Correct answer: Nb5+1.Bd8[A]? zz 1...Kb6 2.Nb5# (C) but 1...Kb8[a]! 1.Bf4[Б]? zz 1...Kb8 2.Nb5# (C) but 1...Kb6[a]! 1.Nb5+!(C) 1...Kb6/[б]/ 2.Bd