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* Pyupgrade to Python 3.9 * updating DIRECTORY.md Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
52 lines
1.2 KiB
Python
52 lines
1.2 KiB
Python
"""
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Problem 119: https://projecteuler.net/problem=119
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Name: Digit power sum
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The number 512 is interesting because it is equal to the sum of its digits
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raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a number
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with this property is 614656 = 28^4. We shall define an to be the nth term of
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this sequence and insist that a number must contain at least two digits to have a sum.
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You are given that a2 = 512 and a10 = 614656. Find a30
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"""
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import math
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def digit_sum(n: int) -> int:
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"""
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Returns the sum of the digits of the number.
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>>> digit_sum(123)
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6
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>>> digit_sum(456)
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15
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>>> digit_sum(78910)
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25
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"""
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return sum(int(digit) for digit in str(n))
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def solution(n: int = 30) -> int:
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"""
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Returns the value of 30th digit power sum.
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>>> solution(2)
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512
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>>> solution(5)
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5832
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>>> solution(10)
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614656
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"""
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digit_to_powers = []
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for digit in range(2, 100):
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for power in range(2, 100):
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number = int(math.pow(digit, power))
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if digit == digit_sum(number):
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digit_to_powers.append(number)
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digit_to_powers.sort()
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return digit_to_powers[n - 1]
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if __name__ == "__main__":
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print(solution())
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