From aaa12e97d61e881a87642a3103265aed4d12ab92 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Wed, 11 Aug 2021 20:29:41 +0200
Subject: [PATCH] Solution to problem 23 in Julia

---
 src/Julia/Problem023.jl | 60 +++++++++++++++++++++++++++++++++++++++++
 1 file changed, 60 insertions(+)
 create mode 100644 src/Julia/Problem023.jl

diff --git a/src/Julia/Problem023.jl b/src/Julia/Problem023.jl
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+++ b/src/Julia/Problem023.jl
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+#=
+Created on 11 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 23 of Project Euler
+https://projecteuler.net/problem=23
+=#
+
+function Problem23()
+    #=
+    A perfect number is a number for which the sum of its proper divisors is
+    exactly equal to the number. For example, the sum of the proper divisors
+    of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect
+    number.
+
+    A number n is called deficient if the sum of its proper divisors is less
+    than n and it is called abundant if this sum exceeds n.
+
+    As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest
+    number that can be written as the sum of two abundant numbers is 24. By
+    mathematical analysis, it can be shown that all integers greater than 28123
+    can be written as the sum of two abundant numbers. However, this upper
+    limit cannot be reduced any further by analysis even though it is known
+    that the greatest number that cannot be expressed as the sum of two
+    abundant numbers is less than this limit.
+
+    Find the sum of all the positive integers which cannot be written as the
+    sum of two abundant numbers.
+    =#
+    LIMIT = 28123
+    divisorsum = zeros(Int32,LIMIT)
+    for i in range(1, LIMIT, step=1)
+        for j in range(2i, LIMIT, step=i)
+            divisorsum[j] += i
+        end
+    end
+    abundantnums = [i for (i, x) in enumerate(divisorsum) if x > i]
+
+    expressible = falses(LIMIT)
+    for i in abundantnums
+        for j in abundantnums
+            if i + j < LIMIT+1
+                expressible[i + j] = true
+            else
+                break
+            end
+        end
+    end
+
+    ans = sum(i for (i, x) in enumerate(expressible) if x == false)
+    return ans
+end
+
+
+println("Time to evaluate Problem 23:")
+@time Problem23()
+println("")
+println("Result for Problem 23: ", Problem23())