"""
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 permutations
fact = {}
for x in xrange(0, 10):
f = 1
for y in xrange(2, x + 1):
f *= y
fact[x] = f
factSum = 0
for i in xrange(10, 1000*100): # found out experimentally
if i == sum([fact[j] for j in permutations.split_into_digits(i)]):
factSum += i
print str(factSum)
The number, 197, is called a circular prime because all rotations of the digits:
197, 971, and 719, are themselves prime.
There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
How many circular primes are there below one million?
"""
from collections import deque
import primes
import permutations
allPrimes = primes.create_primes(1000*1000)
countOfCircularPrimes = 0
for prime in allPrimes:
isCircularPrime = True
digits = deque(permutations.split_into_digits(prime))
for i in xrange(0, len(digits)):
digits.rotate(1)
newPrime = permutations.combine_numbers(digits)
if newPrime not in allPrimes:
isCircularPrime = False
break
if isCircularPrime:
countOfCircularPrimes += 1
print str(countOfCircularPrimes)
