From 20c35a6140364a8148581549db28987e75f88006 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Tue, 14 Sep 2021 18:55:33 +0200
Subject: [PATCH] Solution to problem 44 in Julia

---
 src/Julia/Problem044.jl | 51 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 51 insertions(+)
 create mode 100644 src/Julia/Problem044.jl

diff --git a/src/Julia/Problem044.jl b/src/Julia/Problem044.jl
new file mode 100644
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+++ b/src/Julia/Problem044.jl
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+#=
+Created on 14 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 44 of Project Euler
+https://projecteuler.net/problem=44
+=#
+
+using BenchmarkTools
+using Combinatorics
+
+function pentagonal(n)
+    return Int(n*(3*n-1)/2)
+end
+
+function Problem44()
+    #=
+    Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2.
+    The first ten pentagonal numbers are:
+
+        1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
+
+    It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their
+    difference, 70 − 22 = 48, is not pentagonal.
+
+    Find the pair of pentagonal numbers, Pj and Pk, for which their
+    sum and difference are pentagonal and D = |Pk − Pj| is minimised.
+
+    What is the value of D?
+    =#
+    dif = 0
+    pentagonal_list = [pentagonal(n) for n in 1:2500]
+    pairs = combinations(pentagonal_list, 2)
+    for p in pairs
+        if reduce(+, p) in pentagonal_list && abs(reduce(-, p)) in pentagonal_list
+            dif = (abs(reduce(-,p)))
+            # the first one found would be the smallest
+            break
+        end
+    end
+
+    return dif
+end
+
+
+println("Time to evaluate Problem 44:")
+@btime Problem44()
+println("")
+println("Result for Problem 44: ", Problem44())