Yes. Guess 1st set (1,3,5,7,9,11,13,15) -> If the number is in the set, write down 1 Guess 2nd set (2,3,6,7,10,11,14,15) -> If the number is in the set, write down 2 Guess 3rd set (4,5,6,7,12,13,14,15) -> If the number is in the set, write down 4 Guess 4th set (8,9,10,11,12,13,14,15) -> If the number is in the set, write down 8 After all 4 guesses, add the ...


A simple way is to pick the number $X$ which is half-way through the range and ask Is it less than $X$? From the answer you can discard either the lower half or the upper half of the range. Repeat until you have 1 number left. This method is known as a binary search or binary chop. In general, If you are allowed $q$ queries, you can get an answer from ...


The result of 7@6 is because


To build off @ChrisCudmore’s correct answer (go upvote it!), here is the same answer but with a simpler less technical explanation – hopefully this will be easier to understand… If you are allowed to ask four questions before then saying which number it is, you should use the following 4 questions: Now to find the number, start with 0. You will always get ...


The answer is because the operator x@y is clearly the function: Alternatively:


You can do this by splitting the range in half. If the numbers to guess from (1-16) is less than 2^X where x is the number of guesses you get. Look at the picture below, and each color is 1 guess: Hypothetically, the answer is 16 (I rounded up to split the numbers easier). 1. you guess 1-8, and that isn't it 2. you guess 9-12 and that isn't it 3. you guess ...


The maximum is Reasoning


The answer may be because fits for the given examples.


Since we are talking about multiplication (products grow really fast), and our data gathering is relatively cheap, it stands to reason that the best strategy is to I had this idea yesterday, but it took me this long to get through the final annoying special case. However, I'm pretty convinced now that This is going to be a longish read because of the ...


Maybe would be worth a try to go for:


Slightly simplifying Dark Malthorp's remarkable solution, we get: $$\pi \approx A_k = \left[1 - \left( 2 \cdot \log\lceil \sqrt{3!_k}\rceil + \sqrt{4}\right)\div(\lfloor\sqrt{\lceil\sqrt 5\rceil!_k}\rfloor!)\right]^{\lceil\sqrt{6!_{k-1}}\rceil!} \div \lfloor\sqrt 7\rfloor \cdot \lceil\sqrt{\lceil\sqrt 8\rceil!_k} \rceil!\cdot \lfloor\sqrt{\sqrt{9}!_k}\rfloor!...


Expanding on @Dason's answer:


No, it is not possible. At least not unless you put up more restrictions. E.g. that the numbers have to be integers. None of the solutions shown so far give a way to find $\sqrt \pi$ , which is between 1-15. If you restrict yourself to integers you can use this method: sum = 0 s = 1 while s <= max value: Ask: "Is the number odd?" Yes: Add s to sum ...


Maybe her 4-digit password is: Reason:


So I can’t comment since I don’t have enough reputation yet. I just want to say that this is the coolest thing I have ever seen. I had no idea that binary could be used in such a way that you could guess a number in a certain amount of guesses 100% of the time as long as the amount of guesses was equal to the amount of digits in binary of the highest number ...



There are 2 problems with trying to eliminate half of the numbers between 1-15 1st problem: you eventually narrow it down to 2 numbers at which point you have to ask about 1 specific number which isn’t allowed. 2nd problem: if the questions aren’t answered when they’re asked, but rather all 4 questions are answered after they’re all asked, then ...


7@6 is Because formula is


I think it's unlikely that this is really the solution, since you could make up several similar ways of "solving" the puzzle, but maybe it's worth a try.

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