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2

This is probably not the minimum, but it is an improvement to the answer from @Sriotchilism O'Zaic: with the sizes In 9 parts: In 8 parts: In 7 parts: In 6 parts: In 5 parts: I used the $18$ pieces from the linked puzzle as my starting point and


-2

The problem might be similar to this problem: Given a multiset M of rational numbers where the sum of M equals 1, this is "n-distributable" for an n n∈N, if there exists a partition M1⊔...⊔Mn of M, that the sum of each subset Xi equals 1/n. We would like to find for a fixed k (k∈N) the minimal possible cardinality of a (a1,a2,..,ak) distributable multiset. ...


10

First guess:


0

An attempt: Let us call $a$ the number of parts the cake will be divided into. $a$ must be divisible by $ 2, 3, 4, 5, 6, 7, 8\,$and$\,9$. The above set of divisors can be reduced to $5, 7, 2^{3}$ and $9$. So $a$ must be of the form $$a=k\times 5\times7\times8\times 9$$ where $k$ is a natural number. What is the value of $k$ for which $a$ is minimum? ...


22

(Kind of) analytical solution that only requires small amount of calculations, (potentially) doable by hand. First step: we can safely drop 2, 3 and 7 from the equation as those digits are used in 23 and 17. Now, we need to build a prime from: 5, 11, 13, 17, 19, 23, 29 and 31. Second step: let's try to build the shortest number possible from these numbers. ...


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