10
$\begingroup$

Today, as usual, we were doing all those boring numerical computations in our calculators. It all started when my professor replaced $0.2$ with $\frac{1}{5}$. I got into calculating the unit fractions one by one. But soon, I got indulged in unit fractions made from primes, as other numbers can be decomposed into prime factors (and partially because I've always thought that primes are special). Then, I observed a few things.

  • All unit prime fractions (except $\frac{1}{2}$ & $\frac{1}{5}$) have recurring digits.
  • But, there were a few special fractions. For instance, $\frac{1}{7}$ had a digit frequency of $6$ (i.e) $0.\overline{142857}$ - $6$ recurring digits. $\frac{1}{17}$ had a frequency of $16$, $\frac{1}{19}$ had a frequency of $18$, etc.

Only after an hour or so, I was shocked to notice something. "Most" of these primes had a similar property. If $\mathcal{R}(n)$ is the number of recurring digits, then $\mathcal{R}(n)=(n-1)$ for those special primes (Well, it doesn't work for $13$, but the latter result is still true).

Then, came the pattern. First of all, $\mathcal{R}(n)$ is even for these primes, since all primes are odd. While calculating $\frac{1}{13}$, I saw that when we split the recurring digits $0.0\overline{769230}$ in half and add them ($769+230$), we get $999=10^3-1$.

Then, I did the same for $\frac{1}{17}$ and $\frac{1}{19}$, for which I got $(10^8-1)$ and $(10^9-1)$. For every fraction of this kind, the sum is of the form $10^k-1$ where $k\ \in \mathbb N$.

Soon, I found that there was a condition for this form to appear (after writing it out for a few of these primes $7, 13, 17, 19, 23, 29$, etc.). It happens only when

$$\mathcal{R}(n_i)\geq\mathcal{R}(n_{i-1})\ \forall\ \ n \in \mathbb P$$

For instance, this doesn't happen for $1/11$, but it occurred for $1/7$ and $1/13$, since $$\mathcal{R}(11)=2\ <\ \mathcal{R}(7)=8=\mathcal{R}(13)$$

I'm curious about this result. We're slicing those recurring digits in half, right? I can't quite visualize how the summing up those sliced digits converge to the same form.

Why's this so? Is this true for all these special primes? Or, does this have a limit beyond which this condition breaks down?

$\endgroup$
  • 1
    $\begingroup$ What are recurring digits? One has $\dfrac 1{13}=0.\overline{076923}$ I take it the number of recurring digits is $6$, but $6\neq 13-1$. And what does it mean to split the recurring digits in half? $\endgroup$ – Git Gud Aug 25 '14 at 15:44
  • $\begingroup$ @GitGud: Okay, I took it this way - $\dfrac 1{13}=0.0\overline{769230}$. And, I forgot to mention that $(n-1)$ doesn't work for $13$. It works for $7,17,19$ and so on, in the order of increasing recurring digits. By "splitting into half", I meant, $769+230=999$. $\endgroup$ – Waffle's Crazy Peanut Aug 25 '14 at 15:49
  • 1
    $\begingroup$ Regarding the number of recurring digits, I asked that question a while back and got a good answer: math.stackexchange.com/questions/298844/… $\endgroup$ – Arthur Aug 25 '14 at 16:02
  • $\begingroup$ Another thing to explore is these same fractions in other bases... $\endgroup$ – Thomas Andrews Aug 25 '14 at 16:08
  • 2
    $\begingroup$ This "splitting into half" stuff can be looked-at a little differently. If $x=0.\overline{abcdef}$, then $10^3x = abc.\overline{defabc}$, so that $(1+10^3)x = abc.\overline{[a+d][b+e][c+f]}$, where $[m+n]$ is a single digit (so, I'm assuming no "carries" here); that is, the repeating part of $(1+10^3)x$ is what you get by adding the "halves" of the repeating part of $x$. That you get $999$ as the sum of these halves says that $(1+10^3)x$ is an integer; namely $abc.\overline{999} = abc+1$. (In particular, $1001\cdot\frac{1}{13}=77=076+1$.) $\endgroup$ – Blue Aug 25 '14 at 16:17
10
$\begingroup$

So, here's what's going on here.

First of all, let's take $1/13$ for an example. Writing $1/13 = 0.\overline{076923}$ is the same as saying $\frac{1}{13} = \frac{76923}{999999} = \frac{76923}{10^{6} - 1}$. So we see that, $1/13$ having a period of 6 digits corresponds to the fact that we can write $\frac{1}{13} = \frac{a}{10^6 - 1}$ for some integer $a$. We can see $a = \frac{10^6 - 1}{13}$. So the first observation is the following:

FACT: The number of repeating digits in $1/n$ is the smallest integer $k$ such that $n$ is a factor of $10^k - 1$. (When $n$ has no factors of $2$ or $5$).

I don't know if you've seen modular arithmetic before, but saying that $n$ is a factor of $10^k - 1$ is the same as saying $10^k \equiv 1$ (mod n). The smallest such integer $k$ is called the order of 10 (mod $n$). It is not so hard to prove that, if $n$ is prime (lets call it $p$), the order of any element must divide $p - 1$. For instance, when $p = 13$, the number of repeating digits in $1/13$ must be a divisor of $12$ (and this is true in any base $b$, as long as $b$ has no factors of $13$). In usual base ten, the number of repeating digits is $6$, as you've discovered.

Okay, so now lets talk about the 9999.. part.

Suppose $p$ is a prime number such that $1/p$ has a repeating decimal expansion whose length $k$ is even (for instance, if $k = p - 1$). So we have $p$ divides $10^k - 1$, or $10^k \equiv 1$ (mod p). This implies that $10^{k/2} \equiv -1$ (mod p). Why? Well suppose $10^{k/2} \equiv x$ (mod p). Then it follows that $x^2 \equiv (10^{k/2})^2 \equiv 10^k \equiv 1$ (mod p). But the equation $x^2 \equiv 1$ only has two solutions mod $p$: $x = \pm 1$. $x = 1$ is not possible here, since we said that $k$ was the smallest integer for which $10^k \equiv 1$ (mod p).

[Note that this is not true if $p$ is not prime. For example, $10^6 \equiv 1$ (mod 21), but $10^3 \equiv 13$ (mod 21). One can check that $13$ is a solution to $x^2 \equiv 1$ (mod 21), and $13$ is neither $1$ nor $-1$. But this cannot occur when $p$ is prime.]

Okay. So we've concluded that $10^{k/2} \equiv -1$ (mod p), meaning $p$ divides $10^{k/2} + 1$. Hence we can write:

$$\frac{1}{p} = \frac{a}{10^{k/2} + 1}$$

For some integer $k$. Multiplying the numerator and denominator of the RHS by $10^{k/2} - 1$, we get:

$$\frac{1}{p} = \frac{a(10^{k/2} - 1)}{10^k - 1} = \frac{10^{k/2}(a - 1) + (10^{k/2} - a)}{10^k - 1}$$

This solves the problem! Why? Well, $10^{k/2} - a$ is a number with $k/2$ digits. Similarly, $a - 1$ is a number with $k/2$ digits. The numerator of this fraction will be $a - 1$ followed by $10^{k/2} - a$, since $a - 1$ is being multiplied by $10^{k/2}$ and so shifted $k/2$ digits. Hence the sum of the 'first half' and the 'second half' is $(a - 1) + (10^{k/2} - a) = 10^{k/2} - 1$, which is a sequence of $9$s.

I'll finish with doing $1/13$ as an example. Here $k = 6$, and so $k/2 = 3$. Then $10^{k/2} = 1000 \equiv -1$ (mod 13). Indeed, 1001 is a multiple of 13. So we can write:

$$\frac{1}{13} = \frac{77}{1001}$$

Here, $77$ is our 'a'. So we get:

$$\frac{1}{13} = \frac{77 \cdot 999}{999999} = \frac{10^3(76) + (1000 - 77)}{999999} = \frac{10^3(76) + (923)}{999999}$$

This corresponds to the fact that the first 3 digits are $a - 1 = 76$ (or 076) and the last 3 digits are $10^{k/2} - a = 1000 - 77 = 923$.

$\endgroup$
  • $\begingroup$ Well, you're right that I don't have a good experience with modulo operations. But, I'm certainly convinced by your answer. It was rather charming! Thanks Alex :) $\endgroup$ – Waffle's Crazy Peanut Aug 26 '14 at 13:18
7
$\begingroup$

Something that might blow your mind. Not only is $142+857=999$, but $14+28+57=99$. Similarly, for $1/13=0.\overline{076923}$. You have $076+923=999$ and $07+69+23=99$.

The number of repeating digits for $\frac{1}{n}$ in general, when $n$ is relatively prime to $10$, is the smallest positive integer $q$ such that $n$ is a divisor of $10^q-1$. So, for example, $3\mid 10^1-1$, so the frequency of $1/3$ is $1$. For $7$, $10^6-1$ is divisible by $7$. When $n$ is prime, we can show that $q\mid n-1$.

Note that $10^q-1=999\dots9$, with $q$ digits $9$.

Since $37\mid 999$, we get $1/37$ has frequency $3$, which is rather extreme. Then you don't get your '999' formula, because $1/37 = 0.\overline{027}$ can't be broken into halves of digits.

Similarly, $\frac{1}{31}=0.\overline{032258064516129}$ has frequency $15$. Note that $$03226+80645+16129=99999$$

But when $q$ is even, say $q=2k$, then $n\mid 10^{2k}-1=(10^k-1)(10^k+1)$. When $n$ is prime, since $q$ is the smallest value, we know that $n$ is not a divisor of $10^k-1$, so $n$ must be a divisor of $10^k+1$. (That is why you won't see this necessarily when $n$ is not prime.)

Now, if $$\frac{1}{n}=\frac{A}{10^k}+\frac{B}{10^{2k}}+\frac{A}{10^{3k}}\dots$$ with $0\leq A,B<10^k$, we can multiply by $10^k+1$ and the result must be an integer. So:

$$\begin{align}\frac{10^k+1}{n} &= A+\frac{A+B}{10^k}+\frac{A+B}{10^{2k}}+\frac{A+B}{10^{3k}}\dots\\&=A+ \frac{10^k(A+B)}{10^k-1} \end{align}$$ This means that $A+B$ must be divisible by $10^k-1$. Since $0\leq A,B\leq 10^k-1$, the only way for this to be an integer is if $A+B=10^k-1=99\dots9$.

The cases when $q$ is divisible by $3$, as with $1/31$ and $1/37$, we saw that there is something more going on. In those cases, $q=3k$, $10^{3k}-1=(10^k-1)(10^{2k}+10^k+1)$, and by the same reasoning above, $n$ must divide $10^{2k}+10^k+1$, which means when you multiply $\frac{1}{n}$ by $10^{2k}+10^k+1$, you should get an integer, the sums of the three components, by the above reasoning, will be a multiple of $10^k-1$. (Is it possible for the sum to be $2(10^k-1)$? I haven't ruled that out...)

This can be written more generally. If $n$ is prime and not a factor of $b(b-1)$, then $\frac{1}{n}$ in base $b$ is repeating, and the sum of the (base $b$) digits will be a multiple of $b-1$. When the fraction is repeating $2$ digits, then the sum of the digits will always be $b-1$. Then above, when $q=2k$ in base $10$, this result means that $\frac{1}{n}=0.ABAB\dots$ in base $b=10^k$, and we can show that the sum must be exactly $10^k-1$.

When $n=71$, we have:

$$\frac{1}{71}=0.\overline{01408450704225352112676056338028169}$$

That sequence has length $35$, and when we break it up into set of $7$ digits, we get:

$$01408+45070+42253+52112+67605+63380+28169=299997=3\cdot99999$$

and into groups of seven digits: $$0140845+0704225+3521126+7605633+8028169=19999998= 2\cdot 9999999$$

Finally, breaking $1/7$ into groups of $5$ and $7$, and $11$, we get:

$$\begin{align}299997=&14285+71428+57142+85714+28571+42857\\ 29999997=&1428571+4285714+2857142+8571428+5714285+7142857\\ 299999999997=&14285714285+71428571428+57142857142+\\&85714285714+28571428571+42857142857 \end{align}$$

This is not just true for $n$ prime. If $n\mid 1+b+b^2+\dots b^{d-1}$ then the first $d$ digits base $b$ add up to a multiple of $b-1$. (The frequency of the repetition in $\frac{1}{n}$ will be a factor of $d$.)

For example, when $n=7\cdot 13=91$, and $91\mid 1+10^3$ and $91\mid 1+10^2+10^4$. So:

$$\frac{1}{91}=0.\overline{010989}\\01+09+89=99\\ 010+989=999$$

$\endgroup$
  • $\begingroup$ Wow! That's one deep analysis you've done! Thanks Thomas :) And, you've made your answer precisely, that it wasn't obvious in the first look. It took my time ;-) $\endgroup$ – Waffle's Crazy Peanut Aug 26 '14 at 13:20
  • 1
    $\begingroup$ Thanks. It probably changed a bit since you first read it, though, which might explain why it got clearer on later reading. :) Early drafts were more scattershot. $\endgroup$ – Thomas Andrews Aug 26 '14 at 14:07
3
$\begingroup$

A fraction in lowest terms with a prime denominator other than 2 or 5 (i.e. coprime to 10) always produces a repeating decimal.

The length of the period of the repeating decimal of $1/p$ is equal to the order of $10\mod p$. If 10 is a primitive root modulo $p$, the period of the repeating decimal length is equal to $p − 1$; if not, the period of the repeating decimal length is a factor of $p − 1$. This result can be deduced from Fermat's little theorem, which states that $10^{p−1} = 1 (\mod p)$.

For $13$: $$10^1\equiv 10\pmod{13}\\10^2\equiv9\pmod{13}\\10^3\equiv12\pmod{13}\\10^4\equiv3\pmod{13}\\10^5\equiv4\pmod{13}\\10^6\equiv1\pmod{13}\\10^7\equiv10\pmod{13}$$ Here we see that the period of $10^k$ modulo $13$ is $6$. The remainders in the period, which are $10,9,12,3,4,1$ do not form a rearrangement of all nonzero remainders modulo $13$, implying that 10 is not a primitive root modulo 13.

$1/13$ contains $6$ repeating digits which indeed is a factor of $12(=13-1)$

If the period of the repeating decimal length of $1/p$ for prime $p$ is equal to $p − 1$ then the period of the repeating decimal, expressed as an integer, is called a cyclic number.Examples include $7,17,19,23,29,97,\cdots$

If $p$ is a prime and $10$ is a primitive root modulo $p$, then the length $\mathcal{R}$ of the period of the repeating decimal of $1/p$ is $\phi(p)$, totient's function.


Let suppose we take $1/p$ which is your "special prime number".Now decimal form of it must be of $$1/p=0.\overline{a_{1}a_{2}\cdots a_{\frac{p-1}2}a_{\frac{p-1}2+1}\cdots a_{p-1}}$$ Now $$10^{\frac{p-1}2}/p+1/p=\left(a_{1}a_{2}\cdots a_{\frac{p-1}2}\right).\left(\overline{a_{\frac{p-1}2+1}\cdots a_{p-1}a_{1}a_{2}\cdots a_{\frac{p-1}2}}\right)+0.\left(\overline{a_{1}a_{2}\cdots a_{\frac{p-1}2}a_{\frac{p-1}2+1}\cdots a_{p-1}}\right)$$ We can see that LHS becomes $\frac{10^{\frac{p-1}2}+1}p$ which must be an integer so your result of $999\cdots9$ automatically becomes true, since the decimal part of this number is actually the sum of half parts.

$\endgroup$
  • $\begingroup$ I didn't know that there's a theorem for this thing! Thanks for bringing it up :) $\endgroup$ – Waffle's Crazy Peanut Aug 26 '14 at 13:36

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.