Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 25 May 2017, 07:09

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If you have done a variety of questions, you have come

Author Message
TAGS:

### Hide Tags

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7374
Location: Pune, India
Followers: 2288

Kudos [?]: 15107 [0], given: 224

If you have done a variety of questions, you have come [#permalink]

### Show Tags

03 Mar 2011, 20:45
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (00:50) wrong based on 7 sessions

### HideShow timer Statistics

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 18 Oct 2010 Posts: 91 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: 1666 Location: United States (IN) Concentration: Strategy, Technology Followers: 34 Kudos [?]: 533 [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: 42 Followers: 0 Kudos [?]: 6 [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: 151 Nationality: Indian Concentration: Entrepreneurship, General Management Followers: 7 Kudos [?]: 59 [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: 906 Followers: 14 Kudos [?]: 358 [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: 194 Followers: 4 Kudos [?]: 135 [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: 906 Followers: 14 Kudos [?]: 358 [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: 7374 Location: Pune, India Followers: 2288 Kudos [?]: 15107 [1] , given: 224 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

Veritas Prep Reviews

Manager
Joined: 14 Feb 2011
Posts: 194
Followers: 4

Kudos [?]: 135 [0], given: 3

### Show Tags

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
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: 906
Followers: 14

Kudos [?]: 358 [0], given: 123

### Show Tags

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
Manager
Status: TIME FOR 700+
Joined: 06 Dec 2010
Posts: 204
Schools: Fuqua
WE 1: Research in Neurology
WE 2: MORE research in Neurology
Followers: 3

Kudos [?]: 46 [0], given: 55

### Show Tags

04 Mar 2011, 10:18
tough one
_________________

Back to the grind, goal 700+

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: 906
Followers: 14

Kudos [?]: 358 [0], given: 123

### Show Tags

04 Mar 2011, 18:59
Brilliant ! Makes all the sense now.
SVP
Joined: 16 Nov 2010
Posts: 1666
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 34

Kudos [?]: 533 [0], given: 36

### Show Tags

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
_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7374
Location: Pune, India
Followers: 2288

Kudos [?]: 15107 [0], given: 224

### Show Tags

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Re: Number Properties [#permalink] 05 Mar 2011, 06:00 Similar topics Replies Last post Similar Topics: 2 If you are buying a product by rounding up to a$1 digit, is the sum o 1 24 Feb 2017, 16:24
3 How many questions did Joey attempt in a maths test having 25 question 4 13 May 2016, 05:36
1 Can you showing me the strategy and solving a few DS problem 8 13 Oct 2010, 12:31
1 This is easy, but I want to know if you have a method to 3 02 Sep 2010, 01:06
Multplying by a negative and inequalities 4 04 Aug 2010, 08:38
Display posts from previous: Sort by