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Newly Defined Functions

  1. Feb 28, 2015 #1
    1. The problem statement, all variables and given/known data
    For positive integers m, k, and n , let mkn be defined as mkn = kmn , where [itex] k\frac {m}{n} [/itex] is a mixed fraction. What is the value of 643 + 364 ?

    2. Relevant equations
    I attempt the other few similar questions where the solution are as follow
    832 + 382 = [itex] \frac {169}{24} [/itex]
    641 + 164 = [itex] \frac {85}{6} [/itex]

    The answer for this question.
    643 + 364 = [itex] \frac {21}{2} [/itex]

    But the problem is I have no idea how to get the answer.

    3. The attempt at a solution
    Based on the problem, I can see that

    643 + 364 = 463 + 634
    = 463 + 364
    = 643 + 634
    or more simply

    643 = 463 and 364 = 634

    Notice that 463 and 364 are in reverse position with each other.

    Since it mention that [itex] m\frac {k}{n} [/itex] is a mixed fraction, I suspect that
    643 + 364 = 643 + 634

    I try to relate 643 = [itex] 6 \frac {4}{3} [/itex] and 364 = [itex] 3 \frac {6}{4} [/itex]
    but they are not the same for 463 and 634 if expressed in term of mixed fraction

    Even attempt the way:
    643 = 463 → (24)3 and 364 = 634 → (18)4 by multiply the first 2 numbers and
    643 = 463 = (10)3 and 364 = 634 → (9)4 by adding the first 2 numbers.

    Not sure if mkn = kmn = [itex] k\frac {m}{n} [/itex]

    Please try to give some hints about how to see this problem.
     
  2. jcsd
  3. Feb 28, 2015 #2

    Mark44

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    I don't get it. The sentence above is defining mkn as kmn. How does ##k\frac m n## tie in to either of the expressions shown earlier?
    If there is an actual definition here, I'm not seeing it.
     
  4. Feb 28, 2015 #3
  5. Feb 28, 2015 #4

    haruspex

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  6. Feb 28, 2015 #5
    Is the question not giving enough information to be solved?
     
  7. Feb 28, 2015 #6

    Mark44

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    In my opinion, and apparently that of haruspex, no.
     
  8. Feb 28, 2015 #7

    Ray Vickson

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    Your definition (or, rather, the definition in your source) is vague and almost useless. Do you mean
    [tex] m\underline{k}n = m\, \frac{k}{n} \;(\text{which} \: = m + \frac{k}{n} = \frac{mn + k}{n}) [/tex]
    or do you mean
    [tex] m\underline{k}n = m \times \frac{k}{n} \:(\text{which} \: = \frac{mk}{n} )? [/tex]

    Note that ##a\underline{b}c = b\underline{a}c## for the second definition of ##a\underline{b}c##, etc., but not for the first one.

    The people who set this question should be ashamed, unless they have already defined the notation ##a \frac{b}{c}## somewhere else in the paper.
     
  9. Feb 28, 2015 #8

    SammyS

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    Here is an image of the problem. (It looks as if the website randomizes the the numbers.) It a real puzzle as to what that operation is.
    ?temp_hash=e1642ed97ddabf9570e6574d9ace7c16.png
    The correct answer is (C) . I have no idea why that's the case.

    To view the solution seems to require signing up for a fee. (I'm not curious enough & too cheap for that.) The website is Brilliant.org

    SammyS
     

    Attached Files:

  10. Feb 28, 2015 #9

    HallsofIvy

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    That "definition" is meaningless since it defines "[itex]m\underline{k}n[/itex]" as "[itex]k\underline{m}n[/itex]" which has not been defined!

    This is like defining a number to be "whacky" by "x is 'whacky' if and only if x/2 is 'whacky'. The definition is circular.
     
    Last edited: Feb 28, 2015
  11. Mar 1, 2015 #10
    alright, I think I have found the way to solve it
    Let take 643 + 364
    = 463 + 634

    I got the answer by performing this:

    [tex](\frac {(24)(24)}{(18)(16)} + \frac {9}{18})+8[/tex] = [tex] \frac {21}{2}[/tex]

    Notice that first 2 term 6 x 4 = 4 x 6 = 24
    final 2 term 6 x 3 = 18 and 4 x 4 = 16 respectively
    The fraction consists of
    [tex] \frac {3 x 3}{3 x 6}[/tex] = [tex] \frac {9}{18}[/tex]
    the 8 = 4 + 4 is the addition of last digit of the last number.

    I try this and it applies to all the similar questions. I still cannot find a logical reason how it leads to this, so if you got any insights, can let me know.
     
  12. Mar 1, 2015 #11

    Mark44

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    Five people have responded in this thread, and none of them can make any sense out of this problem. Since the problem is so poorly defined, the best thing to do would be to forget it, and move on to other questions.
     
  13. Mar 1, 2015 #12

    SammyS

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    As we suspected, there was some error in the statement of the problem. - - a typo or some major definition left out. I just looked it up again. Here's an image from today.
    PhysForum_NewlyDefFunc_2.PNG
    So, it's not too interesting a problem after all.

    This doesn't seem to work for some of the examples cited yesterday. ... or do I have brain fade?
     
    Last edited: Mar 1, 2015
  14. Mar 1, 2015 #13
    I can't make any sense of the questions/answers in your post but when i tried the link the most similar looking question i got was
    For positive integers ##m,k,n## and , let ##m\mathbf kn## be defined as ##m\mathbf kn = \mathbf k \frac{m}{n}##, where ##\mathbf k \frac{m}{n}## is a mixed fraction. What is the value of ##6\mathbf 21 + 1\mathbf 26##
    so we get
    ##2\frac{6}{1} + 2\frac{1}{6} = 8 + \frac{13}{3} = \frac{48+13}{6} = \frac{61}{6}##
    which gives me correct answer according to that site. However for the answers in your question i get the wrong answer so maybe I found the wrong question.

    Edit: I guess the problem was the question had an error in it when you tried it before
     
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