I don't have a clue to even start the following question so any pointers would be welcome!
Let n be an integer with n greater than or equal to 2. Show that n! + 2 is even.
Now show that m divides n! + m, for all m such that 2<m<n (here it is a less than or equal to sign) and show why n! +m is composite.
Use this to find a sequence of n-1 consecutive integers (that is a sequence a2, a3,...,an, with ai+1=ai +1 and all ai composite).