Solution of this problem.

Question

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.

Answer 1

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.

Answer 2

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 $$y=2m$$