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

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New post Updated on: 25 Jun 2017, 21:16
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Question Stats:

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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, 08:01.
Last edited by Bunuel on 25 Jun 2017, 21:16, edited 1 time in total.
Renamed the topic and edited the question.
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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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New post 25 Jun 2017, 19:19
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1
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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New post 01 Jul 2017, 22:38
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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 \(y^x\)?

(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 \(y^x\)

Hence, Answer is 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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New post 26 Jul 2017, 17:56
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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New post 26 Jul 2017, 19:39
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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New post 26 Jul 2017, 19:49
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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New post 31 Jul 2017, 19:22
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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Re: If x and y are integers and x^y=256, what is the number of the differe &nbs [#permalink] 09 Aug 2018, 01:54
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