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1/2^10+1/2^11+1/2^12+1/2^12=

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1/2^10+1/2^11+1/2^12+1/2^12=  [#permalink]

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New post 09 Sep 2010, 14:08
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Question Stats:

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\(\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}+\frac{1}{2^{12}}=\)


(A) 1/2^7

(B) 1/2^8

(C) 1/2^9

(D) 1/2^13

(E) 1/2^45
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Re: 1/2^10+1/2^11+1/2^12+1/2^12=  [#permalink]

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New post 09 Sep 2010, 14:15
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Sarah707 wrote:
1/2^10+1/2^11+1/2^12+1/2^12=


(A) 1/2^7

(B) 1/2^8

(C) 1/2^9

(D) 1/2^13

(E) 1/2^45


Please help me with the above problem. Is there any short cut for this??

Thanks


\(\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}+\frac{1}{2^{12}}\)

= \(\frac{1}{2^{10}}*[ 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^2}]\)

= \(\frac{1}{2^{10}}*[ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4}]\)

= \(\frac{1}{2^{10}}*[ 2]\)

= \(\frac{1}{2^{9}}\)

Hence C
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Re: 1/2^10+1/2^11+1/2^12+1/2^12=  [#permalink]

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New post 09 Sep 2010, 15:30
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Great question, and great explanation from Gurpreet.

This question brings up a pretty important strategic point on exponents. Keep in mind that we're not particularly good at adding/subtracting exponents, but we're really good at multiplying/dividing them. Accordingly, whenever you're asked to add/subtract exponents, you should ask what you can do to turn that into multiplication. Typically that involves factoring, as we had to do here.

For exponents in general, I see three guiding principles that can help you attack problems that look intimidating:

1) Multiply - turn addition/subtraction into multiplication (usually by factoring) so that you can apply the exponent rules that you know

2) Find common bases - in order to multiply exponents or set exponents equal, you'll usually need common bases, so you should look to break bases down into prime factors (e.g. 6^4 = (2*3)^4 = 2^4 * 3^4) so that you can create common bases.

3) Look for patterns - exponents are quite pattern driven, so they lend themselves well to number properties both common (units digit; positive/negative) and unique (10^x minus something will give you a difference with digits that are a series of 9s followed by the last few unique digits given by the subtraction).

If you're thinking of these general principles when you see exponent questions, you should always be able to turn the inconvenient problem that they give you into something that you know you do well.
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Re: 1/2^10+1/2^11+1/2^12+1/2^12=  [#permalink]

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New post 20 Jul 2017, 13:16
Sarah707 wrote:
\(\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}+\frac{1}{2^{12}}=\)


(A) 1/2^7

(B) 1/2^8

(C) 1/2^9

(D) 1/2^13

(E) 1/2^45


\(\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}+\frac{1}{2^{12}}\)

\(\frac{1}{2^{10}} (1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^2})\)

\(\frac{1}{2^{10}} (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4})\)

\(\frac{1}{2^{10}} (\frac{4 + 2 + 1 + 1}{4})\) \(= \frac{1}{2^{10}} (\frac{8}{4})\)

\(\frac{1}{2^{10}} (2) =\) \(\frac{2^1}{2^{10}}\)

\(\frac{1}{(2^{-1})(2^{10})} = \frac{1}{2^{(10-1)}} = \frac{1}{2^9}\)

Answer (C)...

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Re: 1/2^10+1/2^11+1/2^12+1/2^12=  [#permalink]

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New post 21 Jul 2017, 03:10
Sarah707 wrote:
\(\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}+\frac{1}{2^{12}}=\)


(A) 1/2^7

(B) 1/2^8

(C) 1/2^9

(D) 1/2^13

(E) 1/2^45



Let 1/2^10=x

Therefore,

1/x+1/2x+1/4x+1/4x
=3/2x+1/2x
=4/2x
=2/x
=2/2^10
=1/2^9

C..
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Re: 1/2^10+1/2^11+1/2^12+1/2^12=  [#permalink]

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New post 05 May 2019, 00:47
Sarah707 wrote:
\(\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}+\frac{1}{2^{12}}=\)


(A) 1/2^7

(B) 1/2^8

(C) 1/2^9

(D) 1/2^13

(E) 1/2^45


Take 1/12 as common from all the terms, the expression will become
= \(\frac{1}{12} ( 2^{2}+2+1+1) = 8/2^{12} = 2^{3}/ 2^{12}= 1/2^{9}\)
Hence, answer is C
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Re: 1/2^10+1/2^11+1/2^12+1/2^12=   [#permalink] 05 May 2019, 00:47
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