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Perfect Square [#permalink]
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21 May 2011, 05:45
For which value of n below is a perfect square (2^8)+(2^11)+(2^n)
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Re: Perfect Square [#permalink]
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21 May 2011, 06:28
2^8(1 + 2^3) + 2^n = 2^8 * (9 + (2)^n8) = 9 + (2)^n8 = 25 then its a perfect square => (2)^n8 = 16 => (2)^n8 = 2^4 => n = 12
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Re: Perfect Square [#permalink]
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21 May 2011, 06:29
subhashghosh wrote: 2^8(1 + 2^3) + 2^n
= 2^8 * (9 + (2)^n8)
= 9 + (2)^n8 = 25 then its a perfect square
=> (2)^n8 = 16
=> (2)^n8 = 2^4
=> n = 12 yup thats the answer
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Re: Perfect Square [#permalink]
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21 May 2011, 06:55
@subhashghosh can you please explain me this part? subhashghosh wrote: = 2^8 * (9 + (2)^n8)
= 9 + (2)^n8 = 25 then its a perfect square
Thanks.
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Re: Perfect Square [#permalink]
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21 May 2011, 07:02
hussi9 wrote: For which value of n below is a perfect square
(2^8)+(2^11)+(2^n) (2^8)+(2^11)+(2^n) =(2^(4*2))+(2. 2^4 .2^6 )+ (2^n)To complete the square n ==12 i.e 6*2 (2^2)^2 + +(2. 2^4 .2^6 ) + (2^6)^2
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Re: Perfect Square [#permalink]
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21 May 2011, 08:23
wish answer options were given would have done answer fitting. 2^8[1+2^3+ 2^(n8)] means 9+2^(n8) = perfect square meaning 2^(n8) = perfect square 16,64 are possible values. n=12 fits.
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Re: Perfect Square [#permalink]
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21 May 2011, 08:32
@jamifahad, consider this : 2^8 * (9 + (2)^n8) is a product of 2^8 (perfect square) and (9 + (2)^n8) So we want (9 + (2)^n8) to be a perfect square. Now 2^any number is even while 9 is odd So we want 9 + even # to be a perfect square, and that will be and odd #. So check for instance 9 + powers of two which can be a perfect square. After a few calculations by checking powers of 2 in increasing order, you can see that 2^4 = 16, and 16 + 9 = 25, a perfect square. Please let me know if you have any other query.
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Re: Perfect Square [#permalink]
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21 May 2011, 09:24
@subhash got it thanks.
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Re: Perfect Square [#permalink]
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21 May 2011, 09:39
(2^8)+(2^11)+(2^n) must be equal to (a+b)^2 form =>(2^8)+(2^11)+(2^n) = a^2 + b^2 + 2ab 2^11 looks like a 2ab form . if we consider 2^8 = a^2 => a = 2^4 => b must be 2^11 / (2*2^4) => b = 6 also n = 2b = 12
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Re: Perfect Square [#permalink]
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21 May 2011, 14:39
\(2^8+2^11+2^n\)
taking 2^8 common we have
(2^8)(1+2^3+2^(n8)) = (2^8)(9+2^(n8))
we know that 2^8 is a perfect square .
so 9+2^(n8) has to be perfect square
minimum power of 2 that can be added to 9 to make it a perfect square is 4.( 9+16 = 25)
=> n 8 =4
Hence n=12.



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Re: Perfect Square [#permalink]
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18 Aug 2012, 02:30
subhashghosh wrote: 2^8(1 + 2^3) + 2^n
= 2^8 * (9 + (2)^n8)
= 9 + (2)^n8 = 25 then its a perfect square
=> (2)^n8 = 16
=> (2)^n8 = 2^4
=> n = 12 just a small query, we are assuming that 2^n > 2^8 , which enables us to take 2^8 as a common factor and we get 2^8 ( 2^ n8 ) but what if 2^n = 4 then n = 2 in which case we cannot take 2^8 as a common factor . any thoughts as to how we can easily assume 2^n > 2^8 ?
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Re: Perfect Square [#permalink]
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18 Aug 2012, 03:22
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hussi9 wrote: For which value of n below is a perfect square
(2^8)+(2^11)+(2^n) Use the formula for \((a+b)^2=a^2+2ab+b^2.\) Since \(2^8+2^{11}=(2^4)^2+2*2^4*2^6\), we need an extra term, that of \((2^6)^2=2^{12}\) to complete the expression to a perfect square, \((2^4+2^6)^2.\) So, \(n\) should be \(12\).
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Re: Perfect Square [#permalink]
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18 Aug 2012, 18:51
EvaJager wrote: hussi9 wrote: For which value of n below is a perfect square
(2^8)+(2^11)+(2^n) Use the formula for \((a+b)^2=a^2+2ab+b^2.\) Since \(2^8+2^{11}=(2^4)^2+2*2^4*2^6\), we need an extra term, that of \((2^6)^2=2^{12}\) to complete the expression to a perfect square, \((2^4+2^6)^2.\) So, \(n\) should be \(12\). Thank you eva I think this approach is more justified, like it better
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Re: Perfect Square [#permalink]
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10 Sep 2012, 15:11
gurpreetsingh wrote: (2^8)+(2^11)+(2^n)
must be equal to (a+b)^2 form
=>(2^8)+(2^11)+(2^n) = a^2 + b^2 + 2ab
2^11 looks like a 2ab form . if we consider 2^8 = a^2 => a = 2^4
=> b must be 2^11 / (2*2^4) => b = 6
also n = 2b = 12 Your approach is absolutely right, except some ambiguity. Let me correct it. Where from did you get n=2b ? Also, b is not equal to 6. Correct expression is \(b=2^6\) As \(2^n = b^2\) So, \(2^n= (2^6)^2=2^{12}\) Therefore, n=12
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Re: Perfect Square
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