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As described in CONTRIBUTING.md > Expand acronyms because gcd() is hard to understand but greatest_common_divisor() is not. Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com>
76 lines
1.5 KiB
Python
76 lines
1.5 KiB
Python
"""
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Project Euler Problem 5: https://projecteuler.net/problem=5
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Smallest multiple
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2520 is the smallest number that can be divided by each of the numbers
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from 1 to 10 without any remainder.
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What is the smallest positive number that is _evenly divisible_ by all
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of the numbers from 1 to 20?
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References:
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- https://en.wiktionary.org/wiki/evenly_divisible
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- https://en.wikipedia.org/wiki/Euclidean_algorithm
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- https://en.wikipedia.org/wiki/Least_common_multiple
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"""
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def greatest_common_divisor(x: int, y: int) -> int:
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"""
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Euclidean Greatest Common Divisor algorithm
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>>> greatest_common_divisor(0, 0)
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0
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>>> greatest_common_divisor(23, 42)
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1
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>>> greatest_common_divisor(15, 33)
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3
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>>> greatest_common_divisor(12345, 67890)
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15
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"""
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return x if y == 0 else greatest_common_divisor(y, x % y)
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def lcm(x: int, y: int) -> int:
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"""
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Least Common Multiple.
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Using the property that lcm(a, b) * greatest_common_divisor(a, b) = a*b
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>>> lcm(3, 15)
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15
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>>> lcm(1, 27)
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27
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>>> lcm(13, 27)
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351
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>>> lcm(64, 48)
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192
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"""
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return (x * y) // greatest_common_divisor(x, y)
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def solution(n: int = 20) -> int:
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"""
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Returns the smallest positive number that is evenly divisible (divisible
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with no remainder) by all of the numbers from 1 to n.
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>>> solution(10)
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2520
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>>> solution(15)
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360360
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>>> solution(22)
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232792560
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"""
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g = 1
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for i in range(1, n + 1):
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g = lcm(g, i)
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return g
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if __name__ == "__main__":
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print(f"{solution() = }")
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