Found problems: 26
Let $({{x}_{n}})$ be an integer sequence such that $0\le {{x}_{0}}<{{x}_{1}}\le 100$ and
$${{x}_{n+2}}=7{{x}_{n+1}}-{{x}_{n}}+280,\text{ }\forall n\ge 0.$$
a) Prove that if ${{x}_{0}}=2,{{x}_{1}}=3$ then for each positive integer $n,$ the sum of divisors of the following number is divisible by $24$ $${{x}_{n}}{{x}_{n+1}}+{{x}_{n+1}}{{x}_{n+2}}+{{x}_{n+2}}{{x}_{n+3}}+2018.$$
b) Find all pairs of numbers $({{x}_{0}},{{x}_{1}})$ such that ${{x}_{n}}{{x}_{n+1}}+2019$ is a perfect square for infinitely many nonnegative integer numbers $n.$
[b]Q[/b] Let $n\geq 2$ be a fixed positive integer and let $d_1,d_2,...,d_m$ be all positive divisors of $n$. Prove that:
$$\frac{d_1+d_2+...+d_m}{m}\geq \sqrt{n+\frac{1}{4}}$$Also find the value of $n$ for which the equality holds.
[i]Proposed by dangerousliri [/i]
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.
Let $X$ be the sum of all divisors of the number $(3\cdot 2009)^{((2\cdot 2009)^{2009}-1)}$ . Find the last digit of $X$.
Prove that for each $n \ge 3$ there exist $n$ distinct positive divisors $d_1,d_2, ...,d_n$ of $n!$ such that $n! = d_1 +d_2 +...+d_n$.
Assuming that the positive integer $a$ is not a perfect square, prove that for any positive integer n, the sum ${S_{n}=\sum_{i=1}^{n}\{a^{\frac{1}{2}}\}^{i}}$ is irrational.
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.
Do there exist infinitely many positive integers $m$ such that the sum of the positive divisors of $m$ (including $m$ itself) is a perfect square?
[i]Proposed by Dylan Toh[/i]
Let $n$ be a positive integer such that $24\mid n+1$. Prove that the sum of the positive divisors of $n$ is divisble by 24.
For every positive integer $n$, $\sigma(n)$ is equal to the sum of all the positive divisors of $n$ (for example, $\sigma(6)=1+2+3+6=12$) . Find the solution of the equation $$\sigma(p^2)=\sigma(q^b)$$ where $p$ and $q$ are primes where $p>q$ and $b$ are positive integers.
[i](gools)[/i]
Given a positive integer $k$. Prove that there are infinitely many positive integers $n$ satisfy the following conditions at the same time:
a) $n$ has at least $k$ distinct prime divisors
b) All prime divisors other than $3$ of $n$ have the form $4t+1$, with $t$ some positive integer.
c) $n | 2^{\sigma(n)}-1$
Here $\sigma(n)$ demotes the sum of the positive integer divisors of $n$.
An integer $n\ge2$ is called [i]resistant[/i], if it is coprime to the sum of all its divisors (including $1$ and $n$).
Determine the maximum number of consecutive resistant numbers.
For instance:
* $n=5$ has sum of divisors $S=6$ and hence is resistant.
* $n=6$ has sum of divisors $S=12$ and hence is not resistant.
* $n=8$ has sum of divisors $S=15$ and hence is resistant.
* $n=18$ has sum of divisors $S=39$ and hence is not resistant.
Determine all integers $n>1$ whose positive divisors add up to a power of $3.$
[i]Andrei Bâra[/i]
Determine all integers $n>1$ whose positive divisors add up to a power of $3.$
[i]Andrei Bâra[/i]
A natural number $n$ is said to be [i]abundant [/i] if the sum of its divisors is greater than $2n$. For example, $18$ is abundant because the sum of its divisors, $1 + 2 + 3 + 6 + 9 + 18$, is greater than $36$. Write every even number greater than $46$ as a sum of two numbers abundant.
Determine all polynomials $P(x)$ with integer coefficients such that $P(n)$ is divisible by $\sigma(n)$ for all positive integers $n$. (As usual, $\sigma(n)$ denotes the sum of all positive divisors of $n$.)
Do there exist infinitely many positive integers $m$ such that the sum of the positive divisors of $m$ (including $m$ itself) is a perfect square?
[i]Proposed by Dylan Toh[/i]
For a positive integer $n$ we denote by $d(n)$ the number of positive divisors of $n$ and by $s(n)$ the sum of these divisors. For example, $d(2018)$ is equal to $4$ since $2018$ has four divisors $(1, 2, 1009, 2018)$ and $s(2018) = 1 + 2 + 1009 + 2018 = 3030$.
Determine all positive integers $x$ such that $s(x) \cdot d(x) = 96$.
(Richard Henner)
Given $N = 2^ap_1p_2...p_m$, $m \ge 1$, $a \in N$ with $p_1, p_2,..., p_m$ are different primes. It is known that $\sigma (N) = 3N $ where $\sigma (N)$ is the sum of all positive integers which are factors of $N$. Show that there exists a prime number $p$ such that $2^p- 1$ is also a prime, and $2^p - 1|N$.
Given two positives integers $m$ and $n$, prove that there exists a positive integer $k$ and a set $S$ of at least $m$ multiples of $n$ such that the numbers $\frac {2^k{\sigma({s})}} {s}$ are odd for every $s \in S$. $\sigma({s})$ is the sum of all positive integers of $s$ (1 and $s$ included).
A positive integer \(n\) is called perfect if the sum of its positive divisors \(\sigma(n)\) is twice \(n\), that is, \(\sigma(n) = 2n\). For example, \(6\) is a perfect number since the sum of its positive divisors is \(1 + 2 + 3 + 6 = 12\), which is twice \(6\). Prove that if \(n\) is a positive perfect integer, then:
\[
\sum_{p|n} \frac{1}{p + 1} < \ln 2 < \sum_{p|n} \frac{1}{p - 1}
\]
where the sums are taken over all prime divisors \(p\) of \(n\).
Given $N = 2^ap_1p_2...p_m$, $m \ge 1$, $a \in N$ with $p_1, p_2,..., p_m$ are different primes. It is known that $\sigma (N) = 3N $ where $\sigma (N)$ is the sum of all positive integers which are factors of $N$. Show that there exists a prime number $p$ such that $2^p- 1$ is also a prime, and $2^p - 1|N$.
For each positive integer $k$ let $a_k$ be the largest divisor of $k$ which is not divisible by $3$. Let $s_n=a_1+a_2+\dots+a_n$. Show that:
(a) The number $s_n$ is divisible by $3$ iff the number of ones in the ternary expansion of $n$ is divisible by $3$.
(b) There are infinitely many $n$ for which $s_n$ is divisible by $3^3$.
Let $\sigma(n)$ denote the sum of positive divisors of a number $n$. A positive integer $N=2^r b$ is given, where $r$ and $b$ are positive integers and $b$ is odd. It is known that $\sigma(N)=2N-1$. Prove that $b$ and $\sigma(b)$ are coprime.
(J. Antalan, J. Dris)
For a positive integer $n$, $\sigma(n)$ denotes the sum of the positive divisors of $n$. Determine $$\limsup\limits_{n\rightarrow \infty} \frac{\sigma(n^{2023})}{(\sigma(n))^{2023}}$$
[b]Note:[/b] Given a sequence ($a_n$) of real numbers, we say that $\limsup\limits_{n\rightarrow \infty} a_n = +\infty$ if ($a_n$) is not upper bounded, and, otherwise, $\limsup\limits_{n\rightarrow \infty} a_n$ is the smallest constant $C$ such that, for every real $K > C$, there is a positive integer $N$ with $a_n < K$ for every $n > N$.