💡Alright, so I've got this problem here where a robot wrote the word "MATHEMATICS," and I need to figure out what number the robot got. Hmm, okay, let's see. I guess the robot must have converted each letter into a number somehow. Maybe it's a cipher or some kind of code where each letter corresponds to a specific number.First, I need to understand how the robot is converting letters to numbers. Is it a simple substitution cipher where each letter is replaced by a unique number? Or maybe it's based on the position of the letter in the alphabet? Let's think about that. In the English alphabet, A is 1, B is 2, and so on up to Z being 26. So, if that's the case, let's try converting "MATHEMATICS" using that method.M is the 13th letter, A is 1, T is 20, H is 8, E is 5, M is 13, A is 1, T is 20, I is 9, C is 3, S is 19. So, putting that together, we'd get 13 1 20 8 5 13 1 20 9 3 19. That seems pretty straightforward, but the problem mentions that the robot wrote the word, so maybe it's not that simple.Wait, the problem also mentions something about the robot writing the word "MATHEMATICS." Maybe it's not just a direct letter-to-number conversion. Perhaps there's a pattern or a specific rule the robot is using to generate the number. Let me think about other possibilities.Maybe the robot is using a different cipher, like a Caesar cipher, where each letter is shifted by a certain number of places. For example, shifting each letter by 1 would make A become B, B become C, and so on. But without knowing the shift value, that might not help.Alternatively, the robot could be using a more complex cipher, like a Vigenère cipher, which uses a keyword to shift letters. But again, without knowing the keyword, that seems difficult.Another thought: maybe the robot is using a numerical cipher where each letter is represented by a two-digit number, like A=01, B=02, ..., Z=26. If that's the case, then "MATHEMATICS" would be 13 01 20 08 05 13 01 20 09 03 19. That seems plausible, but the problem didn't specify that it's a two-digit number for each letter.Wait, maybe the robot is using a combination of letters and numbers in a specific sequence. For example, taking the numerical value of each letter and concatenating them together. So, M=13, A=1, T=20, H=8, E=5, M=13, A=1, T=20, I=9, C=3, S=19. Concatenating these would give 1312085131209319. That's a long number, but it's possible.But I'm not sure if that's what the robot is doing. Maybe there's a different approach. Let's think about the word "MATHEMATICS" itself. It's a 11-letter word, and it contains repeated letters like M, A, T, and E. Maybe the robot is counting the number of times each letter appears or something like that.Wait, another idea: perhaps the robot is assigning a unique number to each letter based on some rule, not necessarily the position in the alphabet. For example, maybe it's using a telephone keypad where letters are mapped to numbers. On a phone keypad, letters are grouped under numbers like 2, 3, 4, etc. Let's see:- 2: ABC- 3: DEF- 4: GHI- 5: JKL- 6: MNO- 7: PQRS- 8: TUV- 9: WXYZSo, using this mapping:- M is on 6- A is on 2- T is on 8- H is on 4- E is on 3- M is on 6- A is on 2- T is on 8- I is on 4- C is on 2- S is on 7So, converting "MATHEMATICS" using this method would give us 6 2 8 4 3 6 2 8 4 2 7. That's 62843628427. That seems like a possible number the robot could have generated.But I'm not entirely sure if that's the correct approach. Maybe the robot is using a different mapping or a combination of methods. Let's think of another possibility.Perhaps the robot is using a cipher where each letter is replaced by its position in the alphabet, but with some modification. For example, adding a constant to each position. If we add 1 to each position, A would be 2, B would be 3, and so on. But without knowing the exact modification, it's hard to say.Another idea: maybe the robot is using a binary representation of each letter. For example, converting each letter to its ASCII value and then to binary. But that would result in a very long string of 1s and 0s, which might not be what the problem is asking for.Wait, the problem says "Write down the number he got." So, it's expecting a single number, not multiple numbers or a binary string. That makes me think it's more likely a substitution cipher where each letter is replaced by a specific number, and then those numbers are concatenated together.Given that, let's go back to the telephone keypad idea. If we map each letter to its corresponding number on the keypad, we get:- M: 6- A: 2- T: 8- H: 4- E: 3- M: 6- A: 2- T: 8- I: 4- C: 2- S: 7So, concatenating these numbers together, we get 6 2 8 4 3 6 2 8 4 2 7, which is 62843628427.But let's double-check if there's another possible mapping. For example, sometimes people use a different mapping where letters are assigned numbers based on their row on the keypad:- First row: 1 (no letters)- Second row: ABC (2)- Third row: DEF (3)- Fourth row: GHI (4)- Fifth row: JKL (5)- Sixth row: MNO (6)- Seventh row: PQRS (7)- Eighth row: TUV (8)- Ninth row: WXYZ (9)Using this mapping, we get the same result as before: M=6, A=2, T=8, H=4, E=3, M=6, A=2, T=8, I=4, C=2, S=7. So, 62843628427.Alternatively, some people might use a different mapping where each letter is assigned a number based on its position in the alphabet, but grouped into sets. For example, A=1, B=2, ..., Z=26, but then grouped into pairs or something. But that seems more complicated, and the problem doesn't specify that.Another thought: maybe the robot is using a cipher where each letter is replaced by the number of segments in its digital display. For example, on a seven-segment display, each digit from 0 to 9 lights up a certain number of segments. But that seems too abstract and not directly related to letters.Wait, perhaps the robot is using a cipher where each letter is converted to its corresponding Roman numeral and then to a number. For example, M=1000, A=1, T=100, H=100, E=5, etc. But that would result in a very large number, and it's not clear how to handle letters like I, C, S, etc.Alternatively, maybe the robot is using a cipher where each letter is assigned a number based on its Scrabble tile value. For example, A=1, B=3, C=3, D=2, E=1, F=4, G=2, H=4, I=1, J=8, K=5, L=1, M=3, N=1, O=1, P=3, Q=10, R=1, S=1, T=1, U=1, V=4, W=4, X=8, Y=4, Z=10. Using this, let's see:- M=3- A=1- T=1- H=4- E=1- M=3- A=1- T=1- I=1- C=3- S=1So, concatenating these, we get 3 1 1 4 1 3 1 1 1 3 1, which is 31141311131. That's another possible number.But I'm not sure if that's the intended method. The problem doesn't specify that it's using Scrabble values, so that might be a stretch.Going back to the telephone keypad idea, it seems more plausible because it's a common method for mapping letters to numbers, especially in contexts like phone keypad input. So, I think that's a good candidate for the solution.To summarize, if we map each letter in "MATHEMATICS" to its corresponding number on a telephone keypad, we get:- M: 6- A: 2- T: 8- H: 4- E: 3- M: 6- A: 2- T: 8- I: 4- C: 2- S: 7Concatenating these numbers together, we get 62843628427.I think that's the most logical approach, given the information provided. Unless there's another specific cipher or method that the problem is referring to, which isn't mentioned here, this seems like the best answer.

💡Alright, so I'm trying to figure out this problem about the knowledge contest with five students: A, B, and three others. The question is about determining the number of possible rankings based on the information given to A and B.First, let's parse the information given:1. The respondent told A, "All 5 of you have different scores, unfortunately, your score is not the best." So, A is not the top scorer.2. The respondent told B, "You are not the last one." So, B is not the last scorer.Our goal is to find the total number of possible rankings for these five people given these constraints.Okay, so we have five students, all with different scores. That means we're dealing with permutations of five distinct elements. Without any constraints, the total number of possible rankings would be 5 factorial, which is 120. But we have some constraints here, so we need to adjust for those.Let's break it down step by step.**Understanding the Constraints:**1. **A is not the best (first place).**2. **B is not the last (fifth place).**So, A can be in positions 2, 3, 4, or 5.B can be in positions 1, 2, 3, or 4.But we also need to consider that these two constraints might overlap. For example, if A is in position 5, then B can still be in positions 1-4, but if A is in position 2, then B can be in positions 1, 3, 4, or 5. Wait, no, B can't be in position 5 regardless of where A is. So, actually, B's position is independent of A's position except for the fact that all scores are different, so their positions must be unique.Hmm, maybe it's better to approach this by considering the possible positions of A and B and then counting the permutations accordingly.**Case 1: A is in the last position (5th place).**If A is last, then B cannot be last, so B must be in positions 1-4. But since A is already in position 5, B can be in positions 1-4, and the remaining three students can be in the remaining three positions.So, the number of permutations in this case would be:- Choose a position for B: 4 choices (positions 1-4).- Then, arrange the remaining three students in the remaining three positions: 3! = 6 ways.So, total permutations for this case: 4 * 6 = 24.**Case 2: A is not in the last position.**This means A can be in positions 2, 3, or 4.For each of these positions of A, we need to consider where B can be.But B cannot be in position 5, so B can be in positions 1, 2, 3, or 4, but not in the same position as A.So, let's break it down further.**Subcase 2.1: A is in position 2.**- A is fixed in position 2.- B can be in positions 1, 3, 4, or 5. But wait, B cannot be in position 5, so B can be in positions 1, 3, or 4.- So, 3 choices for B.- Then, the remaining three students can be arranged in the remaining three positions: 3! = 6 ways.Total permutations for this subcase: 3 * 6 = 18.**Subcase 2.2: A is in position 3.**- A is fixed in position 3.- B can be in positions 1, 2, 4, or 5. Again, B cannot be in position 5, so B can be in positions 1, 2, or 4.- So, 3 choices for B.- Then, the remaining three students can be arranged in the remaining three positions: 3! = 6 ways.Total permutations for this subcase: 3 * 6 = 18.**Subcase 2.3: A is in position 4.**- A is fixed in position 4.- B can be in positions 1, 2, 3, or 5. But B cannot be in position 5, so B can be in positions 1, 2, or 3.- So, 3 choices for B.- Then, the remaining three students can be arranged in the remaining three positions: 3! = 6 ways.Total permutations for this subcase: 3 * 6 = 18.So, adding up all the subcases in Case 2:18 (Subcase 2.1) + 18 (Subcase 2.2) + 18 (Subcase 2.3) = 54.**Total permutations:**Case 1 (A is last): 24Case 2 (A is not last): 54Total = 24 + 54 = 78.So, the total number of possible rankings is 78.Wait, let me double-check to make sure I didn't miss anything.In Case 1, A is last, so B can be in any of the first four positions, and the rest can be arranged freely. That seems right: 4 * 3! = 24.In Case 2, A is in positions 2, 3, or 4. For each of these positions, B has 3 choices (since B can't be in position 5 or the same position as A). Then, the remaining three students are arranged in the remaining three positions, which is 3!.So, 3 positions for A, each with 3 choices for B, and 6 arrangements for the rest: 3 * 3 * 6 = 54.Adding them up: 24 + 54 = 78. That seems consistent.I think that's correct. So, the answer should be 78, which is option B.

💡Okay, so I have this problem here about a sequence {a_n} where the sum of the first n terms is given by S_n = n² - 2n - 1. I need to find a₁ + a₁₇. Hmm, let me think about how to approach this.First, I remember that the sum of the first n terms, S_n, is related to the individual terms of the sequence. Specifically, each term a_n can be found by subtracting the sum of the previous terms from the sum up to n. So, for n ≥ 2, a_n = S_n - S_{n-1}. But for n = 1, a₁ is just S₁ because there are no terms before it.So, let me write that down:- a₁ = S₁- a_n = S_n - S_{n-1} for n ≥ 2Alright, so I need to find a₁ and a₁₇ and then add them together. Let's start with a₁.Calculating a₁:S₁ = (1)² - 2*(1) - 1 = 1 - 2 - 1 = -2So, a₁ = -2Wait, that seems a bit odd. Is a₁ supposed to be negative? Well, the problem didn't specify any constraints on the terms, so I guess it's possible. Let me double-check my calculation:1² is 1, 2*1 is 2, so 1 - 2 is -1, then -1 -1 is -2. Yeah, that's correct.Now, moving on to a₁₇. Since 17 is greater than 1, I'll use the formula a_n = S_n - S_{n-1}.Calculating a₁₇:First, find S₁₇:S₁₇ = (17)² - 2*(17) - 117² is 289, 2*17 is 34, so 289 - 34 is 255, then 255 - 1 is 254.So, S₁₇ = 254Next, find S₁₆:S₁₆ = (16)² - 2*(16) - 116² is 256, 2*16 is 32, so 256 - 32 is 224, then 224 - 1 is 223.So, S₁₆ = 223Therefore, a₁₇ = S₁₇ - S₁₆ = 254 - 223 = 31Wait, that seems straightforward. Let me verify:254 - 223: 254 - 200 is 54, then subtract 23 more, which is 54 - 23 = 31. Yep, that's correct.Now, I need to find a₁ + a₁₇. So, that's (-2) + 31.Calculating that: -2 + 31 = 29So, the sum of a₁ and a₁₇ is 29.Looking back at the options given:A: 31B: 29C: 30D: 39829 is option B. So, the answer should be B.But wait, just to make sure I didn't make any mistakes in my calculations. Let me go through each step again.First, S₁:1² - 2*1 -1 = 1 - 2 -1 = -2. Correct.S₁₇:17² is 289, 2*17 is 34, so 289 - 34 = 255, then 255 -1 = 254. Correct.S₁₆:16² is 256, 2*16 is 32, so 256 -32 = 224, then 224 -1 = 223. Correct.a₁₇ = 254 - 223 = 31. Correct.a₁ + a₁₇ = -2 + 31 = 29. Correct.So, all my steps seem to check out. Therefore, I'm confident that the answer is 29, which is option B.**Final Answer**boxed{B}

💡Alright, so I've got this problem about a pool with five pipes, numbered 1 to 5. Some of these pipes are inlet pipes that fill the pool, and some are outlet pipes that drain it. The problem gives me a table that shows how long it takes to fill the pool when two pipes are opened simultaneously. My task is to figure out which single pipe, when opened alone, fills the pool the fastest. The options are (A) pipe 1, (B) pipe 2, (C) pipe 4, and (D) either pipe 3 or 5.Okay, let's break this down step by step. First, I need to understand the information given in the table. The table lists pairs of pipes and the time it takes to fill the pool when those two pipes are open. Here's the table again for clarity:begin{tabular}{|c|c|c|c|c|c|}hlineOpen Pipe Numbers & (1)(2) & (2)(3) & (3)(4) & (4)(3) & (5)(1) hlineTime to Fill the Pool (hours) & 2 & 15 & 6 & 3 & 10 hlineend{tabular}So, for example, if I open pipes 1 and 2 together, the pool fills in 2 hours. If I open pipes 2 and 3, it takes 15 hours, and so on.My goal is to find out which single pipe can fill the pool the fastest. To do this, I think I need to figure out the individual rates at which each pipe fills or empties the pool. Once I have those rates, I can determine which pipe has the highest filling rate.Let me denote the rate at which each pipe fills or empties the pool as ( Q_i ) for pipe ( i ), where ( i ) is 1 to 5. If ( Q_i ) is positive, it means the pipe is an inlet and fills the pool. If ( Q_i ) is negative, it means the pipe is an outlet and drains the pool. The units for ( Q_i ) would be pool per hour, so a positive ( Q_i ) would mean the pipe fills the pool at that rate, and a negative ( Q_i ) would mean it drains the pool at that rate.Given that, when two pipes are open together, their rates add up. So, for example, if pipes 1 and 2 are open together, their combined rate is ( Q_1 + Q_2 ), and since they fill the pool in 2 hours, their combined rate is ( frac{1}{2} ) pool per hour.Similarly, for pipes 2 and 3, their combined rate is ( Q_2 + Q_3 = frac{1}{15} ) pool per hour because they take 15 hours to fill the pool.Wait, hold on. If pipes 2 and 3 take 15 hours to fill the pool, does that mean their combined rate is ( frac{1}{15} ) pool per hour? Yes, that's correct. Because rate is reciprocal of time when working together.But here's a thought: if pipe 2 is part of both the (1)(2) pair and the (2)(3) pair, and the (1)(2) pair fills the pool much faster than the (2)(3) pair, that suggests that pipe 1 is a strong inlet pipe, while pipe 3 might be an outlet pipe or a weak inlet pipe.Similarly, looking at the (3)(4) and (4)(3) pairs, both take different times: 6 hours and 3 hours. Wait, that seems odd. How can opening pipes 3 and 4 together take 6 hours, and then opening pipes 4 and 3 together take 3 hours? That doesn't make sense because it's the same pair of pipes. Maybe that's a typo in the problem? Or perhaps it's indicating something else.Wait, looking back at the table, it says (3)(4) and then (4)(3). Maybe that's just the same pair listed twice, but with different times? That doesn't make sense because the order shouldn't matter. So, perhaps it's a mistake in the problem statement. Maybe one of them is supposed to be a different pair? Or maybe one of the times is different?Alternatively, maybe the first (3)(4) is 6 hours, and the second (4)(3) is 3 hours. That would mean that depending on which pipe is opened first, the time changes? But that doesn't make sense because the order of opening shouldn't affect the time to fill the pool. So, perhaps it's a mistake in the problem.Alternatively, maybe the first (3)(4) is 6 hours, and the second (4)(3) is 3 hours, but that would imply that pipe 4 is an inlet pipe and pipe 3 is an outlet pipe, or vice versa. Let me think about that.If pipe 3 is an outlet pipe, then opening pipes 3 and 4 together would result in a slower filling time because pipe 3 is draining the pool while pipe 4 is filling it. But if pipe 3 is an inlet pipe, then opening pipes 3 and 4 together would fill the pool faster.Wait, but the times are different: 6 hours and 3 hours. That suggests that in one case, the combined rate is ( frac{1}{6} ) pool per hour, and in the other case, it's ( frac{1}{3} ) pool per hour. That would mean that the combined rate is different depending on which pipe is opened first, which doesn't make sense because the order shouldn't matter.So, perhaps the problem intended to list different pairs, but there was a mistake. Alternatively, maybe it's a trick question where the order does matter, but that seems unlikely.Alternatively, maybe the first (3)(4) is 6 hours, and the second (4)(3) is 3 hours, but that would mean that pipe 4 is a much stronger inlet pipe than pipe 3, or pipe 3 is an outlet pipe.Wait, let's try to proceed with the information given, assuming that it's correct. Maybe the (3)(4) pair takes 6 hours, and the (4)(3) pair takes 3 hours. That would mean that when pipe 4 is opened first, it somehow affects the rate, but that seems unlikely.Alternatively, perhaps the problem meant to list different pairs, like (3)(4) and (4)(5), but that's just speculation.Given that, perhaps I should proceed with the assumption that the (3)(4) pair takes 6 hours, and the (4)(3) pair takes 3 hours, even though that seems contradictory.Alternatively, maybe it's a typo, and the second (4)(3) should be a different pair, like (4)(5). But without more information, it's hard to say.Alternatively, perhaps the problem is correct, and the (3)(4) pair takes 6 hours, and the (4)(3) pair takes 3 hours, meaning that pipe 4 is a much stronger inlet pipe than pipe 3.Wait, but that would mean that when pipe 4 is opened with pipe 3, the combined rate is higher than when pipe 3 is opened with pipe 4, which is the same pair. That doesn't make sense.Therefore, perhaps the problem intended to list different pairs, and the second (4)(3) is actually a different pair, like (4)(5). But since I don't have that information, I'll proceed with the given data, assuming that it's correct, even though it seems contradictory.So, moving forward, let's list out the equations based on the given data.1. Pipes 1 and 2 together fill the pool in 2 hours: ( Q_1 + Q_2 = frac{1}{2} ) pool per hour.2. Pipes 2 and 3 together fill the pool in 15 hours: ( Q_2 + Q_3 = frac{1}{15} ) pool per hour.3. Pipes 3 and 4 together fill the pool in 6 hours: ( Q_3 + Q_4 = frac{1}{6} ) pool per hour.4. Pipes 4 and 3 together fill the pool in 3 hours: ( Q_4 + Q_3 = frac{1}{3} ) pool per hour.5. Pipes 5 and 1 together fill the pool in 10 hours: ( Q_5 + Q_1 = frac{1}{10} ) pool per hour.Wait a minute, now I see that equations 3 and 4 are both about pipes 3 and 4, but with different times. That's the contradiction I was talking about earlier. Equation 3 says that pipes 3 and 4 together fill the pool in 6 hours, so their combined rate is ( frac{1}{6} ). Equation 4 says that pipes 4 and 3 together fill the pool in 3 hours, so their combined rate is ( frac{1}{3} ).But that's impossible because the order of opening shouldn't change the combined rate. Therefore, this suggests that there's a mistake in the problem statement. Perhaps one of the times is incorrect, or the pair is different.Given that, perhaps I should assume that one of the times is correct, and the other is a mistake. For example, maybe the second (4)(3) should be a different pair, like (4)(5), which would make more sense.Alternatively, perhaps the times are correct, but the pairings are different. For example, maybe the first (3)(4) is 6 hours, and the second (4)(3) is 3 hours, but that's the same pair.Alternatively, perhaps the first (3)(4) is 6 hours, and the second (4)(3) is 3 hours, but that would mean that pipe 4 is a much stronger inlet pipe than pipe 3, or vice versa.Wait, but if I take both equations 3 and 4 as given, I can set up the following:From equation 3:( Q_3 + Q_4 = frac{1}{6} )From equation 4:( Q_4 + Q_3 = frac{1}{3} )But that's impossible because ( frac{1}{6} ) is not equal to ( frac{1}{3} ). Therefore, this suggests that there's a mistake in the problem statement.Given that, perhaps I should proceed by assuming that one of the times is correct, and the other is a mistake. For example, perhaps the second (4)(3) is actually a different pair, like (4)(5), and the time is 3 hours.Alternatively, perhaps the first (3)(4) is 6 hours, and the second (4)(3) is 3 hours, but that's the same pair, so that's not possible.Alternatively, perhaps the second (4)(3) is a typo, and it's supposed to be (4)(5), with a time of 3 hours.Given that, perhaps I should proceed by assuming that the second (4)(3) is actually (4)(5), and the time is 3 hours. That would make more sense, and I can proceed with that assumption.So, revising the equations:1. ( Q_1 + Q_2 = frac{1}{2} ) (from pipes 1 and 2)2. ( Q_2 + Q_3 = frac{1}{15} ) (from pipes 2 and 3)3. ( Q_3 + Q_4 = frac{1}{6} ) (from pipes 3 and 4)4. ( Q_4 + Q_5 = frac{1}{3} ) (assuming the second pair is (4)(5))5. ( Q_5 + Q_1 = frac{1}{10} ) (from pipes 5 and 1)Now, with these five equations, I can try to solve for the individual rates ( Q_1 ) to ( Q_5 ).Let me write down the equations again:1. ( Q_1 + Q_2 = frac{1}{2} ) -- Equation (1)2. ( Q_2 + Q_3 = frac{1}{15} ) -- Equation (2)3. ( Q_3 + Q_4 = frac{1}{6} ) -- Equation (3)4. ( Q_4 + Q_5 = frac{1}{3} ) -- Equation (4)5. ( Q_5 + Q_1 = frac{1}{10} ) -- Equation (5)Now, I have five equations with five unknowns. I can solve this system of equations step by step.First, let's express ( Q_2 ) from Equation (1):( Q_2 = frac{1}{2} - Q_1 ) -- Equation (1a)Next, substitute ( Q_2 ) into Equation (2):( (frac{1}{2} - Q_1) + Q_3 = frac{1}{15} )Simplify:( frac{1}{2} - Q_1 + Q_3 = frac{1}{15} )Rearrange:( -Q_1 + Q_3 = frac{1}{15} - frac{1}{2} )Calculate ( frac{1}{15} - frac{1}{2} ):( frac{1}{15} - frac{7.5}{15} = -frac{6.5}{15} = -frac{13}{30} )So:( -Q_1 + Q_3 = -frac{13}{30} )Multiply both sides by -1:( Q_1 - Q_3 = frac{13}{30} ) -- Equation (2a)Now, let's look at Equation (3):( Q_3 + Q_4 = frac{1}{6} )We can express ( Q_4 ) as:( Q_4 = frac{1}{6} - Q_3 ) -- Equation (3a)Now, substitute ( Q_4 ) into Equation (4):( (frac{1}{6} - Q_3) + Q_5 = frac{1}{3} )Simplify:( frac{1}{6} - Q_3 + Q_5 = frac{1}{3} )Rearrange:( -Q_3 + Q_5 = frac{1}{3} - frac{1}{6} )Calculate ( frac{1}{3} - frac{1}{6} = frac{1}{6} )So:( -Q_3 + Q_5 = frac{1}{6} ) -- Equation (4a)Now, let's look at Equation (5):( Q_5 + Q_1 = frac{1}{10} )We can express ( Q_5 ) as:( Q_5 = frac{1}{10} - Q_1 ) -- Equation (5a)Now, substitute ( Q_5 ) from Equation (5a) into Equation (4a):( -Q_3 + (frac{1}{10} - Q_1) = frac{1}{6} )Simplify:( -Q_3 + frac{1}{10} - Q_1 = frac{1}{6} )Rearrange:( -Q_1 - Q_3 = frac{1}{6} - frac{1}{10} )Calculate ( frac{1}{6} - frac{1}{10} ):Convert to common denominator, which is 30:( frac{5}{30} - frac{3}{30} = frac{2}{30} = frac{1}{15} )So:( -Q_1 - Q_3 = frac{1}{15} )Multiply both sides by -1:( Q_1 + Q_3 = -frac{1}{15} ) -- Equation (4b)Now, we have Equation (2a):( Q_1 - Q_3 = frac{13}{30} )And Equation (4b):( Q_1 + Q_3 = -frac{1}{15} )Now, we can solve these two equations simultaneously.Let's add Equation (2a) and Equation (4b):( (Q_1 - Q_3) + (Q_1 + Q_3) = frac{13}{30} + (-frac{1}{15}) )Simplify left side:( 2Q_1 = frac{13}{30} - frac{2}{30} = frac{11}{30} )So:( 2Q_1 = frac{11}{30} )Divide both sides by 2:( Q_1 = frac{11}{60} ) pool per hour.Now, substitute ( Q_1 = frac{11}{60} ) into Equation (1a):( Q_2 = frac{1}{2} - frac{11}{60} )Convert ( frac{1}{2} ) to 30/60:( Q_2 = frac{30}{60} - frac{11}{60} = frac{19}{60} ) pool per hour.Next, substitute ( Q_1 = frac{11}{60} ) into Equation (4b):( frac{11}{60} + Q_3 = -frac{1}{15} )Convert ( -frac{1}{15} ) to -4/60:( Q_3 = -frac{4}{60} - frac{11}{60} = -frac{15}{60} = -frac{1}{4} ) pool per hour.Wait, that's interesting. So, ( Q_3 = -frac{1}{4} ) pool per hour. That means pipe 3 is an outlet pipe, draining the pool at a rate of ( frac{1}{4} ) pool per hour.Now, let's find ( Q_4 ) using Equation (3a):( Q_4 = frac{1}{6} - Q_3 = frac{1}{6} - (-frac{1}{4}) = frac{1}{6} + frac{1}{4} )Convert to common denominator, which is 12:( frac{2}{12} + frac{3}{12} = frac{5}{12} ) pool per hour.So, ( Q_4 = frac{5}{12} ) pool per hour.Finally, let's find ( Q_5 ) using Equation (5a):( Q_5 = frac{1}{10} - Q_1 = frac{1}{10} - frac{11}{60} )Convert ( frac{1}{10} ) to 6/60:( Q_5 = frac{6}{60} - frac{11}{60} = -frac{5}{60} = -frac{1}{12} ) pool per hour.So, ( Q_5 = -frac{1}{12} ) pool per hour, meaning pipe 5 is also an outlet pipe, draining the pool at a rate of ( frac{1}{12} ) pool per hour.Now, let's summarize the rates we've found:- ( Q_1 = frac{11}{60} ) pool per hour (inlet pipe)- ( Q_2 = frac{19}{60} ) pool per hour (inlet pipe)- ( Q_3 = -frac{1}{4} ) pool per hour (outlet pipe)- ( Q_4 = frac{5}{12} ) pool per hour (inlet pipe)- ( Q_5 = -frac{1}{12} ) pool per hour (outlet pipe)Now, to determine which single pipe fills the pool the fastest, we need to look at the individual rates of the inlet pipes, since outlet pipes would drain the pool, not fill it.So, the inlet pipes are pipes 1, 2, and 4, with rates:- Pipe 1: ( frac{11}{60} ) pool per hour- Pipe 2: ( frac{19}{60} ) pool per hour- Pipe 4: ( frac{5}{12} ) pool per hourNow, let's convert these rates to a common denominator to compare them easily.The common denominator for 60 and 12 is 60.- Pipe 1: ( frac{11}{60} )- Pipe 2: ( frac{19}{60} )- Pipe 4: ( frac{5}{12} = frac{25}{60} )So, in terms of pool per hour:- Pipe 1: 11/60 ≈ 0.1833- Pipe 2: 19/60 ≈ 0.3167- Pipe 4: 25/60 ≈ 0.4167Therefore, the rates from fastest to slowest are:1. Pipe 4: 25/60 pool per hour2. Pipe 2: 19/60 pool per hour3. Pipe 1: 11/60 pool per hourSo, pipe 4 has the highest rate, meaning it can fill the pool the fastest when opened alone.To double-check, let's calculate the time it would take for each inlet pipe to fill the pool alone.Time is the reciprocal of the rate.- Pipe 1: ( frac{1}{frac{11}{60}} = frac{60}{11} ) hours ≈ 5.4545 hours- Pipe 2: ( frac{1}{frac{19}{60}} = frac{60}{19} ) hours ≈ 3.1579 hours- Pipe 4: ( frac{1}{frac{25}{60}} = frac{60}{25} = frac{12}{5} = 2.4 ) hoursSo, pipe 4 takes approximately 2.4 hours to fill the pool alone, which is faster than pipe 2's approximately 3.16 hours and pipe 1's approximately 5.45 hours.Therefore, the single pipe that fills the pool the fastest is pipe 4.

💡Okay, so I have this problem here: If ( f(x) = frac{1}{2^{x} - 1} + a ) is an odd function, then what is ( a )? Hmm, let me try to figure this out step by step.First, I remember that an odd function satisfies the condition ( f(-x) = -f(x) ) for all ( x ) in its domain. So, if ( f(x) ) is odd, then plugging in ( -x ) into the function should give me the negative of the original function. That seems like a good starting point.Let me write down the given function again: ( f(x) = frac{1}{2^{x} - 1} + a ). So, to check if it's odd, I need to compute ( f(-x) ) and set it equal to ( -f(x) ).Calculating ( f(-x) ):[f(-x) = frac{1}{2^{-x} - 1} + a]Hmm, ( 2^{-x} ) is the same as ( frac{1}{2^{x}} ), right? So, let me rewrite that:[f(-x) = frac{1}{frac{1}{2^{x}} - 1} + a]To simplify the denominator, I can combine the terms:[frac{1}{frac{1 - 2^{x}}{2^{x}}} = frac{2^{x}}{1 - 2^{x}} = -frac{2^{x}}{2^{x} - 1}]So, substituting back into ( f(-x) ):[f(-x) = -frac{2^{x}}{2^{x} - 1} + a]Now, according to the property of odd functions, this should equal ( -f(x) ). Let me compute ( -f(x) ):[-f(x) = -left( frac{1}{2^{x} - 1} + a right) = -frac{1}{2^{x} - 1} - a]So, setting ( f(-x) = -f(x) ):[-frac{2^{x}}{2^{x} - 1} + a = -frac{1}{2^{x} - 1} - a]Hmm, let's rearrange this equation to solve for ( a ). First, I'll bring all the terms involving ( a ) to one side and the rest to the other side.Adding ( frac{2^{x}}{2^{x} - 1} ) to both sides:[a = -frac{1}{2^{x} - 1} - a + frac{2^{x}}{2^{x} - 1}]Wait, that seems a bit messy. Maybe I should combine the terms on each side first before moving anything.Looking back at the equation:[-frac{2^{x}}{2^{x} - 1} + a = -frac{1}{2^{x} - 1} - a]Let me bring all the ( a ) terms to the left and the other terms to the right:[a + a = -frac{1}{2^{x} - 1} + frac{2^{x}}{2^{x} - 1}]Simplifying both sides:[2a = frac{-1 + 2^{x}}{2^{x} - 1}]Wait, the numerator on the right is ( -1 + 2^{x} ), which is the same as ( 2^{x} - 1 ). So, substituting that in:[2a = frac{2^{x} - 1}{2^{x} - 1}]Oh, that simplifies nicely because the numerator and denominator are the same:[2a = 1]So, solving for ( a ):[a = frac{1}{2}]Let me double-check my steps to make sure I didn't make any mistakes. Starting from ( f(-x) ), I correctly transformed ( 2^{-x} ) into ( frac{1}{2^{x}} ) and simplified the expression. Then, setting ( f(-x) = -f(x) ), I correctly expanded both sides and rearranged the equation. Combining like terms led me to ( 2a = 1 ), so ( a = frac{1}{2} ). That seems right.Just to be thorough, let me plug ( a = frac{1}{2} ) back into the original function and check if it's indeed odd.Original function with ( a = frac{1}{2} ):[f(x) = frac{1}{2^{x} - 1} + frac{1}{2}]Compute ( f(-x) ):[f(-x) = frac{1}{2^{-x} - 1} + frac{1}{2} = frac{1}{frac{1}{2^{x}} - 1} + frac{1}{2}]Simplify the denominator:[frac{1}{frac{1 - 2^{x}}{2^{x}}} = frac{2^{x}}{1 - 2^{x}} = -frac{2^{x}}{2^{x} - 1}]So, ( f(-x) = -frac{2^{x}}{2^{x} - 1} + frac{1}{2} ).Now, compute ( -f(x) ):[-f(x) = -left( frac{1}{2^{x} - 1} + frac{1}{2} right) = -frac{1}{2^{x} - 1} - frac{1}{2}]Let me see if ( f(-x) = -f(x) ):[-frac{2^{x}}{2^{x} - 1} + frac{1}{2} stackrel{?}{=} -frac{1}{2^{x} - 1} - frac{1}{2}]Bring all terms to one side:[-frac{2^{x}}{2^{x} - 1} + frac{1}{2} + frac{1}{2^{x} - 1} + frac{1}{2} = 0]Combine like terms:[left( -frac{2^{x}}{2^{x} - 1} + frac{1}{2^{x} - 1} right) + left( frac{1}{2} + frac{1}{2} right) = 0]Simplify each group:[frac{-2^{x} + 1}{2^{x} - 1} + 1 = 0]Notice that ( -2^{x} + 1 = -(2^{x} - 1) ), so:[frac{-(2^{x} - 1)}{2^{x} - 1} + 1 = -1 + 1 = 0]Yep, that checks out. So, ( f(-x) = -f(x) ) when ( a = frac{1}{2} ), which means the function is indeed odd. Therefore, my answer is correct.I think I've covered all the steps and double-checked my work. It seems solid, so I feel confident that ( a = frac{1}{2} ) is the right answer.**Final Answer**The value of ( a ) is boxed{dfrac{1}{2}}.

💡Alright, so I have this sequence problem here. It says that the sequence {x_n} starts with x₁ = 4, and each subsequent term is defined by x_n = sqrt(2x_{n-1} + 3) for n ≥ 2. I need to prove two things:1. That the absolute difference between x_n and 3 is less than or equal to (2/3) times the absolute difference between x_{n-1} and 3. In symbols, |x_n - 3| ≤ (2/3)|x_{n-1} - 3|.2. That x_n is bounded between 3 - (2/3)^{n-1} and 3 + (2/3)^{n-1}. So, 3 - (2/3)^{n-1} ≤ x_n ≤ 3 + (2/3)^{n-1}.Okay, let's tackle the first part first. I need to show that each term gets closer to 3 by a factor of 2/3. That sounds like a recursive inequality. Maybe I can express |x_n - 3| in terms of |x_{n-1} - 3|.Given that x_n = sqrt(2x_{n-1} + 3), let's try to manipulate this expression to relate it to 3.First, let's write down |x_n - 3|:|x_n - 3| = |sqrt(2x_{n-1} + 3) - 3|Hmm, to simplify this, I remember that expressions like sqrt(a) - b can be rationalized by multiplying by sqrt(a) + b over sqrt(a) + b. Let me try that.So,|sqrt(2x_{n-1} + 3) - 3| = |(sqrt(2x_{n-1} + 3) - 3)(sqrt(2x_{n-1} + 3) + 3)| / |sqrt(2x_{n-1} + 3) + 3|The numerator simplifies because it's a difference of squares:= |(2x_{n-1} + 3) - 9| / |sqrt(2x_{n-1} + 3) + 3|Simplify the numerator:= |2x_{n-1} + 3 - 9| = |2x_{n-1} - 6| = 2|x_{n-1} - 3|So now we have:|x_n - 3| = 2|x_{n-1} - 3| / |sqrt(2x_{n-1} + 3) + 3|Now, I need to bound this expression. Specifically, I need to show that it's less than or equal to (2/3)|x_{n-1} - 3|.So, let's look at the denominator: sqrt(2x_{n-1} + 3) + 3.I need to find a lower bound for this denominator to make the entire fraction as large as possible, which would help me establish the upper bound.What's the minimum value that sqrt(2x_{n-1} + 3) can take? Since x₁ = 4, let's compute x₂:x₂ = sqrt(2*4 + 3) = sqrt(11) ≈ 3.3166x₃ = sqrt(2*sqrt(11) + 3). Let's approximate that:2*sqrt(11) ≈ 6.6332, so 6.6332 + 3 = 9.6332, sqrt(9.6332) ≈ 3.104x₄ = sqrt(2*3.104 + 3) = sqrt(6.208 + 3) = sqrt(9.208) ≈ 3.034x₅ = sqrt(2*3.034 + 3) = sqrt(6.068 + 3) = sqrt(9.068) ≈ 3.011x₆ = sqrt(2*3.011 + 3) ≈ sqrt(6.022 + 3) = sqrt(9.022) ≈ 3.0036x₇ ≈ sqrt(2*3.0036 + 3) ≈ sqrt(6.0072 + 3) = sqrt(9.0072) ≈ 3.0012So, it seems like the sequence is approaching 3 from above. So, all terms x_n are greater than 3, right? Because starting from 4, each subsequent term is the square root of something greater than 9, so it's greater than 3.Wait, actually, let's check x₁ = 4, which is greater than 3. Then x₂ = sqrt(11) ≈ 3.3166, which is still greater than 3. x₃ ≈ 3.104, still greater than 3. So, all x_n are greater than 3.Therefore, sqrt(2x_{n-1} + 3) + 3 is greater than sqrt(6) + 3, since x_{n-1} is at least 3, so 2x_{n-1} is at least 6, so sqrt(6) ≈ 2.45, so sqrt(6) + 3 ≈ 5.45.But actually, since x_{n-1} is greater than 3, 2x_{n-1} + 3 is greater than 9, so sqrt(2x_{n-1} + 3) is greater than 3. Therefore, sqrt(2x_{n-1} + 3) + 3 is greater than 6.Wait, that's not right. If x_{n-1} is greater than 3, then 2x_{n-1} + 3 is greater than 9, so sqrt(2x_{n-1} + 3) is greater than 3, so sqrt(...) + 3 is greater than 6.But in our earlier calculations, x₂ ≈ 3.3166, so sqrt(2x₁ + 3) + 3 = sqrt(11) + 3 ≈ 3.3166 + 3 = 6.3166.Similarly, x₃ ≈ 3.104, so sqrt(2x₂ + 3) + 3 ≈ sqrt(11) + 3 ≈ 6.3166.Wait, no, x₃ is sqrt(2x₂ + 3). So, x₂ is sqrt(11), so 2x₂ + 3 = 2*sqrt(11) + 3 ≈ 6.6332 + 3 = 9.6332, so sqrt(9.6332) ≈ 3.104.So, sqrt(2x₂ + 3) + 3 ≈ 3.104 + 3 = 6.104.Similarly, x₄ = sqrt(2x₃ + 3) ≈ sqrt(6.208 + 3) = sqrt(9.208) ≈ 3.034, so sqrt(2x₃ + 3) + 3 ≈ 3.034 + 3 = 6.034.So, as n increases, sqrt(2x_{n-1} + 3) + 3 approaches 6 from above.Therefore, the denominator sqrt(2x_{n-1} + 3) + 3 is always greater than 6.Wait, but in our earlier step, we had:|x_n - 3| = 2|x_{n-1} - 3| / [sqrt(2x_{n-1} + 3) + 3]Since the denominator is greater than 6, then:|x_n - 3| ≤ (2|x_{n-1} - 3|) / 6 = (1/3)|x_{n-1} - 3|Wait, but the problem statement says it's less than or equal to (2/3)|x_{n-1} - 3|. Hmm, that suggests that my bound is too loose.Wait, maybe I made a mistake in the denominator's lower bound.Wait, let's think again. If x_{n-1} is greater than 3, then 2x_{n-1} + 3 is greater than 9, so sqrt(2x_{n-1} + 3) is greater than 3.Therefore, sqrt(2x_{n-1} + 3) + 3 is greater than 6.But in reality, when x_{n-1} is close to 3, say x_{n-1} = 3, then sqrt(2*3 + 3) = sqrt(9) = 3, so sqrt(2x_{n-1} + 3) + 3 = 6.But as x_{n-1} increases, sqrt(2x_{n-1} + 3) increases, so the denominator increases beyond 6.Wait, but in our earlier calculations, when x_{n-1} is 4, the denominator is sqrt(11) + 3 ≈ 6.3166.When x_{n-1} is 3.3166, denominator is sqrt(2*3.3166 + 3) + 3 ≈ sqrt(9.6332) + 3 ≈ 3.104 + 3 = 6.104.When x_{n-1} is 3.104, denominator is sqrt(2*3.104 + 3) + 3 ≈ sqrt(9.208) + 3 ≈ 3.034 + 3 = 6.034.So, as x_{n-1} approaches 3, the denominator approaches 6.Therefore, the denominator is always greater than or equal to 6, but in reality, it's slightly more than 6 when x_{n-1} is larger.Wait, but if we take the denominator as being at least 6, then:|x_n - 3| = 2|x_{n-1} - 3| / [sqrt(2x_{n-1} + 3) + 3] ≤ 2|x_{n-1} - 3| / 6 = (1/3)|x_{n-1} - 3|But the problem statement says it's ≤ (2/3)|x_{n-1} - 3|, which is a weaker bound than (1/3)|x_{n-1} - 3|.So, perhaps my initial approach is too aggressive in bounding the denominator.Wait, maybe I need to find a better lower bound for the denominator.Let me think. Since x_{n-1} is greater than 3, let's denote x_{n-1} = 3 + d, where d > 0.Then, 2x_{n-1} + 3 = 2*(3 + d) + 3 = 6 + 2d + 3 = 9 + 2d.So, sqrt(2x_{n-1} + 3) = sqrt(9 + 2d) = 3*sqrt(1 + (2d)/9).Using the binomial approximation for sqrt(1 + ε) ≈ 1 + ε/2 when ε is small.But since d is decreasing, maybe for small d, sqrt(9 + 2d) ≈ 3 + (2d)/(2*3) = 3 + d/3.Therefore, sqrt(2x_{n-1} + 3) + 3 ≈ (3 + d/3) + 3 = 6 + d/3.So, the denominator is approximately 6 + d/3, where d = x_{n-1} - 3.Therefore, |x_n - 3| = 2d / (6 + d/3) = (2d) / (6 + d/3).Let me write this as:|x_n - 3| ≈ (2d) / (6 + d/3) = (2d) / (6 + d/3) = (2d) / [6(1 + d/(18))] = (2d)/(6) * 1/(1 + d/18) ≈ (d/3)*(1 - d/18) for small d.So, approximately, |x_n - 3| ≈ (d/3)*(1 - d/18) ≈ d/3 - d²/(54).But since d is positive and decreasing, the term d²/(54) is positive, so |x_n - 3| ≈ d/3 - something positive, which is less than d/3.Wait, but that suggests that |x_n - 3| is less than d/3, which is (1/3)|x_{n-1} - 3|.But the problem statement says it's less than or equal to (2/3)|x_{n-1} - 3|.So, perhaps my initial approach was correct, but the problem statement is giving a weaker bound.Alternatively, maybe I need to consider a different approach.Wait, let's go back to the original expression:|x_n - 3| = 2|x_{n-1} - 3| / [sqrt(2x_{n-1} + 3) + 3]We need to show that this is ≤ (2/3)|x_{n-1} - 3|.So, we need:2 / [sqrt(2x_{n-1} + 3) + 3] ≤ 2/3Which simplifies to:1 / [sqrt(2x_{n-1} + 3) + 3] ≤ 1/3Which implies:sqrt(2x_{n-1} + 3) + 3 ≥ 3But sqrt(2x_{n-1} + 3) + 3 is always greater than 3, since sqrt(2x_{n-1} + 3) is at least sqrt(9) = 3, so sqrt(...) + 3 ≥ 6.Wait, but 6 is greater than 3, so 1/[sqrt(...) + 3] ≤ 1/6, which is less than 1/3.Wait, that suggests that 2/[sqrt(...) + 3] ≤ 2/6 = 1/3, which is less than 2/3.So, actually, the inequality |x_n - 3| ≤ (2/3)|x_{n-1} - 3| is automatically satisfied because the left side is ≤ (1/3)|x_{n-1} - 3|, which is ≤ (2/3)|x_{n-1} - 3|.Therefore, the bound given in the problem is not tight, but it's still valid.So, to summarize, since sqrt(2x_{n-1} + 3) + 3 ≥ 6, we have:|x_n - 3| = 2|x_{n-1} - 3| / [sqrt(2x_{n-1} + 3) + 3] ≤ 2|x_{n-1} - 3| / 6 = (1/3)|x_{n-1} - 3| ≤ (2/3)|x_{n-1} - 3|Because (1/3) ≤ (2/3).Therefore, part (1) is proven.Now, moving on to part (2). We need to show that x_n is bounded between 3 - (2/3)^{n-1} and 3 + (2/3)^{n-1}.Given that we have part (1), which gives us a recursive inequality, we can use induction or iterative application of the inequality.Let's try to use the result from part (1). We have:|x_n - 3| ≤ (2/3)|x_{n-1} - 3|This is a recursive inequality that suggests that the error term |x_n - 3| is decreasing by a factor of 2/3 each time.Starting from x₁ = 4, we have |x₁ - 3| = 1.Then, |x₂ - 3| ≤ (2/3)|x₁ - 3| = (2/3)*1 = 2/3.Similarly, |x₃ - 3| ≤ (2/3)|x₂ - 3| ≤ (2/3)^2 * |x₁ - 3| = (2/3)^2 *1 = 4/9.Continuing this pattern, we get:|x_n - 3| ≤ (2/3)^{n-1} * |x₁ - 3| = (2/3)^{n-1} *1 = (2/3)^{n-1}.Therefore, |x_n - 3| ≤ (2/3)^{n-1}.This implies that:- (2/3)^{n-1} ≤ x_n - 3 ≤ (2/3)^{n-1}Adding 3 to all parts:3 - (2/3)^{n-1} ≤ x_n ≤ 3 + (2/3)^{n-1}Which is exactly what we needed to prove for part (2).So, putting it all together, we've shown both parts using the recursive inequality and induction.