Every day from day 2, neighboring cubes (cubes with common faces) to red cubes also turn red and are numbered with the day number.
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AdventuringHobbit
23.12.2019 01:54
For b, consider the set of all lattice points (x,y,z) such that 0<=x,y,z<=9. Letting (0,0,0) have the number 1, we know in general the point (x,y,z) has number x+y+z+1. So we are looking for the number of ordered pairs (x,y,z) such that x+y+z=6. There are 8 choose 2 = 28 ordered pairs by sticks and stones.
skrublord420
23.12.2019 03:35
In general, for an $n \times n \times n$ cube, the number of cubes marked $k$ is the number of solutions to $x+y+z = k-1$ subject to $0 \le x, y, z \le n-1.$ You can use bagels and dots and donuts and stars and stripes forever to solve this.
totem
23.12.2019 15:47
A specific formula for (c):
In general I believe for even $n$, there will be two numbers that appear equally most often for $n\geq 4$: $\{n+\frac{n-4}{2}+1,n+\frac{n-4}{2}+2\}$ both having frequency $n+2\sum_{k=\frac{n}{2}+1}^{n-1} k$ and thus I propose numbers $3026$ and $3027$ occur $3054243$ times although I've not rigorously confirmed this but rather looking at the six cube below and working out the pattern and then checking it in code for larger $n$ although I suspect that's not allowed in a contest. Was a fun problem to work on!
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Math5000
23.12.2019 16:33
Anything, can you correct me, please! Thanks for the solutions!
An n × n × n cube
Consider the day number $i$.
If $i\le n$, then the number of times it appears is $1+2+...+i=\frac{i(i+1)}{2}$.
If $2n-1\le i\le 3n-2$, then (symmetry) it appears the same number of times as the number $3n-1-i$.
If $n< i<2n-1$, then the number of times it appears is $$(i+1-n)+(i-n)+...(n-1)+n+(n-1)+...+(2n-i).$$
Note that it is the third of these formulae which will give the largest number of appearances and this will be when $n$ is as close as possible to being in the middle of the summation.Then $n-(i+1-n)\approx n-(2n-i)$ i.e. $2i+1\approx 3n$.
$7$ appears $\frac{7\times8}{2}=28$ times.
For $13$, we need $4+...+9+10+9+...+7=73$.
$2i+1\approx 3\times 2018$ when $i=3026$ and $3027$.