According to the VB6/VBA documentation
The modulus, or remainder, operator divides number1 by number2 (rounding floating-point numbers to integers) and returns only the remainder as result. For example, in the following expression, A (result) equals 5. A = 19 Mod 6.7 Usually, the data type of result is a Byte, Byte variant, Integer, Integer variant, Long, or Variant containing a Long, regardless of whether or not result is a whole number. Any fractional portion is truncated. However, if any expression is Null, result is Null. Any expression that is Empty is treated as 0.
Remember, mod returns the remainder of the division. Any integer mod 1 = 0.
debug.print 12 mod 1
'12/1 = 12 r 0
The real culprit here though is that vba truncates (rounds down) the double to an integer before performing the modulo.
?13 mod 10
'==>3
?12.5 mod 10
'==>2
debug.print 12.5 mod 1
'vba truncates 12.5 to 12
debug.print 12 mod 1
'==> 0