🧠:[<think>Okay, so there are three friends: Ben, Dan, and Cam. Each of them makes a statement about the others' truthfulness. The goal is to figure out who is lying. Let me break down each statement and see how they interact.First, Ben says, "Exactly one of Dan and Cam is telling the truth." So if Ben is telling the truth, then between Dan and Cam, one is truthful and the other is lying.Then Dan says, "Exactly one of Ben and Cam is telling the truth." Similarly, if Dan is truthful, then exactly one of Ben or Cam is truthful.Cam says, "Neither Ben nor Dan is telling the truth." So if Cam is truthful, both Ben and Dan are lying.Now, let's consider possible scenarios. These types of logic puzzles often require checking each person's potential truthfulness and seeing if it leads to a contradiction.Let me start by assuming Cam is telling the truth. If Cam is truthful, then neither Ben nor Dan is telling the truth. That means both Ben and Dan are lying. Let's check their statements in that case.If Ben is lying, then his statement "Exactly one of Dan and Cam is telling the truth" is false. The negation of "exactly one" is "both or neither." So if Ben is lying, either both Dan and Cam are truthful or both are lying. But Cam is supposed to be truthful in this scenario, so that would mean Dan must also be truthful. But wait, if Cam is truthful, then Dan is lying. So there's a contradiction here. If Cam is truthful, Dan must be lying, but if Ben is lying, then Dan must be truthful. That's a problem. So this scenario leads to a contradiction. Therefore, Cam cannot be telling the truth. So Cam is lying. That's the first conclusion.Since Cam is lying, his statement "Neither Ben nor Dan is telling the truth" is false. The negation of "neither" is "at least one." So either Ben or Dan or both are telling the truth. So now we know Cam is lying. Let's update that.Now, let's check Ben and Dan. Since Cam is lying, Cam's statement is false, so at least one of Ben or Dan is truthful.Let me consider the possibility that Ben is telling the truth. If Ben is truthful, then exactly one of Dan and Cam is telling the truth. But we already established Cam is lying, so if Ben is truthful, then Dan must be telling the truth. Let's check Dan's statement in that case.Dan says, "Exactly one of Ben and Cam is telling the truth." If Dan is truthful, then exactly one of Ben and Cam is truthful. Since we're assuming Ben is truthful and Cam is lying, that would satisfy Dan's statement. So Dan's statement would be true. But wait, if Ben is truthful, then according to Ben's statement, Dan must be truthful. And if Dan is truthful, his statement checks out because Ben is truthful and Cam is lying. So in this case, both Ben and Dan are telling the truth, and Cam is lying. But hold on, Ben's statement says "Exactly one of Dan and Cam is telling the truth." If both Ben and Dan are truthful, then Cam is lying, so Ben's statement would be "Exactly one of Dan and Cam is telling the truth." But Dan is truthful, Cam is lying, so exactly one (Dan) is truthful. Wait, but Ben is truthful as well. Wait, no, Ben's statement is only about Dan and Cam. So Ben is saying "Exactly one of Dan and Cam is truthful." If Ben is truthful, then exactly one of Dan and Cam is truthful. But Cam is lying, so Dan must be truthful. So that works. So Ben's statement is about Dan and Cam. So if Ben is truthful, Dan is truthful and Cam is lying. Then Dan's statement is "Exactly one of Ben and Cam is telling the truth." But Ben is truthful and Cam is lying, so exactly one (Ben) is truthful. So Dan's statement would be true. But wait, that would mean both Ben and Dan are truthful. But Dan's statement is "Exactly one of Ben and Cam is truthful," which would be false if both Ben and Dan are truthful. Wait, hold on. Wait, this seems contradictory.Let me rephrase. If Ben is truthful, then exactly one of Dan and Cam is truthful. Since Cam is lying, Dan must be truthful. If Dan is truthful, then his statement "Exactly one of Ben and Cam is telling the truth" should be true. But in this case, Ben is truthful and Cam is lying, so exactly one (Ben) is truthful. Therefore, Dan's statement is true. But wait, Dan's statement would be "Exactly one of Ben and Cam is truthful," which is true because Ben is truthful and Cam is lying. So Dan is truthful. So both Ben and Dan are truthful. But then, Ben's original statement is "Exactly one of Dan and Cam is truthful." If Dan is truthful and Cam is lying, then exactly one (Dan) is truthful, so Ben's statement is true. Dan's statement is "Exactly one of Ben and Cam is truthful," which is true because Ben is truthful and Cam is lying. So both Ben and Dan are truthful? But the problem is that if both Ben and Dan are truthful, then Cam's statement "Neither Ben nor Dan is telling the truth" is a lie, which is consistent. But wait, but if both Ben and Dan are telling the truth, then Cam is lying, which is okay. Wait, but does that create a contradiction?Wait, let's check the numbers. If Ben and Dan are both truthful, then:- Ben's statement: Exactly one of Dan and Cam is truthful. But Dan is truthful and Cam is lying. So exactly one (Dan) is truthful. So Ben's statement holds.- Dan's statement: Exactly one of Ben and Cam is truthful. Ben is truthful and Cam is lying. So exactly one (Ben) is truthful. So Dan's statement holds.- Cam's statement: Neither Ben nor Dan is truthful. Which is false because both are truthful.So in this scenario, all statements are consistent. Ben and Dan are truthful, Cam is lying. But wait, but Cam is lying, which we established earlier. So is this possible? But the problem is that the answer options don't include "Cam is lying and Ben and Dan are truthful" as an option. Wait, let me check the options again.The options are:A: Just BenB: Just DanC: Just CamD: Each of Ben and CamE: Each of Ben, Cam and DanSo option C is "Just Cam" is lying, which would mean Ben and Dan are truthful. But according to my previous reasoning, that seems consistent. But wait, in the problem, if Ben and Dan are both telling the truth, then their statements should be consistent.Wait, but let's check again. If Ben is truthful, then exactly one of Dan or Cam is truthful. But in this case, Dan is truthful and Cam is lying. So exactly one, which matches Ben's statement. Then Dan is truthful, so his statement is exactly one of Ben or Cam is truthful. Ben is truthful and Cam is lying, so exactly one (Ben) is truthful. So Dan's statement is correct. So both Ben and Dan can be truthful. But that would mean Cam is lying, so the answer would be C: Just Cam. But let me check again if there's a contradiction.Wait, but if both Ben and Dan are truthful, then Cam's lie is okay. But then, the problem is that Dan's statement would be saying that exactly one of Ben and Cam is truthful, but if Ben is truthful and Cam is lying, then that is exactly one. So Dan's statement is true. Similarly, Ben's statement is true. So they can both be truthful. Therefore, Cam is the only liar. So answer C.But wait, the answer options include C: Just Cam. So that's possible. But hold on, let me check again. Alternatively, maybe there's another scenario where someone else is lying.Wait, let's suppose instead that Ben is lying. If Ben is lying, then his statement "Exactly one of Dan and Cam is telling the truth" is false. Therefore, either both Dan and Cam are telling the truth, or neither is. But Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is telling the truth, then Ben and Dan are lying. But if Cam is truthful, then Dan must be lying. So if Ben is lying, Cam could be truthful or lying. Wait, let's walk through this.Case 1: Suppose Ben is lying. Then his statement is false, so Dan and Cam are either both truthful or both lying. Let's check both possibilities.Subcase 1a: Both Dan and Cam are truthful. If Dan is truthful, his statement is "Exactly one of Ben and Cam is telling the truth." But in this subcase, Cam is truthful, so Dan's statement would require exactly one of Ben and Cam to be truthful. But Cam is truthful, and Ben is lying (as per our initial assumption). So exactly one (Cam) is truthful, which matches Dan's statement. Wait, so Dan's statement would be true. However, Cam's statement is "Neither Ben nor Dan is telling the truth." But Dan is supposed to be truthful in this subcase, which contradicts Cam's statement. Therefore, this subcase is impossible.Subcase 1b: Both Dan and Cam are lying. So if Ben is lying, then both Dan and Cam are lying. Let's check Cam's statement: "Neither Ben nor Dan is telling the truth." If Cam is lying, then the negation is "At least one of Ben or Dan is telling the truth." But in this case, Ben is lying (by our assumption), Dan is lying (as per subcase 1b). So both are lying, which makes Cam's statement a lie. So that's consistent. Now, Dan's statement: "Exactly one of Ben and Cam is telling the truth." Dan is lying, so the negation is that it's not exactly one. So either both or neither. Since Ben is lying and Cam is lying (subcase 1b), then neither is telling the truth, which would mean Dan's statement is false. So that's consistent. So in this scenario: Ben is lying, Dan is lying, Cam is lying. So all three are lying. But Cam's statement is "Neither Ben nor Dan is telling the truth," which is a lie, so at least one of Ben or Dan is telling the truth. But in this case, both are lying. Wait, that's a contradiction. Because if Cam is lying, then the negation of his statement is that at least one of Ben or Dan is truthful. But in this subcase, both are lying, so Cam's lie would mean that at least one is truthful, which is not the case. Therefore, this leads to a contradiction. Therefore, Subcase 1b is invalid.Therefore, if Ben is lying, both Subcases 1a and 1b lead to contradictions. Therefore, Ben cannot be lying. Therefore, Ben must be telling the truth.So going back to the previous scenario where Ben is truthful. Then, as per Ben's statement, exactly one of Dan and Cam is truthful. Since Cam is lying (as established earlier), Dan must be truthful. Then, Dan's statement is "Exactly one of Ben and Cam is telling the truth." Which is true because Ben is truthful and Cam is lying. So exactly one (Ben) is truthful. So Dan's statement is true. Therefore, both Ben and Dan are truthful, Cam is lying. So the answer is C: Just Cam.But wait, the answer options include C: Just Cam. But let me check once more. Let's consider another angle. Suppose that Dan is lying. If Dan is lying, then his statement "Exactly one of Ben and Cam is telling the truth" is false. So either both Ben and Cam are truthful or both are lying.If Dan is lying, let's check possibilities.Subcase 2a: Both Ben and Cam are truthful. If Ben is truthful, then exactly one of Dan and Cam is truthful. But Cam is supposed to be truthful here. So Dan is lying (as per our assumption) and Cam is truthful. Therefore, Ben's statement would require exactly one of Dan and Cam to be truthful. Since Cam is truthful and Dan is lying, that's exactly one, so Ben's statement holds. But Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is truthful, then both Ben and Dan are lying. But we assumed Ben is truthful here. Contradiction. Therefore, Subcase 2a invalid.Subcase 2b: Both Ben and Cam are lying. If Ben is lying, then his statement "Exactly one of Dan and Cam is truthful" is false. So Dan and Cam are both truthful or both lying. But Cam is lying (as per subcase 2b). So Dan must also be lying. But we assumed Dan is lying (from the initial assumption). So both Ben and Cam are lying. Then Cam's statement "Neither Ben nor Dan is telling the truth" is a lie. Therefore, at least one of Ben or Dan is truthful. But Ben is lying, Dan is lying, so contradiction. Therefore, Subcase 2b invalid. Therefore, Dan cannot be lying. Wait, but we initially assumed Dan is lying here, but both subcases lead to contradictions. Therefore, Dan must be truthful.Wait, this is getting confusing. Let me recap.We already determined that Cam is lying. Therefore, Cam's statement is false, so at least one of Ben or Dan is truthful.If Ben is truthful, then exactly one of Dan and Cam is truthful. Cam is lying, so Dan must be truthful. Then Dan's statement holds. So both Ben and Dan are truthful. But does that work? Let me check each statement again.Ben: "Exactly one of Dan and Cam is telling the truth." If Dan is truthful and Cam is lying, this is true.Dan: "Exactly one of Ben and Cam is telling the truth." If Ben is truthful and Cam is lying, this is also true.Cam: "Neither Ben nor Dan is telling the truth." This is false, which is consistent.Therefore, in this case, both Ben and Dan are truthful, and Cam is lying. So the answer should be C: Just Cam is lying. But wait, option C is "Just Cam," which would mean Ben and Dan are truthful. But according to the answer options, that is an option. So perhaps C is the correct answer.But wait, the problem is presented with options, and option C is "Just Cam." So according to our reasoning, that's possible.Alternatively, let's check if there's another possible scenario where more than one is lying.Wait, suppose that both Ben and Dan are lying. If Ben is lying, then his statement "Exactly one of Dan and Cam is truthful" is false. So Dan and Cam are either both truthful or both lying. If Ben is lying, let's see:If both Dan and Cam are truthful, but Cam says "Neither Ben nor Dan is truthful." If Cam is truthful, then Dan must be lying. But Dan is supposed to be truthful here. Contradiction.If both Dan and Cam are lying, then Cam's lie means at least one of Ben or Dan is truthful. But Ben is lying (as per assumption) and Dan is lying (as per this subcase). So contradiction again.Therefore, if Ben is lying, it's impossible. Therefore, Ben must be truthful.Therefore, Ben is truthful, Cam is lying, Dan is truthful. So answer C: Just Cam. But according to the answer options, C is "Just Cam," which is lying. But wait, the answer options list C as "Just Cam," so that would mean Cam is the only one lying. But according to the scenario, Ben and Dan are truthful. Wait, but in the options, answer C is "Just Cam," which is correct. However, according to my previous reasoning, that's possible, but the options also have D: Each of Ben and Cam, and E: All three.Wait, perhaps the error is that Ben and Dan cannot both be truthful because their statements would conflict. Wait, let's check again.Ben says exactly one of Dan and Cam is truthful. If Ben is truthful, then Dan is truthful and Cam is lying. Then Dan says exactly one of Ben and Cam is truthful. Since Ben is truthful and Cam is lying, Dan's statement is true. So both can be truthful. So there's no conflict. Each of their statements is about the other two. Ben's statement is about Dan and Cam, Dan's is about Ben and Cam. Since Cam is lying, both statements hold. Therefore, this is a possible scenario where Ben and Dan are truthful, and Cam is lying.But let me check if there's another scenario where Cam is truthful. If Cam is truthful, then Ben and Dan are lying. If Ben is lying, his statement "Exactly one of Dan and Cam is truthful" is false. Therefore, Dan and Cam are both truthful or both lying. But Cam is supposed to be truthful here, so Dan must also be truthful. But Dan's statement is "Exactly one of Ben and Cam is truthful." If Dan is truthful, then exactly one of Ben and Cam is truthful. But Ben is lying and Cam is truthful, so exactly one (Cam) is truthful. So Dan's statement is true. But we assumed Cam is truthful, which would require Dan to be lying (from Cam's statement). Contradiction. Therefore, Cam cannot be truthful.Thus, Cam is lying. Therefore, at least one of Ben or Dan is truthful. Then, if we suppose Ben is truthful, Dan must be truthful, which is consistent. If we suppose Ben is lying, Dan must be... But we saw that leads to contradictions. Therefore, the only consistent scenario is Ben and Dan truthful, Cam lying. So answer C: Just Cam. Therefore, the answer is C.But wait, let me check the answer options again. Option C is "Just Cam." So yes, Cam is lying. Therefore, the answer should be C. So I think that's the correct answer.Final Answerboxed{C}</think>Let's analyze the statements made by Ben, Dan, and Cam step by step:1. Ben's statement: "Exactly one of Dan and Cam is telling the truth."2. Dan's statement: "Exactly one of Ben and Cam is telling the truth."3. Cam's statement: "Neither Ben nor Dan is telling the truth."Step 1: Assume Cam is telling the truth- If Cam is truthful, then both Ben and Dan are lying.- If Ben is lying, his statement that "Exactly one of Dan and Cam is telling the truth" is false. This means both Dan and Cam are either lying or both are truthful. However, Cam is assumed truthful, so Dan must also be truthful, contradicting Cam's statement. Therefore, Cam cannot be truthful.Step 2: Cam is lying- Since Cam is lying, his statement "Neither Ben nor Dan is telling the truth" is false. Therefore, at least one of Ben or Dan is truthful.Step 3: Check Ben's statement- Assume Ben is truthful. Then, exactly one of Dan and Cam is telling the truth. Since Cam is lying, Dan must be truthful.- Dan's statement: "Exactly one of Ben and Cam is telling the truth." Since Ben is truthful and Cam is lying, Dan's statement is true.Step 4: Verify consistency- Ben is truthful, Dan is truthful, and Cam is lying.- Ben's statement holds (Dan is truthful, Cam is lying).- Dan's statement holds (Ben is truthful, Cam is lying).- Cam's statement is false (Ben and Dan are truthful).Thus, the only consistent scenario is that Cam is lying, while Ben and Dan are telling the truth. Therefore, the answer is:boxed{C},<think>Okay, let's try to figure out this logic puzzle. So, we have three friends: Ben, Dan, and Cam. Each of them makes a statement about the others. Our job is to determine who is lying based on their statements. Let me break down their statements first.Ben says, "Exactly one of Dan and Cam is telling the truth." So, Ben is claiming that between Dan and Cam, one is truthful and the other is lying.Dan says, "Exactly one of Ben and Cam is telling the truth." That's similar to Ben's statement but refers to Ben and Cam instead. So Dan is saying that between Ben and himself, and Cam, one is truthful and the other is lying.Cam says, "Neither Ben nor Dan is telling the truth." So Cam is outright stating that both Ben and Dan are liars.Alright, so we need to figure out which of these statements are true and which are false. Since each person is making a statement about the others, their truths or lies are interconnected. Let's try to approach this systematically.First, let's consider the possibilities. Each person can either be telling the truth (T) or lying (F). There are three people, so there are 2^3 = 8 possible combinations of truth-tellers and liars. But many of these combinations will contradict each other. Let's check each possibility one by one.1. All three are telling the truth (T, T, T): If all are telling the truth, let's check Cam's statement. Cam says neither Ben nor Dan is telling the truth. But if Cam is telling the truth, then both Ben and Dan must be lying. But this contradicts our initial assumption that all three are truthful. So this scenario is impossible.2. Ben and Dan are truthful, Cam is lying (T, T, F): Let's test this. If Ben is truthful, then exactly one of Dan or Cam is truthful. Since we assumed Dan is truthful, that would mean Cam must be lying, which aligns with our assumption. Dan's statement: exactly one of Ben and Cam is truthful. Ben is truthful, Cam is lying, so exactly one (Ben) is truthful. Wait, but Dan says "exactly one of Ben and Cam," which would be true since Ben is T and Cam is F. So Dan's statement is also true. Cam is lying, as per our assumption. However, Cam's lie would mean that at least one of Ben or Dan is telling the truth. But in reality, both Ben and Dan are telling the truth here, which would make Cam's statement false. But wait, Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is lying, then it's not true that neither is telling the truth, meaning at least one of them is telling the truth. But in this case, both are telling the truth, which still makes Cam's statement a lie. So this scenario is possible? Let me double-check.Wait, Ben's statement: Exactly one of Dan and Cam is truthful. If Dan is T and Cam is F, then exactly one (Dan) is truthful, so Ben's statement is true. Dan's statement: Exactly one of Ben and Cam is truthful. Ben is T, Cam is F, so exactly one (Ben) is truthful. So Dan's statement is also true. Cam's statement: Neither Ben nor Dan is truthful. But they are both truthful, so Cam is lying. So in this case, all statements are consistent. But the problem is, if both Ben and Dan are truthful, then Cam is lying. But Cam's lie is okay. However, in this scenario, Ben and Dan both claim that exactly one of the other two is truthful, but if both Ben and Dan are truthful, then each of their statements would require that only one of the others is truthful. Wait, hold on. Wait, Ben's statement is about Dan and Cam. If Ben is truthful, then exactly one of Dan and Cam is truthful. If Dan is truthful, then Cam must be lying. Similarly, Dan's statement is about Ben and Cam. If Dan is truthful, then exactly one of Ben and Cam is truthful. If Ben is truthful, then Cam must be lying. So in this case, both Ben and Dan being truthful requires that Cam is lying. But then, in their statements, Ben and Dan each assert that exactly one of the other two is truthful, but in reality, if both Ben and Dan are truthful, then each of their statements is actually saying that only one of the other two is truthful. Wait, this is a contradiction.Wait, let's parse this again. If Ben is truthful, then exactly one of Dan and Cam is truthful. So if Ben is T, then either Dan is T and Cam is F, or Dan is F and Cam is T. Similarly, if Dan is truthful, then exactly one of Ben and Cam is truthful. So if Dan is T, then either Ben is T and Cam is F, or Ben is F and Cam is T. If both Ben and Dan are truthful, then Ben's statement requires that exactly one of Dan or Cam is truthful. But Dan is T, so Cam must be F. Similarly, Dan's statement requires that exactly one of Ben or Cam is truthful. Ben is T, so Cam must be F. So Cam is F. Then Cam's statement is a lie, which is consistent. So in this scenario, both Ben and Dan are truthful, and Cam is lying, and all statements are consistent. Wait, but in this case, Ben and Dan's statements are both true, even though they each say "exactly one" of the other two. But in reality, each of their statements is referring to the other two people. Wait, Ben's statement is about Dan and Cam, not including himself. Similarly, Dan's statement is about Ben and Cam. So if Ben and Dan are both truthful, then Ben's statement says Dan and Cam have exactly one truthful, but Dan is truthful, so Cam must be lying. Similarly, Dan's statement says Ben and Cam have exactly one truthful, which is Ben, since Cam is lying. So this works. Then Cam, who is lying, says neither Ben nor Dan is truthful, which is false because both are. So this scenario is possible. Hmm, but then according to the answer choices, if Cam is lying, the answer would be C: Just Cam. But let's check other possibilities to see if there's another possible scenario.3. Ben and Cam are truthful, Dan is lying (T, F, T): Let's check this. If Ben is truthful, then exactly one of Dan and Cam is truthful. Since Cam is T, Dan must be F. That works. Dan is lying, so his statement "Exactly one of Ben and Cam is telling the truth" is false. Since Ben and Cam are both T, then Dan's statement is false, which is consistent. Cam's statement: "Neither Ben nor Dan is telling the truth." But Ben is T and Dan is F, so Cam's statement is false. But we assumed Cam is truthful here. Contradiction. So this scenario is invalid.4. Dan and Cam are truthful, Ben is lying (F, T, T): Ben is lying, so his statement "Exactly one of Dan and Cam is telling the truth" is false. Therefore, either both are lying or both are truthful. But we assumed Dan and Cam are both T, so Ben's lie is consistent. Dan's statement: "Exactly one of Ben and Cam is telling the truth." Since Cam is T and Ben is F, then exactly one (Cam) is truthful. So Dan's statement is true. Cam's statement: "Neither Ben nor Dan is telling the truth." But Dan is T, so Cam's statement is false. But we assumed Cam is T. Contradiction. So invalid.5. Only Ben is truthful (T, F, F): Ben says exactly one of Dan or Cam is truthful. If Ben is T, then one of Dan or Cam is T. But in this scenario, both Dan and Cam are F. So Ben's statement is false, which contradicts Ben being T. Invalid.6. Only Dan is truthful (F, T, F): Dan's statement: exactly one of Ben and Cam is truthful. If Dan is T, then one of Ben or Cam must be T. But in this scenario, both are F. So Dan's statement is false. Contradiction. Invalid.7. Only Cam is truthful (F, F, T): Cam's statement: "Neither Ben nor Dan is telling the truth." If Cam is T, then both Ben and Dan are F. Let's check Ben and Dan's statements. Ben's statement: "Exactly one of Dan and Cam is truthful." Since Dan is F and Cam is T, exactly one is truthful. But Ben is F here, so his statement should be false. Wait, Ben's statement would be true in reality, but since Ben is supposed to be lying, this is a contradiction. Therefore, invalid.8. All three are lying (F, F, F): If all are lying, then Cam's statement "Neither Ben nor Dan is telling the truth" is a lie. Therefore, at least one of Ben or Dan is telling the truth. But since all are lying, that contradicts. So invalid.So, from all possible scenarios, the only one that holds without contradiction is scenario 2: Ben and Dan are truthful, Cam is lying. However, wait a minute, when I first thought about scenario 2, I felt that if both Ben and Dan are truthful, each of them is stating that exactly one of the other two is truthful, but in reality, if both Ben and Dan are truthful, then when Ben talks about Dan and Cam, he's saying exactly one is truthful. But Dan is T, so Cam must be F. Similarly, Dan talks about Ben and Cam: Ben is T, Cam is F. So their statements are consistent. But wait, hold on, if Ben is truthful, then his statement is about Dan and Cam. So he's not talking about himself. So Ben's statement is "Exactly one of Dan and Cam is truthful." If Ben is T, then that requires one of Dan or Cam is T. If Dan is T, then Cam must be F, which aligns with Dan being T and Cam F. Similarly, Dan's statement is "Exactly one of Ben and Cam is truthful." If Dan is T, then exactly one of Ben and Cam is T. Since Ben is T, Cam must be F. So both statements are true. Cam's statement is a lie, so that's okay. So in this case, Ben and Dan are truthful, and Cam is lying. But according to the answer options, the choices are A: Just Ben, B: Just Dan, C: Just Cam, D: Each of Ben and Cam, E: All three. So in this scenario, only Cam is lying, so the answer would be C: Just Cam.But wait, let me verify once again. Let me check if there's another possible scenario. For example, if Cam is telling the truth. If Cam is T, then neither Ben nor Dan is telling the truth. So Ben and Dan are lying. If Ben is lying, then his statement "Exactly one of Dan and Cam is truthful" is false. So the negation is that either both Dan and Cam are truthful or both are lying. But Cam is T (per our assumption), so Dan must be T as well for Ben's statement to be false. But if Cam is T, then Dan is lying. Wait, this is a contradiction. If Cam is truthful, then Dan is lying. But if Dan is lying, his statement "Exactly one of Ben and Cam is truthful" is false. So the negation is that either both Ben and Cam are truthful or both are lying. But Cam is T (per assumption), so Ben must be T. But Cam is T, so Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is T, then Ben is F. But if Ben must be T (from Dan's lie), then contradiction. Therefore, Cam cannot be truthful. Therefore, Cam must be lying, which brings us back to scenario 2. Therefore, the only consistent scenario is Ben and Dan telling the truth, Cam lying. So the answer is C: Just Cam.Wait, but the answer options given include D: Each of Ben and Cam, and E: All three. So maybe I made a mistake here.Wait, let me check again. If Cam is lying, then his statement "Neither Ben nor Dan is telling the truth" is false, which means at least one of Ben or Dan is telling the truth. So either Ben is T, Dan is F; Ben is F, Dan is T; or both Ben and Dan are T. Let's explore these possibilities.Case 1: Cam is lying, Ben is T, Dan is F. If Ben is T, then exactly one of Dan or Cam is T. But Dan is F and Cam is F (since Cam is lying, his statement is a lie, but he could still be a liar). Wait, if Cam is lying, he's F. So Ben says exactly one of Dan and Cam is T. If Ben is T, then one of them must be T. But Dan is F and Cam is F. So Ben's statement would be false. Contradiction. So this case is invalid.Case 2: Cam is lying, Ben is F, Dan is T. If Dan is T, then exactly one of Ben and Cam is T. Ben is F, Cam is F (lying). So Dan's statement is "Exactly one of Ben and Cam is truthful." But both are F, so Dan's statement is F, contradicting Dan being T. So invalid.Case 3: Cam is lying, Ben is T, Dan is T. As in scenario 2. Then Ben's statement: Exactly one of Dan and Cam is T. Dan is T, Cam is F. So exactly one (Dan) is T. So Ben's statement is T. Dan's statement: Exactly one of Ben and Cam is T. Ben is T, Cam is F. Exactly one (Ben) is T. So Dan's statement is T. Cam's statement is F. All consistent. So this works.Therefore, the only possible scenario is Ben and Dan are truthful, Cam is lying. Therefore, the answer is C: Just Cam.But wait, looking back at the answer options: C is Just Cam, D is Each of Ben and Cam, E is All three. Wait, but according to our analysis, both Ben and Dan are telling the truth. So only Cam is lying. Therefore, the answer should be C. But let me check again.Wait, in the problem statement: "Which of the three friends is lying?" The options are:A: Just BenB: Just DanC: Just CamD: Each of Ben and CamE: Each of Ben, Cam and DanSo according to the reasoning, only Cam is lying. Therefore, the answer is C. But hold on, I need to verify once more.If Cam is lying, then Cam's statement "Neither Ben nor Dan is telling the truth" is false. Therefore, at least one of Ben or Dan is telling the truth. In our case, both Ben and Dan are telling the truth, so Cam's lie holds. Ben and Dan's statements are both true, as we checked. Therefore, yes, only Cam is lying. So answer C.Alternatively, maybe the answer is D: Each of Ben and Cam. Let me check if there's another scenario where both Ben and Cam are lying.Suppose Ben is lying. Then his statement "Exactly one of Dan and Cam is telling the truth" is false. So either both are lying or both are truthful. If Ben is lying, let's assume both Dan and Cam are lying. Then Dan's statement "Exactly one of Ben and Cam is telling the truth" would be a lie. Since Ben is lying and Cam is lying, Dan's statement is "Exactly one of Ben and Cam is T," which is false, so Dan is lying, which is consistent. Then Cam's statement "Neither Ben nor Dan is telling the truth" would be a lie, meaning at least one of them is truthful. But in this scenario, Ben and Dan are lying, Cam is lying. So all are lying, which contradicts Cam's lie. Wait, if Cam is lying, then "Neither Ben nor Dan is telling the truth" is false, so at least one is truthful. But if all are lying, then Cam's statement is false, which is okay. But then Ben's lie would mean that both Dan and Cam are lying or both are truthful. Since both Dan and Cam are lying, Ben's lie is okay. Dan's lie would mean that it's not exactly one of Ben and Cam. Since both are lying, Dan's statement is a lie. So in this scenario, all three are lying. But earlier, we dismissed this scenario because Cam's lie would mean that at least one of Ben or Dan is truthful, but if all are lying, then Cam's lie is just that it's not the case that both are lying. Wait, actually, if Cam's statement is "Neither Ben nor Dan is telling the truth," which is a lie, then the negation is "At least one of Ben or Dan is telling the truth." But in the all-lying scenario, both Ben and Dan are lying, so Cam's statement is a lie because it's not true that neither is truthful—actually, wait, in reality, neither is truthful, so Cam's statement would be true. Wait, hold on. If all three are lying, then Cam's statement is "Neither Ben nor Dan is telling the truth." If Cam is lying, then the statement is false, meaning at least one of Ben or Dan is telling the truth. But if all are lying, then both Ben and Dan are lying, so Cam's statement is actually true. Therefore, Cam cannot be lying in this scenario. Therefore, the all-lying scenario is invalid because Cam's statement would be true if all are lying, which contradicts the assumption that Cam is lying.Therefore, the only valid scenario is where Ben and Dan are truthful, Cam is lying. Therefore, only Cam is lying. The answer is C.Wait, but looking back, some sources might have different answers. Let me check once again step by step.Let me represent the statements logically.Let B, D, C represent whether Ben, Dan, Cam are telling the truth (T) or lying (F).Ben's statement: Exactly one of D and C is T. So, B ↔ (D ⊕ C).Dan's statement: Exactly one of B and C is T. So, D ↔ (B ⊕ C).Cam's statement: Neither B nor D is T. So, C ↔ (¬B ∧ ¬D).We need to find the truth values of B, D, C that satisfy all these statements.Let's consider possible cases.Case 1: C = T.Then Cam's statement requires that ¬B ∧ ¬D is true. So B = F, D = F.But if B = F, then Ben's statement is false. Ben's statement is (D ⊕ C). Since D = F and C = T, D ⊕ C = T. But Ben's statement is supposed to be false, so contradiction. Therefore, C cannot be T.Case 2: C = F.Then Cam's statement is false. Therefore, ¬(¬B ∧ ¬D) = B ∨ D. So at least one of B or D is T.Now, let's look at Ben and Dan's statements.Ben's statement: B ↔ (D ⊕ C). Since C = F, this simplifies to B ↔ (D ⊕ F) ↔ B ↔ D. So Ben's statement is equivalent to B ↔ D.Dan's statement: D ↔ (B ⊕ C). Similarly, since C = F, this simplifies to D ↔ (B ⊕ F) ↔ D ↔ B. So Dan's statement is also equivalent to D ↔ B.Therefore, Ben and Dan's statements both reduce to B ↔ D. Therefore, Ben and Dan must have the same truth value. Either both are T or both are F.But from Cam's lie, we know that at least one of B or D is T. So possibilities:Subcase 2a: B = T and D = T.Then, from Ben's statement: B ↔ D is T ↔ T, which is T. So Ben's statement is T.From Dan's statement: D ↔ B is T ↔ T, which is T. So Dan's statement is T.So in this subcase, B = T, D = T, C = F. This is consistent because:- Ben's statement: Exactly one of D and C is T. D is T, C is F. Exactly one (D) is T. So Ben's statement is T.- Dan's statement: Exactly one of B and C is T. B is T, C is F. Exactly one (B) is T. So Dan's statement is T.- Cam's statement: Neither B nor D is T. But both are T, so Cam's statement is F. So C = F, consistent.Therefore, this subcase is valid.Subcase 2b: B = F and D = F.But from Cam's lie, at least one of B or D must be T. Therefore, this subcase is invalid.Therefore, the only possible solution is B = T, D = T, C = F. Hence, only Cam is lying. The answer is C.Therefore, after thorough analysis, the answer is C: Just Cam.Final Answerboxed{C}</think>Ben, Dan, and Cam each make statements about the truthfulness of the others. We need to determine who is lying.1. Ben's statement: "Exactly one of Dan and Cam is telling the truth."2. Dan's statement: "Exactly one of Ben and Cam is telling the truth."3. Cam's statement: "Neither Ben nor Dan is telling the truth."We analyze the truth values of each statement systematically:- Cam's statement: If Cam were telling the truth, then both Ben and Dan would be lying. However, if Ben and Dan were lying, their statements would be false, leading to contradictions. Therefore, Cam must be lying.- Cam is lying: This means at least one of Ben or Dan is telling the truth. We now check Ben and Dan's statements under this condition.- Ben's statement: If Ben is telling the truth, then exactly one of Dan or Cam is telling the truth. Since Cam is lying, Dan must be telling the truth.- Dan's statement: If Dan is telling the truth, then exactly one of Ben or Cam is telling the truth. Since Cam is lying, Ben must be telling the truth.Both Ben and Dan's statements are consistent with each other and Cam's lie. Therefore, Ben and Dan are telling the truth, and Cam is lying.Thus, the answer is (boxed{C}).]

🔑:SolutionCIf Cam's statement is true, then both Ben and Dan are lying. But then exactly one of Ben and Cam is telling the truth. So Dan is telling the truth. This is a contradiction.We deduce that Cam is lying.Hence at least one of Ben and Dan is telling the truth.If Ben's statement is true, then exactly one of Ben and Cam is telling the truth. Hence Dan is telling the truth.Similarly, if Dan's statement is true, then Ben is telling the truth.We deduce that Ben and Dan are telling the truth and that Cam is lying.

🧠:[<think>Okay, so I have this problem here about vectors. Let me try to figure it out step by step. The problem says that the angle between vectors a and b is 60 degrees. The magnitude of vector a is 1, and the magnitude of the vector (a - 2b) is sqrt(13). They want to find the magnitude of vector b, and the options are given as A) sqrt(2), B) 3/2, C) 2 sqrt(2), and D) 2.Alright, let's recall some vector properties. When dealing with the magnitude of a vector difference, like |a - 2b|, I remember that the formula involves the dot product. Specifically, the square of the magnitude of a vector difference is equal to the sum of the squares of the magnitudes minus twice the dot product. So, |a - 2b|^2 = |a|^2 + |2b|^2 - 2 * a · (2b). Wait, let me check that formula again. Maybe I should write it out more carefully.The general formula is |u - v|^2 = |u|^2 + |v|^2 - 2u·v. So in this case, u is a and v is 2b. Therefore, |a - 2b|^2 = |a|^2 + |2b|^2 - 2 * a · (2b). Let me compute each term.First, |a| is given as 1, so |a|^2 is 1^2 = 1.Then, |2b|^2 is (2|b|)^2 = 4|b|^2.Now, the dot product term: 2 * a · (2b) = 4 * a · b.So putting it all together, |a - 2b|^2 = 1 + 4|b|^2 - 4(a · b).But the problem says that |a - 2b| = sqrt(13), so squaring both sides gives |a - 2b|^2 = 13. Therefore:1 + 4|b|^2 - 4(a · b) = 13.Now, I need to express a · b in terms of the angle between them. The dot product formula is a · b = |a||b|cos(theta). The angle theta is 60 degrees, and |a| is 1. So a · b = 1 * |b| * cos(60°). Cos(60°) is 0.5, so a · b = 0.5|b|.Substituting back into the equation:1 + 4|b|^2 - 4*(0.5|b|) = 13.Simplify each term:First, 4*(0.5|b|) is 2|b|. So the equation becomes:1 + 4|b|^2 - 2|b| = 13.Now, let's rearrange the terms:4|b|^2 - 2|b| + 1 - 13 = 0.Simplify 1 -13 to get -12:4|b|^2 - 2|b| -12 = 0.Hmm, that's a quadratic equation in terms of |b|. Let me denote x = |b|. Then the equation becomes:4x^2 - 2x -12 = 0.Let me divide all terms by 2 to simplify:2x^2 - x -6 = 0.Now, solving for x using the quadratic formula. The quadratic equation is ax^2 + bx + c = 0, so here a = 2, b = -1, c = -6.The quadratic formula is x = [-b ± sqrt(b^2 -4ac)] / (2a).Plugging in the values:x = [1 ± sqrt( (-1)^2 -4*2*(-6) )]/(2*2).Compute the discriminant first:(-1)^2 is 1.Then, -4*2*(-6) is -4* -12 = +48.So the discriminant is 1 + 48 = 49.Therefore, x = [1 ± sqrt(49)]/4 = [1 ±7]/4.So two solutions:x = (1 +7)/4 = 8/4 = 2.x = (1 -7)/4 = -6/4 = -1.5.But since x represents the magnitude of vector b, it can't be negative. So x = 2 is the only valid solution.Therefore, |b| = 2. So the answer is D) 2.Wait, let me double-check the calculations to be sure I didn't make any mistakes.Starting from |a -2b|^2 = 13. Then expanding:|a|^2 + |2b|^2 -4(a·b) =13.Wait, hold on, actually, when expanding |a -2b|^2, the formula is |a|^2 + |2b|^2 - 2*(a · 2b). So that would be |a|^2 + (2|b|)^2 - 4(a·b). Which is what I did before. So that's correct.Then substituting a·b = |a||b|cos60°, which is 0.5|b|. Therefore, substituting that in gives:1 +4|b|^2 -4*(0.5|b|)=1 +4|b|^2 -2|b|=13.Subtracting 13: 4|b|^2 -2|b| -12=0. Dividing by 2: 2|b|^2 -|b| -6=0.Yes, quadratic in |b|: 2x^2 -x -6=0.Solutions x=(1 ±sqrt(1 +48))/4=(1±7)/4. So 2 and -1.5. So magnitude is 2. That's correct. Therefore, the answer is D) 2.But let me check once more. Suppose |b|=2. Then |a -2b| should be sqrt(13). Let's verify.Compute |a -2b|^2: |a|^2 + |2b|^2 -4(a·b).|a|=1, so |a|^2=1.|2b|=4, so |2b|^2=16.a·b=|a||b|cos60°=1*2*0.5=1.Thus, 1 +16 -4*1=1 +16 -4=13. Which matches the given |a -2b|^2=13. Perfect, that checks out. So the answer is indeed D) 2.I think that's solid. Let me just check if there's another way someone might approach this problem, maybe using the law of cosines?Since vectors a and b have an angle of 60 degrees between them, the vector a -2b can be thought of as the third side of a triangle with sides a, 2b, and angle between them of 60 degrees. Wait, but actually, it's a - 2b, so the angle between a and -2b would be 180 - 60 = 120 degrees? Wait, maybe not exactly, because the angle between a and b is 60 degrees, so the angle between a and -2b is 180 -60=120 degrees. But the triangle formed by vectors a, -2b, and a -2b. So by the law of cosines, |a -2b|^2 = |a|^2 + |2b|^2 -2*|a|*|2b|*cos(theta), where theta is the angle between a and 2b. But since the angle between a and b is 60 degrees, the angle between a and 2b is still 60 degrees because scaling a vector doesn't change the angle. Therefore, theta=60 degrees.Wait, hold on, this seems conflicting with my earlier thought. Let me clarify. If the angle between a and b is 60 degrees, then the angle between a and 2b is still 60 degrees because 2b is just a scalar multiple of b. So when we compute the magnitude of a -2b, using the law of cosines, it's |a|^2 + |2b|^2 -2*|a|*|2b|*cos(60°). Which would give the same result as before.So let's compute that:|a -2b|^2 = |a|^2 + |2b|^2 -2*|a|*|2b|*cos(60°).Which is 1 + (2|b|)^2 -2*1*(2|b|)*0.5.Simplify each term:1 +4|b|^2 -2*1*2|b|*0.5.The last term: 2*1*2|b|*0.5 = 2|b|. So:1 +4|b|^2 -2|b| = 13, same as before. So that leads to the same equation. So solving gives |b|=2. Hence, same answer. So that confirms it again.Therefore, the answer is D) 2.Final Answerboxed{D}</think>Given the angle between vectors (boldsymbol{a}) and (boldsymbol{b}) is (60^circ), (|boldsymbol{a}| = 1), and (|boldsymbol{a} - 2boldsymbol{b}| = sqrt{13}), we need to find (|boldsymbol{b}|).First, we use the formula for the magnitude of the vector difference:[|boldsymbol{a} - 2boldsymbol{b}|^2 = |boldsymbol{a}|^2 + |2boldsymbol{b}|^2 - 2 cdot boldsymbol{a} cdot (2boldsymbol{b})]Substituting the known values:[|boldsymbol{a} - 2boldsymbol{b}|^2 = 1^2 + (2|boldsymbol{b}|)^2 - 4 cdot (boldsymbol{a} cdot boldsymbol{b})]Since the angle between (boldsymbol{a}) and (boldsymbol{b}) is (60^circ), we use the dot product formula:[boldsymbol{a} cdot boldsymbol{b} = |boldsymbol{a}| |boldsymbol{b}| cos 60^circ = 1 cdot |boldsymbol{b}| cdot frac{1}{2} = frac{1}{2} |boldsymbol{b}|]Substituting this into the equation:[13 = 1 + 4|boldsymbol{b}|^2 - 4 left(frac{1}{2} |boldsymbol{b}|right)]Simplifying the equation:[13 = 1 + 4|boldsymbol{b}|^2 - 2|boldsymbol{b}|]Rearranging terms:[4|boldsymbol{b}|^2 - 2|boldsymbol{b}| - 12 = 0]Dividing by 2:[2|boldsymbol{b}|^2 - |boldsymbol{b}| - 6 = 0]Solving this quadratic equation using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}) where (a = 2), (b = -1), and (c = -6):[|boldsymbol{b}| = frac{1 pm sqrt{1 + 48}}{4} = frac{1 pm 7}{4}]This gives two solutions:[|boldsymbol{b}| = frac{8}{4} = 2 quad text{or} quad |boldsymbol{b}| = frac{-6}{4} = -1.5]Since the magnitude cannot be negative, we have (|boldsymbol{b}| = 2).Thus, the final answer is (boxed{D}).,<think>Okay, let's see. I need to find the magnitude of vector b given some information about vectors a and a-2b. The angle between a and b is 60 degrees, the magnitude of a is 1, and the magnitude of a-2b is sqrt(13). The options are given, so maybe I can use the properties of vectors and trigonometry to solve this.First, I remember that the magnitude of a vector difference can be found using the formula similar to the law of cosines. For two vectors u and v, the magnitude of u - v is sqrt(|u|² + |v|² - 2|u||v|cosθ), where θ is the angle between them. But in this case, the vector is a - 2b, so maybe I need to adjust the formula accordingly.Let me write down what I know:- |a| = 1- The angle between a and b is 60°, so θ = 60°- |a - 2b| = sqrt(13)- Need to find |b|Hmm. So, let's denote |b| as x. Then, maybe I can express |a - 2b| in terms of x and use the given angle.Using the formula for the magnitude of the difference of vectors, but here we have a - 2b. Let's consider 2b as a vector. So, the magnitude squared of a - 2b is |a|² + |2b|² - 2 * |a| * |2b| * cosθ. Wait, but actually, the formula for |u - v|² is |u|² + |v|² - 2u·v. So, maybe I should use the dot product here.Yes, that's a good point. The square of the magnitude of a - 2b is equal to (a - 2b)·(a - 2b). Let's compute that.Expanding the dot product:(a - 2b)·(a - 2b) = a·a - 2a·2b + (2b)·(2b)Wait, no, actually, expanding (u - v)·(u - v) is u·u - 2u·v + v·v. So in this case, u is a and v is 2b. So:|a - 2b|² = |a|² - 2(a·2b) + |2b|²Which simplifies to:|a|² - 4(a·b) + 4|b|²Since |2b|² is 4|b|², and the middle term is -2*(a·2b) = -4(a·b).Okay, so we have:|a - 2b|² = |a|² - 4(a·b) + 4|b|²We know |a - 2b| is sqrt(13), so squaring that gives 13.|a| is 1, so |a|² is 1.The term a·b is the dot product of a and b, which is |a||b|cosθ. Since θ is 60°, this becomes |a||b|cos60°. |a| is 1, so it's 1 * x * 0.5 = 0.5x, where x is |b|.Putting all these into the equation:13 = 1 - 4*(0.5x) + 4x²Let me compute each term:First term: 1Second term: -4*(0.5x) = -2xThird term: 4x²So the equation becomes:13 = 1 - 2x + 4x²Let me rearrange that to form a quadratic equation:4x² - 2x + 1 - 13 = 0Simplify the constants:1 - 13 = -12So:4x² - 2x - 12 = 0Hmm, this is a quadratic in terms of x. Let's see if I can simplify this equation. Maybe divide all terms by 2 to make it simpler:2x² - x - 6 = 0So, 2x² - x - 6 = 0Now, solving for x using the quadratic formula. The quadratic formula is x = [-b ± sqrt(b² - 4ac)]/(2a), where here the coefficients are:a = 2, b = -1, c = -6Wait, hold on. Let me check:The quadratic equation is 2x² - x - 6 = 0, so a = 2, b = -1, c = -6Therefore:x = [ -(-1) ± sqrt( (-1)^2 - 4*2*(-6) ) ] / (2*2)Simplify step by step:First, the numerator:-(-1) = 1Inside the square root:(-1)^2 = 14*2*(-6) = -48So, sqrt(1 - (-48)) = sqrt(1 + 48) = sqrt(49) = 7Therefore:x = [1 ± 7]/4So two solutions:x = (1 + 7)/4 = 8/4 = 2x = (1 - 7)/4 = (-6)/4 = -1.5But since x represents the magnitude of vector b, it cannot be negative. Therefore, x = 2.So, the magnitude of vector b is 2. Let's check the options. Option D is 2. So the answer should be D.But let me verify this result to make sure I didn't make any calculation errors.Starting from the beginning:Given angle between a and b is 60°, |a|=1, |a - 2b|=sqrt(13). Find |b|.We used the formula |a - 2b|² = |a|² + |2b|² - 2*|a|*|2b|*cosθWait, actually, hold on. Earlier, I expanded (a - 2b)·(a - 2b) as |a|² -4(a·b) + 4|b|², which seems correct because:(a - 2b)·(a - 2b) = a·a - 2b·a - 2a·b + 4b·b. Wait, no. Wait, actually, when expanding (a - 2b)·(a - 2b):= a·a - 2b·a - 2a·b + (2b)·(2b)But since dot product is commutative, b·a = a·b. So:= |a|² - 4a·b + 4|b|². That's correct.Therefore, the expansion is correct. Then substituting |a|² =1, a·b= |a||b|cos60° = 0.5x, so yes, the equation becomes:13 = 1 - 4*(0.5x) +4x²Which is 13 = 1 -2x +4x², leading to 4x² -2x -12=0, then dividing by 2 gives 2x² -x -6=0. Then quadratic formula gives x=2 or x=-1.5. So x=2 is valid. That seems correct.Alternatively, maybe to double-check, let's compute |a -2b| if |b|=2.Compute |a -2b|. The angle between a and b is 60°, |a|=1, |b|=2. Then |a -2b| can be computed as sqrt(|a|² + |2b|² - 2*|a|*|2b|*cosθ)Which is sqrt(1 + 16 - 2*1*4*cos60°)Because |2b|=2*2=4, and |2b|²=16.Then cos60°=0.5, so:sqrt(1 + 16 - 2*1*4*0.5) = sqrt(17 - 4) = sqrt(13). Which matches the given value. Therefore, correct.So, yes, |b|=2 is the correct answer. Therefore, the answer is D.Final Answerboxed{D}</think>Given the angle between vectors (boldsymbol{a}) and (boldsymbol{b}) is (60^circ), (|boldsymbol{a}| = 1), and (|boldsymbol{a} - 2boldsymbol{b}| = sqrt{13}), we need to find the magnitude of (boldsymbol{b}).First, we use the formula for the magnitude of the vector difference:[|boldsymbol{a} - 2boldsymbol{b}|^2 = |boldsymbol{a}|^2 + |2boldsymbol{b}|^2 - 2 cdot |boldsymbol{a}| cdot |2boldsymbol{b}| cdot cos(60^circ)]Substituting the known values:[13 = 1^2 + (2|boldsymbol{b}|)^2 - 2 cdot 1 cdot 2|boldsymbol{b}| cdot cos(60^circ)]Simplifying each term:[13 = 1 + 4|boldsymbol{b}|^2 - 4|boldsymbol{b}| cdot 0.5][13 = 1 + 4|boldsymbol{b}|^2 - 2|boldsymbol{b}|]Rearranging the equation into a standard quadratic form:[4|boldsymbol{b}|^2 - 2|boldsymbol{b}| - 12 = 0]Dividing through by 2 to simplify:[2|boldsymbol{b}|^2 - |boldsymbol{b}| - 6 = 0]Solving this quadratic equation using the quadratic formula (|boldsymbol{b}| = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 2), (b = -1), and (c = -6):[|boldsymbol{b}| = frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 2 cdot (-6)}}{2 cdot 2}][|boldsymbol{b}| = frac{1 pm sqrt{1 + 48}}{4}][|boldsymbol{b}| = frac{1 pm 7}{4}]This gives two solutions:[|boldsymbol{b}| = frac{8}{4} = 2 quad text{or} quad |boldsymbol{b}| = frac{-6}{4} = -1.5]Since the magnitude cannot be negative, we have (|boldsymbol{b}| = 2).Thus, the answer is (boxed{D}).]

🔑:begin{array}{l}text { Given }|boldsymbol{a}-2 boldsymbol{b}|=sqrt{13} Rightarrow(boldsymbol{a}-2 boldsymbol{b})^{2}=13 Rightarrow|a|^{2}-4|a||boldsymbol{b}| cos theta +4 b^{2}=13 Rightarrow|boldsymbol{b}|=2 text {. }end{array}

🧠:[<think>Okay, let's see. I need to find the area of triangle ABC given some trigonometric conditions. Let me first write down all the given information and then figure out how to approach this step by step.Given:1. In triangle ABC, angles A, B, C are opposite sides a, b, c respectively.2. The equation: sin C * cos(A/2) = (2 - cos C) * sin(A/2)3. cos A = 3/54. a = 45. Find the area of triangle ABC.First, since cos A is given as 3/5, maybe I can find sin A using the Pythagorean identity. Let's do that first.cos A = 3/5, so sin A = sqrt(1 - cos² A) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. So sin A = 4/5.Also, since angle A is known in terms of its sine and cosine, maybe we can find angle A in degrees or radians? But maybe we don't need to. Let's see.The next part is the trigonometric equation: sin C * cos(A/2) = (2 - cos C) * sin(A/2). Hmm. Maybe I can manipulate this equation to express in terms of tan(A/2) or something like that?Let me divide both sides by cos(A/2) to get:sin C = (2 - cos C) * tan(A/2)Alternatively, divide both sides by sin(A/2):sin C * cot(A/2) = 2 - cos CHmm, not sure if that helps. Let's think about using some trigonometric identities here. Maybe express sin C and cos C in terms of angles A and B, since in a triangle, angles sum to π (180 degrees). So C = π - A - B. Maybe that could help, but since we don't know angle B, maybe not directly.Alternatively, perhaps express both sides in terms of angle A/2. Let's recall that there are half-angle formulas. For example, sin C and cos C can be related to angle C, but maybe we can relate angle C to angle A if we can find a relationship from the given equation.Wait, let's see. Let's denote angle A as α, then angle C is something. Let me try to solve the equation sin C * cos(α/2) = (2 - cos C) * sin(α/2).Let me rearrange the equation:sin C * cos(α/2) + cos C * sin(α/2) = 2 sin(α/2)Wait, that's because if I move the cos C * sin(α/2) term to the left, the equation becomes:sin C * cos(α/2) + cos C * sin(α/2) = 2 sin(α/2)But the left side is sin(C + α/2) by the sine addition formula. So:sin(C + α/2) = 2 sin(α/2)But the maximum value of sine is 1, so 2 sin(α/2) must be ≤ 1. Wait, but sin(α/2) is positive since α is an angle in a triangle, so 0 < α < π, so 0 < α/2 < π/2, so sin(α/2) is between 0 and 1. Therefore, 2 sin(α/2) must be ≤ 1, which would mean sin(α/2) ≤ 1/2. So α/2 ≤ π/6, so α ≤ π/3 (60 degrees). But given that cos α = 3/5, let's check if α is less than 60 degrees.cos α = 3/5, so α = arccos(3/5). Let's compute arccos(3/5). Since cos 53.13 degrees ≈ 3/5 (from the 3-4-5 triangle), so α ≈ 53.13 degrees, which is less than 60 degrees. Therefore, 2 sin(α/2) ≈ 2 sin(26.57 degrees) ≈ 2 * 0.4472 ≈ 0.8944, which is less than 1. So it's possible. Therefore, the equation sin(C + α/2) = 2 sin(α/2) is valid because the right-hand side is less than 1.Wait, but sin(C + α/2) = 2 sin(α/2). Let me write that as:sin(C + A/2) = 2 sin(A/2)Hmm. Let's let θ = C + A/2. Then sin θ = 2 sin(A/2). Since the maximum of sin θ is 1, we already confirmed that 2 sin(A/2) ≤ 1, which is true here because sin(A/2) ≈ sin(26.57 degrees) ≈ 0.447, so 2*0.447 ≈ 0.894 < 1. Therefore, θ = arcsin(2 sin(A/2)) or θ = π - arcsin(2 sin(A/2)). But θ = C + A/2, so:C + A/2 = arcsin(2 sin(A/2)) or C + A/2 = π - arcsin(2 sin(A/2))But since C is an angle in the triangle, and A is already known to be approximately 53 degrees, let's see. Let's compute sin(A/2). Given that cos A = 3/5, so A is arccos(3/5). Let's compute sin(A/2).Using the half-angle formula: sin(A/2) = sqrt[(1 - cos A)/2] = sqrt[(1 - 3/5)/2] = sqrt[(2/5)/2] = sqrt(1/5) = 1/√5 ≈ 0.447.Therefore, 2 sin(A/2) ≈ 0.894. Then arcsin(0.894) ≈ 63.43 degrees (since sin 63.43 ≈ 0.894). So θ ≈ 63.43 degrees or 180 - 63.43 = 116.57 degrees.Therefore, C + A/2 ≈ 63.43 degrees or 116.57 degrees.But let's compute A/2 first. Since A ≈ 53.13 degrees, then A/2 ≈ 26.57 degrees. Therefore, C + 26.57 ≈ 63.43 or 116.57.Therefore, C ≈ 63.43 - 26.57 ≈ 36.86 degrees or C ≈ 116.57 - 26.57 ≈ 90 degrees.Wait, so angle C could be approximately 36.86 degrees or 90 degrees. But in a triangle, angles sum to 180 degrees. Let's check both possibilities.First, if C ≈ 36.86 degrees, then angle B = 180 - A - C ≈ 180 - 53.13 - 36.86 ≈ 90.01 degrees, which is approximately 90 degrees.Alternatively, if C ≈ 90 degrees, then angle B ≈ 180 - 53.13 - 90 ≈ 36.87 degrees.So there are two possible solutions? Hmm. But maybe in reality, only one of them satisfies the original equation. Let's check.Wait, but the equation came from rearranging sin C * cos(A/2) = (2 - cos C) * sin(A/2). So perhaps both solutions are valid? Hmm. Let's check with exact expressions instead of approximate degrees.Alternatively, perhaps we can solve the equation more precisely.Let me write the equation again:sin C * cos(A/2) = (2 - cos C) * sin(A/2)Let's divide both sides by sin(A/2):sin C * cot(A/2) = 2 - cos CLet me express cot(A/2) in terms of cos A. Since we know cos A, maybe we can find cot(A/2) using half-angle identities.Recall that cot(A/2) = (1 + cos A)/sin A. Let's verify that:Yes, because tan(A/2) = sin A / (1 + cos A), so cot(A/2) = (1 + cos A)/sin A.Given cos A = 3/5, sin A = 4/5. Therefore, cot(A/2) = (1 + 3/5)/(4/5) = (8/5)/(4/5) = 2. So cot(A/2) = 2.Therefore, our equation becomes:sin C * 2 = 2 - cos CSo:2 sin C + cos C = 2Hmm, this is simpler. So 2 sin C + cos C = 2.Now, we can write this equation as:2 sin C + cos C = 2Let me think about solving this equation for C. Let's denote angle C as x. Then:2 sin x + cos x = 2Let me rearrange:2 sin x = 2 - cos xSquare both sides to eliminate the trigonometric functions? But squaring might introduce extraneous solutions. Alternatively, use auxiliary angle method.Express 2 sin x + cos x as R sin(x + φ) or R cos(x - φ). Let's try that.We can write 2 sin x + cos x = R sin(x + φ). Let's compute R and φ.R = sqrt(2² + 1²) = sqrt(5)φ = arctan(1/2) (since the coefficient of sin x is 2 and cos x is 1, so φ = arctan(1/2))Therefore:sqrt(5) sin(x + φ) = 2Thus:sin(x + φ) = 2 / sqrt(5)But 2 / sqrt(5) ≈ 0.894, which is less than 1, so valid.Therefore:x + φ = arcsin(2 / sqrt(5)) or π - arcsin(2 / sqrt(5))Compute φ = arctan(1/2) ≈ 0.4636 radians ≈ 26.565 degreesarcsin(2 / sqrt(5)) = arcsin(2/sqrt(5)) ≈ 63.4349 degrees ≈ 1.1071 radiansTherefore:x + 0.4636 ≈ 1.1071 or x + 0.4636 ≈ π - 1.1071 ≈ 2.0344First case:x ≈ 1.1071 - 0.4636 ≈ 0.6435 radians ≈ 36.87 degreesSecond case:x ≈ 2.0344 - 0.4636 ≈ 1.5708 radians ≈ 90 degreesSo angle C is either approximately 36.87 degrees or 90 degrees. So two possible solutions for angle C. Therefore, two possible triangles.But we need to check if both are valid. Let's check with the original equation.Case 1: C ≈ 36.87 degrees.Then compute left side: sin C * cos(A/2)We have sin C ≈ sin 36.87° ≈ 0.6cos(A/2): since A ≈ 53.13°, A/2 ≈ 26.57°, cos(26.57°) ≈ 0.8944So left side ≈ 0.6 * 0.8944 ≈ 0.5366Right side: (2 - cos C) * sin(A/2)cos C ≈ cos 36.87° ≈ 0.8sin(A/2) ≈ sin 26.57° ≈ 0.4472Thus, (2 - 0.8) * 0.4472 ≈ 1.2 * 0.4472 ≈ 0.5366, which matches left side.Case 2: C = 90 degrees.Left side: sin 90° * cos(A/2) = 1 * 0.8944 ≈ 0.8944Right side: (2 - cos 90°) * sin(A/2) = (2 - 0) * 0.4472 ≈ 2 * 0.4472 ≈ 0.8944, which also matches.Therefore, both solutions are valid. So angle C can be either 36.87° or 90°, leading to two different triangles.But given that a = 4 (side opposite angle A), we need to find the sides and then compute the area. But since there are two possible triangles, does that mean two possible areas? Wait, but maybe given the information, only one is possible.Wait, in the problem statement, there is no mention of the triangle being acute or obtuse, so both solutions may be possible, but maybe the side lengths would differ. So we need to check both possibilities.Let me first find angle C for each case, then find angle B, then use the Law of Sines to find sides b and c, then compute the area.First, let's find angle A in radians or exact terms. Since cos A = 3/5, angle A is arccos(3/5). Similarly, sin A = 4/5.Case 1: C = arcsin(2/sqrt(5)) - φ. Wait, but maybe better to use degrees for intuition.But let's work with exact expressions.Given angle C is either:1. x = arcsin(2/sqrt(5)) - arctan(1/2)But 2/sqrt(5) is equal to sin θ where θ = arcsin(2/sqrt(5)) ≈ 63.43°, and arctan(1/2) ≈ 26.565°, so x ≈ 63.43 - 26.565 ≈ 36.87°, which is arctan(3/4) because tan(36.87°) ≈ 0.75 = 3/4.Similarly, angle C = 90° in the second case.But let's see:Case 1: C = arctan(3/4) ≈ 36.87°, since tan C = 3/4.Case 2: C = 90°.Let me compute angles for both cases.Case 1:Angles:A ≈ 53.13°, C ≈ 36.87°, so B = 180 - 53.13 - 36.87 ≈ 90°.Case 2:Angles:A ≈ 53.13°, C = 90°, so B ≈ 180 - 53.13 - 90 ≈ 36.87°.So depending on angle C, angle B is 90° or 36.87°, which swaps the roles.Now, using Law of Sines: a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius.Given a = 4, sin A = 4/5.Therefore, 4 / (4/5) = 5 = 2R => R = 5/2.Therefore, sides:In Case 1:B = 90°, so sin B = 1.Therefore, b = 2R * sin B = 5/2 * 1 = 5/2? Wait, no. Wait, the Law of Sines states a/sin A = b/sin B = c/sin C.Given a = 4, sin A = 4/5, so 4 / (4/5) = 5. Therefore, 2R = 5, so R = 2.5.Therefore:- Side a = 4 corresponds to angle A ≈ 53.13°- Side b corresponds to angle B. In Case 1, angle B = 90°, so sin B = 1, so b = 5 * sin B = 5 * 1 = 5.- Side c corresponds to angle C ≈ 36.87°, sin C = 3/5 (since sin(arctan(3/4)) = 3/5), so c = 5 * sin C = 5 * 3/5 = 3.So sides in Case 1: a=4, b=5, c=3. Wait, that's a 3-4-5 triangle! Which makes sense, since angles are approximately 53.13°, 90°, 36.87°, corresponding to sides 3,4,5.In this case, the area would be (1/2)*base*height. If it's a right-angled triangle at B, then the area is (1/2)*a*c = (1/2)*4*3 = 6.Case 2:Angle C = 90°, so it's a right-angled triangle at C.Therefore, angle B ≈ 36.87°, angle A ≈ 53.13°, angle C = 90°.Using Law of Sines again:a/sin A = 4 / (4/5) = 5.Therefore:- Side a = 4 (opposite angle A)- Side b (opposite angle B ≈ 36.87°): sin B = 3/5, so b = 5 * sin B = 5*(3/5) = 3- Side c (opposite angle C = 90°): sin C = 1, so c = 5*1 = 5Therefore, sides are a=4, b=3, c=5, again a 3-4-5 triangle but with the right angle at C. Then the area is (1/2)*a*b = (1/2)*3*4 = 6. Same area.Wait, so in both cases, the area is 6? That's interesting. So regardless of the configuration, the area remains the same. Therefore, the answer should be 6.But let me verify this.Wait, in Case 1: right-angled at B, sides 3,4,5. Area (1/2)*3*4=6.In Case 2: right-angled at C, sides 3,4,5. Area (1/2)*3*4=6.Yes, same area.Alternatively, if it's not a right-angled triangle, but wait in both cases, one of the angles is 90 degrees. Wait, but the problem didn't specify that the triangle is right-angled, but through solving the equation, we arrived at two possible right-angled triangles. So regardless, both cases give the same area.Alternatively, maybe there is a non-right-angled solution? Wait, but according to our solving, the equation leads to angle C being either 36.87° or 90°, which both result in the triangle being right-angled, either at B or at C. So both cases are right-angled, leading to same area.Wait, that's fascinating. So regardless of where the right angle is, the area is the same because the legs are 3 and 4 in both cases. Therefore, the area is 6.But let me check using another formula for the area. For any triangle, area = (1/2)*a*b*sin C. Let's use this formula.In Case 1: angle C = 36.87°, sides a=4, b=5. Wait, no, sides are a=4, b=5, c=3. Wait, no, in Case 1, angle B is 90°, so sides would be a=4 (opposite A), b=5 (opposite B), c=3 (opposite C). Then area is (1/2)*a*c = (1/2)*4*3=6.In Case 2: angle C=90°, sides a=4, b=3, c=5. Then area is (1/2)*a*b= (1/2)*4*3=6.Alternatively, using the formula area = (1/2)*b*c*sin A. Let's try that.In Case 1: sides b=5, c=3, angle A=53.13°, sin A=4/5. So area=(1/2)*5*3*(4/5)= (1/2)*3*4=6.In Case 2: sides b=3, c=5, angle A=53.13°, same calculation: (1/2)*3*5*(4/5)=6.Therefore, regardless of the case, the area is 6. Therefore, the answer is 6.But let me see if there is another way to compute the area without going through the cases. Maybe using the given equation and cos A.Alternatively, perhaps using formula area = (a² sin B sin C)/(2 sin A). Let me recall that formula.But maybe better to use the formula area = (1/2)ab sin C. But we need to find sides b and c.Alternatively, since we know angle A and side a, we can use the Law of Sines to find other sides once we know angles B and C.But since we have two possible configurations, but both give the same area, so the answer is uniquely 6.Alternatively, let's compute the area using the formula:Area = (1/2) * b * c * sin ABut we need to find sides b and c.From the Law of Sines:a/sin A = b/sin B = c/sin C = 5 (as we found before)In Case 1:sin B = 1 (angle B=90°), so b=5*1=5sin C = 3/5 (angle C ≈36.87°), so c=5*(3/5)=3Thus, area = (1/2)*5*3*(4/5)= (1/2)*12=6In Case 2:sin B = 3/5 (angle B≈36.87°), so b=5*(3/5)=3sin C=1 (angle C=90°), so c=5*1=5Thus, area=(1/2)*3*5*(4/5)=6Same result.Alternatively, use Heron's formula. Let's check that.In Case 1: sides 3,4,5. Semiperimeter s=(3+4+5)/2=6Area=sqrt(s(s-a)(s-b)(s-c))=sqrt(6*3*2*1)=sqrt(36)=6In Case 2: sides 3,4,5 as well, same calculation.Thus, Heron's formula confirms the area is 6.Therefore, regardless of the case, the area is 6. Therefore, the answer is 6.But let me check once again if there could be another solution where the triangle is not right-angled, but from the equation solving, we arrived at two solutions both right-angled. So no, there is no other solution. Therefore, the area is 6.Final Answerboxed{6}</think>Given the problem, we need to find the area of triangle ( triangle ABC ) with the following conditions:1. ( sin C cos frac{A}{2} = (2 - cos C) sin frac{A}{2} )2. ( cos A = frac{3}{5} )3. ( a = 4 )First, we find ( sin A ) using the Pythagorean identity:[sin A = sqrt{1 - cos^2 A} = sqrt{1 - left(frac{3}{5}right)^2} = sqrt{frac{16}{25}} = frac{4}{5}]Next, we use the given trigonometric equation:[sin C cos frac{A}{2} = (2 - cos C) sin frac{A}{2}]Dividing both sides by ( sin frac{A}{2} ) and using the half-angle identity for cotangent:[sin C cot frac{A}{2} = 2 - cos C]Given ( cot frac{A}{2} = frac{1 + cos A}{sin A} = frac{1 + frac{3}{5}}{frac{4}{5}} = 2 ), the equation simplifies to:[2 sin C = 2 - cos C]Rearranging and solving:[2 sin C + cos C = 2]Using the auxiliary angle method, we find:[sin left( C + arctan left( frac{1}{2} right) right) = frac{2}{sqrt{5}}]This gives two solutions for ( C ):1. ( C approx 36.87^circ )2. ( C = 90^circ )For both cases, we use the Law of Sines to find the sides and verify the area:- Case 1: ( C approx 36.87^circ ), ( B = 90^circ ) - Sides: ( a = 4 ), ( b = 5 ), ( c = 3 ) - Area: ( frac{1}{2} times 4 times 3 = 6 )- Case 2: ( C = 90^circ ), ( B approx 36.87^circ ) - Sides: ( a = 4 ), ( b = 3 ), ( c = 5 ) - Area: ( frac{1}{2} times 3 times 4 = 6 )Both cases result in the same area. Using Heron's formula for sides 3, 4, 5 confirms the area is 6.Thus, the area of ( triangle ABC ) is:[boxed{6}],<think>Okay, so I need to find the area of triangle ABC given some trigonometric conditions. Let me start by writing down all the given information and see how I can connect them.First, the problem says:In triangle ABC, angles A, B, C are opposite sides a, b, c respectively. We have the equation:sin C * cos(A/2) = (2 - cos C) * sin(A/2)Also, cos A = 3/5 and a = 4. We need to find the area of triangle ABC.Alright, let's take it step by step.1. Given cos A = 3/5. From this, I can find sin A because in any triangle, sin²θ + cos²θ = 1. So, sin A = sqrt(1 - (3/5)²) = sqrt(1 - 9/25) = sqrt(16/25) = 4/5. Since angle A is in a triangle, it must be between 0 and π, so sin A is positive.2. Given a = 4. Side a is opposite angle A. So, in the Law of Sines, a/sin A = 2R, where R is the circumradius. Maybe this will come in handy later.3. The equation sin C * cos(A/2) = (2 - cos C) * sin(A/2). Hmm, this looks like a trigonometric identity that I can manipulate. Let me try to rearrange this equation.First, let's write the equation:sin C * cos(A/2) = (2 - cos C) * sin(A/2)Let me divide both sides by sin(A/2):sin C * cot(A/2) = 2 - cos CAlternatively, maybe express everything in terms of sin and cos. Let me see. Let's recall some trigonometric identities. Maybe express sin C and cos C in terms of other angles? Because in a triangle, angles sum to π, so C = π - A - B. But maybe that's too vague.Alternatively, let's use the identity for sin C and cos C in terms of angle A and B. Wait, but perhaps it's better to express the equation in terms of angle A.Alternatively, maybe use substitution. Let me set θ = A/2, so the equation becomes:sin C * cos θ = (2 - cos C) * sin θDivide both sides by cos θ:sin C = (2 - cos C) * tan θSo, sin C + cos C * tan θ = 2 tan θHmm, not sure if that helps. Let me see. Alternatively, express tan θ in terms of angle A. Since θ = A/2, tan(A/2) can be expressed in terms of sin A and cos A. The formula is tan(A/2) = sin A / (1 + cos A). Let me check that:Yes, tan(θ) = sin(2θ)/(1 + cos(2θ)) so tan(A/2) = sin A / (1 + cos A) = (4/5)/(1 + 3/5) = (4/5)/(8/5) = 4/8 = 1/2. Wait, that's useful! Because cos A = 3/5, so tan(A/2) = (1 - cos A)/sin A. Wait, another formula: tan(θ/2) = (1 - cos θ)/sin θ. So, using θ = A, tan(A/2) = (1 - cos A)/sin A = (1 - 3/5)/(4/5) = (2/5)/(4/5) = 2/4 = 1/2. So tan(A/2) is 1/2. Therefore, tan θ = 1/2 where θ = A/2.So going back, we had:sin C + cos C * tan θ = 2 tan θWhich is:sin C + cos C*(1/2) = 2*(1/2) = 1Therefore:sin C + (1/2)cos C = 1Let me write this equation:sin C + (1/2)cos C = 1Hmm. This is a linear combination of sin C and cos C equal to 1. Maybe we can write this as a single sine function. Remember that a sin x + b cos x = R sin(x + φ), where R = sqrt(a² + b²), and φ is some phase shift. Let me try that.Here, a = 1, b = 1/2. So R = sqrt(1 + (1/2)^2) = sqrt(1 + 1/4) = sqrt(5/4) = sqrt(5)/2. So:sin C + (1/2)cos C = sqrt(5)/2 * sin(C + φ) = 1But then sqrt(5)/2 * sin(C + φ) = 1 => sin(C + φ) = 2/sqrt(5) ≈ 0.894, which is valid because 2/sqrt(5) ≈ 0.894 < 1. Therefore, possible.Alternatively, we can write sin C + (1/2)cos C = 1. Let me solve this equation for C.Let me rearrange:sin C = 1 - (1/2)cos CSquare both sides:sin² C = 1 - cos C + (1/4)cos² CBut sin² C = 1 - cos² C, so:1 - cos² C = 1 - cos C + (1/4)cos² CBring all terms to one side:1 - cos² C -1 + cos C - (1/4)cos² C = 0Simplify:(-cos² C - (1/4)cos² C) + cos C = 0Combine like terms:- (5/4)cos² C + cos C = 0Multiply both sides by -4 to eliminate fractions:5 cos² C - 4 cos C = 0Factor:cos C (5 cos C - 4) = 0Therefore, cos C = 0 or cos C = 4/5.Now, check which solutions are valid.Case 1: cos C = 0. Then C = π/2. Let's check if this satisfies the original equation.Original equation: sin C * cos(A/2) = (2 - cos C) * sin(A/2)If cos C = 0, then sin C = 1 (since C = π/2), so left side: 1 * cos(A/2)Right side: (2 - 0) * sin(A/2) = 2 sin(A/2)So equation becomes cos(A/2) = 2 sin(A/2)Divide both sides by cos(A/2):1 = 2 tan(A/2)But we know tan(A/2) = 1/2, so 2*(1/2) = 1, which is true. Therefore, cos C = 0 is a valid solution. So C = π/2.Case 2: cos C = 4/5. Then sin C = sqrt(1 - (16/25)) = 3/5 (since angle C is between 0 and π, sin C is positive). Let's check if this satisfies the original equation.Left side: sin C * cos(A/2) = (3/5) * cos(A/2)Right side: (2 - 4/5) * sin(A/2) = (6/5) * sin(A/2)So equation: (3/5) cos(A/2) = (6/5) sin(A/2)Multiply both sides by 5:3 cos(A/2) = 6 sin(A/2)Divide by 3:cos(A/2) = 2 sin(A/2)Again, divide both sides by cos(A/2):1 = 2 tan(A/2)Which is 1 = 2*(1/2) = 1, which is true. Therefore, cos C = 4/5 is also a valid solution. So angle C can be either π/2 or an angle with cos C = 4/5.Wait, so both solutions are valid? Hmm. So we have two possible solutions for angle C: either C = π/2, or C = arccos(4/5). Therefore, the triangle is not uniquely determined by the given conditions? That seems odd. But maybe the other given information (a = 4, cos A = 3/5) can help us determine which one is the correct solution.Wait, let's think. Since cos A = 3/5, angle A is arccos(3/5). Let's compute angle A in radians to get an idea, but maybe we can work with fractions.Alternatively, angle A is arccos(3/5), so using the value of cos A = 3/5, angle A is acute. Then angle C can be either π/2 or arccos(4/5). Let me check the sum of angles.Since in a triangle, A + B + C = π. If angle C is π/2, then A + B = π/2. But angle A is arccos(3/5). Let's compute angle A.cos A = 3/5, so A = arccos(3/5). Let me compute this angle. Using a calculator, arccos(3/5) is approximately 53.13 degrees. So if angle C is 90 degrees, then angle B is 90 - 53.13 ≈ 36.87 degrees. Then angle B is arccos(4/5), because cos B = 4/5 (since in this case, angle B would be 36.87 degrees, and cos 36.87° ≈ 4/5). Wait, is that correct? Let me check:If angle C is 90 degrees, then angle B = π/2 - A. So:cos B = cos(π/2 - A) = sin A = 4/5. So cos B = 4/5. So angle B would be arccos(4/5), which is approximately 36.87 degrees. Wait, but sin B would be 3/5 then? Wait, no. If cos B = 4/5, then sin B = 3/5. So in this case, angle B has cos B = 4/5, sin B = 3/5.Alternatively, if angle C is arccos(4/5), then cos C = 4/5, so angle C is approximately 36.87 degrees, then angle A + angle B = π - angle C ≈ 180 - 36.87 ≈ 143.13 degrees. But angle A is already 53.13 degrees, so angle B would be approximately 143.13 - 53.13 ≈ 90 degrees. So angle B would be 90 degrees. Then, sin B = 1, cos B = 0.Therefore, depending on which solution we take for angle C, angle B is either arccos(4/5) or π/2. Therefore, there are two possible triangles? But maybe given the side a = 4, and using the Law of Sines, we can determine which one is possible.Wait, let me check. Let's first assume that angle C is π/2 (Case 1). Then, the triangle is right-angled at C. Then, using Law of Sines:a / sin A = b / sin B = c / sin C = 2RGiven that a = 4, sin A = 4/5, so 4 / (4/5) = 5. Therefore, 2R = 5 => R = 2.5.Therefore, sides:b = 2R sin B = 5 sin BBut angle B = π/2 - A. Since angle A is arccos(3/5), angle B = π/2 - A. So sin B = sin(π/2 - A) = cos A = 3/5. Therefore, b = 5*(3/5) = 3.Similarly, c = 2R sin C = 5*1 = 5.Therefore, sides would be a=4, b=3, c=5. Then the area is (1/2)*b*c*sin A? Wait, but since it's a right-angled triangle at C, the area is (1/2)*a*b. Wait, in a right-angled triangle, the area is (1/2)*leg1*leg2. Here, legs are a and b? Wait, no. Wait, angle C is the right angle, so sides a and b are the legs? Wait, side a is opposite angle A, which is 53.13 degrees, and side b is opposite angle B, which is 36.87 degrees. Side c is opposite the right angle, so c is the hypotenuse.Therefore, sides: legs are a and b, hypotenuse c. Therefore, area is (1/2)*a*b = (1/2)*4*3 = 6.Alternatively, in Case 2, where angle C = arccos(4/5) ≈ 36.87 degrees, then angle B would be π - A - C ≈ 180 - 53.13 - 36.87 ≈ 90 degrees. So angle B is 90 degrees. Then, the triangle is right-angled at B. Then, sides:Using Law of Sines again, 2R = a / sin A = 4 / (4/5) = 5. So sides:b = 2R sin B = 5*sin(π/2) = 5*1 = 5.c = 2R sin C = 5*sin(arccos(4/5)) = 5*(3/5) = 3.So sides would be a=4, b=5, c=3. Then, the area is (1/2)*a*c*sin B. Since angle B is 90 degrees, sin B = 1, so area is (1/2)*4*3 = 6. So same area.Wait, so both cases lead to the same area? Interesting. Therefore, regardless of whether angle C is 90 degrees or angle B is 90 degrees, the area is 6. Therefore, the area is 6. But let me confirm.Wait, but in the first case, legs are a=4, b=3, hypotenuse c=5, area 6. In the second case, legs are b=5, c=3, hypotenuse a=4? Wait, hypotenuse is the side opposite the right angle. Wait, if angle B is 90 degrees, then the hypotenuse is side b, which would be 5, and legs are a and c. Wait, but in this case, Law of Sines gives a=4, which is opposite angle A (53.13 degrees), and c=3, opposite angle C (36.87 degrees). Then the legs would be a=4 and c=3, hypotenuse b=5. Then area is (1/2)*4*3=6. So same as before.Therefore, regardless of which angle is the right angle (C or B), the sides are 3,4,5, leading to area 6. So the area is 6 in both cases.But wait, in the problem statement, they just give general triangle ABC with those conditions. So perhaps both possibilities are allowed, but in both cases, the area is the same.Therefore, the answer is 6. So the area is 6. Therefore, the answer is boxed{6}But let me check once more to ensure I haven't missed anything.Alternative approach: Use the formula for area in terms of sides and angles. The area is (1/2)*b*c*sin A. Wait, but I need to find sides b and c. Alternatively, use the formula area = (1/2)*a*b*sin C, but again, need sides.Alternatively, since we have angles and a side, maybe use Law of Sines.From cos A = 3/5, sin A = 4/5.Law of Sines: a/sin A = b/sin B = c/sin C = 2R.So 4/(4/5) = 5, so 2R = 5. Therefore, R = 5/2.Therefore, sides:b = 2R sin B = 5 sin Bc = 5 sin CBut angles A + B + C = π. So B + C = π - A.But we need to find B and C. However, from the given equation, we have two possibilities for angle C: either π/2 or arccos(4/5). As before.If angle C = π/2, then angle B = π/2 - A. Then sin B = cos A = 3/5. Therefore, b = 5*(3/5)=3, c=5*1=5. So area is (1/2)*3*5=15/2=7.5. Wait, wait, this contradicts earlier.Wait, hold on, if angle C is π/2, then sides are a=4, b=3, c=5. Then area is (1/2)*3*4=6. But according to Law of Sines, area would be (1/2)*a*b*sin C. Wait, angle between a and b is angle C. Wait, no. Wait, in triangle, sides a, b, c opposite angles A, B, C. So angle between sides b and c is A, angle between a and c is B, angle between a and b is C. Therefore, area is (1/2)*b*c*sin A = (1/2)*3*5*(4/5) = (15/2)*(4/5) = (15/2)*(4/5) = (60/10) = 6.Alternatively, (1/2)*a*b*sin C = (1/2)*4*3*1 = 6. Similarly, if angle B is the right angle, then area is (1/2)*a*c*sin B = (1/2)*4*3*1 = 6. So yes, regardless, the area is 6.But wait, earlier when I thought of (1/2)*3*5=7.5, that was a mistake because in that case, angle C is between sides a and b, so the area is (1/2)*a*b*sin C. So with a=4, b=3, sin C=1, area=6. Correct. So area is 6.Therefore, regardless of which angle is the right angle, the area is 6. So the answer is 6. Therefore, the area of triangle ABC is 6.Final Answerboxed{6}</think>Given the problem, we need to find the area of triangle ( triangle ABC ) with the following conditions:1. ( sin C cos frac{A}{2} = (2 - cos C) sin frac{A}{2} )2. ( cos A = frac{3}{5} )3. ( a = 4 )First, we determine ( sin A ) using the given ( cos A ):[sin A = sqrt{1 - cos^2 A} = sqrt{1 - left(frac{3}{5}right)^2} = sqrt{frac{16}{25}} = frac{4}{5}]Next, we use the given trigonometric equation:[sin C cos frac{A}{2} = (2 - cos C) sin frac{A}{2}]Dividing both sides by ( sin frac{A}{2} ):[sin C cot frac{A}{2} = 2 - cos C]We know ( tan frac{A}{2} = frac{1 - cos A}{sin A} = frac{1 - frac{3}{5}}{frac{4}{5}} = frac{1}{2} ), so ( cot frac{A}{2} = 2 ). Substituting this into the equation:[sin C cdot 2 = 2 - cos C]Simplifying:[2 sin C + cos C = 2]Solving this equation, we find two possible solutions for ( cos C ):[cos C = 0 quad text{or} quad cos C = frac{4}{5}]Case 1: ( cos C = 0 ) (i.e., ( C = frac{pi}{2} ))- Triangle is right-angled at ( C ).- Using the Law of Sines: ( frac{a}{sin A} = 2R Rightarrow 2R = frac{4}{frac{4}{5}} = 5 ).- Sides ( b ) and ( c ) are found using ( 2R ): [ b = 5 sin B = 5 cdot frac{3}{5} = 3, quad c = 5 sin C = 5 cdot 1 = 5 ]- Area: ( frac{1}{2} times a times b = frac{1}{2} times 4 times 3 = 6 ).Case 2: ( cos C = frac{4}{5} )- Using the Law of Sines, we find ( sin C = frac{3}{5} ).- Angle ( B ) becomes ( frac{pi}{2} ), making the triangle right-angled at ( B ).- Sides ( b ) and ( c ) are: [ b = 5 sin B = 5 cdot 1 = 5, quad c = 5 sin C = 5 cdot frac{3}{5} = 3 ]- Area: ( frac{1}{2} times a times c = frac{1}{2} times 4 times 3 = 6 ).In both cases, the area of ( triangle ABC ) is ( 6 ).[boxed{6}]]

🔑:Solve:cos A=frac{1-tan ^{2} frac{A}{2}}{1+tan ^{2} frac{A}{2}}=frac{3}{5} Rightarrow tan frac{A}{2}=frac{1}{2}, thus frac{sin C}{2-cos C}=tan frac{A}{2}=frac{1}{2}Rightarrow 2 sin C+cos C=2 Rightarrow 3 cos ^{2} C-4 sin C cos C=0 Rightarrow C=90^{circ} or tan C=frac{3}{4}.When C=90^{circ}, a=4, b=3, c=5 Rightarrow S_{triangle A B C}=6;When tan C=frac{3}{4}, sin C=cos A=sin left(90^{circ}-Aright) Rightarrow B=90^{circ},thus a=4, b=5, c=4 Rightarrow S_{triangle A B C}=6. In summary, S_{triangle A B C}=6.

🧠:[<think>Okay, so I need to find the expected number of composite numbers when selecting 5 numbers randomly from the set {1,2,3,...,100}, and repetition is allowed. Hmm, let's start by recalling what expected value means. The expected value is like the average outcome we would expect if we did an experiment many times. In this case, the experiment is picking 5 numbers from 1 to 100 with replacement and counting how many composites we get each time. The expected value would be the average number of composites over many such trials.To compute the expected value, I remember that for each number we pick, the probability of it being composite contributes to the overall expectation. Since expectation is linear, the expected number of composite numbers in 5 trials is just 5 times the probability that a single number picked is composite. So, maybe I can first figure out the probability that a single number selected is composite, and then multiply that by 5.But wait, first I need to figure out how many composite numbers are there between 1 and 100. Let me think. A composite number is a positive integer that has at least one positive divisor other than 1 and itself. So, numbers greater than 1 that are not prime. Also, note that 1 is neither prime nor composite. So, in the set {1,2,3,...,100}, we have numbers that are 1, primes, and composites.So, first, let's count the number of primes between 1 and 100. Then subtract that from 100, and also subtract 1 for the number 1, to get the number of composite numbers.But how many primes are there between 1 and 100? Let me recall. The primes less than 100 are:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Let me count them. Starting with 2: that's 1, 3 is 2, 5 is 3, 7 is 4, 11 is 5, 13 is 6, 17 is 7, 19 is 8, 23 is 9, 29 is 10, 31 is 11, 37 is 12, 41 is 13, 43 is 14, 47 is 15, 53 is 16, 59 is 17, 61 is 18, 67 is 19, 71 is 20, 73 is 21, 79 is 22, 83 is 23, 89 is 24, 97 is 25. So, 25 primes.So, there are 25 primes between 1 and 100. Then, the number 1 is neither prime nor composite. Therefore, the number of composite numbers should be 100 - 25 -1 = 74.Wait, let me check that. Total numbers from 1 to 100: 100. Subtract primes (25) and 1, which is 26 numbers. So 100 -26 = 74. Yes, that seems right.Therefore, the number of composite numbers is 74. Therefore, the probability that a single randomly selected number is composite is 74/100 = 0.74. Then, since we are selecting 5 numbers with replacement, each selection is independent, right? So, the expected number of composite numbers is 5 * 0.74 = 3.7.Wait, but let me think again. Is that right? So, expectation is linear, even if the events are not independent. But in this case, since each number is selected with replacement, the selections are independent. Therefore, the expectation would just be the sum of the expectations for each trial. Since each trial has an expectation of 0.74, then 5*0.74 is correct. So, 3.7. Therefore, the answer should be 3.7, which is 37/10. So, in box form, boxed{dfrac{37}{10}} or boxed{3.7}.But let me double-check my count of primes. Because if I messed up the number of primes, that would affect the composite count. Let me recount the primes between 1 and 100.Starting from 2:2, 3, 5, 7 – primes between 1 and 10: 4 primes.Then 11, 13, 17, 19 – primes between 11-20: 4 primes. Total so far: 8.21-30: 23, 29 – 2 primes. Total: 10.31-40: 31, 37 – 2 primes. Total: 12.41-50: 41, 43, 47 – 3 primes. Total: 15.51-60: 53, 59 – 2 primes. Total: 17.61-70: 61, 67 – 2 primes. Total: 19.71-80: 71, 73, 79 – 3 primes. Total: 22.81-90: 83, 89 – 2 primes. Total: 24.91-100: 97 – 1 prime. Total: 25.Yes, 25 primes. So that's correct. So 100 -25 -1 = 74 composites.So the probability of picking a composite is 74/100. So 74 divided by 100 is 0.74, so 5 times that is 3.7, which is 37/10. So, the expected number is 3.7.But wait, hold on. Let me confirm once more. The problem says "numbers are randomly selected" with repetition allowed. So, each selection is independent, and each has the same probability. So yes, linearity of expectation applies regardless of dependence, but in this case, since they're independent, the variance would be different, but expectation is just additive.Alternatively, if we model this as a binomial distribution with n=5 trials and probability p=74/100, the expected value is np=5*(74/100)= 370/100=37/10=3.7.Therefore, the answer should be 3.7, or 37/10.But just to be thorough, let me check if my definition of composite numbers is correct. Composite numbers are positive integers greater than 1 that are not prime. So, numbers like 4, 6, 8, etc. And 1 is excluded. So yes, from 1 to 100, 1 is neither prime nor composite. So total composites are 100 -25 -1 =74.Therefore, yes, 74 composites. So probability is 74/100, expectation 5*(74/100)=3.7.Therefore, the answer is boxed{dfrac{37}{10}}.Final AnswerThe expected number of composite numbers is boxed{dfrac{37}{10}}.</think>To find the expected number of composite numbers when selecting 5 numbers randomly from the set ({1, 2, 3, ldots, 100}) with repetition allowed, we start by determining the number of composite numbers in the set.1. Count the prime numbers between 1 and 100: - There are 25 prime numbers between 1 and 100. - The number 1 is neither prime nor composite.2. Calculate the number of composite numbers: - Total numbers from 1 to 100: 100 - Subtract the number of primes (25) and the number 1: (100 - 25 - 1 = 74) - Therefore, there are 74 composite numbers.3. Determine the probability of selecting a composite number: - The probability of selecting a composite number in a single trial is (frac{74}{100} = 0.74).4. Calculate the expected number of composite numbers in 5 trials: - The expected value for each trial is 0.74. - Using linearity of expectation, the expected number of composite numbers in 5 trials is (5 times 0.74 = 3.7).Thus, the expected number of composite numbers is (boxed{dfrac{37}{10}}).,<think>Okay, let's see. The problem is asking for the expected number of composite numbers when we randomly select 5 numbers from the set {1, 2, 3, ..., 100}, and repetition is allowed. Hmm, expected value... right, so I need to find the average number of composite numbers we can expect in such a selection. First, I remember that the expected value of a random variable is the sum of each possible value multiplied by its probability. In this case, each selected number can be either composite or not. Since we are selecting 5 numbers with replacement, each selection is independent. So maybe I can model this using indicator variables. An indicator variable for each selection that is 1 if the number is composite and 0 otherwise. Then the expected number of composite numbers would be the sum of the expected values of these indicators. Because expectation is linear, right? So even if the selections are dependent, the expectation would just be the sum of each individual expectation. Wait, but since repetition is allowed, each selection is independent, so the covariance terms would be zero? Hmm, maybe not necessary here. But the linearity of expectation holds regardless of independence. So even if they were dependent, the expectation would just be 5 times the probability that a single number is composite. Therefore, if I can compute the probability that a single randomly selected number from 1 to 100 is composite, then the expected number in 5 trials would be 5 multiplied by that probability. That seems straightforward. So the key step here is figuring out how many composite numbers are there between 1 and 100. But wait, we have to remember that composite numbers are numbers greater than 1 that are not prime. So first, we need to count how many primes are there between 1 and 100, and then subtract 1 (because 1 is neither prime nor composite) to get the number of composite numbers. Wait, let's verify that. The numbers from 1 to 100 can be categorized into primes, composites, and 1. So total numbers: 100. Number of primes: let's call that P. Number of composites: 100 - P - 1. So composites = 99 - P. Therefore, the probability that a randomly selected number is composite is (99 - P)/100. Then the expected number of composites in 5 trials is 5*(99 - P)/100. So I need to find P, the number of primes between 1 and 100. Let me recall the primes up to 100. Starting from 2: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Let me count them. 2,3,5,7: 4 primes11,13,17,19: 4 (total 8)23,29: 2 (10)31,37: 2 (12)41,43,47: 3 (15)53,59: 2 (17)61,67: 2 (19)71,73,79: 3 (22)83,89: 2 (24)97: 1 (25)So there are 25 primes between 1 and 100. Therefore, P=25. Then composites would be 99 -25 = 74. Wait, 99 -25 is 74? Wait, 100 numbers total, subtract 1 (the number 1), subtract 25 primes, so composites are 74. So 74 composite numbers. Therefore, the probability that a single randomly selected number is composite is 74/100 = 0.74. Then the expected number of composite numbers in 5 trials is 5 * 0.74 = 3.7. So the answer should be 3.7, which as a fraction is 37/10, or 3.7. But let me double-check the number of primes between 1 and 100. Maybe I made a mistake in counting. Let me list them again more carefully:Primes less than 100:2, 3, 5, 7, 11, 13, 17, 19, 23, 29,31, 37, 41, 43, 47, 53, 59, 61, 67, 71,73, 79, 83, 89, 97.Let's count them line by line:First line: 10 primes (from 2 to 29)Second line: 10 primes (31 to 71)Third line: 5 primes (73 to 97). Wait, 73, 79, 83, 89, 97: that's 5 primes. Wait, total 10 + 10 + 5 = 25. So yes, 25 primes. So that's correct. Therefore, composites = 100 - 1 -25 = 74. So probability is 74/100. So the expectation is 5*(74/100) = 370/100 = 3.7. So the answer should be 3.7. But just to make sure, let's verify that 1 is not considered composite. Yes, composite numbers are defined as positive integers that have at least one positive divisor other than 1 and themselves. Since 1 has only one divisor, it's neither prime nor composite. So yes, 1 is excluded. So the count of composite numbers is 100 - primes -1. So 74. Therefore, the expected number is 5*(74/100) = 3.7. So the answer is 3.7. To confirm, 74/100 is 0.74, times 5 is 3.7. Alternatively, if they want it as a fraction, 74*5 = 370, over 100, simplifies to 37/10, which is 3.7. So either way, 3.7 is correct. I think that's solid. The key steps were realizing that expectation can be calculated by linearity, counting the number of primes, subtracting from 100 and accounting for 1, then multiplying by 5. Final AnswerThe expected number of composite numbers is boxed{3.7}.</think>To find the expected number of composite numbers when 5 numbers are randomly selected from the set ({1, 2, 3, ldots, 100}) with repetition allowed, we use the linearity of expectation. First, we need to determine the probability that a single randomly selected number is composite. 1. Count the prime numbers between 1 and 100: There are 25 prime numbers in this range.2. Calculate the number of composite numbers: - Total numbers: 100 - Subtract 1 (since 1 is neither prime nor composite) and the 25 primes: (100 - 1 - 25 = 74)3. Probability of selecting a composite number: (frac{74}{100} = 0.74)Using the linearity of expectation, the expected number of composite numbers in 5 selections is:[5 times 0.74 = 3.7]Thus, the expected number of composite numbers is (boxed{3.7}).]

🔑:(7) frac{37}{10} Hint: Among {1,2,3, cdots, 100}, there are 74 composite numbers. Let xi be the number of composite numbers drawn, thenP(xi=i)=mathrm{C}_{5}^{i}left(frac{74}{100}right)^{i}left(frac{26}{100}right)^{5-i}(0 leqslant i leqslant 5),Thus, xi follows a binomial distribution.Therefore, E xi=5 times frac{74}{100}=frac{37}{10}.

🧠:[<think>Alright, let's tackle this problem step by step. So, the question is: "Ten consecutive natural numbers are written on the board. What is the maximum number of them that can have a digit sum equal to a perfect square?" First, I need to understand what exactly is being asked here. We have ten consecutive numbers, like n, n+1, n+2, ..., n+9. We need to determine the maximum number of these ten numbers where the sum of their digits is a perfect square. A perfect square is a number that's the square of an integer, such as 1, 4, 9, 16, 25, etc. So, for each of these consecutive numbers, we calculate the sum of their digits and check if that sum is one of these perfect squares. The goal is to find the largest possible number of such numbers in a consecutive block of ten.To approach this, I think I should first recall how the digit sums of consecutive numbers behave. When you increment a number by 1, most of the time, only the last digit changes, which increases by 1. However, if the last digit is 9, there will be a carry-over. For example, 199 becomes 200 when incremented, so the digit sum decreases by 9 + 9 - 2 = 16. So, the digit sum can either increase by 1 (if there's no carry-over) or decrease by some amount if there is a carry-over. This is important because the digit sum can fluctuate quite a bit, especially when numbers have trailing 9s.Given that, we need to find a sequence of ten consecutive numbers where as many as possible have digit sums that are perfect squares. Since the digit sums can both increase and decrease, depending on the presence of trailing 9s, it might be possible to have multiple numbers in a row where their digit sums are perfect squares, interrupted by some that are not. The key is to find a starting number n where several of the numbers from n to n+9 have digit sums that hit perfect squares, either through incremental increases or through carry-over resets.Let me try to think of an example. Let's say we start with a number that ends with a lot of 9s. For example, take 1999. The digit sum is 1+9+9+9=28. The next number is 2000, whose digit sum is 2. Then 2001: 3, 2002:4, which is a perfect square. So here, from 1999 to 2002, the digit sums go from 28 to 2, 3, 4. The digit sums drop drastically due to the carry-over. But in this case, only 2002 has a digit sum that's a perfect square (4). So maybe starting near a number that causes a carry-over can reset the digit sum to a lower number, which might be a perfect square. But in this case, only one number in that sequence had a digit sum that's a perfect square.Alternatively, consider numbers that don't have trailing 9s. For example, let's take a number like 100. The digit sum is 1. Then 101: 2, 102:3, 103:4, which is a perfect square. 104:5, 105:6, 106:7, 107:8, 108:9 (another perfect square), 109:10. Here, in the sequence from 100 to 109, we have two numbers (103 and 108) with digit sums equal to perfect squares. That's better, but still only two. However, this is a very small example. Perhaps in a longer sequence, there could be more.Wait, but the question is about ten consecutive numbers. Maybe there's a way to arrange the numbers such that multiple of them have digit sums that are perfect squares. Let's think about how digit sums progress. For numbers without trailing 9s, each subsequent number increases the digit sum by 1. For example, 1000:1, 1001:2, 1002:3, ..., 1009:10. Then 1010:2, because of the trailing 9 turning into 0 and the next digit increments. So here, the digit sum drops from 10 to 2. So, in this case, there's a range where the digit sums go from 1 to 10, then drop to 2. The perfect squares here would be 1 (1000), 4 (1003), 9 (1008), and then 1 again at 1010 (but 1010 is outside the 10-number range). So within 1000 to 1009, the numbers 1000 (1), 1003 (4), 1008 (9) have digit sums as perfect squares. So that's three numbers in ten. Is it possible to get more?Alternatively, maybe if the numbers have more digits. Let's consider numbers in the hundreds. Let's take 990. Digit sum: 9+9+0=18. 991:19, 992:20, 993:21, 994:22, 995:23, 996:24, 997:25 (which is 5²), 998:26, 999:27, 1000:1. So here, 997 has a digit sum of 25, which is a perfect square. So only one in that sequence. Not so good.Wait, maybe a different range. Let's pick numbers where the digit sums are near a perfect square, so that by adding 1 each time (without carry-over), they can reach the next perfect square. For example, if we have a digit sum of 14 (not a square), then adding 1 gives 15 (still not), 16 (which is 4²). So if a number has digit sum 15, the next number (without carry-over) would have 16, which is a square. Similarly, if a number has digit sum 3, the next would be 4. So if we can have a sequence where consecutive numbers increment the digit sum into a perfect square, then that would be good. But carry-over complicates this.Alternatively, if we have a number that ends with several 9s, then incrementing it would cause the digit sum to drop. For example, going from x999 to x+1000, the digit sum drops by 9*3 -1 = 26. So if x999 has digit sum S, then x+1000 has digit sum S -26 +1 = S -25. So this is a significant drop. But how can we use this to our advantage?Suppose we have a number n where n has a digit sum that is a perfect square. Then n+1, if there's no carry-over, has a digit sum one more. If that's also a perfect square, great. If not, maybe after a carry-over. Wait, but carry-over can cause digit sums to drop. For example, if a number is ...a9...9, then adding 1 turns those 9s to 0 and increments the next digit. So the digit sum decreases by 9*k - (next digit increment). So if we can have a number that after a carry-over, the digit sum becomes a perfect square.This seems a bit abstract. Maybe I need to look for actual examples.Let me think of some perfect squares. The possible perfect squares for digit sums. The digit sum of a number can be as low as 1 (for numbers like 10, 100, etc.), and as high as... well, for a number with d digits, the maximum digit sum is 9*d. For example, a 3-digit number can have a maximum digit sum of 27. So, perfect squares up to, say, 25 (5²=25) or 36 (6²=36) might be relevant here.So, perfect squares to consider: 1, 4, 9, 16, 25, 36, 49, etc. But 49 is already quite high. For numbers with more digits, higher digit sums are possible. Let's say, for a 4-digit number, maximum digit sum is 36, so 36 is possible. For a 5-digit number, 45, so 49 is still too high. So possible perfect squares for digit sums in numbers up to, say, 5 digits: 1, 4, 9, 16, 25, 36.So, our target digit sums are these numbers. Now, we need to find ten consecutive numbers where as many as possible have digit sums in this set.Now, let's think about consecutive numbers and their digit sums. If a number doesn't end with a 9, then the next number's digit sum is one more. So, for example, if a number has digit sum 15, the next one is 16, which is a perfect square. So, in such a case, two numbers in a row could have digit sums 15 and 16, but only 16 is a perfect square. However, if you have a number with digit sum 3, the next one is 4, which is a perfect square. So here, if the first number has digit sum 3, the next one is 4. So, in that case, one is a perfect square, the next is also a perfect square. Wait, but 3 is not a perfect square. So if a number has digit sum 3, the next number has digit sum 4. So, the second number is a perfect square. Similarly, if a number has digit sum 8, the next number is 9, which is a perfect square. So, here, if you have a number ending with 8, then the next number ends with 9, digit sum increases by 1, so 8+1=9. Wait, but if the last digit is 8, adding 1 gives 9, so digit sum increases by 1. But if it's 199, adding 1 gives 200, digit sum decreases by 17. So, it's tricky.Wait, so if the number does NOT end with a 9, the digit sum increases by 1. If it ends with one 9, the digit sum decreases by 9 - 1 = 8 (since the last digit goes from 9 to 0, and the next digit increments by 1). If it ends with two 9s, the digit sum decreases by 9 + 9 - 1 = 17. Three 9s, 9*3 -1 = 26, etc.So, the key idea is that trailing 9s cause the digit sum to drop when incremented, while non-trailing 9s just increment the digit sum by 1. Therefore, if we can find numbers where the transition over a trailing 9 causes the digit sum to drop to a perfect square, and then the subsequent numbers (without trailing 9s) have digit sums that increment by 1 each time, possibly hitting more perfect squares.So, to maximize the number of such numbers, we might want a sequence where after a carry-over that drops the digit sum to a perfect square, the next few numbers have digit sums that are consecutive perfect squares. For example, if a number n ends with several 9s, so that n+1 has a much lower digit sum, which is a perfect square, then n+2, n+3, etc., have digit sums increasing by 1 each, which could be the next perfect squares.But perfect squares are spaced quadratically, so the difference between consecutive squares increases as the numbers get larger. For example, 1, 4 (diff 3), 9 (diff 5), 16 (diff 7), 25 (diff 9), etc. So, if you have a digit sum that is a perfect square, the next possible perfect square is at least 3 more. Therefore, if the digit sums are increasing by 1 each time, you can't have two consecutive perfect squares unless they are 0 and 1, but 0 isn't a digit sum for natural numbers. Wait, the digit sum of 0 is only for the number 0, but we are talking about natural numbers, which typically start at 1. So 0 is excluded. Therefore, the only case where two consecutive numbers can have digit sums that are consecutive perfect squares is if one digit sum is 1 (for example, 10, 100, etc.), and the next is 4 (digit sum increases by 3). But that would require a carry-over. Wait, no. If you have a number like 9999999, which has a digit sum of 63. Adding 1 would turn it into 10000000, digit sum 1. So from 63 to 1. That's a big drop. So in this case, you could have 9999999 (63) and 10000000 (1). 1 is a perfect square, 63 isn't. But the next number after 10000000 is 10000001, digit sum 2, then 3, 4, etc. So here, 10000000 (1), 10000003 (4), 10000008 (9), etc. But these are spread out. So within ten numbers, you might have 1, 4, 9, 16, 25, etc., but they are spaced apart.Alternatively, perhaps starting just before a multiple of 100 or 1000 where the digit sum drops to a lower number. For example, let's say a number is 3999. Digit sum is 3 + 9 + 9 + 9 = 30. Next number is 4000, digit sum 4. Then 4001:5, 4002:6, 4003:7, 4004:8, 4005:9. So here, from 4000 to 4005, the digit sums go from 4 to 9. So 4000 (4), 4004 (8), 4005 (9). Wait, 4 is 2², 9 is 3². So in this case, 4000 and 4005 have digit sums as perfect squares. Then 4000 is n+1, and 4005 is n+6. So within ten numbers, maybe more?Wait, let's map this. Starting at 3999:3999: 3+9+9+9=30 (not a square)4000: 4 (2²)4001:5 (not)4002:6 (not)4003:7 (not)4004:8 (not)4005:9 (3²)4006:10 (not)4007:11 (not)4008:12 (not)4009:13 (not)4010:5 (not)So, in this sequence, only 4000 and 4005 have digit sums that are perfect squares. So that's two. Hmm.Alternatively, maybe a different starting point. Let's consider numbers near digit sums of 15 and 16. For example, if a number has digit sum 15, the next number (if no carry-over) has digit sum 16, which is 4². So if we can find a number with digit sum 15, followed by a number with digit sum 16, that's two in a row. Then, if the next number has digit sum 17, which is not a square, and so on. But maybe after that, another carry-over could drop the digit sum again.Alternatively, let's look for a sequence where multiple carry-overs happen, causing digit sums to drop to perfect squares multiple times. For example, if we have numbers with multiple trailing 9s such that incrementing causes multiple drops.Wait, let's try to construct such a sequence. Let's say we have a number like 169999. Let's compute its digit sum: 1+6+9+9+9+9 = 1+6=7, 7+9=16, 16+9=25, 25+9=34, 34+9=43. So digit sum 43. Then, adding 1 gives 170000, digit sum 1+7+0+0+0+0=8. Then, 170001:9, 170002:10, 170003:11, ..., 170008:16, 170009:17, then 170010:8. So in this case, the digit sums go from 43 (169999) to 8 (170000), then 9 (170001), 10, ..., 16 (170008), 17 (170009), and back to 8 (170010). So in this range, 170000 (8, not a square), 170001 (9=3²), 170008 (16=4²). So two perfect squares here. But in ten numbers, starting from 169999 to 170008, the numbers with digit sums as perfect squares are 170001 and 170008. So again, two. Not better.Wait, perhaps if we have numbers where digit sums go from one perfect square to another with small gaps. For example, if we can have a sequence where digit sums go 15, 16, 24, 25. But 15 to 16 is +1, 16 to 24 is +8 (due to carry-over?), 24 to 25 is +1. Hmm, not sure.Alternatively, let's consider numbers that are near transitions where digit sums can hit multiple perfect squares. Let's take numbers where digit sums are 15, 16, 25, etc. For example, consider the number 799. Digit sum: 7+9+9=25 (which is 5²). Then 800:8 (not a square), 801:9 (3²), 802:10, 803:11, 804:12, 805:13, 806:14, 807:15, 808:16 (4²), 809:17, 810:9 (3²). So here, starting at 799:799:25 (5²)800:8801:9 (3²)802:10803:11804:12805:13806:14807:15808:16 (4²)809:17810:9 (3²)So within the numbers 799 to 808 (ten numbers: 799-808), the digit sums that are perfect squares are 799 (25), 801 (9), 808 (16). That's three perfect squares. Then, the next number 809 (17) isn't, but 810 (9) is. However, 810 is outside the ten-number range. So in the ten consecutive numbers from 799 to 808, three of them have digit sums as perfect squares. That's better than the previous examples.Wait, but in this case, 799 is the first number, and then 801, 808. So three in ten. Maybe this is a candidate. Let's check the exact numbers:Numbers:799:25 (5²) – yes800:8 – no801:9 (3²) – yes802:10 – no803:11 – no804:12 – no805:13 – no806:14 – no807:15 – no808:16 (4²) – yesSo that's three numbers: 799, 801, 808. So three out of ten.Is it possible to get more than three? Let's see.Another example: let's take numbers around 9999. 9999 has a digit sum of 36 (which is 6²). Then 10000 has digit sum 1, 10001:2, 10002:3, 10003:4 (2²), 10004:5, 10005:6, 10006:7, 10007:8, 10008:9 (3²), 10009:10. So here, from 9999 to 10008:9999:36 (6²)10000:1 (1²)10001:210002:310003:4 (2²)10004:510005:610006:710007:810008:9 (3²)So within these ten numbers, 9999 (36), 10000 (1), 10003 (4), 10008 (9). That's four numbers with digit sums as perfect squares. Wait, that's four! So starting from 9999 to 10008, there are four numbers with digit sums equal to perfect squares. Let's list them:9999:36 (6²) – yes10000:1 (1²) – yes10003:4 (2²) – yes10008:9 (3²) – yesSo that's four. Is that correct? Let's check:9999: 9+9+9+9=36 (6²)10000:1+0+0+0+0=1 (1²)10001:1+0+0+0+1=210002:310003:4 (2²)10004:510005:610006:710007:810008:9 (3²)Yes, exactly four numbers. That's better. So in this case, starting with 9999, we get four numbers in the next ten numbers with digit sums as perfect squares. So maybe four is possible.But can we get more than four?Let me try to find another example. Let's see. How about numbers around 1999999. Let's compute digit sums.1999999:1+9+9+9+9+9+9 = 1 + 9*6 = 1 +54=55. Not a square.Next number:2000000, digit sum 2. Then 2000001:3, 2000002:4 (2²), 2000003:5, 2000004:6, 2000005:7, 2000006:8, 2000007:9 (3²), 2000008:10, 2000009:11, 2000010:3. So in this case, starting from 1999999 to 2000008:1999999:55 – no2000000:2 – no2000001:3 – no2000002:4 – yes2000003:5 – no2000004:6 – no2000005:7 – no2000006:8 – no2000007:9 – yes2000008:10 – noSo here, two numbers with digit sums as perfect squares. Not as good as the previous example.Wait, going back to the previous example with 9999, four numbers. Let's see if we can extend that.Wait, another example. Let's take 9999999999 (10 9s). But that's probably overkill. Let's take 999999. Digit sum 54. Next number is 1000000, digit sum 1. Then 1000001:2, 1000002:3, 1000003:4 (2²), 1000004:5, 1000005:6, 1000006:7, 1000007:8, 1000008:9 (3²), 1000009:10. So here, 1000003 (4), 1000008 (9). So two. Not better than four.Wait, how about numbers where multiple carries happen? For example, let's take a number like 699999.Digit sum:6 + 9*5 =6 +45=51. Next number:700000, digit sum7. Then 700001:8, 700002:9, 700003:10, 700004:11, 700005:12, 700006:13, 700007:14, 700008:15, 700009:16. So here, 700000:7, 700001:8, 700002:9 (3²), 700008:16 (4²). So two perfect squares. Not great.Hmm. Let's think back to the 9999 example. The reason that worked was because the transition from 9999 (36) to 10000 (1) allowed for digit sums to reset, and then with numbers increasing their digit sums by 1 each time, hitting 1, 4, 9. But since the numbers from 10000 to 10009 have digit sums 1 through 10, and perfect squares in that range are 1,4,9. So starting at 9999 (which is 36), but 36 is also a perfect square, so in the ten numbers from 9999 to 10008, we have 9999 (36), 10000 (1), 10003 (4), 10008 (9). Four numbers. So that's four. Can we find a sequence where five numbers have digit sums as perfect squares?Alternatively, maybe using smaller numbers. Let's think of numbers in the teens or twenties, where digit sums are smaller. For example, numbers from 10 to 19:10:1 (1²)11:212:313:4 (2²)14:515:616:717:818:9 (3²)19:10So in this range, 10 (1), 13 (4), 18 (9). Three numbers. That's three out of ten.Alternatively, numbers around 100:100:1 (1²)101:2102:3103:4 (2²)104:5105:6106:7107:8108:9 (3²)109:10Again, three numbers: 100, 103, 108.Wait, another thought. If a number and some of its neighbors can have digit sums that are the same perfect square. For example, if multiple numbers have digit sum 9. But how?For example, consider numbers like 18, 27, 36, 45, 54, 63, 72, 81, 90, 108, etc., which all have digit sum 9. But these are spread out. However, in consecutive numbers, it's unlikely to have multiple numbers with the same digit sum unless there's some carry-over. For instance, 89:8+9=17, 90:9+0=9. So 89 to 90: digit sum drops from 17 to 9. So if 90 is in the sequence, that's a perfect square. Similarly, 99:18, 100:1. So 100 is a perfect square. But again, these are single instances.Alternatively, maybe a sequence where two numbers have digit sums equal to the same perfect square. For example, numbers like 100 (1), and 1000 (1). But these are not consecutive. Consecutive numbers with the same digit sum would require that when you increment, the digit sum remains the same. For example, 1000 (1), 1001 (2), so no. To have the same digit sum, you need to have a carry-over that reduces the digit sum by 9 in one place and adds 1 in another, keeping the total the same. For example, 199 -> 200: digit sum 19 -> 2, which is not the same. Another example: 2099 -> 2100: digit sum 20 -> 3. Not same.Wait, 9999 -> 10000: 36 ->1. Not same. So it's hard to have consecutive numbers with the same digit sum unless there's a specific arrangement. For example, 325 -> 3+2+5=10, 326 ->3+2+6=11, 327->12, etc. So consecutive numbers with same digit sum require that there's a 9 in the digits that turns into 0 and increments the next digit. But even then, the digit sum changes. For example, 399 -> 400: 3+9+9=21, 4+0+0=4. So different. Therefore, consecutive numbers usually have different digit sums. So unless there's a 9 in the middle digits, like 119 -> 120: 1+1+9=11, 1+2+0=3. No. So, in general, consecutive numbers have digit sums differing by +1 or -9k +1 for some k. Therefore, consecutive numbers can't have the same digit sum unless there's a carry-over that exactly compensates the loss from 9s. But this is rare. For example, 1999 -> 2000: digit sum 28 -> 2. Not the same. 2899 -> 2900: 2+8+9+9=28, 2+9+0+0=11. Still different.Therefore, consecutive numbers usually have different digit sums, so in a ten-number sequence, each number's digit sum is either +1 or some lower number due to carry-over. Therefore, the maximum number of perfect squares we can get is limited by how many perfect squares lie in the possible range of digit sums for those ten numbers.In the previous example, with 9999 to 10008, we had four perfect squares. Let's see if there's a way to get five.Let me try constructing such a sequence. Let's consider a number that when incremented multiple times, crosses multiple perfect squares. For example, starting at a number with digit sum 2, then next numbers go 3,4 (perfect square),5,6,7,8,9 (perfect square),10,11,12. So in this case, two perfect squares. Not better.Alternatively, starting at 15:1+5=6, 16:7, 17:8, 18:9, 19:10, 20:2, 21:3, 22:4, 23:5, 24:6. Here, 18 (9), 22 (4). Two perfect squares.Alternatively, starting at 7999999999. That's 7 followed by nine 9s. Digit sum:7 + 9*9=7+81=88. Next number:8000000000, digit sum8. Then 8000000001:9 (3²), 8000000002:10, 8000000003:11, ..., 8000000008:16 (4²), 8000000009:17, 8000000010:9. So here, from 7999999999 to 8000000008:7999999999:88 – no8000000000:8 – no8000000001:9 (3²)8000000002:108000000003:118000000004:128000000005:138000000006:148000000007:158000000008:16 (4²)So in this ten-number span, two perfect squares: 8000000001 and 8000000008. That's two. Not better.Alternatively, starting at a number with digit sum 24, next number is 25 (5²). Let's see. For example, 6999:6+9+9+9=33. Next number:7000:7. Not helpful.Wait, if I have a number with digit sum 24, then the next number would have digit sum 25 if there's no carry-over. For example, 996:9+9+6=24. Next number:997:25 (5²). So here, two numbers in a row:996 (24), 997 (25). But 24 is not a perfect square, but 25 is. So only one. But if we can have a number with digit sum 24, next 25, then perhaps others.Wait, another example. Let's take 7997:7+9+9+7=32. Next number:7998:7+9+9+8=33. Then 7999:34, 8000:8, 8001:9, 8002:10, 8003:11, 8004:12, 8005:13, 8006:14. So here, 8001 (9) and 8000 (8). Only one perfect square. Not helpful.Alternatively, let's think of a number where incrementing it multiple times causes the digit sum to hit multiple perfect squares. For example, starting at a digit sum that's 3 less than a perfect square, so that the next three numbers each increment the digit sum by 1, reaching the perfect square. For instance, if a number has digit sum 13, then next numbers have 14,15,16 (which is 4²). So if we can have three numbers:14,15,16, but only 16 is a square. Not helpful.Alternatively, if a number has digit sum 1 (which is a square), then the next numbers have 2,3,4 (another square). So in this case, numbers with digit sums 1 and 4. So two squares. For example, 100:1, 101:2, 102:3, 103:4. So here, 100 and 103. Two squares.Wait, going back to the previous example with 9999, which had four squares. Let me verify that again. Starting from 9999:1. 9999: 36 (6²) – yes2. 10000:1 (1²) – yes3. 10001:2 – no4. 10002:3 – no5. 10003:4 (2²) – yes6. 10004:5 – no7. 10005:6 – no8. 10006:7 – no9. 10007:8 – no10.10008:9 (3²) – yesSo indeed, four numbers: 9999,10000,10003,10008. So that's four. Is there a way to get five? Let me think. Maybe if we can have two separate carry-overs that each drop the digit sum into a perfect square, and then have some numbers in between also hit perfect squares.For example, take a number like 199999999. Its digit sum is 1+9*8=1+72=73. Next number:200000000:2. Then 200000001:3, 200000002:4 (2²), 200000003:5, 200000004:6, 200000005:7, 200000006:8, 200000007:9 (3²), 200000008:10, 200000009:11. So in this ten-number span, 200000000 (2), 200000002 (4), 200000007 (9). Three numbers. Not four.Another attempt: 8999999999 (9 followed by nine 9s). Digit sum:9 + 9*9=9+81=90. Next number:9000000000:9. Then 9000000001:10, 9000000002:11, ..., 9000000008:17, 9000000009:18. So here, 9000000000:9 (3²), 9000000001:10, etc. So only one. Not helpful.Wait, another idea: using numbers where digit sum drops to a perfect square and then increases again through increments. For example, a number like 999: digit sum 27. Next number:1000:1. Then 1001:2, 1002:3, 1003:4 (2²), 1004:5, 1005:6, 1006:7, 1007:8, 1008:9 (3²). So here, from 999 to 1008:999:27 – no1000:1 (1²) – yes1001:2 – no1002:3 – no1003:4 (2²) – yes1004:5 – no1005:6 – no1006:7 – no1007:8 – no1008:9 (3²) – yesSo three numbers:1000,1003,1008. Three out of ten.Still not better than four.Wait, maybe combining two transitions. For example, a number that ends with two separate groups of 9s. For example, 399999. Digit sum:3+9*5=3+45=48. Next number:400000:4. Then 400001:5, 400002:6, 400003:7, 400004:8, 400005:9 (3²), 400006:10, 400007:11, 400008:12, 400009:13. Then 400010:5. So here, 400000:4 (2²), 400005:9 (3²). Two numbers. Not better.Alternatively, let's take 6999999. Digit sum:6+9*6=6+54=60. Next number:7000000:7. Then 7000001:8, 7000002:9 (3²), 7000003:10, ..., 7000008:15, 7000009:16 (4²). So here, 7000002 (9), 7000009 (16). Two numbers. Hmm. So it seems challenging to get more than four. Let's check the example with 9999 again. Four numbers. Maybe that's the maximum. Wait, but let's try another similar example.Consider the number 89999. Digit sum:8+9*4=8+36=44. Next number:90000:9. Then 90001:10, 90002:11, 90003:12, 90004:13, 90005:14, 90006:15, 90007:16 (4²), 90008:17, 90009:18. So here, 90000 (9), 90007 (16). Two numbers. Not better.Another example: 999999999 (nine 9s). Digit sum:81. Next number:1000000000:1. Then 1000000001:2, ..., 1000000008:9, 1000000009:10. So in this range, 1000000000 (1), 1000000003 (4), 1000000008 (9). So three numbers.Wait, but the transition from 999999999 to 1000000000 gives us 1, and then we have 4 and 8 in the next numbers. But that's three. Still less than four.Wait, let's try another example where multiple carry-overs happen. For example, 27999999. Digit sum:2+7+9*6=2+7+54=63. Next number:28000000:2+8=10. Then 28000001:11, 28000002:12, 28000003:13, 28000004:14, 28000005:15, 28000006:16 (4²), 28000007:17, 28000008:18. So here, 28000006 (16). Just one. Not helpful.Another idea: start with a number that is a perfect square itself and see if the digit sums can be perfect squares. For example, 100:1 (which is 1²), 121:4 (2²), 144:9 (3²), etc. But these are individual numbers. Consecutive numbers would be 100,101,102,...,109. As before, only three.Alternatively, think of a number where digit sum is 25 (5²), and the next numbers have digit sums 26,27,28,29,30,31,32,33,34. But none of these are perfect squares except 25 and 36. So if a number has digit sum 25, the next numbers won't reach 36 within ten numbers unless there's a carry-over. But 25 + 9 =34, which is still not 36. So not helpful.Wait, another example: let's take the number 999. As before:999:27 – no1000:1 (1²)1001:21002:31003:4 (2²)1004:51005:61006:71007:81008:9 (3²)So here, 1000,1003,1008. Three numbers. Still less than four.Wait, maybe if we can find a starting number where the digit sum is a perfect square, and then through carry-over, another digit sum becomes a perfect square, and then the next numbers also hit some. For example, let's take 9999999999 (10 9s). Digit sum:90. Not a square. Next number:10000000000 (1). Then 10000000001 (2), ..., up to 10000000009 (10). So here, 10000000000 (1), 10000000003 (4), 10000000008 (9). Three numbers.Another example: 4444444444. Digit sum:4*10=40. Next number:4444444445 (40+1=41). Not helpful.Alternatively, start with a number like 198. Digit sum:1+9+8=18. Next number:199:19. Then 200:2, 201:3, 202:4 (2²), 203:5, 204:6, 205:7, 206:8, 207:9 (3²). So from 198 to 207:198:18 – no199:19 – no200:2 – no201:3 – no202:4 (2²) – yes203:5 – no204:6 – no205:7 – no206:8 – no207:9 (3²) – yesSo two numbers:202 and 207.Hmm. It seems challenging to get beyond four. The earlier example with 9999 to 10008 had four numbers. Let's see if we can find a similar case with more.Wait, let's check another similar example. Let's take 999999 (6 digits). Digit sum:54. Next number:1000000 (1). Then:1000000:1 (1²)1000001:21000002:31000003:4 (2²)1000004:51000005:61000006:71000007:81000008:9 (3²)1000009:10So here, 1000000 (1), 1000003 (4), 1000008 (9). Three numbers. Still three.Wait, how about a number like 999999999999 (12 9s). Digit sum:108. Next number:1000000000000 (1). Then:1000000000000:1 (1²)1000000000001:2...1000000000003:4 (2²)...1000000000008:9 (3²)...So, again, three numbers. Another thought: Maybe use a number where the digit sum drops to a perfect square and then there are multiple numbers with digit sums that are perfect squares. For example, if the digit sum drops to 1, then the numbers can go up to 10, but only 1,4,9 are squares. But in ten numbers, you can only hit these three. However, in the previous example with 9999, we had four because the starting number (9999) had digit sum 36 (6²), which is also a perfect square. So in that case, 9999 (36), then 10000 (1), 10003 (4), 10008 (9). That's four.So to get four, you need the transition number (the one that causes the carry-over) to have a digit sum that's a perfect square, and then the following numbers hit the lower perfect squares as they increment.Therefore, to maximize the count, we need a starting number n such that:1. n has a digit sum that's a perfect square.2. n+1, due to a large carry-over, has a digit sum that's a perfect square (like 1).3. n+4 has digit sum 4.4. n+9 has digit sum 9.But wait, in the 9999 example:n = 9999 (36, 6²)n+1 = 10000 (1, 1²)n+4 = 10003 (4, 2²)n+9 = 10008 (9, 3²)So four numbers. Therefore, to get four, you need four numbers where their digit sums are perfect squares at positions that can be hit by the initial drop and then increments.But can we have another perfect square in between? For example, if we have a digit sum that drops to 1, then increments to 4, 9, and also maybe 16 within the next ten numbers. But 1 to 16 requires a difference of 15, which would require 15 increments. But since we're only considering ten numbers, the digit sum can go from 1 up to 10 (if there are no more carry-overs). So 16 is too high. So the maximum digit sum in the next ten numbers would be 1 +9=10. Therefore, the perfect squares we can hit are 1,4,9. So three, but if the starting number also had a perfect square digit sum, that's four.Therefore, the maximum seems to be four. But let's check if there's another scenario where more than four can be achieved.Wait, what if the starting number n has a digit sum that's a perfect square, and the next number n+1 also has a digit sum that's a perfect square without a large carry-over? For example, if n has digit sum 15, n+1 has digit sum 16. But 15 isn't a square, so n itself wouldn't count. However, if n has digit sum 3, n+1 has digit sum 4. Then n has digit sum 3 (not a square), but n+1 is 4. Still only one.Alternatively, if n-1 has digit sum 15, n has 16. But we're only counting from n to n+9.Alternatively, what if we have a number where multiple carry-overs happen in the ten-number span? For example, numbers like 199, 200, 201,...,208. Here, 199 has digit sum 19, 200:2, 201:3, 202:4, ..., 208:10. So here, 202:4, 207:9. Two numbers. Not helpful.Alternatively, a number like 2999:2+9+9+9=29. Next number:3000:3, 3001:4, 3002:5, ..., 3007:10, 3008:11, 3009:12. Here, 3001:4 (2²). Only one.Wait, how about a number like 38999:3+8+9+9+9=38. Next number:39000:3+9=12. Then 39001:13, 39002:14, 39003:15, 39004:16 (4²), 39005:17, 39006:18, 39007:19, 39008:20, 39009:21. So here, 39004:16. One number.Another example: 6999:6+9+9+9=33. Next number:7000:7. Then 7001:8, 7002:9 (3²), 7003:10, ..., 7008:15, 7009:16 (4²). So here, 7002 (9), 7009 (16). Two numbers.Hmm. It's starting to look like four is the maximum possible. The example with 9999 to 10008 gives four, and I can't find any sequence with more than four. Let's verify that example again to make sure.Numbers:1. 9999: 9+9+9+9 = 36 (6²) – yes2. 10000:1+0+0+0+0 =1 (1²) – yes3. 10001:2 – no4. 10002:3 – no5. 10003:4 (2²) – yes6. 10004:5 – no7. 10005:6 – no8. 10006:7 – no9. 10007:8 – no10.10008:9 (3²) – yesYes, four numbers: 1,2,5,10th positions (9999,10000,10003,10008). Each of these has digit sums 36,1,4,9, which are all perfect squares. So four in total.Is there a way to include another perfect square in this sequence? Let's see. The digit sums go from 36 (9999) to 1 (10000), then 2,3,4 (10003),5,6,7,8,9 (10008). So after 10008, the next numbers would be 10009:10, 10010:2, etc. So within the ten-number span, we can't get another perfect square. So four is indeed the maximum here.But wait, another idea: what if the starting number is such that when it rolls over, the digit sum drops to a perfect square, and there's another number later in the sequence that also rolls over a smaller set of 9s, causing another drop to a perfect square. For example, starting with a number like 399999. Digit sum:3+9*5=48. Next number:400000:4. Then 400001:5, 400002:6, 400003:7, 400004:8, 400005:9 (3²), 400006:10, 400007:11, 400008:12, 400009:13. Then next number:400010:5. So here, 400000:4 (2²), 400005:9 (3²). That's two. Not helpful.Alternatively, start with 699999999. Digit sum:6+9*8=6+72=78. Next number:700000000:7. Then 700000001:8, 700000002:9 (3²), 700000003:10, ..., 700000008:15, 700000009:16 (4²). So here, two numbers:700000002 and 700000009. Two.Hmm. It seems very challenging to engineer a sequence with five perfect squares. Therefore, the example with four seems to be the maximum.Wait, but let's think differently. Maybe using numbers where the digit sum can reach multiple perfect squares through incremental increases and decreases. For example, if we have a number like 1000000000000000 (1 followed by zeros). Its digit sum is 1. The next numbers would have digit sums 2,3,4,...,10. So in this case, 1 (1²), 4 (2²), and 9 (3²) are all within the first ten numbers. So starting from 1000000000000000, we have:1. 1000000000000000:1 – yes2. 1000000000000001:2 – no3. 1000000000000002:3 – no4. 1000000000000003:4 – yes5. 1000000000000004:5 – no6. 1000000000000005:6 – no7. 1000000000000006:7 – no8. 1000000000000007:8 – no9. 1000000000000008:9 – yes10.1000000000000009:10 – noSo here, three numbers:1,4,9. Three. However, if the starting number itself is 999...999 with digit sum 36 (6²), then the sequence would be:1. 9999...999:36 (6²) – yes2. 1000...000:1 (1²) – yes3. 1000...001:2 – no4. 1000...002:3 – no5. 1000...003:4 (2²) – yes6. 1000...004:5 – no7. 1000...005:6 – no8. 1000...006:7 – no9. 1000...007:8 – no10.1000...008:9 (3²) – yesSo four numbers. This seems to be the pattern.Therefore, it seems that the maximum number of perfect squares achievable in ten consecutive natural numbers is four. This happens when the sequence includes a number with a digit sum of 36 (like 9999), which is a perfect square, followed by numbers that reset the digit sum to 1 (another perfect square), and then increment to 4 and 9 (also perfect squares). Therefore, the maximum is four.But just to be thorough, let's check if there's any other sequence where five numbers can have digit sums as perfect squares.Suppose we have a number n with digit sum 25 (5²). Then, n+1 has digit sum 26. If n+1 has a digit sum that is not a square, but n+ something else. For example, if n=799, digit sum=25. Then:799:25 (5²)800:8 – no801:9 (3²)802:10 – no803:11 – no804:12 – no805:13 – no806:14 – no807:15 – no808:16 (4²)So here, 799,801,808. Three numbers. Three.Alternatively, if we have a number like 168999. Digit sum:1+6+8+9+9+9=42. Next number:169000:1+6+9+0+0+0=16 (4²). Then 169001:17, 169002:18, 169003:19, 169004:20, 169005:21, 169006:22, 169007:23, 169008:24, 169009:25 (5²). So here, 169000:16 (4²), 169009:25 (5²). Two numbers. Not better.Another example: 9999999999999999 (16 9s). Digit sum:144 (12²). Next number:10000000000000000:1 (1²). Then the next numbers go up to 10. So in the sequence:1. 9999999999999999:144 (12²) – yes2. 10000000000000000:1 (1²) – yes3. 10000000000000001:2 – no4. 10000000000000002:3 – no5. 10000000000000003:4 (2²) – yes6. 10000000000000004:5 – no7. 10000000000000005:6 – no8. 10000000000000006:7 – no9. 10000000000000007:8 – no10.10000000000000008:9 (3²) – yesSo four numbers again. Same pattern.Therefore, based on all these examples and considerations, it appears that the maximum number of digit sums that can be perfect squares in ten consecutive natural numbers is four.Final AnswerThe maximum number is boxed{4}.</think>To determine the maximum number of ten consecutive natural numbers that can have a digit sum equal to a perfect square, we need to analyze how digit sums behave when numbers are incremented. Digit sums can either increase by 1 (if there's no carry-over) or decrease significantly (if there is a carry-over involving trailing 9s).Key steps and considerations:1. Digit Sum Behavior: Incrementing a number without trailing 9s increases the digit sum by 1. Incrementing a number with trailing 9s causes a carry-over, significantly decreasing the digit sum.2. Perfect Squares: The relevant perfect squares for digit sums (considering typical digit sums) are 1, 4, 9, 16, 25, 36, etc.3. Example Analysis: By examining sequences of numbers, particularly those involving carry-overs, we found that starting with a number like 9999 (digit sum 36, a perfect square) and incrementing through the next ten numbers (10000 to 10008) can yield multiple perfect squares in their digit sums.Detailed example:- 9999: Digit sum 36 (6²)- 10000: Digit sum 1 (1²)- 10003: Digit sum 4 (2²)- 10008: Digit sum 9 (3²)This sequence results in four numbers (9999, 10000, 10003, 10008) with digit sums that are perfect squares. Other examples and analyses confirmed that four is the maximum number achievable in such sequences.Thus, the maximum number of ten consecutive natural numbers that can have a digit sum equal to a perfect square is (boxed{4}).,<think>Okay, so the problem is: Ten consecutive natural numbers are written on the board. What is the maximum number of them that can have a digit sum equal to a perfect square?Hmm, let me start by understanding the problem. We have ten consecutive numbers, like n, n+1, n+2, ..., n+9. We need to find the maximum possible number of these where the sum of their digits is a perfect square. So, for each number in this sequence, calculate the sum of its digits, check if that sum is a perfect square, and then figure out the maximum number of such numbers possible in any such sequence of ten consecutive numbers.First, I need to recall that a perfect square is a number that is the square of an integer. So, 0, 1, 4, 9, 16, 25, 36, etc. Since we are dealing with digit sums, the maximum possible digit sum for a number depends on the number of digits. For example, a two-digit number can have a maximum digit sum of 18 (99), a three-digit number can have 27 (999), and so on. So, depending on how large the numbers are, the possible digit sums can vary, but the perfect squares up to, say, 81 (9^2) might be relevant here.But the problem doesn't specify the range of the numbers, so the answer might be a general maximum regardless of where the sequence is. So, maybe there's a configuration where, in ten consecutive numbers, as many as possible have digit sums that are perfect squares. The question is, how to maximize that count.I need to think about how digit sums change when you increment a number. When you add 1 to a number, the digit sum either increases by 1 (if there's no carry-over) or decreases by a certain amount (if there is a carry-over). For example, 199 -> 200: digit sum goes from 1+9+9=19 to 2+0+0=2, so a decrease of 17. Similarly, 999 -> 1000: digit sum from 27 to 1, decrease of 26. So, the digit sum can decrease a lot when there are multiple 9s in the number.Therefore, consecutive numbers can have digit sums that are both increasing and decreasing. The key is to find a sequence where as many of these digit sums hit perfect squares. Since perfect squares are spaced apart quadratically, their frequency decreases as numbers get larger, but digit sums can vary both up and down.So, perhaps the optimal scenario is when there's a lot of numbers where digit sums are perfect squares, and the transitions between them occur via small increments or controlled carry-overs that adjust the digit sums to the next perfect square.Let me try to think of an example. Let's say we have a number ending with several 9s. When we increment it, the digit sum drops significantly. So, maybe if we have a number like 3999, the digit sum is 3+9+9+9=30. Then 4000 has 4+0+0+0=4, which is a perfect square. So, here, the digit sum drops from 30 to 4, which is 26 less. So, 4 is a perfect square, but 30 is not. If we could arrange for the number before the transition to have a digit sum that's a perfect square, and the next number after the transition to also be a perfect square, then we can have two consecutive numbers with perfect square digit sums. But how often can this happen?Alternatively, perhaps sequences where numbers don't have trailing 9s, so that their digit sums increment by 1 each time. For example, 100, 101, 102, ..., 109. Their digit sums are 1, 2, 3, ..., 10. So, in this case, among these, only 1 (1^2), 4 (2^2), and 9 (3^2) would be perfect squares. So, in ten numbers, we might have three with digit sums as perfect squares. But maybe there's a better sequence where more numbers have digit sums that are perfect squares.Alternatively, let's think about numbers where digit sums are clustered around perfect squares. For example, if a number has a digit sum of 15, which is not a perfect square, but adding 1 gives 16, which is a perfect square. So, if we have a number with digit sum 15, the next number's digit sum is 16. But to have that, the number must not end with 9, so that incrementing doesn't cause a carry-over. So, for example, 69 has digit sum 15. Then 70 has digit sum 7, which is not 16. Wait, that's a problem. Wait, no, 69: 6+9=15. Then 70: 7+0=7. So, the digit sum drops by 8. So, that's not helpful. So, perhaps if we avoid numbers ending with 9, so that the digit sum increases by 1 each time. For example, 1000: digit sum 1. 1001: 2, 1002: 3, ..., 1009: 10. Then 1010: 2, 1011: 3, etc. So, here, digit sums cycle through small numbers. But maybe in this case, we can get a few perfect squares.But maybe in other ranges, the digit sums can reach higher perfect squares. For example, numbers with digit sums 16, 25, 36, etc. So, if we can have consecutive numbers where their digit sums are moving from one perfect square to another via increments or controlled carry-overs.Alternatively, think of a number like 9999999, which when incremented becomes 10000000, with digit sum dropping from 63 to 1. But 63 is not a perfect square, and 1 is. So, maybe this is an example where only one number in the transition has a perfect square digit sum.Alternatively, consider numbers that are just below a perfect square digit sum, and then when you add 1, the digit sum becomes a perfect square. For example, digit sum 15, then 16. But as above, this requires that the number does not end with a 9. If the number ends with a 9, adding 1 would cause a carry-over, which would reduce the digit sum by 9 and add 1, so net decrease of 8. So, for example, 199: digit sum 19. Add 1: 200, digit sum 2. So, decrease by 17. So, that's a big jump.Alternatively, if a number ends with a single 9: like 19: digit sum 10. Add 1: 20, digit sum 2. So, decrease by 8.If a number ends with two 9s: 299: digit sum 20. Add 1: 300, digit sum 3. Decrease by 17.So, in order to have a digit sum that's a perfect square followed by another digit sum that's a perfect square after incrementing, we need a way to transition from one perfect square to another. Since the digit sum can either increase by 1 (no carry-over) or decrease by some amount (carry-over). So, if we can have two consecutive numbers where the first has digit sum k^2 - 1, and the next has digit sum k^2, that would be good. But if there's a carry-over, then the digit sum would drop, so maybe from a higher perfect square to a lower one.Alternatively, maybe in some cases, a carry-over can cause the digit sum to decrease to another perfect square. For example, going from 9999 (digit sum 36, which is 6^2) to 10000 (digit sum 1). But 36 is a perfect square, and 1 is also a perfect square. So, here, two consecutive numbers have digit sums that are perfect squares. But only two. Wait, 9999 is 36, then 10000 is 1, so those two. So, if the sequence includes 9999 and 10000, then those two numbers would have perfect square digit sums. So, in a ten-number sequence from 9991 to 10000, how many have digit sums that are perfect squares?Wait, let's check:Take numbers from 9991 to 10000.Calculate digit sums:9991: 9+9+9+1 = 289992: 9+9+9+2 = 299993: 9+9+9+3 = 309994: 9+9+9+4 = 319995: 31 +1=32Wait, no: 9995: 9+9+9+5=329996: 9+9+9+6=339997: 9+9+9+7=349998: 9+9+9+8=359999: 9+9+9+9=36 (perfect square)10000: 1+0+0+0+0=1 (perfect square)So, in this sequence, only 9999 and 10000 have digit sums that are perfect squares. So, two numbers. Not great.But maybe another range. Let's think of numbers around transitions where digit sums go from, say, 15 to 16 (which are 15 and 16, with 16 being a perfect square). If a number has digit sum 15, and the next number, when incremented, doesn't have a carry-over, then the digit sum becomes 16. So, for example, let's take the number 69: digit sum 6+9=15. Next number is 70: 7+0=7. So, that's a decrease, so that's not helpful. Wait, but 69 ends with a 9, so adding 1 causes a carry-over. So, to avoid that, we need a number that doesn't end with 9, right?Wait, if a number ends with, say, 8, then adding 1 would make it 9, so digit sum increases by 1. For example, 88: 8+8=16 (perfect square). Then 89: 8+9=17, not a perfect square. Then 90: 9+0=9 (perfect square). So, here, 88 has digit sum 16, 90 has digit sum 9. So, two numbers in a sequence of three: 88,89,90. So, in three numbers, two have perfect square digit sums. Then, if you have a longer sequence, maybe you can chain these.Wait, let's check:Suppose we have numbers from 88 to 97:88: 16 (perfect square)89: 1790: 9 (perfect square)91: 1092: 1193: 1294: 1395: 1496: 1597: 16 (perfect square)So, in this case, the numbers 88,90,97 have digit sums 16,9,16, which are perfect squares. So, three numbers in a ten-number sequence. Hmm, that's better than the previous example. Wait, but 88 to 97 is ten numbers: 88,89,...,97. Wait, 97 is the tenth number. So, three perfect squares here. But maybe even better.Wait, let's see another example. Let's take numbers around 100, where digit sums are lower.100:1101:2102:3103:4 (perfect square)104:5105:6106:7107:8108:9 (perfect square)109:10So, here, 103 (4), 108 (9) have perfect squares. So, two in ten numbers.Another example, numbers from 10 to 19:10:1 (perfect square)11:212:313:4 (perfect square)14:515:616:717:818:9 (perfect square)19:10So, in this sequence, 10,13,18 have digit sums 1,4,9. So, three numbers. Not bad.Wait, so in the 10-19 range, three numbers. In the 88-97 range, three numbers. But in the 9999-10000 transition, only two. So, maybe three is a common number. But the question is asking for the maximum. So, can we find a ten-number sequence where more than three numbers have digit sums as perfect squares?Wait, perhaps in a higher range where digit sums can reach multiple perfect squares. Let's consider numbers with digit sums like 16, 25, 36, etc. Let me see.Suppose we have a number with digit sum 15. If we can increment it to 16 without a carry-over, that would be good. For example, 69: digit sum 15. But adding 1 gives 70: digit sum 7. That's a problem. So, the number must not end with 9. Let's take a number like 78: 7+8=15. Adding 1: 79: 7+9=16. So, here, 78 has digit sum 15, 79 has digit sum 16. So, two consecutive numbers with digit sums 15 and 16. But 15 is not a perfect square, 16 is. So, only one here. But if we can find a number where the digit sum is 15 (not square) and the next is 16 (square). Then another number in the sequence with digit sum 25, etc.Alternatively, let's think of numbers where multiple transitions happen. For example, a number ending with two 9s: 399. Digit sum: 3+9+9=21. Next number: 400: 4+0+0=4 (perfect square). Then 401: 4+0+1=5, and so on. So, here, only 400 has a perfect square digit sum. But 399's digit sum is 21, which is not. So, only one in that case.Alternatively, let's take a number like 7999: digit sum 7+9+9+9=34. Next number: 8000:8+0+0+0=8. Then 8001:8+0+0+1=9 (perfect square). So, 8001 has digit sum 9. Then 8002:8+0+0+2=10. So, here, 8000:8, 8001:9. So, two numbers: 8000 (not square), 8001 (square). Hmm, not helpful.Wait, maybe consider a sequence where there are multiple numbers with digit sums that are perfect squares, spaced in such a way that their digit sums cycle through squares. For example, numbers with digit sums 1, 4, 9, 16, 25, 36, etc. But how to get multiple of these in ten numbers.Alternatively, think about numbers that have digit sums that are perfect squares, and where their consecutive numbers don't interfere via carry-overs.Wait, here's another idea. Let's consider the numbers from 1000 to 1009:1000:11001:21002:31003:4 (perfect square)1004:51005:61006:71007:81008:9 (perfect square)1009:10So, here, 1003 and 1008 have digit sums 4 and 9. So, two numbers. Not better than previous examples.Wait, maybe go back to the earlier example where 88-97 had three numbers. Let me check again:88: 1689:1790:991:1092:1193:1294:1395:1496:1597:16So, 88 (16), 90 (9), 97 (16). So, three numbers. Hmm. So, 16,9,16. So, 16 is 4^2, 9 is 3^2, 16 again. So, three perfect squares. So, three numbers. Maybe that's the maximum?Wait, let me think of another example. Suppose we have a sequence where multiple numbers have digit sums that are consecutive perfect squares.For example, numbers with digit sums 16, 25, etc. Let's try to find numbers with digit sums 16, 25, 36, 49, etc., but in a consecutive sequence.Wait, suppose we have a number with digit sum 15. Then adding 1 gives 16 if there's no carry-over. So, let's take a number like 79: 7+9=16, wait, that's already 16. Let's take 78:15. Then 79:16. So, two consecutive numbers where digit sums are 15 and 16. But only 16 is a perfect square.Alternatively, if we have a number with digit sum 24, then adding 1 could make it 25 if there's no carry-over. For example, 699:6+9+9=24. Next number:700:7+0+0=7. Not helpful. So, carry-over causes a big drop. So, not helpful.Alternatively, a number like 899:8+9+9=26. Next number:900:9+0+0=9. So, 26 to 9. 9 is a perfect square. So, here, 900 has digit sum 9. So, if we can have 899 and 900: 899 is 26 (not square), 900 is 9 (square). So, only one.Wait, how about numbers that are near multiple perfect squares? For example, numbers with digit sums 1,4,9,16,25,36, etc., within the same sequence. Let's see if we can have a ten-number sequence that includes multiple of these.Wait, here's an idea: Let's take a number like 1000000000 (1 followed by nine zeros). Its digit sum is 1. Then, adding 1 each time:1000000000:1 (perfect square)1000000001:21000000002:31000000003:4 (perfect square)1000000004:51000000005:61000000006:71000000007:81000000008:9 (perfect square)1000000009:10So, here, the numbers 1000000000, 1000000003, 1000000008 have digit sums 1,4,9. So, three numbers in the sequence. Same as previous examples.Alternatively, maybe in a different range. Let's take numbers from 1 to 10:1:1 (perfect square)2:23:34:4 (perfect square)5:56:67:78:89:9 (perfect square)10:1 (perfect square)Wait, 10 is part of the next ten. So, numbers 1-9: digits sums 1-9. So, 1,4,9. Then 10:1. So, in 1-10, we have four numbers:1,4,9,10. Wait, 10's digit sum is 1. So, 1 occurs twice. So, in numbers 1-10, the digit sums that are perfect squares are 1 (1), 4 (4), 9 (9), and 1 (10). So, four numbers? Wait, 1,4,9,10. But 10 is part of the next set. Wait, but if we take numbers 1 through 10, inclusive, then yes, 1,4,9,10. Wait, 10 is the tenth number. So, in 1-10, four numbers have digit sums equal to perfect squares. That's better.Wait, let's check:1:1 (1^2)2:23:34:4 (2^2)5:56:67:78:89:9 (3^2)10:1 (1^2)So, numbers 1,4,9,10. Each of these have digit sums of 1,4,9,1. So, four numbers. So, in the first ten natural numbers (assuming natural numbers start at 1), there are four numbers with digit sums that are perfect squares.That's interesting. So, here, 1,4,9,10. So, four numbers. Wait, but 10 is the tenth number. So, if we take numbers n=1 to n=10, we have four. So, maybe this is an example where the maximum is four.But is four the maximum possible?Wait, let's check numbers 10-19:10:111:212:313:414:515:616:717:818:919:10So, digit sums: 1,2,3,4,5,6,7,8,9,10. So, perfect squares are 1 (10),4 (13),9 (18). So, three numbers. So, similar to the first example.But in the numbers 1-10, we had four. So, maybe sequences starting near 1 can have more?Wait, let's check numbers 0-9 if we consider 0 as a natural number (though usually it's not). If we include 0:0:0 (0 is a perfect square)1:12:23:34:45:56:67:78:89:9So, all numbers from 0-9 have digit sums equal to themselves. So, 0,1,4,9 are perfect squares. So, numbers 0,1,4,9. Four numbers. But since 0 is often excluded from natural numbers, this might not count.But if we do consider 0 as a natural number, then in 0-9, we have four numbers. But again, depends on definition.Assuming natural numbers start at 1, then 1-10 have four numbers. So, that's a candidate.But is there a sequence of ten consecutive numbers where more than four have digit sums as perfect squares?Let me think. Suppose we have a sequence where several numbers have digit sums 1,4,9,16,25, etc., in the same ten-number span.Wait, let's consider numbers in the range where digit sums can reach multiple perfect squares. Let's pick numbers where digit sums can be 1,4,9,16,25, etc. For example, suppose we have a number with digit sum 16, and another with 25, and another with 9, all within ten consecutive numbers.But how?Alternatively, let's think about numbers where digit sums can be 1, 4, 9, 16, 25, 36, etc., and see if multiple of these can occur close together.For example, 10 has digit sum 1, 13 has 4, 18 has 9. So, within 10-19, we have 10,13,18. Then, if we can get another number in this range with a digit sum of 16, but 16 is 7+9, so 79, but 79 is outside 10-19.Wait, maybe another example. Let's take numbers around 100. Suppose we have numbers 100-109:100:1101:2102:3103:4104:5105:6106:7107:8108:9109:10So, here, 100 (1), 103 (4), 108 (9). Three numbers. If we can find a way to get another one, but how? For example, 106:7, not a square. 109:10. So, maybe not.Alternatively, let's go higher. Let's take numbers in the 1000s. 1000-1009:1000:11001:21002:31003:41004:51005:61006:71007:81008:91009:10Same pattern as before. Three numbers.Wait, but in the numbers 1-10, we had four. So, maybe starting at 1 is better.But let's check another example. Let's take numbers 9999999999 (ten 9s). Digit sum is 9*10=90. 90 is not a perfect square. Then, adding 1: 10000000000, digit sum 1. So, from 90 to 1. Not helpful.Alternatively, let's think of numbers that roll over multiple digits. For example, 1999, 2000. Digit sums 1+9+9+9=28 to 2+0+0+0=2. Not helpful.Alternatively, 999:27 to 1000:1. 27 is not a square.Wait, 9999:36 (6^2) to 10000:1 (1^2). So, here, two consecutive numbers with digit sums 36 and 1. Both perfect squares. So, if we have a ten-number sequence that includes 9999 and 10000, then those two. But the rest of the numbers from 9990 to 9999:9990:9+9+9+0=279991:289992:299993:309994:319995:329996:339997:349998:359999:36So, in 9990-9999, only 9999 has a digit sum of 36 (perfect square). Then 10000:1. So, in the sequence 9991-10000, only 9999 and 10000 have digit sums as perfect squares. So, two numbers. Not better than previous examples.Hmm. Let's think differently. Suppose we have a sequence where several numbers have digit sums that are perfect squares because of their digits. For example, numbers like 100, which has digit sum 1; 121, which has digit sum 4; 144, digit sum 9; etc. But these are spread out, not consecutive.Alternatively, maybe in some magic sequence, multiple numbers have digit sums that hit perfect squares. For example, numbers like:- 10:1- 13:4- 18:9- 31:4 (3+1=4)- 40:4- 45:9- 54:9- 63:9- 90:9- 97:16Wait, but these are not consecutive numbers.Wait, maybe if we can find a ten-number consecutive sequence where multiple numbers have digit sums that are perfect squares. Let's try to construct such a sequence.Let me take the example from 1 to 10. As previously noted, numbers 1,4,9,10 have digit sums 1,4,9,1. So, four numbers. If we can find another ten-number sequence with four or more, that would be better.Wait, let's check numbers 10-19:10:111:212:313:414:515:616:717:818:919:10So, here, 10,13,18 have digit sums 1,4,9. So, three numbers.How about numbers 0-9 (if allowed):0:01:12:23:34:45:56:67:78:89:9So, 0,1,4,9. Four numbers. But again, 0 is often excluded.Wait, in some definitions, natural numbers start at 0. If that's the case, then 0-9 have four perfect squares. But the problem says "natural numbers," which can be ambiguous. However, in mathematics, natural numbers usually start at 1, but sometimes include 0. If the problem assumes starting at 1, then 1-10 is four. If starting at 0, then 0-9 is four. So, either way, four.Is there a way to get five?Wait, let's try to think of numbers where multiple numbers in the same ten-number span have digit sums that are perfect squares. For example, starting at 1:1,4,9,10. Maybe starting at another number where there's overlap.Wait, let's consider numbers from 8 to 17:8:89:910:111:212:313:414:515:616:717:8So, in this case, 9 (9),10 (1),13 (4). So, three numbers. Not better.How about numbers from 3 to 12:3:34:45:56:67:78:89:910:111:212:3Here, 4 (4), 9 (9),10 (1). So, three.Alternatively, numbers from 4 to 13:4:45:56:67:78:89:910:111:212:313:4So, here, 4,9,10,13. So, four numbers:4,9,10,13. Each with digit sums 4,9,1,4. So, four numbers. So, this is similar to the 1-10 example. So, numbers 4-13: four perfect squares. Wait, let's check:4:45:56:67:78:89:910:111:212:313:4Yes, digit sums:4,5,6,7,8,9,1,2,3,4. So, perfect squares at positions 4 (4),9 (9),10 (1), and 13 (4). Wait, but 13 is the tenth number in the sequence starting at 4: 4,5,6,7,8,9,10,11,12,13. So, numbers 4,9,10,13. Each with digit sums 4,9,1,4. So, four numbers. So, that's another example with four.Therefore, sequences starting at 1,4, etc., can have four numbers with digit sums as perfect squares. So, four is achievable. But is five possible?Let me try to find a ten-number sequence with five numbers having digit sums as perfect squares.Suppose we have numbers where digit sums are 1,4,9,16,25. But getting five of these in ten numbers.Wait, let's consider numbers where digit sums cycle through multiple perfect squares. For example, numbers that have digit sums 1, 4, 9, 16, 25, etc., but within ten consecutive numbers.Wait, but 1,4,9 are small squares. 16,25 are larger. So, perhaps in numbers where digit sums can reach 16 or 25 within the same ten-number span.Let's take an example. Let's consider numbers in the 100s where digit sums can be higher.Take numbers 199-208:199:1+9+9=19200:2+0+0=2201:2+0+1=3202:2+0+2=4 (perfect square)203:2+0+3=5204:2+0+4=6205:2+0+5=7206:2+0+6=8207:2+0+7=9 (perfect square)208:2+0+8=10So, in this sequence, 202 (4), 207 (9). Two numbers. Not helpful.Another example: 399-408:399:3+9+9=21400:4+0+0=4 (perfect square)401:4+0+1=5402:4+0+2=6403:4+0+3=7404:4+0+4=8405:4+0+5=9 (perfect square)406:4+0+6=10407:4+0+7=11408:4+0+8=12So, here, 400 (4), 405 (9). Two numbers. Still not better.Wait, what about numbers where transitioning from a high digit sum to a low one via multiple carry-overs?For example, 6999999999: digit sum 6+9*9=6+81=87. Next number:7000000000:7+0*9=7. 87 is not a square, 7 isn't. Then 7000000001:7+0+1=8. Not helpful.Alternatively, 8999999999: digit sum 8+9*9=8+81=89. Next number:9000000000:9. So, 89 is not a square, 9 is. Then 9000000001:10, etc. Only one square.Alternatively, let's think of numbers where the digit sum is a square, and due to the way numbers are structured, several numbers in a row can have square digit sums.Wait, for example, let's take numbers 1000000000, which has digit sum 1, and then 1000000001 (2), ..., up to 1000000009 (10). As before, three squares.But if we can find a number where incrementing doesn't change the digit sum much, but the digit sum is a square. For example, 1001: digit sum 2. Not a square. 1002:3. 1003:4. Hmm. So, 1003 is 4.Alternatively, numbers like 1111111111 (digit sum 10), but that's not a square. 1111111113: digit sum 12. Not a square.Alternatively, numbers with digit sums 16. For example, 79:7+9=16. 80:8+0=8. 81:8+1=9. So, 79,80,81: digit sums 16,8,9. So, two squares here:16 and9.So, in three numbers, two squares. If we can chain these.Wait, let's take numbers 79-88:79:1680:881:982:1083:1184:1285:1386:1487:1588:16So, in this ten-number span, 79 (16), 81 (9), 88 (16). So, three numbers. Same as previous examples.Another example: 16-25:16:717:818:919:1020:221:322:423:524:625:7Here, 18 (9), 22 (4). So, two numbers.Wait, what about numbers from 22-31:22:423:524:625:726:827:928:1029:1130:331:4So, here, 22 (4),27 (9),31 (4). So, three numbers. Again, three.Hmm. It seems challenging to get beyond four. Let's go back to the earlier example where starting from 1, there were four numbers. Let me check that again:Numbers 1-10:1:12:23:34:45:56:67:78:89:910:1So, perfect squares at 1,4,9,10. Wait, 10's digit sum is 1, which is a square. So, four numbers.Similarly, numbers 4-13:4:45:56:67:78:89:910:111:212:313:4So, perfect squares at 4,9,10,13. So, four numbers.So, in both these sequences, four numbers. So, maybe four is achievable. But is there a sequence with five?Wait, let's think about numbers where multiple numbers have digit sum 1, which is a perfect square.For example, numbers like 10, 100, 1000, etc. But these are spread out.However, in the sequence 1-10, we have two numbers with digit sum 1:1 and 10.Similarly, in the sequence 4-13, we have two numbers with digit sum 4:4 and 13.So, maybe if we can have a sequence where multiple numbers have the same digit sum, which is a perfect square. For example, three numbers with digit sum 4 and two numbers with digit sum 9. But how?Wait, let's see:Suppose we have a ten-number sequence where two numbers have digit sum 1, two have digit sum 4, and one has digit sum 9. Total of five. Is that possible?For example, numbers like:1:110:14:413:49:9But these are spread out. To have them in a consecutive ten-number span.Wait, numbers 1-10:1,4,9,10. So, two ones (1 and10), one four, one nine. Total four.If we can find a ten-number sequence where two numbers have digit sum 1, two have digit sum 4, and one has digit sum 9, that would be five.Is that possible?Wait, let's consider numbers like 1000000000 to 1000000009. Their digit sums are 1,2,...,10. So, two ones (1000000000 and1000000009's digit sum is 10, not 1). Wait, no. 1000000000:1, 1000000001:2,...,1000000009:10. So, only one 1. So, no.Another example, numbers 9999999999 to 10000000008. Digit sums: 90,91,...,99,1,2,...,8. Only 10000000000 has digit sum 1. So, one.Not helpful.Alternatively, let's think of numbers where multiple numbers can have digit sum 1. For example, 10, 100, 1000, etc. But these are not consecutive.Alternatively, consider numbers like 1000000, 1000001, ..., 1000009. Their digit sums are 1,2,...,10. So, again, only one 1.Alternatively, think of numbers where two numbers have digit sum 1, but these would have to be 1 and 10 in the same ten-number sequence. So, 1-10 includes 1 and10. So, two ones. Similarly, 10-19 includes 10 (digit sum1) and19 (digit sum10). So, only one 1 in each subsequent ten-number span.So, if we have a sequence starting at 1, we get two ones. If we start at 10, only one. So, perhaps sequences starting at 1 can have two ones, plus other squares.In the sequence 1-10, we have four numbers:1,4,9,10. So, two ones (1 and10), one four (4), and one nine (9). Total four.If we can find another number in this sequence with a digit sum that is a perfect square, but how? The other numbers in 1-10 have digit sums from 1 to 10. Only 1,4,9 are perfect squares.So, in this case, maximum four.Another example: let's consider numbers like 1999, 2000, 2001, etc. 1999:1+9+9+9=28, 2000:2, 2001:3, 2002:4, 2003:5, 2004:6, 2005:7, 2006:8, 2007:9, 2008:10, 2009:11. So, here, 2002:4, 2007:9. So, two numbers.Not better.Wait, let's think of a number where digit sum is 25, and then the next number has digit sum 16. For example, 799:7+9+9=25. Next number 800:8+0+0=8. Not helpful. But 25 and8 are not consecutive squares.Alternatively, a number with digit sum 16, next number 25. But how?Wait, 88:16, next number 89:17, not 25. So, no.Alternatively, let's take numbers where digit sums can reach higher squares through carry-over. For example, a number with digit sum 36, then next number 1. But 36 and1 are squares. So, two numbers.But 36 is 6^2, and1 is1^2. So, consecutive numbers with digit sums 36 and1. Like 9999 and10000.But in the ten-number span from 9991 to10000, only 9999 and10000 have digit sums 36 and1. So, two numbers.But this is lower than the four we saw earlier.Hmm. So, based on all these examples, the maximum number of numbers with digit sums as perfect squares in a ten-consecutive-number sequence seems to be four. For example, in the numbers 1-10, four numbers:1,4,9,10.But let me verify this once more. Let's list the numbers from 1 to10:1:1 (1^2)2:23:34:4 (2^2)5:56:67:78:89:9 (3^2)10:1 (1^2)So, numbers 1,4,9,10. Four numbers. Each of these has a digit sum that's a perfect square.If we take another sequence, say, 4-13, as before:4:4 (2^2)5:56:67:78:89:9 (3^2)10:1 (1^2)11:212:313:4 (2^2)So, numbers 4,9,10,13. Four numbers. Same count.Another example, numbers 9-18:9:910:111:212:313:414:515:616:717:818:9So, here, 9,10,13,18. Four numbers. Same count.Therefore, it seems that in various sequences, four is achievable. So, four is a candidate for the maximum.But let's see if we can find a ten-number sequence with five.Wait, let's check numbers from 0-9, if 0 is allowed:0:0 (0^2)1:1 (1^2)2:23:34:4 (2^2)5:56:67:78:89:9 (3^2)So, numbers 0,1,4,9. Four numbers. If 0 is considered a natural number, then 0-9 have four, but not five.Alternatively, suppose we have a sequence like 9-18, which includes 9 (9),10 (1),13 (4),18 (9). Four numbers.Alternatively, numbers 100-109:100:1101:2102:3103:4104:5105:6106:7107:8108:9109:10So, 100 (1),103 (4),108 (9). Three numbers.Hmm.Wait, maybe we need to think of a different range where digit sums can reach more squares. For example, numbers where digit sums can be 16, 25, 36, etc. But in a ten-number span.Suppose we have a number with digit sum 16, and the next number has digit sum 25. But how?For example, 79:7+9=16. Next number 80:8+0=8. Not helpful.Alternatively, 799:7+9+9=25. Next number800:8+0+0=8. Not helpful.Alternatively, 9999:36. Next number10000:1. Both squares. So, two numbers.Alternatively, numbers like 4444:4+4+4+4=16. Next number4445:4+4+4+5=17. Then 4446:18, and so on. So, in this case, 4444 has digit sum16. Then numbers up to4453:4444:164445:174446:184447:194448:204449:214450:134451:144452:154453:16So, in this ten-number span, 4444 (16) and4453 (16). So, two numbers. Only two.Alternatively, numbers where digit sums can reach multiple squares via incrementing without carry-over. For example, numbers like 169:1+6+9=16, 170:1+7+0=8, 171:1+7+1=9. So, in this case, 169 (16), 171 (9). Two squares.Alternatively, numbers like 196:1+9+6=16, 197:17, 198:18, 199:19, 200:2, 201:3, 202:4, 203:5, 204:6, 205:7. So, in this ten-number span, 196 (16),202 (4). Two numbers.Not helpful.Hmm. After going through various examples and scenarios, it seems that the maximum number of numbers in a ten-consecutive-natural-number sequence with digit sums equal to perfect squares is four. This is achieved in sequences like 1-10, 4-13, 9-18, etc., where four numbers have digit sums of 1,4,9,16, etc.Therefore, the answer should be four. However, before finalizing, let me check if there's any possibility of five.Wait, let's take numbers from 1 to 10: four numbers.Is there a sequence where five numbers have digit sums as squares?Suppose we have a number like 10 (digit sum1), then 100 (digit sum1), but these are not consecutive. So, no.Alternatively, numbers where digit sums reset multiple times. For example, numbers transitioning from 99 to 100: digit sums go from 18 to1. Not helpful.Alternatively, numbers with multiple trailing 9s causing multiple resets. For example, 999999999 to 1000000000. But these would have digit sums 81 (not a square) to1 (square). Only one number.Alternatively, think of a sequence where multiple numbers have digit sums 1. For example, 1, 10, 100, 1000, etc., but these are not consecutive.Alternatively, numbers like 1000000000 to1000000009: digit sums1 to10. So, only one 1.Alternatively, maybe numbers where two different carry-overs cause digit sums to hit perfect squares multiple times.Wait, let's consider a number like 3999999999. Digit sum:3 +9*9=3+81=84. Next number:4000000000:4. Then 4000000001:5, ...,4000000009:13. So, in this ten-number span, 4000000000 has digit sum4 (square), and 4000000009 has digit sum13. Only one square.Not helpful.Another example: 6999999999. Digit sum:6+9*9=6+81=87. Next number:7000000000:7. Not a square. Then7000000001:8. Not helpful.Hmm. Alternatively, numbers where digit sums decrease to a square and then increase to another square. For example, 999:27,1000:1,1001:2,1002:3,1003:4. Here,1000 (1) and1003 (4). So, two numbers. Still not five.After exhausting various possibilities, I can't find a ten-number sequence with more than four numbers having digit sums that are perfect squares. Therefore, I conclude that the maximum number is four.Final Answerboxed{4}</think>To determine the maximum number of ten consecutive natural numbers that can have a digit sum equal to a perfect square, we analyzed various sequences and their digit sums. Key steps and examples considered:1. Digit Sum Changes: Incrementing a number can either increase the digit sum by 1 (no carry-over) or decrease it significantly (carry-over).2. Perfect Squares: Relevant perfect squares include 1, 4, 9, 16, 25, 36, etc.3. Example Sequences: - Numbers 1 to 10: Digit sums are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1. Perfect squares at 1 (1), 4 (4), 9 (9), 1 (10). Four numbers. - Numbers 4 to 13: Digit sums are 4, 5, 6, 7, 8, 9, 1, 2, 3, 4. Perfect squares at 4 (4), 9 (9), 1 (10), 4 (13). Four numbers. - Numbers 9 to 18: Digit sums are 9, 10, 11, 12, 13, 14, 15, 16, 17, 9. Perfect squares at 9 (9), 1 (10), 16 (16), 9 (18). Four numbers.After extensive analysis of various sequences and considering the transitions caused by carry-overs and increments, the maximum number of consecutive numbers with digit sums as perfect squares found was four. This is achieved in sequences like 1-10, 4-13, and 9-18, where four numbers have digit sums that are perfect squares (1, 4, 9, and another occurrence of 1 or 4).Thus, the maximum number of ten consecutive natural numbers that can have a digit sum equal to a perfect square is (boxed{4}).]

🔑:Answer: 4.Solution: Note that the sums of the digits of consecutive natural numbers within the same decade are consecutive natural numbers. Since there are 10 numbers, they span two decades. Also note that among ten consecutive natural numbers, there can be no more than 3 perfect squares, and three perfect squares can only occur if there are 9 numbers. Therefore, among the 10 consecutive natural numbers written on the board, there cannot be more than 4 numbers with a sum of digits equal to the square of a natural number. We can provide an example where there are exactly 4 such numbers: {1,2,3,4,5,6,7,8,9,10}.Grading criteria. Full solution - 20 points. Only the estimate - 12 points. Answer + example - 9 points. Only the answer - 3 points.

🧠:[<think>Okay, so I need to find the value of the real number a given that the complex number z, which is defined as z = (a)/(2 + i) + (2 + i)/5, has a sum of its real part and imaginary part equal to 1. The options are A: 0, B: 1, C: 2, D: 3. Hmm, let's start by recalling that any complex number can be written in the form x + yi, where x is the real part and y is the imaginary part. The problem states that x + y = 1. So, I need to first simplify z into its standard form and then set the sum of the real and imaginary parts to 1, then solve for a.First, let's simplify each term in z separately. The first term is a/(2 + i). To simplify this, I should multiply the numerator and denominator by the complex conjugate of the denominator to rationalize it. The complex conjugate of 2 + i is 2 - i. So:a/(2 + i) * (2 - i)/(2 - i) = [a(2 - i)] / [(2 + i)(2 - i)]Let me compute the denominator first: (2 + i)(2 - i) = 2*2 + 2*(-i) + i*2 + i*(-i) = 4 - 2i + 2i - i². Since i² = -1, this becomes 4 - (-1) = 5. So the denominator is 5.Therefore, the first term simplifies to [a(2 - i)] / 5 = (2a)/5 - (a i)/5.Okay, that's the first part. Now the second term in z is (2 + i)/5. That's straightforward; it can be written as 2/5 + i/5.Now, adding both terms together to get z:z = [(2a)/5 - (a i)/5] + [2/5 + i/5]Combine the real parts and the imaginary parts separately:Real part: (2a)/5 + 2/5Imaginary part: (-a/5 + 1/5)iSo, the real part is (2a + 2)/5 and the imaginary part is (-a + 1)/5.According to the problem, the sum of the real part and the imaginary part is 1. So:Real part + Imaginary part = 1Which translates to:(2a + 2)/5 + (-a + 1)/5 = 1Since both fractions have the same denominator, we can combine them:[(2a + 2) + (-a + 1)] / 5 = 1Simplify the numerator:2a + 2 - a + 1 = (2a - a) + (2 + 1) = a + 3So now we have:(a + 3)/5 = 1Multiply both sides by 5:a + 3 = 5Subtract 3 from both sides:a = 2Therefore, the value of a is 2, which is option C. Let me check my steps again to ensure I didn't make a mistake.First, simplifying a/(2 + i) by multiplying numerator and denominator by (2 - i) is correct. The denominator becomes 4 + 1 = 5, so that's right. Then, expanding the numerator gives 2a - ai, so dividing by 5 gives (2a/5) - (ai/5). Then the second term is (2 + i)/5, which splits into 2/5 + i/5. Adding the real parts: (2a/5 + 2/5). Adding the imaginary parts: (-a/5 + 1/5)i. Then, adding the real and imaginary parts together (as per the problem's requirement) is (2a + 2)/5 + (-a + 1)/5. Combining those gives (a + 3)/5 = 1, leading to a = 2. Seems correct. Let me plug a = 2 back into z and check if the sum of real and imaginary parts is indeed 1.First term: 2/(2 + i) = 2*(2 - i)/5 = (4 - 2i)/5. Second term: (2 + i)/5. Adding them: (4 - 2i)/5 + (2 + i)/5 = [4 + 2]/5 + [-2i + i]/5 = 6/5 - i/5. So real part is 6/5, imaginary part is -1/5. Their sum is 6/5 - 1/5 = 5/5 = 1. Yes, that checks out. So a = 2 is indeed the correct answer. Therefore, the answer is C: 2.Final Answerboxed{C}</think>Given the complex number ( z = frac{a}{2+i} + frac{2+i}{5} ), we need to find the value of the real number ( a ) such that the sum of its real part and imaginary part is 1.First, simplify each term in ( z ):1. Simplify ( frac{a}{2+i} ): [ frac{a}{2+i} cdot frac{2-i}{2-i} = frac{a(2-i)}{(2+i)(2-i)} = frac{a(2-i)}{5} = frac{2a}{5} - frac{ai}{5} ]2. Simplify ( frac{2+i}{5} ): [ frac{2+i}{5} = frac{2}{5} + frac{i}{5} ]Combine the terms to form ( z ):[z = left( frac{2a}{5} - frac{ai}{5} right) + left( frac{2}{5} + frac{i}{5} right)]Separate the real and imaginary parts:- Real part: ( frac{2a}{5} + frac{2}{5} )- Imaginary part: ( -frac{a}{5} + frac{1}{5} )According to the problem, the sum of the real part and the imaginary part is 1:[left( frac{2a + 2}{5} right) + left( frac{-a + 1}{5} right) = 1]Combine the fractions:[frac{(2a + 2) + (-a + 1)}{5} = 1]Simplify the numerator:[frac{a + 3}{5} = 1]Solve for ( a ):[a + 3 = 5 implies a = 2]Thus, the value of ( a ) is (boxed{C}).,<think>Okay, let me try to figure out this problem step by step. So, the problem is about a complex number z, which is given by the expression z = (a)/(2 + i) + (2 + i)/5. We are told that the sum of its real part and imaginary part is 1, and we need to find the value of the real number a from the options provided.First, I need to recall that a complex number can be written in the form z = x + yi, where x is the real part and y is the imaginary part. The problem says that the sum of the real part and the imaginary part is 1, so that means x + y = 1.Our goal is to find the value of a such that when we compute z using the given expression, the real part plus the imaginary part equals 1.To do this, I think I need to simplify the expression for z by combining the two terms into a single complex number, then separate the real and imaginary parts, and finally set up the equation x + y = 1 to solve for a.Let me start by simplifying each term separately. The first term is (a)/(2 + i). To simplify this, I remember that dividing by a complex number can be done by multiplying the numerator and denominator by the complex conjugate of the denominator. The complex conjugate of 2 + i is 2 - i. So, let's do that:First term:(a)/(2 + i) = [a * (2 - i)] / [(2 + i)(2 - i)]Let's compute the denominator first: (2 + i)(2 - i) = 2*2 + 2*(-i) + i*2 + i*(-i) = 4 - 2i + 2i - i². The terms -2i and +2i cancel out, and since i² = -1, we have 4 - (-1) = 4 + 1 = 5. So the denominator is 5.So the first term becomes [a * (2 - i)] / 5 = (2a - ai)/5.Now, the second term is (2 + i)/5. Let's write that as (2/5) + (i/5).So, combining both terms:z = (2a - ai)/5 + (2 + i)/5.Since they have the same denominator, we can combine them:z = [(2a - ai) + 2 + i]/5.Let's group the real parts and the imaginary parts:Real parts: 2a + 2Imaginary parts: (-a)i + i = (-a + 1)iTherefore, z = [ (2a + 2) + (-a + 1)i ] / 5.Wait, hold on, actually, when we combine them, the numerator is (2a - ai + 2 + i). Let me verify that step again.Original numerator after expanding both terms:First term: 2a - aiSecond term: 2 + iAdding them together: 2a - ai + 2 + iCombine like terms:Real parts: 2a + 2Imaginary parts: (-a + 1)iYes, that's correct. So z = (2a + 2)/5 + [(-a + 1)/5]i.Therefore, the real part of z is (2a + 2)/5 and the imaginary part is (-a + 1)/5.The problem states that the sum of the real part and the imaginary part is 1. So, we can write:(2a + 2)/5 + (-a + 1)/5 = 1Since both fractions have the same denominator, we can combine them:[ (2a + 2) + (-a + 1) ] / 5 = 1Simplify the numerator:2a + 2 - a + 1 = (2a - a) + (2 + 1) = a + 3So, (a + 3)/5 = 1Multiply both sides by 5:a + 3 = 5Subtract 3 from both sides:a = 2Hmm, so according to this, the answer is 2, which is option C. But let me check my steps again to make sure I didn't make any mistakes.First, simplifying (a)/(2 + i):Multiply numerator and denominator by (2 - i):(a)(2 - i) / (4 + 1) = (2a - ai)/5. That's correct.Second term is (2 + i)/5. Correct.Adding them together:(2a - ai)/5 + (2 + i)/5 = [2a - ai + 2 + i]/5. Correct.Grouping real and imaginary parts:Real: 2a + 2, Imaginary: (-a + 1)i. So yes, then divide by 5. Then the real part is (2a + 2)/5 and the imaginary part is (-a + 1)/5. Summing them gives [(2a + 2) + (-a + 1)]/5 = (a + 3)/5 = 1, so a = 2. That seems correct.Wait, but let me verify by plugging a = 2 back into the original expression to check if the sum of the real and imaginary parts is indeed 1.Compute z when a = 2:First term: 2/(2 + i) = 2*(2 - i)/5 = (4 - 2i)/5.Second term: (2 + i)/5.Adding them together:(4 - 2i)/5 + (2 + i)/5 = (4 + 2)/5 + (-2i + i)/5 = 6/5 - i/5.So the real part is 6/5, the imaginary part is -1/5. The sum is 6/5 - 1/5 = 5/5 = 1. Perfect, that's correct. So a = 2 is indeed the correct answer. Therefore, the answer is C: 2.But wait, the options given are A:0, B:1, C:2, D:3. So C is correct.But let me check if I interpreted the problem correctly. The problem says "the sum of its real part and imaginary part is 1". So, for example, if z = x + yi, then x + y = 1. Wait, is that correct? Or is it Re(z) + Im(z) = 1, which is x + y = 1? Yes, that's how the problem is stated. So, the real part is x, the imaginary part is y (the coefficient of i), and their sum is 1.So, according to my calculation, when a = 2, Re(z) = 6/5 and Im(z) = -1/5, and 6/5 + (-1/5) = 5/5 = 1, which is correct.Therefore, the answer is C:2.Alternatively, maybe I can try solving it another way to confirm. Let's represent z = (a)/(2 + i) + (2 + i)/5. Let's write z as x + yi, where x is the real part and y is the imaginary part. Then x + y = 1.First, compute (a)/(2 + i). Multiply numerator and denominator by (2 - i):(a)(2 - i)/(4 + 1) = (2a - ai)/5. So, real part is 2a/5, imaginary part is -a/5.Second term: (2 + i)/5. Real part is 2/5, imaginary part is 1/5.Therefore, adding both terms:Real part total: (2a/5) + (2/5) = (2a + 2)/5Imaginary part total: (-a/5) + (1/5) = (-a + 1)/5Sum of real and imaginary parts: (2a + 2)/5 + (-a + 1)/5 = (2a + 2 - a + 1)/5 = (a + 3)/5Set equal to 1: (a + 3)/5 = 1 => a + 3 = 5 => a = 2. Yep, same result.Therefore, confident that the answer is C:2.Final Answerboxed{C}</think>Given a complex number ( z = frac{a}{2+i} + frac{2+i}{5} ), we need to find the value of the real number ( a ) such that the sum of its real part and imaginary part is 1.First, simplify each term separately:1. For the term ( frac{a}{2+i} ): [ frac{a}{2+i} = frac{a(2-i)}{(2+i)(2-i)} = frac{a(2-i)}{5} = frac{2a - ai}{5} ]2. For the term ( frac{2+i}{5} ): [ frac{2+i}{5} = frac{2}{5} + frac{i}{5} ]Combine the two terms:[z = frac{2a - ai}{5} + frac{2 + i}{5} = frac{(2a - ai + 2 + i)}{5}]Group the real and imaginary parts:- Real part: ( frac{2a + 2}{5} )- Imaginary part: ( frac{-a + 1}{5} )According to the problem, the sum of the real part and the imaginary part is 1:[frac{2a + 2}{5} + frac{-a + 1}{5} = 1]Combine the fractions:[frac{(2a + 2) + (-a + 1)}{5} = 1]Simplify the numerator:[frac{a + 3}{5} = 1]Solve for ( a ):[a + 3 = 5 implies a = 2]Thus, the value of ( a ) is (boxed{C}).]

🔑:First, we simplify the expression for z by getting rid of the complex denominator in the first fraction and combining the terms.We have z = frac{a}{2+i} + frac{2+i}{5}. To remove the imaginary unit from the denominator of the first fraction, we multiply the numerator and denominator by the conjugate of the denominator: 2 - i.Thus, z becomes:begin{align*}z &= frac{a(2 - i)}{(2+i)(2-i)} + frac{2+i}{5} &= frac{a(2 - i)}{4+1} + frac{2}{5} + frac{i}{5} &= frac{2a - ai}{5} + frac{2}{5} + frac{i}{5} &= frac{2a + 2}{5} + frac{1 - a}{5}i.end{align*}Given that the sum of the real part and the imaginary part is 1, we can now set up the equation:frac{2a+2}{5} + frac{1-a}{5} = 1.Combining the terms and multiplying through by 5 to clear fractions yields:(2a+2) + (1-a) = 5.Simplifying further, we get:2a + 2 + 1 - a = 5 implies a + 3 = 5.Hence, a = 5 - 3 = 2.So, the correct answer is boxed{C}.