From 23fa61009ddc93191208a28fb25635252733f834 Mon Sep 17 00:00:00 2001
From: Mattrixwv <m_ellison@ymail.com>
Date: Sun, 7 Feb 2021 17:26:20 -0500
Subject: [PATCH] Added solution to problem 33

---
 Problem33.m | 99 +++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 99 insertions(+)
 create mode 100644 Problem33.m

diff --git a/Problem33.m b/Problem33.m
new file mode 100644
index 0000000..09443a8
--- /dev/null
+++ b/Problem33.m
@@ -0,0 +1,99 @@
+function [] = Problem33()
+%ProjectEuler/Octave/Problem33.m
+%Matthew Ellison
+% Created: 02-07-21
+%Modified: 02-07-21
+%{
+The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s
+We shall consider fractions like, 30/50 = 3/5, to be trivial examples
+There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator
+If the product of these four fractions is given in its lowest common terms, find the value of the denominator
+%}
+%{
+	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
+MIN_NUMERATOR = 10;
+MAX_NUMERATOR = 98;
+MIN_DENOMINATOR = 11;
+MAX_DENOMINATOR = 99;
+NUMERATORS = [];
+DENOMINATORS = [];
+prodDenominator = 1;
+
+%Start the timer
+startTime = clock();
+
+%Search every possible numerator/denominator pair
+for denominator = MIN_DENOMINATOR : MAX_DENOMINATOR
+	for numerator = MIN_NUMERATOR : (denominator - 1)
+		num = num2str(numerator);
+		denom = num2str(denominator);
+		tempNum = 0;
+		tempDenom = 0;
+
+		%Check that this isn't a trivial example
+		if((num(2) == '0') && (denom(2) == '0'))
+			continue;
+		elseif(num(1) == denom(1))
+			tempNum = str2num(num(2));
+			tempDenom = str2num(denom(2));
+		elseif(num(1) == denom(2))
+			tempNum = str2num(num(2));
+			tempDenom = str2num(denom(1));
+		elseif(num(2) == denom(1))
+			tempNum = str2num(num(1));
+			tempDenom = str2num(denom(2));
+		elseif(num(2) == denom(2))
+			tempNum = str2num(num(1));
+			tempDenom = str2num(denom(1));
+		end
+		if(tempDenom == 0)
+			continue;
+		end
+
+		%Test if the new fraction is the same as the old one
+		if((tempNum / tempDenom) == (numerator / denominator))
+			numerators(end + 1) = numerator;
+			denominators(end + 1) = denominator;
+		end
+	end
+end
+
+%Get the product of the numbers
+numProd = prod(numerators);
+denomProd = prod(denominators);
+%Get the gcd to reduce to lowest terms
+GCD = gcd(numProd, denomProd);
+%Save the denominator
+prodDenominator = denomProd / GCD;
+
+
+%Stop the timer
+endTime = clock();
+
+%Print the results
+printf("The denominator of the product is %d\n", prodDenominator)
+printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
+
+end
+
+%{
+Results:
+The denominator of the product is 100
+It took 4.728714 seconds to run this algorithm
+%}