Removing Floors - Part IV

The following inequalities hold true:

$0 \geq \lfloor {\frac{n}{6}} \rfloor - \frac{n}{6} > -1 \\ 0 \geq \lfloor {\frac{n}{3}} \rfloor - \frac{n}{3} > -1 \\ 0 \geq \lfloor {\frac{n}{2}} \rfloor - \frac{n}{2} > -1 \\ \text{Add 'em up, you get:} \\ 0 \geq \lfloor {\frac{n}{6}} \rfloor + \lfloor {\frac{n}{3}} \rfloor + \lfloor {\frac{n}{2}} \rfloor - n > -3 \\ \text{Which comes to:} \\ n-0 \geq \lfloor {\frac{n}{6}} \rfloor + \lfloor {\frac{n}{3}} \rfloor + \lfloor {\frac{n}{2}} \rfloor > n-3 \\ \text{Since } m \text { is bounded above by 0 and below by 3, the only possibilities for } m \text{ are 0,1,2:} \\ \lfloor {\frac{6}{6}} \rfloor + \lfloor {\frac{6}{3}} \rfloor + \lfloor {\frac{6}{2}} \rfloor = 6 = 6-0 \\ \lfloor {\frac{5}{6}} \rfloor + \lfloor {\frac{5}{3}} \rfloor + \lfloor {\frac{5}{2}} \rfloor = 3 = 5-2 \\ \lfloor {\frac{4}{6}} \rfloor + \lfloor {\frac{4}{3}} \rfloor + \lfloor {\frac{4}{2}} \rfloor = 3 = 4-1 \\ \text{Since all possibilities work for certain } n \text{, sum of all } m = 0+1+2=3$