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Re: n is an integer. [#permalink]
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Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT \(2n + 1\) but it is (if you notice very carefully) that 2 (in the first quantity) is \(2^{n+1}\)
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Re: n is an integer. [#permalink]
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Carcass wrote:
Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT \(2n + 1\) but it is (if you notice very carefully) that 2 (in the first quantity) is \(2^{n+1}\)


Ok, yes the forum wasn't showing it right.

Thanks

I edit my answer then. The exponent in quantity A can be rewritten as \(2^{n+1} = 2n+2\). Thus, the exponent is always even and -1 to an even exponent becomes 1, so the answer would be C, not D. Sorry for bothering you but may you provide the OE?
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Re: n is an integer. [#permalink]
Shouldnt the answer be "B" ?
becuase A is always -1 and B is always 1
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



n is an integer.

Quantity A
Quantity B
\((-1)^2^^{n+1}\)
\((1)^n\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Re: n is an integer. [#permalink]
Thank you now I get it
IlCreatore wrote:
Carcass wrote:
Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT \(2n + 1\) but it is (if you notice very carefully) that 2 (in the first quantity) is \(2^{n+1}\)


Ok, yes the forum wasn't showing it right.

Thanks

I edit my answer then. The exponent in quantity A can be rewritten as \(2^{n+1} = 2n+2\). Thus, the exponent is always even and -1 to an even exponent becomes 1, so the answer would be C, not D. Sorry for bothering you but may you provide the OE?
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Re: n is an integer. [#permalink]
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



n is an integer.

Quantity A
Quantity B
\((-1)^{2}^{n+1}\)
\((1)^n\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Carcass, you are still writing it (D)? I agree that it's (C).
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Re: n is an integer. [#permalink]
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



n is an integer.

Quantity A
Quantity B
\((-1)^{2}^{n+1}\)
\((1)^n\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

ANswer IS D
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Re: n is an integer. [#permalink]
1
IlCreatore wrote:
Carcass wrote:
Sorry, sometimes the forum does not show properly the formatting (or maybe I am wrong).

It is NOT \(2n + 1\) but it is (if you notice very carefully) that 2 (in the first quantity) is \(2^{n+1}\)


Ok, yes the forum wasn't showing it right.

Thanks

I edit my answer then. The exponent in quantity A can be rewritten as \(2^{n+1} = 2n+2\). Thus, the exponent is always even and -1 to an even exponent becomes 1, so the answer would be C, not D. Sorry for bothering you but may you provide the OE?


You did right in considering the exponent as 2n+2. Since n is an integer, if for example you consider n=-1, then 2n+2 = 0. Then answer A would be equal to answer B. Thus you cannot get to a definite conclusion. Answer is D.

Regards
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Re: n is an integer. [#permalink]
1
since n is a integer. if the n =1 then the value in A is -1 while in B is 1. if the n=2 then the both values are equal.
So the answer is D.
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Re: n is an integer. [#permalink]
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I believe the answer is C.

the question is (-1) to the power of 2 to the power of (n+1)

So to solve first we have to designate different values for n (integer) including +ve and -ve values.

if n = +1 => the last power will be (1+1). so it will be (-1) to the power of (2) to the power of (2) which is (-1) to the power of 4 and thus equals to 1.
Answer is C

if n = -1 => the last power will be (0). So it will be (-1) to the power of (2) to the power of (0), which is (-1) to the power of (0) which is also equal to 1.
Answer is C

if n = 2 => the last power will be (3). So it will be (-1) to the power of (2) to the power of (3), which is (-1) to the power of (6) which is also equal to 1.
Answer is C.

if n=-2 => the last power will be (-1). So it will be (-1) to the power of (2) to the power of (-1), which is (-1) to the power of (-2), which is synonymous to the 1/(-1)^2 which equals to 1 eventually.

Answer is still C.

I think the idea over here in this question is that we have two exponents, last of them is n+1, whatever value we put its going to be multiplied by (2) and thus becoming even, although it might be negative (the -ve sign in the exponent signals a reciprocal) the reciprocal of 1 is going to be 1. So since the value - or reciprocal- of (-1)^even it is always going to give us a positive signed 1.

Thus the answer MUST be C.
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Re: n is an integer. [#permalink]
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Question of the day on Instagram. Promoted

Thanks!!!
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Re: n is an integer. [#permalink]
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Since its (-1)^(2^(n+1)) and n is an integer, if we take n to be say -2
the equation would become
(-1)^(2^(-1))
= (-1)^(1/2)
= sqrt(-1)
which is not defined hence answer is D
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Re: n is an integer. [#permalink]
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I found this problem easiest to considering some scenarios that might yield interesting results.

n=1 --> A = 1, B = 1 --> A=B
n=0 --> A = 1, B = 1 --> A=B
n=-1 --> A = -1, B = 1 --> A<B

Hence, the answer is D.
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Re: n is an integer. [#permalink]
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