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Easy Proof

  1. Feb 16, 2004 #1
    I'm getting confused and can't seem to wrap my head around this problem. Prove that the sum of the squares of any 3 consecutive odd numbers when divided by 12 gives a remainder of 11.

    I'm not sure how to set this up or proceed I figured that

    (n^2 + (n +2)^2 + (n+4)^2)/12 = x + 11

    Where n is any odd integer

    then I got

    3n^2 +12n + 20 = 12x +123
    3n(n+4) = 12x + 103
    I'm not sure where to take it from here
    Can someone start out on an alternate solution or set it up differently for me?

  2. jcsd
  3. Feb 16, 2004 #2

    matt grime

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    you will get a lot further if you let n=2k+1 which is always what you should do when asked to consider odd numbers

    the bit where you write x+11 is wrong, as that isn't what 'has remainder 11 on dividing by 12' means.

    omit the x entirely

    look at the line 3n^2 +12n + 20

    this has the same remainder as 3n^2 +20, now put n=2k+, and remember 12 = 0 mod 12.
  4. Feb 16, 2004 #3
    (2n+1)^2 + (2n+3)^2 + (2n+5)^2
    after expanding and simplifying
    12n^2 + 36n + 35

    as you see, 12 goes into 12n^2 evenly, as does 12 into 36...so far so good

    and then we come to 35...35/12 = 2 + 11R....there is your 11


    in general terms f[unction](n)=d[ivisor](n)q[uotient](n) + r[emainder](n)
    where n is a positive integer

    so the division statement would be:
    12n^2 + 36n + 35 = (12)(n^2 + 3n + 2) + 11
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