💡To determine how many minutes longer it takes Cheenu to walk a mile now compared to when he was biking as a young man, we need to calculate the time per mile for both biking and walking.1. **Calculate the time per mile when biking:** - Cheenu biked 20 miles in 80 minutes. - Time per mile biking = (frac{80 text{ minutes}}{20 text{ miles}} = 4 text{ minutes per mile}).2. **Calculate the time per mile when walking:** - Cheenu walks 8 miles in 160 minutes. - Time per mile walking = (frac{160 text{ minutes}}{8 text{ miles}} = 20 text{ minutes per mile}).3. **Determine the difference in time per mile:** - Difference = (20 text{ minutes per mile} - 4 text{ minutes per mile} = 16 text{ minutes per mile}).Thus, it takes Cheenu 16 minutes longer to walk a mile now compared to when he was biking as a young man.

💡Okay, so I have this problem where a teacher tells Xu Jun, "Two years ago, my age was three times your age." Then Xu Jun says to the teacher, "In eight years, your age will be twice my age." I need to figure out how old Xu Jun is right now.Alright, let's break this down. First, I'll assign some variables to make it easier. Let me call Xu Jun's current age "x" and the teacher's current age "y." That seems straightforward.Now, the teacher says that two years ago, their age was three times Xu Jun's age. So, two years ago, the teacher's age would have been y - 2, and Xu Jun's age would have been x - 2. According to the teacher, y - 2 equals three times (x - 2). So, I can write that as an equation:y - 2 = 3(x - 2)Okay, that's the first equation. Now, Xu Jun says that in eight years, the teacher's age will be twice his age. So, in eight years, the teacher's age will be y + 8, and Xu Jun's age will be x + 8. According to Xu Jun, y + 8 equals twice (x + 8). So, that gives me the second equation:y + 8 = 2(x + 8)Alright, now I have two equations:1. y - 2 = 3(x - 2)2. y + 8 = 2(x + 8)I need to solve these two equations to find the value of x, which is Xu Jun's current age.Let me start with the first equation:y - 2 = 3(x - 2)I'll expand the right side:y - 2 = 3x - 6Now, I'll solve for y:y = 3x - 6 + 2y = 3x - 4Okay, so that's equation one simplified: y = 3x - 4.Now, let's look at the second equation:y + 8 = 2(x + 8)Again, I'll expand the right side:y + 8 = 2x + 16Now, solve for y:y = 2x + 16 - 8y = 2x + 8So, equation two simplified: y = 2x + 8.Now I have two expressions for y:1. y = 3x - 42. y = 2x + 8Since both equal y, I can set them equal to each other:3x - 4 = 2x + 8Now, solve for x:3x - 2x = 8 + 4x = 12So, Xu Jun is 12 years old right now.Wait, let me double-check that to make sure I didn't make any mistakes.If x = 12, then y = 3(12) - 4 = 36 - 4 = 32.So, the teacher is currently 32 years old.Let's check the first statement: Two years ago, the teacher was 32 - 2 = 30, and Xu Jun was 12 - 2 = 10. Is 30 equal to three times 10? Yes, 30 = 3 * 10. That checks out.Now, the second statement: In eight years, the teacher will be 32 + 8 = 40, and Xu Jun will be 12 + 8 = 20. Is 40 equal to twice 20? Yes, 40 = 2 * 20. That also checks out.Okay, so my answer seems correct. Xu Jun is 12 years old.

💡Okay, so I'm trying to figure out this problem about making all the streets in a hexagonal grid one-way, such that you can still get from any corner to any other corner. Let me start by understanding the problem better.First, the city is made up of regular hexagonal blocks. That means each block is a six-sided figure, and all the blocks are identical. From any corner to any other corner, there are two different routes. So, if I pick any two corners, there are at least two distinct paths connecting them. That seems important because it suggests there's some redundancy in the network, which might help in making the streets one-way without losing connectivity.Now, the question is whether we can assign a one-way direction to each street such that it's still possible to travel from any corner to any other corner following these directions. This is essentially asking if the directed graph formed by the streets is strongly connected.I remember that in graph theory, a directed graph is strongly connected if there's a directed path from every vertex to every other vertex. So, the problem is asking if this particular graph (the hexagonal grid) can be oriented as a strongly connected directed graph.Let me think about the properties of a hexagonal grid. Each intersection (corner) in a hexagonal grid has three streets meeting at it. So, each vertex has degree three. In an undirected graph, this is a 3-regular graph. Now, when we make the streets one-way, we're essentially assigning a direction to each edge. The question then becomes: can we assign directions to the edges such that the resulting directed graph is strongly connected?I recall that for a directed graph to be strongly connected, it must be possible to traverse from any vertex to any other vertex following the directed edges. One way to ensure this is to have a directed cycle that covers all vertices, but in a hexagonal grid, which is an infinite graph, that's not feasible. However, since the problem doesn't specify the size of the city, I assume it's a finite section of the hexagonal grid.Wait, actually, the problem says "in a city," which might imply a finite number of blocks. But it doesn't specify how many. Hmm, maybe it's a general question about any finite hexagonal grid.But regardless, the key point is that from any corner to any other corner, there are two different routes. That suggests that the graph is 2-connected, meaning there are at least two distinct paths between any pair of vertices. In graph theory, a 2-connected undirected graph can be oriented as a strongly connected directed graph. Is that true?I think there's a theorem related to this. Maybe Robbins' theorem? Let me recall. Robbins' theorem states that a graph is strongly orientable (i.e., it can be oriented as a strongly connected directed graph) if and only if it is 2-edge-connected. Wait, 2-edge-connected means that the graph remains connected whenever fewer than two edges are removed. So, if the graph is 2-edge-connected, it can be oriented as a strongly connected directed graph.In our case, the hexagonal grid is 3-regular and, I believe, 3-edge-connected. That means it's certainly 2-edge-connected. So, according to Robbins' theorem, it should be possible to orient the edges such that the directed graph is strongly connected.But let me double-check. Robbins' theorem says that a graph has a strong orientation if and only if it is 2-edge-connected. Since the hexagonal grid is 3-edge-connected, it is also 2-edge-connected, so it should satisfy the condition.Wait, but the problem mentions that from any corner to any other corner, there are two different routes. That seems to imply 2-vertex-connectedness, not necessarily 2-edge-connectedness. Is there a difference?Yes, 2-vertex-connectedness (or 2-connectedness) means that the graph remains connected whenever fewer than two vertices are removed. 2-edge-connectedness is about edges instead of vertices. So, the problem states that there are two different routes between any two corners, which suggests that the graph is 2-connected (vertex-connected). But Robbins' theorem is about edge-connectedness.Hmm, so maybe I need to think differently. If the graph is 2-connected, does that imply it can be strongly oriented? I think there's another theorem related to this. Maybe it's about ear decompositions or something else.Alternatively, perhaps I can think about the hexagonal grid as a planar graph. Since it's a grid, it's planar. Planar graphs have certain properties that might help. For example, in planar graphs, the max-flow min-cut theorem applies, but I'm not sure if that's directly relevant here.Wait, maybe I can approach this by trying to construct such an orientation. If I can find a way to assign directions to the streets so that the resulting graph is strongly connected, that would answer the question affirmatively.One common method to make a graph strongly connected is to find an ear decomposition, which is a way of building up the graph by adding paths (ears) that connect back to the existing graph. But I'm not sure how to apply that here.Alternatively, I could think about using a spanning tree and then adding back edges to make it strongly connected. But since the graph is 2-connected, it has multiple cycles, which might help in creating the necessary directed paths.Wait, another idea: if I can find a directed cycle that covers all the vertices, then the graph is strongly connected. But in a hexagonal grid, which is a planar graph, finding such a cycle might not be straightforward.Alternatively, maybe I can use the fact that the hexagonal grid is bipartite. Wait, is it? Let me think. In a hexagonal grid, each vertex can be colored with two colors such that no two adjacent vertices share the same color. So, yes, it's bipartite. But how does that help?In a bipartite graph, any cycle must be of even length. But I'm not sure if that's useful here.Wait, maybe I can use a checkerboard pattern to assign directions. For example, in a chessboard, you can assign directions based on the color of the squares. Maybe something similar can be done here.But in a hexagonal grid, the coloring is a bit different. Each hexagon can be colored in a way that no two adjacent hexagons share the same color, using two colors. But I'm not sure if that helps with assigning directions.Alternatively, maybe I can use a consistent orientation, like always directing the streets in a clockwise or counterclockwise manner around each hexagon. But would that make the entire graph strongly connected?Let me visualize a single hexagon. If I direct all the streets clockwise around the hexagon, then within that hexagon, you can go around in a cycle. But if I do this for all hexagons, would that create a strongly connected graph?Wait, no, because if all hexagons are directed clockwise, then the entire graph would have a consistent orientation, but you might not be able to move freely between different hexagons. For example, moving from one hexagon to an adjacent one might require going against the direction of some streets.Hmm, maybe a better approach is to use a more complex orientation that allows for movement in both directions across the grid.Wait, another idea: since the graph is 2-connected, it has an ear decomposition starting from a cycle. So, if I can orient the initial cycle in a consistent direction, and then orient each ear in a way that maintains strong connectivity, that might work.But I'm not sure how to formalize that.Alternatively, maybe I can use the fact that the graph is 3-regular and 2-connected, and apply some known result about such graphs being strongly orientable.Wait, I think there is a theorem that says that every 2-connected graph has a strong orientation. Is that true?Let me check. Robbins' theorem says that a graph has a strong orientation if and only if it is 2-edge-connected. But 2-edge-connectedness is a different condition from 2-vertex-connectedness.So, if the graph is 2-vertex-connected, does it imply it's 2-edge-connected? Not necessarily. For example, consider a graph that is 2-vertex-connected but has a bridge (an edge whose removal disconnects the graph). Such a graph is not 2-edge-connected.But in our case, the hexagonal grid is 3-edge-connected, which is stronger than 2-edge-connected. So, it is 2-edge-connected as well.Therefore, by Robbins' theorem, since the hexagonal grid is 2-edge-connected, it can be oriented as a strongly connected directed graph.Wait, but the problem states that from any corner to any other corner, there are two different routes. That suggests that the graph is 2-connected (vertex-connected), but as I thought earlier, 2-connectedness doesn't necessarily imply 2-edge-connectedness.But in the case of the hexagonal grid, which is 3-regular and 3-edge-connected, it is indeed 2-edge-connected. So, Robbins' theorem applies, and it can be strongly oriented.Therefore, the answer should be yes, it's possible to make all streets one-way such that any corner can be reached from any other corner.But let me try to think of a concrete example. Suppose I have a small hexagonal grid, say, a single hexagon. If I direct all the streets clockwise, then within that hexagon, you can go around in a cycle. But if I have multiple hexagons, how do I direct the streets between them?Wait, in a larger grid, each internal street is shared by two hexagons. So, if I direct the streets around each hexagon in the same direction, say clockwise, then the streets between hexagons would have conflicting directions. For example, the street between two adjacent hexagons would be directed clockwise in one hexagon and counterclockwise in the other, which is impossible.Therefore, directing all hexagons in the same rotational direction doesn't work for the entire grid.So, maybe a different approach is needed. Perhaps using a checkerboard-like pattern where directions alternate in some fashion.Alternatively, maybe using a consistent global orientation, like directing all streets from west to east or something, but that might not work because the grid is hexagonal and doesn't have a clear global direction.Wait, another idea: since the hexagonal grid is bipartite, maybe I can assign directions based on the two partitions. For example, direct all streets from partition A to partition B. Then, since the graph is bipartite, you can only go from A to B, not the other way around, which would not make it strongly connected.So, that approach wouldn't work.Alternatively, maybe I can use a more sophisticated orientation where directions are assigned in a way that allows for cycles that cover the entire graph.Wait, perhaps using a concept similar to a feedback arc set, where we remove certain edges to make the graph acyclic, but then add them back in a way that creates cycles.But I'm not sure.Wait, maybe I can use the fact that the hexagonal grid is a planar graph and apply some planar graph orientation techniques.Alternatively, perhaps I can use a depth-first search approach to assign directions, ensuring that the graph remains strongly connected.But I'm not sure how to formalize that.Wait, going back to Robbins' theorem, since the hexagonal grid is 2-edge-connected, it can be oriented as a strongly connected directed graph. So, theoretically, it's possible.But how to construct such an orientation?Maybe by finding an ear decomposition and orienting each ear appropriately.Alternatively, perhaps using a spanning tree and then orienting the remaining edges to create back edges that allow for strong connectivity.Wait, let's try that. Suppose I pick a root vertex and build a spanning tree from it. Then, I can orient all the tree edges away from the root. Then, for the remaining edges (the non-tree edges), I can orient them towards the root or in a way that creates cycles.But I'm not sure if that would ensure strong connectivity.Alternatively, maybe I can use a bidirectional spanning tree approach, where each edge is assigned a direction such that both the original and reverse directions are present in the spanning tree.But that might not be necessary.Wait, perhaps a better approach is to use a consistent orientation based on a global property, like the orientation of the hexagons.Wait, another idea: since the hexagonal grid can be embedded in the plane, maybe I can use a consistent orientation based on the clockwise or counterclockwise order around each face.But as I thought earlier, if I do that for each hexagon, the directions on the shared edges would conflict.So, maybe instead, I can choose a global orientation, like directing all streets in a particular global direction, such as from the center outwards or something.But in an infinite grid, that's not possible, but since we're talking about a city, it's probably a finite grid.Wait, but the problem doesn't specify the size, so maybe it's a general question.Alternatively, maybe I can use a spiral-like orientation, directing streets in a way that allows movement in both directions across the grid.But I'm not sure.Wait, perhaps I can use the fact that the hexagonal grid is a Cayley graph for the free abelian group on two generators, which is known to be strongly connected.But that might be too abstract.Alternatively, maybe I can think of the hexagonal grid as a torus, but that's not necessarily the case.Wait, perhaps I can use a known result that all 2-edge-connected graphs have strong orientations, and since the hexagonal grid is 3-edge-connected, it satisfies this condition.Therefore, the answer is yes, it's possible.But to be more concrete, maybe I can think of an explicit orientation.Wait, here's an idea: choose a spanning tree of the hexagonal grid, and orient all edges away from the root. Then, for each non-tree edge, orient it in the direction that creates a cycle. This way, the graph remains strongly connected because you can go from any node to the root via the tree, and from the root to any node via the reverse tree, and the cycles provide alternative paths.But I'm not sure if this works for all cases.Alternatively, maybe I can use a more systematic approach. For example, in a hexagonal grid, each vertex has three neighbors. Assign directions to the edges such that each vertex has at least one incoming and one outgoing edge. This would ensure that the graph is strongly connected.Wait, but just having at least one incoming and one outgoing edge per vertex doesn't necessarily make the graph strongly connected. For example, you could have multiple disconnected components each satisfying that condition.So, that's not sufficient.Wait, perhaps I can use a more careful assignment. For example, in a hexagonal grid, you can assign directions to the edges such that they form a consistent cycle covering the entire grid.But in a finite grid, that might be possible, but in an infinite grid, it's not.Wait, but the problem is about a city, which is finite, so maybe it's possible.Alternatively, maybe I can use a checkerboard pattern where directions alternate in a way that allows for movement in both directions across the grid.Wait, another idea: since the hexagonal grid is bipartite, I can divide the vertices into two sets, say, black and white, such that no two adjacent vertices share the same color. Then, I can direct all edges from black to white. But this would make the graph a DAG (directed acyclic graph), which is not strongly connected because you can't go back from white to black.So, that's not helpful.Alternatively, maybe I can direct edges in a way that creates cycles that cover the entire graph.Wait, perhaps using a concept similar to a Hamiltonian cycle, but in a directed version, called a directed Hamiltonian cycle. If I can find such a cycle, then the graph is strongly connected.But finding a directed Hamiltonian cycle is a hard problem in general, but for a hexagonal grid, maybe it's possible.Alternatively, maybe I can use the fact that the hexagonal grid is highly symmetric and find a systematic way to assign directions.Wait, here's a possible approach: choose a starting vertex and perform a depth-first search, assigning directions as you go. When you backtrack, assign directions to allow returning.But I'm not sure if that would cover all edges.Alternatively, maybe I can use a consistent orientation based on the coordinates of the vertices.Wait, in a hexagonal grid, each vertex can be assigned axial coordinates, which are two coordinates (q, r) such that q + r + s = 0, where s is the third coordinate. Using these coordinates, I can define a consistent orientation, like directing edges in the positive q direction, or something like that.But I'm not sure if that would ensure strong connectivity.Wait, another idea: since the hexagonal grid is a tiling of the plane with hexagons, maybe I can use a consistent orientation based on the orientation of the hexagons. For example, direct all edges in a way that follows the right-hand rule around each hexagon.But as I thought earlier, this would cause conflicts on shared edges between hexagons.Wait, perhaps instead of orienting all hexagons in the same direction, I can alternate the orientation in a checkerboard pattern. For example, orient the streets clockwise in some hexagons and counterclockwise in others, in a way that resolves the conflicts on shared edges.But I'm not sure if that's possible.Wait, maybe I can use a 3-coloring of the hexagons and assign directions based on that. Since the hexagonal grid can be 3-colored, maybe I can direct edges from one color to another in a cyclic manner.But I'm not sure.Wait, perhaps I can use the fact that the hexagonal grid is dual to the triangular lattice, and use properties from there.But that might be too abstract.Wait, going back to Robbins' theorem, since the hexagonal grid is 2-edge-connected, it can be oriented as a strongly connected directed graph. So, theoretically, it's possible. Therefore, the answer should be yes.But I want to make sure I'm not missing something. The problem states that from any corner to any other corner, there are two different routes. That suggests that the graph is 2-connected, which is a bit different from 2-edge-connected. But since the hexagonal grid is 3-edge-connected, it is also 2-edge-connected, so Robbins' theorem applies.Therefore, the answer is yes, it's possible to make all streets one-way such that any corner can be reached from any other corner.But to be more concrete, maybe I can think of an explicit example. Suppose I have a small hexagonal grid, like a single hexagon. If I direct all the streets clockwise, then within that hexagon, you can go around in a cycle. But if I have multiple hexagons, how do I direct the streets between them?Wait, in a larger grid, each internal street is shared by two hexagons. So, if I direct the streets around each hexagon in the same direction, say clockwise, then the streets between hexagons would have conflicting directions. For example, the street between two adjacent hexagons would be directed clockwise in one hexagon and counterclockwise in the other, which is impossible.Therefore, directing all hexagons in the same rotational direction doesn't work for the entire grid.So, maybe a different approach is needed. Perhaps using a checkerboard-like pattern where directions alternate in some fashion.Alternatively, maybe using a consistent global orientation, like directing all streets from west to east or something, but that might not work because the grid is hexagonal and doesn't have a clear global direction.Wait, another idea: since the hexagonal grid is bipartite, maybe I can assign directions based on the two partitions. For example, direct all streets from partition A to partition B. Then, since the graph is bipartite, you can only go from A to B, not the other way around, which would not make it strongly connected.So, that approach wouldn't work.Alternatively, maybe I can use a more sophisticated orientation where directions are assigned in a way that allows for cycles that cover the entire graph.Wait, perhaps using a concept similar to a feedback arc set, where we remove certain edges to make the graph acyclic, but then add them back in a way that creates cycles.But I'm not sure.Wait, perhaps I can use the fact that the hexagonal grid is a planar graph and apply some planar graph orientation techniques.Alternatively, maybe I can use a depth-first search approach to assign directions, ensuring that the graph remains strongly connected.But I'm not sure how to formalize that.Wait, going back to Robbins' theorem, since the hexagonal grid is 2-edge-connected, it can be oriented as a strongly connected directed graph. So, theoretically, it's possible.But how to construct such an orientation?Maybe by finding an ear decomposition and orienting each ear appropriately.Alternatively, perhaps using a spanning tree and then orienting the remaining edges to create back edges that allow for strong connectivity.Wait, let's try that. Suppose I pick a root vertex and build a spanning tree from it. Then, I can orient all the tree edges away from the root. Then, for the remaining edges (the non-tree edges), I can orient them towards the root or in a way that creates cycles.But I'm not sure if that would ensure strong connectivity.Alternatively, maybe I can use a bidirectional spanning tree approach, where each edge is assigned a direction such that both the original and reverse directions are present in the spanning tree.But that might not be necessary.Wait, perhaps a better approach is to use a consistent orientation based on a global property, like the orientation of the hexagons.Wait, another idea: since the hexagonal grid can be embedded in the plane, maybe I can use a consistent orientation based on the clockwise or counterclockwise order around each face.But as I thought earlier, if I do that for each hexagon, the directions on the shared edges would conflict.So, maybe instead, I can choose a global orientation, like directing all streets in a particular global direction, such as from the center outwards or something.But in an infinite grid, that's not possible, but since we're talking about a city, it's probably a finite grid.Wait, but the problem doesn't specify the size, so maybe it's a general question.Alternatively, maybe I can use a known result that all 2-edge-connected graphs have strong orientations, and since the hexagonal grid is 3-edge-connected, it satisfies this condition.Therefore, the answer is yes, it's possible.But to be more concrete, maybe I can think of an explicit orientation.Wait, here's an idea: choose a spanning tree of the hexagonal grid, and orient all edges away from the root. Then, for each non-tree edge, orient it in the direction that creates a cycle. This way, the graph remains strongly connected because you can go from any node to the root via the tree, and from the root to any node via the reverse tree, and the cycles provide alternative paths.But I'm not sure if this works for all cases.Alternatively, maybe I can use a more systematic approach. For example, in a hexagonal grid, each vertex has three neighbors. Assign directions to the edges such that each vertex has at least one incoming and one outgoing edge. This would ensure that the graph is strongly connected.Wait, but just having at least one incoming and one outgoing edge per vertex doesn't necessarily make the graph strongly connected. For example, you could have multiple disconnected components each satisfying that condition.So, that's not sufficient.Wait, perhaps I can use a more careful assignment. For example, in a hexagonal grid, you can assign directions to the edges such that they form a consistent cycle covering the entire grid.But in a finite grid, that might be possible, but in an infinite grid, it's not.Wait, but the problem is about a city, which is finite, so maybe it's possible.Alternatively, maybe I can use a checkerboard pattern where directions alternate in a way that allows for movement in both directions across the grid.Wait, another idea: since the hexagonal grid is bipartite, I can divide the vertices into two sets, say, black and white, such that no two adjacent vertices share the same color. Then, I can direct all edges from black to white. But this would make the graph a DAG (directed acyclic graph), which is not strongly connected because you can't go back from white to black.So, that's not helpful.Alternatively, maybe I can direct edges in a way that creates cycles that cover the entire graph.Wait, perhaps using a concept similar to a Hamiltonian cycle, but in a directed version, called a directed Hamiltonian cycle. If I can find such a cycle, then the graph is strongly connected.But finding a directed Hamiltonian cycle is a hard problem in general, but for a hexagonal grid, maybe it's possible.Alternatively, maybe I can use the fact that the hexagonal grid is highly symmetric and find a systematic way to assign directions.Wait, here's a possible approach: choose a starting vertex and perform a depth-first search, assigning directions as you go. When you backtrack, assign directions to allow returning.But I'm not sure if that would cover all edges.Alternatively, maybe I can use a consistent orientation based on the coordinates of the vertices.Wait, in a hexagonal grid, each vertex can be assigned axial coordinates, which are two coordinates (q, r) such that q + r + s = 0, where s is the third coordinate. Using these coordinates, I can define a consistent orientation, like directing edges in the positive q direction, or something like that.But I'm not sure if that would ensure strong connectivity.Wait, another idea: since the hexagonal grid is a tiling of the plane with hexagons, maybe I can use a consistent orientation based on the orientation of the hexagons. For example, direct all edges in a way that follows the right-hand rule around each hexagon.But as I thought earlier, this would cause conflicts on shared edges between hexagons.Wait, perhaps instead of orienting all hexagons in the same direction, I can alternate the orientation in a checkerboard pattern. For example, orient the streets clockwise in some hexagons and counterclockwise in others, in a way that resolves the conflicts on shared edges.But I'm not sure if that's possible.Wait, maybe I can use a 3-coloring of the hexagons and assign directions based on that. Since the hexagonal grid can be 3-colored, maybe I can direct edges from one color to another in a cyclic manner.But I'm not sure.Wait, perhaps I can use the fact that the hexagonal grid is dual to the triangular lattice, and use properties from there.But that might be too abstract.Wait, going back to Robbins' theorem, since the hexagonal grid is 2-edge-connected, it can be oriented as a strongly connected directed graph. So, theoretically, it's possible. Therefore, the answer should be yes.But I want to make sure I'm not missing something. The problem states that from any corner to any other corner, there are two different routes. That suggests that the graph is 2-connected, which is a bit different from 2-edge-connected. But since the hexagonal grid is 3-edge-connected, it is also 2-edge-connected, so Robbins' theorem applies.Therefore, the answer is yes, it's possible to make all streets one-way such that any corner can be reached from any other corner.

💡First, I need to evaluate the expression ( x^5 - 5x ) when ( x = 4 ).I'll start by substituting ( x = 4 ) into the expression:[x^5 - 5x = 4^5 - 5 times 4]Next, I'll calculate ( 4^5 ). Since ( 4^5 = 4 times 4 times 4 times 4 times 4 ), this equals 1024.Then, I'll calculate ( 5 times 4 ), which equals 20.Finally, I'll subtract 20 from 1024 to find the result:[1024 - 20 = 1004]

💡Alright, so I've got this problem here about choosing the right sampling methods for two different surveys. Let me try to break it down step by step. First, I need to understand what each survey is about and what kind of sampling method would be most appropriate for each.Starting with the first survey: It says we need to select 100 households from a community that has 125 high-income families, 280 middle-income families, and 95 low-income families. The goal is to investigate a certain indicator of social purchasing power. Okay, so this seems like a survey where the population is divided into different income groups. High-income, middle-income, and low-income. Since these groups might have different purchasing behaviors, it makes sense that we want to ensure that each group is adequately represented in our sample.Now, thinking about sampling methods. There are several types like random sampling, stratified sampling, systematic sampling, cluster sampling, and so on. For the first survey, since the population is divided into distinct subgroups (strata) based on income, stratified sampling seems like a good fit. Stratified sampling involves dividing the population into homogeneous subgroups and then randomly selecting samples from each subgroup. This ensures that each subgroup is proportionally represented in the sample, which is important if we suspect that income level affects the purchasing power indicator we're investigating.So, for the first survey, I'm leaning towards stratified sampling. It allows us to capture the diversity within the community and ensures that each income group is represented, which should give us more accurate and reliable results.Moving on to the second survey: We need to select 3 students from 12 high school freshmen with sports specialties to investigate their study load situation. Hmm, this seems like a smaller population, and we're selecting a small sample size from it. Since the population is relatively small and homogeneous (all are high school freshmen with sports specialties), random sampling might be the most straightforward and appropriate method here.Random sampling means that every individual in the population has an equal chance of being selected. This method is simple and ensures that there's no bias in the selection process. Given that the population is small and the sample size is also small, random sampling should suffice and provide a fair representation of the study load situation among these students.So, for the second survey, random sampling seems appropriate.Now, looking back at the options provided:A: ① Random sampling ② Systematic sampling B: ① Stratified sampling ② Random sampling C: ① Systematic sampling ② Stratified sampling D: ① and ② both use Stratified samplingFrom my analysis, for the first survey, stratified sampling is suitable, and for the second survey, random sampling is appropriate. That matches option B.But just to double-check, let's consider why the other options might not be as suitable.Option A suggests random sampling for the first survey. However, since the population is divided into distinct income groups, random sampling might not ensure proportional representation of each group, which could lead to biased results.Option C suggests systematic sampling for the first survey and stratified sampling for the second. Systematic sampling involves selecting every nth individual from a population list, which might not account for the different income groups in the first survey. For the second survey, since it's a small population, stratified sampling might be overkill, and random sampling is simpler and sufficient.Option D suggests using stratified sampling for both surveys. While stratified sampling is good for the first survey, it's unnecessary for the second survey because the population is small and homogeneous. Using stratified sampling there might complicate things unnecessarily.Therefore, after considering all options, option B seems to be the most appropriate choice, matching my initial analysis.Another thing to consider is the purpose of each survey. For the first survey, understanding social purchasing power likely requires insights across different income levels, which stratified sampling can provide by ensuring each income group is represented. For the second survey, since we're looking at study load among sports-specialized freshmen, random sampling should give a fair snapshot of the study load without needing to stratify further.Also, thinking about the sample sizes: 100 households out of a total of 125 + 280 + 95 = 490 households. That's a significant sample size, and ensuring each income group is proportionally represented is crucial. For the second survey, selecting 3 out of 12 is manageable with random sampling without needing more complex methods.In conclusion, using stratified sampling for the first survey and random sampling for the second survey aligns with the nature and requirements of each survey, making option B the correct choice.

💡Okay, so I have this equation: ( xy - 10x + 3y = 670 ). I need to find positive integers ( x ) and ( y ) that satisfy this equation and then determine the smallest possible value of ( |x - y| ). Hmm, that sounds a bit tricky, but let me try to figure it out step by step.First, I remember that sometimes equations like this can be factored by rearranging terms. Let me see if I can manipulate this equation to make it factorable. The equation is ( xy - 10x + 3y = 670 ). I notice that both ( x ) and ( y ) are present in the terms, so maybe I can group them somehow.Let me try grouping the ( x ) terms and the ( y ) terms together. So, ( xy - 10x + 3y ) can be rewritten as ( x(y - 10) + 3y ). That gives me ( x(y - 10) + 3y = 670 ). Hmm, that's a bit better, but I still have both ( x ) and ( y ) in the equation.I wonder if I can factor this further. Maybe I can factor out a common term from both parts. Let's see, if I factor out ( (y - 10) ) from the first part, but the second part is ( 3y ), which isn't directly a multiple of ( (y - 10) ). Maybe I can adjust the equation to make it factorable.I recall a technique where you add a constant to both sides to make the equation factorable. Let me try that. If I add 30 to both sides, the equation becomes ( xy - 10x + 3y + 30 = 700 ). Now, let's see if I can factor this.Looking at ( xy - 10x + 3y + 30 ), I can group the terms as ( x(y - 10) + 3(y - 10) ). Oh, that works! So, factoring out ( (y - 10) ), I get ( (x + 3)(y - 10) = 700 ). Great, now the equation is factored into two binomials multiplied together equaling 700.Now, since ( x ) and ( y ) are positive integers, ( x + 3 ) and ( y - 10 ) must also be positive integers. That means I need to find pairs of positive integers that multiply to 700. Let me list all the factor pairs of 700.Starting with the smallest factor, 1: ( 1 times 700 = 700 )Then, 2: ( 2 times 350 = 700 )4: ( 4 times 175 = 700 )5: ( 5 times 140 = 700 )7: ( 7 times 100 = 700 )10: ( 10 times 70 = 700 )14: ( 14 times 50 = 700 )20: ( 20 times 35 = 700 )25: ( 25 times 28 = 700 )I think that's all the factor pairs. Now, each of these pairs corresponds to ( (x + 3, y - 10) ). So, for each pair, I can solve for ( x ) and ( y ).Let's go through each pair:1. ( (1, 700) ): ( x + 3 = 1 ) implies ( x = -2 ), which is not positive. So, discard this pair.2. ( (2, 350) ): ( x + 3 = 2 ) implies ( x = -1 ), still not positive. Discard.3. ( (4, 175) ): ( x + 3 = 4 ) implies ( x = 1 ). Then, ( y - 10 = 175 ) implies ( y = 185 ). So, ( (x, y) = (1, 185) ).4. ( (5, 140) ): ( x + 3 = 5 ) implies ( x = 2 ). Then, ( y - 10 = 140 ) implies ( y = 150 ). So, ( (x, y) = (2, 150) ).5. ( (7, 100) ): ( x + 3 = 7 ) implies ( x = 4 ). Then, ( y - 10 = 100 ) implies ( y = 110 ). So, ( (x, y) = (4, 110) ).6. ( (10, 70) ): ( x + 3 = 10 ) implies ( x = 7 ). Then, ( y - 10 = 70 ) implies ( y = 80 ). So, ( (x, y) = (7, 80) ).7. ( (14, 50) ): ( x + 3 = 14 ) implies ( x = 11 ). Then, ( y - 10 = 50 ) implies ( y = 60 ). So, ( (x, y) = (11, 60) ).8. ( (20, 35) ): ( x + 3 = 20 ) implies ( x = 17 ). Then, ( y - 10 = 35 ) implies ( y = 45 ). So, ( (x, y) = (17, 45) ).9. ( (25, 28) ): ( x + 3 = 25 ) implies ( x = 22 ). Then, ( y - 10 = 28 ) implies ( y = 38 ). So, ( (x, y) = (22, 38) ).I should also consider the reverse pairs, where the first element is larger than the second, but since we're dealing with positive integers, those would just give the same solutions with ( x ) and ( y ) swapped, but since we're looking for ( |x - y| ), it doesn't matter which is larger.Now, let's list all the valid ( (x, y) ) pairs and calculate ( |x - y| ) for each:1. ( (1, 185) ): ( |1 - 185| = 184 )2. ( (2, 150) ): ( |2 - 150| = 148 )3. ( (4, 110) ): ( |4 - 110| = 106 )4. ( (7, 80) ): ( |7 - 80| = 73 )5. ( (11, 60) ): ( |11 - 60| = 49 )6. ( (17, 45) ): ( |17 - 45| = 28 )7. ( (22, 38) ): ( |22 - 38| = 16 )Looking at these differences, the smallest one is 16, which comes from the pair ( (22, 38) ).Wait, let me double-check if I missed any factor pairs. I considered all the pairs from 1 up to 25, but 700 is 2^2 * 5^2 * 7, so the factors should be covered. I think I got them all.Also, I need to make sure that ( x ) and ( y ) are positive integers in each case. For example, when ( x + 3 = 1 ), ( x = -2 ), which is invalid, so I correctly discarded those. All other pairs give positive ( x ) and ( y ).Therefore, the smallest possible value of ( |x - y| ) is 16.