Composite numbers

**mitch_nufc** · January 5th 2009, 04:34 AM

let n be an integer with n> or equal 2. Show that n! + 2 is even. Now show that m|n! + m, for all m such that 2(</=)m(</=)n and explain why n! + m is composite. Use this to find a sequence of n - 1 consecutive compositive integers. (the sequence is an a_2,a_3,...,a_n , with a_i+1 = a_i +1 and all a_i composite.

**Grandad** · January 5th 2009, 04:41 AM

Hello mitch_nufc

Originally Posted by mitch_nufc

let n be an integer with n> or equal 2. Show that n! + 2 is even. Now show that m|n! + m, for all m such that 2(</=)m(</=)n and explain why n! + m is composite. Use this to find a sequence of n - 1 consecutive compositive integers. (the sequence is an a_2,a_3,...,a_n , with a_i+1 = a_i +1 and all a_i composite.

Look at the definition of $n!$ for a positive integer $n$ . It consists of the product of all integers from $1$ to $n$ , inclusive. Can you see why this is divisible by $m$ , provided $2 \le m \le n$ ?

Can you now see why $n!+m$ is also divisible by $m$ ?

So what about the numbers $n!+2$ , $n!+3$ , ..., $n!+n$ ?

Grandad

**mitch_nufc** · January 5th 2009, 06:02 AM

i still dont get it

**Grandad** · January 5th 2009, 10:19 AM

Hello mitch_nufc

$n! = 1 \times 2 \times 3 \times \dots \times n$

So $n!$ is divisible by all integers from $2$ to $n$ . OK so far?

In particular $n!$ is divisible by $2$ . So if you add $2$ to $n!$ , you get another number divisible by $2$ , because an even number plus an even number is always even.

That was the first thing you were asked to prove.

Next, note that if two integers, $a$ and $b$ , say, are each divisible by another integer $m$ , then $(a+b)$ is also divisible by $m$ . (Can you see why?)

Now we've already said that $n!$ is divisible by all integers between $2$ and $n$ . In other words, it's divisible by $m$ if $2 \le m \le n$ . And $m$ is (obviously) also divisible by $m$ . So $n! + m$ is also ...?

Now think again about the integers $n! + 2, n! + 3, \dots, n! + n$ . What can you say about them?

Grandad

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