Answer
See below.
Work Step by Step
(a) Let $n, n+1$ be any two consecutive integers, based on the quotient-remainder theorem, we have:
case1: $n=3m+0$ and $n(n+1)=3m(3m+1)=3(3m^2+m)$,
thus $k=3m^2+m$;
case2: $n=3m+1$ and $n(n+1)=(3m+1)(3m+2)=9m^2+9m+2=3(3m^2+3m)+2$,
thus $k=3m^2+3m$;
case3: $n=3m+2$ and $n(n+1)=(3m+2)(3m+3)=9m^2+15m+6=3(3m^2+5m+2)$, thus $k=3m^2+5m+2$;
In any case, we can write the product in the form of $3k$ or $3k+2$.
(b) Let $n, n+1$ be any two consecutive integers, use the mod form, we have:
case1: $n$ mod $3=0$ and $n(n+1)=3m(3m+1)=3(3m^2+m)$,
thus $n(n+1)$ mod $3=0$;
case2: $n$ mod $3=1$ and $n(n+1)=(3m+1)(3m+2)=9m^2+9m+2=3(3m^2+3m)+2$,
thus $n(n+1)$ mod $3=2$;
case3: $n$ mod $3=2$ and $n(n+1)=(3m+2)(3m+3)=9m^2+15m+6=3(3m^2+5m+2)$, thus $n(n+1)$ mod $3=2$;
In any case, we have the product satisfy $n(n+1)$ mod $3=0$ or $n(n+1$ mod $3=2$.
