The Lakeside School, Seattle, Washington

From: "Leslie Robertson" (leslie_robertson@morris.lakeside.sea.wa.us) We have decided that for the variable 'n' units to the side of a cube, there will be 8 cubes with three surfaces, 12n-24 cubes with 2 surfaces, (n-2)squared *6 cubes with 1 surface, and (n-2)cubed cubes with no surfaces covered with paint. This was gathered from the properties of a cube. It always has 8 corners(3 sides). To know how many contain no sides that are painted you must realize that they are one layer deep from all sides. So you subtract 2 from it to account for the one on either side of it and then cube it because we need to calculate the 3 demensions. To calculate the number of cubes with 2 sides painted you must realize again that they cannot be the ones on the corners but must be the ones on the edges. So you subtract 2 from the length of an edge and then multiply it by 12(the number of edges). Next, you can easily calculate the amount of 1 paint sided ones by knowing that they must be seen, and cannot be on a edge or corner of the whole cube. Thus we know that you must take one off for both edges in any dimention. So, in conclusion, you can subtract 2 from the length in units for one side, square that and then multiply it by 6 because that is how many sides there are to a cube. (TA DA!)

Germantown Academy, Fort Washington, Pennsylvania

From: "Kandee Linden" (Kandee_Linden@msn.com) If a cube is n units long and painted on all sides, then cut into 1 unit cubes, how many cubes have paint on various sides? We're obviously looking for some equations here. So, we know that no matter how big the cube is, only eight of the 1 unit cubes will have paint on three sides because they are the corners and each cube has exactly the same number of them. Now, 12(n-2) cubes will have paint on two sides. We are saying that you take 2 less than however many 1 u cubes are in one row of the total cube (to exclude the corners) and times that by the twelve edges that every cube has(only the edges contain cubes with two sides painted). For only 1 side painted, you need the number of unit cubes on the outside of the larger one excluding the edges and corners. So, you need (n-2) squared, and that gets the surface area of one side with no corners or edges. You also need to times that by six so you have the total paint-on-one-side area on all six faces of the larger cube. So, it's 6{(n-2)(n-2)} Now, if you want to know how many don't have any paint at all, you need to find the area of the cube inside the larger cube. The equation is (n-2) - that excludes the entire outside - cubed. That is, (n-2)(n-2)(n-2).

Germantown Academy, Fort Washington, Pennsylvania

From: ruth@mathforum.org (Ruth Carver) The first thing that I did in solving this problem was to draw a cube with sides labeled "N". By drawing a picture it really help understand the puzzle. First, you painted a cube on all sides and then cut it into smaller one unit cubes. Therefore, there will always be eight cubes who have three surfaces painted because these are the corners and there are always eight corners in any cube. Second, I know that all the blocks located on the edges of the cube have two surfaces painted. To find the number, just subtract two from "N" and multiply it by twelve. There are twelve edges to any cube and when you subtract the corner block you will get the remainder blocks that are located on that edge. Next, to calculate the blocks who have only one surface painted (these would be all the leftover blocks on the outside of the cube), look at one side of cube. If you multiply "N" minus two times "N" minus two you will get all the blocks with one surface painted. So to find all the blocks with one surface painted multiply that by six (there are six sides on a cube). Finally, the blocks with no surfaces painted would be all the blocks located within the cube. It is a cube within a cube. The formula for a cube is height times length time width. So, the solution to find the inside cube would be "N" minus two times "N" minus two times "N" minus two. 3 surfaces- 8 (the corners!) 2 surfaces- 12(N-2) 1 surface- 6(N-2)(N-2) no surfaces- (N-2) (N-2) (N-2)

Germantown Academy, Fort Washington, Pennsylvania

From: MJfan11463@aol.com Here is my answer: I figured out that for a),3 sides painted, for every cube that you make,the answer is always 8 because there are 8 corners on every cue and there are only 3 corners painted on the corners of the cube and the number of corners is a constant. For b),2 sides painted, I figured out that the euation for this is 12(n-2), n being the number of units the cube is. I figured this out because I knew that there were 12 sections on each cube where the edges meet and I subtracted out the corners because we know that all of the corners are painted 3 on three sides. There are only 12 sections because 4 of the sections are overlapping with eachother. So I took n-2 and multiplied it by 12 so get the number of cubes that were painted on 2 sides. For c), 1 side painted, I came up with the equation, {(n-2)^2} that is n minus 2 to the second power. N again being the number of units the cube is. Again, I subtracted out the corners and there is an exact suare in the middle of the cube itself. You square this because it is an exact square and there are 6 faces to each cube so you multiply your answer by 6. For d), 0 faces painted, I came up with the equation (n-2)^3, that is n minus 2 to the third power. N being the number of units the cube is. You know that this has to be the interior of the cube because all of the exterior cubes are painted on at least one side. You have to subtract 2 because the cube inside is 1 unit in on both sides of the cube. You have to cube the answer because the interior cubes is their own, unique cube, just like the full cube of the outside.

Germantown Academy, Fort Washington, Pennsylvania

From: ruth@mathforum.org (Ruth Carver) The problem states that there is a cube with n sides long that is painted on all sides and that it is then cut into cubes with sides 1 unit long. I started out by making a chart with the titles of each column being: cube size, 3 surfaces, 2 surfaces, 1 surface, and no surfaces. When I started to fill the chart in I used numbers in order to find a pattern, which I did. In the end I used the variable n. So, under the cube size I used a 2x2x2 up to a 10x10x10. Then next to that I put what those numbers equal if you multiply them out. For 2x2x2 it was 8 and for 10x10x10 it was 1000. Using n you could say nxnxn or n^3. Now I had to do the three surfaces. I have allergies so I have a tissue box sitting next to my computer and I was staring right at it when I did this problem. My tissue box helped me with this. The only little cubes that could have 3 surfaces with paint on them can be the corners of the cube, in my case, tissue box. This is true because the corners have surfaces that help to make up 3 surfaces of the big cube itself. I then counted the corners on my tissue box and came up with 8. So 8 it is for every cube. You could give me any cube and ask how many of the little cubes that make it up have 3 surfaces with paint on them and it would always be 8. There's not exactly a method for finding a cube with 3 surfaces of paint on them it's just always 8. For 2 surfaces I'm going to use a 5x5x5to explain it. Say you are looking for 2 surface owith paint on them, the logical place to look then is on the edge of the cube because that is the only place on the cube where a little cube shares 2 surfaces with the big cube. On that edge there are 2 little cubes that have been used, the top one and the bottom one. They were used under the 3 surfaces column. So if your cube is a 5x5x5 but you have to take 2 away from each edge. A 3x3x3 would not be correct that would be finding the area of a 3x3x3. You have to count the number of edges on the cube not repeting any which is 12 for all cubes as well and multiply that by the 5-2 which is 3. A method using n for finding a cube with 2 surfaces of paint on would be: (n-2)x12. Now I have 1 surface to do. First I counted how many faces were on the cube which is 6 and is always 6 for any cube. Then I figured out that the only surfaces with 1 surface of pait would be the little cubes inside the edge. Not the cubes inside the outer layer of cubes. If you were looking for cubes with 1 surface of paint on them at just one surface of the cube it would be (n-2)(n-2). Using the 5x5x5 cube it would be (5-2)(5-2)which is 9. But that is just one face and in the beginning I counted 6 faces so you have to multiply that number by 6 giving you 54 little cubes with 1 surface of paint on them for the 5x5x5 cube. So a method for finding a cube with 1 surface of paint on it would be (n-2)(n-2)x6. Now we have to find a method for finding the number of cubes with no surfaces that have paint. using a 5x5x5 cube whose area equals 125 you know that there are 8 cubes with 3 surfaces that have paint, 36 cubes with 2 surfaces of paint and 54 cubes with 1 surface ofd paint. So you could just say that 125-8-36-54 equals the number of subes with no surfaces of paint which is 27. f you add up the cubes with 3 surfaces, 2 surfaces, 1 surface and no surfaces you should get the area of the big cube . You could do that if you wanted to check yourself. Using n you could find the number of cubes with no surfaces of paint on them by doing (n^3)-(8)-{(n-2)x12}-{(n-2)x6}. When i first read this problem it sounded rather hard but in the end it was fun and not too dificult either . the only hard thing was explaining to my mom why my tissue box had lines and squares all over it !!

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