Divisible Product

• If $m\equiv 0(mod\quad 3)$ , then $3|(m+3)$ , so the given expression is divisible by $3$ .

• If $m\equiv 1(mod\quad 3)$ , then $3|(m+5)$ , and the result follows.

• If $m\equiv 2(mod\quad 3)$ , then $3|(m+1)$ , anew.

Thus, the given product must be divisible by 3.

A single counterexample is enough to show that the expression is not necessary divisible by the other options.

Choose $m\equiv 1(mod\quad 315)$ , then the given product is not divisible by neither 5, 7 nor 9.

Product of any consecutive term in the sequence 2n or 2n+1 ( (2n+1 were,N>=2) and (2n were N>=3) N=no of terms) is divisible by 3.

=(2016!-2015!)/2017=2015! (2016-1)/2017=2015! 2015

here 2015! Is divisible with 2017. then the reminder must be 2015.