D : 5 = r + 1 r
D = 5×(r +1) + r
D = 5r + 5 + r
D = 6r + 5
avem conditia r < 5
ptr r = 4
D = 6 × 4 + 5 = 29
ptr r = 3
D = 6 × 3 + 5 = 23
ptr r = 2
D = 6 × 2 + 5 = 17
ptr r = 1
D = 6 × 1 + 5 = 11
deci avem urmatoarele numere:
29 ; 23 ; 17 ; 11 .
29 : 5 = 5 rest 4
23 : 5 = 4 rest 3
17 : 5 = 3 rest 2
11 : 5 = 2 rest 1