Bunuel wrote:
2. If n is a positive integer and p is a prime number, is p a factor of n!?
(1) p is a factor of (n+2)!-n! --> if \(n=2\) then \((n+2)!-n!=22\) and for \(p=2\) then answer will be YES but for \(p=11\) the answer will be NO. Not sufficient.
(2) p is a factor of (n+2)!/n! --> \(\frac{(n+2)!}{n!}=(n+1)(n+2)\) --> if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) then answer will be YES but for \(p=3\) the answer will be NO. Not sufficient.
(1)+(2) \((n+2)!-n!=n!((n+1)(n+2)-1)\). Now, \((n+1)(n+2)-1\) and \((n+1)(n+2)\) 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.
I didnot understand why 1 is not sufficient, can you elaborate on that ?
what i did is - I took n! common * coprime number
So if we have n! everytime the factors for n! and n!* coprime should always be same.. What am i missing?
Will Vote for A
Given
n = +ve Int
P = Prime
(A)
p is a factor of (n+2)!-n!
Let n = 4
= 6! - 4!
= 4![(6*5) - 1] = 4! (29)
therefore n! = 4! will have sufficient factors as 4! (29)
data sufficient
Let n = 17
= (19)! - 17!
= 17! [(19 * 18) - 1]
data sufficient
Let n = 11
= (13!) - 11!
= 11! [(13*12) - 1]
data sufficient
(B)
p is a factor of (n+2)!/n!
Let n = 2
= (4!) / 2! = 12 = 2*2*3 = data sufficient
Let n = 4
= (6!) / 4! = 6 * 5 = 2*3*5 = data not sufficient