In 1/6=1.6666..., repeating 6 is called repeatend ( or reptend ) . I learn from https://en.wikipedia.org/wiki/Repeating_decimal, the reptend in the decimal form of 1/97 is a 96-digit string. Find fraction(s) having longer reptend string(s)?

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Answer 1 · 2016-10-16T02:59:49

We can find the fraction for an arbitrary repeating string with the following method:

Let $x = 0.bar(a_1a_2...a_n)$ , where $bar(a_1a_2...a_n)$ denotes the repeating $n$ -digit string $a_1a_2...a_n$ .

$=> 10^nx = a_1a_2...a_n.bar(a_1a_2...a_n)$

$=> 10^nx - x = a_1a_2...a_n.bar(a_1a_2...a_n)-0.bar(a_1a_2...a_n)$

$=> (10^n-1)x = a_1a_2...a_n$

$:. x = (a_1a_2...a_n)/(10^n-1)$

So, for example, if we wanted a 1000-digit repeating string, we could let $a_1 = a_2 = ... = a_999 = 0$ and $a_1000 = 1$ , and then we would have

$(000...01)/(10^1000-1) = 1/(10^1000-1)$

give a repeating sequence of $999$ zeros followed by a $1$ . Note that this method works to generate any repeating sequence.

If we let $a_i$ represent the $i^"th"$ digit of $pi$ , then

$(a_1a_2...a_10000)/(10^10000-1)$

would generate a a repeating string of the first $10000$ digits of $pi$ .

We can also use a similar method to find the fraction for any rational value with a repeating string.

Given a general real number with a repeating string

$c_1c_2...c_j.b_1b_2...b_kbar(a_1a_2...a_n) , a_i, b_i, c_i in {0, 1, 2, ..., 9}$ .

let
$a = a_1a_2...a_n$
$b = b_1b_2...b_k$
$c = c_1c_2...c_j$

Note that by the above work, we have $0.bar(a) = (a_1a_2...a_n)/(10^n-1)$

Then we can rewrite our number as

$c_1c_2...c_j.b_1b_2...b_kbar(a_1a_2...a_n) = c+10^(-k)b+10^(-k)*0.bar(a)$

$=c+b/10^k+a/(10^k(10^n-1))$

$=((10^kc+b)(10^n-1)+a)/(10^k(10^n-1))$