Note that for any integer \(k\),
\(n\equiv\begin{cases}0\mod6&\text{if }n=6k\\1\mod6&\text{if }n=6k+1\\2\mod6&\text{if }n=6k+2\\3\mod6&\text{if }n=6k+3\\4\mod6&\text{if }n=6k+4\\5\mod6&\text{if }n=6k+5\end{cases}\)
and if \(n=6k+6=6(k+1)\), well \(n\) 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 \(0+1+2+3+4+5=15\).
