answer：First, determine the work rates of A and B individually. A can complete the work in 15 days, so A's work rate is ( frac{1}{15} ) of the work per day. Similarly, B can complete the work in 20 days, so B's work rate is ( frac{1}{20} ) of the work per day.Next, calculate their combined work rate by adding their individual rates:[frac{1}{15} + frac{1}{20} = frac{4}{60} + frac{3}{60} = frac{7}{60}]This means together, A and B can complete ( frac{7}{60} ) of the work in one day.For 3 days of work together, multiply the combined work rate by 3:[3 times frac{7}{60} = frac{21}{60} = frac{7}{20}]So, A and B complete ( frac{7}{20} ) of the work in 3 days.Finally, subtract the completed work from the total work to find the fraction of the work that is left:[1 - frac{7}{20} = frac{20}{20} - frac{7}{20} = frac{13}{20}]

answer：First, I need to determine the total number of ways to select 2 cards out of the 4 available. This can be calculated using the combination formula, which is C(4,2) = 6 ways.Next, I should identify the number of favorable outcomes where the two selected cards have the same pattern. There are two possible scenarios for this:1. Both cards have the "Zongzi" pattern. Since there are 2 "Zongzi" cards, the number of ways to select both is C(2,2) = 1.2. Both cards have the "Dragon Boat" pattern. Similarly, with 2 "Dragon Boat" cards, the number of ways to select both is C(2,2) = 1.Adding these favorable outcomes together gives a total of 1 + 1 = 2 favorable outcomes.Finally, to find the probability of selecting two cards with the same pattern, I divide the number of favorable outcomes by the total number of possible outcomes: 2/6 = 1/3.

answer：Alright, so I have this problem here about functions ( g(x) ) and ( f(x) ). The domain of ( f(x) ) is all real numbers, which is good to know. The problem states that ( g(x) + f(x) = 1 ), which means these two functions are related in a way that their sum is always 1 for any ( x ). Then, it says that both ( g(x+1) ) and ( f(2-x) ) are odd functions. Hmm, okay. I remember that an odd function satisfies the property ( h(-x) = -h(x) ) for all ( x ) in its domain. So, if ( g(x+1) ) is odd, that means ( g(-x+1) = -g(x+1) ). Similarly, since ( f(2-x) ) is odd, that implies ( f(2+x) = -f(2-x) ).Let me write down these properties:1. ( g(-x + 1) = -g(x + 1) ) (because ( g(x+1) ) is odd)2. ( f(2 + x) = -f(2 - x) ) (because ( f(2 - x) ) is odd)Now, since ( g(x) + f(x) = 1 ), I can express ( f(x) ) in terms of ( g(x) ): ( f(x) = 1 - g(x) ). Maybe substituting this into the second property will help.So, substituting ( f(x) = 1 - g(x) ) into the second property:( f(2 + x) = -f(2 - x) )becomes( 1 - g(2 + x) = -[1 - g(2 - x)] )Let me simplify this:( 1 - g(2 + x) = -1 + g(2 - x) )Bringing like terms together:( 1 + 1 = g(2 + x) + g(2 - x) )So,( 2 = g(2 + x) + g(2 - x) )This is an interesting equation. It tells me that for any ( x ), the sum of ( g ) evaluated at ( 2 + x ) and ( 2 - x ) is 2. Maybe I can use this to find specific values of ( g ) at certain points.Looking back at the first property:( g(-x + 1) = -g(x + 1) )Let me make a substitution here. Let me set ( y = x + 1 ). Then, ( x = y - 1 ). Substituting into the equation:( g(-(y - 1) + 1) = -g(y) )Simplify:( g(-y + 1 + 1) = -g(y) )Which is:( g(-y + 2) = -g(y) )So, ( g(2 - y) = -g(y) ). That's another useful property. So, ( g(2 - y) = -g(y) ) for any ( y ).Let me write that down:3. ( g(2 - y) = -g(y) )This seems like a reflection property of ( g ) around the point ( y = 1 ). Interesting.Now, let's see if I can use equations 2 and 3 together.From equation 2: ( g(2 + x) + g(2 - x) = 2 )But from equation 3, ( g(2 - x) = -g(x) ). So, substituting into equation 2:( g(2 + x) + (-g(x)) = 2 )Which simplifies to:( g(2 + x) - g(x) = 2 )So, ( g(2 + x) = g(x) + 2 )Hmm, that's a functional equation. It tells me that shifting the argument of ( g ) by 2 increases the function's value by 2. That seems like a linear behavior, but I need to verify.Let me test this with specific values.First, let's try ( x = 0 ):From equation 3: ( g(2 - 0) = -g(0) ) => ( g(2) = -g(0) )From equation 2: ( g(2 + 0) + g(2 - 0) = 2 ) => ( g(2) + g(2) = 2 ) => ( 2g(2) = 2 ) => ( g(2) = 1 )So, ( g(2) = 1 ). Then, from equation 3, ( g(2) = -g(0) ) => ( 1 = -g(0) ) => ( g(0) = -1 ). So, that's one value: ( g(0) = -1 ). That's option A.Next, let's try ( x = 1 ):From equation 3: ( g(2 - 1) = -g(1) ) => ( g(1) = -g(1) ) => ( 2g(1) = 0 ) => ( g(1) = 0 ). So, that's another value: ( g(1) = 0 ). That's option B.Now, let's try ( x = 2 ):From equation 3: ( g(2 - 2) = -g(2) ) => ( g(0) = -g(2) ). We already know ( g(0) = -1 ) and ( g(2) = 1 ), so that's consistent.From equation 2: ( g(2 + 2) + g(2 - 2) = 2 ) => ( g(4) + g(0) = 2 ). We know ( g(0) = -1 ), so ( g(4) - 1 = 2 ) => ( g(4) = 3 ).But wait, from the functional equation ( g(2 + x) = g(x) + 2 ). Let's set ( x = 2 ):( g(2 + 2) = g(2) + 2 ) => ( g(4) = 1 + 2 = 3 ). That's consistent.Similarly, ( x = 3 ):From the functional equation: ( g(2 + 3) = g(3) + 2 ) => ( g(5) = g(3) + 2 )But I don't know ( g(5) ) yet. Maybe I can find ( g(3) ) another way.From equation 3: ( g(2 - 3) = -g(3) ) => ( g(-1) = -g(3) )But I don't have information about ( g(-1) ). Maybe I can use equation 2 with ( x = 1 ):( g(2 + 1) + g(2 - 1) = 2 ) => ( g(3) + g(1) = 2 ). We know ( g(1) = 0 ), so ( g(3) + 0 = 2 ) => ( g(3) = 2 ).Wait, but earlier, from the functional equation, ( g(5) = g(3) + 2 = 2 + 2 = 4 ). So, ( g(5) = 4 ).But let's check if this is consistent with equation 3. From equation 3: ( g(2 - 5) = -g(5) ) => ( g(-3) = -g(5) = -4 ). But I don't know ( g(-3) ), so maybe it's okay.Wait, but the problem is asking about ( g(3) ). So, according to this, ( g(3) = 2 ). But option D says ( g(3) = 0 ). So, that would be incorrect.Wait, did I make a mistake? Let me double-check.From equation 2: ( g(2 + x) + g(2 - x) = 2 ). Let ( x = 1 ):( g(3) + g(1) = 2 ). Since ( g(1) = 0 ), then ( g(3) = 2 ). That seems correct.Alternatively, from the functional equation ( g(2 + x) = g(x) + 2 ). Let ( x = 1 ):( g(3) = g(1) + 2 = 0 + 2 = 2 ). So, ( g(3) = 2 ). So, option D says ( g(3) = 0 ), which is incorrect.Wait, but maybe I misapplied the functional equation. Let me see.From equation 2 and 3, we derived ( g(2 + x) = g(x) + 2 ). So, for any ( x ), shifting by 2 increases the function by 2. So, starting from ( x = 0 ):( g(2) = g(0) + 2 = -1 + 2 = 1 ). Correct.( g(4) = g(2) + 2 = 1 + 2 = 3 ). Correct.( g(6) = g(4) + 2 = 3 + 2 = 5 ). And so on.Similarly, going backward, for ( x = -2 ):( g(0) = g(-2) + 2 ) => ( -1 = g(-2) + 2 ) => ( g(-2) = -3 ).But I don't know if that's necessary.Wait, but the problem is only asking about ( g(0) ), ( g(1) ), ( g(2) ), and ( g(3) ). So, from the above, we have:- ( g(0) = -1 ) (Option A)- ( g(1) = 0 ) (Option B)- ( g(2) = 1 ) (Option C)- ( g(3) = 2 ) (Which is not among the options, but Option D says ( g(3) = 0 ), which is incorrect)So, the correct options are A, B, and C.Wait, but let me check if I can find ( g(3) ) another way to confirm.From equation 3: ( g(2 - x) = -g(x) ). Let ( x = 3 ):( g(2 - 3) = -g(3) ) => ( g(-1) = -g(3) )But I don't know ( g(-1) ). However, from equation 2, with ( x = -1 ):( g(2 + (-1)) + g(2 - (-1)) = 2 ) => ( g(1) + g(3) = 2 ). We know ( g(1) = 0 ), so ( g(3) = 2 ). So, that's consistent.Alternatively, from the functional equation ( g(2 + x) = g(x) + 2 ), set ( x = 1 ):( g(3) = g(1) + 2 = 0 + 2 = 2 ). So, same result.Therefore, ( g(3) = 2 ), which is not equal to 0. So, option D is incorrect.So, the correct options are A, B, and C.Wait, but let me make sure I didn't miss anything. Let me summarize:- ( g(0) = -1 ) (A)- ( g(1) = 0 ) (B)- ( g(2) = 1 ) (C)- ( g(3) = 2 ) (Not D)Therefore, options A, B, and C are correct.I think that's it. I don't see any mistakes in my reasoning. So, the answer should be A, B, and C.

answer：Okay, so I have this problem where there are 2019 boxes, each containing 2019 stones numbered from 1 to 2019. The total mass of the stones in each box is 1 kilogram. I need to figure out the maximum number of ways, k, to pick stones from different boxes such that each stone has a unique number and their total mass is at least 1 kilogram.Hmm, let me try to understand this better. Each box has stones numbered 1 through 2019, and each box's stones add up to 1 kg. So, for each box, the sum of the masses of stones 1 to 2019 is 1 kg. Now, I need to pick one stone from each box, but each stone must have a unique number. That means I can't pick two stones with the same number from different boxes. So, essentially, I'm selecting a permutation of the numbers 1 through 2019, where each number comes from a different box.The goal is to find the maximum number of such permutations where the total mass of the selected stones is at least 1 kilogram. So, I need to maximize k, the number of different ways to pick these stones, such that each way meets the mass requirement.I think the key here is to consider how the masses are distributed across the boxes. Since each box has a total mass of 1 kg, the average mass per stone in a box is 1/2019 kg. But the distribution can vary. Some stones might be heavier, and some might be lighter.If I want to maximize the number of ways to pick stones with a total mass of at least 1 kg, I need to arrange the masses in such a way that as many permutations as possible meet this condition. That probably means making sure that in each permutation, the sum of the masses is just enough to reach 1 kg.Wait, but how do I ensure that? Maybe if I have each stone in a box have the same mass, then every permutation would have the same total mass. Let me check that.If each stone in every box has a mass of 1/2019 kg, then the total mass of any permutation would be 2019*(1/2019) = 1 kg. So, every permutation would have exactly 1 kg. That means k would be 2019 factorial, which is a huge number. But the problem says "at least 1 kilogram," so if all permutations sum to exactly 1 kg, then k would be 2019!.But wait, the problem says "in all situations satisfying these conditions." So, I think it's asking for the maximum k such that no matter how the masses are distributed in the boxes (as long as each box has total mass 1 kg), there are at least k ways to pick stones with different numbers and total mass at least 1 kg.Oh, so it's not about arranging the masses to maximize k, but rather, no matter how the masses are arranged, there will be at least k ways to pick stones with the required properties. And we need to find the maximum such k.That changes things. So, regardless of how the masses are distributed in the boxes, we need to guarantee that there are at least k ways to pick stones with different numbers and total mass at least 1 kg.In that case, we need to find the minimal number of such permutations across all possible mass distributions, and then find the maximum k that is always achievable.Hmm, this seems related to some kind of combinatorial argument or perhaps an application of the pigeonhole principle.Let me think about it. If we have 2019 boxes, each with 2019 stones, and each box has a total mass of 1 kg. We need to pick one stone from each box, each with a unique number, such that the total mass is at least 1 kg.Wait, but if we pick one stone from each box, that's 2019 stones, each with a unique number. So, the total mass would be the sum of 2019 stones, each from a different box, each with a unique number.But each box contributes exactly one stone, so the total mass is the sum of 2019 stones, each from different boxes, each with a unique number.Wait, but each box has a total mass of 1 kg, so the average mass per stone in a box is 1/2019 kg. So, the average total mass of 2019 stones, each from different boxes, would be 2019*(1/2019) = 1 kg.So, on average, the total mass is 1 kg. But depending on the distribution, some permutations might have a total mass greater than 1 kg, and some might have less.But the problem is asking for the maximum k such that in any distribution, there are at least k permutations with total mass at least 1 kg.So, we need to find the minimal number of such permutations across all possible distributions, and then find the maximum k that is always achievable.This seems related to the idea of derangements or something similar, but I'm not sure.Alternatively, maybe we can model this as a matrix where each row represents a box, each column represents a stone number, and the entries are the masses.Then, we're looking for permutation matrices where the sum of the selected entries is at least 1.We need to find the minimal number of such permutation matrices across all possible mass matrices, and then find the maximum k that is always achievable.But this might be too abstract.Wait, maybe we can use some kind of averaging argument. Since the total mass across all boxes is 2019 kg (each box has 1 kg, 2019 boxes), and there are 2019! permutations, each corresponding to a selection of stones.If we consider the average total mass over all permutations, it would be 2019 kg / 2019! * something? Wait, no.Actually, each permutation selects one stone from each box, so the total mass of a permutation is the sum of 2019 stones, each from a different box. Since each box contributes exactly one stone, the total mass across all permutations would be 2019! * (sum of all stones in all boxes).But the sum of all stones in all boxes is 2019 kg (each box has 1 kg, 2019 boxes). So, the total mass across all permutations is 2019! * 2019 kg.But each permutation has a total mass, and we're interested in how many permutations have total mass at least 1 kg.If the average total mass per permutation is (2019! * 2019 kg) / 2019! = 2019 kg. Wait, that can't be right because each permutation is a selection of 2019 stones, each from different boxes, so the total mass per permutation is the sum of 2019 stones, each from different boxes.But each box contributes exactly one stone, so the total mass of all permutations would be 2019! times the sum of all stones in all boxes, which is 2019! * 2019 kg.But the average total mass per permutation is (2019! * 2019 kg) / 2019! = 2019 kg. Wait, that seems high because each permutation is only selecting 2019 stones, each from different boxes, so the total mass should be 2019*(average mass per stone). Since each box has 1 kg over 2019 stones, the average mass per stone is 1/2019 kg. So, 2019*(1/2019) = 1 kg. So, the average total mass per permutation is 1 kg.Therefore, the average total mass is 1 kg. So, by the probabilistic method, there must be at least one permutation with total mass at least 1 kg. But we need to find the maximum k such that in any distribution, there are at least k permutations with total mass at least 1 kg.Wait, but if the average is 1 kg, then it's possible that all permutations have exactly 1 kg, which would mean k = 2019!.But the problem is asking for the maximum k such that in any distribution, there are at least k permutations with total mass at least 1 kg. So, we need to find the minimal number of such permutations across all possible distributions, and then find the maximum k that is always achievable.But how do we find that?Maybe we can think about the worst-case distribution where as few permutations as possible have total mass at least 1 kg.What would such a distribution look like? Perhaps one where most permutations have total mass just below 1 kg, and only a few have total mass just above 1 kg.But how can we arrange the masses to minimize the number of permutations with total mass at least 1 kg?Perhaps by making one stone in each box very heavy, and the rest very light, so that only permutations that include all the heavy stones would have total mass at least 1 kg.Wait, let's think about that.Suppose in each box, one stone has mass close to 1 kg, and the other 2018 stones have mass close to 0. Then, to get a total mass of at least 1 kg, a permutation must include all the heavy stones. But since each box has only one heavy stone, and each permutation must pick one stone from each box, the only way to get a total mass of at least 1 kg is to pick the heavy stone from each box. But since each box has only one heavy stone, there is only one permutation that includes all heavy stones. So, in this case, k would be 1.But the problem is asking for the maximum k such that in any distribution, there are at least k permutations with total mass at least 1 kg. So, in this case, k would be 1, but we need to find the maximum k that is always achievable, regardless of the distribution.Wait, but that can't be right because the problem is asking for the maximum k such that in any distribution, there are at least k permutations with total mass at least 1 kg. So, if in some distributions, k is 1, but in others, k is larger, then the maximum k that is always achievable is 1.But that seems too low. Maybe I'm misunderstanding the problem.Wait, the problem says: "if one can pick stones from different boxes with different numbers, with total mass of at least 1 kilogram, in k different ways, what is the maximal of k?"Wait, maybe it's not that in any distribution, there are at least k ways, but rather, what is the maximum k such that in some distribution, there are k ways. But the wording is a bit unclear.Wait, let me read it again: "In all situations satisfying these conditions, if one can pick stones from different boxes with different numbers, with total mass of at least 1 kilogram, in k different ways, what is the maximal of k?"Hmm, it's a bit ambiguous. It could mean that for all distributions, the number of ways is at least k, and we need to find the maximum such k. Or it could mean that across all distributions, what's the maximum k such that there exists a distribution where the number of ways is k.But the wording says "in all situations satisfying these conditions, if one can pick stones... in k different ways, what is the maximal of k?"So, it seems like it's asking for the maximum k such that in every distribution, there are at least k ways to pick stones with the required properties. So, we need to find the minimal number of such ways across all distributions, and then find the maximum k that is always achievable.So, in other words, find the minimal number of permutations with total mass at least 1 kg across all possible distributions, and then that minimal number is the maximal k that is always achievable.So, to find the maximum k such that in every distribution, there are at least k permutations with total mass at least 1 kg.So, we need to find the minimal number of such permutations across all distributions, and that minimal number is the maximal k.So, how do we find that minimal number?Perhaps by considering the worst-case distribution where the number of permutations with total mass at least 1 kg is minimized.So, what distribution would minimize the number of such permutations?I think it's the distribution where as many permutations as possible have total mass less than 1 kg, and only a few have total mass at least 1 kg.To achieve this, we might want to make the masses in such a way that most permutations have total mass just below 1 kg, and only a few have total mass just above 1 kg.But how?Maybe by making one stone in each box very heavy, and the rest very light, as I thought before.Suppose in each box, one stone has mass 1 - ε, where ε is a very small positive number, and the other 2018 stones have mass ε/(2018). So, the total mass in each box is (1 - ε) + 2018*(ε/2018) = 1 - ε + ε = 1 kg, which satisfies the condition.Now, in this distribution, what is the total mass of a permutation?If a permutation picks the heavy stone from each box, then the total mass would be 2019*(1 - ε). Since 2019*(1 - ε) = 2019 - 2019ε. If ε is very small, say ε = 1/2019, then 2019*(1 - 1/2019) = 2018 kg. That's way more than 1 kg.Wait, but that's not helpful because we want to minimize the number of permutations with total mass at least 1 kg.Wait, maybe I need to adjust the distribution.Suppose in each box, one stone has mass just over 1/2019 kg, and the rest have mass just under 1/2019 kg. Then, the total mass of a permutation would be the sum of 2019 stones, each just over or under 1/2019 kg.But how does that affect the total?Wait, if each stone in a box is just over 1/2019, then the total mass of a permutation would be just over 1 kg. If each stone is just under, then the total would be just under 1 kg.But we need to arrange it so that only a few permutations have total mass at least 1 kg.Wait, maybe if we have one box where one stone is very heavy, and the rest are light, and the other boxes have their stones arranged so that only when you pick the heavy stone from that one box and the light stones from the others, the total mass is just over 1 kg.But I'm not sure.Alternatively, maybe we can use the concept of derangements or something similar.Wait, perhaps we can think of this as a matrix where rows are boxes and columns are stone numbers, and entries are masses. We need to find permutation matrices where the sum of the selected entries is at least 1.We need to find the minimal number of such permutation matrices across all possible mass matrices, and then find the maximum k that is always achievable.But this is getting too abstract.Wait, maybe we can use the probabilistic method. If we consider the average total mass of a permutation, which is 1 kg, as we calculated before, then by Markov's inequality, the number of permutations with total mass at least 1 kg is at least something.But Markov's inequality gives an upper bound on the probability that a random variable exceeds a certain value, but we need a lower bound on the number of permutations with total mass at least 1 kg.Wait, maybe we can use the fact that the average is 1 kg, so the number of permutations with total mass at least 1 kg is at least something.But I'm not sure.Alternatively, maybe we can use the pigeonhole principle. Since the total mass across all permutations is 2019! * 2019 kg, and each permutation has a total mass, then the average total mass per permutation is 2019 kg. Wait, that doesn't make sense because each permutation is only selecting 2019 stones, each from different boxes, so the total mass per permutation should be 1 kg on average.Wait, no, the total mass across all permutations is 2019! * (sum of all stones in all boxes). Since each box has 1 kg, and there are 2019 boxes, the total mass is 2019 kg. So, the total mass across all permutations is 2019! * 2019 kg.But each permutation has a total mass, so the average total mass per permutation is (2019! * 2019 kg) / 2019! = 2019 kg. Wait, that can't be right because each permutation is only selecting 2019 stones, each from different boxes, so the total mass per permutation should be 1 kg on average.Wait, I'm confused. Let me recast this.Each box has 2019 stones, each with some mass, summing to 1 kg. There are 2019 boxes, so the total mass across all boxes is 2019 kg.Each permutation selects one stone from each box, so the total mass of a permutation is the sum of 2019 stones, each from a different box. The total mass across all permutations is the sum over all permutations of the sum of their selected stones.But each stone is selected in exactly (2018)! permutations because for each stone, there are 2018! ways to permute the other stones.So, the total mass across all permutations is (2018)! * sum of all stones in all boxes = (2018)! * 2019 kg.Therefore, the average total mass per permutation is (2018)! * 2019 kg / 2019! = (2018)! * 2019 / 2019! = (2018)! * 2019 / (2019 * 2018!) ) = 1 kg.So, the average total mass per permutation is 1 kg.Therefore, by the probabilistic method, there must be at least one permutation with total mass at least 1 kg. But we need to find the maximum k such that in any distribution, there are at least k permutations with total mass at least 1 kg.Given that the average is 1 kg, and we're dealing with a large number of permutations, perhaps we can use some concentration inequality to bound the number of permutations with total mass at least 1 kg.But I'm not sure.Alternatively, maybe we can think about the problem in terms of linear algebra or combinatorics.Wait, perhaps the answer is 2018! because that's the number of cyclic permutations or something.Wait, let me think differently.Suppose we fix the numbers of the stones we're picking. For example, we could pick stone 1 from box 1, stone 2 from box 2, and so on, up to stone 2019 from box 2019. That's one permutation.But we need to pick stones with different numbers from different boxes, so each permutation corresponds to a rearrangement of the numbers.Wait, maybe the key is to consider that for each cyclic shift of the numbers, we can get a different permutation.For example, if we shift all numbers by 1, so stone 1 goes to box 2, stone 2 goes to box 3, ..., stone 2019 goes to box 1. That's another permutation.If we do this for all possible cyclic shifts, we get 2019 different permutations.But how does that help us?Wait, if we consider all cyclic permutations, there are 2018! such permutations because cyclic permutations are considered the same up to rotation.Wait, no, the number of cyclic permutations of 2019 elements is (2019 - 1)! = 2018!.So, there are 2018! cyclic permutations.Now, if we consider each cyclic permutation as a way to pick stones, then perhaps each of these cyclic permutations corresponds to a different way of picking stones with different numbers from different boxes.But how does that relate to the total mass?Wait, if we have 2018! cyclic permutations, and each corresponds to a different way of picking stones, then perhaps in the worst-case distribution, only these 2018! permutations have total mass at least 1 kg.But why?Wait, maybe if we arrange the masses such that only the cyclic permutations have total mass at least 1 kg, and all other permutations have total mass less than 1 kg.But how?Suppose we arrange the masses so that in each box, the mass of stone i is 1/2019 + ε if i is in a certain cyclic order, and 1/2019 - ε otherwise.But I'm not sure.Alternatively, maybe if we have each box arranged so that the masses are in a certain cyclic pattern, then only the cyclic permutations would sum up to at least 1 kg.But I'm not sure.Wait, maybe the answer is 2018! because that's the number of cyclic permutations, and in the worst case, only these permutations would have total mass at least 1 kg.But I need to think more carefully.Suppose we have a distribution where each box has one stone with mass 1 - ε and the rest with mass ε/(2018). Then, as I thought before, only the permutation that picks the heavy stone from each box would have total mass 2019*(1 - ε). If ε is very small, this total mass is much larger than 1 kg.But in this case, there's only one such permutation, so k would be 1. But we need to find the maximum k such that in any distribution, there are at least k permutations with total mass at least 1 kg.So, in this case, k would be 1, but we need to find the maximum k that is always achievable, regardless of the distribution.Wait, but maybe the answer is 2018! because that's the number of cyclic permutations, and in any distribution, there must be at least that many permutations with total mass at least 1 kg.But I'm not sure.Wait, let me think about it differently. Suppose we have 2019 boxes, each with 2019 stones. Each box has total mass 1 kg.We need to pick one stone from each box, each with a unique number, such that the total mass is at least 1 kg.What's the minimal number of such permutations across all distributions?I think it's 2018! because that's the number of cyclic permutations, and in the worst case, only these permutations would have total mass at least 1 kg.But I'm not sure.Wait, maybe I can think of it as a Latin square problem. Each permutation corresponds to a transversal in a Latin square. But I'm not sure how that helps.Alternatively, maybe we can use Hall's Marriage Theorem. If we consider the boxes as one set and the stone numbers as another set, and we need to find a matching where the total mass is at least 1 kg.But I'm not sure.Wait, perhaps the answer is 2018! because that's the number of cyclic permutations, and in any distribution, there must be at least that many permutations with total mass at least 1 kg.But I need to verify this.Suppose we have a distribution where the masses are arranged such that only cyclic permutations have total mass at least 1 kg. Then, the number of such permutations would be 2018!.But is this the minimal number across all distributions?I think so because if you arrange the masses in a way that only cyclic permutations have total mass at least 1 kg, then the number of such permutations is 2018!.Therefore, the maximal k is 2018!.So, the answer is 2018!.**Final Answer**The maximal value of ( k ) is (boxed{2018!}).

answer：First, I need to determine the unknown dimension of the room, which I'll call "x" feet.The room has four walls. Two of these walls have a length of 25 feet and a height of 12 feet, while the other two walls have a length of "x" feet and the same height of 12 feet.Calculating the area of the four walls:- Area of the two 25 feet walls: 2 * (25 * 12) = 600 square feet- Area of the two "x" feet walls: 2 * (x * 12) = 24x square feet- Total area of the four walls: 600 + 24x square feetNext, I need to subtract the areas of the door and the three windows since they are not to be whitewashed:- Area of the door: 6 feet * 3 feet = 18 square feet- Area of one window: 4 feet * 3 feet = 12 square feet- Area of three windows: 3 * 12 = 36 square feet- Total area to be subtracted: 18 + 36 = 54 square feetThe net area to be whitewashed is:600 + 24x - 54 = 546 + 24x square feetThe cost of whitewashing per square foot is Rs. 3, so the total cost is:Cost = (546 + 24x) * 3Given that the total cost is Rs. 2718, I can set up the equation:(546 + 24x) * 3 = 2718Dividing both sides by 3 to simplify:546 + 24x = 906Subtracting 546 from both sides:24x = 360Dividing both sides by 24 to find x:x = 15 feetSo, the unknown dimension of the room is 15 feet.

answer：First, I need to determine how many cups of vegetables and broth are required for one serving of soup. I know that one serving contains 1 cup of vegetables and an unknown amount of broth, which I'll denote as ( x ) cups.For 8 servings, the total amount of vegetables needed is ( 8 times 1 = 8 ) cups. The total amount of broth needed is ( 8 times x = 8x ) cups.The combined total of vegetables and broth for 8 servings is given as 14 pints. Since 1 pint equals 2 cups, 14 pints is equivalent to ( 14 times 2 = 28 ) cups.Now, I can set up an equation to represent the total combined volume:[8 text{ cups (vegetables)} + 8x text{ cups (broth)} = 28 text{ cups (total)}]To find the value of ( x ), I'll subtract the amount of vegetables from the total:[8x = 28 - 8][8x = 20][x = frac{20}{8}][x = 2.5]Therefore, each serving of soup contains 2.5 cups of broth.