Co-prime integer (high level question) : GMAT Problem Solving (PS)
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# Co-prime integer (high level question)

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Co-prime integer (high level question) [#permalink]

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09 Jan 2012, 12:45
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Given:

2012*2010*2008*2006 - 9
----------------------------
2014*2010*2008*2004 + 135

to be simplified to x/y.

Last edited by adgir on 11 Jan 2012, 10:02, edited 1 time in total.
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Re: Co-prime integer (high level question) [#permalink]

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10 Jan 2012, 00:20
Given:

2012*2010*2008*2006 - 9
----------------------------
2014*2010*2008*2004 + 135

can be presented as x/y, where x and y are co-prime.

Find x + y.

I started with 2009 = a,

(a+3)(a-3)(a+1)(a-1) - 9 = (a2-9)(a2-1) - 9

Then on the bottom I've got

(a2-25)(a2-1) + 135

Stuck from this point...

Thanks!

In such questions, the options are very important. Off hand, I would guess that it might have something to do with odd-even properties. Both numerator and denominator will be odd so there sum will be even and possibly only one option will be even. We will only know when you give the options.
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Re: Co-prime integer (high level question) [#permalink]

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10 Jan 2012, 09:49
This question is beyond GMAT and doesn't come with options for an answer...

Anybody?

It seems that nobody can crack this one.
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Re: Co-prime integer (high level question) [#permalink]

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10 Jan 2012, 13:42
Given:

2012*2010*2008*2006 - 9
----------------------------
2014*2010*2008*2004 + 135

can be presented as x/y, where x and y are co-prime.

Find x + y.

I started with 2009 = a,

(a+3)(a-3)(a+1)(a-1) - 9 = (a2-9)(a2-1) - 9

Then on the bottom I've got

(a2-25)(a2-1) + 135

Stuck from this point...

Thanks!

Nothing wrong with how you started the problem. If a = 2009, the fraction is equal to:

[ (a^2 - 9)(a^2 - 1) - 9 ]/ [ (a^2 - 25)(a^2 - 1) + 135 ]

Now if you multiply out the products on the top and bottom, you get:

= [a^4 - 10a^2 + 9 - 9] / [a^4 - 26a^2 + 25 + 135]

= (a^4 - 10a^2) / (a^4 - 26a^2 + 160)

and we can now factor on top and bottom:

(a^2)(a^2 - 10) / (a^2 - 16)(a^2 - 10)

Now one factor cancels, leaving us with:

a^2/(a^2 - 16)

Now, plugging in a = 2009, this is equal to:

2009^2 / 2009^2 - 16 = 2009 / 2013*2005

Neither factor in the denominator could possibly share a divisor with 2009 (they're too close together), so this fraction is completely reduced. So the answer is 2009 + 2013*2005, which is a number in the millions.

This is a really awkward question. There's nothing mathematically interesting about it unless the answer turns out to be something very simple. It's not very satisfying to do all this work only to find that the answer is 4,038,074. Where are these questions from? They aren't GMAT-like at all.
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Re: Co-prime integer (high level question) [#permalink]

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11 Jan 2012, 10:01
Thanks! it was from my college professor, I was just really stuck on it, even though it is beyond GMAT.
Thank you again for explanation! I completely didn't think of multiplying top and bottom, this is where I was stuck.
Re: Co-prime integer (high level question)   [#permalink] 11 Jan 2012, 10:01
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