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20/2^5=1/2^m+1/2^n,where m and n are integers, mn= [#permalink]
 15 Aug 2008, 20:26
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20/2^5=1/2^m+1/2^n,where m and n are integers, mn=?
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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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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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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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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 = 3/4Kindly 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, 23:01, edited 1 time in total.
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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+2Then 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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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 = 3/4Kindly 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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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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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
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