harey

I had this solution in mind, unfortunately, it fails because of repeated situations. Starting situation: 6 lanterns, only lantern 6 is lit. First round: lantern 1: no action lantern 2: devil turns it on remaining lanterns: angel does not do anything (If the devil turns a lantern on before the angel has turned any off, then the angel should just do nothing on that circuit. ) Situation: Lanterns 2 and 6 are lit 2nd round: lantern 1: no action lantern 2: angel turns it off (on each circuit, turn off all lit lanterns until the devil turns an unlit lantern on) Now, only the lantern 6 is lit: the game ends if the lanterns return to a previously encountered state What am I missing?

Here's an example from my code's 6 lantern output: 111110 <one lit lantern (interpret 0 as "on" and 1 as "off") 111101 <one lit lantern You suppose the devil turns the lantern 5 on. But what if he turns the lantern 4 and not 5 on? 111110 <one lit lantern (interpret 0 as "on" and 1 as "off") 111010 your turn P.S. I am not touchy (did not even think it could be interpreted this way), but I think things over. In the last time, I wrote a lot of nonsense, i.e. my first answer.

@EventHorizonMy post is unrelated to yours. The number of lamps is not important, just a little bit more that those I quote. "Let's start with one lantern lit, i.e. 33.": This is not a binary number, it is the 33rd lantern assuming the starting position is the 1st lantern (conveniently chosen), following the lantern 32 and followed by the lantern 34.

&EventHorizon

OK, so here the solution.

It seems you need a hint:

To decide who will do the dishes, we put in a jar tickets with prime numbers up to 111. The one who draws the ticket with a larger number wins. I look at my ticket: 3. Maybe you have a low number like 11... So I offer to exchange our tickets before showing them. What is the probability I will do the dishes?

I get 13. Take one coin of each chest. a) You get two Gold, one Silver: Silver is identified, remains to distinguish all_Gold and Gold_Silver. Take 10 more coins from the first G: - if they are all gold, you identified all_Gold, the remaining is Gold_Silver - if you get a silver coin, you identified Gold_Silver, the remaining is all_Gold b) You get one Gold, two Silver: Same proceeding, permute Gold/Silver.

A good question would be "How many steps on average until I win?"

There is a strip of N (N>13) squares. In the middle is the SILVER DOLLAR. Two players alternatively place a penny on an empty square. Then, at each turn, a player must: - move one coin one or more squares to the left, observing: - the coin cannot move out of the strip - the coin cannot jump on another coin - the coin cannot jump over another coin - or - - pocket the leftmost coin Best strategy?

@Plasmid I translated your notation into mine, as far as we are gone, we have same results (excepted some doublets). I thought about interpolation, too, but I did not go this way estimating there is not enough data.

I fear there is no solution.

Same as rocdocmac Combinatorics:

Thanks for finding the problem. I was so sure that if X contradicts Y, they must be in different categories - it did not occur to me that can be both liars.

Do not worry, on my first attempt, I did not manage it to write it clearly enough that I myself could read and understand it. When you asked your question, I checked my solution written about a year ago and wondered whether it would not be easier to start from the beginning. I have rewritten it and found a kind of notation: P.S. Can someone change in the title ONE in ONCE? Thanks in advance.

Bonus question: Answer your question. Why don't you post your solution?

On an island, every statement is true if the islander is aged less than L and false if he is at least L years old. Find their ages. [1] A: "B is more than 20 years old." [2] B: "C is more than 18 years old." [3] C: "D is less than 22 years old." [4] D: "E is not 17 years old." [5] E: "A is more than 21 years old." [6] A: "D is more than 16 years old." [7] B: "E is less than 20 years old." [8] C: "A is 19 years old." [9] D: "B is 20 years old." [A] E: "C is less than 18 years old."

There be sixty-and-four flowers-de-luce (in a grid 8x8), and the riddle is to show how I may remove six of these so that there may yet be an even number of the flowers in every row and every column. I am not able to remove 6 of them: interactive version What am I missing? Solution. that does not help me. I got it now, to late to delete.

I did some research... At first, the drawing Nick made is somewhat confusing. What he calls "forward force" should be "thrust" and what he calls "thrust" should be "drag". Just google "drag thrust weight lift". If thrust=drag and lift=weight, the plane will fly at constant speed at constant height (plenty of pages). Now, the question is how much thrust we need to generate lift=weight (or a little better). On the end of the article https://en.wikipedia.org/wiki/Lift-to-drag_ratio, there is a table: latest air-crafts have a coefficient over 19.

I fear there are multiple solutions. Babysnoot is correct, just the only reason to proclaim Jester truthteller is that there is no further contradiction in the system. For Bear=truthteller, Drummer=liar, others=mix (it does not matter whether what they say is true or false), there is no contradiction in the system, neither. Does someone agree?

A table with English words by the set of all combinations (with repetition) of 7 letters. However, even with some tricks like sorting the letters, I do not think my computer would give me the answer overnight. Remains the Monte-Carlo method. However, I am pretty sure it is not the expected answer. A hint?

Read "passenger 2-99" instead of "2-98".