'''
Name:		Problem 38: Pandigital multiples (30445)

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

192 x 1 = 192
192 x 2 = 384
192 x 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?

Let m be the multiplicand and n be the multiplier
If m is 9 digits long, n = 1, so no 9-digit numbers fulfill the requirement
In fact, m must have 4 or fewer digits to meet the requirement
Only multiplicands with unique digits need be considered.

'''
from itertools import permutations
perms=[]
panTest=[]
panNum=[]
numbers=[1,2,3,4,5,6,7,8,9]
largestPandigital=0
for d in range(1,5):
	perms=set(permutations(numbers,d))
for p in perms:
	number=""
	n=1
	m=int("".join(str(q) for q in p))
	while (len(number)<9):
		number=number + str(m*n)
		if len(number)==9:
			if "0" not in number:
				panTest.append(number)
		n=n+1
for pt in panTest:
	usedNumbers=[]
	for c in pt:
		if int(c) not in usedNumbers and int(c) in numbers:
			usedNumbers.append(int(c))
	if len(usedNumbers)==9:
		if int(pt) > largestPandigital:
			largestPandigital = int(pt)
print(largestPandigital)