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abcd divizibil cu 5 => d=0 si d=5
b+c
I) Cazul d=0 => a= -----
3
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a=1 => 1bc0 unde (b+c)/3 trebuie sa fie 1, adica sa aiba valoarea => (1,2) abcd=1120, (2,1)=> abcd=1210
a=2 => 2bc0 unde (b+c)/3 = 2 => (0,6)=> abcd=2060, (6,0), (1,5), (5,1), (2,4), (4,2), (3,3)
a=3 => 3bc0 unde (b+c)/3 = 3 => (0,9)=> abcd=3090, (9,0), (1,8), (8,0), (2,7), (7,2), (8,1), (1,8)
Similar se trateaza si cazul in care d=5
