Prove that $2005|\underbrace{55 \ldots5}_{800\text{ digits}}$

Question

Prove that $2005|\underbrace{55 \ldots 5}_{800\text{ digits}}$

I know that $2005=5\cdot 401$ since $55 \ldots 5$ is divisibility with $5$ i only need to prove that $55 \ldots 5$ is divisibility with 401.

$55 \ldots 5=5(10^{799}+10^{798}+\cdots+10+1)$, then I can find a remainder for example $10^3\equiv198 \pmod {401}$ for $10^5\equiv151 \pmod {401}$, and put in sum and prove that sum $(10^{799}+10^{798}+\cdots+10+1)$ is divisibility with number $401$ but I it seem like bad idea, do you have something?

Answer 1

Since $401$ is a prime number, then, by Fermat's little theorem, $10^{400}\equiv1\pmod{401}$. So, $10^{800}\equiv1\pmod{401}$, which means that $401\mid10^{800}-1$. But $9(=10-1)\mid10^{88}-1$. so$$401\mid\frac{10^{800}-1}9\times9;$$since $\gcd(401,9)=1$, it follows from this that $401\mid\frac{10^{800}-1}9$. This is equivalent to the assertion $2\,005\mid\overbrace{55\ldots5}^{800\text{ times}}$.

Answer 2

Note that $5^{800}$ is divisible only by $5^k$ with $0\le k \le 800$ but

$$2005=5\cdot 401$$

and $401$ is prime.

Answer 3

Note that your number is $5(10^{800}-1)/9 =5(10^{400}-1)(10^{400}+1)/9$ and then apply Fermat.

Answer 4

Hint: $$2005\,\big|\,\underbrace{55 \ldots 5}_{800\text{ digits}}$$ $$401\cdot5\,\big|\,(\underbrace{11 \ldots 1}_{800\text{ digits}})\cdot5$$