I want to do a program in python (3.6.5) that tell the length of e.g. 1/7. The output should be for this example something like: "length: 6, repeated numbers: 142857". I got this so far:
n = int(input("numerator: "))
d = int(input("denominator: "))
def t(n, d):
x = n * 9
z = x
k = 1
while z % d:
z = z * 10 + x
k += 1
print ("length:", k)
print ("repeated numbers:", t)
return k, z / d
t(n, d)
Doing print ("repeated numbers:", t) prints the representation of the t function itself, not its output.
Here's a repaired version of your code. I use a Python 3.6+ f-string to convert the repeating digits to a string, and add zeros to the front to make it the correct length.
def find_period(n, d):
z = x = n * 9
k = 1
while z % d:
z = z * 10 + x
k += 1
digits = f"{z // d:0{k}}"
return k, digits
# Test
num, den = 1, 7
period, digits = find_period(num, den)
print('num:', num, 'den:', den, 'period:', period, 'digits:', digits)
num, den = 1, 17
period, digits = find_period(num, den)
print('num:', num, 'den:', den, 'period:', period, 'digits:', digits)
output
num: 1 den: 7 period: 6 digits: 142857
num: 1 den: 17 period: 16 digits: 0588235294117647
This line may be a bit mysterious:
f"{z // d:0{k}}"
It says: Find the largest integer less than or equal to z divided by d, convert it to a string, and pad it on the left with zeroes (if necessary) to give it a length of k.
As Goyo points out in the comments, this algorithm is not perfect. It gets stuck in a loop if the decimal contains any non-repeating part, that is, if the denominator has any factors of 2 or 5. See if you can figure out a way to deal with that.