Solution to problem 48 in Julia

src/Julia/Problem048.jl

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+#=
+Created on 18 Sep 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 48 of Project Euler
+https://projecteuler.net/problem=48
+=#
+
+using BenchmarkTools
+
+function Problem48()
+    #=
+    The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317.
+
+    Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000.
+    =#
+    series = sum(big(i)^i for i in 1:1000)
+    return string(series)[end-9:end]
+end
+
+
+println("Time to evaluate Problem 48:")
+@btime Problem48()
+println("")
+println("Result for Problem 48: ", Problem48())