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Math Help - Algebra2 homework urgent please help.

  1. #1
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    Algebra2 homework urgent please help.

    what is the least positive integer k for which 12 k is the cube of an integer?
    Last edited by Schatzle06; April 22nd 2007 at 02:53 PM. Reason: was typed wrong
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Schatzle06 View Post
    what is the least positive integer k for which 12k is the cube of an integer?
    We want 12k = m^3 for integers m and k
    => k = (m^3)/12

    so we want the smallest m for which (m^3)/12 gives an integer. The smallest m for which this happens is 0, but that would mean k = 0, we want k positive. The next smallest m for which this happens is 6. So k = (6^3)/12 = 18
    Last edited by Jhevon; April 22nd 2007 at 03:22 PM. Reason: made an error in calculating k
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    Hello, Schatzle06!

    What is the least positive integer k for which 12k is the cube of an integer?
    Sorry, JHevon . . . your reasoning was correct; your answer wasn't.


    If 12k = m, then the left side must be cube.

    The left side is: .23k

    We want the exponents to be a multiple of 3 (smallest).
    . . This happens when k = 23

    Then: .12k .= .(23)(23) .= .23 .= .6

    Therefore: .k = 18

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Schatzle06!

    Sorry, JHevon . . . your reasoning was correct; your answer wasn't.


    If 12k = m, then the left side must be cube.

    The left side is: .23k

    We want the exponents to be a multiple of 3 (smallest).
    . . This happens when k = 23

    Then: .12k .= .(23)(23) .= .23 .= .6

    Therefore: .k = 18

    Thanks for looking out Soroban. I saw that my answer was wrong a little before your post and corrected it. Apparently I was thinking of something else at the time not sure what
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