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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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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.  +1Kudos if the answer helped



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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13 Apr 2014, 06:01



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



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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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]
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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 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]
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04 Oct 2015, 17:30
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 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/gmatdatasufficiency?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 nonprime integer (aka a "composite integer") REPHRASED target question: Is integer k a nonprime integer? Statement 1: k > 4! In other words, k > 24 This does not help us determine whether or not k is a nonprime integer? No. Consider these two conflicting cases: Case a: k = 25, in which case k is a nonprime integerCase b: k = 29, in which case k is a prime integerSince 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 nonprime integerk = 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 nonprime integerk = 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 nonprime integerAs you can see, this pattern can be repeated all the way up to k = 13! + 13. In EVERY case, k is a nonprime integerSince we can answer the target question with certainty, statement 2 is SUFFICIENT Answer = B Cheers, Brent
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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28 Mar 2016, 17:21
Here is a visual that should help.
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Screen Shot 20160328 at 6.20.15 PM.png [ 133.71 KiB  Viewed 1391 times ]
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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14 May 2016, 13:39
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 St1. we an have a prime and no prime, hence > not suff. St2. We need to find only 1 p that fits the discription. 13!+2=[2(13!/2+1)] > so 2 fits the description. > Suff.



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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04 Dec 2016, 20:07
"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?



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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05 Dec 2016, 00:16



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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08 Jul 2017, 12:07
[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



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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09 Jul 2017, 01:16
rocko911 wrote: 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 Forget about p. The question asks does a positive integer k, has a factor which is greater than 1 and less than k itself. As explained on previous pages this is the same as asking is k a prime number. If k is a prime number it won't have any such factor but if k is not a prime number, then it'll have such factor(s).
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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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09 Jul 2017, 06:08
Bunuel  i have subtracted 13! from each side of the inequality and got that k is equal or between 0 to 11. From this inequality i can't got the answer. Where am i wrong? By the way  i looked at the math book for inequality theory but haven't find something relevant. Is there any post that covers this topic? Thanks in advance!



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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12 Nov 2017, 08:22
Hi Bunuel, I have the same question as oryahalom. Cant we subtract 13! from both sides of the inequality? Thanks, RD



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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12 Nov 2017, 08:28



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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12 Nov 2017, 08:48
Thanks Bunuel for the immediate reply! : )



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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18 Nov 2017, 05:10
I did something wrong, which I would like to bring for everyone's attention:When I saw this question in QP1, I hurriedly cancelled 13! from both sides, reducing the expression to:
13! + 2 ≤ k ≤ 13! + 13 Thereafter, I went on to establish that since the expression is left as 2 ≤ k ≤ 13, so we can have both primes (2, 3, 5, 7, 11, 13) and nonprimes (4, 6, 8, 10, 12) in the range. So I picked insufficient for this statement.
Now I realize that the my method was not right. What I should have realized is that we take 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13 common from 2 + 13!, 3 + 13!, 4 + 13!, 5 + 13!, 6 + 13!, 7 + 13!, 8 + 13!, 9 + 13!, 10 + 13!, 11 + 13!, 12 + 13!, and 13 + 13!. And by doing so, we have proven that none of the numbers in the given range is prime.Now I realize that my method was incorrect. Maybe my concept of crossing out terms from an inequality is flawed and I need to work on it. Bunuel: If you don't mind, could you please let me know why we cannot crossout from an inequality expression like I tried when attempting this question?



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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18 Nov 2017, 09:05
1
This post received KUDOS
See Bunuel's last post above
you would have to cross out 13! in all 3 parts of the inequality, while it is only present in 2
13!+2≤k≤13!+13 =
13!+2≤k AND k≤13!+13



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Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
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19 Dec 2017, 06:46
gmatcracker2010 wrote: Does the integer k have a factor p such that 1<p<k?
(1) k > 4!
(2) \(13! + 2 \leq k \leq 13!+13\) We need to determine whether k has a factor p such that 1<p<k, or in other words, whether k is a prime number. If it is, then it doesn’t have a factor between 1 and itself. If it isn’t, then it does. Statement One Alone: k > 4! Since there are prime numbers greater than 4! and composite (nonprime) numbers greater than 4!, statement one alone is not sufficient to answer the question. Statement Two Alone: 13! + 2 ≤ k ≤ 13! + 13 13! will have a factor of any integers from 2 to 13 inclusive. For any number k is, in the range of integers from 13! + 2 to 13! + 13 inclusive, k will have a factor from 2 to 13 inclusive. In particular, if k = 13! + n where (2 ≤ n ≤ 13), k will have n as a factor. For example, if k = 13! + 5, then k will have a factor of 5, since 5 divides into 13! and 5. If k = 13! + 8, then k will have a factor of 8 and hence a factor of 2. Thus, statement two is sufficient to answer the question. Answer: B
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