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10 Jun 2018, 05:19
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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
(1) x/y = 2
(2) n < y < x
10 Jun 2018, 06:45
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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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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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10 Jun 2018, 07:34
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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.
Thus, answer is A.
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.
(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.
Thus, answer is A.
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.
