For integers x, y, and z, if ((2^x)^y)^z = 131072 which of

Question

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 ?

KarishmaB · Accepted Answer

((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)

Qoofi · Answer

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 .

blueseas · Answer

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

KarishmaB · Answer

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

blueseas · Answer

joy ,
please refer to attached article.it might help you.

joylive · Answer

Thanks @Karishma, my confusion was  the product xyz , thought it is asking x*y*z = odd, thanks to you all, it's resolved now.

mvictor · Answer

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..

KarishmaB · Answer

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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/11 ... -the-gmat/

mvictor · Answer

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..

KarishmaB · Answer

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.

GMATYoda · Answer

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 4th-plus-one 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).

JiddiHuMain · Answer

Hey,

Can someone please explain what is wrong with option E "The product yz is positive".

Responses are much appreciated.

TheNightKing · Answer

I took almost 4 minutes in the test but got it right. I guess that's a good thing? There are 2 ways to look at it assuming you don't realize that it could be solved using cyclicity. 1: Just divide by 2. You get 65536. If you ever spent time for powers of 2 beyond 10 you will realise this is 2^16. So now you know that number is 2^17. (Side Note : If you are an engineer then you probably are very familiar with 65536

) 2. Try a higher power of 2. Don't stick to 2. Try 8. You will see it works. Try 16. Try 32. It will take few minutes but you will reach the end result. Just thought to share my 2 cents

Basshead · Answer

Cyclicity of 2:  2, 4, 8, 6

We're told  2^{xyz} = 131,072

Since 131,072 ends with 2, we can conclude that xyz is an odd number. Answer is B.

bumpbot · Answer

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For integers x, y, and z, if ((2^x)^y)^z = 131072 which of

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GMAT Problem Solving (PS) Questions

For integers x, y, and z, if ((2^x)^y)^z = 131072 which of

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