solution plz

[brick2009]

I could not understand this problem...help plz


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

[Stuart@KaplanGMAT]

I could not understand this problem...help plz


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

Step 1 of the Kaplan Method for DS: focus on the question stem.

The first word that should jump out is "does"; the answer to a "does" question is either "yes" or "no", so we're dealing with a yes/no question.

The next thing we need to understand is the question itself. We know that 1 and x are definitely factors of x, the question is are there any other factors of x? Well, what numbers don't have any factors other than 1 and themselves? Primes! So, the question is really asking:

Is x NOT a prime number?

which we can just rethink as:

Is x a prime number?

since we don't really care what the answer is, we just care whether we can get a definite answer.

Now that we've greatly simplified the question, let's move on to:

Step 2 of the Kaplan Method for DS: consider each statement by itself.

(1) looks much simpler, so let's start here.

x > 3!

or

x > 3*2*1

or

x > 6

If x is greater than 6, could it be prime? YES
If x is greater than 6, could it be non-prime? YES

Since x may or may not be prime, (1) is insufficient.

(2) 15! + 2 ≤ x ≤ 15! + 15

This statement is much trickier - we really have to understand factorials and factoring.

15! is simply 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15

Therefore, 15! is a multiple of every integer from 1 to 15.

Well, if 15! is a multiple of 2, then 15! + 2 is also a multiple of 2.
If 15! is a multiple of 3, then 15! + 3 is also a multiple of 3.
If 15! is a multiple of 4, then 15! + 4 is also a multiple of 4.
.
.
.
If 15! is a multiple of 15, then 15! + 15 is also a multiple of 15.

So, if x is an integer (info from the question stem) and is in the range provided (info from statement (2)), then x is DEFINITELY NOT a prime number.

Step 3 of the Kaplan Method for DS: if necessary, combine the statements.

One of the statements was sufficient alone, so no need to combine!

(2) is sufficient alone: choose B.

[November Rain]

Hi

I think the answer is B. Could you please confirm to me the OA?

The question is asking if X is a prime number.

(1) Only says that X is bigger than 6. But you have several primes bigger than 6, so it's insufficient.

(2) Says that X is a number that has, at least, a factor of 2 (15! + 2) or a factor of 3 (15! + 3), or a factor of 4 (15! + 4) etc... or a factor of 15 (15!+15). So whatever the real number, X will always have at least one factor, so X is not prime. Sufficient

[mehravikas]

Let's rephrase the statement 2 as 4! + 2 <= x <= 4! + 15

Therefore 26 <= x <= 39

Using this we still can't confirm that x is not a prime number? What's the difference if 4! is replaced by 15!?

I could not understand this problem...help plz


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

Step 1 of the Kaplan Method for DS: focus on the question stem.

The first word that should jump out is "does"; the answer to a "does" question is either "yes" or "no", so we're dealing with a yes/no question.

The next thing we need to understand is the question itself. We know that 1 and x are definitely factors of x, the question is are there any other factors of x? Well, what numbers don't have any factors other than 1 and themselves? Primes! So, the question is really asking:

Is x NOT a prime number?

which we can just rethink as:

Is x a prime number?

since we don't really care what the answer is, we just care whether we can get a definite answer.

Now that we've greatly simplified the question, let's move on to:

Step 2 of the Kaplan Method for DS: consider each statement by itself.

(1) looks much simpler, so let's start here.

x > 3!

or

x > 3*2*1

or

x > 6

If x is greater than 6, could it be prime? YES
If x is greater than 6, could it be non-prime? YES

Since x may or may not be prime, (1) is insufficient.

(2) 15! + 2 ≤ x ≤ 15! + 15

This statement is much trickier - we really have to understand factorials and factoring.

15! is simply 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15

Therefore, 15! is a multiple of every integer from 1 to 15.

Well, if 15! is a multiple of 2, then 15! + 2 is also a multiple of 2.
If 15! is a multiple of 3, then 15! + 3 is also a multiple of 3.
If 15! is a multiple of 4, then 15! + 4 is also a multiple of 4.
.
.
.
If 15! is a multiple of 15, then 15! + 15 is also a multiple of 15.

So, if x is an integer (info from the question stem) and is in the range provided (info from statement (2)), then x is DEFINITELY NOT a prime number.

Step 3 of the Kaplan Method for DS: if necessary, combine the statements.

One of the statements was sufficient alone, so no need to combine!

(2) is sufficient alone: choose B.

[Stuart@KaplanGMAT]

Let's rephrase the statement 2 as 4! + 2 <= x <= 4! + 15

Therefore 26 <= x <= 39

Using this we still can't confirm that x is not a prime number? What's the difference if 4! is replaced by 15!?

The difference is the number you're adding at the end; 15 is more than 4.

4! must be a multiple of 2, 3 and 4, but doesn't have to be a multiple of 5, 6, 7, ..., 15, so we can't use the same reasoning.

Each of the following would be analagous to (2):

4! + 2 <= x <= 4! + 4
9! + 2 <= x <= 9! + 9
100! + 2 <= x <= 100! + 100

or, in more general terms:

n! + 2 <= x <= n! + n

and would be sufficient to conclude that x isn't a prime number.

[mehravikas]

I get it now... !!

Thanks so much.

Let's rephrase the statement 2 as 4! + 2 <= x <= 4! + 15

Therefore 26 <= x <= 39

Using this we still can't confirm that x is not a prime number? What's the difference if 4! is replaced by 15!?

The difference is the number you're adding at the end; 15 is more than 4.

4! must be a multiple of 2, 3 and 4, but doesn't have to be a multiple of 5, 6, 7, ..., 15, so we can't use the same reasoning.

Each of the following would be analagous to (2):

4! + 2 <= x <= 4! + 4
9! + 2 <= x <= 9! + 9
100! + 2 <= x <= 100! + 100

or, in more general terms:

n! + 2 <= x <= n! + n

and would be sufficient to conclude that x isn't a prime number.