Someone must have discovered this...

(It seems trivial to me, so I don't know if this is correct. Can anyone help me write a program to check this?)

So for many (yes, this is the reason I ask for a program) fractions $\frac { a }{ b } =0.\overline { { x }_{ 1 }{ x }_{ 2 }...{ x }_{ n } } $ in which $b>2$ is a prime and $(a,b)=1$, I found out that $b-1$ is divisible by $n$. I've checked it for about 50 numbers and they all seem correct. However, nothing is correct without a proof, so I'm stuck. Can anyone please explain why this works???

#NumberTheory

Easy Math Editor

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Comments

Your claim is wrong if $b = 5$ .

If $b \neq 2, 5$ , then $10^{b-1} \equiv 1 \pmod b$ by Fermat's little theorem, so $10^{b-1} - 1 \equiv 0 \pmod b$ . Thus $\frac{a}{b} = \frac{x}{10^{b-1} - 1}$ for some integer $x$ , so clearly its decimal representation will be $x$ repeated every $b-1$ digits. Thus the period $n$ must divide $b-1$ .

Ivan Koswara - 3 years, 9 months ago

Thanks for the proof. Anyways, I think that 1/5=0.200... so it should have a period of 1. Where was my mistake?

Steven Jim - 3 years, 9 months ago

It's not in the form $0.\overline{x_1 x_2 \ldots x_n}$ as you described. It's in the form $0.y_1 y_2 \ldots y_m \overline{x_1 x_2 \ldots x_n}$ . (In the latter case, all $b \neq 0$ works, not necessarily those relatively prime to 10.)

Ivan Koswara - 3 years, 9 months ago

@Ivan Koswara – Oh yeah. Thanks a lot.

Steven Jim - 3 years, 9 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}`	$\boxed{123}$