New answers tagged

0

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.


0

The answer is: Proof of uniqueness:


5

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 ...


8

The maximum is Reasoning


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Maybe would be worth a try to go for:


-1

Another way to answer the question is the following: 7@6=11342 (2x7)-(7+6)=1 (2x7)+(6-7)=13 [ (6+1)/2 + (6-1)/2 ] 7=42


0

7@6 the answer is 7-6=1 7+6=13 7X6=42 The answer is 11342


1

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 ...


2

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 ...


11

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 ...


54

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 ...


2

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 ...


23

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 ...


70

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 ...


0

Write the general case as a@b The answer seems to be Some question are, do we limit a and b? For instance do we require a > b, what about a=b? This answer seems to require that a and b be positive because otherwise you get weird results for the 2nd and 3rd terms.


1

7@6 is Because formula is


4

Expanding on @Dason's answer:


7

The answer may be because fits for the given examples.


13

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


2

Maybe her 4-digit password is: Reason:


42

The result of 7@6 is because


4

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!...


7

I believe this is the highest score possible with 100 or fewer operations: 99 operations; Score > $10^{10^{10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}}}}$ As the rather absurd score above suggests, arbitrarily large scores are possible. In particular, we can construct a sequence of approximations $A_k$ that converge extremely rapidly to $\pi$, giving scores ...


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