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Show that for any n>0, there are n consecutive composite numbers.

User Pirijan
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Final answer:

To show that there are n consecutive composite numbers for any n>0, consider the sequence ((n+1)! + 1, (n+1)! + 2, ..., (n+1)! + n). Each term is composite as 1 <= k <= n divides (n+1)! + k.

Step-by-step explanation:

The question asks to show that for any positive integer n, there exists a sequence of n consecutive composite numbers. To prove this, we can construct an example using factorials.

Consider the sequence given by (n+1)!, (n+1)! + 2, (n+1)! + 3, …, (n+1)! + n. Here, (n+1)! is the factorial of (n+1), which is the product of all positive integers up to (n+1). Each term in the sequence can be written as (n+1)! + k where 1 <= k <= n. Since k divides (n+1)! + k and k is a positive integer less than or equal to n, each term is composite.

Therefore, this sequence provides n consecutive composite numbers.

User Necreaux
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