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For the positive integers a, b (OG16)

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boomgoesthegmat Senior | Next Rank: 100 Posts Default Avatar
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For the positive integers a, b (OG16)

Post Thu May 19, 2016 3:43 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    For the positive integers a, b, and k, a^k || b means that a^k is a divisor of b, but a^(k+1) is not a divisor of b. If k is a positive integer and 2^k ||72, then k is equal to

    A) 2
    B) 3
    C) 4
    D) 8
    E) 18

    OA: B

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    OptimusPrep Master | Next Rank: 500 Posts
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    Post Thu May 19, 2016 7:58 pm
    boomgoesthegmat wrote:
    For the positive integers a, b, and k, a^k || b means that a^k is a divisor of b, but a^(k+1) is not a divisor of b. If k is a positive integer and 2^k ||72, then k is equal to

    A) 2
    B) 3
    C) 4
    D) 8
    E) 18

    OA: B
    2^k ||72 means 2^k is a divisor of 72, but 2^(k+1) is not a divisor of 72

    72 = 2^3*3^2

    The maximum powers of 2 in 72 = 3
    Hence 2^3 is a divisor of 72 and 2^4 is not a divisor of 72
    k = 3

    Correct Option: B

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    Post Mon Oct 09, 2017 7:57 am
    Any positive integer can be expressed as the product of prime factors - we call this PRIME FACTORIZATION. For example, the prime factorization of 140 is (2^2)(5^1)(7^1), and 108 = (2^2)(3^3). This prime factorization is really helpful when you're solving questions with variable exponents. For example:

    12 = (2^x)(3^y)
    In order to solve for x and y here, we need to break 12 down to its prime factors:
    12 = (2^2)(3)
    So, (2^2)(3) = (2^x)(3^y)
    Therefore x = 2 and y = 1.

    In #110, we have "k" and "k + 1" as exponents, which is a big clue that we probably want to think in terms of prime factorization.

    The hardest part about this question is simply figuring out what it's asking - breaking down "a^k is a divisor of b, but a^(k + 1) is not." How do we translate that? So "a" to a certain exponent goes evenly into "b," but "a" to the next highest exponent does not. That would mean that a^k was the maximum number of a's that go into "b." In other words... all the a's!

    So, 2^k || 72 would be the maximum number of factors of 2 that go into 72 - all the 2's in 72. If we factor 72, we find that the prime factorization is (2^3)(3^2). There are 3 factors of 2 in 72 (that's what 2^3 tells us), so k must equal 3.

    There's more on prime factorization here: http://www.beatthegmat.com/2-x-2-x-2-3-2-13-what-is-x-t178975.html#578272

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