PDQ 4/30By Ryan Smolenski

Problem Statement

In this problem, we have a cube that measures N by N. The outside of the cube is painted a different

color than the inside, so that the surface area is easily distinguishable (if the cube were to be taken

apart). Our job was to divide the cube into even sections so that the length, width, and height of each

smaller cube was 1. Depending on what value N was assigned, we were to find out how many cubes

had exactly 6, 5, 4, 3, 2, or 1 painted side(s), or at least 6, 5, 4, 3, 2, or 1 painted side(s).

Process

To start off, I drew four cubes to give me an idea of what the situation would look like. They were a

1x1x1, 2x2x2, 3x3x3, and 4x4x4. I noticed that the first two cubes had unique properties compared to

the following ones, so I decided to take note. All the sides on the 1x1x1 were coated, so I didn’t need to

evaluate further on that. All the subcubes on the 2x2x2 had three sides coated on them (8), so I didn’t

need to evaluate on that. However, the 3x3x3 and subsequent cubes had a pattern for all types of sides.

Solution

Assuming you are asking for the exact amount of sides painted on a

subcube

Let’s start with the 1x1x1. This cannot be broken down any further, so there are no subcubes, and all

the sides were coated in paint. That means no sides had only 5, 4, 3, 2, or 1 side coated. Let’s move on

to the rest of the cubes (2x2x2, 3x3x3...NxNxN). Since the whole cube is...a cube, none of the sides will

ever have 6, 5, or 4 sides coated on them. This is true for all whole numbers N greater than 1.

Let’s imagine we have a whole cube now (a 3x3x3 or greater). When the subcubes are divided, they

can only be in three positions: In the direct middle of a side (where only one side of the subcube is

painted), in the corners of a side (where three sides of the subcube are painted), and the middle of an

edge of a side (where two sides of the subcube are painted). Anywhere else would lead to petrusion,

which I assume does not happen. Since there are multiple sides, this assumption must hold true for all

six sides.

Okay, now for some patterns. Let’s start with the subcubes in the middle of a side. As a cube grows in

incriments of one, that mean there will be one additional row, column, and depth of subcubes. If we

start with a 2x2x2, there are zero subcubes in the direct middle, where only one side is painted. In a

3x3x3, there is one side like this from a two dimensional standpoint, and for a 4x4x4, there are four

sides like this. This pattern continues and looks like this: 0, 1, 4, 9, 16, 25... We notice this is a pattern

of squares. There are six sides on a cube, so a formula to find the number is [(N - 2)^2] x 6, where N is

no less than 2.

Now we will look at the corner of a cube, where three sides of a subcube are painted. On a cube, there

can only be four corners, so there are always four subcubes painted with three sides, as long as N is

more than 1.

Last, we have the subcubes that are in a middle of an edge that are only painted on two sides. I pic-

tured this in an odd way, but try to visualise this. You have a cube of NxNxN. You divided itup into

subcubes, but you still retained the original shape. Now, take divide them into sections with a length of

N and a height of N, and a width of one. You should have N amounts of those in total. Look at a 3x3x3,

if it’s easier. It’s like cutting a Rubik’s cube into three equal parts, 3x3x1, where there are 3 groups.

Starting with a 3x3x3, I will list the amount of subcubes that are found for each cube: 12, 24, 36, 48,

60... These are just multiples of 12, no? Each time you add a 1x1x1 to a whole cube, you add 12 more

to the edges that are only painted twice. The formula for this is equal to (N - 2) x 12, where N is no less

than 2.

There you have it. All of these formulas work on numbers larger than two. While a 1x1x1 and 2x2x2 are

unique because they don’t have any sides that have two sides coated on a subcube, the following sizes

do, and that’s why they follow the formulas. You can try to figure out on your own if you want to, if you

have a big enough Rubik’s cube...

5/8/13 C_pop Ryan Smolenski

1 of 3Printed by Wolfram Mathematica Student Edition

Let’s start with the 1x1x1. This cannot be broken down any further, so there are no subcubes, and all

the sides were coated in paint. That means no sides had only 5, 4, 3, 2, or 1 side coated. Let’s move on

to the rest of the cubes (2x2x2, 3x3x3...NxNxN). Since the whole cube is...a cube, none of the sides will

ever have 6, 5, or 4 sides coated on them. This is true for all whole numbers N greater than 1.

Let’s imagine we have a whole cube now (a 3x3x3 or greater). When the subcubes are divided, they

can only be in three positions: In the direct middle of a side (where only one side of the subcube is

painted), in the corners of a side (where three sides of the subcube are painted), and the middle of an

edge of a side (where two sides of the subcube are painted). Anywhere else would lead to petrusion,

which I assume does not happen. Since there are multiple sides, this assumption must hold true for all

six sides.

Okay, now for some patterns. Let’s start with the subcubes in the middle of a side. As a cube grows in

incriments of one, that mean there will be one additional row, column, and depth of subcubes. If we

start with a 2x2x2, there are zero subcubes in the direct middle, where only one side is painted. In a

3x3x3, there is one side like this from a two dimensional standpoint, and for a 4x4x4, there are four

sides like this. This pattern continues and looks like this: 0, 1, 4, 9, 16, 25... We notice this is a pattern

of squares. There are six sides on a cube, so a formula to find the number is [(N - 2)^2] x 6, where N is

no less than 2.

Now we will look at the corner of a cube, where three sides of a subcube are painted. On a cube, there

can only be four corners, so there are always four subcubes painted with three sides, as long as N is

more than 1.

Last, we have the subcubes that are in a middle of an edge that are only painted on two sides. I pic-

tured this in an odd way, but try to visualise this. You have a cube of NxNxN. You divided itup into

subcubes, but you still retained the original shape. Now, take divide them into sections with a length of

N and a height of N, and a width of one. You should have N amounts of those in total. Look at a 3x3x3,

if it’s easier. It’s like cutting a Rubik’s cube into three equal parts, 3x3x1, where there are 3 groups.

Starting with a 3x3x3, I will list the amount of subcubes that are found for each cube: 12, 24, 36, 48,

60... These are just multiples of 12, no? Each time you add a 1x1x1 to a whole cube, you add 12 more

to the edges that are only painted twice. The formula for this is equal to (N - 2) x 12, where N is no less

than 2.

There you have it. All of these formulas work on numbers larger than two. While a 1x1x1 and 2x2x2 are

unique because they don’t have any sides that have two sides coated on a subcube, the following sizes

do, and that’s why they follow the formulas. You can try to figure out on your own if you want to, if you

have a big enough Rubik’s cube...

Assuming I’m finding the sides with at least a certain amount of sides

painted

Here, I assumed that the subcubes had exactly the number of painted sides asked for. If I were to solve

for the total amount, as long as there are at least ___ amount of painted sides, I would simply add my

results. The results for the corners (three subsides painted) will always be eight, since there aren’t any

greater number of sides painted to that.

Now for the subcubes that have two sides painted on them, the formula to the exact amount would be

used, but the final result would involve adding the ones with three sides painted on them. This would

make it (N - 2) x 12 + 8, or (N - 2) x 122

3.

To find the grand total of subcubes with at least one painted side, we would add our results for just the

ones with three and just the ones with two, or we could add all the ones with at least two. The formula

to find solely the ones with one side remains true, so the equation would be: ([(N - 2)^2] x 6) + ((N - 2) x

12) + 8, or ([(N - 2)^2] x 6) + ((N - 2) x 122

3). The 1x1x1 has all one cube that has at least one side

painted on it, so the answer would be one for that question. The 2x2x2 would have subsides with 1-3

painted be eight. The others follow a strange pattern. The very center of the cube is not being painted.

A formula can be found to find those unpainted sides, based on this. I’ll start with the number in a

3x3x3. 1, 4, 9, 16, 25... Another squares signal! Since it is a cube, you can also use this formula to find

the number of subcubes with at least one pained side: (N^3) - (N - 2)^2. You take the total amount of

subcubes and subtract the amount of unpainted subcubes.

5/8/13 C_pop Ryan Smolenski

2 of 3Printed by Wolfram Mathematica Student Edition

Here, I assumed that the subcubes had exactly the number of painted sides asked for. If I were to solve

for the total amount, as long as there are at least ___ amount of painted sides, I would simply add my

results. The results for the corners (three subsides painted) will always be eight, since there aren’t any

greater number of sides painted to that.

Now for the subcubes that have two sides painted on them, the formula to the exact amount would be

used, but the final result would involve adding the ones with three sides painted on them. This would

make it (N - 2) x 12 + 8, or (N - 2) x 122

3.

To find the grand total of subcubes with at least one painted side, we would add our results for just the

ones with three and just the ones with two, or we could add all the ones with at least two. The formula

to find solely the ones with one side remains true, so the equation would be: ([(N - 2)^2] x 6) + ((N - 2) x

12) + 8, or ([(N - 2)^2] x 6) + ((N - 2) x 122

3). The 1x1x1 has all one cube that has at least one side

painted on it, so the answer would be one for that question. The 2x2x2 would have subsides with 1-3

painted be eight. The others follow a strange pattern. The very center of the cube is not being painted.

A formula can be found to find those unpainted sides, based on this. I’ll start with the number in a

3x3x3. 1, 4, 9, 16, 25... Another squares signal! Since it is a cube, you can also use this formula to find

the number of subcubes with at least one pained side: (N^3) - (N - 2)^2. You take the total amount of

subcubes and subtract the amount of unpainted subcubes.

Generalizations

The variable N represents any whole positive number.

I assumed that all of the squares were attached to each other in the form of a cube when the cube was

being painted. All of the suface area of the cube was painted and nothing inside the cube was painted.

N is a whole number, so there are no subcubes that have a length of 1/2, for example. Also, The 1x1x1

would not work with any of the formulas, since it only had one square to begin with. Last, the cube could

be any positive whole number, so it is not limited in size. None of the subcubes can be petruding out in

any way when the cube is dipped in, and it everything must be in the shape of a cube.

I also assumed that there could be two possible outcumes to the answer, since it was not specified

whether or not cubes already accounted for can be repeated.

Self-Assessment

Overall, I think I did an okay job with this Mathematica PDQ. To be honest, I’m not sure if I covered all

the generalizations, but that was all I could think of. I tried to be creative with my answers, and I tried

my best to elaborate on them. I covered all the requirements, but I did draw a few diagrams of 3x3x3

and 4x4x4 Rubik’s cubes to help me out with my thinking. I did not include them in Mathematica.

Joe Mielinski did help me with the problem, but I didn’t really ask for his help. He gave me ideas how to

find the formulas for the exact amount of cubes, but didn’t tell me the formulas. I did everything else on

my own.

Overall, I have learned that there can be multiple correct answers to large math problems, and that

there are multiple methods to find those answers. I just worked at the most logical way I could think of,

and I achieved my goals because of that. I feel that my work was diligent and not last minute, like how

some of my older assignments were.

5/8/13 C_pop Ryan Smolenski

3 of 3Printed by Wolfram Mathematica Student Edition