Another intuitive, no-math (and I believe best) strategy could be as follows: This ensures that The majority won't vote 1 (Steaks) when they would be wrong. The majority will vote 1 the first night they all had steak.


9 days. Every day, each prisoner with an unknown key passes it to the next cell to the left (or to the next guy in alphabetic order, sentence length, age... w/e, just agree on the order on that first day). Since they know where in the order the owner of their own key is (relative to them), they know on which day to expect their key - they get to keep that, ...


Ilmari's answer covers the logic of the answer perfectly, but in case anyone's still confused, here's a version of the diagram I find more intuitive (I made it myself while solving the puzzle). The Pink squares are rooms where the princess could be on the given day, the blue squares are where the prince knocks that day, and the black squares are rooms in ...


Bob can get it in Here's how: What I mean by the term in quotes:


Yes, you should definitely accept the deal. The puzzle clearly states that pushing a button labeled "Poof!" will remove a coin (emphasis mine) from box $i$, and cause two coins to magically appear in another box. By means of swapping and poofing you can keep increasing the amount of money inside the machine to over one trillion coins, but this process takes ...


Stay put for about 45 days, after which the pirates would have circumnavigated the globe and returned to your current position.


Maybe this diagram will help you visualize the solution: In this diagram, the vertical axis shows the room number (1–17) and the horizontal axis shows the day (1–30). The $\color{red}{\text{red}}$ dots show the rooms on whose door the prince knocks on each day, and the $\color{darkgreen}{\text{green}}$ dots and arrows show the possible ...


The lockets and coffins are always found in loops. open a coffin look at the number of the locket in the coffin go to the coffin with the same number as the locket open the coffin repeat from 2 At one point you will definitely find the locket of the coffin you first opened To prove this lets look at the other possibility and try to enter a loop without ...


That's easy


Here is the strategy:   It works because:


There is another way to do this problem, that doesn't involve any sort of conditional branching at all. It is in fact possible to set a fixed weighing schedule beforehand and still determine which ball is lighter or heavier in just 3 weighings. I'll explain how below. The gist of problems like these is, how much information can you get from the procedure ...


It doesn't matter which option you choose, because Your probability of survival if you're one of n players left is as follows: Informal proof It was established in the question that if there are only 2 players left, they must annihilate each other. So if there are 3 left, your only hope of survival is to annihilate both your opponents, which can only ...


If we assume the ocean is flat and extends indefinitely in all directions, there is a strategy that guarantees we can catch the pirates in at most 800,000 years. Put our current location as the origin of a coordinate system. We will describe our position in polar coordinates, as a function of time: $(r(t),\theta(t))$ (where we have arbitrarily chosen a ...


Satan should stick to fiddling. You will win, and here is a simple proof. Consider the game $n$ turns at a time. After each cycle of $n$ turns, all the coins are in their original position (though not necessarily flipped the same way). Replace $H$ with $0$ and $T$ with $1$. In each cycle, you flip all $1$'s to $0$'s, until Satan flips a $0$ to a $1$. Once ...




You can make arbitrarily large sets of dice with this property. Start with Efron's dice: A: 4, 4, 4, 4, 0, 0 B: 3, 3, 3, 3, 3, 3 C: 6, 6, 2, 2, 2, 2 D: 5, 5, 5, 1, 1, 1 A beats B, B beats C, C beats D, and D beats A with probability $\frac{2}{3}>60\%$. Now add many copies of die B, each using a different value between 2 and 4. For example, one of ...


The winning strategy for the first player is to put their coin in the dead center of the table. Then whatever move their opponent makes, they exactly mirror it, around the center. e.g. If the second player puts their first coin 1 inch to the left of the center coin, the first player mirrors this by putting their coin 1 inch to the right of the center coin. ...


The current president should choose the numbers 9($L$), 32($n_1$), 8($n_2$), 5($n_3$), 5($n_4$), 5($n_5$), 5($n_6$), 5($n_7$), 5($n_8$), 5($n_9$). In the lowest sub-sub-...-subgroup, the president should place voters in such a way that his supporters outnumber his opposers in the ratio of 3:5, till he runs out of supporters, and he can let the remaining ...


The volume filled is:


I have a hunch that the answer is Explanation: Continuing this way, we see that


A dwarf with a red dot will A dwarf with a blue dot will Therefore, on the $N$th day, all dwarves with red dots are present, and all dwarves with blue dots are not.


The prisoners agree that: Each prisoner with a blue nose will: Each prisoner with a red nose will:


Yes, the policeman can catch the thief, although it may take a very long time. In particular, the following strategy works: The policeman starts in the center. First, he makes a counterclockwise loop around the top right quadrant. Then, he makes a counterclockwise loop around the top left quadrant. Then, he makes a counterclockwise loop around the bottom ...


Please ignore my painting skill. i am not good at it


This: Just to break down the first version of this: So we have with 3 eggs:


There is an easy way to solve it in the minimum number of moves. The reason this works is:


The answer, as I'm sure most people are guessing, is... Here's my solution: Also, I apologise for the formatting. I'm new to SE.



You could count upwards from 1, because none of the numbers less than a billion contains the letter "b". However, this might be repeating words if "101" is considered to contain "one". (The version I've seen before is "Provide 100 words that don't contain the letter 'a'.") If you happen to have the system memorized, you can use polygon names instead.

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