answer：Okay, so I have this problem where Alex is painting cells inside a 6x6 square on graph paper. After he paints some cells, he marks the nodes (which are the intersections of the graph paper lines) that are adjacent to an equal number of painted and unpainted squares. The question is asking for the maximum number of nodes that can be marked.First, I need to visualize the grid. A 6x6 grid means there are 6 rows and 6 columns of cells. Each cell is a square, and the intersections of the grid lines are the nodes. So, for a 6x6 grid, there are 7 horizontal lines and 7 vertical lines, making a total of 7x7 = 49 nodes.Now, each node is adjacent to some number of cells. The number of adjacent cells depends on the position of the node:- Corner nodes are adjacent to only 1 cell.- Edge nodes (but not corners) are adjacent to 2 cells.- Internal nodes are adjacent to 4 cells.The problem states that a node is marked if it is adjacent to an equal number of painted and unpainted cells. So, for a node to be marked, the number of painted cells adjacent to it must equal the number of unpainted cells adjacent to it.Let's break this down for each type of node:1. **Corner Nodes**: These are adjacent to only 1 cell. Since 1 is an odd number, it's impossible to have an equal number of painted and unpainted cells adjacent to them. Therefore, corner nodes cannot be marked. 2. **Edge Nodes (excluding corners)**: These are adjacent to 2 cells. To have an equal number of painted and unpainted cells, each edge node must be adjacent to 1 painted and 1 unpainted cell. This is possible, so edge nodes can potentially be marked.3. **Internal Nodes**: These are adjacent to 4 cells. To have an equal number of painted and unpainted cells, each internal node must be adjacent to 2 painted and 2 unpainted cells. This is also possible, so internal nodes can potentially be marked.Now, let's count how many nodes there are of each type:- **Corner Nodes**: There are 4 corners in the grid.- **Edge Nodes (excluding corners)**: Each side of the grid has 6 nodes, but we exclude the corners. So, each side has 6 - 2 = 4 edge nodes. Since there are 4 sides, the total number of edge nodes is 4 x 4 = 16.- **Internal Nodes**: The total number of nodes is 49. Subtracting the 4 corner nodes and 16 edge nodes gives 49 - 4 - 16 = 29 internal nodes.So, corner nodes cannot be marked, but edge and internal nodes can potentially be marked. However, just because they can be marked doesn't mean they will be. We need to find the maximum number of nodes that can be marked, which means we need to find a way to paint the cells such that as many edge and internal nodes as possible meet the condition of being adjacent to an equal number of painted and unpainted cells.Let's think about how to maximize the number of marked nodes. For edge nodes, each needs to be adjacent to 1 painted and 1 unpainted cell. For internal nodes, each needs to be adjacent to 2 painted and 2 unpainted cells.One strategy might be to alternate painted and unpainted cells in such a way that each node meets the required condition. For example, if we paint every other cell in a checkerboard pattern, each internal node would be adjacent to 2 painted and 2 unpainted cells, satisfying the condition. Similarly, each edge node would be adjacent to 1 painted and 1 unpainted cell, also satisfying the condition.But wait, let's test this idea. If we paint a checkerboard pattern on the 6x6 grid, each internal node would indeed be adjacent to 2 painted and 2 unpainted cells. However, edge nodes would also be adjacent to 1 painted and 1 unpainted cell, as each edge node is adjacent to two cells, and in a checkerboard pattern, these two cells would alternate between painted and unpainted.But hold on, in a checkerboard pattern, the corner cells would be either all painted or all unpainted, depending on where you start. Since corner nodes are adjacent to only 1 cell, and if that cell is painted, the corner node would be adjacent to 1 painted and 0 unpainted cells, which doesn't satisfy the condition. Similarly, if the corner cell is unpainted, the corner node would be adjacent to 0 painted and 1 unpainted cell, which also doesn't satisfy the condition. But as we established earlier, corner nodes cannot be marked because they are adjacent to an odd number of cells, so this doesn't affect our count.So, in a checkerboard pattern, all edge nodes and all internal nodes would be marked. That would give us 16 edge nodes + 29 internal nodes = 45 marked nodes. But wait, is this possible? Let's think about the total number of painted cells.In a 6x6 grid, a checkerboard pattern would result in 18 painted cells and 18 unpainted cells (since 6x6 = 36, and half of 36 is 18). However, each internal node is adjacent to 4 cells, and each edge node is adjacent to 2 cells. If we mark all edge and internal nodes, we need to ensure that the painting of the cells satisfies the condition for all these nodes simultaneously.But is this possible? Let's consider the constraints. Each internal node needs 2 painted and 2 unpainted adjacent cells. Each edge node needs 1 painted and 1 unpainted adjacent cell. If we paint the grid in a checkerboard pattern, this condition is satisfied for all nodes except the corners, which we already determined cannot be marked.Therefore, in this case, the maximum number of marked nodes would indeed be 45. However, I need to verify if this is achievable or if there's a limitation I'm missing.Wait a minute, let's think about the parity of the grid. A 6x6 grid has even dimensions, so a perfect checkerboard pattern is possible without any conflicts. Each row alternates between painted and unpainted cells, and each column does the same. This ensures that every internal node is adjacent to 2 painted and 2 unpainted cells, and every edge node is adjacent to 1 painted and 1 unpainted cell.Therefore, it seems that the maximum number of marked nodes is indeed 45. However, I need to make sure that this is the case and that there isn't a configuration where more nodes can be marked.Is there a way to mark more than 45 nodes? Well, since there are only 49 nodes in total and 4 of them are corners which cannot be marked, the maximum possible number of marked nodes is 45. Therefore, 45 is indeed the maximum.But just to be thorough, let's consider if there's any other configuration that might allow more nodes to be marked. For example, what if we paint all cells? Then, every node would be adjacent to only painted cells, so no nodes would be marked. Similarly, if we paint no cells, no nodes would be marked either. So, the checkerboard pattern seems to be the optimal configuration.Another thought: what if we paint more cells in certain areas to try to satisfy the condition for more nodes? But since each node's condition depends on its adjacent cells, painting more cells in one area might disrupt the balance in another area. For example, if we paint an extra cell near an edge, it might cause some edge nodes to have more painted cells than unpainted, thus making them ineligible for marking.Therefore, the checkerboard pattern, which evenly distributes painted and unpainted cells, seems to be the best approach to maximize the number of marked nodes.In conclusion, the maximum number of nodes that can be marked is 45.

answer：First, I need to determine the relationship between the front and back wheels of the tandem bicycle. Since there is no slippage, the distance traveled by both wheels must be the same.I'll start by calculating the circumference of each wheel, which is the distance it covers in one full revolution. The circumference of a circle is given by the formula ( C = pi times text{diameter} ).For the front wheel with a diameter of 40 inches:[C_{text{front}} = pi times 40 = 40pi text{ inches}]For the back wheel with a diameter of 20 inches:[C_{text{back}} = pi times 20 = 20pi text{ inches}]Next, I'll find the total distance traveled by the front wheel when it makes 150 revolutions:[text{Distance} = 150 times 40pi = 6000pi text{ inches}]Since the back wheel must cover the same distance without slippage, I'll calculate the number of revolutions it makes:[text{Revolutions}_{text{back}} = frac{6000pi}{20pi} = 300]Therefore, the back wheel makes 300 revolutions while the front wheel makes 150 revolutions.

answer：Okay, so I have this problem about a hyperbola, and I need to find the equations of its asymptotes. Let me try to understand what's given and what I need to do.First, the hyperbola is given by the equation (frac{x^2}{a^2} - y^2 = 1), where (a > 0). I know that the standard form of a hyperbola centered at the origin opening horizontally is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1). Comparing this to the given equation, it looks like (b^2 = 1), so (b = 1). The asymptotes of a hyperbola are the lines that the hyperbola approaches but never touches. For a hyperbola of the form (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), the equations of the asymptotes are (y = pm frac{b}{a}x). So in this case, the asymptotes should be (y = pm frac{1}{a}x). That means if I can find the value of (a), I can determine the equations of the asymptotes.Now, the problem mentions a point (P) on the hyperbola. Through this point (P), two lines are drawn parallel to the asymptotes. These lines intersect the asymptotes at points (A) and (B). The figure formed by these intersections and the origin (O) is a parallelogram (OBPA), and its area is given as 1. So, I need to use this information to find (a).Let me try to visualize this. The hyperbola is centered at the origin, opening to the left and right. Point (P) is somewhere on the hyperbola. From (P), I draw two lines: one parallel to one asymptote and the other parallel to the other asymptote. These lines will intersect the asymptotes at points (A) and (B). Connecting these points with the origin (O) and point (P) forms a parallelogram. The area of this parallelogram is 1.I think I need to find the coordinates of point (P), then find the equations of the lines through (P) parallel to the asymptotes, find their intersection points (A) and (B) with the asymptotes, and then calculate the area of the parallelogram (OBPA). Setting this area equal to 1 should give me an equation to solve for (a).Let me denote point (P) as ((m, n)). Since (P) lies on the hyperbola, it must satisfy the equation (frac{m^2}{a^2} - n^2 = 1).The asymptotes of the hyperbola are (y = frac{1}{a}x) and (y = -frac{1}{a}x). So, the slopes of these asymptotes are (frac{1}{a}) and (-frac{1}{a}).Now, I need to find the equations of the lines through (P(m, n)) that are parallel to these asymptotes. 1. The line parallel to the asymptote (y = frac{1}{a}x) will have the same slope, (frac{1}{a}). The equation of this line can be written using point-slope form: (y - n = frac{1}{a}(x - m)).2. Similarly, the line parallel to the asymptote (y = -frac{1}{a}x) will have a slope of (-frac{1}{a}). Its equation is (y - n = -frac{1}{a}(x - m)).These two lines intersect the asymptotes at points (A) and (B). Let me find the coordinates of these intersection points.First, let's find point (A), which is the intersection of the line through (P) parallel to (y = frac{1}{a}x) and the asymptote (y = -frac{1}{a}x).Wait, hold on. If the line is parallel to one asymptote, it should intersect the other asymptote. So, the line parallel to (y = frac{1}{a}x) should intersect (y = -frac{1}{a}x), and the line parallel to (y = -frac{1}{a}x) should intersect (y = frac{1}{a}x). That makes sense because if they were parallel to the same asymptote, they wouldn't intersect.So, let's correct that:- The line through (P) with slope (frac{1}{a}) will intersect the asymptote (y = -frac{1}{a}x) at point (A).- The line through (P) with slope (-frac{1}{a}) will intersect the asymptote (y = frac{1}{a}x) at point (B).Let me find the coordinates of point (A):We have two equations:1. (y - n = frac{1}{a}(x - m)) (the line through (P) with slope (frac{1}{a}))2. (y = -frac{1}{a}x) (the asymptote)Substitute equation 2 into equation 1:(-frac{1}{a}x - n = frac{1}{a}(x - m))Multiply both sides by (a) to eliminate denominators:(-x - a n = x - m)Bring all terms to one side:(-x - a n - x + m = 0)Combine like terms:(-2x + m - a n = 0)Solve for (x):(-2x = -m + a n)(x = frac{m - a n}{2})Now, substitute this back into equation 2 to find (y):(y = -frac{1}{a} cdot frac{m - a n}{2} = -frac{m - a n}{2a} = frac{a n - m}{2a})So, point (A) has coordinates (left( frac{m - a n}{2}, frac{a n - m}{2a} right)).Similarly, let's find point (B), the intersection of the line through (P) with slope (-frac{1}{a}) and the asymptote (y = frac{1}{a}x).We have two equations:1. (y - n = -frac{1}{a}(x - m))2. (y = frac{1}{a}x)Substitute equation 2 into equation 1:(frac{1}{a}x - n = -frac{1}{a}(x - m))Multiply both sides by (a):(x - a n = - (x - m))Simplify:(x - a n = -x + m)Bring all terms to one side:(x + x - a n - m = 0)Combine like terms:(2x - a n - m = 0)Solve for (x):(2x = a n + m)(x = frac{a n + m}{2})Substitute back into equation 2 to find (y):(y = frac{1}{a} cdot frac{a n + m}{2} = frac{a n + m}{2a})So, point (B) has coordinates (left( frac{a n + m}{2}, frac{a n + m}{2a} right)).Now, I have points (O(0, 0)), (B), (P(m, n)), and (A). The figure (OBPA) is a parallelogram. To find its area, I can use the shoelace formula or vector cross product.But since it's a parallelogram, the area can also be calculated as the magnitude of the cross product of vectors (overrightarrow{OB}) and (overrightarrow{OA}).Wait, actually, in a parallelogram, the area is the base times the height, but since it's defined by vectors, the area is the magnitude of the cross product of the vectors defining two adjacent sides.Alternatively, since (OBPA) is a parallelogram, the area can be found using the determinant of a matrix formed by vectors (overrightarrow{OB}) and (overrightarrow{OP}), but I need to make sure I'm using the correct vectors.Wait, maybe it's simpler to use coordinates and the shoelace formula.Let me list the coordinates of the parallelogram:- (O(0, 0))- (Bleft( frac{a n + m}{2}, frac{a n + m}{2a} right))- (P(m, n))- (Aleft( frac{m - a n}{2}, frac{a n - m}{2a} right))But wait, is this the correct order? Let me make sure the points are in order, either clockwise or counterclockwise.Starting from (O), moving to (B), then to (P), then to (A), and back to (O). Let me verify if this forms a parallelogram.Alternatively, perhaps it's better to consider vectors.Let me think of vectors from the origin:- Vector (overrightarrow{OB}) is from (O) to (B): (left( frac{a n + m}{2}, frac{a n + m}{2a} right))- Vector (overrightarrow{OA}) is from (O) to (A): (left( frac{m - a n}{2}, frac{a n - m}{2a} right))But in a parallelogram, the vectors defining the sides from a common vertex should be equal. So, from (O), the sides are (OB) and (OA), but actually, in the parallelogram (OBPA), the sides are (OB) and (OP), but I need to be careful.Wait, maybe I should consider the vectors defining the sides of the parallelogram. Since it's (OBPA), the sides are (OB) and (PA), but (PA) is equal to (OB), and (BP) is equal to (OA). Hmm, this is getting a bit confusing.Alternatively, perhaps I can use the formula for the area of a parallelogram given by four points. The shoelace formula can be used for polygons, so maybe that's the way to go.The shoelace formula for a quadrilateral with vertices ((x_1, y_1)), ((x_2, y_2)), ((x_3, y_3)), ((x_4, y_4)) is:[text{Area} = frac{1}{2} |x_1 y_2 - x_2 y_1 + x_2 y_3 - x_3 y_2 + x_3 y_4 - x_4 y_3 + x_4 y_1 - x_1 y_4|]But since it's a parallelogram, I can also use the fact that the area is twice the area of triangle (OBP) or something similar. Wait, no, because it's a four-sided figure.Alternatively, since it's a parallelogram, the area can be calculated as the magnitude of the cross product of vectors (overrightarrow{OB}) and (overrightarrow{OP}). Wait, is that correct?Wait, no. In a parallelogram, the area is the magnitude of the cross product of two adjacent sides. So, if I consider vectors (overrightarrow{OB}) and (overrightarrow{OA}), then the area is (|overrightarrow{OB} times overrightarrow{OA}|).But let me verify this.Given points (O(0,0)), (B), (P), and (A), the vectors from (O) are (overrightarrow{OB}) and (overrightarrow{OA}). The area of the parallelogram formed by these vectors is indeed (|overrightarrow{OB} times overrightarrow{OA}|).But in our case, the parallelogram is (OBPA), so the sides are (OB) and (BP), but (BP) is equal to (OA). Hmm, maybe I need to think differently.Alternatively, perhaps it's better to use coordinates and compute the area using the shoelace formula.Let me list the coordinates in order:1. (O(0, 0))2. (Bleft( frac{a n + m}{2}, frac{a n + m}{2a} right))3. (P(m, n))4. (Aleft( frac{m - a n}{2}, frac{a n - m}{2a} right))Now, applying the shoelace formula:Compute the sum of (x_i y_{i+1}) and subtract the sum of (y_i x_{i+1}), then take half the absolute value.Let me compute each term step by step.First, list the coordinates in order, repeating the first at the end:1. (O(0, 0))2. (Bleft( frac{a n + m}{2}, frac{a n + m}{2a} right))3. (P(m, n))4. (Aleft( frac{m - a n}{2}, frac{a n - m}{2a} right))5. (O(0, 0))Compute the sum (S1 = x_1 y_2 + x_2 y_3 + x_3 y_4 + x_4 y_5):- (x_1 y_2 = 0 cdot frac{a n + m}{2a} = 0)- (x_2 y_3 = frac{a n + m}{2} cdot n = frac{n(a n + m)}{2})- (x_3 y_4 = m cdot frac{a n - m}{2a} = frac{m(a n - m)}{2a})- (x_4 y_5 = frac{m - a n}{2} cdot 0 = 0)So, (S1 = 0 + frac{n(a n + m)}{2} + frac{m(a n - m)}{2a} + 0)Simplify (S1):First term: (frac{n(a n + m)}{2})Second term: (frac{m(a n - m)}{2a} = frac{a n m - m^2}{2a} = frac{n m}{2} - frac{m^2}{2a})So, (S1 = frac{n(a n + m)}{2} + frac{n m}{2} - frac{m^2}{2a})Expand the first term:(frac{a n^2 + m n}{2} + frac{n m}{2} - frac{m^2}{2a})Combine like terms:(frac{a n^2}{2} + frac{m n}{2} + frac{m n}{2} - frac{m^2}{2a})Simplify:(frac{a n^2}{2} + m n - frac{m^2}{2a})Now, compute the sum (S2 = y_1 x_2 + y_2 x_3 + y_3 x_4 + y_4 x_5):- (y_1 x_2 = 0 cdot frac{a n + m}{2} = 0)- (y_2 x_3 = frac{a n + m}{2a} cdot m = frac{m(a n + m)}{2a})- (y_3 x_4 = n cdot frac{m - a n}{2} = frac{n(m - a n)}{2})- (y_4 x_5 = frac{a n - m}{2a} cdot 0 = 0)So, (S2 = 0 + frac{m(a n + m)}{2a} + frac{n(m - a n)}{2} + 0)Simplify (S2):First term: (frac{a n m + m^2}{2a} = frac{n m}{2} + frac{m^2}{2a})Second term: (frac{n m - a n^2}{2})So, (S2 = frac{n m}{2} + frac{m^2}{2a} + frac{n m}{2} - frac{a n^2}{2})Combine like terms:(frac{n m}{2} + frac{n m}{2} + frac{m^2}{2a} - frac{a n^2}{2})Simplify:(n m + frac{m^2}{2a} - frac{a n^2}{2})Now, the area is (frac{1}{2} |S1 - S2|):Compute (S1 - S2):[left( frac{a n^2}{2} + m n - frac{m^2}{2a} right) - left( n m + frac{m^2}{2a} - frac{a n^2}{2} right)]Simplify term by term:First, expand the subtraction:(frac{a n^2}{2} + m n - frac{m^2}{2a} - n m - frac{m^2}{2a} + frac{a n^2}{2})Combine like terms:- (frac{a n^2}{2} + frac{a n^2}{2} = a n^2)- (m n - n m = 0)- (-frac{m^2}{2a} - frac{m^2}{2a} = -frac{m^2}{a})So, (S1 - S2 = a n^2 - frac{m^2}{a})Therefore, the area is:[frac{1}{2} |a n^2 - frac{m^2}{a}| = 1]So,[frac{1}{2} |a n^2 - frac{m^2}{a}| = 1]Multiply both sides by 2:[|a n^2 - frac{m^2}{a}| = 2]Since (P(m, n)) lies on the hyperbola (frac{m^2}{a^2} - n^2 = 1), we can express (m^2) in terms of (n^2):[frac{m^2}{a^2} = 1 + n^2 implies m^2 = a^2(1 + n^2)]Substitute (m^2) into the area equation:[|a n^2 - frac{a^2(1 + n^2)}{a}| = 2]Simplify the expression inside the absolute value:[|a n^2 - a(1 + n^2)| = |a n^2 - a - a n^2| = |-a| = | -a | = a]So, we have:[a = 2]Therefore, (a = 2). Now, recall that the asymptotes of the hyperbola are (y = pm frac{1}{a}x). Substituting (a = 2), the equations of the asymptotes are:[y = pm frac{1}{2}x]Looking back at the options provided:A: (y = pm frac{1}{2}x)B: (y = pm frac{sqrt{2}}{2}x)C: (y = pm frac{1}{3}x)D: (y = pm frac{sqrt{3}}{3}x)So, the correct answer is option A.I think I followed the steps correctly, but let me just double-check if I made any mistakes in the calculations.Starting from the area expression, I used the shoelace formula, which should be correct. I found the coordinates of points (A) and (B) by solving the intersections, which seems right. Then, I set up the area equation and used the hyperbola equation to substitute (m^2). The simplification led me to (a = 2), which makes sense. So, the asymptotes are (y = pm frac{1}{2}x), which is option A.**Final Answer**boxed{A}

answer：Okay, so I have this problem about finite subsets of integers and two functions d₂(S) and d₃(S). Let me try to understand what these functions represent.For part (a), S is a set of consecutive integers from m to m+2019. I need to show that d₂(S) is greater than (13/7) times d₃(S). Hmm, okay. Let me recall what d₂(S) and d₃(S) mean.d₂(S) counts how many elements in S can be written as a difference of two squares. So, for each a in S, there exist integers x and y such that x² - y² = a. Similarly, d₃(S) counts how many elements in S can be written as a difference of two cubes, meaning x³ - y³ = a for some integers x and y.I think I remember that not all integers can be expressed as a difference of squares. There's some condition on the form of a. Let me think. If a number is a difference of squares, it can be factored as (x - y)(x + y). So, both (x - y) and (x + y) must be integers, and they must have the same parity because x and y are integers. That means either both are even or both are odd. So, their product will be either odd or divisible by 4. Therefore, numbers that are 2 mod 4 cannot be expressed as a difference of squares. So, numbers of the form 4k + 2 are excluded.So, in any set of consecutive integers, roughly 1/4 of them are of the form 4k + 2, which cannot be expressed as a difference of squares. Therefore, d₂(S) should be about 3/4 of the size of S. Since S has 2020 elements, d₂(S) should be roughly 1515.Now, for d₃(S), which counts the numbers expressible as a difference of cubes. I know that cubes modulo some number have certain properties. Maybe I can use modular arithmetic to find restrictions on a.Let me think about cubes modulo 7. The cubes modulo 7 are 0, 1, and 6. So, the difference of two cubes modulo 7 can be 0, 1, 2, 5, or 6. Similarly, modulo 9, the cubes are 0, 1, and 8. So, the difference of two cubes modulo 9 can be 0, 1, 2, 7, or 8.Using the Chinese Remainder Theorem, the possible residues modulo 63 (since 7 and 9 are coprime) would be combinations of these residues. So, each residue modulo 7 can pair with each residue modulo 9, giving 5 x 5 = 25 possible residues modulo 63. Therefore, the density of numbers expressible as a difference of cubes is at most 25/63, which is roughly 0.3968.So, in a set of 2020 numbers, d₃(S) should be roughly 2020 x 25/63 ≈ 804. So, d₂(S) is about 1515, and d₃(S) is about 804. Then, 1515 / 804 ≈ 1.885, which is greater than 13/7 ≈ 1.857. So, that shows d₂(S) > (13/7) d₃(S).Wait, but I need to make sure that these counts are exact or at least tight enough. Maybe I should compute the exact number of elements of the form 4k + 2 in S. Since S has 2020 elements, the number of elements congruent to 2 mod 4 is either 505 or 504, depending on m. So, d₂(S) is 2020 - number of 4k + 2 elements, which is at least 2020 - 505 = 1515. So, d₂(S) ≥ 1515.For d₃(S), since the density is 25/63, the maximum possible d₃(S) is 2020 x 25/63 ≈ 804. So, d₃(S) ≤ 804. Therefore, d₂(S) ≥ 1515 > (13/7) x 804 ≈ 1514. So, 1515 > 1514, which is just barely true. Hmm, maybe I need to adjust for exact counts.Alternatively, maybe the exact number of differences of cubes is less than 804, so the inequality holds.Okay, moving on to part (b). Here, S_n is {1, 2, ..., n}, and I need to show that for sufficiently large n, d₂(S_n) > 4 d₃(S_n).From part (a), I know that d₂(S_n) is roughly 3n/4, since only numbers of the form 4k + 2 are excluded. So, d₂(S_n) ≈ 3n/4.For d₃(S_n), I need to estimate how many numbers up to n can be expressed as a difference of cubes. I think the number of such numbers grows slower than linearly. Maybe it's roughly proportional to n^(2/3), since for each x, y, x³ - y³ can be up to n, so y is roughly up to n^(1/3), and x is up to n^(1/3) as well. So, the number of pairs (x, y) is roughly (n^(1/3))² = n^(2/3). But each difference can be repeated, so the actual number of distinct differences might be less.Wait, but in part (a), I used modular restrictions to bound d₃(S). Maybe I can use a similar approach here. The density of numbers expressible as differences of cubes is limited by the modular constraints. Earlier, I found that modulo 63, only 25 residues are possible. So, the density is 25/63, which is about 0.3968. So, d₃(S_n) is roughly 0.3968 n.But wait, that can't be right because in part (a), with n=2020, d₃(S) was about 804, which is roughly 0.3968 x 2020 ≈ 804. So, if I apply the same density, d₃(S_n) ≈ 0.3968 n.But then, d₂(S_n) ≈ 0.75 n, and 0.75 / 0.3968 ≈ 1.885, which is greater than 4? No, 1.885 is less than 4. So, that approach doesn't help.Wait, maybe I need a better estimate for d₃(S_n). Perhaps the number of differences of cubes up to n is much smaller. Let me think about how many distinct differences x³ - y³ there are for x, y ≤ n^(1/3). Each x can pair with y < x, so the number of differences is roughly (n^(1/3))² / 2, which is n^(2/3)/2. So, d₃(S_n) is O(n^(2/3)).Therefore, d₂(S_n) is Θ(n), and d₃(S_n) is Θ(n^(2/3)). So, as n grows, d₂(S_n) grows much faster than d₃(S_n). Therefore, for sufficiently large n, d₂(S_n) will be greater than 4 d₃(S_n).To make this precise, let me formalize it. Let me denote d₂(S_n) ≈ 3n/4 and d₃(S_n) ≈ C n^(2/3) for some constant C. Then, 3n/4 > 4 C n^(2/3). Dividing both sides by n^(2/3), we get 3/4 n^(1/3) > 4 C. Since n^(1/3) grows without bound, for sufficiently large n, this inequality will hold.Therefore, there exists an N such that for all n > N, d₂(S_n) > 4 d₃(S_n).I think that makes sense. So, the key idea is that d₂(S_n) grows linearly, while d₃(S_n) grows much more slowly, like n^(2/3), so eventually, d₂(S_n) will dominate.**Final Answer**(a) boxed{d_2(S) > frac{13}{7} d_3(S)}(b) boxed{d_2(S_n) > 4 d_3(S_n)}

answer：Alright, so I have this problem here about propositions and their negations. Let me try to wrap my head around it step by step. The proposition given is p: "There exists an x in real numbers such that x squared plus x minus one is less than zero." Okay, so in logical terms, that's like saying ∃x ∈ ℝ, x² + x - 1 < 0.Now, I need to find the negation of this proposition, which is ¬p. From what I remember, the negation of an existential quantifier (∃) is a universal quantifier (∀). So, if p is saying that there's at least one real number x that satisfies the inequality, then ¬p should be saying that for all real numbers x, the inequality does not hold. That makes sense because if it's not true that there's at least one x that makes the inequality true, then it must be that for every x, the inequality is false.So, applying that, the negation of p would be: For every x in real numbers, x squared plus x minus one is greater than or equal to zero. In symbols, that's ∀x ∈ ℝ, x² + x - 1 ≥ 0.Let me double-check that. The original statement is about existence, so the negation flips it to universality. Also, the inequality flips from less than to greater than or equal to. Yeah, that seems right.Looking at the options given:A: There exists an x in real numbers such that x squared plus x minus one is greater than zero.Hmm, that's still an existential statement, but with a different inequality. Not sure if that's the negation.B: For every x not in real numbers, x squared plus x minus one is greater than zero.Wait, x not in real numbers? That doesn't make sense because the original statement is about real numbers. The negation should still be about real numbers, right?C: There exists an x not in real numbers, x squared plus x minus one is greater than or equal to zero.Again, talking about x not in real numbers. That seems off because the original proposition was about real numbers. The negation should pertain to the same domain.D: For every x in real numbers, x squared plus x minus one is greater than or equal to zero.This matches what I thought earlier. It's a universal statement about all real numbers, and the inequality is flipped to greater than or equal to. So, this must be the correct negation.Just to be thorough, let me think about why the other options aren't correct. Option A is still an existence statement, which doesn't negate the original proposition properly. It's not saying that for all x, the inequality doesn't hold; it's just saying there's some x where it's greater than zero. That doesn't cover all cases.Option B and C are talking about x not in real numbers, which is irrelevant because the original statement was specifically about real numbers. The negation should still be within the same domain to be logically consistent.So, yeah, D is definitely the right choice. It correctly applies the rules of negating an existential quantifier and adjusts the inequality appropriately.

answer：Okay, so I have this problem here: A company is buying new uniforms for its employees. They bought some uniforms total for their 32 stores. Each store should get about 4 uniforms. How many uniforms did the company buy in total?Alright, let's break this down. First, there are 32 stores. Each store needs about 4 uniforms. So, to find out the total number of uniforms the company bought, I think I need to multiply the number of stores by the number of uniforms each store needs.So, that would be 32 stores multiplied by 4 uniforms per store. Let me write that out: 32 * 4.Hmm, 32 times 4. Let me do the multiplication step by step to make sure I don't make a mistake. 30 times 4 is 120, and 2 times 4 is 8. So, adding those together, 120 + 8 equals 128.Wait, does that make sense? If each store gets 4 uniforms, and there are 32 stores, then 4 times 32 should give the total number of uniforms. Yeah, that seems right.But just to double-check, maybe I can think of it differently. If one store gets 4 uniforms, then two stores would get 8 uniforms, right? So, 4 uniforms per store times the number of stores gives the total uniforms.So, 4 uniforms/store * 32 stores = 128 uniforms.I think that's correct. It seems straightforward, but I want to make sure I didn't miss anything in the problem. The company bought uniforms for 32 stores, each needing about 4 uniforms. So, multiplying those two numbers should give the total uniforms bought.Yeah, I'm pretty confident that the total number of uniforms the company bought is 128.