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# Indices

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Manager
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Indices [#permalink]

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25 Mar 2009, 22:07
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Which of the following is equal to (2^12 – 2^6) / (2^6 – 2^3)?
A. 2^6 + 2^3
B. 2^6 - 2^3
C. 2^9
D. 2^3
E. 2
Manager
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Re: Indices [#permalink]

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26 Mar 2009, 01:50
A
[2^6(2^6-1)]/[2^3(2^3-1)] simplify to [2^3(2^6-1)]/[(2^3-1)]
Since 2^6-1=(2^3+1)(2^3-1), you can replace
[2^3(2^3+1)(2^3-1)]/[(2^3-1)] , the 2^3-1 cancels
2^3(2^3+1) = 2^6+2^3
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Re: Indices [#permalink]

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26 Mar 2009, 02:35
(2^12 – 2^6) / (2^6 – 2^3)
=2^6(2^6-1)/2^3(2^3-1)
=2^6(2^3-1)(2^3+1)/2^3(2^3-1)
=2^3(2^3+1)
=2^6+2^3

Answer : A
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Lahoosaher

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Re: Indices [#permalink]

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14 Oct 2009, 07:50
amolsk11 wrote:
(2^12 – 2^6) / (2^6 – 2^3)
=2^6(2^6-1)/2^3(2^3-1)
=2^6(2^3-1)(2^3+1)/2^3(2^3-1)
=2^3(2^3+1)
=2^6+2^3

Answer : A

we can do it as (2^6 -2^3)(2^6 + 2^3)/(2^6 – 2^3) which simplifies to A
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Re: Indices [#permalink]

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14 Oct 2009, 16:52
knowing $$a^2 - b^2 = (a+b)(a-b)$$ helps solve all such problems, this is infact straight forward we can see both $$a^2 - b^2$$ and a-b in the problem
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Re: Indices [#permalink]

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22 Oct 2009, 13:00
I got stuck on this question for well over two minutes. What would you say is the level of this question?
Re: Indices   [#permalink] 22 Oct 2009, 13:00
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