6

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


6

Let $R(n)$ be the number of codes of total width $n$, so the question asks for $R(14)$.


6

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


5

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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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 ...


3

I can do it in Basically, working from the front


3

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


3



3

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 $N$,


3

There are: Proof: Now:


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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:


1

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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