Added solution to problem 34

Problem34.m

@@ -0,0 +1,68 @@
+function [] = Problem34()
+%ProjectEuler/ProjectEulerOctave/Problem34.lua
+%Matthew Ellison
+% Createe: 06-01-21
+%Modified: 06-01-21
+%Find the sum of all numbers which are equal to the sum of the factorial of their digits
+%{
+	Copyright (C) 2021  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 the variables
+
+%Start the timer
+startTime = clock();
+MAX_NUM = 1499999;	%The largest num that can be the sum of its own digits
+numberFactorials = [];	%Holds the pre-computed factorials of the numbers 0-9
+totalSum = 0;	%Holds the sum of all numbers equal to the sum of their digit's factorials
+
+%Pre-compute the possible factorials from 0! to 9!
+for cnt = 1 : 9
+	numberFactorials(cnt) = factorial(cnt);
+end
+%Run through all possible numbers from 3-MAX_NUM and see if they equal the sum of their digit's factorials
+for cnt = 3 : MAX_NUM
+	%Split the number into its digits and add each one to the sum
+	numString = num2str(cnt);
+	currentSum = 0;
+	for numCnt = 1 : size(numString)(2)
+		num = str2num(numString(numCnt));
+		if(num != 0)
+			currentSum += numberFactorials(num);
+		else
+			currentSum += 1;
+		end
+	end
+	%If the number is equal to the sum add the sum to the running sum
+	if(currentSum == cnt)
+		totalSum += currentSum;
+	end
+end
+
+%Stop the timer
+endTime = clock();
+
+%Print the results
+printf("The sum of all numbers that are the sum of their digit's factorials is %d\n", totalSum);
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+
+end
+
+%{
+Results:
+The sum of all numbers that are the sum of their digit's factorials is 40730
+It took 1988.403809 seconds to run this algorithm
+%}