surendar26 wrote:

Does the prime number p divide n!?

1. Prime number p divides n! + (n+2)!.

2. The prime number p divides (n+2)!/n!.

The question should be:

is the prime number p a factor of n!, where n is a nonnegative integer.

(Basically the question asks whether \(p\leq{n}\))

(1) p is a factor of \(n!+(n+2)!\) --> \(p\) is a factor of \(n!(1+(n+1)(n+2))\).

Now, if \(p=3\) and \(n=0\) (\(n!+(n+2)!=1+2=3\)) then the answer will be NO: \(p=3\) is not a factor of \(n!=0!=1\).

But if \(p=2\) and \(n=2\) (\(n!+(n+2)!=2+24=26\)) then the answer will be YES: \(p=2\) is a factor of \(n!=2!=2\).

Two different answer, hence not sufficient.

(2) p is a factor of \(\frac{(n+2)!}{n!}\) --> \(p\) is a factor of \((n+1)(n+2)\).

Now, if \(p=2\) and \(n=0\) then the answer will be NO but if \(p=2\) and \(n=2\) then the answer will be YES. Not sufficient.

(1)+(2) Now, \((n+1)(n+2)\) and \((n+1)(n+2)+1\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)+1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)+1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.

Answer: C.