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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.

enter image description here

$n_{B}$ 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:

enter image description here

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 $n_A+\dfrac{n_{B}}{2}$

enter image description here

Puzzle is to compute $n_C$

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  • $\begingroup$ Sorry I'm not a native speaker, what does 'plans cross this cube by at least 3 vertices' mean? $\endgroup$ – newbie Apr 4 at 8:09
  • $\begingroup$ @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 :) $\endgroup$ – JKHA Apr 4 at 8:14
  • $\begingroup$ 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)? $\endgroup$ – newbie Apr 4 at 8:14
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    $\begingroup$ Oh ok, I would call that planes, not plans :) @JKHA $\endgroup$ – newbie Apr 4 at 8:15
  • $\begingroup$ @newbie, yes, each face is independent and a move is for one face only :) $\endgroup$ – JKHA Apr 4 at 8:15
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Part A:

$n_A=6+6+8=20$.
Six faces, six other axis-paralleling ones, eight non-axis-paralleling ones.
ugly drawing

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.
such a configuration
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. final part Therefore, $n_C=3\times 8+20=44$.

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

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