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Kangaroo Maths Contest 2010 Grades 7-8 Question #25 (1 Viewer)

TuePhuong

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A Kangaroo has a large collection of small cubes . Each cube is a single colour. Kangaroo want to use 27 small cubes to make a cube so that any two cubes with at least one common vertex are of different colours. At least how many colours have to be used?
 
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Paradoxica

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The central cube must be different to all other cubes. +1

Consider the vertex cubes. None of them touch so they can all have the same colour. +1

The Consider the center cube of each face.

Each one does not touch the opposite face, so 3 more colours are required, as anymore would not be possible, and they are all in contact with the center and 4 vertex cubes. +3

Lastly, consider the edge cubes. We can classify each one into three "rings" consisting of four cubes that do not touch each other, in a manner similar to the vertex cubes. Three colours are thus required for these three groups of four cubes. +3

1+1+3+3=8

So 8 colours minimum are required in order for the condition to be satisfied.

A lot of details are missing as this is not a proof, only based on intuition. (basically a greedy algorithm)

Perhaps you could assume you only need 7 colours and arrive at a contradiction of some sort.
 

TuePhuong

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Lastly, consider the edge cubes. We can classify each one into three "rings" consisting of four cubes that do not touch each other, in a manner similar to the vertex cubes. Three colours are thus required for these three groups of four cubes. +3
Can you explain more clearly? Thanks a lot.
 

Paradoxica

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Can you explain more clearly? Thanks a lot.
"Slice" the cube into 3 3X3 slices.

The middle slice (i.e. the one least exposed to the air) has four pieces that do not touch each other. Then they can be assigned a colour without trouble.

Repeat along the other two axes.
 

TuePhuong

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"Slice" the cube into 3 3X3 slices.

The middle slice (i.e. the one least exposed to the air) has four pieces that do not touch each other. Then they can be assigned a colour without trouble.

Repeat along the other two axes.
I can understand. Thank you so much! I colour as following
Q25.jpg
 
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RealiseNothing

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A much easier method is to notice that any 2x2x2 cube has a vertex (at the centre) which is common for all 8 cubes. So clearly you need at least 8 colours.

You could go ahead and find an 8-colouring, but for an a more satisfying proof that 8 colours is sufficient:

Start with a 2x2x2 cube where all smaller cubes are distinctly coloured. Use the following algorithm:

1) Split your entire shape in half.

2) Fix one half, and reflect the other half about the fixed half. For example, this is what it would look like for a 2x2 square:

blue green

red yellow

Step 1) Fix the RHS half, so green and yellow are fixed.

Step 2) Reflect the LHS over the RHS:

blue green blue

red yellow red

As you can see, this process allows us to colour more cubes without the need for extra colours. This works because the line we are reflecting about will always separate cubes of the same colour (in this case the fixed green/yellow is separating the blue's and red's).

So starting with your 2x2x2 cube, we can do this process to obtain a 3x3x3 cube with only 8 colours, as so:

2x2x2 ---> 2x2x3 ---> 2x3x3 --->3x3x3 (try visualise this process)
 

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