Color Cubed (July '24 Challenge Question)

You can paint the 1st side 3 colors, the second 3 colors, ...or a total of 3⁶ = 729. I am assuming that, for example, if you have 1 red side that it is different if it is on the top, bottom, front, back, left, or right side.

Suppose you can rotate the cube so you can have duplicates...

You can paint side #1, 3 different colors. Now you have 5 sides left which you can paint any one of 3 colors. Then you have 4 sides left one of which you can paint a color, and so on.

Number of combinations: 6x3 for first side, 5x3 for second side, 4x3 for 3rd side, ..., 3 choices for the 6th side, 18 x 15 x 12 x 9 x 6 x 3 = 3 x 6! = 3 x 720 = 2160 combinations.

Now you have to sort out the duplicates that are rotations. The cube can be rotated about 3 perpendicular axes in 4 90 degree positions, one of six sides facing up in 4 positions = 4 x 6 = 24 combinations. 2160/24 = 90 unique painted cubes.

You have 3 colors and white, that's 4 possibilities, and 6 sides and 36 combination possibilities...4*6*36 = 864

The question states that you have to paint all sides, so you're only dealing with three colours.

53

If the cube was embossed with the numbers 1 to 6 like a dice, so that the numbers are still visible after painting then the answer would clearly be 3 raised to the power six 729.

But when you remove the numbers trying to eliminate all the rotations which are similar seems impossible; so, perhaps it's better to approach this from the point of view of one colour, say: red; avoid duplicates (hopefully) then multiply by three at the end.

All red____________________________________________________________________1

Five red plus 1 blue or 1 yellow________________________________________________2

How many different ways can you arrange 4 red around a cube: easier to look at

the other two faces. They can be either opposite or adjacent: so

Four red plus: 2 blue or; 2 yellow or; 1 blue and 1 yellow = 2 x 3 ______________________6

How many different ways can you arrange 3 red around a cube: start with two opposite,

the third must form a saddle together with the complementary saddle of other colours;

or, start with two adjacent, the third must form a saddle as above, or, a "complete"

corner together with the "complimentary" corner of other colours: so,

saddle with bby, byb, byy, yby, or, corner with bby or byy {PLUS symmetrical **)_________6

----

That's 15 x 3 = 45 so far____________________________________________________45

Plus symmetrical ** 3 red saddle plus bbb or yyy, OR, 3 red corner plus bbb or yyy_______4

Note that we don't need to consider 2 red with more than 2 of any other colour

because we've already taken care of those above: so,

red opposite red with blue opposite blue and yellow opposite yellow__________________1

r opposite r with bbyy, b opposite b with rryy y opposite y with rrbb___________________3

Total 53

Please tell me if I've missed something.

I hope that formatting come out on other screens: it looks like this on mine:-

Bingo, bango, bongo.

57 is right! And thanks for explaining how to get there!

I can paint with only one color, or two colors, or all three colors.

With one color, there is 1 pattern and 3 colors, for 1x3=3 choices.

For the rest, lets label the sides, starting with the first one painted (F for front). The others are L (left), R (right), U (up), D (down), or O (opposite).

With two colors I can either paint 1,5 sides or 2,4 sides or 3,3 sides.

For 1,5 sides all combinations are rotations so with one pattern and 3x2 color choices I have 1x3x2=6 choices, to total 9 so far.

For 2,4 sides and two colors I can paint two sides with one and the other 4 sides with the other. If the first color's second side is L, R, U, or D, all possible rotations are equal, so I have one pattern. If the second side is O, again I have only one pattern. Therefore with painting two sides I have two patterns with 3x2 color combinations. This gives 2x3x2=12 choices, with a total of 21 so far.

For 3,3 sides and two colors the first color can be F, U, D or F, L, R, giving only one pattern because all others are rotations. If the first color is on sides that share a corner, other possible patterns are rotations. Therefore for 3,3 sides painted we have a total of 2 patterns and 3x2 color combinations we add 2x3x2=12 choices, to total 33 so far.

If I paint using all three colors, I can paint 1,1,4 or 2,3,4 or 2,2,2 sides with the colors.

For 1,1,4 the second color can either be on any adjacent side (L, U, D, or R), for which all are rotations giving 1 pattern for the third color, If on the opposite side (O), again all rotations give 1 pattern for the third color. So we have add (1+1)x3=6 choices, to total of 39 so far.

For 1,2,3 if the second color is on two adjacent sides these are either adjacent to each other, with rotations giving 1 pattern (L,U becomes U,R or R,D or D,L); of if opposite each other rotations allow only 1 pattern (L,R becomes U,D). If the second color is on opposite and adjacent sides, then there is only one pattern. The third color occupies the remaining sides. We have 3 colors for the single side, 2 colors for the two sides, and a total of 3 possible patterns, this is 3x2x3=18 possible choices, with a total of 57 so far.

For 2,2,2 sides painted, if all colors are on opposite sides we have only 1 choice (total is 58 so far) since all are rotations without mirror images.

If the first color is on opposite sides then the other two are on adjacent sides with all patterns being the same via rotations and no mirror images. Therefore with a color choice for the opposite sides we have 3x1=3 choices, and a total of 61.

If the first color is on the front and an adjacent side (such as F, L), and the second color is on two opposite sides, we revert back to the above with no new patterns. If the second color is on adjacent sides (R & D or R & U), the third can only be O and the remaining adjacent side (D or U). All possible patterns are rotations but with mirror images. This gives 2 more patterns, and our total is 63 choices.

--JMM

GA.

But

"For 3,3 sides and two colors the first color can be F, U, D or F, L, R, giving only one pattern because all others are rotations. If the first color is on sides that share a corner, other possible patterns are rotations. Therefore for 3,3 sides painted we have a total of 2 patterns and 3x2 color combinations we add 2x3x2=12 choices, to total 33 so far."

You are effectively claiming that there are 6 ways of choosing two from three colours.

(This kind of works if there is a way to distinguish between the two colours: e.g. for a 1/5 or 2/4 split; but for a 3/3 split both patterns: two "saddles" or two "whole corners" are symmetrical. So there are only two patterns times three choices of colours.)

If we reduce your total by 6, and, increase mine by 4 as above: we both get 57.

Perhaps Heinz should start repackaging their products in coloured cubes.

Good catch. Thanks for the correction. Heinz it is.

--JMM

OK, I agree. I put it together in a spreadsheet, separated into groups which have the 28 combinations of colored sides, e.g., "1,2,3" is the group with 1 red side, 2 yellow sides, and 3 blue sides, etc. Zero indicates color not used.

I came up with the Heinz number

Total = 57

I think this is correct, but I need to make some actual pictures to verify, especially the 3 color cubes.

Corrected table:

..."If you were to paint each face of this cube in a single color, how many different variations could you paint on this cube?".... = 0

Ha !

To make semantic fun of the wording, I'd suggest infinity possibilities with vast amounts of paint. Thin coat of one colour on day one, another coating (any colour) day 2 etc. Separate coatings are not exactly mixing. Then there's the issue of can we paint the inside of the cube.

"If you were to paint each face of this cube in a single color, how many different variations could you paint on this cube?".... = 0"

Wouldn't it be 3?

If each face(6 faces) are painted in a single color(all faces the same color) there would be 3 possible options for the 3 colors of paint

<Disclaimer - other sauces are available.>

There is no time limit stated in the original posting,so wait for all colors to dry then repaint over them,so the answer is :INFINITE

K.I.S.S.!