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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].

Answer:

Step-by-step explanation:

Six consecutive integers are:

x,x+1,x+2,x+3,x+4,x+5,

Hence,sum of their remainders are 1+2+3+4+5=15

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