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Difficult Math Problem #72 - Number Theory

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #72 - Number Theory

Post Wed Dec 06, 2006 7:35 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

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    projase Junior | Next Rank: 30 Posts Default Avatar
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    Post Thu Dec 07, 2006 11:12 am
    n*2^7*3^2*7^3 should be divisible by 5^2 and3^3.

    Then the minimun value of n is 5^2*3=75.

    Please comment,

    Regards,

    PR

    800guy wrote:
    If both 5^2 and 3^3 are factors of n x (2^5) x (6^2) x (7^3), what is the smallest possible positive value of n?

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Fri Dec 08, 2006 10:30 am
    OA:

    Write down n x (2^5) x (6^2) x (7^3) as
    = n x (2^5) x (3^2) x (2^2) x (7^3),
    = n x (2^7) x (3^2) x (7^3)

    now at a minimum 5^2 and a 3 is missing from this to make it completely divisible by 5^2 x 3^3

    Hence answer = 5^2 x 3 = 75

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