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# If n is an integer, xy does not equal zero, and x^n = y^n, what is

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If n is an integer, xy does not equal zero, and x^n = y^n, what is  [#permalink]

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10 Jun 2018, 05:19
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55% (hard)

Question Stats:

48% (02:08) correct 52% (01:57) wrong based on 33 sessions

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If n is an integer, xy does not equal zero, and $$x^n = y^n$$, what is the value of n?

(1) x/y = 2
(2) n < y < x

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Re: If n is an integer, xy does not equal zero, and x^n = y^n, what is  [#permalink]

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10 Jun 2018, 06:45
From statement 1 we know that x is not equal to y. Thus, for the above situation to be true, either x and y are equal or x and y being unequal with n being zero. Note: n is an integer so not fraction or decimal value can be considered. Thus, n = 0.

Similar is the case in statement 2, we know n is less than x and y and x and y are unequal numbers. Thus, n = 0.

Thus both statements are sufficient. IMO D.

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Re: If n is an integer, xy does not equal zero, and x^n = y^n, what is  [#permalink]

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10 Jun 2018, 07:34
2
If n is an integer, xy does not equal zero, and xn=ynxn=yn, what is the value of n?

(1) x/y = 2
(2) n < y < x

So a few things. Firstly, we know that neither x nor y is 0. Then, let's figure it out.

Statement (1): x/y = 2, so then x = 2y, and we have (2y)^n = y^n. The only way this could happen is if n = 0. So as such, (1) is sufficient.

Statement (2): n < y < x. To try and break this, let's have y be a negative number. Say -2. Now, let's have x be a positive number, say 2. The reason why we choose for y and x be the negative and positive forms of a number, is because we want the possibility that a specific power will result in the same results. So, we now have (-2)^n = 2^n. Now since n is a negative number (because y is), can we still solve the equation? The answer is yes! So long as n is an negative even integer, we will have the final fractions be equal. Since we can have multiple negative even integers for n however... statement (2) is insufficient.

P.S. If we want to input a few numbers in for statement (2).

Let n = -4, y = -2, x = 2. We have (-2)^-4 = (2)^-4.
<=> 1/(-2)^4 = 1/2^4
<=> 1/16 = 1/16. It works!

Let n = -6, y = -2, x = 2. We have (-2)^-6 = (2)^-6.
<=> 1/(-2)^6 = 1/2^6
<=> 1/64 = 1/64. It also works!

Disclaimer: I have not mathed for a very long time. Please excuse me if I figured statement (1) out wrong.
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Not a professional entity or a quant/verbal expert or anything. So take my answers with a grain of salt.

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Re: If n is an integer, xy does not equal zero, and x^n = y^n, what is  [#permalink]

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10 Jun 2018, 07:54
I think it should be A.
From Statement 1 we can deduce that (x/y)^n=1
=> n should be zero.

From statement two, we could create cases.
(x/y)^n=1
for example n could be 0, y could be 2, x could be three. Thus n =0.
Another case could be
n=-6, y=-1 and x=1, I think that would yield 1 also. Thus n=-6 here.
Inconsistent means insuffiecient

Thus A.
Re: If n is an integer, xy does not equal zero, and x^n = y^n, what is &nbs [#permalink] 10 Jun 2018, 07:54
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