# Show stretch of numbers are composite, n! + 2, n! + 3 etc

Let $n\in\mathbb{N}$ with $n\geq2$. Consider the numbers $n!+2,n!+3,...,n!+ n$. Show that none of them is prime. Deduce that for each positive integer $N$ there is a stretch of $N$ consecutive composite numbers.

What I don't understand is what happens when n=2 and you work out n! + 3=5 which is prime.

• I understand that 2 divides n! + 2 when n=2, but what about when the n is less than the number you add to it. – teddyt55 Nov 3 '13 at 18:40
• When $n=2$ the last number in your list is $2!+2$ ;) – N. S. Nov 3 '13 at 18:41

If $n = 2$, then the first number on your list is

$$n! + 2 = 2! + 2 = 4$$

which is not prime. On the other hand, the last number is also

$$n! + n = 2! + 2 = 4$$

which is still not prime.

• Thanks very much. It is clear now! – teddyt55 Nov 3 '13 at 18:59

n!=1*2*3*4*5*...*(n-1)*n so if n>=2 n!+2 =1*2*3*4*5*...*(n-1)*n +2 =2(1*3*4*...n) it is composite (not prime) n!+3= 1*2*3*4*5*...*(n-1)*n +3 =3(1*2*4*...n) not prime n!+4 = factor 2 or 4 not prime . . . n!+n= 1*2*3*4*5*...*(n-1)*n +n =n(1*2*3*4*5*...*(n-1) +1) not prime

• n!+3 =1*2*3*4*5*...*(n-1)*n +3 =3(1*2*4 ...n+1) when you can do it if only n>=3 to factor 3 – Khosrotash Nov 3 '13 at 18:45
• when p is prime number 1*2*3*4*5*...*(n-1)*n +p is prime when you can factor p from both of them ,so n>=p – Khosrotash Nov 3 '13 at 18:48
• Thanks very much. It is clear now! – teddyt55 Nov 3 '13 at 18:59

The "etc" here should mean the numbers go on until $n!+n$, not to infinity. (As you explained in the body of your question), so with $n=2$ we should reform the question to: Show that $2!+2$ is prime, which is obvious ($2!+2=2+2=4$).