💡Let the two initial numbers be ( x ) and ( y ). Their sum is given by[ x + y = S. ]After adding 5 to each number, the numbers become ( x + 5 ) and ( y + 5 ). Tripling each of these results in[ 3(x + 5) quad text{and} quad 3(y + 5). ]The sum of these tripled numbers is:[ 3(x + 5) + 3(y + 5) = 3x + 15 + 3y + 15. ]Combine like terms and simplify:[ 3x + 3y + 30. ]Since ( x + y = S ), replace ( x + y ) with ( S ):[ 3(x + y) + 30 = 3S + 30. ]Thus, the sum of the final two numbers is ( 3S + 30 ).The final answer is (boxed{textbf{(D)} 3S + 30}).

💡First, I need to determine the time it takes for the first worker to load the truck alone. Let's denote this time as ( T ) hours.The first worker's loading rate is ( frac{1}{T} ) trucks per hour, and the second worker's rate is ( frac{1}{8} ) trucks per hour since they can load the truck in 8 hours.When both workers work together, their combined loading rate is the sum of their individual rates. According to the problem, together they can load the truck in approximately 3.0769230769230766 hours. Therefore, their combined rate is ( frac{1}{3.0769230769230766} ) trucks per hour.I can set up the equation:[frac{1}{T} + frac{1}{8} = frac{1}{3.0769230769230766}]To solve for ( T ), I'll first find a common denominator for the fractions on the left side of the equation, which is ( 8T ):[frac{8}{8T} + frac{T}{8T} = frac{1}{3.0769230769230766}][frac{8 + T}{8T} = frac{1}{3.0769230769230766}]Next, I'll cross-multiply to solve for ( T ):[8 + T = 8T times frac{1}{3.0769230769230766}][8 + T = frac{8T}{3.0769230769230766}]Simplifying the right side:[8 + T = 2.6T]Now, I'll solve for ( T ):[8 = 2.6T - T][8 = 1.6T][T = frac{8}{1.6}][T = 5]So, it takes the first worker 5 hours to load the truck alone.

💡Okay, so I have this problem where there are 36 points equally spaced around the perimeter of a square. Each side of the square is divided into eight equal segments, meaning each side has nine points, including the vertices. I need to find how many distinct possible positions there are for the centroid of a triangle formed by any three non-collinear points among these 36 points.First, let me visualize the square. Let's assume it's a square with side length 8 units, so each side has points at every integer coordinate from 0 to 8. That makes sense because 8 segments mean 9 points per side, including the corners. So, the total number of points is 4 sides × 9 points = 36 points, which matches the problem statement.Now, the centroid of a triangle is the average of the coordinates of its three vertices. So, if I have three points A, B, and C with coordinates (x_A, y_A), (x_B, y_B), and (x_C, y_C), then the centroid G will have coordinates:G_x = (x_A + x_B + x_C) / 3G_y = (y_A + y_B + y_C) / 3Since all the points are on the perimeter of the square, their coordinates are either on the x-axis, y-axis, or the lines x=8 or y=8. So, each coordinate is an integer between 0 and 8.Therefore, the sums x_A + x_B + x_C and y_A + y_B + y_C will be integers between 0 and 24 (since 3×8=24). When we divide these sums by 3, the centroid coordinates will be fractions with denominators of 3. So, G_x and G_y can be written as p/3 and q/3 where p and q are integers between 0 and 24.But wait, since A, B, and C are not collinear, the centroid can't be just any point. It has to be the centroid of a valid triangle. So, I need to figure out all possible (p/3, q/3) points that can be formed by the centroids of such triangles.Let me think about the possible values of p and q. Since each x-coordinate of the points is between 0 and 8, the sum x_A + x_B + x_C can be between 0 and 24. Similarly, the sum y_A + y_B + y_C can be between 0 and 24. Therefore, p and q can range from 0 to 24, but since we're dividing by 3, the centroid coordinates can range from 0 to 8.However, not all combinations of p and q will be achievable because the points are constrained to the perimeter of the square. For example, if all three points are on the same side, the centroid will lie along that side, but since the points are not collinear, they can't all be on a straight line. Wait, actually, the problem states that A, B, and C are not collinear, so they can't all lie on the same straight line, which includes the sides of the square.So, the centroid can't lie on the edges of the square because that would require all three points to be on the same side, which would make them collinear. Therefore, the centroid must lie strictly inside the square.This means that the centroid coordinates must be strictly greater than 0 and strictly less than 8. So, p and q must be integers such that p/3 and q/3 are between 0 and 8, but not equal to 0 or 8. Therefore, p and q must be integers from 1 to 23, inclusive.So, p and q can each take 23 possible values, leading to 23×23 = 529 possible centroid positions. But wait, is that actually the case? Are all these positions achievable?I need to verify if every point of the form (p/3, q/3) where p and q are integers from 1 to 23 can indeed be the centroid of some triangle formed by three non-collinear points on the square's perimeter.Let me consider how the coordinates of the centroid are formed. Since each coordinate is the average of three integers, the centroid coordinates will be multiples of 1/3. So, the grid of possible centroids is a 23×23 grid of points spaced 1/3 units apart, starting from (1/3, 1/3) up to (23/3, 23/3).But I need to ensure that for every such point, there exist three points on the perimeter whose coordinates average to that point. Let's think about how to construct such triangles.For example, if I want a centroid at (1/3, 1/3), I can take three points: (0,0), (0,0), and (1,1). Wait, but points can't be repeated because we need three distinct points. So, maybe (0,0), (0,1), and (1,0). Let's check the centroid:G_x = (0 + 0 + 1)/3 = 1/3G_y = (0 + 1 + 0)/3 = 1/3Yes, that works. So, (1/3, 1/3) is achievable.Similarly, for (23/3, 23/3), which is approximately (7.666, 7.666), I can take points near the top-right corner. For example, (8,8), (8,7), and (7,8). The centroid would be:G_x = (8 + 8 + 7)/3 = 23/3G_y = (8 + 7 + 8)/3 = 23/3So, that works too.What about a point in the middle, say (4,4)? That would require the sum of x-coordinates to be 12 and the sum of y-coordinates to be 12. Let's pick three points: (0,0), (8,8), and (4,4). Wait, (4,4) is a point on the perimeter? No, (4,4) is the center of the square, which is not on the perimeter. So, I need three points on the perimeter whose x and y coordinates sum to 12 each.Let me think of three points: (0,0), (8,8), and (4,4). But again, (4,4) isn't on the perimeter. Maybe (0,8), (8,0), and (4,4). Again, (4,4) isn't on the perimeter. Hmm, maybe I need three points on different sides.Let's try (0,0), (8,4), and (4,8). Let's calculate the centroid:G_x = (0 + 8 + 4)/3 = 12/3 = 4G_y = (0 + 4 + 8)/3 = 12/3 = 4Yes, that works. So, (4,4) is achievable.What about a point like (1/3, 2/3)? Let's see. I need three points whose x-coordinates sum to 1 and y-coordinates sum to 2. Since all x and y coordinates are integers, the sum must be integers, so 1 and 2 are possible.For example, take (0,0), (0,0), and (1,2). Wait, but again, points can't be repeated. So, maybe (0,0), (0,1), and (1,1). Let's check:G_x = (0 + 0 + 1)/3 = 1/3G_y = (0 + 1 + 1)/3 = 2/3Yes, that works. So, (1/3, 2/3) is achievable.Similarly, for a point like (2/3, 1/3), I can take (0,0), (1,0), and (1,1):G_x = (0 + 1 + 1)/3 = 2/3G_y = (0 + 0 + 1)/3 = 1/3Yes, that works.What about a point like (7/3, 5/3)? Let's see. I need x-coordinates summing to 7 and y-coordinates summing to 5.Let me pick three points: (0,0), (7,0), and (0,5). Wait, (0,5) is on the left side, (7,0) is on the bottom side, and (0,0) is the corner. Let's calculate the centroid:G_x = (0 + 7 + 0)/3 = 7/3G_y = (0 + 0 + 5)/3 = 5/3Yes, that works. So, (7/3, 5/3) is achievable.It seems like for any p and q between 1 and 23, I can find three points on the perimeter whose coordinates sum to p and q respectively, leading to the centroid (p/3, q/3).But wait, is there any restriction? For example, can I always find three points on the perimeter such that their x-coordinates sum to any integer p between 1 and 23, and similarly for y-coordinates?Let me think about the possible sums. Since each x-coordinate can be 0,1,2,...,8, the sum of three x-coordinates can range from 0+0+0=0 to 8+8+8=24. Similarly for y-coordinates.But since the points are on the perimeter, not all combinations of x and y are independent. For example, a point on the top side has y=8, and x can be 0 to 8. Similarly, a point on the right side has x=8, and y can be 0 to 8, etc.So, the x and y coordinates are not entirely independent because a point can only be on one side at a time. Therefore, when choosing three points, their x and y coordinates are constrained based on which sides they are on.This might complicate things because the sums of x and y coordinates are not entirely independent. For example, if two points are on the top side (y=8), then their y-coordinates are fixed, which affects the sum of y-coordinates.But wait, the centroid's x and y coordinates are determined independently by the sums of the x and y coordinates of the three points. So, as long as I can find three points whose x-coordinates sum to p and three points whose y-coordinates sum to q, regardless of their positions on the square, then (p/3, q/3) is achievable.But the problem is that the same three points must have x-coordinates summing to p and y-coordinates summing to q. So, the sums are linked through the same three points.Therefore, I need to ensure that for any p and q, there exist three points on the perimeter such that the sum of their x-coordinates is p and the sum of their y-coordinates is q.This might not always be possible because the x and y coordinates are linked for each point. For example, if a point is on the top side, its y-coordinate is 8, but its x-coordinate can vary. Similarly, a point on the right side has x=8 and y can vary.So, perhaps some combinations of p and q are not achievable because the x and y sums are constrained by the sides the points are on.Let me think of an example. Suppose I want p=1 and q=1. That would require the sum of x-coordinates to be 1 and the sum of y-coordinates to be 1. How can I achieve that?I need three points whose x-coordinates add up to 1 and y-coordinates add up to 1. Since each x and y coordinate is at least 0, the only way to get a sum of 1 is to have two points with x=0 and one point with x=1, and similarly for y.But let's see if that's possible. Let's say two points are at (0,0) and one point is at (1,0). Then, the sum of x-coordinates is 0+0+1=1, and the sum of y-coordinates is 0+0+0=0. But we need the sum of y-coordinates to be 1. So, maybe one of the points has y=1.Wait, if I take two points at (0,0) and one point at (1,1). Then, the sum of x-coordinates is 0+0+1=1, and the sum of y-coordinates is 0+0+1=1. That works. So, the centroid would be (1/3, 1/3), which is achievable.But wait, (1,1) is on the perimeter? No, (1,1) is inside the square, not on the perimeter. So, I can't use that point. Hmm, that complicates things.So, I need three points on the perimeter such that their x-coordinates sum to 1 and y-coordinates sum to 1. Let's see.The only way to get x-coordinates summing to 1 is to have two points with x=0 and one point with x=1. Similarly, for y-coordinates summing to 1, two points with y=0 and one point with y=1.But since the points are on the perimeter, the point with x=1 must be on the bottom side (y=0) or the top side (y=8). Similarly, the point with y=1 must be on the left side (x=0) or the right side (x=8).Wait, let's try to find such points.Let me take two points at (0,0) and one point at (1,0). Then, the sum of x-coordinates is 0+0+1=1, and the sum of y-coordinates is 0+0+0=0. Not enough.Alternatively, take two points at (0,0) and one point at (0,1). Then, the sum of x-coordinates is 0+0+0=0, which is less than 1. Not good.Wait, maybe take one point at (0,0), one point at (1,0), and one point at (0,1). Then, the sum of x-coordinates is 0+1+0=1, and the sum of y-coordinates is 0+0+1=1. Perfect! So, the centroid would be (1/3, 1/3).But wait, are these points on the perimeter? (0,0) is a corner, (1,0) is on the bottom side, and (0,1) is on the left side. Yes, all three points are on the perimeter and not collinear. So, this works.Therefore, (1/3, 1/3) is achievable.Similarly, for (23/3, 23/3), as I thought earlier, I can take points near the top-right corner.What about a point like (2/3, 2/3)? Let's see. I need x-coordinates summing to 2 and y-coordinates summing to 2.Let me take two points at (0,0) and one point at (2,0). Then, the sum of x-coordinates is 0+0+2=2, and the sum of y-coordinates is 0+0+0=0. Not enough.Alternatively, take one point at (0,0), one at (1,0), and one at (1,1). Wait, (1,1) is not on the perimeter. So, maybe take one point at (0,0), one at (1,0), and one at (0,2). Then, the sum of x-coordinates is 0+1+0=1, which is less than 2. Not good.Wait, maybe take one point at (0,0), one at (2,0), and one at (0,2). Then, the sum of x-coordinates is 0+2+0=2, and the sum of y-coordinates is 0+0+2=2. So, the centroid would be (2/3, 2/3). But are these points on the perimeter? (0,0) is a corner, (2,0) is on the bottom side, and (0,2) is on the left side. Yes, all three are on the perimeter and not collinear. So, this works.Therefore, (2/3, 2/3) is achievable.What about a point like (3/3, 3/3) = (1,1)? Let's see. I need x-coordinates summing to 3 and y-coordinates summing to 3.Let me take three points: (0,0), (1,0), and (2,0). Then, the sum of x-coordinates is 0+1+2=3, and the sum of y-coordinates is 0+0+0=0. Not enough.Alternatively, take one point at (0,0), one at (1,1), and one at (2,2). But (1,1) and (2,2) are not on the perimeter. Hmm.Wait, maybe take one point at (0,0), one at (3,0), and one at (0,3). Then, the sum of x-coordinates is 0+3+0=3, and the sum of y-coordinates is 0+0+3=3. So, the centroid is (1,1). But are these points on the perimeter? (0,0) is a corner, (3,0) is on the bottom side, and (0,3) is on the left side. Yes, all three are on the perimeter and not collinear. So, this works.Therefore, (1,1) is achievable.Wait, but (1,1) is inside the square, not on the perimeter. But the centroid can be inside the square, right? The centroid is the average of the three points, so it doesn't have to be on the perimeter.Yes, that's correct. The centroid can be anywhere inside the square, depending on the positions of the three points.So, going back, it seems like for any p and q between 1 and 23, I can find three points on the perimeter such that their x-coordinates sum to p and y-coordinates sum to q, leading to the centroid (p/3, q/3).But I need to make sure that for every p and q, such points exist. Let me think of a more challenging case, like p=23 and q=23.As I thought earlier, taking points near the top-right corner: (8,8), (8,7), and (7,8). The sum of x-coordinates is 8+8+7=23, and the sum of y-coordinates is 8+7+8=23. So, the centroid is (23/3, 23/3). That works.Another challenging case: p=22 and q=22. Let's see. I need x-coordinates summing to 22 and y-coordinates summing to 22.Let me take two points at (8,8) and one point at (6,8). Then, the sum of x-coordinates is 8+8+6=22, and the sum of y-coordinates is 8+8+8=24. Not quite 22.Alternatively, take two points at (8,8) and one point at (6,7). Then, the sum of x-coordinates is 8+8+6=22, and the sum of y-coordinates is 8+8+7=23. Still not 22.Wait, maybe take one point at (8,8), one at (8,7), and one at (7,7). Then, the sum of x-coordinates is 8+8+7=23, which is too much. Hmm.Alternatively, take one point at (8,8), one at (8,6), and one at (6,8). Then, the sum of x-coordinates is 8+8+6=22, and the sum of y-coordinates is 8+6+8=22. Perfect! So, the centroid is (22/3, 22/3). Are these points on the perimeter? (8,8) is a corner, (8,6) is on the right side, and (6,8) is on the top side. Yes, all three are on the perimeter and not collinear. So, this works.Therefore, (22/3, 22/3) is achievable.What about p=1 and q=23? Let's see. I need x-coordinates summing to 1 and y-coordinates summing to 23.Wait, x-coordinates summing to 1 would require two points with x=0 and one point with x=1. But y-coordinates summing to 23 would require two points with y=8 and one point with y=7, since 8+8+7=23.But can I have a point with x=1 and y=7? Yes, (1,7) is on the left side. So, let's take two points at (0,8) and one point at (1,7). Then, the sum of x-coordinates is 0+0+1=1, and the sum of y-coordinates is 8+8+7=23. So, the centroid is (1/3, 23/3). Are these points on the perimeter? (0,8) is a corner, (0,8) is the same point, but we need three distinct points. So, maybe take two different points at (0,8) and (0,8), but that's not allowed. Wait, I need three distinct points.So, let's take one point at (0,8), one at (0,8), and one at (1,7). But again, points can't be repeated. So, maybe take one point at (0,8), one at (0,7), and one at (1,8). Then, the sum of x-coordinates is 0+0+1=1, and the sum of y-coordinates is 8+7+8=23. So, the centroid is (1/3, 23/3). Are these points on the perimeter? (0,8) is a corner, (0,7) is on the left side, and (1,8) is on the top side. Yes, all three are on the perimeter and not collinear. So, this works.Therefore, (1/3, 23/3) is achievable.Similarly, for p=23 and q=1, I can take one point at (8,0), one at (8,1), and one at (7,0). The sum of x-coordinates is 8+8+7=23, and the sum of y-coordinates is 0+1+0=1. So, the centroid is (23/3, 1/3). These points are on the perimeter: (8,0) is a corner, (8,1) is on the right side, and (7,0) is on the bottom side. Not collinear. So, this works.Therefore, it seems like for any combination of p and q between 1 and 23, I can find three points on the perimeter such that their x-coordinates sum to p and y-coordinates sum to q, leading to the centroid (p/3, q/3).But wait, is there any case where p and q are such that it's impossible to find such points? For example, p=2 and q=23.Let me try p=2 and q=23. I need x-coordinates summing to 2 and y-coordinates summing to 23.For x-coordinates summing to 2, I can have two points with x=0 and one point with x=2. For y-coordinates summing to 23, I need two points with y=8 and one point with y=7.So, let's take two points at (0,8) and one point at (2,7). Then, the sum of x-coordinates is 0+0+2=2, and the sum of y-coordinates is 8+8+7=23. So, the centroid is (2/3, 23/3). Are these points on the perimeter? (0,8) is a corner, (0,8) is the same point, but we need three distinct points. So, maybe take one point at (0,8), one at (0,7), and one at (2,8). Then, the sum of x-coordinates is 0+0+2=2, and the sum of y-coordinates is 8+7+8=23. So, the centroid is (2/3, 23/3). These points are on the perimeter: (0,8) is a corner, (0,7) is on the left side, and (2,8) is on the top side. Not collinear. So, this works.Therefore, (2/3, 23/3) is achievable.Another tricky case: p=24 and q=24. Wait, p and q can only go up to 23 because the centroid coordinates must be strictly less than 8. So, p and q are from 1 to 23.Wait, actually, p and q can be up to 24, but since the centroid must be strictly inside the square, p and q must be less than 24, so p and q go up to 23. So, p and q are from 1 to 23.But let me confirm: if p=24, then G_x=24/3=8, which is on the edge, but the centroid must be strictly inside, so p=24 is not allowed. Similarly, p=0 would give G_x=0, which is on the edge, so p=0 is not allowed. Therefore, p and q must be from 1 to 23.So, in conclusion, every point of the form (p/3, q/3) where p and q are integers from 1 to 23 is achievable as the centroid of some triangle formed by three non-collinear points on the perimeter of the square.Therefore, the number of distinct possible positions for the centroid is 23×23=529.Wait, but let me think again. Is there any overlap? Could different sets of three points lead to the same centroid? For example, could two different triangles have the same centroid?Yes, definitely. For example, different triangles can have the same centroid. But the question is asking for the number of distinct possible positions, not the number of distinct triangles. So, even if multiple triangles lead to the same centroid, we only count that position once.But in my previous reasoning, I concluded that every (p/3, q/3) with p and q from 1 to 23 is achievable, so the number of distinct positions is 23×23=529.But wait, let me think about whether all these points are indeed inside the square. Since p and q are from 1 to 23, G_x and G_y are from 1/3 to 23/3, which is approximately 0.333 to 7.666, which is strictly inside the square (since the square goes from 0 to 8). So, all these centroids are inside the square, not on the edges.Therefore, the number of distinct possible positions for the centroid is 23×23=529.But wait, I just realized something. The problem says "any three of these points which are not collinear." So, we need to ensure that the three points are not collinear. But in my previous examples, I was careful to choose points that are not collinear. So, as long as the three points are not all on the same straight line, which includes the sides of the square, the centroid is valid.But does this restriction affect the number of achievable centroids? I don't think so, because even if some sets of three points are collinear, we can always find other sets that are not collinear to achieve the same centroid. For example, if three points on a side give a centroid on that side, but since we exclude collinear points, we need to find other points not on that side to achieve the same centroid.But wait, in my earlier examples, I was able to achieve centroids on the edges by using points from different sides. For example, (1/3, 1/3) is achieved by points on the bottom and left sides, not on the same side. So, even though the centroid is near the corner, it's not on the edge because the points are not collinear.Wait, actually, the centroid can't be on the edge because that would require the three points to be collinear, which is not allowed. So, all centroids must be strictly inside the square, which aligns with p and q being from 1 to 23, leading to G_x and G_y from 1/3 to 23/3, which is inside the square.Therefore, my conclusion remains that there are 23×23=529 distinct possible positions for the centroid.But let me check if there's any mathematical reason why some points might not be achievable. For example, in modular arithmetic, sometimes certain sums are not possible. But since we're dealing with sums of integers, and the possible sums range from 0 to 24, and we're dividing by 3, which is a divisor of 24, I think all residues modulo 3 are covered.Wait, actually, the sums p and q can be any integer from 0 to 24, so when divided by 3, they can be any multiple of 1/3 from 0 to 8. But since p and q are from 1 to 23, we cover all multiples of 1/3 in the interior of the square.Therefore, I think my reasoning holds, and the number of distinct possible positions for the centroid is 529.

💡Okay, so I have this problem where I'm given two vectors, **a** and **b**, and I need to find a vector **p** such that when both **a** and **b** are projected onto some vector **v**, they both result in **p**. Additionally, **a**, **b**, and **p** are collinear. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, let me recall what it means for vectors to be collinear. Collinear vectors lie on the same straight line, which means they are scalar multiples of each other. So, if **a**, **b**, and **p** are collinear, then **p** must lie somewhere along the line that passes through both **a** and **b**. That gives me a starting point.Next, the projection part. I remember that the projection of a vector **u** onto another vector **v** is given by the formula:[text{proj}_{mathbf{v}} mathbf{u} = left( frac{mathbf{u} cdot mathbf{v}}{mathbf{v} cdot mathbf{v}} right) mathbf{v}]This projection results in a vector that's in the direction of **v**. Since both **a** and **b** project to the same vector **p**, that means:[text{proj}_{mathbf{v}} mathbf{a} = text{proj}_{mathbf{v}} mathbf{b} = mathbf{p}]So, both projections are equal, which implies that:[left( frac{mathbf{a} cdot mathbf{v}}{mathbf{v} cdot mathbf{v}} right) mathbf{v} = left( frac{mathbf{b} cdot mathbf{v}}{mathbf{v} cdot mathbf{v}} right) mathbf{v}]Since the denominators are the same and non-zero (assuming **v** is not the zero vector), we can set the numerators equal:[mathbf{a} cdot mathbf{v} = mathbf{b} cdot mathbf{v}]Which simplifies to:[(mathbf{a} - mathbf{b}) cdot mathbf{v} = 0]So, the vector (**a** - **b**) is orthogonal to **v**. That's an important condition.Now, let me write down the vectors **a** and **b**:[mathbf{a} = begin{pmatrix} 1 -1 2 end{pmatrix}, quad mathbf{b} = begin{pmatrix} 2 3 0 end{pmatrix}]Calculating (**a** - **b**):[mathbf{a} - mathbf{b} = begin{pmatrix} 1 - 2 -1 - 3 2 - 0 end{pmatrix} = begin{pmatrix} -1 -4 2 end{pmatrix}]So, (**a** - **b**) is **(-1, -4, 2)**. This vector must be orthogonal to **v**. That means their dot product is zero:[begin{pmatrix} -1 -4 2 end{pmatrix} cdot mathbf{v} = 0]Let me denote **v** as **v** = (v₁, v₂, v₃). Then the dot product is:[-1 cdot v₁ - 4 cdot v₂ + 2 cdot v₃ = 0]So, that's one equation that **v** must satisfy. But I don't know **v**, so maybe I can find another condition.Wait, since **p** is the projection of both **a** and **b** onto **v**, and **p** is collinear with **a** and **b**, that means **p** must lie on the line passing through **a** and **b**. So, **p** can be expressed as a linear combination of **a** and (**b** - **a**). Let me write that down.Let me parameterize the line passing through **a** and **b**. A general point on this line can be written as:[mathbf{p} = mathbf{a} + t (mathbf{b} - mathbf{a})]Where t is a scalar parameter. Let me compute (**b** - **a**):[mathbf{b} - mathbf{a} = begin{pmatrix} 2 - 1 3 - (-1) 0 - 2 end{pmatrix} = begin{pmatrix} 1 4 -2 end{pmatrix}]So, the parametric equation for **p** is:[mathbf{p} = begin{pmatrix} 1 -1 2 end{pmatrix} + t begin{pmatrix} 1 4 -2 end{pmatrix} = begin{pmatrix} 1 + t -1 + 4t 2 - 2t end{pmatrix}]Now, since **p** is the projection of both **a** and **b** onto **v**, and projections are vectors in the direction of **v**, **p** must be a scalar multiple of **v**. So, **p** is parallel to **v**, which means **v** can be written as a scalar multiple of **p**. Let me denote **v** = k**p**, where k is a scalar.But since **p** is already a vector on the line through **a** and **b**, and **v** is a direction vector, I can use this relationship to find **p**.Wait, but earlier I had that (**a** - **b**) is orthogonal to **v**, which is equivalent to (**a** - **b**) being orthogonal to **p** because **v** is a scalar multiple of **p**. So, (**a** - **b**) is orthogonal to **p**.Let me write that condition:[(mathbf{a} - mathbf{b}) cdot mathbf{p} = 0]Substituting the expressions for (**a** - **b**) and **p**:[begin{pmatrix} -1 -4 2 end{pmatrix} cdot begin{pmatrix} 1 + t -1 + 4t 2 - 2t end{pmatrix} = 0]Calculating the dot product:[(-1)(1 + t) + (-4)(-1 + 4t) + (2)(2 - 2t) = 0]Let me compute each term:First term: (-1)(1 + t) = -1 - tSecond term: (-4)(-1 + 4t) = 4 - 16tThird term: (2)(2 - 2t) = 4 - 4tNow, adding them all together:(-1 - t) + (4 - 16t) + (4 - 4t) = (-1 + 4 + 4) + (-t -16t -4t) = 7 - 21tSo, 7 - 21t = 0Solving for t:21t = 7t = 7 / 21 = 1/3Okay, so t is 1/3. Now, plugging this back into the parametric equation for **p**:[mathbf{p} = begin{pmatrix} 1 + frac{1}{3} -1 + 4 times frac{1}{3} 2 - 2 times frac{1}{3} end{pmatrix} = begin{pmatrix} frac{4}{3} frac{1}{3} frac{4}{3} end{pmatrix}]So, **p** is (4/3, 1/3, 4/3). Let me just verify if this makes sense.First, check if (**a** - **b**) is orthogonal to **p**:Dot product:(-1)(4/3) + (-4)(1/3) + (2)(4/3) = (-4/3) + (-4/3) + (8/3) = (-8/3) + (8/3) = 0Good, that works.Also, since **p** is collinear with **a** and **b**, it should lie on the line connecting them, which it does because we parameterized it that way.Additionally, since **p** is the projection of both **a** and **b** onto **v**, and **p** is collinear with **a** and **b**, this makes sense because the projection would essentially be the point where the line intersects the direction of **v**.Wait, but let me think again. If both **a** and **b** project to the same **p**, then **p** must be the point where the line through **a** and **b** is closest to the origin along the direction of **v**. Since **p** is collinear, it's like the point where the line intersects the line defined by **v**.But in this case, since **p** is on the line through **a** and **b**, and **v** is in the direction of **p**, it's consistent.Let me also check the projection formula. Let's compute the projection of **a** onto **p** and see if it equals **p**.Using the projection formula:[text{proj}_{mathbf{p}} mathbf{a} = left( frac{mathbf{a} cdot mathbf{p}}{mathbf{p} cdot mathbf{p}} right) mathbf{p}]Compute **a** ⋅ **p**:(1)(4/3) + (-1)(1/3) + (2)(4/3) = 4/3 - 1/3 + 8/3 = (4 - 1 + 8)/3 = 11/3Compute **p** ⋅ **p**:(4/3)^2 + (1/3)^2 + (4/3)^2 = 16/9 + 1/9 + 16/9 = 33/9 = 11/3So, the projection is:(11/3) / (11/3) * **p** = 1 * **p** = **p**Similarly, projecting **b** onto **p**:**b** ⋅ **p** = (2)(4/3) + (3)(1/3) + (0)(4/3) = 8/3 + 1 + 0 = 11/3**p** ⋅ **p** is still 11/3, so the projection is again **p**.Perfect, that checks out. So, **p** is indeed the projection of both **a** and **b** onto itself, which makes sense because **p** is collinear with **v**.Therefore, after all these steps and verifications, I'm confident that **p** is (4/3, 1/3, 4/3).

💡Okay, so I have this problem here: I need to find the value of x that gives the minimum value for the expression 3x² + 9x - 7. Hmm, quadratic expressions, right? I remember that quadratic functions graph as parabolas, and since the coefficient of x² is positive (which is 3 in this case), the parabola opens upwards. That means the vertex of the parabola will be its lowest point, so the minimum value of the expression occurs at the vertex. Alright, so I need to find the vertex of this quadratic function. I think there are a couple of ways to do this. One method is completing the square, and another is using the vertex formula. I'm a bit fuzzy on completing the square, so maybe I'll try the vertex formula first. The standard form of a quadratic equation is ax² + bx + c, and the vertex occurs at x = -b/(2a). Let me write that down. In this case, a is 3 and b is 9. Plugging those into the formula, x = -9/(2*3) = -9/6 = -3/2. So, x = -3/2. That should be the value where the expression reaches its minimum. Wait, let me double-check that. Maybe I should try completing the square to see if I get the same result. Starting with the expression 3x² + 9x - 7. First, I can factor out the coefficient of x² from the first two terms. So, that would be 3(x² + 3x) - 7. Now, to complete the square inside the parentheses, I need to take half of the coefficient of x, which is 3, so half of that is 3/2, and then square it, which gives (3/2)² = 9/4. So, I add and subtract 9/4 inside the parentheses to maintain equality. So, it becomes 3[(x² + 3x + 9/4) - 9/4] - 7. That simplifies to 3(x + 3/2)² - 3*(9/4) - 7. Calculating the constants, 3*(9/4) is 27/4, and 27/4 + 7 is... wait, 7 is 28/4, so 27/4 + 28/4 is 55/4. But since it's minus 27/4 and then minus 7, it's -27/4 - 28/4 = -55/4. So, the expression becomes 3(x + 3/2)² - 55/4. Since the square term is always non-negative, the minimum value occurs when (x + 3/2)² = 0, which is when x = -3/2. That matches the result I got earlier using the vertex formula. Hmm, seems like both methods give me the same answer. Maybe I should also check by plugging in x = -3/2 into the original expression to see what value I get. Let's do that. Calculating 3*(-3/2)² + 9*(-3/2) - 7. First, (-3/2) squared is 9/4. So, 3*(9/4) is 27/4. Then, 9*(-3/2) is -27/2. So, putting it all together: 27/4 - 27/2 - 7. Converting everything to quarters, 27/4 is 27/4, -27/2 is -54/4, and -7 is -28/4. Adding them up: 27/4 - 54/4 - 28/4 = (27 - 54 - 28)/4 = (-55)/4. So, the minimum value is -55/4, which occurs at x = -3/2. Just to make sure I didn't make any calculation errors, let me go through the steps again. Starting with the vertex formula: x = -b/(2a). Here, a = 3, b = 9. So, x = -9/(2*3) = -9/6 = -3/2. Yep, that's correct. Completing the square: 3x² + 9x - 7. Factor out 3: 3(x² + 3x) - 7. Complete the square by adding and subtracting (3/2)² = 9/4: 3[(x² + 3x + 9/4) - 9/4] - 7. That becomes 3(x + 3/2)² - 27/4 - 7. Converting 7 to quarters: 28/4. So, -27/4 - 28/4 = -55/4. So, the expression is 3(x + 3/2)² - 55/4, confirming the minimum at x = -3/2. Plugging x = -3/2 back into the original expression: 3*(-3/2)² + 9*(-3/2) - 7. Calculating each term: (-3/2)² = 9/4, so 3*(9/4) = 27/4. 9*(-3/2) = -27/2. Then, -7 is just -7. Converting all to quarters: 27/4 - 54/4 - 28/4 = (27 - 54 - 28)/4 = (-55)/4. So, the minimum value is indeed -55/4 at x = -3/2. I think I've checked it thoroughly. Both methods give the same result, and plugging it back in confirms the minimum value. So, I'm confident that x = -3/2 is the correct answer.

💡Alright, so I have this problem here where I need to find a positive integer ( N ) such that the function ( f(n) = frac{d(n)}{sqrt[3]{n}} ) is maximized. Here, ( d(n) ) represents the number of positive divisors of ( n ). The question also mentions that this ( N ) is unique, meaning there's only one such number where ( f(N) ) is greater than ( f(n) ) for any other positive integer ( n ). Once I find this ( N ), I need to calculate the sum of its digits, and the options given are from 5 to 9.First, I need to understand what ( d(n) ) means. It's the number of positive divisors of ( n ). For example, ( d(1) = 1 ) because 1 only has one divisor, which is itself. ( d(2) = 2 ) because 2 has two divisors: 1 and 2. Similarly, ( d(12) = 6 ) because 12 has six divisors: 1, 2, 3, 4, 6, and 12.So, ( f(n) ) is taking the number of divisors of ( n ) and dividing it by the cube root of ( n ). The goal is to find the ( N ) where this ratio is the highest possible, and it's unique.I remember that the number of divisors function, ( d(n) ), is related to the prime factorization of ( n ). If ( n ) can be expressed as ( n = p_1^{e_1} p_2^{e_2} dots p_k^{e_k} ), where ( p_i ) are prime numbers and ( e_i ) are their respective exponents, then the number of divisors is given by ( d(n) = (e_1 + 1)(e_2 + 1) dots (e_k + 1) ).So, for example, if ( n = 12 = 2^2 times 3^1 ), then ( d(n) = (2+1)(1+1) = 3 times 2 = 6 ), which matches the given example.Given that, ( f(n) = frac{d(n)}{sqrt[3]{n}} ) can be rewritten in terms of the prime factors. Let me try to express ( f(n) ) using the prime factorization:[ f(n) = frac{(e_1 + 1)(e_2 + 1) dots (e_k + 1)}{sqrt[3]{p_1^{e_1} p_2^{e_2} dots p_k^{e_k}}} ]This simplifies to:[ f(n) = prod_{i=1}^{k} frac{e_i + 1}{p_i^{e_i/3}} ]So, each prime factor contributes a term ( frac{e_i + 1}{p_i^{e_i/3}} ) to the product. Therefore, to maximize ( f(n) ), we need to choose primes and their exponents such that each term ( frac{e_i + 1}{p_i^{e_i/3}} ) is as large as possible.I think the strategy here is to consider small primes because larger primes raised to any power will make the denominator too large, reducing the overall value of the term. So, small primes like 2, 3, 5, etc., are more promising.Let me start by considering the prime 2. Let's see how the term ( frac{e + 1}{2^{e/3}} ) behaves as ( e ) increases.For ( e = 1 ):[ frac{1 + 1}{2^{1/3}} = frac{2}{2^{1/3}} approx frac{2}{1.26} approx 1.587 ]For ( e = 2 ):[ frac{2 + 1}{2^{2/3}} = frac{3}{2^{2/3}} approx frac{3}{1.587} approx 1.89 ]For ( e = 3 ):[ frac{3 + 1}{2^{3/3}} = frac{4}{2} = 2 ]For ( e = 4 ):[ frac{4 + 1}{2^{4/3}} = frac{5}{2.519} approx 1.984 ]For ( e = 5 ):[ frac{5 + 1}{2^{5/3}} = frac{6}{3.1748} approx 1.89 ]For ( e = 6 ):[ frac{6 + 1}{2^{6/3}} = frac{7}{4} = 1.75 ]So, it seems that for the prime 2, the term ( frac{e + 1}{2^{e/3}} ) peaks at ( e = 3 ) with a value of 2, and then starts decreasing. So, the optimal exponent for prime 2 is 3.Now, let's do the same for the next prime, which is 3.For ( e = 1 ):[ frac{1 + 1}{3^{1/3}} = frac{2}{1.442} approx 1.386 ]For ( e = 2 ):[ frac{2 + 1}{3^{2/3}} = frac{3}{2.080} approx 1.442 ]For ( e = 3 ):[ frac{3 + 1}{3^{3/3}} = frac{4}{3} approx 1.333 ]For ( e = 4 ):[ frac{4 + 1}{3^{4/3}} = frac{5}{4.326} approx 1.156 ]So, for prime 3, the term peaks at ( e = 2 ) with approximately 1.442.Next, let's check prime 5.For ( e = 1 ):[ frac{1 + 1}{5^{1/3}} = frac{2}{1.710} approx 1.169 ]For ( e = 2 ):[ frac{2 + 1}{5^{2/3}} = frac{3}{2.924} approx 1.026 ]For ( e = 3 ):[ frac{3 + 1}{5^{3/3}} = frac{4}{5} = 0.8 ]So, for prime 5, the term is maximized at ( e = 1 ) with approximately 1.169.Similarly, for prime 7:For ( e = 1 ):[ frac{1 + 1}{7^{1/3}} = frac{2}{1.913} approx 1.046 ]For ( e = 2 ):[ frac{2 + 1}{7^{2/3}} = frac{3}{3.659} approx 0.820 ]So, even lower. Thus, for primes larger than 3, the optimal exponent is 1, and even then, their contribution is lower compared to primes 2 and 3.Therefore, it seems that the optimal ( N ) should be composed of the primes 2 and 3, with exponents 3 and 2 respectively, as these give the highest contributions.So, let's compute ( N = 2^3 times 3^2 = 8 times 9 = 72 ).Wait, but hold on. The options given are 5, 6, 7, 8, 9. 72 is not among them. So, perhaps I made a mistake.Wait, the problem says "the sum of the digits of ( N )". So, if ( N = 72 ), the sum of its digits is 7 + 2 = 9, which is one of the options. Hmm, so maybe 9 is the answer? But wait, let me verify.Is 72 indeed the unique ( N ) that maximizes ( f(n) )?Wait, let's compute ( f(72) ).First, ( d(72) ). The prime factorization of 72 is ( 2^3 times 3^2 ). So, ( d(72) = (3+1)(2+1) = 4 times 3 = 12 ).Then, ( sqrt[3]{72} approx 4.160 ).Thus, ( f(72) = frac{12}{4.160} approx 2.885 ).Now, let's check ( f(36) ). ( 36 = 2^2 times 3^2 ). So, ( d(36) = (2+1)(2+1) = 9 ). ( sqrt[3]{36} approx 3.301 ). Thus, ( f(36) = frac{9}{3.301} approx 2.726 ), which is less than ( f(72) ).What about ( f(48) )? ( 48 = 2^4 times 3^1 ). So, ( d(48) = (4+1)(1+1) = 5 times 2 = 10 ). ( sqrt[3]{48} approx 3.634 ). Thus, ( f(48) = frac{10}{3.634} approx 2.752 ), still less than ( f(72) ).How about ( f(24) )? ( 24 = 2^3 times 3^1 ). ( d(24) = (3+1)(1+1) = 4 times 2 = 8 ). ( sqrt[3]{24} approx 2.884 ). So, ( f(24) = frac{8}{2.884} approx 2.773 ), still less than 72.What about ( f(64) )? ( 64 = 2^6 ). ( d(64) = 6 + 1 = 7 ). ( sqrt[3]{64} = 4 ). So, ( f(64) = frac{7}{4} = 1.75 ), which is much less.What about ( f(144) )? ( 144 = 2^4 times 3^2 ). ( d(144) = (4+1)(2+1) = 5 times 3 = 15 ). ( sqrt[3]{144} approx 5.241 ). So, ( f(144) = frac{15}{5.241} approx 2.862 ), which is still less than 72's 2.885.Wait, so 72 is higher than 144. So, perhaps 72 is the maximum.But let's check ( f(36) ) again, which was 2.726, and ( f(72) ) is 2.885. So, 72 is higher.What about ( f(216) )? ( 216 = 2^3 times 3^3 ). ( d(216) = (3+1)(3+1) = 16 ). ( sqrt[3]{216} = 6 ). So, ( f(216) = frac{16}{6} approx 2.666 ), which is less than 72.So, 72 seems higher.Wait, but let me check ( f(54) ). ( 54 = 2^1 times 3^3 ). ( d(54) = (1+1)(3+1) = 2 times 4 = 8 ). ( sqrt[3]{54} approx 3.78 ). So, ( f(54) = frac{8}{3.78} approx 2.116 ), which is less.How about ( f(18) )? ( 18 = 2^1 times 3^2 ). ( d(18) = (1+1)(2+1) = 2 times 3 = 6 ). ( sqrt[3]{18} approx 2.620 ). So, ( f(18) = frac{6}{2.620} approx 2.290 ), still less.Wait, is there a number smaller than 72 that might have a higher ( f(n) )?Let me check ( f(36) ) again: 2.726.What about ( f(24) ): 2.773.Wait, 24 is 2^3 * 3^1, so exponents 3 and 1. Maybe if I adjust exponents, I can get a higher value.Wait, let's try ( N = 2^3 * 3^2 = 72 ), which gives us 12 divisors and cube root ~4.16, so 12 / 4.16 ~2.885.If I try ( N = 2^4 * 3^2 = 16 * 9 = 144 ), which gives 15 divisors, cube root ~5.24, so 15 / 5.24 ~2.862, which is less.If I try ( N = 2^2 * 3^3 = 4 * 27 = 108 ). ( d(108) = (2+1)(3+1) = 3 * 4 = 12 ). ( sqrt[3]{108} approx 4.762 ). So, ( f(108) = 12 / 4.762 approx 2.520 ), which is less.What if I try ( N = 2^3 * 3^1 = 24 ), which we already saw gives 2.773.Alternatively, ( N = 2^3 * 3^2 * 5^1 = 72 * 5 = 360 ). ( d(360) = (3+1)(2+1)(1+1) = 4 * 3 * 2 = 24 ). ( sqrt[3]{360} approx 7.106 ). So, ( f(360) = 24 / 7.106 approx 3.38 ). Wait, that's higher than 72's 2.885.Wait, that can't be. Because 360 is a larger number, but it's getting a higher ratio? Hmm, maybe I made a mistake.Wait, let me recalculate ( f(360) ). ( d(360) = 24 ). ( sqrt[3]{360} approx 7.106 ). So, 24 / 7.106 ≈ 3.38. That's actually higher than 72's 2.885.Wait, but that contradicts my earlier assumption that 72 is the maximum. So, perhaps I need to consider more primes.Wait, but earlier, when I considered primes beyond 3, like 5, the contribution was lower. But in this case, adding 5^1 to 72 gives a higher ( f(n) ). So, maybe I need to explore more.Wait, let me check ( f(360) ). Is that correct? 360 is 2^3 * 3^2 * 5^1. So, ( d(360) = (3+1)(2+1)(1+1) = 4*3*2 = 24 ). ( sqrt[3]{360} approx 7.106 ). So, 24 / 7.106 ≈ 3.38.But wait, let me check ( f(720) ). 720 is 2^4 * 3^2 * 5^1. ( d(720) = (4+1)(2+1)(1+1) = 5*3*2 = 30 ). ( sqrt[3]{720} approx 8.96 ). So, ( f(720) = 30 / 8.96 ≈ 3.35 ), which is slightly less than 360's 3.38.Wait, so 360 is higher than 720.What about ( N = 2^3 * 3^2 * 5^1 * 7^1 = 72 * 5 * 7 = 2520 ). ( d(2520) = (3+1)(2+1)(1+1)(1+1) = 4*3*2*2 = 48 ). ( sqrt[3]{2520} ≈ 13.6 ). So, ( f(2520) = 48 / 13.6 ≈ 3.529 ), which is higher than 360.Wait, so adding more primes seems to increase ( f(n) ). But that can't be, because as we add more primes, the cube root of ( n ) increases more rapidly than the number of divisors.Wait, but in the case of 2520, it's 48 / 13.6 ≈ 3.529, which is higher than 360's 3.38. So, perhaps adding more primes can actually increase ( f(n) ).Wait, let's try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 = 2520 * 11 = 27720 ). ( d(27720) = (3+1)(2+1)(1+1)(1+1)(1+1) = 4*3*2*2*2 = 96 ). ( sqrt[3]{27720} ≈ 30.2 ). So, ( f(27720) = 96 / 30.2 ≈ 3.18 ), which is less than 2520's 3.529.Hmm, so adding 11^1 actually decreased the ratio.Wait, so maybe there is an optimal number of primes to include. Let's see.Wait, 2520 gives 3.529, which is higher than 360's 3.38. Let's check if adding another prime beyond 7 would help.Wait, let's try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1 = 27720 * 13 = 360360 ). ( d(360360) = (3+1)(2+1)(1+1)(1+1)(1+1)(1+1) = 4*3*2*2*2*2 = 192 ). ( sqrt[3]{360360} ≈ 71.1 ). So, ( f(360360) = 192 / 71.1 ≈ 2.70 ), which is less than 2520's 3.529.So, adding more primes beyond 7 actually decreases the ratio.Wait, so maybe 2520 is better than 360, but 2520 is 2^3 * 3^2 * 5 * 7. Let's see if we can adjust the exponents to get a higher ratio.Wait, let's try ( N = 2^4 * 3^3 * 5^1 * 7^1 ). So, exponents 4,3,1,1. Then, ( d(N) = (4+1)(3+1)(1+1)(1+1) = 5*4*2*2 = 80 ). ( N = 16 * 27 * 5 * 7 = 16*27=432, 432*5=2160, 2160*7=15120 ). ( sqrt[3]{15120} ≈ 24.7 ). So, ( f(N) = 80 / 24.7 ≈ 3.24 ), which is less than 2520's 3.529.Wait, so increasing exponents beyond 3 for 2 and 2 for 3 actually decreases the ratio.Wait, let's try ( N = 2^3 * 3^3 * 5^1 * 7^1 ). So, exponents 3,3,1,1. ( d(N) = (3+1)(3+1)(1+1)(1+1) = 4*4*2*2 = 64 ). ( N = 8 * 27 * 5 * 7 = 8*27=216, 216*5=1080, 1080*7=7560 ). ( sqrt[3]{7560} ≈ 19.6 ). So, ( f(N) = 64 / 19.6 ≈ 3.265 ), still less than 2520's 3.529.Wait, so maybe 2520 is better. Let's see if we can adjust exponents differently.Wait, let's try ( N = 2^3 * 3^2 * 5^2 * 7^1 ). So, exponents 3,2,2,1. ( d(N) = (3+1)(2+1)(2+1)(1+1) = 4*3*3*2 = 72 ). ( N = 8 * 9 * 25 * 7 = 8*9=72, 72*25=1800, 1800*7=12600 ). ( sqrt[3]{12600} ≈ 23.2 ). So, ( f(N) = 72 / 23.2 ≈ 3.103 ), still less than 2520's 3.529.Hmm, so 2520 seems better. Let me check if 2520 is indeed the maximum.Wait, let's try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 ). Wait, that's 27720, which we saw earlier gives ( f(N) ≈ 3.18 ), which is less than 2520's 3.529.Wait, so 2520 is better. Let me check if there's a number between 72 and 2520 that might have a higher ratio.Wait, let's try ( N = 2^3 * 3^2 * 5^1 * 7^1 = 2520 ). So, that's the same as before.Wait, perhaps if I remove one prime, like 7, and see what happens.Wait, ( N = 2^3 * 3^2 * 5^1 = 360 ). ( f(360) ≈ 3.38 ), which is less than 2520's 3.529.Wait, so 2520 is better.Wait, but the problem says that ( N ) is unique. So, if 2520 is the maximum, but the options are 5,6,7,8,9, which are single-digit sums, but 2520's digit sum is 2+5+2+0=9, which is one of the options. So, maybe 9 is the answer.But wait, earlier I thought 72 was the maximum, but 2520 is higher. So, perhaps I need to check if 2520 is indeed the maximum.Wait, let me check ( f(5040) ). 5040 is 2^4 * 3^2 * 5^1 * 7^1. ( d(5040) = (4+1)(2+1)(1+1)(1+1) = 5*3*2*2 = 60 ). ( sqrt[3]{5040} ≈ 17.15 ). So, ( f(5040) = 60 / 17.15 ≈ 3.5 ), which is slightly less than 2520's 3.529.Wait, so 2520 is higher than 5040.Wait, let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1 ). That's 360360, which we saw earlier gives ( f(N) ≈ 2.70 ), which is less.Wait, so 2520 is higher than 5040 and 360360.Wait, let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 ). That's 27720, which gives ( f(N) ≈ 3.18 ), which is less than 2520.Wait, so 2520 seems to be the maximum. But let me check if there's a number with more primes but higher exponents.Wait, let's try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 ). That's a very large number, but let's compute ( f(N) ).( d(N) = (3+1)(2+1)(1+1)(1+1)(1+1)(1+1)(1+1) = 4*3*2*2*2*2*2 = 192 ).( N = 8 * 9 * 5 * 7 * 11 * 13 * 17 ). Let's compute that:8*9=7272*5=360360*7=25202520*11=2772027720*13=360360360360*17=6126120So, ( sqrt[3]{6126120} ≈ 183 ). So, ( f(N) = 192 / 183 ≈ 1.049 ), which is way less than 2520's 3.529.So, adding more primes beyond 7 actually decreases the ratio.Wait, so 2520 seems to be the maximum. But let me check if there's a number with exponents higher than 3 for 2 or 3.Wait, let's try ( N = 2^4 * 3^3 * 5^1 * 7^1 ). So, exponents 4,3,1,1. ( d(N) = (4+1)(3+1)(1+1)(1+1) = 5*4*2*2 = 80 ). ( N = 16 * 27 * 5 * 7 = 16*27=432, 432*5=2160, 2160*7=15120 ). ( sqrt[3]{15120} ≈ 24.7 ). So, ( f(N) = 80 / 24.7 ≈ 3.24 ), which is less than 2520's 3.529.Wait, so increasing exponents beyond 3 for 2 and 2 for 3 actually decreases the ratio.Wait, let me try ( N = 2^3 * 3^3 * 5^1 * 7^1 ). So, exponents 3,3,1,1. ( d(N) = (3+1)(3+1)(1+1)(1+1) = 4*4*2*2 = 64 ). ( N = 8 * 27 * 5 * 7 = 8*27=216, 216*5=1080, 1080*7=7560 ). ( sqrt[3]{7560} ≈ 19.6 ). So, ( f(N) = 64 / 19.6 ≈ 3.265 ), still less than 2520's 3.529.Wait, so 2520 is still higher.Wait, let me check ( N = 2^3 * 3^2 * 5^2 * 7^1 ). So, exponents 3,2,2,1. ( d(N) = (3+1)(2+1)(2+1)(1+1) = 4*3*3*2 = 72 ). ( N = 8 * 9 * 25 * 7 = 8*9=72, 72*25=1800, 1800*7=12600 ). ( sqrt[3]{12600} ≈ 23.2 ). So, ( f(N) = 72 / 23.2 ≈ 3.103 ), still less than 2520's 3.529.Wait, so 2520 is still the maximum.Wait, but let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 ). That's 27720, which gives ( f(N) ≈ 3.18 ), which is less than 2520's 3.529.Wait, so 2520 is higher.Wait, but let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1 ). That's 360360, which gives ( f(N) ≈ 2.70 ), which is less.Wait, so 2520 is the highest so far.Wait, but let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 ). That's 6126120, which gives ( f(N) ≈ 1.049 ), which is way less.So, it seems that 2520 is the maximum.Wait, but the problem says that ( N ) is unique. So, is 2520 the unique maximum? Or is there another number with the same ( f(n) )?Wait, let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 ). That's 2520, which gives ( f(N) ≈ 3.529 ).Is there another number with the same ( f(n) )?Wait, let's try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 ). That's 27720, which gives ( f(N) ≈ 3.18 ), which is less.Wait, let me try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 13^1 ). That's 2520 * 13 = 32760. ( d(32760) = (3+1)(2+1)(1+1)(1+1)(1+1) = 4*3*2*2*2 = 96 ). ( sqrt[3]{32760} ≈ 31.9 ). So, ( f(N) = 96 / 31.9 ≈ 3.01 ), which is less than 2520's 3.529.Wait, so 2520 is unique in that it's the highest.Wait, but the problem says that ( N ) is unique, so 2520 is the maximum. But the options are 5,6,7,8,9. So, the sum of the digits of 2520 is 2+5+2+0=9, which is option E.But wait, earlier I thought 72 was the maximum, but 2520 is higher. So, maybe the answer is 9.Wait, but let me check if there's a smaller number than 2520 that gives a higher ( f(n) ).Wait, let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 ). That's 27720, which gives ( f(N) ≈ 3.18 ), which is less than 2520's 3.529.Wait, let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1 ). That's 360360, which gives ( f(N) ≈ 2.70 ), which is less.Wait, so 2520 is the maximum.But wait, let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1 * 17^1 ). That's 6126120, which gives ( f(N) ≈ 1.049 ), which is way less.So, 2520 is the maximum.But wait, the problem says that ( N ) is unique. So, is there another number with the same ( f(n) )?Wait, let me check ( N = 2^3 * 3^2 * 5^1 * 7^1 ). That's 2520, which gives ( f(N) ≈ 3.529 ).Is there another number with the same ( f(n) )?Wait, let me try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 ). That's 27720, which gives ( f(N) ≈ 3.18 ), which is less.Wait, let me try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 13^1 ). That's 32760, which gives ( f(N) ≈ 3.01 ), which is less.Wait, let me try ( N = 2^3 * 3^2 * 5^1 * 7^1 * 17^1 ). That's 2520 * 17 = 42840. ( d(42840) = (3+1)(2+1)(1+1)(1+1)(1+1) = 4*3*2*2*2 = 96 ). ( sqrt[3]{42840} ≈ 35.0 ). So, ( f(N) = 96 / 35 ≈ 2.74 ), which is less than 2520's 3.529.Wait, so 2520 is unique in that it's the highest.Therefore, the sum of the digits of ( N = 2520 ) is 2 + 5 + 2 + 0 = 9.So, the answer is E) 9.