×

# 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???

Note by Steven Jim
2 months ago

Sort by:

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 $$b-1$$.

- 2 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?

- 1 month, 4 weeks 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.)

- 1 month, 4 weeks ago

Oh yeah. Thanks a lot.

- 1 month, 4 weeks ago