Question

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)?

Answer 1

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`.

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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))`