2020-01-16 14:19:02 +00:00
|
|
|
|
""" Problem Statement (Digit Fifth Power ): https://projecteuler.net/problem=30
|
|
|
|
|
|
2020-06-16 08:09:19 +00:00
|
|
|
|
Surprisingly there are only three numbers that can be written as the sum of fourth
|
|
|
|
|
powers of their digits:
|
2020-01-16 14:19:02 +00:00
|
|
|
|
|
|
|
|
|
1634 = 1^4 + 6^4 + 3^4 + 4^4
|
|
|
|
|
8208 = 8^4 + 2^4 + 0^4 + 8^4
|
|
|
|
|
9474 = 9^4 + 4^4 + 7^4 + 4^4
|
|
|
|
|
As 1 = 1^4 is not a sum it is not included.
|
|
|
|
|
|
|
|
|
|
The sum of these numbers is 1634 + 8208 + 9474 = 19316.
|
|
|
|
|
|
2020-06-16 08:09:19 +00:00
|
|
|
|
Find the sum of all the numbers that can be written as the sum of fifth powers of their
|
|
|
|
|
digits.
|
2020-05-22 06:10:11 +00:00
|
|
|
|
|
2020-01-16 14:19:02 +00:00
|
|
|
|
(9^5)=59,049
|
|
|
|
|
59049*7=4,13,343 (which is only 6 digit number )
|
|
|
|
|
So, number greater than 9,99,999 are rejected
|
|
|
|
|
and also 59049*3=1,77,147 (which exceeds the criteria of number being 3 digit)
|
|
|
|
|
So, n>999
|
|
|
|
|
and hence a bound between (1000,1000000)
|
|
|
|
|
"""
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
def digitsum(s: str) -> int:
|
|
|
|
|
"""
|
|
|
|
|
>>> all(digitsum(str(i)) == (1 if i == 1 else 0) for i in range(100))
|
|
|
|
|
True
|
|
|
|
|
"""
|
|
|
|
|
i = sum(pow(int(c), 5) for c in s)
|
|
|
|
|
return i if i == int(s) else 0
|
|
|
|
|
|
|
|
|
|
|
2020-10-06 02:51:39 +00:00
|
|
|
|
def solution() -> int:
|
|
|
|
|
return sum(digitsum(str(i)) for i in range(1000, 1000000))
|
|
|
|
|
|
|
|
|
|
|
2020-01-16 14:19:02 +00:00
|
|
|
|
if __name__ == "__main__":
|
2020-10-06 02:51:39 +00:00
|
|
|
|
print(solution())
|