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Simplest answer is


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The sequence is: and we want the coefficient of $x^{10}$. The sequence equals: which is the; We therefore want:


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The number of ways to achieve $10$ ounces is: To make it general: So: Note that:


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Generalizing my comment on Gareth's solution, we can arrange Pascal's triangle as a right triangular array and ignore the right half ($n < 2k$) to obtain something like this: 1 1 1 2 1 3 1 4 6 ... We then, for any $N$,


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Gareth has found the optimal solutions, but here is an R script if anyone wants to mess around with the upper bounds for n, just change the value of the variable"UpperBound". require(pracma) UpperBound<-500 n<-rep(1:UpperBound,each=UpperBound) k<-rep(1:UpperBound,times=UpperBound) data<-as.data.frame(cbind(n,k)) colnames(data)<-c("n","k") ...


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Choosing gets to within about of the desired answer. I think this is best possible with <= 100 cards. Found with the help of a computer, but purely as an aid to calculation. My approach was to [EDITED to add:] Out of curiosity, I also ran a more automated search for the larger bound of n=500 mentioned in the OP. For this, The automated search also ...


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An initial lower Bound Finding a better lower bound Doing a bit better with some brute force The code used : https://pastebin.com/8nQ9UgBP Notes about the code:


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Edit: as @DarkThunder pointed out, this is incorrect. My most perfect soup contains


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since they are 1/64th increments, you can be left with a minimum of 1/64th of a cup. Add 1 cup, remove 1/2 cup, remove 1/4 cup, remove 1/8 cup, remove 1/16 cup, remove 1/32 cup, remove 1/64 cup, there is 1/64th of a cup left.


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An upper bound This can surely be improved substantially because Another (smaller) upper-bound Applying a bit of brute force, A smaller upper bound still And smaller still


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0k I'll start with my intuition which probably is not the best one, but you can outdo me now ;) . Idea: Java code: Result: Note: Other idea:


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Let’s start with notation: I just tried to


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I can do it in Basically, working from the front


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