💡To solve the problem, we need to determine the number of months A kept his oxen in the pasture. We'll use the concept of oxen-months to calculate each person's share of the rent.1. **Calculate the total oxen-months contributed by each person:** - A: (10 text{ oxen} times x text{ months}) - B: (12 text{ oxen} times 5 text{ months} = 60 text{ oxen-months}) - C: (15 text{ oxen} times 3 text{ months} = 45 text{ oxen-months})2. **Set up the equation for the total oxen-months:** [ 10x + 60 + 45 = 10x + 105 ]3. **Determine the total rent and C's share:** - Total rent: (210 text{ Rs}) - C's share: (54 text{ Rs})4. **Calculate the total oxen-months using C's share:** [ 54 = frac{45}{10x + 105} times 210 ] [ 54 = frac{9450}{10x + 105} ] [ 10x + 105 = 175 ] [ 10x = 70 ] [ x = 7 ]Therefore, A kept his oxen in the pasture for 7 months.

💡Okay, so I have this function f(x) defined piecewise. It's x squared plus b times x plus c when x is less than or equal to zero, and it's just 2 when x is greater than zero. The problem gives me two conditions: f(-4) equals f(0), and f(-2) equals -2. I need to find the number of solutions to the equation f(x) equals x.Alright, let's break this down. First, I think I need to figure out the values of b and c. Since f(-4) equals f(0), I can set up an equation using those points. Also, f(-2) equals -2 gives me another equation. So, I should have two equations with two variables, b and c, which I can solve.Let me write down what f(0) is. Since 0 is less than or equal to zero, f(0) is 0 squared plus b times 0 plus c, which simplifies to just c. So, f(0) is c.Now, f(-4) is also equal to f(0), which is c. Let's compute f(-4). Plugging x = -4 into the first piece of the function: (-4)^2 + b*(-4) + c. That's 16 minus 4b plus c. So, f(-4) is 16 - 4b + c, and that's equal to c. So, I can set up the equation 16 - 4b + c = c.Hmm, if I subtract c from both sides, I get 16 - 4b = 0. That simplifies to 16 = 4b, so b must be 4. Okay, so b is 4.Now, let's use the other condition: f(-2) equals -2. Again, since -2 is less than or equal to zero, we use the first piece of the function. So, f(-2) is (-2)^2 + b*(-2) + c. That's 4 minus 2b plus c. We know this equals -2, so 4 - 2b + c = -2.We already found that b is 4, so plug that in: 4 - 2*(4) + c = -2. That's 4 - 8 + c = -2, which simplifies to -4 + c = -2. Adding 4 to both sides gives c = 2.So, now we have b = 4 and c = 2. Therefore, the function f(x) when x is less than or equal to zero is x squared plus 4x plus 2.Now, the problem asks for the number of solutions to the equation f(x) = x. So, we need to solve f(x) = x for all x.Since f(x) is defined piecewise, we should consider two cases: when x is less than or equal to zero and when x is greater than zero.First, let's handle the case when x is less than or equal to zero. In this case, f(x) is x squared + 4x + 2. So, we set up the equation:x squared + 4x + 2 = x.Let's subtract x from both sides to bring everything to one side:x squared + 4x + 2 - x = 0.Simplify that:x squared + 3x + 2 = 0.Now, we can factor this quadratic equation. Looking for two numbers that multiply to 2 and add up to 3. That would be 1 and 2. So, factoring:(x + 1)(x + 2) = 0.Setting each factor equal to zero gives the solutions x = -1 and x = -2. Both of these solutions are less than or equal to zero, so they are valid in this case.Next, let's consider the case when x is greater than zero. In this case, f(x) is just 2. So, we set up the equation:2 = x.So, solving for x gives x = 2. Since 2 is greater than zero, this solution is valid in this case.Therefore, we have three solutions in total: x = -2, x = -1, and x = 2.Wait, let me double-check to make sure I didn't make any mistakes. For the quadratic equation, I had x squared + 3x + 2 = 0, which factors to (x + 1)(x + 2) = 0, giving x = -1 and x = -2. That seems correct.And for x > 0, f(x) = 2, so setting that equal to x gives x = 2, which is valid. So, yes, three solutions.I think that's it. So, the number of solutions is three.**Final Answer**The number of solutions to the equation ( f(x) = x ) is boxed{3}.

💡First, I need to determine the total amount of work required to mow the lawn. If four people can mow the lawn in 6 hours, the total work can be represented as 4 people multiplied by 6 hours, which equals 24 person-hours.Next, I want to find out how many people are needed to complete the same amount of work in 3 hours. Using the total work of 24 person-hours, I can set up the equation:[text{Number of people} times 3 text{ hours} = 24 text{ person-hours}]Solving for the number of people gives:[text{Number of people} = frac{24}{3} = 8]Finally, to determine how many more people are needed beyond the original four, I subtract the original number of people from the required number:[8 text{ people} - 4 text{ people} = 4 text{ additional people}]

💡Okay, so I have this problem here with two parts: a system of inequalities involving x and a fractional equation involving y. I need to find all integer values of m that satisfy both conditions and then sum them up. Hmm, let me take it step by step.First, let me look at the system of inequalities about x:[left{begin{array}{l}frac{x-1}{3} leq 1 2(x-1) > mend{array}right.]Alright, so I need to solve these inequalities to find the range of x, and then figure out what m can be based on the number of integer solutions.Starting with the first inequality: (frac{x-1}{3} leq 1). To solve for x, I can multiply both sides by 3:(x - 1 leq 3)Then add 1 to both sides:(x leq 4)Okay, so x is less than or equal to 4.Now the second inequality: (2(x - 1) > m). Let me solve for x here as well. Divide both sides by 2:(x - 1 > frac{m}{2})Then add 1 to both sides:(x > frac{m}{2} + 1)So combining both inequalities, we have:(frac{m}{2} + 1 < x leq 4)This means x is greater than (frac{m}{2} + 1) and less than or equal to 4.Now, the problem states that this system has solutions and at most 3 integer solutions. So, I need to find the range of m such that the interval (left( frac{m}{2} + 1, 4 right]) contains at most 3 integers.Let me think about how many integers are in this interval. The number of integers in the interval depends on the lower bound (frac{m}{2} + 1).Suppose the lower bound is between k and k+1, then the number of integers in the interval would be 4 - k. Since we want at most 3 integer solutions, 4 - k ≤ 3, which implies k ≥ 1.Wait, maybe another approach. Let me consider the smallest integer greater than (frac{m}{2} + 1). Let's denote this integer as n. Then the number of integers in the interval is 4 - n + 1 = 5 - n. We need 5 - n ≤ 3, so n ≥ 2.Therefore, the smallest integer n must be at least 2. So, (frac{m}{2} + 1 < 2), which implies (frac{m}{2} < 1), so m < 2. But wait, that might not be the right way.Alternatively, maybe I should find the range of m such that the interval (left( frac{m}{2} + 1, 4 right]) includes at most 3 integers. Let's list the possible integer values of x: 2, 3, 4. So, if the lower bound is less than 2, then x can be 2,3,4, which is 3 integers. If the lower bound is between 2 and 3, then x can be 3,4, which is 2 integers. If the lower bound is between 3 and 4, then x can be 4, which is 1 integer. If the lower bound is greater than 4, there are no solutions, but the problem says there are solutions, so the lower bound must be less than 4.So, to have at most 3 integer solutions, the lower bound must be less than 2, because if it's greater than or equal to 2, the number of integer solutions would be less than 3. Wait, no. If the lower bound is less than 2, x can be 2,3,4, which is 3 integers. If the lower bound is between 2 and 3, x can be 3,4, which is 2 integers. If the lower bound is between 3 and 4, x can be 4, which is 1 integer. So, to have at most 3 integer solutions, the lower bound can be anywhere, but since the problem says "at most 3", which includes 1,2,3. But since the system must have solutions, the lower bound must be less than 4.But I think the key is that the number of integer solutions is at most 3, so the lower bound must be such that the interval includes no more than 3 integers. So, if the lower bound is less than 2, we have 3 integer solutions. If it's between 2 and 3, we have 2. If it's between 3 and 4, we have 1. So, the maximum number of integer solutions is 3, which occurs when the lower bound is less than 2.But the problem says "has solutions and at most 3 integer solutions". So, to ensure that the system has solutions, the lower bound must be less than 4, which it already is because x ≤4. So, the main condition is that the number of integer solutions is at most 3, which is always true because the maximum is 3. Wait, no, because if the lower bound is less than 2, it's 3 solutions. If it's between 2 and 3, it's 2 solutions, etc. So, the condition is automatically satisfied as long as the system has solutions.But maybe I'm overcomplicating. Let's think about the range of m such that the system has solutions. The system has solutions if (frac{m}{2} + 1 < 4), which is always true because m is an integer, and if m is too large, the lower bound might exceed 4, making no solutions. So, to have solutions, we need (frac{m}{2} + 1 < 4), which simplifies to m < 6.So, m must be less than 6. Also, since the number of integer solutions is at most 3, which is already satisfied as long as m < 6 because the maximum number of integer solutions is 3 when m is small.Wait, maybe I need to ensure that the number of integer solutions is exactly 3. Because "at most 3" includes 1,2,3, but the problem says "has solutions and at most 3 integer solutions". So, it's okay if it's 1,2, or 3. So, m can be such that the number of integer solutions is 1,2, or 3.But perhaps the key is that the system must have solutions, so m must be such that (frac{m}{2} + 1 < 4), which is m < 6. Also, since the number of integer solutions is at most 3, which is already covered because the maximum is 3.Wait, maybe I need to ensure that the lower bound is such that the interval includes at least one integer. So, (frac{m}{2} + 1 < 4) and (frac{m}{2} + 1 < x leq 4). So, x must be an integer in that interval. So, to have at least one integer solution, the lower bound must be less than 4, which is already satisfied.But perhaps the problem is more about the number of integer solutions being at most 3, which is always true because the maximum is 3. So, maybe the main condition is just m < 6.But let me check. If m is 6, then (frac{6}{2} + 1 = 4), so the interval is (4,4], which is empty. So, m must be less than 6.So, m can be up to 5.But let's think about the fractional equation now.The fractional equation is:[frac{y}{2y - 4} - 1 = frac{m - 4}{y - 2}]I need to solve this equation for y and ensure that the solutions are non-negative integers.First, let me simplify the equation.Note that 2y - 4 is 2(y - 2), so the left side can be rewritten:[frac{y}{2(y - 2)} - 1 = frac{m - 4}{y - 2}]Let me combine the terms on the left side:[frac{y}{2(y - 2)} - frac{2(y - 2)}{2(y - 2)} = frac{m - 4}{y - 2}]Simplify the left side:[frac{y - 2(y - 2)}{2(y - 2)} = frac{m - 4}{y - 2}]Simplify the numerator:[y - 2y + 4 = -y + 4]So, the left side becomes:[frac{-y + 4}{2(y - 2)} = frac{m - 4}{y - 2}]Now, multiply both sides by 2(y - 2) to eliminate denominators:[(-y + 4) = 2(m - 4)]Simplify:[-y + 4 = 2m - 8]Bring all terms to one side:[-y + 4 - 2m + 8 = 0]Combine like terms:[-y - 2m + 12 = 0]Solve for y:[-y = 2m - 12 y = 12 - 2m]So, the solution is y = 12 - 2m.But we need to ensure that y is a non-negative integer and also that y ≠ 2, because in the original equation, the denominator y - 2 cannot be zero.So, y = 12 - 2m must be a non-negative integer, and y ≠ 2.Therefore:1. (12 - 2m geq 0) → (12 geq 2m) → (m leq 6)2. (12 - 2m neq 2) → (12 - 2m ≠ 2) → (10 ≠ 2m) → (m ≠ 5)So, m must be an integer less than or equal to 6 and not equal to 5.But from the inequalities earlier, m must be less than 6. So, combining both conditions, m must be less than 6 and m ≤ 6, so m < 6, and m ≠ 5.Wait, but m is an integer, so m can be 0,1,2,3,4,5, but m <6, so m can be 0,1,2,3,4,5. But m ≠5, so m can be 0,1,2,3,4.Wait, but earlier, from the inequalities, m must be less than 6, so m can be up to 5, but m must be less than 6, so m can be 0,1,2,3,4,5. But from the fractional equation, m must be ≤6 and m ≠5. So, combining both, m must be less than 6 and m ≠5, so m can be 0,1,2,3,4.Wait, but earlier, from the inequalities, m must be less than 6, so m can be up to 5, but m must be less than 6, so m can be 0,1,2,3,4,5. But from the fractional equation, m must be ≤6 and m ≠5. So, combining both, m must be less than 6 and m ≠5, so m can be 0,1,2,3,4.Wait, but let me double-check. From the inequalities, we have that m must be less than 6 because if m=6, the lower bound is 4, which makes the interval empty. So, m must be less than 6.From the fractional equation, m must be ≤6 and m ≠5. So, combining both, m must be less than 6 and m ≠5, so m can be 0,1,2,3,4.But wait, m can be 5 as long as it's less than 6? No, because m must be less than 6, so m can be up to 5, but m ≠5. So, m can be 0,1,2,3,4.Wait, but m=5 is allowed in the fractional equation? No, because m=5 would make y=12-10=2, which is not allowed because y≠2. So, m=5 is excluded.So, the possible integer values of m are 0,1,2,3,4.Now, let me check if these values satisfy both conditions.For m=0:From the inequalities: (frac{0}{2} +1=1), so x >1 and x ≤4. So, integer solutions are 2,3,4. That's 3 solutions, which is acceptable.From the fractional equation: y=12-0=12, which is a non-negative integer and y≠2. So, acceptable.m=1:Inequalities: (frac{1}{2} +1=1.5), so x>1.5 and x≤4. Integer solutions:2,3,4. 3 solutions.Fractional equation: y=12-2=10. Non-negative integer, y≠2. Good.m=2:Inequalities: (frac{2}{2} +1=2), so x>2 and x≤4. Integer solutions:3,4. 2 solutions.Fractional equation: y=12-4=8. Non-negative integer, y≠2. Good.m=3:Inequalities: (frac{3}{2} +1=2.5), so x>2.5 and x≤4. Integer solutions:3,4. 2 solutions.Fractional equation: y=12-6=6. Non-negative integer, y≠2. Good.m=4:Inequalities: (frac{4}{2} +1=3), so x>3 and x≤4. Integer solutions:4. 1 solution.Fractional equation: y=12-8=4. Non-negative integer, y≠2. Good.m=5:Wait, m=5 is excluded because y=12-10=2, which is not allowed. So, m=5 is out.So, the valid integer values of m are 0,1,2,3,4.Now, summing them up: 0+1+2+3+4=10.So, the sum of all integer values of m that satisfy the conditions is 10.

💡Okay, so I have this problem where I need to figure out for which positive integers ( n ) it's possible to label the elements of a set ( B ) with either 0 or 1. The set ( B ) has subsets ( A_1, A_2, ldots, A_{2n+1} ) each containing exactly ( 2n ) elements. Additionally, any two subsets ( A_i ) and ( A_j ) intersect in exactly one element, and every element of ( B ) is in at least two subsets. The goal is to label each element of ( B ) such that each subset ( A_i ) has exactly ( n ) elements labeled 0.Hmm, let me break this down. Each subset has ( 2n ) elements, and they all intersect in exactly one element. So, if I pick any two subsets, they share precisely one common element. Also, every element is in at least two subsets, which means there are no elements that are unique to a single subset.I think this structure is similar to something in combinatorics, maybe a block design? Specifically, it reminds me of a projective plane or something like that, where each pair of lines intersects in exactly one point. But I'm not entirely sure. Maybe I can model this as a graph?Let me try to visualize this. If I consider each subset ( A_i ) as a vertex in a graph, then the intersection condition implies that every pair of vertices is connected by an edge, which represents the unique common element. So, the graph would be a complete graph ( K_{2n+1} ), since there are ( 2n+1 ) subsets, each connected to every other subset.In this graph, each edge corresponds to an element of ( B ), and each vertex corresponds to a subset ( A_i ). So, the problem reduces to labeling the edges of ( K_{2n+1} ) with 0s and 1s such that each vertex is incident to exactly ( n ) edges labeled 0.Wait, that sounds like a kind of edge labeling problem. Each vertex has degree ( 2n ) because it's connected to all other ( 2n ) vertices. We need exactly ( n ) of those edges to be labeled 0, and the remaining ( n ) to be labeled 1. So, essentially, we need a way to partition the edges of ( K_{2n+1} ) into two sets: one set with ( n ) edges per vertex labeled 0, and the other set with the remaining ( n ) edges labeled 1.Is this always possible? I know that in graph theory, certain regular graphs can be decomposed into perfect matchings or other regular subgraphs. For example, a complete graph with an odd number of vertices can be decomposed into Hamiltonian cycles or something similar. But in this case, we need a decomposition into two ( n )-regular graphs.Wait, each vertex needs to have exactly ( n ) edges labeled 0, so the subgraph induced by the 0-labeled edges must be an ( n )-regular graph on ( 2n+1 ) vertices. Similarly, the 1-labeled edges would form another ( n )-regular graph.But hold on, ( 2n+1 ) is an odd number. In a regular graph, the degree must divide the number of vertices minus one. Here, each vertex has degree ( n ), and the number of vertices is ( 2n+1 ). So, ( n ) must divide ( 2n ), which it does because ( 2n = 2 times n ). But is that sufficient?I recall that a regular graph of degree ( d ) on ( v ) vertices exists if and only if ( v times d ) is even, because the total number of edges must be an integer. In our case, the total number of edges in the 0-labeled subgraph would be ( frac{(2n+1) times n}{2} ). For this to be an integer, ( (2n+1) times n ) must be even.Let's check: ( (2n+1) times n = 2n^2 + n ). The term ( 2n^2 ) is always even, and ( n ) is either even or odd. If ( n ) is even, then ( 2n^2 + n ) is even. If ( n ) is odd, ( 2n^2 ) is even and ( n ) is odd, so the sum is odd. Therefore, ( (2n+1) times n ) is even only if ( n ) is even.So, this suggests that such a labeling is only possible when ( n ) is even. If ( n ) is odd, the total number of edges would not be an integer, which is impossible because you can't have a fraction of an edge.Let me test this with a small example. Let's take ( n = 1 ). Then, we have ( 2n+1 = 3 ) subsets, each with ( 2 ) elements. Each pair of subsets intersects in exactly one element, and every element is in at least two subsets.So, the set ( B ) must have 3 elements, each appearing in exactly two subsets. Let's denote the subsets as ( A_1 = {a, b} ), ( A_2 = {a, c} ), and ( A_3 = {b, c} ). Now, we need to label each element with 0 or 1 such that each subset has exactly 1 element labeled 0.If I label ( a = 0 ), then ( A_1 ) and ( A_2 ) each have one 0. But ( A_3 ) has neither ( a ) nor ( b ) labeled 0, so we need to label either ( b ) or ( c ) as 0. Suppose we label ( b = 0 ). Then ( A_1 ) has two 0s, which is not allowed. If we label ( c = 0 ), then ( A_2 ) and ( A_3 ) each have one 0, but ( A_1 ) only has ( a = 0 ). Wait, that works? ( A_1 ) has ( a = 0 ) and ( b ) unlabeled, but actually, all elements must be labeled either 0 or 1. So, if ( a = 0 ), ( b ) must be 1, and ( c ) must be 0. Then, ( A_1 = {0, 1} ), ( A_2 = {0, 0} ), and ( A_3 = {1, 0} ). But ( A_2 ) has two 0s, which violates the condition. Alternatively, if ( a = 1 ), ( b = 0 ), ( c = 0 ), then ( A_1 = {1, 0} ), ( A_2 = {1, 0} ), ( A_3 = {0, 0} ). Again, ( A_3 ) has two 0s. It seems impossible for ( n = 1 ).Wait, maybe I made a mistake in constructing the subsets. For ( n = 1 ), each subset has 2 elements, and each pair intersects in one element. So, the structure is like a triangle where each edge is an element, and each vertex is a subset. So, the elements are the edges, and each subset corresponds to a vertex containing the two edges incident to it. So, in this case, the elements are the edges of ( K_3 ), which are three elements: ( ab, ac, bc ). Each subset ( A_i ) corresponds to a vertex and contains the two edges incident to it. So, ( A_1 = {ab, ac} ), ( A_2 = {ab, bc} ), ( A_3 = {ac, bc} ).Now, we need to label each edge (element) with 0 or 1 such that each subset (vertex) has exactly one 0. So, each vertex has two edges, and exactly one of them must be labeled 0. So, essentially, we need a perfect matching in the line graph of ( K_3 ), which is ( K_3 ) itself. Wait, but in ( K_3 ), a perfect matching would consist of edges that don't share a vertex, but ( K_3 ) has three edges, which is odd, so a perfect matching isn't possible.Alternatively, maybe I'm overcomplicating it. If each vertex needs exactly one 0 edge, then we need to choose one edge incident to each vertex to label 0. However, since each edge is incident to two vertices, if we label an edge 0, it satisfies the condition for both of its endpoints. But in ( K_3 ), each edge is connected to two vertices, so if I label one edge 0, it counts towards two vertices. But we have three vertices, each needing one 0 edge. So, labeling one edge 0 would satisfy two vertices, and the third vertex would need another edge labeled 0, but that edge is already connected to another vertex, which would then have two 0 edges, which is not allowed. Hence, it's impossible for ( n = 1 ).This supports the earlier conclusion that ( n ) must be even. Let me check ( n = 2 ). Then, we have ( 5 ) subsets, each with ( 4 ) elements. Each pair of subsets intersects in exactly one element, and every element is in exactly two subsets. So, the structure is similar to ( K_5 ), where each edge represents an element, and each vertex represents a subset containing four edges (since each subset has four elements). Wait, no, each subset corresponds to a vertex, and each element corresponds to an edge. So, each subset ( A_i ) has four elements, which are the four edges incident to vertex ( i ) in ( K_5 ).Wait, no, in ( K_5 ), each vertex has degree 4, so each subset ( A_i ) would correspond to the four edges incident to vertex ( i ). But in our problem, each subset has ( 2n = 4 ) elements, and each pair of subsets intersects in exactly one element, which corresponds to the unique edge connecting two vertices in ( K_5 ). So, yes, this structure is exactly ( K_5 ), where each subset is a vertex, and each element is an edge.Now, we need to label each edge (element) with 0 or 1 such that each vertex (subset) has exactly two edges (elements) labeled 0. So, we need a 2-regular labeling on the edges of ( K_5 ). This is equivalent to finding a 2-factor in ( K_5 ), which is a spanning 2-regular subgraph. In ( K_5 ), a 2-factor would consist of cycles covering all vertices. Since ( K_5 ) is a complete graph with an odd number of vertices, it can be decomposed into cycles. For example, a 5-cycle is a 2-regular graph, but that only covers one cycle. However, we need a 2-factor, which can be a combination of cycles.Wait, actually, in ( K_5 ), a 2-factor can consist of a single 5-cycle or a combination of smaller cycles. But since we need each vertex to have degree 2, the 2-factor must cover all vertices with cycles. For ( K_5 ), it's possible to decompose it into a 5-cycle and a 4-cycle, but that would leave one vertex uncovered, which isn't allowed. Alternatively, we can have two edge-disjoint 5-cycles, but that would require more edges than available.Wait, no, a 2-factor in ( K_5 ) must cover all edges with cycles. But ( K_5 ) has 10 edges, and a 5-cycle has 5 edges. So, two edge-disjoint 5-cycles would cover all 10 edges, but that's not possible because each edge can only be in one cycle. Alternatively, a 2-factor can be a single 5-cycle, but that only uses 5 edges, leaving the other 5 edges unused, which isn't a 2-factor.I think I'm getting confused here. Let me recall that a 2-factor is a spanning 2-regular subgraph, meaning it covers all vertices with cycles, but doesn't necessarily cover all edges. So, in ( K_5 ), a 2-factor can be a single 5-cycle, which uses 5 edges, leaving the other 5 edges uncovered. But in our problem, we need to label edges with 0 or 1 such that each vertex has exactly two 0 edges. So, the 0-labeled edges must form a 2-regular graph, which is a union of cycles covering all vertices.But in ( K_5 ), can we find a 2-regular subgraph that covers all vertices? Yes, a 5-cycle is such a subgraph. So, if we label the edges of a 5-cycle as 0, then each vertex is incident to exactly two 0 edges. The remaining edges can be labeled 1. This satisfies the condition.Wait, but in ( K_5 ), a 5-cycle uses 5 edges, and each vertex has degree 2 in the 5-cycle. So, yes, each vertex would have exactly two 0 edges, and the remaining two edges (since each vertex has degree 4 in ( K_5 )) would be labeled 1. That works.So, for ( n = 2 ), it's possible. This supports the idea that even ( n ) works. Let me see if I can generalize this.In general, for ( K_{2n+1} ), we need to find an ( n )-regular subgraph that covers all vertices. Such a subgraph is called an ( n )-factor. A theorem in graph theory states that a regular graph of degree ( d ) on ( v ) vertices has a ( k )-factor if certain conditions are met. In our case, we need an ( n )-factor in ( K_{2n+1} ).I recall that a necessary condition for a ( k )-factor to exist is that ( k ) divides the number of vertices, but in our case, ( n ) doesn't necessarily divide ( 2n+1 ). Wait, actually, the condition is that the total number of edges in the ( k )-factor must be an integer. The total number of edges in an ( n )-factor of ( K_{2n+1} ) would be ( frac{(2n+1) times n}{2} ). For this to be an integer, ( (2n+1) times n ) must be even, as I thought earlier.Since ( 2n+1 ) is odd, ( n ) must be even for the product to be even. Therefore, ( n ) must be even for such a labeling to exist.To summarize, the problem reduces to finding an ( n )-regular subgraph in ( K_{2n+1} ), which is possible if and only if ( n ) is even. This is because the total number of edges in such a subgraph must be an integer, which only happens when ( n ) is even.So, the answer is that ( n ) must be an even number.

💡Okay, so I have this problem here about sequences and sums. Let me try to understand it step by step. The problem says that ( S_n ) is the sum of the first ( n ) terms of the sequence ( {a_n} ). We're given that ( a_1 = 1 ), and for ( n geq 2 ), ( a_n = n + (-1)^{n-1} ). We need to find the ratio ( frac{S_{24}}{S_{23}} ).First, let me write down what I know. The sequence starts with ( a_1 = 1 ). For each subsequent term, ( a_n ) is equal to ( n ) plus ( (-1)^{n-1} ). So, let me compute the first few terms to see the pattern.For ( n = 2 ):( a_2 = 2 + (-1)^{2-1} = 2 + (-1)^1 = 2 - 1 = 1 ).For ( n = 3 ):( a_3 = 3 + (-1)^{3-1} = 3 + (-1)^2 = 3 + 1 = 4 ).For ( n = 4 ):( a_4 = 4 + (-1)^{4-1} = 4 + (-1)^3 = 4 - 1 = 3 ).For ( n = 5 ):( a_5 = 5 + (-1)^{5-1} = 5 + (-1)^4 = 5 + 1 = 6 ).Hmm, so it seems like for even ( n ), ( a_n = n - 1 ), and for odd ( n geq 3 ), ( a_n = n + 1 ). Let me check that.If ( n ) is even, say ( n = 2k ), then ( (-1)^{n-1} = (-1)^{2k - 1} = (-1)^{odd} = -1 ). So, ( a_n = n - 1 ).If ( n ) is odd, say ( n = 2k + 1 ), then ( (-1)^{n-1} = (-1)^{2k} = 1 ). So, ( a_n = n + 1 ).Okay, that makes sense. So, the sequence alternates between adding 1 and subtracting 1 starting from ( n = 2 ). So, the sequence goes: 1, 1, 4, 3, 6, 5, 8, 7, and so on.Now, I need to find ( S_{24} ) and ( S_{23} ). Since ( S_n ) is the sum of the first ( n ) terms, ( S_{24} = a_1 + a_2 + dots + a_{24} ) and ( S_{23} = a_1 + a_2 + dots + a_{23} ).So, ( S_{24} = S_{23} + a_{24} ). Therefore, ( frac{S_{24}}{S_{23}} = frac{S_{23} + a_{24}}{S_{23}} = 1 + frac{a_{24}}{S_{23}} ).But I don't know ( S_{23} ) yet, so maybe I should compute ( S_{24} ) and ( S_{23} ) separately.Let me think about how to compute ( S_n ). Since the sequence alternates between adding 1 and subtracting 1, maybe I can pair the terms or find a pattern.Looking at the sequence:( a_1 = 1 )( a_2 = 1 )( a_3 = 4 )( a_4 = 3 )( a_5 = 6 )( a_6 = 5 )( a_7 = 8 )( a_8 = 7 )And so on.So, starting from ( a_2 ), each pair of terms is ( (2k - 1) ) and ( 2k ), where ( a_{2k-1} = 2k ) and ( a_{2k} = 2k - 1 ). Wait, let me check that.For ( k = 1 ), ( a_1 = 1 ), which is given.For ( k = 2 ), ( a_2 = 1 ), ( a_3 = 4 ).For ( k = 3 ), ( a_4 = 3 ), ( a_5 = 6 ).For ( k = 4 ), ( a_6 = 5 ), ( a_7 = 8 ).So, actually, for each ( k geq 2 ), the odd-indexed term ( a_{2k-1} = 2k ) and the even-indexed term ( a_{2k} = 2k - 1 ).Wait, no. Let me correct that. For ( n = 2k ), ( a_n = n - 1 = 2k - 1 ). For ( n = 2k - 1 ), ( a_n = n + 1 = 2k ). So, yes, each pair ( a_{2k-1} ) and ( a_{2k} ) adds up to ( (2k) + (2k - 1) = 4k - 1 ).But wait, let me compute ( a_{2k-1} + a_{2k} ):( a_{2k-1} = (2k - 1) + 1 = 2k )( a_{2k} = 2k + (-1)^{2k - 1} = 2k - 1 )So, ( a_{2k-1} + a_{2k} = 2k + (2k - 1) = 4k - 1 ).Hmm, so each pair sums to ( 4k - 1 ).But let's see for ( k = 1 ), ( a_1 = 1 ), which is alone.For ( k = 2 ), ( a_2 + a_3 = 1 + 4 = 5 ), which is ( 4*2 - 1 = 7 ). Wait, that doesn't match. Wait, maybe my indexing is off.Wait, for ( k = 2 ), ( a_3 + a_4 = 4 + 3 = 7 ), which is ( 4*2 - 1 = 7 ). Okay, that works.Similarly, ( a_5 + a_6 = 6 + 5 = 11 ), which is ( 4*3 - 1 = 11 ). Okay, that works.So, starting from ( a_3 ) and ( a_4 ), each pair sums to ( 4k - 1 ), where ( k ) starts at 2.But ( a_1 ) is alone, and ( a_2 ) is also alone? Wait, no, ( a_2 ) is 1, which is part of the first pair.Wait, let's think differently. Maybe group the terms as ( a_1 ), ( a_2 ), ( a_3 + a_4 ), ( a_5 + a_6 ), etc.So, ( S_n ) can be broken down as:- ( a_1 = 1 )- ( a_2 = 1 )- ( (a_3 + a_4) = 4 + 3 = 7 )- ( (a_5 + a_6) = 6 + 5 = 11 )- ( (a_7 + a_8) = 8 + 7 = 15 )- And so on.So, each pair after ( a_2 ) adds up to an increasing sequence: 7, 11, 15, etc., each time increasing by 4.Wait, 7, 11, 15, 19,... That's an arithmetic sequence with common difference 4.So, starting from ( a_3 ) and ( a_4 ), each pair sums to 7, then 11, then 15, etc.So, if I can find how many such pairs are there in ( S_{24} ) and ( S_{23} ), I can compute their sums.Let me see. For ( S_{24} ), which is the sum of the first 24 terms.We have:- ( a_1 = 1 )- ( a_2 = 1 )- Then, from ( a_3 ) to ( a_{24} ), which is 22 terms, so 11 pairs.Each pair sums to 7, 11, 15,... up to the 11th term.Similarly, for ( S_{23} ), it's the sum of the first 23 terms.- ( a_1 = 1 )- ( a_2 = 1 )- Then, from ( a_3 ) to ( a_{23} ), which is 21 terms, so 10 pairs and one extra term.Wait, 21 terms would be 10 pairs (20 terms) plus 1 term.But let me confirm:From ( a_3 ) to ( a_{24} ), that's 22 terms, which is 11 pairs.From ( a_3 ) to ( a_{23} ), that's 21 terms, which is 10 pairs and one term.So, perhaps I can compute ( S_{24} ) and ( S_{23} ) by summing these pairs.Let me formalize this.For ( S_n ):- If ( n = 1 ), ( S_1 = 1 ).- If ( n = 2 ), ( S_2 = 1 + 1 = 2 ).- For ( n geq 3 ), ( S_n = 2 + sum_{k=2}^{m} (a_{2k-1} + a_{2k}) ), where ( m ) is such that ( 2k leq n ).Wait, maybe another approach. Let me think about how many pairs are in ( S_{24} ) and ( S_{23} ).In ( S_{24} ):- ( a_1 = 1 )- ( a_2 = 1 )- Then, ( a_3 ) to ( a_{24} ) is 22 terms, which is 11 pairs.Similarly, in ( S_{23} ):- ( a_1 = 1 )- ( a_2 = 1 )- Then, ( a_3 ) to ( a_{23} ) is 21 terms, which is 10 pairs and one term.Wait, but 21 terms would be 10 pairs (20 terms) plus 1 term. So, the last term is ( a_{23} ).But ( a_{23} ) is an odd index, so ( a_{23} = 23 + 1 = 24 ).So, ( S_{23} = 1 + 1 + sum_{k=2}^{11} (a_{2k-1} + a_{2k}) - a_{24} ). Wait, no, because ( S_{23} ) stops at ( a_{23} ), which is the 23rd term.Wait, maybe it's better to compute ( S_{24} ) first.Let me try to compute ( S_{24} ):( S_{24} = a_1 + a_2 + (a_3 + a_4) + (a_5 + a_6) + dots + (a_{23} + a_{24}) )So, that's 1 + 1 + 7 + 11 + 15 + ... up to 11 pairs.Wait, how many pairs are there? From ( a_3 ) to ( a_{24} ), that's 22 terms, so 11 pairs.Each pair sums to 7, 11, 15,..., which is an arithmetic sequence with first term 7 and common difference 4.So, the sum of these 11 pairs is the sum of an arithmetic series with 11 terms, first term 7, last term ( 7 + (11-1)*4 = 7 + 40 = 47 ).Sum of pairs = ( frac{11}{2} times (7 + 47) = frac{11}{2} times 54 = 11 times 27 = 297 ).Then, ( S_{24} = 1 + 1 + 297 = 299 ).Now, ( S_{23} ):( S_{23} = a_1 + a_2 + (a_3 + a_4) + dots + (a_{21} + a_{22}) + a_{23} )So, that's 1 + 1 + 7 + 11 + 15 + ... up to 10 pairs, plus ( a_{23} ).The sum of the first 10 pairs is the sum of an arithmetic series with 10 terms, first term 7, last term ( 7 + (10-1)*4 = 7 + 36 = 43 ).Sum of pairs = ( frac{10}{2} times (7 + 43) = 5 times 50 = 250 ).Then, ( a_{23} = 23 + 1 = 24 ).So, ( S_{23} = 1 + 1 + 250 + 24 = 2 + 250 + 24 = 276 ).Therefore, ( frac{S_{24}}{S_{23}} = frac{299}{276} ).Simplify this fraction:Divide numerator and denominator by GCD(299,276). Let's compute GCD(299,276).299 ÷ 276 = 1 with remainder 23.276 ÷ 23 = 12 with remainder 0.So, GCD is 23.Thus, ( frac{299}{276} = frac{299 ÷ 23}{276 ÷ 23} = frac{13}{12} ).So, the ratio is ( frac{13}{12} ).Therefore, the answer is A: ( frac{13}{12} ).