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Veritas Prep GMAT Instructor
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If you have done a variety of questions, you have come [#permalink]

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03 Mar 2011, 20:45
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If you have done a variety of questions, you have come across this concept somewhere before. The point is to recognize which concept I am talking about.

Ques. Given that x and y are positive integers, is x prime?

I. $$(y + 1)! <= x <= (y + 1)(y! + 1)$$

II. $$(y + 1)! + 1$$ has five factors
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Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 18 Oct 2010 Posts: 92 Followers: 2 Kudos [?]: 8 [0], given: 0 Re: Number Properties [#permalink] Show Tags 03 Mar 2011, 21:43 this question is really tough and time-consuming. my answer is E and it took me > 5 mins to solve it without knowing if i am right. statement 2 is insufficient cos we dont know x statement 1: (y+1)!=1*2*...*y*(y+1) =1.2.....y.(y+1)= =1.2.....y.y +1.2.3....y=y.y!+y! (y+1)*(y!+1)=y.y!+y!+y+1=(y+1)!+(y+1) => (y+1)!<=x<=(y+1)!+(y+1) whatever y is, prime or not prime, (y+1)! is never a prime if y=1 so 2!=2<x<4 > x=3 y=2 so 3!=6<x<6+2+1=9 so x coule be 7,8,9 so insufficient statement 1+2 (y+1)!+1<=x<=(y+1)!+1+(y+1) A<=x<=A+y+1 ( consider A to be (y+1)!+1) insufficient E ( still confused) SVP Joined: 16 Nov 2010 Posts: 1673 Location: United States (IN) Concentration: Strategy, Technology Followers: 34 Kudos [?]: 481 [0], given: 36 Re: Number Properties [#permalink] Show Tags 03 Mar 2011, 21:44 I think the answer is C. _________________ Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant) GMAT Club Premium Membership - big benefits and savings Intern Joined: 06 Sep 2010 Posts: 45 Followers: 0 Kudos [?]: 5 [0], given: 0 Re: Number Properties [#permalink] Show Tags 03 Mar 2011, 22:06 Looks like this is pretty time consuming. On the D Day, I'll move on to the next question after spending 30 seconds and understanding the complexity. The answer looks like E. Experts ? VeritasPrepKarishma wrote: If you have done a variety of questions, you have come across this concept somewhere before. The point is to recognize which concept I am talking about. Ques. Given that x and y are positive integers, is x prime? I. $$(y + 1)! <= x <= (y + 1)(y! + 1)$$ II. $$(y + 1)! + 1$$ has five factors Manager Joined: 26 Sep 2010 Posts: 155 Nationality: Indian Concentration: Entrepreneurship, General Management GMAT 1: 760 Q V Followers: 7 Kudos [?]: 58 [0], given: 18 Re: Number Properties [#permalink] Show Tags 03 Mar 2011, 23:19 (1) implies (y+1)! <= x <= (y+1)! + (y+1) Lets replace y+1 by a => a! <= x <= a! + a For all, a! + i where i<=a, there will be a common factor (a+i) in a! + a =>x is not prime. (2) No result derivable. Answer should be A. OA please. _________________ You have to have a darkness...for the dawn to come. Director Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing. Affiliations: University of Chicago Booth School of Business Joined: 03 Feb 2011 Posts: 920 Followers: 14 Kudos [?]: 321 [0], given: 123 Re: Number Properties [#permalink] Show Tags 03 Mar 2011, 23:54 Agreed ! 100% it has to be A. a will have factors between a! + a and a!. x is NOT prime. I think this pattern is called "LONE WOLF" trap. it should be A. http://gmatclub.com/wiki/GMAT_math_tips IndigoIntentions wrote: (1) implies (y+1)! <= x <= (y+1)! + (y+1) Lets replace y+1 by a => a! <= x <= a! + a For all, a! + i where i<=a, there will be a common factor (a+i) in a! + a =>x is not prime. (2) No result derivable. Answer should be A. OA please. Manager Joined: 14 Feb 2011 Posts: 196 Followers: 4 Kudos [?]: 113 [0], given: 3 Re: Number Properties [#permalink] Show Tags 04 Mar 2011, 00:59 IndigoIntentions wrote: (1) implies (y+1)! <= x <= (y+1)! + (y+1) Lets replace y+1 by a => a! <= x <= a! + a For all, a! + i where i<=a, there will be a common factor (a+i) in a! + a =>x is not prime. (2) No result derivable. Answer should be A. OA please. The statement above is not true. For e.g. for a = 2 x lies between 2 and 4, lets say 3 and 3 is prime Director Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing. Affiliations: University of Chicago Booth School of Business Joined: 03 Feb 2011 Posts: 920 Followers: 14 Kudos [?]: 321 [0], given: 123 Re: Number Properties [#permalink] Show Tags 04 Mar 2011, 03:08 From statement 2) can I infer that (y+1)! + 1 is perfect square. some power of 4. ie. z^4 where z > 0. So far, I am not able to find a single integer z, that satisfies this criteria. I wonder if there is a problem with second statement. II. (y + 1)! + 1 has five factors beyondgmatscore wrote: The statement above is not true. For e.g. for a = 2 x lies between 2 and 4, lets say 3 and 3 is prime Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6910 Location: Pune, India Followers: 1990 Kudos [?]: 12362 [1] , given: 221 Re: Number Properties [#permalink] Show Tags 04 Mar 2011, 06:05 1 This post received KUDOS Expert's post VeritasPrepKarishma wrote: If you have done a variety of questions, you have come across this concept somewhere before. The point is to recognize which concept I am talking about. Ques. Given that x and y are positive integers, is x prime? I. $$(y + 1)! <= x <= (y + 1)(y! + 1)$$ II. $$(y + 1)! + 1$$ has five factors Let's analyze the question. The question stem just tells us that x and y are positive integers. The information was provided mainly to rule out a decimal value for x and since we are using factorials for y. Question: Is x prime? I. $$(y + 1)! <= x <= (y + 1)(y! + 1)$$ This needs to be modified a little. Why? because right now, there is no apparent relation between (y+1)! and (y + 1)(y! + 1). (y+1)! makes sense to me, (y! +1) does not. Can I bring everything in terms of (y+1)? I see that I have a (y+1) multiplied with y! and with 1. Recognize that (y+1)*y! = (y+1)! So, (y + 1)(y! + 1) = (y+1)y! + (y+1) = (y+1)! + (y+1) How will you know that this is how you would like to split it? Use (y+1)! on left side as a clue. The right side should make sense with respect to the left side which is in its simplest form. $$(y + 1)! <= x <= (y+1)! + (y+1)$$ Now think about it. x can take any of the following values (general case): (y+1)! - Not prime except if y is 1 (y+1)! + 1 - Cannot say whether it is prime or not. If y = 1, this is prime. If y is 2, this is prime. If y is 3 it is not. (y+1)! + 2 - Has 2 as a factor. Not prime (y+1)! + 3 - Has 3 as a factor. Not prime . . (y+1)! + (y+1) - Has (y+1) as a factor. Not prime Hence x may or may not be prime. II. $$(y + 1)! + 1$$ has five factors Not sufficient on its own. No mention of x. Together: There were two exceptions we found above. 1. If y = 1, then (y+1)! is prime. From stmnt II, since (1+1)! + 1 = 3 does not have 5 factors (It only has 2), y is not 1. Hence y is not 1 and (y+1)! is not prime 2. We don't know whether (y+1)! + 1 is prime Since (y+1)! + 1 has 5 factors, it is definitely not prime. Prime numbers have only 2 factors. Since both exceptions have been dealt with using both statements together, we can say that in every case, x is not prime. Answer (C). The question is not time consuming if you quickly see the pattern. It's all about getting exposed to the various concepts. The explanation seems long but only because I have written out everything my mind thinks in a few seconds. This is a relatively tough question but not beyond GMAT. Expect such questions if you are shooting for 49-51 in Quant. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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04 Mar 2011, 06:20
gmat1220 wrote:
From statement 2) can I infer that (y+1)! + 1 is perfect square. some power of 4. ie. z^4 where z > 0.

So far, I am not able to find a single integer z, that satisfies this criteria. I wonder if there is a problem with second statement.
II. (y + 1)! + 1 has five factors

beyondgmatscore wrote:

The statement above is not true. For e.g. for a = 2 x lies between 2 and 4, lets say 3 and 3 is prime

I now think that above conclusion can also be reached as any number that has more than 2 factors can be written in the form of a^p*b^q.. and so on where a and b are distinct prime numbers and the number of factors for such a number would be given by (p+1)*(q+1). Therefore, any number that has 5 factors would necessarily have the form Z^4 where z is a prime.
However, the second statement in the question is a red herring, as we would get really confused if we try and grapple with the problem of finding a value for y for which (y + 1)! + 1 has five factors. All we need to be concerned here with is the fact that because (y + 1)! + 1 has five factors (y + 1)! + 1 is not prime.

Superb question Karishma +1
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04 Mar 2011, 09:19
Sorry cant read those y+1s. Lets substitute y+1 with a.
I a! <= x <= a!+a
II a! + 1 has five factors.

1) Insufficient. We dont know if a is prime. We also dont know if a! is prime.

2) Insufficient. a!+1 has five factors. We can infer a! is not prime. And a is not prime. We don't know x.

Combine 1) + 2) Sufficient.

The pattern is [a! + Integer] has atleast 3 factors namely 1, (a! + Integer) and F such that 2 <= F <= a
Here Integer <= a

Agree?

beyondgmatscore wrote:
I now think that above conclusion can also be reached as any number that has more than 2 factors can be written in the form of a^p*b^q.. and so on where a and b are distinct prime numbers and the number of factors for such a number would be given by (p+1)*(q+1). Therefore, any number that has 5 factors would necessarily have the form Z^4 where z is a prime.
However, the second statement in the question is a red herring, as we would get really confused if we try and grapple with the problem of finding a value for y for which (y + 1)! + 1 has five factors. All we need to be concerned here with is the fact that because (y + 1)! + 1 has five factors (y + 1)! + 1 is not prime.

Superb question Karishma +1
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04 Mar 2011, 10:18
tough one
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04 Mar 2011, 18:59
Brilliant ! Makes all the sense now.
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05 Mar 2011, 05:12
Hi Karishma

How is :

(y+1)! + 3 - Has 3 as a factor. Not prime

if y = 1, then (2! +3) = 5, right ?

Regards,
Subhash
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05 Mar 2011, 06:00
subhashghosh wrote:
Hi Karishma

How is :

(y+1)! + 3 - Has 3 as a factor. Not prime

if y = 1, then (2! +3) = 5, right ?

Regards,
Subhash

We are considering all terms from (y+1)! to (y+1)! + (y+1)
If y = 1, then we are only considering terms 2!, 2! + 1 and 2! + 2.

I have given the general case above where y is some greater number say 6.
In that case
6!
6! + 1
6! + 2 - factor 2
6! + 3 - factor 3
6! + 4 - factor 4
6! + 5 - factor 5
6! + 6 - factor 6
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Re: Number Properties   [#permalink] 05 Mar 2011, 06:00
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