def solve(nthpow):
"""returns sum of numbers that can be written as sum of the nthpow of
their digits
"""
from functools import partial
from utils import sum_digits
power = partial(pow, exponent=nthpow)
return filter(lambda n: n == sum_digits(n, power), xrange(2, upper(nthpow)))
def curious_numbers():
"""yields all curious numbers. Upper bound can be found by taking a
n-digit number, then the max sum of its factorials is n*9!. So we
want to find the limit on n. This happens when 10**(n-1) > n*9!,
or when n~=7.
"""
for curr in range(3, 10**6):
if curr == sum_digits(curr, math.factorial):
yield curr
def main():
total = 0
for num in range(2, 100):
if num not in {4, 9, 16, 25, 36, 49, 64, 81}:
root = isqrt(num * (10**2)**99)
total += sum_digits(root)
return total
def main():
total = 0
for num in range(2, 100):
if num not in {4, 9, 16, 25, 36, 49, 64, 81}:
root = isqrt(num * (10**2)**99)
total += sum_digits(root)
return total
def main():
cnt = 0
for num_digits in count(2):
max_num = 10**num_digits
min_num = max_num // 10
for digit_sum in range(2, num_digits * 9 + 1):
num = digit_sum * digit_sum
p = 2
while num < max_num:
if min_num <= num < max_num and num == sum_digits(num)**p:
cnt += 1
if cnt == 30:
return num
num *= digit_sum
p += 1
def main():
cnt = 0
for num_digits in count(2):
max_num = 10**num_digits
min_num = max_num//10
for digit_sum in range(2, num_digits*9 + 1):
num = digit_sum * digit_sum
p = 2
while num < max_num:
if min_num <= num < max_num and num == sum_digits(num)**p:
cnt += 1
if cnt == 30:
return num
num *= digit_sum
p += 1
def test_sum_digits(self):
self.assertEqual(sum_digits(0), 0)
self.assertEqual(sum_digits(1), 1)
self.assertEqual(sum_digits(1000), 1)
self.assertEqual(sum_digits(1234567890), 45)
''' n! means n x (n - 1) x ... x 3 x 2 x 1 Find the sum of the digits in the number 100! ''' from time import time from math import factorial from utils import sum_digits start=time() f=factorial(10000) result=sum_digits(f) print 'The answer is: '+str(result) print 'Calculated in '+str(round(time()-start,5))+'s.'
''' 2^(15) = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. What is the sum of the digits of the number 2^(1000)? ''' from time import time from utils import sum_digits start=time() result = sum_digits(pow(2,1000)) print 'The answer is: '+str(result) print 'Calculated in '+str(round(time()-start,5))+'s.'
def test_sum_digits(self):
self.assertEqual(sum_digits(0), 0)
self.assertEqual(sum_digits(1), 1)
self.assertEqual(sum_digits(1000), 1)
self.assertEqual(sum_digits(1234567890), 45)
from utils import factorial, sum_digits print(sum_digits(factorial(100)))