Question #1753

Number theory..<br>Prove or disprove that for every integer "n" there is a sequence of "n" consecutive composite numbers.(example- five consecutive composite numbers are 32,33,34,35,36)

Expert's answer

Recall the denition n! = 1 × 2 × 3 × ... × n for any positive integer n. Consider the consecutive positive integers (n+1)!+2, (n+1)!+3, . . . (n+1)!+(n+1).

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.

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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