'''
Take the number 192 and multiply it by each of 1, 2, and 3:

192 × 1 = 192
192 × 2 = 384
192 × 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
'''

def conLength(n):
    i = 1
    p = len(str(n))
    while p < 9:
        i += 1
        p += len(str(i * n))
    
    if p == 9:
        return True
    
    return False

def isPandigital(n):
    digits = {1, 2, 3, 4, 5, 6, 7, 8, 9}
    if set(map(int, str(n))) == digits:
        return True
    return False

def concatProduct(n):
    i = 1
    d = n

    while len(str(d)) < 9:
        i += 1
        d = int(str(d) + str(n * i))
    
    return d

def main():
    print("Hello this is Patrick")

    m = 0

    for i in range(1, 54321):
        if conLength(i):
            d = concatProduct(i)

            if isPandigital(d) and d > m:
                m = d
    
    print(m)

if __name__ == "__main__":
    main()