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1 0 Clue
TAKE ONE COIN FROM THE FIRST BAG, TWO FROM THE SECOND, THREE FROM THE THIRD, AND SO ON, UNTIL YOU HAVE TEN COINS FROM THE TENTH. THEN STACK AND WEIGH THE 55 COINS (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10). SINCE EACH GOLD COIN WEIGH TEN GRAMS, IF ALL HAD BEEN GOLD, THE SCALE WOULD HAVE READ 550 GRAMS. THE AMOUNT BY WHICH THE WEIGHT WAS TOO LIGHT INDICATED THE NUMBER OF SILVER COINS AND THE NUMBER OF THE THE SILVER BAG. FOR INSTANCE, IF THE WEIGHT WAS 543 GRAMS, IT WOULD INDICATE THAT 7 SILVER COINS (550 - 543 = 7) HAD BEEN WEIGHED WITH THE GOLD COINS AND THAT THE REST OF THE SILVER COINS WERE IN THE SEVENTH BAG NUMBER THE BUCKETS 1 THROUGH 20. TAKE ONE BLOCK FROM BUCKET 1, TWO FROM BUCKET 2, THREE FROM BUCKET 3, AND SO ON, UNTIL YOU HAVE 20 BLOCKS FROM BUCKET 20. THEN STACK AND WEIGH THE 210 BLOCKS (1 + 2 + 3 + ... + 18 + 19 + 20 = 210). IF ALL THE BLOCKS WEIGH 16OZ EACH, THEN THE SCALE WOULD READ 3360OZ. HOWEVER, THE SCALE WILL ACTUALLY READ OVER THIS BY A NUMBER OF OUNCES THAT CORRESPONDS TO THE NUMBERED BUCKET WITH THE HEAVIER BLOCKS SINCE EACH HEAVY BLOCK ADDS 1 OUNCE. FOR EXAMPLE, IF THE HEAVIER BLOCKS ARE IN BUCKET 10, THE SCALE WOULD READ 3370OZ SINCE THERE ARE 10 BLOCKS THAT WEIGHT 17OZ EACH. SUBTRACTING THE WEIGHT ON THE SCALE WITH 3360OZ WILL GIVE THE BUCKET NUMBER: 3370 - 3360 = 10 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 STEP 1: DIVIDE IT INTO SETS OF THREE. IF YOU PUT ONE SET ON EITHER SIDE, IT'LL EITHER EQUAL THE SAME, OR YOU'LL SEE WHICH SET IS LIGHTER. EITHER WAY, YOU NARROW IT DOWN TO THREE BALLS. STEP 2: PUT TWO OF THE BALLS ON THE SCALE. IF THEY'RE EQUAL, THE LAST BALL IS DEFECTIVE, BUT IF NOT, THE SCALE WILL TELL YOU WHICH ONE IS LIGHTER