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Does the integer k have a factor p such that 1 < p < k ?
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19 Jun 2010, 02:49
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Does the integer k have a factor p such that 1<p<k? (1) k > 4! (2) \(13! + 2 \leq k \leq 13!+13\)
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Does the integer k have a factor p such that 1 < p < k ?
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19 Jun 2010, 03:51
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: http://gmatclub.com/forum/factorfactorials100670.htmlHope it's clear.
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Re: Does the integer k have a factor p such that 1 < p < k ?
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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: factorfactorials100670.htmlHope 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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Re: Does the integer k have a factor p such that 1 < p < k ?
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03 Nov 2010, 14:51
metallicafan wrote: 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: factorfactorials100670.htmlHope 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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Re: Does the integer k have a factor p such that 1 < p < k ?
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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: factorfactorials100670.htmlHope 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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Re: Does the integer k have a factor p such that 1 < p < k ?
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10 Feb 2011, 17:19
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: factorfactorials100670.htmlHope 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: mathnumbertheory88376.htmlHope it's clear.
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Re: Does the integer k have a factor p such that 1 < p < k ?
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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 ?
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14 Feb 2012, 15:15
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 ToolKitQuestion 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 ?
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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: ifxisanintegerdoesxhaveafactornsuchthat100670.htmlHope 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 ?
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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: ifxisanintegerdoesxhaveafactornsuchthat100670.htmlHope 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 ?
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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 Essentially this is about prime and nonprime. 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!



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Does the integer k have a factor p such that 1 < p < k ?
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Re: Does the integer k have a factor p such that 1 < p < k ?
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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 ?
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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 18461 times ]



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Re: Does the integer k have a factor p such that 1 < p < k ?
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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 ?
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05 Dec 2016, 00:16
ruchitd wrote: "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? A prime number is a positive integer with only two factors 1 and itself. So, a prime number does not have a factor which is more that 1 and less than itself.
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Re: Does the integer k have a factor p such that 1 < p < k ?
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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 ?
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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 ?
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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 ?
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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




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