A bunch of unit cubes are put together to form a larger cube. Some of the faces of the larger cube are painted. Then the large cube is taken apart and 24 of the small cubes have no paint on them. How big was the large cube and which of its faces were painted?
Annie says: Solutions
This is a really nice problem to reason out. The best example is provided by Justin Lam from Sequoia Middle School. His solution is very clear and easy to read and exactly right.
I'm not going to talk about how to work out the problem, because you only need to read the five solutions by Justin Lam, Becky Dunlap, Katie Quinn-Kerins, Peter Liang, and Ken Duisenberg to see how it might be done.
A list of all the people who got this problem right and most of the solutions are also available.
Justin Lam Grade: 8 School: Sequoia Middle School, Pleasant Hill, California 24 = 12x1x1 = 8x3x1 = 6x4x1 = 4x3x2 are the only possible factorings of 24 into 3 factors. All these factorings can be viewed as how these 24 small cubes form a rectangular solid. For example, 8x3x1 can be viewed as a rectangular solid with sides 8, 3, and 1. The problem can be restated as: how to add layers of small cubes to this solid to form a perfect large cube. Since these layers of cubes must be painted on at least one side, we cannot add more than two layers of small cubes in each dimension (one layer on each side). Therefore,the first 3 factorings (combinations) can be ruled out right away. Let's look at the last combination - 4x3x2. We can make a perfect large cube of 4x4x4 out of this rectangular solid by: (1) Adding two layers of small cubes of 4x3=12 on each side of the rectangular solid. That would increase the side 2 to side 4. The resulting solid will have dimension of 4x3x4. (2) Adding another layer of small cubes of 4x4=16. This would make the resulting solid from (1) into a perfect 4x4x4 cube. So, the total number of painted cubes from this large cube is (4x3)+(4x3)+(4x4) = 40. And, in fact, 64-40=24 unpainted small cubes. Another way to look at this problem is to consider the possible ways we can paint the side of a 4x4x4 cube. |TOTAL # | # OF PAINTED | # OF UNPAINTED | POSSIBILITIES |OF CUBES| CUBES | CUBES | ------------------------|--------|--------------|----------------| | | | | Painted ONE side | 64 | 4(4)=16 | 64-16=48 | | | | | ------------------------|--------|--------------|----------------| | | | | Painted 2 adjacent sides| 64 | 4(4)+3(4)=28 | 64-28=36 | | | | | ------------------------|--------|--------------|----------------| | | | | Painted 2 opposite sides| 64 | 4(4)+4(4)=32 | 64-32=23 | | | | | ------------------------|--------|--------------|----------------| | | | | Painted 3 sides - | 64 |4(4)+3(4)+3(4)| | 2 adjacent & 1 opposite| | = 40 | 64-40=24 *** | | | | | ------------------------|--------|--------------|----------------| | | | | Painted 3 sides - | 64 |4(4)+3(4)+3(3)| | 3 adjacent sides with | | = 37 | 64-37=29 | a common point | | | | | | | | ------------------------|--------|--------------|----------------| Therefore, *** represents the solution which is equivalent to the solution from above.
From: ruth@mathforum.org Becky Dunlap Grade: 9 School: Germantown Academy, Fort Washington, Pennsylvania First, I located unit cubes that I could use to try to figure out the answer. Then, with the cubes, I made a 3-by-3-by-3 cube. I right away realized that the large cube couldn't be a 3-by-3-by-3 because if only one side is painted, the minimum, then there are only 18 cubes have no paint on them, which is already too few. Then I made a 5-by-5-by-5, but that didn't work either because if every side is painted than there were 27 cubes in the middle that weren't painted, which is too many. Now that I knew that only a 4-by-4-by-4 would work, I had to determine which sides were painted. I "painted" one side, and 48 cubes were left unpainted. I painted a side next to it, and 36 were left unpainted. I painted a side next to both of them, and there were 27 unpainted. When I painted a fourth side, there were 18 left unpainted, which is too few. I realized that I would need to try painting a different combination of sides. I "unpainted" the last two sides and painted the side opposite the first side, so I had one cube painted with two sides painted on either side of it that weren't next to each other. When I did that, there were 24 left unpainted
Katie Quinn-Kerins Grade: 10 School: Germantown Academy, Fort Washington, Pennsylvania Mrs. Ruth Carver's Geometry Honors Class Possible number of small cubes to make one large cube: 1^3 = 1 cube (large) 2^3 = 8 cubes 3^3 = 27 cubes 4^3 = 64 cubes 5^3 = 125 cubes 2 units^3 = 8 small cubes. This is too small because the problem states that there will be 24 small cubes left. 3 units^3 = 27 small cubes. This is too small also because if one side is painted 9 of the small cubes will have paint on them leaving 18 unpainted small cubes. 4units^3 = 64 smaller cubes. - If any one side or face is painted, 16 small cubes will show paint leaving 48 unpainted small cubes. - If the opposite side or face is also painted, an additional 16 small cubes will show paint leaving 32 unpainted small cubes and 32 painted small cubes. - If ANY third side or face is painted, 16 small cubes will get paint on them, but 8 of those 16 already have paint on them. This would therefore add an additional 8 small cubes to the painted small cubes giving us 40 small painted cubes and 24 small cubes with no paint. In conclusion: the large cube was 64 cubic units. Two opposite faces and any third face were painted so that when the large cube was taken apart, 24 of the smaller cubes would have no paint on them while 40 of the small cubes would have paint on at least one side. Actually, of the cubes that had paint on them, 8 of them had paint on two faces and 32 small cubes had paint on only one face.
Peter Liang Grade: 10 School: Newport High School, Bellevue, Washington Mr. Art Mabbott The large cube is formed by 64 unit cubes. Three sides must be painted in order for the problem to work. I didn't find an equation for this problem, but relied more on the trial and error method of reasoning. The large cube must fit several rules. First, its total number of unit cubes must exceed 24. Also, when the length of a side is raised to the third power and then subtracted by 24, it must be divisible (whole numbers) by the length of the side and must be greater than the length of a side squared. Finally, the difference of the side cubed and 24 must have a quotient less than 6 when divided by the side squared. This is because a cube has 6 faces. So, the length of any side of the 64 unit cube is 4 units. This becomes our variable, x. x^3 minus 24 equals 40. x goes into 40 and 40 is greater than x^2. Plus, 40 divided by x^2 equals 2.5, less than 6. Let's say 2 non-adjacent or non-consecutive faces are painted in the large cube. The number of unit cubes not painted becomes 64-32=32. Then paint any one of the four remaining faces. Since 8 cubes of this face are shared by the previous two faces, the number of new cubes painted is really just 8. It's easy to see that 32-8=24, the number of cubes not painted.
Ken Duisenberg Grade: Hewlett Packard Engineer School: Stanford University '93 Answer: 4 This is similar to the problem used at SMSU Problem Corner 4/8/96: http://science.smsu.edu/~les/POTW.html No.Sides w/paint SmallCubesPainted SmallCubesNotPainted N(for 24) 0 0 N^3 2.8845 1 N^2 N^3 - N^2 3.26 2(adjacent) 2*N^2 - N N^3 - 2*N^2 + N 3.6 2(opposite) 2*N^2 N^3 - 2*N^2 3.7 3(corner) 3*N^2 - 3N + 1 N^3 - 3*N^2 + 3N - 1 3.86 3(U) 3*N^2 - 2N N^3 - 3*N^2 + 2N 4 4(adjacent) 4*N^2 - 5N + 2 N^3 - 4*N^2 + 5N - 2 4.25 4(circle) 4*N^2 - 4N N^3 - 4*N^2 + 4N 4.35 5 5*N^2 - 8N + 4 N^3 - 5*N^2 + 8N - 4 4.6 6 6*N^2 - 12N + 8 (N-2)^3 4.8845 The only configuration leading to N as an integer is three sides painted, not adjacent around a vertex, but shaped like a U around the cube. For this case, the side length is 4, for a total of 4^3 = 64 unit cubes.
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