answer：First, I need to understand the problem. We have two sequences: the Heavenly Stems with 10 characters and the Earthly Branches with 12 characters. They are arranged in a cyclic manner, and I need to find the column number where "Jia" and "Zi" appear together for the second time.To solve this, I'll determine the cycle at which both sequences align again. The Heavenly Stems repeat every 10 columns, and the Earthly Branches repeat every 12 columns. The least common multiple (LCM) of 10 and 12 is 60, meaning the sequences will realign at column 60.Since "Jia Zi" first appears at column 1, the second occurrence will be at column 60 + 1 = 61.Therefore, the column number where "Jia" and "Zi" are in the same column for the second time is 61.

answer：Alright, so I came across this interesting geometry problem today. It goes like this: An astronomical projector illuminates an octant, which is a three-dimensional angle where all plane angles are right angles. The projector is placed at the center of a cube. The question is, can it be rotated in such a way that it doesn't illuminate any vertices of the cube?Hmm, okay. Let me try to visualize this. A cube has eight vertices, right? And the projector is at the center, so it's equidistant from all the vertices. An octant is like one-eighth of the space around a point, bounded by three mutually perpendicular planes. So, if the projector is shining an octant, it's illuminating a region that's one-eighth of the entire space around the center.Now, the cube has its vertices at the corners, each one unit away from the center if we consider a unit cube. The projector is shining an octant, so it's illuminating a region that's like a three-dimensional right angle. The question is whether we can rotate this octant such that none of the cube's vertices fall within the illuminated region.Let me think about the cube's structure. Each vertex can be represented by coordinates (±1, ±1, ±1) if we consider the cube centered at the origin with side length 2. The center is at (0,0,0). The projector is shining an octant, which can be thought of as the region where x, y, z are all positive, or any combination of signs depending on the orientation.But wait, the projector can be rotated. So, the octant doesn't have to align with the cube's edges. It can be oriented in any direction. The key is whether we can rotate the octant such that none of the eight vertices lie within it.Let me consider the cube's vertices. Each vertex is at a corner, so each has coordinates with all components either +1 or -1. The octant, when aligned with the coordinate axes, would contain the vertex (1,1,1). But if we rotate the octant, maybe we can avoid all these vertices.But how does rotation affect the octant? Rotating the octant would change the direction in which it's pointing, but the shape remains the same—it's still a three-dimensional right angle. The challenge is to position this octant such that none of the cube's vertices lie inside it.I think it might help to consider the cube's diagonals. The space diagonal of the cube goes from (1,1,1) to (-1,-1,-1), passing through the center. If the projector is shining along this diagonal, would it illuminate any vertices? Well, the octant is a three-dimensional angle, not a line, so it's not just along the diagonal but in a region around it.Wait, maybe I should think about the angles involved. The octant is 90 degrees in each plane, so it's quite a large region. But the cube's vertices are all at the maximum distance from the center in each axis. So, if the projector is shining an octant, it's illuminating a region that's 1/8th of the sphere around the center.But the cube's vertices are all on the sphere of radius sqrt(3) (since the distance from the center to a vertex is sqrt(1^2 + 1^2 + 1^2) = sqrt(3)). So, the projector is shining an octant on this sphere. The question is whether we can rotate the octant such that none of the eight points (the vertices) lie within the illuminated region.Hmm, another approach: perhaps considering the cube's dual, the octahedron. But I'm not sure if that's helpful here.Wait, maybe I can think about the cube's vertices in terms of their position relative to the octant. Each vertex is at a corner, so each is in a different octant if we consider the standard coordinate system. But the projector's octant can be rotated, so it's not fixed to the coordinate axes.I think the key is whether the octant can be oriented such that it doesn't contain any of the cube's vertices. Since the cube has eight vertices, each in a different octant, but the projector's octant is just one octant. So, if we can rotate the projector's octant such that it doesn't align with any of the cube's octants, maybe it can avoid all vertices.But wait, the cube's octants are fixed relative to the cube. If the projector's octant is rotated, it might still overlap with some of the cube's octants. Hmm, this is getting a bit confusing.Let me try to think in terms of vectors. Each vertex can be represented as a vector from the center. The projector's octant can be thought of as the set of vectors where each coordinate is either positive or negative, depending on the octant. But if we rotate the octant, the signs of the coordinates change accordingly.Wait, maybe it's helpful to consider the angles between the projector's octant and the cube's vertices. If the projector's octant is rotated such that none of the vertices lie within it, that means each vertex must lie in one of the other seven octants relative to the projector's orientation.But is that possible? Since the cube has eight vertices, each in a different octant, if the projector's octant is rotated, it might still contain one of the vertices. Or maybe not, if the rotation is such that the octant is positioned between the vertices.Wait, another thought: the cube's vertices are all at the corners, so they are all at 45 degrees from the center along each axis. If the projector's octant is rotated such that its boundaries are not aligned with these 45-degree points, maybe it can avoid illuminating any vertices.But how exactly? Let me try to formalize this. Suppose we represent the projector's octant as a region in 3D space. The octant is defined by three planes intersecting at the center, each at 90 degrees to the others. If we rotate these planes, the octant will rotate accordingly.The cube's vertices are at (±1, ±1, ±1). For the projector's octant not to illuminate any vertex, each vertex must lie outside the octant. That means, for each vertex, at least one of the coordinates in the rotated coordinate system must be negative.Wait, but if we rotate the coordinate system, the signs of the coordinates change. So, if we can rotate the octant such that for every vertex, at least one of the coordinates in the rotated system is negative, then the vertex is not in the octant.But is this possible? Let me think about the cube's symmetry. The cube is symmetric, so any rotation that avoids one vertex would have to avoid all others. But since the vertices are in all octants, it's tricky.Wait, maybe it's impossible because no matter how you rotate the octant, it will always align with one of the cube's octants, thereby illuminating one vertex. But I'm not sure.Alternatively, maybe you can rotate the octant so that it's positioned between the vertices, not pointing directly at any of them. Since the octant is a region, not a line, perhaps it can be oriented such that it doesn't contain any vertex.Wait, let's think about the angles. The octant is 90 degrees in each plane, so it's a fairly large region. The cube's vertices are each at 45 degrees from the center along each axis. If the projector's octant is rotated by 45 degrees, maybe it can avoid the vertices.But how? Let me try to imagine rotating the octant so that its boundaries are not aligned with the cube's edges. For example, if we rotate the octant by 45 degrees around one axis, the planes defining the octant would be at 45 degrees relative to the cube's faces.In that case, the octant would extend into regions that are not directly aligned with the cube's vertices. Maybe this way, none of the vertices would lie within the octant.But I'm not entirely sure. Let me try to do some calculations. Suppose we rotate the octant by 45 degrees around the x-axis. Then, the y and z coordinates would be transformed. The vertices of the cube would have their y and z coordinates altered accordingly.Wait, maybe it's better to use a rotation matrix. Let's consider a rotation around the x-axis by θ degrees. The rotation matrix would be:[1, 0, 0][0, cosθ, -sinθ][0, sinθ, cosθ]If we apply this rotation to the octant, which is initially aligned with the coordinate axes, the new boundaries of the octant would be defined by the rotated planes.But I'm not sure how to determine if a vertex lies within the rotated octant. Maybe I can check for each vertex whether, after rotation, all its coordinates are positive.Wait, but the octant is defined by three inequalities: x > 0, y > 0, z > 0 in the original coordinate system. After rotation, these inequalities would change.Alternatively, perhaps it's easier to consider the inverse transformation: instead of rotating the octant, we can rotate the cube and check if any vertex lies within the original octant.That might be a simpler approach. So, if we rotate the cube such that none of its vertices lie in the positive octant (x > 0, y > 0, z > 0), then the projector can be placed at the center and the octant won't illuminate any vertices.But is such a rotation possible? Let me think about the cube's vertices. Each vertex has coordinates (±1, ±1, ±1). If we can rotate the cube such that for each vertex, at least one of the coordinates becomes negative in the rotated system, then none of the vertices would lie in the positive octant.But is that possible? Since the cube is symmetric, any rotation that affects one vertex will affect others in a symmetric way. So, if we rotate the cube such that one vertex moves into the negative octant, others might still remain in the positive.Wait, maybe it's impossible because the cube's vertices are too symmetric. No matter how you rotate it, at least one vertex will end up in the positive octant.But I'm not entirely sure. Let me try to think of a specific example. Suppose we rotate the cube by 45 degrees around the x-axis. Then, the y and z coordinates of the vertices will change.For example, the vertex (1,1,1) after rotation would have y' = y cosθ - z sinθ and z' = y sinθ + z cosθ. If θ is 45 degrees, then cosθ = sinθ = √2/2 ≈ 0.707.So, y' = 1*(√2/2) - 1*(√2/2) = 0z' = 1*(√2/2) + 1*(√2/2) = √2 ≈ 1.414So, the vertex (1,1,1) becomes (1, 0, √2). So, in the rotated coordinate system, x is still positive, y is zero, and z is positive. So, this vertex is still in the positive octant (since x > 0, z > 0, and y = 0 is on the boundary). Hmm, so it's still illuminated.Wait, but the octant is defined by x > 0, y > 0, z > 0. So, if y is zero, it's on the boundary, not inside the octant. So, maybe this vertex is not illuminated.But wait, the octant is a closed set, so it includes the boundaries. So, if y is zero, it's on the boundary of the octant. So, does that count as being illuminated? The problem says "illuminate," which might mean the interior. Hmm, the problem statement is a bit ambiguous.But let's assume that the octant includes the boundaries. Then, the vertex (1,1,1) after rotation would have y' = 0, so it's on the boundary of the octant. So, it's still illuminated.But maybe if we rotate it by a different angle, we can make sure that all vertices have at least one coordinate negative.Wait, let's try a different approach. Let's consider the cube's vertices in terms of their angles from the center. Each vertex is at a point where all coordinates are either +1 or -1, so their direction vectors are (±1, ±1, ±1). The angle between any two vertices is the same, due to the cube's symmetry.If we can rotate the octant such that none of these direction vectors lie within the octant, then we're good. But is that possible?Wait, the octant is a region that's 1/8th of the sphere. The cube has eight vertices, each in a different octant. So, if the projector's octant is rotated, it might still contain one of the cube's vertices.But wait, the cube's vertices are in the standard octants, but the projector's octant can be rotated to any orientation. So, maybe it's possible to rotate the projector's octant such that it doesn't align with any of the cube's octants, thereby not containing any vertices.But I'm not sure. Let me think about the angles again. The octant is 90 degrees in each plane, so it's quite a large region. The cube's vertices are each at 45 degrees from the center along each axis. So, if the projector's octant is rotated by 45 degrees, maybe it can avoid the vertices.Wait, let's consider a specific rotation. Suppose we rotate the octant by 45 degrees around the x-axis. Then, the y and z coordinates are transformed. The vertex (1,1,1) becomes (1, 0, √2), as before. The vertex (1,1,-1) becomes (1, -√2, 0). The vertex (1,-1,1) becomes (1, -√2, 0). Wait, no, let me recalculate.Actually, the rotation matrix around the x-axis by θ is:x' = xy' = y cosθ - z sinθz' = y sinθ + z cosθSo, for θ = 45 degrees, cosθ = sinθ = √2/2.So, for vertex (1,1,1):x' = 1y' = 1*(√2/2) - 1*(√2/2) = 0z' = 1*(√2/2) + 1*(√2/2) = √2So, (1,0,√2). So, x' > 0, z' > 0, y' = 0. So, it's on the boundary of the octant.Similarly, for vertex (1,1,-1):x' = 1y' = 1*(√2/2) - (-1)*(√2/2) = √2z' = 1*(√2/2) + (-1)*(√2/2) = 0So, (1, √2, 0). Again, x' > 0, y' > 0, z' = 0. So, on the boundary.Similarly, for vertex (1,-1,1):x' = 1y' = (-1)*(√2/2) - 1*(√2/2) = -√2z' = (-1)*(√2/2) + 1*(√2/2) = 0So, (1, -√2, 0). Here, x' > 0, y' < 0, z' = 0. So, this vertex is not in the positive octant.Wait, so in this case, some vertices are on the boundary, and some are outside. But the problem is whether the projector illuminates any vertices. If the octant includes the boundaries, then the vertices on the boundary are illuminated. If not, then maybe not.But the problem statement says "illuminate an octant," which is typically considered a closed set, including the boundaries. So, if any vertex lies on the boundary, it's illuminated.But in this rotation, some vertices are on the boundary, so they are illuminated. So, this rotation doesn't solve the problem.Hmm, maybe a different rotation. What if we rotate the octant by some angle other than 45 degrees? Let's say we rotate it by θ degrees around the x-axis, and choose θ such that none of the vertices lie in the positive octant.Wait, but how can we ensure that? Each vertex has coordinates (±1, ±1, ±1). After rotation, their coordinates become (x', y', z') where:x' = xy' = y cosθ - z sinθz' = y sinθ + z cosθWe need to choose θ such that for all vertices, at least one of x', y', z' is negative.But since x is either +1 or -1, if we rotate around the x-axis, x' remains the same. So, for vertices with x = 1, x' = 1, which is positive. For vertices with x = -1, x' = -1, which is negative.So, for the vertices with x = 1, we need to ensure that either y' or z' is negative. Similarly, for vertices with x = -1, since x' is already negative, they are not in the positive octant.So, the problem reduces to ensuring that for all vertices with x = 1, either y' or z' is negative.Let's consider a vertex with x = 1, y = 1, z = 1. After rotation, y' = cosθ - sinθ, z' = sinθ + cosθ.We need either y' < 0 or z' < 0.Similarly, for vertex (1,1,-1): y' = cosθ + sinθ, z' = sinθ - cosθ.We need either y' < 0 or z' < 0.For vertex (1,-1,1): y' = -cosθ - sinθ, z' = -sinθ + cosθ.We need either y' < 0 or z' < 0.For vertex (1,-1,-1): y' = -cosθ + sinθ, z' = -sinθ - cosθ.We need either y' < 0 or z' < 0.So, let's analyze each case.1. Vertex (1,1,1): y' = cosθ - sinθ z' = sinθ + cosθ We need y' < 0 or z' < 0. Let's see when y' < 0: cosθ - sinθ < 0 => cosθ < sinθ => tanθ > 1 => θ > 45 degrees. When θ > 45 degrees, y' < 0. When θ < 45 degrees, y' > 0. Similarly, z' = sinθ + cosθ. Since sinθ and cosθ are both positive for θ between 0 and 90 degrees, z' is always positive in this range. So, for θ > 45 degrees, y' < 0, so this vertex is not in the positive octant.2. Vertex (1,1,-1): y' = cosθ + sinθ z' = sinθ - cosθ We need y' < 0 or z' < 0. y' = cosθ + sinθ is always positive for θ between 0 and 90 degrees, since both cosθ and sinθ are positive. z' = sinθ - cosθ < 0 when sinθ < cosθ => tanθ < 1 => θ < 45 degrees. So, for θ < 45 degrees, z' < 0, so this vertex is not in the positive octant.3. Vertex (1,-1,1): y' = -cosθ - sinθ z' = -sinθ + cosθ We need y' < 0 or z' < 0. y' = - (cosθ + sinθ) is always negative for θ between 0 and 90 degrees, since cosθ + sinθ > 0. So, y' < 0, so this vertex is not in the positive octant regardless of θ.4. Vertex (1,-1,-1): y' = -cosθ + sinθ z' = -sinθ - cosθ We need y' < 0 or z' < 0. z' = - (sinθ + cosθ) is always negative for θ between 0 and 90 degrees. So, z' < 0, so this vertex is not in the positive octant regardless of θ.So, summarizing:- For vertex (1,1,1): Not in positive octant if θ > 45 degrees.- For vertex (1,1,-1): Not in positive octant if θ < 45 degrees.- For vertices (1,-1,1) and (1,-1,-1): Always not in positive octant.So, if we choose θ = 45 degrees, then:- For vertex (1,1,1): y' = 0, z' = √2 > 0. So, it's on the boundary.- For vertex (1,1,-1): y' = √2 > 0, z' = 0. So, it's on the boundary.- The other two vertices are already outside.But if we choose θ slightly more than 45 degrees, say θ = 46 degrees, then:- For vertex (1,1,1): y' < 0, so it's not in the positive octant.- For vertex (1,1,-1): z' = sinθ - cosθ. Let's compute sin46 ≈ 0.7193, cos46 ≈ 0.6947. So, z' ≈ 0.7193 - 0.6947 ≈ 0.0246 > 0. So, z' > 0, and y' = cos46 + sin46 ≈ 0.6947 + 0.7193 ≈ 1.414 > 0. So, both y' and z' are positive, so this vertex is in the positive octant.Wait, that's a problem. So, if θ > 45 degrees, vertex (1,1,1) is not in the positive octant, but vertex (1,1,-1) is in the positive octant.Similarly, if θ < 45 degrees, vertex (1,1,-1) is not in the positive octant, but vertex (1,1,1) is in the positive octant.So, it seems that no matter what θ we choose, either vertex (1,1,1) or (1,1,-1) will be in the positive octant.Therefore, it's impossible to rotate the octant around the x-axis such that none of the vertices are illuminated.But wait, maybe we can rotate around a different axis or combine rotations. Let's consider rotating around another axis, say the y-axis or z-axis, or even a combination of axes.Alternatively, maybe we can rotate the octant such that it's not aligned with any of the cube's edges, but positioned in a way that all vertices are outside.Wait, another idea: if we rotate the octant by 45 degrees around the line x=y=z, which is the space diagonal. This might spread the octant's boundaries in such a way that none of the vertices are inside.But I'm not sure. Let me try to think about this.The space diagonal is the line from (1,1,1) to (-1,-1,-1). Rotating around this axis might change the orientation of the octant such that it's not aligned with any of the cube's vertices.But I'm not sure how to calculate this. Maybe it's too complex for my current understanding.Alternatively, perhaps the answer is no, it's not possible to rotate the octant such that it doesn't illuminate any vertices. Because no matter how you rotate it, due to the cube's symmetry, at least one vertex will end up in the octant.But wait, earlier when I considered rotating around the x-axis, I found that depending on the angle, either (1,1,1) or (1,1,-1) would be in the positive octant. So, maybe it's impossible.But I'm not entirely certain. Maybe there's a way to rotate the octant such that none of the vertices are in it.Wait, another approach: consider the cube's vertices as points on the unit sphere (scaled by sqrt(3)). The octant is a region on the sphere. The question is whether we can rotate the octant such that none of the eight points lie within it.But the sphere is divided into eight octants, each containing one vertex. So, if we rotate the octant, it's still an octant, just in a different orientation. But the cube's vertices are still in their respective octants. So, if the projector's octant is rotated, it might overlap with one of the cube's octants, thereby illuminating a vertex.Wait, but the cube's octants are fixed relative to the cube. If the projector's octant is rotated, it's relative to the cube's coordinate system. So, maybe it's possible to rotate the projector's octant such that it doesn't align with any of the cube's octants, thereby not containing any vertices.But I'm not sure. It seems like due to the cube's symmetry, any octant would align with one of the cube's octants, thereby illuminating a vertex.Wait, but the cube's octants are just the standard ones, but the projector's octant can be rotated arbitrarily. So, maybe it's possible to position the projector's octant such that it's between the cube's octants, not containing any vertices.But how? Let me think about the angles again. The octant is 90 degrees in each plane. The cube's vertices are at 45 degrees from the center along each axis. So, if the projector's octant is rotated by 45 degrees, maybe it can avoid the vertices.Wait, but earlier when I tried rotating around the x-axis by 45 degrees, the vertex (1,1,1) ended up on the boundary, which is still illuminated if the octant includes boundaries.But if we rotate it by a little more than 45 degrees, maybe the vertex (1,1,1) moves outside the octant, but then another vertex moves inside.Alternatively, maybe we can rotate the octant such that all vertices are outside. But I'm not sure.Wait, another idea: the octant is a convex region. The cube's vertices are all on the sphere. If we can find a rotation such that the octant is entirely contained within a hemisphere that doesn't contain any vertices, then it's possible.But the cube's vertices are spread out, so any hemisphere would contain at least one vertex.Wait, no. A hemisphere is half of the sphere, so it contains four vertices. But the octant is 1/8th of the sphere. So, maybe it's possible to position the octant such that it doesn't contain any of the eight vertices.But I'm not sure. It's getting a bit too abstract.Wait, maybe I can think about the cube's dual, the octahedron. The octahedron has faces that correspond to the cube's vertices. If the projector's octant is rotated such that it doesn't align with any of the octahedron's faces, maybe it can avoid the vertices.But I'm not sure if that helps.Alternatively, maybe the answer is yes, it's possible. Because the octant can be rotated such that it's positioned between the cube's vertices, not pointing directly at any of them.But I'm not entirely certain. I think I need to look for a more rigorous approach.Let me consider the cube's vertices in terms of their position relative to the octant. Each vertex is at a point where all coordinates are ±1. The octant is defined by three inequalities, say x > 0, y > 0, z > 0 in some rotated coordinate system.If we can find a rotation such that for all vertices, at least one of the coordinates in the rotated system is negative, then none of the vertices will be in the octant.But is that possible? Let's consider the cube's vertices. Each vertex has coordinates (±1, ±1, ±1). If we can rotate the coordinate system such that for each vertex, at least one coordinate becomes negative, then we're done.But how? Let's think about the rotation as a change of basis. We need a new basis where for each vertex, at least one of the new coordinates is negative.But since the cube is symmetric, any rotation that affects one vertex will affect others in a symmetric way. So, if we rotate the coordinate system such that one vertex has a negative coordinate, others might still have positive coordinates.Wait, maybe it's impossible because the cube's vertices are too symmetric. No matter how you rotate the coordinate system, at least one vertex will have all coordinates positive in the new system.But I'm not sure. Let me try to think of a specific example. Suppose we rotate the coordinate system by 45 degrees around the x-axis, as before. Then, the vertex (1,1,1) becomes (1, 0, √2), which is on the boundary. The vertex (1,1,-1) becomes (1, √2, 0), which is on the boundary. The vertex (1,-1,1) becomes (1, -√2, 0), which is outside. The vertex (1,-1,-1) becomes (1, -√2, -√2), which is outside.So, in this case, two vertices are on the boundary, and two are outside. But if the octant includes the boundaries, then those two vertices are illuminated. So, this rotation doesn't solve the problem.Wait, but what if we rotate by a different angle? Let's say we rotate by θ degrees around the x-axis, and choose θ such that for all vertices with x = 1, either y' or z' is negative.From earlier analysis, for vertex (1,1,1), y' = cosθ - sinθ. For this to be negative, θ > 45 degrees.For vertex (1,1,-1), z' = sinθ - cosθ. For this to be negative, θ < 45 degrees.So, it's impossible to choose θ such that both y' < 0 and z' < 0 for these vertices. Therefore, no matter what θ we choose, either (1,1,1) or (1,1,-1) will have both y' and z' positive, thereby being in the positive octant.Therefore, it's impossible to rotate the octant around the x-axis such that none of the vertices are illuminated.But what if we rotate around a different axis, say the y-axis or z-axis, or a combination? Maybe that would help.Wait, let's try rotating around the y-axis. The rotation matrix around the y-axis by θ is:[cosθ, 0, sinθ][0, 1, 0][-sinθ, 0, cosθ]So, for a vertex (1,1,1), after rotation:x' = cosθ + sinθy' = 1z' = -sinθ + cosθWe need x' < 0 or z' < 0.x' = cosθ + sinθ. For θ between 0 and 90 degrees, cosθ + sinθ is always positive. So, x' > 0.z' = -sinθ + cosθ. For θ < 45 degrees, cosθ > sinθ, so z' > 0. For θ > 45 degrees, cosθ < sinθ, so z' < 0.So, for θ > 45 degrees, z' < 0, so vertex (1,1,1) is not in the positive octant.But what about vertex (1,1,-1):x' = cosθ + (-1)*sinθy' = 1z' = sinθ + (-1)*cosθWe need x' < 0 or z' < 0.x' = cosθ - sinθ. For θ > 45 degrees, cosθ < sinθ, so x' < 0.z' = sinθ - cosθ. For θ > 45 degrees, sinθ > cosθ, so z' > 0.So, for θ > 45 degrees, x' < 0, so vertex (1,1,-1) is not in the positive octant.Similarly, vertex (1,-1,1):x' = cosθ + sinθy' = -1z' = -sinθ + cosθSince y' = -1 < 0, this vertex is not in the positive octant.Vertex (1,-1,-1):x' = cosθ - sinθy' = -1z' = sinθ - cosθAgain, y' = -1 < 0, so not in the positive octant.So, for θ > 45 degrees, vertices (1,1,1) and (1,1,-1) are not in the positive octant, and the other two vertices are already outside.Wait, does that mean that by rotating around the y-axis by more than 45 degrees, we can make sure that none of the vertices are in the positive octant?Wait, let's check vertex (1,1,1):x' = cosθ + sinθy' = 1z' = -sinθ + cosθFor θ > 45 degrees, z' < 0, so vertex (1,1,1) is not in the positive octant.Vertex (1,1,-1):x' = cosθ - sinθy' = 1z' = sinθ - cosθFor θ > 45 degrees, x' = cosθ - sinθ < 0, so vertex (1,1,-1) is not in the positive octant.The other two vertices have y' = -1, so they are not in the positive octant.So, in this case, by rotating the octant around the y-axis by more than 45 degrees, we can make sure that none of the vertices are in the positive octant.Wait, that seems promising. So, if we rotate the octant by, say, 60 degrees around the y-axis, then both (1,1,1) and (1,1,-1) would have either x' < 0 or z' < 0, thus not being in the positive octant.But let me verify with θ = 60 degrees.cos60 = 0.5, sin60 ≈ 0.866.For vertex (1,1,1):x' = 0.5 + 0.866 ≈ 1.366 > 0y' = 1z' = -0.866 + 0.5 ≈ -0.366 < 0So, z' < 0, so vertex is not in positive octant.For vertex (1,1,-1):x' = 0.5 - 0.866 ≈ -0.366 < 0y' = 1z' = 0.866 - 0.5 ≈ 0.366 > 0So, x' < 0, so vertex is not in positive octant.For vertex (1,-1,1):x' = 0.5 + 0.866 ≈ 1.366 > 0y' = -1 < 0z' = -0.866 + 0.5 ≈ -0.366 < 0So, y' < 0, so vertex is not in positive octant.For vertex (1,-1,-1):x' = 0.5 - 0.866 ≈ -0.366 < 0y' = -1 < 0z' = 0.866 - 0.5 ≈ 0.366 > 0So, x' < 0, so vertex is not in positive octant.Wait, so in this case, none of the vertices are in the positive octant. So, the projector can be rotated around the y-axis by more than 45 degrees, say 60 degrees, and then the octant would not illuminate any vertices.But wait, is this correct? Because earlier when I rotated around the x-axis, it didn't work, but rotating around the y-axis seems to work.Wait, but the cube is symmetric, so rotating around any axis should be similar. But in this case, rotating around the y-axis by more than 45 degrees seems to avoid illuminating any vertices.But wait, let me check another vertex. What about vertex (-1,1,1)?After rotation around the y-axis by 60 degrees:x' = (-1)*cosθ + 1*sinθ = -0.5 + 0.866 ≈ 0.366 > 0y' = 1z' = (-1)*sinθ + 1*cosθ = -0.866 + 0.5 ≈ -0.366 < 0So, z' < 0, so vertex (-1,1,1) is not in the positive octant.Similarly, vertex (-1,1,-1):x' = (-1)*cosθ + (-1)*sinθ = -0.5 - 0.866 ≈ -1.366 < 0y' = 1z' = (-1)*sinθ + (-1)*cosθ = -0.866 - 0.5 ≈ -1.366 < 0So, x' < 0, so vertex is not in positive octant.Vertex (-1,-1,1):x' = (-1)*cosθ + 1*sinθ = -0.5 + 0.866 ≈ 0.366 > 0y' = -1 < 0z' = (-1)*sinθ + 1*cosθ = -0.866 + 0.5 ≈ -0.366 < 0So, y' < 0, so vertex is not in positive octant.Vertex (-1,-1,-1):x' = (-1)*cosθ + (-1)*sinθ = -0.5 - 0.866 ≈ -1.366 < 0y' = -1 < 0z' = (-1)*sinθ + (-1)*cosθ = -0.866 - 0.5 ≈ -1.366 < 0So, all vertices are either x' < 0, y' < 0, or z' < 0. Therefore, none of the vertices are in the positive octant.Wow, so it seems that by rotating the octant around the y-axis by more than 45 degrees, we can make sure that none of the cube's vertices are illuminated.But wait, earlier when I rotated around the x-axis, it didn't work because one vertex would always end up in the positive octant. But rotating around the y-axis seems to avoid that.Is this because rotating around the y-axis affects different vertices in a way that allows us to avoid all of them?Alternatively, maybe it's because when rotating around the y-axis, we're not just affecting the x and z coordinates, but also the z coordinate in a way that allows us to push both (1,1,1) and (1,1,-1) out of the positive octant.So, in conclusion, it seems that it is possible to rotate the projector's octant such that it doesn't illuminate any vertices of the cube. Specifically, by rotating the octant around the y-axis by more than 45 degrees, we can position it in such a way that none of the cube's vertices fall within the illuminated region.Therefore, the answer should be yes, it can be rotated in such a way.

answer：First, I need to solve each inequality separately.For the first inequality, 2x + 4 ≥ 0, I'll subtract 4 from both sides to get 2x ≥ -4. Then, dividing both sides by 2 gives x ≥ -2.Next, for the second inequality, 6 - x > 3, I'll subtract 6 from both sides to get -x > -3. Multiplying both sides by -1 (and remembering to reverse the inequality sign) gives x < 3.Combining both inequalities, the solution is all real numbers x such that -2 ≤ x < 3.Now, I'll list all the integer values within this range: -2, -1, 0, 1, and 2.Finally, I'll calculate the sum of these integers: -2 + (-1) + 0 + 1 + 2 = 0.

answer：Okay, so I have this problem about an arithmetic sequence. It says that in the arithmetic sequence {a_n}, a_4 + a_8 = 16. I need to find the sum of the first 11 terms, S_11. Hmm, arithmetic sequences, right? Let me recall what I know about them.An arithmetic sequence is a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is a_1 and the common difference is d, then the nth term can be written as a_n = a_1 + (n - 1)d. That formula seems familiar.So, the problem gives me a_4 + a_8 = 16. Let me write expressions for a_4 and a_8 using the formula for the nth term.a_4 = a_1 + (4 - 1)d = a_1 + 3da_8 = a_1 + (8 - 1)d = a_1 + 7dAdding these together: a_4 + a_8 = (a_1 + 3d) + (a_1 + 7d) = 2a_1 + 10dAnd this is equal to 16, so 2a_1 + 10d = 16. Hmm, maybe I can simplify this equation.Dividing both sides by 2: a_1 + 5d = 8. Okay, so that's one equation.Now, I need to find S_11, the sum of the first 11 terms. The formula for the sum of the first n terms of an arithmetic sequence is S_n = n/2 * (2a_1 + (n - 1)d). Alternatively, it can also be written as S_n = n*(a_1 + a_n)/2. Maybe I can use either formula.Let me try the first one: S_11 = 11/2 * (2a_1 + 10d). Wait, because n is 11, so (n - 1) is 10, so 2a_1 + 10d. Hmm, interesting. But from earlier, I have 2a_1 + 10d = 16. So, substituting that into the sum formula, S_11 = 11/2 * 16.Calculating that: 11/2 * 16. Let's see, 16 divided by 2 is 8, so 11 * 8 = 88. So, S_11 is 88.Wait, that seems straightforward. Let me double-check using the other formula for the sum. S_n = n*(a_1 + a_n)/2. So, for S_11, it would be 11*(a_1 + a_11)/2.What's a_11? Using the nth term formula: a_11 = a_1 + (11 - 1)d = a_1 + 10d.So, S_11 = 11*(a_1 + a_1 + 10d)/2 = 11*(2a_1 + 10d)/2. Which simplifies to 11*(2a_1 + 10d)/2. But 2a_1 + 10d is 16, so again, 11*16/2 = 11*8 = 88. Same result. Okay, that seems consistent.Alternatively, maybe I can think about the average of the first and the 11th term multiplied by the number of terms. Since the average of the first and last term is (a_1 + a_11)/2, which is equal to (a_1 + (a_1 + 10d))/2 = (2a_1 + 10d)/2 = a_1 + 5d. From earlier, I found that a_1 + 5d = 8. So, the average is 8, and the number of terms is 11, so the sum is 11*8 = 88. Yep, same answer.So, all methods lead me to 88. That gives me confidence that 88 is the correct answer.I think I've covered all the bases here. I used the formula for the nth term to express a_4 and a_8, added them together to get an equation involving a_1 and d, then used that equation in the sum formula. I checked it using two different expressions for the sum, and both gave me the same result. So, I think I'm good.**Final Answer**The sum of the first 11 terms is boxed{88}.

answer：Alright, so I have this integral to solve:int_{frac{pi}{2}}^{2 arctan 2} frac{dx}{sin x(1+sin x)}Hmm, okay. Let me try to figure out how to approach this. The integrand is frac{1}{sin x(1+sin x)}. That looks a bit complicated, but maybe I can simplify it somehow. First, I remember that when dealing with integrals involving trigonometric functions, sometimes substitution or trigonometric identities can help. Let me see if I can rewrite the denominator in a more manageable form.The denominator is sin x(1+sin x). Maybe I can split this into partial fractions? Let me try that. So, I want to express frac{1}{sin x(1+sin x)} as frac{A}{sin x} + frac{B}{1+sin x}. To find A and B, I'll set up the equation:1 = A(1 + sin x) + B sin xExpanding this, I get:1 = A + A sin x + B sin xCombine like terms:1 = A + (A + B) sin xFor this equation to hold for all x, the coefficients of like terms must be equal on both sides. So, the coefficient of sin x must be zero, and the constant term must be 1. That gives me two equations:1. A + B = 0 (coefficient of sin x)2. A = 1 (constant term)From the second equation, I immediately get A = 1. Plugging this into the first equation:1 + B = 0 implies B = -1So, the partial fraction decomposition is:frac{1}{sin x(1+sin x)} = frac{1}{sin x} - frac{1}{1+sin x}Great, that simplifies the integrand. Now, the integral becomes:int left( frac{1}{sin x} - frac{1}{1+sin x} right) dxWhich can be split into two separate integrals:int frac{1}{sin x} dx - int frac{1}{1+sin x} dxI know that int frac{1}{sin x} dx is a standard integral, which is ln | tan frac{x}{2} | + C. But let me double-check that. Yes, because frac{1}{sin x} = csc x, and the integral of csc x is indeed ln | tan frac{x}{2} | + C.Now, the second integral is int frac{1}{1+sin x} dx. Hmm, this one is a bit trickier. I remember that multiplying the numerator and denominator by 1 - sin x can sometimes help. Let me try that:int frac{1}{1+sin x} dx = int frac{1 - sin x}{(1+sin x)(1 - sin x)} dx = int frac{1 - sin x}{1 - sin^2 x} dxSince 1 - sin^2 x = cos^2 x, this simplifies to:int frac{1 - sin x}{cos^2 x} dx = int frac{1}{cos^2 x} dx - int frac{sin x}{cos^2 x} dxI recognize these integrals. The first one is int sec^2 x dx = tan x + C, and the second one can be solved by substitution. Let me set u = cos x, then du = -sin x dx. So, the second integral becomes:- int frac{sin x}{cos^2 x} dx = - int frac{-du}{u^2} = int frac{du}{u^2} = - frac{1}{u} + C = - frac{1}{cos x} + CPutting it all together, the second integral is:tan x - frac{1}{cos x} + CSo, going back to the original split integral, we have:int frac{1}{sin x} dx - int frac{1}{1+sin x} dx = ln | tan frac{x}{2} | - left( tan x - frac{1}{cos x} right) + CSimplifying, the integral becomes:ln | tan frac{x}{2} | - tan x + frac{1}{cos x} + CNow, let me write that as the antiderivative:F(x) = ln | tan frac{x}{2} | - tan x + sec x + COkay, so now I need to evaluate this from frac{pi}{2} to 2 arctan 2. Let's denote the upper limit as b = 2 arctan 2 and the lower limit as a = frac{pi}{2}.So, the definite integral is:F(b) - F(a) = left[ ln | tan frac{b}{2} | - tan b + sec b right] - left[ ln | tan frac{a}{2} | - tan a + sec a right]Let me compute each part step by step.First, let's compute F(b):1. Compute tan frac{b}{2} where b = 2 arctan 2.2. Compute tan b.3. Compute sec b.Similarly, compute F(a) where a = frac{pi}{2}:1. Compute tan frac{a}{2}.2. Compute tan a.3. Compute sec a.Starting with F(b):1. Compute tan frac{b}{2} where b = 2 arctan 2.Let me set theta = arctan 2, so that b = 2theta. Then, tan theta = 2, and we need to find tan frac{b}{2} = tan theta = 2. Wait, that's interesting. Because if b = 2theta, then frac{b}{2} = theta, so tan frac{b}{2} = tan theta = 2.So, ln | tan frac{b}{2} | = ln 2.2. Compute tan b = tan 2theta.Using the double-angle formula for tangent:tan 2theta = frac{2 tan theta}{1 - tan^2 theta}We know that tan theta = 2, so:tan 2theta = frac{2 times 2}{1 - 2^2} = frac{4}{1 - 4} = frac{4}{-3} = -frac{4}{3}So, tan b = -frac{4}{3}.3. Compute sec b = sec 2theta.We can use the identity sec 2theta = frac{1}{cos 2theta}. Alternatively, since we know tan theta = 2, we can find cos 2theta using the double-angle formula:cos 2theta = frac{1 - tan^2 theta}{1 + tan^2 theta} = frac{1 - 4}{1 + 4} = frac{-3}{5}Therefore, sec 2theta = frac{1}{cos 2theta} = frac{1}{-3/5} = -frac{5}{3}.So, putting it all together, F(b) is:ln 2 - left( -frac{4}{3} right) + left( -frac{5}{3} right) = ln 2 + frac{4}{3} - frac{5}{3} = ln 2 - frac{1}{3}Now, moving on to F(a) where a = frac{pi}{2}:1. Compute tan frac{a}{2} = tan frac{pi}{4} = 1.So, ln | tan frac{a}{2} | = ln 1 = 0.2. Compute tan a = tan frac{pi}{2}. Hmm, wait a second. tan frac{pi}{2} is undefined; it approaches infinity. But in the context of the integral, we need to evaluate the limit as x approaches frac{pi}{2} from below.Looking back at the antiderivative:F(x) = ln | tan frac{x}{2} | - tan x + sec xAs x approaches frac{pi}{2} from the left, tan x approaches infinity, and sec x also approaches infinity. But let's see how they behave together.Let me write sec x as frac{1}{cos x} and tan x as frac{sin x}{cos x}. So, sec x - tan x = frac{1 - sin x}{cos x}.As x approaches frac{pi}{2}, sin x approaches 1, so 1 - sin x approaches 0, and cos x approaches 0 as well. Let me see if this limit exists.Let me set x = frac{pi}{2} - epsilon, where epsilon approaches 0 from the positive side.Then, sin x = sin left( frac{pi}{2} - epsilon right) = cos epsilon approx 1 - frac{epsilon^2}{2}cos x = cos left( frac{pi}{2} - epsilon right) = sin epsilon approx epsilon - frac{epsilon^3}{6}So, 1 - sin x approx 1 - (1 - frac{epsilon^2}{2}) = frac{epsilon^2}{2}Thus, frac{1 - sin x}{cos x} approx frac{frac{epsilon^2}{2}}{epsilon} = frac{epsilon}{2} to 0 as epsilon to 0.Therefore, sec x - tan x approaches 0 as x approaches frac{pi}{2} from the left.So, in the antiderivative F(x), as x approaches frac{pi}{2}, the term ln | tan frac{x}{2} | approaches ln | tan frac{pi}{4} | = ln 1 = 0, and the term - tan x + sec x approaches 0. Therefore, F(a) = 0 + 0 = 0.Wait, but let me verify that. Because if I plug x = frac{pi}{2} into F(x), I get:ln | tan frac{pi}{4} | - tan frac{pi}{2} + sec frac{pi}{2}Which is ln 1 - infty + infty. That seems indeterminate. But from the limit, as x approaches frac{pi}{2} from below, the expression - tan x + sec x approaches 0, as we saw earlier. So, the limit of F(x) as x approaches frac{pi}{2} from below is 0.Therefore, F(a) = 0.Putting it all together, the definite integral is:F(b) - F(a) = left( ln 2 - frac{1}{3} right) - 0 = ln 2 - frac{1}{3}So, the value of the integral is ln 2 - frac{1}{3}.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Partial fraction decomposition: Correct, A=1, B=-1.2. Integrals of csc x and the second integral: Correct, used substitution and identities.3. Evaluating F(b): Correct, used substitution theta = arctan 2, computed tan frac{b}{2} = 2, tan b = -4/3, sec b = -5/3. Then, substituted into F(b): ln 2 - (-4/3) + (-5/3) = ln 2 + 4/3 -5/3 = ln 2 -1/3.4. Evaluating F(a): Correct, recognized that as x approaches pi/2, tan x and sec x go to infinity, but their combination sec x - tan x approaches 0, so F(a) = 0.Therefore, the final result is ln 2 - frac{1}{3}.I think that's correct. Let me just write it neatly.

answer：Okay, so I have this problem where I need to find the number of positive integers ( x ) less than 100 such that the sum ( 3^{x} + 5^{x} + 7^{x} + 11^{x} + 13^{x} + 17^{x} + 19^{x} ) is a prime number. Hmm, that sounds a bit challenging, but let me try to break it down step by step.First off, I know that prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, for the given expression to be prime, it must evaluate to a prime number when ( x ) is a positive integer less than 100.Let me start by testing small values of ( x ) because sometimes patterns emerge or specific properties can be observed.**Case 1: ( x = 1 )**Let me compute each term:- ( 3^1 = 3 )- ( 5^1 = 5 )- ( 7^1 = 7 )- ( 11^1 = 11 )- ( 13^1 = 13 )- ( 17^1 = 17 )- ( 19^1 = 19 )Adding them all up: ( 3 + 5 + 7 + 11 + 13 + 17 + 19 ).Let me compute this step by step:- ( 3 + 5 = 8 )- ( 8 + 7 = 15 )- ( 15 + 11 = 26 )- ( 26 + 13 = 39 )- ( 39 + 17 = 56 )- ( 56 + 19 = 75 )So, the sum is 75. Is 75 a prime number? No, because 75 is divisible by 5 and 3 (since 7 + 5 = 12, which is divisible by 3). So, 75 is not prime. Therefore, ( x = 1 ) doesn't work.**Case 2: ( x = 2 )**Compute each term:- ( 3^2 = 9 )- ( 5^2 = 25 )- ( 7^2 = 49 )- ( 11^2 = 121 )- ( 13^2 = 169 )- ( 17^2 = 289 )- ( 19^2 = 361 )Adding them up: ( 9 + 25 + 49 + 121 + 169 + 289 + 361 ).Let me add step by step:- ( 9 + 25 = 34 )- ( 34 + 49 = 83 )- ( 83 + 121 = 204 )- ( 204 + 169 = 373 )- ( 373 + 289 = 662 )- ( 662 + 361 = 1023 )So, the sum is 1023. Is 1023 prime? Let me check. 1023 divided by 3 is 341, so it's divisible by 3. Therefore, 1023 is not prime. So, ( x = 2 ) is out.**Case 3: ( x = 3 )**Compute each term:- ( 3^3 = 27 )- ( 5^3 = 125 )- ( 7^3 = 343 )- ( 11^3 = 1331 )- ( 13^3 = 2197 )- ( 17^3 = 4913 )- ( 19^3 = 6859 )Adding them up: ( 27 + 125 + 343 + 1331 + 2197 + 4913 + 6859 ).Let me compute step by step:- ( 27 + 125 = 152 )- ( 152 + 343 = 495 )- ( 495 + 1331 = 1826 )- ( 1826 + 2197 = 4023 )- ( 4023 + 4913 = 8936 )- ( 8936 + 6859 = 15795 )So, the sum is 15795. Is this prime? Let's see. The number ends with a 5, so it's divisible by 5. Hence, 15795 is not prime. So, ( x = 3 ) doesn't work.**Case 4: ( x = 4 )**Compute each term:- ( 3^4 = 81 )- ( 5^4 = 625 )- ( 7^4 = 2401 )- ( 11^4 = 14641 )- ( 13^4 = 28561 )- ( 17^4 = 83521 )- ( 19^4 = 130321 )Adding them up: ( 81 + 625 + 2401 + 14641 + 28561 + 83521 + 130321 ).Let me compute step by step:- ( 81 + 625 = 706 )- ( 706 + 2401 = 3107 )- ( 3107 + 14641 = 17748 )- ( 17748 + 28561 = 46309 )- ( 46309 + 83521 = 129830 )- ( 129830 + 130321 = 260151 )So, the sum is 260,151. Is this prime? Let's check. The sum is 260,151. I can test divisibility by small primes. Let's see, does 260,151 divide by 3? Adding the digits: 2+6+0+1+5+1 = 15, which is divisible by 3. So, 260,151 is divisible by 3, hence not prime. So, ( x = 4 ) is out.**Case 5: ( x = 5 )**Compute each term:- ( 3^5 = 243 )- ( 5^5 = 3125 )- ( 7^5 = 16807 )- ( 11^5 = 161051 )- ( 13^5 = 371293 )- ( 17^5 = 1419857 )- ( 19^5 = 2476099 )Adding them up: ( 243 + 3125 + 16807 + 161051 + 371293 + 1419857 + 2476099 ).This is getting quite large. Let me compute step by step:- ( 243 + 3125 = 3368 )- ( 3368 + 16807 = 20175 )- ( 20175 + 161051 = 181,226 )- ( 181,226 + 371,293 = 552,519 )- ( 552,519 + 1,419,857 = 1,972,376 )- ( 1,972,376 + 2,476,099 = 4,448,475 )So, the sum is 4,448,475. Is this prime? Well, it's an odd number, so not divisible by 2. Let's check divisibility by 5: it ends with a 5, so yes, it's divisible by 5. Therefore, 4,448,475 is not prime. So, ( x = 5 ) doesn't work.Hmm, so far, for ( x = 1 ) to ( x = 5 ), none of the sums are prime. Maybe I should look for a pattern or think about properties that could help me determine if the sum is prime without computing each time.Let me consider the parity of the sum. All the bases except 3 are odd primes, so each term ( p^x ) where ( p ) is odd will be odd if ( x ) is any positive integer. 3 is also odd, so all terms are odd. Now, how many terms are there? Let's count: 3, 5, 7, 11, 13, 17, 19. That's 7 terms. So, adding 7 odd numbers together.An odd number plus another odd number is even, but adding an odd number of odd numbers results in an odd number. So, 7 is odd, so the sum will be odd. So, the sum is odd, which is a good start because primes except 2 are odd. But that doesn't necessarily mean it's prime.Another thought: maybe considering modulo operations to see if the sum is divisible by some small primes, which would help us eliminate possible ( x ) values.Let me try modulo 3. Since 3 is one of the bases, perhaps that can help.Compute each term modulo 3:- ( 3^x mod 3 = 0 ) because 3 is divisible by 3.- ( 5 mod 3 = 2 ), so ( 5^x mod 3 = 2^x mod 3 ).- ( 7 mod 3 = 1 ), so ( 7^x mod 3 = 1^x = 1 mod 3 ).- ( 11 mod 3 = 2 ), so ( 11^x mod 3 = 2^x mod 3 ).- ( 13 mod 3 = 1 ), so ( 13^x mod 3 = 1 mod 3 ).- ( 17 mod 3 = 2 ), so ( 17^x mod 3 = 2^x mod 3 ).- ( 19 mod 3 = 1 ), so ( 19^x mod 3 = 1 mod 3 ).So, adding all these up modulo 3:Sum modulo 3 = ( 0 + 2^x + 1 + 2^x + 1 + 2^x + 1 mod 3 ).Simplify:Sum modulo 3 = ( 3 times 2^x + 3 mod 3 ).Which is ( 0 + 0 mod 3 ) because 3 is 0 mod 3. So, the sum is congruent to 0 modulo 3.Wait, that means the sum is divisible by 3. Therefore, unless the sum is 3 itself, it can't be prime because it would have 3 as a divisor.So, the sum is divisible by 3. Therefore, the only way the sum can be prime is if the sum is equal to 3. But let's see if that's possible.Looking back at ( x = 1 ), the sum was 75, which is way larger than 3. For ( x = 0 ), which isn't considered here since we're only looking at positive integers, the sum would be 7 (since any number to the power of 0 is 1, and there are 7 terms). But 7 is prime, but ( x = 0 ) isn't a positive integer, so we can't consider that.Therefore, for all positive integers ( x ), the sum is divisible by 3 and greater than 3, hence composite. So, the sum can never be prime for any positive integer ( x ).Wait, but hold on. Let me verify this because I might have made a mistake in my modulo calculation.Let me re-examine the modulo 3 computation.Each term:- ( 3^x mod 3 = 0 ).- ( 5^x mod 3 = (2)^x mod 3 ).- ( 7^x mod 3 = (1)^x = 1 mod 3 ).- ( 11^x mod 3 = (2)^x mod 3 ).- ( 13^x mod 3 = (1)^x = 1 mod 3 ).- ( 17^x mod 3 = (2)^x mod 3 ).- ( 19^x mod 3 = (1)^x = 1 mod 3 ).So, adding them up:( 0 + 2^x + 1 + 2^x + 1 + 2^x + 1 mod 3 ).So, that's ( 3 times 2^x + 3 times 1 mod 3 ).Which simplifies to ( 0 + 0 = 0 mod 3 ). So, yes, the sum is divisible by 3.Therefore, the sum is always divisible by 3, and since the sum is greater than 3 for all positive integers ( x ), it can never be prime.Wait, but hold on a second. Let me check for ( x = 0 ). If ( x = 0 ), each term is 1, so the sum is 7, which is prime. But ( x ) has to be a positive integer, so ( x = 0 ) isn't considered here. Therefore, for all ( x geq 1 ), the sum is divisible by 3 and greater than 3, hence composite.Therefore, there are no positive integers ( x ) less than 100 for which the given sum is prime.But wait, let me test ( x = 1 ) again just to be sure. Earlier, I got 75, which is divisible by 3 and 5. So, that's composite. ( x = 2 ) gave 1023, which is divisible by 3. ( x = 3 ) gave 15795, which is divisible by 5 and 3. ( x = 4 ) gave 260,151, which is divisible by 3. ( x = 5 ) gave 4,448,475, which is divisible by 5 and 3. So, seems consistent.Is there any possible ( x ) where the sum is 3? Well, the sum is 7 when ( x = 0 ), which is prime, but ( x ) must be positive. For ( x = 1 ), it's 75, which is way larger. So, no, the sum can't be 3 for any positive ( x ).Therefore, my conclusion is that there are no positive integers ( x ) less than 100 for which the given expression is prime. So, the number of such ( x ) is zero.**Final Answer**The number of positive integers ( x ) less than 100 is boxed{0}.