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factorial dilemma

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Source: — Data Sufficiency |
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by goelmaya » Fri Feb 24, 2012 12:00 am
Statement 1: x > 3!
Consider x=7, A prime number. Insufficient(Because there are only two factors)

Statement 2: 15! + 2 ≤ x ≤ 15! + 15

15! = 15.14.13.12.........1
x=15!+2, 2 is a factor
x=15!+3, 3 is a factor
x=15!+4, 4 is a factor
x=15!+5, 5 is a factor
x=15!+6, 6 is a factor
x=15!+7, 7 is a factor
x=15!+8, 8 is a factor
x=15!+9, 9 is a factor
x=15!+10, 10 is a factor
x=15!+11, 11 is a factor
x=15!+12, 12 is a factor
x=15!+13, 13 is a factor
x=15!+14, 14 is a factor
x=15!+15, 15 is a factor

Statement 2 is sufficient

IMO, Statement B alone is sufficient.
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by Brent@GMATPrepNow » Fri Feb 24, 2012 7:09 am
nafiul9090 wrote: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
This is a great candidate for rephrasing the target question.

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
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by nafiul9090 » Fri Feb 24, 2012 7:17 am
Brent@GMATPrepNow wrote:
nafiul9090 wrote: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
This is a great candidate for rephrasing the target question.

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
thanks Brent for your feedback but i have a question, what if the limit is 15!+2<x<15!+17 or any prime greater than 15??
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by Brent@GMATPrepNow » Fri Feb 24, 2012 7:39 am
nafiul9090 wrote:
Thanks Brent for your feedback but i have a question, what if the limit is 15!+2<x<15!+17 or any prime greater than 15??
Great question!

15! + P (where P is a prime greater than 15) may or may not be prime. The only conclusion we can draw is that 15! + P is not divisible by P

Take 15! + 17
We know that the prime factorization of 15! does not contain a 17, so 15! is not divisible by 17.
However, we do know that 17 is divisible by 17.
As such, we know that 15! + 17 is not divisible by 17.

Side rule: If X is divisible by d, and Y is not divisible by d, then X+Y is not divisible by d (where X, Y and d are positive integers)

What about determining whether or not 15! + P is prime (where P is a prime greater than 15)?
Unfortunately, 15! is a very big number. To show why we can't make any conclusions about whether or not 15! + P is prime (where P is a prime greater than 15), let's examine an easier case .

Is 4! + P prime (where P is a prime greater than 4)?
Well, 4! + P may or may not be prime.

case a) if P=5 then 4! + P = 29. which is prime
case b) if P=11 then 4! + P = 35. which is not prime

Similar logic can be applied to 15! + P (where P is a prime greater than 15)

Cheers,
Brent
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