Solution to problem 43 in Julia
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src/Julia/Problem043.jl
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src/Julia/Problem043.jl
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#=
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Created on 13 Sep 2021
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for Problem 43 of Project Euler
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https://projecteuler.net/problem=43
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=#
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using BenchmarkTools
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using Combinatorics
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function Problem43()
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#=
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The number, 1406357289, is a 0 to 9 pandigital number because
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it is made up of each of the digits 0 to 9 in some order, but
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it also has a rather interesting sub-string divisibility property.
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Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this
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way, we note the following:
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d2d3d4=406 is divisible by 2
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d3d4d5=063 is divisible by 3
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d4d5d6=635 is divisible by 5
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d5d6d7=357 is divisible by 7
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d6d7d8=572 is divisible by 11
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d7d8d9=728 is divisible by 13
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d8d9d10=289 is divisible by 17
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Find the sum of all 0 to 9 pandigital numbers with this property.
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=#
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ans = []
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pandigital = join(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'])
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for n in permutations(pandigital)
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n_ =join(n)
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if n_[1] != 0 && join(sort(n)) == pandigital
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if parse(Int, n_[8:end]) % 17 == 0
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if parse(Int, n_[7:9]) % 13 == 0
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if parse(Int, n_[6:8]) % 11 == 0
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if parse(Int, n_[5:7]) % 7 == 0
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if parse(Int, n_[4:6]) % 5 == 0
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if parse(Int, n_[3:5]) % 3 == 0
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if parse(Int, n_[2:4]) % 2 == 0
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push!(ans, n_)
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end
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end
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end
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end
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end
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end
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end
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end
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end
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return sum(parse(Int, num) for num in ans)
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end
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println("Time to evaluate Problem 43:")
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@btime Problem43()
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println("")
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println("Result for Problem 43: ", Problem43())
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