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# M04 DS # 11

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M04 DS # 11 [#permalink]  13 Dec 2008, 17:40
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If $$j\neq{0}$$, what is the value of $$j$$ ?

(1) $$|j| = j^{-1}$$
(2) $$j^j = 1$$

Source: GMAT Club Tests - hardest GMAT questions

REVISED VERSION OF THIS QUESTION IS HERE: m04-ds-74542-20.html#p1250063
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Re: M04 DS # 11 [#permalink]  09 Mar 2012, 06:54
Expert's post
topmbaseeker wrote:
What is the value of integer $$J$$ ?

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

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

This question needs revision, it should state that J does not equal zero.
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Re: M04 DS # 11 [#permalink]  12 Mar 2012, 04:08
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Expert's post
ENAFEX wrote:
|J| = J^{-1}

This gives us two equations

eqn 1
J=1/J

i.e, J^2 = 1
Therefore, J= +1 or - 1

eqn 2
-J= 1/J
i.e, J^2= -1
Therefore J= SQRT(-1)

How to proceed?
Is SQRT(-1)=1? Even if this value is 1 we will still have another value -1 that we got from eqn 1 right?

If $$J\neq{0}$$, what is the value of $$J$$ ?

(1) $$|J| = J^{-1}$$
(2) $$J^J = 1$$

Two reasons why should the stem state that $$J\neq{0}$$:
For statement (1) if $$J=0$$ then we'll have $$0^{-1}=\frac{1}{0}=undefined$$. Remember you can't raise zero to a negative power.
For statement (2) if $$J=0$$ then we'll have $$0^0$$. 0^0, in some sources equals to 1, some mathematicians say it's undefined. Anyway you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT. So on the GMAT the possibility of 0^0 is always ruled out.

Also notice that saying in the stem that J is an integer is a redundant.

AS FOR THE SOLUTION:
If $$J\neq{0}$$, what is the value of $$J$$ ?

(1) $$|J| = J^{-1}$$ --> $$|J|*J=1$$ --> $$J=1$$ (here J can no way be a negative number, since in this case we would have $$|J|*J=positive*negative=negative\neq{1}$$). Sufficient.

(2) $$J^J = 1$$ --> again only one solution: $$J=1$$. Sufficient.

If you consider $$J<0$$ for (1) you'll have: $$-J=\frac{1}{J}$$ --> $$J^2=-1$$. This equation has no real roots (square root of negative number is undefined for the GMAT) and since GMAT is only dealing with real numbers then for the GMAT it has no roots ($$\sqrt{-1}=i$$, where $$i$$ is complex number/imaginary unit).

Hope it's clear.
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Re: M04 DS # 11 [#permalink]  25 Jul 2012, 14:27
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I would like to highlight a common "gotcha" when dealing with solutions involving absolute values.
|J| = 1/J (By the way (J ^ -1) = 1/J)
This equation may seem to have 2 solutions
J = 1/J (OR) -J = 1/J
J^2 = 1 (OR) J^2 = -1

Of the above, only J^2 = 1 makes sense.
This equation has 2 solutions, i.e J = +1 or -1, HOWEVER the uniqueness of absolute value equations is that
some of the solutions may NOT satisfy the original equation. One has to substitute the options BACK
into the original equation to check the feasibility.
In this case, J = -1 doesn't satisfy the original equation (which is |J| = 1/J) , hence J = +1 is the only solution.

Be wary of equations involving absolute values ! Always substitute the answer choices back into the equation to evaluate
the feasibility
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Re: M04 DS # 11 [#permalink]  25 Jul 2012, 17:20
Good one. I have read the explanations above and found them to be plausible. I rejected 0^0 as invalid and hence got 1 as the value of J for both the statements. hence chose D.

A few guys have said that GMAC will not test 0^0 and I am of the same opinion.
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Re: M04 DS # 11 [#permalink]  25 Jul 2013, 05:52
Expert's post
BELOW IS REVISED VERSION OF THIS QUESTION:

If $$j\neq{0}$$, what is the value of $$j$$ ?

(1) $$|j| = j^{-1}$$
(2) $$j^j = 1$$

(1) $$|j| = j^{-1}$$ --> $$|j|*j=1$$ --> $$j=1$$ (here $$j$$ can no way be a negative number, since in this case we would have $$|j|*j=positive*negative=negative\neq{1}$$). Sufficient.

(2) $$j^j = 1$$ --> again only one solution: $$j=1$$. Sufficient.

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Re: M04 DS # 11 [#permalink]  10 Jul 2014, 05:28
Just elaborating from Statement 1:

|J| = 1/J

J cannot be negative - Because if it is negative:

|-J| = J
1/(-J) = -1/J

J = -1/J (Which is not possible for any integer value - we already ruled out J=0)

So we get |J|*J = 1

J^2 - 1 = 0

This equation leads to 2 values of J
J = 1
J= -1

But J is not negative so J = 1.

Thanks
Re: M04 DS # 11   [#permalink] 10 Jul 2014, 05:28
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# M04 DS # 11

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