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Questions tagged [repunit-numbers]

For questions about repunit numbers, that is, numbers that contain only the digit 1.

Repunit numbers are the numbers of the form: $$R_n=\frac{10^{n}-1}{9}=\underbrace{111...111}_{n\text{ times}}$$ So $$R_1=1, R_2=11, R_3=111, R_4=1111,...$$

If $R_n$ is a prime number, $n$ is a prime number.
The first Repunit primes are:

$R_2,\ R_{19},\ R_{23},\ R_{317},\ R_{1031},\ R_{49081}, R_{86453},\ ...$ sequence A004023 in OEIS

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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 makes $\frac{a^p-1}{a-1}$ composite and $a \not…
arbashn
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Prove that none of $\{11, 111, 1111,\dots \}$ is the perfect square of an integer

Please help me with solving this : prove that none of $\{11, 111, 1111 \ldots \}$ is the square of any $x\in\mathbb{Z}$ (that is, there is no $x\in\mathbb{Z}$ such that $x^2\in\{11, 111, 1111, \ldots\}$).
Pechenka
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Why do repunit primes have only a prime number of consecutive $1$s?

Repunit primes are primes of the form $\frac{10^n - 1}{9} = 1111\dots11 \space (n-1 \space ones)$. Each repunit prime is denoted by $R_i$, where $i$ is the number of consecutive $1$s it has. So far, very few of these have been found: $R_2, R_{19},…
Gerard
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Prove that every number ending in a $3$ has a multiple which consists only of ones.

Prove that every number ending in a $3$ has a multiple which consists only of ones. Eg. $3$ has $111$, $13$ has $111111$. Also, is their any direct way (without repetitive multiplication and checking) of obtaining such multiple of any given…
Eight
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Is $1111111111111111111111111111111111111111111111111111111$ ($55$ $1$'s) a composite number?

This is an exercise from a sequence and series book that I am solving. I tried manipulating the number to make it easier to work with: $$111...1 = \frac{1}9(999...) = \frac{1}9(10^{55} - 1)$$ as the number of $1$'s is $55$. The exercise was under…
user69284
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Prove that 17 divides 1111111111111111 (16 1's) and doesn't divide 11111111

I need to prove that $17$ divides $\underbrace{1111111111111111}_{\text{16 1's}}$ and doesn't divide $\underbrace{11111111}_{\text{8 1's}}$ by using congruence. I know that $\underbrace{1111111111111111}_{\text{16 1's}}= \frac{10^{16}-1}{9}$ and…
ematth7
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Can $7$ be the smallest prime factor of a repunit?

Repunits are numbers whose digits are all $1$. In general, finding the full prime factorization of a repunit is nontrivial. Sequence A067063 in the OEIS gives the smallest prime factor of repunits. There are no $7$'s (just the number $7$, not as a…
Vincent Luo
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What is $\underbrace{555\cdots555}_{1000\ \text{times}} \ \text{mod} \ 7$ without a calculator

It can be calculated that $\frac{555555}{7} = 79365$. What is the remainder of the number $5555\dots5555$ with a thousand $5$'s, when divided by $7$? I did the following: $$\begin{array} & 5 \ \text{mod} \ 7=& &5 \\ 55 \ \text{mod} \ 7= & &6…
Adnan
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Proving that none of these elements 11, 111, 1111, 11111...can be a perfect square

How can i prove that no number in set S S = {11, 111, 1111, 11111...} Is a perfect square. I have absolutely no idea how to tackle this problem i tried rewriting it in powers of 10 but that didn't really get me anywhere... Thanks in advance.
user2396852
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Prove that $11,111,111.....$ is not a perfect square

I know there are many other same related questions, but as my profile says I don't have a mentor, I just wanted you guys to check my solution. First I noticed that, $11$ = $10 + 1$, $111$ = $10^2+10+1$, Similarly,…
Kritesh Dhakal
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Why primes $\neq 2,5$ divide infinitely many repunits? (cyclic permutation orbits)

I first noticed this is true for the integers in the sequence $9, 99, 999, 9999,\dots$, since for some term $a_n=10^n-1$ in the sequence and $p$ a prime other than $2$ or $5$, we have $a_n\equiv 0 \pmod p\!\iff\! 10^n\equiv 1 \pmod p$ and we can…
yunone
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Repunit Divisibility

We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$. Lemma: Let $n$ be a positive integer and $GCD(10,n) = 1$. Then, there exists a $k$ such that $R(k) \equiv 0$ $ $ $ $ $mod$ $(n)$. I 'm asking this question because I…
aeyalcinoglu
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When is the number $11111\cdots1$ a prime number?

For which $n$ is the sum: $$\sum_{k=0}^{n}10^k$$ a prime number? Are they finite?
user59671
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Is the repunit $R_{3^k}$ divisible by $3^k$?

By the repunit $R_n$ is meant the base ten positive integer consisting of $n$ digits all 1. My question is whether when $n=3^k$ we have the repunit divisible by $3^k$. I checked up to $R_{243}$, a moderate check I realize, but wasn't sure of my…
coffeemath
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How to find the values of $n$ for which $\displaystyle\underbrace{111\cdots1111}_{n}\equiv0 \pmod {41}$?

What are the values of $n$ satisfying $\displaystyle\underbrace{111\cdots1111}_{n}\equiv0 \pmod {41}$? I think $n=5k$, with $k=1,2,\cdots$, but I can't prove it.
math110
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