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Added solution to problem 36
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@@ -23,7 +23,6 @@ function [] = Problem35()
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%Setup the variables
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MAX_NUM = 999999;
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%MAX_NUM = 100;
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circularPrimes = [];
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%Start the timer
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83
Problem36.m
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83
Problem36.m
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@@ -0,0 +1,83 @@
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function [] = Problem36()
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%ProjectEuler/ProjectEulerOctave/Problem36.lua
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%Matthew Ellison
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% Created: 06-29-21
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%Modified: 06-29-21
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%Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
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%{
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Copyright (C) 2021 Matthew Ellison
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This program is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as published by
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the Free Software Foundation, either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>.
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%}
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%Setup the variables
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MAX_NUM = 999999; %The largest number that will be checked
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palindromes = []; %All numbers that are palindromes in base 10 and 2
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sumOfPal = 0; %The sum of all elements in the list of palindromes
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%Start the timer
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startTime = clock();
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%Start with 1, check if it is a palindrome in base 10 and 2, and continue to MAX_NUM
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for num = 1 : MAX_NUM
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%Check if num is a palindrome
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if(isPalindrome(num2str(num)))
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%Convert num to base 2 and see if that is a palindrome
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binNum = toBin(num);
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if(isPalindrome(binNum))
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%Add num to the list of palindromes
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palindromes(end + 1) = num;
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end
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end
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end
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%Get the sum of all palindromes in the list
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sumOfPal = sum(palindromes);
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%Stop the timer
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endTime = clock();
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%Print the results
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printf("The sum of all base 10 and base 2 palindromic numbers < %d is %d\n", MAX_NUM, sumOfPal);
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printf("It took %f seconds to run this algorithm\n", etime(endTime, startTime))
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end
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function [revStr] = reverse(str)
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counter = size(str)(2); %Set the counter to the last element in string
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%Loop until the counter reaches 0
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while(counter > 0)
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%Add the current element of string to rString
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revStr(end + 1) = str(counter);
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--counter;
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end
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end
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function [isPal] = isPalindrome(str)
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rev = reverse(str);
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if(str == rev)
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isPal = true;
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else
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isPal = false;
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end
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end
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function [binStr] = toBin(num)
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binStr = dec2bin(num);
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end
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%{
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Results:
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The sum of all base 10 and base 2 palindromic numbers < 999999 is 872187
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It took 630.539528 seconds to run this algorithm
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%}
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