# stuff

• Jul 18th 2009, 11:42 AM
billym
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...
• Jul 18th 2009, 01:22 PM
Plato
Quote:

Originally Posted by billym
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.
$\displaystyle \begin{gathered} xyz = \left( {2^{\alpha _1 } \cdot 3^{\beta _1 } } \right)\left( {2^{\alpha _2 } \cdot 3^{\beta _2 } } \right)\left( {2^{\alpha _3 } \cdot 3^{\beta _3 } } \right) \hfill \\ \alpha _1 + \alpha _2 + \alpha _3 = 13\;\& \,\beta _1 + \beta _2 + \beta _3 = 6 \hfill \\ \end{gathered}$

Can you use it to finish?
• Jul 18th 2009, 01:25 PM
CaptainBlack
Quote:

Originally Posted by billym

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 $\displaystyle x,y$ and $\displaystyle z$ must each be of the form $\displaystyle 3^n 2^m$, such that the powers of 3 sum to 6 and those of 2 sum to 13.

CB
• Jul 18th 2009, 01:50 PM
billym
So:

C(15,2)*C(8,2) = 2940
• Jul 18th 2009, 01:55 PM
Plato
Yes. Well done.
• Jul 18th 2009, 01:57 PM
Soroban
~
Hello, billym!

Quote:

a) How many solutions exist to the equation $\displaystyle x + y + z \:=\: 15$
where $\displaystyle x, y, z$ 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: .$\displaystyle \_\:O\_\:O\_\:O\_\:O\_\:O \_\:O\_\:O\_\:O\_\:O\_\:O\_\: O\_\:O\_\:O\_\:O\_\:O\_$

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: .$\displaystyle 17^2 \:=\:{\color{blue}289}$ solutions.

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

. . . . . . . . Examples

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

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

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

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

• Jul 18th 2009, 02:04 PM
Plato
Quote:

Originally Posted by Soroban
There are: .$\displaystyle 17^2 \:=\:{\color{blue}289}$ solutions.

Actually billym is correct in the OP.
The correct answer is $\displaystyle \binom{17}{2}=135$.
• Jul 18th 2009, 02:22 PM
billym
I thought so. If I didn't include the zero then it would have to be (?) :

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

then:

$\displaystyle \binom{14}{2}=91$
• Jul 18th 2009, 02:36 PM
Plato
Quote:

Originally Posted by billym
I thought so. If I didn't include the zero then it would have to be (?) :
(x - 1) + (y - 1) + (z - 1) = 12
then:
$\displaystyle \binom{14}{2}=91$

That answer is correct but the concept is not.
That is the number of positive solutions.

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