💡Okay, so I have this problem where I need to determine the number of ways to select a sequence of 8 sets ( A_{1}, A_{2}, ldots, A_{8} ), each of which is a subset of ({1, 2}). The key condition here is that ( A_{m} ) must contain ( A_{n} ) if ( m ) divides ( n ). First, I need to understand what this condition really means. If ( m ) divides ( n ), then ( A_{m} ) has to be a superset of ( A_{n} ). So, for example, ( A_{1} ) must contain all the other sets because 1 divides every number. Similarly, ( A_{2} ) must contain ( A_{4} ), ( A_{6} ), and ( A_{8} ), since 2 divides those numbers. Given that each ( A_i ) is a subset of ({1, 2}), each set can be one of four possibilities: the empty set, ({1}), ({2}), or ({1, 2}). So, without any constraints, there would be (4^8) possible sequences. But the constraints here significantly reduce that number.I think the best way to approach this is to consider each element in ({1, 2}) separately. Since the elements are independent, the total number of valid sequences will be the square of the number of valid sequences for a single element. So, if I can figure out how many valid sequences there are for just the element 1, I can square that number to get the total.Let's focus on the element 1 first. For each set ( A_i ), we can represent whether element 1 is in ( A_i ) or not. Let's denote this as a binary variable, where 1 means the element is present, and 0 means it's absent. So, for each ( A_i ), we have a binary value indicating the presence of element 1.Now, the condition that ( A_{m} ) contains ( A_{n} ) if ( m ) divides ( n ) translates to: if ( m ) divides ( n ), then the binary value for ( A_{m} ) must be greater than or equal to the binary value for ( A_{n} ). In other words, if ( m ) divides ( n ), then if element 1 is in ( A_{n} ), it must also be in ( A_{m} ).This seems similar to a problem where we have to assign binary values to the numbers 1 through 8 such that if ( m ) divides ( n ), then the value assigned to ( m ) is at least as large as the value assigned to ( n ). This structure reminds me of posets (partially ordered sets) and order ideals. Specifically, the divisibility relation on the set ({1, 2, ldots, 8}) forms a poset, and we're looking for the number of order ideals in this poset. However, since we're dealing with binary assignments, it might be simpler to model this as a graph where each node represents a number, and edges represent the divisibility relation.But maybe I'm overcomplicating it. Let's think recursively. For each number ( n ), the presence of element 1 in ( A_n ) depends on the presence of element 1 in all its divisors. So, if I know the assignments for the divisors of ( n ), I can determine the possible assignments for ( n ).Starting from the smallest number, which is 1. Since 1 divides every number, the presence of element 1 in ( A_1 ) affects all other sets. If element 1 is not in ( A_1 ), then it cannot be in any other set. If it is in ( A_1 ), then it can be in any of the other sets, but with the condition that if a set ( A_m ) contains element 1, then all multiples of ( m ) must also contain it.Wait, no. Actually, the condition is that if ( m ) divides ( n ), then ( A_m ) must contain ( A_n ). So, if ( A_n ) contains element 1, then ( A_m ) must also contain it. But if ( A_m ) does not contain element 1, then ( A_n ) cannot contain it either.This means that the presence of element 1 in any set ( A_n ) implies its presence in all divisors of ( n ). So, the presence of element 1 propagates upwards through the divisibility chain.Therefore, to count the number of valid sequences for element 1, I need to count the number of ways to assign 0s and 1s to the numbers 1 through 8 such that if a number ( n ) is assigned 1, then all its divisors are also assigned 1.This is equivalent to choosing a subset of numbers from 1 to 8 such that if a number is in the subset, all its divisors are also in the subset. Such subsets are called "order ideals" or "downward-closed sets" in poset terminology.So, the number of valid sequences for element 1 is equal to the number of order ideals in the poset defined by divisibility on ({1, 2, ldots, 8}).Now, I need to compute the number of order ideals in this poset. This might be a bit involved, but perhaps I can break it down by considering the structure of the poset.First, let's list the numbers from 1 to 8 and their prime factorizations:1: 12: 23: 34: 2²5: 56: 2×37: 78: 2³Looking at this, the poset under divisibility can be visualized as a forest of trees, where each tree is rooted at a prime number, and the nodes are the multiples of that prime.But actually, since 1 is a divisor of every number, it's connected to all other numbers. So, the poset is a single tree with root 1, and branches corresponding to the prime factors.Wait, no. Actually, 1 is connected to all numbers, but numbers can have multiple parents if they have multiple prime factors. For example, 6 is divisible by both 2 and 3, so in the poset, 6 would have edges from both 2 and 3.This makes the poset a lattice rather than a tree. Specifically, it's the lattice of divisors of 8!, but restricted to numbers up to 8.To count the number of order ideals, I can use the fact that the number of order ideals in a poset is equal to the product of (1 + the number of order ideals of each connected component). But in this case, the poset is connected because 1 is connected to everyone.Alternatively, I can use the fact that the number of order ideals in a poset is equal to the Dedekind number for that poset, but Dedekind numbers are known only for small posets.Alternatively, I can try to compute it recursively. Let's consider the poset and try to count the number of order ideals.Let me try to list all the order ideals. An order ideal is a set of elements such that if an element is in the set, all elements below it (in the divisibility order) are also in the set.So, starting with the minimal elements, which are the primes: 2, 3, 5, 7.But actually, in the poset, 1 is the minimal element, and then the primes are the next level.Wait, no. In the poset, 1 is the minimal element, and then the primes are the atoms (elements covering 1). Then, the composites are above the primes.So, the poset has 1 at the bottom, then 2, 3, 5, 7 above it, then 4, 6 above 2 and 3 respectively, 8 above 4, etc.Let me try to visualize the poset:- Level 0: 1- Level 1: 2, 3, 5, 7- Level 2: 4 (divisible by 2), 6 (divisible by 2 and 3)- Level 3: 8 (divisible by 4)So, the poset has four levels: 1, primes, composites, and higher composites.Now, to count the number of order ideals, I can use the fact that the number of order ideals in a poset is equal to the product over all elements of (1 + the number of order ideals of the subposet generated by the elements above it).But this might be complicated. Alternatively, I can use the fact that the number of order ideals in a poset is equal to the number of antichains, but that's not directly helpful.Wait, actually, the number of order ideals is equal to the number of antichains, because each order ideal is uniquely determined by its maximal elements, which form an antichain.But counting antichains is also non-trivial.Alternatively, I can use the fact that the poset is a distributive lattice, and the number of order ideals is equal to the product of (1 + the number of order ideals of each connected component). But in this case, the poset is connected.Alternatively, I can use the fact that the poset is a tree, but it's not a tree because 6 is divisible by both 2 and 3, creating a diamond shape.Wait, maybe I can use the inclusion-exclusion principle or some recursive formula.Let me try to approach it recursively. Let's consider the poset and see how adding each element affects the number of order ideals.Start with just 1. The number of order ideals is 1 (only the empty set and the set containing 1, but since 1 is the minimal element, any order ideal must include 1 if it includes anything above it. Wait, actually, in the poset, an order ideal must include all elements below any element it includes. So, if we include 1, we can include any subset of the elements above it, but if we don't include 1, we can't include anything.Wait, no. Actually, an order ideal is a subset ( I ) such that if ( x in I ) and ( y leq x ), then ( y in I ). So, if we don't include 1, we can't include anything. If we include 1, we can include any subset of the elements above 1, but with the condition that if we include an element, we must include all elements below it.Wait, this is getting confusing. Maybe I should think in terms of generating functions or something.Alternatively, perhaps I can model this as a graph where each node represents a number, and edges go from a number to its multiples. Then, the number of order ideals is the number of independent sets in this graph, but actually, it's more specific than that.Wait, no. An order ideal is a set of elements such that if an element is in the set, all elements below it are also in the set. So, it's like a downward-closed set.In this case, since the poset is graded by the number of prime factors, maybe I can compute the number of order ideals level by level.Let me try to list all possible order ideals.Starting with the minimal element 1. If we include 1, we can choose to include any combination of the elements above it, but with the condition that if we include an element, we must include all elements below it.Wait, no. If we include 1, we can include any subset of the elements above it, but if we include an element, we must include all elements below it. But since 1 is already included, including an element just requires that all its divisors are included, which they are because 1 is included.Wait, actually, no. If we include an element, we must include all elements below it, but since 1 is already included, we don't need to worry about that. So, if we include 1, we can include any subset of the elements above it, but with the condition that if we include an element, we must include all elements below it in the poset.But since 1 is already included, including an element just requires that all its divisors are included, which they are because 1 is included. So, actually, if we include 1, we can include any subset of the elements above it, without any further restrictions, because the only divisor of any element is 1, which is already included.Wait, that can't be right because, for example, if we include 2, we must include all divisors of 2, which is just 1. Since 1 is already included, including 2 is fine. Similarly, including 3 requires only 1, which is already included.But what about including 4? If we include 4, we must include all divisors of 4, which are 1 and 2. So, if we include 4, we must also include 2. Similarly, including 6 requires including 2 and 3.Including 8 requires including 4, which in turn requires including 2.So, the dependencies are:- Including 2 requires including 1.- Including 3 requires including 1.- Including 4 requires including 2 and 1.- Including 5 requires including 1.- Including 6 requires including 2, 3, and 1.- Including 7 requires including 1.- Including 8 requires including 4, 2, and 1.So, to count the number of order ideals, we can think of it as choosing a subset of the elements above 1 (i.e., 2, 3, 4, 5, 6, 7, 8) such that if we include an element, we must include all its required dependencies.This is similar to a dependency graph where each element has certain prerequisites.To count the number of such subsets, we can use the principle of inclusion-exclusion or model it as a graph and count the number of independent sets with certain constraints.Alternatively, we can model this as a graph where each node represents an element, and edges represent dependencies (i.e., if you include a node, you must include its neighbors). Then, the number of order ideals is the number of subsets of nodes that include all neighbors of included nodes.This is equivalent to the number of closed sets in the graph, which is a well-known problem but doesn't have a straightforward formula.However, since the poset is small, maybe we can compute it manually.Let me try to list all possible order ideals.First, the minimal case: not including 1. But since 1 is the minimal element, if we don't include 1, we can't include anything else. So, that's one order ideal: the empty set.Now, if we include 1, we can choose to include any subset of the elements above it, but with the dependencies.So, let's consider the elements above 1: 2, 3, 4, 5, 6, 7, 8.We need to count the number of subsets of these elements such that if an element is included, all its divisors (other than 1, which is already included) are also included.So, let's consider the dependencies:- 2: no dependencies (other than 1)- 3: no dependencies- 4: depends on 2- 5: no dependencies- 6: depends on 2 and 3- 7: no dependencies- 8: depends on 4, which depends on 2So, to include 4, we must include 2.To include 6, we must include 2 and 3.To include 8, we must include 4 and 2.So, let's think of this as a graph where:- 2 is connected to 4 and 6 and 8- 3 is connected to 6- 4 is connected to 8- 5 and 7 are isolatedSo, the dependencies form a graph with connected components:- Component 1: 2, 4, 6, 8- Component 2: 3- Component 3: 5- Component 4: 7Wait, no. Actually, 3 is connected to 6, which is connected to 2, which is connected to 4 and 8. So, actually, 2, 3, 4, 6, 8 are all connected in one component.5 and 7 are isolated.So, the poset has two connected components: one large component containing 2, 3, 4, 5, 6, 7, 8, and one isolated element 1.Wait, no. 5 and 7 are only connected to 1, not to anything else. So, in terms of dependencies, 5 and 7 can be included independently.So, the number of order ideals can be computed as the product of the number of order ideals in each connected component.But since 1 is connected to all, it's a bit tricky. Wait, actually, 1 is the minimal element, so it's part of all order ideals except the empty set.But since we're considering order ideals that include 1, we can factor out the 1 and consider the rest.So, the number of order ideals is 1 (the empty set) plus the number of order ideals that include 1.The number of order ideals that include 1 is equal to the number of order ideals in the poset excluding 1, but with the condition that if an element is included, all its divisors are included.But since we've already included 1, the dependencies are only among the elements above 1.So, the number of order ideals that include 1 is equal to the number of order ideals in the poset induced by the elements {2, 3, 4, 5, 6, 7, 8} with the divisibility relation.This poset is connected as follows:- 2 is connected to 4, 6, 8- 3 is connected to 6- 4 is connected to 8- 5 and 7 are isolatedSo, to count the number of order ideals in this poset, we can consider the connected components:- Component 1: 2, 3, 4, 6, 8- Component 2: 5- Component 3: 7Since the poset is a forest of trees, the number of order ideals is the product of the number of order ideals in each component.So, let's compute the number of order ideals for each component.Starting with Component 1: 2, 3, 4, 6, 8This component has the following structure:- 2 is connected to 4, 6, 8- 3 is connected to 6- 4 is connected to 8So, it's a bit complex. Let's try to compute the number of order ideals for this component.We can model this as a graph and use the principle of inclusion-exclusion or recursion.Let me try to approach it recursively. Let's consider the elements in order of increasing number of dependencies.First, consider 8. To include 8, we must include 4, which must include 2.Similarly, to include 6, we must include 2 and 3.So, let's think of the possible choices:1. Do not include 2: Then, we cannot include 4, 6, or 8. So, the only elements we can include are 3. But including 3 doesn't require anything else except 1, which is already included. So, in this case, we can choose to include or exclude 3 and 5 and 7 independently.Wait, no. Wait, we're only considering Component 1, which is 2, 3, 4, 6, 8. So, in this component, if we don't include 2, we can include 3 or not, but 3 doesn't depend on anything else in this component. So, the number of order ideals when not including 2 is 2 (include or exclude 3).2. Include 2: Then, we can choose to include or exclude 4, 6, and 8, but with dependencies.- If we include 4, we must include 2 (already included).- If we include 6, we must include 2 and 3.- If we include 8, we must include 4 and 2.So, let's break it down:- Including 2: Now, we can choose to include or exclude 4, 6, and 8, but with the following conditions: - Including 4 requires including 2 (already included). - Including 6 requires including 3. - Including 8 requires including 4 (and thus 2).So, let's consider the possibilities:a. Do not include 4, 6, or 8: Then, we can choose to include or exclude 3 independently. So, 2 possibilities.b. Include 4: Then, we can choose to include or exclude 6 and 8. - If we include 4, we can include or exclude 6 and 8. - Including 6 requires including 3. - Including 8 requires including 4 (already included). So, for each inclusion of 4, we have: - Include 4 only: 1 possibility. - Include 4 and 6: requires including 3. - Include 4 and 8: requires including 4 (already included). - Include 4, 6, and 8: requires including 3. Wait, this is getting tangled. Maybe a better approach is to consider the choices step by step. Let's think of it as a tree: - Start with 2. - From 2, we can go to 4, 6, or 8. - From 4, we can go to 8. - From 6, we can go to 3. So, the structure is: 2 -> 4 -> 8 2 -> 6 -> 3 So, to count the number of order ideals in this component, we can consider the possible subsets that include 2 and any combination of 4, 6, 8, 3, with the dependencies. Let's consider the possible cases: 1. Include 2 only: 1 possibility. 2. Include 2 and 4: Then, we can choose to include or exclude 8. - Include 2 and 4 only: 1 possibility. - Include 2, 4, and 8: 1 possibility. 3. Include 2 and 6: Then, we must include 3. - Include 2, 6, and 3: 1 possibility. 4. Include 2, 4, and 6: Then, we must include 3 and can choose to include or exclude 8. - Include 2, 4, 6, 3: 1 possibility. - Include 2, 4, 6, 3, 8: 1 possibility. 5. Include 2, 4, 8, and 6: Then, we must include 3. - Include 2, 4, 8, 6, 3: 1 possibility. 6. Include 2, 6, 3, and 4: Then, we can include or exclude 8. - Include 2, 6, 3, 4: 1 possibility. - Include 2, 6, 3, 4, 8: 1 possibility. Wait, this is getting too detailed and I might be double-counting. Maybe a better approach is to use the principle of multiplication. Let's consider the choices for each element, considering dependencies. Starting from the top: - 8 depends on 4, which depends on 2. - 6 depends on 2 and 3. So, the choices are: - Whether to include 2 or not. If we include 2, then: - Whether to include 4 or not. - Whether to include 6 or not. - If we include 6, we must include 3. - If we include 4, we can choose to include 8 or not. So, let's structure it: If we include 2: - Choose to include or exclude 4: 2 choices. - Choose to include or exclude 6: 2 choices. - If we include 6, we must include 3: 1 choice. - If we include 4, choose to include or exclude 8: 2 choices. So, the total number of order ideals when including 2 is: (choices for 4) * (choices for 6) * (choices for 8 if 4 is included) But this is a bit tangled. Alternatively, we can think of it as: - Including 2: 1 choice. - For 4: 2 choices (include or exclude). - For 6: 2 choices (include or exclude). - For 8: If 4 is included, 2 choices; else, 0. - For 3: If 6 is included, must include 3; else, 0. So, the total number is: (1) * [ (1 + 1) for 4 ] * [ (1 + 1) for 6 ] * [ (1 + 1) for 8 if 4 is included ] * [ (1 + 1) for 3 if 6 is included ] Wait, this is getting too convoluted. Maybe a better way is to use the fact that the number of order ideals in a poset is equal to the product of (1 + the number of order ideals of each connected component). But in this case, the component is connected, so that doesn't help. Alternatively, perhaps I can use the fact that the number of order ideals in a poset is equal to the number of antichains, but counting antichains is also non-trivial. Wait, maybe I can use the fact that the poset is a tree, but it's not a tree because 6 is connected to both 2 and 3. Alternatively, perhaps I can use the principle of inclusion-exclusion. Let me try to list all possible order ideals in Component 1: 1. {} (empty set) 2. {2} 3. {2, 4} 4. {2, 4, 8} 5. {2, 6} 6. {2, 6, 3} 7. {2, 4, 6} 8. {2, 4, 6, 3} 9. {2, 4, 6, 3, 8} 10. {2, 4, 8, 6} 11. {2, 4, 8, 6, 3} 12. {2, 6, 3, 4} 13. {2, 6, 3, 4, 8} Wait, this seems like I'm listing all possible combinations, but I might be missing some or double-counting. Alternatively, perhaps I can think of it as follows: - The number of order ideals in Component 1 is equal to the number of subsets of {2, 3, 4, 6, 8} that are closed under taking divisors. So, let's list all such subsets: 1. {} 2. {2} 3. {2, 4} 4. {2, 4, 8} 5. {2, 6} 6. {2, 6, 3} 7. {2, 4, 6} 8. {2, 4, 6, 3} 9. {2, 4, 6, 3, 8} 10. {2, 4, 8, 6} 11. {2, 4, 8, 6, 3} 12. {2, 6, 3, 4} 13. {2, 6, 3, 4, 8} Wait, this seems like 13 order ideals, but I'm not sure if I've covered all possibilities. Alternatively, perhaps I can use the fact that the number of order ideals in a poset is equal to the product of (1 + the number of order ideals of each connected component). But in this case, the component is connected, so that doesn't help. Alternatively, perhaps I can use the fact that the number of order ideals in a poset is equal to the number of antichains, but counting antichains is also non-trivial. Wait, maybe I can use the fact that the poset is a tree, but it's not a tree because 6 is connected to both 2 and 3. Alternatively, perhaps I can use the principle of inclusion-exclusion. Let me try to think differently. Since the poset is small, maybe I can compute the number of order ideals by considering the elements in a specific order and multiplying the possibilities. Let's consider the elements in the order 2, 3, 4, 6, 8. For each element, decide whether to include it or not, considering the dependencies. - Start with 2: If we include 2, we can proceed; if not, we can't include 4, 6, or 8. - If we include 2, then: - For 3: Can include or exclude independently. - For 4: Can include or exclude. - For 6: If we include 6, we must include 3. - For 8: If we include 8, we must include 4. So, let's break it down: Case 1: Do not include 2. - Then, we cannot include 4, 6, or 8. - We can choose to include or exclude 3. - So, 2 possibilities: {} and {3}. Case 2: Include 2. - Now, we can choose to include or exclude 3, 4, 6, 8, but with dependencies. Subcase 2a: Do not include 4 or 6. - Then, we can choose to include or exclude 3. - So, 2 possibilities: {2} and {2, 3}. Subcase 2b: Include 4. - Then, we can choose to include or exclude 8. - And we can choose to include or exclude 6, but if we include 6, we must include 3. So, Subsubcase 2b1: Do not include 6. - Then, we can choose to include or exclude 8. - So, 2 possibilities: {2, 4} and {2, 4, 8}. Subsubcase 2b2: Include 6. - Then, we must include 3. - We can choose to include or exclude 8. - So, 2 possibilities: {2, 4, 6, 3} and {2, 4, 6, 3, 8}. Subcase 2c: Include 6. - Then, we must include 3. - We can choose to include or exclude 4 and 8. - If we include 4, we can choose to include or exclude 8. So, Subsubcase 2c1: Do not include 4. - Then, we cannot include 8. - So, 1 possibility: {2, 6, 3}. Subsubcase 2c2: Include 4. - Then, we can choose to include or exclude 8. - So, 2 possibilities: {2, 6, 3, 4} and {2, 6, 3, 4, 8}. Subcase 2d: Include both 4 and 6. - Then, we must include 3. - We can choose to include or exclude 8. - So, 2 possibilities: {2, 4, 6, 3} and {2, 4, 6, 3, 8}. Wait, this seems like some overlap with Subcase 2b2 and 2c2. Maybe I'm overcomplicating it. Let's try to count systematically: When including 2: - We can choose to include or exclude 3, 4, 6, 8. - With the constraints: - If we include 4, we can include 8. - If we include 6, we must include 3. So, let's consider the possibilities: 1. Do not include 3, 4, 6, 8: {2} 2. Include 3, do not include 4, 6, 8: {2, 3} 3. Include 4, do not include 3, 6, 8: {2, 4} 4. Include 4 and 8, do not include 3, 6: {2, 4, 8} 5. Include 6, which requires including 3: {2, 6, 3} 6. Include 4 and 6, which requires including 3: {2, 4, 6, 3} 7. Include 4, 6, and 8, which requires including 3: {2, 4, 6, 3, 8} 8. Include 6 and 4, which requires including 3: {2, 6, 3, 4} 9. Include 6, 4, and 8, which requires including 3: {2, 6, 3, 4, 8} Wait, this seems like 9 possibilities when including 2. But earlier, when not including 2, we had 2 possibilities. So, total order ideals in Component 1: 2 (without 2) + 9 (with 2) = 11. Hmm, but earlier I thought it was 13. Maybe I'm missing some. Let me try to list them again: Without 2: 1. {} 2. {3} With 2: 3. {2} 4. {2, 3} 5. {2, 4} 6. {2, 4, 8} 7. {2, 6} 8. {2, 6, 3} 9. {2, 4, 6} 10. {2, 4, 6, 3} 11. {2, 4, 6, 3, 8} 12. {2, 4, 8, 6} 13. {2, 4, 8, 6, 3} 14. {2, 6, 3, 4} 15. {2, 6, 3, 4, 8} Wait, now I'm getting 15, which seems too high. Maybe I'm overcounting because some of these sets are the same. For example, {2, 4, 6, 3} and {2, 6, 3, 4} are the same set. Similarly, {2, 4, 6, 3, 8} and {2, 6, 3, 4, 8} are the same. So, actually, the unique sets are: 1. {} 2. {3} 3. {2} 4. {2, 3} 5. {2, 4} 6. {2, 4, 8} 7. {2, 6} 8. {2, 6, 3} 9. {2, 4, 6} 10. {2, 4, 6, 3} 11. {2, 4, 6, 3, 8} So, 11 unique order ideals in Component 1. Now, moving on to Component 2: {5} This is a single element with no dependencies except 1, which is already included. So, the number of order ideals in Component 2 is 2: {} and {5}. Similarly, Component 3: {7} The number of order ideals is also 2: {} and {7}. Therefore, the total number of order ideals in the poset excluding 1 is: Number of order ideals in Component 1 * Number of order ideals in Component 2 * Number of order ideals in Component 3 = 11 * 2 * 2 = 44. But wait, this counts the number of order ideals that include 1, right? Because we factored out 1 and considered the rest. So, the total number of order ideals in the entire poset is: 1 (the empty set) + 44 (the order ideals that include 1) = 45. Therefore, the number of valid sequences for element 1 is 45. Since the elements 1 and 2 are independent, the total number of valid sequences for both elements is 45 * 45 = 2025. So, the final answer is 2025.

💡Okay, so I have this trigonometry problem here: If cos(π/2 + α) = 1/3, find the value of cos(2α). Hmm, let me try to figure this out step by step. First, I remember that there are some trigonometric identities that relate functions of angles shifted by π/2. Specifically, I think cos(π/2 + α) is related to sine or something like that. Let me recall... Oh yeah, I think it's one of the co-function identities. Wait, co-function identities relate sine and cosine of complementary angles. So, cos(π/2 - α) is equal to sin(α), right? But in this case, it's cos(π/2 + α). Hmm, that's not exactly the same. Maybe there's another identity for that. I think there's a formula for cos(A + B). Let me write that down: cos(A + B) = cos A cos B - sin A sin B. So, if I apply that to cos(π/2 + α), it should be cos(π/2)cos(α) - sin(π/2)sin(α). Let me compute each part. Cos(π/2) is 0, and sin(π/2) is 1. So substituting those in, I get 0 * cos(α) - 1 * sin(α), which simplifies to -sin(α). So, cos(π/2 + α) = -sin(α). But the problem says that this is equal to 1/3. Therefore, -sin(α) = 1/3. If I solve for sin(α), I can multiply both sides by -1, which gives sin(α) = -1/3. Okay, so the sine of alpha is negative one-third. Now, the question asks for cos(2α). I remember there are double-angle formulas for cosine. Let me recall them. The double-angle formula for cosine can be written in a few ways: 1. cos(2α) = cos²α - sin²α2. cos(2α) = 2cos²α - 13. cos(2α) = 1 - 2sin²αHmm, since I already know sin(α), maybe the third formula would be the most straightforward to use because it only involves sin²α, which I can compute from sin(α) = -1/3. So, using the third formula: cos(2α) = 1 - 2sin²α. Let's plug in sin(α) = -1/3. First, compute sin²α: (-1/3)² = 1/9. Then, multiply that by 2: 2 * (1/9) = 2/9. Now, subtract that from 1: 1 - 2/9. To compute 1 - 2/9, I can think of 1 as 9/9, so 9/9 - 2/9 = 7/9. Therefore, cos(2α) = 7/9. Wait, let me double-check my steps to make sure I didn't make a mistake. 1. I started with cos(π/2 + α) = 1/3.2. I used the identity cos(π/2 + α) = -sin(α), which seems correct because cos(π/2 + α) is indeed -sin(α).3. So, -sin(α) = 1/3 implies sin(α) = -1/3. That makes sense.4. Then, I used the double-angle formula for cosine, specifically the one involving sin²α, because I already had sin(α).5. Plugging in sin(α) = -1/3, I squared it to get 1/9, multiplied by 2 to get 2/9, and subtracted that from 1 to get 7/9.Everything seems to check out. I don't think I made any calculation errors. So, I'm confident that cos(2α) is 7/9.**Final Answer**The value of cos(2α) is boxed{dfrac{7}{9}}.

💡Alright, so I have this problem here: Given θ in the interval (0, π/2), I need to find the maximum value of the expression (2 sinθ cosθ) / [(sinθ + 1)(cosθ + 1)]. Hmm, okay. Let me try to figure this out step by step.First, I remember that 2 sinθ cosθ is equal to sin2θ. So maybe I can rewrite the numerator as sin2θ. That might make things simpler. So the expression becomes sin2θ / [(sinθ + 1)(cosθ + 1)]. Hmm, that seems a bit better, but I'm not sure if that's the most helpful step yet.Let me look at the denominator: (sinθ + 1)(cosθ + 1). If I expand that, it would be sinθ cosθ + sinθ + cosθ + 1. So the denominator is sinθ cosθ + sinθ + cosθ + 1. Okay, so the entire expression is sin2θ divided by that.Wait, so sin2θ is 2 sinθ cosθ, which is actually twice the sinθ cosθ term. So maybe I can write the numerator as 2 sinθ cosθ, which is the same as sin2θ. So the expression is 2 sinθ cosθ / [sinθ cosθ + sinθ + cosθ + 1]. Hmm, maybe I can factor something out here or find a substitution.I've heard that sometimes substituting t = sinθ + cosθ can be useful. Let me try that. So let t = sinθ + cosθ. Then, I remember that t² = sin²θ + 2 sinθ cosθ + cos²θ. Since sin²θ + cos²θ = 1, that means t² = 1 + 2 sinθ cosθ. So, 2 sinθ cosθ = t² - 1. That's interesting because the numerator is 2 sinθ cosθ, which is t² - 1.So, substituting back into the expression, the numerator becomes t² - 1, and the denominator is sinθ cosθ + sinθ + cosθ + 1. Wait, sinθ cosθ is (t² - 1)/2, right? Because 2 sinθ cosθ = t² - 1, so sinθ cosθ = (t² - 1)/2.So, substituting that into the denominator, we get [(t² - 1)/2] + t + 1. Let me write that out: denominator = (t² - 1)/2 + t + 1. Let me combine those terms. First, let's write 1 as 2/2 to have a common denominator. So, denominator = (t² - 1)/2 + t + 2/2 = (t² - 1 + 2t + 2)/2 = (t² + 2t + 1)/2.Wait, t² + 2t + 1 is (t + 1)². So the denominator simplifies to (t + 1)² / 2. So now, the entire expression becomes (t² - 1) / [(t + 1)² / 2]. Which is the same as 2(t² - 1) / (t + 1)².Hmm, that seems manageable. Let's see, t² - 1 is (t - 1)(t + 1). So, 2(t - 1)(t + 1) / (t + 1)². The (t + 1) terms cancel out, leaving 2(t - 1) / (t + 1). So, the expression simplifies to 2(t - 1)/(t + 1).Now, I need to find the maximum value of this expression with respect to t. But what is the range of t? Since θ is in (0, π/2), t = sinθ + cosθ. I remember that sinθ + cosθ can be written as √2 sin(θ + π/4). The maximum value of sin is 1, so the maximum value of t is √2, and the minimum value is... Well, when θ approaches 0, sinθ approaches 0 and cosθ approaches 1, so t approaches 1. Similarly, when θ approaches π/2, sinθ approaches 1 and cosθ approaches 0, so t approaches 1 again. So t is in (1, √2].So, t ranges from just above 1 to √2. So, now I have the expression 2(t - 1)/(t + 1), and I need to find its maximum value for t in (1, √2].Let me think about how to maximize this expression. Maybe I can take the derivative with respect to t and set it to zero to find critical points. Alternatively, since it's a function of t, I can analyze its behavior.Let me denote f(t) = 2(t - 1)/(t + 1). Let's find its derivative f’(t). Using the quotient rule: f’(t) = [2(t + 1) - 2(t - 1)(1)] / (t + 1)^2. Simplifying the numerator: 2(t + 1) - 2(t - 1) = 2t + 2 - 2t + 2 = 4. So, f’(t) = 4 / (t + 1)^2.Since t > 1, the denominator is always positive, so f’(t) is positive. That means f(t) is increasing on the interval (1, √2]. Therefore, the maximum occurs at t = √2.So, plugging t = √2 into f(t): f(√2) = 2(√2 - 1)/(√2 + 1). Let me simplify this expression. Multiply numerator and denominator by (√2 - 1) to rationalize the denominator:f(√2) = 2(√2 - 1)(√2 - 1) / [(√2 + 1)(√2 - 1)].The denominator becomes (√2)^2 - 1^2 = 2 - 1 = 1. So, denominator is 1.The numerator is 2(√2 - 1)^2. Let's expand (√2 - 1)^2: (√2)^2 - 2√2*1 + 1^2 = 2 - 2√2 + 1 = 3 - 2√2.So, numerator is 2*(3 - 2√2) = 6 - 4√2.Therefore, f(√2) = 6 - 4√2.So, the maximum value of the original expression is 6 - 4√2.Let me double-check my steps to make sure I didn't make a mistake. I used substitution t = sinθ + cosθ, which is a common technique. Then, I expressed sinθ cosθ in terms of t, which is correct. Then, I rewrote the denominator in terms of t and simplified it correctly to (t + 1)^2 / 2. Then, I simplified the entire expression to 2(t - 1)/(t + 1), which seems right.Then, I found the derivative and saw that it's always positive, meaning the function is increasing on the interval, so the maximum is at t = √2. Plugging that in and simplifying gives 6 - 4√2. That seems correct.I think I did everything correctly. So, the maximum value is 6 - 4√2.

💡Okay, so I need to find the smallest "good number," which is defined as an integer that has exactly 8 positive divisors and the sum of these divisors is 3240. The example given is 2006, but I need to find the smallest one. First, I remember that the number of divisors of a number relates to its prime factorization. If a number n has a prime factorization of ( n = p_1^{a_1} times p_2^{a_2} times dots times p_k^{a_k} ), then the number of divisors is ( (a_1 + 1)(a_2 + 1)dots(a_k + 1) ). Since we need exactly 8 divisors, I need to find all possible combinations of exponents such that their increments multiply to 8.The possible combinations for the exponents are:1. One prime factor raised to the 7th power: ( (7 + 1) = 8 )2. One prime factor squared and another prime factor cubed: ( (2 + 1)(3 + 1) = 3 times 4 = 12 ) (Wait, that's 12, not 8. Hmm, maybe I got that wrong.)3. Three distinct prime factors each raised to the first power: ( (1 + 1)(1 + 1)(1 + 1) = 2 times 2 times 2 = 8 )Wait, actually, the correct combinations for 8 divisors are:- ( 8 = 8 times 1 ) which corresponds to a single prime raised to the 7th power.- ( 8 = 4 times 2 ) which corresponds to a prime cubed times another prime.- ( 8 = 2 times 2 times 2 ) which corresponds to three distinct primes each to the first power.So, these are the three cases I need to consider.Now, the sum of the divisors is given by the formula for each case:1. For ( n = p^7 ), the sum of divisors is ( 1 + p + p^2 + p^3 + p^4 + p^5 + p^6 + p^7 ).2. For ( n = p^3 times q ), the sum of divisors is ( (1 + p + p^2 + p^3)(1 + q) ).3. For ( n = p times q times r ), the sum of divisors is ( (1 + p)(1 + q)(1 + r) ).I need each of these sums to equal 3240. Let's tackle each case one by one.**Case 1: ( n = p^7 )**The sum of divisors is ( 1 + p + p^2 + p^3 + p^4 + p^5 + p^6 + p^7 = 3240 ). Let's see if such a prime p exists.Trying p=2:Sum = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255, which is way less than 3240.Trying p=3:Sum = 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 = Let's compute step by step:1 + 3 = 44 + 9 = 1313 + 27 = 4040 + 81 = 121121 + 243 = 364364 + 729 = 10931093 + 2187 = 3280. Hmm, that's 3280, which is more than 3240. So p=3 is too big.p=5:Sum would be even larger, so this case doesn't seem possible. So, no solution in this case.**Case 2: ( n = p^3 times q )**Sum of divisors is ( (1 + p + p^2 + p^3)(1 + q) = 3240 ).Let me factorize 3240 to see possible values for ( (1 + p + p^2 + p^3) ) and ( (1 + q) ).3240 factors: 3240 = 2^3 * 3^4 * 5.So, possible factor pairs (a, b) where a*b=3240, and a = 1 + p + p^2 + p^3, b = 1 + q.Since p and q are primes, q must be at least 2, so b = 1 + q >= 3.Similarly, a must be at least 1 + 2 + 4 + 8 = 15 (if p=2), but let's check.Let me list possible a values:Start with p=2:a = 1 + 2 + 4 + 8 = 15Then b = 3240 / 15 = 216So, 1 + q = 216 => q=215. Is 215 prime? Let's check. 215 divided by 5 is 43, so 215=5*43, not prime. So invalid.Next p=3:a = 1 + 3 + 9 + 27 = 40b = 3240 / 40 = 811 + q =81 => q=80, which is not prime.p=5:a=1 +5 +25 +125=156b=3240 /156=20.769... Not integer. So invalid.p=7:a=1 +7 +49 +343=400b=3240 /400=8.1, not integer.p=11:a=1 +11 +121 +1331=1464b=3240 /1464≈2.213, not integer.p=13:a=1 +13 +169 +2197=2380b=3240 /2380≈1.36, not integer.So, none of these p give integer b where q is prime. Maybe p=2 with a different a?Wait, maybe I missed some factor pairs. Let's list all possible factor pairs of 3240 where a >=15 and b >=3.Factor pairs of 3240:1 * 32402 * 16203 * 10804 * 8105 * 6486 * 5408 * 4059 * 36010 * 32412 * 27015 * 21618 * 18020 * 16224 * 13527 * 12030 * 10836 * 9040 * 8145 * 7254 * 60Now, a must be 1 + p + p^2 + p^3, which for p=2 is 15, p=3 is 40, p=5 is 156, p=7 is 400, etc.Looking at the factor pairs, a could be 15, 40, 156, 400, etc.From the list above, a=15 corresponds to b=216, which we saw didn't work.a=40 corresponds to b=81, which also didn't work.a=156: 3240 /156=20.769, not integer.a=400: 3240 /400=8.1, not integer.So, no valid solutions in this case either.**Case 3: ( n = p times q times r )**Sum of divisors is ( (1 + p)(1 + q)(1 + r) = 3240 ).We need to find three primes p < q < r such that the product of (1+p)(1+q)(1+r)=3240.To minimize n, we should choose the smallest possible primes p, q, r.Let me factorize 3240 again: 3240 = 2^3 * 3^4 * 5.We need to express 3240 as a product of three integers greater than 1 (since p, q, r are primes, so 1+p, 1+q, 1+r are at least 3, 4, 6, etc.)Let me list possible triplets (a, b, c) such that a*b*c=3240, and a <= b <= c, and a >=3, b >=4, c >=6.We need to find such triplets and then check if a-1, b-1, c-1 are primes.Let me start with the smallest possible a.a=3:Then b*c=3240/3=1080.Now, find b and c such that b <= c, b >=4, c >=b.Factor pairs of 1080:1 * 10802 * 5403 * 3604 * 2705 * 2166 * 1808 * 1359 * 12010 * 10812 * 9015 * 7218 * 6020 * 5424 * 4527 * 4030 * 36Now, b must be >=4, so starting from b=4:b=4, c=270: Check if 4-1=3 (prime), 270-1=269 (prime). So p=3, q=269, r=?Wait, n = p*q*r, but in this case, a=3 corresponds to p=2 (since 1+p=3 => p=2), b=4 corresponds to q=3 (1+q=4 => q=3), c=270 corresponds to r=269 (1+r=270 => r=269). So n=2*3*269=1614.Is 269 prime? Let's check. 269 is not divisible by 2,3,5,7,11,13,17. 17^2=289>269, so yes, 269 is prime.So n=2*3*269=1614.Wait, but let's see if there are smaller n.Continuing with a=3:b=5, c=216: p=2, q=4 (but 4 is not prime). So invalid.b=6, c=180: q=5, r=179. 179 is prime. So n=2*5*179=1790, which is larger than 1614.b=8, c=135: q=7, r=134 (134 is not prime).b=9, c=120: q=8 (not prime).b=10, c=108: q=9 (not prime).b=12, c=90: q=11, r=89. 89 is prime. So n=2*11*89=1958, which is larger.b=15, c=72: q=14 (not prime).b=18, c=60: q=17, r=59. Both primes. So n=2*17*59=2006, which is the example given.b=20, c=54: q=19, r=53. Both primes. n=2*19*53=2014.b=24, c=45: q=23, r=44 (not prime).b=27, c=40: q=26 (not prime).b=30, c=36: q=29, r=35 (not prime).So the smallest n in this case is 1614.But let's check other a values to see if we can get a smaller n.a=4:Then b*c=3240/4=810.Factor pairs of 810:1*8102*4053*2705*1626*1359*9010*8115*5418*4527*30Now, a=4 corresponds to p=3 (1+p=4). So b and c must be >=5.Check b=5, c=162: q=4 (not prime).b=6, c=135: q=5, r=134 (not prime).b=9, c=90: q=8 (not prime).b=10, c=81: q=9 (not prime).b=15, c=54: q=14 (not prime).b=18, c=45: q=17, r=44 (not prime).b=27, c=30: q=26 (not prime).So no valid solutions here.a=5:b*c=3240/5=648.Factor pairs of 648:1*6482*3243*2164*1626*1088*819*7212*5418*3624*27a=5 corresponds to p=4 (not prime). So invalid.a=6:b*c=3240/6=540.Factor pairs of 540:1*5402*2703*1804*1355*1086*909*6010*5412*4515*3618*30a=6 corresponds to p=5. So b and c must be >=7.Check b=9, c=60: q=8 (not prime).b=10, c=54: q=9 (not prime).b=12, c=45: q=11, r=44 (not prime).b=15, c=36: q=14 (not prime).b=18, c=30: q=17, r=29. Both primes. So n=5*17*29=2465, which is larger than 1614.a=8:b*c=3240/8=405.Factor pairs of 405:1*4053*1355*819*4515*27a=8 corresponds to p=7. So b and c must be >=9.b=9, c=45: q=8 (not prime).b=15, c=27: q=14 (not prime).No valid solutions.a=9:b*c=3240/9=360.Factor pairs of 360:1*3602*1803*1204*905*726*608*459*4010*3612*3015*2418*20a=9 corresponds to p=8 (not prime). So invalid.a=10:b*c=3240/10=324.Factor pairs of 324:1*3242*1623*1084*816*549*3612*2718*18a=10 corresponds to p=9 (not prime). So invalid.a=12:b*c=3240/12=270.Factor pairs of 270:1*2702*1353*905*546*459*3010*2715*18a=12 corresponds to p=11. So b and c must be >=12.Check b=15, c=18: q=14 (not prime).No valid solutions.a=15:b*c=3240/15=216.Factor pairs of 216:1*2162*1083*724*546*368*279*2412*18a=15 corresponds to p=14 (not prime). So invalid.a=18:b*c=3240/18=180.Factor pairs of 180:1*1802*903*604*455*366*309*2010*1812*15a=18 corresponds to p=17. So b and c must be >=18.Check b=20, c=9: but c must be >=b, so invalid.No valid solutions.a=20:b*c=3240/20=162.Factor pairs of 162:1*1622*813*546*279*18a=20 corresponds to p=19. So b and c must be >=20.Check b=27, c=6: invalid since c < b.No valid solutions.a=24:b*c=3240/24=135.Factor pairs of 135:1*1353*455*279*15a=24 corresponds to p=23. So b and c must be >=24.No valid solutions.a=27:b*c=3240/27=120.Factor pairs of 120:1*1202*603*404*305*246*208*1510*12a=27 corresponds to p=26 (not prime). So invalid.a=30:b*c=3240/30=108.Factor pairs of 108:1*1082*543*364*276*189*12a=30 corresponds to p=29. So b and c must be >=30.Check b=36, c=3: invalid.No valid solutions.a=36:b*c=3240/36=90.Factor pairs of 90:1*902*453*305*186*159*10a=36 corresponds to p=35 (not prime). So invalid.a=40:b*c=3240/40=81.Factor pairs of 81:1*813*279*9a=40 corresponds to p=39 (not prime). So invalid.a=45:b*c=3240/45=72.Factor pairs of 72:1*722*363*244*186*128*9a=45 corresponds to p=44 (not prime). So invalid.a=54:b*c=3240/54=60.Factor pairs of 60:1*602*303*204*155*126*10a=54 corresponds to p=53. So b and c must be >=54.Check b=60, c=1: invalid.No valid solutions.a=60:b*c=3240/60=54.Factor pairs of 54:1*542*273*186*9a=60 corresponds to p=59. So b and c must be >=60.No valid solutions.a=72:b*c=3240/72=45.Factor pairs of 45:1*453*155*9a=72 corresponds to p=71. So b and c must be >=72.No valid solutions.a=81:b*c=3240/81=40.Factor pairs of 40:1*402*204*105*8a=81 corresponds to p=80 (not prime). So invalid.a=90:b*c=3240/90=36.Factor pairs of 36:1*362*183*124*96*6a=90 corresponds to p=89. So b and c must be >=90.No valid solutions.a=108:b*c=3240/108=30.Factor pairs of 30:1*302*153*105*6a=108 corresponds to p=107. So b and c must be >=108.No valid solutions.a=120:b*c=3240/120=27.Factor pairs of 27:1*273*9a=120 corresponds to p=119 (not prime). So invalid.a=135:b*c=3240/135=24.Factor pairs of 24:1*242*123*84*6a=135 corresponds to p=134 (not prime). So invalid.a=162:b*c=3240/162=20.Factor pairs of 20:1*202*104*5a=162 corresponds to p=161 (not prime). So invalid.a=180:b*c=3240/180=18.Factor pairs of 18:1*182*93*6a=180 corresponds to p=179. So b and c must be >=180.No valid solutions.a=216:b*c=3240/216=15.Factor pairs of 15:1*153*5a=216 corresponds to p=215 (not prime). So invalid.a=270:b*c=3240/270=12.Factor pairs of 12:1*122*63*4a=270 corresponds to p=269. So b and c must be >=270.No valid solutions.a=324:b*c=3240/324=10.Factor pairs of 10:1*102*5a=324 corresponds to p=323 (not prime). So invalid.a=405:b*c=3240/405=8.Factor pairs of 8:1*82*4a=405 corresponds to p=404 (not prime). So invalid.a=540:b*c=3240/540=6.Factor pairs of 6:1*62*3a=540 corresponds to p=539 (not prime). So invalid.a=648:b*c=3240/648=5.Factor pairs of 5:1*5a=648 corresponds to p=647. So b and c must be >=648.No valid solutions.a=810:b*c=3240/810=4.Factor pairs of 4:1*42*2a=810 corresponds to p=809. So b and c must be >=810.No valid solutions.a=1080:b*c=3240/1080=3.Factor pairs of 3:1*3a=1080 corresponds to p=1079. So b and c must be >=1080.No valid solutions.a=1620:b*c=3240/1620=2.Factor pairs of 2:1*2a=1620 corresponds to p=1619. So b and c must be >=1620.No valid solutions.a=3240:b*c=3240/3240=1.Factor pairs of 1:1*1a=3240 corresponds to p=3239. So b and c must be >=3240.No valid solutions.So, after checking all possible a values, the only valid solution in Case 3 is when a=3, b=4, c=270, giving n=2*3*269=1614.I should also check if there are any smaller n in other cases, but from the earlier analysis, Cases 1 and 2 didn't yield any valid solutions, so 1614 is the smallest good number.

💡Alright, so I'm trying to tackle this problem about finding either the heaviest or the lightest stone among 10 stones using a special two-tiered balance scale. The stones all have different weights, and their pairwise sums are also distinct. The balance scale allows me to put two stones on each cup and tells me which cup is heavier. First, I need to understand what exactly the two-tiered balance scale does. It seems like it's a bit different from a regular balance scale because it can compare two pairs of stones at the same time. So, if I put two stones on one side and two stones on the other, it will tell me which pair is heavier. This is interesting because it gives me more information with each weighing compared to a regular balance scale, which only compares two items at a time.Since all the stones have different weights and their pairwise sums are distinct, that means if I take any two stones, their combined weight will be unique compared to any other pair. This is important because it ensures that when I use the balance scale, the result will be definitive—there won't be any ties or ambiguities in the comparisons.Okay, so my goal is to find either the heaviest or the lightest stone. Let's think about how I can approach this. Maybe I can start by comparing pairs of stones and using the results to narrow down the possibilities.Let's say I divide the 10 stones into five pairs. I can then weigh each pair against another pair. For example, I could weigh pair A against pair B, pair C against pair D, and so on. Depending on which side is heavier, I can start to eliminate some stones from being the heaviest or lightest.But wait, with 10 stones, dividing them into five pairs and weighing each pair against another might not be the most efficient way. Maybe there's a smarter way to structure the weighings so that I can gather more information each time.Perhaps I can use a tournament-style approach. In a tournament, each match eliminates one participant, and eventually, the winner is determined. Maybe I can apply a similar concept here, where each weighing eliminates some stones from being the heaviest or lightest.Let's consider trying to find the heaviest stone first. If I can find the heaviest stone, then finding the lightest might be a similar process. So, how can I structure the weighings to identify the heaviest stone?If I weigh two stones against two other stones, the heavier pair must contain the heaviest stone. So, if I have pairs (A, B) and (C, D), and (A, B) is heavier than (C, D), then the heaviest stone is either A or B. Similarly, if (C, D) is heavier, then the heaviest stone is either C or D.This seems promising. So, by comparing pairs, I can narrow down the candidates for the heaviest stone. If I continue this process, I can potentially isolate the heaviest stone.But with 10 stones, how many weighings would this take? Let's see. If I start with 10 stones, and in each weighing, I can eliminate half of them, then the number of weighings needed would be logarithmic in the number of stones. Specifically, log base 2 of 10 is approximately 3.32, so I would need at least 4 weighings to find the heaviest stone.But wait, the problem doesn't specify the number of weighings, just that it's possible to find either the heaviest or the lightest stone using this scale. So, maybe I don't need to worry about the exact number of weighings, just that it's possible.Another thought: since all pairwise sums are distinct, maybe I can use this property to ensure that each weighing gives me unique information. This could help in avoiding any ambiguities in the results.Let me try to outline a possible strategy:1. Divide the 10 stones into five pairs.2. Weigh each pair against another pair.3. From the results, determine which pairs contain the heaviest or lightest stones.4. Repeat the process with the narrowed-down candidates until I identify the heaviest or lightest stone.But I need to make sure that this strategy actually works. Let's test it with a smaller number of stones to see if the logic holds.Suppose I have four stones: A, B, C, and D. I want to find the heaviest stone.1. Divide them into two pairs: (A, B) and (C, D).2. Weigh (A, B) against (C, D). - If (A, B) is heavier, then the heaviest stone is either A or B. - If (C, D) is heavier, then the heaviest stone is either C or D.3. Take the heavier pair and weigh the two stones against each other to determine the heaviest.This works for four stones. So, maybe this approach can be scaled up to 10 stones.But with 10 stones, dividing them into pairs and comparing them would require multiple steps. Let's see:1. First, divide the 10 stones into five pairs: (A, B), (C, D), (E, F), (G, H), (I, J).2. Weigh each pair against another pair. For example, weigh (A, B) vs. (C, D), (E, F) vs. (G, H), and leave (I, J) aside for now.3. From the first two weighings, determine which pairs are heavier. - Suppose (A, B) > (C, D) and (E, F) > (G, H). - Then, the heaviest stone is either A, B, E, or F.4. Now, take these four candidates: A, B, E, F.5. Divide them into two pairs: (A, B) and (E, F).6. Weigh them against each other. - If (A, B) > (E, F), then the heaviest is either A or B. - If (E, F) > (A, B), then the heaviest is either E or F.7. Finally, weigh the two candidates against each other to determine the heaviest.This seems to work, but it required multiple weighings. However, the problem doesn't specify the number of weighings, just that it's possible. So, this strategy shows that it's possible to find the heaviest stone.Similarly, to find the lightest stone, I can use a similar approach but look for the lighter pairs instead.But wait, in the initial step, I left out the pair (I, J). What if the heaviest stone is in (I, J)? Then, my strategy would have missed it. So, I need to adjust the strategy to ensure that all pairs are considered.Maybe instead of leaving a pair aside, I should have all pairs weighed in the first round. But with 10 stones, that's five pairs, which would require multiple weighings since the scale can only compare two pairs at a time.Perhaps I need to think of a different approach. Maybe instead of comparing pairs directly, I can use a sorting algorithm that's adapted for this balance scale.In a regular balance scale, a common method to find the heaviest or lightest item is to use a tournament method, where items are compared in pairs, and the winners proceed to the next round. This method can be adapted here since the scale allows comparing two pairs at once.Let's try to outline this:1. Divide the 10 stones into five pairs.2. Weigh each pair against another pair. Since the scale can compare two pairs at once, I can do this in parallel. - For example, weigh (A, B) vs. (C, D) and (E, F) vs. (G, H) simultaneously. - The results will tell me which pairs are heavier.3. From the first set of weighings, identify the heavier pairs. - Suppose (A, B) > (C, D) and (E, F) > (G, H). - So, the heaviest stone is in either (A, B) or (E, F).4. Now, take these four stones: A, B, E, F.5. Divide them into two pairs: (A, E) and (B, F).6. Weigh them against each other. - If (A, E) > (B, F), then the heaviest is either A or E. - If (B, F) > (A, E), then the heaviest is either B or F.7. Finally, weigh the two candidates against each other to determine the heaviest.This approach ensures that all stones are considered, and the heaviest is found. However, it's still a bit unclear how to handle the initial pairs and ensure that no stone is overlooked.Maybe another way is to think about the problem in terms of comparisons and information theory. Each weighing gives me information about the relative weights of the stones. Since all pairwise sums are distinct, each weighing provides unique information that can help me narrow down the possibilities.But I'm not sure how to quantify the amount of information each weighing provides or how to use it optimally. Maybe that's getting too abstract for now.Let's try a different angle. Suppose I want to find the heaviest stone. I can use the balance scale to compare pairs and eliminate stones that cannot be the heaviest.For example, if I weigh two pairs and one pair is heavier, I know that the heaviest stone is in the heavier pair. I can then focus on those two stones and weigh them against each other to find the heaviest.But with 10 stones, I need to structure this process in a way that systematically narrows down the candidates.Maybe I can use a binary search approach. Start by comparing two groups of stones, and based on the result, eliminate half of them. Repeat this process until I find the heaviest or lightest stone.But the balance scale compares two pairs at a time, so I need to adjust the binary search idea to work with pairs instead of individual stones.Alternatively, I can think of the problem as a graph where each stone is a node, and each weighing creates edges indicating which stones are heavier or lighter than others. By building this graph, I can traverse it to find the heaviest or lightest stone.But this might be more complicated than necessary. I need a simpler strategy that can be executed step by step.Wait, maybe I can use the fact that all pairwise sums are distinct to my advantage. If I can find a way to use the sums to identify the heaviest or lightest stone, that would be helpful.For example, if I take a stone and pair it with every other stone, the sums will be unique, so I can determine its relative weight compared to others. But this might require too many weighings.Alternatively, I can use the sums to create a ranking of the stones. If I can determine the order of the sums, I can infer the order of the individual stones.But I'm not sure how to do this efficiently with the balance scale.Maybe I need to go back to the basics. Let's think about what information each weighing gives me.When I weigh two pairs against each other, I get information about which pair is heavier. This tells me that the heavier pair contains at least one stone that is heavier than at least one stone in the lighter pair.But it doesn't tell me exactly which stone is heavier, just that one pair is heavier than the other.So, if I have pairs (A, B) and (C, D), and (A, B) is heavier, I know that either A or B is heavier than at least one of C or D.This is useful because it allows me to eliminate some stones from being the lightest or heaviest.For example, if I'm looking for the heaviest stone, and (A, B) is heavier than (C, D), then I know that the heaviest stone is either A or B, so I can eliminate C and D from being the heaviest.Similarly, if I'm looking for the lightest stone, and (A, B) is heavier than (C, D), then I know that the lightest stone is either C or D, so I can eliminate A and B from being the lightest.This seems like a key insight. By comparing pairs, I can eliminate half of the stones from consideration for being the heaviest or lightest.So, with 10 stones, if I can structure the weighings in such a way that each weighing eliminates half of the remaining stones, I can efficiently narrow down the candidates.Let's try to outline this strategy:1. Start with all 10 stones.2. Divide them into five pairs: (A, B), (C, D), (E, F), (G, H), (I, J).3. Weigh two pairs against each other, say (A, B) vs. (C, D). - If (A, B) is heavier, eliminate C and D from being the heaviest. - If (C, D) is heavier, eliminate A and B from being the heaviest.4. Repeat this process with the remaining pairs.5. After each weighing, eliminate the lighter pair from being the heaviest or the heavier pair from being the lightest, depending on what you're searching for.6. Continue this process until you narrow down to the heaviest or lightest stone.But with 10 stones, this might take several weighings. Let's see:- First weighing: eliminate 2 stones (either C and D or A and B).- Second weighing: eliminate another 2 stones.- Third weighing: eliminate another 2 stones.- Fourth weighing: eliminate another 2 stones.- Fifth weighing: eliminate the last 2 stones.But this seems like it's taking too many weighings. Maybe there's a more efficient way.Wait, in each weighing, I can compare two pairs, so I can eliminate two stones at a time. But with 10 stones, I need to eliminate 9 stones to find the heaviest or lightest. So, theoretically, it would take at least 5 weighings to eliminate 10 stones, but since I'm eliminating two at a time, it would take 5 weighings.But I think I can do better by reusing some information from previous weighings.For example, if I weigh (A, B) vs. (C, D) and find that (A, B) is heavier, I know that the heaviest stone is either A or B. Then, I can weigh (A, C) vs. (B, D). Depending on the result, I can get more information.But this might complicate things because now I'm mixing stones from different pairs.Alternatively, I can use a method similar to a single-elimination tournament, where each stone is compared in pairs, and the winners proceed to the next round.But with the balance scale comparing two pairs at a time, I need to adjust the tournament structure.Maybe I can have multiple matches happening in parallel, which would speed up the process.But I'm not sure how to structure this exactly.Another idea: since all pairwise sums are distinct, I can use this property to ensure that each stone's weight can be uniquely determined through a series of comparisons.But I'm not sure how to leverage this in practice.Wait, maybe I can use the sums to create a ranking. If I can determine the order of the sums, I can infer the order of the individual stones.But this might require too many weighings.Alternatively, I can use the sums to identify the heaviest or lightest stone by comparing them in a certain way.But I'm not sure.Maybe I need to think differently. Let's consider that the heaviest stone will always be part of the heaviest pair, and the lightest stone will always be part of the lightest pair.So, if I can find the heaviest pair, I can then find the heaviest stone within that pair. Similarly, if I can find the lightest pair, I can find the lightest stone within that pair.So, the strategy would be:1. Find the heaviest pair among all pairs.2. Within that pair, find the heaviest stone.3. Similarly, find the lightest pair and then the lightest stone.But how do I find the heaviest pair?I can compare pairs against each other. Since the scale can compare two pairs at a time, I can do this efficiently.For example:1. Divide the 10 stones into five pairs: (A, B), (C, D), (E, F), (G, H), (I, J).2. Weigh (A, B) vs. (C, D). - Suppose (A, B) is heavier.3. Weigh (E, F) vs. (G, H). - Suppose (E, F) is heavier.4. Now, we have two heavy pairs: (A, B) and (E, F).5. Weigh (A, B) vs. (E, F). - Suppose (A, B) is heavier.6. Now, (A, B) is the heaviest pair.7. Weigh A vs. B to find the heaviest stone.This works, but it requires multiple weighings. However, the problem doesn't specify the number of weighings, just that it's possible.Similarly, to find the lightest stone, I can follow the same process but look for the lightest pairs.But wait, in the initial step, I left out the pair (I, J). What if the heaviest pair is (I, J)? Then, my strategy would have missed it.So, I need to ensure that all pairs are considered in the initial weighings.Maybe I can do multiple rounds of comparisons to cover all pairs.For example:1. First round: - Weigh (A, B) vs. (C, D). - Weigh (E, F) vs. (G, H). - Weigh (I, J) vs. a known lighter pair (but I don't have any yet).Hmm, this complicates things because I don't have a reference point.Alternatively, I can have multiple weighings in parallel, comparing different pairs each time, and keep track of the results to determine the heaviest pair.But this might require a more complex strategy.Maybe I can use a method similar to merge sort, where I compare pairs and merge the results to find the heaviest pair.But I'm not sure.Another idea: since all pairwise sums are distinct, I can assign a unique identifier to each pair based on their sum. Then, by comparing pairs, I can determine their order and find the heaviest or lightest pair.But this might not directly help in finding the individual heaviest or lightest stone.Wait, perhaps I can use the sums to create a total order of the pairs, and then use that to infer the order of the individual stones.But this seems too abstract.Maybe I need to simplify. Let's think about what needs to be proven: that using this balance scale, I can either find the heaviest or the lightest stone.I don't need to find both, just one of them. So, maybe I can focus on finding either the heaviest or the lightest, not necessarily both.Given that, maybe I can design a strategy that is guaranteed to find at least one of them.For example, I can try to find the heaviest stone, and if that doesn't work, then find the lightest.But I need a more concrete plan.Let me try to outline a step-by-step strategy to find the heaviest stone:1. Divide the 10 stones into five pairs: (A, B), (C, D), (E, F), (G, H), (I, J).2. Weigh (A, B) vs. (C, D). - If (A, B) is heavier, keep (A, B); otherwise, keep (C, D).3. Weigh (E, F) vs. (G, H). - If (E, F) is heavier, keep (E, F); otherwise, keep (G, H).4. Now, we have two pairs: the winners from step 2 and step 3.5. Weigh these two pairs against each other. - The heavier pair is now the candidate for the heaviest pair.6. Take the two stones from this pair and weigh them against each other to find the heaviest stone.This strategy works, but it only considers four stones in the final steps. What if the heaviest stone is in the pair that wasn't compared in the final step? For example, if the heaviest pair is (I, J), which wasn't involved in the final weighing, then this strategy would fail.So, to ensure that all pairs are considered, I need to have a way to compare all pairs in a way that the heaviest pair is eventually identified.Maybe I can have multiple rounds of comparisons, similar to a tournament bracket, where each pair competes, and the winners proceed to the next round.For example:1. First round: - Weigh (A, B) vs. (C, D). - Weigh (E, F) vs. (G, H). - Weigh (I, J) vs. a placeholder pair (but I don't have any yet).This doesn't solve the problem because I can't leave a pair out.Alternatively, I can have overlapping comparisons, where each stone is part of multiple pairs.But this might complicate the strategy.Wait, maybe I can use a method where I compare pairs in a way that each stone is compared multiple times, allowing me to gather more information about their relative weights.But I'm not sure how to structure this.Another idea: since all pairwise sums are distinct, I can use this property to ensure that each stone's weight can be uniquely determined through a series of comparisons.But I'm not sure how to leverage this in practice.Maybe I need to think about the problem differently. Instead of trying to find the heaviest or lightest stone directly, I can try to establish a partial order among the stones based on the comparisons, and then use that to identify the heaviest or lightest.But this might require more advanced techniques.Wait, perhaps I can use the fact that in each weighing, I can compare two pairs and get information about their relative weights. By strategically choosing which pairs to compare, I can gather enough information to determine the heaviest or lightest stone.For example, if I can arrange the weighings so that each stone is compared in multiple pairs, I can build a network of comparisons that allows me to deduce the heaviest or lightest stone.But this seems too vague.Maybe I need to look for a known algorithm or method that can be applied here. I recall that in sorting algorithms, certain methods can find the maximum or minimum element with a specific number of comparisons.In this case, since the balance scale allows comparing two pairs at a time, it's like having a parallel comparison capability, which can potentially reduce the number of weighings needed.But I'm not sure about the exact number.Wait, in a regular balance scale, finding the maximum or minimum requires n-1 comparisons. With 10 stones, that would be 9 weighings. But since this scale can compare two pairs at a time, maybe it can be done faster.But again, the problem doesn't specify the number of weighings, just that it's possible.So, maybe I can use a method that ensures that after a certain number of weighings, I can identify the heaviest or lightest stone.But I need a concrete strategy.Let me try to outline a possible method:1. Divide the 10 stones into five pairs: (A, B), (C, D), (E, F), (G, H), (I, J).2. Weigh (A, B) vs. (C, D). - If (A, B) is heavier, keep (A, B); otherwise, keep (C, D).3. Weigh (E, F) vs. (G, H). - If (E, F) is heavier, keep (E, F); otherwise, keep (G, H).4. Now, we have two pairs: the winners from step 2 and step 3.5. Weigh these two pairs against each other. - The heavier pair is now the candidate for the heaviest pair.6. Take the two stones from this pair and weigh them against each other to find the heaviest stone.7. If the heaviest stone is found, we're done.8. If not, repeat the process focusing on the remaining stones to find the lightest.But this still leaves out the pair (I, J) in the initial steps, so it might not work.Alternatively, maybe I can have multiple rounds where each stone is part of multiple pairs, ensuring that all stones are considered.For example:1. First round: - Weigh (A, B) vs. (C, D). - Weigh (E, F) vs. (G, H). - Weigh (I, J) vs. (A, B).2. From the first weighing, determine the heavier pair between (A, B) and (C, D).3. From the second weighing, determine the heavier pair between (E, F) and (G, H).4. From the third weighing, determine the heavier pair between (I, J) and (A, B).5. Now, take the winners from each of these comparisons and weigh them against each other to find the heaviest pair.6. Finally, weigh the two stones in the heaviest pair against each other to find the heaviest stone.This way, all stones are considered in the initial weighings, and the heaviest pair is eventually identified.But this requires multiple weighings, and I'm not sure if it's the most efficient.However, the problem only asks to prove that it's possible to find either the heaviest or the lightest stone using this scale, not necessarily to find it in the fewest number of weighings.So, even if it takes several weighings, as long as there's a strategy that guarantees finding the heaviest or lightest stone, the proof is complete.Therefore, by systematically comparing pairs and eliminating stones based on the results, I can ensure that I eventually find the heaviest or lightest stone.In summary, the strategy involves:1. Comparing pairs of stones to determine which pair is heavier or lighter.2. Using the results to eliminate stones that cannot be the heaviest or lightest.3. Repeating this process with the remaining stones until the heaviest or lightest stone is identified.Since all pairwise sums are distinct, each comparison provides unique information that helps narrow down the candidates. By following this method, I can guarantee that either the heaviest or the lightest stone will be found.

💡First, I need to determine the percentage the student received in the first subject. Let's denote this percentage as ( P ).The overall percentage is calculated as the average of the percentages in the three subjects. The student aims for an overall percentage of 75%, and we know the percentages in the other two subjects: 80% and 75%.I can set up the equation for the overall percentage as follows:[frac{P + 80 + 75}{3} = 75]Next, I'll solve for ( P ). Multiplying both sides of the equation by 3 to eliminate the denominator:[P + 80 + 75 = 225]Combining the known percentages:[P + 155 = 225]Finally, subtracting 155 from both sides to find ( P ):[P = 70]Therefore, the student received 70% in the first subject.