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Number System

Expert replies
by knight247 » Wed Aug 03, 2011 5:26 am
Does the prime number p divide N!
(1) The prime number p divides N!+(N+2)!
(2) The prime number p divides (N+2)!/N!

Don't have the OA on this one. According to me it is A.Because when two multiples of a number are added the resultant sum is also a multiple of that number. For statement two, I plugged in numbers and managed to figure that it is insufficient. Am I correct? Detailed explanations would be appreciated
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Source: — Data Sufficiency |

by Frankenstein » Wed Aug 03, 2011 6:59 am
Hi,
Because when two multiples of a number are added the resultant sum is also a multiple of that number.
True. But you are assuming the converse is true. If a number is a factor of the sum of two numbers, each number doesn't necessarily be a multiple of the factor.
From(1):
Let n = 3
n!+(n+2)! = 3! + 5! = 126
p = 7 divides n!+(n+2)!, but it doesn't divide n! = 6 -> No
p = 3 divides n!+(n+2)!, it also divides n! = 6 -> Yes
Not sufficient

From(2):
Not sufficient(As you have already checked)

Both(1) and (2):
p is prime
p is a factor of (n+1)(n+2), i.e p is factor of either (n+1) or(n+2)
p is factor of n!+(n+2)! = n![1+(n+1)(n+2)]
So, p is factor of n! or [1+(n+1)(n+2)].
Since, 'p' is factor of (n+1)(n+2), it cannot be factor of 1+(n+1)(n+2). So, p has to be factor of n!.

Hence, C
Cheers!

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