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Does the integer k have a factor p such that 1<p<k? [#permalink]
 19 Jun 2010, 03:49
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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.htmlHope it's clear.
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Thanks..can not be clearer than this.
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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.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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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: factor-factorials-100670.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 can not 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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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.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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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.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: math-number-theory-88376.htmlHope it's clear.
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153. 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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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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Re: Does the integer k have a factor p such that 1<p<k? [#permalink]
22 Jan 2013, 03:40
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]
22 Jan 2013, 03:40
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