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The operator a# is defined on a number a as equal to (–1)^m*a, where

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The operator a# is defined on a number a as equal to (–1)^m*a, where  [#permalink]

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New post 15 Jul 2019, 00:20
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The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?

(1) a# + a = 0
(2) (a#)*a + a^2 = 0
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Re: The operator a# is defined on a number a as equal to (–1)^m*a, where  [#permalink]

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New post 15 Jul 2019, 10:30
Bunuel wrote:
The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?

(1) a# + a = 0
(2) (a#)*a + a^2 = 0


giving a try though not sure
#1
a=1 and m=1
so a# + a = 0 ;
a#'(–1)^m*a ; -1 and a=1 so -1+1 ; 0
sufficient
(a#)# – a ; -1-1 ; -2
#2
(a#)*a + a^2 = 0
a=1 so a# ; -1 and a^2 = 1 ; -1+1 ; 0
sufficient
IMO D
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Re: The operator a# is defined on a number a as equal to (–1)^m*a, where  [#permalink]

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New post 15 Jul 2019, 13:56
Archit3110 wrote:
Bunuel wrote:
The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?

(1) a# + a = 0
(2) (a#)*a + a^2 = 0


giving a try though not sure
#1
a=1 and m=1
so a# + a = 0 ;
a#'(–1)^m*a ; -1 and a=1 so -1+1 ; 0
sufficient
(a#)# – a ; -1-1 ; -2
#2
(a#)*a + a^2 = 0
a=1 so a# ; -1 and a^2 = 1 ; -1+1 ; 0
sufficient
IMO D


Why didn't you checked with a=-1, because that option will give you different results

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Re: The operator a# is defined on a number a as equal to (–1)^m*a, where  [#permalink]

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New post 28 Nov 2019, 10:12
Bump up.

Couldn't figure out how to approach this question.

If I take "a" common from equation in St 2, I get the equation in St 1 multiplied by "a". But couldn't figure out next steps
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Re: The operator a# is defined on a number a as equal to (–1)^m*a, where  [#permalink]

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New post 28 Nov 2019, 14:20
Bunuel wrote:
The operator a# is defined on a number a as equal to (–1)^m*a, where m is an integer. If a is not equal to zero, what is (a#)# – a ?

(1) a# + a = 0
(2) (a#)*a + a^2 = 0


depending on the value of m, a# can be either -a or a. if we can derive the nature of m (odd or even or -ve or +ve), that will suffice.

(1) a# + a =0 , a# = -a, that means m was odd in this case. Lets say m= 1, a=4, so a# = (-1)^1*4 = -4, (a#)#= (-1)(-4)=4, (a#)# -a =0. Again if we take a = -4, a# will be 4 and (a#)# will be -4. so (a#)# - a = -4 -(-4) =0. Sufficient.

(2) a {(a#) + a} = 0. since a can not be 0, {(a#) + a} =0, so, a# = -a, which is just like the previous data. Sufficient.
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Re: The operator a# is defined on a number a as equal to (–1)^m*a, where   [#permalink] 28 Nov 2019, 14:20
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