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# If x and y are integers and x^y=256, what is the number of the differe

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Intern
Joined: 28 Dec 2014
Posts: 9
GMAT 1: 710 Q47 V41
If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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Updated on: 25 Jun 2017, 22:16
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Difficulty:

95% (hard)

Question Stats:

29% (01:54) correct 71% (02:12) wrong based on 252 sessions

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If x and y are integers and $$x^y=256$$, what is the number of the different values of $$y^x$$?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Originally posted by aazt on 25 Jun 2017, 09:01.
Last edited by Bunuel on 25 Jun 2017, 22:16, edited 1 time in total.
Renamed the topic and edited the question.
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Joined: 02 Aug 2009
Posts: 7991
Re: If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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25 Jun 2017, 20:19
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aazt wrote:
If x and y are integers and $$x^y$$=256, what is the number of the different values of $$y^x$$?

Options:
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

Hi,
256 has only 2 as a factor.
256 =$$2^8=4^4=16^2=256^1$$...
Those with even power can have x as NEGATIVE too..
So values can be $$2^8....(-2)^8....4^4....(-4)^4.....16^2....(-16)^2.....256^1$$..
So $$y^x$$ can be
8^2=64
8^{-2}=1/64
4^4=256..
4^(-4)=1/256
2^16
2^(-16)
1^256=1

So 7 values..
D
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Re: If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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01 Jul 2017, 23:38
2
aazt wrote:
If x and y are integers and $$x^y=256$$, what is the number of the different values of $$y^x$$?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

This is a very good question!

$$x^y = 256$$

Lets check all the values of x and y that satisfy the equation.

$$2^8 = 256$$

$$4^4 = 256$$

$$16^2 = 256$$

$$256^1 = 256$$

For all those numbers which are raised to even powers we know that we can get the same answer with a negative number as well.

$$2^8 = 256$$

$${-2}^8 = 256$$

$$4^4 = 256$$

$${-4}^4 = 256$$

$$16^2 = 256$$

$${-16}^2 = 256$$

$$256^1 = 256$$

We are looking for different values for $$y^x$$ with unique values:

$$8^2 = 16$$

$$8^{-2} = \frac{1}{8^2} = \frac{1}{16}$$

$$4^4 = 256$$

$$4^{-4} = \frac{1}{4^4} = \frac{1}{256}$$

$$2^16 =$$ no need to calculate as you know the value is going to be greater than 256

$$2^{-16} =$$$$\frac{1}{2^{16}}$$ = Once again no need to calculate as you know the value is going to be different from all above values

$$1^256 = 1$$

As we can see that we are getting$$7$$ different values for $$y^x$$

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Re: If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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26 Jul 2017, 18:56
Can someone explain what happens if we account for negative values of y?
For example, let x = 65536 (=256^2), and y = -2. Then -2^65536 gives another unique value for y^x, does it not?
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Re: If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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26 Jul 2017, 20:39
aazt wrote:
If x and y are integers and $$x^y=256$$, what is the number of the different values of $$y^x$$?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

$$x^y = 256$$

$$256 = 2^8 = 4^4 = 16^2 = 256^1$$, but since $$x$$ and $$y$$ can be both positive and negative integers, $$(-2)^8, (-4)^4, (-16)^2$$ also fits the condition.

So, $$y^x$$ can be $$1^{256}, 2^{16}, 8^{-2}, 4^{-4}, 2^{-16}, 4^{4}, 8^{2}$$ = 7 values. Ans - D.
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Re: If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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26 Jul 2017, 20:49
rocheng wrote:
Can someone explain what happens if we account for negative values of y?
For example, let x = 65536 (=256^2), and y = -2. Then -2^65536 gives another unique value for y^x, does it not?

Hi,

It's given that $$x^y = 256$$, but if we assume negative values for $$y$$, then $$x^y$$ will not equal $$256$$, for e.g., if $$x^y = 256$$, then $$x$$ and $$y$$ satisfy the value $$256$$ and $$1$$, respectively, but when we assign $$x = 256$$ and $$y = -1$$, we get $$x^y = 256^{-1} = \frac{1}{256}$$, which breaches the initial condition that $$x^y = 256$$. The given conditions must hold true throughout, for this reason $$y$$ cannot be a negative integer.
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Re: If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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31 Jul 2017, 20:22
TimeTraveller wrote:
rocheng wrote:
Can someone explain what happens if we account for negative values of y?
For example, let x = 65536 (=256^2), and y = -2. Then -2^65536 gives another unique value for y^x, does it not?

Hi,

It's given that $$x^y = 256$$, but if we assume negative values for $$y$$, then $$x^y$$ will not equal $$256$$, for e.g., if $$x^y = 256$$, then $$x$$ and $$y$$ satisfy the value $$256$$ and $$1$$, respectively, but when we assign $$x = 256$$ and $$y = -1$$, we get $$x^y = 256^{-1} = \frac{1}{256}$$, which breaches the initial condition that $$x^y = 256$$. The given conditions must hold true throughout, for this reason $$y$$ cannot be a negative integer.

Hi -- thanks for the response. You said that "if we assume negative values for $$y$$, then $$x^y$$ will not equal $$256$$." What about my example? $$x = 65536$$, $$y = -2$$, $$x^y = 256$$
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Re: If x and y are integers and x^y=256, what is the number of the differe  [#permalink]

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09 Aug 2018, 02:54
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Re: If x and y are integers and x^y=256, what is the number of the differe   [#permalink] 09 Aug 2018, 02:54
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