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Difficult Math Question #23 - Algebra

This topic has 5 member replies
800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Question #23 - Algebra

Post Wed Oct 04, 2006 8:38 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    OA coming after few people have answerd--see attachment
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    gamemaster Senior | Next Rank: 100 Posts Default Avatar
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    Posted:
    34 messages
    Post Thu Oct 05, 2006 12:29 am
    x = 13!

    a = x^16 - x^8 / x^8 + x ^ 4 =

    (x^8+x^4)(x^8-x^4)/(x^8+x^4) =

    x^8 - x^4

    a/x^4 =

    x^4 - 1

    digit number of 13! must be 0 because its like (1*2*3...*11*12*13) * 10

    so the final answer would be 1

    sanjeevs50 Newbie | Next Rank: 10 Posts Default Avatar
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    2 messages
    Post Thu Oct 05, 2006 2:26 pm
    (E) is the anwer

    0 of unit digit - 1 = 9

    shail_8206 Newbie | Next Rank: 10 Posts Default Avatar
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    Post Thu Oct 05, 2006 9:35 pm
    Let (13!)^4 = x
    (x^4 - x^2)/ (x^2 - x) = a
    by solving further we get:
    x^3 + x^2 = ax
    or a/x = x + 1
    a/(13! ^ 4 ) = (13!^ 4) + 1

    as 13! ends with 0, so is 13!^ 4; so (13!^ 4) + 1 ends with 1.
    Unit's digit is 1.

    gamemaster Senior | Next Rank: 100 Posts Default Avatar
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    Post Fri Oct 06, 2006 8:29 am
    the answer is 9

    i had a typo...

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Oct 09, 2006 4:38 pm
    here's the oa, look at the attachment
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