Question

Aflati cel mai micc numar natural care impartit la numerele 15,30 si 45 da, de fiecare data ,un cat diferit de 0 si catul 13.

Answer 1

a : 15 = x rest 13 ==> a = 15x+13
a : 30 = y rest 13 ==> a = 30y+13
a : 45 = z rest 13 ==> a = 45z+13

c.m.m.m.c [15;30;45] = 2×3^2×5 = 90

a = 90+13 = 103 (numarul cautat)

Facem si verificarea:

103 : 15 = 6 rest 13
103 : 30 = 3 rest 13
103 : 45 = 2 rest 13