def is_curious(n):
return sum([factorial(x) for x in get_digits(n)]) == n
def is_npower_number(i, n=4):
"""Return True if sum of `n`th power of digits in `i` equates `i`"""
assert i > 1
return i == sum(a**n for a in get_digits(i))
def truncatable_from_left(n):
d = get_digits(n, integer=False)
m = [is_prime(int("".join(d[i:]))) for i in range(len(d))]
return reduce(lambda x, y: x and y, m)
def truncatable_form_right(n):
d = get_digits(n, integer=False)
m = [is_prime(int("".join(d[:i]))) for i in range(1, len(d))] + [is_prime(n)]
return reduce(lambda x, y: x and y, m)
#!/usr/bin/python
# coding: UTF-8
"""
@author: CaiKnife
Powerful digit sum
Problem 56
A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.
Considering natural numbers of the form, ab, where a, b 100, what is the maximum digital sum?
"""
from euler import get_digits
m = 0
for a in xrange(1, 100):
for b in xrange(1, 100):
digit_sum = sum(get_digits(a ** b))
if m < digit_sum:
m = digit_sum
print m
