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

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20/2^5=1/2^m+1/2^n,where m and n are integers, mn= [#permalink] New post 15 Aug 2008, 19:26
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20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?
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Re: PS: Exponents [#permalink] New post 15 Aug 2008, 19:49
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pmenon wrote:
20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?



20/2^5 = 5/8

5/8= 4/8 + 1/8
= 1/2 + 1/8

m=1
n=3

mn = 3
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Re: PS: Exponents [#permalink] New post 15 Aug 2008, 19:52
pmenon wrote:
20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?

mn=3, is it right?
Assume that m>n
5 x 2^(m+n)= 8 x 2^n x[2^(m-n)+1]
So 2^(m-n)+1=5
So m-n=2 => m=n+2
Then 5*26(2n+2)=8*2^n*5
=> 2n+2=n+3
Then n=1, m=2
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Re: PS: Exponents [#permalink] New post 15 Aug 2008, 20:57
pmenon wrote:
20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?


20/2^5=5/8=1/2^m+1/2^n= 2^m+ 2^n/ 2^M+N
m=1 and N=2 then 2^m+ 2^n/ 2^M+N = 5/8
m=2 and N=1 then 2^m+ 2^n/ 2^M+N = 5/8

mn=2
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Re: PS: Exponents [#permalink] New post 15 Aug 2008, 21:50
x2suresh wrote:
pmenon wrote:
20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?


20/2^5=5/8=1/2^m+1/2^n= 2^m+ 2^n/ 2^M+N
m=1 and N=2 then 2^m+ 2^n/ 2^M+N = 5/8
m=2 and N=1 then 2^m+ 2^n/ 2^M+N = 5/8

mn=2


if m=1 and N=2 then 2^m+ 2^n/ 2^M+N = 5/8

therefore, 2^1 + 2^2/ 2^3 = 2 + 4/8 = 6/8 = 3/4


if m=2 and N=1 then 2^m+ 2^n/ 2^M+N = 5/8

therefore 2^2 + 2^1/ 2^3 = 4+2/8 = 3/4



Kindly correct if I am wrong. If your approach is right, kindly help me finding fault in my approach.

Last edited by rahulgoyal1986 on 15 Aug 2008, 22:01, edited 1 time in total.
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Re: PS: Exponents [#permalink] New post 15 Aug 2008, 21:59
DavidArchuleta wrote:
pmenon wrote:
20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?

mn=3, is it right?
Assume that m>n
5 x 2^(m+n)= 8 x 2^n x[2^(m-n)+1]
So 2^(m-n)+1=5
So m-n=2 => m=n+2
Then 5*26(2n+2)=8*2^n*5
=> 2n+2=n+3
Then n=1, m=2


the bold equation is inconsistent with ur answer.

m= n+2 -------------- (1)
thn m has to be 3.

2n+2=n+3
Then n=1 but m = 3 (from 1)

Correct me if u find me wrong
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Re: PS: Exponents [#permalink] New post 16 Aug 2008, 06:03
rahulgoyal1986 wrote:
x2suresh wrote:
pmenon wrote:
20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?


20/2^5=5/8=1/2^m+1/2^n= 2^m+ 2^n/ 2^M+N
m=1 and N=2 then 2^m+ 2^n/ 2^M+N = 5/8
m=2 and N=1 then 2^m+ 2^n/ 2^M+N = 5/8

mn=2


if m=1 and N=2 then 2^m+ 2^n/ 2^M+N = 5/8

therefore, 2^1 + 2^2/ 2^3 = 2 + 4/8 = 6/8 = 3/4


if m=2 and N=1 then 2^m+ 2^n/ 2^M+N = 5/8

therefore 2^2 + 2^1/ 2^3 = 4+2/8 = 3/4



Kindly correct if I am wrong. If your approach is right, kindly help me finding fault in my approach.




sorry buddy.. you are right. I took 2^m = 2^1 = 1 instead of 2.

it should 1 and 3

2^m+ 2^n/ 2^M+N = 2+8 / (16) = 5/8

Thanks for pointing out.
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Re: PS: Exponents [#permalink] New post 16 Aug 2008, 14:34
The OA is 3.

Here is my question. I simplified it down to 2^m + 2^n / 2^m+n = 5/8 ... then i equated the numerators to each other, and the denominators to each other, i.e.:

2^m + 2^n = 5
2^m+n = 8

this gave me two different answers .... where did i go wrong ?
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Re: PS: Exponents [#permalink] New post 16 Aug 2008, 18:46
20*2^(-5)=2^(-m) + 2^(-n)
=>5 = 2^(3-m ) + 2^(3 -n)

One of the terms 2^(3-m ) or 2^(3 -n) has to be 1

so , if n= 3 => 3-m= 2 => m=1 so mn= 3

If we consider m=3 so n=1

So, mn= 3
Re: PS: Exponents   [#permalink] 16 Aug 2008, 18:46
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20/2^5=1/2^m+1/2^n,where m and n are integers, mn=

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