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# Perfect Square

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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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21 May 2011, 06:28
2^8(1 + 2^3) + 2^n

= 2^8 * (9 + (2)^n-8)

= 9 + (2)^n-8 = 25 then its a perfect square

=> (2)^n-8 = 16

=> (2)^n-8 = 2^4

=> n = 12
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21 May 2011, 06:29
subhashghosh wrote:
2^8(1 + 2^3) + 2^n

= 2^8 * (9 + (2)^n-8)

= 9 + (2)^n-8 = 25 then its a perfect square

=> (2)^n-8 = 16

=> (2)^n-8 = 2^4

=> n = 12

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21 May 2011, 06:55
@subhashghosh can you please explain me this part?

subhashghosh wrote:
= 2^8 * (9 + (2)^n-8)

= 9 + (2)^n-8 = 25 then its a perfect square

Thanks.
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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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21 May 2011, 08:23

2^8[1+2^3+ 2^(n-8)] means

9+2^(n-8) = perfect square meaning 2^(n-8) = perfect square

16,64 are possible values.

n=12 fits.
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21 May 2011, 08:32

2^8 * (9 + (2)^n-8) is a product of 2^8 (perfect square) and (9 + (2)^n-8)

So we want (9 + (2)^n-8) 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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21 May 2011, 09:24
@subhash got it thanks.
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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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21 May 2011, 14:39
$$2^8+2^11+2^n$$

taking 2^8 common we have

(2^8)(1+2^3+2^(n-8))
= (2^8)(9+2^(n-8))

we know that 2^8 is a perfect square .

so 9+2^(n-8) 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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18 Aug 2012, 02:30
subhashghosh wrote:
2^8(1 + 2^3) + 2^n

= 2^8 * (9 + (2)^n-8)

= 9 + (2)^n-8 = 25 then its a perfect square

=> (2)^n-8 = 16

=> (2)^n-8 = 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^ n-8 )

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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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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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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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   [#permalink] 10 Sep 2012, 15:11
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