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# If e is a nonzero integer, is 1/2^e greater than or less than 1?

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If e is a nonzero integer, is 1/2^e greater than or less than 1?  [#permalink]

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05 Mar 2019, 05:16
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65% (hard)

Question Stats:

41% (01:31) correct 59% (01:45) wrong based on 48 sessions

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If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0

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Joined: 17 May 2018
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Re: If e is a nonzero integer, is 1/2^e greater than or less than 1?  [#permalink]

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05 Mar 2019, 05:24
If e is less than 0 the value of 1/2^e can be greater than 1
1. E>-1
So e>=1
1/2^e greater than 1--> no
So statement 1 is giving us the answer to the question
2. E^2>0
E can be either negative or positive
So statement 2 is not sufficient to ans the question

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Re: If e is a nonzero integer, is 1/2^e greater than or less than 1?  [#permalink]

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05 Mar 2019, 10:27
Bunuel wrote:
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0

#1
-e<1
means e=-ve integer
so
1/2^e will always be >1
sufficient
#2
e^2 > 0
e can be +ve or -ve so 1/2^e can be > 1 or <1 ; not sufficient
IMO A
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If e is a nonzero integer, is 1/2^e greater than or less than 1?  [#permalink]

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Updated on: 05 Mar 2019, 10:58
Bunuel wrote:
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0

(1) −e < 1
(2) e^2 > 0[/quote]

Ok (1/2) ^ e, where e does not equal 0 and e is an integer will be greater than 1 if e <0 (a fraction to a negative exponent gives us 1 divided by some number less than 1, and thus we get a number greater than 1). So the question simplifies to is e <0 subject to the constraint e is an integer >0 Yes/No?

(1) -e <1 gives us e >-1 |e is an integer not equal to 0 --> This means that e = 1,2,3, thus e>0, thus we can mark sufficient.

(2) e^2 >0. Therefore either e>0 or e <0 | e is an integer not equal to 0. First consider e >0. Then e = 1,2,3... thus we get a YES answer. Now lets consider e <0 | e is an integer not equal to 0. Then e = -1,-2,-3...., thus e <0 giving us a NO answer. Thus (2) is NS

Originally posted by ocelot22 on 05 Mar 2019, 10:46.
Last edited by ocelot22 on 05 Mar 2019, 10:58, edited 1 time in total.
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Re: If e is a nonzero integer, is 1/2^e greater than or less than 1?  [#permalink]

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05 Mar 2019, 10:50
ocelot22

Please see question : e is a nonzero integer
in #1 you have considered e as a fraction....

ocelot22 wrote:
Bunuel wrote:
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0

Ok (1/2) ^ e, where e does not equal 0 and e is an integer will be greater than 1 if e <0 (a fraction to a negative exponent gives us 1 divided by some number less than 1, and thus we get a number greater than 1). So the question simplifies to is e <0 Yes/No?

(1) -e <1 gives us e >-1 Lete =-1/2. Then (1/2)^(-1/2) = 1/((1/2)^1/2) >1 so we get a YES. Let e =2, then (1/2)^2 = 1/4 and we get a NO. Since we get a Yes and a No, we can mark NS

(2) e^2 >0. Therefore either e>0 or e <0. Let e =-1, then we get a YES answer (since (1/2)^-1 = 2). Let e = 1. then we get (1/2)^1 = 1/2 and we get a No answer NS

(1) and (2) e>-1 and either e>0 or e<0. First lets consider e >-1 subject to e>0. Suppose e = 1. then we get a No answer (since 1/2^1 = 1/2). Now lets consider e >-1 subject to e<0. This gives -1<e<0. Let e = -1/2. Then We get a Yes answer (since 1/2^(-1/2) = 1/((1/2)^1/2) > 0). Since we can get a Yes answer from one case and a no answer from the other case, statements (1) and (2) combined are Not Sufficient.

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If e is a nonzero integer, is 1/2^e greater than or less than 1?  [#permalink]

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05 Mar 2019, 10:56
Archit3110 wrote:
ocelot22

Please see question : e is a nonzero integer
in #1 you have considered e as a fraction....

ocelot22 wrote:
Bunuel wrote:
If e is a nonzero integer, is 1/2^e greater than or less than 1?

(1) −e < 1
(2) e^2 > 0

Ok (1/2) ^ e, where e does not equal 0 and e is an integer will be greater than 1 if e <0 (a fraction to a negative exponent gives us 1 divided by some number less than 1, and thus we get a number greater than 1). So the question simplifies to is e <0 subject to the constraint e is an integer >0 Yes/No?

(1) -e <1 gives us e >-1 |e is an integer not equal to 0 --> This means that e = 1,2,3, thus e>0, thus we can mark sufficient.

(2) e^2 >0. Therefore either e>0 or e <0 | e is an integer not equal to 0. First consider e >0. Then e = 1,2,3... thus we get a YES answer. Now lets consider e <0 | e is an integer not equal to 0. Then e = -1,-2,-3...., thus e <0 giving us a NO answer. Thus (2) is NS

Fixed
If e is a nonzero integer, is 1/2^e greater than or less than 1?   [#permalink] 05 Mar 2019, 10:56
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