Question

aflati numerele naturale care impartite la 9 dau catul si restul doua numere naturale consecutive ( catul mai mic decat restul )

Answer 1

c = 1  r = 2
a : 9 = 1 rest 2
a = 9 x1 + 2
a = 9 + 2
a = 11

c = 2 , r = 3
a : 9 = 2 rest 3
a = 9 x2 + 3
a = 18 + 3
a = 21

c = 3  , r = 4
a : 9 = 3 trest 4
a = 9 x 3 + 4
a = 27 + 4
a = 31

c = 4 , r = 5
a : 9 = 4 rest 5
a = 9 x4 + 5
a = 36 + 5
a = 41

c = 5  r= 6
a : 9 = 5 rest 6
a = 9 x5 + 6
a = 45 + 6
a = 51

c = 6 , r = 7
a: 9 = 6 rest 7
a = 9 x6 + 7
a = 54 + 7
a = 61

c = 7 , r = 8
a : 9 = 7 rest 8
a = 9 x7 + 8
a = 63 + 8
a = 71