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Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 04:11
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Does positive integer x have a factor f such that 1<f<x ? (1) x > 19! (2) 19! +2<=x<=19! +22 Thank you!
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Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 04:24
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The stem can be rephrased as: Is x prime? With this information we can start looking at the statements: Statement 1: This is obviously not enough, as there is an infinite number of primes greater than 19!. E.G. 19!+1. As consecutive integers are mutually coprime, 19!+1 must be a prime number. Statement 2: Sufficient. This statement explicitly excluded 19!+1, so we need to check if any value x is coprime with 19!, in which case x would be prime, since 19! contains all factors<=19. Because of the same reason, we do not need to actively check all values x <=19! + 19, for which x cannot be prime. One example for this: x=19! + 19: \(x=19! + 19=18!*19+19=(18!+1)*19\) Therefore, x=19!+19 cannot be prime. The same goes for all other x<=19!+19 Now we need to exclude x=19!+20, x=19!+21 and x=19!+22. Since all of the summands added to 19! are nonprime numbers with factors between 1 and 19, they share factors as 19!. Therefore statement 2 is sufficient to conclude that x cannot be prime. Answer B.
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 05:02
littlewarthog wrote: The stem can be rephrased as: Is x prime? With this information we can start looking at the statements: Statement 1: This is obviously not enough, as there is an infinite number of primes greater than 19!. E.G. 19!+1. As consecutive integers are mutually coprime, 19!+1 must be a prime number. 19!+1 is not a prime no. it is divisible by 71. though, the highlighted portion is correct.



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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 05:13
In deed, you are right. Thanks for the heads up. The conclusion stays valid though. For the numbers 2 to 22, it can easily be determined that they are not prime, so statement 2 is still sufficient.
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 07:07
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 09:22
littlewarthog wrote: The stem can be rephrased as: Is x prime? With this information we can start looking at the statements: Statement 1: This is obviously not enough, as there is an infinite number of primes greater than 19!. E.G. 19!+1. As consecutive integers are mutually coprime, 19!+1 must be a prime number. Statement 2: Sufficient. This statement explicitly excluded 19!+1, so we need to check if any value x is coprime with 19!, in which case x would be prime, since 19! contains all factors<=19. Because of the same reason, we do not need to actively check all values x <=19! + 19, for which x cannot be prime. One example for this: x=19! + 19: \(x=19! + 19=18!*19+19=(18!+1)*19\) Therefore, x=19!+19 cannot be prime. The same goes for all other x<=19!+19 Now we need to exclude x=19!+20, x=19!+21 and x=19!+22. Since all of the summands added to 19! are nonprime numbers with factors between 1 and 19, they share factors as 19!. Therefore statement 2 is sufficient to conclude that x cannot be prime. Answer B. Why are you saying that we are looking for a prime number? For instance 6 has a factor greater than 1 and less than 6..



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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 09:38
Dear gmatmania, You are right, actually the stem translates to "Is x NOT a prime number?" not "Is x a prime number?" as I have written mistakenly. For your example, as 6 does have factors greater than 1 and smaller than itself, it is not a prime. However, this doesn't have an impact on the strategy for this question, or even its outcome. As this is a DS question, the question is "Do the statements provide another information to conclude that x is NOT a prime?", which is equivalent to "Do the statements provide enough information to conclude that x is a prime?" Hope that helps.
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 10:03
I can'understand what kind of number are19!+2 and 19!+22 without calculating them.. I can't understand statement 2
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 10:20
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If all summands of a sum are divisible by a factor, then the sim is also divisible by that factor. Take for example 19!+6. 19! Is divisible by 6, as 6 is one of its factors and 6 is of course divisible by 6, the sum 19!+6 is also divisible by 6 and therefore no prime number. With that approach, you can conclude that all numbers from 19!+2 to 19!+22 are not prime numbers, as they all have at least one factor other than 1 and itself.
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 11:02
Will i have to consider only integers between 19! +2 and 19!+22? Can i consider 19!+2.1?
Thank you!
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 11:27
The question stem says that x is a positive integer. Happy to help. If this is helpful to you, please consider giving me Kudos.
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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29 Dec 2014, 11:31
gmatmania17 wrote: I can'understand what kind of number are19!+2 and 19!+22 without calculating them.. I can't understand statement 2
Posted from my mobile device lets consider a simple example lets say we have 5!+2. now 5! = 1*2*3*4*5. thus 5!+2= 1*2*3*4*5 +2 = 2(1*3*4*5 +1). now since the whole expression is a multiple of 2. thus 5!+2 is not a prime number. you can apply the same analogy for 19!+2. also, for 19!+22. we know that 19! contains both 11 and 2. hence we can easily take out 22 as common factor. thus 19!+22 will not be a prime gmatmania17 wrote: Will i have to consider only integers between 19! +2 and 19!+22? Can i consider 19!+2.1?
Thank you!
Posted from my mobile device NO. since x is an integer. therefore fractional values are not allowed. i hope it helps.



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Does positive integer x have a factor f such that 1<f<x? [#permalink]
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06 Jun 2016, 12:48
Does positive integer x have a factor f such that 1<f<x?
1) x> 19!
2) 19! + 2<x<19! + 22



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Re: Does positive integer x have a factor f such that 1<f<x? [#permalink]
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06 Jun 2016, 14:47
pepo wrote: Does positive integer x have a factor f such that 1<f<x?
1) x> 19!
2) 19! + 2<x<19! + 22 Dear pepo, Please do not start a new thread for a problem already posted. Please search extensively before beginning a brand new thread for a math question, because most GMAT practice math questions in existence have already been posted. You can find this one here: dosepositiveintegerxhaveafactorfsuchthat1fx190858.htmlI will ask Bunuel to merge these topics. Mike
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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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06 Jun 2016, 22:58



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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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06 Jun 2016, 23:57
gmatmania17 wrote: Does positive integer x have a factor f such that 1<f<x ? (1) x > 19! (2) 19! +2<=x<=19! +22
Thank you! Let us take statement 2 First We had enough explanation for statements such as 19! +2<=x<=19! +22. x will always have a factor (other than 1 and itself). In other words x is NOT prime. [say x = 19!+2 > we can take 2 out. This means that x is divisible by 2. Similarly we can take out 3, 4, 5, .......upto 19. Bottomline: x is composite number] B is sufficient. Statement 1 x > 19! x can be a prime number. In such a case we do not have any factor f. Or it can be a composite number. In which case there will be some f Not sufficient. B is the answer.



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Re: Dose positive integer x have a factor f such that 1<f<x ? [#permalink]
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17 Aug 2017, 19:18
the answer is B, 20,21,22 share factors with 19!




Re: Dose positive integer x have a factor f such that 1<f<x ?
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17 Aug 2017, 19:18






