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Created problem_34 in project_euler #2305

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Aug 13, 2020
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1 change: 1 addition & 0 deletions project_euler/problem_34/__init__.py
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#
76 changes: 76 additions & 0 deletions project_euler/problem_34/sol1.py
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"""
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.

"""


def factorial(n: int) -> int:
"""Return the factorial of n.

>>> factorial(5)
120
>>> factorial(1)
1
>>> factorial(0)
0
>>> factorial(-1)
Traceback (most recent call last):
...
ValueError: n must be >= 0

>>> factorial(1.1)
Traceback (most recent call last):
...
ValueError: n must be exact integer
"""

import math

if not n >= 0:
raise ValueError("n must be >= 0")
if math.floor(n) != n:
raise ValueError("n must be exact integer")
if n + 1 == n: # catch a value like 1e300
raise OverflowError("n too large")
result = 1
factor = 2
while factor <= n:
result *= factor
factor += 1
return result


def sum_of_digit_factorial(n: int) -> int:
"""
Returns the sum of the digits in n
>>> sum_of_digit_factorial(15)
121
>>> sum_of_digit_factorial(0)
1
"""
digits = list(map(int, str(n)))
summ = sum(factorial(digit) for digit in digits)
return summ


def compute() -> int:
"""
Returns the sum of all numbers whose
sum of the factorials of all digits
add up to the number itself.
>>> compute()
40730
"""
the_list = []
for num in range(3, 7 * factorial(9) + 1):
if sum_of_digit_factorial(num) == num:
the_list.append(num)
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Ok sir. Will do it

return sum(the_list)


if __name__ == "__main__":
print(compute())