The sum of five consecutive integers is $$2015$$. What is the smallest of these integers? 
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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 $$403-2=401$$. 