You appear to be assuming that information about the remainder is available while doing integer division, or at least in its decimal approximation. That doesn't have to be true; if the division is done by shifts&subtracts, then the remainder will be left over in the other register, but no one is going to bother doing a comparison between twice the remainder and the divisor to see whether to add one more, as I would expect that would add significantly to the total time to take the division.