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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

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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 - 2 months ago

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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 - 1 month, 4 weeks ago

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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 - 1 month, 4 weeks ago

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@Ivan Koswara Oh yeah. Thanks a lot.

Steven Jim - 1 month, 4 weeks ago

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