Prove: $2005|\underbrace{5 5\cdots 5}_{800\text{ digits}}$

Question

$2005=5\cdot 401$

Divisibility by $5$ follows from the last digit of $\underbrace{5 5\cdots 5}_{800\text{ digits}}$.

How toprove divisibility by $401$?

http://math.stackexchange.com/questions/2014958/proove-that-2005-devides-55555-dots-with-800-5s — lab bhattacharjee, Nov 23 '16 at 10:00

score 3 · Accepted Answer · edited Jun 12 '20 at 10:38

3

Your number is $5\left(\dfrac{10^{800} - 1}{9} \right)$

$401$ is a prime number $\implies 10^{400} \equiv 1 \pmod{401} \implies 10^{800} \equiv 1 \pmod{40}$

$\implies 5\left( \dfrac{10^{800} - 1}{9} \right) \equiv 0 \pmod{401}$

edited Jun 12 '20 at 10:38

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answered Nov 23 '16 at 10:08

kotomord

1,814

I understand how you get that $10^{800}-1 \equiv 0 (mod 401)$ but I don't understand how you get that $\frac{10^{800}-1}{9} \equiv 0 (mod 401)$? – Postskjerm Nov 23 '16 at 10:20
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@stenvic 9x is divisible to 401 => x is divisible to 401 – kotomord Nov 23 '16 at 10:21
do you mean $401\mid 9x \Rightarrow 401\mid x$? – Postskjerm Nov 23 '16 at 10:25
You have to prove that the numerator is divisible by $9$. – Piquito Nov 23 '16 at 10:56

Prove: $2005|\underbrace{5 5\cdots 5}_{800\text{ digits}}$

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