2020-01-16 14:19:02 +00:00
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""" Problem Statement (Digit Fifth Power ): https://projecteuler.net/problem=30
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Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
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1634 = 1^4 + 6^4 + 3^4 + 4^4
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8208 = 8^4 + 2^4 + 0^4 + 8^4
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9474 = 9^4 + 4^4 + 7^4 + 4^4
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As 1 = 1^4 is not a sum it is not included.
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The sum of these numbers is 1634 + 8208 + 9474 = 19316.
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Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
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(9^5)=59,049
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59049*7=4,13,343 (which is only 6 digit number )
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So, number greater than 9,99,999 are rejected
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and also 59049*3=1,77,147 (which exceeds the criteria of number being 3 digit)
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So, n>999
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and hence a bound between (1000,1000000)
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"""
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def digitsum(s: str) -> int:
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"""
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>>> all(digitsum(str(i)) == (1 if i == 1 else 0) for i in range(100))
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True
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"""
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i = sum(pow(int(c), 5) for c in s)
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return i if i == int(s) else 0
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if __name__ == "__main__":
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2020-01-18 12:24:33 +00:00
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count = sum(digitsum(str(i)) for i in range(1000, 1000000))
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2020-01-16 14:19:02 +00:00
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print(count) # --> 443839
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