First commit with 3 problems

Problem1.c

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+//ProjectEuler/C/Problem1.c
+//Matthew Ellison
+// Created: 03-08-19
+//Modified: 03-28-19
+//What is the sum of all the multiples of 3 or 5 that are less than 1000
+//Unless otherwise listed all non-standard includes are my own creation and available from https://bibucket.org/Mattrixwv/myHelpers
+/*
+	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/>.
+*/
+
+
+#include <stdio.h>
+#include <inttypes.h>
+#include "Stopwatch.h"
+
+const int MAX_NUMBER = 1000;
+
+
+int main(){
+	//Setup the stopwatch
+	struct Stopwatch timer;
+	initStopwatch(&timer);
+	//Setup the variables you will need
+
+	uint64_t fullSum = 0;	//For the sum of all the numbers
+
+	//Start the timer
+	startStopwatch(&timer);
+
+	//Step through ever number < 1000 and see if either 3 or 5 divides it evenly
+	for(uint64_t cnt = 1;cnt < MAX_NUMBER;++cnt){
+		if((cnt % 3) == 0){
+			fullSum += cnt;
+		}
+		else if((cnt % 5) == 0){
+			fullSum += cnt;
+		}
+	}
+
+	//Stop the timer
+	stopStopwatch(&timer);
+	char* timerStr = getStrStopwatch(&timer);
+
+	//Print the results
+	printf("The sum of all the numbers < %d that are divisible by 3 or 5 is %lu\n", MAX_NUMBER, fullSum);
+	printf("It took %s to run this algorithm\n", timerStr);
+
+	//Free the memory taken up by the string
+	free(timerStr);
+	timerStr = NULL;
+
+	return 0;
+}
+
+/* Results:
+The sum of all the numbers < 1000 that are divisible by 3 or 5 is 233168
+It took 3.000 microseconds to run this algorithm
+*/