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MAKE 2 PILES WITH EQUAL NUMBER OF COINS. NOW, FLIP ALL THE COINS IN ONE OF THE PILE. HOW THIS WILL WORK? LET'S TAKE A LOOK AT AN EXAMPLE. SO INITIALLY THERE ARE 5 HEADS, SO SUPPOSE YOU DIVIDE IT IN 2 PILES. P1 : H H T T T P2 : H H H T T NOW WHEN PILE P1 IS FLIPPED: P1 : T T H H H THE NUMBER OF HEADS IN P1 AND P2 ARE EQUAL NOW YOU SPLIT THE COINS INTO A GROUP OF NINETY AND A GROUP OF TEN. YOU THEN FLIP ALL OF THE COINS IN THE GROUP OF TEN. WHEN THE LIGHTS ARE TURNED ON YOU'LL FIND THAT THERE ARE AN EQUAL NUMBER OF HEADS IN BOTH GROUPS. THIS METHOD WILL ALWAYS WORK. IF IT'S DIFFICULT TO COMPREHEND HOW THIS WORKS, LET'S LOOK AT AN EXAMPLE. ASSUME THAT THERE ARE 3 COINS THAT HAVE HEADS IN THE GROUP OF NINETY. SO THE GROUP OF TEN WILL HAVE 7 COINS WITH HEADS AND 3 COINS WITH TAILS. WHEN ALL THE COINS ARE FLIPPED IN THE GROUP OF TEN, THE NUMBER OF COINS WITH HEADS WILL BECOME THREE YOU WILL HAVE TO SAY, "YOU WILL GIVE ME NEITHER COPPER NOR SILVER COIN." IF IT IS TRUE, THEN YOU WILL GET THE GOLD COIN. IF IT IS A LIE, THEN THE NEGATION MUST BE TRUE, SO THE STATEMENT BECOMES: "YOU WILL GIVE ME EITHER COPPER OR SILVER COIN". THIS WOULD BREAK THE GIVEN CONDITIONS THAT YOU GET NO COIN WHEN LYING. SO THE FIRST SENTENCE MUST BE TRUE SEPARATE THE COINS INTO 2 GROUPS OF 10 AND 16, THEN FLIP OVER ALL THE COINS IN THE GROUP WITH 10 COINS. TO ILLUSTRATE, LET'S GO THROUGH AN EXAMPLE. IN THE GROUP OF 10, IF THERE ARE 4 HEADS, THEN THERE WOULD BE 6 HEADS IN THE GROUP OF 16. WHEN YOU FLIP ALL THE COINS IN THE GROUP OF 10, YOU WILL GET 6 HEADS