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

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i dont have any idea to solve the attached problem. Please provide methodology to solve such problems.

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

(1) k > 4!
(2) 13! + 2<= k <= 13!+13
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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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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: factor-factorials-100670.html

Hope it's clear.
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Re: gmat prep ds [#permalink]

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New post 19 Jun 2010, 04:46
Thanks..can not be clearer than this.
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Re: gmat prep ds [#permalink]

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New post 03 Nov 2010, 14:20
Bunuel wrote:
Does the integer k have a factor p such that 1<p<k?

Question basically asks is \(k\) 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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Answer: B.

Check similar question: factor-factorials-100670.html

Hope it's clear.


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

Question basically asks is \(k\) 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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Answer: B.

Check similar question: factor-factorials-100670.html

Hope it's clear.


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!


So basically we should determine whether \(13!+1\) is a prime number (as shown above all other possible values of k are not are not prime), which cannot be done without a computer. There are some particular values of \(k=n!+1\) for which we can say whether it's a prime or not with help of Wilson's theorem, but again it's out of the scope of GMAT.

By the way: \(13!+1\) is not a prime number, it has two distinct prime factors: \(13!+1=83*75,024,347=6,227,020,801\), so the answer still will be B.
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New post 10 Feb 2011, 17:04
Bunuel wrote:
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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Answer: B.

Check similar question: factor-factorials-100670.html

Hope it's clear.


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 ...
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ChenggongMAS wrote:
Bunuel wrote:
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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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 \(13!+x\)). Sufficient.

Answer: B.

Check similar question: factor-factorials-100670.html

Hope it's clear.


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 ...


k=13!+8 means k=2*3*4*5*6*7*8*9*10*11*12*13+8, now factor out 8: k=8*(2*4*5*6*7*9*10*11*12*13+1) --> k is a multiple of 8 as k=8*something --> as k is a multiple of 8 it can not be a prime number. You can have any number from 2 to 13 inclusive instead of 8, and you'll be able to factor out this number the same way as you did with 8, so any number of a type 13!+x, wher x is from 2 to 13 inclusive will be a multiple of x, thus not a prime number.

Check the link in my first post for similar problem. Also check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it's clear.
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New post 13 Dec 2011, 14:37
Answer can't be E, it has to be B. I don't have OG with me, but I need to check.

1. By itself not sufficient. K could be prime.
2. 13! + 2 <= K <= 13! + 13 ensures that you have ever number from 2 - 13 dividing k.
That is:
13! + 2 -> div by 2
13! + 3 -> div by 3
... so on till 13.

So, answer should be B.
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Does the integer k have a factor p such that 1 < p < k ?

(1) k > 4!
(2) 13! + 2 <= k <= 13! + 13

[Reveal] Spoiler:
So considering statement 1

K > 24

Lets say k = 25, so the factors are 1, 5 and 25. P will be 1 as 1 is the factor of every number. Therefore, insufficient.

Considering Statement 2

13!+2<= k<=13!+13

simplifying this expression will give

2<=K<=13, so can be anything from 2...............13. As p can and cannot be greater than 1 this statement is insufficient.

Combining the two, I think will not give an answer too. Therefore, for me its E. Any thoughts anyone as I don't have an OA?

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Last edited by Bunuel on 16 Sep 2014, 12:56, edited 4 times in total.
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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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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 29 Jan 2012, 15:22
Thanks for the great explanation.
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 14 Feb 2012, 13:27
Hi bunuel
I don't understand the definition of a prime ( 1<p<k ). I know what a prime is but this drives me nuts.
For an example I take 10 for k. I have factors 2x5 that are greater than 1 and smaller than 10. :?:

Or is it asking for one factor? I see "a factor" in the question, which is singular. Then I understand this. It is really tricky.

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M3tm4n wrote:
Hi bunuel
I don't understand the definition of a prime ( 1<p<k ). I know what a prime is but this drives me nuts.
For an example I take 10 for k. I have factors 2x5 that are greater than 1 and smaller than 10. :?:

Or is it asking for one factor? I see "a factor" in the question, which is singular. Then I understand this. It is really tricky.

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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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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 01 Sep 2012, 07:32
Got this one wrong. Can somebody explain how to solve this?
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 18 Nov 2012, 02:18
Bunuel wrote:
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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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.



Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time)
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 18 Nov 2012, 04:01
mneeti wrote:
Bunuel wrote:
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 \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), 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.



Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time)


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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Re: Does the integer k have a factor p such that 1<p<k? [#permalink]

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New post 22 Jan 2013, 02:40
gmatcracker2010 wrote:
i dont have any idea to solve the attached problem. Please provide methodology to solve such problems.

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

(1) k > 4!
(2) 13! + 2<= k <= 13!+13


1.
if k=29 p=29= k NO!
if k =35 p could be 7 or 5 YES!
INSUFFICIENT!

2.
13! + 2 = 2(13!/2 + 1) = 2*odd
13! + 3 = 3(13!/3 + 1) = 3*odd
...
13! + 13 = 13 (12! +1) = 13*odd
Always there is a 1< p < 13! + n

Answer: B
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 27 Jan 2013, 20:25
So basically in the second statement we can factor out a number from 2-13 but if its higher than that we can't say for sure it would be a prime or not?
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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crazypriya wrote:
Does the integer k have a factor p such that 1<p<k?
1.k>4!
2.13!+2<=k<=13!+13

[Reveal] Spoiler:
OA B


The question is basically asking whether K is prime or not. If it is prime then P can be 1 or k else it will follow 1<p<k.

Statement 1. k>4!
k>24;
let's say k=25, then p=5; we get a Yes to the question.
if k=29; then p=1 or 29; we get a No to the question.
Thus, insufficient.


Statement 2: 13!+2<=k<=13!+13
k can take any value from 13!+2 to 13!+13.
All the values in this range has one thing in common. They all can have one common value:
for e.g. 13!+2= 2[(13*12*...except 2)+1]
13!+10=10[(13*12*...(except 10)+1]
No matter whether the value in the bracket is prime or not; the common value will make it non prime and will always give a yes to the question.
Thus, sufficient.
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]

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New post 29 May 2013, 22:17
crazypriya wrote:
Does the integer k have a factor p such that 1<p<k?
1.k>4!
2.13!+2<=k<=13!+13

[Reveal] Spoiler:
OA B

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!
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Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 29 May 2013, 22:17

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