Prove that the number $ \underbrace{11 \ldots 1}_{(p-1) \mathrm{l}^{\prime} \mathrm{s}} $ is divisible by $p$

Question

Question -

Example 1.30. Let $p \geq 7$ be a prime. Prove that the number $ \underbrace{11 \ldots 1}_{(p-1) \mathrm{l}^{\prime} \mathrm{s}} $ is divisible by $p$

Proof: We have $ \underbrace{11 \ldots 1}_{p-1^{\prime} \mathrm{s}}=\frac{10^{p-1}-1}{9} $ and the conclusion follows from Fermat's little theorem.

But by FLT we get ${10^{p-1}-1} \equiv 0 \pmod{p}$ how we will get that $\frac{10^{p-1}-1}{9} \equiv 0 \pmod{p}$?

Answer 1

If $ak \equiv 0 \pmod n$ and $\gcd(k,n) =1$ then $a \equiv 0\pmod n$.

Pf: If $\gcd(k,n) =1$ then by Bezouts Lemma there are $x,y$ so that $kx + ny = 1$. So $kx \equiv 1 \pmod n$. So $ak \equiv 0 \pmod n$ then $akx \equiv 0x\pmod n$ but $akx\equiv a(kx) \equiv a\pmod n$.

.....

So $p$ is prime and $p \ge 7$ so $\gcd(p,9) = 1$.

And $9|10^{p-1} -1$. And $10^{p-1} \equiv 1 \pmod p$. So......

$10^{p-1}-1= 9*\underbrace{111...1}\equiv 0 \pmod p$.

So $\underbrace{111...1}\equiv 0 \pmod p$.

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Alternatively.

$9|10^p-1 = 9999999....9$ and $p|10^p-1$. So the least common multiple of $9$ and $p$ will also divide $10^9-1$.

The least common multiple of $9$ and $p$ is $9p$ as $\gcd(p,9) =1$.

So $9p|10^p-1$ and so $p|\frac {10^p - 1}9$.

Answer 2

Since $p$ is not divisible by $3$, modulo $p$ we have that division by $9$ corresponds to multiplication by some constant. Which constant will depend on $p$, but for instance, for $p=7$ we have $\frac19\equiv 4$, and for $p=11$ we have $\frac19\equiv 5$.

This gives us $10^{p-1}-1\equiv 0\implies \frac{10^{p-1}-1}{9}\equiv 0$.

Answer 3

Well, simply $$9 | (10^{p-1}-1) \ \ \text{and} \ \ p|(10^{p-1}-1)$$ So, $$10^{p-1}-1=9pk$$ where $\gcd(9,p)=\gcd(9,k)=\gcd(p,k)=1$ for some $k \in \mathbb{Z^{+}}$. So $$\frac{10^{p-1}-1}{9}=pk \implies p|\left(\frac{10^{p-1}-1}{9}\right) \ \Box.$$

Note: $9 | (10^{p-1}-1)$ because of the factoring formula $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+ \dots + xy^{n-2}+y^{n-1})$$ Plugging in $x=10$, $y=1$ and $n=p-1$ gives the conclusion.

Answer 4

By the geometric series sum formula $$ \begin{aligned} \underbrace{11 \ldots 1}_{p-1^{\prime} \mathrm{s}}&=1+10+\dotsb+10^{p-2} =\frac{10^{p-1}-1}{10-1} \end{aligned} $$ Now this is clearly an integer, which means $9\mid 10^{p-1}-1$. Since $p\geq7$, $p\nmid9$, so for $p$ to divide $\underbrace{11 \ldots 1}_{p-1^{\prime} \mathrm{s}} $ we need $p\mid10^{p-1}-1$, which follows since $\gcd(10,p)=1$, and so we may use FLT, $10^{p-1}\equiv1\pmod{p}$. Note it does not matter to us that $9$ divides into $10^{p-1}-1$, since $\gcd(p,9)=1$, so $p$ divides into some part of $\underbrace{11 \ldots 1}_{p-1^{\prime} \mathrm{s}} $ relatively prime to $9$.