New answers tagged

11

Not sure about the possibility of a dictionnary, but maybe this can do what you want : divide X by 500 and cap the result between 0 and 21, then use an array dict = [1,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,5]; dict[min(21,max(0,(x/500)))]; the part with the array is kinda bad because it means a very big array depending on the unequal ranges, but in your ...


7

Considering hints:


6

Here is a solution with a constant product of which I think is the minimum possible: Some partial progress for a lower bound on the product: This leaves only a few possibilities for improvement:


3

I found a solution trying to minimize prime factors. And finding a balance between the minimum value and minimum number of factors. I noticed that some fields are connected, in that multiplying one of them results in the multiplication of fixed other fields. There are three such patterns: -C, D, H, L, and M (and any mirror image of that) -A, E, G, L, M (...


6

Edit: In an effort to find the minumum, here is a much smaller solution in which the mutual product is Solution As MKBakker pointed out we could further reduce this by dividing each of the entries 4,8,16,96 and 192 by 2 to get a mutual product of although they have subsequently improved on this.


14

There are a number of ways to do this. An easy strategy:


2

For N=4: Addendum (@JaapScherphuis)


0

We think we have a partial answer. For the case N=1, you will find the treasure in the first night For N=2, you just have to search the same room twice and you will have the treasure in the second night (worst case scenario) For N=3, same thing with the middle room For N=4 or more, we don't see any strategy that can allow you to get the treasure in a ...


11

And here is the number you are probably thinking of: It works only for $ab$ where $a \le b$. I suppose that it is a mistake in the problem statement. Others have proven that as it is, the problem is unsolvable. And here is how I came to that number. PS: I have been playing with this problem. You can extend it to $ab$ with $a > b$ with the ...


7

I am going to prove that Indeed Let me dump here previous thoughts that turned out not to be useful but might be in the future. First, Second


4

I revisited the question, and ran a brute force over all possible cases, and found that there are two optimal solutions. I used the following github code and modified it a bit to fit my task. Other than the $1,4,6,7$ digit set that holds the optimal record of $53$ that Oray found, we can also accomplish the same record with digit case $2,3,8,9$: $$\begin{...


9

To do this, Finally, I can give you the Python code if you want. Not sure it can be spoilered (I've never learned how to have multi-line spoilers). Python Code


6

Start at the number 1. You should now be able to find how to fill in the missing value.


1

The players ante a sequence of odd numbers, so if a round ends after $x$ pairs of cards are drawn then a player's payoff is $\pm x^2$. Since the players break even after three rounds we must have $x^2 + y^2 = z^2$, where $x,y,z$ are the number of pairs drawn in each round, but not necessarily in that order. The tuple $(x,y,z)$ is therefore a Pythagorean ...


3

Bob must've won the last round, resulting in fewer black cards in the remaining deck, otherwise he wouldn't call for a reshuffle. Sum of the won money is $\left(\frac{n}{2}\right)^2$, where $n$ is the number of rounds played. So we have to solve: $\left(\frac{n_{1}}{2}\right)^2 + \left(\frac{n_{2}}{2}\right)^2 = \left(\frac{n_{3}}{2}\right)^2$, where $n_{1},...


3

I think it is Because:


1

Perhaps you are looking for We note that Then So the final answer is


14

First, we need to figure what is the minimum amount of superconductor that we'll lose when making the cut. Optimally, we can cut the large cube net into two smaller cube nets without any waste at all. Let's start by trying to find such a cube net: The blue and yellow parts are pretty simple cube wraps, so after the cut, we can easily make the required two ...


4

Not a full answer (I don't think so at least), but maybe something someone can build off of:


18

The answer is 7, It's a simple extension of the Magic Calculator trick. Each card represents a bit, with the upper left index value being the value of that bit. To perform it as a trick, you ask the volunteer to pick a number between 1 and 64, and hand you the cards on which their number appears. Add up the index values on the selected cards, and you have ...


20

The answer is: Here's the way you get that number:


4

I am a 4 digit number The digit in the ones place is 6 The value of the digit in the hundreds places is 500 The digits in the tens and thousands places are the same and they add up to 8 Maybe some traps in the question...?


6

Found it, thanks to number 48734


5

452


3

First, the answer to the actual question in the test: This makes step 2: Step 3: Final output:


0

Possible for 452 I'm not sure whether this is allowed or not but here goes:


0

The last RHS factors out as


6



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