# Problem 38: Pandigital multiples
# 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?
pand = '123456789'
best = 0
for value in range(1,1000000):
  for N in range(2,10):
    v = []
    for n in range(1,N+1):
      v.extend([s for s in str(n*value)])
    if len(v)>9:
      break
    s = ''.join(v)
    if ''.join(sorted(s))==pand and int(s)>best:
      best = int(s)
print(best)
      