Solution to problem 5 in Julia
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src/Julia/Problem005.jl
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src/Julia/Problem005.jl
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Created on 20 Jun 2021
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for Problem 5 of Project Euler
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https://projecteuler.net/problem=5
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=#
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#=
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The LCM of two natural numbers x and y is given by:
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def lcm(x, y):
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return x * y // math.gcd(x, y)
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It is possible to compute the LCM of more than two numbers by iteratively
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computing the LCM of two numbers, i.e. LCM(a, b, c) = LCM(a, LCM(b, c))
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=#
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function Problem5()
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#=
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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 of
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the numbers from 1 to 20?
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=#
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ans = 1
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for i in 1:21
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ans *= i ÷ gcd(i, ans)
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end
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return ans
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end
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println("Time to evaluate Problem 5:")
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@time Problem5()
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println("")
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println("Result for Problem 5: ", Problem5())
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