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  1. #1
    Member billym's Avatar
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    stuff

    2 Questions, I think I got the first one:


    a) How many solutions exist to the equation x + y + z = 15, where x, y, and z have to be non-negative integers?

    C(17,2) = 136 ... correct?

    b) How many solutions exist to the equation xyz = 3^6 * 2 ^ 13, where x, y, and z have to be positive integers?

    No idea...
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  2. #2
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    Quote Originally Posted by billym View Post
    b) How many solutions exist to the equation xyz = 3^6 * 2 ^ 13, where x, y, and z have to be positive integers?
    Here is a model of part b.
    \begin{gathered}<br />
  xyz = \left( {2^{\alpha _1 }  \cdot 3^{\beta _1 } } \right)\left( {2^{\alpha _2 }  \cdot 3^{\beta _2 } } \right)\left( {2^{\alpha _3 }  \cdot 3^{\beta _3 } } \right) \hfill \\<br />
  \alpha _1  + \alpha _2  + \alpha _3  = 13\;\& \,\beta _1  + \beta _2  + \beta _3  = 6 \hfill \\ <br />
\end{gathered}

    Can you use it to finish?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by billym View Post

    b) How many solutions exist to the equation xyz = 3^6 * 2 ^ 13, where x, y, and z have to be positive integers?

    No idea...
    The fundamental theorem of arithmetic tells you that x,y and z must each be of the form 3^n 2^m, such that the powers of 3 sum to 6 and those of 2 sum to 13.

    CB
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  4. #4
    Member billym's Avatar
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    So:

    C(15,2)*C(8,2) = 2940
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  5. #5
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    Yes. Well done.
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  6. #6
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    ~

    Hello, billym!

    a) How many solutions exist to the equation x + y + z \:=\: 15
    where x, y, z are non-negative integers?
    You have the right idea, but your answer is slightly off.
    ["Non-negative integers" include the zero.]


    We have 15 objects in a row: . \_\:O\_\:O\_\:O\_\:O\_\:O \_\:O\_\:O\_\:O\_\:O\_\:O\_\: O\_\:O\_\:O\_\:O\_\:O\_

    There are 17 spaces before, after, and between them.

    We will place two "dividers" in any of the 17 spaces.

    There are 17 choices for the first divider, and 17 choices for the second.

    There are: . 17^2 \:=\:{\color{blue}289} solutions.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    . . . . . . . . Examples

    . . OO{\color{red}|}OOO{\color{red}|}OOOOOOOOOO \quad\Rightarrow\quad 2 + 3 + 10

    . . OOOO{\color{red}||}OOOOOOOOOOO \quad\Rightarrow\quad 4+0+11

    . . {\color{red}||}OOOOO OOOOO OOOOO \quad\Rightarrow\quad 0 + 0 + 15

    . . {\color{red}|}OOOOO OOOOO OOOOO{\color{red}|} \quad\Rightarrow\quad 0 + 15 + 0

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  7. #7
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    Quote Originally Posted by Soroban View Post
    There are: . 17^2 \:=\:{\color{blue}289} solutions.
    Actually billym is correct in the OP.
    The correct answer is \binom{17}{2}=135.
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  8. #8
    Member billym's Avatar
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    I thought so. If I didn't include the zero then it would have to be (?) :

    (x - 1) + (y - 1) + (z - 1) = 12

    then:

    <br /> <br />
\binom{14}{2}=91<br />
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  9. #9
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    Quote Originally Posted by billym View Post
    I thought so. If I didn't include the zero then it would have to be (?) :
    (x - 1) + (y - 1) + (z - 1) = 12
    then:
    \binom{14}{2}=91
    That answer is correct but the concept is not.
    That is the number of positive solutions.

    The concept is (x + 1) + (y + 1) + (z + 1) = 12
    Think about putting a one into each variable. Leaving 12 to put in any of the three.
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