33

The answer should be Here is how you do it.


28

This sounds straightforward if


15

The same idea as APrough's answer, but presented visually instead of with words: πŸ“… 🏠 🏜️ 🏜️ 🏜️ 🏜️ 🏜️ 🏁 0 πŸ€ πŸ—πŸ—πŸ—πŸ—πŸ™‚πŸ—πŸ—πŸ—πŸ—πŸ™‚πŸ—πŸ—πŸ—πŸ— 1 πŸ€ πŸ—πŸ—πŸ—πŸ™‚πŸ—πŸ—πŸ—πŸ™‚πŸ—πŸ—πŸ— 1 πŸ€ πŸ—πŸ—πŸ—πŸ—πŸ™‚πŸ—πŸ—πŸ—πŸ—πŸ™‚πŸ— 2 πŸ™‚ πŸ€ πŸ—πŸ—πŸ—πŸ™‚πŸ—πŸ—πŸ— 2 πŸ™‚ πŸ€ πŸ—πŸ—πŸ—πŸ—πŸ™‚πŸ—πŸ— 3 πŸ™‚ πŸ™‚πŸ— πŸ€ πŸ—πŸ—πŸ— 4 πŸ™‚πŸ™‚ πŸ€ πŸ—πŸ— 5 πŸ™‚πŸ™‚ πŸ€ πŸ— 6 πŸ™‚πŸ™‚ 🀠


14

Alternative solution: Then here's how that works Here is the total number of rations and helpers required


11

If you are unsure about the puzzle, you are probably misquoting a classical puzzle as Rand Al'Thor mentions. If you cannot distinguish the type of the coins, regardless of how you make two piles, there is always a way to swap a heads-up and a tails-up coin between the two piles, changing the heads-up counts. So you cannot guarantee the counts are the same. ...


9

Here is one way to do it


6



6

You should:


4

A solution to this problem exists because there are an even number of both heads-up coins and tails-up coins, and thus they can be split evenly into two piles. If a solution exists, then all you would need to do is arrange the coins into two piles, and randomly swap one coin from each pile, over and over. Given enough time, you would generate all possible ...


4

This is 33-14-16-73-7-117 translates to As Si S Ta N Ts - assistants. ??-15-92-22-99 translates to ?? P U Ti Es - __puties, which I guessed as "deputies". Given that this was about chemistry, and I know $\text{n}^0$ means neutrons in that context, I looked up "assistant deputy neutron" and came across something called However, seeing ...


3

Not sure if this is worthy of an answer. By colour coding the letters and varying the grid size, This happens... Maybe someone else can build up on this...


3

He could do it as follows


3

I think the answer might be Because


2

It seems like you are talking about Line them up in order, And into teams as well. Each of them’s a character, Under someone’s spell. Some have foreign accents And put on different things: A cute hat, something grave, Tails and dots and rings. The leader might be biggest, Or they’re equal in one case. Some will often go alone. And neighbour only ...


2

Based on @Florian-F's out-of-the-box approach, I would Then, I would randomly split the coins into two piles. Any two piles will have the same number of heads facing up, which would be exactly


2

I don't see a solution without some kind of 'cheating'. My first idea was to stand the coins on their edge, however this is not allowed but perhaps one of the following might be allowed: 1.) Fix coins and turn the table by 90 degrees upwards Make two piles. It doesn't matter how many coins are in the first or second pile. Now use some sticky tape (scotch) to ...


1



1

I think,


1

Possibly wrong but I'll put this out there as a guess as it hasn't been specifically denied. If we are looking for someone who has 'done the dirty', it is interesting that


1

Minimum food & helper time wasted, but requires reliable communication across the desert: 4 man-days (both food and time) wasted for helpers. (If both helpers start with you, the minimum is 6 wasted man-days. The same if both helpers start from the opposite side.)


1

My first thought is to work backwards...


1

I'm not sure how to score this but I've found a neat little formula using the As follows


1

Not sure if this fits the requirements (it probably doesn't). I kind of got lost in them. But here goes. A more general approach


1

Well, I have absolutely no idea how this may or may not score, but here are the operations I'll be using: "concatenate some of the original numbers in the order they were given, with an optional decimal point" (a common operation in formation-of-numbers puzzles) square root subtraction division the "round to nearest integer" operation (...


1

Another answer where


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