You can use the floor function:

result = floor(13/3)

This function always rounds down to the lower integer

Why not play with mod? See what it does? Lets see what we can do. K = 1:10; Now, pick some number that we want to use to test for divisibility. p = 3; So, which elements of K will P divide exactly? We would expect 3, 6, and 9 for this vector of integers K and divisor P. So if you understand how mod works, what does this tell you? mod(K,p) ans = 1 2 0 1 2 0 1 2 0 1 find(mod(K,p) == 0) ans = 3 6 9 So when mod(K,p) is zero, that tells us that p divides K, or the corresponding element thereof for vector K. Similarly, does 17 divide 34? mod(34,17) ans = 0 Yes, it does. But not 35. mod(35,17) ans = 1 As far as your assignment goes, you need to spend some time working it out. But this should get you started.

Just use a conditional. It will not meaningfully affect efficiency.

inline double realmod (x, y)
{
  result = fmod(x, y);
  return result >= 0 ? result : result + y;
}

fmod() calls assembly instruction FPREM which takes 16-64 cycles (according to the Pentium manual, http://www.intel.com/design/pentium/manuals/24143004.pdf). The jump instructions for the conditional and the floating point addition only amount to 5 or so.

When your code has floating point division, you don't need to sweat the small stuff.

b = mod(a,m) returns the remainder after division of a by m, where a is the dividend and m is the divisor. From here https://www.mathworks.com/help/matlab/ref/mod.html