Does the integer k have a factor p such that 1<p<k?


(1) k > 4!

(2) \(13! + 2 \leq k \leq 13!+13\)

Thanks Bunuel. I have a question, what would happen if the second statement said this?

(2) \(13!+1\leq{k}\leq{13!+13}\)

I have seen some similar problems, in which they add 1. Could you post some links about it?

Thanks!
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Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters?
I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...

Does the integer k have a factor p such that 1<p<k?

Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number).

(1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient.

(2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*3*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient.

Answer: B.

Check similar question: if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html

Hope it's clear.
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Question asks whether some number k has a factor p which is more than 1 but less than k. For example if k=10 then the answer is yes, since both 2 and 5 are factors of 10 and are more than 1 and less than 10. But if for example k=7=prime then the answer is no, since 7 has no factor which is more than 1 and less than 7.

Now, look at the definition of a prime number: a prime number is a positive integer with exactly two factors: 1 and itself. So, we can say that the questions asks whether k is a prime number, because if it is then it won't have a factor which is more than 1 and less than k.

Hope it's clear.
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If \(13!+2\leq{k}\leq{13!+13}\), then we can factor out \(x\) out of \(13!+x\) (where \(2\leq{x}\leq{13}\)), which means that k is not a prime.

But you cannot apply the same logic with 4!+2<k<4!+6, since if k=4!+5, then you cannot factor out 5 out of it.
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Essentially this is about prime and non-prime.
Option 1: k>4!
Now assume k=5. then is there any factor of 5 which lies between 1 and 5? NO.
Take k=6. then there is 2,3 which lies between 1 and 6.
So 1 alone insufficient.

Option 2: 13!+2<=k<=13!+13
Take say 13!+3. now it can be rewritten as 3*{(13.12.11.....4.2.1)+1}
So there is 3 as factor between 1 and k. Similarly you can take the common factor out of all nos. in option 2.
Therefore option 2 is sufficient!
Add Kudos if this helps!

Target question: Does the integer k have a factor p such that 1 < p < k ?

This question is a great candidate for rephrasing the target question. (We have a free video with tips on rephrasing the target question: http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1100)

Let's look at a few cases to get a better idea of what the target question is asking.
- Try k = 6. Since 2 is a factor of 6, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 10 Since 5 is a factor of 10, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 16. Since 4 is a factor of 14, we can see that k DOES have a factor p such that 1<p<k.
- Try k = 5. Since 1 and 5 are the ONLY factors of 5, we can see that k does NOT have a factor p such that 1<p<k.
Aha, so if k is a prime number, then it CANNOT satisfy the condition of having a factor p such that 1 < p < k
In other words, the target question is really asking us whether k is a non-prime integer (aka a "composite integer")

REPHRASED target question: Is integer k a non-prime integer?

Statement 1: k > 4!
In other words, k > 24
This does not help us determine whether or not k is a non-prime integer? No.
Consider these two conflicting cases:
Case a: k = 25, in which case k is a non-prime integer
Case b: k = 29, in which case k is a prime integer
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: 13! + 2 ≤ k ≤ 13! + 13
Let's examine a few possible values for k.

k = 13! + 2
= (13)(12)(11)....(5)(4)(3)(2)(1) + 2
= 2[(13)(12)(11)....(5)(4)(3)(1) + 1]
Since k is a multiple of 2, k is a non-prime integer

k = 13! + 3
= (13)(12)(11)....(5)(4)(3)(2)(1) + 3
= 3[(13)(12)(11)....(5)(4)(2)(1) + 1]
Since k is a multiple of 3, k is a non-prime integer

k = 13! + 4
= (13)(12)(11)....(5)(4)(3)(2)(1) + 4
= 4[(13)(12)(11)....(5)(3)(2)(1) + 1]
Since k is a multiple of 4, k is a non-prime integer

As you can see, this pattern can be repeated all the way up to k = 13! + 13. In EVERY case, k is a non-prime integer

Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent
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"Question basically asks whether k is a prime number. If it is, then it won't have a factor psuch that 1<p<k

(definition of a prime number)."

Hi,

I just don't understand this. Can you please explain?

[quote="Bunuel"][quote="mneeti"][quote="Bunuel"]


Bunuel , Please make me understand one thing.

The question tells us that 1<p<k

which means p can be anything smaller than k , and question asks if p is a factor of K?

it means if k is 25 then is p a factor of 25? or we can assume p be any number less than 25 and check if its factor of 25 or not

My question is in option B,

If 13! + 2 <= k <= 13! + 13 , it means k is a very large digit and lies between those equalities.
SO NOW , p must be any integer less than k and we need to see if p is a factor of k?

How about 17? lets take p as 17

1<=17<=(13!+2,13!+3,13!+4........13!+13)
as we know from B that k has a factor , yes we can see 2, 3 ,4 ,5 till 13 it has factor but hat about 17?

17 satisfies the criteria and it should also be a factor and if its not then it must be uncertain if p is factor of k or not?


PLEASE HELP