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TS
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Old Qn revisited 07 Apr 2005, 18:21
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Please provide ans with explanation.

Find an integer F.

(1) F=1/F
(2) |F|=1/F



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kapslock
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Old Qn revisited 07 Apr 2005, 23:24
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TS
Please provide ans with explanation.

Find an integer F.

(1) F=1/F
(2) |F|=1/F
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Is this a DS question TS?

(1) suggests F^2 = 1 => F = 1 or -1. No single solution.

(2) suggests
either F = 1/F which implies F^2 = 1 => F = +/-1 (as above)
Or F = -1/F which implies F^2 = -1 => F= i (iota) which is non-integer anyway. No solution.

Thus again a non unique solution.

Combine both, and we still have F = +/-1

Thus no unique solution even when we combine both. Therefore E.

Hope that helps.
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thearch
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Old Qn revisited 08 Apr 2005, 02:02
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B for me
from statement 2 you know that |F|*F=1
so, since |F| is integer and positive, F can't be negative. So F is 1.
a basic rule of inequalities is that a positive number can be obtained only by multiplying 2 numbers both positive or negative. Since |F| is positive, to obtain 1, the only satisfying number is 1.
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kapslock
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Old Qn revisited 08 Apr 2005, 03:37
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thearch
B for me
from statement 2 you know that |F|*F=1
so, since |F| is integer and positive, F can't be negative. So F is 1.
a basic rule of inequalities is that a positive number can be obtained only by multiplying 2 numbers both positive or negative. Since |F| is positive, to obtain 1, the only satisfying number is 1.
Show more


Oh I see it there - I was wrong in the first statement of
"(2) suggests..."
There's only F = 1 solution available there, not +/-1 because
|F|xF = 1 does NOT reduce to F^2 = 1.

I too change my solution from E to B. Learnt a valuable lesson - pay more attention to questions :(

However I saw some erroneous statement in your solution too. Just pointing it out.

----------------------------------------------------------------------
Firstly, when you say |F| is an integer and positive, its alright, but not the conclusion that F can't be negative.

Here's why.

Suppose F = -1. Thus, |F| = 1 is an integer and positive, but F still isn't a positive. It still remains negative.

Now in this speciific case I cited,
|F| x F = 1 can be either F = 1. (As I agreed above)

Secondly, its satisfied if F = -i where i = sqrt(-1), the complex number.
Thats coz LHS = i x -i = -i^2 = -(-1) = 1 = RHS.

Of course its not an integral solution, hence inadmissible, and so you end up staying with the solution F = 1.

Hope that helps.
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Old Qn revisited 08 Apr 2005, 04:58
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kapslock, you're going too far for me :!: :!:
I always assume that we're talking about real numbers, otherwise I'd give up my GMAT fight, 'cause I'm a stupid Italian studying "law and business administration" :oops:



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Hi there,
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