All Questions
38
questions
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61
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Subset of natural numbers with largest amount of divisors
Let $n, k \in \mathbb{N}$, with $k \le n$.
Which $k$ natural numbers not greater than $n$ have the largest amount of divisors altogether?
Formally, let $D(x)$ be the set of positive divisors of some $...
0
votes
0
answers
146
views
How do we get the number of prime divisors?
We have a natural square-free number $n$ such that $2^5\cdot 3^6\cdot 5^4\equiv 0 \pmod {\tau(n)}$.
Which is the maximum number of different primes that can divide $n$ ?
$$$$
We have that $\tau(n)$ is ...
- 14k
8
votes
3
answers
505
views
How many numbers are there such that its number of decimal digits equals to the number of its distinct prime factors?
Problem
A positive integer is said to be balanced if the number of its decimal digits equals the number of its distinct prime factors. For instance, $15$ is balanced, while $49$ is not. How many ...
- 3,966
1
vote
1
answer
57
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When is the number of Divisors of a Number equivalent to one of its Factors?
My math teacher asked me this problem for homework and I am unsure how to solve it.
Which numbers contain a number of factors equivalent to the value of one of their divisors?
I found that 8 works, ...
- 171
-1
votes
2
answers
50
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Prove that $a|p+1$ if and only if ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m} }$ such that $n,m \in N$ and $p$ is prime [duplicate]
Let $p$ be a prime and let $n$ be a natural number such that $n \gt a$ .
Prove that $a|p+1$ if and only if exists integers $n,m$
Such that ${\frac{a}{p} = \frac{1}{n} + \frac{1}{m}}$
- 9
-2
votes
2
answers
159
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Condition in Type of Prime Factors of Consecutive Integers
We define a odd-prime $p$ as $i$-type prime if $p \equiv - i \pmod q$ where $ 1 \leq i \leq q-1$ (see similar definition on page 24, CHAPTER 2, of the book "Summing It Up" by Avner Ash ...
12
votes
3
answers
462
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On the diophantine equation $x^{m-1}(x+1)=y^{n-1}(y+1)$ with $x>y$, over integers greater or equal than two
I don't know if the following diophantine equation (problem) is in the literature. We consider the diophantine equation $$x^{m-1}(x+1)=y^{n-1}(y+1)\tag{1}$$ over integers $x\geq 2$ and $y\geq 2$ with $...
8
votes
1
answer
635
views
How many integers are there that are not divisible by any prime larger than 20 and not divisible by the square of any prime?
I tackled the problem in the following way but i'm not sure if i'm correct.
I need the count of the numbers that have in their prime factorization only primes p such that $p \lt 20$ and those ...
0
votes
1
answer
196
views
Calculate all prime numbers $x$, where $x^{18} - 1$ is divisible by $28728$
Question: Calculate all prime numbers $x$, where $x^{18} - 1$ is divisible by $28728$
Apparently, the answer is all prime numbers except $2, 3, 7,$ and $19.$ I did some prime factorisation and found ...
- 315
5
votes
1
answer
195
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Find all $n$ such that $n/d(n) = p$, a prime, where $d(n)$ is the number of positive divisors of $n$
Let $d(n)$ denote the number of positive divisors of $n$. Find all $n$ such that $n/d(n) = p$, a prime.
I tried this, but only I could get two solutions.
I proceeded like this -
Suppose
$$n = p^r \...
user636268
3
votes
3
answers
185
views
Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime
Let $p>3$ be a prime number. How to prove that $\left(p^2+1\right)^2$ has no divisors congruent to $1 \bmod p$, except the trivial ones $1$, $p^2+1$, and $\left(p^2+1\right)^2$?
When $p=3$, you ...
- 51
1
vote
1
answer
58
views
Are there exactly $d$ distinct remainders when $x^{\frac{p-1}{d}}$ is divided by $p$?
Q1: Is it true that if $d$, if $d|p-1$, where $p$ a prime, then there for all $(x,p) = 1$, are exactly $d$ distinct remainders when $x^{\frac{p-1}{d}}$ is divided by $p$?
Q2: For what composite ...
- 20.8k
2
votes
1
answer
142
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The Chinese hypothesis revisited
In the past I tried to get different variations of the so-called Chinese hypothesis, see this Wikipedia (a disproven conjecture).
Today I wanted to combine in an artificious way also Wilson-Lagrange ...
user243301
0
votes
2
answers
66
views
Fractions of the form $\frac{a}{k}\cdot\frac{b}{k}\cdot\frac{c}{k}\cdots=\frac{n}{k}$
For any $n,k\in\mathbb{Z}^+$, $n\gt k$. I found that the majority of the time, either there are no positive integers, $a,b,c\dots,$ such that
$$\frac{a}{k}\cdot\frac{b}{k}\cdot\frac{c}{k}\cdots=\...
- 3,557
1
vote
0
answers
48
views
When $f(x)$ divides $d$ $f(x)=d(c+2ax+dx^2)\mod{N}$
Given $f(0)$ divides $d$ and $f(1)$ not, how to find other $x$ values that make $f(x)$ divisible by $d$?
$$f(x)=d(c+2ax+dx^2)\mod{N}$$
$a,c,d,x,N$ are positive integers
$c$ is a small number
$d$ is ...
- 1,450