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THE MINIMUM NUMBER OF WEIGHTS REQUIRED IS FIVE AND THESE SHOULD WEIGHT 1, 3, 9, 27 AND 81 POUNDS. THE MERCHANT HAS TO USE A BALANCE WEIGHING SCALE TO DO THE JOB. TO WEIGH 2 POUNDS, HE'LL HAVE TO PUT THE 3 POUND WEIGHT ON ONE PAN AND 1 POUND WEIGHT ON THE OTHER PAN. TO WEIGH 5 POUNDS, HE'LL HAVE TO PUT THE 9 POUND WEIGHT ON ONE PAN AND 1 AND 3 POUND WEIGHTS ON THE OTHER PAN HE WOULD NEED TO PLACE THE 3-POUND WEIGHT AND THE 1-POUND WEIGHT ON THE SAME SIDE AS THE OBJECT AND BALANCE IT WITH THE 9-POUND WEIGHT. IF THE 2 GROUPS WEIGH THE SAME, THEN THE WEIGHT OF THE OBJECT IS 5 POUNDS. WHEN YOU PLACE A WEIGHT ON THE SAME PAN OF THE SCALE AS THE OBJECT YOU ARE WEIGHING, YOU SUBTRACT THAT WEIGHT FROM THE TOTAL OF THE WEIGHTS ON THE PAN ON THE OTHER SIDE THE FOUR WEIGHTS ARE: 1, 3, 9 AND 17. THE FOURTH WEIGHT IS NOT 27 BECAUSE BY USING A 27 POUND WEIGHT, WE WOULD ABLE TO MEASURE UP TO 40 POUNDS. BUT SINCE WE NEED TO MEASURE ONLY UP TO 30 POUNDS, WE CAN SUBTRACT 10 FROM 27 TO GET 17. LET A=1, B=3, C=9 AND D=17. 1LBS = A 2LBS + A = B 3LBS = C 4LBS = A + B 5LBS + A + B = C 6LBS + B = C 7LBS + B = A + C 8LBS + A = C 9LBS = C 10LBS = A + C ... 20LBS = B + D ... 30LBS = A + B + C + D IF THERE WAS ONLY ONE BLUE-EYED PERSON ON THE ISLAND, THEN THAT PERSON WOULD LOOK AROUND AND SEE THAT THERE IS NO OTHER BLUE-EYED PERSON. SO HE REALIZES THAT HE IS THE ONLY PERSON WITH BLUE EYES ON THE ISLAND AND LEAVES ON THE DAY OF THE ANNOUNCEMENT. IF THERE ARE 2 BLUE-EYED PEOPLE, THEN THEY LOOK AT EACH OTHER. EACH ONE EXPECTS THE OTHER TO LEAVE ON THE DAY OF THE ANNOUNCEMENT. HOWEVER, ON THE NEXT DAY, WHEN THEY REALIZE THAT NEITHER OF THEM LEFT THE ISLAND, THEY WOULD BE ABLE TO DEDUCE THAT BOTH OF THEM HAVE BLUE EYES. THEY BOTH LEAVE THE ISLAND ON THE SECOND DAY. THROUGH MATHEMATICAL INDUCTION, THIS LOGIC CAN BE APPLIED TO THE 100 BLUE-EYED PEOPLE ON THE ISLAND. SO ON THE 100TH DAY, ALL THE 100 BLUE-EYED PEOPLE LEAVE THE ISLAND