Let us call the sum of the first half and the latter half of the cyclic numbers of an irreducible fraction 'a division sum' when the period of a repeating decimal is even. Also, let $\lambda(l)$ be the length of the repeating digits of $\frac 1l$ in decimal expansion.

Example 1 : The division sum of $\frac 17$ is $142+857=999$ and $\lambda(7)=6$ because $\frac 17=0.\dot 142\ 85\dot 7$.

Then, here is my question.

Question : Is the following true?

"Suppose that $p,q$ are distinct prime numbers other than $2$ or $5$. Then, either $(1)$ or $(2)$ holds.

$(1)$ $\lambda(p)=2^h\mu, \lambda(q)=2^h\nu\ (h\ge 1, \mu,\nu$ are odd$)\Rightarrow$ the division sum of $\frac{1}{pq}$ can be represented as $99\cdots 9$.

$(2)$ $\lambda(p)=2^e\mu, \lambda(q)=2^f\nu\ (0\le e\lt f, \mu,\nu$ are odd$)\Rightarrow$ the division sum of $\frac{1}{pq}$ is in the form of repeating the cyclic number of $\frac rp$ where $r$ is the minimum natural number such that $qr\equiv 2\ $(mod $p).$"

Example 2 : The division sum of $\frac 1{77}=\frac{1}{7\cdot 11}=0.\dot 012\ 98\dot 7$ is $012+987=999$. Note that $\lambda(7)=6=2^1\cdot 3,\lambda(11)=2^1\cdot 1$.

Example 3 : The division sum of $\frac 1{21}=\frac{1}{3\cdot 7}=0.\dot 047\ 61\dot 9$ is $047+619=666$. Since we know that $\lambda(3)=1=2^0\cdot 1,\lambda(7)=6=2^1\cdot 3$, $7r\equiv 2\ $(mod$3)$ leads $r=2$. Note that $\frac 23=0.\dot 6$.

Example 4 : The division sum of $\frac 1{949}=\frac{1}{13\cdot 73}=0.\dot 001053740779\ 76817702845\dot 1$ is $1053740779+768177028451=769230\ 769230$. Since we know that $\lambda(13)=6=2^1\cdot 3,\lambda(73)=8=2^3\cdot 1$, $73r\equiv 2\ $(mod$13)$ leads $r=10$. Note that $\frac{10}{13}=0.\dot 76923\dot 0$.

Motivation : We know that a fraction in lowest terms with a prime denominator other than $2$ or $5$ (i.e. coprime to $10$) always produces a repeating decimal. We can prove that the division sum of $\frac 1p$ can be represented as $99\cdots 9$ when $p$ is a prime other than $2$ or $5$. I've been thinking about its generalization. Then, I reached the above expectation. This expectation seems true, but I can't prove it. Can anyone help?

By the way, if my expectation is true, the following follows.

"Suppose that $p\ge 7$ is a prime number and that the period of the repeating decimal $\frac{1}{3p}$ is even. If $p\equiv 1\ ($mod $3)$, then the division sum of $\frac{1}{3p}$ can be represented as $66\cdots 6$. If $p\equiv 2\ ($mod $3)$, then the division sum of $\frac{1}{3p}$ can be represented as $33\cdots 3.$"

Update : I crossposted to MO.


If $n$ is coprime with $10$, then $\lambda(n)$ is the order of the class of $10$ in $(\Bbb Z/n \Bbb Z)^*$.

If $\lambda(n) = 2k$, then $10^k$ is a nontrivial square root of $1$ mod $n$ (there may be several of them), and if $a,b$ are the first and second half of the repeating pattern of $\frac 1n$, then :

$0 \le a,b \lt 10^k$ ; $a \neq b$ ; $10^ka+b = (10^{2k}-1)\frac 1n > 0$.
From this equality we get $a+b = (10^k-1)((10^k+1)\frac 1n - a)$.
Also, since $a = \lfloor 10^k \frac 1n \rfloor$, we get $a+b \le (10^k-1)(1+ \frac 1n)$

If $10^k = -1 \pmod n$, then $a+b$ is a positive multiple of $10^k-1$, so it must be $10^k-1$.

Suppose $n=pq$. By the chinese remainder theorem, $\lambda(n) = lcm(\lambda(p),\lambda(q))$.
If both $\lambda(p)$ and $\lambda(q)$ are even, say $\lambda(p)=2k_p, \lambda(q)=2k_q$, then $k = lcm(k_p,k_q)$.

Then, $10^k \equiv -1 \pmod n \iff 10^k \equiv -1 \pmod {p,q} \iff k$ is an odd multiple of $k_p$ and $k_q$. And this happen if and only if the exposant of $2$ in the prime factorisation of $k_p$ and $k_q$ are equal. So we have done your case $(1)$.

If we are not in case $(1)$ (and $\lambda(pq)$ is even), then we are in case $(2)$ : $\lambda(p)=2^e\mu, \lambda(q) = 2^f \nu$, with $e \neq f$ (we can assume $e<f$ by switching $p$ with $q$ if it's not the case). We then have $\lambda(pq) = 2^f lcm(\mu,\nu)$, so $k$ is an odd multiple of $k_q$ and a multiple of $\lambda(p)$. Hence $10^k \equiv 1 \pmod p$ and $10^k \equiv -1 \pmod q$. Looking back at our formula for $a+b$, $(10^k+1)\frac 1n$ simplifies into $u/p$ for some integer $u$. And so, $a+b = (10^k-1)\frac rp$ for some positive integer $r$. From our upper bound on $a+b$, we have $r \le p(1+ \frac 1n) = p + \frac 1q$, hence $1 \le r \le p$.

So to determine $r$ it is enough to determine it modulo $p$. We know that modulo $p$, $r \equiv u = (10^k+1)/q \equiv 2/q$, hence $qr \equiv 2 \pmod p$, which is what you wanted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.