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How will you find the number of positive integral solutions of $5x + 4y + 3z + 2m + k = 20 $ . I could find out by hit and trial but I want to get the general solutions to such equations. I tried to connect with dividing n ( which is 20 here ) to 15 groups where 5 groups are of same size but couldnt get any general solution. Please help me out.

Jasmine
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  • this is a linear Diophantine equation – Dr. Sonnhard Graubner Mar 21 '18 at 18:29
  • I got it but cant we have any general solution to it – Jasmine Mar 21 '18 at 18:30
  • see here file:///F:/FULLTEXT01.pdf Jasmine – Dr. Sonnhard Graubner Mar 21 '18 at 18:33
  • or here https://math.stackexchange.com/questions/20717/how-to-find-solutions-of-linear-diophantine-ax-by-c – Dr. Sonnhard Graubner Mar 21 '18 at 18:34
  • With two variables you can pick a value for $y$ and see if there is a solution for $x$. For instance with $y = 1$ you have $2x = 17$ (no solution) with $y = 2$ you have $2x = 14$ ($x = 7$). – Trevor Gunn Mar 21 '18 at 18:34
  • The file is not related to math – Jasmine Mar 21 '18 at 18:35
  • A Diophantine equation is simply an equation for which we are only interested in integer solutions (or non-negative integer solutions depending on desire). Linear refers to that the terms are all of the form a scalar times a single variable and nothing more exotic than that. As for an approach, you could describe this with generating functions, looking at the coefficient of $x^{20}$ in the expansion of $(1+x^2+x^4+x^6+x^8+\dots)(1+x^3+x^6+x^9+\dots)$ (or without the 1's if strictly +) but given the low values and low number of variables here, brute force is perfectly easy and acceptable. – JMoravitz Mar 21 '18 at 18:36
  • Here i am not interested in solutions but only the number of solutions – Jasmine Mar 21 '18 at 18:37
  • And, in Diophantine equations, in some contexts you concern yourself with describing the properties of each solution. In other contexts you concern yourself only with counting how many solutions there are. Its a very general term. Your problem is still well described as a Diophantine equation problem. – JMoravitz Mar 21 '18 at 18:38
  • Please xheck my edit that was the actual question but i wanted to know the concept and apply it on my own but not getting desirable solution – – Jasmine Mar 21 '18 at 18:45
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    In your new edit, if you are still looking for strictly positive integral solutions, once you set each variable to $1$, the total is merely $5$ away from your desired total of $20$, which is still a small enough number brute force is easy to work with. Otherwise, generating functions again are convenient. The number of positive integer solutions of $c_1x_1+c_2x_2+c_3x_3+\dots+c_kx_k=N$ where $c_1,c_2,\dots$ are constants and $x_1,x_2,\dots$ are variables is the coefficient of $x^N$ in expansion of $\prod_{n=1}^k\frac{x^{c_n}}{1-x^{c_n}}$. If non-negative solutions, then clear the numerators – JMoravitz Mar 21 '18 at 18:49
  • The later one is more complicated can we do it by simple concept of grouping? – Jasmine Mar 21 '18 at 18:54
  • Considering that I don't know what you consider "grouping" I'm going to expect the answer is no. – JMoravitz Mar 21 '18 at 18:55
  • Grouping is simple considering the whole 20 to be distributed among 15 groups where 5 groups are of same size – Jasmine Mar 21 '18 at 18:57
  • Why should that have to be the case? There are plenty of distributions where they aren't in equal sized groups – JMoravitz Mar 21 '18 at 19:05
  • @JMoravitz can you explain how did you write that result regarding the the number of positive integral solution of $c_{1}x_{1}+c_{2}x_{2}+..c_{n}x_{n}=N$ – Jasmine Aug 10 '18 at 08:20
  • The short answer is that I used generating functions and well known Taylor series. The long answer might be too long to explain here, but it is explained in many introductory textbooks on combinatorics and gone into in great length in the book Generatingfunctionology, available online for free. – JMoravitz Aug 10 '18 at 13:51
  • @JMoravitz thank you it really helps! – Jasmine Aug 10 '18 at 15:45

2 Answers2

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2x+3y=20

As y is positive $2x\leq17 \implies x\leq8.5$

Add x to both sides.

$3(x+y)=20+x$

We observe that $20+x$ is a multiple of $3$ so, the only possible values of $x$ are $1,4,$ and $ 7$.

Answer After edit: coefficient of $x^20$ in the following product gives the number of positive integral solutions

$(x^5+x^{10}+x^{15}+...)(x^4+x^8+x^{12}+...)(x^3+x^6+x^9+...)(x^2+x^4+x^8+...)(x+x^2+x^3+...)$

This expression can be simplified by using formula for summation of geometric series. $x^{15}\frac{1}{1-x^5}\frac{1}{1-x^4}\frac{1}{1-x^3}\frac{1}{1-x^2}\frac{1}{1-x}$

So, the required number is just the coefficient of $x^5$ in $\frac{1}{1-x^5}\frac{1}{1-x^4}\frac{1}{1-x^3}\frac{1}{1-x^2}\frac{1}{1-x}$

Which can we found by breaking this product into partial fractions and expanding each fraction.

420
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  • Please xheck my edit that was the actual question but i wanted to know the concept and apply it on my own but not getting desirable solution – Jasmine Mar 21 '18 at 18:44
  • By hit and trial there are only 7 solutions. But if we do this way, it will get more complicated.I wanted to get an easier way of solving this. – Jasmine Mar 21 '18 at 19:01
  • This is a general method, hit and trial can be done only when the number of solutions is not very large. – 420 Mar 21 '18 at 19:04
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from the given equation we get $$x=10-\frac{3}{2}y$$ with $$y=2m$$ where $m$ is an integer number we obtain $$x=10-m$$ $$y=2m$$ that's all! from $$2x=20-3y$$ we get by dividing by $2$ $$x=10-\frac{3}{2}x$$ since $x$ must be a integer number, then $y$ must be even, for example $$y=2m$$

Dr. Sonnhard Graubner
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