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

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Does the integer k have a factor p such that 1 < p < k ? [#permalink]  29 Jan 2012, 15:15
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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?
[Reveal] Spoiler: 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 ? [#permalink]  29 Jan 2012, 15:21
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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.

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]  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]  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.

Posted from GMAT ToolKit
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  14 Feb 2012, 15:15
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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.

Posted from GMAT ToolKit

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]  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]  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.

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]  18 Nov 2012, 04:01
Expert's post
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.

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]  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]  29 May 2013, 22:06
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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]  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]  12 Apr 2014, 17:25
Does the integer g have a factor f such that 1 < f < g ?

1) g>3!
2) 11!+11≥g≥11!+2
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  13 Apr 2014, 01:10
Mountain14 wrote:
Does the integer g have a factor f such that 1 < f < g ?

1) g>3!
2) 11!+11≥g≥11!+2

What we need to figure out in this question is if g is a prime number or not.
If g is a prime number, it will not have a factor f, else it will have a factor f.

1) g > 3! means g is any number greater than 6. It can be a prime number, we do not know for sure.
2) 11!+11≥g≥11!+2 means g = 11! + k where 2 <= k <= 11, as all the numbers between 2 and 11 is present in 11!. g = k *(1+ 1*2*....)
g will always have x as factor, so it is not prime.

Hence, 2 is sufficient to answer the question. So option B.

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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  13 Apr 2014, 06:01
Expert's post
Mountain14 wrote:
Does the integer g have a factor f such that 1 < f < g ?

1) g>3!
2) 11!+11≥g≥11!+2

Merging similar topics. Please refer to the solutions above.

Similar questions to practice:
if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.html

for-any-integer-n-greater-than-1-n-denotes-the-product-of-168575.html
does-integer-n-have-2-factors-x-y-such-that-1-x-y-n-165983.html
if-z-is-an-integer-is-z-prime-128732.html

Hope this helps.
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  04 Jul 2014, 17:45
thank you so much, this is great!
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  19 Jul 2014, 03:23
Just as an FYI, this is a GMATPrep question. I saw it in the GMATPrep software.
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  16 Sep 2014, 12:49
enigma123 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:
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?

Great question. I answered it in 47 secs. Thanks for posting , keep it up.
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  06 Dec 2014, 05:06
Bunuel wrote:
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.

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.

Hi Bunuel- I thought the question asked for single value of p that satisfied the condition 1<p<k. What did I miss/ How avoid such a mistake again?
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Does the integer k have a factor p such that 1 < p < k ? [#permalink]  06 Feb 2015, 20:25
Pretty sure there's a 3 missing in the otherwise brilliant explanation provided by Bunuel.

13!+8= 8(2*3*4*5*6*7*9*10*11*12*13+1)
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]  03 Mar 2015, 09:29
enigma123 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:
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?

Question is asking whether K is prime ?
Option A : K>4! ; if K=25 then there exists 5 ; If K=29 ; there does not exists any integer between 1 and 29 which is factor of 29. Not Sufficient.
option B: 13!+2<= k<=13!+13 ; note here that 13! is multiple of all the numbers from 2 to 13 (inclusive) ; adding any number between 2 and 13 will just give another Even number and we know that there is only one even Prime Number 2 so clearly all integers in this range 13!+2 to 13!+13 will be Even numbers , so B is sufficient in telling that K is NOT PRIME.
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Re: Does the integer k have a factor p such that 1 < p < k ?   [#permalink] 03 Mar 2015, 09:29
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