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Re: Veritas Prep PS 17 - For integers x, y, and z [#permalink]

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03 Oct 2013, 23:33

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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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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.

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 [#permalink]

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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.

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

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26 Nov 2014, 09:04

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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: for-integers-x-y-and-z-if-2-x-y-z-131072-which-of-160992.html#p1274052 If you are not sure about cyclicity, check out this post: http://www.veritasprep.com/blog/2015/11 ... -the-gmat/ _________________

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

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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: for-integers-x-y-and-z-if-2-x-y-z-131072-which-of-160992.html#p1274052 If you are not sure about cyclicity, check out this post: http://www.veritasprep.com/blog/2015/11 ... -the-gmat/

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..

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/ odd-even / 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 [#permalink]

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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.
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