the riddle at 12 is also solvable. 1a 2a 3a 1b 2b 3b We know, it's impossible for 3b to be a mimic. Let's assume 3a would be a mimic. booth red boxes accuse one of the black boxes to be a mimic. One would have to tell the truth and we would have 3 mimics. So wie know booth blue ones are save and tell the truth ? ? S ? ? S Let's assume booth red ones would be mimics - we would have 4 Mimics, and black ones would have to be mimics too. Let's assume booth black ones would be mimics, 2a says the the bottom row contains no mimic, but the other black box is on the bottom row. So we know we have a mix bunch. 1b tells as there is no mimic among the blue boxes - which is true. it's the mimic. - 2b tells us, correctly that ab is a mimic, which means it's also telling the truth and with that, also a mimic. S S S M M S L L T T T T