Correct answer: Kd31.Ra4[A]? (2.Kd3#[Б]) 1...N:b3[a] 2.K:b3#[C] 1...N:c2/[б]/ 2.K:c2#[D] 1...B:c4[c] 2.K:c4#[E] but 1...a:b5! 1.Kd3[Б]! (2.Ra4#[A])