Which is longer?

Number Theory Level 1

We see that the fractional part of $\frac{1}{3}$ is 0.333... It repeats after one digit. Let us say that $\frac{1}{3}$ has "period" one. We can also say that $\frac{1}{7} = 0.142857142857142857...$ has "period' six. Which of the following has the longest "period'?

\frac{1}{23}

\frac{1}{17}

\frac{1}{19}

\frac{1}{11}

1 solution

Lixin Zheng
May 13, 2017

Fermat's little theorem states that if $a$ is not divisible by $p$ , then $a^{p-1}\equiv 1 \pmod{p}$ . The period, $k$ of a fraction, $\dfrac1{q}$ , can be modeled as $10^{k}\equiv 1 \pmod{q}$ . This means that we are looking for a number $k$ , such that $10^{k}\equiv 1 \pmod{q}$ . Applying Fermat's little theorem and noticing that 10 is coprime to all of these denominators, we obtain that $k$ is 1 less than each of the denominators. Thus $k$ is the biggest for the fraction $\boxed{\frac{1}{23}}$

I'm not sure that solution is fully correct, considering that it implies 1/11 has a period of 10, when it actually has a period of 2.

Tyler Friedman - 4 years ago

Thanks for pointing that out, however, it has a period of two because 10^2 is congruent to 1 mod 11, so in this case, 2 would be the smallest number that satisfies 10^k is congruent to 1 mod 11.

Lixin Zheng - 4 years ago

Your argument is wrong. 1/37 has a period length of 3 and it is coprime to 10.

Stephan Hensel - 3 years, 1 month ago

7 is not divisible by 3, but 7^(3-1) = 2 (mod 3), what did I get wrong about Fermat's little theorem?

Johannes H - 2 years, 4 months ago

Which is longer?

1 solution

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