💡First, I need to calculate the dot product of the two vectors:[begin{pmatrix} 4 -5 6 -3 end{pmatrix} text{ and } begin{pmatrix} -2 8 -7 4 end{pmatrix}]The dot product is found by multiplying corresponding components and then summing the results.1. Multiply the first components: (4 times (-2) = -8)2. Multiply the second components: (-5 times 8 = -40)3. Multiply the third components: (6 times (-7) = -42)4. Multiply the fourth components: (-3 times 4 = -12)Now, add all these products together:[-8 + (-40) + (-42) + (-12) = -102]Therefore, the dot product of the two vectors is (-102).

💡First, I need to determine the possible number of members per row that divides evenly into the total number of choir members, which is 90.I'll start by listing all the divisors of 90. These are the numbers that can multiply by another integer to result in 90. The divisors of 90 are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.Next, I need to identify which of these divisors fall within the specified range of 6 to 15 members per row. From the list of divisors, the numbers that meet this criterion are 6, 9, 10, and 15.Therefore, there are 4 possible row-lengths that satisfy the given conditions.

💡First, I need to calculate the values of (6!) and (8!).Calculating (6!):[6! = 6 times 5 times 4 times 3 times 2 times 1 = 720]Calculating (8!):[8! = 8 times 7 times 6! = 8 times 7 times 720 = 40320]Next, I add 2 to each of these factorial values:[6! + 2 = 720 + 2 = 722][8! + 2 = 40320 + 2 = 40322]Now, I need to find the greatest common divisor (GCD) of 722 and 40322. To do this, I can use the Euclidean algorithm.First, subtract 722 from 40322:[40322 - 722 = 39600]Now, find the GCD of 722 and 39600. I can continue this process until I find a common divisor.Subtracting 722 from 39600 multiple times:[39600 - 722 times 54 = 39600 - 39168 = 432]Now, find the GCD of 722 and 432.Subtracting 432 from 722:[722 - 432 = 290]Now, find the GCD of 432 and 290.Subtracting 290 from 432:[432 - 290 = 142]Now, find the GCD of 290 and 142.Subtracting 142 from 290:[290 - 142 = 148]Now, find the GCD of 142 and 148.Subtracting 142 from 148:[148 - 142 = 6]Now, find the GCD of 142 and 6.Dividing 142 by 6:[142 div 6 = 23 text{ with a remainder of } 4]Now, find the GCD of 6 and 4.Subtracting 4 from 6:[6 - 4 = 2]Now, find the GCD of 4 and 2.Dividing 4 by 2:[4 div 2 = 2 text{ with a remainder of } 0]Since the remainder is 0, the GCD is 2.Therefore, the greatest common divisor of (6! + 2) and (8! + 2) is (boxed{2}).

💡Alright, so I have this problem about two factories producing clothing sets. Let me try to understand what's going on here. Both Factory A and Factory B make the same type of clothing, which I assume means they produce both tops and pants. The goal is to find out the maximum number of clothing sets they can produce together in one month.First, let's break down the information given:- Factory A produces 2700 sets per month. The time ratio for producing tops to pants is 2:1. So, for every 2 units of time spent on tops, they spend 1 unit on pants.- Factory B produces 3600 sets per month. Their time ratio for tops to pants is 3:2. That means for every 3 units of time on tops, they spend 2 units on pants.I need to figure out how they can work together to maximize the number of clothing sets produced. I think the key here is to optimize their production by possibly having one factory focus more on tops and the other on pants, or some combination of that, to balance the production and avoid bottlenecks.Let me start by understanding how much time each factory spends on tops and pants.For Factory A:- The time ratio is 2:1 for tops to pants. So, the total parts are 2 + 1 = 3 parts.- Tops take 2/3 of the time, and pants take 1/3 of the time.For Factory B:- The time ratio is 3:2 for tops to pants. So, the total parts are 3 + 2 = 5 parts.- Tops take 3/5 of the time, and pants take 2/5 of the time.Now, if they work together, maybe they can specialize in the part they are more efficient at. That is, Factory A, which spends more time on tops, could focus more on tops, and Factory B, which also spends more time on tops, could also focus more on tops. But wait, both factories are spending more time on tops than pants. That might not be the most efficient way to balance the production.Alternatively, maybe Factory A should focus on pants since they spend less time on them, and Factory B can focus on tops since they spend more time on them. Let me think about that.But first, let's figure out how much time each factory spends on each part.For Factory A:- Produces 2700 sets per month.- Tops take 2/3 of the time, so the number of tops produced is (2/3) * 2700 = 1800 tops.- Pants take 1/3 of the time, so the number of pants produced is (1/3) * 2700 = 900 pants.Wait, but a clothing set consists of one top and one pair of pants. So, if Factory A produces 1800 tops and 900 pants in a month, they can only make 900 complete sets because they have only 900 pants. That doesn't make sense because the problem states they produce 2700 sets per month. Hmm, maybe I'm misunderstanding the time ratio.Let me re-examine the problem. It says the time ratio of producing tops to pants is 2:1 for Factory A. So, for every 2 units of time spent on tops, they spend 1 unit on pants. That doesn't necessarily mean the number of tops to pants is 2:1. It's about the time spent, not the quantity produced.So, perhaps I need to calculate the time taken to produce one top and one pair of pants for each factory.Let's denote:- For Factory A: - Time to produce one top: t1 - Time to produce one pair of pants: t2 - Given the ratio t1:t2 = 2:1, so t1 = 2*t2- For Factory B: - Time to produce one top: t3 - Time to produce one pair of pants: t4 - Given the ratio t3:t4 = 3:2, so t3 = (3/2)*t4Now, let's assume that Factory A works for T hours in a month. Then, the total time spent on tops is 2*t2 per set, and on pants is t2 per set. Since they produce 2700 sets, the total time spent is 2700*(2*t2 + t2) = 2700*3*t2 = 8100*t2 = T.Similarly, for Factory B, the total time spent is 3600*(3*t4 + 2*t4) = 3600*5*t4 = 18000*t4 = T.So, for both factories, T = 8100*t2 = 18000*t4.Therefore, t2 = (18000/8100)*t4 = (20/9)*t4 ≈ 2.222*t4.So, t2 is approximately 2.222 times t4.Now, if they work together, they can pool their resources. Let's see how much time they can spend on tops and pants.Total time available from both factories: T + T = 2*T.But wait, actually, each factory has its own time T, so combined, they have 2*T.But we need to figure out how to allocate this combined time to produce as many sets as possible.Each set requires one top and one pair of pants. So, the production of tops and pants needs to be balanced.Let me denote:- x = number of tops produced- y = number of pants producedWe need x = y to have complete sets.But since both factories can produce both tops and pants, we need to decide how much time each factory spends on tops and pants to maximize x = y.Let me denote:- For Factory A: - Time spent on tops: a - Time spent on pants: b - So, a + b = T - Number of tops produced: a / t1 = a / (2*t2) - Number of pants produced: b / t2- For Factory B: - Time spent on tops: c - Time spent on pants: d - So, c + d = T - Number of tops produced: c / t3 = c / (1.5*t4) - Number of pants produced: d / t4Total tops produced: (a / (2*t2)) + (c / (1.5*t4)) = xTotal pants produced: (b / t2) + (d / t4) = yWe need x = y.Also, a + b = T and c + d = T.We need to maximize x = y.This seems a bit complicated. Maybe I can express everything in terms of t4, since we have t2 = (20/9)*t4.So, t2 = (20/9)*t4.Therefore, t1 = 2*t2 = (40/9)*t4.And t3 = 1.5*t4.So, let's rewrite the equations:For Factory A:- Number of tops: a / t1 = a / (40/9*t4) = (9a)/(40*t4)- Number of pants: b / t2 = b / (20/9*t4) = (9b)/(20*t4)For Factory B:- Number of tops: c / t3 = c / (1.5*t4) = (2c)/(3*t4)- Number of pants: d / t4Total tops: (9a)/(40*t4) + (2c)/(3*t4) = xTotal pants: (9b)/(20*t4) + (d)/t4 = yWe need x = y.Also, a + b = T and c + d = T.We can express b = T - a and d = T - c.Substituting:Total tops: (9a)/(40*t4) + (2c)/(3*t4) = xTotal pants: (9(T - a))/(20*t4) + (T - c)/t4 = ySet x = y:(9a)/(40*t4) + (2c)/(3*t4) = (9(T - a))/(20*t4) + (T - c)/t4Multiply both sides by t4 to eliminate denominators:(9a)/40 + (2c)/3 = (9(T - a))/20 + (T - c)Let's multiply everything by 120 to eliminate denominators:120*(9a)/40 + 120*(2c)/3 = 120*(9(T - a))/20 + 120*(T - c)Simplify:3*9a + 40*2c = 6*9(T - a) + 120(T - c)27a + 80c = 54(T - a) + 120(T - c)27a + 80c = 54T - 54a + 120T - 120cCombine like terms:27a + 80c + 54a + 120c = 54T + 120T81a + 200c = 174TNow, we have one equation: 81a + 200c = 174TWe need another equation to solve for a and c. Since we want to maximize x = y, we can express x in terms of a and c.From total tops:x = (9a)/(40*t4) + (2c)/(3*t4)We can factor out 1/t4:x = (9a/40 + 2c/3)/t4Similarly, from total pants:y = (9(T - a))/(20*t4) + (T - c)/t4= (9(T - a)/20 + (T - c))/t4But since x = y, we already used that to get the equation above.So, we need to maximize x, which is (9a/40 + 2c/3)/t4, subject to 81a + 200c = 174T.Let me express c in terms of a from the equation:81a + 200c = 174T200c = 174T - 81ac = (174T - 81a)/200Now, substitute c into x:x = (9a/40 + 2*(174T - 81a)/200/3)/t4Simplify:First, simplify 2*(174T - 81a)/200/3:= (2*(174T - 81a))/(200*3)= (348T - 162a)/600= (58T - 27a)/100So, x = (9a/40 + (58T - 27a)/100)/t4Combine the terms:Find a common denominator for 40 and 100, which is 200.= (45a/200 + (116T - 54a)/200)/t4= (45a + 116T - 54a)/200/t4= (-9a + 116T)/200/t4So, x = (116T - 9a)/200/t4To maximize x, we need to minimize a, since x decreases as a increases.But a cannot be negative, so the minimum value of a is 0.If a = 0, then c = (174T - 0)/200 = 174T/200 = 87T/100So, c = 0.87TNow, check if c <= T, which it is since 0.87T < T.So, with a = 0 and c = 0.87T, we can calculate x:x = (116T - 0)/200/t4 = 116T/(200*t4) = 29T/(50*t4)Now, let's find T in terms of t4.From Factory A:T = 8100*t2 = 8100*(20/9)*t4 = 8100*(20/9)*t4 = 900*20*t4 = 18000*t4Wait, that can't be right. Wait, earlier we had T = 8100*t2 and T = 18000*t4.So, 8100*t2 = 18000*t4Therefore, t2 = (18000/8100)*t4 = (20/9)*t4 ≈ 2.222*t4So, T = 8100*t2 = 8100*(20/9)*t4 = 900*20*t4 = 18000*t4Wait, that's the same as Factory B's T.So, T = 18000*t4Therefore, x = 29T/(50*t4) = 29*(18000*t4)/(50*t4) = 29*18000/50 = 29*360 = 10440Wait, that seems too high because Factory A alone produces 2700 sets and Factory B produces 3600 sets, so together they should produce more than 6300, but 10440 seems excessive.Maybe I made a mistake in the calculations.Let me go back.We had:x = (116T - 9a)/200/t4With a = 0, x = 116T/(200*t4)But T = 18000*t4So, x = 116*18000*t4/(200*t4) = 116*18000/200 = 116*90 = 10440Hmm, that's the same result.But considering that Factory A can only produce 2700 sets in a month, and Factory B 3600, together they can't produce more than 6300 unless they specialize.Wait, but if they specialize, maybe they can produce more.Wait, let's think differently.If Factory A focuses entirely on pants, how many pants can they produce?Factory A's time ratio is 2:1 for tops to pants.So, if they spend all their time on pants, the time per pair of pants is t2.Total time T = 8100*t2So, number of pants = T / t2 = 8100*t2 / t2 = 8100 pants.Similarly, if Factory B focuses entirely on tops, how many tops can they produce?Factory B's time ratio is 3:2 for tops to pants.So, time per top is t3 = 1.5*t4Total time T = 18000*t4Number of tops = T / t3 = 18000*t4 / (1.5*t4) = 12000 tops.So, if Factory A makes 8100 pants and Factory B makes 12000 tops, the number of sets they can make is limited by the smaller number, which is 8100.But wait, Factory A can only produce 2700 sets in a month, which includes both tops and pants. If they focus entirely on pants, they can produce more pants, but does that mean they can produce 8100 pants in a month?Wait, no. Because Factory A's total production capacity is 2700 sets per month, which is 2700 tops and 2700 pants if they were balanced.But since they have a time ratio, their actual production is skewed.Wait, maybe I need to calculate how many pants Factory A can produce if they focus entirely on pants.Given that the time ratio is 2:1 for tops to pants, the time to produce one top is twice the time to produce one pair of pants.So, if Factory A spends all their time on pants, they can produce more pants.Let me denote:Let t be the time to produce one pair of pants.Then, the time to produce one top is 2t.Factory A's total time per month is T = 2700*(2t + t) = 2700*3t = 8100tIf they focus entirely on pants, the number of pants produced is T / t = 8100t / t = 8100 pants.Similarly, for Factory B:Time ratio 3:2 for tops to pants.Let t' be the time to produce one pair of pants.Then, time to produce one top is (3/2)t'Factory B's total time per month is T = 3600*( (3/2)t' + t' ) = 3600*( (5/2)t' ) = 9000t'If they focus entirely on tops, the number of tops produced is T / ( (3/2)t' ) = 9000t' / ( (3/2)t' ) = 9000*(2/3) = 6000 tops.So, if Factory A produces 8100 pants and Factory B produces 6000 tops, the number of sets they can make is limited by the smaller number, which is 6000.But wait, Factory A can produce 8100 pants, and Factory B can produce 6000 tops. So, they can make 6000 sets, using 6000 tops and 6000 pants, leaving 2100 pants unused.But Factory A could also produce some tops while producing pants, and Factory B could produce some pants while producing tops, to balance the production.Wait, but if Factory A focuses entirely on pants, they can produce 8100 pants, and Factory B focuses entirely on tops, producing 6000 tops. Then, they can make 6000 sets, which is more than the sum of their individual productions (2700 + 3600 = 6300). Wait, 6000 is less than 6300, so that's not better.Wait, maybe I'm missing something.Alternatively, if they both focus on the part they are more efficient at.Factory A is more efficient at pants since they spend less time on them (time ratio 2:1, so more time on tops, meaning less efficient at pants? Wait, no, actually, the time ratio is tops to pants, so higher ratio means more time on tops, meaning less efficient at pants.Wait, actually, the time ratio is the time spent on tops compared to pants. So, a higher ratio means they spend more time on tops, implying they are less efficient at producing pants.Similarly, Factory B has a time ratio of 3:2, meaning they spend more time on tops than pants, so they are less efficient at pants as well.So, both factories are less efficient at pants, but Factory A is more skewed towards tops (2:1) compared to Factory B (3:2).So, maybe Factory A should focus more on tops, and Factory B should focus more on pants? Wait, but Factory B is also less efficient at pants.This is getting confusing.Let me try another approach.Let me assume that both factories can work together to produce tops and pants, and we need to find the optimal allocation of their time to maximize the number of sets.Each set requires one top and one pair of pants.Let me denote:Let x be the number of sets produced.Then, we need x tops and x pants.The time required by Factory A to produce x tops and x pants is:For Factory A:- Time to produce x tops: x * t1 = x * 2*t2- Time to produce x pants: x * t2- Total time: x*(2*t2 + t2) = 3x*t2But Factory A's total time is T = 8100*t2So, 3x*t2 <= 8100*t2 => 3x <= 8100 => x <= 2700Similarly, for Factory B:- Time to produce x tops: x * t3 = x * 1.5*t4- Time to produce x pants: x * t4- Total time: x*(1.5*t4 + t4) = 2.5x*t4Factory B's total time is T = 18000*t4So, 2.5x*t4 <= 18000*t4 => 2.5x <= 18000 => x <= 7200But since both factories are working together, the total time they can spend is T + T = 2*T.Wait, but actually, each factory has its own time T, so combined, they have 2*T.But we need to consider the time each factory spends on tops and pants.Let me denote:For Factory A:- a = time spent on tops- b = time spent on pants- a + b = TFor Factory B:- c = time spent on tops- d = time spent on pants- c + d = TTotal tops produced: (a / t1) + (c / t3) = xTotal pants produced: (b / t2) + (d / t4) = xWe need to maximize x.Given that t1 = 2*t2 and t3 = 1.5*t4, and T = 8100*t2 = 18000*t4.So, t2 = (18000/8100)*t4 = (20/9)*t4 ≈ 2.222*t4Therefore, t1 = 2*t2 = (40/9)*t4 ≈ 4.444*t4t3 = 1.5*t4So, let's express everything in terms of t4.For Factory A:- a / t1 = a / (40/9*t4) = (9a)/(40*t4)- b / t2 = b / (20/9*t4) = (9b)/(20*t4)For Factory B:- c / t3 = c / (1.5*t4) = (2c)/(3*t4)- d / t4Total tops: (9a)/(40*t4) + (2c)/(3*t4) = xTotal pants: (9b)/(20*t4) + (d)/t4 = xAlso, a + b = T and c + d = T.Express b = T - a and d = T - c.Substitute into pants equation:(9(T - a))/(20*t4) + (T - c)/t4 = xNow, set tops equal to pants:(9a)/(40*t4) + (2c)/(3*t4) = (9(T - a))/(20*t4) + (T - c)/t4Multiply both sides by t4:(9a)/40 + (2c)/3 = (9(T - a))/20 + (T - c)Multiply everything by 120 to eliminate denominators:27a + 80c = 54(T - a) + 120(T - c)Expand:27a + 80c = 54T - 54a + 120T - 120cCombine like terms:27a + 80c + 54a + 120c = 54T + 120T81a + 200c = 174TNow, we have:81a + 200c = 174TWe need to maximize x, which is:x = (9a)/(40*t4) + (2c)/(3*t4)Express x in terms of a and c:x = (9a + (80c)/3)/40*t4But we have 81a + 200c = 174TLet me express c in terms of a:200c = 174T - 81ac = (174T - 81a)/200Substitute into x:x = (9a + (80*(174T - 81a)/200)/3)/40*t4Simplify:First, simplify the term with c:80*(174T - 81a)/200 = (80/200)*(174T - 81a) = (2/5)*(174T - 81a) = (348T - 162a)/5Now, divide by 3:(348T - 162a)/15So, x = (9a + (348T - 162a)/15)/40*t4Combine terms:Multiply numerator and denominator appropriately to combine:= (135a + 348T - 162a)/15 /40*t4= (-27a + 348T)/15 /40*t4= (348T - 27a)/600*t4To maximize x, we need to minimize a, since x decreases as a increases.The minimum value of a is 0.So, set a = 0:x = (348T - 0)/600*t4 = 348T/(600*t4) = 29T/(50*t4)Now, recall that T = 18000*t4So, x = 29*(18000*t4)/(50*t4) = 29*18000/50 = 29*360 = 10440Wait, that's the same result as before, but it seems too high because individually they produce 2700 and 3600, totaling 6300.But if they specialize, Factory A can produce 8100 pants and Factory B can produce 6000 tops, making 6000 sets, which is less than 10440.This discrepancy suggests that my approach might be flawed.Perhaps I need to consider that the time spent on tops and pants by each factory cannot exceed their total time T.Let me try a different method.Let me assume that both factories work together to produce x sets.Each set requires one top and one pair of pants.So, total tops needed: xTotal pants needed: xNow, let's calculate the time required by each factory to produce their share of tops and pants.For Factory A:- Let a be the number of tops produced by Factory A- Let b be the number of pants produced by Factory A- So, a + b = 2700 (since they produce 2700 sets per month)- Time ratio: 2:1 for tops to pants- So, time spent on tops: 2t per top- Time spent on pants: t per pair- Total time: a*2t + b*t = TSimilarly, for Factory B:- Let c be the number of tops produced by Factory B- Let d be the number of pants produced by Factory B- So, c + d = 3600- Time ratio: 3:2 for tops to pants- So, time spent on tops: 1.5t' per top- Time spent on pants: t' per pair- Total time: c*1.5t' + d*t' = TBut since both factories are working together, their total time is T + T = 2T.But I'm getting confused again.Maybe I should use linear programming.Let me define variables:Let x be the number of sets produced.Each set requires one top and one pair of pants.Total tops produced: xTotal pants produced: xNow, let's calculate how much time each factory spends on tops and pants.For Factory A:- Let a be the time spent on tops- Let b be the time spent on pants- a + b = T- Number of tops produced: a / t1 = a / (2*t2)- Number of pants produced: b / t2- So, a / (2*t2) + c / (1.5*t4) = x- And b / t2 + d / t4 = xWait, this is getting too tangled.Maybe I should look for a simpler approach.Let me consider the time each factory takes to produce one set.For Factory A:- Time per set: t1 + t2 = 2t2 + t2 = 3t2- So, time per set = 3t2- Therefore, t2 = T / (3*2700) = T / 8100For Factory B:- Time per set: t3 + t4 = 1.5t4 + t4 = 2.5t4- So, time per set = 2.5t4- Therefore, t4 = T / (2.5*3600) = T / 9000Now, t2 = T / 8100t4 = T / 9000So, t2 = (T / 8100) = (1/8100)Tt4 = (1/9000)TNow, if they work together, the total time available is 2T.But we need to find how to allocate this time to produce x sets.Each set requires one top and one pair of pants.Let me denote:- Let a be the time Factory A spends on tops- Let b be the time Factory A spends on pants- Let c be the time Factory B spends on tops- Let d be the time Factory B spends on pantsSo, a + b = T (Factory A's total time)c + d = T (Factory B's total time)Total tops produced:= (a / t1) + (c / t3)= (a / (2*t2)) + (c / (1.5*t4))= (a / (2*(T/8100))) + (c / (1.5*(T/9000)))= (a * 8100 / (2T)) + (c * 9000 / (1.5T))= (a * 4050 / T) + (c * 6000 / T)Total pants produced:= (b / t2) + (d / t4)= (b / (T/8100)) + (d / (T/9000))= (b * 8100 / T) + (d * 9000 / T)We need total tops = total pants = xSo,(4050a + 6000c)/T = (8100b + 9000d)/TSimplify:4050a + 6000c = 8100b + 9000dDivide both sides by 150:27a + 40c = 54b + 60dNow, we also have:a + b = Tc + d = TSo, b = T - ad = T - cSubstitute into the equation:27a + 40c = 54(T - a) + 60(T - c)Expand:27a + 40c = 54T - 54a + 60T - 60cCombine like terms:27a + 40c + 54a + 60c = 54T + 60T81a + 100c = 114TNow, we have:81a + 100c = 114TWe need to maximize x, which is:x = (4050a + 6000c)/TExpress x in terms of a and c:x = (4050a + 6000c)/TWe can express c in terms of a from the equation:81a + 100c = 114T100c = 114T - 81ac = (114T - 81a)/100Substitute into x:x = (4050a + 6000*(114T - 81a)/100)/TSimplify:First, simplify the term with c:6000*(114T - 81a)/100 = 60*(114T - 81a) = 6840T - 4860aSo, x = (4050a + 6840T - 4860a)/T= (6840T - 810a)/T= 6840 - 810a/TTo maximize x, we need to minimize a.The minimum value of a is 0.So, set a = 0:x = 6840 - 0 = 6840But this can't be right because Factory A alone produces 2700 sets, and Factory B 3600, so together they can't produce 6840 sets unless they are working together efficiently.Wait, but 6840 is more than the sum of their individual productions (2700 + 3600 = 6300). So, it's possible if they specialize.But earlier, when I considered specializing, I got 6000 sets, which is less than 6840.This suggests that there's a more efficient way to allocate their time to produce more sets.Let me check the calculations again.From the equation:81a + 100c = 114TWhen a = 0:c = 114T / 100 = 1.14TBut c cannot exceed T, since c + d = T.So, c = 1.14T is not possible because c <= T.Therefore, the maximum c can be is T, which would require a to be:81a + 100T = 114T81a = 14Ta = (14/81)T ≈ 0.1728TSo, a ≈ 0.1728TThen, c = TNow, calculate x:x = (4050a + 6000c)/T= (4050*0.1728T + 6000*T)/T= (4050*0.1728 + 6000)≈ (700 + 6000)= 6700Ah, so x ≈ 6700Therefore, the maximum number of sets they can produce together is 6700.This makes sense because it's more than the sum of their individual productions (6300) but less than the theoretical maximum when a = 0 which was 6840, but that was not feasible due to c exceeding T.So, the optimal allocation is:Factory A spends approximately 17.28% of its time on tops and the rest on pants.Factory B spends all its time on tops.This results in 6700 sets produced.Therefore, the maximum number of sets they can produce together in one month is 6700.

💡Okay, so I have this functional equation to solve: ( f(m + f(n)) = f(f(m)) + f(n) ) for all non-negative integers ( m ) and ( n ). I need to find all functions ( f ) from ( mathbb{N} ) to ( mathbb{N} ) that satisfy this condition. First, I should probably start by plugging in some specific values for ( m ) and ( n ) to see if I can get some useful information about ( f ). A common strategy is to set one of the variables to zero and see what happens. Let me try that.Let’s set ( m = 0 ) and ( n = 0 ). Then the equation becomes:[ f(0 + f(0)) = f(f(0)) + f(0) ]Simplifying, that's:[ f(f(0)) = f(f(0)) + f(0) ]Hmm, if I subtract ( f(f(0)) ) from both sides, I get:[ 0 = f(0) ]So, ( f(0) = 0 ). That's a good start. It tells me that the function ( f ) maps zero to zero.Next, maybe I can set ( n = 0 ) in the original equation and see what happens. Let's do that:[ f(m + f(0)) = f(f(m)) + f(0) ]But since ( f(0) = 0 ), this simplifies to:[ f(m) = f(f(m)) ]Interesting. So, applying ( f ) twice is the same as applying it once. This suggests that ( f ) is idempotent, meaning that ( f ) restricted to its image is the identity function. In other words, for any ( k ) in the image of ( f ), ( f(k) = k ).Let me denote the image of ( f ) as ( E ), so ( E = { f(n) mid n in mathbb{N} } ). If ( E ) only contains zero, then ( f ) is the zero function. That is, ( f(n) = 0 ) for all ( n ). Let me check if this works.If ( f(n) = 0 ) for all ( n ), then the left-hand side of the original equation becomes:[ f(m + f(n)) = f(m + 0) = f(m) = 0 ]And the right-hand side becomes:[ f(f(m)) + f(n) = f(0) + 0 = 0 + 0 = 0 ]So, yes, the zero function satisfies the equation. But I need to check if there are other solutions as well.Assuming ( E ) is not just {0}, let's explore further. Let me take ( m ) and ( n ) from ( E ). Since ( f(m) = m ) and ( f(n) = n ) for ( m, n in E ), substituting into the original equation gives:[ f(m + n) = f(m) + n = m + n ]So, ( f(m + n) = m + n ) for all ( m, n in E ). This suggests that ( E ) is closed under addition. If ( E ) is closed under addition, it must be a subsemigroup of ( mathbb{N} ). The simplest non-trivial subsemigroups of ( mathbb{N} ) are those generated by a single element, say ( d ), meaning ( E = { kd mid k in mathbb{N} } ) for some fixed ( d in mathbb{N} ). Let me assume that ( E ) is generated by some ( d ), so ( E = { 0, d, 2d, 3d, ldots } ). Then, for any ( n ), ( f(n) ) must be a multiple of ( d ). Let me denote ( f(n) = k_n d ) where ( k_n ) is a non-negative integer.Now, let's see if this assumption holds. If ( f(n) = k_n d ), then the original equation becomes:[ f(m + k_n d) = f(f(m)) + k_n d ]But ( f(f(m)) = f(k_m d) = k_{k_m d} d ). Since ( k_{k_m d} ) is just another integer, let's denote it as ( k'_m ). So, the equation becomes:[ f(m + k_n d) = k'_m d + k_n d ]But ( f(m + k_n d) ) should also be a multiple of ( d ), say ( k_{m + k_n d} d ). Therefore:[ k_{m + k_n d} d = (k'_m + k_n) d ]Dividing both sides by ( d ), we get:[ k_{m + k_n d} = k'_m + k_n ]This suggests that the function ( k ) (which maps ( n ) to ( k_n )) satisfies:[ k(m + k_n d) = k(f(m)) + k_n ]But since ( f(m) = k_m d ), we have:[ k(m + k_n d) = k(k_m d) + k_n ]This seems a bit abstract. Maybe I need to make a more specific assumption about ( f ). Let me suppose that ( f(n) = c cdot n ) for some constant ( c ). Let's test this.If ( f(n) = c n ), then substituting into the original equation:[ f(m + f(n)) = f(m + c n) = c(m + c n) = c m + c^2 n ]On the other hand, the right-hand side is:[ f(f(m)) + f(n) = f(c m) + c n = c(c m) + c n = c^2 m + c n ]So, equating both sides:[ c m + c^2 n = c^2 m + c n ]Rearranging terms:[ c m - c^2 m = c n - c^2 n ][ c(1 - c) m = c(1 - c) n ]For this to hold for all ( m ) and ( n ), either ( c(1 - c) = 0 ) or ( m = n ). Since ( m ) and ( n ) are arbitrary, the only possibility is ( c(1 - c) = 0 ). Therefore, ( c = 0 ) or ( c = 1 ).If ( c = 0 ), then ( f(n) = 0 ) for all ( n ), which is the zero function we already considered. If ( c = 1 ), then ( f(n) = n ) for all ( n ). Let's check if this works.If ( f(n) = n ), then the left-hand side becomes:[ f(m + f(n)) = f(m + n) = m + n ]And the right-hand side becomes:[ f(f(m)) + f(n) = f(m) + n = m + n ]So, yes, the identity function also satisfies the equation.Wait a minute, so both the zero function and the identity function satisfy the equation. Are there any other functions that could satisfy it? Maybe functions that are zero for some inputs and identity for others?Let me consider a function that is zero for some numbers and identity for others. For example, suppose ( f(n) = 0 ) if ( n ) is even, and ( f(n) = n ) if ( n ) is odd. Let's test this.Take ( m = 1 ) and ( n = 1 ). Then:Left-hand side: ( f(1 + f(1)) = f(1 + 1) = f(2) = 0 )Right-hand side: ( f(f(1)) + f(1) = f(1) + 1 = 1 + 1 = 2 )But ( 0 neq 2 ), so this function doesn't work. Hmm, maybe such piecewise functions don't work because they disrupt the consistency required by the equation. Let me think differently.Earlier, I considered that the image ( E ) is a subsemigroup generated by ( d ). So, perhaps ( f(n) ) is a multiple of ( d ) for some fixed ( d ). Let me formalize this.Suppose ( f(n) = d cdot g(n) ) where ( g(n) ) is some function from ( mathbb{N} ) to ( mathbb{N} ). Then, substituting into the original equation:[ f(m + f(n)) = f(m + d g(n)) = d cdot g(m + d g(n)) ]And the right-hand side:[ f(f(m)) + f(n) = d cdot g(d g(m)) + d cdot g(n) = d (g(d g(m)) + g(n)) ]So, equating both sides:[ d cdot g(m + d g(n)) = d (g(d g(m)) + g(n)) ]Dividing both sides by ( d ):[ g(m + d g(n)) = g(d g(m)) + g(n) ]This seems similar to the original equation but with ( g ) instead of ( f ). Maybe I can find a function ( g ) that satisfies this.If I set ( g(n) = k ) for some constant ( k ), then:Left-hand side: ( g(m + d k) = k )Right-hand side: ( g(d k) + k = k + k = 2k )So, ( k = 2k ) implies ( k = 0 ). Thus, ( g(n) = 0 ) for all ( n ), which gives ( f(n) = 0 ), the zero function.Alternatively, if ( g(n) = n ), then:Left-hand side: ( g(m + d n) = m + d n )Right-hand side: ( g(d m) + n = d m + n )So, ( m + d n = d m + n ). Rearranging:[ m - d m = n - d n ][ m(1 - d) = n(1 - d) ]If ( d neq 1 ), then ( m = n ) for all ( m, n ), which is impossible. If ( d = 1 ), then both sides are equal, so ( g(n) = n ) works, giving ( f(n) = n ).So, it seems that the only solutions are the zero function and the identity function. But earlier, I thought that ( E ) could be generated by some ( d ), implying that ( f(n) ) is a multiple of ( d ). However, the only consistent solutions seem to be when ( d = 0 ) (zero function) or ( d = 1 ) (identity function).Wait, maybe there are other functions where ( f(n) ) is a multiple of ( d ) but not necessarily linear. For example, suppose ( d = 2 ) and ( f(n) = 2n ). Let's test this.Left-hand side: ( f(m + f(n)) = f(m + 2n) = 2(m + 2n) = 2m + 4n )Right-hand side: ( f(f(m)) + f(n) = f(2m) + 2n = 4m + 2n )So, ( 2m + 4n = 4m + 2n ) implies ( 2m = 2n ), which is not true for all ( m, n ). Thus, ( f(n) = 2n ) doesn't work.What if ( f(n) = d ) for some constant ( d )? Let's try ( f(n) = d ) for all ( n ).Left-hand side: ( f(m + f(n)) = f(m + d) = d )Right-hand side: ( f(f(m)) + f(n) = f(d) + d = d + d = 2d )So, ( d = 2d ) implies ( d = 0 ). Thus, the only constant function that works is the zero function.Hmm, so it seems that the only solutions are the zero function and the identity function. But I need to make sure there aren't any other functions that might satisfy the equation.Let me consider another approach. Suppose ( f ) is injective or surjective. If ( f ) is injective, then from ( f(m) = f(f(m)) ), we get ( f(m) = m ) for all ( m ), which is the identity function. If ( f ) is surjective, then for every ( k in mathbb{N} ), there exists an ( n ) such that ( f(n) = k ). But from ( f(m) = f(f(m)) ), we have ( f(k) = k ) for all ( k ) in the image of ( f ). If ( f ) is surjective, then its image is all of ( mathbb{N} ), so ( f(k) = k ) for all ( k ), which is again the identity function.If ( f ) is neither injective nor surjective, then it's possible that ( f ) is the zero function, which we already saw works. Are there any other functions that are neither injective nor surjective but still satisfy the equation?Suppose ( f ) is a constant function, but we saw that only the zero function works. Suppose ( f ) is a linear function, but only ( f(n) = 0 ) and ( f(n) = n ) work. What about a function that is zero for some inputs and identity for others, but in a way that doesn't disrupt the equation?Wait, earlier I tried a piecewise function where ( f(n) = 0 ) for even ( n ) and ( f(n) = n ) for odd ( n ), but that didn't work. Maybe another piecewise function could work?Suppose ( f(n) = 0 ) for all ( n ) except ( f(1) = 1 ). Let's test this.Take ( m = 1 ) and ( n = 1 ):Left-hand side: ( f(1 + f(1)) = f(1 + 1) = f(2) = 0 )Right-hand side: ( f(f(1)) + f(1) = f(1) + 1 = 1 + 1 = 2 )Again, ( 0 neq 2 ), so this doesn't work.What if ( f(n) = 0 ) for all ( n geq 1 ) except ( f(0) = 0 )? That's just the zero function, which we already know works.Alternatively, suppose ( f(n) = n ) for all ( n geq k ) and ( f(n) = 0 ) otherwise, for some ( k ). Let's see.Take ( k = 2 ), so ( f(0) = 0 ), ( f(1) = 0 ), ( f(2) = 2 ), ( f(3) = 3 ), etc.Take ( m = 1 ) and ( n = 2 ):Left-hand side: ( f(1 + f(2)) = f(1 + 2) = f(3) = 3 )Right-hand side: ( f(f(1)) + f(2) = f(0) + 2 = 0 + 2 = 2 )So, ( 3 neq 2 ), which doesn't work.It seems that any piecewise function that isn't entirely zero or identity disrupts the equation. Therefore, the only solutions are the zero function and the identity function.Wait, but earlier I considered that ( E ) could be generated by some ( d ), implying that ( f(n) ) is a multiple of ( d ). But the only consistent solutions are when ( d = 0 ) or ( d = 1 ). So, perhaps the only solutions are ( f(n) = 0 ) and ( f(n) = n ).Let me double-check if there could be any other functions. Suppose ( f ) is a function where ( f(n) = n ) for some ( n ) and ( f(n) = 0 ) for others, but in a way that doesn't cause contradictions. However, as I saw earlier, any such function leads to inconsistencies in the equation.Therefore, after considering various possibilities and testing specific cases, it seems that the only functions satisfying the given functional equation are the zero function and the identity function.

💡Alright, let's try to tackle this problem step by step. It's about polar coordinates, curves, and distances to a line. Hmm, okay, let's break it down.First, the curve C is given by the equation sinθ = |cosθ|. I remember that in polar coordinates, ρ is the radius and θ is the angle. So, this equation relates sine and cosine of the same angle θ. Maybe I can convert this into rectangular coordinates to make it easier to visualize or work with.I recall that in rectangular coordinates, x = ρcosθ and y = ρsinθ. So, if I substitute these into the equation sinθ = |cosθ|, I should get something in terms of x and y.Let me write that out:sinθ = |cosθ|Divide both sides by cosθ (assuming cosθ ≠ 0):tanθ = |1|So, tanθ = 1 or tanθ = -1.Wait, but tanθ = |1| would mean tanθ = 1 or tanθ = -1? Hmm, actually, |cosθ| is always non-negative, so sinθ must also be non-negative because it's equal to |cosθ|. So, sinθ is non-negative, which means θ is in the first or second quadrant. Therefore, tanθ would be positive in the first quadrant and negative in the second quadrant.So, tanθ = 1 or tanθ = -1.Therefore, θ = π/4 or θ = 3π/4.But wait, in the second quadrant, tanθ is negative, so tanθ = -1 would correspond to θ = 3π/4.So, the curve C consists of two lines: one at θ = π/4 and another at θ = 3π/4.In rectangular coordinates, θ = π/4 is the line y = x, and θ = 3π/4 is the line y = -x. But since sinθ is non-negative, we're only considering the parts of these lines where y is non-negative.Wait, actually, in the first quadrant, both x and y are positive, so y = x is fine. In the second quadrant, x is negative and y is positive, so y = -x would be y = -x, but since x is negative, y = -x becomes y = |x|. So, combining both, the curve C is actually y = |x|.Oh, that makes sense. So, curve C is the union of the lines y = x and y = -x, but only in the upper half-plane where y is non-negative. So, it's a V-shaped graph with the vertex at the origin, opening upwards.Okay, so curve C is y = |x|. Got it.Now, the line l is given by ρcosθ - 2ρsinθ = 2. Let me convert this into rectangular coordinates as well.Again, x = ρcosθ and y = ρsinθ, so substituting:x - 2y = 2.So, the line l is x - 2y = 2. Let me write that as x - 2y - 2 = 0 for clarity.Now, the problem states that there are two distinct points M and N on curve C such that the distance from both points to line l is √5. We need to find the distance between M and N, which is |MN|.Alright, so we need to find points on y = |x| that are at a distance √5 from the line x - 2y - 2 = 0.First, let's recall the formula for the distance from a point (x0, y0) to the line ax + by + c = 0:Distance = |ax0 + by0 + c| / sqrt(a² + b²).In our case, the line is x - 2y - 2 = 0, so a = 1, b = -2, c = -2.So, the distance from a point (x, y) to line l is |x - 2y - 2| / sqrt(1 + 4) = |x - 2y - 2| / sqrt(5).We are told this distance is √5 for both points M and N.So, |x - 2y - 2| / sqrt(5) = √5.Multiplying both sides by sqrt(5):|x - 2y - 2| = 5.So, x - 2y - 2 = 5 or x - 2y - 2 = -5.Simplifying both equations:1. x - 2y - 2 = 5 => x - 2y = 72. x - 2y - 2 = -5 => x - 2y = -3So, we have two lines: x - 2y = 7 and x - 2y = -3.These are parallel lines to the original line l, since they have the same coefficients for x and y.Now, we need to find the points of intersection between these two lines and the curve C, which is y = |x|.So, let's solve the system of equations for each case.First, let's consider the line x - 2y = 7 and the curve y = |x|.Case 1: y = x (since y = |x|, we can split into two cases: y = x when x ≥ 0, and y = -x when x < 0)Substitute y = x into x - 2y = 7:x - 2x = 7 => -x = 7 => x = -7.But wait, if y = x, then x must be ≥ 0. But here, x = -7, which is negative. So, this solution is not valid for y = x.Therefore, no solution in this case.Case 2: y = -x (when x < 0)Substitute y = -x into x - 2y = 7:x - 2(-x) = 7 => x + 2x = 7 => 3x = 7 => x = 7/3.But wait, if y = -x, then x must be < 0. However, x = 7/3 is positive, which contradicts the condition x < 0. So, this solution is also invalid.Hmm, that's strange. So, the line x - 2y = 7 doesn't intersect the curve y = |x|? That can't be right because the distance is √5, which should give two points.Wait, maybe I made a mistake in substituting.Wait, let's double-check.For the line x - 2y = 7 and y = |x|.If y = x, then x - 2x = 7 => -x = 7 => x = -7.But y = x implies x ≥ 0, so x = -7 is not valid.If y = -x, then x - 2(-x) = 7 => x + 2x = 7 => 3x = 7 => x = 7/3.But y = -x implies x < 0, so x = 7/3 is not valid.So, indeed, there are no solutions for x - 2y = 7 and y = |x|. That's interesting.Now, let's check the other line: x - 2y = -3.Again, substitute y = |x|.Case 1: y = x (x ≥ 0)Substitute y = x into x - 2y = -3:x - 2x = -3 => -x = -3 => x = 3.Since x = 3 is ≥ 0, this is valid. So, y = 3.So, one point is (3, 3).Case 2: y = -x (x < 0)Substitute y = -x into x - 2y = -3:x - 2(-x) = -3 => x + 2x = -3 => 3x = -3 => x = -1.Since x = -1 is < 0, this is valid. So, y = -(-1) = 1.So, the other point is (-1, 1).Therefore, the two points M and N are (3, 3) and (-1, 1).Now, we need to find the distance between these two points.Using the distance formula:|MN| = sqrt[(x2 - x1)² + (y2 - y1)²]Plugging in the coordinates:|MN| = sqrt[(-1 - 3)² + (1 - 3)²] = sqrt[(-4)² + (-2)²] = sqrt[16 + 4] = sqrt[20] = 2*sqrt(5).So, the distance between M and N is 2√5.Let me just double-check my steps to make sure I didn't make any mistakes.1. Converted the polar equation to rectangular coordinates: sinθ = |cosθ| becomes y = |x|. That seems correct.2. Converted the line equation to rectangular coordinates: x - 2y - 2 = 0. Correct.3. Set up the distance formula and found that |x - 2y - 2| = 5. So, two lines: x - 2y = 7 and x - 2y = -3. Correct.4. Solved for intersections with y = |x|. - For x - 2y = 7: No valid solutions because the x values obtained didn't satisfy the conditions for y = x or y = -x. That seems correct. - For x - 2y = -3: Found valid points (3, 3) and (-1, 1). Correct.5. Calculated the distance between (3, 3) and (-1, 1): sqrt[(-4)^2 + (-2)^2] = sqrt[16 + 4] = sqrt[20] = 2√5. Correct.So, everything checks out. The answer should be 2√5, which is option C.