# Math Help - Composite numbers

1. ## Composite numbers

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.

2. ## Divisibility of factorials

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 $n!+2$, $n!+3$, ..., $n!+n$?

3. i still dont get it

4. ## Divisibility of factorials

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?