Please help with the method of these questions...

**aab300** · April 18th 2012, 05:43 AM

(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?

**Soroban** · April 18th 2012, 10:36 AM

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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