💡Okay, so I need to find the greatest prime factor of 15! + 17!. Hmm, factorials can get really big, so I don't think I want to calculate them directly. Maybe there's a way to simplify the expression first.Let me recall that 17! is 17 × 16 × 15!. So, 17! = 17 × 16 × 15!. That means I can factor out 15! from both terms in the expression 15! + 17!. Let me try that:15! + 17! = 15! + 17 × 16 × 15! = 15!(1 + 17 × 16).Okay, so that simplifies to 15! multiplied by (1 + 17 × 16). Let me calculate what's inside the parentheses:17 × 16 is 272, right? So, 1 + 272 is 273. Therefore, the expression becomes 15! × 273.Now, I need to find the prime factors of this product. Since 15! is the product of all integers from 1 to 15, its prime factors include all the prime numbers up to 15. Let me list those primes: 2, 3, 5, 7, 11, and 13. So, the prime factors of 15! are 2, 3, 5, 7, 11, and 13.Next, I need to factorize 273 to see if it introduces any new prime factors. Let me divide 273 by small primes:273 ÷ 3 = 91. So, 3 is a factor. Now, 91 can be divided by 7: 91 ÷ 7 = 13. So, 273 factors into 3 × 7 × 13.Wait, those are all primes, and they are already included in the prime factors of 15!. So, does that mean the greatest prime factor is still 13? But hold on, 17! includes 17 as a factor, right? But in the expression 15! + 17!, we factored out 15!, so the 17 is still present in the 17! term.Wait, no. Because when we factored out 15!, the 17 was part of the 17! which became 17 × 16 × 15!. So, when we factored out 15!, the 17 was part of the coefficient 273. But 273 is 3 × 7 × 13, so 17 isn't a factor of 273. Hmm, that's confusing.Let me think again. The original expression is 15! + 17!. We factored out 15! to get 15!(1 + 17 × 16). So, 1 + 17 × 16 is 273, which is 3 × 7 × 13. So, 17 isn't a factor of 273. Therefore, 17 isn't a factor of the entire expression 15! + 17!.But wait, 17! is divisible by 17, and 15! is not. So, when we add 15! and 17!, does 17 divide the sum? Let me check:15! is not divisible by 17, but 17! is divisible by 17. So, 15! + 17! = 15! + 17 × 16 × 15! = 15!(1 + 17 × 16). Since 17 divides 17!, it doesn't necessarily divide 15! + 17! unless 17 divides 1 + 17 × 16. Let me check if 17 divides 1 + 17 × 16.17 × 16 is 272, so 1 + 272 is 273. Does 17 divide 273? Let me divide 273 by 17: 17 × 16 is 272, so 273 - 272 is 1. So, 273 = 17 × 16 + 1, which means 273 ≡ 1 mod 17. Therefore, 17 does not divide 273, so 17 is not a factor of 15! + 17!.Wait, that contradicts my earlier thought. So, if 17 doesn't divide 273, then 17 isn't a factor of the entire expression. Therefore, the prime factors are still the primes up to 13.But that doesn't seem right because 17! is part of the sum, and 17 is a prime number. Maybe I made a mistake in my reasoning.Let me try another approach. Let's consider the expression 15! + 17! = 15!(1 + 17 × 16). So, the prime factors are the union of the prime factors of 15! and the prime factors of 273. Since 273 is 3 × 7 × 13, which are all already in 15!, the greatest prime factor is still 13.But wait, 17! includes 17, but when we factor out 15!, the 17 is part of the coefficient 273, which doesn't include 17 as a factor. So, 17 isn't a factor of the entire expression. Therefore, the greatest prime factor is 13.But I'm still confused because 17! is part of the sum, and 17 is a prime. Maybe I need to check if 17 divides the sum 15! + 17!.Let me think about it modulo 17. Since 17 is prime, 15! is not divisible by 17, but 17! is divisible by 17. So, 15! + 17! ≡ 15! + 0 ≡ 15! mod 17. Is 15! congruent to 0 mod 17? No, because 17 is prime and greater than 15, so 15! doesn't include 17 as a factor. Therefore, 15! + 17! ≡ 15! mod 17, which is not 0. So, 17 does not divide the sum.Therefore, the greatest prime factor is indeed 13.Wait, but in the original problem, the user's assistant concluded that the greatest prime factor is 17. So, maybe I'm missing something. Let me double-check.The expression is 15! + 17! = 15!(1 + 17 × 16) = 15! × 273. The prime factors of 15! are up to 13, and 273 is 3 × 7 × 13. So, the prime factors are 2, 3, 5, 7, 11, 13. Therefore, the greatest prime factor is 13.But the user's assistant said 17. Maybe they thought that since 17! is part of the sum, 17 is a factor. But as I just reasoned, 17 doesn't divide the sum because 15! + 17! ≡ 15! mod 17, which isn't 0. So, 17 isn't a factor.Therefore, the greatest prime factor should be 13.Wait, but maybe I made a mistake in the modulo calculation. Let me check 15! mod 17.By Wilson's theorem, (p-1)! ≡ -1 mod p for prime p. So, 16! ≡ -1 mod 17. Therefore, 15! = 16! / 16 ≡ (-1)/16 mod 17. Since 16 ≡ -1 mod 17, so 15! ≡ (-1)/(-1) ≡ 1 mod 17. Therefore, 15! ≡ 1 mod 17.So, 15! + 17! ≡ 1 + 0 ≡ 1 mod 17. Therefore, 17 does not divide the sum. So, the greatest prime factor is indeed 13.But the user's assistant concluded 17. Maybe they made a mistake. So, I think the correct answer is 13.Wait, but let me check another way. Let me compute 15! + 17! = 15!(1 + 17×16) = 15!×273. Now, 273 is 3×7×13, so the prime factors are 2,3,5,7,11,13. So, the greatest is 13.Yes, so I think the correct answer is 13, not 17. The user's assistant might have made a mistake by assuming that 17 is a factor, but it's not.Wait, but let me check if 17 divides 273. 273 ÷ 17 is 16.058... So, no, it's not divisible. So, 17 isn't a factor. Therefore, the greatest prime factor is 13.I think I'm confident now that the greatest prime factor is 13.

💡Okay, so I have this problem about quadratic trinomials. Let me try to understand it step by step. We have 100 quadratic trinomials, labeled f₁(x), f₂(x), up to f₁₀₀(x). Each of these trinomials has the same coefficient for x² and the same coefficient for x, but they have different constant terms. So, if I write them out, they would look like:f₁(x) = ax² + bx + c₁ f₂(x) = ax² + bx + c₂ ... f₁₀₀(x) = ax² + bx + c₁₀₀ Here, 'a' and 'b' are the same for all trinomials, but each cᵢ is different. Each of these trinomials has two roots, which means each quadratic equation fᵢ(x) = 0 has two solutions.Now, for each trinomial fᵢ(x), one root is chosen and denoted as xᵢ. So, x₁ is a root of f₁(x), x₂ is a root of f₂(x), and so on, up to x₁₀₀, which is a root of f₁₀₀(x).The problem asks for the possible values that the sum f₂(x₁) + f₃(x₂) + ... + f₁₀₀(x₉₉) + f₁(x₁₀₀) can take.Hmm, okay. So, let me try to break this down. First, since each xᵢ is a root of fᵢ(x), that means fᵢ(xᵢ) = 0. So, for example, f₁(x₁) = 0, f₂(x₂) = 0, etc.But in the sum we're asked about, we're plugging x₁ into f₂, x₂ into f₃, and so on, wrapping around at the end so that x₁₀₀ is plugged into f₁.So, let's write out the sum:Sum = f₂(x₁) + f₃(x₂) + f₄(x₃) + ... + f₁₀₀(x₉₉) + f₁(x₁₀₀)Each term in this sum is a quadratic trinomial evaluated at a root of another trinomial.Let me consider one term, say f₂(x₁). Since x₁ is a root of f₁(x), we know that f₁(x₁) = 0. But f₂(x) is similar to f₁(x), except for the constant term. So, f₂(x) = ax² + bx + c₂, whereas f₁(x) = ax² + bx + c₁.Therefore, f₂(x₁) = a(x₁)² + b(x₁) + c₂. But from f₁(x₁) = 0, we have a(x₁)² + b(x₁) + c₁ = 0. So, a(x₁)² + b(x₁) = -c₁.Substituting this into f₂(x₁), we get:f₂(x₁) = (-c₁) + c₂ = c₂ - c₁Ah, interesting! So, f₂(x₁) is just the difference between the constant terms of f₂ and f₁.Let me check another term to see if this pattern holds. Take f₃(x₂). Since x₂ is a root of f₂(x), we have f₂(x₂) = 0, which means a(x₂)² + b(x₂) + c₂ = 0. Therefore, a(x₂)² + b(x₂) = -c₂.Then, f₃(x₂) = a(x₂)² + b(x₂) + c₃ = (-c₂) + c₃ = c₃ - c₂Yes, the same pattern! Each term f_{i+1}(x_i) = c_{i+1} - c_i.So, if I apply this to all terms in the sum:Sum = (c₂ - c₁) + (c₃ - c₂) + (c₄ - c₃) + ... + (c₁₀₀ - c₉₉) + (c₁ - c₁₀₀)Now, let's write this out:Sum = c₂ - c₁ + c₃ - c₂ + c₄ - c₃ + ... + c₁₀₀ - c₉₉ + c₁ - c₁₀₀Looking at this, each positive c term cancels with the negative of the same c term in the next part. For example, +c₂ cancels with -c₂, +c₃ cancels with -c₃, and so on, all the way up to +c₁₀₀ cancels with -c₁₀₀. The only terms that don't cancel are the very first -c₁ and the very last +c₁.So, Sum = (-c₁) + (c₁) = 0Wait, that's interesting. All the terms cancel out, and the sum is zero. So, regardless of the specific values of the constant terms c₁, c₂, ..., c₁₀₀, as long as each xᵢ is a root of fᵢ(x), the sum will always be zero.Let me double-check this reasoning. Each term f_{i+1}(x_i) is equal to c_{i+1} - c_i because x_i is a root of f_i(x), so when we plug x_i into f_{i+1}(x), the quadratic part cancels out to -c_i, and adding c_{i+1} gives the difference. Then, when we sum all these differences, it's a telescoping series where all intermediate terms cancel, leaving only the first negative and the last positive, which also cancel each other. So, yes, the sum is zero.Therefore, the sum f₂(x₁) + f₃(x₂) + ... + f₁₀₀(x₉₉) + f₁(x₁₀₀) must equal zero.**Final Answer**boxed{0}

💡Okay, so I have this geometry problem here. It says: In triangle ABC, AB equals AC, angle C is 70 degrees, and I need to find angle B. Hmm, let me try to visualize this. Since AB equals AC, that means triangle ABC is isosceles with AB and AC as the two equal sides. In an isosceles triangle, the angles opposite the equal sides are also equal. So, that would mean angle B is equal to angle C, right? Wait, no, hold on. If AB equals AC, then the angles opposite those sides would be equal. So, angle opposite AB is angle C, and angle opposite AC is angle B. So, actually, angle B equals angle C. But angle C is given as 70 degrees, so angle B should also be 70 degrees. But wait, that doesn't seem right because in a triangle, the sum of all angles should be 180 degrees. If angle B and angle C are both 70 degrees, then angle A would be 180 minus 70 minus 70, which is 40 degrees. That seems possible, but I'm not sure if I'm applying the properties correctly. Maybe I should draw the triangle to get a better idea.Alright, let's sketch triangle ABC where AB equals AC. So, point A is at the top, and BC is the base. Since AB equals AC, the triangle is symmetric along the altitude from A to BC. Now, angle C is given as 70 degrees. Since angle C is at point C, which is one of the base angles, and because AB equals AC, angle B should also be equal to angle C. So, angle B is 70 degrees, and angle A is 40 degrees. Let me check if this adds up to 180 degrees. 70 plus 70 is 140, plus 40 is 180. Okay, that works. So, maybe my initial thought was correct. But I'm still a bit confused because sometimes in isosceles triangles, people get confused about which angles are equal. Let me double-check.In triangle ABC, sides AB and AC are equal, so the angles opposite these sides, which are angle C and angle B respectively, should be equal. So, angle B equals angle C, which is 70 degrees. Therefore, angle B is 70 degrees. Yeah, that seems consistent. I think I was overcomplicating it at first, but now I'm pretty sure that angle B is 70 degrees. To be absolutely certain, let me recall the properties of isosceles triangles. In an isosceles triangle, the angles opposite the equal sides are equal. Since AB equals AC, angles at B and C are equal. Given that angle C is 70 degrees, angle B must also be 70 degrees. So, I think I've got it right.

💡First, determine how many math and science problems Sarah has left after Saturday.She started with 35 math problems and completed 15, leaving her with 20 math problems.She started with 25 science problems and completed 10, leaving her with 15 science problems.In total, she has 35 problems left to do.Next, calculate how many problems are on each of the 5 pages.35 problems ÷ 5 pages = 7 problems per page.Now, find the time Sarah spent per math problem on Saturday.She spent 2 hours on 15 math problems, so the time per math problem is 2 hours ÷ 15 problems = 0.1333 hours per problem.Similarly, calculate the time per science problem.She spent 1.5 hours on 10 science problems, so the time per science problem is 1.5 hours ÷ 10 problems = 0.15 hours per problem.Since all problems take approximately the same amount of time, average the time per problem.(0.1333 hours/problem + 0.15 hours/problem) ÷ 2 = 0.14165 hours per problem.Finally, calculate the total time needed to complete the remaining 35 problems.35 problems × 0.14165 hours/problem = 4.95775 hours.So, it will take Sarah approximately 4.96 hours to finish the remaining problems.

💡Okay, so I have this question about Proposition A and Proposition B. Proposition A is that the derivative of a function f at a point x is zero, which is written as f'(x) = 0. Proposition B is that the point x is an extremum of the differentiable function f(x). The question is asking how Proposition A relates to Proposition B in terms of being a sufficient or necessary condition. The options are A, B, C, or D, where A is a sufficient but not necessary condition, B is a necessary but not sufficient condition, C is both necessary and sufficient, and D is neither sufficient nor necessary.Alright, let me start by recalling what extremum points are. An extremum is a maximum or minimum point of a function. So, if x is an extremum, that means f(x) is either higher or lower than the points around it. Now, I remember from calculus that if a function has an extremum at a point x and it's differentiable there, then the derivative at that point must be zero. So, that would mean that if x is an extremum, then f'(x) = 0. So, in this case, Proposition B implies Proposition A. That makes Proposition A a necessary condition for Proposition B because without f'(x) = 0, x can't be an extremum.But wait, is the converse true? If f'(x) = 0, does that necessarily mean that x is an extremum? Hmm, I think not. I remember something about points where the derivative is zero but they aren't extrema. For example, consider the function f(x) = x³. The derivative is f'(x) = 3x², which is zero at x = 0. But at x = 0, the function doesn't have a maximum or minimum; it's actually a point of inflection. So, in this case, f'(0) = 0, but x = 0 isn't an extremum. That means Proposition A doesn't necessarily lead to Proposition B. So, f'(x) = 0 is not a sufficient condition for x being an extremum.So, putting this together: If x is an extremum, then f'(x) must be zero. So, f'(x) = 0 is a necessary condition for x being an extremum. However, just because f'(x) = 0 doesn't mean x is an extremum; it could be a saddle point or something else. Therefore, f'(x) = 0 is necessary but not sufficient for x being an extremum.Let me think if there are any other cases or exceptions. Well, in some cases, if the derivative changes sign around x, then f'(x) = 0 would imply an extremum. But the question is about the relationship between Proposition A and Proposition B in general. Since Proposition A can be true without Proposition B being true, it's not sufficient. But Proposition B can't be true without Proposition A being true, so it's necessary.I think that's it. So, the answer should be that Proposition A is a necessary but not sufficient condition for Proposition B, which corresponds to option B.

💡Alright, let me try to figure out this problem step by step. So, Millie starts on Monday by putting a quart of seeds into the bird feeder, and 25% of those seeds are millet. That means on Monday, there's 0.25 quarts of millet and 0.75 quarts of other seeds. Each day after that, she adds another quart of the same mix—25% millet and 75% other seeds. But here's the catch: every day, the birds eat 25% of the millet and all of the other seeds. So, the non-millet seeds get eaten completely each day, but only a quarter of the millet gets eaten, leaving three-quarters of it for the next day.The question is asking on which day, just after Millie has added the seeds, will the birds find that more than half the seeds in the feeder are millet. So, we need to figure out when the amount of millet exceeds 50% of the total seeds in the feeder.Let me break this down day by day.**Monday (Day 1):**- Millie adds 1 quart of seeds.- 25% is millet: 0.25 quarts.- 75% is other seeds: 0.75 quarts.- Birds eat 25% of millet: 0.25 * 0.25 = 0.0625 quarts eaten.- Remaining millet: 0.25 - 0.0625 = 0.1875 quarts.- All other seeds (0.75 quarts) are eaten.- Total seeds left: 0.1875 quarts (all millet).**Tuesday (Day 2):**- Millie adds another quart: 0.25 millet + 0.75 other seeds.- Total before birds eat: 0.1875 + 0.25 = 0.4375 millet; 0.75 other seeds.- Birds eat 25% of millet: 0.4375 * 0.25 = 0.109375 quarts eaten.- Remaining millet: 0.4375 - 0.109375 = 0.328125 quarts.- All other seeds (0.75 quarts) are eaten.- Total seeds left: 0.328125 quarts (all millet).**Wednesday (Day 3):**- Millie adds another quart: 0.25 millet + 0.75 other seeds.- Total before birds eat: 0.328125 + 0.25 = 0.578125 millet; 0.75 other seeds.- Birds eat 25% of millet: 0.578125 * 0.25 = 0.144703125 quarts eaten.- Remaining millet: 0.578125 - 0.144703125 = 0.433421875 quarts.- All other seeds (0.75 quarts) are eaten.- Total seeds left: 0.433421875 quarts (all millet).**Thursday (Day 4):**- Millie adds another quart: 0.25 millet + 0.75 other seeds.- Total before birds eat: 0.433421875 + 0.25 = 0.683421875 millet; 0.75 other seeds.- Birds eat 25% of millet: 0.683421875 * 0.25 = 0.17085546875 quarts eaten.- Remaining millet: 0.683421875 - 0.17085546875 = 0.51256640625 quarts.- All other seeds (0.75 quarts) are eaten.- Total seeds left: 0.51256640625 quarts (all millet).**Friday (Day 5):**- Millie adds another quart: 0.25 millet + 0.75 other seeds.- Total before birds eat: 0.51256640625 + 0.25 = 0.76256640625 millet; 0.75 other seeds.- Birds eat 25% of millet: 0.76256640625 * 0.25 = 0.1906416015625 quarts eaten.- Remaining millet: 0.76256640625 - 0.1906416015625 = 0.5719248046875 quarts.- All other seeds (0.75 quarts) are eaten.- Total seeds left: 0.5719248046875 quarts (all millet).Wait, so after each day, the amount of millet is increasing, but only by a fraction each time. It seems like it's approaching 1 quart, but we need to find when it exceeds half of the total seeds, which is 0.5 quarts.Looking at the totals:- After Monday: 0.1875 quarts (18.75%)- After Tuesday: 0.328125 quarts (32.8125%)- After Wednesday: 0.433421875 quarts (43.3421875%)- After Thursday: 0.51256640625 quarts (51.256640625%)- After Friday: 0.5719248046875 quarts (57.19248046875%)So, on Thursday, just after Millie adds the seeds, the total seeds in the feeder are 0.683421875 quarts of millet and 0.75 quarts of other seeds, making a total of 1.433421875 quarts. Wait, that doesn't make sense because each day Millie adds 1 quart, so the total seeds should be 1 quart after she adds them, right?Wait, maybe I made a mistake in calculating the total seeds. Let me rethink this.Each day, Millie adds 1 quart, so the total seeds before the birds eat are 1 quart plus whatever is left from the previous day. But the birds eat all the non-millet seeds each day, so the only seeds that remain are the uneaten millet.So, actually, each day, the total seeds in the feeder just after Millie adds them are 1 quart (the new seeds) plus the remaining millet from the previous day.Therefore, the total seeds just after adding are always 1 quart plus some amount of millet. But the problem states that each day, Millie adds another quart without removing any seeds. So, the total seeds in the feeder each day just after she adds them would be the previous day's remaining millet plus the new quart.But the birds eat 25% of the millet and all of the other seeds each day. So, the remaining seeds after the birds eat are 75% of the millet from the previous day.Wait, maybe I need to model this differently.Let me try to model the amount of millet each day.Let’s denote M_n as the amount of millet just after Millie adds the seeds on day n.On day 1 (Monday):- M_1 = 0.25 quarts (25% of 1 quart)- Birds eat 25% of M_1: 0.25 * 0.25 = 0.0625 quarts eaten- Remaining millet: 0.25 - 0.0625 = 0.1875 quartsOn day 2 (Tuesday):- Millie adds another 0.25 quarts of millet- Total millet before birds eat: 0.1875 + 0.25 = 0.4375 quarts- Birds eat 25% of 0.4375: 0.4375 * 0.25 = 0.109375 quarts eaten- Remaining millet: 0.4375 - 0.109375 = 0.328125 quartsOn day 3 (Wednesday):- Millie adds another 0.25 quarts of millet- Total millet before birds eat: 0.328125 + 0.25 = 0.578125 quarts- Birds eat 25% of 0.578125: 0.578125 * 0.25 = 0.144703125 quarts eaten- Remaining millet: 0.578125 - 0.144703125 = 0.433421875 quartsOn day 4 (Thursday):- Millie adds another 0.25 quarts of millet- Total millet before birds eat: 0.433421875 + 0.25 = 0.683421875 quarts- Birds eat 25% of 0.683421875: 0.683421875 * 0.25 = 0.17085546875 quarts eaten- Remaining millet: 0.683421875 - 0.17085546875 = 0.51256640625 quartsOn day 5 (Friday):- Millie adds another 0.25 quarts of millet- Total millet before birds eat: 0.51256640625 + 0.25 = 0.76256640625 quarts- Birds eat 25% of 0.76256640625: 0.76256640625 * 0.25 = 0.1906416015625 quarts eaten- Remaining millet: 0.76256640625 - 0.1906416015625 = 0.5719248046875 quartsWait, so each day, the total seeds just after Millie adds them are 1 quart (the new seeds) plus the remaining millet from the previous day. But the birds eat all the non-millet seeds each day, so the only seeds that remain are the uneaten millet.Therefore, the total seeds just after Millie adds them on day n are 1 quart (the new seeds) plus the remaining millet from day n-1.But the problem is asking for when more than half the seeds are millet just after Millie has placed the seeds. So, we need to find the day when the amount of millet is more than half of the total seeds in the feeder.Let’s calculate the total seeds and the proportion of millet each day just after Millie adds the seeds.**Monday (Day 1):**- Total seeds: 1 quart- Millet: 0.25 quarts- Proportion: 0.25 / 1 = 25%**Tuesday (Day 2):**- Total seeds: 1 quart (new) + 0.1875 quarts (remaining millet) = 1.1875 quarts- Millet: 0.25 + 0.1875 = 0.4375 quarts- Proportion: 0.4375 / 1.1875 ≈ 36.84%**Wednesday (Day 3):**- Total seeds: 1 quart (new) + 0.328125 quarts (remaining millet) = 1.328125 quarts- Millet: 0.25 + 0.328125 = 0.578125 quarts- Proportion: 0.578125 / 1.328125 ≈ 43.53%**Thursday (Day 4):**- Total seeds: 1 quart (new) + 0.433421875 quarts (remaining millet) = 1.433421875 quarts- Millet: 0.25 + 0.433421875 = 0.683421875 quarts- Proportion: 0.683421875 / 1.433421875 ≈ 47.71%**Friday (Day 5):**- Total seeds: 1 quart (new) + 0.51256640625 quarts (remaining millet) = 1.51256640625 quarts- Millet: 0.25 + 0.51256640625 = 0.76256640625 quarts- Proportion: 0.76256640625 / 1.51256640625 ≈ 50.43%So, on Friday, just after Millie adds the seeds, the proportion of millet is approximately 50.43%, which is just over 50%. Therefore, the answer should be Friday.Wait, but let me double-check the calculations because the proportion on Thursday was 47.71%, which is still below 50%, and on Friday, it's just over 50%. So, Friday is the first day when more than half the seeds are millet.Alternatively, maybe I should model this using a formula instead of day-by-day to see if it aligns.The amount of millet each day can be modeled as a geometric series because each day, 75% of the previous day's millet remains, and 0.25 quarts are added each day.The formula for the amount of millet after n days is:M_n = 0.25 * (1 - (0.75)^n) / (1 - 0.75) = 1 - (0.75)^nWe want M_n / (1 + M_n - 0.25) > 0.5Wait, that might be more complicated. Alternatively, since each day the total seeds are 1 quart plus the remaining millet, and we want millet to be more than half of that total.So, M_n > 0.5 * (1 + M_n - 0.25)Wait, maybe it's better to set up the inequality:M_n > 0.5 * (1 + M_n)Because the total seeds are 1 quart (new) plus the remaining millet from previous days, which is M_n - 0.25 (since 0.25 is added each day). Wait, this is getting confusing.Alternatively, let's denote T_n as the total seeds just after adding on day n, which is 1 quart plus the remaining millet from day n-1, which is 0.75 * M_{n-1}.But M_n = 0.25 + 0.75 * M_{n-1}So, T_n = 1 + 0.75 * M_{n-1}We want M_n / T_n > 0.5Substituting M_n = 0.25 + 0.75 * M_{n-1}So, (0.25 + 0.75 * M_{n-1}) / (1 + 0.75 * M_{n-1}) > 0.5Let’s solve this inequality:(0.25 + 0.75 * M_{n-1}) > 0.5 * (1 + 0.75 * M_{n-1})0.25 + 0.75 * M_{n-1} > 0.5 + 0.375 * M_{n-1}0.75 * M_{n-1} - 0.375 * M_{n-1} > 0.5 - 0.250.375 * M_{n-1} > 0.25M_{n-1} > 0.25 / 0.375 = 2/3 ≈ 0.6667So, we need M_{n-1} > 2/3 quarts.From our earlier calculations:- M_1 = 0.25- M_2 = 0.4375- M_3 = 0.578125- M_4 = 0.683421875- M_5 = 0.76256640625So, M_4 ≈ 0.6834 > 0.6667, which means on day 5 (Friday), the condition is met.Therefore, the answer is Friday.