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

This topic has 3 expert replies and 0 member replies

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

Post 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?

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    Post 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

    Answer: B

    Cheers,
    Brent

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    Post 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...

    Final Answer: B

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

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    Contact Rich at Rich.C@empowergmat.com

    Post 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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    Our average student improves 98 points.

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