(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?
(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?
Hello, aab300!
(a) How many positive factors are there of 2009?
$\displaystyle 2009 \:=\:7^2\cdot41$
There are 6 positive factors: .$\displaystyle 1,\,7,\,41,\,49,\,287,\,2009$
(b) How many solutions in positive integers are there to the equation: .$\displaystyle x+y+z\:=\:2009\,?$
Place 2009 objects in a row, leaving one space between them.
. . . . . $\displaystyle \circ\,\_\,\circ\,\_\,\circ\,\_\,\circ \,\_\,\hdots\,\_\;\circ $
Select two of the 2008 spaces and insert "dividers".
. . So that:. $\displaystyle \circ\,\_\,\circ\,\_\,\circ\,|\,\circ\,\_\;\circ \,| \,\circ\,\_\, \hdots \,\_\;\circ \;\text{ represents }\,(3,\,2,\,2004)$
There are: .$\displaystyle {2008\choose2} \,=\,2,\!015,\!028$ possible partitions.
Therefore, there are 2,015,028 solutions to $\displaystyle x+ y +z \:=\:2009$
(c) How many solutions in positive integers x,y,z are there to the equation: .$\displaystyle xyz \:=\:2009\,?$
The prime factorization 2009 is: .$\displaystyle 7^2\cdot41$
We have:
. . $\displaystyle \begin{array}{cc}\text{Factors} & \text{Permutations} \\ \hline 1, 7, 287 & 6 \\ 1, 41, 49 & 6 \\ 7,7,41 & 3 \\ \hline \text{Total:} & 15 \end{array}$
There are 15 solutions to $\displaystyle xyz = 2009$