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I could be wrong, but I think it floors. I'm actually quite sure about this, because if you do modulo arithmetic, you always get the remainder.
So, 6/4 would give you 1 remainder 2. The remainder can be calculated using 6%4 (=2). So, in general a%b = r (a,b integers).
The integer division and modulus operators are defined to perform truncation towards zero. (In C89, it was implementation-defined whether truncation was done towards zero or -infinity. This is (obviously) important only if one or both operands are negative. Consider:
-22 / 7 = -3
-22 % 7 = -1
truncation towards zero
-22 / 7 = -4
-22 % 7 = 6
truncation towards -infinity
Both satisfy the required equation (a/b)*b + a%b == a. The second has the advantage that the modulus is always positive -- but they decided on the other (more Fortran-like, less Pascal-like) variant...)