💡Okay, so I have this problem where I need to prove that in a graph with 20 vertices, where each vertex has a degree of at least 10, there must be a Hamiltonian path. Hmm, let me think about how to approach this.First, I remember that a Hamiltonian path is a path that visits every vertex exactly once. So, I need to show that such a path exists in this graph. The graph has 20 vertices, and each vertex is connected to at least 10 others. That seems like a pretty dense graph, but I need to be precise.Maybe I can use Dirac's theorem or something similar? Wait, Dirac's theorem is about Hamiltonian cycles, not paths. It states that if every vertex has degree at least n/2 in a graph with n vertices, then the graph is Hamiltonian. But here, n is 20, so n/2 is 10, and each vertex has degree at least 10. So, does that mean the graph is Hamiltonian? But Dirac's theorem is for cycles, not paths. Hmm.Wait, if the graph is Hamiltonian, meaning it has a Hamiltonian cycle, then it certainly has a Hamiltonian path, right? Because a cycle can be broken into a path by removing one edge. So, if I can show that the graph has a Hamiltonian cycle, then it must have a Hamiltonian path.But is Dirac's theorem applicable here? Let me double-check. Dirac's theorem requires that the graph is simple and that every vertex has degree at least n/2. In this case, n is 20, so n/2 is 10, and each vertex has degree at least 10. So, yes, Dirac's theorem applies, and the graph must have a Hamiltonian cycle, which implies a Hamiltonian path.Wait, but the problem specifically asks for a Hamiltonian path, not necessarily a cycle. Maybe there's a more straightforward way without invoking Dirac's theorem. Maybe I can use induction or some other method.Let me think about the structure of the graph. Each vertex has degree at least 10, which is more than half of 20. So, it's a pretty well-connected graph. Maybe I can use the concept of connectivity or something related to paths.I remember that in a graph where each vertex has degree at least k, the graph is k-connected. But I'm not sure if that's directly applicable here.Alternatively, maybe I can use the pigeonhole principle. Since each vertex is connected to at least 10 others, there must be some overlap in connections that can be used to form a long path.Wait, another idea: if I can show that the graph is traceable, meaning it has a Hamiltonian path, then I'm done. There are some theorems related to traceable graphs, but I don't recall the exact conditions.Let me try to construct a Hamiltonian path step by step. Start with any vertex, say v1. It has at least 10 neighbors. Choose one, say v2. Now, v2 has at least 10 neighbors, but one of them is v1, so it has at least 9 new neighbors. Continue this process, always choosing a new vertex that hasn't been visited yet.But wait, what if I get stuck before visiting all 20 vertices? I need to ensure that I can always extend the path. Maybe the high degree ensures that there are enough connections to keep extending the path.Alternatively, suppose I have a path that doesn't include all vertices. Then, there must be some vertex outside the path that is connected to at least 10 vertices on the path. Maybe I can use that to extend the path.Wait, that sounds similar to the argument used in Dirac's theorem. Let me try to formalize it.Assume, for contradiction, that the graph does not have a Hamiltonian path. Then, there exists a longest path P with fewer than 20 vertices. Let’s say P has k vertices, where k < 20. The end vertex of P, say v, has degree at least 10. Since P is the longest path, v cannot be connected to any vertex outside P, otherwise, we could extend P. Therefore, all neighbors of v must be within P.But P has k vertices, and v is connected to at least 10 of them. So, k must be at least 11, because v is connected to 10 others in P. Wait, but k could be larger. Hmm, maybe I need a different approach.Alternatively, consider that if the graph is 10-connected, then it's highly connected, which would imply the existence of a Hamiltonian path. But I'm not sure about the exact relationship between connectivity and Hamiltonian paths.Wait, another thought: in a graph where each vertex has degree at least (n-1)/2, the graph is Hamiltonian. Here, n=20, so (n-1)/2 is 9.5, and each vertex has degree at least 10, which is more than 9.5. So, maybe this is another theorem that applies.Yes, I think that's Chvásal's theorem or something similar. It states that if for every pair of non-adjacent vertices, the sum of their degrees is at least n, then the graph is Hamiltonian. Let me check.If two vertices are non-adjacent, their degrees sum to at least 10 + 10 = 20, which is equal to n. So, by Chvásal's theorem, the graph is Hamiltonian. Therefore, it has a Hamiltonian cycle, which implies a Hamiltonian path.But wait, Chvásal's theorem requires that for every pair of non-adjacent vertices, the sum of their degrees is at least n. In this case, since each vertex has degree at least 10, any two non-adjacent vertices would have degrees summing to at least 20, which is exactly n. So, Chvásal's condition is satisfied, and thus the graph is Hamiltonian.Therefore, the graph must have a Hamiltonian cycle, and hence a Hamiltonian path.Wait, but I'm not entirely sure if Chvásal's theorem directly applies here. Let me double-check the exact statement. Chvásal's theorem says that if for every pair of non-adjacent vertices u and v, deg(u) + deg(v) ≥ n, then the graph is Hamiltonian. In our case, deg(u) + deg(v) ≥ 10 + 10 = 20 = n, so the condition is satisfied. Therefore, the graph is Hamiltonian.So, yes, the graph must have a Hamiltonian cycle, which implies a Hamiltonian path. Therefore, we've proven that such a graph must have a Hamiltonian path.Alternatively, if I don't want to use Chvásal's theorem, maybe I can use a more elementary approach. Let's try induction.Base case: For n=2, trivial. Assume it's true for all graphs with fewer than 20 vertices where each vertex has degree at least 10. Now, consider a graph with 20 vertices. If I can remove a vertex and still maintain the degree condition, then by induction, the remaining graph has a Hamiltonian path, and I can add the removed vertex back in somehow.But I'm not sure if removing a vertex would maintain the degree condition. Each vertex has degree at least 10, but after removing one, the degrees of its neighbors decrease by 1, so they might have degree 9, which is below the required 10. So, induction might not work directly.Another idea: use the fact that in a graph with minimum degree δ, the graph is connected if δ ≥ n/2. Wait, in our case, δ=10 and n=20, so δ=10= n/2. So, the graph is connected. But being connected doesn't necessarily mean it has a Hamiltonian path.Wait, but if the graph is connected and has high enough connectivity, maybe it has a Hamiltonian path. I think there's a theorem that says that if a graph is k-connected and satisfies certain degree conditions, then it's Hamiltonian.Alternatively, maybe I can use the concept of ear decomposition or something related to building up the graph.Wait, perhaps I can use the fact that in a graph with minimum degree δ, the length of the longest path is at least δ + 1. But in our case, δ=10, so the longest path would be at least 11 vertices. But we need a path of 20 vertices.Hmm, not directly helpful.Wait, another approach: consider the complement graph. The complement of our graph would have each vertex with degree at most 9, since the original graph has degree at least 10. Maybe properties of the complement graph can help.But I'm not sure how. Maybe if the complement graph has certain properties, like being disconnected or having certain structures, that could imply something about the original graph.Alternatively, maybe I can use the pigeonhole principle. Since each vertex is connected to at least 10 others, and there are 20 vertices, there must be some overlap in connections that allows for a long path.Wait, let me think about the number of edges. Each vertex has at least 10 edges, so the total number of edges is at least (20*10)/2 = 100 edges. That's a lot of edges for a graph with 20 vertices. The complete graph has 190 edges, so 100 is about half of that. Maybe that's enough to ensure a Hamiltonian path.But I need a more precise argument. Maybe I can use the fact that in such a dense graph, any two large sets of vertices have many connections between them, which can be used to extend paths.Wait, another idea: use the concept of a "greedy" algorithm for constructing a Hamiltonian path. Start at any vertex, and at each step, choose a neighbor that hasn't been visited yet. Since each vertex has at least 10 neighbors, and we're only visiting one at a time, we should be able to continue until we've visited all vertices.But wait, what if we get stuck before visiting all vertices? For example, suppose we have a path of length k, and the current vertex has all its neighbors already in the path. Then we can't extend the path further. But in our case, since each vertex has degree at least 10, and the path has k vertices, the current vertex has at least 10 neighbors, but only k-1 are already in the path. So, as long as k-1 < 10, there must be a neighbor outside the path.Wait, let's see. If k-1 < 10, then k < 11. So, when k reaches 11, the current vertex could have all 10 neighbors already in the path, which would prevent us from extending the path further. But we need to reach k=20.Hmm, so this approach might not work because we could get stuck at k=11. So, maybe the greedy algorithm isn't sufficient.Wait, but maybe we can use a different strategy. Instead of always choosing any neighbor, maybe choose a neighbor that has many connections to the remaining vertices. That way, we can ensure that we can continue extending the path.Alternatively, maybe use the fact that the graph is highly connected to reroute the path if we get stuck.Wait, another approach: consider that the graph is 10-connected. If a graph is k-connected, then it has k internally disjoint paths between any two vertices. So, in our case, the graph is 10-connected, which is a very high level of connectivity.But how does that help with finding a Hamiltonian path? Maybe it implies that the graph is highly resilient to disconnections, so there must be a way to traverse all vertices without getting stuck.Alternatively, maybe use Menger's theorem, which relates connectivity to the number of disjoint paths. But I'm not sure how to apply that here.Wait, going back to the original idea with the longest path. Suppose P is the longest path in the graph, with k vertices. If k=20, we're done. Otherwise, consider the end vertex v of P. Since v has degree at least 10, it must be connected to at least 10 vertices in P. But P has k vertices, so if k < 20, then v has at least 10 neighbors in P, which implies that k must be at least 11.But we need to show that k=20. So, suppose k < 20. Then, the end vertex v has at least 10 neighbors in P. Let’s say v is connected to vertices v1, v2, ..., v10 in P. Now, consider the vertex v10. It is connected to v, and also to its neighbors in P. If v10 is connected to any vertex outside P, then we can extend P by going from v10 to that vertex, contradicting the maximality of P.Therefore, v10 cannot be connected to any vertex outside P. Similarly, v9 cannot be connected to any vertex outside P, because otherwise, we could extend P by going from v9 to that vertex. Wait, but v9 is connected to v10, which is connected to v, so maybe there's a way to rearrange the path.Wait, perhaps we can use the fact that v has many neighbors in P to find a way to extend P. For example, if one of the neighbors of v in P is connected to a vertex outside P, then we can extend P. But if none of them are connected to outside P, then all their neighbors are within P, which would mean that the subgraph induced by P has high connectivity.But I'm not sure how to formalize this. Maybe I need to use the concept of a "dead end" and show that such a situation is impossible given the high minimum degree.Alternatively, maybe consider that if P is the longest path, then all vertices not in P must be connected only to vertices in P. But each vertex not in P has degree at least 10, so they must be connected to at least 10 vertices in P. But P has k vertices, so the number of vertices not in P is 20 - k. Each of these vertices has at least 10 connections to P, so the total number of edges from P to non-P vertices is at least 10*(20 - k).On the other hand, each vertex in P can have at most (20 - k) connections to non-P vertices, but since each vertex in P has degree at least 10, and they are already connected to other vertices in P, the number of connections to non-P vertices is limited.Wait, let's calculate. Each vertex in P has degree at least 10. Suppose P has k vertices. Each vertex in P can be connected to at most (k - 1) vertices within P, and the rest must be connected to non-P vertices. So, the number of edges from P to non-P is at least 10*k - (k*(k - 1))/2. Wait, no, that's not quite right.Actually, the total degree of all vertices in P is at least 10*k. The number of edges within P is at most (k*(k - 1))/2, because that's the maximum number of edges in a complete graph of k vertices. Therefore, the number of edges from P to non-P is at least 10*k - (k*(k - 1))/2.But the number of edges from P to non-P is also equal to the sum of degrees of non-P vertices, which is at least 10*(20 - k). So, we have:10*k - (k*(k - 1))/2 ≥ 10*(20 - k)Let me write that down:10k - (k(k - 1))/2 ≥ 10(20 - k)Multiply both sides by 2 to eliminate the fraction:20k - k(k - 1) ≥ 20(20 - k)Expand both sides:20k - k² + k ≥ 400 - 20kCombine like terms:(20k + k) - k² ≥ 400 - 20k21k - k² ≥ 400 - 20kBring all terms to one side:21k - k² - 400 + 20k ≥ 0Combine like terms:41k - k² - 400 ≥ 0Rearrange:-k² + 41k - 400 ≥ 0Multiply both sides by -1 (which reverses the inequality):k² - 41k + 400 ≤ 0Now, solve the quadratic inequality k² - 41k + 400 ≤ 0.Find the roots of the equation k² - 41k + 400 = 0.Using the quadratic formula:k = [41 ± sqrt(41² - 4*1*400)] / 2Calculate discriminant:41² = 16814*1*400 = 1600So, sqrt(1681 - 1600) = sqrt(81) = 9Thus, k = [41 ± 9]/2So, k = (41 + 9)/2 = 50/2 = 25k = (41 - 9)/2 = 32/2 = 16So, the quadratic is ≤ 0 between k=16 and k=25. But k is the length of the path, which is at most 20. So, the inequality holds for 16 ≤ k ≤ 25, but since k ≤ 20, it holds for 16 ≤ k ≤ 20.But we assumed that k < 20, so k ≤ 19. Therefore, the inequality k² - 41k + 400 ≤ 0 holds for k=16 to k=20. But if k=16, then the inequality is satisfied, but we need to see if this leads to a contradiction.Wait, let's think about this. If k < 20, then k ≤ 19. But the inequality k² - 41k + 400 ≤ 0 holds for k=16 to k=25. So, for k=16,17,18,19,20, the inequality holds. But we need to see if this leads to a contradiction when k < 20.Wait, maybe I made a mistake in the earlier steps. Let me go back.We had:10k - (k(k - 1))/2 ≥ 10(20 - k)Which simplifies to:20k - k(k - 1) ≥ 400 - 20kWhich becomes:20k - k² + k ≥ 400 - 20kThen:21k - k² ≥ 400 - 20kThen:21k - k² - 400 + 20k ≥ 0Which is:41k - k² - 400 ≥ 0Then:-k² + 41k - 400 ≥ 0Multiply by -1:k² - 41k + 400 ≤ 0So, the quadratic is ≤ 0 between its roots, which are 16 and 25. So, for k between 16 and 25, the inequality holds. But since k is the length of the path, which is at most 20, we have that for k=16 to 20, the inequality holds.But we assumed that k < 20, so k=16 to 19. Therefore, the inequality holds for these values. But how does this help us?Wait, maybe I need to find a contradiction. If k < 20, then the inequality holds, but we need to show that this leads to a contradiction, implying that k must be 20.But I'm not sure how to proceed from here. Maybe I need to consider the number of edges and see if it's possible for k to be less than 20.Alternatively, maybe I can use the fact that if k=16, then the number of edges from P to non-P is at least 10*(20 - 16)=40. But the number of edges from P to non-P is also equal to the sum of degrees of non-P vertices, which is at least 10*(20 - k). Wait, but we already used that.Hmm, maybe I need to think differently. Let's consider that if the longest path has k vertices, then the subgraph induced by these k vertices must have certain properties.Wait, another idea: use the fact that in a graph with minimum degree δ, the diameter is at most something. But I'm not sure.Wait, going back to the original problem, maybe I can use the theorem by Erdős and Gallai, which gives conditions for the existence of Hamiltonian paths based on degree sequences. But I don't remember the exact conditions.Alternatively, maybe use the fact that in a graph where every vertex has degree at least n/2, the graph is Hamiltonian, which is Dirac's theorem. Wait, but Dirac's theorem is for cycles, not paths. However, if the graph is Hamiltonian, meaning it has a Hamiltonian cycle, then it certainly has a Hamiltonian path.So, if I can show that the graph has a Hamiltonian cycle, then it has a Hamiltonian path. Since each vertex has degree at least 10, which is n/2, Dirac's theorem applies, and the graph is Hamiltonian, meaning it has a Hamiltonian cycle, hence a Hamiltonian path.Therefore, the conclusion is that the graph must have a Hamiltonian path.Wait, but I'm not sure if Dirac's theorem directly applies to paths. Let me check.Dirac's theorem states that if n ≥ 3 and every vertex has degree at least n/2, then the graph is Hamiltonian, meaning it has a Hamiltonian cycle. So, yes, if the graph has a Hamiltonian cycle, it certainly has a Hamiltonian path.Therefore, the graph must have a Hamiltonian path.Alternatively, if I don't want to rely on Dirac's theorem, maybe I can use a more elementary approach.Suppose the graph does not have a Hamiltonian path. Then, there exists a longest path P with k vertices, where k < 20. Let v be the end vertex of P. Since v has degree at least 10, it must be connected to at least 10 vertices in P. Let’s say v is connected to vertices v1, v2, ..., v10 in P.Now, consider the vertex v10. It is connected to v9 and v11 in P, and also to v. If v10 is connected to any vertex outside P, say u, then we can extend P by going from v10 to u, contradicting the maximality of P. Therefore, v10 cannot be connected to any vertex outside P.Similarly, v9 cannot be connected to any vertex outside P, because otherwise, we could extend P by going from v9 to that vertex. Wait, but v9 is connected to v8 and v10 in P, and also to v. If v9 is connected to any vertex outside P, we could extend P.Therefore, all vertices from v1 to v10 must be connected only within P, which means that the subgraph induced by P is highly connected. But since each of these vertices has degree at least 10, and P has k vertices, the number of connections within P must be at least 10 for each vertex.But if k < 20, then the number of vertices outside P is 20 - k. Each vertex outside P has degree at least 10, so they must be connected to at least 10 vertices in P. But if all vertices in P are already connected only within P, then the vertices outside P cannot have enough connections, leading to a contradiction.Therefore, our assumption that k < 20 must be false, meaning that the longest path P must have k=20 vertices, i.e., a Hamiltonian path.So, in conclusion, the graph must have a Hamiltonian path.

💡Okay, so I have this problem where I need to compute the length of the longest interval [a, b] such that the function f(x) = sum from i=1 to 2014 of |x - i| is constant on that interval. Hmm, that sounds a bit tricky, but let me try to break it down.First, I know that f(x) is the sum of absolute values. Each term |x - i| is a V-shaped graph that has a minimum at x = i. So, when I add up all these V-shaped graphs, the resulting function f(x) should have some interesting properties.I remember that the sum of absolute values like this has a minimum at the median of the points. Since we're summing from 1 to 2014, the median should be around the middle of these numbers. Let me calculate that. There are 2014 terms, so the median would be between the 1007th and 1008th terms. That would be between 1007 and 1008. So, the median is 1007.5.Wait, so f(x) should be minimized at x = 1007.5. That makes sense because the median minimizes the sum of absolute deviations. But the question is about where f(x) is constant. Hmm, how does that work?I think that around the median, the function f(x) might be flat, meaning it's constant over some interval. Let me think about why that would happen. Each absolute value term |x - i| has a slope of -1 when x < i and +1 when x > i. So, when we add up all these slopes, the total slope of f(x) changes at each integer i.At the median, the number of terms where x < i is equal to the number of terms where x > i. So, the total slope from the left would be equal to the total slope from the right, resulting in a flat region. But how long is this flat region?Let me consider the slopes. For x < 1007.5, the number of terms where x < i is more than the number of terms where x > i. So, the slope would be negative because the sum of the slopes from the left terms would dominate. Similarly, for x > 1007.5, the slope would be positive because the right terms would dominate.But wait, if the number of terms where x < i is equal to the number of terms where x > i at x = 1007.5, then the slope is zero there. But does this mean that the function is flat only at that exact point, or over an interval?I think it's over an interval because when x is between two integers, say between 1007 and 1008, the number of terms where x < i and x > i remains the same. So, the slope doesn't change in that interval, meaning the function is linear with a slope of zero, hence constant.So, the interval where f(x) is constant should be between 1007 and 1008. Therefore, the length of this interval is 1008 - 1007 = 1.Wait, but let me double-check. If I take x = 1007, how many terms are less than x and greater than x? For x = 1007, the number of terms less than x is 1006, and the number of terms greater than x is 2014 - 1007 = 1007. Hmm, so the slopes would be -1006 and +1007, resulting in a total slope of +1. So, at x = 1007, the slope is +1.Similarly, at x = 1008, the number of terms less than x is 1007, and the number of terms greater than x is 2014 - 1008 = 1006. So, the slopes would be -1007 and +1006, resulting in a total slope of -1.Wait, that's interesting. So, at x = 1007, the slope is +1, and at x = 1008, the slope is -1. But at x = 1007.5, the slope is zero. So, does that mean that the function is increasing before 1007.5 and decreasing after 1007.5? But that contradicts my earlier thought that it's constant between 1007 and 1008.Hmm, maybe I made a mistake. Let me think again. The slope of f(x) is given by the difference between the number of terms where x > i and the number of terms where x < i. So, when x is less than all the i's, the slope is -2014. As x increases, each time x crosses an integer i, the slope increases by 2 because one more term switches from negative to positive.So, starting from x = 0, the slope is -2014. As x increases, each time x passes an integer, the slope increases by 2. So, when x is between 1007 and 1008, how many terms are less than x and greater than x?For x between 1007 and 1008, the number of terms less than x is 1007, and the number of terms greater than x is 2014 - 1007 = 1007. So, the slope is (1007 - 1007) = 0. Wait, that's different from what I thought earlier.Wait, no. Let me clarify. The slope is (number of terms where x > i) - (number of terms where x < i). So, for x between 1007 and 1008, the number of terms where x > i is 1007 (since i goes up to 1007), and the number of terms where x < i is 2014 - 1007 = 1007. So, the slope is 1007 - 1007 = 0. So, the slope is zero in that interval.But earlier, when I considered x = 1007, I thought the slope was +1, but maybe that's not correct. Let me recast that.At x = 1007, the number of terms where x > i is 1007 (i from 1 to 1007), and the number of terms where x < i is 2014 - 1007 = 1007. So, the slope is 1007 - 1007 = 0. Wait, that contradicts my earlier calculation. Hmm.Wait, no. At x = 1007, the term |x - 1007| is zero, so it doesn't contribute to the slope. So, the number of terms where x > i is 1006 (i from 1 to 1006), and the number of terms where x < i is 2014 - 1007 = 1007. So, the slope is 1006 - 1007 = -1.Similarly, at x = 1008, the number of terms where x > i is 1007 (i from 1 to 1007), and the number of terms where x < i is 2014 - 1008 = 1006. So, the slope is 1007 - 1006 = +1.So, at x = 1007, the slope is -1, and at x = 1008, the slope is +1. But in between, for x between 1007 and 1008, the slope is zero because the number of terms where x > i and x < i are equal.Wait, that makes sense. So, the function f(x) is decreasing with slope -1 up to x = 1007, then becomes flat (slope 0) between x = 1007 and x = 1008, and then starts increasing with slope +1 after x = 1008.Therefore, the interval where f(x) is constant is [1007, 1008], and the length of this interval is 1.But wait, let me confirm this with a smaller example to make sure I'm not making a mistake. Suppose I have f(x) = |x - 1| + |x - 2|. Then, the median is between 1 and 2, which is 1.5. Let's see where f(x) is constant.For x < 1, f(x) = (1 - x) + (2 - x) = 3 - 2x, slope -2.For 1 < x < 2, f(x) = (x - 1) + (2 - x) = 1, which is constant.For x > 2, f(x) = (x - 1) + (x - 2) = 2x - 3, slope +2.So, indeed, f(x) is constant between 1 and 2, which is an interval of length 1.Similarly, if I take f(x) = |x - 1| + |x - 2| + |x - 3|. The median is 2. Let's see.For x < 1, f(x) = (1 - x) + (2 - x) + (3 - x) = 6 - 3x, slope -3.For 1 < x < 2, f(x) = (x - 1) + (2 - x) + (3 - x) = (x - 1) + (2 - x) + (3 - x) = (x - 1 + 2 - x) + (3 - x) = (1) + (3 - x) = 4 - x, slope -1.For 2 < x < 3, f(x) = (x - 1) + (x - 2) + (3 - x) = (x - 1 + x - 2) + (3 - x) = (2x - 3) + (3 - x) = x, slope +1.For x > 3, f(x) = (x - 1) + (x - 2) + (x - 3) = 3x - 6, slope +3.So, in this case, the function is not constant over an interval, but rather has different slopes in different regions. The minimum is at x = 2, but it's not flat around it.Wait, that's different from the previous case. So, in the case of an even number of terms, we had a flat interval, but with an odd number, we don't. So, maybe when the number of terms is even, the function is flat between the two middle points, but when it's odd, it's not.In our original problem, we have 2014 terms, which is even. So, the median is between 1007 and 1008, and the function is flat between those two points, making the interval [1007, 1008], length 1.But wait, in the smaller example with 2 terms, it was flat between 1 and 2, which is length 1. With 3 terms, it wasn't flat. So, seems consistent.So, in our problem, since 2014 is even, the function is flat between 1007 and 1008, which is an interval of length 1.Therefore, the length of the longest interval where f(x) is constant is 1.But just to make sure, let me think about another example. Suppose f(x) = |x - 1| + |x - 2| + |x - 3| + |x - 4|. The median is between 2 and 3. Let's compute the slopes.For x < 1: slope = -4.For 1 < x < 2: slope = -2.For 2 < x < 3: slope = 0.For 3 < x < 4: slope = +2.For x > 4: slope = +4.So, between 2 and 3, the slope is zero, meaning f(x) is constant there. So, interval length is 1.So, again, for even number of terms, the function is flat between the two middle points, length 1.Therefore, in our problem, the answer should be 1.

💡Okay, so I have this problem here: "Which of the following integers is closest to 0?" And then there are five options: (A) -1, (B) 2, (C) -3, (D) 4, and (E) -5. Alright, let's see how to approach this.First, I need to understand what it means for an integer to be closest to 0. I think it means which number is nearest to zero on the number line. So, numbers can be positive or negative, but their distance from zero is what matters here, right? So, for example, -1 is just one unit away from zero, whereas 2 is two units away, and so on.Maybe I should list out all the numbers and their distances from zero to compare them. Let's see:- For option A, which is -1, the distance from zero is 1.- For option B, which is 2, the distance from zero is 2.- For option C, which is -3, the distance from zero is 3.- For option D, which is 4, the distance from zero is 4.- For option E, which is -5, the distance from zero is 5.So, if I list these distances: 1, 2, 3, 4, 5. Clearly, 1 is the smallest distance here. That means -1 is the closest to zero.Wait, but I want to make sure I'm not missing anything. Is there a chance that maybe a negative number could be closer in some other way? Hmm, no, because distance from zero is always positive, regardless of the direction. So, whether the number is negative or positive doesn't affect the distance; it's just how far it is from zero.Let me think about the number line. Zero is in the middle, and then to the right are positive numbers increasing, and to the left are negative numbers decreasing. So, -1 is just one step to the left of zero, and 2 is two steps to the right. So, visually, -1 is closer to zero than 2 is.Similarly, -3 is three steps to the left, which is farther away than -1. And 4 is four steps to the right, which is even farther. -5 is five steps to the left, which is the farthest from zero among all the options.So, based on this, -1 is definitely the closest to zero.But wait, let's double-check. Maybe I should consider the actual values without thinking about the number line. If I take the absolute value of each number, that gives me the distance from zero. So, |-1| = 1, |2| = 2, |-3| = 3, |4| = 4, and |-5| = 5. Comparing these absolute values: 1, 2, 3, 4, 5. Again, 1 is the smallest, so -1 is the closest to zero.Is there any other way to interpret the question? Maybe considering the integers themselves rather than their distances? But that doesn't make much sense because the question is about being closest to zero, which inherently involves distance.Alternatively, could there be a trick where, for example, negative numbers are considered differently? But no, in terms of proximity to zero, negative numbers are treated the same as positive numbers because distance is always positive.So, I think I'm confident that -1 is the closest to zero among the given options. It's only one unit away, whereas the others are two or more units away. Therefore, the answer should be option A, which is -1.Just to make sure, let's recap:- Option A: -1 → Distance from zero: 1- Option B: 2 → Distance from zero: 2- Option C: -3 → Distance from zero: 3- Option D: 4 → Distance from zero: 4- Option E: -5 → Distance from zero: 5Clearly, 1 is the smallest distance, so option A is the correct answer.I don't think I've missed anything here. It seems straightforward once you consider the absolute values or visualize the number line. So, yeah, I'm pretty sure -1 is the closest to zero.

💡Okay, so I have this problem here where I need to find the arithmetic mean of a set of numbers. Let me read it again to make sure I understand.There's a set of ( n + 1 ) numbers, where ( n > 1 ). In this set, one number is ( 1 - frac{1}{n} ), another number is ( 1 + frac{1}{n} ), and the remaining ( n - 1 ) numbers are all 1. I need to calculate the arithmetic mean of this set.Alright, let's break this down step by step. First, let's figure out how many numbers we have in total. It says ( n + 1 ) numbers. So, if ( n ) is, say, 3, then we have 4 numbers in total. But since ( n ) is a variable, I need to keep it general.Now, among these ( n + 1 ) numbers, one is ( 1 - frac{1}{n} ), another is ( 1 + frac{1}{n} ), and the rest ( n - 1 ) are just 1s. So, the set looks like this:- ( 1 - frac{1}{n} )- ( 1 + frac{1}{n} )- 1, 1, 1, ..., 1 (total of ( n - 1 ) times)To find the arithmetic mean, I need to calculate the sum of all these numbers and then divide by the total number of numbers, which is ( n + 1 ).Let me write down the sum:Sum = ( left(1 - frac{1}{n}right) + left(1 + frac{1}{n}right) + underbrace{1 + 1 + ldots + 1}_{n - 1 text{ times}} )Hmm, let's compute each part separately.First, let's compute ( left(1 - frac{1}{n}right) + left(1 + frac{1}{n}right) ). If I add these two together:( 1 - frac{1}{n} + 1 + frac{1}{n} = 1 + 1 - frac{1}{n} + frac{1}{n} )Oh, look at that! The ( -frac{1}{n} ) and ( +frac{1}{n} ) cancel each other out. So, that simplifies to:( 1 + 1 = 2 )Okay, so the sum of those two numbers is 2.Now, the remaining ( n - 1 ) numbers are all 1s. So, their sum is just ( (n - 1) times 1 = n - 1 ).So, the total sum of all the numbers in the set is:Sum = 2 + (n - 1) = 2 + n - 1 = n + 1Alright, so the total sum is ( n + 1 ).Now, the arithmetic mean is the total sum divided by the number of numbers. There are ( n + 1 ) numbers in total.So, Mean = ( frac{n + 1}{n + 1} )Simplify that, and it's just 1.Wait, so the arithmetic mean is 1? That seems straightforward. Let me double-check to make sure I didn't make a mistake.Starting over, the two special numbers are ( 1 - frac{1}{n} ) and ( 1 + frac{1}{n} ). When I add them together, the ( frac{1}{n} ) terms cancel, giving me 2. Then, the rest of the numbers are all 1s, and there are ( n - 1 ) of them, so their sum is ( n - 1 ). Adding that to the 2 gives me ( n + 1 ). Dividing by ( n + 1 ) gives me 1.Yes, that seems correct. So, the arithmetic mean is 1.Looking back at the options given:A) 1B) ( frac{n}{n+1} )C) ( 1 - frac{1}{n+1} )D) ( 1 - frac{1}{n^2} )So, the correct answer is A) 1.I think that makes sense because the two numbers ( 1 - frac{1}{n} ) and ( 1 + frac{1}{n} ) balance each other out, effectively contributing 2 to the sum, and the rest are all 1s. So, the average ends up being 1.Let me think if there's another way to approach this problem. Maybe by considering deviations from the mean?If I assume the mean is 1, then each number's deviation from 1 would be:- For ( 1 - frac{1}{n} ), the deviation is ( -frac{1}{n} )- For ( 1 + frac{1}{n} ), the deviation is ( +frac{1}{n} )- For the rest, the deviation is 0So, the total deviation is ( -frac{1}{n} + frac{1}{n} + 0 + 0 + ldots + 0 = 0 )Since the total deviation is 0, the mean must be 1. That's another way to see it.Yeah, that confirms it. The mean is indeed 1.

💡First, determine the number of lollipop sticks Felicity has collected so far by calculating 72% of the total required sticks.72% of 825 sticks = 0.72 × 825 = 594 sticksNext, calculate how many sticks Felicity collects each week. Since she visits the store four times a week and receives one stick per visit:4 sticks per weekFinally, divide the total number of sticks collected by the number of sticks collected each week to find the number of weeks:594 sticks ÷ 4 sticks per week = 148.5 weeksSince Felicity cannot collect a fraction of a week, round down to the nearest whole number, indicating she has been collecting for 148 full weeks.

💡Okay, so I have this complex number problem here: If the complex number ( z ) satisfies ( frac{z}{1 - i} = i^{2016} + i^{2017} ), then what is ( z )? The options are A: -2, B: 2, C: 2i, D: -2i.Hmm, let me try to figure this out step by step. I remember that ( i ) is the imaginary unit, where ( i^2 = -1 ). So, powers of ( i ) cycle every four exponents: ( i^1 = i ), ( i^2 = -1 ), ( i^3 = -i ), ( i^4 = 1 ), and then it repeats. So, ( i^5 = i ), ( i^6 = -1 ), and so on.First, I need to simplify ( i^{2016} ) and ( i^{2017} ). Let me see, 2016 divided by 4 is 504 with no remainder, right? So, ( i^{2016} = (i^4)^{504} = 1^{504} = 1 ). That makes sense because every multiple of 4 for the exponent of ( i ) gives 1.Now, ( i^{2017} ) is just one more than 2016, so that's ( i^{2016} times i = 1 times i = i ). So, ( i^{2017} = i ).So, putting it back into the equation, ( i^{2016} + i^{2017} = 1 + i ). Therefore, the equation becomes ( frac{z}{1 - i} = 1 + i ).Now, I need to solve for ( z ). To do that, I can multiply both sides of the equation by ( 1 - i ). So, ( z = (1 + i)(1 - i) ).Wait, that looks like the difference of squares formula. Remember, ( (a + b)(a - b) = a^2 - b^2 ). So, applying that here, ( (1 + i)(1 - i) = 1^2 - (i)^2 ).Calculating that, ( 1^2 = 1 ) and ( i^2 = -1 ), so ( 1 - (-1) = 1 + 1 = 2 ). Therefore, ( z = 2 ).Looking back at the options, B is 2, so that should be the answer.But wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning:1. Simplify ( i^{2016} ): Since 2016 is divisible by 4, ( i^{2016} = 1 ).2. Simplify ( i^{2017} ): That's just ( i^{2016} times i = 1 times i = i ).3. So, ( i^{2016} + i^{2017} = 1 + i ).4. Then, ( frac{z}{1 - i} = 1 + i ).5. Multiply both sides by ( 1 - i ): ( z = (1 + i)(1 - i) ).6. Expand the product: ( 1 times 1 = 1 ), ( 1 times (-i) = -i ), ( i times 1 = i ), ( i times (-i) = -i^2 ).7. Combine like terms: ( 1 - i + i - i^2 ). The ( -i ) and ( +i ) cancel out, leaving ( 1 - i^2 ).8. Since ( i^2 = -1 ), this becomes ( 1 - (-1) = 1 + 1 = 2 ).Yep, that all checks out. So, ( z = 2 ), which is option B.Just to make sure, let me consider if there's another way to approach this. Maybe by rationalizing the denominator or something? Let's see.Starting from ( frac{z}{1 - i} = 1 + i ). If I multiply both sides by ( 1 - i ), that's the same as before, so I think that's the right way.Alternatively, I could express ( z ) as ( (1 + i)(1 - i) ) directly, which is the same as ( 1 - i^2 ), which is 2. So, same result.I think I'm confident that the answer is B: 2.**Final Answer**boxed{B}