diff --git a/Problem1.m b/Problem1.m
index 27068d4..4d56d56 100644
--- a/Problem1.m
+++ b/Problem1.m
@@ -1,10 +1,11 @@
+function [] = Problem1()
 %ProjectEuler/Octave/Problem1.m
 %Matthew Ellison
-% Created: 
-%Modified: 03-28-19
+% Created: 03-28-19
+%Modified: 10-26-20
 %What is the sum of all the multiples of 3 or 5 that are less than 1000
 %{
-	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
@@ -21,42 +22,32 @@
 %}
 
 
-%Setup your variables
-fullSum = 0;	%To hold the sum of all the numbers
-numbers = 0;	%To hold all of the numbers
-counter = 0;	%The number. It must stay below 1000
+	%Setup your variables
+	fullSum = 0;	%To hold the sum of all the numbers
+	numbers = 0;	%To hold all of the numbers
+	counter = 0;	%The number. It must stay below 1000
 
-%Start the timer
-startTime = clock();
+	%Start the timer
+	startTime = clock();
 
-%When done this way it removes the possibility of duplicate numbers
-while(counter < 1000)
-	%See if the number is a multiple of 3
-	if(mod(counter, 3) == 0)
-		numbers(end + 1) = counter;
-	%See if the number is a multiple of 5
-	elseif(mod(counter, 5) == 0)
-		numbers(end + 1) = counter;
-	end
+	%Get the sum of the progressions of 3 and 5 and remove the sum of progressions of the overlap
+	sumOfMultiples = sumOfProgression(3) + sumOfProgression(5) - sumOfProgression(3 * 5);
 
-	%Increment the number
-	++counter;
-end
+	%Stop the timer
+	endTime = clock();
 
-%Stop the timer
-endTime = clock();
+	%Print the results
+	printf("The sum of all the numbers less than 1000 that is divisibly by 3 or 5 is: %d\n", sumOfMultiples)
+	printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
 
-%Print the results
-printf("The sum of all the numbers less than 1000 that is divisibly by 3 or 5 is: %d\n", sum(numbers))
-printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+end %End of Problem1()
 
-%Cleanup your variables
-clear fullSum;
-clear numbers;
-clear counter;
-clear startTime;
-clear endTime;
-clear ans;
+%Gets the sum of the progression of the multiple
+function [sum] = sumOfProgression(multiple)
+	numTerms = floor(999 / multiple);	%This gets the number of multiples of a particular number that is < MAX_NUMBER
+	%The sum of progression formula is (n / 2)(a + l). n = number of terms, a = multiple, l = last term
+	sum = ((numTerms / 2) * (multiple + (numTerms * multiple)));
+end %End of sumOfProgression
 
 %{
 Results: