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Bunuel
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Vyshak
St1: (-1)^odd = -1 --> x^2 = odd --> x = odd
odd/2 --> Rem = 1
Sufficient

St2: n^x = n^(2x - 1)
If n = 1; x can be odd or even.
Not sufficient

Answer: A
I think its D ..

Statemnt 1 is right as u said.
But in st.2

As base are same so we can equalise expo.
x= 2x-1
x=1
So sufficient...

Answer is D..

Correct me if i am wrong...

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Vyshak
St1: (-1)^odd = -1 --> x^2 = odd --> x = odd
odd/2 --> Rem = 1
Sufficient

St2: n^x = n^(2x - 1)
If n = 1; x can be odd or even.
Not sufficient

Answer: A
I think its D ..

Statemnt 1 is right as u said.
But in st.2

As base are same so we can equalise expo.
x= 2x-1
x=1
So sufficient...

Answer is D..

Correct me if i am wrong...

Sent from my BND-AL10 using GMAT Club Forum mobile app

Hi

I think we cannot always equate the exponents if we dont know anything about the base. What if n = -1.
(-1)^3 is same as (-1)^5, but that doesn't mean 3 is equal to 5
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Bunuel

Fresh GMAT Club Tests' Question



If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\)

(2) \(n^x = n^{(2x - 1)}\)

I think the answer must be D.

Statement 1, for (−1)^(x^2)=−1, x^2 must be an odd number and odd number when divided by 2 always give a remainder 1. Hence sufficient.

Statement 2, considering n=-1, (-1)^x = (-1)^(2x-1) only when both the exponents are either odd or even. When x is even, 2x-1 cannot be even. Therefore x is odd and the remainder again is 1. Hence sufficient.

Correct me if I am wrong.
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I think ur right coz when I take X=2 then 2x-1 comes to be odd. When I take X as odd 2x-1 is odd. So second option is correct too
Ans Is D

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Thanks for the explanation. Understand the key take away here is that exponents can be equated only when base is not equal to 0, 1 or -1.

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Bunuel
OFFICIAL SOLUTION:


If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\).

This implies that x^2 is odd (if it were even, then (-1)^even = 1). x^2 = odd, on the other hand means that x is odd (since given that x is an integer). Any odd number divided by 2 gives the remainder of 1. Sufficient.


(2) \(n^x = n^{(2x - 1)}\).

Be careful not to fall into the trap. Remember we can automatically equate the exponents of equal bases when that base does not equal 0, 1 or -1:

\(1^x = 1^y\), for any values of x and y (they are not necessarily equal);
\((-1)^x = (-1)^y\), for any even values of x and y (they are not necessarily equal);
\(0^x = 0^y\), for any non-zero x and y (they are not necessarily equal).

Thus, for \(n^x = n^{(2x - 1)}\), we could equate the exponents if we knew that n is not 0, 1 or -1. In this case we'd have: x = 2x - 1, which would give x = 1. But if n is 0, 1 or -1, then we cannot equate the exponents. For example, if n = 1, then x can be even as well as odd. Not sufficient.

Answer: A.

Thank you "BUNUEL"...for nice expalination...

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Bunuel

Fresh GMAT Club Tests' Question



If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\)

(2) \(n^x = n^{(2x - 1)}\)


OFFICIAL SOLUTION IS HERE.

Par of GMAT CLUB'S New Year's Quantitative Challenge Set

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Bunuel

Fresh GMAT Club Tests' Question



If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\)

(2) \(n^x = n^{(2x - 1)}\)


OFFICIAL SOLUTION IS HERE.

Asked: If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\)
x^2 is odd
x is odd
the remainder when x is divided by 2 is 1
SUFFICIENT

(2) \(n^x = n^{(2x - 1)}\)
If n = 1, any value of x will satisfy the relation.
If n = 0, any value of x will satisfy the relation.
If n = -1, odd values of x will satisfy the relation.
Otherwise, x = 2x-1
x = 1
NOT SUFFICIENT

IMO A
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For example, if n = 1, then x can be even as well as odd. Not sufficient.
I understand that if numbers are -1;1 or 0 then the exponents do not have a role in the equation. But considering the second statement, if we have x in both sides of the equation should not they equal the same value? Whic even number can satisfy the equation? Just want to understand.


Bunuel
OFFICIAL SOLUTION:


If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\).

This implies that x^2 is odd (if it were even, then (-1)^even = 1). x^2 = odd, on the other hand means that x is odd (since given that x is an integer). Any odd number divided by 2 gives the remainder of 1. Sufficient.


(2) \(n^x = n^{(2x - 1)}\).

Be careful not to fall into the trap. Remember we can automatically equate the exponents of equal bases when that base does not equal 0, 1 or -1:

\(1^x = 1^y\), for any values of x and y (they are not necessarily equal);
\((-1)^x = (-1)^y\), for any even values of x and y (they are not necessarily equal);
\(0^x = 0^y\), for any non-zero x and y (they are not necessarily equal).

Thus, for \(n^x = n^{(2x - 1)}\), we could equate the exponents if we knew that n is not 0, 1 or -1. In this case we'd have: x = 2x - 1, which would give x = 1. But if n is 0, 1 or -1, then we cannot equate the exponents. For example, if n = 1, then x can be even as well as odd. Not sufficient.

Answer: A.
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Sevil92
For example, if n = 1, then x can be even as well as odd. Not sufficient.
I understand that if numbers are -1;1 or 0 then the exponents do not have a role in the equation. But considering the second statement, if we have x in both sides of the equation should not they equal the same value? Whic even number can satisfy the equation? Just want to understand.


Bunuel
OFFICIAL SOLUTION:


If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\).

This implies that x^2 is odd (if it were even, then (-1)^even = 1). x^2 = odd, on the other hand means that x is odd (since given that x is an integer). Any odd number divided by 2 gives the remainder of 1. Sufficient.


(2) \(n^x = n^{(2x - 1)}\).

Be careful not to fall into the trap. Remember we can automatically equate the exponents of equal bases when that base does not equal 0, 1 or -1:

\(1^x = 1^y\), for any values of x and y (they are not necessarily equal);
\((-1)^x = (-1)^y\), for any even values of x and y (they are not necessarily equal);
\(0^x = 0^y\), for any non-zero x and y (they are not necessarily equal).

Thus, for \(n^x = n^{(2x - 1)}\), we could equate the exponents if we knew that n is not 0, 1 or -1. In this case we'd have: x = 2x - 1, which would give x = 1. But if n is 0, 1 or -1, then we cannot equate the exponents. For example, if n = 1, then x can be even as well as odd. Not sufficient.

Answer: A.

Sorry, but I cannot follow what you mean at all... Can you please elaborate what you mean ?
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-------
What we know right away
-------

x an integer, x > 0


------------------------
What Statement 1 says
------------------------

This is saying that x^2 is odd; i.e. there does not exist some integer k such that x^2 = 2k

This is sufficient. Why?

Assert that x^2 is odd; we'll arrive at a contradiction quickly
--> Suppose x is even. Then we can write x = 2r for some integer r
--> Then x^2 = 4r^2
--> This contradicts x^2 odd, because we can set k = 2r^2

-------------------------
What Statement 2 says
-------------------------

n^x = n^(2x-1)

Unfortunately we don't know anything about n. If we suppose it's a real number, then we can set it equal to 0.

As a result, we get 0^x = 0^(2x-1), in which case x could be any natural number.

So this isn't sufficient
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Bunuel
Sevil92
For example, if n = 1, then x can be even as well as odd. Not sufficient.
I understand that if numbers are -1;1 or 0 then the exponents do not have a role in the equation. But considering the second statement, if we have x in both sides of the equation should not they equal the same value? Whic even number can satisfy the equation? Just want to understand.


Bunuel
OFFICIAL SOLUTION:


If x is a positive integer, what is the remainder when x is divided by 2?

(1) \((-1)^{(x^2)} = -1\).

This implies that x^2 is odd (if it were even, then (-1)^even = 1). x^2 = odd, on the other hand means that x is odd (since given that x is an integer). Any odd number divided by 2 gives the remainder of 1. Sufficient.


(2) \(n^x = n^{(2x - 1)}\).

Be careful not to fall into the trap. Remember we can automatically equate the exponents of equal bases when that base does not equal 0, 1 or -1:

\(1^x = 1^y\), for any values of x and y (they are not necessarily equal);
\((-1)^x = (-1)^y\), for any even values of x and y (they are not necessarily equal);
\(0^x = 0^y\), for any non-zero x and y (they are not necessarily equal).

Thus, for \(n^x = n^{(2x - 1)}\), we could equate the exponents if we knew that n is not 0, 1 or -1. In this case we'd have: x = 2x - 1, which would give x = 1. But if n is 0, 1 or -1, then we cannot equate the exponents. For example, if n = 1, then x can be even as well as odd. Not sufficient.

Answer: A.

Sorry, but I cannot follow what you mean at all... Can you please elaborate what you mean ?

Already understood what I haven't seen before. Thank you!
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