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# 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 $$10^r \equiv 1 \pmod{p}$$.

Note by Paramjit Singh
3 years, 10 months ago

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

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?

- 3 years, 10 months ago

Where are using the fact that p is a prime?

- 3 years, 10 months ago

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

- 3 years, 10 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.)

- 3 years, 10 months ago

Well, it still is valid though.

- 3 years, 10 months ago

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

- 3 years, 10 months ago

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

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

- 3 years, 10 months ago

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

- 3 years, 10 months ago