How to solve the hardest logic puzzle ever in two questions GABRIEL UZQUIANO Rabern and Rabern (2008) have noted the need to modify 'the hardest logic puzzle ever' as presented in Boolos 1996 in order to avoid trivialization. Their paper ends with a two-question solution to the original puzzle, which does not carry over to the amended puzzle. The purpose of this note is to offer a two-question solution to the latter puzzle, which is, after all, the one with a claim to being the hardest logic puzzle ever. Recall, first, Boolos's statement of the puzzle: Three gods A, B and C are called, in some order, True, False and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for 'yes' and 'no' are 'da' and 'ja', in some order. You don't know which word means which. (Boolos 1996: 62) And remember his guidelines: It could be that some god gets asked more than one question. What the second question is, and to which god it's put, may depend on the answer to the first question. Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely. Random will answer 'da' or 'ja' when asked any yes-no question. (Boolos 1996: 62) Since the stipulation that Random should set out to speak truly or lie, albeit randomly, trivializes the puzzle, Rabern and Rabern (2008) have proposed