[email protected] wrote:Hi shibsriz,
This DS question should be posted in the DS Forum, but I'll answer it here.
This question provides a great opportunity to TEST Values, but the real "secret" to it is in understanding how factorials and division "work." This would be a rare issue on the GMAT and the explanation is going to be a bit long-winded, but you asked....
We're told that N > 5 and we're asked for the remainder when (N! + N + 1) is divided by (N+1).
Fact 1: (N+2) is prime.
Let's TEST N = 9
(9! +9+1) / (9+1) = (9! + 10)/10
This can be "split" into...
9!/10 + 10/10
9! = 9x8x7x6x5x4x3x2x1 = (2x5)x(9x8x7x6x4x3x1)
So 9!/10 has no remainder and 10/10 has no remainder, so (9! + 10)/10 has a remainder of 0
With the limitations provided by Fact 1, the remainder will ALWAYS be 0. Here's why:
Since N + 2 = prime and N > 5, then...
N MUST be odd....
N+1 MUST be even.....
N! MUST be even....(because there's a "2" in the sequence)
N! MUST be a multiple of (N+1)....(because there's a "2" and the "odd number" that you need for 2(odd) = (N+1))
(N! + N+1)/(N+1) can always be split into this....
N!/(N+1) + (N+1)/(N+1)
So... N!/(N+1) has no remainder and (N+1)/(N+1) = 1 and has no remainder.
Fact 1 ALWAYS provides a remainder of 0
Fact 1 is SUFFICIENT.
Fact 2: (N-2) is prime
This tells us that N = odd, which creates the same circumstances as in Fact 1.
Fact 2 is SUFFICIENT.
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
