Question 286600
9 consecutive integers:
n,n+1,n+2,n+3,n+4,n+5,n+6,n+7,n+8

So their sum is 9n+36=9(n+4)

So, we want 9(n+4)=9  this implies n+4=1 and n=-3.

So, -3+-2+-1+0+1+2+3+4+5=9.

Now, one should note that 0 is included, so the product of any numbers with 0 has to be 0.