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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 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 THE CORRECT COMBINATION IS 65292. SINCE THE THIRD DIGIT IS THREE LESS THAN THE SECOND, AND THE FOURTH IS FOUR GREATER THAN THE SECOND, THERE ARE ONLY THREE POSSIBLE COMBINATIONS FOR THE SECOND, THIRD AND FOURTH DIGITS. THESE ARE -307-, -418-, AND -529-. WITH THE FIRST DIGIT THREE TIMES THE FIFTH, THE ONLY POSSIBLE COMBINATIONS FOR THE FIRST AND FIFTH DIGITS ARE 0 0,3 1, 6 2, AND 9 3. THE SOLUTION ARISES FROM COMBINING THESE TWO SETS OF POSSIBILITIES, WITH THE ADDED CRITERIA THAT THERE ARE THREE COMBINATIONS OF TWO DIGITS THAT THAT EACH SUM TO 11 THE WEIGHT OF THE 5 RINGS ARE 1, 2, 4, 8 AND 16 GRAMS. USING THE COMBINATION OF THE 5 TYPE OF WEIGHTS YOU CAN REWARD FROM 1 TO 31 GRAMS IN WEIGHT TO THE WISE MAN. FOR EXAMPLE: FOR THE 3RD DAY, YOU CAN GIVE HIM THE 1 AND 2 GRAM RINGS. FOR THE 15TH DAY, YOU CAN GIVE HIM THE 1, 2, 4, 8 GRAM RINGS. FOR THE 30TH DAY, YOU CAN GIVE HIM THE 2, 4, 8, 16 GRAMS