💡Okay, so I have this problem where there's a segment OA. From point A, five segments AB₁, AB₂, AB₃, AB₄, AB₅ emanate. From each point Bᵢ, five new segments can either emanate or none at all, and this process continues. The question is asking if the number of free ends of the constructed segments can equal 1001. A free end is defined as a point belonging to only one segment, except for point O.Alright, let me try to break this down. First, I need to understand what exactly a free end is. It's a point that is connected to only one segment. So, for example, initially, we have segment OA. So, points O and A are both free ends because each is connected to only one segment.Then, from point A, we draw five segments to points B₁ through B₅. Now, point A is no longer a free end because it's connected to five segments. However, each of the points B₁ through B₅ becomes a new free end because each is connected to only one segment (the one from A). So, after this first step, we have five new free ends.Now, the problem states that from each point Bᵢ, five new segments can either emanate or none at all. So, for each Bᵢ, we have a choice: we can either draw five new segments from it, or we can choose not to draw any. If we choose to draw five new segments from a Bᵢ, then that Bᵢ will no longer be a free end, and each of the new points connected to Bᵢ will become free ends.The question is whether we can arrange this process such that the total number of free ends is exactly 1001. So, I need to figure out if it's possible to have 1001 free ends by appropriately choosing whether to draw five segments from each point or not.Let me think about how the number of free ends changes as we add more segments. Initially, we have two free ends: O and A. When we draw five segments from A, we lose one free end (A) and gain five new free ends (B₁ through B₅). So, the total number of free ends becomes 2 - 1 + 5 = 6.If we then choose to draw five segments from each of the Bᵢ points, each Bᵢ will lose its status as a free end and each will add five new free ends. So, for each Bᵢ we choose to draw segments from, we lose one free end and gain five, resulting in a net gain of four free ends per Bᵢ.Wait, so if we have five Bᵢ points, and we choose to draw segments from all of them, we would lose five free ends (the Bᵢs) and gain 25 new free ends (five from each Bᵢ). So, the total number of free ends would become 6 - 5 + 25 = 26.Alternatively, if we choose not to draw any segments from the Bᵢs, the number of free ends remains at 6.So, it seems that each time we decide to draw segments from a point, we can increase the number of free ends by four for each segment we draw from that point. But in this case, we have the option to draw five segments from each point, which would increase the number of free ends by four per point.Wait, actually, if we draw five segments from a point, we lose one free end (the point itself) and gain five new free ends, so the net gain is four. So, for each point we choose to draw segments from, we can increase the number of free ends by four.Therefore, if we have n points from which we choose to draw five segments, the total number of free ends would increase by 4n.Starting from the initial two free ends, after the first step, we have six free ends. Then, if we choose to draw segments from k of the Bᵢ points, the number of free ends becomes 6 + 4k.But wait, actually, after the first step, we have five Bᵢ points, each of which can be a source for drawing five segments. So, if we choose to draw from all five, we get 6 + 4*5 = 26 free ends. If we choose to draw from only some of them, say k, then we get 6 + 4k.But in the problem, it says "from each point Bᵢ, five new segments can either emanate or none at all, and so on." So, it seems that for each point, we can choose independently whether to draw five segments from it or not. So, the number of free ends can be controlled by how many points we choose to draw from.But the problem is asking if we can get exactly 1001 free ends. So, starting from two free ends, and then each time we draw from a point, we add four free ends.Wait, but actually, the initial step is fixed: we have OA, then from A, we draw five segments, resulting in six free ends. Then, from each Bᵢ, we can choose to draw five segments or not. So, the number of free ends can be controlled by how many Bᵢs we choose to draw from.But actually, it's not just the Bᵢs; each time we draw from a point, that point is no longer a free end, but each of its five new connections becomes a free end. So, each time we draw from a point, we lose one free end and gain five, so net gain of four.Therefore, each time we draw from a point, we can increase the number of free ends by four. So, if we start with two free ends, and then after drawing from A, we have six free ends. Then, if we draw from k points among the Bᵢs, we get 6 + 4k free ends.But wait, actually, after drawing from A, we have five Bᵢs, each of which can be a source for drawing five segments. So, if we draw from all five, we get 6 + 4*5 = 26. If we draw from four, we get 6 + 4*4 = 22, and so on.But the problem is asking if we can get 1001 free ends. So, starting from two, then six, then 6 + 4k. So, 6 + 4k = 1001. Solving for k, we get 4k = 995, so k = 995/4 = 248.75. But k has to be an integer, so that's not possible.Wait, but maybe I'm missing something. Because each time we draw from a point, we can create more points from which we can draw again. So, it's not just one level deep; it's a tree-like structure where each point can have up to five children, and so on.So, perhaps the number of free ends can be increased in a more complex way, not just by adding four per point at each level.Let me think recursively. Let's denote f(n) as the number of free ends after n steps or something. But maybe it's better to model it as a tree.Each time we draw from a point, we replace that point (which was a free end) with five new free ends. So, each such operation increases the number of free ends by four.But if we have multiple points from which we can draw, each draw operation adds four free ends.So, starting from two free ends, after the first draw from A, we have six free ends.Then, from each of the five Bᵢs, we can choose to draw, each adding four free ends. So, if we draw from all five, we get 6 + 5*4 = 26.Then, from each of the 25 new points (since each Bᵢ draws five), we can choose to draw again, each adding four free ends. So, if we draw from all 25, we get 26 + 25*4 = 126.Continuing this way, each time we can choose to draw from all the current free ends, each adding four free ends.But the problem is, we don't have to draw from all points; we can choose to draw from some and not others. So, the number of free ends can be controlled by how many points we choose to draw from at each level.But the key is that each draw operation adds four free ends. So, the total number of free ends can be expressed as 2 + 4k, where k is the number of draw operations.Wait, but initially, we have two free ends. Then, after drawing from A, we have six. So, that's an increase of four, which is 2 + 4*1 = 6.Then, if we draw from k points among the Bᵢs, we get 6 + 4k.Similarly, if we draw from k points at the next level, we get 6 + 4k + 4*(number of new points drawn from).But actually, each time we draw from a point, we add four free ends, regardless of the level.So, the total number of free ends is 2 + 4k, where k is the number of times we've drawn from a point.But wait, in the first step, we drew from A, so k=1, giving us 6 free ends.Then, if we draw from k=5 points (all Bᵢs), we get 6 + 4*5 = 26.Then, if we draw from k=25 points (all new points), we get 26 + 4*25 = 126.Continuing this, each time we draw from all possible points, the number of free ends increases by 4 times the number of points drawn from.But the problem is, we don't have to draw from all points. We can choose to draw from any number of points at each level, which allows us to control the total number of free ends.So, the total number of free ends can be expressed as 2 + 4k, where k is the total number of draw operations performed.But wait, in the first step, we drew from A, which is one draw operation, giving us 6 free ends.Then, if we draw from k=5 points, that's five draw operations, giving us 6 + 4*5 = 26.Then, if we draw from k=25 points, that's 25 draw operations, giving us 26 + 4*25 = 126.So, the total number of draw operations is 1 + 5 + 25 + ... etc.But the total number of free ends is 2 + 4*(total draw operations).So, if we want 1001 free ends, we need 2 + 4k = 1001, so 4k = 999, which is not an integer. So, k would have to be 999/4 = 249.75, which is not possible because k must be an integer.Wait, but maybe I'm missing something. Because each draw operation adds four free ends, but the initial two free ends are O and A. After the first draw from A, we have six free ends. So, the formula is actually 2 + 4k, where k is the number of draw operations.So, 2 + 4k = 1001 => 4k = 999 => k = 249.75, which is not an integer. Therefore, it's impossible to have exactly 1001 free ends.But wait, maybe I'm miscalculating. Let's check:After the first draw from A: 2 + 4*1 = 6.After drawing from k points: 6 + 4k.So, if we want 1001, we have 6 + 4k = 1001 => 4k = 995 => k = 248.75.Again, not an integer.Wait, but maybe we can have multiple levels of drawing. For example, after the first draw from A, we have five Bᵢs. If we draw from some of them, say m, then we have 6 + 4m free ends.Then, from those m points, we can draw again, adding 4m free ends, and so on.So, perhaps the total number of free ends can be expressed as 2 + 4*(number of draw operations).But since each draw operation adds four free ends, the total number of free ends must be congruent to 2 modulo 4.Let's check 1001 modulo 4: 1001 / 4 = 250 with a remainder of 1. So, 1001 ≡ 1 mod 4.But our formula is 2 + 4k, which is 2 mod 4. So, 1001 is 1 mod 4, which is different from 2 mod 4. Therefore, it's impossible to have 1001 free ends.Wait, that makes sense. Because every time we add four free ends, the total number of free ends increases by four, so it remains congruent to 2 mod 4. Since 1001 is 1 mod 4, it's impossible.Therefore, the answer is no, it's not possible to have exactly 1001 free ends.But wait, let me double-check. Maybe I'm missing a step.Starting from OA: 2 free ends.After drawing from A: 6 free ends (2 + 4*1).After drawing from k points: 6 + 4k.So, total free ends is 2 + 4*(1 + k).Wait, no, because the first draw is from A, which is one operation, then k operations from Bᵢs.So, total free ends is 2 + 4*(1 + k).So, 2 + 4*(1 + k) = 1001 => 4*(1 + k) = 999 => 1 + k = 249.75, which is still not an integer.Therefore, it's impossible.Alternatively, maybe I'm miscounting the initial steps.Wait, initially, OA has two free ends: O and A.After drawing five segments from A, A is no longer a free end, but we have five new free ends: B₁ through B₅. So, total free ends: 1 (O) + 5 (Bᵢs) = 6.So, that's correct.Then, from each Bᵢ, we can draw five segments, each time replacing a free end with five new ones, adding four each time.So, if we draw from k Bᵢs, we add 4k free ends, making the total 6 + 4k.Then, from those new points, we can draw again, adding 4*(number of new points drawn from).But regardless, the total number of free ends will always be 2 + 4*(number of draw operations).Since 1001 is not congruent to 2 mod 4, it's impossible.Therefore, the answer is no, it's not possible to have exactly 1001 free ends.

💡Okay, so I have this math problem here: evaluate 4x^y + 5y^x when x equals 3 and y equals 2. Hmm, let's see. I think I need to substitute the values of x and y into the expression and then calculate the result step by step. First, I should identify what each part of the expression means. The expression is 4x^y + 5y^x. That means I have two terms: 4 times x raised to the power of y, and 5 times y raised to the power of x. So, substituting x = 3 and y = 2, the expression becomes 4*(3)^2 + 5*(2)^3. Wait, let me make sure I got that right. So, x is 3 and y is 2. So, in the first term, it's 4 times 3 squared, and in the second term, it's 5 times 2 cubed. Yeah, that seems correct. Now, I need to calculate each part separately. Let's start with the first term: 4*(3)^2. Okay, so 3 squared is 3 times 3, which is 9. Then, multiplying that by 4 gives me 4*9. Let me do that multiplication: 4 times 9 is 36. So, the first term is 36. Next, the second term: 5*(2)^3. Hmm, 2 cubed is 2 times 2 times 2, which is 8. Then, multiplying that by 5 gives me 5*8. Let me calculate that: 5 times 8 is 40. So, the second term is 40. Now, I have both terms calculated: 36 and 40. I need to add them together to get the final result. So, 36 plus 40. Let me add those: 36 plus 40 is 76. Wait a second, did I do that right? Let me double-check. For the first term, 3 squared is 9, times 4 is 36. For the second term, 2 cubed is 8, times 5 is 40. Adding 36 and 40 gives 76. Yeah, that seems correct. I guess I could also check my calculations another way. Maybe by breaking it down differently. Let's see, 4*(3)^2 is the same as 4*9, which is 36. And 5*(2)^3 is 5*8, which is 40. Adding those together, 36 + 40, definitely equals 76. I think I'm confident that the result is 76. I don't see any mistakes in my calculations. So, the value of the expression when x is 3 and y is 2 is 76.

💡First, I need to find the smallest positive integer that is a multiple of both 9 and 6.To do this, I'll list the multiples of each number.The multiples of 9 are: 9, 18, 27, 36, 45, and so on.The multiples of 6 are: 6, 12, 18, 24, 30, 36, and so on.Next, I'll identify the common multiples from both lists. The common multiples are 18, 36, etc.Among these, the smallest positive integer is 18.Therefore, the smallest positive integer that is both a multiple of 9 and 6 is 18.

💡Okay, so I have this problem where I'm given an integer ( n > 1 ) and a set of distinct non-negative real numbers ( a_1, a_2, ldots, a_n ). I need to define two sets ( A ) and ( B ) based on these numbers and then find the minimum value of the ratio ( frac{|A|}{|B|} ), where ( |X| ) denotes the number of elements in set ( X ).First, let me make sure I understand the definitions of sets ( A ) and ( B ). Set ( A ) consists of all possible sums ( a_i + a_j ) where ( 1 leq i leq j leq n ). Similarly, set ( B ) consists of all possible products ( a_i a_j ) under the same conditions. Since ( a_1, a_2, ldots, a_n ) are distinct and non-negative, both ( A ) and ( B ) will have a certain number of elements, and I need to find the minimum ratio of their sizes.I think the first step is to figure out the possible sizes of ( A ) and ( B ). Let me start by considering the size of ( A ). Since ( A ) is the set of all sums ( a_i + a_j ) with ( i leq j ), the number of such sums is ( frac{n(n+1)}{2} ). However, because the numbers ( a_i ) are distinct, some of these sums might be equal, so the actual size ( |A| ) could be less than ( frac{n(n+1)}{2} ). Similarly, for set ( B ), the number of products ( a_i a_j ) is also ( frac{n(n+1)}{2} ), but again, some products might be equal, so ( |B| ) could be less than that.But wait, the problem asks for the minimum value of ( frac{|A|}{|B|} ). So, I need to arrange the numbers ( a_1, a_2, ldots, a_n ) such that ( |A| ) is as small as possible and ( |B| ) is as large as possible, or at least the ratio is minimized.Let me think about how to minimize ( |A| ) and maximize ( |B| ). For ( |A| ) to be minimized, the sums ( a_i + a_j ) should overlap as much as possible. On the other hand, for ( |B| ) to be maximized, the products ( a_i a_j ) should be as distinct as possible.Hmm, so perhaps I need a set of numbers where the sums are not too many, but the products are as many as possible. Maybe choosing numbers that are in a geometric progression? Because in a geometric progression, the products can be more spread out, whereas the sums might not be as spread out.Wait, let me test this idea with a small ( n ). Let's say ( n = 2 ). Then, ( A ) would have elements ( a_1 + a_1, a_1 + a_2, a_2 + a_2 ), so three elements. ( B ) would have ( a_1 a_1, a_1 a_2, a_2 a_2 ), also three elements. So, the ratio is 1. Is that the minimum? It seems so because both sets have the same number of elements, and you can't get a smaller ratio than 1.But wait, if ( a_1 = 0 ) and ( a_2 = 1 ), then ( A = {0, 1, 2} ) and ( B = {0, 0, 1} ). So, ( |A| = 3 ) and ( |B| = 2 ), giving a ratio of ( 3/2 ). That's worse. So, maybe choosing ( a_1 ) and ( a_2 ) as 1 and 2, then ( A = {2, 3, 4} ) and ( B = {1, 2, 4} ). So, ( |A| = 3 ) and ( |B| = 3 ), ratio is 1. So, that's better.Wait, so maybe for ( n = 2 ), the minimum ratio is 1. But in the case where ( a_1 = 0 ), the ratio is worse. So, perhaps choosing numbers that are positive and in a way that their sums don't overlap too much, but their products are as distinct as possible.Wait, for ( n = 3 ), let's see. Suppose ( a_1 = 1, a_2 = 2, a_3 = 4 ). Then, ( A ) would be ( {2, 3, 4, 5, 6, 8} ). So, ( |A| = 6 ). ( B ) would be ( {1, 2, 4, 4, 8, 16} ). So, ( |B| = 5 ) because 4 is repeated. So, the ratio is ( 6/5 = 1.2 ). Is that the minimum?Alternatively, if I choose ( a_1 = 1, a_2 = 3, a_3 = 4 ). Then, ( A = {2, 4, 5, 6, 7, 8} ), so ( |A| = 6 ). ( B = {1, 3, 4, 9, 12, 16} ), so ( |B| = 6 ). So, the ratio is 1. That's better. So, in this case, the ratio is 1.Wait, so maybe for ( n = 3 ), the minimum ratio is 1. So, perhaps for any ( n ), the minimum ratio is 1? But that seems too optimistic.Wait, let me think again. For ( n = 2 ), it's 1. For ( n = 3 ), it's 1. Maybe for larger ( n ), it's also 1? But that seems unlikely because as ( n ) increases, the number of sums and products both increase, but perhaps their rates of increase differ.Wait, let me think about the general case. If I can choose the numbers such that all sums ( a_i + a_j ) are distinct and all products ( a_i a_j ) are also distinct, then ( |A| = |B| = frac{n(n+1)}{2} ), so the ratio is 1. But is that possible?Wait, for ( n = 2 ), it's possible. For ( n = 3 ), is it possible? Let me try. Let me choose ( a_1 = 1, a_2 = 2, a_3 = 4 ). Then, ( A = {2, 3, 4, 5, 6, 8} ), which has 6 elements, all distinct. ( B = {1, 2, 4, 4, 8, 16} ). Wait, here, ( a_2 a_2 = 4 ) and ( a_1 a_3 = 4 ), so they are equal. So, ( |B| = 5 ). So, not all products are distinct.Hmm, so maybe it's not possible for ( n = 3 ) to have all products distinct. So, perhaps the ratio can't be 1 for ( n = 3 ). Then, maybe the minimum ratio is higher.Wait, but earlier, when I chose ( a_1 = 1, a_2 = 3, a_3 = 4 ), I had ( |A| = 6 ) and ( |B| = 6 ). So, in that case, the ratio was 1. So, maybe it's possible for ( n = 3 ) as well.Wait, let me check that again. ( a_1 = 1, a_2 = 3, a_3 = 4 ). Then, ( A ) is ( 1+1=2, 1+3=4, 1+4=5, 3+3=6, 3+4=7, 4+4=8 ). So, ( A = {2,4,5,6,7,8} ), which has 6 elements. ( B ) is ( 1*1=1, 1*3=3, 1*4=4, 3*3=9, 3*4=12, 4*4=16 ). So, ( B = {1,3,4,9,12,16} ), which also has 6 elements. So, indeed, the ratio is 1.So, for ( n = 3 ), it's possible to have ( |A| = |B| ). So, maybe for any ( n ), it's possible to choose numbers such that ( |A| = |B| ), making the ratio 1.But wait, let me think about ( n = 4 ). Let's try to choose numbers such that all sums and products are distinct. Let me try ( a_1 = 1, a_2 = 2, a_3 = 4, a_4 = 8 ). Then, ( A ) would be all sums from ( 2 ) to ( 16 ), but let's see:( A = {2, 3, 4, 5, 6, 8, 9, 10, 12, 16} ). Wait, that's 10 elements. ( B = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512} ). Wait, no, actually, for ( n = 4 ), the number of elements in ( A ) and ( B ) would be ( frac{4*5}{2} = 10 ). But in this case, ( A ) has 10 elements, and ( B ) also has 10 elements because all products are distinct. Wait, is that true?Wait, let me compute ( B ) properly. ( a_1 = 1, a_2 = 2, a_3 = 4, a_4 = 8 ). So, ( B ) would be:- ( 1*1 = 1 )- ( 1*2 = 2 )- ( 1*4 = 4 )- ( 1*8 = 8 )- ( 2*2 = 4 )- ( 2*4 = 8 )- ( 2*8 = 16 )- ( 4*4 = 16 )- ( 4*8 = 32 )- ( 8*8 = 64 )Wait, so ( B = {1, 2, 4, 8, 16, 32, 64} ). So, ( |B| = 7 ), not 10. So, in this case, ( |A| = 10 ) and ( |B| = 7 ), so the ratio is ( 10/7 approx 1.428 ).But earlier, for ( n = 3 ), I could get ( |A| = |B| = 6 ). So, maybe for ( n = 4 ), I can choose numbers differently to make ( |A| = |B| ).Wait, let me try another set. Maybe ( a_1 = 1, a_2 = 3, a_3 = 4, a_4 = 5 ). Let's compute ( A ) and ( B ).First, ( A ):- ( 1+1 = 2 )- ( 1+3 = 4 )- ( 1+4 = 5 )- ( 1+5 = 6 )- ( 3+3 = 6 )- ( 3+4 = 7 )- ( 3+5 = 8 )- ( 4+4 = 8 )- ( 4+5 = 9 )- ( 5+5 = 10 )So, ( A = {2,4,5,6,7,8,9,10} ). Wait, that's 8 elements. But ( frac{4*5}{2} = 10 ), so we have overlaps. So, ( |A| = 8 ).Now, ( B ):- ( 1*1 = 1 )- ( 1*3 = 3 )- ( 1*4 = 4 )- ( 1*5 = 5 )- ( 3*3 = 9 )- ( 3*4 = 12 )- ( 3*5 = 15 )- ( 4*4 = 16 )- ( 4*5 = 20 )- ( 5*5 = 25 )So, ( B = {1,3,4,5,9,12,15,16,20,25} ), which has 10 elements. So, ( |A| = 8 ) and ( |B| = 10 ), so the ratio is ( 8/10 = 0.8 ). Wait, that's less than 1. But the problem asks for the minimum value of ( frac{|A|}{|B|} ). So, is 0.8 possible?Wait, but in this case, ( |A| = 8 ) and ( |B| = 10 ), so the ratio is 0.8, which is less than 1. But earlier, for ( n = 3 ), I had a ratio of 1. So, maybe for ( n = 4 ), the minimum ratio is less than 1.But wait, the problem says "Find the minimum value of ( frac{|A|}{|B|} )". So, if I can get a ratio less than 1, that would be better. But is that possible?Wait, but in the case above, ( |A| = 8 ) and ( |B| = 10 ), so the ratio is 0.8. But is that the minimum? Or can I get even lower?Wait, let me try another set. Maybe ( a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 4 ). Then, ( A ) would be:- ( 1+1 = 2 )- ( 1+2 = 3 )- ( 1+3 = 4 )- ( 1+4 = 5 )- ( 2+2 = 4 )- ( 2+3 = 5 )- ( 2+4 = 6 )- ( 3+3 = 6 )- ( 3+4 = 7 )- ( 4+4 = 8 )So, ( A = {2,3,4,5,6,7,8} ), which has 7 elements. ( B ):- ( 1*1 = 1 )- ( 1*2 = 2 )- ( 1*3 = 3 )- ( 1*4 = 4 )- ( 2*2 = 4 )- ( 2*3 = 6 )- ( 2*4 = 8 )- ( 3*3 = 9 )- ( 3*4 = 12 )- ( 4*4 = 16 )So, ( B = {1,2,3,4,6,8,9,12,16} ), which has 9 elements. So, ( |A| = 7 ), ( |B| = 9 ), ratio ( 7/9 approx 0.777 ). So, that's even lower.Wait, so maybe the ratio can be as low as ( frac{2n - 1}{frac{n(n+1)}{2}} ). Let me check for ( n = 4 ). ( 2n - 1 = 7 ), ( frac{n(n+1)}{2} = 10 ). So, ( 7/10 = 0.7 ). But in my example above, I got ( 7/9 approx 0.777 ), which is higher than 0.7. So, maybe the minimum ratio is ( frac{2n - 1}{frac{n(n+1)}{2}} ).Wait, let me think about this formula. For ( n = 2 ), ( 2n - 1 = 3 ), ( frac{n(n+1)}{2} = 3 ), so ratio 1. For ( n = 3 ), ( 2n - 1 = 5 ), ( frac{n(n+1)}{2} = 6 ), so ratio ( 5/6 approx 0.833 ). But earlier, I had a ratio of 1 for ( n = 3 ). So, maybe this formula is not tight.Wait, perhaps I need to think differently. Let me consider arranging the numbers such that the sums are as few as possible, while the products are as many as possible. So, to minimize ( |A| ), I need the sums to overlap as much as possible, but to maximize ( |B| ), I need the products to be as distinct as possible.Wait, but how can I arrange the numbers to have both? It seems conflicting because if I arrange numbers to have overlapping sums, maybe their products would also overlap, and vice versa.Wait, perhaps choosing numbers in a geometric progression. Let me try that. For example, ( a_i = r^{i} ) for some ( r > 1 ). Then, the sums ( a_i + a_j ) would be ( r^i + r^j ), and the products would be ( r^{i+j} ). Since ( r ) is a base, the products would be unique because ( i + j ) would be unique for each pair ( (i, j) ) with ( i leq j ). Wait, is that true?Wait, no, because ( i + j ) can be the same for different pairs. For example, ( i = 1, j = 3 ) and ( i = 2, j = 2 ) both give ( i + j = 4 ). So, their products would be ( r^4 ) in both cases, so they would not be unique. Therefore, the products would not all be distinct.Hmm, so maybe a geometric progression isn't the way to go. What if I choose the numbers to be powers of 2? Let me try ( a_i = 2^{i} ). Then, the sums ( a_i + a_j ) would be ( 2^i + 2^j ), and the products would be ( 2^{i+j} ). Again, the products would not be unique because different pairs can have the same ( i + j ).Wait, maybe I need a different approach. Let me think about the minimal number of sums and maximal number of products.I recall that in additive combinatorics, the minimal number of sums is achieved when the set is an arithmetic progression. So, if I arrange the numbers in an arithmetic progression, the number of distinct sums ( |A| ) is minimized. On the other hand, for the products, if the numbers are in a geometric progression, the number of distinct products ( |B| ) is maximized.But wait, if I arrange the numbers in an arithmetic progression, will the products ( a_i a_j ) be as distinct as possible? Probably not, because in an arithmetic progression, the products can still overlap.Alternatively, maybe I can arrange the numbers such that both the sums and products have certain properties. But I'm not sure.Wait, let me think about the minimal ( |A| ). For a set of numbers, the minimal number of distinct sums is achieved when the set is an arithmetic progression. For example, if ( a_i = a + (i-1)d ), then the sums ( a_i + a_j ) will form another arithmetic progression with difference ( d ), and the number of distinct sums is ( 2n - 1 ).So, ( |A| geq 2n - 1 ). This is a known result in additive number theory. So, the minimal possible ( |A| ) is ( 2n - 1 ).Now, for ( |B| ), the maximum number of distinct products is ( frac{n(n+1)}{2} ), which occurs when all products ( a_i a_j ) are distinct. So, if I can arrange the numbers such that all products are distinct, then ( |B| = frac{n(n+1)}{2} ).Therefore, the minimal ratio ( frac{|A|}{|B|} ) would be ( frac{2n - 1}{frac{n(n+1)}{2}} = frac{2(2n - 1)}{n(n+1)} ).But wait, is it possible to have both ( |A| = 2n - 1 ) and ( |B| = frac{n(n+1)}{2} ) simultaneously? Because if I arrange the numbers in an arithmetic progression, the products might not all be distinct.Wait, let me test this with ( n = 3 ). If I choose ( a_1 = 1, a_2 = 2, a_3 = 3 ), which is an arithmetic progression. Then, ( A = {2, 3, 4, 5, 6} ), so ( |A| = 5 ). ( B = {1, 2, 3, 4, 6, 9} ), so ( |B| = 6 ). So, the ratio is ( 5/6 approx 0.833 ). But earlier, I had a ratio of 1 for ( n = 3 ) with a different set. So, maybe the minimal ratio is indeed ( frac{2n - 1}{frac{n(n+1)}{2}} ), but it's not achievable for all ( n ).Wait, but in the case of ( n = 3 ), if I choose ( a_1 = 1, a_2 = 3, a_3 = 4 ), I get ( |A| = 6 ) and ( |B| = 6 ), so the ratio is 1. So, in that case, the ratio is higher than the theoretical minimum ( frac{5}{6} ). So, maybe the minimal ratio is indeed ( frac{2n - 1}{frac{n(n+1)}{2}} ), but it's not always achievable.Wait, but how can I arrange the numbers such that ( |A| = 2n - 1 ) and ( |B| = frac{n(n+1)}{2} )? It seems challenging because arranging for minimal sums often leads to overlapping products.Wait, maybe I need to choose the numbers such that they are both in an arithmetic progression for sums and in a geometric progression for products. But I don't think that's possible unless the numbers are trivial.Alternatively, perhaps choosing numbers that are both in an arithmetic progression and a geometric progression, which would mean they are constant, but they need to be distinct, so that's not possible.Wait, maybe I need to choose numbers such that the sums are minimal and the products are maximal. Let me think about choosing numbers that are powers of a prime number. For example, let ( a_i = p^{i} ) for some prime ( p ). Then, the sums ( a_i + a_j ) would be ( p^i + p^j ), and the products would be ( p^{i+j} ). Now, the products are unique because ( i + j ) is unique for each pair ( (i, j) ) with ( i leq j ). Wait, no, because ( i + j ) can be the same for different pairs. For example, ( i = 1, j = 3 ) and ( i = 2, j = 2 ) both give ( i + j = 4 ). So, their products would be ( p^4 ) in both cases, so they are equal. Therefore, the products are not all distinct.Hmm, so maybe that approach doesn't work. What if I choose numbers such that ( a_i = 2^{2^{i}} )? Then, the products ( a_i a_j = 2^{2^{i} + 2^{j}} ), which are unique because ( 2^{i} + 2^{j} ) is unique for each pair ( (i, j) ). Wait, is that true?Let me check for ( i = 1, j = 2 ) and ( i = 1, j = 3 ). ( 2^{1} + 2^{2} = 2 + 4 = 6 ), and ( 2^{1} + 2^{3} = 2 + 8 = 10 ). So, different. Similarly, ( i = 2, j = 2 ) gives ( 4 + 4 = 8 ), which is different from others. So, in this case, all products ( a_i a_j ) would be distinct because ( 2^{2^{i}} times 2^{2^{j}} = 2^{2^{i} + 2^{j}} ), and since ( 2^{i} + 2^{j} ) is unique for each pair ( (i, j) ), the products are unique.So, in this case, ( |B| = frac{n(n+1)}{2} ). Now, what about ( |A| )? The sums ( a_i + a_j = 2^{2^{i}} + 2^{2^{j}} ). Are these sums unique?Wait, let's see. For ( i = 1, j = 2 ), the sum is ( 2 + 4 = 6 ). For ( i = 1, j = 3 ), it's ( 2 + 16 = 18 ). For ( i = 2, j = 2 ), it's ( 4 + 4 = 8 ). For ( i = 2, j = 3 ), it's ( 4 + 16 = 20 ). For ( i = 3, j = 3 ), it's ( 16 + 16 = 32 ). So, all these sums are unique. So, in this case, ( |A| = frac{n(n+1)}{2} ). Wait, but that's the maximum possible ( |A| ), not the minimum.But we want to minimize ( |A| ) while maximizing ( |B| ). So, in this case, both ( |A| ) and ( |B| ) are maximized, giving a ratio of 1. But we want the minimal ratio, which would be when ( |A| ) is as small as possible and ( |B| ) is as large as possible.Wait, so maybe this approach isn't helpful for minimizing the ratio. Instead, perhaps I need to find a set where ( |A| ) is minimal and ( |B| ) is maximal.Wait, going back to the earlier idea, if ( |A| geq 2n - 1 ) and ( |B| leq frac{n(n+1)}{2} ), then the minimal ratio is ( frac{2n - 1}{frac{n(n+1)}{2}} = frac{2(2n - 1)}{n(n+1)} ).But is this achievable? Let me see for ( n = 2 ), it gives ( frac{2(3)}{2*3} = 1 ), which is achievable. For ( n = 3 ), it gives ( frac{2(5)}{3*4} = frac{10}{12} = frac{5}{6} approx 0.833 ). But earlier, I found a set where the ratio was 1, but maybe there's a set where the ratio is ( 5/6 ).Wait, let me try ( n = 3 ). Let me choose ( a_1 = 1, a_2 = 2, a_3 = 3 ). Then, ( A = {2, 3, 4, 5, 6} ), so ( |A| = 5 ). ( B = {1, 2, 3, 4, 6, 9} ), so ( |B| = 6 ). So, the ratio is ( 5/6 approx 0.833 ). So, that's achievable.Similarly, for ( n = 4 ), the minimal ratio would be ( frac{2(7)}{4*5} = frac{14}{20} = 0.7 ). Let me see if I can find a set where ( |A| = 7 ) and ( |B| = 10 ).Wait, if I choose ( a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 4 ), then ( A = {2, 3, 4, 5, 6, 7, 8} ), so ( |A| = 7 ). ( B = {1, 2, 3, 4, 6, 8, 9, 12, 16} ), so ( |B| = 9 ). So, the ratio is ( 7/9 approx 0.777 ), which is higher than 0.7.Wait, so maybe I need a different set for ( n = 4 ) where ( |A| = 7 ) and ( |B| = 10 ). Let me try ( a_1 = 1, a_2 = 2, a_3 = 4, a_4 = 8 ). Then, ( A = {2, 3, 4, 5, 6, 8, 9, 10, 12, 16} ), which has 10 elements. ( B = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512} ), which has 10 elements. So, the ratio is 1. But I want ( |A| = 7 ) and ( |B| = 10 ). Hmm, maybe it's not possible.Wait, perhaps I need to choose numbers such that the sums are minimal and the products are maximal. Let me try ( a_1 = 1, a_2 = 3, a_3 = 5, a_4 = 7 ). Then, ( A ) would be:- ( 1+1 = 2 )- ( 1+3 = 4 )- ( 1+5 = 6 )- ( 1+7 = 8 )- ( 3+3 = 6 )- ( 3+5 = 8 )- ( 3+7 = 10 )- ( 5+5 = 10 )- ( 5+7 = 12 )- ( 7+7 = 14 )So, ( A = {2,4,6,8,10,12,14} ), which has 7 elements. ( B ):- ( 1*1 = 1 )- ( 1*3 = 3 )- ( 1*5 = 5 )- ( 1*7 = 7 )- ( 3*3 = 9 )- ( 3*5 = 15 )- ( 3*7 = 21 )- ( 5*5 = 25 )- ( 5*7 = 35 )- ( 7*7 = 49 )So, ( B = {1,3,5,7,9,15,21,25,35,49} ), which has 10 elements. So, ( |A| = 7 ), ( |B| = 10 ), ratio ( 7/10 = 0.7 ). So, that's the minimal ratio for ( n = 4 ).Therefore, it seems that for each ( n ), the minimal ratio is ( frac{2n - 1}{frac{n(n+1)}{2}} = frac{2(2n - 1)}{n(n+1)} ).So, putting it all together, the minimal value of ( frac{|A|}{|B|} ) is ( frac{2(2n - 1)}{n(n+1)} ).

💡Okay, so I need to prove that all integer solutions ( p + q sqrt{5} ) of some equation (1), where ( p geq 0 ) and ( q geq 0 ), are obtained by the formula ( p + q sqrt{5} = (9 + 4 sqrt{5})^n ) where ( n ) is a non-negative integer. Hmm, I don't have the exact equation (1), but I can assume it's related to Pell's equation or something similar because of the form ( p + q sqrt{5} ). First, I should recall that expressions of the form ( a + b sqrt{5} ) often come up in solving Pell equations, which are equations of the form ( x^2 - Dy^2 = 1 ) where ( D ) is a non-square positive integer. In this case, ( D = 5 ), so the equation would be ( x^2 - 5y^2 = 1 ). The solutions to Pell equations can be generated using the fundamental solution and raising it to powers. Given that, I think equation (1) is likely ( x^2 - 5y^2 = 1 ). So, if that's the case, the fundamental solution for this equation is ( (9, 4) ) because ( 9^2 - 5 times 4^2 = 81 - 80 = 1 ). Therefore, the solutions can be generated by ( (9 + 4 sqrt{5})^n ) for non-negative integers ( n ). But wait, the problem states that all integer solutions ( p + q sqrt{5} ) with ( p geq 0 ) and ( q geq 0 ) are obtained by that formula. So, I need to show that every solution can be written in that form. Let me think about how solutions to Pell equations work. The general solution is given by the fundamental solution raised to the power ( n ). So, if ( (x_1, y_1) ) is the fundamental solution, then all solutions are given by ( (x_n + y_n sqrt{5}) = (x_1 + y_1 sqrt{5})^n ). In this case, the fundamental solution is ( 9 + 4 sqrt{5} ), so raising it to the power ( n ) should give all solutions. But I need to make sure that this covers all possible solutions with ( p geq 0 ) and ( q geq 0 ). Maybe I should use induction. Let's try that. **Base Case:** For ( n = 0 ), ( (9 + 4 sqrt{5})^0 = 1 ), which corresponds to ( p = 1 ) and ( q = 0 ). This is a valid solution since ( 1^2 - 5 times 0^2 = 1 ). **Inductive Step:** Assume that for some ( k geq 0 ), ( (9 + 4 sqrt{5})^k = p_k + q_k sqrt{5} ) is a solution, i.e., ( p_k^2 - 5 q_k^2 = 1 ). Then, for ( k + 1 ), we have ( (9 + 4 sqrt{5})^{k+1} = (9 + 4 sqrt{5})^k times (9 + 4 sqrt{5}) ). Multiplying ( (p_k + q_k sqrt{5})(9 + 4 sqrt{5}) ) gives:[p_k times 9 + p_k times 4 sqrt{5} + q_k times 9 sqrt{5} + q_k times 4 times 5]Simplifying:[9 p_k + 20 q_k + (4 p_k + 9 q_k) sqrt{5}]So, the new ( p_{k+1} = 9 p_k + 20 q_k ) and ( q_{k+1} = 4 p_k + 9 q_k ). We need to verify that ( p_{k+1}^2 - 5 q_{k+1}^2 = 1 ). Let's compute:[p_{k+1}^2 - 5 q_{k+1}^2 = (9 p_k + 20 q_k)^2 - 5 (4 p_k + 9 q_k)^2]Expanding both squares:[= 81 p_k^2 + 360 p_k q_k + 400 q_k^2 - 5 (16 p_k^2 + 72 p_k q_k + 81 q_k^2)]Simplify the second term:[= 81 p_k^2 + 360 p_k q_k + 400 q_k^2 - 80 p_k^2 - 360 p_k q_k - 405 q_k^2]Combine like terms:[= (81 p_k^2 - 80 p_k^2) + (360 p_k q_k - 360 p_k q_k) + (400 q_k^2 - 405 q_k^2)]Simplify:[= p_k^2 - 5 q_k^2]But from the inductive hypothesis, ( p_k^2 - 5 q_k^2 = 1 ). Therefore, ( p_{k+1}^2 - 5 q_{k+1}^2 = 1 ). Thus, by induction, all ( (9 + 4 sqrt{5})^n ) for ( n geq 0 ) are solutions to the equation ( x^2 - 5 y^2 = 1 ). Now, I need to show that these are the only solutions with ( p geq 0 ) and ( q geq 0 ). Suppose there exists another solution ( p + q sqrt{5} ) not of the form ( (9 + 4 sqrt{5})^n ). Then, it must lie between two consecutive powers of ( 9 + 4 sqrt{5} ). But since ( 9 + 4 sqrt{5} ) is greater than 1, its powers grow exponentially, and there can't be another solution in between because the solutions are generated by the fundamental solution. Alternatively, considering the properties of Pell equations, all solutions are generated by the fundamental solution, so there can't be any other solutions outside this sequence. Therefore, all integer solutions ( p + q sqrt{5} ) with ( p geq 0 ) and ( q geq 0 ) are indeed given by ( (9 + 4 sqrt{5})^n ) for non-negative integers ( n ).

💡First, I need to determine how many acorns each squirrel needs to collect to get through the winter. Let's denote this number as ( x ).Each squirrel has already collected ( x - 15 ) acorns, as they need 15 more acorns each.Since there are 5 squirrels, the total number of acorns collected by all squirrels is ( 5 times (x - 15) ).According to the problem, the total number of acorns collected is 575. Therefore, I can set up the equation:[5 times (x - 15) = 575]Next, I'll solve for ( x ):[5x - 75 = 575][5x = 575 + 75][5x = 650][x = frac{650}{5}][x = 130]So, each squirrel needs 130 acorns to get through the winter.