Finding all values of $n$ for which $1/n$ has a specific decimal period in different bases












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I wanted to know if there is a way to easily determine the lowest value Natural number $n$ to where $1/n$ decimal actual period is purely repeating to $5$ digits in base $5$ and purely repeating to $4$ digits in base $6$?



What I learned so far



I had seen where $n$’s decimal representation independent of the base if the fraction is in lowest terms (is why I choose $1/n$) the period is limited to some $n-1$ or an integer $n-1$ is divisible by. It does not give the way to actually determine the period and it states the period may well be different for other bases.



For example $1/33$ could have periods of ($1,2,4,8,16,$ or $32$).



So my guessing point is lowest $n$ would need to be some multiple of $20$ then adding one also not having $2,3$ or $5$ as a prime factor to insure purely repeating for all chosen bases.



However I also saw for the denominator of $33$ that a period length of $10$ was given even though $33-1$ has no factors of $5$ in it so am left wondering if that theorem is actually true.






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    0














    I wanted to know if there is a way to easily determine the lowest value Natural number $n$ to where $1/n$ decimal actual period is purely repeating to $5$ digits in base $5$ and purely repeating to $4$ digits in base $6$?



    What I learned so far



    I had seen where $n$’s decimal representation independent of the base if the fraction is in lowest terms (is why I choose $1/n$) the period is limited to some $n-1$ or an integer $n-1$ is divisible by. It does not give the way to actually determine the period and it states the period may well be different for other bases.



    For example $1/33$ could have periods of ($1,2,4,8,16,$ or $32$).



    So my guessing point is lowest $n$ would need to be some multiple of $20$ then adding one also not having $2,3$ or $5$ as a prime factor to insure purely repeating for all chosen bases.



    However I also saw for the denominator of $33$ that a period length of $10$ was given even though $33-1$ has no factors of $5$ in it so am left wondering if that theorem is actually true.






    number-theory







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      I wanted to know if there is a way to easily determine the lowest value Natural number $n$ to where $1/n$ decimal actual period is purely repeating to $5$ digits in base $5$ and purely repeating to $4$ digits in base $6$?



      What I learned so far



      I had seen where $n$’s decimal representation independent of the base if the fraction is in lowest terms (is why I choose $1/n$) the period is limited to some $n-1$ or an integer $n-1$ is divisible by. It does not give the way to actually determine the period and it states the period may well be different for other bases.



      For example $1/33$ could have periods of ($1,2,4,8,16,$ or $32$).



      So my guessing point is lowest $n$ would need to be some multiple of $20$ then adding one also not having $2,3$ or $5$ as a prime factor to insure purely repeating for all chosen bases.



      However I also saw for the denominator of $33$ that a period length of $10$ was given even though $33-1$ has no factors of $5$ in it so am left wondering if that theorem is actually true.






      number-theory







      share|cite|improve this question















      I wanted to know if there is a way to easily determine the lowest value Natural number $n$ to where $1/n$ decimal actual period is purely repeating to $5$ digits in base $5$ and purely repeating to $4$ digits in base $6$?



      What I learned so far



      I had seen where $n$’s decimal representation independent of the base if the fraction is in lowest terms (is why I choose $1/n$) the period is limited to some $n-1$ or an integer $n-1$ is divisible by. It does not give the way to actually determine the period and it states the period may well be different for other bases.



      For example $1/33$ could have periods of ($1,2,4,8,16,$ or $32$).



      So my guessing point is lowest $n$ would need to be some multiple of $20$ then adding one also not having $2,3$ or $5$ as a prime factor to insure purely repeating for all chosen bases.



      However I also saw for the denominator of $33$ that a period length of $10$ was given even though $33-1$ has no factors of $5$ in it so am left wondering if that theorem is actually true.






      number-theory




      number-theory






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      edited Nov 19 '18 at 3:20









      Tianlalu

      3,09621038




      3,09621038










      asked Nov 9 '18 at 23:49









      Toni Stack

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          For your first problem, the answer is that no such $n$ exists. Indeed, the base $6$ expansion having period $4$ tells you $n$ is a factor of $6^4-1=5times 7times 37$, and the base $5$ having period $5$ gives $n$ divides $5^5-1=2^2times 11times 71$. You can see these are incompatible constraints.



          Now your confusion with $n=33$ seems to come from applying a theorem outside its domain




          Theorem 1 Let $p$ be prime. The fundamental period of $1/p$ in base $b$ expansion, $b$ not divisible by $p$, is a factor of $p-1$.




          This need not work for $n$ composite. Indeed, the above is derived from




          Theorem 2 The minimal period of $1/n$ in base $b$ expansion, $gcd(n,b)=1$, is the multiplicative order of $b$ modulo $n$.




          Since the multiplicative group $(mathbb{Z}/p)^times$ has order $phi(p)=p-1$, this gives the case $n=p$ is prime.






          share|cite|improve this answer





















          • What is then the period constraints for an unreducable fraction (non factors of 3,11)/33 The table I thought was referring to fractions already in lowest terms
            – Toni Stack
            Nov 10 '18 at 1:35






          • 1




            In general, the order has to be a divisor of $phi(n)$, where $phi$ is the Euler's totient function. But it is possible to give a slightly stronger bound -- it has to be a divisor of $operatorname{lcm}(phi(p_1^{r_1}),phi(p_2^{r_2}),dots,phi(p_k^{r_k}))$, where $n=p_1^{r_1}p_2^{r_2}dots p_k^{r_k}$. In your case, 33=3*11 and $phi(3)=2$, $phi(11)=10$ gives lcm =10.
            – user10354138
            Nov 10 '18 at 2:44











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          For your first problem, the answer is that no such $n$ exists. Indeed, the base $6$ expansion having period $4$ tells you $n$ is a factor of $6^4-1=5times 7times 37$, and the base $5$ having period $5$ gives $n$ divides $5^5-1=2^2times 11times 71$. You can see these are incompatible constraints.



          Now your confusion with $n=33$ seems to come from applying a theorem outside its domain




          Theorem 1 Let $p$ be prime. The fundamental period of $1/p$ in base $b$ expansion, $b$ not divisible by $p$, is a factor of $p-1$.




          This need not work for $n$ composite. Indeed, the above is derived from




          Theorem 2 The minimal period of $1/n$ in base $b$ expansion, $gcd(n,b)=1$, is the multiplicative order of $b$ modulo $n$.




          Since the multiplicative group $(mathbb{Z}/p)^times$ has order $phi(p)=p-1$, this gives the case $n=p$ is prime.






          share|cite|improve this answer





















          • What is then the period constraints for an unreducable fraction (non factors of 3,11)/33 The table I thought was referring to fractions already in lowest terms
            – Toni Stack
            Nov 10 '18 at 1:35






          • 1




            In general, the order has to be a divisor of $phi(n)$, where $phi$ is the Euler's totient function. But it is possible to give a slightly stronger bound -- it has to be a divisor of $operatorname{lcm}(phi(p_1^{r_1}),phi(p_2^{r_2}),dots,phi(p_k^{r_k}))$, where $n=p_1^{r_1}p_2^{r_2}dots p_k^{r_k}$. In your case, 33=3*11 and $phi(3)=2$, $phi(11)=10$ gives lcm =10.
            – user10354138
            Nov 10 '18 at 2:44
















          0














          For your first problem, the answer is that no such $n$ exists. Indeed, the base $6$ expansion having period $4$ tells you $n$ is a factor of $6^4-1=5times 7times 37$, and the base $5$ having period $5$ gives $n$ divides $5^5-1=2^2times 11times 71$. You can see these are incompatible constraints.



          Now your confusion with $n=33$ seems to come from applying a theorem outside its domain




          Theorem 1 Let $p$ be prime. The fundamental period of $1/p$ in base $b$ expansion, $b$ not divisible by $p$, is a factor of $p-1$.




          This need not work for $n$ composite. Indeed, the above is derived from




          Theorem 2 The minimal period of $1/n$ in base $b$ expansion, $gcd(n,b)=1$, is the multiplicative order of $b$ modulo $n$.




          Since the multiplicative group $(mathbb{Z}/p)^times$ has order $phi(p)=p-1$, this gives the case $n=p$ is prime.






          share|cite|improve this answer





















          • What is then the period constraints for an unreducable fraction (non factors of 3,11)/33 The table I thought was referring to fractions already in lowest terms
            – Toni Stack
            Nov 10 '18 at 1:35






          • 1




            In general, the order has to be a divisor of $phi(n)$, where $phi$ is the Euler's totient function. But it is possible to give a slightly stronger bound -- it has to be a divisor of $operatorname{lcm}(phi(p_1^{r_1}),phi(p_2^{r_2}),dots,phi(p_k^{r_k}))$, where $n=p_1^{r_1}p_2^{r_2}dots p_k^{r_k}$. In your case, 33=3*11 and $phi(3)=2$, $phi(11)=10$ gives lcm =10.
            – user10354138
            Nov 10 '18 at 2:44














          0












          0








          0






          For your first problem, the answer is that no such $n$ exists. Indeed, the base $6$ expansion having period $4$ tells you $n$ is a factor of $6^4-1=5times 7times 37$, and the base $5$ having period $5$ gives $n$ divides $5^5-1=2^2times 11times 71$. You can see these are incompatible constraints.



          Now your confusion with $n=33$ seems to come from applying a theorem outside its domain




          Theorem 1 Let $p$ be prime. The fundamental period of $1/p$ in base $b$ expansion, $b$ not divisible by $p$, is a factor of $p-1$.




          This need not work for $n$ composite. Indeed, the above is derived from




          Theorem 2 The minimal period of $1/n$ in base $b$ expansion, $gcd(n,b)=1$, is the multiplicative order of $b$ modulo $n$.




          Since the multiplicative group $(mathbb{Z}/p)^times$ has order $phi(p)=p-1$, this gives the case $n=p$ is prime.






          share|cite|improve this answer












          For your first problem, the answer is that no such $n$ exists. Indeed, the base $6$ expansion having period $4$ tells you $n$ is a factor of $6^4-1=5times 7times 37$, and the base $5$ having period $5$ gives $n$ divides $5^5-1=2^2times 11times 71$. You can see these are incompatible constraints.



          Now your confusion with $n=33$ seems to come from applying a theorem outside its domain




          Theorem 1 Let $p$ be prime. The fundamental period of $1/p$ in base $b$ expansion, $b$ not divisible by $p$, is a factor of $p-1$.




          This need not work for $n$ composite. Indeed, the above is derived from




          Theorem 2 The minimal period of $1/n$ in base $b$ expansion, $gcd(n,b)=1$, is the multiplicative order of $b$ modulo $n$.




          Since the multiplicative group $(mathbb{Z}/p)^times$ has order $phi(p)=p-1$, this gives the case $n=p$ is prime.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 10 '18 at 0:30









          user10354138

          7,4042824




          7,4042824












          • What is then the period constraints for an unreducable fraction (non factors of 3,11)/33 The table I thought was referring to fractions already in lowest terms
            – Toni Stack
            Nov 10 '18 at 1:35






          • 1




            In general, the order has to be a divisor of $phi(n)$, where $phi$ is the Euler's totient function. But it is possible to give a slightly stronger bound -- it has to be a divisor of $operatorname{lcm}(phi(p_1^{r_1}),phi(p_2^{r_2}),dots,phi(p_k^{r_k}))$, where $n=p_1^{r_1}p_2^{r_2}dots p_k^{r_k}$. In your case, 33=3*11 and $phi(3)=2$, $phi(11)=10$ gives lcm =10.
            – user10354138
            Nov 10 '18 at 2:44


















          • What is then the period constraints for an unreducable fraction (non factors of 3,11)/33 The table I thought was referring to fractions already in lowest terms
            – Toni Stack
            Nov 10 '18 at 1:35






          • 1




            In general, the order has to be a divisor of $phi(n)$, where $phi$ is the Euler's totient function. But it is possible to give a slightly stronger bound -- it has to be a divisor of $operatorname{lcm}(phi(p_1^{r_1}),phi(p_2^{r_2}),dots,phi(p_k^{r_k}))$, where $n=p_1^{r_1}p_2^{r_2}dots p_k^{r_k}$. In your case, 33=3*11 and $phi(3)=2$, $phi(11)=10$ gives lcm =10.
            – user10354138
            Nov 10 '18 at 2:44
















          What is then the period constraints for an unreducable fraction (non factors of 3,11)/33 The table I thought was referring to fractions already in lowest terms
          – Toni Stack
          Nov 10 '18 at 1:35




          What is then the period constraints for an unreducable fraction (non factors of 3,11)/33 The table I thought was referring to fractions already in lowest terms
          – Toni Stack
          Nov 10 '18 at 1:35




          1




          1




          In general, the order has to be a divisor of $phi(n)$, where $phi$ is the Euler's totient function. But it is possible to give a slightly stronger bound -- it has to be a divisor of $operatorname{lcm}(phi(p_1^{r_1}),phi(p_2^{r_2}),dots,phi(p_k^{r_k}))$, where $n=p_1^{r_1}p_2^{r_2}dots p_k^{r_k}$. In your case, 33=3*11 and $phi(3)=2$, $phi(11)=10$ gives lcm =10.
          – user10354138
          Nov 10 '18 at 2:44




          In general, the order has to be a divisor of $phi(n)$, where $phi$ is the Euler's totient function. But it is possible to give a slightly stronger bound -- it has to be a divisor of $operatorname{lcm}(phi(p_1^{r_1}),phi(p_2^{r_2}),dots,phi(p_k^{r_k}))$, where $n=p_1^{r_1}p_2^{r_2}dots p_k^{r_k}$. In your case, 33=3*11 and $phi(3)=2$, $phi(11)=10$ gives lcm =10.
          – user10354138
          Nov 10 '18 at 2:44


















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