February 18, 2019 February 18, 2019 10:00 PM PST 11:00 PM PST We don’t care what your relationship status this year  we love you just the way you are. AND we want you to crush the GMAT! February 18, 2019 February 18, 2019 10:00 PM PST 11:00 PM PST Buy "AllInOne Standard ($149)", get free Daily quiz (2 mon). Coupon code : SPECIAL
Author 
Message 
TAGS:

Hide Tags

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8883
Location: Pune, India

Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
03 Mar 2011, 19:45
Question Stats:
40% (02:32) correct 60% (02:26) wrong based on 118 sessions
HideShow timer Statistics
Given that x and y are positive integers, is x prime? (1) \((y + 1)! <= x <= (y + 1)(y! + 1)\) (2) \((y + 1)! + 1\) has five factors 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.
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 18 Oct 2010
Posts: 70

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
03 Mar 2011, 20:43
this question is really tough and timeconsuming. 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)



Retired Moderator
Joined: 16 Nov 2010
Posts: 1408
Location: United States (IN)
Concentration: Strategy, Technology

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
03 Mar 2011, 20:44
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: 28

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
03 Mar 2011, 21:06
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: 126
Nationality: Indian
Concentration: Entrepreneurship, General Management

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
03 Mar 2011, 22:19
(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: 725

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
03 Mar 2011, 22:54
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_tipsIndigoIntentions 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: 174

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
03 Mar 2011, 23:59
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: 725

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
04 Mar 2011, 02:08
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: 8883
Location: Pune, India

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
04 Mar 2011, 05:05
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 4951 in Quant.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 14 Feb 2011
Posts: 174

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
04 Mar 2011, 05:20
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: 725

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
04 Mar 2011, 08:19
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



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: 725

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
04 Mar 2011, 17:59
Brilliant ! Makes all the sense now.



Retired Moderator
Joined: 16 Nov 2010
Posts: 1408
Location: United States (IN)
Concentration: Strategy, Technology

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
05 Mar 2011, 04:12
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: 8883
Location: Pune, India

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
05 Mar 2011, 05:00
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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 27 Mar 2014
Posts: 74

Re: Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
29 Sep 2017, 08:11
VeritasPrepKarishma wrote: 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 4951 in Quant. Experts help me understand the combination part.



GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 732

Given that x and y are positive integers, is x prime?
[#permalink]
Show Tags
18 Oct 2018, 07:19
VeritasKarishma wrote: Given that x and y are positive integers, is x prime?
(1) \((y + 1)! <= x <= (y + 1)(y! + 1)\)
(2) \((y + 1)! + 1\) has five positive factors
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.
Dear Karishma, Very beautiful problem, congrats! Let me contribute with a different wording. I recognize the famous result: for any integer n greater than 1, we don´t have primes in the interval [ n!+2 , n!+n ] (*) . (Because j is a factor of n!+j , where j is 2,3, ..., n.) \(x,y\,\, \ge 1\,\,{\rm{ints}}\) \(x\,\,\mathop {\rm{ = }}\limits^? \,\,{\rm{prime}}\) \(\left( 1 \right)\,\,\,n!\,\, \le x\,\, \le \,\,n!\, + n\,\,\,,\,\,{\rm{where}}\,\,\,n = y + 1\,\,\,\,\,\,\,\left[ {\,n!\, + n\,\, = \,\,\left( {y + 1} \right) \cdot y!\, + \left( {y + 1} \right) = \left( {y + 1} \right)\left( {y!\, + 1} \right)\,} \right]\) \(\left\{ \matrix{ \,{\rm{Take}}\,\,y = 1\,\,\left( {n = 2} \right)\,\,{\rm{and}}\,\,x\,{\rm{ = }}\,{\rm{2}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,y = 1\,\,\left( {n = 2} \right)\,\,{\rm{and}}\,\,x\,{\rm{ = }}\,4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\) \(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{ \,n!\,\, \le x\,\, \le \,\,n!\, + n \hfill \cr \,n!\,\, + \,\,1\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\, \Rightarrow \,\,n \ge 3\,\,\left( {y \ge 2} \right)\,\,\, \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\{ \matrix{ \,x = n!\,\,\,{\rm{not}}\,\,{\rm{prime}} \,\,\, (n \ge 3) \hfill \cr \,x = n!\,\, + \,\,1\,\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\left( {{\rm{statement}}\,\,\left( 2 \right)} \right) \hfill \cr \,n!\, + 2\,\, \le x\,\, \le \,\,n!\, + n\,\,\,,\,\,x\,\,\,{\rm{not}}\,\,{\rm{prime}}\,\,\,\left( * \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle\) Kind Regards, Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT) Our highlevel "quant" preparation starts here: https://gmath.net




Given that x and y are positive integers, is x prime?
[#permalink]
18 Oct 2018, 07:19






