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For integers x, y, and z, if ((2^x)^y)^z = 131072 which of
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Updated on: 03 Oct 2013, 22:57
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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 ?
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Originally posted by joylive on 03 Oct 2013, 22:03.
Last edited by Bunuel on 03 Oct 2013, 22:57, edited 1 time in total.
Renamed the topic and edited the question.




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Re: Veritas Prep PS 17  For integers x, y, and z
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03 Oct 2013, 22:37
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)
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Re: Veritas Prep PS 17  For integers x, y, and z
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03 Oct 2013, 22:32
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 .
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Re: Veritas Prep PS 17  For integers x, y, and z
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03 Oct 2013, 22:33
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
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Re: Veritas Prep PS 17  For integers x, y, and z
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03 Oct 2013, 22:47
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
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Re: Veritas Prep PS 17  For integers x, y, and z
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03 Oct 2013, 22:51
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.
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Re: Veritas Prep PS 17  For integers x, y, and z
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03 Oct 2013, 22:57
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.



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Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of
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31 Jan 2016, 17:42
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..



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Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of
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31 Jan 2016, 19:20
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/
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Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of
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01 Feb 2016, 07:06
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..



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Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of
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01 Feb 2016, 07:35
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.
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Re: For integers x, y, and z, if ((2^x)^y)^z = 131072 which of
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22 Oct 2018, 08:35
Veritas Prep Official Solution: Simplifying the given expression, the question really provides that 2(xyz) ends in a 2. And finding the pattern for exponential units digits, one can see that the cycle goes: \(2^1=2\) \(2^2=4\) \(2^3=8\) \(2^4=16(ends−in−6)\) \(2^5=32(ends−in−2)\) And because the ending digit of 2 is the first repeating digit, it is known that every fourth exponent ends in 6, and that every 4thplusone exponent will yield a units digit of 2. So the exponent xyz must be odd, meaning that the each of x, y, and z must be odd, for if any one term was even the entire term would be even. That eliminates A and C (and guarantees B). D is wrong because, whatever the odd value (it's 17) that yields 2x=131072 , y and z could each be 1 leaving all the value to come from x, so the values must not yield a prime product. And, similarly, E is incorrect because both y and z could be 1, still allowing for xyz to be 17 (which you don't need to know  you can just know that all the positive value could come from one term).
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