"I dont understand whats wrong with my code"
Let's go line-by-line.
odd = rem(x, 2) ~= 0
Odd numbers are whole numbers. rem(2.1, 2) equals 0.1 which is not equal to 0 and would be misclassified as odd. I guess that you want to force the user to provide an integer for x in which case you could throw an error if x is not an integer
assert(mod(x,1)==0, 'x must be an integer.')
or you could add an integer test to the condition,
odd = rem(x,2)~=0 & mod(x,1)==0;
while x > sqrt(132) ... end
- What happens when x is less than sqrt(132)?
- Your while-loop never changes the value of x so the while-loop will be infinite.
if x / 11
Is x scalar? Unless x=0, this condition will always be true. If-statements should contain a binary condition that tests for true or false. Any scalar other than 0 will satisfy this condition. If x is a vector or array the if-statement is even more problematic and should be replaced by indexing. If you want to test whether x is divisible by 11, use
min x
This is not correct matlab syntax. Look at the documentation for examples.
" Im writing a code to find the lowest value of x which is dividable by 11 and bigger than sqrt(132), x is also a odd number."
Assuming x is a vector, here are some test you can combine using indexing, Now combine those indices to pull out values of x that satisfy all tests and apply min().
