Solution to problem 21 in Julia

src/Julia/Problem021.jl

@@ -0,0 +1,55 @@
+#= 
+Created on 05 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 21 of Project Euler
+https://projecteuler.net/problem=21 =#
+
+
+function divisors(n)
+    divisors = Int64[1]
+    m = round(Int, n / 2)
+    for i in 2:m
+        if n % i == 0
+            push!(divisors, i)
+        end
+    end
+    return divisors
+end
+
+function Problem21()
+    #= 
+    Let d(n) be defined as the sum of proper divisors of n (numbers 
+    less than n which divide evenly into n).
+    If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable
+    pair and each of a and b are called amicable numbers.
+
+    For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 
+    44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 
+    1, 2, 4, 71 and 142; so d(284) = 220.
+
+    Evaluate the sum of all the amicable numbers under 10000 =#
+    
+    n = 9999
+    s = zeros(Int, n)
+    amicable = Int64[]
+    for i in 2:n
+        s[i] = sum(divisors(i))
+    end
+    
+    for i in 2:n
+        if s[i] <= n && i != s[i] && i == s[s[i]]
+            push!(amicable, i)
+        end
+    end
+
+    return sum(amicable)
+end
+
+
+println("Time to evaluate Problem 21:")
+@time Problem21()
+println("")
+println("Result for Problem 21: ", Problem21())