While loop problems finding lowesst value

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. I dont understand whats wrong with my code under. Any help is appreciated.

odd = rem(x, 2) ~= 0
while x > sqrt(132)
    if x / 11  
        min x  
    else
        fprintf('Try again/n') 
    end 
end

[S. Walter]

Your if statement is incomplete - right now it just says: "If the array of x is divided by 11, give the minimum value of x". It doesn't even get to the else part.

You want your if statement to say something like "If x divided by 11 has no remainder, give me that value of that x", then check to make sure it's odd and bigger than sqrt(132).

Answer 1

[Adam Danz]

"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

mod(x,11)==0

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,

x = 1:1000; 
idx0 = mod(x,1)==0;   % is integer
idx1 = mod(x,11)==0;  % divisible by 11
idx2 = x < sqrt(132); % less than sqrt 132
idx3 = mod(1:10,2)==1 % is odd

Now combine those indices to pull out values of x that satisfy all tests and apply min().