From bfab0b1af2423768ba1bf25942953f982637ca81 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Mon, 30 Aug 2021 15:22:38 +0200
Subject: [PATCH] Solution to problem 32 in Julia

---
 src/Julia/Problem032.jl | 47 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 47 insertions(+)
 create mode 100644 src/Julia/Problem032.jl

diff --git a/src/Julia/Problem032.jl b/src/Julia/Problem032.jl
new file mode 100644
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+++ b/src/Julia/Problem032.jl
@@ -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())