Data Sufficiency

Does the prime number P divides n!

(i) The prime number P divides n!+(n+2)!

n!(1+(n+1)(n+2))
n! can be a prime number multiple. or (1+(n+1)(n+2)) can be a prime number multiple.
so cannot be determined.

Example:
n= 2
2!(4!) = 2!(1+3*4)
when P = 2, 2! is divisible by 2.
when P = 13, 2! is not divisible by 13.

Hence INSUFF

(ii) The Prime number P divides (n+2)!/n!
P divides (n+1)(n+2)
Let n= 2
P = 3 divides 3*4 = 12
P =3 does not divide 2! =2...NO
but P =2 divides 2! =2..YES
hence INSUFF

Combining:
P divides n!(1+(n+1)(n+2)) and (n+1)(n+2)
(1+(n+1)(n+2)) and (n+1)(n+2) are consecutive numbers and hence cannot have a common prime factor.

Hence P should be a factor of n!

pick C.

What is the source and OA? Seems tough question.

narik11 wrote:Does the prime number P divides n!

(i) The prime number P divides n!+(n+2)!

(ii) The Prime number divides (n+2)!/n!

Is n! = P x, where n, x are positive integers and P being a prime?

[1] It reads n! + (n + 2)! = P y, where y is a positive integer.

Or, n! + (n + 2) (n + 1) n! = P y

Or, n! (n^2 + 3 n + 3) = P y.


Obscurity what's left…P could be either a factor n! or a factor of (n^2 + 3 n + 3), or may be of both. Insufficient

[2] It reads that P is a factor of (n^2 + 3 n + 3), or we accidentally get a gem that if n is a positive integer, then (n^2 + 3 n + 3) is a prime. Now, if a prime is a factor of a prime, then it’s the prime itself, hence, P = (n^2 + 3 n + 3), take an example for n like if n = 2, then P is 19. Here, 2! Is not containing 19 as one factor, and this is pretty sure that (n^2 + 3 n + 3) > n, hence the prime P would be greater than any element in the flag march of the factorial of n, a positive integer; hence the answer is always NO. Sufficient


[spoiler]B[/spoiler]

sanju09 wrote:

narik11 wrote:Does the prime number P divides n!

(i) The prime number P divides n!+(n+2)!

(ii) The Prime number divides (n+2)!/n!

Is n! = P x, where n, x are positive integers and P being a prime?

[1] It reads n! + (n + 2)! = P y, where y is a positive integer.

Or, n! + (n + 2) (n + 1) n! = P y

Or, n! (n^2 + 3 n + 3) = P y.


Obscurity what's left…P could be either a factor n! or a factor of (n^2 + 3 n + 3), or may be of both. Insufficient

[2] It reads that P is a factor of (n^2 + 3 n + 3), or we accidentally get a gem that if n is a positive integer, then (n^2 + 3 n + 3) is a prime. Now, if a prime is a factor of a prime, then it’s the prime itself, hence, P = (n^2 + 3 n + 3), take an example for n like if n = 2, then P is 19. Here, 2! Is not containing 19 as one factor, and this is pretty sure that (n^2 + 3 n + 3) > n, hence the prime P would be greater than any element in the flag march of the factorial of n, a positive integer; hence the answer is always NO. Sufficient


[spoiler]B[/spoiler]

Not Quite Right..
Firstly, stmnt B reduces to n^2 + 3 n + 2 and not n^2 + 3 n + 3...
Secondly, Even if u use n^2 + 3 n + 3 , still u cant generalize n^2 + 3 n + 3 as a prime number..take n = 3 , it reduces the expression to 21 which is not a prime.... So the P = n^2 + 3 n + 3 is not correct

rohan_vus wrote:

sanju09 wrote:

narik11 wrote:Does the prime number P divides n!

(i) The prime number P divides n!+(n+2)!

(ii) The Prime number divides (n+2)!/n!

Is n! = P x, where n, x are positive integers and P being a prime?

[1] It reads n! + (n + 2)! = P y, where y is a positive integer.

Or, n! + (n + 2) (n + 1) n! = P y

Or, n! (n^2 + 3 n + 3) = P y.


Obscurity what's left…P could be either a factor n! or a factor of (n^2 + 3 n + 3), or may be of both. Insufficient

[2] It reads that P is a factor of (n^2 + 3 n + 3), or we accidentally get a gem that if n is a positive integer, then (n^2 + 3 n + 3) is a prime. Now, if a prime is a factor of a prime, then it’s the prime itself, hence, P = (n^2 + 3 n + 3), take an example for n like if n = 2, then P is 19. Here, 2! Is not containing 19 as one factor, and this is pretty sure that (n^2 + 3 n + 3) > n, hence the prime P would be greater than any element in the flag march of the factorial of n, a positive integer; hence the answer is always NO. Sufficient


[spoiler]B[/spoiler]

Not Quite Right..
Firstly, stmnt B reduces to n^2 + 3 n + 2 and not n^2 + 3 n + 3...
Secondly, Even if u use n^2 + 3 n + 3 , still u cant generalize n^2 + 3 n + 3 as a prime number..take n = 3 , it reduces the expression to 21 which is not a prime.... So the P = n^2 + 3 n + 3 is not correct

When we try factoring n! + (n + 2)!, it unfolds like


= n! + (n + 2) (n + 1) n!

= n! (1 + n^2 + 3 n + 2)

= n! (n^2 + 3 n + 3), so that part is correct.

For second part, I must confess that it was some EUREKA fever spread by Mr. Archimedes, I didn't even realized that when n is a multiple of 3, (n^2 + 3 n + 3) cannot be a prime, and I carelessly jumped out of the tub and started shouting it a gem. Heart changed to



[spoiler]C, now[/spoiler]

Stmnt B is not n! + (n+2)! but (n+2)!/n! , so still it wont be n^2 + 3n+3 .....

rohan_vus wrote:Stmnt B is not n! + (n+2)! but (n+2)!/n! , so still it wont be n^2 + 3n+3 .....

What is (n + 2)!/n! equal to? Yeah yo're right! Thanks