Answer on Discrete Mathematics Question for scott sugg
By denition of factorial, all integers from 2 to (n+1) divide (n+1)!, so 2|((n+1)!+2),
3|((n + 1)! + 3), and so on up to
(n + 1)|((n + 1)! + (n + 1))
Thus all members of this sequence are composite, making n consecutive composite numbers. This sequence can be generated for any n, so for all n there exists at least n consecutive composite numbers.
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