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"""

Problem:

The fraction 49/98 is a curious fraction, as an inexperienced

mathematician in attempting to simplify it may incorrectly believe

that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction,

less than one in value, and containing two digits in the numerator

and denominator.

If the product of these four fractions is given in its lowest common

terms, find the value of the denominator.

"""

def isDigitCancelling(num, den):

if num != den:

if num % 10 == den // 10:

if (num // 10) / (den % 10) == num / den:

return True

def solve(digit_len: int) -> str:

"""

>>> solve(2)

'16/64 , 19/95 , 26/65 , 49/98'

>>> solve(3)

'16/64 , 19/95 , 26/65 , 49/98'

>>> solve(4)

'16/64 , 19/95 , 26/65 , 49/98'

>>> solve(0)

>>> solve(5)

'16/64 , 19/95 , 26/65 , 49/98'

"""

solutions = []

den = 11

last_digit = int("1" + "0" * digit_len)

for num in range(den, last_digit):

while den <= 99:

if (num != den) and (num % 10 == den // 10) and (den % 10 != 0):

if isDigitCancelling(num, den):

solutions.append(f"{num}/{den}")

den += 1

num += 1

den = 10

solutions = " , ".join(solutions)

return solutions

if __name__ == "__main__":

print(solve(2))

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