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Math Help - Please help with the method of these questions...

  1. #1
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    Please help with the method of these questions...

    (i) How many positive factors are there of 2009?

    (ii) How many solutions in positive integers x,y,z are there to the equation x+y+z=2009?

    (iii) How many solutions in positive integers x,y,z are there to the equation x*y*z=2009?
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  2. #2
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    Re: Please help with the method of these questions...

    Hello, aab300!

    (a) How many positive factors are there of 2009?

    2009 \:=\:7^2\cdot41

    There are 6 positive factors: . 1,\,7,\,41,\,49,\,287,\,2009




    (b) How many solutions in positive integers are there to the equation: . x+y+z\:=\:2009\,?

    Place 2009 objects in a row, leaving one space between them.
    . . . . . \circ\,\_\,\circ\,\_\,\circ\,\_\,\circ \,\_\,\hdots\,\_\;\circ

    Select two of the 2008 spaces and insert "dividers".

    . . So that:. \circ\,\_\,\circ\,\_\,\circ\,|\,\circ\,\_\;\circ \,| \,\circ\,\_\, \hdots \,\_\;\circ \;\text{ represents }\,(3,\,2,\,2004)

    There are: . {2008\choose2} \,=\,2,\!015,\!028 possible partitions.

    Therefore, there are 2,015,028 solutions to x+ y +z \:=\:2009




    (c) How many solutions in positive integers x,y,z are there to the equation: . xyz \:=\:2009\,?

    The prime factorization 2009 is: . 7^2\cdot41

    We have:

    . . \begin{array}{cc}\text{Factors} & \text{Permutations} \\ \hline 1, 7, 287 & 6 \\ 1, 41, 49 & 6 \\ 7,7,41 & 3 \\ \hline \text{Total:} & 15 \end{array}

    There are 15 solutions to xyz = 2009
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