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1+1/3)(1+1/4)(1+1/5)(1+1/6)(1+1/7) divided by

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1+1/3)(1+1/4)(1+1/5)(1+1/6)(1+1/7) divided by [#permalink] New post 09 Aug 2007, 12:06
(1+1/3)(1+1/4)(1+1/5)(1+1/6)(1+1/7) divided by
(1/3)+(1/4)+(1/5)+(1/6)+(1/7)

anyone know a fast and easy way to solve this?
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 [#permalink] New post 09 Aug 2007, 12:51
EDIT: nevermind, just noticed the terms in the denominator are added to one another, not multiplied
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Re: arithmatic challenge [#permalink] New post 09 Aug 2007, 13:24
madcowudub wrote:
(1+1/3)(1+1/4)(1+1/5)(1+1/6)(1+1/7) divided by
(1/3)+(1/4)+(1/5)+(1/6)+(1/7)

anyone know a fast and easy way to solve this?


= (1+1/3)(1+1/4)(1+1/5)(1+1/6)(1+1/7) / (1/3)+(1/4)+(1/5)+(1/6)+(1/7)
= (4x5x6x7x8) / [(4x5x6x7) + (3x5x6x7) + (3x4x6x7) + (3x4x5x7) + (3x4x5x6)]
= (4x5x6x7x8) / [6(4x5x7) + (3x5x7) + (3x4x7) + (2x5x7) + (3x4x5)]
= (4x5x7x8) / [(4x5x7) + (3x5x7) + (3x4x7) + (2x5x7) + (3x4x5)]
= 1120 / (140 + 105 + 84 + 70 + 60)
= 1120 / 459
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Re: arithmatic challenge [#permalink] New post 09 Aug 2007, 16:26
madcowudub wrote:
(1+1/3)(1+1/4)(1+1/5)(1+1/6)(1+1/7) divided by
(1/3)+(1/4)+(1/5)+(1/6)+(1/7)

anyone know a fast and easy way to solve this?


Top part = (4/3)*(5/4)*(6/5)*(7/6)*(8/7) = 8/3
Below = [(1/3) + (1/6)] + (1/4) + (1/5) + (1/7) = [[(1/2)] + (1/4)] + [(1/5) + (1/7)] = [3/4] + [12/35]
At this point, if the answer is some sort of decimal, I would estimate 12/35 to 12/36 ~= 1/3
Then move on.
Otherwise, I would just solve the three fractions I got.
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Re: arithmatic challenge [#permalink] New post 09 Aug 2007, 16:38
bkk145 wrote:
madcowudub wrote:
(1+1/3)(1+1/4)(1+1/5)(1+1/6)(1+1/7) divided by
(1/3)+(1/4)+(1/5)+(1/6)+(1/7)

anyone know a fast and easy way to solve this?


Top part = (4/3)*(5/4)*(6/5)*(7/6)*(8/7) = 8/3

Below = [(1/3) + (1/6)] + (1/4) + (1/5) + (1/7) = [[(1/2)] + (1/4)] + [(1/5) + (1/7)] = [3/4] + [12/35]

At this point, if the answer is some sort of decimal, I would estimate 12/35 to 12/36 ~= 1/3

Then move on.
Otherwise, I would just solve the three fractions I got.


i like your approach. excellent. we are almost there.
Re: arithmatic challenge   [#permalink] 09 Aug 2007, 16:38
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