💡First, I need to determine the time it takes the tailor to make one shirt, one pair of pants, and one jacket based on the given ratio of 1:2:3. Let's assume the time to make one shirt is ( x ) hours. Therefore, the time to make a pair of pants is ( 2x ) hours, and the time to make a jacket is ( 3x ) hours.Next, I'll set up an equation based on the information that the tailor can make 2 shirts, 3 pairs of pants, and 4 jackets in 10 working hours. This gives the equation:[2x + 3 times (2x) + 4 times (3x) = 10]Simplifying this equation:[2x + 6x + 12x = 10 20x = 10 x = 0.5]So, the time to make one shirt is 0.5 hours, one pair of pants is ( 2x = 1 ) hour, and one jacket is ( 3x = 1.5 ) hours.Finally, I'll calculate the total time required to make 14 shirts, 10 pairs of pants, and 2 jackets:[14 times 0.5 + 10 times 1 + 2 times 1.5 = 7 + 10 + 3 = 20 text{ hours}]

💡Okay, so I have this problem about pairing letters, and I need to prove a recurrence relation for the number of different ways to perform the pairing. Let me try to understand the problem first.We have n distinct letters, and each letter can be used twice. So, for each letter, we have two copies. We need to pair these letters into n pairs. Each pair can consist of the same letter or different letters. The number of different ways to do this is denoted by u_n. The key point is that only the order of the pairs or the order within the pairs matters. So, if two pairings are the same except for the order of the pairs or the order within a pair, they are considered the same.The recurrence relation we need to prove is:u_{n+1} = (n + 1) u_n - [n(n - 1)/2] u_{n - 2}Alright, so I need to figure out how to relate u_{n+1} to u_n and u_{n-2}. Let me think about how adding an additional letter affects the number of pairings.Suppose we have n letters, and we add an (n+1)-th letter. Let's call this new letter 'b'. Now, we need to form n+1 pairs using the existing n letters and the new letter 'b'. Each of the existing letters can be used twice, and the new letter 'b' can also be used twice.One approach is to consider how the new letter 'b' can be paired. There are two possibilities: either 'b' is paired with itself, or 'b' is paired with one of the existing letters.Case 1: 'b' is paired with itself. In this case, the remaining n letters need to be paired among themselves. The number of ways to do this is u_n.Case 2: 'b' is paired with one of the existing letters. Let's say 'b' is paired with letter 'a'. Then, we have one 'a' left, which needs to be paired with another letter. But wait, since each letter can be used twice, pairing 'b' with 'a' uses one 'a' and one 'b', leaving one 'a' and one 'b' still available. Hmm, but we need to form n+1 pairs, so perhaps this isn't the right way to think about it.Wait, maybe I need to think differently. If we have n+1 letters, each appearing twice, that's a total of 2(n+1) letters. We need to pair them into n+1 pairs. So, adding the new letter 'b' adds two more letters to the pool, making it 2n + 2 letters. We need to pair these into n+1 pairs.Let me try another approach. Suppose we have a pairing for n letters, which gives us u_n ways. If we add a new letter 'b', we can insert 'b' into the existing pairings in some way. But I'm not sure exactly how.Alternatively, maybe I can use the concept of derangements or some combinatorial argument. Let me think about the total number of ways to pair 2n letters, which is (2n - 1)!!, but in this case, letters can be repeated, so it's different.Wait, no, in our problem, each letter is distinct but can be used twice. So, it's like having two copies of each letter, making a total of 2n letters. We need to pair them into n pairs, where each pair can be two of the same letter or two different letters.So, the total number of pairings is similar to the number of perfect matchings in a multigraph where each vertex has degree 2, but with labeled vertices. Hmm, but I'm not sure if that helps directly.Let me try to think recursively. Suppose I have n+1 letters. How can I relate the number of pairings for n+1 letters to the number of pairings for n letters and n-2 letters.Maybe I can consider fixing the pair that includes the new letter 'b'. There are two possibilities: either 'b' is paired with itself, or 'b' is paired with one of the existing n letters.If 'b' is paired with itself, then the remaining 2n letters (which are two copies of each of the original n letters) need to be paired, which can be done in u_n ways.If 'b' is paired with one of the existing letters, say 'a', then we have one 'a' and one 'b' left. But wait, we have two copies of each letter, so pairing 'b' with 'a' uses one 'a' and one 'b', leaving one 'a' and one 'b' still available. But we need to pair all letters, so we have to pair the remaining letters.Wait, but if we pair 'b' with 'a', then we have one 'a' and one 'b' left, but we also have the other letters. So, perhaps we need to adjust the pairings accordingly.Alternatively, maybe I need to consider that pairing 'b' with an existing letter affects the pairing of that letter. For example, if 'b' is paired with 'a', then the other 'a' must be paired with something else, which could be another letter or itself.This seems complicated. Maybe I should think about the recurrence relation given and see if I can derive it by considering the different ways the new letter can be incorporated.The recurrence is u_{n+1} = (n + 1) u_n - [n(n - 1)/2] u_{n - 2}So, the idea is that u_{n+1} is related to u_n multiplied by (n + 1), minus some term involving u_{n - 2}.Perhaps the (n + 1) factor comes from choosing which letter to pair with the new letter 'b'. There are n existing letters, so n choices, plus the choice of pairing 'b' with itself, making n + 1 choices.But then why subtract [n(n - 1)/2] u_{n - 2}?Maybe because when we pair 'b' with an existing letter, we might be overcounting some configurations. Specifically, if we pair 'b' with two different letters, we might be creating duplicate pairings.Wait, let me think carefully. Suppose we have n letters, and we add 'b'. If we pair 'b' with itself, that's one case, contributing u_n.If we pair 'b' with one of the n letters, say 'a', then we have one 'a' and one 'b' left. But we also have the other letters. So, perhaps the number of ways to pair 'b' with 'a' is equal to the number of ways to pair the remaining letters, which would be u_{n - 1}, since we've used up one 'a' and one 'b'.But wait, no, because we still have another 'a' and another 'b' left. So, actually, pairing 'b' with 'a' doesn't reduce the number of letters by one, but rather by two, since we're using one 'a' and one 'b'.Wait, that doesn't make sense. Let me clarify.We have 2n + 2 letters in total when we add 'b'. Each letter is present twice. So, when we pair 'b' with 'a', we're using one 'a' and one 'b', leaving us with one 'a', one 'b', and the other 2n - 2 letters (which are two copies of each of the remaining n - 1 letters).So, the remaining letters to pair are: one 'a', one 'b', and two copies of each of the remaining n - 1 letters. That's a total of 2n letters.Wait, but we need to pair them into n pairs. So, the remaining letters are: one 'a', one 'b', and two copies of each of the remaining n - 1 letters. That's 2n letters, which need to be paired into n pairs.But how does this relate to u_{n - 2}?Wait, maybe I'm overcomplicating it. Let me try a different approach.Suppose we have n+1 letters. The total number of pairings is u_{n+1}. Let's consider how many ways there are to pair the new letter 'b'.As I thought earlier, 'b' can either be paired with itself or with one of the existing n letters.If 'b' is paired with itself, the remaining 2n letters (two copies of each of the original n letters) can be paired in u_n ways.If 'b' is paired with one of the existing letters, say 'a', then we have one 'a' and one 'b' left. But we also have the other letters. So, the number of ways to pair 'b' with 'a' is equal to the number of ways to pair the remaining letters, which is u_{n}.Wait, but that would suggest that pairing 'b' with 'a' contributes u_n ways, and since there are n choices for 'a', the total contribution would be n u_n.But then, the total number of pairings would be u_n (for pairing 'b' with itself) plus n u_n (for pairing 'b' with each of the n letters), giving (n + 1) u_n. But this can't be right because the recurrence also subtracts a term.So, where is the overcounting happening?Ah, I see. When we pair 'b' with two different letters, say 'a' and 'c', we might be creating configurations that are being counted multiple times. Specifically, if we pair 'b' with 'a' and then 'b' with 'c', we might be double-counting some pairings.Wait, no, because each pairing is unique. If 'b' is paired with 'a', that's a specific pairing, and 'b' can't be paired with both 'a' and 'c' in the same pairing.Wait, perhaps the issue is that when we pair 'b' with 'a', we might be creating a situation where two 'a's are left, which could lead to overcounting.Alternatively, maybe the problem is that when we pair 'b' with 'a', we're effectively removing one 'a' and one 'b' from the pool, but since each letter is present twice, we still have another 'a' and another 'b' left. So, perhaps the number of ways to pair the remaining letters is not u_{n - 1}, but something else.Wait, let's think about it step by step.When we pair 'b' with 'a', we use one 'a' and one 'b', leaving one 'a', one 'b', and two copies of each of the remaining n - 1 letters. So, the remaining letters are: one 'a', one 'b', and two copies of each of the remaining n - 1 letters. That's a total of 2n letters.Now, how many ways are there to pair these 2n letters into n pairs?This seems similar to the original problem but with n letters, except that we have one 'a' and one 'b' instead of two copies each. So, it's not exactly u_n, because in u_n, each letter has two copies.Hmm, so maybe we need to adjust for this.Alternatively, perhaps we can think of it as u_{n} minus some term. But I'm not sure.Wait, maybe I can model this as a recurrence where pairing 'b' with 'a' affects the count in a way that requires subtracting some overcounted cases.Alternatively, perhaps I should consider the inclusion-exclusion principle. When we count all possible pairings by considering pairing 'b' with itself and pairing 'b' with each existing letter, we might be overcounting cases where 'b' is paired with multiple letters, but since each pairing is unique, that shouldn't happen.Wait, no, because in each pairing, 'b' can only be paired once. So, each pairing is either pairing 'b' with itself or with one other letter, and each of these is unique.So, why does the recurrence have a subtraction term?Maybe the issue is that when we pair 'b' with an existing letter, we might be creating configurations that are equivalent to pairing 'b' with itself in some way, leading to overcounting.Alternatively, perhaps the subtraction term accounts for cases where two different letters are paired with 'b', but since 'b' can only be paired once, those cases are invalid and need to be subtracted.Wait, that doesn't make sense because in each pairing, 'b' is only paired once.Wait, maybe the problem is that when we pair 'b' with an existing letter, we might be creating a situation where two letters are both paired with 'b', which is impossible, so we need to subtract those cases.But that can't happen because in a pairing, each element is in exactly one pair. So, 'b' can only be paired once.Hmm, I'm getting confused. Let me try to think of a small example to see what's happening.Let's take n = 1. Then, u_1 is the number of ways to pair two letters, which is 1 (just the pair (a, a)).For n = 2, we have two letters, a and b, each appearing twice. The possible pairings are:1. (a, a) and (b, b)2. (a, b) and (a, b)So, u_2 = 2.Now, let's compute u_3 using the recurrence.According to the recurrence, u_3 = (3) u_2 - [2(2 - 1)/2] u_{1} = 3*2 - [2*1/2]*1 = 6 - 1 = 5.Let's check if u_3 is indeed 5.For n = 3, we have letters a, b, c, each appearing twice. The possible pairings are:1. (a, a), (b, b), (c, c)2. (a, a), (b, c), (b, c)3. (b, b), (a, c), (a, c)4. (c, c), (a, b), (a, b)5. (a, b), (a, c), (b, c)Wait, that's 5 pairings, so u_3 = 5, which matches the recurrence.But how does this help me understand the recurrence?When we go from n=2 to n=3, we add the letter 'c'. The number of pairings increases from 2 to 5.According to the recurrence, u_3 = 3*u_2 - [2*1/2]*u_1 = 6 - 1 = 5.So, the 3*u_2 term is accounting for adding 'c' in different ways, and the subtraction term is correcting for overcounting.But how exactly?When we add 'c', we can either pair 'c' with itself or pair it with one of the existing letters.If we pair 'c' with itself, we get u_2 pairings for the remaining letters, which is 2.If we pair 'c' with an existing letter, say 'a', then we have one 'a' and one 'c' left, and the other letters. But in this case, pairing 'c' with 'a' affects the pairing of 'a's.Wait, in the case of n=2, when we add 'c', pairing 'c' with 'a' would leave one 'a' and one 'c', but we also have another 'a' and another 'c'. So, the remaining letters are one 'a', one 'c', and two 'b's.Wait, no, for n=2, we have letters a, b, each appearing twice. Adding 'c' gives us a, b, c, each appearing twice. So, when we pair 'c' with 'a', we use one 'a' and one 'c', leaving one 'a', one 'c', and two 'b's.So, the remaining letters are one 'a', one 'c', and two 'b's. How many ways can we pair these?We have to pair one 'a', one 'c', and two 'b's into two pairs.The possible pairings are:1. (a, c) and (b, b)2. (a, b) and (c, b)So, that's two ways.Similarly, if we pair 'c' with 'b', we would have one 'b', one 'c', and two 'a's, leading to two pairings.And if we pair 'c' with itself, we have two 'c's, and the remaining letters are two 'a's and two 'b's, which can be paired in two ways.So, total pairings when adding 'c' are:- Pair 'c' with itself: 2 pairings- Pair 'c' with 'a': 2 pairings- Pair 'c' with 'b': 2 pairingsTotal: 2 + 2 + 2 = 6But wait, u_3 is 5, not 6. So, where is the overcounting?Ah, I see. When we pair 'c' with 'a' and then pair the remaining letters, we get two pairings, but one of them might be equivalent to pairing 'c' with itself in some way.Wait, no, in the case of n=2, when we pair 'c' with 'a', the remaining letters are one 'a', one 'c', and two 'b's. Pairing these can be done in two ways: (a, c) and (b, b), or (a, b) and (c, b). Similarly for pairing 'c' with 'b'.But when we count all these, we get 6, but u_3 is 5. So, one of these pairings is being double-counted.Looking back at the list of u_3 pairings:1. (a, a), (b, b), (c, c)2. (a, a), (b, c), (b, c)3. (b, b), (a, c), (a, c)4. (c, c), (a, b), (a, b)5. (a, b), (a, c), (b, c)So, when we pair 'c' with 'a', we get pairings 2 and 5. Similarly, pairing 'c' with 'b' gives pairings 3 and 5. Pairing 'c' with itself gives pairings 1, 4, and 5? Wait, no.Wait, pairing 'c' with itself gives:- (c, c), and then the remaining letters are two 'a's and two 'b's, which can be paired as (a, a) and (b, b), or (a, b) and (a, b). So, that's two pairings: 1 and 4.Wait, but in the list, pairing 1 is (a, a), (b, b), (c, c), and pairing 4 is (c, c), (a, b), (a, b). So, that's two pairings.Then, pairing 'c' with 'a' gives:- (c, a), and then the remaining letters are one 'a', one 'c', and two 'b's. Pairing these as (a, c) and (b, b) gives pairing 2, and pairing as (a, b) and (c, b) gives pairing 5.Similarly, pairing 'c' with 'b' gives:- (c, b), and then the remaining letters are one 'b', one 'c', and two 'a's. Pairing these as (b, c) and (a, a) gives pairing 3, and pairing as (b, a) and (c, a) gives pairing 5.So, pairing 5 is being counted twice: once when pairing 'c' with 'a' and once when pairing 'c' with 'b'. Therefore, the total count is 6, but pairing 5 is counted twice, so we need to subtract one to get the correct count of 5.Therefore, the recurrence relation must account for this overcounting by subtracting the number of such duplicate pairings.In the case of n=2, the number of duplicate pairings is 1, which is equal to [2(2 - 1)/2] u_{0}. Wait, u_{0} is 1, since there's one way to pair zero letters. So, [2(2 - 1)/2] * 1 = 1, which matches the subtraction term.So, in general, when we add a new letter 'b', pairing 'b' with each existing letter can lead to overcounting when two different pairings result in the same overall pairing. Specifically, when we pair 'b' with two different letters, we might end up with the same configuration, leading to duplicates.Therefore, the number of such duplicates is equal to the number of ways to choose two letters from the existing n letters, which is C(n, 2) = n(n - 1)/2, multiplied by the number of ways to pair the remaining letters, which is u_{n - 2}.Hence, the recurrence relation is:u_{n+1} = (n + 1) u_n - [n(n - 1)/2] u_{n - 2}This makes sense because the first term counts all possible ways to add the new letter 'b' either by pairing it with itself or with one of the existing n letters, and the second term subtracts the overcounted cases where pairing 'b' with two different letters results in the same overall pairing.So, putting it all together, the recurrence relation is proven by considering the different ways to incorporate the new letter and correcting for overcounting.

💡Alright, so I've got this problem about a sequence. Let me try to figure it out step by step. First, the problem says that in the sequence {a_n}, the first term a₁ is 1/2. Then, it mentions that the point (a_n, a_{n+1}) lies on the line y = x + 1/2 for any natural number n. Okay, so part (I) is asking for the general term formula for the sequence {a_n}. Hmm, let's see. If the point (a_n, a_{n+1}) is on the line y = x + 1/2, that means when x is a_n, y must be a_{n+1}. So, substituting into the equation, we get:a_{n+1} = a_n + 1/2Oh, so this is a recurrence relation. It tells me that each term is equal to the previous term plus 1/2. That means the sequence is arithmetic, right? Because in an arithmetic sequence, each term increases by a constant difference, which in this case is 1/2.Given that a₁ is 1/2, and the common difference d is 1/2, I can use the formula for the nth term of an arithmetic sequence:a_n = a₁ + (n - 1)dPlugging in the values:a_n = 1/2 + (n - 1)(1/2)Let me compute that:a_n = 1/2 + (n - 1)/2Combine the terms:a_n = (1 + n - 1)/2Simplify numerator:a_n = n/2Wait, that seems straightforward. So, the general term is a_n = n/2. Let me check with the first few terms to make sure.For n = 1: a₁ = 1/2, which matches.For n = 2: a₂ = 2/2 = 1. Let's see if that fits the recurrence. a₂ should be a₁ + 1/2 = 1/2 + 1/2 = 1. Yep, that works.For n = 3: a₃ = 3/2. Using the recurrence, a₃ = a₂ + 1/2 = 1 + 1/2 = 3/2. Perfect.Okay, so part (I) seems done. The general term is a_n = n/2.Now, moving on to part (II). It defines another sequence {b_n} where each term is 1 divided by (a_n times a_{n+1}). So,b_n = 1 / (a_n * a_{n+1})From part (I), we know a_n = n/2 and a_{n+1} = (n + 1)/2. Let's plug those in:b_n = 1 / [(n/2) * ((n + 1)/2)]Multiply the denominators:b_n = 1 / [(n(n + 1))/4]Which is the same as:b_n = 4 / [n(n + 1)]Hmm, okay. So, b_n simplifies to 4 divided by n(n + 1). That looks like it might be a telescoping series when we sum it up.The problem asks for the sum of the first n terms of {b_n}, denoted as T_n. So, T_n = b₁ + b₂ + ... + b_n.Given that b_n = 4 / [n(n + 1)], let's see if we can express this as a telescoping series. I remember that fractions like 1/[n(n + 1)] can be broken down using partial fractions.Let me try that. Let's write:1 / [n(n + 1)] = A/n + B/(n + 1)Multiplying both sides by n(n + 1):1 = A(n + 1) + BnExpanding:1 = An + A + BnCombine like terms:1 = (A + B)n + ASince this must hold for all n, the coefficients of like terms must be equal on both sides. So,A + B = 0 (coefficient of n)A = 1 (constant term)From A = 1, substitute into A + B = 0:1 + B = 0 => B = -1So, the partial fractions decomposition is:1 / [n(n + 1)] = 1/n - 1/(n + 1)Therefore, b_n = 4 / [n(n + 1)] = 4(1/n - 1/(n + 1))Alright, so each term b_n can be written as 4 times (1/n - 1/(n + 1)). Now, let's write out the sum T_n:T_n = Σ (from k=1 to n) [4(1/k - 1/(k + 1))]Factor out the 4:T_n = 4 Σ (from k=1 to n) [1/k - 1/(k + 1)]This is a telescoping series because when we expand the sum, most terms will cancel out. Let's write out the first few terms to see the pattern:For k=1: 1/1 - 1/2For k=2: 1/2 - 1/3For k=3: 1/3 - 1/4...For k=n: 1/n - 1/(n + 1)When we add all these up, the -1/2 from the first term cancels with the +1/2 from the second term, the -1/3 from the second term cancels with the +1/3 from the third term, and so on. This cancellation continues all the way to the last term.So, after cancellation, we're left with the first term of the first expression and the last term of the last expression:T_n = 4 [1 - 1/(n + 1)]Simplify inside the brackets:1 - 1/(n + 1) = (n + 1)/(n + 1) - 1/(n + 1) = n/(n + 1)Therefore, T_n = 4 * (n/(n + 1)) = 4n/(n + 1)Let me double-check with n=1 to make sure. If n=1, T₁ = b₁ = 4/(1*2) = 2. Plugging into 4n/(n + 1): 4*1/(1 + 1) = 4/2 = 2. Correct.For n=2, T₂ = b₁ + b₂ = 4/(1*2) + 4/(2*3) = 2 + 4/6 = 2 + 2/3 = 8/3. Using the formula: 4*2/(2 + 1) = 8/3. Correct.For n=3, T₃ = 8/3 + 4/(3*4) = 8/3 + 1/3 = 9/3 = 3. Formula: 4*3/(3 + 1) = 12/4 = 3. Correct.Looks good. So, the sum T_n is 4n/(n + 1).**Final Answer**(I) The general term is boxed{dfrac{n}{2}}.(II) The sum of the first ( n ) terms is boxed{dfrac{4n}{n + 1}}.

💡Alright, so I have this problem here about a sociological survey in the 7th "Ya" class. It says that 20% of students interested in mathematics are also interested in physics, and 25% of students interested in physics are also interested in mathematics. Only Pete and Vasya are not interested in either subject. The total number of students is more than 20 but less than 30. I need to find out how many students are in the class.Okay, let's break this down. First, I need to understand what the percentages mean. If 20% of math-interested students are also interested in physics, that means there's some overlap between the two groups. Similarly, 25% of physics-interested students are also into math, so there's a reciprocal overlap.Let me define some variables to make this clearer. Let's say:- M = number of students interested in mathematics- P = number of students interested in physics- B = number of students interested in both mathematics and physicsFrom the problem, I know that 20% of M are also in P, so:B = 0.2 * MSimilarly, 25% of P are also in M, so:B = 0.25 * PSince both expressions equal B, I can set them equal to each other:0.2 * M = 0.25 * PLet me solve for one variable in terms of the other. Let's solve for M:M = (0.25 / 0.2) * PM = 1.25 * PHmm, so M is 1.25 times P. That means the number of math-interested students is 25% more than the number of physics-interested students.Now, I also know that only Pete and Vasya are not interested in either subject. So, the total number of students in the class is the number of students interested in math, physics, or both, plus these two.Let me denote the total number of students as T. So:T = (M + P - B) + 2I subtract B because those students are counted twice when I add M and P.From earlier, I have B = 0.2 * M and B = 0.25 * P. Let me substitute B in the equation for T.T = (M + P - 0.2 * M) + 2T = (0.8 * M + P) + 2But I also know that M = 1.25 * P. Let me substitute that in:T = (0.8 * 1.25 * P + P) + 2T = (1 * P + P) + 2T = 2 * P + 2So the total number of students is twice the number of physics-interested students plus 2.Now, the problem says the total number of students is more than 20 but less than 30. So:20 < T < 30Substituting T from above:20 < 2 * P + 2 < 30Let me solve this inequality for P. First, subtract 2 from all parts:18 < 2 * P < 28Then, divide all parts by 2:9 < P < 14So P must be an integer between 9 and 14. Let's list the possible values for P: 10, 11, 12, 13.Now, let's find the corresponding T for each P:- If P = 10, then T = 2*10 + 2 = 22- If P = 11, then T = 2*11 + 2 = 24- If P = 12, then T = 2*12 + 2 = 26- If P = 13, then T = 2*13 + 2 = 28So possible total students are 22, 24, 26, or 28.But I also need to check if these values make sense with the original percentages. Remember, B = 0.2 * M and B = 0.25 * P. Since B must be an integer (you can't have a fraction of a student), let's check each case.Starting with P = 10:- M = 1.25 * 10 = 12.5But M must be an integer, so this is not possible.Next, P = 11:- M = 1.25 * 11 = 13.75Again, not an integer. So this is invalid.Next, P = 12:- M = 1.25 * 12 = 15- B = 0.2 * 15 = 3- B must also equal 0.25 * P = 0.25 * 12 = 3So B = 3, which is consistent. Therefore, T = 26 is a valid solution.Next, P = 13:- M = 1.25 * 13 = 16.25Not an integer, so invalid.So the only valid total number of students is 26.Let me double-check:- P = 12- M = 15- B = 3- Total students = 15 + 12 - 3 + 2 = 26Yes, that adds up correctly. So the class has 26 students.

💡Alright, so I've got this problem here about a 200x200 square that's colored like a chessboard. The question is whether we can make all the cells the same color by using a specific move: taking every 2x3 rectangle and flipping the colors of all its cells. Hmm, interesting.First off, let me visualize the chessboard coloring. It's alternating black and white squares, right? So, every cell has a color opposite to its neighbors. Now, the move involves selecting a 2x3 rectangle and flipping all the colors within it. That means if a cell is white, it becomes black, and vice versa.I wonder, what does flipping a 2x3 rectangle do to the overall color distribution? Well, in a 2x3 rectangle, there are 6 cells. On a chessboard, these 6 cells would consist of 3 white and 3 black squares, right? So flipping them would turn 3 white into black and 3 black into white. That means the number of white and black squares remains the same globally, but their positions change.Wait, but the goal is to make all cells the same color. So, we need to somehow eliminate one color entirely. But if each move just swaps 3 white and 3 black squares, it seems like the total number of white and black squares remains balanced. That makes me think it might not be possible to make all cells the same color because we can't actually change the total count of each color.But maybe I'm missing something. Perhaps there's a way to strategically flip rectangles to gradually shift the color distribution. Let me think about smaller boards first to see if it's possible there.Consider a 2x3 board. If I flip the entire board, which is a 2x3 rectangle, it would flip all 6 cells. Since it's a chessboard, flipping it would just invert the colors, but it's still a chessboard pattern. So, I can't make all cells the same color in a 2x3 board with this move.What about a 4x4 board? If I try to flip 2x3 rectangles within it, I can overlap them in different ways. But each flip still affects 3 white and 3 black squares. It seems like no matter how I flip, I can't get rid of one color entirely because the number of white and black squares remains balanced.Going back to the 200x200 board, it's much larger, but the principle should be the same. Each flip affects an equal number of white and black squares, so the total count remains balanced. Therefore, it's impossible to make all cells the same color because we can't change the overall balance of white and black squares.Wait, but maybe there's a way to use multiple flips to create a larger effect. For example, flipping overlapping rectangles might have a cumulative effect. Let me think about that.If I flip one 2x3 rectangle, it flips 3 white and 3 black squares. If I flip another overlapping 2x3 rectangle, it might flip some of the same squares again, effectively flipping them back. This could potentially create a pattern where certain areas are flipped multiple times, but I'm not sure if that helps in making all cells the same color.Another approach could be to consider the problem in terms of parity. Each cell's color can be represented as a binary state, where 0 is white and 1 is black. Flipping a cell changes its state from 0 to 1 or from 1 to 0. The move of flipping a 2x3 rectangle is equivalent to adding 1 modulo 2 to each of the 6 cells in that rectangle.If we think of the entire board as a system of linear equations over the field GF(2), where each equation represents the desired change in a cell's color, we can analyze whether it's possible to reach a state where all cells are the same color.However, setting up such a system for a 200x200 board would be extremely complex, and I'm not sure if it's feasible to solve manually. But perhaps there's a pattern or invariant that can be identified.Considering the chessboard pattern, each 2x3 rectangle contains an equal number of black and white squares. Therefore, flipping it doesn't change the overall balance of black and white squares on the board. This suggests that any sequence of such moves will preserve the balance between black and white squares.Since the initial board has an equal number of black and white squares (or as close as possible, given that 200x200 is even), and each move preserves this balance, it's impossible to reach a state where all squares are the same color because that would require an imbalance in the number of black and white squares.Therefore, based on the parity argument and the balance of colors, it's impossible to make all cells of the 200x200 square the same color using the described move.

💡First, I need to calculate the price after a 20% discount on the original price of 250 yuan.20% of 250 yuan is 50 yuan. Subtracting this discount from the original price gives me 200 yuan.Next, I need to increase this discounted price by 20%. 20% of 200 yuan is 40 yuan. Adding this increase to the discounted price results in 240 yuan.Finally, I compare the final price of 240 yuan to the original price of 250 yuan. Since 240 yuan is less than 250 yuan, the current price has been lowered.