Solution to problem 43 in Julia

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David Doblas Jiménez 2021-09-13 20:54:00 +02:00
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#=
Created on 13 Sep 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem 43 of Project Euler
https://projecteuler.net/problem=43
=#
using BenchmarkTools
using Combinatorics
function Problem43()
#=
The number, 1406357289, is a 0 to 9 pandigital number because
it is made up of each of the digits 0 to 9 in some order, but
it also has a rather interesting sub-string divisibility property.
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this
way, we note the following:
d2d3d4=406 is divisible by 2
d3d4d5=063 is divisible by 3
d4d5d6=635 is divisible by 5
d5d6d7=357 is divisible by 7
d6d7d8=572 is divisible by 11
d7d8d9=728 is divisible by 13
d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
=#
ans = []
pandigital = join(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'])
for n in permutations(pandigital)
n_ =join(n)
if n_[1] != 0 && join(sort(n)) == pandigital
if parse(Int, n_[8:end]) % 17 == 0
if parse(Int, n_[7:9]) % 13 == 0
if parse(Int, n_[6:8]) % 11 == 0
if parse(Int, n_[5:7]) % 7 == 0
if parse(Int, n_[4:6]) % 5 == 0
if parse(Int, n_[3:5]) % 3 == 0
if parse(Int, n_[2:4]) % 2 == 0
push!(ans, n_)
end
end
end
end
end
end
end
end
end
return sum(parse(Int, num) for num in ans)
end
println("Time to evaluate Problem 43:")
@btime Problem43()
println("")
println("Result for Problem 43: ", Problem43())