Prove that there is prime number and natural so $n\times q$ digits are $1$.

Question

I need your helping to prove that there is a prime number $ 2\lt n\in \mathbb N$ and $ 5\neq q\in \mathbb N$ so that the digits of $n\times q$ are only $1$.

for example:if $n=3$ then $3\times 37=111$

if $n=7$ then $7\times 15873=111111 $

also if $2<n\in \mathbb N$ is odd and not divisible by $5$, $q\in \mathbb N$* , so $n\times q$ are only $1$ digits.

Answer 1

Here is an alternative solution to the problem

Given a prime number $p \notin \lbrace 2, 5 \rbrace$, there exists a mutliple $np$ which consists only of ones.

Consider the first $p+1$ postive integers which consit only of ones, i.e. the integers

$$1,11, 111,\dots, \underbrace{1 \cdots 1}_{p+1} $$

There are two among these, whose remainders upon divison by $p$ are equal (since there are only $p$ remainders, but $p+1$ numbers). Hence the diffrence of two such numbers is divisible by $p$ and looks like $$11\cdots 1100\cdots 00 = (11\cdots 11) \cdot 10^k$$ where $k$ is the number of zeros. As $p$ is not equal to 2 or 5, $p$ does not divide $10^k$. But $p$ divides the product $(11\cdots 11) \cdot 10^k$, hence $p$ divides $11\cdots11$.

Answer 2

$\underbrace{11\cdots11}_{( d\text{ digits})}=\dfrac{10^d-1}9$ where $d$ is any positive integer

We need $\dfrac{10^d-1}9\equiv0\pmod p$

For $p=3, 111=3\cdot37$

For $p>5,\dfrac{10^d-1}9\equiv0\pmod p\iff10^d\equiv1\pmod p$

The sufficient condition is $d$ is divisible by $(p-1)$ (from Fermat's Little Theorem)