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Problem23.m

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+function [] = Problem23()
+%ProjectEuler/Octave/Problem23.m
+%Matthew Ellison
+% Created: 03-22-19
+%Modified: 03-28-19
+%Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers
+%{
+	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/>.
+%}
+
+
+MAX_NUM = 28123;	%The highest number that will be evaluated
+
+
+%Setup the variables
+divisorSums = [];
+%Make sure every element has a 0 in it's location
+for cnt = 1 : MAX_NUM
+	divisorSums(end + 1) = 0;
+end
+
+%Start the timer
+startTime = clock();
+
+%Get the sum of the divisors of all numbers < MAX_NUM
+for cnt = 1 : MAX_NUM
+	div = getDivisors(cnt);
+	if(size(div)(2) > 1)
+		div(end) = [];
+	end
+	divisorSums(cnt) = sum(div);
+end
+
+%Get the abundant numbers
+abund = [];
+for cnt = 1 : size(divisorSums)(2)
+	if(divisorSums(cnt) > cnt)
+		abund(end + 1) = cnt;
+	end
+end
+
+%Check if each number can be the sum of 2 abundant numbers and add to the sum if no
+sumOfNums = 0;
+for cnt = 1 : MAX_NUM
+	if(~isSum(abund, cnt))
+		printf("Added %d to sum\n", cnt)
+		sumOfNums += cnt;
+	end
+end
+
+%Stop the timer
+endTime = clock();
+
+%Print the results
+printf("The answer is %d\n", sumOfNums)
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+
+end	%Problem23
+
+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;
+	end
+
+	%Start at 3 and loop through all numbers < sqrt(goalNumber) looking for a number that divides it evenly
+	topPossibleDivisor = ceil(sqrt(goalNumber));
+	possibleDivisor = 1;
+	while(possibleDivisor <= 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
+			%Take care of a few occations where a number was added twice
+			if(divisors(end) == (possibleDivisor + 1))
+				possibleDivisor += 1;
+			end
+		end
+		possibleDivisor += 1;
+	end
+
+	%Sort the list before returning for neatness
+	divisors = sort(divisors);
+end
+
+function [answer] = isSum(abund, num)
+	sumOfNums = 0;
+	%Pick a number for the first part of the sum
+	for firstNum = 1 : size(abund)(2)
+		%Pick a number for the second part of the sum
+		for secondNum = firstNum : size(abund)(2)
+			sumOfNums = abund(firstNum) + abund(secondNum);
+			if(sumOfNums == num)
+				answer = true;
+				return;
+			elseif(sumOfNums > num)
+				break;
+			end
+		end
+	end
+	answer = false;
+end
+
+%{
+Results: I let this run for 11 hours and did not come up with a result. I added some tracking and tested again and it seems like it is doing what it is supposed to and will come to the correct answer... eventually
+However, for now, this remains unproven.
+%}