ProjectEulerOctave/Problem1.m

%ProjectEuler/Octave/Problem1.m
%Matthew Ellison
% Created: 
%Modified: 03-28-19
%What is the sum of all the multiples of 3 or 5 that are less than 1000
%{
	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
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();

%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

	%Increment the number
	++counter;
end

%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", sum(numbers))
printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))

%Cleanup your variables
clear fullSum;
clear numbers;
clear counter;
clear startTime;
clear endTime;
clear ans;

%{
Results:
The sum of all the numbers less than 1000 that is divisibly by 3 or 5 is: 233168
It took 0.014717 seconds to run this algorithm
%}