Quiz#3 NOT(Rubik's Cube)

[drsarao]

A cube is painted Red, Green and Blue with opposing faces having same colour. The cube is now sawed with 3 cuts in each plane resulting in 64 equal cubes.

How many of following would you find?

Cubes with at-least TWO sides of different color.
Cubes with paint on ONE face only.
Cubes without ANY paint.
Cubes with ONE side without paint.
Cubes with THREE painted surfaces.
Cubes with RED and BLUE paints only.

Bonus question:
In the original uncut cube, how many SQUARES can you count?

Quiz#1 The-age-of-three-daughters

Quiz#2 Johnny-come-early

 

[BiocideJ]

Here is what I came up with, although I have some reservations on the accuracy of some of my responses (as noted)

Cubes with at-least TWO sides of different color: 32 (2*12 edges on the opposite sides + 4*2 non-corner edges on each of the adjacent sides [24+8])
Cubes with paint on ONE face only: 24 (the 4 interior non-edge cubes from all 6 sides [4x6])
Cubes without ANY paint: 8 (the 2x2x2 cube in the center)
Cubes with ONE side without paint: 64 (all of the cubes would have at least one side without paint so perhaps I don't understand the question)
Cubes with THREE painted surfaces: 8 (the corners of the original cube [1*8])
Cubes with RED and BLUE paints only: 8 (the non-corner edges of the 2 common sides [2+2+2+2])

Bonus question:
In the original uncut cube, how many SQUARES can you count?
I would think 6 (each of the sides of the cube), however, that seems too simplistic so perhaps I don't understand the question.

 

[drsarao]

Biocide,
You are right on all counts.

Two questions were wrongly worded by me. (Hence the ambiguity). Should be read as:
Cubes with exactly ONE side without paint.
In the original uncut cube (after the cut lines are marked), how many SQUARES can you count?

(I typed instead of copy-pasting to make it more difficult for a google-solution! But then Lost in Translation)

 

[BiocideJ]

Cubes with exactly ONE side without paint: 0 (at most a cube could have 3 sides painted if it was a corner of the original cube)
In the original uncut cube (after the cut lines are marked), how many SQUARES can you count?
I believe it would be 126. (16:1x1 squares + 9:2x2 squares + 4:3x3 squares + 1:4x4 square = 21 squares on a face * 6 faces to a cube = 126.

 

[Joe4]

I have moved this quiz to the Lounge. Please post all the quizzes there.
(With less traffic in the Lounge too, they will stay up on the front page longer)

 

[drsarao]

Thanks Joe!

I believe it would be 126. (16:1x1 squares + 9:2x2 squares + 4:3x3 squares + 1:4x4 square = 21 squares on a face * 6 faces to a cube = 126.

Biocide,
I can't believe, you were done in by simple sum in the end! The answer is 30*6=180.

I created a general formula using SUMPRODUCT() with segments for chessboard filled-in: (This is for ONE face only)
(Not at all elegant and limited to SUMPRODUCT() range)
Next iteration will be better.
Anybody else give it a try?
For other conditions also?

Excel Workbook
	A	B	C	D	E	F	G	H	I	J
1
2	Segments on one side	8	7	6	5	4	3	2	1	0
3
4	Total Nos of Squares	204
Sheet12

Excel 2003

Cell Formulas
Range	Formula
C2	=IF(B2>1,B2-1,0)
D2	=IF(C2>1,C2-1,0)
E2	=IF(D2>1,D2-1,0)
F2	=IF(E2>1,E2-1,0)
G2	=IF(F2>1,F2-1,0)
H2	=IF(G2>1,G2-1,0)
I2	=IF(H2>1,H2-1,0)
J2	=IF(I2>1,I2-1,0)
B4	=SUMPRODUCT(B2:J2,B2:J2)

 

[BiocideJ]

Oops... Not exactly sure how I did that. especially since I had the right numbers in my explanation... 16+9+4+1 = 30 like you said.
I think there is a life lesson here. When you focus so hard on the difficult parts you are prone to overlook the obvious, simple things that are staring you right in the face.

 

[drsarao]

A better formula for 'squares within squares':

Excel Workbook
	A	B
1	Segments on one side	200
2	Total Nos of Squares	2686700
Sheet12

Excel 2003

Cell Formulas
Range	Formula
B2	=SUMPRODUCT(ROW(INDIRECT("$A$1:$A$" & B1)),ROW(INDIRECT("$A$1:$A$" & B1)))

 

[taurean]

Array Entered:
=SUM(POWER(ROW($A$1:INDEX(A:A,B1)),2))
OR
=SUM(ROW($A$1:INDEX(A:A,B1))^2)

 

[drsarao]

Taurean,
Crisp!
I could also shorten my formula based on it:

=SUMPRODUCT(ROW(INDIRECT("$A$1:$A$" & B1))^2)

(I had thought of it. But there was a mental block to putting only one argument in SUMPRODUCT()!)

Here is the complete generalized Excel solution. (I think it is correct. Submitted for peer review.)

Excel Workbook
	A	B
1	CUBE Quiz
2	No of cuts	2
3	No of segments	3
4	Total resulting cubes	27
5
6	Atleast 2 different colours	20
7	Exactly 2 different colours	8
8	Exactly 2 specific colours	4
9	Atleast 3 different colours	8
10	Only one face painted	6
11	No paint at all	1
12	Total Nos of Squares	84
Sheet12

Excel 2003

Cell Formulas
Range	Formula
B3	=B2+1
B4	=B3^3
B6	=B4-B11-B10
B7	=(B3-2)*8
B8	=(B3-2)*4
B9	=1*8
B10	=(B3-2)*(B3-2)*6
B11	=(B3-2)^3
B12	=SUMPRODUCT(ROW(INDIRECT("$A$1:$A$" & B3))^2)*6