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## 4+2^2+2^3+2^4+2^5+2^6+2^7=

tagged by: Brent@GMATPrepNow

This topic has 3 expert replies and 0 member replies

### Top Member

Vincen Master | Next Rank: 500 Posts
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#### 4+2^2+2^3+2^4+2^5+2^6+2^7=

Fri Sep 29, 2017 6:37 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
4+2^2+2^3+2^4+2^5+2^6+2^7=

A) 2^7
B) 2^8
C) 2^16
D) 2^28
E) 2^29

The OA is B.

Is there any way to make this PS question without using a calculator?

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### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Fri Sep 29, 2017 7:29 pm
Vincen wrote:
4+2^2+2^3+2^4+2^5+2^6+2^7=

A) 2^7
B) 2^8
C) 2^16
D) 2^28
E) 2^29
Let's look for a pattern...

We want: 4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = ?

4 + 2^2 = 4 + 4 = 8 = 2^3
So, 4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = 2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7

2^3 + 2^3 = 2(2^3) = 2^4
So, 2^3 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = 2^4 + 2^4 + 2^5 + 2^6 + 2^7

2^4 + 2^4 = 2(2^4) = 2^5
So, 2^4 + 2^4 + 2^5 + 2^6 + 2^7 = 2^5 + 2^5 + 2^6 + 2^7

Continuing the pattern, we get: 4 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = 2^8

Cheers,
Brent

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Rich.C@EMPOWERgmat.com Elite Legendary Member
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Sat Sep 30, 2017 3:52 pm
Hi Vincen,

With a little math and a bit of logic, you can get to the correct answer here without too much trouble. To start, you have to recognize what it really means to raise 2 to higher and higher exponents. With each increase of 1 to the exponent, the total DOUBLES.

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32
Etc.

Thus, at higher and higher exponents, the numbers are NOT close together. From the sum of terms given to us, we can see that the largest term will be 2^7 and (when we word 'down' the list) we're adding smaller and smaller numbers to it. Clearly the sum will be GREATER than 2^7, but not that much greater (relatively speaking). Looking at the answer choices, there's only one option that makes sense...

GMAT assassins aren't born, they're made,
Rich

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### GMAT/MBA Expert

EconomistGMATTutor GMAT Instructor
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Tue Oct 10, 2017 1:20 pm
Vincen wrote:
4+2^2+2^3+2^4+2^5+2^6+2^7=

A) 2^7
B) 2^8
C) 2^16
D) 2^28
E) 2^29

The OA is B.

Is there any way to make this PS question without using a calculator?
Hi Vincen,
Let's take a look at your question.

4+2^2+2^3+2^4+2^5+2^6+2^7
We can write the first term of the sum 4 as 2^2,
=(2^2+2^2)+2^3+2^4+2^5+2^6+2^7
= 2(2^2)+2^3+2^4+2^5+2^6+2^7

We know that 2(2^2) = 2^3, so
= (2^3+2^3)+2^4+2^5+2^6+2^7

Add 2^3 + 2^3 = 2(2^3)
= 2(2^3) + 2^4+2^5+2^6+2^7
= 2^4 + 2^4+2^5+2^6+2^7

Add 2^4 + 2^4 = 2(2^4)
= 2(2^4) +2^5+2^6+2^7
=( 2^5 + 2^5)+2^6+2^7

Add 2^5 + 2^5 = 2(2^5)
= 2(2^5) + 2^6+2^7
=( 2^6+2^6)+2^7

Add 2^6 + 2^6 = 2(2^6)
= 2(2^6)+2^7
= 2^7+2^7
= 2(2^7)
= 2^8

Therefore, Option B is correct.

I am available if you'd like any follow up.

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