Question 1109417
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<pre>
r + j + c + L = 30    (L -cLassical; the remaining notations are clear)

2r = 3c

j = L

================>

r + j + {{{2r/3}}} + j = 30,

3r + 3j + 2r + 3j = 90

5r + 6j = 90

j = {{{(90-5r)/6}}}


You need to choose the "free" value of r in a way 

    a)  r  is integer >= 0,   and

    b)  {{{(90-5r)/6}}}  is  integer >= 0,   and

    c)  {{{2r/3}}} is integer >= 0.


Table

    r    j =(90-5r)/6    c = {{{(2r)/3}}}   L = j
---------------------------------------------------
    0      15              0          15
    6      10              4          10
    12      5              8           5    
    18      0             12           0
</pre>

Your answers are ALL listed in the table.