# The Big Cube and Squares Puzzle

The following logical puzzle is mainly inspired by a mathematical contest I went to.

# Part A

You are given a cube. How many planes cross this cube by at least 3 vertices? This number will be named $$n_{A}$$.

# Part B

You are given a cube. Here is one of its faces. $$n_{B}$$ will be its surface area.

# Part C

You are given a 5x5x5 cube with a 25 blank squares grid on each of its 6 faces. Here is an example: A move on one of its face consists of switching colors of 3 squares in a row — horizontally or vertically — from white to black or from black to white.

$$n_{C}$$ is the minimum number of moves needed so half of your cube faces is a checkerboard — like in the next diagram — and with the constraint that the total number of black squares in your cube is greater than $$n_A+\dfrac{n_{B}}{2}$$ Puzzle is to compute $$n_C$$

• Sorry I'm not a native speaker, what does 'plans cross this cube by at least 3 vertices' mean? Apr 4 '20 at 8:09
• @newbie, I'm not a native too, writing to you from France :) I tried to ask how many different plans cross, or, "touch" this cube, by exactly 3 vertices. I recall there are 8 vertices in a cube :)
– JKHA
Apr 4 '20 at 8:14
• Also, in part C, does a move only affect the colors of squares in that one face (so we can treat a cube as 6 different non-interacting faces)? Apr 4 '20 at 8:14
• Oh ok, I would call that planes, not plans :) @JKHA Apr 4 '20 at 8:15
• @newbie, yes, each face is independent and a move is for one face only :)
– JKHA
Apr 4 '20 at 8:15

Part A:

$$n_A=6+6+8=20$$.
Six faces, six other axis-paralleling ones, eight non-axis-paralleling ones. Part B:

$$NL^2=(NT+UL)^2+UT^2=50$$.
Surface area of one face: $$S=NL^2/2=25$$.
$$n_B=6S=150$$.

Part C:

We'll first focus on making a face checkerboard-like. Notice that whenever we make a move, the parity of black cells will be changed, therefore we need to make even number of moves.
An lowerbound of number of moves is $$6$$. This is because for a move of length $$3$$, you can 'cover' at most $$2$$ black squares and there're $$12$$ of them.
However, $$6$$ is not achieveable, because we can show by contradiction that there must be at least one 'move' that is not touching other 'move's, so that can't be a valid move (the middle cell will be black instead of white).
On the other hand, $$8$$ moves are enough. Enough for the three checkboard faces. $$n_A+n_B/2=20+75=95$$, so the total number of black squares must be at least $$96$$ and the total number of black squares in three faces of checkboards is $$36$$. For the other three faces, there must be $$60$$ black cells. We can make $$3$$ black cells in one move, so at least we need $$20$$ moves. Not surprisingly, it's indeed achieveable. Therefore, $$n_C=3\times 8+20=44$$.

• Quite a fast and correct answer :)
– JKHA
Apr 4 '20 at 9:30
• $n_C$ is $44$ which is quite linked to squares :)
– JKHA
Apr 4 '20 at 9:30