Previously, I gave you a puzzle on prime numbers.

Let F(n) be (1st prime) * (2nd prime) * … * (nth prime)

The question was to describe the group of numbers between 2 and F(n)-1 that F(n) cannot divide.

And the answer is… **all composite numbers between 2 and F(n)-1 with repeated prime factors**. Hmm… I guess that doesn’t quite fit the “as plain an English as possible”, but it *is* concise.

For example, F(n) cannot divide 4, because 4 = 2*2 (repeated 2).

But F(n) divides 6, because 6 = 2*3 (no repeated primes).

F(n) cannot divide 18, because 18 = 2*3*3 (repeated 3).

### The proof (sort of)

If a number A in [ 2, F(n)-1 ), is prime, then F(n) divides A by definition because F(n) is a product of primes.

Let A be a composite number in [ 2, F(n)-1 ) with no repeated prime factors. Then F(n) divides A because F(n) is a product of primes where the prime factors form a superset of the prime factors of A.

### Can you complete the proof?

Let A be a composite number in [ 2, F(n)-1 ) with repeated prime factors. Let *p* be a repeated prime.

…

[*Can you complete the proof?*]