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Dose positive integer x have a factor f such that 1<f<x ?

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Dose positive integer x have a factor f such that 1<f<x ?  [#permalink]

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New post 29 Dec 2014, 05:11
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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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New post 29 Dec 2014, 05:24
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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 co-prime, 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 co-prime 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 non-prime 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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New post 29 Dec 2014, 06:02
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 co-prime, 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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New post 29 Dec 2014, 06:13
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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New post 29 Dec 2014, 08:07
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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!


Similar questions to practice:
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does-p-have-a-factor-x-where-1-x-p-and-x-and-p-are-positive-integers-188773.html

Hope this helps.
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Re: Dose positive integer x have a factor f such that 1<f<x ?  [#permalink]

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New post 29 Dec 2014, 10:22
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 co-prime, 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 co-prime 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 non-prime 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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New post 29 Dec 2014, 10:38
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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New post 29 Dec 2014, 11:03
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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New post 29 Dec 2014, 11:20
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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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New post 29 Dec 2014, 12:02
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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New post 29 Dec 2014, 12:27
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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New post 29 Dec 2014, 12:31
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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New post 06 Jun 2016, 13:48
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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New post 06 Jun 2016, 15:47
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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:
dose-positive-integer-x-have-a-factor-f-such-that-1-f-x-190858.html
I 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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Re: Dose positive integer x have a factor f such that 1<f<x ?  [#permalink]

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New post 07 Jun 2016, 00:57
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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New post 17 Aug 2017, 20:18
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 ? &nbs [#permalink] 17 Aug 2017, 20:18
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