answer：First, I'll express ( b ) and ( c ) in terms of ( a ) using the given division conditions.For ( b ) divided by ( a ), the quotient is 3 and the remainder is 3. This can be written as:[ b = 3a + 3 ]For ( c ) divided by ( a ), the quotient is 9 and the remainder is 11. This can be written as:[ c = 9a + 11 ]Next, I'll substitute the expression for ( b ) into the expression for ( c ) to find ( c ) in terms of ( b ):[ c = 9a + 11 = 3 times (3a + 3) + 2 = 3b + 2 ]Finally, to find the remainder when ( c ) is divided by ( b ), I observe that:[ c = 3b + 2 ]So, the remainder is 2.

answer：Alright, so I have this problem about algae covering a pond. It says that the amount of algae doubles every day, and by day 30, the pond is completely covered in algae. The question is asking on which day the pond was 75% algae-free. Hmm, okay, let me try to wrap my head around this.First, let me rephrase the problem in my own words. The algae coverage doubles each day, so it's growing exponentially. By day 30, the entire pond is covered. So, if I think about it, on day 29, the pond must have been half covered because it doubled on day 30 to become fully covered. Similarly, on day 28, it would have been a quarter covered, right? Because doubling a quarter gives half, and doubling half gives the full coverage on day 30.Wait, so if on day 28 it's 25% covered, then on day 29 it's 50%, and day 30 it's 100%. So, the coverage goes like 25%, 50%, 100% on days 28, 29, 30 respectively. So, the problem is asking when was the pond 75% algae-free. That means the pond was 25% covered by algae, right? Because if 75% is free, then 25% is covered.So, if 25% coverage is on day 28, then 75% algae-free would also be day 28. That seems straightforward, but let me make sure I'm not missing anything.Let me think about the exponential growth. If the algae doubles every day, then the coverage on day n is 2^(n) times the coverage on day 0. But wait, actually, it's better to model this as the coverage on day n being 2^(n - 30) because on day 30, it's 100% covered. So, working backwards, each day before day 30, the coverage halves.So, on day 29, it's 50%, day 28, 25%, day 27, 12.5%, and so on. So, if we want the coverage to be 25%, that's day 28. Therefore, 75% algae-free would be day 28.But let me make sure I'm interpreting the question correctly. It says "75% of the pond algae-free," which means 25% covered. So, yes, that would be day 28. Hmm, that seems consistent.Alternatively, maybe I can model this with an equation. Let's let A(n) be the area covered by algae on day n. Since it doubles every day, we can write A(n) = A0 * 2^n, where A0 is the initial coverage on day 0. But we know that on day 30, the pond is completely covered, so A(30) = 1 (representing 100%). Therefore, 1 = A0 * 2^30, so A0 = 1 / 2^30.Therefore, the coverage on day n is A(n) = (1 / 2^30) * 2^n = 2^(n - 30). So, on day n, the coverage is 2^(n - 30). We want to find the day when the coverage is 25%, which is 0.25. So, set up the equation:2^(n - 30) = 0.25We know that 0.25 is equal to 2^(-2), so:2^(n - 30) = 2^(-2)Therefore, n - 30 = -2, so n = 28.So, that confirms it. Day 28 is when the pond was 25% covered, meaning 75% algae-free.Wait, just to double-check, let me think about the progression:- Day 30: 100% covered- Day 29: 50% covered- Day 28: 25% covered- Day 27: 12.5% covered- Day 26: 6.25% covered- Etc.So, yes, each day going back halves the coverage. So, 25% is two days before full coverage, which is day 28.Therefore, the answer should be day 28.But just to make sure, let's think about it another way. If on day 30, it's 100%, then on day 29, it's 50%, and on day 28, it's 25%. So, 25% coverage is day 28, meaning 75% is free. So, that's consistent.Alternatively, maybe someone could think that 75% algae-free is 75% covered, but that would be incorrect because "algae-free" means the part without algae. So, 75% algae-free implies 25% covered. So, yeah, that's correct.I think I've thought this through multiple ways and each time I get day 28 as the answer. So, I feel confident about that.**Final Answer**The pond was 75% algae-free on day boxed{28} of the month.

answer：Okay, so I have this problem about cards with numbers on them, either 1 or -1. There are 100 more cards with -1 than with 1. If I take the product of the numbers on each pair of different cards and sum all these products, the result is 1000. I need to find how many cards have the number 1.Hmm, let me break this down. Let's denote the number of cards with 1 as ( n ). Then, the number of cards with -1 would be ( n + 100 ) because there are 100 more -1 cards than 1 cards. So, the total number of cards is ( n + (n + 100) = 2n + 100 ).Now, the problem mentions the sum of the products of each pair of different cards. That sounds like a combination problem where I have to consider all possible pairs. The number of such pairs is given by the combination formula ( C(total, 2) ), which is ( frac{(2n + 100)(2n + 99)}{2} ). But wait, actually, the sum of the products isn't just the number of pairs; it's the sum of each product. Each product is either 1*1, 1*(-1), or (-1)*(-1). Let me think. If I have two 1s, their product is 1. If I have a 1 and a -1, their product is -1. If I have two -1s, their product is 1. So, the total sum of all these products will depend on how many pairs of each type there are.Let me denote:- ( a ) as the number of 1s, which is ( n ).- ( b ) as the number of -1s, which is ( n + 100 ).Then, the total number of pairs is ( C(a + b, 2) ). The number of pairs with two 1s is ( C(a, 2) ), the number of pairs with two -1s is ( C(b, 2) ), and the number of mixed pairs (1 and -1) is ( a times b ).So, the sum of all products would be:- For two 1s: Each contributes 1, so total is ( C(a, 2) times 1 ).- For two -1s: Each contributes 1, so total is ( C(b, 2) times 1 ).- For mixed pairs: Each contributes -1, so total is ( a times b times (-1) ).Therefore, the total sum ( S ) is:[ S = C(a, 2) + C(b, 2) - a times b ]We are given that this sum is 1000. So,[ C(a, 2) + C(b, 2) - a times b = 1000 ]Let me express ( C(a, 2) ) and ( C(b, 2) ) in terms of ( a ) and ( b ):[ C(a, 2) = frac{a(a - 1)}{2} ][ C(b, 2) = frac{b(b - 1)}{2} ]Substituting these into the equation:[ frac{a(a - 1)}{2} + frac{b(b - 1)}{2} - a b = 1000 ]Let me combine the terms:First, combine the two fractions:[ frac{a(a - 1) + b(b - 1)}{2} - a b = 1000 ]Multiply numerator:[ frac{a^2 - a + b^2 - b}{2} - a b = 1000 ]Let me write this as:[ frac{a^2 + b^2 - a - b}{2} - a b = 1000 ]Multiply both sides by 2 to eliminate the denominator:[ a^2 + b^2 - a - b - 2 a b = 2000 ]Hmm, let's see. The left-hand side can be rewritten. Notice that ( a^2 + b^2 - 2ab = (a - b)^2 ), so:[ (a - b)^2 - a - b = 2000 ]We know that ( b = a + 100 ), so let's substitute ( b ) with ( a + 100 ):[ (a - (a + 100))^2 - a - (a + 100) = 2000 ]Simplify ( (a - a - 100)^2 ):[ (-100)^2 - a - a - 100 = 2000 ]Calculate ( (-100)^2 = 10000 ):[ 10000 - 2a - 100 = 2000 ]Simplify:[ 9900 - 2a = 2000 ]Subtract 9900 from both sides:[ -2a = 2000 - 9900 ][ -2a = -7900 ]Divide both sides by -2:[ a = frac{-7900}{-2} ][ a = 3950 ]Wait, that can't be right. If ( a = 3950 ), then ( b = 3950 + 100 = 4050 ). The total number of cards is 3950 + 4050 = 8000. That seems quite large, but let me check the calculations.Wait, let's go back step by step.We had:[ (a - b)^2 - a - b = 2000 ]Since ( b = a + 100 ), substitute:[ (a - (a + 100))^2 - a - (a + 100) = 2000 ]Which simplifies to:[ (-100)^2 - 2a - 100 = 2000 ]So:[ 10000 - 2a - 100 = 2000 ]Which is:[ 9900 - 2a = 2000 ]Subtract 9900:[ -2a = 2000 - 9900 ][ -2a = -7900 ]Divide by -2:[ a = 3950 ]Hmm, seems consistent. So, the number of 1s is 3950. Let me verify if this makes sense.Total number of cards is 3950 + 4050 = 8000. The number of pairs is ( C(8000, 2) = frac{8000 times 7999}{2} ), which is a huge number, but the sum of the products is 1000, which is small compared to the total number of pairs. Hmm, that seems a bit odd because the sum is positive, but most of the products are 1 or -1.Wait, let me think about the sum again. The sum is 1000, which is positive, so the number of pairs with product 1 must exceed the number of pairs with product -1 by 1000.But if there are so many cards, how does the sum stay at 1000? Maybe my initial approach is correct, but let me see.Alternatively, perhaps I can model this problem using variables and see if I can get another equation.Let me denote ( a = n ) (number of 1s) and ( b = n + 100 ) (number of -1s). Then, the sum of the products is 1000.But another way to compute the sum of the products is to consider that the sum over all pairs ( (i, j) ) where ( i < j ) of ( x_i x_j ). This is equal to ( frac{(x_1 + x_2 + dots + x_{2n+100})^2 - (x_1^2 + x_2^2 + dots + x_{2n+100}^2)}{2} ).Since each ( x_i ) is either 1 or -1, ( x_i^2 = 1 ). So, the sum of squares is equal to the total number of cards, which is ( 2n + 100 ).Let me denote ( S = x_1 + x_2 + dots + x_{2n+100} ). Then, the sum of the products is ( frac{S^2 - (2n + 100)}{2} ).We are given that this equals 1000:[ frac{S^2 - (2n + 100)}{2} = 1000 ]Multiply both sides by 2:[ S^2 - (2n + 100) = 2000 ]So,[ S^2 = 2000 + 2n + 100 ][ S^2 = 2100 + 2n ]Now, let's compute ( S ). ( S ) is the sum of all the numbers on the cards. Each 1 contributes +1 and each -1 contributes -1. So,[ S = (a) times 1 + (b) times (-1) ][ S = a - b ]But ( b = a + 100 ), so:[ S = a - (a + 100) ][ S = -100 ]Therefore, ( S = -100 ). Plugging back into the equation:[ (-100)^2 = 2100 + 2n ][ 10000 = 2100 + 2n ]Subtract 2100:[ 7900 = 2n ]Divide by 2:[ n = 3950 ]So, the number of 1s is 3950. That matches my earlier result. So, despite the large number of cards, the sum of the products is 1000 because the total sum squared is 10000, and subtracting the total number of cards (8000) gives 2000, which when divided by 2 gives 1000.Therefore, the number of cards with 1 is 3950.**Final Answer**The number of cards with the number 1 is boxed{3950}.

answer：Okay, so I have this problem where I need to find the possible values of (2n + m) given three inequalities involving integers (n) and (m). The inequalities are:1. (3n - m < 5)2. (n + m > 26)3. (3m - 2n < 46)Hmm, let me try to break this down step by step. Maybe I can solve these inequalities to find the range of possible values for (n) and (m), and then see what (2n + m) can be.Starting with the first inequality: (3n - m < 5). I can rearrange this to express (m) in terms of (n). Let's see:(3n - m < 5)Subtract (3n) from both sides:(-m < -3n + 5)Multiply both sides by (-1), which reverses the inequality:(m > 3n - 5)Okay, so (m) is greater than (3n - 5). Got that.Moving on to the second inequality: (n + m > 26). This one is straightforward. It tells me that the sum of (n) and (m) is more than 26.Third inequality: (3m - 2n < 46). Let me rearrange this as well. Maybe express (m) in terms of (n):(3m - 2n < 46)Add (2n) to both sides:(3m < 2n + 46)Divide both sides by 3:(m < frac{2n + 46}{3})So, (m) is less than (frac{2n + 46}{3}).Now, let me summarize the inequalities I have:1. (m > 3n - 5)2. (n + m > 26)3. (m < frac{2n + 46}{3})Since (n) and (m) are integers, I can try to find integer solutions that satisfy all three inequalities. Maybe I can express all inequalities in terms of (m) and then find the overlapping ranges.From inequality 1: (m > 3n - 5)From inequality 2: (m > 26 - n)From inequality 3: (m < frac{2n + 46}{3})So, combining inequalities 1 and 2, (m) must be greater than the maximum of (3n - 5) and (26 - n). And it must be less than (frac{2n + 46}{3}).Therefore, the range for (m) is:(max(3n - 5, 26 - n) < m < frac{2n + 46}{3})Since (m) must be an integer, the number of possible (m) values depends on how these bounds overlap.I think the next step is to find the values of (n) for which the lower bound is less than the upper bound. That is, when:(max(3n - 5, 26 - n) < frac{2n + 46}{3})This will give me the range of (n) where such (m) exists.Let me consider two cases for the maximum function:Case 1: (3n - 5 geq 26 - n)Case 2: (3n - 5 < 26 - n)Let me solve for (n) in each case.Case 1: (3n - 5 geq 26 - n)Adding (n) to both sides:(4n - 5 geq 26)Adding 5:(4n geq 31)Dividing by 4:(n geq frac{31}{4}), which is (n geq 7.75). Since (n) is an integer, (n geq 8).Case 2: (3n - 5 < 26 - n)Adding (n) to both sides:(4n - 5 < 26)Adding 5:(4n < 31)Dividing by 4:(n < 7.75), so (n leq 7) since (n) is an integer.So, depending on whether (n) is 8 or higher, or 7 or lower, the maximum is different.Now, let's handle each case separately.**Case 1: (n geq 8)**Here, (max(3n - 5, 26 - n) = 3n - 5). So, the inequality becomes:(3n - 5 < frac{2n + 46}{3})Multiply both sides by 3 to eliminate the denominator:(9n - 15 < 2n + 46)Subtract (2n) from both sides:(7n - 15 < 46)Add 15:(7n < 61)Divide by 7:(n < frac{61}{7}), which is approximately 8.714.Since (n) is an integer and (n geq 8), the only possible value is (n = 8).So, in this case, (n = 8). Let's find the corresponding (m).From inequality 1: (m > 3(8) - 5 = 24 - 5 = 19)From inequality 2: (m > 26 - 8 = 18)From inequality 3: (m < frac{2(8) + 46}{3} = frac{16 + 46}{3} = frac{62}{3} approx 20.666)So, (m) must satisfy:(19 < m < 20.666)Since (m) is an integer, the possible values are (m = 20).Thus, when (n = 8), (m = 20). Let's compute (2n + m):(2(8) + 20 = 16 + 20 = 36)So, 36 is a possible value.**Case 2: (n leq 7)**Here, (max(3n - 5, 26 - n) = 26 - n). So, the inequality becomes:(26 - n < frac{2n + 46}{3})Multiply both sides by 3:(78 - 3n < 2n + 46)Add (3n) to both sides:(78 < 5n + 46)Subtract 46:(32 < 5n)Divide by 5:(6.4 < n)Since (n) is an integer, (n geq 7). But in this case, (n leq 7), so (n = 7).So, (n = 7). Let's find (m).From inequality 1: (m > 3(7) - 5 = 21 - 5 = 16)From inequality 2: (m > 26 - 7 = 19)From inequality 3: (m < frac{2(7) + 46}{3} = frac{14 + 46}{3} = frac{60}{3} = 20)So, (m) must satisfy:(19 < m < 20)But (m) is an integer, so there are no integer values of (m) that satisfy this. Therefore, (n = 7) doesn't yield any solution.Hmm, so in Case 2, (n = 7) doesn't work. Let me check if I did that correctly.Wait, when (n = 7), the lower bound from inequality 2 is (m > 19), and the upper bound is (m < 20). So, (m) must be greater than 19 but less than 20, which is impossible for integers. So, indeed, no solution for (n = 7).Therefore, in Case 2, there are no solutions.Wait, but hold on. Maybe I should check for (n) less than 7? Because in Case 2, (n leq 7), but when (n < 7), the maximum is still (26 - n), but perhaps the inequality (26 - n < frac{2n + 46}{3}) could have solutions for (n < 7). Let me check.So, in Case 2, (n leq 7). Let's solve the inequality again:(26 - n < frac{2n + 46}{3})Multiply both sides by 3:(78 - 3n < 2n + 46)Add (3n) to both sides:(78 < 5n + 46)Subtract 46:(32 < 5n)Divide by 5:(6.4 < n)So, (n > 6.4), which for integers is (n geq 7). Therefore, even in Case 2, the only possible (n) is 7, which doesn't yield a solution. So, indeed, no solutions for (n leq 7).Therefore, the only solution is (n = 8) and (m = 20), giving (2n + m = 36).Wait, but that seems too restrictive. Maybe I missed something. Let me think again.Perhaps I should consider that (m) must be an integer, so even if (n) is 8, maybe (m) can take other values? Let me double-check.For (n = 8):- From inequality 1: (m > 19)- From inequality 2: (m > 18)- From inequality 3: (m < 20.666)So, (m) must be greater than 19 and less than 20.666. Since (m) is integer, only (m = 20) is possible.So, that's correct, only (m = 20) works.But wait, maybe (n) can be higher than 8? Because in Case 1, I got (n < 8.714), so (n = 8) is the only integer. So, (n) can't be 9 or higher because (n < 8.714), which would mean (n = 8) is the maximum.But let me check for (n = 9):If (n = 9), let's see if there exists an (m) that satisfies all three inequalities.From inequality 1: (m > 3(9) - 5 = 27 - 5 = 22)From inequality 2: (m > 26 - 9 = 17)From inequality 3: (m < frac{2(9) + 46}{3} = frac{18 + 46}{3} = frac{64}{3} approx 21.333)So, (m) must satisfy:(22 < m < 21.333)Which is impossible because 22 is greater than 21.333. So, no solution for (n = 9).Similarly, for (n = 10):Inequality 1: (m > 3(10) -5 = 25)Inequality 2: (m > 26 -10 = 16)Inequality 3: (m < frac{2(10) +46}{3} = frac{20 +46}{3} = frac{66}{3} = 22)So, (m) must be greater than 25 and less than 22, which is impossible.So, indeed, (n) cannot be higher than 8.Wait, but what about (n = 6)? Let me check if (n = 6) could yield a solution.For (n = 6):Inequality 1: (m > 3(6) -5 = 18 -5 =13)Inequality 2: (m > 26 -6 =20)Inequality 3: (m < frac{2(6) +46}{3} = frac{12 +46}{3} = frac{58}{3} approx19.333)So, (m) must satisfy:(20 < m <19.333)Which is impossible. So, no solution.Similarly, (n = 5):Inequality 1: (m > 10)Inequality 2: (m >21)Inequality 3: (m < frac{10 +46}{3} = frac{56}{3} approx18.666)So, (m) must be greater than 21 and less than 18.666, which is impossible.Same with (n = 4):Inequality 1: (m >7)Inequality 2: (m >22)Inequality 3: (m < frac{8 +46}{3} = frac{54}{3}=18)So, (m) must be greater than 22 and less than 18, impossible.Similarly, (n = 3):Inequality 1: (m >4)Inequality 2: (m >23)Inequality 3: (m < frac{6 +46}{3} = frac{52}{3} approx17.333)Again, impossible.(n = 2):Inequality 1: (m >1)Inequality 2: (m >24)Inequality 3: (m < frac{4 +46}{3} = frac{50}{3} approx16.666)Impossible.(n = 1):Inequality 1: (m > -2)Inequality 2: (m >25)Inequality 3: (m < frac{2 +46}{3} = frac{48}{3}=16)Impossible.(n = 0):Inequality 1: (m > -5)Inequality 2: (m >26)Inequality 3: (m < frac{0 +46}{3} approx15.333)Impossible.Negative (n) values:Let me check (n = -1):Inequality 1: (m > -8)Inequality 2: (m >27)Inequality 3: (m < frac{-2 +46}{3} = frac{44}{3} approx14.666)Impossible.So, it seems that for (n <8), there are no solutions, except for (n =8), which gives (m =20).Wait, but hold on. Maybe I should check (n =8) again.For (n =8):Inequality 1: (m >19)Inequality 2: (m >18)Inequality 3: (m <20.666)So, (m) must be 20.But wait, is (m =20) satisfying all inequalities?Let me plug into the original inequalities:1. (3n - m =24 -20=4 <5) ✔️2. (n + m =8 +20=28 >26) ✔️3. (3m -2n=60 -16=44 <46) ✔️Yes, all are satisfied.So, only (n =8) and (m =20) satisfy all three inequalities.Therefore, (2n + m =16 +20=36).Wait, but the problem says "list all possible options". So, is 36 the only possible value?But let me think again. Maybe I missed some other (n) and (m) pairs.Wait, perhaps I should approach this differently. Maybe express (2n + m) in terms of the inequalities.Let me denote (k =2n + m). I need to find all possible integer values of (k).Given the inequalities:1. (3n - m <5) → (m >3n -5)2. (n + m >26)3. (3m -2n <46) → (3m <2n +46) → (m < frac{2n +46}{3})So, combining inequalities 1 and 3:(3n -5 < m < frac{2n +46}{3})Also, from inequality 2: (m >26 -n)So, (m) must satisfy both (m > max(3n -5, 26 -n)) and (m < frac{2n +46}{3})So, for each (n), we can find the range of (m), and then compute (k =2n +m).But since (n) and (m) are integers, the possible (k) values depend on the possible (m) for each (n).Alternatively, maybe I can express (k) in terms of (n):(k =2n +m). From inequality 1: (m >3n -5), so (k =2n +m >2n +3n -5 =5n -5)From inequality 3: (m < frac{2n +46}{3}), so (k =2n +m <2n + frac{2n +46}{3} = frac{6n +2n +46}{3} = frac{8n +46}{3})So, (5n -5 <k < frac{8n +46}{3})Also, from inequality 2: (m >26 -n), so (k =2n +m >2n +26 -n =n +26)Therefore, combining all:( max(5n -5, n +26) <k < frac{8n +46}{3})So, for each (n), (k) must be between (max(5n -5, n +26)) and (frac{8n +46}{3})Let me analyze when (5n -5) is greater than (n +26):(5n -5 >n +26)Subtract (n):(4n -5 >26)Add 5:(4n >31)Divide by 4:(n >7.75)So, for (n geq8), (5n -5) is greater than (n +26), so (max(5n -5, n +26) =5n -5)For (n leq7), (max(5n -5, n +26) =n +26)So, let's split into two cases again.**Case 1: (n geq8)**Here, (5n -5 <k < frac{8n +46}{3})But from earlier, we saw that for (n geq8), the only possible (n) is 8, giving (k =36). Let me confirm.For (n =8):Lower bound: (5(8) -5 =40 -5=35)Upper bound: (frac{8(8) +46}{3} = frac{64 +46}{3}= frac{110}{3} approx36.666)So, (k) must be greater than 35 and less than 36.666. Since (k) is integer, possible (k =36)Which matches our earlier result.**Case 2: (n leq7)**Here, (n +26 <k < frac{8n +46}{3})So, let's compute for each (n) from 1 to7 (since (n) must be positive? Wait, no, (n) can be negative as well, but earlier checks showed that for (n <8), no solutions exist except (n=8). But let me check for (n) from 1 to7.Wait, but in the earlier analysis, for (n =7), (m) had to be between 19 and20, which was impossible. So, perhaps even for (n leq7), there are no solutions.But let me proceed.For (n =7):Lower bound: (7 +26=33)Upper bound: (frac{8(7) +46}{3}= frac{56 +46}{3}= frac{102}{3}=34)So, (k) must be greater than33 and less than34. So, (k =34)But wait, does (k =34) correspond to some (m)?Since (k =2n +m =14 +m =34), so (m =20)But for (n =7), (m =20). Let's check if this satisfies all inequalities.1. (3n -m =21 -20=1 <5) ✔️2. (n +m =7 +20=27 >26) ✔️3. (3m -2n =60 -14=46 <46)? Wait, 46 is not less than46. It's equal. So, this violates the third inequality.Therefore, (m =20) for (n =7) doesn't satisfy the third inequality. So, (k =34) is not a valid solution.Wait, but according to the upper bound, (k <34), so (k) must be less than34, but (k >33). So, (k =34) is not allowed, since it's not less than34. So, no solution for (n =7).Similarly, for (n =6):Lower bound: (6 +26=32)Upper bound: (frac{8(6) +46}{3}= frac{48 +46}{3}= frac{94}{3} approx31.333)So, (k) must be greater than32 and less than31.333, which is impossible.Similarly, (n =5):Lower bound:31Upper bound: (frac{40 +46}{3}= frac{86}{3} approx28.666)Impossible.(n =4):Lower bound:30Upper bound: (frac{32 +46}{3}= frac{78}{3}=26)Impossible.(n =3):Lower bound:29Upper bound: (frac{24 +46}{3}= frac{70}{3} approx23.333)Impossible.(n =2):Lower bound:28Upper bound: (frac{16 +46}{3}= frac{62}{3} approx20.666)Impossible.(n =1):Lower bound:27Upper bound: (frac{8 +46}{3}= frac{54}{3}=18)Impossible.So, indeed, for (n leq7), there are no solutions.Therefore, the only possible value of (k =2n +m) is36.But wait, let me check (n =8) and (m =20) again.(3n -m =24 -20=4 <5) ✔️(n +m =28 >26) ✔️(3m -2n =60 -16=44 <46) ✔️Yes, all inequalities are satisfied.Is there any other (n) and (m) that could satisfy all three inequalities?Wait, let me think differently. Maybe I can express (2n +m) as a variable and try to find its bounds.Let me denote (k =2n +m). Then, from the inequalities:1. (3n -m <5) → (m >3n -5)2. (n +m >26)3. (3m -2n <46) → (3m <2n +46) → (m < frac{2n +46}{3})So, combining these:(3n -5 <m < frac{2n +46}{3})Also, from inequality 2: (m >26 -n)So, (m) must satisfy both (m > max(3n -5, 26 -n)) and (m < frac{2n +46}{3})Let me express (k =2n +m). So, (m =k -2n)Substitute into the inequalities:1. (3n - (k -2n) <5) → (3n -k +2n <5) → (5n -k <5) → (k >5n -5)2. (n + (k -2n) >26) → (-n +k >26) → (k >n +26)3. (3(k -2n) -2n <46) → (3k -6n -2n <46) → (3k -8n <46) → (3k <8n +46) → (k < frac{8n +46}{3})So, combining:( max(5n -5, n +26) <k < frac{8n +46}{3})Which is the same as before.So, for each (n), (k) must lie in that interval.We saw that for (n geq8), (k >5n -5) and (k < frac{8n +46}{3})For (n =8):(5(8) -5=35 <k < frac{64 +46}{3}= frac{110}{3}≈36.666)So, (k =36)For (n =9):(5(9) -5=40 <k < frac{72 +46}{3}= frac{118}{3}≈39.333)But (k =2n +m), and for (n =9), (m) must satisfy (m >22) and (m <21.333), which is impossible. So, no solution.Similarly, for (n =7):(5(7) -5=30 <k < frac{56 +46}{3}= frac{102}{3}=34)But (k >30) and (k <34). So, possible (k =31,32,33)But does any of these correspond to integer (m)?For (k =31):(m =31 -2(7)=31 -14=17)Check inequalities:1. (3(7) -17=21 -17=4 <5) ✔️2. (7 +17=24 >26)? No, 24 <26. So, invalid.For (k =32):(m =32 -14=18)Check:1. (21 -18=3 <5) ✔️2. (7 +18=25 >26)? No.For (k =33):(m =33 -14=19)Check:1. (21 -19=2 <5) ✔️2. (7 +19=26 >26)? No, it's equal, which doesn't satisfy the strict inequality.So, no solutions for (k =31,32,33) when (n =7).Therefore, even though (k) is in the range, the corresponding (m) doesn't satisfy all inequalities.Similarly, for other (n), even if (k) is in the range, (m) might not satisfy all inequalities.Thus, the only valid solution is (k =36).Wait, but let me check (n =8) and (m =20) again.Yes, all inequalities are satisfied.Is there a possibility of another (n) and (m) where (k =36)?Let me see.Suppose (k =36), so (2n +m =36). Then, (m =36 -2n)Substitute into the inequalities:1. (3n - (36 -2n) <5) → (3n -36 +2n <5) → (5n -36 <5) → (5n <41) → (n <8.2). So, (n leq8)2. (n + (36 -2n) >26) → (-n +36 >26) → (-n >-10) → (n <10)3. (3(36 -2n) -2n <46) → (108 -6n -2n <46) → (108 -8n <46) → (-8n < -62) → (8n >62) → (n >7.75). So, (n geq8)Therefore, combining these:From 1: (n leq8)From 3: (n geq8)So, (n =8)Therefore, the only solution is (n =8), (m =20), giving (k =36)Thus, the only possible value of (2n +m) is36.**Final Answer**The only possible value is boxed{36}.

answer：Okay, so I have this problem where a farmer wants to build a rectangular region using a river as one side and fencing for the other three sides. He has 1200 feet of fencing. The goal is to find the dimensions (length L and width W) that will enclose the greatest area, and then find L + W. Hmm, let me think about how to approach this.First, since the farmer is using the river as one side, he doesn't need fencing along that side. So, he only needs fencing for the other three sides. Let me visualize this: imagine the river is along one length of the rectangle, so the farmer needs fencing for the other length and the two widths. So, the total fencing required would be L + 2W. Is that right? Let me confirm: if the river is one length, then the fencing is needed for the opposite length and the two widths. Yes, that makes sense.He has 1200 feet of fencing, so the equation for the fencing is:L + 2W = 1200Okay, so that's our constraint. Now, we need to maximize the area. The area A of a rectangle is given by:A = L * WSo, we need to express A in terms of one variable and then find its maximum. Since we have the fencing constraint, we can solve for one variable in terms of the other.Let me solve the fencing equation for L:L = 1200 - 2WNow, plug this into the area formula:A = (1200 - 2W) * WSimplify that:A = 1200W - 2W^2So, now we have A as a function of W: A(W) = -2W^2 + 1200WThis is a quadratic equation, and since the coefficient of W^2 is negative (-2), the parabola opens downward, meaning the vertex is the maximum point. So, the maximum area occurs at the vertex of this parabola.The vertex of a parabola given by A = aW^2 + bW + c is at W = -b/(2a). In this case, a = -2 and b = 1200.So, let's compute W:W = -1200 / (2 * -2) = -1200 / (-4) = 300So, W = 300 feet.Now, plug this back into the equation for L:L = 1200 - 2W = 1200 - 2*300 = 1200 - 600 = 600So, L = 600 feet.Therefore, the dimensions that maximize the area are L = 600 feet and W = 300 feet. The problem asks for L + W, so let's compute that:L + W = 600 + 300 = 900So, L + W is 900 feet.Wait, let me double-check my calculations to make sure I didn't make any mistakes. The fencing is L + 2W = 1200, so if W is 300, then 2W is 600, and L is 600, which adds up to 1200. That seems correct. The area would be 600 * 300 = 180,000 square feet. Is that the maximum?Let me think, if I choose a different width, say W = 250, then L would be 1200 - 2*250 = 700. The area would be 700 * 250 = 175,000, which is less than 180,000. Similarly, if W = 350, then L = 1200 - 700 = 500, and area is 500*350 = 175,000. So, yes, 180,000 is indeed the maximum. Therefore, my calculations seem correct.Alternatively, I can use calculus to confirm. Taking the derivative of A with respect to W:A = -2W^2 + 1200WdA/dW = -4W + 1200Set derivative equal to zero for maximum:-4W + 1200 = 0-4W = -1200W = 300So, same result. Therefore, L = 600, W = 300, so L + W = 900.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**The value of ( L + W ) is boxed{900}.

answer：Alright, so I've got this problem here: there are three prime numbers m, n, and p. Their product equals five times the sum of these three primes. I need to find the value of m² + n² + p². Hmm, okay. Let me think about how to approach this.First, let's write down the equation given in the problem. It says that the product of m, n, and p is equal to five times their sum. So, in mathematical terms, that would be:m * n * p = 5 * (m + n + p)Alright, so we have m, n, p as primes, and this equation holds. My goal is to find m² + n² + p². So, I need to figure out what m, n, and p are first.Since m, n, p are primes, they can be 2, 3, 5, 7, 11, etc. Let me think about the possible primes that could satisfy this equation.First, let's note that 2 is the only even prime number. All other primes are odd. So, if any of m, n, or p is 2, that might make the product even, which could affect the equation.Let me consider the case where one of the primes is 2. Let's say m = 2. Then the equation becomes:2 * n * p = 5 * (2 + n + p)So, simplifying that:2np = 5(2 + n + p)Let me write that as:2np = 10 + 5n + 5pHmm, okay. Let's rearrange this equation to bring all terms to one side:2np - 5n - 5p = 10Hmm, this looks a bit complicated. Maybe I can factor this somehow. Let me see.Let me factor n from the first two terms:n(2p - 5) - 5p = 10Hmm, not sure if that helps. Alternatively, maybe I can add 25/2 to both sides to complete the rectangle or something. Wait, that might complicate things.Alternatively, maybe I can express this as:2np - 5n - 5p = 10Let me add 25 to both sides:2np - 5n - 5p + 25 = 35Hmm, does that factor into something? Let me see:2np -5n -5p +25. Hmm, maybe factor by grouping.Let me group terms:(2np -5n) - (5p -25) = 35Wait, that would be:n(2p -5) -5(p -5) = 35Hmm, not sure if that helps. Alternatively, maybe factor differently.Wait, perhaps I can write it as:(2n -5)(2p -5) = something. Let me try that.Let me consider the equation:2np -5n -5p = 10Multiply both sides by 2:4np -10n -10p = 20Now, add 25 to both sides:4np -10n -10p +25 = 45Now, factor the left side:(2n -5)(2p -5) = 45Ah, that seems promising. So, (2n -5)(2p -5) = 45.Now, since n and p are primes, let's think about possible values for (2n -5) and (2p -5). These must be positive integers because n and p are primes greater than 2 (since m is already 2, and primes are at least 2, but n and p could be 2 as well, but let's see).Wait, actually, n and p could be 2 as well, but let's check if that's possible.If n = 2, then 2n -5 = 4 -5 = -1. Similarly, p =2 would give 2p -5 = -1. But since (2n -5)(2p -5) =45, which is positive, so both factors must be positive or both negative. But since 45 is positive, both factors must be positive or both negative. However, if n and p are primes greater than 2, then 2n -5 and 2p -5 are positive. If n or p is 2, then 2n -5 is negative. So, if both n and p are 2, then (2n -5)(2p -5) = (-1)(-1)=1, which is not 45. If one is 2 and the other is greater, then one factor is negative and the other is positive, resulting in a negative product, which is not 45. So, both n and p must be primes greater than 2, so that both (2n -5) and (2p -5) are positive integers.Therefore, (2n -5) and (2p -5) are positive integer factors of 45. Let's list the positive integer factor pairs of 45:1 * 453 * 155 * 99 * 515 * 345 * 1So, these are the possible pairs. Now, since 2n -5 and 2p -5 are positive integers, let's solve for n and p in each case.First pair: 1 and 45.So, 2n -5 =1 => 2n=6 => n=32p -5=45 => 2p=50 => p=25But 25 is not a prime number. So, this pair is invalid.Second pair: 3 and 15.2n -5=3 => 2n=8 => n=4But 4 is not a prime number. So, invalid.Third pair: 5 and 9.2n -5=5 => 2n=10 => n=52p -5=9 => 2p=14 => p=7Okay, both 5 and 7 are primes. So, this is a valid solution.Fourth pair: 9 and 5.2n -5=9 => 2n=14 => n=72p -5=5 => 2p=10 => p=5Same as above, just swapped. So, n=7, p=5. Still valid.Fifth pair:15 and 3.2n -5=15 => 2n=20 => n=1010 is not prime. Invalid.Sixth pair:45 and1.2n -5=45 => 2n=50 => n=25, not prime.2p -5=1 => p=3, which is prime, but n=25 is not. So, invalid.So, the only valid factor pair is 5 and 9, giving n=5 and p=7, or n=7 and p=5.Therefore, in this case, m=2, n=5, p=7.So, the primes are 2,5,7.Let me check if this satisfies the original equation.Product: 2*5*7=70Sum:2+5+7=145 times sum:5*14=70Yes, 70=70. So, that works.Therefore, the primes are 2,5,7.So, now, m² +n² +p²=2² +5² +7²=4 +25 +49=78.Wait, that seems straightforward. But just to make sure, are there any other possibilities where m is not 2?Let me consider that case. Suppose none of m, n, p is 2. Then all three primes are odd. So, their product would be odd*odd*odd=odd. The sum would be odd+odd+odd=odd. So, 5 times the sum would be 5*odd=odd. So, the product is odd, and 5 times the sum is odd, so that's possible.But let's see if such primes exist.So, if m, n, p are all odd primes, then m*n*p=5*(m +n +p)Let me see if such primes exist.Let me try small primes: 3,5,7.Compute product:3*5*7=105Sum:3+5+7=155 times sum:75105≠75, so that doesn't work.Next, try 3,5,11.Product:3*5*11=165Sum:3+5+11=195*19=95165≠95Not equal.Next, 3,5,13.Product:195Sum:215*21=105195≠105Nope.How about 3,7,11.Product:231Sum:215*21=105231≠105No.Wait, maybe smaller primes.Wait, 3,3, something.But primes can be repeated? The problem says three prime numbers, doesn't specify distinct. So, maybe m, n, p can be same primes.So, let's try m=n=3.Then, equation becomes 3*3*p=5*(3+3+p)So, 9p=5*(6 +p)=30 +5pSo, 9p -5p=304p=30p=7.5Not an integer, so invalid.Similarly, m=n=5.5*5*p=5*(5+5+p)25p=5*(10 +p)=50 +5p25p -5p=5020p=50p=2.5, not integer.Similarly, m=n=7.7*7*p=5*(7+7+p)49p=5*(14 +p)=70 +5p49p -5p=7044p=70p=70/44=35/22≈1.59, not integer.So, no solution when two primes are same.Alternatively, maybe all three primes are same.So, m=n=p.Then, m³=5*(3m)So, m³=15mDivide both sides by m (m≠0):m²=15m=√15≈3.87, not integer. So, no solution.So, seems like if all primes are odd, we can't get a solution. Because when we tried small primes, the product was way larger than 5 times the sum.Wait, but let me think again. Maybe with larger primes, but 5 times the sum would be too small compared to the product.Wait, for example, let me take m=3, n=5, p= something.Compute 3*5*p=15pSum=3+5+p=8 +p5*(8 +p)=40 +5pSo, equation:15p=40 +5p15p -5p=4010p=40p=4, which is not prime.So, that doesn't work.Similarly, m=3, n=7.3*7*p=21pSum=3+7+p=10 +p5*(10 +p)=50 +5pEquation:21p=50 +5p16p=50p=50/16=25/8≈3.125, not integer.Not prime.Similarly, m=5, n=7.5*7*p=35pSum=5+7+p=12 +p5*(12 +p)=60 +5pEquation:35p=60 +5p30p=60p=2Ah, p=2. So, primes are 5,7,2.Wait, but that's the same solution as before, just rearranged. So, m=5, n=7, p=2.So, regardless, the primes are 2,5,7.So, even if I assume m is not 2, I still end up with one of the primes being 2.Therefore, the only solution is 2,5,7.Hence, m² +n² +p²=4 +25 +49=78.Wait, just to make sure, let me think if there are any other possibilities where m is 2, but n or p is 2 as well.Wait, if m=2, n=2, then the equation becomes:2*2*p=5*(2 +2 +p)So, 4p=5*(4 +p)=20 +5pSo, 4p -5p=20-p=20p=-20, which is not a prime. So, invalid.Similarly, if m=2, p=2, then n would be:2*n*2=5*(2 +n +2)4n=5*(n +4)=5n +204n -5n=20-n=20n=-20, invalid.So, no solution where two primes are 2.Therefore, the only solution is m=2, n=5, p=7.Thus, the value of m² +n² +p² is 4 +25 +49=78.**Final Answer**The value of (m^2 + n^2 + p^2) is boxed{78}.