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 \(10^r \equiv 1 \pmod{p}\).

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.)

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TopNewest\(\displaystyle \frac{1}{p} = 0.\overline{a_1a_2\ldots a_r}\)

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?

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Where are using the fact that p is a prime?

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

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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.)

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Well, it still is valid though.

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

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

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

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

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