Are There Divisibility Rules For Primes?

This problem can be solved by directly applying Fermat's little theorem .

The given number can be written as

$\dfrac19 \left( \underbrace{999999\ldots 9}_{(p-1) \text{ 9's}} \right) = \dfrac19 \left(10^{p-1} - 1 \right)$

Because $p\geq 7$ and $\gcd(10,p) = 1$ , we can apply Fermat's little theorem:

$\begin{aligned} && 10^{p-1} \equiv 1 \pmod p \\ &\Rightarrow& 10^{p-1} - 1 \equiv 0 \pmod p \\ &\Rightarrow& \dfrac19 \left( 10^{p-1} - 1 \right) \equiv 0 \pmod p \\ &\Rightarrow& \underbrace{111111\ldots 1}_{(p-1) \text{ 1's}} \equiv 0 \pmod p \\ \end{aligned}$

From the last congruence above, we know that $\underbrace{111111\ldots 1}_{(p-1) \text{ 1's}}$ gives a remainder of 0 when divided by $p$ , thus the expression in question is divisible by $p$ .

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Food for thought : What primes number(s) smaller than 7 satisfy this condition as well? And which one doesn't? Why?