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Math Help - algebra question (squares?)

  1. #1
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    algebra question (squares?)

    1 Write down the first ten square numbers (perfect squares) [can do this]
    2 Show that when each of these squares are divided by 3, the remainder is either 0 or 1
    3 Write down three consecutive numbers the first of which is divisible by three[can do up to this part]
    4 Prove that the square of any number is of the form of 3n or 3n+1
    5 Use modular arithmetic(what's that) to prove the same result
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  2. #2
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    Quote Originally Posted by requal View Post
    1 Write down the first ten square numbers (perfect squares) [can do this]
    2 Show that when each of these squares are divided by 3, the remainder is either 0 or 1
    3 Write down three consecutive numbers the first of which is divisible by three[can do up to this part]
    4 Prove that the square of any number is of the form of 3n or 3n+1
    5 Use modular arithmetic(what's that) to prove the same result
    In part 2 you showed that when a square is divided by 3, the remainder is either 0 or 1.

    This means that the number can be written in the form:

    k = 3n+1
    or
    k = 3n.
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  3. #3
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    Okay i got the worked solutions of the question, but I don't understand it, someone please clarify it for me.

    iii, 3n, 3n+1, 3n+2
    iv Any number must be of the form of 3k, 3k+1, 3k+2

    Hence, (3k)^2=9k^2=3(3k^2)=3n
    And, (3k+1)^2=9k^2+6k+1=3(3k+2)+1=3n+1
    And, (3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1=3n+1

    Can someone explain to me how they got the solution?? I mean how can n is equal to three randomly different terms
    Last edited by requal; January 20th 2009 at 11:18 PM. Reason: clarifying the question
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  4. #4
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    Quote Originally Posted by requal View Post
    Can someone explain to me how they got the solution?? I mean how can n is equal to three randomly different terms
    Well Its another way of saying every number when divided by 3 leaves a remainder of either 0,1 or 2.
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