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Problem12.m

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+function [] = Problem12()
+%ProjectEuler/Octave/Problem12.m
+%Matthew Ellison
+% Created: 
+%Modified: 03-28-19
+%What is the value of the first triangle number to have over five hundred divisors?
+%{
+	Copyright (C) 2019  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
+	the Free Software Foundation, either version 3 of the License, or
+	(at your option) any later version.
+
+	This program is distributed in the hope that it will be useful,
+	but WITHOUT ANY WARRANTY; without even the implied warranty of
+	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+	GNU Lesser General Public License for more details.
+
+	You should have received a copy of the GNU Lesser General Public License
+	along with this program.  If not, see <https://www.gnu.org/licenses/>.
+%}
+
+
+%Setup your variables
+number = 1;		%To hold the current triangle number
+nextNumber = 2;	%To hold the current number you have counted up to
+found = false;	%To tell when the answer has been found
+numDivisors = 0;
+
+%Start the timer
+startTime = clock();
+
+%Keep checking to find the correct numbers until you 
+while((~found) && (number > 0))
+	%See how many divisors the number has
+	numDivisors = size(getDivisors(number))(2);
+
+	%If it is enough set the flag to stop the loop
+	if(numDivisors > 500)
+		found = true;
+	else
+		number += nextNumber;
+		nextNumber += 1;
+	end
+end
+
+%Stop the timer
+endTime = clock();
+
+%Print your result
+printf("The first triangular number with more than 500 divisors is %d\n", number)
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+
+end %End of Problem12()
+
+function [divisors] = getDivisors(goalNumber)
+	divisors = [];
+	%Start by checking that the number is positive
+	if(goalNumber <= 0)
+		return;
+	%If the number is 1 return just itself
+	elseif(goalNumber == 1)
+		divisors(end + 1) = 1;
+		return;
+	%Otherwise add 1 and itself to the list
+	else
+		divisors(end + 1) = 1;
+		divisors(end + 1) = goalNumber;
+	end
+
+	%Start at 3 and loop through all numbers < sqrt(goalNumber) looking for a number that divides it evenly
+	topPossibleDivisor = ceil(sqrt(goalNumber));
+	for possibleDivisor = 2 : topPossibleDivisor
+		%If you find one add it and the number it creates to the list
+		if(mod(goalNumber, possibleDivisor) == 0)
+			divisors(end + 1) = possibleDivisor;
+			%Account for the possibility sqrt(goalNumber) being a divisor
+			if(possibleDivisor != topPossibleDivisor)
+				divisors(end + 1) = goalNumber / possibleDivisor;
+			end
+		end
+	end
+
+	%Sort the list before returning for neatness
+	divisors = sort(divisors);
+end
+
+%{
+Results:
+The first triangular number with more than 500 divisors is 76576500
+It took 344.606964 seconds to run this algorithm
+%}