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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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Last edited by Carcass on 03 Oct 2019, 12:32, edited 1 time in total.
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Re: n is an integer. [#permalink]
20 Oct 2017, 07:23
Whatever is n, 2n+1 is always odd. Thus, quantity A is always 1, while quantity B is always 1 because it does not matter if the exponent is even or odd.
How can it be D? What am I losing?



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Re: n is an integer. [#permalink]
20 Oct 2017, 16:20
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]
21 Oct 2017, 07:37
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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]
04 Mar 2018, 02:19
Shouldnt the answer be "B" ? becuase A is always 1 and B is always 1 Carcass wrote: 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]
20 Mar 2018, 11:54
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]
09 Jul 2018, 01:50
Carcass wrote: 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]
09 Jul 2018, 18:32
Carcass wrote: 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]
03 Oct 2019, 12:08
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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]
03 Oct 2019, 12:47
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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.




Re: n is an integer.
[#permalink]
03 Oct 2019, 12:47





