012 021 102 120 201 210
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3,
4, 5, 6, 7, 8 and 9?
"""
import euler_utils as utils
digits = [chr(i+48) for i in range(0,10)]
count = 0
permutation = []
for position in range(9,0,-1):
#we could have just divided here, it will straightaway give us the index
factorial = utils.factorial_of(position)
for i in range(len(digits)):
count += factorial
if count >= 1000000:
count -= factorial
permutation.append(digits[i])
digits.remove(digits[i])
break
permutation = permutation + digits
print ''.join(permutation)
"""
Project Euler Problem 34
========================
145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
Find the sum of all numbers which are equal to the sum of the factorial of
their digits.
Note: as 1! = 1 and 2! = 2 are not sums they are not included.
"""
import euler_utils as utils
factorials = [utils.factorial_of(i) for i in range(0,10)]
total_sum = 0
for i in range(10,7*utils.factorial_of(9)):
if i == sum([factorials[int(d)] for d in str(i)]):
total_sum += i
print total_sum