Answer to Question #1753 in Discrete Mathematics 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.
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!