Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 18 Apr 2011
Posts: 46
Location: United States
WE: Information Technology (Computer Software)

For integers x, y, and z, if ((2^x)^y)^z = 131072 which of [#permalink]
Show Tags
03 Oct 2013, 23:03
1
This post received KUDOS
13
This post was BOOKMARKED
Question Stats:
65% (01:00) correct 35% (00:59) wrong based on 257 sessions
HideShow timer Statistics
For integers x, y, and z, if \(((2^x)^y)^z = 131072\) which of the following must be true ? A. The product xyz is even B. The product xyz is odd C. The product xy is even D. The product yz is prime E. The product yz is positive I can understand that the last digit will be odd (i.e. z = odd) but how to determine the rest, x & y ?
Official Answer and Stats are available only to registered users. Register/ Login.
Last edited by Bunuel on 03 Oct 2013, 23:57, edited 1 time in total.
Renamed the topic and edited the question.



Manager
Joined: 18 Dec 2012
Posts: 95
Location: India
Concentration: General Management, Strategy
GMAT 1: 660 Q49 V32 GMAT 2: 530 Q37 V25
GPA: 3.32
WE: Manufacturing and Production (Manufacturing)

Re: Veritas Prep PS 17  For integers x, y, and z [#permalink]
Show Tags
03 Oct 2013, 23:32
2
This post received KUDOS
It is a PS problem. The problems comes down to 2^(xyz) = 131072 2^10 = 1024 ( Good to remember this ) So if you count the powers 11 = 2***, 12 = 4***. 13 = 8***, so on We find 2^17 = 131072, which imples xyz = 17 Hence the best possible solution is B. xyz is odd .
_________________
I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?



Director
Joined: 14 Dec 2012
Posts: 823
Location: India
Concentration: General Management, Operations
GPA: 3.6

Re: Veritas Prep PS 17  For integers x, y, and z [#permalink]
Show Tags
03 Oct 2013, 23:33
1
This post received KUDOS
2
This post was BOOKMARKED
joylive wrote: For integers x, y, and z, if \(((2^x)^y)^z = 131072\) which of the following must be true ?  The product xyz is even
 The product xyz is odd
 The product xy is even
 The product yz is prime
 The product yz is positive
I can understand that the last digit will be odd (i.e. z = odd) but how to determine the rest, x & y ? The number 2 has the cyclicity of 4 means: 2^1 =last digit = 2 2^2 = last digit =4 2^3 =last digit = 8 2^4 = last digit = 6 now after this again this will repeat. so every 4k + 1 term will have the unit digit = 2 now in our question last digit is 2 so x*y*z will be of form 4k + 1 = ODD hence B
_________________
When you want to succeed as bad as you want to breathe ...then you will be successfull....
GIVE VALUE TO OFFICIAL QUESTIONS...
GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabularylistforgmatreadingcomprehension155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmatanalyticalwritingassessment : http://www.youtube.com/watch?v=APt9ITygGss



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7988
Location: Pune, India

Re: Veritas Prep PS 17  For integers x, y, and z [#permalink]
Show Tags
03 Oct 2013, 23:37
4
This post received KUDOS
Expert's post
4
This post was BOOKMARKED
joylive wrote: For integers x, y, and z, if \(((2^x)^y)^z = 131072\) which of the following must be true ?  The product xyz is even
 The product xyz is odd
 The product xy is even
 The product yz is prime
 The product yz is positive
I can understand that the last digit will be odd (i.e. z = odd) but how to determine the rest, x & y ? \(((2^x)^y)^z = 2^{xyz} =\) (the powers get multiplied) Now look at the last digit of 131072 and think of the cyclicity of 2. \(2^1 = 2\) \(2^2 = 4\) \(2^3 = 8\) \(2^4 = 16\) \(2^5 = 32\) \(2^6 = 64\) Since 131072 ends with a 2, the power xyz must be of the form (4n + 1) i.e. it must be 1 more than a multiple of 4 (since the cyclicity of 2 is 4). Since xyz is one more than a multiple of 4, it must be odd. Answer (B)
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7988
Location: Pune, India

Re: Veritas Prep PS 17  For integers x, y, and z [#permalink]
Show Tags
03 Oct 2013, 23:47
joylive wrote: blueseas wrote: joylive wrote: For integers x, y, and z, if \(((2^x)^y)^z = 131072\) which of the following must be true ?  The product xyz is even
 The product xyz is odd
 The product xy is even
 The product yz is prime
 The product yz is positive
I can understand that the last digit will be odd (i.e. z = odd) but how to determine the rest, x & y ? The number 2 has the cyclicity of 4 means: 2^1 =last digit = 2 2^2 = last digit =4 2^3 =last digit = 8 2^4 = last digit = 6 now after this again this will repeat. so every 4k + 1 term will have the unit digit = 2 now in our question last digit is 2 so x*y*z will be of form 4k + 1 = ODD hence B Thanks for replying  still not getting the question. I'm aware of Last Digit cycle of 2 , that is how z is odd, but the product is 2 ^ (xyz), we know z = odd, but how do we figure out x and y to answer the question, finding out the the powers of 2 till 131072 is definitely not an option in GMAT, i suppose Ok, why don't you try this: Say x = 2, y = 3, z = 5 (z is odd) \(((2^2)^3)^5 = 2^{30}\) What digit do you think this will end with? I hope you will agree it will end with 4. Notice that the last digit is decided by the entire power of 2. Only z being odd is not enough. If either one of x or y is even, the entire power will become even and then it will end with 4 or 6
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Director
Joined: 14 Dec 2012
Posts: 823
Location: India
Concentration: General Management, Operations
GPA: 3.6

Re: Veritas Prep PS 17  For integers x, y, and z [#permalink]
Show Tags
03 Oct 2013, 23:51
joylive wrote: For integers x, y, and z, if \(((2^x)^y)^z = 131072\) which of the following must be true ?  The product xyz is even
 The product xyz is odd
 The product xy is even
 The product yz is prime
 The product yz is positive
I can understand that the last digit will be odd (i.e. z = odd) but how to determine the rest, x & y ? joy , please refer to attached article.it might help you.
_________________
When you want to succeed as bad as you want to breathe ...then you will be successfull....
GIVE VALUE TO OFFICIAL QUESTIONS...
GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabularylistforgmatreadingcomprehension155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmatanalyticalwritingassessment : http://www.youtube.com/watch?v=APt9ITygGss



Intern
Joined: 18 Apr 2011
Posts: 46
Location: United States
WE: Information Technology (Computer Software)

Re: Veritas Prep PS 17  For integers x, y, and z [#permalink]
Show Tags
03 Oct 2013, 23:57
VeritasPrepKarishma wrote: The number 2 has the cyclicity of 4 means: 2^1 =last digit = 2 2^2 = last digit =4 2^3 =last digit = 8 2^4 = last digit = 6
Ok, why don't you try this: Say x = 2, y = 3, z = 5 (z is odd)
\(((2^2)^3)^5 = 2^{30}\) What digit do you think this will end with? I hope you will agree it will end with 4.
Notice that the last digit is decided by the entire power of 2. Only z being odd is not enough. If either one of x or y is even, the entire power will become even and then it will end with 4 or 6 Thanks @Karishma, my confusion was the product xyz, thought it is asking x*y*z = odd, thanks to you all, it's resolved now.



Board of Directors
Joined: 17 Jul 2014
Posts: 2752
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of [#permalink]
Show Tags
31 Jan 2016, 18:42
it is somehow way too time consuming to be an actual gmat type question... it all comes down to identify how many powers of 2 we have..



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7988
Location: Pune, India

Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of [#permalink]
Show Tags
31 Jan 2016, 20:20
mvictor wrote: it is somehow way too time consuming to be an actual gmat type question... it all comes down to identify how many powers of 2 we have.. No. If the question wanted you to find the powers of 2, then the number on the right hand side would have been something like 512 or 1024... May be 2048 or 4096  something you are reasonably expected to know or figure out quickly. The question expects you to understand the concept of cyclicity and exponents multiplication. Then it takes no more than a few secs to arrive at the answer. Look at my solution here: forintegersxyandzif2xyz131072whichof160992.html#p1274052If you are not sure about cyclicity, check out this post: http://www.veritasprep.com/blog/2015/11 ... thegmat/
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Board of Directors
Joined: 17 Jul 2014
Posts: 2752
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of [#permalink]
Show Tags
01 Feb 2016, 08:06
VeritasPrepKarishma wrote: mvictor wrote: it is somehow way too time consuming to be an actual gmat type question... it all comes down to identify how many powers of 2 we have.. No. If the question wanted you to find the powers of 2, then the number on the right hand side would have been something like 512 or 1024... May be 2048 or 4096  something you are reasonably expected to know or figure out quickly. The question expects you to understand the concept of cyclicity and exponents multiplication. Then it takes no more than a few secs to arrive at the answer. Look at my solution here: forintegersxyandzif2xyz131072whichof160992.html#p1274052If you are not sure about cyclicity, check out this post: http://www.veritasprep.com/blog/2015/11 ... thegmat/You are right...nevertheless, with such a big number, I started to think that xyz is a fraction. Dumb me, did not pay attention to the answer choices.. I started to find the prime factorization...and it took way too much time..



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7988
Location: Pune, India

Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of [#permalink]
Show Tags
01 Feb 2016, 08:35
mvictor wrote: You are right...nevertheless, with such a big number, I started to think that xyz is a fraction. Dumb me, did not pay attention to the answer choices.. I started to find the prime factorization...and it took way too much time..
For future reference, remember that in GMAT Quant you are not expected to mess around with such an unwieldy number. If they want you to prime factorise it, they will not go beyond a 3 digit number. If it is larger, then the number would have factors that you can instantly see (a number such as 10000 etc). In case of such large numbers, focus on last digit/ oddeven / simplification (such as 9999 = 10000  1) etc.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



NonHuman User
Joined: 09 Sep 2013
Posts: 6527

Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of [#permalink]
Show Tags
25 Sep 2017, 05:06
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of
[#permalink]
25 Sep 2017, 05:06






