Initial commit with existing files

Problem16.m

@@ -0,0 +1,90 @@
+%ProjectEuler/Octave/Problem16.m
+%Matthew Ellison
+% Created: 
+%Modified: 03-28-19
+%What is the sum of the digits of the number 2^1000?
+%{
+	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
+numberToPower = 2;
+POWER = 1000;
+nums = [1];
+carry = 0;
+currentPower = 0;
+
+%Start the timer
+startTime = clock();
+
+%Loop until you reach the correct power to raise the number to
+while(currentPower < POWER)
+	counter = 1;
+	%Loop through every element in the list and multiply by 2
+	while(counter <= size(nums)(2))
+		nums(counter) = (nums(counter) * numberToPower) + carry;
+		carry = 0;
+		%If one of the elements is >= 10 you need to carry to the next element
+		if(nums(counter) >= 10)
+			nums(counter) -= 10;
+			++carry;
+		end
+		++counter;
+	end
+	%If you have something to carry after everything has been multiplied you need to add a new column
+	if(carry > 0)
+		nums(end + 1) = carry;
+		carry = 0;
+	end
+	++currentPower;
+end
+
+%Stop the timer
+endTime = clock();
+
+%Print the results
+nums = fliplr(nums);	%Flip the results so it is easier to move to a string
+counter = 1;
+number = [''];
+while(counter <= size(nums)(2))
+	number(counter) = num2str(nums(counter));	%Convert each number to a character
+	++counter;
+end
+
+%Print the results
+printf("2^1000 = %s\n", number)
+printf("The sum of the digits is: %d\n", sum(nums))
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+
+%Cleanup your variables
+clear numberToPower;
+clear POWER;
+clear nums;
+clear carry;
+clear currentPower;
+clear counter;
+clear startTime;
+clear endTime;
+clear number;
+clear ans;
+
+%{
+Results:
+2^1000 = 10715086071862673209484250490600018105614048117055336074437503883703510511249361224931983788156958581275946729175531468251871452856923140435984577574698574803934567774824230985421074605062371141877954182153046474983581941267398767559165543946077062914571196477686542167660429831652624386837205668069376
+The sum of the digits is: 1366
+It took 3.203796 seconds to run this algorithm
+%}