A cube painted green on all faces

Contents • 1 Problem • 2 Solution 1 • 3 Solution 2 • 4 Solution 3 (Similar to solution 1) • 5 See also Problem Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color? Solution 1 (Alcumus solution) If the orientation of the cube is fixed, there are possible arrangements of colors on the faces. There are arrangements in which all six faces are the same color and arrangements in which exactly five faces have the same color. In each of these cases the cube can be placed so that the four vertical faces have the same color. The only other suitable arrangements have four faces of one color, with the other color on a pair of opposing faces. Since there are three pairs of opposing faces, there are such arrangements. The total number of suitable arrangements is therefore , and the probability is Solution 2 Label the six sides of the cube by numbers to as on a classic dice. Then the "four vertical faces" can be: , , or . Let be the set of colorings where are all of the same color, similarly let and be the sets of good colorings for the other two sets of faces. There are possible colorings, and there are good colorings. Thus the result is . We need to compute . Using the Clearly , as we have two possibilities for the common color of the four vertical faces, and two possibili...

Six faces of identical cubes are painted 6 distinct colors. How many different cubes can be formed? The answer is surprisingly small but that may be because I don't know how to express the different ways properly. I just incorrectly did $6!$ because if there are 6 ways to color the first face, there are now 5 colors left so 5 ways to color the second face. And so on. Why is this wrong? Perhaps I need to make use of "identical" and "distinct" from the question. If the cubes are identical, I must divide the answer by something, perhaps $6!$ for the 6 identical faces? And there should be some "choosing" happening but I can see where this fits in but it must be for the "distinct" part. Choose colors as in $6C1, 5C1 ...etc$ but this is the same as $6!$ Please help $\begingroup$ Ok. But why do you consider sides to be the same if the question stipulates that there are 6 distinct colors? The worked solutions say "If all the 6 sides were different it could be arranged in 6! Ways" and then process with your logic. I'm confused - what would a cube with "all 6 sides different mean"? $\endgroup$ This is perhaps a more concrete way to arrive at the answer. Say we will color the faces with the colors, $R,O,Y,G,B,V$. We have to color some face $R$. Place the cube on the table, with that face down. Now we have to color some face $O$. We can paint the top face, or one of the lateral faces. Suppose first that we paint the top face. Now we have to paint one of the lateral faces $Y$. Whicheve...

People Also Read: The number of the smaller cubes have 3 faces painted is 8 cubes. Given: A cube is painted green on all surfaces and is cut into 1000 identical cubes. What is Painted Cube Concept, Tricks, and Shortcuts – Hitbullseye Cubes on 4 edges of the unpainted side of the cube will have 1 side painted (due to the unpainted side). Therefore, total cubes with 2 sides painted= 8*10 + 4= 84 cubes. Three 2-faces painted cube on every edge; ⇒ Number of cubes with 2 face painted = 3 × 1 2 = 3 6. Nine single face painted cube on side; ⇒ Number of cubes with 1 face painted = 9 × 6 = 5 4. How to use A cube of side 4 cm is painted on all its sides. If it is sliced in 1 In each side of the larger cube, the smaller cubes In the edges will have more than one face painted. The cubes which are situated at the corners of the big cube, have three faces painted. A cube painted yellow on all faces is cut into 27 small cubes of equal sizes. Green face should be adjacent to the silver face. If a cube is painted on all of its surfaces with the same colour, and then divided into smaller cubes of equal size, then the number of smaller cubes so obtained will be calculated as follows: Number of cubes with no face painted = (n – 2) 3. Number of cubes with one face painted = 6 (n – 2) 2. Answer (1 of 7): Exactly the cubes on the edges (except the corners) have two faces colored. Each edge is subdivided in 6 pieces (since 6 cubed is 216), two of them are corners, so four are genuine edge blocks....

Hint: The given problem is easy if you have some imagination of cubes. For example, Rubik’s cube has alignment \[3 \times 3 \times 3\]. In Rubik’s cube if you count the number of cubes you will find 27 cubes of equal size. That is exactly the case in our problem. We need to find the number cubes painted on one side only. Complete step-by-step answer: Given, a cube is painted yellow on all six faces and then the cube cuts into 27 small cubes of equal part. So it has been cut into an \[3 \times 3 \times 3\] arrangement. The image that comes to mind is a Rubik’s cube. There is 1 cube in every centre (middle of 3in each axis direction i.e., longitudinal lateral and vertical) which has no colour. On each of the 6 sides of the cube, there is a central smaller cube that is painted on one face. There are 8 corners of the cube, which are painted on 3 surfaces. Remaining 12 cubes are painted on 2 sides only. Thus, only 6 sides of the cube are painted on one face. So, the correct answer is “Option B”. Note: In cube all sides are equal. 12 cubes are painted on 2 sides because on each side of the cube, there are 4 cubes at the middle of the edges. These are shared with one other side of the cube. A cube has 6 faces, 8 vertices and 12 edges. Imagine a cube and follow the explanation which is done above.

Question Download Solution PDF A cube is coloured red on one face, green on the opposite face, yellow on another face and blue on a face adjacent to the yellow face. The other two faces are left uncoloured. It is then cut into 125 smaller cubes of equal size. How many cubes are uncoloured on all the faces? Given: The pairs of opposite faces of a large cube are painted red, yellow and green . Acube is made using 125smaller cubes of equal size. N = 125 n = 5 Shortcut Trick We make the following observations: n = 5because there will be 5rows of 125pieces (small cubes) each. If a cube has been painted red, green, and yellowon pairs of opposite faces and divide it into X parts. 1) Total number of cubes = X 3 2) Cubes that have three faces painted = 8 (Because there are only 8 corners in the cube) 3) Cubes which has only two faces painted = (X - 2) × No of Edges (12) 4) Cubes that have only one face painted = (X - 2) 2× 6 5) Cubes which have no face painted = (X - 2) 3 The cubes will be uncoloured : Let the side of the bigger cube be 5 cm. So, one side of smaller cube is 1 cm. ∴ n = \(\) Number of Smaller Cubes having all faces uncoloured = Number of Smaller Cubes at 2 Surfaces + Number of Smaller Cubes at 1 edge + Number of Smaller Cubes from inside the Cube = 2× (n - 2) 2+ (n - 2)× 1 + (n - 2) 3 = 2× (5 - 2) 2+ (5 - 2)× 1 + (5 - 2) 3 = 2× 3 2+ 3 + 3 3 = 2× 9+ 3 + 27 = 18 + 3 + 27 = 48 Hence, the correct answer is "48".

Cube root of 64 is 4 therefore cube is 4 x 4 x 4. Each side of the cube has 16 visible small cube faces, 4 of which are corners (3 green sides) and 8 more are edges (2 green sides), so there are only four small cubes with one green face on each of the six sides of the large cube. Your answer is therefore 24. A board game comes with 9 white and green number cubes and there are twice as many white cubes as green cubes How many cubes of each color are there? There are 6 white cubes and 3 green cubes. Algebraically, the equations are W = white cubes, G= green cubes W+G = 9 and W = 2G so substituting we have 2G + G = 9, and 3G = 9, so G=3 and W=6.

The correct option is C 1 The cuboid can be cut into 27 smaller identical pieces by applying 2 cuts parallel to each face of the cube, hence will be having 3 layers of pieces, the front and back layer would be painted, similarly top and bottom layer, left side and right side layer would be painted, so number of layer of no face painted pieces would be (3-2) x (3-2) x (3-2) = 1 x 1 x 1 = 1 piece.

The correct option is C 1 The cuboid can be cut into 27 smaller identical pieces by applying 2 cuts parallel to each face of the cube, hence will be having 3 layers of pieces, the front and back layer would be painted, similarly top and bottom layer, left side and right side layer would be painted, so number of layer of no face painted pieces would be (3-2) x (3-2) x (3-2) = 1 x 1 x 1 = 1 piece.

Cube root of 64 is 4 therefore cube is 4 x 4 x 4. Each side of the cube has 16 visible small cube faces, 4 of which are corners (3 green sides) and 8 more are edges (2 green sides), so there are only four small cubes with one green face on each of the six sides of the large cube. Your answer is therefore 24. A board game comes with 9 white and green number cubes and there are twice as many white cubes as green cubes How many cubes of each color are there? There are 6 white cubes and 3 green cubes. Algebraically, the equations are W = white cubes, G= green cubes W+G = 9 and W = 2G so substituting we have 2G + G = 9, and 3G = 9, so G=3 and W=6.

Contents • 1 Problem • 2 Solution 1 • 3 Solution 2 • 4 Solution 3 (Similar to solution 1) • 5 See also Problem Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color? Solution 1 (Alcumus solution) If the orientation of the cube is fixed, there are possible arrangements of colors on the faces. There are arrangements in which all six faces are the same color and arrangements in which exactly five faces have the same color. In each of these cases the cube can be placed so that the four vertical faces have the same color. The only other suitable arrangements have four faces of one color, with the other color on a pair of opposing faces. Since there are three pairs of opposing faces, there are such arrangements. The total number of suitable arrangements is therefore , and the probability is Solution 2 Label the six sides of the cube by numbers to as on a classic dice. Then the "four vertical faces" can be: , , or . Let be the set of colorings where are all of the same color, similarly let and be the sets of good colorings for the other two sets of faces. There are possible colorings, and there are good colorings. Thus the result is . We need to compute . Using the Clearly , as we have two possibilities for the common color of the four vertical faces, and two possibili...