Answer
See below.
Work Step by Step
(a) Let the three consecutive integers be $n, n+1, n+2$, we have:
case1: if $n=3k+0$, then $n$ is divisible by 3;
case2: if $n=3k+1$, then $n+2=3k+3$ is divisible by 3;
case3: if $n=3k+2$, then $n+1=3k+3$ is divisible by 3;
Since the remainder can be $0,1,2$ only, the above cases should cover all the possibilities, thus the product of $n(n+1)(n+2)$ is divisible by 3 in any case.
(b) Let the three consecutive integers be $n, n+1, n+2$, we have:
case1: if $n$ mod $3=0$, then $n$ is divisible by 3;
case2: if $n$ mod $3=1$, then $(n+2)$ mod $3=0$ and $n+2$ is divisible by 3;
case3: if $n$ mod $3=2$, then $(n+1)$ mod $3=0$ and $n+1$ is divisible by 3;
Since the remainder can be $0,1,2$ only, the above cases should cover all the possibilities, thus we have $n(n+1)(n+2)$ mod $3=0$ in any case.
