Răspuns:
n1= 7*c1+r1 n2=7*c2+r2 n3=7*c3+r3 r<7 r= {1,2,3,4,5,6} dar
r1+r2+r3=9 => 2+3+4=9 r1=2, r2=3, r3=4 (resturile)
n2=n1+1 n3=n2+1=n1+2
c1+c2+c3=87 ( daca numerele sunt consecutive catul este acelasi)
87:3= 29 (catul)
n1=7*29+2= 205 n1=205
n2=7*29+3= 206 n2=206
n3=7*29+4=207 n3=207