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YOU WILL NEED TO MAKE THREE CUTS. CUT ALL THREE LINKS ON ONE CHAIN AND SEPARATE THEM. NOW YOU HAVE 3 CHAINS AND 3 LINKS. THEN, USE THESE 3 INDIVIDUAL LINKS TO JOIN THE OTHER THREE CHAINS TOGETHER POUR THE LENTILS INTO THE INNKEEPER'S SACK, BIND IT AND TURN INSIDE OUT. POUR IN THE PEAS. THEN UNBIND THE SACK AND POUR THE LENTILS BACK TO YOUR SACK THE SHEEP WILL SURVIVE. IF THERE WERE 1 LION AND 1 SHEEP, THEN THE LION WOULD SIMPLY EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 2 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IF ONE OF THEM WOULD, IT WOULD SURELY BE EATEN BY THE OTHER LION AFTERWARDS. THE SHEEP WILL SURVIVE. IF THERE WERE 3 LIONS AND 1 SHEEP, THEN ONE OF THE LIONS COULD SAFELY EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 2 LIONS, WHERE NO ONE CAN EAT THE SHEEP. THE SHEEP WILL NOT SURVIVE. IF THERE WERE 4 LIONS AND 1 SHEEP, THEN NO LION WOULD EAT THE SHEEP, BECAUSE IT WOULD TURN INTO THE SCENARIO WITH 3 LIONS. THE SHEEP WILL SURVIVE. CONTINUING THIS ARGUMENT, THE CONCLUSION IS AS FOLLOWS: IF THERE IS AN EVEN NUMBER OF LIONS, THEN NOTHING HAPPENS AND THE SHEEP SURVIVES. IF THERE IS AN ODD NUMBER OF LIONS, THEN ANY LION COULD SAFELY EAT THE SHEEP AND THE SHEEP WILL NOT SURVIVE. THIS IS SIMILAR TO THE UNEXPECTED HANGING PARADOX 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