From e7c2e60c25b3f2d504961335169ab00f7ce67c39 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Wed, 8 Sep 2021 21:00:32 +0200
Subject: [PATCH] Solution to problem 38 in Julia

---
 src/Julia/Problem038.jl | 56 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 56 insertions(+)
 create mode 100644 src/Julia/Problem038.jl

diff --git a/src/Julia/Problem038.jl b/src/Julia/Problem038.jl
new file mode 100644
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+++ b/src/Julia/Problem038.jl
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+#=
+Created on 08 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 38 of Project Euler
+https://projecteuler.net/problem=38
+=#
+
+using BenchmarkTools
+
+function Problem38()
+    #=
+    Take the number 192 and multiply it by each of 1, 2, and 3:
+
+      192 × 1 = 192
+      192 × 2 = 384
+      192 × 3 = 576
+
+    By concatenating each product we get the 1 to 9 pandigital,
+    192384576. We will call 192384576 the concatenated product of
+    192 and (1,2,3)
+
+    The same can be achieved by starting with 9 and multiplying by
+    1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is
+    the concatenated product of 9 and (1,2,3,4,5).
+
+    What is the largest 1 to 9 pandigital 9-digit number that can
+    be formed as the concatenated product of an integer with
+    (1,2, ... , n) where n > 1?
+    =#
+
+    results = []
+    pandigital = join(['1', '2', '3', '4', '5', '6', '7', '8', '9'])
+    # Number must 4 digits (exactly) to be pandigital
+    # if n > 1
+    for i in 1:10_000
+        integer = 1
+        number = ""
+        while length(number) < 9
+            number *= string(integer * i)
+            if join(sort(collect(number))) == pandigital
+                push!(results,number)
+            end
+            integer += 1
+        end
+    end
+    return maximum(results)
+end
+
+
+println("Time to evaluate Problem 38:")
+@btime Problem38()
+println("")
+println("Result for Problem 38: ", Problem38())