Remainder

[manik11]

Given n>5,when (n!+n+1)is divided by (n+1),what is the remainder ?

(1) (n+2) is a prime number.

(2) (n−2) is a prime number.

Hi Experts...I could really use some help on this one. I simplified the question down to [n!/(n+1) +1],but after that I have got no clue how to proceed further.

Would really appreciate if you guys could post links to some problems, which use similar concept.

OA : D
Source : Veritas Prep

[GMATGuruNY]

Given n>5,when (n!+n+1)is divided by (n+1),what is the remainder ?

(1) (n+2) is a prime number.

(2) (n−2) is a prime number.

Statement 1: (n+2) is a prime number
Case 1: n+2 = 11, implying that n=9
Plugging n=9 into (n! + n+1)/(n+1), we get:
(9! + 10)/10 = (multiple of 10 + 10)/10 = (multiple of 10)/10.
Since 10 can divide evenly into any multiple of 10, the remainder will be 0.

Case 2: n+2 = 13, implying that n=11
Plugging n=11 into (n! + n+1)/(n+1), we get:
(11! + 12)/12 = (multiple of 12 + 12)/12 = (multiple of 12)/12.
Since 12 can divide evenly into any multiple of 12, the remainder will be 0.

Case 3: n+2 = 37, implying that n=35
Plugging n=35 into (n! + n+1)/(n+1), we get:
(35! + 36)/35 = (multiple of 36 + 36)/36 = (multiple of 36)/36.
Since 36 can divide evenly into any multiple of 36, the remainder will be 0.

In every case, the remainder is 0.
SUFFICIENT.

Statement 2: (n-2) is a prime number
Case 1: n-2 = 11, implying that n=13
Plugging n=13 into (n! + n+1)/(n+1), we get:
(13! + 14)/14 = (multiple of 14 + 14)/14 = (multiple of 14)/14.
Since 14 can divide evenly into any multiple of 14, the remainder will be 0.

Case 2: n-2 = 13, implying that n=15
Plugging n=15 into (n! + n+1)/(n+1), we get:
(15! + 16)/16 = (multiple of 16 + 16)/16 = (multiple of 16)/16.
Since 16 can divide evenly into any multiple of 16, the remainder will be 0.

Case 3: n-2 = 37, implying that n=39
Plugging n=39 into (n! + n+1)/(n+1), we get:
(39! + 40)/40 = (multiple of 40 + 40)/40 = (multiple of 40)/40.
Since 40 can divide evenly into any multiple of 40, the remainder will be 0.

In every case, the remainder is 0.
SUFFICIENT.

The correct answer is D.