Even stranger commons!

Just like the method in previous problem, we can say that for common solutions, the variables $j,k,l,m,n,o,p,q$ can be replaced by $a,b,c,d,e,f,g,h$ respectively.

This transforms the equations into $a+4b+c+3d+3e+3f+6g+6h=20.................(i)$ $a+2b+c+3d+e+f+g+6h=20.................(ii)$

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Subtract (ii) from (i), then we get

$2b+2e+2f+5g=0$

As we want non-negative integer solutions, $b=e=f=g=0$ .

Hence we just want solutions to the equation

$a+c+3d+6h=20$

(See that the variables whose co-efficient corresponding to the variable in 2nd equation is same , for them only the values will be non-zero)

So, this is same as finding non-negative integer solutions to the equation $a+c+d' + h'=20$ , with the condition $d'$ is multiple of 3 and $h'$ is multiple of $6$ .

This will have the generating function for $a$ as $\displaystyle \sum_{k=0}^{20} x^k$

Generating function for $c$ as $\displaystyle \sum_{k=0}^{20} x^k$

Generating function for $d'$ as $\displaystyle \sum_{k=0}^{6} x^{3k}$

Generating function for $h'$ as $\displaystyle \sum_{k=0}^{3} x^{6k}$

Thus, the generating function for our sum, to be 20, it'll be the co-efficient of $x^{20}$ in the function

$\displaystyle \biggl( \sum_{k=0}^{20} x^k \biggr) ^2 \biggl( \sum_{k=0}^{6} x^{3k} \biggr) \biggl( \sum_{k=0}^{3} x^{6k} \biggr)$

And it is $\boxed{150}$

A way without generating functions (casework wink wink)

We subtract the two equations to get 2b+2e+2f+5g=0.

Therefore, b=e=f=g=0.

Now we must solve the equation a+c+3d+6h=20.

We do casework on 6h (not that bad since there are only 4 cases).

Case 1: 6h=18.

Then, a+c+3d=2 which there are only 2 solutions for (d=0).

Case 2: 6h=12.

subcasework on 3d.

When 3d=6, there are 3 solutions.

3d=3, 6 solutions.

3d=0, 9 solutions.

3+6+9=18 for this case.

Case 3: 6h=6

Similarly, 3+6+9+12+15=45.

Case 4: 6h=0

3+6+9+12+15+18+21=84.

We add up all of these to get $3+18+45+84=\boxed{150}$