Updated problems to be more in line with conventions

Problem21.lua

@@ -1,11 +1,11 @@
 --ProjectEuler/lua/Problem21.lua
 --Matthew Ellison
 -- Created: 03-19-19
---Modified: 03-28-19
+--Modified: 06-19-20
 --Evaluate the sum of all the amicable numbers under 10000
 --All of my requires, unless otherwise listed, can be found at https://bitbucket.org/Mattrixwv/luaClasses
 --[[
-	Copyright (C) 2019  Matthew Ellison
+	Copyright (C) 2020  Matthew Ellison
 
 	This program is free software: you can redistribute it and/or modify
 	it under the terms of the GNU Lesser General Public License as published by
@@ -29,26 +29,26 @@ LIMIT = 10000;	--The top number that will be evaluated
 
 
 --Setup the timer
-timer = Stopwatch:create();
+local timer = Stopwatch:create();
 
 --Setup the variables
-divisorSum = {};	--Holds the sum of the factors of the subscript number
+local divisorSum = {};	--Holds the sum of the factors of the subscript number
 
 --Start the timer
 timer:start();
 
 --Generate the factors of all the numbers < 10000, get their sum, and add it to the list
 for cnt = 1, LIMIT do
-	divisors = getDivisors(cnt);	--Get all the divisors of a number
+	local divisors = getDivisors(cnt);	--Get all the divisors of a number
 	if(#divisors > 1) then
 		table.remove(divisors);	--Remove the last entry because it will be the number itself
 	end
 	divisorSum[#divisorSum + 1] = getSum(divisors);	--Add the sum of the divisors to the vector
 end
 --Check every sum of divisors in the list for a matching sum
-amicable = {};
+local amicable = {};
 for cnt = 1, #divisorSum do
-	sum = divisorSum[cnt];
+	local sum = divisorSum[cnt];
 	--If the sum is greater than the number of divisors then it is impossible to be amicable. Skip the number and continue
 	if(sum >= #divisorSum) then
 	--We know that divisorSum[cnt] == sum, do if divisorSum[sum] == cnt we have found an amicable number