How to enter this program?

Hello, I have tried unsuccessfully for a couple hours to enter a program that loops until a number that is even, divisible by both 13 and 16, and whose square root is greater than 120 is found. I have reached a wall, so I have come to ask for help. Would using a while statement be the first step? Please don't solve this problem, I only ask for help getting it going. Yes, this is basic stuff, but I still have little understanding on it.

Answer 1

[Roger Stafford]

You need to translate each of those conditions into a matlab logical proposition. For example if n is an integer you are investigating, try dividing it by 13, then comparing the 'round' of the quotient with the original quotient to see if it is still an integer after the division. And yes, the 'while' function is the right thing to use here. Keep it looping until all the conditions turn out to be true for the same integer, for which you can make great use of the "&" operation.

[Rick Rosson]

You may also find that the mod function is helpful in this particular case. This function returns the modulus of two integers, which is the remainder that is left after dividing the first integer by the second integer. So, for example:

   mod(45,13)

returns 6, whereas

   mod(52,13)

returns 0.

This provides a very easy way to test for divisibility. For example:

   if mod(n,13)
      disp('n is not divisible by 13');
   else
      disp('n is divisible by 13');
   end

[Roger Stafford]

Yes, that's a little easier than writing

   round(n/13) == n/13

[Colton]

Okay, so I said

a = 2;
while sqrt(a) <= 120 && mod(a,13) && mod(a,16)
    a = a + 2;
end

But this returns a = 16, satisfying only the final condition. Could you point me in the right direction? Thank you so much for your help so far though.

[Rick Rosson]

Try using || instead of &&.

[Colton]

@Rick, thank you. That fixed my issues and I was able to find the correct value! And thanks everyone else for the help!

Answer 2

[Rick Rosson]

Here is a faster and simpler approach:

   a = 0;
   while a < 120^2
        a = a + 13*16;
   end

[per isakson]

Note &nbsp "square root is greater than 120"

Answer 3

[per isakson]

Try

    >> factor( cssm )
    ans =
         2     2     2     2     2     5     7    13

where cssm is

    function ix = cssm()
        % chose a start value, which is larger than 120^2 and divisible by 16
        ix = 120*120+16; 
        while  ne( mod(ix,13), 0 )
            ix = ix + 16;
        end
    end