Prove that a number consisting only of $7's$ is prime one time

Question

$7$ is prime.

However, $77$ is not prime. $777$ is also not prime.

Prove that a number consisting only of $7's$ is only prime one time.

I think this amounts to proving that $7$ is always a factor of a number entirely consisting of $7's,$ for at least two $7's$ in said number.

Answer 1

$7$ is obviously prime.

However, the successive numbers are obviously not. If the number is a repdigit of $n$ $7$'s, for $n\ge 2$, then the number factors into $7 \times r$, where $r$ is a repdigit of $n$ $1$'s.

For example:

• $77 = 7 \times 11$
• $777 = 7 \times 111$
• $7777 = 7 \times 1111$

Answer 2

Notice that the number

$$\begin{align*} \underbrace{7\cdot \cdots\cdot 7}_{\color{red}{n\text{ times}}} &=7\cdot10^{n-1}+7\cdot10^{n-2}+\cdots+7\cdot10^{0}\\\\ &=7\cdot\left(10^{n-1}+\cdots+10^{0}\right)\\\\ &=7k \end{align*}$$

for some $k\in\mathbb{N}$

Answer 3

Not only is a string of 7's not prime ( divisibly by 7). Even if it were, it would need a prime length. Because, if not, divide up the number into substrings of length a factor of its total length. This substring divides into each substring cut once, making it divide the whole number.