🧠:<think>First, I'll calculate the total runtime of all the MCU movies by adding up each movie's runtime.Next, I'll convert the total runtime from minutes to hours by dividing by 60. Then, to find out how many days Tony needs to watch all the movies if he watches for 8 hours each day, I'll divide the total hours by 8 and round up to the nearest whole number.For the second part, I'll determine the number of ways to order the movies within each phase. Since the order of movies within each phase matters, I'll use factorial calculations for each phase. Then, I'll multiply the number of ways for each phase together to get the total number of distinct ways Tony can schedule his marathon.</think>

🧠:<think>Alright, so I have this problem where a stylist is creating a new collection with specific constraints on the aesthetic scores of the outfits. The goal is to figure out how many unique combinations of jacket, trousers, and shirt result in a prime total aesthetic score. Then, there's a second part where adding another shirt changes things, and I need to find how many additional combinations become possible.First, let me parse the problem step by step.We have three categories of clothing items:1. Jackets: 5 different ones with aesthetic scores ( J = {2, 3, 5, 7, 11} ).2. Trousers: 4 different ones with scores ( T = {3, 5, 7, 11} ).3. Shirts: 6 different ones with scores ( S = {2, 3, 5, 7, 11, 13} ).Each outfit is a combination of one jacket, one trouser, and one shirt. The total aesthetic score is the sum of the three individual scores. The stylist wants this total to be a prime number.So, the first task is to find how many such combinations result in a prime total.Let me think about how to approach this.First, I know that prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. So, the total score must be a prime number.Given that the scores are all positive integers, the total will be at least (2 + 3 + 2 = 7) and at most (11 + 11 + 13 = 35). So, the possible prime totals lie between 7 and 35.I can list all the prime numbers in this range to have a reference. Let me jot them down:Primes between 7 and 35: 7, 11, 13, 17, 19, 23, 29, 31.So, these are the primes we need to consider.Now, the problem is to count all possible triples (jacket, trouser, shirt) such that their sum is one of these primes.Given that each category has a different number of items, I can't just multiply the counts directly. I need to consider each combination and check if the sum is prime.But since the numbers are manageable, maybe I can find a systematic way to compute this without enumerating all possible combinations, which would be 5*4*6=120 combinations. That's a lot, but perhaps manageable.Alternatively, maybe I can find a pattern or use some mathematical properties to simplify the counting.Let me think about the parity of the numbers because primes (except 2) are odd. So, the sum must be odd or 2, but since the minimum sum is 7, which is odd, all primes we are considering are odd.So, the sum of jacket + trouser + shirt must be odd.Now, the sum of three numbers is odd if:- All three are odd, or- Exactly one is odd and the other two are even.But looking at the scores:Jackets: ( J = {2, 3, 5, 7, 11} ). So, 2 is even, others are odd.Trousers: ( T = {3, 5, 7, 11} ). All odd.Shirts: ( S = {2, 3, 5, 7, 11, 13} ). 2 is even, others are odd.So, let's analyze the parity of each component:- Jacket: 1 even (2), 4 odd (3,5,7,11)- Trouser: All 4 are odd- Shirt: 1 even (2), 5 odd (3,5,7,11,13)So, the sum's parity is determined by the number of even components.Since Trouser is always odd, the sum's parity depends on Jacket and Shirt.Let me denote:- E for even, O for odd.So, the sum is:- If Jacket is E (2) and Shirt is E (2): E + O + E = E + E + O = O (since E + E = E, E + O = O)Wait, no, let's think again.Wait, the sum is Jacket + Trouser + Shirt.Trouser is always O.So, the sum is (Jacket + Shirt) + Trouser.Since Trouser is O, the sum's parity is determined by (Jacket + Shirt) + O.But O + O = E, E + O = O.Wait, no, let me think differently.Let me consider the parity of each component:- Jacket: E or O- Trouser: O- Shirt: E or OSo, the total sum's parity is:E (Jacket) + O (Trouser) + E (Shirt) = E + O + E = (E + E) + O = E + O = OSimilarly,E (Jacket) + O (Trouser) + O (Shirt) = E + O + O = E + (O + O) = E + E = EO (Jacket) + O (Trouser) + E (Shirt) = O + O + E = (O + O) + E = E + E = EO (Jacket) + O (Trouser) + O (Shirt) = O + O + O = (O + O) + O = E + O = OSo, in summary:- If Jacket and Shirt are both E: sum is O- If Jacket is E and Shirt is O: sum is E- If Jacket is O and Shirt is E: sum is E- If Jacket and Shirt are both O: sum is OBut since the sum must be prime and greater than 2, it must be odd. So, the sum must be odd.Therefore, the sum is odd only in two cases:1. Jacket is E and Shirt is E: sum is O2. Jacket is O and Shirt is O: sum is OSo, only these two cases can result in an odd sum, which is necessary (but not sufficient) for being prime.Therefore, to get a prime total, the combination must be either:- E jacket + E shirt + O trouser, which is E + E + O = O- O jacket + O shirt + O trouser, which is O + O + O = OSo, these are the only two scenarios where the total sum is odd, hence potentially prime.Therefore, we can split the problem into two parts:1. Count combinations where Jacket is E and Shirt is E.2. Count combinations where Jacket is O and Shirt is O.Then, for each of these counts, we need to check how many of them result in a prime total.So, let's compute the number of combinations in each case.First, case 1: E jacket and E shirt.E jackets: only 1 (score 2)E shirts: only 1 (score 2)So, number of combinations: 1 (jacket) * 4 (trousers) * 1 (shirt) = 4 combinations.Each of these combinations has a total score of 2 (jacket) + 3,5,7,11 (trousers) + 2 (shirt).So, let's compute the totals:- 2 + 3 + 2 = 7- 2 + 5 + 2 = 9- 2 + 7 + 2 = 11- 2 + 11 + 2 = 15Now, check which of these totals are prime:7: prime9: not prime (divisible by 3)11: prime15: not prime (divisible by 3 and 5)So, out of 4 combinations, 2 result in prime totals (7 and 11).Therefore, case 1 contributes 2 valid combinations.Case 2: O jacket and O shirt.O jackets: 4 (scores 3,5,7,11)O shirts: 5 (scores 3,5,7,11,13)Number of combinations: 4 * 4 * 5 = 80 combinations.But we need to check how many of these 80 combinations result in a prime total.This is a bit more involved. Let me think about how to approach this.Each combination is (jacket, trouser, shirt) where jacket is O, trouser is O, shirt is O.So, the total is O + O + O = O, which is odd, so potentially prime.But we need to check for each combination whether the sum is prime.Given that there are 80 combinations, it's impractical to compute each one manually. So, perhaps we can find a pattern or use some mathematical properties.Alternatively, maybe we can precompute the possible sums and count how many times each prime occurs.But let's see:First, let's note the ranges:Jacket: 3,5,7,11Trouser: 3,5,7,11Shirt: 3,5,7,11,13So, the minimum total is 3+3+3=9Maximum total is 11+11+13=35So, the possible primes in this range are: 11,13,17,19,23,29,31Wait, earlier I had primes from 7 to 35, but since the minimum here is 9, the primes are 11,13,17,19,23,29,31.So, we need to find how many combinations result in each of these primes.Alternatively, perhaps we can fix two variables and compute the third.But that might be complicated.Alternatively, maybe we can compute the frequency of each possible sum.But given the time constraints, perhaps a better approach is to note that for each combination, the sum is jacket + trouser + shirt.Given that all three are odd, their sum is odd, so we can focus on the primes in the range.But perhaps another approach is to note that for each prime p in 11,13,...,31, we can compute the number of solutions to jacket + trouser + shirt = p, where jacket ∈ {3,5,7,11}, trouser ∈ {3,5,7,11}, shirt ∈ {3,5,7,11,13}.So, for each prime p, we can compute the number of triples (j,t,s) such that j + t + s = p.This seems feasible.So, let's list the primes and compute the number of solutions for each.Primes to consider: 11,13,17,19,23,29,31.Let's go one by one.1. Prime = 11We need j + t + s = 11j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}So, possible combinations:Let me think about the minimum sum: 3+3+3=9, so 11 is just 2 more than the minimum.So, possible combinations:- j=3, t=3, s=5 (3+3+5=11)- j=3, t=5, s=3- j=5, t=3, s=3Are there any others?Wait, let's check:j=3, t=3, s=5: 3+3+5=11j=3, t=5, s=3: 3+5+3=11j=5, t=3, s=3: 5+3+3=11Is there another combination?j=3, t=7, s=1: but s=1 is not available, s starts at 3.Similarly, j=3, t=11, s=-2: invalid.Similarly, j=5, t=5, s=1: invalid.So, only 3 combinations.But wait, s can be 5, but in the case above, s=5 is allowed.Wait, in the first case, s=5 is allowed, so that's valid.So, total 3 combinations.But let me check if there are more.Wait, j=3, t=3, s=5j=3, t=5, s=3j=5, t=3, s=3Is there a j=3, t=7, s=1? No, s=1 isn't available.Similarly, j=3, t=11, s=-2: invalid.j=5, t=5, s=1: invalid.j=5, t=7, s=-1: invalid.j=7, t=3, s=1: invalid.So, only 3 combinations.Therefore, for p=11, we have 3 combinations.2. Prime =13We need j + t + s =13Again, j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Let me find all possible triples.Start with j=3:Then t + s =10t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s such that t + s=10:t=3, s=7t=5, s=5t=7, s=3t=11, s=-1: invalidSo, for j=3, we have 3 combinations: (3,3,7), (3,5,5), (3,7,3)Next, j=5:Then t + s =8t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=5t=5, s=3t=7, s=1: invalidt=11, s=-3: invalidSo, 2 combinations: (5,3,5), (5,5,3)Next, j=7:Then t + s =6t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=3t=5, s=1: invalidt=7, s=-1: invalidt=11, s=-5: invalidSo, only 1 combination: (7,3,3)j=11:Then t + s =2But t and s are at least 3, so no solutions.Therefore, total combinations for p=13:From j=3: 3From j=5: 2From j=7:1Total: 3+2+1=6 combinations.3. Prime=17j + t + s=17Again, j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Let me approach this by fixing j and finding t and s.Start with j=3:Then t + s=14t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=11t=5, s=9: s=9 not availablet=7, s=7t=11, s=3So, valid combinations:(3,3,11), (3,7,7), (3,11,3)So, 3 combinations.j=5:t + s=12t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=9: invalidt=5, s=7t=7, s=5t=11, s=1: invalidSo, valid combinations:(5,5,7), (5,7,5)2 combinations.j=7:t + s=10t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=7t=5, s=5t=7, s=3t=11, s=-1: invalidSo, valid combinations:(7,3,7), (7,5,5), (7,7,3)3 combinations.j=11:t + s=6t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=3t=5, s=1: invalidt=7, s=-1: invalidt=11, s=-5: invalidSo, only 1 combination: (11,3,3)Therefore, total combinations for p=17:From j=3:3From j=5:2From j=7:3From j=11:1Total: 3+2+3+1=9 combinations.4. Prime=19j + t + s=19Again, j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Start with j=3:t + s=16t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=13t=5, s=11t=7, s=9: invalidt=11, s=5So, valid combinations:(3,3,13), (3,5,11), (3,11,5)3 combinations.j=5:t + s=14t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=11t=5, s=9: invalidt=7, s=7t=11, s=3So, valid combinations:(5,3,11), (5,7,7), (5,11,3)3 combinations.j=7:t + s=12t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=9: invalidt=5, s=7t=7, s=5t=11, s=1: invalidSo, valid combinations:(7,5,7), (7,7,5)2 combinations.j=11:t + s=8t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=5t=5, s=3t=7, s=1: invalidt=11, s=-3: invalidSo, valid combinations:(11,3,5), (11,5,3)2 combinations.Therefore, total combinations for p=19:From j=3:3From j=5:3From j=7:2From j=11:2Total: 3+3+2+2=10 combinations.5. Prime=23j + t + s=23j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Start with j=3:t + s=20t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=7, s=13t=11, s=9: invalidt=5, s=15: invalidt=3, s=17: invalidSo, only 1 combination: (3,7,13)j=5:t + s=18t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=5, s=13t=7, s=11t=11, s=7t=3, s=15: invalidSo, valid combinations:(5,5,13), (5,7,11), (5,11,7)3 combinations.j=7:t + s=16t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=13t=5, s=11t=7, s=9: invalidt=11, s=5So, valid combinations:(7,3,13), (7,5,11), (7,11,5)3 combinations.j=11:t + s=12t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=3, s=9: invalidt=5, s=7t=7, s=5t=11, s=1: invalidSo, valid combinations:(11,5,7), (11,7,5)2 combinations.Therefore, total combinations for p=23:From j=3:1From j=5:3From j=7:3From j=11:2Total:1+3+3+2=9 combinations.6. Prime=29j + t + s=29j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Start with j=3:t + s=26t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=13, s=13: but t only goes up to 11, so no solution.Wait, t can be at most 11, so t=11, s=15: invalid.Similarly, t=7, s=19: invalid.t=5, s=21: invalid.t=3, s=23: invalid.So, no solutions for j=3.j=5:t + s=24t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=11, s=13: 11+13=24t=7, s=17: invalidt=5, s=19: invalidt=3, s=21: invalidSo, only 1 combination: (5,11,13)j=7:t + s=22t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=9, s=13: invalidt=7, s=15: invalidt=11, s=11t=5, s=17: invalidt=3, s=19: invalidSo, only 1 combination: (7,11,11)j=11:t + s=18t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=5, s=13t=7, s=11t=11, s=7t=3, s=15: invalidSo, valid combinations:(11,5,13), (11,7,11), (11,11,7)3 combinations.Therefore, total combinations for p=29:From j=3:0From j=5:1From j=7:1From j=11:3Total:0+1+1+3=5 combinations.7. Prime=31j + t + s=31j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Start with j=3:t + s=28t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=15, s=13: invalidt=11, s=17: invalidt=7, s=21: invalidt=5, s=23: invalidt=3, s=25: invalidNo solutions.j=5:t + s=26t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=13, s=13: invalidt=11, s=15: invalidt=7, s=19: invalidt=5, s=21: invalidt=3, s=23: invalidNo solutions.j=7:t + s=24t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=11, s=13t=7, s=17: invalidt=5, s=19: invalidt=3, s=21: invalidSo, only 1 combination: (7,11,13)j=11:t + s=20t ∈ {3,5,7,11}, s ∈ {3,5,7,11,13}Possible t and s:t=7, s=13t=11, s=9: invalidt=5, s=15: invalidt=3, s=17: invalidSo, only 1 combination: (11,7,13)Therefore, total combinations for p=31:From j=3:0From j=5:0From j=7:1From j=11:1Total:0+0+1+1=2 combinations.Now, let's summarize the number of combinations for each prime:- p=11:3- p=13:6- p=17:9- p=19:10- p=23:9- p=29:5- p=31:2Adding these up: 3+6=9, 9+9=18, 18+10=28, 28+9=37, 37+5=42, 42+2=44.Wait, let me add them step by step:Start with 3 (p=11)+6 (p=13) =9+9 (p=17)=18+10 (p=19)=28+9 (p=23)=37+5 (p=29)=42+2 (p=31)=44So, total combinations in case 2 (O jacket and O shirt) that result in a prime total:44.But wait, earlier in case 1, we had 2 combinations.So, total unique combinations:44 +2=46.But hold on, I need to make sure that these are unique combinations.Wait, in case 1, the combinations are E jacket and E shirt, which are specific items: jacket=2 and shirt=2.So, these are distinct from case 2, which involves O jacket and O shirt.Therefore, the total is indeed 44 +2=46.But let me double-check my counts for case 2.Wait, when I counted for each prime, I added up the combinations and got 44.But let me recount the counts for each prime:p=11:3p=13:6p=17:9p=19:10p=23:9p=29:5p=31:2Adding these:3+6=9, 9+9=18, 18+10=28, 28+9=37, 37+5=42, 42+2=44. Yes, that's correct.So, case 2 contributes 44 combinations.Case 1 contributes 2 combinations.Therefore, total unique combinations:44+2=46.Wait, but hold on, let me think again.In case 1, we had 4 combinations, but only 2 resulted in primes.In case 2, we have 80 combinations, but only 44 resulted in primes.So, total is 2+44=46.But let me think about whether there could be any overlap between case 1 and case 2.But case 1 is E jacket and E shirt, while case 2 is O jacket and O shirt. So, no overlap.Therefore, the total number of unique combinations is 46.Wait, but let me think about whether I might have double-counted any combinations.No, because case 1 and case 2 are mutually exclusive: a combination cannot be both E jacket and E shirt and O jacket and O shirt at the same time.Therefore, 46 is the correct total.But let me cross-verify.Total possible combinations:5*4*6=120.Number of combinations where sum is prime:46.Therefore, the answer to the first sub-problem is 46.Now, moving on to the second sub-problem.The stylist adds one additional vintage shirt with an aesthetic score of 17.So, the shirts now are S = {2,3,5,7,11,13,17}.We need to find how many additional combinations result in a prime total aesthetic score.So, the additional shirt is 17, which is an odd number.Therefore, the new shirt is O.So, in terms of parity, it's O.Therefore, when considering combinations, this shirt will only be part of case 2 (O jacket and O shirt), since E jacket + O shirt would result in an even sum, which cannot be prime (except 2, but the minimum sum is 7).Wait, actually, adding an O shirt, when combined with E jacket and O trouser, would result in E + O + O = E, which is even, hence not prime (except 2, which is too low). So, the only way this new shirt can contribute is in case 2: O jacket + O shirt + O trouser.Therefore, the additional combinations will be those where the shirt is 17, and jacket and trouser are O.So, we need to compute how many combinations with shirt=17 result in a prime total.So, let's compute the number of combinations where shirt=17, jacket ∈ {3,5,7,11}, trouser ∈ {3,5,7,11}, and jacket + trouser +17 is prime.So, the total is jacket + trouser +17.We need this total to be prime.Let me denote the total as p = j + t +17.Given that j ∈ {3,5,7,11}, t ∈ {3,5,7,11}, so j + t ranges from 3+3=6 to 11+11=22.Therefore, p ranges from 6+17=23 to 22+17=39.So, p ∈ {23,24,...,39}.But p must be prime.Primes in this range:23,29,31,37.So, p can be 23,29,31,37.Therefore, we need to find the number of (j,t) pairs such that j + t +17 is one of these primes.So, for each prime p in {23,29,31,37}, compute the number of (j,t) pairs where j + t = p -17.So:For p=23: j + t =6For p=29: j + t=12For p=31: j + t=14For p=37: j + t=20Now, let's compute the number of (j,t) pairs for each:1. p=23: j + t=6j ∈ {3,5,7,11}, t ∈ {3,5,7,11}Possible pairs:j=3, t=3: 3+3=6j=5, t=1: invalidj=7, t=-1: invalidj=11, t=-5: invalidSo, only 1 pair: (3,3)2. p=29: j + t=12Possible pairs:j=3, t=9: invalidj=5, t=7j=7, t=5j=11, t=1: invalidSo, valid pairs: (5,7), (7,5)Total:2 pairs.3. p=31: j + t=14Possible pairs:j=3, t=11j=5, t=9: invalidj=7, t=7j=11, t=3So, valid pairs: (3,11), (7,7), (11,3)Total:3 pairs.4. p=37: j + t=20Possible pairs:j=3, t=17: invalidj=5, t=15: invalidj=7, t=13: invalidj=11, t=9: invalidSo, no valid pairs.Therefore, total additional combinations:For p=23:1For p=29:2For p=31:3For p=37:0Total:1+2+3+0=6 combinations.Therefore, adding the shirt with score 17 adds 6 new combinations that result in a prime total.Therefore, the answer to the second sub-problem is 6 additional combinations.But let me double-check my calculations.For p=23: j + t=6, only (3,3)Yes.For p=29: j + t=12, (5,7) and (7,5)Yes.For p=31: j + t=14, (3,11), (7,7), (11,3)Yes.For p=37: j + t=20, no solutions.Yes.So, total 6 additional combinations.Therefore, the answers are:1. 46 combinations.2. 6 additional combinations.Final Answer1. The number of unique combinations is boxed{46}.2. The number of additional combinations is boxed{6}.</think>

🧠:<think>Okay, so I have this differential equation modeling the population of an endangered species. The equation is:[ frac{dP(t)}{dt} = rP(t) left(1 - frac{P(t)}{K}right) - frac{A}{B + P(t)} ]Professor Jane is looking at how non-lethal research methods affect the population. The problem has two parts. First, when A is zero, find the general solution and discuss the long-term behavior. Second, when A isn't zero, analyze the equilibrium points and their stability.Starting with part 1: A = 0. So the differential equation simplifies to:[ frac{dP(t)}{dt} = rP(t) left(1 - frac{P(t)}{K}right) ]Hmm, that looks familiar. Isn't that the logistic growth model? Yeah, the standard logistic equation. So, I remember that the logistic equation has an analytical solution. Let me recall how to solve it.The logistic equation is:[ frac{dP}{dt} = rPleft(1 - frac{P}{K}right) ]To solve this, I can use separation of variables. Let me write it as:[ frac{dP}{Pleft(1 - frac{P}{K}right)} = r dt ]Integrating both sides. The left side integral can be done using partial fractions. Let me set up partial fractions for the integrand.Let me denote:[ frac{1}{Pleft(1 - frac{P}{K}right)} = frac{A}{P} + frac{B}{1 - frac{P}{K}} ]Multiplying both sides by ( Pleft(1 - frac{P}{K}right) ):[ 1 = Aleft(1 - frac{P}{K}right) + BP ]Expanding:[ 1 = A - frac{A P}{K} + BP ]Grouping terms:[ 1 = A + Pleft(-frac{A}{K} + Bright) ]This must hold for all P, so the coefficients must be equal on both sides. Therefore:For the constant term: 1 = AFor the P term: 0 = -A/K + BSo, since A = 1, then:0 = -1/K + B => B = 1/KTherefore, the partial fractions decomposition is:[ frac{1}{Pleft(1 - frac{P}{K}right)} = frac{1}{P} + frac{1/K}{1 - frac{P}{K}} ]So, the integral becomes:[ int left( frac{1}{P} + frac{1/K}{1 - frac{P}{K}} right) dP = int r dt ]Integrating term by term:First integral: ∫(1/P) dP = ln|P| + CSecond integral: ∫(1/K)/(1 - P/K) dP. Let me make a substitution. Let u = 1 - P/K, then du = -1/K dP, so -du = (1/K) dP.Therefore, the second integral becomes:∫(1/K)/u * (-K du) = -∫(1/u) du = -ln|u| + C = -ln|1 - P/K| + CPutting it all together:ln|P| - ln|1 - P/K| = r t + CCombine the logs:ln| P / (1 - P/K) | = r t + CExponentiate both sides:P / (1 - P/K) = e^{r t + C} = e^C e^{r t}Let me denote e^C as another constant, say, C1.So,P / (1 - P/K) = C1 e^{r t}Solving for P:Multiply both sides by denominator:P = C1 e^{r t} (1 - P/K)Expand the right side:P = C1 e^{r t} - (C1 e^{r t} P)/KBring the P term to the left:P + (C1 e^{r t} P)/K = C1 e^{r t}Factor P:P [1 + (C1 e^{r t}) / K] = C1 e^{r t}Therefore,P = [C1 e^{r t}] / [1 + (C1 e^{r t}) / K]Multiply numerator and denominator by K to simplify:P = [C1 K e^{r t}] / [K + C1 e^{r t}]Now, apply the initial condition P(0) = P0.At t = 0:P0 = [C1 K e^{0}] / [K + C1 e^{0}] = [C1 K] / [K + C1]Solve for C1:Multiply both sides by (K + C1):P0 (K + C1) = C1 KExpand:P0 K + P0 C1 = C1 KBring terms with C1 to one side:P0 K = C1 K - P0 C1 = C1 (K - P0)Therefore,C1 = (P0 K) / (K - P0)So, substitute back into P(t):P(t) = [ (P0 K / (K - P0)) * K e^{r t} ] / [ K + (P0 K / (K - P0)) e^{r t} ]Simplify numerator and denominator:Numerator: (P0 K^2 / (K - P0)) e^{r t}Denominator: K + (P0 K / (K - P0)) e^{r t} = K [1 + (P0 / (K - P0)) e^{r t} ]So,P(t) = [ (P0 K^2 / (K - P0)) e^{r t} ] / [ K (1 + (P0 / (K - P0)) e^{r t} ) ]Cancel K:P(t) = [ (P0 K / (K - P0)) e^{r t} ] / [ 1 + (P0 / (K - P0)) e^{r t} ]Let me factor out (P0 / (K - P0)) e^{r t} in the denominator:Denominator: 1 + (P0 / (K - P0)) e^{r t} = [ (K - P0) + P0 e^{r t} ] / (K - P0)Therefore,P(t) = [ (P0 K / (K - P0)) e^{r t} ] / [ (K - P0 + P0 e^{r t}) / (K - P0) ) ]Simplify:Multiply numerator and denominator:P(t) = (P0 K e^{r t}) / (K - P0 + P0 e^{r t})We can factor out P0 in the denominator:P(t) = (P0 K e^{r t}) / [ P0 e^{r t} + K - P0 ]Alternatively, factor K:Wait, maybe it's better to write it as:P(t) = (K P0 e^{r t}) / (K + P0 (e^{r t} - 1))But I think the standard form is:P(t) = K / (1 + (K / P0 - 1) e^{-r t})Let me check that.Starting from P(t) = (P0 K e^{r t}) / (K - P0 + P0 e^{r t})Divide numerator and denominator by e^{r t}:P(t) = (P0 K) / ( (K - P0) e^{-r t} + P0 )So,P(t) = (P0 K) / ( P0 + (K - P0) e^{-r t} )Which can be written as:P(t) = K / [ 1 + ( (K - P0)/P0 ) e^{-r t} ]Yes, that's the standard logistic growth solution.So, the general solution when A = 0 is:[ P(t) = frac{K}{1 + left( frac{K - P_0}{P_0} right) e^{-r t}} ]Now, discussing the long-term behavior. As t approaches infinity, e^{-r t} approaches zero, so:P(t) approaches K / (1 + 0) = K.So, the population approaches the carrying capacity K, which is the expected behavior for the logistic model. The population grows exponentially at first, then slows down as it approaches K, where it stabilizes.Okay, that was part 1. Now, moving on to part 2: when A ≠ 0. We need to analyze the equilibrium points and their stability.First, equilibrium points occur when dP/dt = 0. So, set the differential equation equal to zero:[ rP left(1 - frac{P}{K}right) - frac{A}{B + P} = 0 ]So,[ rP left(1 - frac{P}{K}right) = frac{A}{B + P} ]This is a nonlinear equation in P. Let's denote this as:[ f(P) = rP left(1 - frac{P}{K}right) - frac{A}{B + P} = 0 ]We need to find the roots of f(P) = 0.First, let's analyze the possible number of equilibrium points. The left side is a quadratic in P, and the right side is a hyperbola. Depending on the parameters, there can be one or two equilibrium points.But let's try to find them.Multiply both sides by (B + P):[ rP left(1 - frac{P}{K}right)(B + P) = A ]Expand the left side:First, expand (1 - P/K)(B + P):= (1)(B + P) - (P/K)(B + P)= B + P - (B P)/K - P^2/KSo,rP [ B + P - (B P)/K - P^2/K ] = AMultiply through:rP [ B + P - (B P)/K - P^2/K ] = ALet me compute each term:First term: rP * B = r B PSecond term: rP * P = r P^2Third term: rP * (-B P / K) = - r B P^2 / KFourth term: rP * (-P^2 / K) = - r P^3 / KSo, combining all terms:r B P + r P^2 - (r B / K) P^2 - (r / K) P^3 = ACombine like terms:- The P term: r B P- The P^2 terms: r P^2 - (r B / K) P^2 = r (1 - B / K) P^2- The P^3 term: - (r / K) P^3So, the equation becomes:- (r / K) P^3 + r (1 - B / K) P^2 + r B P - A = 0Multiply both sides by -1 to make the leading coefficient positive:(r / K) P^3 - r (1 - B / K) P^2 - r B P + A = 0So, the equation is:[ frac{r}{K} P^3 - r left(1 - frac{B}{K}right) P^2 - r B P + A = 0 ]This is a cubic equation in P. Solving cubic equations analytically can be complicated, but perhaps we can analyze the number of real roots and their approximate values.Alternatively, we can consider the function f(P) = rP(1 - P/K) - A/(B + P) and analyze its behavior.Let me think about the behavior of f(P):As P approaches 0:f(P) ≈ rP - A/(B + 0) = rP - A/BSo, f(0) = -A/B < 0 (since A and B are positive constants).As P approaches infinity:f(P) ≈ rP(1 - P/K) - A/P ≈ - r P^2 / K - 0 → -∞So, f(P) tends to -∞ as P approaches infinity.At P = K:f(K) = r K (1 - K/K) - A/(B + K) = 0 - A/(B + K) = -A/(B + K) < 0So, f(K) is negative.Wait, but in the logistic model without the A term, the equilibrium is at P=K. Now, with the A term, the equilibrium might be lower.But let's check the derivative of f(P) to see if there are any maxima or minima.Compute f'(P):f'(P) = d/dP [ rP(1 - P/K) - A/(B + P) ]= r(1 - P/K) + rP(-1/K) + A/(B + P)^2Simplify:= r(1 - P/K - P/K) + A/(B + P)^2= r(1 - 2P/K) + A/(B + P)^2Set f'(P) = 0 to find critical points:r(1 - 2P/K) + A/(B + P)^2 = 0This equation is also non-linear and might not have an analytical solution, but it can help us understand the number of critical points.Alternatively, let's analyze f(P):We know f(0) = -A/B < 0As P increases, initially, f(P) increases because the first term rP(1 - P/K) is increasing for P < K/2.But at some point, the term -A/(B + P) becomes less negative as P increases, so f(P) might have a maximum somewhere.Wait, let me think again.Wait, f(P) is the left-hand side of the equilibrium equation. So, f(P) = 0.We can plot f(P) to understand the number of roots.At P=0: f(0) = -A/B < 0As P increases, f(P) increases because the first term rP(1 - P/K) increases (since P is increasing and 1 - P/K is decreasing, but the product initially increases until P=K/2).At P=K/2: f(K/2) = r*(K/2)*(1 - (K/2)/K) - A/(B + K/2) = r*(K/2)*(1 - 1/2) - A/(B + K/2) = r*(K/2)*(1/2) - A/(B + K/2) = r K /4 - A/(B + K/2)Depending on the values of r, K, A, B, this could be positive or negative.If f(K/2) > 0, then since f(P) starts negative at P=0, increases to a maximum, then decreases to f(K) = -A/(B + K) < 0, then f(P) must cross zero twice: once between P=0 and P=K/2, and once between P=K/2 and P=K.If f(K/2) = 0, then it's tangent at P=K/2.If f(K/2) < 0, then f(P) remains negative throughout, so no equilibrium points except maybe at P=0, but P=0 is already considered.Wait, but f(P) tends to -∞ as P approaches infinity, so if f(P) starts negative at P=0, increases to a maximum, and then decreases to -∞, depending on whether the maximum is above zero or not, we can have one or two positive roots.Wait, let me correct myself. At P=0, f(P) = -A/B < 0. As P increases, f(P) increases because the first term is increasing (since P is increasing and 1 - P/K is still positive for P < K). The second term, -A/(B + P), becomes less negative as P increases, so overall f(P) increases.At P=K, f(K) = -A/(B + K) < 0.So, f(P) starts negative, increases to some maximum, then decreases back to negative at P=K. Therefore, if the maximum of f(P) is positive, then f(P) crosses zero twice: once between P=0 and the maximum, and once between the maximum and P=K.If the maximum is exactly zero, then f(P) touches zero at one point (tangent). If the maximum is negative, then f(P) never crosses zero, so no equilibrium points except P=0, but P=0 is a trivial solution.But in our case, the population P(t) is positive, so we are interested in positive equilibrium points.Therefore, the number of positive equilibrium points depends on whether the maximum of f(P) is positive or not.To find the maximum, set f'(P) = 0:r(1 - 2P/K) + A/(B + P)^2 = 0Let me denote this as:r(1 - 2P/K) = - A/(B + P)^2Since the left side is r(1 - 2P/K) and the right side is negative (because A and (B + P)^2 are positive), so 1 - 2P/K must be negative. Therefore:1 - 2P/K < 0 => P > K/2So, the critical point (maximum) occurs at P > K/2.Now, let's evaluate f(P) at this critical point. If f(P) at this point is positive, then there are two equilibrium points: one between 0 and K/2, and one between K/2 and K.If f(P) at the critical point is zero, then only one equilibrium point at that P.If f(P) at the critical point is negative, then no positive equilibrium points except P=0, which is trivial.But in our case, the population can't be zero because it's an endangered species, so we are looking for positive equilibria.Therefore, the presence of the non-lethal research method (term A/(B + P)) reduces the equilibrium population size compared to the logistic model without A.In the logistic model without A, the equilibrium is at P=K. With A, the equilibrium is less than K.But depending on the parameters, there might be two equilibria: one stable and one unstable, or just one stable equilibrium.Wait, actually, in the logistic model, the equilibrium at P=K is stable. When we add the term -A/(B + P), it introduces another equilibrium point.Wait, let me think again.In the logistic model, dP/dt = rP(1 - P/K). The equilibria are at P=0 and P=K. P=0 is unstable, P=K is stable.When we add the term -A/(B + P), the equilibria are solutions to rP(1 - P/K) = A/(B + P).Depending on the parameters, this can have one or two positive solutions.If there are two positive solutions, one will be stable and the other unstable.To determine stability, we can look at the derivative of dP/dt at the equilibrium points.The derivative is f'(P) = r(1 - 2P/K) + A/(B + P)^2Wait, no. Wait, the derivative of dP/dt with respect to P is:d/dP [ rP(1 - P/K) - A/(B + P) ] = r(1 - 2P/K) + A/(B + P)^2So, at an equilibrium point P*, if f'(P*) < 0, then it's a stable equilibrium. If f'(P*) > 0, it's unstable.So, let's suppose we have two equilibrium points: P1 and P2, with P1 < P2.At P1: Since P1 < K/2, 1 - 2P1/K > 0, so the first term is positive. The second term A/(B + P1)^2 is positive. Therefore, f'(P1) > 0, so P1 is unstable.At P2: Since P2 > K/2, 1 - 2P2/K < 0, so the first term is negative. The second term is positive. So, whether f'(P2) is positive or negative depends on the balance between the two terms.If f'(P2) < 0, then P2 is stable.Therefore, typically, in such models, the lower equilibrium is unstable, and the higher one is stable, but in this case, since the term -A/(B + P) is subtracted, it might be the opposite.Wait, let me think carefully.In the logistic model, the derivative at P=K is f'(K) = r(1 - 2K/K) + 0 = r(1 - 2) = -r < 0, so stable.In our case, with the added term, at P=K, f'(K) = r(1 - 2K/K) + A/(B + K)^2 = -r + A/(B + K)^2.So, if A is small, f'(K) is still negative, so P=K is stable. But if A is large enough, f'(K) could become positive, making P=K unstable.Wait, but in our case, the equilibrium points are not necessarily at P=K anymore.Wait, no. The equilibrium points are solutions to rP(1 - P/K) = A/(B + P). So, P=K is not necessarily an equilibrium unless A=0.Wait, that's correct. When A=0, P=K is an equilibrium. When A≠0, P=K is not an equilibrium anymore.So, the equilibria are at P1 and P2, where P1 < P2.At P1: f'(P1) = r(1 - 2P1/K) + A/(B + P1)^2Since P1 < K/2, 1 - 2P1/K > 0, so f'(P1) is positive (since both terms are positive). Therefore, P1 is unstable.At P2: f'(P2) = r(1 - 2P2/K) + A/(B + P2)^2Since P2 > K/2, 1 - 2P2/K < 0, so the first term is negative. The second term is positive. So, whether f'(P2) is positive or negative depends on which term dominates.If f'(P2) < 0, then P2 is stable.If f'(P2) > 0, then P2 is unstable.But in our case, since the term -A/(B + P) is subtracted, it's likely that P2 is stable.Wait, let me think of it another way. The original logistic model has a stable equilibrium at P=K. Adding a harvesting term (which is what this term is, similar to harvesting) can create a new equilibrium below K, which is stable, and the original K becomes unstable.But in our case, the term is -A/(B + P), which is a type of harvesting that decreases as P increases.Wait, actually, the term is subtracted, so it's like a harvesting effort that increases as P decreases. So, when P is small, the harvesting impact is higher, and as P increases, the impact diminishes.This is different from constant harvesting or proportional harvesting.In this case, the effect is more pronounced when the population is small.So, perhaps the equilibrium P2 is stable, and P1 is unstable.But let me test with specific values.Suppose r=1, K=100, A=10, B=10.Then, f(P) = P(1 - P/100) - 10/(10 + P)We can plot this function or compute f(P) at various points.At P=0: f(0) = 0 - 10/10 = -1 < 0At P=50: f(50) = 50*(1 - 0.5) - 10/(60) = 25 - 0.1667 ≈ 24.833 > 0At P=100: f(100) = 100*(1 - 1) - 10/110 ≈ 0 - 0.0909 ≈ -0.0909 < 0So, f(P) crosses zero between P=0 and P=50, and between P=50 and P=100.Therefore, two equilibrium points: P1 ~ between 0 and 50, P2 ~ between 50 and 100.Now, compute f'(P) at P1 and P2.At P1 ~ say, 10:f'(10) = 1*(1 - 20/100) + 10/(10 + 10)^2 = 0.8 + 10/400 = 0.8 + 0.025 = 0.825 > 0, so P1 is unstable.At P2 ~ say, 90:f'(90) = 1*(1 - 180/100) + 10/(100)^2 = 1*(-0.8) + 0.001 = -0.8 + 0.001 = -0.799 < 0, so P2 is stable.Therefore, in this case, there are two equilibrium points: one unstable at P1 and one stable at P2.So, in general, when A ≠ 0, the model can have two equilibrium points: one unstable (lower) and one stable (higher). The stable equilibrium is less than K, the carrying capacity without the harvesting term.Therefore, the presence of the non-lethal research method reduces the equilibrium population size from K to some P2 < K.But wait, in the example above, P2 is still close to K, but in reality, depending on A, P2 could be significantly lower.If A is very large, the term -A/(B + P) could dominate, pulling the equilibrium much lower.For example, if A is extremely large, even at P=K, the term -A/(B + K) could be very negative, making f(K) = -A/(B + K) very negative, but since f(P) tends to -∞ as P approaches infinity, it's possible that the equilibrium P2 is significantly lower.But in general, the model will have one stable equilibrium at some P2 < K, and one unstable equilibrium at P1 < P2.Therefore, the equilibrium points are solutions to the cubic equation:(r / K) P^3 - r (1 - B / K) P^2 - r B P + A = 0But solving this analytically is complex. Instead, we can state that there are typically two positive equilibrium points when A is not zero, with one being stable and the other unstable.Therefore, the presence of the non-lethal research method reduces the equilibrium population size from K to a lower value P2, and introduces an unstable equilibrium at P1.In summary:1. When A=0, the solution is the logistic growth model, approaching K as t→∞.2. When A≠0, there are typically two equilibrium points: an unstable lower equilibrium P1 and a stable higher equilibrium P2 < K. The stable equilibrium is reduced compared to K due to the impact of the research method.I think that's the analysis.</think>

🧠:<think>Okay, so I have these two questions about optimizing a toy collection management app. Let me try to break them down one by one.Starting with the first question: The developer is using an n-dimensional array where each dimension is an attribute like name, type, year, etc. They want to implement a multi-dimensional binary search. Hmm, binary search is usually O(log n) in one dimension, but how does that scale in multiple dimensions?I remember that in a multi-dimensional array, if each dimension is sorted, you can perform a binary search along each dimension. But wait, is it just a simple extension? Or does it get more complicated? Maybe it's a matter of performing a binary search on each dimension sequentially. So if there are n dimensions, each with size k, then for each dimension, the binary search would take O(log k) time. So for n dimensions, it would be O(n log k). But wait, is that accurate?Alternatively, if the data is structured in a way that allows for a single binary search across all dimensions, maybe the time complexity is different. But I think in reality, each dimension would need to be searched separately, so it's additive. So if each dimension has size k_i, then the time complexity would be the sum of log k_i for each dimension. But if all dimensions are roughly the same size, say k, then it's O(n log k). Wait, but the question says "derive the time complexity in terms of n and the size of each dimension." So maybe it's better to express it as O(log k_1 + log k_2 + ... + log k_n), which is the same as O(log (k_1 * k_2 * ... * k_n)) if we consider the product. But actually, log(a) + log(b) = log(ab), so it's O(log (product of all k_i)). But the product of all k_i would be the total number of elements in the n-dimensional array, right? So if the total number of toys is M, then the time complexity is O(log M). But wait, that can't be right because in multi-dimensional binary search, the complexity isn't necessarily just log of the total elements.Wait, maybe I'm confusing it with something else. Let me think again. In a 2D array, if you perform a binary search on each dimension, it's O(log k1 + log k2). So for n dimensions, it's O(n log k), assuming each dimension has size k. Or if each dimension has a different size, it's O(log k1 + log k2 + ... + log kn). So in terms of n and the size of each dimension, it's the sum of the logs of each dimension's size. So the time complexity would be O(log k1 + log k2 + ... + log kn). Alternatively, if the search is done in a way that each step reduces the search space across all dimensions, maybe it's O(log(max(k1, k2, ..., kn)))? But I don't think that's how it works. I think each dimension is handled separately, so it's additive.So, for the first part, the time complexity is O(log k1 + log k2 + ... + log kn), which can also be written as O(log(k1 * k2 * ... * kn)) if we consider the product, but since the question specifies in terms of n and the size of each dimension, it's better to express it as the sum of logs.Moving on to the second question: The recommendation system uses cosine similarity. I remember that cosine similarity between two vectors is the dot product divided by the product of their magnitudes. So, formula-wise, it's (a · b) / (||a|| ||b||). Now, computing the time complexity. For each pair of toys, we need to compute the cosine similarity. If there are m toys, then the number of pairs is m choose 2, which is O(m^2). For each pair, computing the dot product is O(d), since you multiply each corresponding component and sum them up. Then, computing the magnitudes is also O(d) for each vector. So for each pair, the time is O(d + d + d) = O(d). So overall, the time complexity is O(m^2 * d). But wait, if m is the number of toys and d is the dimensionality of the feature vectors, then yes, it's O(m^2 d). However, if the number of toys is large, this can become computationally expensive. Maybe there are optimizations, like using dimensionality reduction or approximate methods, but the question doesn't ask for optimizations, just the time complexity. So, the time complexity scales as O(m^2 d).Wait, but sometimes people compute cosine similarity in a different way. If you precompute the magnitudes of all vectors first, then for each pair, you only need the dot product and can reuse the magnitudes. So precomputing magnitudes would take O(m d) time, and then for each pair, it's O(d) for the dot product. So total time would still be O(m d + m^2 d) = O(m^2 d), since m^2 d dominates for large m.Alternatively, if you use matrix operations, you can compute all pairwise similarities more efficiently, but in terms of time complexity, it's still O(m^2 d). So yeah, the time complexity is O(m^2 d).Wait, but sometimes people use techniques like locality-sensitive hashing or tree-based methods to reduce the number of comparisons, but again, the question is about the time complexity of computing cosine similarity, not about optimizations. So I think it's safe to say O(m^2 d).So, to recap:1. For the multi-dimensional binary search, the time complexity is the sum of the logarithms of each dimension's size, which is O(log k1 + log k2 + ... + log kn).2. For the cosine similarity, the formula is (a · b)/(||a|| ||b||), and the time complexity is O(m^2 d).I think that's it. Let me just make sure I didn't miss anything.For the first part, is there a more efficient way? Like, if the data is structured in a way that allows for a single binary search across all dimensions, but I don't think that's standard. Usually, multi-dimensional binary search is done by handling each dimension separately, so the time complexity is additive.For the second part, yeah, computing all pairwise similarities is O(m^2 d), which is quadratic in the number of toys and linear in the dimensionality. That makes sense.Final Answer1. The time complexity of the multi-dimensional binary search algorithm is boxed{O(log k_1 + log k_2 + cdots + log k_n)}.2. The formula for cosine similarity is boxed{frac{mathbf{a} cdot mathbf{b}}{|mathbf{a}| |mathbf{b}|}}, and the time complexity of computing cosine similarity for all pairs of toys is boxed{O(m^2 d)}.</think>