## Consecutive Composites

#### Problem

Although there are infinitely many primes it is a most remarkable fact that there can always be found a sequence of $n$ consecutive composite numbers. For example, there are thirteen consecutive composite numbers between the primes 113 and 127.

Prove that there exists a sequence of $n$ consecutive composite numbers for any finite value $n$.

#### Solution

Given that $n$ is a positive integer it is clear that $n! = n \times (n - 1) \times (n - 2) \times ... \times 3 \times 2 \times 1$, is divisible by all the integers $1, 2, 3, ... , n$.

So it follows that $n! + 2$ is divisible by 2, $n! + 3$ is divisible by 3, ... , $n! + n$ is divisible by $n$.

Therefore $(n+1)! + 2, (n+1)! + 3, ... , (n+1)! + n, (n+1)! + (n+1)$ is a sequence of $n$ consecutive composite numbers.

Note that although this proves the existence of a sequence of $n$ consecutive composites it does not find the first such sequence. For example, using this method $n$ = 5, 6! = 720, so 722, 723, 724, 725, and 726 are all composite, but the first sequence of five consecutive composites is 24, 25, 26, 27, and 28.

Also be aware that as $n$ gets larger, the size of the list of consecutive composite numbers increases. However, as $n$ always remains finite, the size of the set remains finite and so this proof does not demonstrate an end to primes.

Problem ID: 325 (26 Jun 2007)     Difficulty: 3 Star

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