28. A large cube has an edge of length 7 cm. On each of its six faces, the two diagonals are drawn in red. The large cube is then cut into small cubes with edges of length of 1 cm. How many small cubes will have at least one red line drawn on it? Consider the face of the cube as a 7x7 grid, and the slashes and backslashes are the two diagonals that are drawn in red. How many cubes have this line through them? We have 13 cubes. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. Each diagonal has 7 cubes, but this one is double-counted. Maybe we can set up a formula to get rid of the double-counted ones later. Which ones are the ones that are counted multiple times? Well, these ones that are in the center of the faces are counted multiple times. They're counted twice, one for each diagonal. What else is counted multiple times? Well, the vertices of the cube are actually counted multiple times. This vertex, this cube, has one line here, but if we also imagine the top face of the cube, then that face also has a diagonal which marks up this smaller cube over here. So this small cube will have one line on this face. It'll also have one line on this face, but it'll also have a third line on this face that's directly to the left. So this small cube, if we do it this way, will be triple-counted. And so how many of these small cubes are there? Well, the small cubes are just the vertices of the big cube. So the vertices of the big cube are triple-counted, and there's one cube per face of the cube is double-counted. So the length of our diagonals is seven, and then we have two diagonals on each face, and we have six faces of a cube. That tells us seven times two times six. Now we have to remove the double-counted ones. Now we remove one for each double-counted one because we want to keep the one that we want at least, we want to count at least once. How many double-counted ones are there? Well, there's one per face in the middle of each face. So there are six faces of a cube minus one times six. How many triple-counted cubes are there? Well, there's a triple-counted cube for every vertex in the big cube. So there's eight vertices of a cube, one, two, three, four. Then if we extend the cube out, we'll see one, two, three, four. If you can visualize the back face of the cube over here. And we're going to remove two of those counted times for this, for these vertex cube because they're triple-counted. We only want one of them to remain. So we have seven times two times six, which is 84, minus six, minus 16. And that gives us an answer of 62 small cubes with at least one red line drawn on it. So that gives us to our answer, which is B.

cube

diagonals

overlaps

calculation

small cubes