💡First, I need to determine how much of the prize will be distributed to 3/5 of the winners. Calculating 2/5 of 500 gives me 200, which will be distributed to 3/5 of the 20 winners, totaling 12 winners.Next, I ensure that each of the 20 winners receives at least 20. This amounts to a total of 400.Subtracting this from the total prize, I have 100 left to distribute among the remaining winners.There are 8 winners left, so I can allocate the remaining 100 to one of them to maximize their award.Finally, adding the initial 20 to the additional 100, the greatest possible individual award is 120.

💡Alright, so I've got this problem about regular n-gons and distances from a point to all the vertices. It's divided into three parts: a, b, and c. Let me try to tackle each part step by step.Starting with part a: It says that if you have a regular n-gon inscribed in a circle of radius R, and you pick any point M on that circle, the sum of the squares of the distances from M to each vertex of the n-gon is equal to 2nR², and it doesn't depend on where M is on the circle.Hmm, okay. So, first, I need to visualize this. Imagine a regular polygon, like a hexagon, inscribed in a circle. If I pick any point on the circumference, say M, and then measure the distance from M to each of the six vertices, square those distances, and add them up, the total should be the same no matter where M is on the circle.I think I remember something about using complex numbers or vectors to represent points on a circle. Maybe I can model the n-gon as points on the unit circle in the complex plane. Let me think: if the polygon is regular, each vertex can be represented as a complex number with magnitude R and angles that are multiples of 2π/n.So, if I let the center of the circle be at the origin, each vertex A_k can be represented as R * e^(iθ_k), where θ_k = 2π(k-1)/n for k = 1, 2, ..., n.Similarly, the point M can be represented as R * e^(iα), where α is some angle. The distance between M and A_k would then be the modulus of the difference between these two complex numbers.So, the distance squared between M and A_k is |M - A_k|². Let's compute that:|M - A_k|² = |R e^(iα) - R e^(iθ_k)|² = R² |e^(iα) - e^(iθ_k)|².Expanding that, |e^(iα) - e^(iθ_k)|² = (e^(iα) - e^(iθ_k))(e^(-iα) - e^(-iθ_k)) = 2 - 2 cos(α - θ_k).So, |M - A_k|² = R² (2 - 2 cos(α - θ_k)) = 2R² (1 - cos(α - θ_k)).Therefore, the sum over all k from 1 to n of |M - A_k|² is 2R² sum_{k=1}^n (1 - cos(α - θ_k)).Simplifying, that's 2R² [n - sum_{k=1}^n cos(α - θ_k)].Now, the key part is evaluating the sum of cos(α - θ_k) over all k. Since θ_k = 2π(k-1)/n, the angles α - θ_k are equally spaced around the circle, starting from α and decreasing by 2π/n each time.I recall that the sum of cosines of equally spaced angles around a circle is zero. Specifically, sum_{k=0}^{n-1} cos(α + 2πk/n) = 0. That's because these points are symmetrically distributed, and their vectors cancel out.Therefore, sum_{k=1}^n cos(α - θ_k) = 0.So, the entire sum becomes 2R² [n - 0] = 2nR². That proves part a.Moving on to part b: Now, the point M is arbitrary in the plane, not necessarily on the circumcircle. We need to show that the sum of the squares of the distances from M to all vertices depends only on the distance l from M to the center O, and is equal to n(R² + l²).Hmm, okay. So, let's consider M somewhere in the plane, at a distance l from the center O. Let's denote the center as O, and the position vectors of the vertices A_k relative to O as vectors a_k, each with magnitude R.The position vector of M relative to O is m, with |m| = l.The distance squared from M to A_k is |m - a_k|². So, the sum we're interested in is sum_{k=1}^n |m - a_k|².Expanding this, |m - a_k|² = |m|² + |a_k|² - 2 m · a_k.So, sum_{k=1}^n |m - a_k|² = sum_{k=1}^n (|m|² + |a_k|² - 2 m · a_k) = n|m|² + sum_{k=1}^n |a_k|² - 2 m · sum_{k=1}^n a_k.Now, |m|² is l², and |a_k|² is R² for each k. So, sum_{k=1}^n |a_k|² = nR².What about sum_{k=1}^n a_k? Since the polygon is regular and centered at O, the vectors a_k are symmetrically distributed around the center. Therefore, their vector sum is zero. So, sum_{k=1}^n a_k = 0.Therefore, the entire expression simplifies to n l² + n R² - 0 = n(R² + l²). That proves part b.For part c: Now, M doesn't have to lie in the plane of the n-gon. We need to show that the same formula holds, i.e., the sum of the squares of the distances from M to all vertices is still n(R² + l²), where l is the distance from M to the center O.Hmm, okay. So, now M is in 3D space, not necessarily in the plane of the n-gon. Let's model this.Let me consider the n-gon lying in a plane, say the xy-plane, with center at the origin O. The point M has coordinates (x, y, z), where z ≠ 0 in general.The distance from M to each vertex A_k is sqrt[(x - a_kx)² + (y - a_ky)² + z²], where (a_kx, a_ky, 0) are the coordinates of A_k.So, the square of the distance is (x - a_kx)² + (y - a_ky)² + z².Therefore, the sum over all k is sum_{k=1}^n [(x - a_kx)² + (y - a_ky)² + z²].This can be rewritten as sum_{k=1}^n [(x - a_kx)² + (y - a_ky)²] + n z².But (x - a_kx)² + (y - a_ky)² is just the square of the distance from the projection of M onto the plane (which is (x, y, 0)) to A_k.From part b, we know that the sum of the squares of the distances from (x, y, 0) to all A_k is n(R² + l²), where l is the distance from (x, y, 0) to O, which is sqrt(x² + y²).Therefore, sum_{k=1}^n [(x - a_kx)² + (y - a_ky)²] = n(R² + (x² + y²)).Adding the n z² term, the total sum becomes n(R² + x² + y²) + n z² = n(R² + x² + y² + z²).But x² + y² + z² is the square of the distance from M to O, which is l². Therefore, the total sum is n(R² + l²).So, even when M is not in the plane of the n-gon, the formula still holds. That proves part c.**Final Answer**a) boxed{2nR^2}b) boxed{n(R^2 + l^2)}c) boxed{n(R^2 + l^2)}

💡First, I need to determine the number of student tickets sold. I'll start by defining variables for the unknowns.Let ( S ) represent the number of student tickets and ( N ) represent the number of non-student tickets.From the problem, I know two key pieces of information:1. The total number of tickets sold is 2000: [ S + N = 2000 ]2. The total revenue from ticket sales is 20,960: [ 9S + 11N = 20960 ]I can solve this system of equations to find ( S ).First, solve the first equation for ( N ):[N = 2000 - S]Next, substitute this expression for ( N ) into the second equation:[9S + 11(2000 - S) = 20960]Simplify and solve for ( S ):[9S + 22000 - 11S = 20960][-2S + 22000 = 20960][-2S = -1040][S = 520]Therefore, 520 student tickets were sold.

💡Okay, so I have this problem: "How many ways are there to put 6 balls in 2 boxes if the balls are distinguishable but the boxes are not?" Hmm, let me try to understand what this means.First, the balls are distinguishable, which means each ball is unique. Maybe they have different colors, numbers, or something that sets them apart from each other. On the other hand, the boxes are not distinguishable, so swapping the contents of the two boxes doesn't create a new arrangement. That makes sense because if the boxes were different, like one red and one blue, then putting different balls in each would count as a different arrangement. But since they're the same, it doesn't matter which box is which.So, I need to figure out how many distinct ways I can distribute these 6 unique balls into 2 identical boxes. Let me think about how to approach this.I remember that when dealing with distinguishable objects and indistinct containers, the problem is related to set partitions. Specifically, it's about partitioning the set of balls into two subsets, where the order of the subsets doesn't matter. So, in this case, I need to find the number of ways to split 6 distinguishable balls into two groups, without caring which group is which.Wait, but isn't this similar to counting the number of subsets? If the boxes were labeled, like Box A and Box B, then the number of ways would be 2^6, since each ball can go into either Box A or Box B. But since the boxes are indistinct, some of these configurations are duplicates. For example, putting Ball 1 in Box A and Balls 2-6 in Box B is the same as putting Ball 1 in Box B and Balls 2-6 in Box A because the boxes are identical.So, to adjust for this overcounting, I need to consider that each distribution is counted twice in the 2^6 total, except for the cases where all balls are in one box or the other. Wait, no, actually, if all balls are in one box, swapping the boxes doesn't change anything, so those cases are only counted once. Hmm, this is getting a bit confusing.Let me try to break it down. If I consider all possible distributions where the boxes are labeled, that's 2^6 = 64 ways. But since the boxes are indistinct, I need to divide by 2 to account for the fact that swapping the boxes doesn't create a new distribution. However, this only works if the distributions are symmetric, meaning that each distribution is counted twice except when the two boxes have the same number of balls.Wait, but in this case, since we have 6 balls, which is an even number, there is a case where each box has 3 balls. In that case, swapping the boxes would result in the same distribution, so those cases are only counted once in the 64 total. Therefore, I can't simply divide the total by 2 because some distributions are unique.This seems complicated. Maybe there's a better way to approach this. I recall something called the Stirling numbers of the second kind, which count the number of ways to partition a set of n objects into k non-empty subsets. In this case, since the boxes are indistinct and can be empty, I need to consider both cases where one box is empty and both boxes have balls.Wait, so if I use the Stirling numbers of the second kind, S(n, k), which counts the number of ways to partition n distinguishable objects into k non-empty indistinct subsets, then for k=2, it would be S(6,2). But since the boxes can be empty, I also need to include the cases where all balls are in one box. So, actually, the total number of ways would be S(6,1) + S(6,2).But S(6,1) is just 1, because there's only one way to put all 6 balls into a single box. And S(6,2) is the number of ways to partition the 6 balls into two non-empty indistinct subsets. I think S(n,2) = 2^(n-1) - 1. Wait, is that right?Let me check. For n=2, S(2,2)=1, and 2^(2-1)-1=1, which matches. For n=3, S(3,2)=3, and 2^(3-1)-1=3, which also matches. So, yes, S(n,2)=2^(n-1)-1. Therefore, S(6,2)=2^5 -1=32-1=31.So, the total number of ways is S(6,1)+S(6,2)=1+31=32. But wait, earlier I thought it was 42. Did I make a mistake?Wait, no. I think I'm confusing something here. Let me go back. The total number of ways to distribute 6 distinguishable balls into 2 indistinct boxes, allowing empty boxes, is equal to the sum of S(6,1) and S(6,2). S(6,1)=1, and S(6,2)=31, so total is 32.But earlier, I thought it was 42. Where did I get that number? Maybe I was considering something else. Let me think again.If I consider labeled boxes, it's 2^6=64. If I consider indistinct boxes, I need to adjust for overcounting. The number of ways where the boxes are non-empty is S(6,2)=31, and the number of ways where one box is empty is 1 (all balls in one box). So total is 31+1=32.But wait, another way to think about it is that for each non-empty subset of the balls, we can pair it with its complement. Since the boxes are indistinct, each pair is counted once. So, the number of ways is equal to the number of subsets divided by 2, but since subsets come in pairs (a subset and its complement), and the total number of subsets is 2^6=64, then the number of distinct pairs is 64/2=32. But we have to be careful because the empty set and the full set are their own complements. So, in this case, the empty set corresponds to putting all balls in one box, which is one way, and the full set is the same as the empty set in the other box, which is the same as the empty set. So, actually, the number of distinct distributions is (2^6)/2=32, which matches the previous result.So, the total number of ways is 32. But wait, earlier I thought it was 42. Did I make a mistake in my initial reasoning?Wait, no. I think the confusion comes from whether the boxes can be empty or not. In the problem statement, it just says "put 6 balls in 2 boxes," which doesn't specify that the boxes must be non-empty. So, allowing empty boxes, the total number of ways is 32.But in the initial problem, the user mentioned 42. Let me check their reasoning. They considered cases where the number of balls in each box is 6-0, 5-1, 4-2, 3-3. For each case, they calculated the number of ways as binomial coefficients: 1, 6, 15, 20, totaling 42.Wait, but that approach assumes that the boxes are labeled, right? Because when you choose 5 balls for one box, you're assuming that the other box gets the remaining 1, but since the boxes are indistinct, choosing 5 for one box is the same as choosing 1 for the other. So, in that case, the total number of ways would be the sum of binomial coefficients divided by 2, except for the case where the distribution is symmetric (i.e., 3-3), which is only counted once.So, let's see: for 6-0, it's 1 way. For 5-1, it's 6 ways, but since the boxes are indistinct, this is the same as 1-5, so we have to divide by 2, giving 3 unique ways. Similarly, for 4-2, it's 15 ways, but again, since 4-2 is the same as 2-4, we divide by 2, giving 7.5, which doesn't make sense because the number of ways should be an integer. Hmm, that's a problem.Wait, actually, the binomial coefficient for 4-2 is 15, which is the number of ways to choose 4 balls out of 6. But since choosing 4 for one box is the same as choosing 2 for the other, the number of unique distributions is 15/2=7.5, which is not possible. So, this approach is flawed.Alternatively, maybe we should consider that for each case where the distribution is asymmetric (i.e., not equal numbers), the number of unique distributions is half the binomial coefficient, and for the symmetric case, it's just the binomial coefficient divided by 2.Wait, let's think differently. The total number of ways to distribute the balls into labeled boxes is 2^6=64. Since the boxes are indistinct, each distribution is counted twice except when the distribution is symmetric (i.e., when the two boxes have the same number of balls). So, the number of unique distributions is (2^6 + S)/2, where S is the number of symmetric distributions.In this case, the only symmetric distribution is when each box has 3 balls. The number of ways to choose 3 balls out of 6 is C(6,3)=20. But since the boxes are indistinct, each such distribution is counted once, not twice. So, the total number of unique distributions is (64 + 20)/2=84/2=42.Ah, that makes sense now. So, the total number of ways is 42. Therefore, the initial reasoning was correct, and my confusion came from trying to apply the Stirling numbers approach without considering the empty boxes properly.So, to summarize, the total number of ways to distribute 6 distinguishable balls into 2 indistinct boxes is 42.

💡Alright, so I have this problem here: given the function f(x) = ax - ln x, and it's supposed to be greater than 1 for all x in the interval (1, +∞). I need to find the range of the real number a. The options are A: (-∞, 1), B: (-∞, 1], C: (1, +∞), and D: [1, +∞). Hmm, okay.First, let me understand the function f(x) = ax - ln x. It's a combination of a linear term ax and a logarithmic term -ln x. Since x is greater than 1, ln x is positive, so -ln x is negative. So, f(x) is ax minus something positive. We need this to be greater than 1 for all x > 1.So, f(x) > 1 implies ax - ln x > 1. Let me rearrange this inequality to solve for a. If I move the ln x to the other side, I get ax > 1 + ln x. Then, dividing both sides by x (since x is positive, the inequality direction doesn't change), I get a > (1 + ln x)/x.So, a has to be greater than (1 + ln x)/x for all x > 1. That means a must be greater than the maximum value that (1 + ln x)/x can take in the interval (1, +∞). Therefore, I need to find the maximum of the function g(x) = (1 + ln x)/x on (1, +∞), and then a must be at least that maximum.To find the maximum of g(x), I should take its derivative and find critical points. Let's compute g'(x). Using the quotient rule: if g(x) = numerator/denominator, then g'(x) = (num’ * denom - num * denom’) / denom².So, numerator is 1 + ln x, so its derivative is 1/x. Denominator is x, so its derivative is 1. Therefore, g'(x) = ( (1/x)*x - (1 + ln x)*1 ) / x². Simplify numerator: (1 - (1 + ln x)) = -ln x. So, g'(x) = (-ln x)/x².Now, since x > 1, ln x is positive, so g'(x) is negative. That means g(x) is decreasing on (1, +∞). So, its maximum occurs at the left endpoint, which is x = 1.Let me compute g(1): (1 + ln 1)/1 = (1 + 0)/1 = 1. So, the maximum value of g(x) on (1, +∞) is 1. Since g(x) is decreasing, for all x > 1, g(x) < 1.Therefore, to satisfy a > g(x) for all x > 1, a must be greater than or equal to 1. Because if a is exactly 1, then at x approaching 1 from the right, f(x) approaches 1 - 0 = 1, but since x is strictly greater than 1, f(x) will be slightly greater than 1. For a greater than 1, f(x) will definitely be greater than 1 for all x > 1.Wait, let me check that. If a is exactly 1, then f(x) = x - ln x. Let's see what happens as x approaches 1 from the right. f(1) would be 1 - 0 = 1, but since x is in (1, +∞), f(x) is slightly more than 1. Actually, let's compute the limit as x approaches 1 from the right: lim x→1+ f(x) = 1 - 0 = 1. But since x is strictly greater than 1, f(x) will be slightly more than 1. However, for a = 1, is f(x) > 1 for all x > 1?Let me test x = 2. f(2) = 2 - ln 2 ≈ 2 - 0.693 ≈ 1.307 > 1. Okay, that's good. What about x approaching infinity? f(x) = x - ln x. As x grows, ln x grows much slower than x, so f(x) tends to infinity. So, f(x) is increasing for large x when a = 1.But wait, is f(x) = x - ln x always increasing? Let's compute its derivative: f'(x) = 1 - 1/x. For x > 1, f'(x) = 1 - 1/x > 0, since 1/x < 1. So, f(x) is increasing on (1, +∞). Therefore, the minimum value of f(x) on (1, +∞) is the limit as x approaches 1 from the right, which is 1. But since x is strictly greater than 1, f(x) is always greater than 1. So, a = 1 is acceptable.If a is greater than 1, say a = 2, then f(x) = 2x - ln x. The derivative is 2 - 1/x, which is even larger, so f(x) is increasing faster, and certainly f(x) > 1 for all x > 1.If a is less than 1, say a = 0.5, then f(x) = 0.5x - ln x. Let's see what happens as x approaches 1 from the right: f(x) approaches 0.5 - 0 = 0.5 < 1. So, f(x) is not greater than 1 near x = 1. Therefore, a cannot be less than 1.Wait, but the options include a = 1. So, the range of a is [1, +∞). That would be option D.Let me double-check. If a = 1, f(x) = x - ln x. We saw that as x approaches 1 from the right, f(x) approaches 1, but since x is strictly greater than 1, f(x) is slightly more than 1. And since f(x) is increasing for x > 1, it will always be greater than 1. So, a = 1 is included.If a is greater than 1, f(x) is even larger, so it's definitely greater than 1. If a is less than 1, near x = 1, f(x) dips below 1, which violates the condition. Therefore, the correct range is [1, +∞), which is option D.I think that makes sense. I don't see any mistakes in my reasoning. So, I'll go with option D.

💡Alright, so I've got this problem here: I need to find the range of values for ( x^2 + y^2 ) given that ( x geq 0 ), ( y geq 0 ), and ( x + y = 1 ). Hmm, okay, let's break this down step by step.First off, since both ( x ) and ( y ) are non-negative and their sum is 1, it means that both ( x ) and ( y ) must be between 0 and 1. That makes sense because if one of them is 0, the other has to be 1 to satisfy the equation ( x + y = 1 ). So, ( x ) and ( y ) are constrained within the interval [0, 1].Now, the expression I need to analyze is ( x^2 + y^2 ). I wonder if there's a way to express this in terms of a single variable since I know that ( x + y = 1 ). Maybe I can substitute ( y ) with ( 1 - x ) or vice versa. Let me try that.Let's substitute ( y ) with ( 1 - x ). So, ( y = 1 - x ). Then, ( x^2 + y^2 ) becomes ( x^2 + (1 - x)^2 ). Let's expand that:( x^2 + (1 - x)^2 = x^2 + (1 - 2x + x^2) = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1 ).Okay, so now I have the expression ( 2x^2 - 2x + 1 ). This is a quadratic function in terms of ( x ). Quadratic functions have either a minimum or maximum value depending on the coefficient of ( x^2 ). In this case, the coefficient is 2, which is positive, so the parabola opens upwards, meaning it has a minimum point.To find the minimum value of this quadratic function, I can use the vertex formula. The vertex of a parabola given by ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). Plugging in the values from my equation:( a = 2 ), ( b = -2 ), so ( x = -frac{-2}{2 times 2} = frac{2}{4} = frac{1}{2} ).So, the vertex is at ( x = frac{1}{2} ). Let's find the value of the function at this point:( fleft(frac{1}{2}right) = 2left(frac{1}{2}right)^2 - 2left(frac{1}{2}right) + 1 = 2 times frac{1}{4} - 1 + 1 = frac{1}{2} - 1 + 1 = frac{1}{2} ).So, the minimum value of ( x^2 + y^2 ) is ( frac{1}{2} ).Now, what about the maximum value? Since the parabola opens upwards, the function will increase as we move away from the vertex. Given that ( x ) is constrained between 0 and 1, I should check the endpoints of this interval to find the maximum value.Let's evaluate the function at ( x = 0 ):( f(0) = 2(0)^2 - 2(0) + 1 = 0 - 0 + 1 = 1 ).And at ( x = 1 ):( f(1) = 2(1)^2 - 2(1) + 1 = 2 - 2 + 1 = 1 ).So, at both endpoints, the function equals 1. Therefore, the maximum value of ( x^2 + y^2 ) is 1.Putting it all together, the range of ( x^2 + y^2 ) is from ( frac{1}{2} ) to 1. That means ( x^2 + y^2 ) can take any value between ( frac{1}{2} ) and 1, inclusive.Just to double-check, let's consider some specific cases:1. When ( x = 0 ), ( y = 1 ): ( x^2 + y^2 = 0 + 1 = 1 ).2. When ( x = 1 ), ( y = 0 ): ( x^2 + y^2 = 1 + 0 = 1 ).3. When ( x = frac{1}{2} ), ( y = frac{1}{2} ): ( x^2 + y^2 = frac{1}{4} + frac{1}{4} = frac{1}{2} ).These results align with what I found earlier, so I'm confident that the range is correct.**Final Answer**The range of values for ( x^2 + y^2 ) is boxed{left[ dfrac{1}{2}, 1 right]}.