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)?

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))#