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FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12. TO SOLVE THIS PROBLEM, WE SHALL HAVE TO START FROM THE END. WE HAVE BEEN TOLD THAT AFTER ALL THE TRANSPOSITIONS, THE NUMBER OF MATCHES IN EACH HEAP IS THE SAME. LET US PROCEED FROM THIS FACT. SINCE THE TOTAL NUMBER OF MATCHES HAS NOT CHANGED IN THE PROCESS, AND THE TOTAL NUMBER BEING 48, IT FOLLOWS THAT THERE WERE 16 MATCHES IN EACH HEAP. AND SO, IN THE END WE HAVE: FIRST HEAP: 16, SECOND HEAP: 16, THIRD HEAP: 16 IMMEDIATELY BEFORE THIS WE HAVE ADDED TO THE FIRST HEAP AS MANY MATCHES AS THERE WERE IN IT, I.E. WE HAD DOUBLED THE NUMBER. SO, BEFORE THE FINAL TRANSPOSITION, THERE ARE ONLY 8 MATCHES IN THE FIRST HEAP. NOW, IN THE THIRD HEAP, FROM WHICH WE TOOK THESE 8 MATCHES, THERE WERE: 16 + 8 = 24 MATCHES. WE NOW HAVE THE NUMBERS AS FOLLOWS: FIRST HEAP: 8, SECOND HEAP: 16, THIRD HEAP: 24. WE KNOW THAT WE TOOK FROM THE SECOND HEAP AS MANY MATCHES AS THERE WERE IN THE THIRD HEAP, WHICH MEANS 24 WAS DOUBLE THE ORIGINAL NUMBER. FROM THIS WE KNOW HOW MANY MATCHES WE HAD IN EACH HEAP AFTER THE FIRST TRANSPOSITION: FIRST HEAP: 8, SECOND HEAP: 16 + 12 = 28, THIRD HEAP: 12. NOW WE CAN DRAW THE FINAL CONCLUSION THAT BEFORE THE FIRST TRANSPOSITION THE NUMBER OF MATCHES IN EACH HEAP WAS: FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12 LET'S ASSUME THAT THERE IS ONLY 1 CHEATING HUSBAND. THEN HIS WIFE DOESN'T SEE ANYBODY CHEATING, SO SHE KNOWS HE CHEATS, AND SHE WILL KILL HIM THAT VERY DAY. NOW, LET'S SAY THAT THERE ARE 2 CHEATING HUSBANDS. THERE WILL BE 98 WOMEN IN THE TOWN WHO KNOW WHO THE 2 CHEATERS ARE. THE 2 WIVES, WHO ARE BEING CHEATED ON, WOULD THINK THAT THERE IS ONLY 1 CHEATER IN THE TOWN. SINCE NEITHER OF THESE 2 WOMEN KNOW THAT THEIR HUSBANDS ARE CHEATERS, THEY BOTH DO NOT REPORT THEIR HUSBANDS IN ON THE DAY OF THE ANNOUNCEMENT. THE NEXT DAY, WHEN THE 2 WOMEN SEE THAT NO HUSBAND WAS EXECUTED, THEY REALIZE THAT THERE COULD ONLY BE ONE EXPLANATION - BOTH THEIR HUSBANDS ARE CHEATERS. THUS, ON THE SECOND DAY, 2 HUSBANDS ARE EXECUTED. THROUGH MATHEMATICAL INDUCTION, IT CAN BE PROVED THAT WHEN THIS LOGIC IS APPLIED TO N CHEATING HUSBANDS, THEY ALL DIE ON THE N TH DAY AFTER THE QUEEN'S ANNOUNCEMENT. SO WITH 100 CHEATING HUSBANDS, ALL OF THEM WILL BE EXECUTED ON THE 100TH DAY THE KING KEEPS POISON 12 SO THAT HE CAN NEUTRALIZE ANY POISON THAT THE CLOWN GIVES HIM. HE POURS POISON 10 (HIS SECOND STRONGEST POISON) AND GIVES IT TO THE CLOWN TO DRINK. CLOWN CAN DEDUCE THIS SO HE KEEPS POISON 11 FOR HIMSELF TO NEUTRALIZE POISON 10. THAT'S HOW HE SURVIVES. HE ALSO KNOWS THAT THE KING, AFTER DRINKING THE CUP HE GAVE HIM, WILL DRINK POISON 12 TO NEUTRALIZE THE POISON. SO THE CLOWN POURS PLAIN WATER INSTEAD OF POISON INTO THE CUP FOR THE KING. THE KING DRINKS THE WATER (WATER IS NOT POISON) AND RIGHT AFTER THAT HE DRINKS THE STRONGEST POISON POSSIBLE (12) AND DIES THE MERCHANT HAS 32 GOLD COINS. TO VERIFY THIS, DIVIDE THE 32 COINS INTO TWO UNEQUAL NUMBERS, SAY, 27 AND 5. THEN: 32 (27 - 5) = (272) - (52). 704 = 729 - 25 IF THE TWO UNEQUAL NUMBERS ARE 22 AND 10, THEN: 32 (22 - 10) = (222) - (102). 384 = 484 - 100 ALTERNATIVELY, LET'S SAY THAT THE 2 NUMBERS ARE X AND Y. WE CAN THEN COME UP WITH THE FOLLOWING EQUATION: 32(X - Y) = X2 - Y2 X2 - Y2 CAN BE EXPANDED TO: (X - Y)(X + Y) SO, NOW THE EQUATION WILL BECOME: 32 = X + Y (X + Y) WILL GIVE THE TOTAL NUMBER OF GOLD COINS