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Hard Question

by beatrix kiddo » Tue Mar 19, 2013 1:55 am
I was struggling with this question for a few days, and couldn't find a solution. Please help.

The expression (2012*2010*2008*2006-9)/(2014*2010*2008*2004+135)can be written in the form a/b, with a and b relatively prime positive integers. Find a and b.
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by Anju@Gurome » Tue Mar 19, 2013 2:12 am
beatrix kiddo wrote:The expression (2012*2010*2008*2006-9)/(2014*2010*2008*2004+135)can be written in the form a/b, with a and b relatively prime positive integers. Find a and b.
Let us assume 2009 = x.
Hence, the expression = [(x + 3)(x + 1)(x - 1)(x - 3) - 9]/[(x + 5)(x + 1)(x - 1)(x - 5) + 135]
= [(x² - 1)(x² - 9) - 9]/[(x² - 1)(x² - 25) + 135]
= [(x� - 10x² + 9) - 9]/[(x� - 26x² + 25) + 135]
= (x� - 10x²)/(x� - 26x² + 160)
= x²(x² - 10)/(x� - 10x² - 16x² + 160)
= x²(x² - 10)/[(x² - 10)(x² - 16)]
= x²/(x² - 16)
= x²/[(x - 4)(x + 4)]

Hence, the given expression = (2009*2009)/(2005*2013)

As 2009 has no common factor with either 2005 or 2013, a = 2009*2009 and b = 2005*2013
Anju Agarwal
Quant Expert, Gurome

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