Step 1 of the Kaplan Method for DS: focus on the question stem.brick2009 wrote: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
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.
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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.
!!
