Note that for any integer [tex]k[/tex],
[tex]n\equiv\begin{cases}0\mod6&\text{if }n=6k\\1\mod6&\text{if }n=6k+1\\2\mod6&\text{if }n=6k+2\\3\mod6&\text{if }n=6k+3\\4\mod6&\text{if }n=6k+4\\5\mod6&\text{if }n=6k+5\end{cases}[/tex]
and if [tex]n=6k+6=6(k+1)[/tex], well [tex]n[/tex] is just another multiple of 6, so we're back to the first case.
For any six consecutive integers, then, the sum of the remainders upon division by 6 is [tex]0+1+2+3+4+5=15[/tex].
