Explicație pas cu pas:
D:I = 3 + rest,
unde
D = deimpartitul,
I = impartitorul si
rest ∈ {0, 1, 2, ..., I-1}
D = 3I + rest
D si I < 20 si rest < I
Avem solutiile:
19:6, 19 = 3x6 + 1, (D, I) = (19, 6)
19:5, 19 = 3x5 + 4, (D, I) = (19, 5)
18:6, 18 = 3x6 + 0, (D, I) = (18, 6)
18:5, 18 = 3x5 + 3, (D, I) = (18, 5)
17:5, 17 = 3x5 + 2, (D, I) = (17, 5)
16:5, 16 =3x5 + 1, (D, I) = (16, 5)
15:5, 15 = 3x5 + 0, (D, I) = (15, 5)
14:4, 14 = 3x4 + 2, (D, I) = (14, 4)
13:4, 13 = 3x4 + 1, (D, I) = (13, 4)
12:4, 12 = 3x4 + 0, (D, I) = (12, 4)
11:3, 11 = 3x3 + 2, (D, I) = (11, 3)
10:3, 10 = 3x3 + 1, (D, I) = (10, 3)
9:3, 9 = 3x3 + 0, (D, I) = (9, 3)
7:2, 7 = 3x2 + 1, (D, I) = (7, 2)
6:2, 6 = 3x2 + 0, (D, I) = (6,2)
3:1, 3 = 3x1 + 0, (D, I) = (3,1)
Daca le consideram NUMAI pe cele care se impart exact, fara rest, deci rest = 0, atunci avem:
(18, 6), (15, 5), (12, 4), (9, 3), (6,2) si (3,1).
