Statement 1) x > 6 ... can be 7 which is prime(no factor between 1 and 7) or 8 which is not prime(has factors between 1 and 8) Not Sufficient!!!
Statement 2) 15! + 2 ≤ x ≤ 15! + 15, x can be (15! + 3; 15! + 4; 15! + 5; or...15! + 14) and from each of these a common factor can be taken out, so no value of x is prime. Sufficient!!!
Answer B.
If x is an integer, does x have a factor n such that 1
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puneetkhurana2000
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This is a great candidate for rephrasing the target question.varun289 wrote:278. If x is an integer, does x have a factor n such that 1 < n < x?
(1) x > 3!
(2) 15! + 2 ≤ x ≤ 15! + 15
What kinds of integers have a factor n such that 1 < n < x? Non-prime integers. So, we're really just asking whether or not n is a non-prime integer. Let's make it even easier on ourselves and ask . . .
Rephrased target question: Is x prime?
Statement 1: x > 3!
In other words, x > 6
case a) x = 7, in which case x is prime
case b) x = 8, in which case x is not prime
Statement 1 is NOT SUFFICIENT
Statement 2: 15! + 2 ≤ x ≤ 15! + 15
This is saying that x can have one of 14 different possible values. So, let's begin checking some values.
Is 15! + 2 prime? No.
Notice that 15! = (15)(14)(13)...(3)(2)(1)
So, we can factor a 2 out of 15! + 2, to get:
15! + 2 = 2[(15)(14)(13)...(3)(1) + 1]
This means that 2 is a factor of 15! + 2, which means it is not prime.
Next, 15! + 3 prime? No.
Notice that 15! = (15)(14)(13)...(4)(3)(2)(1)
So, we can factor a 3 out of 15! + 3, to get:
15! + 3 = 3[(15)(14)(13)...(4)(2)(1) + 1]
This means that 3 is a factor of 15! + 3, which means it is not prime.
We can continue this process to show that none of the 14 possible values of x are prime.
As such, statement 2 is SUFFICIENT and the answer is B
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
