Recurring places...

Suppose for a prime $p > 5$, the decimal representation of $\frac{1}{p} = 0.\overline{a_1a_2\cdots a_r}$, where the over line represents a recurring decimal. Prove that $10^r \equiv 1 \pmod{p}$.

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Comments

$\displaystyle \frac{1}{p} = 0.\overline{a_1a_2\ldots a_r}$

Thus,
$\displaystyle \frac{10^r}{p}= 0.\overline{a_1a_2\ldots a_r}\times 10^r = a_1a_2\ldots a_r.\overline{a_1a_2\ldots a_r}$

Subtracting the two equations,
$\displaystyle \frac{10^r-1}{p} = a_1a_2\ldots a_r$
$\displaystyle\Rightarrow 10^r-1\equiv 0\pmod{p}$
$\displaystyle\Rightarrow 10^r\equiv 1\pmod{p}$

Hence, Proved. I dont know why there has to be $\displaystyle p>5$ ..Could somebody explain that to me?

Anish Puthuraya - 7 years, 4 months ago

As choice of p=2,5 yields 1/p as non-recurring decimals

Anirban Mandal - 7 years, 4 months ago

Where are using the fact that p is a prime?

Rahul Saha - 7 years, 4 months ago

I think that it has to be $p \neq 2,5$ . I tried 3 and it still works.

Samuraiwarm Tsunayoshi - 7 years, 4 months ago

It's pointless to consider an prime under five as each will either never repeat, or repeat continuously and is trivial. 2 and 5 will never repeat, and checking 3 if easy as it is just .3 repeating. (Sorry, awful at using LaTeX, not gonna try.)

Nikhil Pandya - 7 years, 4 months ago

Well, it still is valid though.

Anish Puthuraya - 7 years, 4 months ago

Here $(a_1 a_2\cdot\cdot\cdot a_r)$ denotes place value form.

$\dfrac 1 p = 0.\overline{a_1 a_2 \cdot\cdot\cdot a_r} = \dfrac{(a_1 a_2 \cdot\cdot\cdot a_r)}{10^r-1} ~~~\Longrightarrow ~ 10^r-1=p\times (a_1 a_2 \cdot\cdot\cdot a_r)\equiv 0 \pmod{p}$

Jubayer Nirjhor - 7 years, 4 months ago

Exactly, it's trivial if you use formula for infinite G.P.

A Brilliant Member - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$