## Factorials, prime numbers and a puzzle

There is this interesting math tidbit about composite numbers and factorials by Ned Batchelder. Now prime numbers never appear consecutively (except for 2 and 3). Ned then answered this question: **how many composites can appear consecutively?**

His explanation involves the use of factorials, and you can read about it using the link above. His explanation also gave me something to think about…

Now the factorial of n, denoted by n! is

1*2*3*4* … *(n-1)*n

which is a product of 1 through n.

Let’s define a function F such that F(n) is the product of

(1st prime)*(2nd prime)* … * (nth prime)

For example, the first few prime numbers are 2, 3, 5, 7, 11, 13. So

F(1)=2,

F(2)=2*3=6, and

F(5)=2*3*5*7*11=2310.

This is different from factorial primes (I was actually going to name this special function “prime factorial”).

Now, n! is divisible by 2, n! is divisible by 3 and n! is divisible by 4.

F(n) is divisible by 2, F(n) is divisible by 3, **but F(n) is not divisible by 4**!

My question: **Describe the group of numbers where F(n) cannot divide, in as plain an English as possible.** This group of numbers will necessarily be between 2 and F(n)-1.

Your knee-jerk answer could be “all composite numbers between 2 and F(n)-1!”. Ahh, but F(n) is divisible by 10, and 10 is a composite number (assumption, n is a fairly large number, say greater than 5). This puzzle should be easy to figure out. Articulation of the solution into a couple of sentences might be harder…

[Vincent is currently away on vacation. He asked me, the blog, to take over for a while. Using a proprietary algorithm involving language semantics and neural networks (written by me), I came up with the blog post you’ve just read. It even seems coherent! I mean, uh, of course it makes sense. Oh, the things I do for my master… He’d better come back with lots of pictures for me to post, or he and I are going to have words…]