Răspuns:
Explicație pas cu pas:
ab+ac+ad=189, ⇒ 10a+b+10a+c+10a+d=189, ⇒30·a+(b+c+d)=189
deoarece a,b,c,d distincte, a≠0, ⇒ b+c+d=9 sau b+c+d=19
1) pentru b+c+d=9, ⇒a=6. (b,c,d)∈{(0,1,8),(0,2,7),(0,4,5),(0,5,4),(0,7,2),(0,8,1),(1,0,8),(1,3,5),(1,5,3),(1,8,0),(2,0,7),(2,3,4),(2,4,3),(2,7,0),(3,1,5),(3,2,4),(3,4,2),(3,5,1),(4,0,5),(4,2,3),(4,3,2),(4,5,0),(5,0,4),(5,1,3),(5,3,1),(5,4,0),(7,0,2),(7,2,0),(8,0,1),(8,1,0)}
Deci (ab,ac,ad)∈{(60,61,68),(60,62,67),(60,64,65),(60,65,64),(60,67,62), (60,68,61),(61,60,68),(61,63,65),(61,65,63),(61,68,60),(62,60,67),(62,63,64), (62,64,63),(62,67,60),(63,61,65),(63,62,64),(62,64,63),(62,67,60),(63,61,65), (63,62,64),(63,64,62),(63,65,61),(64,60,65),(64,62,63),(64,63,62),(64,65,60), (65,60,64),(65,61,63),(65,63,61),(65,64,60),(67,60,62),(67,62,60),(68,60,61), (68,61,60).
2) b+c+d=19, nu e valabil, deoarece 30·a+19=189, ⇒30·a=170, a∉N