Hi,
I am hoping that anyone could explaing this question in an easy way...
Dose the integer K has a factor p such that 1<p<K?
1) K>4!
2) 13!+2<k<13!+13
The answer is B but I only was able to understand A k>24 would not be sufficient to answer the question but I have no idea as to 2).
Thanks!
R.
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
DS question
This topic has expert replies
-
magical cook
- Master | Next Rank: 500 Posts
- Posts: 484
- Joined: Sun Jul 30, 2006 7:01 pm
- Thanked: 2 times
- Followed by:1 members
-
amitamit2020
- Junior | Next Rank: 30 Posts
- Posts: 17
- Joined: Wed Jan 24, 2007 3:35 am
Good question.
The question askes - if we can find an number which is lesser than k and can devide k without remainder.
Let us examine statement (2), it says that k belongs to [13!+2, 13!+13].
take k = 13! + 2 -- we are now to find a number (les than k) which can devide this without remainder clearly p = 2 can do that.
go to k = 13! + 3 -- it is clear that p = 3 can answer the question.
... similarly till 13! + 13 ... p = 4,5,6,7,8,9,10,11,12 and 13 can devide k without remainder and obviously it will be lesser than 13! + any positive integer.
hope it helps...
The question askes - if we can find an number which is lesser than k and can devide k without remainder.
Let us examine statement (2), it says that k belongs to [13!+2, 13!+13].
take k = 13! + 2 -- we are now to find a number (les than k) which can devide this without remainder clearly p = 2 can do that.
go to k = 13! + 3 -- it is clear that p = 3 can answer the question.
... similarly till 13! + 13 ... p = 4,5,6,7,8,9,10,11,12 and 13 can devide k without remainder and obviously it will be lesser than 13! + any positive integer.
hope it helps...
GMAT/MBA Expert
-
Stacey Koprince
- GMAT Instructor
- Posts: 2228
- Joined: Wed Dec 27, 2006 3:28 pm
- Location: Montreal, Canada
- Thanked: 639 times
- Followed by:694 members
- GMAT Score:780
Amit is right about this process of examining statement 2. I just want to take one step back and look at the question itself b/c there's a very easy way to rephrase it. This is basically asking us whether k is prime.
A prime number has exactly two factors: itself and 1. By definition, it cannot have a factor p that is between itself and 1. A non-prime number, on the other hand, that is greater than 1 will have a factor between itself and 1 (again, by definition).
So I can answer this question if I know whether k is prime.
(1) just tells me that k>24. There are prime numbers and non-prime numbers greater than 24, so that's not useful.
(2) for any sum, if the two numbers in that sum have a common factor, that factor will also be a factor of the sum. E.g., 2 + 4 = 6. 2 is a factor of 2 and 2 is a factor of 4. Therefore, 2 will also be a factor of 6.
Now, to follow amit's logic:
13! + 2 = 13*12*11*10*9*8*7*6*5*4*3*2 + 2. 2 is a factor of both of the numbers, so the sum will also have 2 as a factor. Therefore, the first possibility is not prime. Follow this logic all the way through 13! + 13; you will always have at least one factor, so that number is not prime. All of the possibilities given by statement 2 are not prime so I can answer the question: No, the integer k does not have a factor p such that 1<p<k.
A prime number has exactly two factors: itself and 1. By definition, it cannot have a factor p that is between itself and 1. A non-prime number, on the other hand, that is greater than 1 will have a factor between itself and 1 (again, by definition).
So I can answer this question if I know whether k is prime.
(1) just tells me that k>24. There are prime numbers and non-prime numbers greater than 24, so that's not useful.
(2) for any sum, if the two numbers in that sum have a common factor, that factor will also be a factor of the sum. E.g., 2 + 4 = 6. 2 is a factor of 2 and 2 is a factor of 4. Therefore, 2 will also be a factor of 6.
Now, to follow amit's logic:
13! + 2 = 13*12*11*10*9*8*7*6*5*4*3*2 + 2. 2 is a factor of both of the numbers, so the sum will also have 2 as a factor. Therefore, the first possibility is not prime. Follow this logic all the way through 13! + 13; you will always have at least one factor, so that number is not prime. All of the possibilities given by statement 2 are not prime so I can answer the question: No, the integer k does not have a factor p such that 1<p<k.
Please note: I do not use the Private Messaging system! I will not see any PMs that you send to me!!
Stacey Koprince
GMAT Instructor
Director of Online Community
Manhattan GMAT
Contributor to Beat The GMAT!
Learn more about me
Stacey Koprince
GMAT Instructor
Director of Online Community
Manhattan GMAT
Contributor to Beat The GMAT!
Learn more about me
