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## DS question

magical cook Master | Next Rank: 500 Posts
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#### DS question

Wed Feb 14, 2007 8:57 am
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
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.

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amitamit2020 Junior | Next Rank: 30 Posts
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Wed Feb 14, 2007 11:02 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...

### GMAT/MBA Expert

Stacey Koprince GMAT Instructor
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Wed Feb 14, 2007 8:13 pm
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.

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.

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