Let the five consecutive integers be $$n-2$$, $$n-1$$, $$n$$, $$n + 1$$ and $$n + 2$$. These have a sum of $$5n$$. Hence $$5n=2015$$ and therefore $$n=403$$. Therefore, the smallest integer is $$403-2=401$$. 
The sum of five consecutive integers is $$2015$$. What is the smallest of these integers? 
[1] $$401$$ 
[2] $$403$$ 
[3] $$405$$ 
[4] $$407$$ 
[5] $$409$$ 