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# 20/2^5=1/2^m+1/2^n,where m and n are integers, mn=

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Manager
Joined: 01 Nov 2007
Posts: 101
20/2^5=1/2^m+1/2^n,where m and n are integers, mn= [#permalink]

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10 Mar 2008, 02:03
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20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?
Director
Joined: 05 Jan 2008
Posts: 687
Re: PS (exponents) [#permalink]

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10 Mar 2008, 02:38
Is it mn=3

20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?
=> 5/8= 1/2^m + 1/2^n solving for m & n we get m=1, n=3

thus mn=3
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Director
Joined: 01 Jan 2008
Posts: 622
Re: PS (exponents) [#permalink]

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10 Mar 2008, 06:52
prasannar wrote:
Is it mn=3

20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?
=> 5/8= 1/2^m + 1/2^n solving for m & n we get m=1, n=3

thus mn=3

i agree. 20/2^5 = 5/2^3 = (4+1)/2^3 = 1/2^1 + 1/2^3 -> mn = 3
Re: PS (exponents)   [#permalink] 10 Mar 2008, 06:52
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# 20/2^5=1/2^m+1/2^n,where m and n are integers, mn=

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