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integers

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Source: — Data Sufficiency |

by ajith » Fri Feb 19, 2010 4:46 am
daretodream wrote:What is the value of integer J?

1. |J| = J^{-1}
2. J^J = 1
1) |J| = J^{-1}

|J| is always +ve
J^{-1} should be +ve always
J is positive

j=1

Sufficient

2) J = +/- 1 ; insufficient

A
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by thephoenix » Fri Feb 19, 2010 10:04 am
daretodream wrote:What is the value of integer J?

1. |J| = J^{-1}
2. J^J = 1

s1)first statement says |J| = 1/J

the only value that is possible is1, so SUFF

s2)second statement

J^J = 1

J could be 0 or 1. 1^1 = 1 , 0^0=1

so INSUFF

hence A
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by November Rain » Fri Feb 19, 2010 10:48 am
IMO B

1. can be either (-1/1 = -1) or (1/1 = 1)

2. only positive 1 can satisfy this equation, because 0^0 is undeffined and it can't be -1 because it has an odd exponent[/spoiler]
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by shashank.ism » Sat Feb 20, 2010 1:53 pm
November Rain wrote:IMO B

1. can be either (-1/1 = -1) or (1/1 = 1)

2. only positive 1 can satisfy this equation, because 0^0 is undeffined and it can't be -1 because it has an odd exponent[/spoiler]
Yeah I am of the same view that 0^0 is indeterminate....
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by girish3131 » Tue Mar 09, 2010 4:34 am
good ques...!


IMO D

wat's OA..
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by outreach » Tue Mar 09, 2010 5:17 am
option 1
|J| = J^{-1}
=> J*|J|=1
J can only have one value that is 1. Hence suff

option 2
J^J = 1
=>j=+1 or -1
insufficient

OA is A
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by schumi_gmat » Tue Mar 09, 2010 9:58 pm
A is sufficient everbody agrees

B I was thinking that it was J raise to J and not sqr(J).

J^J = J^0 implies J=0.

when we resubstitute, 0^0 is indeterminate.

Hence A

Good Problem !!
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by girish3131 » Tue Mar 09, 2010 10:51 pm
Mr outreach N Schumi


i know

J^J = J^0 implies J=0.

when we resubstitute, 0^0 is indeterminate.



AND



option 2
J^J = 1
=>j=+1 or -1
insufficient # wch is wrong ofcourse



BUT consider this thing

J^J = 1 ONLY TRUE In case of when J=1

when u put J=-1 then it become (-1)^(-1) IS NOT EQUAL TO 1 (BECAUSE THIS IS NOT LIKE (-1)(-1) )SO WE CAN NOT CONSIDER THIS SCENARIO EQUAL TO ONE...

HENCE ANS IS D

N outreach

wen u put yr ans then put IMO not OA....

ta

HENCE WE CAN SAY THAT
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by raisethebar » Wed Mar 10, 2010 1:51 am
Agree with Ans D

J^J =1

For second statement concider 1 values for J=1 and J=-1
When J=1
1^1 =1

when J=-1
it will be (-1)^ -1 = 1/(-1) =-1 this is not true.

so we can conclude the 2nd statement itself is sufficient.
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by ShakeM » Sun Mar 14, 2010 1:46 am
Agree with D.

To satisfy the 1st equation J can only be 1.

(-1)^(-1) is not abs(-1), hence -1 cannot be the integer.


To satisfy the second equation J can only be 1.

(-1)^(-1) does not satisfy the second equation.
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by Phirozz » Sun Mar 14, 2010 7:45 am
daretodream wrote:What is the value of integer J?

1. |J| = J^{-1}
2. J^J = 1
IMO D. Both r independently sufficient
whats the OA ?
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