Questions tagged [repunit-numbers]
For questions about repunit numbers, that is, numbers that contain only the digit 1.
65 questions
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Classification of perfect squares with digits strictly less than 7 such that adding 3 to each digit yields another square
Here is the problem I am investigating:I am looking for all perfect squares $N$ (in base 10) that satisfy two conditions:Digit Constraint: Every digit of $N$ is strictly less than $7$ (i.e., digits $\...
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For $a > 0$, $b > 0$, $p \neq 3$ and $gcd(a,b) = 1$, how to prove than $((a^2)^{p}+(ab)^p+(b^2)^{p})/(a^2+ab+b^2)$ is prime only if $p$ is prime?
This is something I noticed while playing with some factors of repunits.
For $a > 0$, $b > 0$, $p \neq 3$ or $p$ different of a multiple of $3$, $gcd(a,b) = 1$ and $a > b$, how to prove than $...
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The chunking aspect of repunit prime factors [closed]
While others have already mentioned the divisibility of decimal repunits,
$$m := \frac{{10}^n - 1}{9}.$$
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Is there a repdigit that is powerful number(Achilles numbers)?
Is there a repdigit that is powerful number(Achilles numbers)?
Powerful numbers(Achilles Numbers) are the numbers whose exponents in prime factors is greater than $1$, but not possible to write it as ...
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Proof of a divisibility property of repunit products
A repunit is a number that contains only the digit 1 under base 10. Define the $k$-th repunit to be $r(k):=\sum_{i=1}^{k} 10^{i-1}$. Show that
$$\left(\prod_{k=1}^m r(k)\right)\left(\prod_{k=1}^n r(k)\...
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Proving 111...111 is divisible by $3^n$ [duplicate]
Problem: Show that the number $11...11$ with $3^n$ digits is divisible by $3^n$.
My Attempt I tried solving this problem on base 3. It is easy to see that
$$3^n = 100....00$$ (upto $n$ 0's)
And
$$ 11.....
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About repunit and Mersenne numbers
Let $r_n=11..11$ be the repunit of 1 repeated n times. In base 10, we have $r_n= (11..11)_{10}=\frac 19 (10^n-1)$ with a repetition of 1 n times. I would like to prove that if n is composite, then $...
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Divisibility of 100 digits number [duplicate]
prove that the $100$ digits number $1.....11$ is divisible by $101$
To show that the $100$-digit number consisting of $1$'s only, i.e. $11...1$ (with $100$ digits), is divisible by $101$, we can use ...
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Basic Repunit divisibility proof [duplicate]
Let $R_n$ denote the repunit with $n$ ones. So $R_1=1, R_2=11, R_3=111, ...$.
I know that $R_n = (10^n - 1)/9$ and I want to show that
$R_n$ divides $R_m$ iff $n$ divides $m$.
How would I go about ...
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On some conjectures regarding repunits
While researching the topic of Descartes numbers, I came across the following seemingly related subproblem:
PROBLEM: Determine conditions on $n$ such that
$$\frac{{10}^n - 1}{9}$$
is squarefree.
MY ...
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For any fixed integer $ a \gt 1 $, how do you prove that $\frac{a^p-1}{a-1}$ is not always prime given prime $ p \not \mid a-1$?
I assumed this would be easy to prove but it turned out to be quite hard since the go to methods don't work on this problem.
Once we fix any $a\gt 1$, we need an algorithm to produce a prime $p$ that ...
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$n$- relatively prime with $10$, then show that there exists another natural number $m$ such that all its digits are $1'$s and $m$ is divisible by $n$ [duplicate]
If there is a natural number $n$ relatively prime with $10$, then show that there exists another natural number $m$ such that all its digits are $1'$s and $m$ is divisible by $n$.
Approach:
Let the ...
user907745
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Prime Repunit Numbers [duplicate]
proof that if a repunit number is prime n has to be prime
So a repunit number is a number that it's all digits are 1. For example $R_{2} = 11$ $R_{7} = 1111111$ and so on. Repunit numbers can be ...
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A similar (little) Fermat's Theorem result and repunit multiple
I got this exercise in arithmetic class (I'm a french student but let me translate the problem)
In this thread I only talk about questions from question 2) on the paper.
Let n and p be 2 integers ...
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Let $a_n = 1 . . . 1 $ with $3^n$ digits. Prove that $a_n$ is divisible by $3a_{n−1}$. [closed]
Let $a_n = 1 . . . 1 $ with $3^n$ digits. Prove that $a_n$ is divisible by $3a_{n−1}$.
Is there any way to solve this question without mathematical induction?
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