answer：First, I need to determine B's contribution to the capital based on the profit-sharing ratio and the time each partner invested.A invested Rs. 3500 for the entire year, which is 12 months. B joined after 6 months, so B's investment period is 6 months.The profit is divided in the ratio 2:3, meaning A's share is 2 parts and B's share is 3 parts.To find the relationship between their investments and the profit ratio, I'll calculate the product of the investment amount and the time period for each partner.For A:Investment × Time = 3500 × 12 = 42000For B:Investment × Time = x × 6 = 6xAccording to the profit-sharing ratio:42000 / 6x = 2 / 3Solving for x:42000 × 3 = 2 × 6x126000 = 12xx = 126000 / 12x = 10500Therefore, B's contribution to the capital is Rs. 10,500.

answer：Alright, so I've got this problem here: The set {-2, 1} is equal to which of the given options? The options are labeled A, B, C, and D, each representing different sets. I need to figure out which one of these options corresponds to the set containing -2 and 1. Let me go through each option step by step.Starting with option A: {(x - 1)(x + 2) = 0}. Hmm, okay. So this is an equation set. I remember that when you have an equation like this, the solutions are the values of x that make the equation true. So, if I solve (x - 1)(x + 2) = 0, I can find the roots. To solve this, I know that if the product of two factors is zero, then at least one of the factors must be zero. So, setting each factor equal to zero:1. x - 1 = 0 ⇒ x = 12. x + 2 = 0 ⇒ x = -2So, the solutions are x = 1 and x = -2. Therefore, the set represented by option A is {1, -2}, which is the same as {-2, 1}. That seems to match the given set. But wait, let me check the other options just to be sure.Moving on to option B: {y | y = x + 1, x ∈ ℤ}. Okay, this is a set defined by a rule. It says that y is equal to x + 1 where x is any integer. So, this set would include all integers shifted by 1. For example, if x is 0, y is 1; if x is 1, y is 2; if x is -1, y is 0; and so on. Essentially, this set is just all integers because for every integer y, there exists an integer x such that y = x + 1. So, this set is actually equal to ℤ, the set of all integers. That's definitely not the same as {-2, 1}, so option B is out.Next is option C: {x | (x + 1)(x - 2) = 0}. Again, this is another equation set. Let me solve this equation to find the elements of the set.Setting each factor equal to zero:1. x + 1 = 0 ⇒ x = -12. x - 2 = 0 ⇒ x = 2So, the solutions are x = -1 and x = 2. Therefore, the set is {-1, 2}, which is different from {-2, 1}. So, option C is not the correct answer.Finally, option D: {x | (x - 1)(x + 2) = 0}. This looks similar to option A, but let me double-check. Solving the equation:1. x - 1 = 0 ⇒ x = 12. x + 2 = 0 ⇒ x = -2So, the solutions are x = 1 and x = -2, which gives the set {1, -2} or {-2, 1}. That's exactly the set we're looking for. Wait a second, both options A and D seem to give the same set. Let me look back at the original problem to make sure I didn't misread anything. The set is {-2, 1}, and both A and D are equation sets that solve to this. But looking back at the problem statement, option A is written as {(x - 1)(x + 2) = 0}, while option D is {x | (x - 1)(x + 2) = 0}. Hmm, actually, option A is written without specifying the variable, just the equation. In set notation, sometimes people write equations inside the set brackets to denote the solutions. So, in that case, both A and D might represent the same set. But in the problem, the options are labeled A, B, C, D, and each is supposed to be a different representation. Wait, maybe I misread option A. Let me check again. It says {(x - 1)(x + 2) = 0}. That's a bit ambiguous because it's just an equation inside the set brackets without specifying the variable. In standard set notation, you usually specify the variable, like in option D: {x | (x - 1)(x + 2) = 0}. So, perhaps option A is not correctly written, or it's interpreted differently. If I consider option A as a set containing the equation itself, rather than the solutions, then it's not a set of numbers but a set containing the equation. That would make option A incorrect because we're looking for a set of numbers, not an equation. So, in that case, only option D correctly represents the set {-2, 1}.Therefore, after carefully going through each option, I conclude that the correct answer is option D.

answer：First, I will identify the multiplications in the expression: 6 × 5 and 3 × 2.Calculating these gives 30 and 6, respectively.Next, I will substitute these results back into the original expression: 8 - 7 + 30 + 4 - 6 + 1 - 0.Then, I will perform the additions and subtractions from left to right:8 - 7 equals 1.1 + 30 equals 31.31 + 4 equals 35.35 - 6 equals 29.29 + 1 equals 30.Finally, 30 - 0 remains 30.Therefore, the final answer is 30.

answer：Okay, so I have this logarithmic equation to solve: log base 81 of (2r - 1) equals -1/2. Hmm, logarithms can be tricky, but I think I remember that a logarithm equation like log_b(a) = c means that b raised to the power of c equals a. So, applying that here, it should mean that 81 raised to the power of -1/2 equals (2r - 1). Let me write that down to make it clearer.So, 81^(-1/2) = 2r - 1. Now, I need to figure out what 81 raised to the power of -1/2 is. I know that a negative exponent means the reciprocal, so 81^(-1/2) is the same as 1 divided by 81^(1/2). And 81^(1/2) is the square root of 81, which is 9. So, 81^(-1/2) is 1/9. That simplifies the equation to 1/9 = 2r - 1.Alright, now I have a simple linear equation to solve for r. Let me add 1 to both sides to isolate the term with r. So, 1/9 + 1 equals 2r. To add these, I need a common denominator. 1 is the same as 9/9, so adding 1/9 and 9/9 gives me 10/9. Therefore, 10/9 equals 2r.Now, to solve for r, I need to divide both sides by 2. Dividing 10/9 by 2 is the same as multiplying by 1/2, so 10/9 * 1/2 equals 10/18, which simplifies to 5/9. So, r equals 5/9.Wait a second, let me double-check my steps to make sure I didn't make any mistakes. Starting from the logarithmic equation, I converted it correctly to its exponential form. Then, I simplified 81^(-1/2) correctly to 1/9. Solving the linear equation step by step, adding 1 to both sides, converting 1 to 9/9, adding 1/9 and 9/9 to get 10/9, and then dividing by 2 to get 5/9. Everything seems to check out.Just to be thorough, maybe I should plug the value of r back into the original equation to verify. If r is 5/9, then 2r - 1 would be 2*(5/9) - 1, which is 10/9 - 9/9, which equals 1/9. So, log base 81 of 1/9 should equal -1/2. Let me confirm that.Since 81 is 9 squared, 81^(1/2) is 9, and 81^(-1/2) is 1/9. So, log base 81 of 1/9 is indeed -1/2. That matches the original equation, so my solution seems correct.I think I've got it. The value of r that satisfies the equation is 5/9.

answer：Okay, so I have this problem here: the function f(x) = log₂x + x - k, where k is a natural number, has only one zero in the interval (2, 3). I need to find what k is, and the options are A: 0, B: 2, C: 4, D: 6.Alright, let me start by understanding the problem. A zero of the function means a value of x where f(x) = 0. So, I need to find k such that the equation log₂x + x - k = 0 has exactly one solution between 2 and 3.First, let me think about the function f(x) = log₂x + x - k. It's a combination of a logarithmic function and a linear function. Since log₂x is defined for x > 0, and in this case, we're only concerned with x between 2 and 3, that's fine.I remember that for continuous functions, the Intermediate Value Theorem can be useful. If a function is continuous on [a, b] and takes values f(a) and f(b) at each end of the interval, then it takes any value between f(a) and f(b) at some point in (a, b). But here, we're told there's only one zero in (2, 3). So, maybe I can use the Intermediate Value Theorem to find conditions on k such that f(2) and f(3) have opposite signs, which would imply a zero crossing in between.But wait, before that, maybe I should check if the function is increasing or decreasing in the interval (2, 3). If the function is strictly increasing or decreasing, then it can have at most one zero in that interval. So, let me find the derivative of f(x) to check its monotonicity.The derivative f’(x) would be the derivative of log₂x plus the derivative of x minus the derivative of k. The derivative of log₂x is 1/(x ln 2), and the derivative of x is 1. The derivative of a constant k is 0. So, f’(x) = 1/(x ln 2) + 1.Since x is in (2, 3), which is positive, and ln 2 is positive, so 1/(x ln 2) is positive. Adding 1, which is also positive, so f’(x) is definitely positive in (2, 3). That means the function is strictly increasing on this interval. So, if the function is strictly increasing, it can cross the x-axis at most once. Therefore, if it does cross, it will do so exactly once.Therefore, for there to be exactly one zero in (2, 3), the function must cross from negative to positive or positive to negative in that interval. But since it's strictly increasing, it can only cross once, so f(2) must be less than 0 and f(3) must be greater than 0, or vice versa. But since it's increasing, if f(2) < 0 and f(3) > 0, then there is exactly one zero in (2, 3). If both f(2) and f(3) are positive or both negative, then there might be no zeros or more than one, but since it's strictly increasing, it can only cross once.Wait, actually, if f(2) < 0 and f(3) > 0, then by the Intermediate Value Theorem, there must be exactly one zero in (2, 3). If f(2) and f(3) are both positive or both negative, then depending on the function's behavior, but since it's strictly increasing, if both are positive, then f(x) is always positive in (2, 3), and if both are negative, f(x) is always negative in (2, 3). So, to have exactly one zero, f(2) must be negative and f(3) must be positive.So, let's compute f(2) and f(3):f(2) = log₂2 + 2 - k. Since log₂2 is 1, so f(2) = 1 + 2 - k = 3 - k.f(3) = log₂3 + 3 - k. Log₂3 is approximately 1.58496, so f(3) ≈ 1.58496 + 3 - k ≈ 4.58496 - k.So, for f(2) < 0 and f(3) > 0:3 - k < 0 ⇒ k > 3and4.58496 - k > 0 ⇒ k < 4.58496Since k is a natural number, which is a positive integer (assuming natural numbers start at 1), so k must be greater than 3 and less than approximately 4.58496. The only integer in that range is 4.Therefore, k must be 4.Wait, let me double-check. If k = 4:f(2) = 3 - 4 = -1 < 0f(3) ≈ 4.58496 - 4 ≈ 0.58496 > 0So, f(2) is negative and f(3) is positive, so by the Intermediate Value Theorem, there is at least one zero in (2, 3). Since the function is strictly increasing, there is exactly one zero.If k were 3:f(2) = 3 - 3 = 0, which is at x=2, but the interval is (2,3), so x=2 is not included. So, f(2) = 0 is not in the open interval, so f(x) would be positive at x=2, but since it's increasing, it would stay positive, so no zero in (2,3). Wait, no, if k=3, f(2)=0, but since the interval is open, (2,3), so x=2 is excluded. So, f(x) at x just above 2 would be slightly positive? Wait, let me compute f(2.1) when k=3:f(2.1) = log₂2.1 + 2.1 - 3 ≈ 1.07 + 2.1 - 3 ≈ 0.17 > 0So, f(x) is positive just above 2, and it's increasing, so it would stay positive in (2,3), so no zero in (2,3). So, k=3 is not acceptable.Similarly, if k=5:f(2) = 3 - 5 = -2 < 0f(3) ≈ 4.58496 - 5 ≈ -0.415 < 0So, both f(2) and f(3) are negative, so the function is negative throughout (2,3), so no zero crossing.If k=4:f(2) = -1 < 0f(3) ≈ 0.58496 > 0So, crosses from negative to positive, exactly one zero.Similarly, if k=2:f(2) = 3 - 2 = 1 > 0f(3) ≈ 4.58496 - 2 ≈ 2.58496 > 0So, both positive, function is increasing, so no zero in (2,3).If k=1:f(2) = 3 - 1 = 2 > 0f(3) ≈ 4.58496 - 1 ≈ 3.58496 > 0Same as above.If k=0:f(2) = 3 - 0 = 3 > 0f(3) ≈ 4.58496 - 0 ≈ 4.58496 > 0Same.So, only when k=4, f(2) is negative and f(3) is positive, so exactly one zero in (2,3).Therefore, the answer should be C: 4.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**boxed{C}

answer：Alright, so I have this problem about a math Olympiad with 30 students and 8 problems each. The scoring rule is a bit tricky. Let me try to understand it step by step.First, each problem is worth ( k ) points if it wasn't solved by exactly ( k ) students. Wait, that's a bit confusing. The remark clarifies it: if a problem is solved by exactly ( k ) students, then each of those ( k ) students gets ( 30 - k ) points for that problem. So, if a problem is solved by ( k ) students, each solver gets ( 30 - k ) points. If it's solved by any other number, they get ( k ) points? Hmm, no, actually, the first rule says each problem is worth ( k ) points if it wasn't solved by exactly ( k ) students. So, maybe if a problem is solved by exactly ( k ) students, it's worth ( 30 - k ) points, otherwise, it's worth ( k ) points. That seems to make more sense.So, to rephrase: For each problem, if exactly ( k ) students solved it, then each of those ( k ) students gets ( 30 - k ) points. If a problem was solved by any number other than ( k ), then each student who solved it gets ( k ) points. Wait, that still doesn't make complete sense. Maybe the problem's total points are ( k ) if it wasn't solved by exactly ( k ) students, and if it was solved by exactly ( k ) students, then each solver gets ( 30 - k ) points. So, the total points for that problem would be ( k times (30 - k) ) if it was solved by exactly ( k ) students, otherwise, it's just ( k ) points total? Or is it that each student gets ( k ) points if the problem wasn't solved by exactly ( k ) students?Wait, the remark says: "if the problem was solved by exactly ( k ) students, then each of them got ( 30 - k ) points in it." So, if a problem is solved by exactly ( k ) students, each solver gets ( 30 - k ) points. If it's solved by any other number, each solver gets ( k ) points? Or is it that the problem is worth ( k ) points if it wasn't solved by exactly ( k ) students, meaning that each solver gets ( k ) points? Hmm, the wording is a bit confusing.Let me try to parse it again. Rule 1: Each problem was worth ( k ) points, if it wasn't solved by exactly ( k ) students. So, if a problem was solved by exactly ( k ) students, then it's not worth ( k ) points. Instead, the remark says that if it was solved by exactly ( k ) students, each solver gets ( 30 - k ) points. So, maybe the problem's total points are ( k ) if it wasn't solved by exactly ( k ) students, and if it was solved by exactly ( k ) students, then each solver gets ( 30 - k ) points, making the total points ( k times (30 - k) ). That seems more consistent.So, to clarify:- For each problem, if the number of students who solved it is not equal to ( k ), then the problem is worth ( k ) points in total, meaning each solver gets ( k ) points.- If the number of students who solved it is exactly ( k ), then each solver gets ( 30 - k ) points, so the total points for that problem would be ( k times (30 - k) ).Wait, but ( k ) is defined as the number of students who didn't solve the problem? Or is ( k ) arbitrary? Hmm, the problem says "Each problem was worth ( k ) points, if it wasn't solved by exactly ( k ) students." So, ( k ) is the number of students who didn't solve the problem. So, if a problem was not solved by exactly ( k ) students, then it's worth ( k ) points. But if it was solved by exactly ( k ) students, then each solver gets ( 30 - k ) points.Wait, that makes more sense. So, ( k ) is the number of students who didn't solve the problem. So, if a problem was not solved by exactly ( k ) students, then the problem is worth ( k ) points, meaning each solver gets ( k ) points. If a problem was solved by exactly ( k ) students, then each solver gets ( 30 - k ) points.Wait, no, the remark says: "if the problem was solved by exactly ( k ) students, then each of them got ( 30 - k ) points in it." So, regardless of ( k ), if a problem is solved by ( k ) students, each gets ( 30 - k ) points. And if it's not solved by exactly ( k ) students, then each solver gets ( k ) points. Wait, that seems inconsistent because ( k ) is defined as the number of students who didn't solve it.I think I need to re-express this.Let me denote:- Let ( s ) be the number of students who solved a particular problem.- Then, according to rule 1, if the problem was not solved by exactly ( k ) students, it's worth ( k ) points. So, if ( s neq k ), then the problem is worth ( k ) points, meaning each solver gets ( k ) points.- If the problem was solved by exactly ( k ) students, then each solver gets ( 30 - k ) points.Wait, but ( k ) is the number of students who didn't solve it. So, if ( s = 30 - k ), then each solver gets ( 30 - (30 - k) = k ) points. That seems redundant.Wait, maybe I'm overcomplicating. Let's try to think differently.Suppose for a problem, ( s ) students solved it. Then:- If ( s = k ), each solver gets ( 30 - k ) points.- If ( s neq k ), each solver gets ( k ) points.But what is ( k )? The problem says "Each problem was worth ( k ) points, if it wasn't solved by exactly ( k ) students." So, perhaps ( k ) is a parameter for each problem, meaning that for each problem, we choose a ( k ) such that if the number of solvers is not equal to ( k ), then the problem is worth ( k ) points, otherwise, it's worth ( 30 - k ) points per solver.Wait, that might make sense. So, for each problem, we can choose a ( k ) such that:- If the number of students who solved the problem is not equal to ( k ), then the problem is worth ( k ) points, meaning each solver gets ( k ) points.- If the number of students who solved the problem is exactly ( k ), then each solver gets ( 30 - k ) points.So, ( k ) is a parameter set per problem, not necessarily related to the number of solvers.But then, how do we choose ( k ) for each problem? The problem doesn't specify, so perhaps we can choose ( k ) strategically to maximize the minimum score, which is Lasha's score.Wait, but the problem says "Each problem was worth ( k ) points, if it wasn't solved by exactly ( k ) students." So, perhaps ( k ) is determined by the number of students who didn't solve it. That is, if a problem was not solved by exactly ( k ) students, then it's worth ( k ) points. So, ( k ) is the number of students who didn't solve the problem.So, for a problem, let ( u ) be the number of students who didn't solve it. Then, if ( u = k ), the problem is worth ( k ) points, meaning each solver gets ( k ) points. Otherwise, if ( u neq k ), then each solver gets ( 30 - k ) points? Wait, that doesn't make sense because ( k ) is defined as the number of students who didn't solve it.Wait, perhaps the problem is worth ( k ) points if it wasn't solved by exactly ( k ) students, meaning that if ( u neq k ), the problem is worth ( k ) points, and if ( u = k ), then the problem is worth ( 30 - k ) points per solver.I think I need to formalize this.Let me denote:- For each problem, let ( u ) be the number of students who didn't solve it.- If ( u neq k ), then the problem is worth ( k ) points, so each solver gets ( k ) points.- If ( u = k ), then each solver gets ( 30 - k ) points.But then, ( k ) is a parameter for each problem. So, for each problem, we can choose ( k ) such that if the number of unsolved students is not equal to ( k ), then each solver gets ( k ) points, otherwise, each solver gets ( 30 - k ) points.But the problem statement says "Each problem was worth ( k ) points, if it wasn't solved by exactly ( k ) students." So, perhaps ( k ) is the number of students who didn't solve it, and if that number is not equal to ( k ), then the problem is worth ( k ) points. Wait, that seems circular.Alternatively, perhaps ( k ) is a fixed value for all problems, but that doesn't make sense because each problem could have a different number of solvers.Wait, maybe ( k ) is the number of students who didn't solve the problem, and if that number is exactly ( k ), then each solver gets ( 30 - k ) points, otherwise, each solver gets ( k ) points. But that would mean ( k ) is both the number of unsolved students and a parameter, which is confusing.I think I need to look at the remark again: "if the problem was solved by exactly ( k ) students, then each of them got ( 30 - k ) points in it." So, if a problem is solved by exactly ( k ) students, each solver gets ( 30 - k ) points. If it's solved by any other number, each solver gets ( k ) points. Wait, but ( k ) is defined as the number of students who didn't solve it? Or is ( k ) arbitrary?Wait, the remark says "if the problem was solved by exactly ( k ) students, then each of them got ( 30 - k ) points in it." So, regardless of ( k ), if a problem is solved by ( k ) students, each gets ( 30 - k ) points. If it's solved by any other number, each solver gets ( k ) points? That can't be, because ( k ) is not defined in the remark.Wait, perhaps the initial rule is that each problem is worth ( k ) points if it wasn't solved by exactly ( k ) students, and if it was solved by exactly ( k ) students, then each solver gets ( 30 - k ) points. So, ( k ) is the number of students who didn't solve it. So, if ( u = k ), where ( u ) is the number of unsolved students, then each solver gets ( 30 - k ) points. Otherwise, each solver gets ( k ) points.So, for each problem:- Let ( u ) be the number of students who didn't solve it.- If ( u = k ), then each solver gets ( 30 - k ) points.- If ( u neq k ), then each solver gets ( k ) points.But then, ( k ) is a parameter for each problem. So, for each problem, we can choose ( k ) such that if the number of unsolved students is ( k ), then each solver gets ( 30 - k ) points, otherwise, each solver gets ( k ) points.But the problem doesn't specify how ( k ) is chosen for each problem. It just says "Each problem was worth ( k ) points, if it wasn't solved by exactly ( k ) students." So, perhaps ( k ) is the number of students who didn't solve it, and if that number is exactly ( k ), then each solver gets ( 30 - k ) points, otherwise, each solver gets ( k ) points.Wait, that seems consistent. So, for each problem:- Let ( u ) be the number of students who didn't solve it.- If ( u = k ), then each solver gets ( 30 - k ) points.- If ( u neq k ), then each solver gets ( k ) points.But then, ( k ) is the number of students who didn't solve it, so ( u = k ). Therefore, if ( u = k ), each solver gets ( 30 - k ) points, otherwise, each solver gets ( k ) points.Wait, that seems redundant because ( u ) is always equal to ( k ) if ( k ) is defined as the number of students who didn't solve it. So, perhaps ( k ) is a parameter set for each problem, not necessarily equal to the number of unsolved students.Wait, I'm getting confused. Let me try to think of an example.Suppose a problem is solved by 15 students. Then, according to the remark, each solver gets ( 30 - 15 = 15 ) points.If a problem is solved by 10 students, each solver gets ( 30 - 10 = 20 ) points.If a problem is solved by 20 students, each solver gets ( 30 - 20 = 10 ) points.Wait, so regardless of ( k ), if a problem is solved by ( s ) students, each solver gets ( 30 - s ) points. So, the initial rule might have been misworded. Maybe it's supposed to say that each problem is worth ( 30 - s ) points if it was solved by ( s ) students, otherwise, it's worth ( k ) points. But the problem says "Each problem was worth ( k ) points, if it wasn't solved by exactly ( k ) students."Wait, perhaps the correct interpretation is:- For each problem, if the number of students who solved it is not equal to ( k ), then the problem is worth ( k ) points, meaning each solver gets ( k ) points.- If the number of students who solved it is exactly ( k ), then each solver gets ( 30 - k ) points.So, ( k ) is a parameter for each problem, and we can choose ( k ) such that if the number of solvers is ( k ), then each solver gets ( 30 - k ) points, otherwise, each solver gets ( k ) points.But since ( k ) is arbitrary, we can choose it to maximize the minimum score. So, to maximize Lasha's score, we need to set ( k ) such that the minimum total score is as high as possible.Wait, but how do we choose ( k ) for each problem? The problem doesn't specify, so perhaps ( k ) is fixed for all problems, or we can choose it per problem.I think the key is that for each problem, we can choose ( k ) such that if the number of solvers is ( k ), then each solver gets ( 30 - k ) points, otherwise, each solver gets ( k ) points.So, to maximize the minimum total score, we need to set ( k ) for each problem such that the distribution of points is as balanced as possible.Let me think about how to distribute the problems among the students to maximize the minimum score.Suppose we have 30 students and 8 problems. Each student solves some number of problems, and for each problem, depending on how many students solved it, the points per solver are determined.Our goal is to arrange the solving of problems such that the student with the least total points (Lasha) has as high a score as possible.This sounds like an optimization problem where we need to maximize the minimum score.To approach this, perhaps we can consider that each problem should be solved by as many students as possible to minimize the points per solver, but we also need to balance it so that no student ends up with too few points.Wait, actually, if a problem is solved by more students, each solver gets fewer points because ( 30 - s ) decreases as ( s ) increases. So, to maximize the points per problem, we want each problem to be solved by as few students as possible. But since we have 8 problems, we need to distribute the solving among the students so that each student solves enough problems to get a decent score.But since we want to maximize the minimum score, we need to ensure that even the student with the least points has as high a score as possible.Perhaps the optimal scenario is when all students have roughly the same number of points, but since Lasha has the least, we need to arrange it so that even the least is as high as possible.Let me consider that each problem is solved by exactly 15 students. Then, each solver gets ( 30 - 15 = 15 ) points per problem. If each student solves 4 problems, they would get ( 4 times 15 = 60 ) points. Since there are 8 problems, each solved by 15 students, the total number of solves is ( 8 times 15 = 120 ). Since there are 30 students, each student solves ( 120 / 30 = 4 ) problems. So, in this case, every student would have exactly 60 points, meaning Lasha would have 60 points.But is this the maximum possible? Let's see if we can make Lasha have more than 60 points.Suppose we try to have each problem solved by 14 students. Then, each solver gets ( 30 - 14 = 16 ) points. If each student solves 4 problems, they would get ( 4 times 16 = 64 ) points. But wait, the total number of solves would be ( 8 times 14 = 112 ), so each student would solve ( 112 / 30 approx 3.73 ) problems, which isn't possible since students can't solve a fraction of a problem. So, we'd have to distribute the solves such that some students solve 4 problems and others solve 3. But then, the students who solve 3 problems would get ( 3 times 16 = 48 ) points, which is less than 60. So, Lasha's score would be 48, which is worse than the previous case.Alternatively, if we have each problem solved by 16 students, each solver gets ( 30 - 16 = 14 ) points. Total solves ( 8 times 16 = 128 ), so each student solves ( 128 / 30 approx 4.27 ) problems. Again, not an integer, so some solve 4, some solve 5. Those solving 4 get ( 4 times 14 = 56 ), those solving 5 get ( 5 times 14 = 70 ). So, Lasha would have 56, which is still less than 60.Wait, so solving each problem by 15 students gives each student exactly 4 problems solved, each worth 15 points, totaling 60. This seems to be the most balanced scenario where everyone has the same score, so Lasha can't have more than 60 without someone else having less.But let's check if we can have some problems solved by more students and some by fewer, such that the minimum score is higher than 60.Suppose we have some problems solved by 14 students and some by 16 students. Let's say 4 problems solved by 14 students and 4 problems solved by 16 students.Total solves: ( 4 times 14 + 4 times 16 = 56 + 64 = 120 ). So, each student solves ( 120 / 30 = 4 ) problems.Now, for the 4 problems solved by 14 students, each solver gets ( 30 - 14 = 16 ) points.For the 4 problems solved by 16 students, each solver gets ( 30 - 16 = 14 ) points.So, each student solves 4 problems, some combination of the two types.To maximize the minimum score, we need to distribute the solves such that each student has as many high-point problems as possible.But since each student solves 4 problems, and there are 4 of each type, we can arrange it so that each student solves 2 of each type. Then, each student would get ( 2 times 16 + 2 times 14 = 32 + 28 = 60 ) points. So, again, Lasha would have 60 points.Alternatively, if some students solve more of the 16-point problems and fewer of the 14-point ones, their total would be higher, but others would have lower. But since we're trying to maximize the minimum, we need to ensure that the least is as high as possible. So, distributing evenly gives everyone 60.What if we have some problems solved by 13 students and others by 17?Total solves: Let's say 4 problems solved by 13 and 4 by 17.Total solves: ( 4 times 13 + 4 times 17 = 52 + 68 = 120 ). Each student solves 4 problems.Points per problem:- 13 solvers: ( 30 - 13 = 17 ) points each.- 17 solvers: ( 30 - 17 = 13 ) points each.If each student solves 2 of each, they get ( 2 times 17 + 2 times 13 = 34 + 26 = 60 ) points.Same as before.Wait, so regardless of how we distribute the number of solvers per problem, as long as the total solves per student is 4, and the points per problem are set such that each problem solved by ( s ) students gives ( 30 - s ) points, then the total per student is ( 4 times ) average points per problem.But in the case where all problems are solved by 15 students, each problem gives 15 points, so total is 60.In the mixed cases, like 14 and 16, or 13 and 17, the average points per problem is still 15, so total is 60.So, it seems that 60 is the maximum possible minimum score, because any deviation from equal distribution of solves per problem would cause some students to have higher or lower scores, but the minimum can't exceed 60 without someone else having lower.Wait, but what if we have some problems solved by fewer students, giving higher points, and others solved by more, giving lower points, but arrange it so that each student solves a mix that gives them at least 60.But as we saw, if we have some problems with higher points and some with lower, the average per problem is still 15, so total is 60.Alternatively, if we have some problems solved by very few students, giving high points, but then those students would have higher total scores, but others might have lower.Wait, let's try an extreme case. Suppose one problem is solved by only 1 student, giving them ( 30 - 1 = 29 ) points. Then, the other 7 problems are solved by 15 students each, giving 15 points each.Total solves: 1 + 7*15 = 1 + 105 = 106. So, each student solves ( 106 / 30 approx 3.53 ) problems. Not an integer, so some solve 3, some solve 4.But the student who solved the 29-point problem would have at least 29 points plus whatever else. But the students who didn't solve that problem would have to solve more of the 15-point problems.Wait, but the total solves are 106, so 30 students solving an average of ~3.53 problems. So, some solve 3, some solve 4.But the student who solved the 29-point problem could have, say, 29 + 15*3 = 29 + 45 = 74 points.Others who solved 4 problems all of 15 points would have 60 points.But Lasha, who solved only 3 problems of 15 points, would have 45 points, which is worse than 60.So, this arrangement actually lowers the minimum score.Therefore, to maximize the minimum score, we need to ensure that all students solve the same number of problems, and each problem is solved by the same number of students, so that the points are evenly distributed.Hence, the optimal scenario is when each problem is solved by 15 students, each student solves 4 problems, and each problem gives 15 points, leading to a total of 60 points per student, including Lasha.Therefore, the maximal number of points Lasha could have is 60.