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https://bitbucket.org/Mattrixwv/octavefunctions.git
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Added code to cleanup the variables at the end of the script
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@@ -22,4 +22,9 @@ end
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%When done this way it removes the possibility of duplicate numbers
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fullSum = sum(numbers);
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fullSum
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ans = fullSum
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%Cleanup your variables
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clear fullSum;
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clear numbers;
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clear counter;
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@@ -6,7 +6,6 @@
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fib = [1, 1, 2]; %Holds the Fibonacci numbers
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currentFib = fib(end) + fib(end - 1); %The current Fibonacci number to be tested
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evenFib = [2]; %A subset of the even Fibonacci numbers
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finalSum = 0;
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while(currentFib < 4000000)
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%Add the number to the list
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@@ -20,5 +19,9 @@ while(currentFib < 4000000)
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currentFib = fib(end) + fib(end - 1);
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end
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finalSum = sum(evenFib);
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finalSum
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sum(evenFib)
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%Cleanup your variables
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clear fib;
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clear currentFib;
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clear evenFib;
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@@ -36,12 +36,13 @@ end
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%When the last number is not divisible by a prime number it must be a prime number
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factors(end + 1) = tempNum;
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%Remove the variables
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%Print the answer
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max(factors)
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%Cleanup your variables
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clear counter;
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clear tempNum;
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clear answer;
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clear number;
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clear primeNums;
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%Print the answer
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max(factors)
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clear factors;
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@@ -28,12 +28,14 @@ while(outerCounter < size(numbers)(2))
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++outerCounter; %Increment
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end
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max(palindromes)
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%Cleanup your variables
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clear outerCounter;
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clear innerCounter;
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clear answer;
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clear numbers;
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max(palindromes)
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clear palindromes;
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%This way is slow. I would like to find a faster way
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%{
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@@ -3,7 +3,6 @@
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%What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
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%Create your variables
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value = 0; %The value that is evenly divisible
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nums = [1:20];
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factors = [1]; %The factors that are already in the number
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list = []; %For a temperary list of the factors of the current number
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@@ -33,13 +32,12 @@ while(counter <= size(nums)(2))
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++counter;
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end
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prod(factors)
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%Cleanup your variables
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clear counter;
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clear factorCnt;
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clear listCnt;
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clear list;
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clear nums;
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clear ans; %Appears whenever you use increment
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value = prod(factors);
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value
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clear factors;
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