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ABEL HAS 50, BILL HAS 20 AND CLARK HAS 30. ABEL ON HIS FIRST TURN OBVIOUSLY DOESN'T KNOW WHETHER HIS NUMBER IS 50 OR 10. SIMILARLY NEITHER BILL NOR CLARK CAN IMMEDIATELY FIGURE OUT THEIR NUMBERS. HOWEVER, ON HIS SECOND TURN ABEL CAN REASON: IF MINE IS A 10, THEN CLARK WOULD KNOW HIS NUMBER IS EITHER 10 OR 30. IF IT IS 10, BILL WOULD IMMEDIATELY KNOW HIS NUMBER IS 20. BUT HE DIDN'T KNOW. SO CLARK SHOULD KNOW HIS NUMBER IS 30. NOW SINCE CLARK DIDN'T KNOW, MY NUMBER MUST BE 50. WITH THIS KIND OF REASONING WE CAN ALSO RULE OUT ALL OTHER COMBINATIONS. SO [50, 20, 30] IS THE ONLY SOLUTION TO THIS PUZZLE 7 DAYS. ONE COP HAS TO SEARCH CLOCKWISE AND THE OTHER COP HAS TO SEARCH ANTI CLOCKWISE. SO, THE COPS START SEARCHING AT: CAVE 13 AND CAVE 1 ON THE 1ST DAY CAVE 12 AND CAVE 2 ON 2ND DAY CAVE 11 AND CAVE 3 ON 3RD DAY CAVE 10 AND CAVE 4 ON 4TH DAY CAVE 9 AND CAVE 5 ON 5TH DAY CAVE 8 AND CAVE 6 ON 6TH DAY CAVE 7 ON 7TH DAY THE WORST CASE IS WHEN THE THIEF STAYS IN CAVE 7 AND DOES NOT MOVE FIRST CUT THE CAKE INTO 4 EQUAL PIECES WITH 2 CUTS ON THE TOP - ONE HORIZONTALLY DOWN THE CENTER OF THE CAKE AND THE OTHER VERTICALLY DOWN THE CENTER OF THE CAKE LIKE A CROSS ('+'). NEXT IS TO CUT THE 4 PIECES INTO HALF WITH A HORIZONTAL FINAL CUT ON THE SIDE. THIS WILL GIVE YOU 4 PIECES OF CAKE ON TOP AND 4 PIECES OF CAKE UNDERNEATH SHE WILL HAVE FIVE PIECES OF CHAINS AFTER THE 2 CUTS. THE CHAINS WILL HAVE LOOPS OF 1, 1, 3, 6 AND 12. THE CHAIN IS MADE UP OF INTERLOCKING LOOPS. SO WHEN YOU MAKE 1 CUT IN THE CHAIN IN THE MIDDLE, YOU WOULD END UP WITH 2 CHAINS AND 1 CUT LOOP. SO 2 CUTS IN THE MIDDLE WOULD LEAVE YOU WITH 3 CHAINS AND 2 CUT SINGLE LOOPS. SO THE JEWELER MAKES ONE CUT ON THE 4TH LOOP MAKING A 1 CUT LOOP, A 3 LOOP CHAIN, AND A 19 LOOP CHAIN. THEN A CUT ON THE 7TH LOOP OF THE 19 LOOP CHAIN LEAVING A 1 CUT LOOP, A 6 LOOP CHAIN, AND A 12 LOOP CHAIN. SO AT THE END SHE HAS 1 CUT LOOP, ANOTHER 1 CUT LOOP, A 3 LOOP CHAIN, A 6 LOOP CHAIN, AND A 12 LOOP CHAIN