At a minimum, one logician - the final one - must be drunk. The first two can reasonably say they don't know if they cannot drive but they think one of the later logicians might The last one ...

For the warmup: Should do it. For the challenge: This statement can only work iff the speaker is a knight, as otherwise it will lead to a logical paradox, which is neither true nor false.

Assuming Gerhard's edit reflects what the OP meant to ask, then yes.

As you say in the question: there should be exactly one interpretation which fits each clue given. However hard the riddle a supermajority of fluent English (or whatever your target language may be) ...

Yes. Each guess eliminates one number as well as dividing the remaining numbers into 2. One guess can pick a number from 3 (is your number 2?). 2 guesses can do 7. N guesses can pick a number from $2^... View answer Accepted answer 20 votes The accepted answer has a subtle error. If we randomly select a 2-sibling family with at least one male, the odds of the other being female are indeed 2/3. If we randomly select a male who is part of ... View answer Accepted answer 17 votes It is not possible. Let v be the set of separations between adjacent vertical lines. Let h be the set of separations between adjacent horizontal lines. Every time an element of v matches an element ... View answer Accepted answer 15 votes I believe it is: View answer 14 votes Rebus #2 Because Rebus #3 Because View answer 14 votes This is impossible Assume 2 houses are fully connected to 3 wells. There must be an 'inner' well which is surrounded by the connections to the other 2 wells. For the third house to be connected to ... View answer 13 votes Take the example square below: -3 2 1 4 0 -4 -1 -2 3 To generate a new square, simply multiply each element of this square by any positive integer. As there are an infinite number of positive ... View answer Accepted answer 13 votes Yes. Start with 0,2,1,3,6,4,8,10,5,7,9,11,22,20,18,16,14,12,24,26,13,15,17,19,21,23,25,27... The pattern is Start with 0,2,1,3 Double Decrement until you cannot Double Increment Divide Increment ... View answer Accepted answer 13 votes$p$is a random variable, chosen with uniform probability over the interval [0,1]. The probability of two heads is$\int_0^1 p^2=\frac{1}{3}$, and by symmetry so is the probability of two tails. View answer Accepted answer 12 votes Set the compass to some size, less than the distance from the 5ft mark to the ceiling. Use the bob to find two points on the circle, one directly above the other. Then use the awl to punch holes at ... View answer Accepted answer 12 votes I don't know if this is optimal play for Mary (though I suspect it is), but she can guarantee winning 2 Euros if she initially plays at (5,5) and copies Ursula's moves by translation through (5,5). ... View answer 12 votes Walk along the river until the rod, held up by my friend at a distance equal to the length of the rope, has the same apparent size as the bridge. Then measure the distance to the bridge in rope ... View answer Accepted answer 11 votes Professor Kafka is Because View answer 11 votes There might be a problem. It looks like Quark's answer of is correct, given this plot for up to 1000 rooms (excess of lights on over lights off) But if we extend that range a little... View answer Accepted answer 10 votes Explanation: Clue: View answer Accepted answer 10 votes You appear to need 3 weighings. Split your 1023 balls into 3 equal groups of 341. Compare 2 of these, and you know which group the heavier ball is in. Call the heavier group H and one of the normal ... View answer 10 votes Edit: as noted by others, this answer is incorrect. It assumes (incorrectly) that the distance to the modal latitude is the modal distance.$\frac{d}{\sqrt{2}}$Assume, wlog, that Romeo is at the ... View answer Accepted answer 10 votes First, each die must have a 0 face. We can see this by noting that we need two 0s to make 001, and a third die with {1,2,3,4,5,6} only gets us to 006. Having this fact, we need 1-8 at least once on ... View answer Accepted answer 10 votes We can make$17 \times 8 = 136$and any number above this by replacing a$17$with$2 \times 9$, to increment the number by$1$. By the time we run out of$17$s to replace ($9 \times 16\$), we continue ...

Sven will Proof:

It is Proof: Forget where the pointer is telling you the treasure is, look at where it can't be. For every pointing, there is a 120 degree region (centred opposite the pointing) that the treasure ...

Step 1: Make 3 groups of 18 coins: A,B,C. Compare A to B and B to C. If A≠B and B≠C then A=C. Select the 'odd' group (assume C) and note if it is lighter or heavier than the others. (assume ...

Noting that 3*E gives a different result in the first two columns, there must be an overflow i.e. E is greater than 3. This overflow can only be 1 or 2 (you cannot get more than 27 from tripling a ...