him1985 wrote:
Bunuel wrote:
goodyear2013 wrote:
If x is a positive integer greater than 1, is x! + x + 1 a prime number?
(1) x < 10
(2) x is odd
OE
(1) INSUFFICIENT
If x = 2: 2! + (2 + 1) = 5, which is prime.
If x = 3: 3! + (3 + 1) = 6 + (3 + 1) = 10, which is not prime.
(2) SUFFICIENT:
If x = 3: 3! + (3 + 1) = 10, which is not prime.
If x = 5: 5! + (5 + 1) = (5 × 4 × 3 × 2 × 1) + 6.
This expression must be divisible by 3, since both of its terms are divisible by 3.
Furthermore it must be even, because both terms are even.
Therefore, it is not a prime number.
If x is a positive integer greater than 1, is x! + x + 1 a prime number?(1) x < 10.
If x=2, then x! + x + 1 = 5 = prime.
If x=3, then x! + x + 1 = 10, not a prime.
Not sufficient.
(2) x is odd. So, x is an odd integer greater than 1: 3, 5, 7, ... In this case x! + x + 1 = even + odd + odd = even > 2, not a prime. Sufficient.
Answer: B.
Is there any other way , apart from testing number. I solved it correctly but took more than 2 min.
The main takeaways from this question is this:
2 is the ONLY number (apart from 1) that, when expressed as a factorial, will give you a prime # (when you add "x" and "1"). Why, you ask? B/c
any number > 2 will include "2" in its prime factorization. Meaning, if you have 3! = 3x2; or 5! = 5x4x3x2.
ALSO, any number multiplied by an even will be EVEN. and the ONLY EVEN PRIME = 2.
Therefore, (1) = insufficient b/c x<10 includes 2 (prime) as well as 3, 4, 5, 6, 7, 8, 9 (which are all NOT prime).
(2) is sufficient b/c we know "1" cannot be included here, so the only numbers to play with are: 3, 5, 7, 9 -- which are all NOT prime. This is sufficient b/c we can safely say "x! + x + 1 is NOT PRIME" under these conditions