslant k$ such that $d_i > d_{i-1}+\cdots+d_1+d_0+1$. Prove that there are infinitely many positive integers $n{}$ such that $n,n+1,\ldots,n+2019$ are all discontinuous.

Let $d_1, d_2, ... , d_k$ be the positive divisors of $n = 1990!$. Show that $\sum \frac{d_i}{\sqrt{n}} = \sum \frac{\sqrt{n}}{d_i}$.

Given a natural number $n$. Prove that you can choose $ \phi (n)+1 $ (not necessarily different) divisors $n$ with the sum $n$. Here $ \phi (n)$ denotes the number of natural numbers less than $n$ that are coprime with $n$. (Fedir Yudin)

Determine all pairs $(n, k)$ of distinct positive integers such that there exists a positive integer $s$ for which the number of divisors of $sn$ and of $sk$ are equal.

A natural number $A$ is given. One may add to it one of its divisors $d$ ($1 < d < A$). One may then repeat this operation with the new number $A + d$ and so on. Prove that starting from $A = 4$ one can get any composite number by these operations. (M Vyalyi)