Solution to problem 32 in Julia
This commit is contained in:
parent
ade9db2153
commit
bfab0b1af2
47
src/Julia/Problem032.jl
Normal file
47
src/Julia/Problem032.jl
Normal file
@ -0,0 +1,47 @@
|
||||
#=
|
||||
Created on 30 Aug 2021
|
||||
|
||||
@author: David Doblas Jiménez
|
||||
@email: daviddoji@pm.me
|
||||
|
||||
Solution for Problem 32 of Project Euler
|
||||
https://projecteuler.net/problem=32
|
||||
=#
|
||||
|
||||
using BenchmarkTools
|
||||
|
||||
function Problem32()
|
||||
#=
|
||||
We shall say that an n-digit number is pandigital if it makes use of all
|
||||
the digits 1 to n exactly once; for example, the 5-digit number, 15234, is
|
||||
1 through 5 pandigital.
|
||||
|
||||
The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
|
||||
multiplicand, multiplier, and product is 1 through 9 pandigital.
|
||||
|
||||
Find the sum of all products whose multiplicand/multiplier/product identity
|
||||
can be written as a 1 through 9 pandigital.
|
||||
HINT: Some products can be obtained in more than one way so be sure to only
|
||||
include it once in your sum.
|
||||
=#
|
||||
|
||||
ans = Set()
|
||||
pandigital = join(['1', '2', '3', '4', '5', '6', '7', '8', '9'])
|
||||
|
||||
for x in 1:100
|
||||
for y in 100:10_000
|
||||
# product = x * y
|
||||
if join(sort(collect(string(x) * string(y) * string(x * y)))) == pandigital
|
||||
push!(ans, x * y)
|
||||
end
|
||||
end
|
||||
end
|
||||
|
||||
return sum(ans)
|
||||
end
|
||||
|
||||
|
||||
println("Time to evaluate Problem 32:")
|
||||
@btime Problem32()
|
||||
println("")
|
||||
println("Result for Problem 32: ", Problem32())
|
Loading…
x
Reference in New Issue
Block a user