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Problem1.m

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+%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
+%}