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B _ T T _ R _ _ S Clue
YOU WOULD ONLY NEED TO TAKE OUT ONE MARBLE BECAUSE WE KNOW THAT ALL OF THE LABELS ARE INCORRECT. SO YOU PULL ONE MARBLE OUT OF THE BOX LABELED "MIXED." IF RED COMES OUT, YOU KNOW THAT HAS TO BE THE ALL-RED BOX, SO YOU PUT THE RED LABEL ON IT. THE BOX LABELLED "BLUE" MUST THEN BE LABELLED "MIXED" BECAUSE YOU KNOW IT IS ALSO LABELED INCORRECTLY, AND THEREFORE CAN'T BE BLUE. YOU WOULD LABEL THE LAST BOX "BLUE" BECAUSE THAT IS THE ONLY COLOR/BOX COMBO LEFT. IF THE FIRST MARBLE YOU PULLED OUT FROM THE BOX LABELED "MIXED" IS A BLUE MARBLE, THEN YOU SOLVE THE PROBLEM IN THE SAME GENERAL WAY ZOE'S SMALLEST POSSIBLE NUMBER IS 6. BASED ON THE FIRST STATEMENT OF ALI, IT INDICATES THAT HE HAS NEITHER 1 NOR 9. IF HE HAD EITHER 1 OR 9 THEN HE WOULD KNOW THAT ZOE MUST HAVE A BIGGER OR SMALLER NUMBER. NOW ZOE, BASED ON ALI'S FIRST STATEMENT, KNOWS THAT ALI DOESN'T HAVE 1 OR 9. ZOE'S FIRST STATEMENT INDICATES THAT SHE DOES NOT HAVE 2 OR 8 (NEITHER 1 NOR 9). IF SHE HAD 1, 2, 8 OR 9, THEN SHE COULD HAVE CONCLUDED THAT ALI HAS A BIGGER OR SMALLER NUMBER. NOW ALI KNOWS THAT ZOE DOESN'T HAVE 1, 2, 8 OR 9. ALI'S SECOND STATEMENT INDICATES THAT HE DOES NOT HAVE 3 OR 7 AND ALSO NOT 1, 2, 8 OR 9. ZONE CAN CONCLUDE THAT ALI DOESN'T HAVE 1, 2, 3, 7, 8 OR 9. IN SHORT, ALI MUST HAVE EITHER 4, 5 OR 6. NOW WHEN ZOE SAYS THAT SHE HAS A BIGGER NUMBER THEN IT MUST BE EITHER 6, 7, 8 OR 9 AND ALI HAVING 4 OR 5. ZOE CAN'T SAY CONFIDENTLY THAT SHE HAS A BIGGER NUMBER IF SHE HAD A 4 OR 5, AS IT COULD BE SMALLER THAN WHAT ALI COULD HAVE. SO ZOE'S SMALLEST POSSIBLE NUMBER IS A 6 YOU WILL NEED TO MAKE THREE CUTS. CUT ALL THREE LINKS ON ONE CHAIN AND SEPARATE THEM. NOW YOU HAVE 3 CHAINS AND 3 LINKS. THEN, USE THESE 3 INDIVIDUAL LINKS TO JOIN THE OTHER THREE CHAINS TOGETHER YOU WILL NEED A MAXIMUM OF 7 ATTEMPTS TO FIND 2 WORKING BATTERIES. BREAK THE BATTERIES INTO 3 GROUPS: TWO GROUPS OF 3 AND ONE GROUP OF 2. BY DOING THIS YOU GUARANTEE THAT AT LEAST ONE OF THE GROUPS HAS 2 WORKING BATTERIES. BOTH OF THE GROUPS OF 3 HAVE 3 POSSIBLE COMBINATIONS OF 2 BATTERIES AND THE GROUP OF 2 ONLY HAS 1 COMBINATION. SO, 3 + 3 + 1 = 7 TRIES AT MOST TO FIND TWO WORKING BATTERIES