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

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Intern
Joined: 25 Aug 2010
Posts: 2

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09 Sep 2010, 14:08
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Difficulty:

15% (low)

Question Stats:

83% (01:16) correct 17% (01:29) wrong based on 46 sessions

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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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Status: Nothing comes easy: neither do I want.
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GMAT 1: 670 Q49 V31
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09 Sep 2010, 14:15
1
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

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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Joined: 26 Jul 2010
Posts: 399

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09 Sep 2010, 15:30
2
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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Joined: 04 Dec 2015
Posts: 750
Location: India
Concentration: Technology, Strategy
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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

$$\frac{1}{2^{10}}+\frac{1}{2^{11}}+\frac{1}{2^{12}}+\frac{1}{2^{12}}$$

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

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

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

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

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

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Intern
Joined: 09 May 2017
Posts: 8
Location: India

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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..
Intern
Joined: 27 Nov 2018
Posts: 34
GMAT 1: 510 Q33 V28

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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
= $$\frac{1}{12} ( 2^{2}+2+1+1) = 8/2^{12} = 2^{3}/ 2^{12}= 1/2^{9}$$
Re: 1/2^10+1/2^11+1/2^12+1/2^12=   [#permalink] 05 May 2019, 00:47
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