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?

You should edit your post for clarity. Presumably you meant $5\times (10^{800}-1)$ but it really isn't clear. — lulu, Nov 16 '18 at 14:08
Note that $5^{800}$ is divisible only for $5^k$ with $0\le k \le 800$. — user, Nov 16 '18 at 14:10
No problem. To your question, note that $400$ divides $800$ so any divisor of $400$ is also a divisor of $800$...now ask yourself what the possible orders of $10$ are $\pmod {401}$. — lulu, Nov 16 '18 at 14:10
Use the geometric series formula, the fact that 9 is coprime to 401, and Fermat's Little Theorem. — Matthew Towers, Nov 16 '18 at 14:18
Possible duplicate: https://math.stackexchange.com/questions/2027055/prove-2005-underbrace5-5-cdots-5-800-text-digits?rq=1 — Darío A. Gutiérrez, Nov 16 '18 at 14:34

José Carlos Santos · Accepted Answer · 2018-11-16T14:38:25.573

3

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 $2\,005\mid\overbrace{55\ldots5}^{800\text{ times}}$.

edited Nov 16 '18 at 14:38

answered Nov 16 '18 at 14:13

José Carlos Santos

427,504

2

$55...5 = 5\frac{10^{800}-1}{10-1}$ so $9*55...5/5 = (10^{800}-1)$. You have to use the fact that $401$ is prime to get the assertion. – Stockfish Nov 16 '18 at 14:19
@Stockfish Sure. I have been most helpful today. Thank you a lot. – José Carlos Santos Nov 16 '18 at 14:38

score 2 · Answer 2 · answered Nov 16 '18 at 14:18

2

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

answered Nov 16 '18 at 14:18

Darío A. Gutiérrez

3,863

marty cohen · Answer 3 · 2018-11-16T15:32:48.907

1

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

edited Nov 16 '18 at 15:32

answered Nov 16 '18 at 14:15

marty cohen

107,799

1

Thanks. Did it from my phone and didn't check as much as I should. – marty cohen Nov 16 '18 at 15:33

user · Answer 4 · 2018-11-16T14:15:13.367

0

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.

edited Nov 16 '18 at 14:15

answered Nov 16 '18 at 14:07

user

154,566

I write that in my answer – Marko Škorić Nov 16 '18 at 14:13

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

4 Answers4