Posers and Puzzles

Posers and Puzzles

  1. Joined
    30 Sep '12
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    17 May '14 17:13
    Days ago I gave an easy and medium problem from a book of math puzzles. Here are two more, an easy and a difficult.

    III.
    For every triple (a, b, c) of non-zero real numbers, form the number
    a/|a| + b/|b| + c/|c| + abc/|abc|.
    (I am using my keyboard's | symbol to denote absolute value.)

    The set of all numbers formed is what?


    IV.
    Let g(x) = x^5 + x^4 + x^3 + x^2 + x + 1. What is the remainder (as a constant number or some polynomial function of x, whichever the case may turn out to be) when the polynomial g(x^12) is divided by the polynomial g(x)?

    (Here I use my keyboard's ^ symbol to denote a power.)
  2. Standard membertalzamir
    Art, not a Toil
    60.13N / 25.01E
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    17 May '14 19:57
    III. Looks like Reveal Hidden Content
    zero, 4, -4
    on a brief glance.

    IV. A cursory look would give Reveal Hidden Content
    (x^72 - 1) * (x - 1) / ((x^12 -1) * (x^6 - 1))
    but this is not a satisfying answer yet, and needs more work. ^^
  3. Joined
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    18 May '14 00:271 edit
    The book agrees with talzamir's answer to III.

    Talzamir is still working on IV.
  4. Joined
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    18 May '14 20:131 edit
    IV.
    Let g(x) = x^5 + x^4 + x^3 + x^2 + x + 1. What is the remainder (as a constant number or some polynomial function of x, whichever the case may turn out to be) when the polynomial g(x^12) is divided by the polynomial g(x)?

    6
  5. Joined
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    18 May '14 23:07
    Originally posted by iamatiger
    Hope that's right, was worked out on pencil & paper - technique on request
  6. Joined
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    19 May '14 01:311 edit
    iamatiger's answer of 6 matches the book.

    The book gives two different lines of derivation. One starts with using the identity
    (x-1)(x^n + x^(n-1) + .. + x + 1) = x^(n+1) - 1,

    and ends with
    g(x^12) = (x-1) g(x) P(x) + 6.


    The other starts with
    g(x^12) = g(x) Q(x) + R(x),
    where Q is a polynomial and R is the remainder being sought. It makes use of the complex sixth roots of unity.

    Do either of those sound like your route to the answer?
  7. Joined
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    19 May '14 07:454 edits
    Not really the same as either of those, after trying long division I noticed that

    (x^(k) - x^(k-1)) * (x^5 + x^4 + x^3 + x^2 + x + 1) = x^(k+5) - x^(k-1)

    carrying that 1 step further
    (x^(k) - x^(k-1) + x^(k-6) - x^(k-7) )*(x^5 + x^4 + x^3 + x^2 + x +1)=
    x^(k+5) - x^(k-7) {equation a}

    for the solution we have to divide
    x^(5*12) + x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12) + x^(0*12)
    by (x^5 + x^4 + x^3 + x^2 + x + 1)

    so setting k to 55, we will first have on the top of the long division:
    x^55 - x ^54 + x^49 - x^48
    {equation a} says this will be equal to x^60 - x^48 = x^(5*12) - x(4*12)
    so the remainder, on the bottom of the long division, at this point, will be:
    2x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12) + x^(0*12)

    by extension of the above we can see the remainder will follow the sequence:
    1x^(5+12) + x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12)
    2x^(4*12) + x^(3*12) + x^(2*12) + x^(1*12)
    3x^(3*12) + x^(2*12) + x^(1*12)
    4x^(2*12) + x^(1*12)
    5x^(1*12)
    6

    so we will stop the long division with the remainder 6x^(0*12), i.e. 6

    the top of the long division will be the long looking polynomial:
    +0
    +1(x^55 - x^54 + x^49 - x^48)
    +2(x^43 - x^42 + x ^37 - x^36)
    +3(x^31 - x^30 + x^25 - x^24)
    +4(x^19 - x^18 + x^13 - x^12)
    +5(x^7 - x^6 + x - 1)
  8. Joined
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    20 May '14 01:25
    Lots of work--I am impressed!

    What sorts of math classes have you taken, if you don't mind saying?
  9. Joined
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    22 May '14 22:461 edit
    A level Maths, Degree in Chemistry, I am now a radar systems designer
  10. Joined
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    24 May '14 04:22
    Originally posted by iamatiger
    A level Maths, Degree in Chemistry, I am now a radar systems designer
    I read a book on the history of radar about three years back. That was one of the most fruitful collaborations between the UK and the USA.
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