answer：First, determine how many food boxes Jamie packed last Saturday, which is 12 boxes.Next, calculate the number of cans of soup in each box, which is 8 cans.Then, find out how many food boxes Jamie plans to pack this Saturday by adding 5 more boxes to last Saturday's total: 12 + 5 = 17 boxes.Finally, multiply the number of boxes this Saturday by the number of cans per box to find the total number of cans needed: 17 boxes × 8 cans = 136 cans.

answer：Okay, so I have this problem about weighted voting systems, and I need to figure out the number of winning coalitions and then calculate the Banzhaf power indices. Let me try to break this down step by step.First, the committee has three members: A, B, and C. Each has a weight, and the total weight is 100. So, ( w_A + w_B + w_C = 100 ). The decision is made if a coalition's combined weight exceeds 50. The first part is to determine the number of winning coalitions possible, assuming each member must be part of the coalition to make it winning. Hmm, so a coalition is any subset of the members, right? But it's only winning if their total weight is more than 50.Wait, but the question says "assuming that each member must be part of the coalition to make it winning." Hmm, maybe I misread that. Let me check again. It says, "assuming that each member must be part of the coalition to make it winning." So, does that mean that a coalition is only winning if all members are included? Or does it mean that each member must be part of some coalition to make it winning? Maybe it's the former. So, perhaps only the grand coalition (all three members) is considered a winning coalition? But that doesn't make much sense because the total weight is 100, so any coalition that includes at least one member with weight over 50 would also be a winning coalition.Wait, maybe I'm overcomplicating. Let me reread the question."A decision is made if a coalition of members has a combined voting weight of more than 50, determine the number of winning coalitions possible, assuming that each member must be part of the coalition to make it winning. A coalition is winning if the sum of the weights of its members exceeds 50. Assume that the weights are strictly positive integers."Hmm, maybe it's saying that each member must be part of a coalition to make it winning, meaning that a coalition is only winning if all members are included? That would mean only the grand coalition is a winning coalition, but that seems too restrictive because, for example, if one member has a weight over 50, then a coalition of just that member would also be a winning coalition.Wait, perhaps the wording is that each member must be part of some winning coalition, not necessarily that a coalition must include all members. So, each member is part of at least one winning coalition. But the question is about the number of winning coalitions possible. Hmm, this is a bit confusing.Alternatively, maybe it's saying that a coalition is winning only if it includes all members, but that doesn't make sense because the total is 100, so any subset with a total over 50 is a winning coalition.Wait, maybe the key is that each member must be part of the coalition to make it winning, meaning that a coalition is only winning if it includes all members? But that can't be, because if a single member has more than 50, then a coalition of just that member is winning.Wait, perhaps the parent is saying that in order for a coalition to be winning, each member must be part of it. So, the only winning coalition is the grand coalition. But that seems odd because usually, in weighted voting systems, a coalition is any subset of members whose weights sum to more than half the total.Wait, maybe the question is that each member must be part of the coalition to make it winning, meaning that a coalition is only winning if all members are included. So, the only winning coalition is the grand coalition. But that would mean only one winning coalition, which seems too restrictive.Alternatively, perhaps the question is that each member must be part of at least one winning coalition, but that's a different concept. It's about the coverage of the members in the winning coalitions, not the definition of a winning coalition.I think I need to clarify this. Let me look up the definition of a winning coalition. A winning coalition is any subset of members whose combined weights exceed the quota, which in this case is 50. So, any subset (including single members, pairs, or the entire committee) that sums to more than 50 is a winning coalition.Therefore, the number of winning coalitions depends on the weights of A, B, and C. Since the weights are positive integers adding up to 100, we need to find all possible subsets of {A, B, C} whose total weight exceeds 50.But without knowing the exact weights, we can't determine the exact number of winning coalitions. Wait, but the problem says "assuming that each member must be part of the coalition to make it winning." Hmm, perhaps that means that each member is necessary for some winning coalition, but that's more about the power indices.Wait, maybe I misread the first part. Let me read it again."1. If the parent explains that a decision is made if a coalition of members has a combined voting weight of more than 50, determine the number of winning coalitions possible, assuming that each member must be part of the coalition to make it winning. A coalition is winning if the sum of the weights of its members exceeds 50. Assume that the weights are strictly positive integers."Wait, maybe it's a translation issue. The original might have been in another language. Maybe "assuming that each member must be part of the coalition to make it winning" is supposed to mean that each member is part of at least one winning coalition. But that's a different concept, called "coverage" in power indices.Alternatively, maybe it's saying that a coalition is only winning if it includes all members, but that can't be because the total is 100, so any subset with over 50 is winning.Wait, perhaps the key is that each member must be part of a winning coalition, meaning that each member is critical in at least one winning coalition. That relates to the Banzhaf power index, which counts the number of times a member is critical in a winning coalition.But the first part is just about the number of winning coalitions, regardless of the members' criticality. So, perhaps the number of winning coalitions is the number of subsets of {A, B, C} whose total weight exceeds 50.Given that, we need to find how many subsets (coalitions) have a total weight >50. Since the total weight is 100, the quota is 50.5, so any subset with total weight ≥51 is a winning coalition.But without knowing the exact weights, we can't determine the exact number. Wait, but the problem says "assuming that each member must be part of the coalition to make it winning." Hmm, maybe that means that each member is necessary for the winning coalition, so the only winning coalition is the grand coalition. But that can't be, because if a single member has more than 50, then that single member is a winning coalition.Wait, perhaps the parent is setting up a scenario where each member is necessary for the winning coalition, meaning that no single member can win alone, and no pair can win without the third. So, the only winning coalition is the grand coalition. But that would require that each member's weight is less than or equal to 50, and the sum of any two members is less than or equal to 50 as well. But since the total is 100, if each member is ≤50, then the sum of any two would be ≤100 - the third member. So, if each member is ≤50, then the sum of any two would be ≥50? Wait, no.Wait, if each member is ≤50, then the sum of any two would be ≤100 - the third. So, if the third is ≥50, then the sum of the other two is ≤50. So, if each member is ≤50, then any two members sum to ≤50, meaning that the only winning coalition is the grand coalition.But in that case, the number of winning coalitions is 1, which is the grand coalition.But is that the case? Let me think. If each member is ≤50, then any single member cannot exceed 50, so no single member can form a winning coalition. Any pair would sum to ≤50, because if each is ≤50, then their sum is ≤100 - the third member. But since the third member is ≥1, the sum of the pair is ≤99. Wait, no, that's not necessarily ≤50.Wait, let me take an example. Suppose each member is exactly 33.333... but since they have to be integers, let's say 33, 33, 34. Then, the sum of any two would be 66, 67, or 67, which are all above 50. So, in that case, even if each member is ≤50, pairs can still sum to over 50.Wait, so my earlier reasoning was wrong. So, if each member is ≤50, it doesn't necessarily mean that pairs can't sum to over 50. For example, if A=40, B=40, C=20, then A+B=80>50, so the pair A and B is a winning coalition. Similarly, A+C=60>50, and B+C=60>50. So, in this case, all pairs and the grand coalition are winning.But if we have A=50, B=25, C=25, then A alone is 50, which is not over 50, so A alone is not a winning coalition. The pairs: A+B=75>50, A+C=75>50, B+C=50, which is not over 50. So, in this case, the winning coalitions are {A,B}, {A,C}, and {A,B,C}. So, three winning coalitions.Wait, so the number of winning coalitions depends on the distribution of weights. So, without knowing the exact weights, we can't determine the exact number of winning coalitions. But the problem says "assuming that each member must be part of the coalition to make it winning." Hmm, maybe that's a mistranslation or misstatement.Alternatively, perhaps the parent is saying that each member is part of some winning coalition, meaning that each member is critical in at least one winning coalition, which relates to the Banzhaf index. But the first part is just about the number of winning coalitions.Wait, maybe the problem is that each member must be part of a winning coalition, meaning that each member is in at least one winning coalition. But that doesn't necessarily restrict the number of winning coalitions. For example, in the case where A=50, B=25, C=25, each member is in at least one winning coalition: A is in {A,B} and {A,C}, B is in {A,B} and {A,B,C}, and C is in {A,C} and {A,B,C}. So, all are covered.But the number of winning coalitions is three in that case.Wait, but the problem is asking for the number of winning coalitions possible, given that each member must be part of the coalition to make it winning. Hmm, maybe it's saying that each member is necessary for the winning coalition, meaning that the only winning coalition is the grand coalition. But as I saw earlier, that's not necessarily the case unless each member's weight is such that no subset of two can exceed 50.Wait, let's think about that. If the sum of any two members is ≤50, then the only winning coalition is the grand coalition. So, in that case, the number of winning coalitions is 1.But how can the sum of any two members be ≤50? Since the total is 100, if each member is ≤50, then the sum of any two would be ≤100 - the third member. So, if the third member is ≥50, then the sum of the other two is ≤50. Therefore, if each member is ≤50, then the sum of any two is ≤50, meaning that the only winning coalition is the grand coalition.But wait, if each member is ≤50, then the sum of any two is ≤100 - the third. So, if the third is ≥50, then the sum of the other two is ≤50. Therefore, if each member is ≤50, then the sum of any two is ≤50, meaning that the only winning coalition is the grand coalition.But in that case, the number of winning coalitions is 1.But is that the case? Let me take an example. Suppose A=50, B=25, C=25. Then, the sum of A and B is 75>50, so {A,B} is a winning coalition. Similarly, {A,C} is also winning. So, in this case, even though each member is ≤50, the sum of two can exceed 50.Wait, so my earlier reasoning was wrong. So, if each member is ≤50, it doesn't necessarily mean that the sum of any two is ≤50. It depends on the distribution.Wait, so to have the sum of any two members ≤50, we need that for all pairs, their sum ≤50. So, for three members, we need:w_A + w_B ≤50,w_A + w_C ≤50,w_B + w_C ≤50.But since w_A + w_B + w_C =100, if each pair sums to ≤50, then adding all three inequalities:2(w_A + w_B + w_C) ≤150,Which is 2*100=200 ≤150? That's not possible. So, it's impossible for all pairs to sum to ≤50. Therefore, it's impossible for the only winning coalition to be the grand coalition. Therefore, the number of winning coalitions must be more than 1.Wait, so that means that the initial assumption that each member must be part of the coalition to make it winning is impossible, because it's impossible for all pairs to sum to ≤50. Therefore, the number of winning coalitions must be at least 3, as in the case where A=50, B=25, C=25.Wait, but in that case, the winning coalitions are {A,B}, {A,C}, and {A,B,C}. So, three winning coalitions.Alternatively, if A=60, B=30, C=10, then the winning coalitions are {A}, {A,B}, {A,C}, {A,B,C}. So, four winning coalitions.Wait, so the number of winning coalitions can vary depending on the weights. Therefore, without knowing the exact weights, we can't determine the exact number of winning coalitions. But the problem says "assuming that each member must be part of the coalition to make it winning." Hmm, maybe that's a misstatement, and it's supposed to mean that each member is part of at least one winning coalition, which is always true unless a member has weight 0, which is not the case here since weights are strictly positive.Wait, but in the case where a member has weight 1, and the others have 49 and 50, then the member with 1 is only in the grand coalition as a winning coalition. So, in that case, each member is part of at least one winning coalition.Therefore, the number of winning coalitions can vary from 3 to 4, depending on the weights.Wait, but in the case where all three have weights 34, 33, 33, then the winning coalitions are all pairs and the grand coalition, so 4 winning coalitions.Wait, let me check:If A=34, B=33, C=33.Then, the single members: A=34 <50, B=33 <50, C=33 <50. So, no single member is a winning coalition.Pairs: A+B=67>50, A+C=67>50, B+C=66>50. So, all three pairs are winning.Grand coalition: 100>50.So, total winning coalitions: 3 pairs +1 grand coalition =4.Alternatively, if A=50, B=25, C=25.Single members: A=50 is not >50, so not winning. B and C are 25 each, also not winning.Pairs: A+B=75>50, A+C=75>50, B+C=50, which is not >50. So, two pairs are winning.Grand coalition: 100>50.So, total winning coalitions: 2 pairs +1 grand coalition=3.Alternatively, if A=60, B=30, C=10.Single members: A=60>50, so {A} is winning.Pairs: A+B=90>50, A+C=70>50, B+C=40<50.Grand coalition: 100>50.So, total winning coalitions: {A}, {A,B}, {A,C}, {A,B,C} =>4.So, depending on the weights, the number of winning coalitions can be 3 or 4.But the problem says "assuming that each member must be part of the coalition to make it winning." Hmm, maybe that's a misstatement, and it's supposed to mean that each member is part of at least one winning coalition, which is always true as long as no member has weight 0, which they don't.Therefore, the number of winning coalitions can be 3 or 4, depending on the weights.But the problem is asking for the number of winning coalitions possible, given that each member must be part of the coalition to make it winning. Hmm, perhaps it's 4, because in the case where all pairs are winning, plus the grand coalition, that's 4. But in other cases, it's 3.Wait, but the problem is asking for the number of winning coalitions possible, so it's the maximum possible number. So, the maximum number of winning coalitions is 4, which occurs when all pairs are winning coalitions, plus the grand coalition.But wait, in the case where all pairs are winning, that would require that each pair sums to >50. So, let's see:If A+B>50,A+C>50,B+C>50.Given that A+B+C=100.So, adding all three inequalities:2(A+B+C) >150,Which is 200>150, which is true. So, it's possible for all pairs to be winning coalitions.For example, A=34, B=33, C=33.As above, all pairs sum to 67, 67, 66, all >50.So, in that case, the winning coalitions are the three pairs and the grand coalition, totaling 4.Therefore, the number of winning coalitions possible is 4.But wait, in the case where A=50, B=25, C=25, the number of winning coalitions is 3.So, the number can vary between 3 and 4.But the problem is asking for the number of winning coalitions possible, assuming that each member must be part of the coalition to make it winning. Hmm, maybe it's 4, because that's the maximum, but I'm not sure.Alternatively, maybe the answer is 4, because each member is part of at least one winning coalition, and the maximum number of winning coalitions is 4.But I'm not entirely sure. Maybe I should proceed to the second part and see if that helps.The second part says: If member A has twice the voting power of member C, and member B has the same voting power as member C, calculate all possible integer values for ( w_A ), ( w_B ), and ( w_C ) such that the voting power indices reflect this relationship. Additionally, calculate the Banzhaf power index for each member.So, let's denote ( w_C = x ). Then, ( w_A = 2x ), and ( w_B = x ). So, total weight is ( 2x + x + x = 4x =100 ). Therefore, ( x=25 ). So, ( w_A=50 ), ( w_B=25 ), ( w_C=25 ).Wait, but 50+25+25=100, which is correct.So, the weights are A=50, B=25, C=25.Now, let's calculate the Banzhaf power index for each member.The Banzhaf power index is calculated by counting the number of times a member is critical in a winning coalition. A member is critical in a coalition if removing them causes the coalition to lose.First, let's list all possible coalitions and determine which are winning.The possible coalitions are:1. {A}: 50>50? No, it's equal, so not winning.2. {B}:25<50, not winning.3. {C}:25<50, not winning.4. {A,B}:50+25=75>50, winning.5. {A,C}:50+25=75>50, winning.6. {B,C}:25+25=50, not winning.7. {A,B,C}:100>50, winning.So, the winning coalitions are {A,B}, {A,C}, {A,B,C}.Now, let's determine the critical members in each winning coalition.1. {A,B}: If we remove A, the coalition becomes {B}=25<50, so A is critical. If we remove B, the coalition becomes {A}=50, which is not >50, so B is also critical.2. {A,C}: Similarly, removing A leaves {C}=25<50, so A is critical. Removing C leaves {A}=50, which is not >50, so C is critical.3. {A,B,C}: Removing A leaves {B,C}=50, which is not >50, so A is critical. Removing B leaves {A,C}=75>50, so B is not critical. Removing C leaves {A,B}=75>50, so C is not critical.So, in terms of criticality:- A is critical in {A,B}, {A,C}, {A,B,C} => 3 times.- B is critical in {A,B}, and not in {A,B,C} => 1 time.- C is critical in {A,C}, and not in {A,B,C} =>1 time.Therefore, the Banzhaf power indices are:- A: 3- B:1- C:1But since Banzhaf indices are usually expressed as fractions of the total, the total criticalities are 3+1+1=5. So, the Banzhaf indices are:- A: 3/5- B:1/5- C:1/5But sometimes, they are expressed as counts, so A has 3, B and C have 1 each.But the problem says "calculate the Banzhaf power index for each member," so probably as fractions.Therefore, the Banzhaf power indices are A: 3/5, B:1/5, C:1/5.So, going back to the first part, the number of winning coalitions is 3, as in the case where A=50, B=25, C=25, which is the case we're dealing with in the second part.Therefore, maybe the first part is referring to this specific case, where the weights are A=50, B=25, C=25, and the number of winning coalitions is 3.But the first part didn't specify the weights, so I'm confused.Wait, the first part is a general question, and the second part is a specific case. So, in the first part, without knowing the weights, we can't determine the exact number of winning coalitions, but in the second part, we have specific weights, so we can calculate the Banzhaf indices.But the first part says "assuming that each member must be part of the coalition to make it winning." Hmm, maybe it's a general question, but given that in the second part, the weights are A=50, B=25, C=25, which is a specific case, maybe the first part is also referring to that case.Wait, but the first part is question 1, and the second part is question 2. So, question 1 is general, and question 2 is specific.Therefore, for question 1, the number of winning coalitions possible is variable, depending on the weights. But the problem says "assuming that each member must be part of the coalition to make it winning." Hmm, maybe it's a specific case where each member is part of at least one winning coalition, but that's always true unless a member has weight 0.Wait, maybe the answer is 4, as in the maximum number of winning coalitions, which is 4, when all pairs and the grand coalition are winning.But in the specific case of the second part, the number of winning coalitions is 3.Wait, I'm getting confused. Maybe I should answer question 1 as 4, the maximum possible, and question 2 as 3, but I'm not sure.Alternatively, maybe the first part is referring to the specific case where each member is part of at least one winning coalition, which is always true, so the number of winning coalitions is variable, but the maximum is 4.But the problem says "determine the number of winning coalitions possible," so maybe it's asking for the maximum possible, which is 4.But I'm not entirely sure. Maybe I should proceed with the second part, as it's more concrete.In the second part, we have A=50, B=25, C=25. The Banzhaf power indices are A:3/5, B:1/5, C:1/5.Therefore, the possible integer values for the weights are A=50, B=25, C=25.So, summarizing:1. The number of winning coalitions possible is 4, assuming that each member is part of at least one winning coalition, which is always true, but the maximum number is 4.2. The weights are A=50, B=25, C=25, and the Banzhaf power indices are A:3/5, B:1/5, C:1/5.But I'm not entirely confident about the first part. Maybe the answer is 3, as in the specific case of the second part.Alternatively, perhaps the first part is asking for the number of winning coalitions in the specific case where each member is part of at least one winning coalition, which is always true, so the number is variable, but the maximum is 4.But since the problem is about teaching a child, maybe it's simpler. If each member must be part of the coalition to make it winning, meaning that the only winning coalition is the grand coalition, but as we saw earlier, that's impossible because the sum of two members can exceed 50. Therefore, the number of winning coalitions must be at least 3.But in the specific case of the second part, it's 3. So, maybe the answer is 3.I think I'll go with 3 for the first part, as in the specific case, and 4 as the maximum possible, but since the problem says "assuming that each member must be part of the coalition to make it winning," which might imply that each member is part of at least one winning coalition, which is always true, so the number is variable, but in the specific case, it's 3.But I'm not entirely sure. Maybe I should look up the definition of "winning coalitions" and "each member must be part of the coalition to make it winning."Wait, "each member must be part of the coalition to make it winning" might mean that each member is necessary for the winning coalition, meaning that the only winning coalition is the grand coalition. But as we saw, that's impossible because the sum of two members can exceed 50. Therefore, the number of winning coalitions must be more than 1.But in the specific case of the second part, it's 3.Therefore, maybe the answer is 3.But I'm still not sure. Maybe I should proceed with the second part, as it's more concrete.In the second part, the weights are A=50, B=25, C=25, and the Banzhaf indices are A:3/5, B:1/5, C:1/5.So, to answer the first part, given that each member must be part of the coalition to make it winning, which might mean that each member is part of at least one winning coalition, which is always true, so the number of winning coalitions is variable, but in the specific case, it's 3.Therefore, the number of winning coalitions possible is 3.But I'm not entirely confident. Maybe the answer is 4, as the maximum possible.I think I'll go with 4 for the first part, as the maximum number of winning coalitions possible, and 3 for the specific case in the second part.But since the problem is about teaching a child, maybe it's simpler. If each member must be part of the coalition to make it winning, meaning that each member is necessary for the winning coalition, which is impossible, so the number of winning coalitions is 3, as in the case where each member is part of at least one winning coalition, but not necessarily all.Wait, I'm overcomplicating. Maybe the answer is 4, as the maximum number of winning coalitions possible, which is when all pairs and the grand coalition are winning.Therefore, I think the answer to the first part is 4, and the second part is 3.But I'm not sure. Maybe I should look up the definition of "each member must be part of the coalition to make it winning." It might mean that each member is part of at least one winning coalition, which is always true, so the number of winning coalitions is variable, but the maximum is 4.Therefore, the answer is 4.But in the specific case of the second part, it's 3.So, to sum up:1. The number of winning coalitions possible is 4.2. The weights are A=50, B=25, C=25, and the Banzhaf power indices are A:3/5, B:1/5, C:1/5.But I'm still not entirely confident. Maybe I should proceed with this answer.

answer：Alright, let me try to figure out these two sub-problems. I'm a bit new to this, so I'll take it step by step.Starting with Sub-problem 1. The collector is using a strategy where they bid a value proportional to the base value raised to the power of 1.5, but not more than 50% above the base value. Hmm, okay. So, for each bottle, the bid is proportional to V_i^1.5, but it can't exceed 1.5 times the base value. First, I need to understand what "proportional" means here. I think it means that the bid is some multiple of V_i^1.5. But since it's proportional, maybe there's a constant k such that bid = k * V_i^1.5. However, the maximum bid allowed is 1.5 * V_i. So, k must be chosen such that k * V_i^1.5 ≤ 1.5 * V_i for each bottle.Wait, but if it's proportional, maybe the same k applies to all bottles? Let me check. If that's the case, then k should be the minimum value that satisfies k * V_i^1.5 ≤ 1.5 * V_i for all i. Alternatively, maybe each bottle has its own k, but that seems less likely since the problem says "proportional" without specifying different constants.Let me think again. The problem says "bid for each bottle a value that is proportional to the base value raised to the power of 1.5, but no more than 50% above the base value." So, perhaps the bid is the minimum between (proportional value) and (1.5 * V_i). But without knowing the exact proportionality constant, how can we compute it?Wait, maybe the proportionality is such that the bid is exactly V_i^1.5, but scaled so that it doesn't exceed 1.5 * V_i. So, perhaps the bid is the minimum of V_i^1.5 and 1.5 * V_i. But that might not make sense because V_i^1.5 is likely larger than V_i, especially for higher V_i.Wait, let's compute V_i^1.5 for each bottle. For V1 = 15,000: 15,000^1.5. Let me compute that. 15,000^1.5 is sqrt(15,000^3). Wait, no, 1.5 is 3/2, so it's the square root of (15,000)^3. Alternatively, 15,000 * sqrt(15,000). Let me compute sqrt(15,000). sqrt(15,000) = sqrt(100 * 150) = 10 * sqrt(150) ≈ 10 * 12.247 ≈ 122.47. So, 15,000 * 122.47 ≈ 1,837,050. That's way more than 1.5 * 15,000 = 22,500. So, the bid would be 22,500.Similarly, for V2 = 20,000: 20,000^1.5 = 20,000 * sqrt(20,000). sqrt(20,000) = sqrt(100 * 200) = 10 * sqrt(200) ≈ 10 * 14.142 ≈ 141.42. So, 20,000 * 141.42 ≈ 2,828,400, which is way above 1.5 * 20,000 = 30,000. So, the bid would be 30,000.For V3 = 25,000: 25,000^1.5 = 25,000 * sqrt(25,000). sqrt(25,000) = 158.11. So, 25,000 * 158.11 ≈ 3,952,750, which is way above 1.5 * 25,000 = 37,500. So, the bid would be 37,500.Wait, but this seems odd because the bids are just 1.5 times the base values. But the problem says "proportional to the base value raised to the power of 1.5, but no more than 50% above the base value." So, maybe the bid is the lesser of (k * V_i^1.5) and (1.5 * V_i). But without knowing k, how do we determine the bid?Alternatively, perhaps the proportionality is such that the bid is (V_i^1.5) divided by some scaling factor so that it doesn't exceed 1.5 * V_i. But without knowing the scaling factor, we can't compute it. Hmm, maybe I'm overcomplicating it.Wait, perhaps the bid is simply V_i^1.5, but if that exceeds 1.5 * V_i, then it's capped at 1.5 * V_i. So, for each bottle, compute V_i^1.5 and if it's more than 1.5 * V_i, set the bid to 1.5 * V_i. Otherwise, use V_i^1.5.But as I computed earlier, V_i^1.5 is way higher than 1.5 * V_i for all three bottles. So, the bids would be 22,500, 30,000, and 37,500 respectively.Wait, but that seems too straightforward. Maybe I'm missing something. Let me check the math again.Wait, 15,000^1.5 is 15,000 * sqrt(15,000). Let me compute sqrt(15,000) more accurately. 15,000 is 15 * 1000, so sqrt(15,000) = sqrt(15) * sqrt(1000) ≈ 3.87298 * 31.62277 ≈ 122.474. So, 15,000 * 122.474 ≈ 1,837,110. That's way more than 22,500. So, the bid is 22,500.Similarly, for 20,000: sqrt(20,000) ≈ 141.421, so 20,000 * 141.421 ≈ 2,828,420, which is way more than 30,000. So, bid is 30,000.For 25,000: sqrt(25,000) ≈ 158.114, so 25,000 * 158.114 ≈ 3,952,850, which is way more than 37,500. So, bid is 37,500.So, the bids are 22,500, 30,000, and 37,500.Wait, but the problem says "proportional to the base value raised to the power of 1.5." So, maybe the bid is k * V_i^1.5, and k is chosen such that k * V_i^1.5 ≤ 1.5 * V_i for all i. So, k ≤ (1.5 * V_i) / V_i^1.5 = 1.5 / V_i^0.5.So, k must be less than or equal to 1.5 / sqrt(V_i) for each i. To find the maximum possible k that works for all i, we take the minimum of 1.5 / sqrt(V_i) across all i.Compute 1.5 / sqrt(15,000), 1.5 / sqrt(20,000), and 1.5 / sqrt(25,000).sqrt(15,000) ≈ 122.474, so 1.5 / 122.474 ≈ 0.01225.sqrt(20,000) ≈ 141.421, so 1.5 / 141.421 ≈ 0.0106.sqrt(25,000) ≈ 158.114, so 1.5 / 158.114 ≈ 0.00949.So, the minimum k is approximately 0.00949. So, the bid for each bottle would be k * V_i^1.5.Compute for V1: 0.00949 * (15,000)^1.5 ≈ 0.00949 * 1,837,110 ≈ 17,450.Similarly, for V2: 0.00949 * (20,000)^1.5 ≈ 0.00949 * 2,828,420 ≈ 26,850.For V3: 0.00949 * (25,000)^1.5 ≈ 0.00949 * 3,952,850 ≈ 37,500.Wait, that's interesting. So, for V3, the bid is exactly 37,500, which is 1.5 * V3. For V1 and V2, the bids are less than 1.5 * V_i.Wait, but this seems a bit more involved. So, the collector is using a proportional strategy where the same k is applied to all bottles, ensuring that no bid exceeds 1.5 * V_i. So, the maximum k is determined by the smallest 1.5 / sqrt(V_i), which is for V3. So, k ≈ 0.00949.Thus, the bids are approximately 17,450, 26,850, and 37,500.Wait, but let me check if this makes sense. If k is 0.00949, then for V1: 0.00949 * 15,000^1.5 ≈ 0.00949 * 1,837,110 ≈ 17,450, which is less than 22,500. For V2: 0.00949 * 2,828,420 ≈ 26,850, which is less than 30,000. For V3: 0.00949 * 3,952,850 ≈ 37,500, which is exactly 1.5 * V3.So, the bids are 17,450, 26,850, and 37,500.Wait, but the problem says "proportional to the base value raised to the power of 1.5, but no more than 50% above the base value." So, perhaps the bids are the minimum of (proportional value) and (1.5 * V_i). But without knowing the exact proportionality, we can't compute it unless we assume that the proportionality constant is chosen such that the highest possible bid is 1.5 * V_i for the highest V_i, and then scaled down for others.Wait, maybe the collector sets the bid for each bottle as the minimum between (V_i^1.5) and (1.5 * V_i). But as we saw, V_i^1.5 is way higher than 1.5 * V_i, so the bids would just be 1.5 * V_i for all. But that seems too simplistic, and the problem mentions "proportional," which suggests a scaling factor.Alternatively, perhaps the collector uses a proportionality constant k such that for each bottle, bid = k * V_i^1.5, and k is chosen so that for the highest V_i, the bid is 1.5 * V_i. Then, for lower V_i, the bid would be less than 1.5 * V_i.So, for V3 = 25,000, bid = k * (25,000)^1.5 = 1.5 * 25,000 = 37,500.So, k = 37,500 / (25,000)^1.5.Compute (25,000)^1.5 = 25,000 * sqrt(25,000) ≈ 25,000 * 158.114 ≈ 3,952,850.So, k = 37,500 / 3,952,850 ≈ 0.00949.Then, for V1: bid = 0.00949 * (15,000)^1.5 ≈ 0.00949 * 1,837,110 ≈ 17,450.For V2: bid = 0.00949 * (20,000)^1.5 ≈ 0.00949 * 2,828,420 ≈ 26,850.So, the bids are approximately 17,450, 26,850, and 37,500.That seems consistent. So, the collector's bids are:Bottle 1: ~17,450Bottle 2: ~26,850Bottle 3: 37,500But let me check if these bids are indeed proportional to V_i^1.5. So, 17,450 / 15,000^1.5 ≈ 17,450 / 1,837,110 ≈ 0.00949.Similarly, 26,850 / 2,828,420 ≈ 0.00949.And 37,500 / 3,952,850 ≈ 0.00949.Yes, so the same k is applied to all, ensuring proportionality.So, that's Sub-problem 1.Now, moving on to Sub-problem 2. There's a competitor with an exponential interest function f(x) = e^{0.0001x}, where x is the base value. The competitor's bids are 10% above their exponential interest value for each bottle. We need to determine which bottles the collector will acquire, assuming the collector's bids are the only competition.So, first, let's compute the competitor's bid for each bottle.For each bottle, competitor's bid = 1.1 * f(x) = 1.1 * e^{0.0001x}.Compute f(x) for each V_i:For V1 = 15,000:f(15,000) = e^{0.0001 * 15,000} = e^{1.5} ≈ 4.4817.So, competitor's bid = 1.1 * 4.4817 ≈ 4.9298 ≈ 4.93.Wait, that can't be right. 4.93 is way below the collector's bid of ~17,450. That seems odd. Did I make a mistake?Wait, maybe I misinterpreted the function. The function is f(x) = e^{0.0001x}, where x is the base value in dollars. So, for x = 15,000, f(x) = e^{0.0001 * 15,000} = e^{1.5} ≈ 4.4817. So, the competitor's bid is 1.1 * 4.4817 ≈ 4.93 dollars. That seems way too low. Maybe the function is f(x) = e^{0.0001 * x} dollars, but that would make the bids extremely low.Alternatively, perhaps the function is f(x) = e^{0.0001x} * some scaling factor? The problem doesn't specify, so maybe I need to assume it's in dollars. But that would make the competitor's bids negligible compared to the collector's bids.Wait, let me double-check. The problem says the competitor's bids are 10% above their exponential interest value. So, if f(x) is in dollars, then the competitor's bid is 1.1 * f(x). But if f(x) is e^{0.0001x}, then for x=15,000, f(x)=e^{1.5}≈4.48, so bid≈4.93. That's way below the collector's bid of ~17,450.Similarly, for V2=20,000:f(20,000)=e^{0.0001*20,000}=e^{2}≈7.389. Competitor's bid=1.1*7.389≈8.128≈8.13.For V3=25,000:f(25,000)=e^{0.0001*25,000}=e^{2.5}≈12.182. Competitor's bid=1.1*12.182≈13.40.So, the competitor's bids are approximately 4.93, 8.13, and 13.40 for the three bottles.Comparing these to the collector's bids:Bottle 1: Collector bids ~17,450 vs Competitor ~4.93 → Collector wins.Bottle 2: Collector bids ~26,850 vs Competitor ~8.13 → Collector wins.Bottle 3: Collector bids 37,500 vs Competitor ~13.40 → Collector wins.Wait, that can't be right because the competitor's bids are way too low. Maybe I misinterpreted the function. Perhaps the function f(x) is in thousands or some other scale? Or maybe it's f(x) = e^{0.0001x} * x? Let me check the problem statement again.The problem says: "the competitor's bids are 10% above their exponential interest value for each bottle." So, f(x) is the exponential interest value, and the bid is 1.1 * f(x). So, unless f(x) is scaled differently, the bids are as I computed.Alternatively, maybe the function is f(x) = e^{0.0001x} * 1000 or something. But the problem doesn't specify, so I have to go with what's given.Alternatively, perhaps the function is f(x) = e^{0.0001x} dollars, but that would make the bids very low. Alternatively, maybe it's f(x) = e^{0.0001x} * 1000, making the bids more reasonable.Wait, let me think. If f(x) = e^{0.0001x}, then for x=15,000, f(x)=e^{1.5}≈4.48, which is about 4.48. That's too low. Maybe the function is f(x) = e^{0.0001x} * 1000, so f(x)=4.48*1000≈4,480. Then, the competitor's bid would be 1.1*4,480≈4,928.Similarly, for x=20,000: f(x)=e^{2}≈7.389*1000≈7,389. Bid≈8,128.For x=25,000: f(x)=e^{2.5}≈12.182*1000≈12,182. Bid≈13,400.That makes more sense. So, perhaps the function is f(x) = e^{0.0001x} * 1000. But the problem doesn't specify that. It just says f(x) = e^{0.0001x}, where x is the base value in dollars.Alternatively, maybe the function is f(x) = e^{0.0001x} dollars, but that would make the bids too low. Alternatively, maybe f(x) is in thousands, so f(x)=e^{0.0001x} thousand dollars. So, for x=15,000, f(x)=e^{1.5}≈4.48 thousand dollars, so 4,480. Then, competitor's bid=1.1*4,480≈4,928.Similarly, for x=20,000: f(x)=e^{2}≈7.389 thousand≈7,389. Bid≈8,128.For x=25,000: f(x)=e^{2.5}≈12.182 thousand≈12,182. Bid≈13,400.That seems more reasonable. So, perhaps the function is in thousands. But the problem doesn't specify, so I'm not sure. Alternatively, maybe the function is f(x) = e^{0.0001x} * 1000, making the bids in the thousands.Alternatively, perhaps the function is f(x) = e^{0.0001x} * 1000, so f(x) is in dollars. Let me assume that for a moment.So, f(x) = e^{0.0001x} * 1000.Then, for x=15,000: f(x)=e^{1.5} * 1000≈4.4817*1000≈4,481.70. Competitor's bid=1.1*4,481.70≈4,929.87.For x=20,000: f(x)=e^{2}*1000≈7.389*1000≈7,389. Competitor's bid≈8,127.90.For x=25,000: f(x)=e^{2.5}*1000≈12.182*1000≈12,182. Competitor's bid≈13,400.20.Now, comparing to the collector's bids:Bottle 1: Collector bids ~17,450 vs Competitor ~4,930 → Collector wins.Bottle 2: Collector bids ~26,850 vs Competitor ~8,128 → Collector wins.Bottle 3: Collector bids 37,500 vs Competitor ~13,400 → Collector wins.So, the collector would acquire all three bottles.But wait, the problem says "assuming the collector's bids are the only competition for the competitor." So, does that mean the collector is the only other bidder, and the competitor is the only other competitor? So, the auction is between the collector and the competitor.In a sealed-bid auction, typically the highest bidder wins. So, for each bottle, the collector and the competitor submit their bids. The higher bid wins.So, if the competitor's bids are as I computed (assuming f(x) is in thousands), then the collector's bids are higher for all three, so the collector wins all.But if f(x) is not scaled, and the competitor's bids are only ~4.93, then the collector would win all as well.But that seems odd because the competitor's bids would be negligible. Maybe I'm misinterpreting the function.Alternatively, perhaps the function f(x) = e^{0.0001x} is in dollars, but that would make the bids too low. Alternatively, maybe it's f(x) = e^{0.0001x} * 1000, making the bids in the thousands.Alternatively, perhaps the function is f(x) = e^{0.0001x} * V_i, meaning the competitor's interest is proportional to both the exponential function and the base value. But the problem doesn't specify that.Wait, the problem says: "the competitor's bids are 10% above their exponential interest value for each bottle." So, the exponential interest value is f(x) = e^{0.0001x}, and the bid is 1.1 * f(x). So, unless f(x) is scaled, the bids are in dollars as computed.But given that the collector's bids are in the tens of thousands, and the competitor's bids are only a few dollars, it's likely that the competitor's bids are much lower, so the collector would win all.But that seems too straightforward. Maybe I'm missing a scaling factor. Alternatively, perhaps the function is f(x) = e^{0.0001x} * 1000, making the bids more competitive.Alternatively, perhaps the function is f(x) = e^{0.0001x} * 1000, so f(x) is in dollars. Let me compute that.For x=15,000: f(x)=e^{1.5} * 1000≈4.4817*1000≈4,481.70. Bid=1.1*4,481.70≈4,929.87.Similarly, for x=20,000: f(x)=e^{2}*1000≈7.389*1000≈7,389. Bid≈8,127.90.For x=25,000: f(x)=e^{2.5}*1000≈12.182*1000≈12,182. Bid≈13,400.20.So, the competitor's bids are ~4,930, ~8,128, and ~13,400.Now, comparing to the collector's bids:Bottle 1: Collector ~17,450 vs Competitor ~4,930 → Collector wins.Bottle 2: Collector ~26,850 vs Competitor ~8,128 → Collector wins.Bottle 3: Collector 37,500 vs Competitor ~13,400 → Collector wins.So, the collector would acquire all three bottles.But wait, if the competitor's bids are only ~4,930 for the first bottle, which is way below the collector's bid of ~17,450, then the collector would win all.Alternatively, if the function is f(x) = e^{0.0001x} without scaling, the competitor's bids are negligible, so the collector wins all.But perhaps I'm overcomplicating it. Let me stick with the initial interpretation: f(x) = e^{0.0001x}, and the competitor's bid is 1.1 * f(x). So, for x=15,000, f(x)=e^{1.5}≈4.48, so bid≈4.93. Similarly for others.Thus, the competitor's bids are way below the collector's bids, so the collector wins all three.But that seems too easy. Maybe the function is f(x) = e^{0.0001x} * 1000, making the bids more competitive.Alternatively, perhaps the function is f(x) = e^{0.0001x} * V_i, so the competitor's interest is both exponential and proportional to the base value.Wait, let me check that. If f(x) = e^{0.0001x} * V_i, then for x=15,000:f(x)=e^{1.5} * 15,000≈4.4817 * 15,000≈67,225.50. Bid=1.1*67,225.50≈73,948.05.Similarly, for x=20,000: f(x)=e^{2} * 20,000≈7.389 * 20,000≈147,780. Bid≈162,558.For x=25,000: f(x)=e^{2.5} * 25,000≈12.182 * 25,000≈304,550. Bid≈335,005.In this case, the competitor's bids are much higher than the collector's bids. So, the competitor would win all three bottles.But the problem says "the collector's bids are the only competition for the competitor." So, if the competitor's bids are higher, the competitor would win, but the collector's bids are lower, so the collector would lose all.But this depends on how we interpret the function. Since the problem doesn't specify scaling, it's ambiguous. However, given that the collector's bids are in the tens of thousands, and the competitor's bids without scaling are only a few dollars, it's more likely that the competitor's bids are negligible, so the collector wins all.Alternatively, perhaps the function is f(x) = e^{0.0001x} * 1000, making the bids in the thousands, which are still lower than the collector's bids.Wait, let me compute f(x) = e^{0.0001x} * 1000 for each bottle:V1=15,000: f(x)=e^{1.5}*1000≈4.4817*1000≈4,481.70. Bid=1.1*4,481.70≈4,929.87.V2=20,000: f(x)=e^{2}*1000≈7.389*1000≈7,389.00. Bid≈8,127.90.V3=25,000: f(x)=e^{2.5}*1000≈12.182*1000≈12,182.00. Bid≈13,400.20.So, the competitor's bids are ~4,930, ~8,128, and ~13,400.Comparing to the collector's bids:Bottle 1: Collector ~17,450 vs Competitor ~4,930 → Collector wins.Bottle 2: Collector ~26,850 vs Competitor ~8,128 → Collector wins.Bottle 3: Collector 37,500 vs Competitor ~13,400 → Collector wins.So, the collector would acquire all three bottles.But if the function is f(x) = e^{0.0001x} without scaling, the competitor's bids are only a few dollars, so the collector wins all.Alternatively, if f(x) is in thousands, as I did earlier, the competitor's bids are still lower than the collector's bids.Therefore, regardless of scaling, the collector's bids are higher, so the collector would acquire all three bottles.But wait, in the first sub-problem, the collector's bids are 17,450, 26,850, and 37,500. The competitor's bids, assuming f(x) is in thousands, are ~4,930, ~8,128, and ~13,400. So, the collector's bids are higher for all.Therefore, the collector would acquire all three bottles.But let me check if the competitor's bids could be higher than the collector's bids for any bottle.For V1: Collector's bid ~17,450 vs Competitor ~4,930 → Collector wins.For V2: Collector's bid ~26,850 vs Competitor ~8,128 → Collector wins.For V3: Collector's bid 37,500 vs Competitor ~13,400 → Collector wins.So, yes, the collector would acquire all three bottles.But wait, if the competitor's function is f(x) = e^{0.0001x} * V_i, then the competitor's bids would be much higher, as I computed earlier, and the collector would lose all. But the problem doesn't specify that scaling, so I think it's safer to assume that f(x) is as given, without scaling, making the competitor's bids negligible.Therefore, the collector would acquire all three bottles.But let me think again. If the function is f(x) = e^{0.0001x}, then for x=15,000, f(x)=e^{1.5}≈4.48, so bid≈4.93. That's way below the collector's bid of ~17,450.Similarly, for x=20,000: bid≈8.13, which is way below 26,850.For x=25,000: bid≈13.40, which is way below 37,500.So, the collector would win all three.Therefore, the answer to Sub-problem 2 is that the collector will successfully acquire all three bottles.

answer：First, I need to determine 25% of the total talent pool of 120 athletes. Calculating 25% of 120 gives me 30 athletes.Next, the manager wants to add 10 more athletes who have demonstrated exceptional skills recently. Adding these 10 athletes to the initial 30 brings the total to 40 athletes.Therefore, the manager should focus on 40 athletes for potential recruitment.

answer：First, I need to calculate the total effective cost of the project after applying the 10% tax rebate. The base construction cost is 5,000,000.Next, I'll determine the expected property value after a 5% increase once the project is completed.Finally, I'll find the difference between the expected property value and the total effective cost to understand the financial outcome of the project.

answer：First, I'll calculate the total number of animals in each section of the zoo.In the reptile house, there are 12 snakes and 8 lizards. Adding these together gives 20 reptiles.Next, in the bird aviary, there are 15 parrots and 9 flamingos. Adding these gives 24 birds.Then, at the mammal exhibit, there are 5 lions and 13 monkeys. Adding these together results in 18 mammals.Finally, I'll add up the totals from each section: 20 reptiles + 24 birds + 18 mammals equals 62 animals in total.