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

**rickrishav** · October 2nd 2011, 04:49 AM

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

**Plato** · October 2nd 2011, 08:38 AM

Originally Posted by 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 $0\in\mathbb{N}$ .
To solve $3x+4y+5z=200$ , consider the coefficient of $x^{200}$ in the expansion of
$\left( {\sum\limits_{k = 0}^{66} {x^{3k} } } \right)\left( {\sum\limits_{k = 0}^{50} {x^{4k} } } \right)\left( {\sum\limits_{k = 0}^{40} {x^{5k} } } \right)$

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

**rickrishav** · October 3rd 2011, 11:15 PM

But finding the coefficeint of $x^{200}$ 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?

**Plato** · October 4th 2011, 03:40 AM

Originally Posted by rickrishav

But finding the coefficeint of $x^{200}$ 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.

**alexmahone** · October 4th 2011, 03:54 AM

Originally Posted by 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 $x^{200}$ in your formula. Perhaps you could complete the solution and find the answer.

**Plato** · October 4th 2011, 04:06 AM

Originally Posted by alexmahone

I think the OP doesn't understand how he's supposed to find the coefficient of $x^{200}$ 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.

**roninpro** · October 6th 2011, 08:05 PM

Originally Posted by Plato

I take it that $0\in\mathbb{N}$ .
To solve $3x+4y+5z=200$ , consider the coefficient of $x^{200}$ in the expansion of
$\left( {\sum\limits_{k = 0}^{66} {x^{3k} } } \right)\left( {\sum\limits_{k = 0}^{50} {x^{4k} } } \right)\left( {\sum\limits_{k = 0}^{40} {x^{5k} } } \right)$

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 $(x,y,z)$ and then attempt to count solutions from there.

**Plato** · October 7th 2011, 02:32 AM

Originally Posted by 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 $(x,y,z)$ and then attempt to count solutions from there.

Please, do show us your method.

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