Number of solutions in Natural Numbers to linear equation in three variables

What are the number of solutions in natural Numbers to the following linear equation in three variables...

*3x+4y+5z=200*

Can there be a general form for the number of solutions for this type of equation:

*ax+by+cz=d*

*a,b,c,d,x,y,z* belong to Natural Numbers

Re: Number of solutions in Natural Numbers to linear equation in three variables

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**rickrishav** What are the number of solutions in natural Numbers to the following linear equation in three variables...

*3x+4y+5z=200*

Can there be a general form for the number of solutions for this type of equation:

*ax+by+cz=d*

*a,b,c,d,x,y,z* belong to Natural Numbers

I take it that $\displaystyle 0\in\mathbb{N}$.

To solve $\displaystyle 3x+4y+5z=200$, consider the coefficient of $\displaystyle x^{200}$ in the expansion of

$\displaystyle \left( {\sum\limits_{k = 0}^{66} {x^{3k} } } \right)\left( {\sum\limits_{k = 0}^{50} {x^{4k} } } \right)\left( {\sum\limits_{k = 0}^{40} {x^{5k} } } \right)$

If zero is not a natural number, begin the sums at one.

Re: Number of solutions in Natural Numbers to linear equation in three variables

But finding the coefficeint of $\displaystyle x^{200}$ by counting, is equivalent to listing down all the solutions of the original equation and counting them up. How can I proceed to find the coefficient in a more efficient manner?

Re: Number of solutions in Natural Numbers to linear equation in three variables

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**rickrishav** But finding the coefficeint of $\displaystyle x^{200}$ by counting, is equivalent to listing down all the solutions of the original equation and counting them up. How can I proceed to find the coefficient in a more efficient manner?

I truly mean you no offence, but I don't think you understand this problem. There is no more *efficient manner* other than the one I gave you.

Re: Number of solutions in Natural Numbers to linear equation in three variables

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**Plato** I truly mean you no offence, but I don't think you understand this problem. There is no more *efficient manner* other than the one I gave you.

I think the OP doesn't understand how he's supposed to find the coefficient of $\displaystyle x^{200}$ in your formula. Perhaps you could complete the solution and find the answer.

Re: Number of solutions in Natural Numbers to linear equation in three variables

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**alexmahone** I think the OP doesn't understand how he's supposed to find the coefficient of $\displaystyle x^{200}$ in your formula. Perhaps you could complete the solution and find the answer.

To do that one either uses a computer algebra system or does a course of study in *generating function theory*.

BTW. I disagree that is what the OP means.

Re: Number of solutions in Natural Numbers to linear equation in three variables

Quote:

Originally Posted by

**Plato** I take it that $\displaystyle 0\in\mathbb{N}$.

To solve $\displaystyle 3x+4y+5z=200$, consider the coefficient of $\displaystyle x^{200}$ in the expansion of

$\displaystyle \left( {\sum\limits_{k = 0}^{66} {x^{3k} } } \right)\left( {\sum\limits_{k = 0}^{50} {x^{4k} } } \right)\left( {\sum\limits_{k = 0}^{40} {x^{5k} } } \right)$

If zero is not a natural number, begin the sums at one.

I'm a little confused by this approach. Is it really necessary to use generating functions to count up the number of desired solutions? It seems more conceptually straightforward to find the parameterisation of the solutions $\displaystyle (x,y,z)$ and then attempt to count solutions from there.

Re: Number of solutions in Natural Numbers to linear equation in three variables

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**roninpro** I'm a little confused by this approach. Is it really necessary to use generating functions to count up the number of desired solutions? It seems more conceptually straightforward to find the parameterisation of the solutions $\displaystyle (x,y,z)$ and then attempt to count solutions from there.

Please, do show us your method.