🧠:Okay, let's try to tackle this problem step by step. It's a bit complex with multiple moving parts, so I need to break it down carefully. First, let me restate the problem to make sure I understand it correctly. Paramon left point A heading towards point B. At 12:00 PM, when Paramon had already traveled halfway to B, two other people start moving: Agafon starts running from A to B, and Solomon starts moving from B to A. Then, at 1:20 PM, Agafon meets Solomon. Later, at 2:00 PM, Agafon catches up with Paramon. We need to find out when Paramon and Solomon met each other.Hmm, okay. Let's start by visualizing the timeline and the movements.Let me note down the key events:1. Paramon starts from A to B at some time before 12:00 PM.2. At 12:00 PM, Paramon is halfway between A and B. So, he has already covered half the distance.3. At this same time (12:00 PM), Agafon starts running from A to B, and Solomon starts moving from B to A.4. They meet each other (Agafon and Solomon) at 1:20 PM. That's 1 hour and 20 minutes after they started, which is 1.333... hours (since 20 minutes is 1/3 of an hour).5. Then, at 2:00 PM, which is 2 hours after 12:00 PM, Agafon catches up with Paramon. So, Agafon takes 2 hours to catch up to Paramon from the time he started.We need to find the time when Paramon and Solomon met.Let me denote the total distance between A and B as D. That might help in setting up equations.First, let's consider Paramon's speed. Since he was halfway to B at 12:00 PM, let's figure out how long he had been traveling before 12:00 PM. However, the problem doesn't mention when Paramon started his journey. Wait, actually, Paramon left point A for point B, and at 12:00 PM, he had traveled halfway. So, we need to know the time Paramon started. Wait, but the problem doesn't specify when Paramon left A. Hmm. Maybe we can figure it out based on the other information.Wait, maybe we don't need the exact starting time of Paramon, but rather his speed relative to the others. Let's see.Let me denote:- Let D be the total distance between A and B.- Let v_p be Paramon's speed.- Let v_a be Agafon's speed.- Let v_s be Solomon's speed.At 12:00 PM, Paramon is halfway, so he has covered D/2. Since he started at some time before 12:00 PM, let's say he started t hours before 12:00 PM. Then, the time from his start to 12:00 PM is t hours, during which he traveled D/2. Therefore, v_p = (D/2) / t = D/(2t).But we don't know t yet. Maybe we can find t from the other information?Alternatively, maybe we can express everything in terms of speeds and times relative to 12:00 PM.Wait, perhaps it's better to set 12:00 PM as time zero for the problem, and then track the times from there. Let's try that.Let me reset the timeline:- Let 12:00 PM be time t = 0.- At t = 0: Paramon is at D/2 from A, moving towards B. Agafon starts from A towards B, and Solomon starts from B towards A.- At t = 1 hour 20 minutes = 4/3 hours (since 1:20 PM is 1 and 1/3 hours after 12:00 PM), Agafon and Solomon meet.- At t = 2 hours (2:00 PM), Agafon catches up with Paramon.We need to find the time t when Paramon and Solomon meet.So, let me assign variables:Let’s denote:- Let D be the total distance between A and B.- Let v_p be Paramon's speed (from A to B). At t = 0, he is at D/2, so his position at time t (after 12:00 PM) is D/2 + v_p * t.- Let v_a be Agafon's speed. He starts from A at t = 0, so his position at time t is v_a * t.- Let v_s be Solomon's speed. He starts from B at t = 0, moving towards A, so his position at time t is D - v_s * t.First, at t = 4/3 hours (1:20 PM), Agafon and Solomon meet. So, their positions are equal:v_a * (4/3) = D - v_s * (4/3)So, (v_a + v_s) * (4/3) = DThus, D = (4/3)(v_a + v_s) ...(1)Second, at t = 2 hours (2:00 PM), Agafon catches up with Paramon. Their positions are equal:v_a * 2 = D/2 + v_p * 2So, 2v_a = D/2 + 2v_pThus, D = 4v_a - 4v_p ...(2)So now, we have equations (1) and (2):From (1): D = (4/3)(v_a + v_s)From (2): D = 4v_a - 4v_pTherefore, equate them:(4/3)(v_a + v_s) = 4v_a - 4v_pMultiply both sides by 3 to eliminate denominator:4(v_a + v_s) = 12v_a - 12v_pExpand left side:4v_a + 4v_s = 12v_a - 12v_pBring all terms to left side:4v_a + 4v_s - 12v_a + 12v_p = 0Combine like terms:-8v_a + 4v_s + 12v_p = 0Divide both sides by 4:-2v_a + v_s + 3v_p = 0So,v_s = 2v_a - 3v_p ...(3)Now, we need another equation to relate these variables. Let's see.We need to find the time when Paramon and Solomon meet. Let's denote that time as t_m.At time t_m, Paramon's position is D/2 + v_p * t_mSolomon's position is D - v_s * t_mWhen they meet, their positions are equal:D/2 + v_p * t_m = D - v_s * t_mRearranged:v_p * t_m + v_s * t_m = D - D/2(v_p + v_s) * t_m = D/2So,t_m = (D/2) / (v_p + v_s) ...(4)Our goal is to find t_m.But we need to express t_m in terms of known quantities or find relations between the variables.We have equations (1), (2), and (3). Let's see if we can find another relation.Wait, perhaps we can find the relation between Paramon's speed and Agafon's speed from equation (2):From (2): D = 4v_a - 4v_pBut from (1): D = (4/3)(v_a + v_s)So, 4v_a - 4v_p = (4/3)(v_a + v_s)Divide both sides by 4:v_a - v_p = (1/3)(v_a + v_s)Multiply both sides by 3:3v_a - 3v_p = v_a + v_sBring terms to left side:3v_a - 3v_p - v_a - v_s = 0Simplify:2v_a - 3v_p - v_s = 0But from equation (3): v_s = 2v_a - 3v_pSubstitute into above:2v_a - 3v_p - (2v_a - 3v_p) = 0Simplify:2v_a -3v_p -2v_a +3v_p = 0 => 0 = 0Hmm, this is just confirming the consistency of the equations, but doesn't give us new information.Therefore, we need another approach. Let's see.We have three variables: v_a, v_s, v_p. But only two equations from (1) and (2), and relation (3). However, equation (3) is derived from (1) and (2). Therefore, we need a third equation. Wait, but perhaps we can use the fact that Paramon had already traveled D/2 by 12:00 PM, which can relate to his speed.But wait, Paramon started at some time before 12:00 PM. Let's denote the time he started as T hours before 12:00 PM. Then, he traveled D/2 in T hours. Therefore, his speed is (D/2)/T = D/(2T).But we don't know T. However, maybe we can express T in terms of other variables.Alternatively, since we have variables in terms of D, perhaps we can set D as a constant (like 1 unit) to simplify the equations. Let's try that.Let’s assume the distance D = 1 (unit). Then, the equations become:From (1): 1 = (4/3)(v_a + v_s)From (2): 1 = 4v_a - 4v_pSo, equation (1): v_a + v_s = 3/4Equation (2): v_a - v_p = 1/4Also, from equation (3): v_s = 2v_a - 3v_pSo now, we can substitute equation (3) into equation (1):v_a + (2v_a - 3v_p) = 3/4Simplify:3v_a - 3v_p = 3/4Divide both sides by 3:v_a - v_p = 1/4But this is exactly equation (2). So again, we still have two equations and three variables. Hmm, need another equation.Wait, perhaps considering Paramon's speed. Since D = 1, and he traveled D/2 = 1/2 by 12:00 PM. If he started T hours before 12:00 PM, then his speed is (1/2)/T = 1/(2T). So, v_p = 1/(2T). But we don't know T. However, perhaps we can relate T to other variables by considering the time when Agafon catches up with Paramon.Wait, Agafon catches up with Paramon at t = 2 hours (2:00 PM). At that time, Agafon has traveled 2 hours, so his distance covered is v_a * 2. Paramon, who had a head start of D/2 and was traveling for 2 hours after 12:00 PM, so his total distance from A is D/2 + v_p * 2. Since Agafon catches up, their positions are equal:v_a * 2 = 1/2 + v_p * 2Which is equation (2): 2v_a = 1/2 + 2v_p => 2v_a - 2v_p = 1/2 => v_a - v_p = 1/4, same as before.So, we still need another relation. Perhaps we need to use the fact that Solomon and Paramon are moving towards each other after 12:00 PM, and we need to find when they meet. But that's what we need to find, so maybe we need to express t_m in terms of T?Alternatively, maybe we can express T in terms of other variables. Since v_p = 1/(2T), perhaps.From equation (2): v_a - v_p = 1/4 => v_a = v_p + 1/4From equation (3): v_s = 2v_a - 3v_p = 2(v_p + 1/4) - 3v_p = 2v_p + 1/2 - 3v_p = -v_p + 1/2From equation (1): v_a + v_s = 3/4 => (v_p + 1/4) + (-v_p + 1/2) = 3/4Simplify:v_p +1/4 -v_p +1/2 = 3/4 => (1/4 + 1/2) = 3/4Which is 3/4 = 3/4, which checks out. So, no new information.Therefore, all equations are consistent, but we need another relation to find the values.Wait, perhaps we can express T in terms of v_p. Since v_p = 1/(2T), then T = 1/(2v_p). So, Paramon started T hours before 12:00 PM, which is 1/(2v_p) hours before 12:00 PM. But how does that help?Alternatively, maybe we need to find the meeting time between Paramon and Solomon in terms of v_p and v_s, but since we can express v_s in terms of v_p (from equation (3)), we can write t_m in terms of v_p.From equation (4):t_m = (D/2)/(v_p + v_s)Since D = 1,t_m = (1/2)/(v_p + v_s)But v_s = -v_p + 1/2 (from earlier, when D=1, v_s = -v_p + 1/2)So,t_m = (1/2)/(v_p + (-v_p + 1/2)) = (1/2)/(1/2) = 1 hour.Wait, that's interesting. So, t_m = 1 hour. That would mean that Paramon and Solomon meet at 12:00 PM + 1 hour = 1:00 PM. But that seems too straightforward. Let me check this again.Wait, let's go step by step.We have:v_s = 2v_a - 3v_p (equation 3)But from equation (2): v_a = v_p + 1/4Substitute into equation (3):v_s = 2(v_p + 1/4) - 3v_p = 2v_p + 1/2 - 3v_p = -v_p + 1/2Thus, v_s = -v_p + 1/2Then, v_p + v_s = v_p + (-v_p + 1/2) = 1/2Therefore, t_m = (D/2)/(v_p + v_s) = (1/2)/(1/2) = 1 hour.So, regardless of the value of v_p, the meeting time between Paramon and Solomon is 1 hour after 12:00 PM, which is 1:00 PM.Wait, but this seems conflicting with the fact that Agafon and Solomon met at 1:20 PM, but Paramon and Solomon met at 1:00 PM? How is that possible?Wait, if Paramon and Solomon met at 1:00 PM, which is 20 minutes before Agafon and Solomon met at 1:20 PM. That seems possible if Solomon continued moving towards A after meeting Paramon, and Agafon was moving towards B and met Solomon later. But let's confirm.Wait, but according to the calculation, t_m = 1 hour after 12:00 PM, which is 1:00 PM, so that's when Paramon and Solomon meet. Then, Agafon and Solomon meet at 1:20 PM.But how does this fit with the positions?Wait, let's verify with actual numbers. Let me assign some values based on the equations.From earlier, with D = 1:From equation (2): v_a = v_p + 1/4From equation (3): v_s = -v_p + 1/2Let’s pick a value for v_p. Let's say, for example, v_p = 1/4. Then,v_a = 1/4 + 1/4 = 1/2v_s = -1/4 + 1/2 = 1/4So, Paramon's speed is 1/4 units per hour, Agafon's speed is 1/2 units per hour, Solomon's speed is 1/4 units per hour.Check equation (1): v_a + v_s = 1/2 + 1/4 = 3/4, which matches D = (4/3)(v_a + v_s) => 1 = (4/3)(3/4) = 1. Correct.Now, let's check the positions at different times.At t = 1 hour (1:00 PM):- Paramon's position: D/2 + v_p * t = 0.5 + (1/4)(1) = 0.5 + 0.25 = 0.75 units from A.- Solomon's position: D - v_s * t = 1 - (1/4)(1) = 0.75 units from A. So, they meet at 0.75 units from A at 1:00 PM. Correct.Agafon and Solomon meet at t = 4/3 hours (1:20 PM):- Agafon's position: v_a * t = (1/2)(4/3) = 2/3 ≈ 0.6667 units from A.- Solomon's position: 1 - (1/4)(4/3) = 1 - 1/3 = 2/3 ≈ 0.6667 units from A. Correct, they meet at 2/3 units from A at 1:20 PM.At t = 2 hours (2:00 PM):- Agafon's position: (1/2)(2) = 1 unit from A, which is point B. Wait, but Paramon is moving towards B. His position at t = 2 hours is 0.5 + (1/4)(2) = 0.5 + 0.5 = 1 unit from A, so he also reaches B at 2:00 PM. So Agafon catches up with Paramon exactly at B at 2:00 PM.But in this case, Paramon reaches B at 2:00 PM. However, according to the problem statement, Agafon catches up with Paramon at 2:00 PM. So, this example works.But in this scenario, Paramon and Solomon meet at 1:00 PM, which is 1 hour after 12:00 PM. So, the answer would be 1:00 PM. But the problem seems to suggest that maybe it's a different time. Wait, but according to the calculations, it's 1:00 PM. Let me check with another example.Let's choose a different v_p. Suppose v_p = 1/8.Then, v_a = 1/8 + 1/4 = 3/8.v_s = -1/8 + 1/2 = 3/8.Check equation (1): v_a + v_s = 3/8 + 3/8 = 6/8 = 3/4. Correct.Now, positions:At t = 1 hour:- Paramon: 0.5 + (1/8)(1) = 0.5 + 0.125 = 0.625.- Solomon: 1 - (3/8)(1) = 5/8 = 0.625. Meet at 0.625.Agafon and Solomon meet at t = 4/3:- Agafon: (3/8)(4/3) = 0.5.- Solomon: 1 - (3/8)(4/3) = 1 - 0.5 = 0.5. Correct.At t = 2 hours:- Agafon: (3/8)(2) = 6/8 = 0.75.- Paramon: 0.5 + (1/8)(2) = 0.5 + 0.25 = 0.75. So they meet at 0.75. Wait, but according to this, Paramon hasn't reached B yet. But if D = 1, then Paramon is at 0.75, which is three-quarters of the way. But according to the previous equations, D = 4v_a - 4v_p. Let's check:4v_a -4v_p = 4*(3/8) -4*(1/8) = 12/8 -4/8 = 8/8 = 1. Correct. So, at t = 2, Agafon catches Paramon at 0.75 from A, which is not B. Wait, but in this case, the total distance D is 1, so point B is at 1.0. So, Paramon hasn't reached B yet. Hmm, this contradicts my previous example. Wait, so in this case, at t = 2 hours, Agafon catches Paramon at 0.75 from A, meaning that Paramon is still en route to B. But in the first example, when v_p = 1/4, they meet at B.So, depending on Paramon's speed, the meeting point varies. However, in both cases, the meeting time between Paramon and Solomon is at t = 1 hour, regardless of the speed. Because t_m = 1 hour in both cases. So, even though the meeting position changes, the time remains the same.Therefore, regardless of Paramon's speed, the meeting time between Paramon and Solomon is 1 hour after 12:00 PM, which is 1:00 PM.Wait, but this seems counterintuitive. How can the meeting time be fixed regardless of their speeds? Let me think.The key here is that the problem gives us two specific meeting times: Agafon and Solomon meet at 1:20 PM, and Agafon catches Paramon at 2:00 PM. These two events constrain the speeds such that the meeting time between Paramon and Solomon becomes fixed.In the equations, when we derived t_m = (D/2)/(v_p + v_s), and through substitution found that v_p + v_s = D/2, leading to t_m = 1 hour, regardless of D. Wait, let me check that.Wait, with D = 1, we had:t_m = (1/2)/(v_p + v_s)But we found that v_p + v_s = 1/2, so t_m = (1/2)/(1/2) = 1 hour.But why does v_p + v_s = 1/2?From equation (3): v_s = 2v_a - 3v_pFrom equation (2): v_a = v_p + 1/4Therefore, substituting into equation (3):v_s = 2(v_p + 1/4) -3v_p = 2v_p + 1/2 -3v_p = -v_p + 1/2Therefore, v_p + v_s = v_p + (-v_p +1/2) =1/2Therefore, regardless of v_p, the sum v_p + v_s is always 1/2 (when D=1). Hence, t_m = (1/2)/(1/2) =1 hour.Therefore, no matter what the individual speeds are, the meeting time between Paramon and Solomon is always 1 hour after 12:00 PM, which is 1:00 PM.Therefore, the answer should be 1:00 PM.But let me double-check with the example where D is not 1. Let's take D as 12 km for ease of calculation.Let’s suppose D = 12 km.From equation (1): D = (4/3)(v_a + v_s) => 12 = (4/3)(v_a + v_s) => v_a + v_s = 9 km/h.From equation (2): D =4v_a -4v_p =>12=4v_a -4v_p =>3= v_a -v_p =>v_a =v_p +3.From equation (3):v_s=2v_a -3v_p=2(v_p +3) -3v_p=2v_p +6 -3v_p= -v_p +6.So, v_a +v_s=(v_p +3)+(-v_p +6)=9, which matches equation (1).Now, let's compute the meeting time between Paramon and Solomon:t_m = (D/2)/(v_p +v_s)=6/(v_p + (-v_p +6))=6/6=1 hour.Again, 1 hour after 12:00 PM, so 1:00 PM.Another example. Let’s take D = 24 km.From equation (1):24=(4/3)(v_a +v_s) =>v_a +v_s=18.From equation (2):24=4v_a -4v_p =>6=v_a -v_p =>v_a =v_p +6.From equation (3):v_s=2v_a -3v_p=2(v_p +6) -3v_p=2v_p +12 -3v_p= -v_p +12.Then, v_p +v_s= v_p + (-v_p +12)=12. So t_m=(24/2)/12=12/12=1 hour.Same result. So, regardless of D, the meeting time is 1 hour after 12:00 PM.Therefore, the answer is 1:00 PM.But wait, the problem says "at 1:20 PM, Agafon met Solomon, and at 2:00 PM, he caught up with Paramon." So, according to our calculations, even though Agafon and Solomon meet at 1:20 PM, Paramon and Solomon meet earlier at 1:00 PM. That seems okay because they are different pairs.But let me check with the example where D=12 km, v_a =v_p +3, v_s= -v_p +6.Let’s take v_p=1 km/h. Then, v_a=4 km/h, v_s=5 km/h.Check meeting between Agafon and Solomon at t=4/3 hours (1:20 PM):Agafon's distance:4*(4/3)=16/3≈5.333 kmSolomon's distance from B:5*(4/3)=20/3≈6.666 km. Since D=12 km, Solomon's position from A is 12 -6.666≈5.333 km. So they meet at 5.333 km from A. Correct.At t=2 hours (2:00 PM):Agafon's distance:4*2=8 kmParamon's distance from A:6 +1*2=8 km. So they meet at 8 km from A. Correct.Meeting between Paramon and Solomon at t=1 hour:Paramon's position:6 +1*1=7 kmSolomon's position:12 -5*1=7 km. Correct.So, yes, everything checks out. Therefore, the answer is 1:00 PM.I think this is consistent across different examples, so the answer should be 1:00 PM.

✅:The meeting between Paramon and Solomon occurred at 1:00 PM. Step-by-Step Explanation:1. Define Variables and Timeline: - Let ( D ) be the total distance between points A and B. - At 12:00 PM (time ( t = 0 )): - Paramon is halfway (( D/2 )) from A to B. - Agafon starts from A towards B at speed ( v_a ). - Solomon starts from B towards A at speed ( v_s ). - At 1:20 PM (( t = frac{4}{3} ) hours), Agafon and Solomon meet. - At 2:00 PM (( t = 2 ) hours), Agafon catches Paramon.2. Key Equations from Meetings: - Agafon-Solomon Meeting: [ v_a cdot frac{4}{3} + v_s cdot frac{4}{3} = D quad Rightarrow quad v_a + v_s = frac{3D}{4} tag{1} ] - Agafon-Paramon Catchup: [ v_a cdot 2 = frac{D}{2} + v_p cdot 2 quad Rightarrow quad 2v_a - 2v_p = frac{D}{2} quad Rightarrow quad 4v_a - 4v_p = D tag{2} ]3. Relate Speeds: - From (1) and (2), solve for ( v_s ): [ v_s = 2v_a - 3v_p tag{3} ] - Substitute ( v_s ) into (1) to confirm consistency.4. Paramon-Solomon Meeting Time (( t_m )): - Positions equate when: [ frac{D}{2} + v_p t_m = D - v_s t_m quad Rightarrow quad t_m = frac{D/2}{v_p + v_s} ] - Substitute ( v_s = 2v_a - 3v_p ) and simplify using (2): [ v_p + v_s = frac{D}{2} quad Rightarrow quad t_m = frac{D/2}{D/2} = 1 text{ hour} ]5. Conclusion: - The meeting occurs 1 hour after 12:00 PM, i.e., 1:00 PM.Answer: The meeting between Paramon and Solomon occurred at 1:00 PM.

🔑:1. Let ( S ) be the total distance between point ( A ) and point ( B ).2. Let the speeds of Paramon, Solomon, and Agafon be ( x ) km/h, ( y ) km/h, and ( z ) km/h, respectively.3. At 12:00, Paramon has travelled half the distance ( S ). Thus, the time taken by Paramon to travel half the distance is: [ frac{S/2}{x} ]4. At 12:00, Agafon starts chasing Paramon. At 14:00, Agafon catches up with Paramon. This means that during the time from 12:00 to 14:00, Agafon travels the same distance that Paramon does plus the half distance that Paramon already travelled. Hence, [ 2 text{ hours} times z = frac{S}{2} + 2 text{ hours} times x ]5. Simplifying the equation: [ 2z = frac{S}{2} + 2x ] [ 4z = S + 4x ] [ S = 4z - 4x ]6. Solomon starts at B at 12:00 and meets Agafon at 13:20. The meeting time between them is: [ 80 text{ minutes} = frac{4}{3} text{ hours} ]So, they meet at:[frac{4}{3} (y + z) = S]7. Therefore, [ z + y = frac{S}{4/3} ] [ z + y = frac{3}{4}S ]8. Solomon and Paramon moved towards each other, meeting in ( t ) hours. Since Solomon starts at 12:00 and they meet in one hour, Paramon excepts 12:00.9. The equation becomes: [ frac{S/2}{x} = 1 ] [ frac{S/2}{x} = 1 implies S times x = frac{S}{2} ] Thus, at they meet: Conclusion: They meet at 13:00.( boxed{13} )

🧠:Okay, so I have this problem: There are n vectors in space such that any pair of them forms an obtuse angle. What is the maximum possible value of n? Hmm, interesting. Let me try to figure this out step by step. First, I need to recall some concepts about vectors and angles between them. The angle between two vectors can be determined using the dot product formula: if u and v are vectors, then the cosine of the angle θ between them is equal to (u · v)/(||u|| ||v||). Since the problem states that any pair of vectors forms an obtuse angle, that means θ > 90 degrees for every pair. Now, cosine of an angle greater than 90 degrees is negative because cos(θ) is negative in the second quadrant (angles between 90 and 180 degrees). Therefore, for all pairs of vectors u and v, the dot product u · v must be negative. So the condition is that for any two distinct vectors in the set, their dot product is negative. The question is, what's the maximum number n of such vectors that can exist in space? The problem doesn't specify the dimension of the space, but I think it's asking for the maximum n in general, so maybe we have to consider vectors in any dimension? Or maybe it's specifically in 3D space? Hmm, the problem says "in space," which typically refers to 3-dimensional space. But I should confirm if that's the case here. Wait, sometimes "in space" just means in some Euclidean space, not necessarily 3D. Hmm. Wait, the problem might be in any dimension, but the answer might be dimension-dependent. Wait, but the problem is asking for the maximum possible n regardless of the dimension. Wait, no, that can't be. Because if you have higher dimensions, you can have more vectors. Wait, but maybe there's a constraint even in higher dimensions? Hmm. Let me think.Wait, actually, in higher dimensions, it's possible to have more vectors with pairwise negative dot products. But maybe there's a maximum regardless of the dimension? That seems unlikely. Wait, maybe the problem is specifically in 3D space? The term "in space" is a bit ambiguous. In math problems, sometimes "space" refers to 3D, but sometimes it can be general. Hmm. Let me check the original problem again. It says: "There are n vectors in space such that any pair of them forms an obtuse angle. What is the maximum possible value of n?" The term "in space" might indeed mean 3-dimensional space. But maybe not. Let me assume first that it's 3-dimensional, and if that leads me somewhere, but also keep in mind higher dimensions just in case.But wait, even if it's in higher dimensions, there might be a bound on n. Let me recall some related concepts. If we have vectors in d-dimensional space, what is the maximum number of vectors such that the dot product between any two is negative? That seems related to the concept of mutually obtuse vectors. Alternatively, perhaps this is connected to the concept of the maximum number of equiangular lines, but in this case, the condition is stricter: not just that all angles are equal, but that all angles are obtuse. Wait, maybe I can think in terms of geometry. Suppose all vectors are in some d-dimensional space. If I have a set of vectors where every pair has a negative dot product, then perhaps these vectors lie in a hemisphere or something? Wait, maybe not. Let me try to visualize in 2D. If in 2D, can I have more than two vectors such that every pair has an obtuse angle? Let's see. Suppose I have three vectors in 2D. If each pair forms an obtuse angle, then each pair has a dot product less than zero. Let's try to arrange three vectors in the plane such that the angle between any two is greater than 90 degrees. Take the first vector along the positive x-axis. The second vector should make an angle greater than 90 degrees with the first, so it would be in the left half-plane. Similarly, the third vector should make an angle greater than 90 degrees with both the first and the second. But in 2D, if the first is along the x-axis, the second is in the left half-plane, say at 150 degrees from the first. Then where can the third vector be? It needs to be in the region that is more than 90 degrees from both the first and the second. The first vector (along x-axis) has angles greater than 90 degrees in the left half-plane. The second vector at 150 degrees (which is 180 - 30 degrees) would have angles greater than 90 degrees in the region opposite to it, which would be the right half-plane shifted by 30 degrees. Wait, the angle opposite to 150 degrees is 30 degrees on the other side. So the regions where a vector would form an angle greater than 90 degrees with both the first and the second vector would be the intersection of the left half-plane (for the first vector) and the region opposite to the second vector. But in 2D, the intersection of these two regions might be empty. Let me check with coordinates.Let’s define the first vector as u = (1, 0). The second vector v is at 150 degrees, so its coordinates would be (cos(150°), sin(150°)) = (-√3/2, 1/2). The third vector w needs to have a negative dot product with both u and v. Let's denote w = (a, b). Then:u · w = a < 0 (since the dot product with u is a*1 + b*0 = a)v · w = (-√3/2)a + (1/2)b < 0Since a < 0, let's let a = -c where c > 0. Then the second inequality becomes:(-√3/2)(-c) + (1/2)b < 0 ⇒ (√3/2)c + (1/2)b < 0 ⇒ √3 c + b < 0 ⇒ b < -√3 cBut since c > 0, b has to be negative and its magnitude has to be greater than √3 c. However, the vector w = (-c, b) must have components such that ||w|| is positive. But even with b being negative and large in magnitude, this is possible. However, the angle between w and u is determined by the dot product, which is -c, so it's negative, so angle is obtuse. The angle between w and v is also obtuse. But is there a third vector that can satisfy this?Wait, but let's see if such a vector exists. Let's pick c = 1, so a = -1. Then b < -√3*1 ≈ -1.732. Let's choose b = -2. Then w = (-1, -2). Let's check the angle between w and u. The dot product is (-1)(1) + (-2)(0) = -1 < 0, so angle is obtuse. The dot product with v is (-√3/2)(-1) + (1/2)(-2) = (√3/2) - 1 ≈ 0.866 - 1 = -0.134 < 0, which is also obtuse. So in this case, there exists a third vector. Wait, so in 2D, we can have three vectors with pairwise obtuse angles. Wait, but can we have four vectors?Let me try to add a fourth vector. Let’s denote the fourth vector as z = (d, e). It needs to have a negative dot product with u, v, and w. Dot product with u: d < 0.Dot product with v: (-√3/2)d + (1/2)e < 0.Dot product with w: (-1)d + (-2)e < 0 ⇒ -d -2e < 0 ⇒ d + 2e > 0.So we have the following inequalities:1. d < 02. (-√3/2)d + (1/2)e < 0 ⇒ -√3 d + e < 0 ⇒ e < √3 d3. d + 2e > 0 ⇒ e > -d/2But since d < 0, let's substitute d = -f where f > 0. Then:1. d = -f < 0 ⇒ f > 02. e < √3*(-f) ⇒ e < -√3 f3. e > -(-f)/2 ⇒ e > f/2But combining inequalities 2 and 3:f/2 < e < -√3 fBut since f > 0, the right-hand side of inequality 2 is negative (because -√3 f < 0), and the left-hand side of inequality 3 is positive (f/2 > 0). Therefore, there is no e that satisfies both e > f/2 (positive) and e < -√3 f (negative). Contradiction. Therefore, in 2D, we cannot have four vectors with pairwise obtuse angles. So in 2D, the maximum n is 3.But the original problem says "in space," which might be 3D. Let me check for 3D.In 3D, can we have more than 3 vectors with pairwise obtuse angles? Let's see.Let me recall that in d-dimensional space, the maximum number of vectors such that any pair has a negative dot product is d+1. Is that a known result? Wait, actually, I might be confusing with other results. Let me think.If we have vectors in d-dimensional space, and we require that each pair has a negative dot product, then we can consider the following approach. Suppose we have vectors v_1, v_2, ..., v_n in R^d. The condition is v_i · v_j < 0 for all i ≠ j.Now, suppose we normalize the vectors so that each has norm 1. Then the dot product is equal to the cosine of the angle between them. So if we have unit vectors, the condition is that cosθ < 0 for all pairs, which means all angles are obtuse.There is a result in linear algebra that gives the maximum number of equiangular lines in a space, but here the angles are not required to be equal, just obtuse. But perhaps there is a bound on the number of vectors with all pairwise dot products negative.I remember that in d-dimensional space, you can have at most d + 1 vectors such that any pair has a negative dot product. Let me see if that makes sense.Consider the case of d = 1. In 1D, vectors are just scalars. If you have two vectors (scalars) pointing in opposite directions, their product is negative. So n = 2. Which is 1 + 1 = 2.In 2D, we saw that n = 3 is possible. 2 + 1 = 3. So that matches. In 3D, it would be 4. Let me try to construct 4 vectors in 3D with pairwise negative dot products.How about the vertices of a regular tetrahedron inscribed in the unit sphere? Wait, in a regular tetrahedron, the angle between any two vectors from the center to the vertices is arccos(1/3) ≈ 70.5 degrees, which is acute. So that's not good. But if we take the vectors pointing to the vertices of a regular tetrahedron but with a different scaling or orientation, maybe we can get obtuse angles? Wait, no. The dot product between vectors in a regular tetrahedron is positive. So that's not helpful.Alternatively, perhaps if we take vectors pointing to the vertices of a regular octahedron. Wait, in a regular octahedron, the vectors are pointing to the positive and negative axes. So for example, (1,0,0), (-1,0,0), (0,1,0), (0,-1,0), (0,0,1), (0,0,-1). But the dot product between (1,0,0) and (0,1,0) is zero, which is not obtuse. So that doesn't work.Wait, maybe another approach. Let's try to construct four vectors in 3D with all pairwise dot products negative.Let me start with the standard basis vectors but with some modifications. Suppose we take vectors that are pointing towards the negative orthants. For example, vectors like (-1, -1, -1), but normalized. Let me see:Let’s define four vectors:v1 = (-1, -1, -1)v2 = (-1, 1, 1)v3 = (1, -1, 1)v4 = (1, 1, -1)Now, let's compute the dot products between each pair.v1 · v2 = (-1)(-1) + (-1)(1) + (-1)(1) = 1 -1 -1 = -1 < 0v1 · v3 = (-1)(1) + (-1)(-1) + (-1)(1) = -1 +1 -1 = -1 < 0v1 · v4 = (-1)(1) + (-1)(1) + (-1)(-1) = -1 -1 +1 = -1 < 0v2 · v3 = (-1)(1) + (1)(-1) + (1)(1) = -1 -1 +1 = -1 < 0v2 · v4 = (-1)(1) + (1)(1) + (1)(-1) = -1 +1 -1 = -1 < 0v3 · v4 = (1)(1) + (-1)(1) + (1)(-1) = 1 -1 -1 = -1 < 0So all pairwise dot products are -1, which is negative. Therefore, these four vectors in 3D space satisfy the condition that every pair forms an obtuse angle. Each vector is in a different orthant, and their dot products are all negative. So n = 4 is possible in 3D.But can we have five vectors? Let me try to see. Suppose we add a fifth vector. Let's say the fifth vector has components (a, b, c). It needs to have a negative dot product with all four existing vectors. Let's write down the conditions:For vector v1 = (-1, -1, -1): (-1)a + (-1)b + (-1)c < 0 ⇒ -a - b - c < 0 ⇒ a + b + c > 0For vector v2 = (-1, 1, 1): (-1)a + (1)b + (1)c < 0 ⇒ -a + b + c < 0For vector v3 = (1, -1, 1): (1)a + (-1)b + (1)c < 0 ⇒ a - b + c < 0For vector v4 = (1, 1, -1): (1)a + (1)b + (-1)c < 0 ⇒ a + b - c < 0So the fifth vector (a, b, c) must satisfy:1. a + b + c > 02. -a + b + c < 03. a - b + c < 04. a + b - c < 0Let me try to analyze these inequalities. Let me denote them as:1. a + b + c > 02. -a + b + c < 0 ⇒ a > b + c3. a - b + c < 0 ⇒ a + c < b4. a + b - c < 0 ⇒ a + b < cFrom inequality 3: a + c < bFrom inequality 4: a + b < cIf I substitute a + c < b into inequality 4: Let's replace b with something larger than a + c in inequality 4.Wait, from 3: b > a + cFrom 4: c > a + bBut substituting b > a + c into c > a + b gives c > a + (a + c) ⇒ c > 2a + c ⇒ 0 > 2a ⇒ a < 0Similarly, from 3: b > a + cBut since a < 0, from inequality 1: a + b + c > 0. Let's express this as (a + c) + b > 0. But from 3, b > a + c, so substituting, we get (a + c) + b > (a + c) + (a + c) = 2(a + c) > 0 ⇒ a + c > 0.But since a < 0, then c must be greater than |a|.From inequality 4: c > a + b. But since b > a + c (from 3), then c > a + (a + c) ⇒ c > 2a + c ⇒ 0 > 2a ⇒ a < 0, which we already have. So this is consistent.But we need to find real numbers a, b, c that satisfy all these inequalities. Let me attempt to assign some values.Let’s assume a = -1 (since a < 0). Then from inequality 1: -1 + b + c > 0 ⇒ b + c > 1From inequality 2: -(-1) + b + c < 0 ⇒ 1 + b + c < 0 ⇒ b + c < -1But this contradicts inequality 1, which requires b + c > 1. Therefore, no solution exists when a = -1.Wait, that's a problem. Let's check if this contradiction is inherent.From inequalities 1 and 2:1. a + b + c > 02. -a + b + c < 0 ⇒ b + c < aBut from 1: a + (b + c) > 0. From 2: b + c < a. Therefore, substituting into 1: a + (something less than a) > 0 ⇒ (a + (b + c)) > 0 ⇒ (a + (something less than a)) > 0 ⇒ (something less than 2a) > 0 ⇒ 2a > something, but since b + c < a, and a is part of the term, but given that a is part of both, this seems tricky.Wait, actually, substituting b + c < a from inequality 2 into inequality 1:a + (b + c) > 0 ⇒ a + (something less than a) > 0 ⇒ (a + (b + c)) > 0 ⇒ (something less than 2a) > 0But if a is negative, then 2a is negative, so something less than 2a (which is negative) would be even more negative, which cannot be greater than 0. Contradiction. Therefore, there is no solution for a, b, c that satisfies all four inequalities. Therefore, in 3D, it's impossible to have five vectors with pairwise negative dot products. Hence, the maximum n in 3D is 4.But wait, the problem says "in space," which is 3D. So the answer would be 4. But wait, before confirming, let me check if there isn't another configuration where you can have more vectors. Maybe not using the orthant approach. Let me think.Suppose we have vectors in 3D space such that each pair forms an obtuse angle. Each vector can be represented as a point on the unit sphere, with the center of the sphere at the origin. The condition is that the dot product between any two points is negative, which means the angle between them is greater than 90 degrees. So, on the unit sphere, this means that the two points are not in the same hemisphere as each other. Wait, no. If two points are in opposite hemispheres, the angle between them can be greater than 90 degrees. Wait, but actually, any two points on the sphere will have an angle between 0 and 180 degrees. If the dot product is negative, the angle is between 90 and 180 degrees.So, in order to have all pairwise angles greater than 90 degrees, each pair of vectors must be in opposite hemispheres with respect to any hyperplane through the origin. Wait, but how can multiple vectors all pairwise be in opposite hemispheres?Actually, if you have a set of points on the sphere where each pair is separated by an angle greater than 90 degrees, then no two points can lie in the same closed hemisphere. Because if two points were in the same closed hemisphere, their angle would be at most 90 degrees. So the condition is that no two points lie in the same closed hemisphere. But by the theorem called the "Hemisphere theorem" or something similar... Wait, there's a theorem by Gohberg and Markus which states that the maximum number of points on a sphere in d-dimensional space such that no two lie in a closed hemisphere is d + 1. Yes! That seems relevant. So in d-dimensional space, the maximum number of vectors (points on the unit sphere) such that no two lie in the same closed hemisphere is d + 1. Since if you have more than d + 1 points, by some theorem like Carathéodory's or Helly's, there must be two in the same hemisphere. Therefore, in 3D space (d = 3), the maximum n is 4. Which aligns with the previous construction. So, according to this theorem, the maximum n is d + 1 for d-dimensional space. Since the problem mentions "in space," which is 3D, the answer should be 4. But wait, let me confirm if the theorem is applicable here. The theorem states that the maximum number of points on a sphere in R^d such that no two lie in a closed hemisphere is d + 1. So in our case, the condition that all angles are greater than 90 degrees is equivalent to all pairs of points not lying in the same closed hemisphere. Because if two points were in the same closed hemisphere, their angle would be at most 90 degrees. Therefore, the maximum n is d + 1. Therefore, in 3D space (d = 3), n = 4. If the problem is in 3D, then the answer is 4. But if "in space" allows for higher dimensions, then n could be higher. But the problem is probably referring to 3D space. However, maybe the answer is d + 1 for d-dimensional space, but the problem didn't specify the dimension, so maybe the answer is that in d-dimensional space, it's d + 1, but since the problem says "in space," which is 3D, the answer is 4. Alternatively, maybe the problem is in general, and the answer is that in any dimension, the maximum n is 4. Wait, that can't be. For example, in 4D, you could have 5 vectors with pairwise negative dot products. Wait, according to the theorem, in d-dimensional space, it's d + 1. So in 4D, it's 5, etc. But the problem didn't specify the dimension. Wait, but the term "in space" usually refers to 3D. Unless it's a translation from another language where "space" could mean any Euclidean space. Hmm. But since the user hasn't specified, perhaps the answer is 4. Alternatively, maybe the problem is intended to be in general, and the answer is n = d + 1 for d-dimensional space. But since it's unspecified, perhaps the answer is 4. Let me check some references. Wait, in the book "Problems and Theorems in Linear Algebra" by Prasolov, there is a problem stating that the maximum number of vectors in R^n such that the pairwise inner products are negative is n + 1. So in 3D, it's 4. Therefore, I think the answer is 4. Therefore, the maximum possible value of n is 4.Final AnswerThe maximum possible value of ( n ) is boxed{4}.

✅:To determine the maximum number ( n ) of vectors in space such that any pair forms an obtuse angle, we start by noting that an obtuse angle between two vectors implies their dot product is negative. For vectors in ( d )-dimensional space, the problem reduces to finding the maximum number of vectors such that all pairwise dot products are negative. This is related to the geometric property that no two vectors can lie in the same closed hemisphere. A known result in linear algebra states that the maximum number of vectors in ( d )-dimensional space with pairwise negative dot products is ( d + 1 ). This is derived from the theorem that the maximum number of points on a sphere in ( mathbb{R}^d ) such that no two lie in a closed hemisphere is ( d + 1 ).For 3-dimensional space (referred to as "in space" in the problem), the maximum number ( n ) is ( 3 + 1 = 4 ). This can be verified by constructing four vectors in 3D space, such as:- ( v_1 = (-1, -1, -1) )- ( v_2 = (-1, 1, 1) )- ( v_3 = (1, -1, 1) )- ( v_4 = (1, 1, -1) )These vectors have pairwise negative dot products, confirming that they form obtuse angles with each other. Attempting to add a fifth vector leads to contradictions in the required inequalities, proving that four is the maximum.Thus, the maximum possible value of ( n ) is (boxed{4}).

🔑:1. We are given n vectors vec{a}_{1}, vec{a}_{2}, ldots, vec{a}_{n} in a certain space such that the angle between any pair of these vectors is obtuse.2. Direct the z-axis of the coordinate system along the vector vec{a}_{n}. Consequently, the z-coordinate of all the remaining vectors must be negative (this follows from the formula for the cosine of the angle between two vectors using their scalar product).3. Let [ vec{a}_{1} = (x_{1}, y_{1}, z_{1}), ldots, vec{a}_{n-1} = (x_{n-1}, y_{n-1}, z_{n-1}) ] be the coordinates of the given vectors, and let [ vec{b}_{1} = (x_{1}, y_{1}), ldots, vec{b}_{n-1} = (x_{n-1}, y_{n-1}) ] be the projections of these vectors onto the xy-plane.4. The angle between any pair of the vectors vec{a}_{i} and vec{a}_{j} is obtuse; hence, the same holds for their projections on the xy plane. Particularly, [ cos(theta_{ij}) = frac{x_{i}x_{j} + y_{i}y_{j}}{left|vec{b}_{i}right| left|vec{b}_{j}right|} < 0 ] because [ x_{i}x_{j} + y_{i}y_{j} + z_{i}z_{j} < 0 ] (since z_{i} and z_{j} are negative, the dot product z_{i}z_{j} is positive and hence subtracts from the left-hand side).5. On the xy-plane, the vectors vec{b}_{1}, ldots, vec{b}_{n-1} also form obtuse angles between one another.6. We need to show that n-1 leq 3. To do this, align vectors from the same origin and sum the angles between neighboring vectors around the circle. Their sum must be 360^circ, and given each angle is obtuse (greater than 90^circ), it can be shown that no more than three such angles can sum to 360^circ.Therefore, n-1 leq 3, and solving this inequality gives: [ n leq 4 ] 7. An example to illustrate when there are exactly 4 vectors is given by: [ vec{a}_1 = (1, 0, -epsilon), vec{a}_2 = left(cos frac{2pi}{3}, sin frac{2pi}{3}, -epsilonright), vec{a}_3 = left(cos frac{4pi}{3}, sin frac{4pi}{3}, -epsilonright), vec{a}_4 = (0, 0, 1) ] where epsilon is a sufficiently small positive number, e.g., epsilon=0.1. Conclusion:(boxed{4})

🧠:Okay, so I have this polynomial problem here. Let me try to figure it out step by step. The problem states that there's a polynomial ( p(x) ) of degree ( 2n ). The conditions given are that it has zeros at all the even integers from 0 up to ( 2n ). So, ( p(0) = p(2) = cdots = p(2n) = 0 ). Then, at all the odd integers from 1 up to ( 2n-1 ), the polynomial takes the value 2. So, ( p(1) = p(3) = cdots = p(2n-1) = 2 ). Finally, when we plug in ( 2n+1 ) into the polynomial, we get ( -30 ). The goal is to find the value of ( n ).First, since the polynomial has roots at 0, 2, 4, ..., up to ( 2n ), which are ( n+1 ) points, I can express ( p(x) ) in terms of these roots. Since it's a polynomial of degree ( 2n ), which is even, and it has ( n+1 ) roots, each of multiplicity 1, but the degree is ( 2n ), that suggests that maybe there's another factor involved. Wait, hold on. If there are ( n+1 ) distinct roots, each contributing a linear factor, then the polynomial would be of degree ( n+1 ). But since the polynomial is of degree ( 2n ), there must be another factor. Maybe a quadratic factor or some higher degree factor. Hmm, but the problem doesn't specify any other roots. Maybe all the roots are the even integers from 0 to ( 2n ), but each with multiplicity 2? That would make the degree ( 2(n+1) ). Wait, but the given polynomial is degree ( 2n ). So that would be conflicting.Wait, let me check again. The roots are at 0, 2, 4, ..., 2n. That's ( n+1 ) roots. Each root is a single root, so the polynomial would be of degree ( n+1 ). But the problem says it's degree ( 2n ). Therefore, there's a contradiction here. Unless the polynomial has those roots but with some multiplicities. Maybe each root is a double root? If each root is a double root, then the total degree would be ( 2(n+1) ). But again, the polynomial is given as degree ( 2n ). So that's still a problem. Hmm.Wait, perhaps I need to consider that the polynomial has roots at 0, 2, ..., 2n, but the degree is ( 2n ). That would mean there are ( 2n+1 ) roots? But no, 0, 2, ..., 2n is ( n+1 ) roots. So unless there are multiplicities. Wait, if each root except for one has multiplicity 1 and one root has multiplicity ( 2n - n = n ). But that seems arbitrary. Maybe there's a different approach.Alternatively, maybe the polynomial is constructed as ( p(x) = q(x) cdot prod_{k=0}^{n} (x - 2k) ), where ( q(x) ) is another polynomial. Since ( p(x) ) is degree ( 2n ), and the product ( prod_{k=0}^{n} (x - 2k) ) is degree ( n+1 ), so ( q(x) ) must be a polynomial of degree ( 2n - (n+1) = n - 1 ). That makes sense. So ( p(x) = left( prod_{k=0}^{n} (x - 2k) right) cdot q(x) ), where ( q(x) ) is a degree ( n - 1 ) polynomial.Then, the other conditions are that at each odd integer ( 1, 3, ..., 2n-1 ), the polynomial ( p(x) ) equals 2. So, for each ( m = 1, 3, ..., 2n-1 ), we have ( p(m) = 2 ). Also, ( p(2n+1) = -30 ).So, since ( p(x) = left( prod_{k=0}^{n} (x - 2k) right) cdot q(x) ), then for each odd integer ( m ), we have:( 2 = p(m) = left( prod_{k=0}^{n} (m - 2k) right) cdot q(m) ).Similarly, ( -30 = p(2n+1) = left( prod_{k=0}^{n} (2n+1 - 2k) right) cdot q(2n+1) ).So, if I can find ( q(x) ), then I can use these equations to find ( n ).Since ( q(x) ) is a polynomial of degree ( n - 1 ), and we have ( n ) conditions (the values at the ( n ) odd integers ( 1, 3, ..., 2n-1 )), and one more condition at ( 2n+1 ). Wait, so in total, there are ( n + 1 ) conditions for ( q(x) ), which is a degree ( n - 1 ) polynomial. That suggests that if we can set up a system of equations, but perhaps there's a smarter way.Alternatively, since ( q(x) ) is a degree ( n - 1 ) polynomial, and we know its values at ( n ) points (the odd integers from 1 to ( 2n - 1 )), we can use interpolation to find ( q(x) ). Once we have ( q(x) ), we can use the additional condition at ( 2n + 1 ) to find ( n ).So let me try to proceed.First, let's define ( Q(x) = q(x) ). Then, ( Q(m) = frac{2}{prod_{k=0}^{n} (m - 2k)} ) for each odd ( m ).But since ( Q(x) ) is a polynomial of degree ( n - 1 ), and we have ( n ) points, we can use Lagrange interpolation to express ( Q(x) ).Alternatively, note that the product ( prod_{k=0}^{n} (x - 2k) ) evaluated at ( x = m ), where ( m ) is odd, would be the product of ( (m - 0)(m - 2)(m - 4)...(m - 2n) ). Since ( m ) is an odd integer between 1 and ( 2n - 1 ), let's think about this product.For example, let's take ( n = 1 ). Then, the roots are 0, 2. So the product is ( x(x - 2) ). Then, for ( x = 1 ), the product is ( 1 times (-1) = -1 ). Then, ( Q(1) = 2 / (-1) = -2 ). Then, the polynomial ( Q(x) ) is a degree ( 0 ) polynomial (constant), so ( Q(x) = -2 ). Then, ( p(x) = -2x(x - 2) ). Then, check ( p(3) ). But ( 3 is beyond the roots (0 and 2), and ( p(3) = -2*3*(3 - 2) = -6 ), which is not 2. So ( n=1 ) is not a solution.Wait, but the problem states ( p(2n+1) = -30 ). For ( n=1 ), ( 2n+1 = 3 ), and we just saw ( p(3) = -6 ne -30 ). So ( n=1 ) is invalid. Let's try ( n=2 ).For ( n=2 ), degree of polynomial is 4. The roots are 0, 2, 4. So, ( p(x) = x(x - 2)(x - 4)Q(x) ), where Q(x) is degree ( 2 - 1 = 1 ). So Q(x) is linear. Then, the polynomial must satisfy p(1) = p(3) = 2. Let's compute.First, compute the product ( prod_{k=0}^{2} (m - 2k) ) at m=1: (1 - 0)(1 - 2)(1 - 4) = 1*(-1)*(-3) = 3. So Q(1) = 2 / 3. Similarly, at m=3: (3 - 0)(3 - 2)(3 - 4) = 3*1*(-1) = -3. So Q(3) = 2 / (-3) = -2/3.So Q(x) is a linear polynomial passing through (1, 2/3) and (3, -2/3). The slope is (-2/3 - 2/3)/(3 - 1) = (-4/3)/2 = -2/3. So the equation is Q(x) = -2/3*(x - 1) + 2/3 = -2/3 x + 2/3 + 2/3 = -2/3 x + 4/3. Wait, let's check:Using the two points (1, 2/3) and (3, -2/3):The linear polynomial can be written as Q(x) = a x + b.At x=1: a + b = 2/3.At x=3: 3a + b = -2/3.Subtracting the first equation from the second: 2a = -4/3 => a = -2/3.Then, from the first equation: -2/3 + b = 2/3 => b = 4/3.So Q(x) = -2/3 x + 4/3.Therefore, the polynomial is p(x) = x(x - 2)(x - 4)(-2/3 x + 4/3).Simplify: Let's factor out -2/3:p(x) = -2/3 x(x - 2)(x - 4)(x - 2).Wait, wait, Q(x) is -2/3 x + 4/3, which is -2/3 (x - 2). So:p(x) = x(x - 2)(x - 4) * (-2/3)(x - 2) = -2/3 x(x - 2)^2(x - 4).So p(x) is a quartic polynomial. Now, let's check p(5) since 2n+1 = 5 when n=2.p(5) = -2/3 *5*(5 - 2)^2*(5 - 4) = -2/3 *5*9*1 = -2/3 *45 = -30. Hey, that's exactly the condition given! So p(5) = -30. So for n=2, all conditions are satisfied.Wait, so maybe the answer is n=2. But let me check if there's another possible n.But before that, let's just recap. For n=2, the polynomial p(x) is constructed as above, and it satisfies all the given conditions. So that seems to work. Let me check n=3 to see if maybe that also works, but probably not.But maybe I should check with n=3.So, n=3. Then, the polynomial p(x) is of degree 6. The roots are 0,2,4,6. So p(x) = x(x - 2)(x - 4)(x - 6)Q(x), where Q(x) is a polynomial of degree 2 (since 6 - 4 = 2; wait, no, wait: original polynomial is degree 2n=6, the product x(x - 2)(x - 4)(x - 6) is degree 4, so Q(x) must be degree 2.Then, the conditions are p(1)=p(3)=p(5)=2 and p(7) = -30.So first, compute Q(1), Q(3), Q(5).At x=1:Product = 1*(1 - 2)*(1 - 4)*(1 - 6) = 1*(-1)*(-3)*(-5) = -15. So Q(1) = 2 / (-15) = -2/15.At x=3:Product = 3*(3 - 2)*(3 - 4)*(3 - 6) = 3*1*(-1)*(-3) = 9. So Q(3) = 2 / 9.At x=5:Product = 5*(5 - 2)*(5 - 4)*(5 - 6) = 5*3*1*(-1) = -15. So Q(5) = 2 / (-15) = -2/15.So Q(x) is a quadratic polynomial passing through (1, -2/15), (3, 2/9), (5, -2/15). Let's see if we can find such a polynomial.Let me set up the equations. Let Q(x) = a x^2 + b x + c.At x=1: a + b + c = -2/15.At x=3: 9a + 3b + c = 2/9.At x=5: 25a + 5b + c = -2/15.Now, subtract the first equation from the second:(9a + 3b + c) - (a + b + c) = 8a + 2b = 2/9 - (-2/15) = 2/9 + 2/15 = (10/45 + 6/45) = 16/45.Similarly, subtract the second equation from the third:(25a + 5b + c) - (9a + 3b + c) = 16a + 2b = -2/15 - 2/9 = (-2/15 - 10/45) = (-6/45 - 10/45) = -16/45.So now we have:8a + 2b = 16/45,16a + 2b = -16/45.Subtract the first equation from the second:8a = (-16/45 - 16/45) = -32/45 => a = -4/45.Then, substitute a = -4/45 into 8a + 2b = 16/45:8*(-4/45) + 2b = 16/45 => -32/45 + 2b = 16/45 => 2b = 48/45 => b = 24/45 = 8/15.Then, substitute a and b into the first equation:(-4/45) + (8/15) + c = -2/15.Convert to common denominator:(-4/45) + (24/45) + c = (-6/45).Thus, 20/45 + c = -6/45 => c = -26/45.So Q(x) = (-4/45)x^2 + (8/15)x - 26/45.Now, check if this quadratic passes through all three points:At x=1: (-4/45) + (8/15) -26/45 = (-4 -26)/45 + 8/15 = (-30)/45 + 24/45 = (-6)/45 = -2/15. Correct.At x=3: (-4/45)*9 + (8/15)*3 -26/45 = (-36/45) + (24/15) -26/45 = (-36/45 -26/45) + 72/45 = (-62/45) + 72/45 = 10/45 = 2/9. Correct.At x=5: (-4/45)*25 + (8/15)*5 -26/45 = (-100/45) + (40/15) -26/45 = (-100 -26)/45 + 120/45 = (-126/45) + 120/45 = (-6)/45 = -2/15. Correct.So Q(x) is correctly determined. Now, compute p(7) where 2n+1=7 (since n=3):p(7) = 7*(7 - 2)*(7 - 4)*(7 - 6)*Q(7).Compute each term:7*5*3*1 = 105.Q(7) = (-4/45)*49 + (8/15)*7 -26/45 = (-196/45) + (56/15) -26/45.Convert to 45 denominator:= (-196/45) + (168/45) -26/45 = (-196 + 168 -26)/45 = (-54)/45 = -6/5.Thus, p(7) = 105 * (-6/5) = -126. But the problem states p(2n+1) = -30. So here, p(7) = -126 ≠ -30. So n=3 does not satisfy the condition.Therefore, n=3 is invalid. Let's check n=2 worked. So perhaps the answer is n=2.Wait, but just to be thorough, let me check n=4. Maybe the pattern continues.But before that, let's check n=0. Wait, n=0 would mean the polynomial is degree 0, but with roots at 0? That doesn't make sense. So n must be at least 1.But n=1 we saw didn't work. n=2 worked. n=3 didn't. Maybe n=2 is the answer.Alternatively, perhaps there's a general approach. Let's try to think about the general case.Given that p(x) is a degree 2n polynomial with roots at 0,2,...,2n, so:p(x) = C x(x - 2)(x - 4)...(x - 2n).Wait, but this product is of degree n+1. Therefore, to make p(x) degree 2n, we need another factor of degree 2n - (n+1) = n-1, which is q(x). So p(x) = q(x) * x(x - 2)...(x - 2n).Now, the conditions at the odd integers give us that for each m = 1,3,...,2n-1, p(m)=2. So:q(m) = 2 / [m(m - 2)(m - 4)...(m - 2n)].Similarly, at x=2n+1, p(2n+1) = -30. So:q(2n+1) = -30 / [(2n+1)(2n+1 - 2)...(2n+1 - 2n)].So, we can model q(x) as the interpolating polynomial that takes those values at the odd integers and also at x=2n+1. Wait, but q(x) is of degree n-1, and we have n points (the odd integers) plus one more point at 2n+1. So in total, n+1 points. Since q(x) is degree n-1, this is over-constrained unless the additional condition is compatible.But in our earlier example with n=2, q(x) was degree 1, and we had 2 points (n=2, so n=2 odd integers: 1 and 3) plus x=5. But in that case, the system was solvable because the value at x=5 gave a consistent equation. So for n=2, with q(x) degree 1, the three points (1,3,5) must lie on a linear polynomial. But in reality, we only used the two points (1 and 3) to determine q(x), and then checked the third point (5) which worked out. So perhaps, for n=2, it's a coincidence. For higher n, the over-constrained system may not have a solution unless the value at 2n+1 is compatible with the interpolating polynomial.Therefore, the key is that for the correct n, the value at 2n+1 will match -30, given the interpolating polynomial q(x). So how do we find n in general?Alternatively, perhaps we can model q(x) as a constant multiple, but given that q(x) is a polynomial, maybe we can use the concept of constructing q(x) such that it takes specific values at those points.Alternatively, since q(x) is a polynomial of degree n−1, and we have n conditions (the odd points), which uniquely defines q(x). Then, the condition at x=2n+1 must also hold, which will fix the value of n. So perhaps there's a general formula.Alternatively, notice that in the case n=2, the value of q(x) at x=5 was -2/3 *5 + 4/3 = (-10 +4)/3 = -6/3 = -2. Then, p(5) = product * q(5) = 5*3*1*(-1) * (-2) = 5*3*1*(-1)*(-2) = 30. Wait, but earlier calculation said p(5) was -30. Wait, maybe I miscalculated.Wait, no. Wait, let me recheck:For n=2, p(x) = x(x - 2)(x - 4) * Q(x), where Q(x) is linear. Wait, earlier, we had Q(x) = -2/3 x + 4/3, so Q(5) = -2/3 *5 + 4/3 = (-10 +4)/3 = -6/3 = -2.Then, p(5) = 5*(5 - 2)*(5 - 4)*Q(5) = 5*3*1*(-2) = -30. Yes, that's correct.But in this case, the product at x=5 is 5*3*1 = 15, multiplied by Q(5)=-2 gives -30. So for general n, p(2n+1) = [product from k=0 to n of (2n+1 - 2k)] * Q(2n+1) = -30.But Q(2n+1) is determined by the interpolating polynomial Q(x) which satisfies Q(m) = 2 / [product from k=0 to n of (m - 2k)] for each odd m from 1 to 2n-1.This seems complicated. Maybe there's a better approach.Alternatively, consider the polynomial p(x) - 2. It has zeros at all the odd integers from 1 to 2n-1, since p(m) - 2 = 0 for m=1,3,...,2n-1. But p(x) is degree 2n, so p(x) - 2 is also degree 2n, and it has n zeros. However, unless multiplicities are involved, this would not account for all roots. But maybe we can factor p(x) - 2 as some product involving the odd terms.But p(x) is already constructed as having roots at the even integers, so combining these two might lead to a system.Alternatively, since p(x) has roots at 0,2,...,2n, we can write p(x) = x(x - 2)(x - 4)...(x - 2n) * Q(x), where Q(x) is a degree n - 1 polynomial. Then, p(x) - 2 has roots at 1,3,...,2n-1. Therefore,x(x - 2)...(x - 2n) * Q(x) - 2 = (x - 1)(x - 3)...(x - (2n - 1)) * S(x),where S(x) is another polynomial. But this might not be straightforward.Alternatively, consider that the polynomial p(x) interpolates the given values. Since it's a polynomial of degree 2n, passing through 2n+1 points. However, we have n+1 roots (0,2,...,2n) and n additional points (1,3,...,2n-1) where it equals 2, plus the point at 2n+1. So total points: (n+1) + n +1 = 2n +2 points. Wait, but a polynomial of degree 2n is determined uniquely by 2n+1 points. Therefore, having 2n+2 conditions would usually be over-constrained. But in our problem, the roots are given as 0,2,...,2n, which are n+1 points, and then the values at n+1 other points (1,3,...,2n-1,2n+1). Therefore, total of 2n+2 conditions. But a degree 2n polynomial has 2n+1 coefficients, so unless the conditions are dependent, there might be no solution. However, the problem states that such a polynomial exists, so the conditions must be compatible. Therefore, the key is that with the given roots and the given values, there's a unique n for which this is possible.In the case of n=2, we found that it works. Let's see if this is the only solution.Alternatively, think about the structure of the polynomial. Since p(x) has zeros at 0,2,...,2n, we can write:p(x) = x(x - 2)(x - 4)...(x - 2n) * Q(x)Then, at x=1,3,...,2n-1, we have:2 = p(m) = m(m - 2)(m - 4)...(m - 2n) * Q(m)Therefore, Q(m) = 2 / [m(m - 2)(m - 4)...(m - 2n)]Similarly, at x=2n+1:-30 = (2n+1)(2n+1 - 2)(2n+1 - 4)...(2n+1 - 2n) * Q(2n+1)Note that the product (2n+1 - 2k) for k from 0 to n is:When k=0: 2n+1 -0 =2n+1k=1: 2n+1 -2=2n-1k=2: 2n+1 -4=2n-3...k=n: 2n+1 -2n=1So the product is (2n+1)(2n -1)(2n -3)...1 = (2n+1)!!Therefore, the product term at x=2n+1 is (2n+1)!!, so:-30 = (2n+1)!! * Q(2n+1)Hence, Q(2n+1) = -30 / (2n+1)!!But Q(x) is a polynomial of degree n -1 determined by its values at the n points m=1,3,...,2n-1.So perhaps there's a way to relate Q(2n+1) to these values.Alternatively, consider that Q(x) is the interpolating polynomial passing through the points (m, 2 / [m(m - 2)...(m - 2n)]) for m=1,3,...,2n-1.Therefore, using Lagrange interpolation, Q(x) can be written as:Q(x) = sum_{j=1}^{n} frac{2}{prod_{k=0}^{n} (m_j - 2k)} cdot prod_{substack{i=1 i neq j}}^{n} frac{x - m_i}{m_j - m_i}where ( m_j = 2j -1 ).But this seems quite complicated. Maybe we can use the concept of finite differences or another method.Alternatively, think of Q(x) as a function defined on the odd integers and extended to a polynomial. But I'm not sure.Alternatively, notice that for n=2, we found that Q(5) = -2, and (2n+1)!! = 5!! = 5*3*1=15. So -30 = 15*(-2). Similarly, for n=3, we saw that Q(7) = -6/5, and (2n+1)!!=7!!=7*5*3*1=105. So -30 = 105*(-6/5)= -126, which didn't work. But in the case of n=2, it worked.So perhaps the relation is that -30 = (2n+1)!! * Q(2n+1). For n=2, 15*(-2) = -30. So Q(2n+1) must be -2. But how is Q(2n+1) related to the other values?In the case of n=2, Q(x) was linear, so Q(5) = -2, which matched. For n=3, Q(x) is quadratic, and Q(7) = -6/5. But how do we express Q(2n+1) in terms of n?Alternatively, notice that in the case of n=2, the product terms in the denominator for each Q(m) were:At m=1: product was 1*(-1)*(-3) = 3, so Q(1) = 2/3.At m=3: product was 3*1*(-1) = -3, so Q(3) = -2/3.Then, Q(x) was linear, passing through (1, 2/3) and (3, -2/3), leading to Q(5) = -2.Similarly, for general n, perhaps there's a pattern in the product terms.Notice that for each odd m=2j-1, the product ( prod_{k=0}^n (m - 2k) ) can be written as ( prod_{k=0}^n (2j -1 - 2k) ).Let's substitute i = j -k. When k=0: i=j; k=1: i=j-1; ... k=n: i=j -n.But not sure.Alternatively, the product is:For m=2j-1 (an odd number between 1 and 2n-1), the product becomes:(2j -1)(2j -1 -2)(2j -1 -4)...(2j -1 -2n).= (2j -1)(2j -3)...(2j -1 -2n).This product can be split into two parts: terms where 2j -1 -2k >0 and terms where it is <0. But perhaps there's a factorial involved.Alternatively, note that the product is similar to a double factorial.For example, when m=1 (j=1), the product is 1*(-1)*(-3)*...*(-(2n -1)).Which is 1*(-1)*(-3)*...*(-(2n -1)) = (-1)^n * (1*3*5*...*(2n -1)) = (-1)^n * (2n -1)!!.Similarly, when m=3 (j=2), the product is 3*1*(-1)*...*(-(2n -3)) = 3*1*(-1)*...*(-(2n -3)).Which is 3*1*(-1)^{n-1} * (1*3*5*...*(2n -3)) = (-1)^{n-1} * 3!! * (2n -3)!! ? Not sure.Alternatively, let's compute the product for general m=2j-1:The terms are (2j-1), (2j-3), (2j-5), ..., (2j-1-2n).This is an arithmetic sequence decreasing by 2, starting at 2j-1 and having n+1 terms.The product is the product of an arithmetic sequence. If we can express this in terms of factorials or double factorials.But perhaps not straightforward. Alternatively, note that the product will include both positive and negative terms, depending on the value of j relative to k.Alternatively, notice that when m=2j-1 is in the middle of the roots 0,2,...,2n, the product will have some symmetry.But maybe it's easier to consider specific cases.For n=2 and m=1:Product is 1*(1-2)*(1-4)*(1-6)=1*(-1)*(-3)*(-5)= -15.Wait, but earlier I thought it was 1*(-1)*(-3) for n=2, but actually, for n=2, the roots are 0,2,4, so wait, no. Wait, original problem states that p(x) is degree 2n, so for n=2, roots are 0,2,4, right? But wait, in the problem statement, it's written that p(0)=p(2)=...=p(2n)=0. So for n=2, 2n=4, so roots at 0,2,4. So the product for x=1 is (1-0)(1-2)(1-4) = 1*(-1)*(-3)=3. So Q(1)=2/3. For x=3: (3-0)(3-2)(3-4)=3*1*(-1)=-3. Q(3)= -2/3. Then, the polynomial Q(x) is linear, and at x=5, which is 2n+1=5, Q(5)= -2.Therefore, for general n, perhaps Q(2n+1) can be found by extrapolating the polynomial Q(x) from the points m=1,3,...,2n-1.Since Q(x) is a polynomial of degree n-1, and we're evaluating it at x=2n+1, there might be a formula for extrapolation.Alternatively, consider that the values of Q(m) alternate between positive and negative, and perhaps the extrapolation to x=2n+1 follows a certain pattern.But this is getting too vague. Let's try to consider the general case.Given that Q(x) is a polynomial of degree n−1 interpolating the points (m, 2 / D(m)), where D(m) = prod_{k=0}^n (m - 2k) for m=1,3,...,2n−1.Then, to find Q(2n+1), we can use the Lagrange interpolation formula:Q(2n+1) = sum_{j=1}^n [2 / D(m_j)] * prod_{substack{i=1 i neq j}}^n (2n+1 - m_i) / (m_j - m_i)But this formula is complex. Maybe there's a pattern or simplification.Alternatively, notice that D(m) = prod_{k=0}^n (m - 2k) = (m)(m - 2)...(m - 2n). For m=2j−1, this product is:(2j−1)(2j−3)...(2j−1−2n). Let's denote this as P(j).Let’s substitute variables for clarity. Let’s set t = j - (k + 1). Not sure.Alternatively, note that the product P(j) can be expressed as:For m=2j−1, which is an odd integer between 1 and 2n−1, the terms in P(j) are:2j−1, 2j−3, ..., 2j−1−2k, ..., 2j−1−2n.The last term is 2j−1−2n = 2(j - n) -1.Depending on the value of j, this product will have a certain number of negative terms.Alternatively, perhaps there's a relationship between Q(2n+1) and the product terms.But this seems too abstract. Let's try with n=2 again to see the pattern.For n=2, Q(5) = -2. Then, (2n+1)!! = 5!! = 15. So -30 = 15*(-2). So Q(2n+1) = -2. Similarly, for n=1, though it didn't work, let's see:For n=1, polynomial is quadratic. Roots at 0,2. Then p(x) = x(x - 2)Q(x), Q(x) is a constant (degree 0). Then, p(1) = 1*(-1)*Q(1) = -Q(1) = 2 ⇒ Q(1) = -2. So Q(x) = -2 for all x. Then p(3) = 3*1*(-2) = -6, which is not 2. But Q(3) is still -2, since it's constant. Then p(3) = -6, which doesn't fit. So n=1 doesn't work. But if we consider the value at x=3 (2n+1=3 for n=1), p(3) = -6, but the problem requires p(2n+1)=-30. So even if we tried, n=1 doesn't work.But for n=2, it worked. Let's try to generalize.Suppose that for general n, Q(2n+1) = (-1)^{n} * 2. Then, p(2n+1) = (2n+1)!! * (-1)^n * 2 = -30.Therefore, (2n+1)!! * (-1)^n * 2 = -30 ⇒ (2n+1)!! * (-1)^n = -15 ⇒ (2n+1)!! * (-1)^n = -15.Since (2n+1)!! is positive for all n, so (-1)^n must be negative, which requires n odd. But 15 is 5*3*1=15. So (2n+1)!!=15 implies 2n+1=5 ⇒n=2. Which matches our previous result. So this suggests that the only solution is n=2.Therefore, the answer is n=2.Final AnswerThe value of ( n ) is boxed{2}.

✅:Given a polynomial ( p(x) ) of degree ( 2n ) with roots at ( 0, 2, 4, ldots, 2n ), and values ( p(1) = p(3) = cdots = p(2n-1) = 2 ) and ( p(2n+1) = -30 ), we need to find ( n ).1. Roots and Polynomial Structure: - The polynomial ( p(x) ) has roots at ( 0, 2, 4, ldots, 2n ), so it can be written as: [ p(x) = x(x - 2)(x - 4) cdots (x - 2n) cdot q(x) ] - Here, ( q(x) ) is a polynomial of degree ( n - 1 ).2. Conditions at Odd Integers: - For each odd integer ( m = 1, 3, 5, ldots, 2n-1 ), we have: [ 2 = p(m) = m(m - 2)(m - 4) cdots (m - 2n) cdot q(m) ] - Solving for ( q(m) ): [ q(m) = frac{2}{m(m - 2)(m - 4) cdots (m - 2n)} ]3. Condition at ( 2n+1 ): - For ( x = 2n+1 ): [ -30 = p(2n+1) = (2n+1)(2n-1)(2n-3) cdots 1 cdot q(2n+1) ] - The product ( (2n+1)(2n-1)(2n-3) cdots 1 ) is the double factorial ( (2n+1)!! ).4. Solving for ( n ): - Using the value at ( 2n+1 ): [ -30 = (2n+1)!! cdot q(2n+1) ] - For ( n = 2 ): - The polynomial ( p(x) = x(x - 2)(x - 4) cdot q(x) ) with ( q(x) ) determined by interpolation. - The product ( (2n+1)!! = 5!! = 15 ). - Solving for ( q(5) ): [ q(5) = frac{-30}{15} = -2 ] - This satisfies all conditions, confirming ( n = 2 ).5. Verification for Other ( n ): - Testing ( n = 3 ) and higher values shows inconsistencies with the given conditions. - Only ( n = 2 ) satisfies all conditions.Thus, the value of ( n ) is (boxed{2}).

🔑:To find ( n ), let's delve into the details provided and use the relationship given by the polynomial properties and difference operator. Here is the step-by-step solution:1. We know that ( p(x) ) is a polynomial of degree ( 2n ) and the provided conditions for ( p(x) ) are: [ p(0) = p(2) = cdots = p(2n) = 0 ] [ p(1) = p(3) = cdots = p(2n-1) = 2 ] [ p(2n+1) = -30 ]2. Consider the finite difference (Delta^{2n+1} p(0)). This operator is formulated based on the alternating sum of polynomial evaluations: [ Delta^{2n+1} p(0) = sum_{k=0}^{2n+1} (-1)^k binom{2n+1}{k} p(k) ]3. Substituting the given values of ( p(k) ): [ Delta^{2n+1} p(0) = sum_{k=0}^{2n+1} (-1)^k binom{2n+1}{k} p(2n+1 - k) ]4. We replace ( p(2n+1 - k) ) with the given values. Only ( p(2n+1) ) is (-30) and for the other terms: [ Delta^{2n+1} p(0) = (-1)^{0} binom{2n+1}{0}(-30) + sum_{j=1}^{n} binom{2n+1}{2j} (2) ]5. We observe that the terms ( p(2j) = 0 ) and ( p(2j-1) = 2 ): [ Delta^{2n+1} p(0) = -30 + 2 sum_{j=1}^{n} binom{2n+1}{2j} ]6. The critical point here is the sum of the binomial coefficients for even indices: [ sum_{j=1}^{n} binom{2n+1}{2j} ] This sum can be derived from the binomial expansion properties: [ sum_{k=0}^{2n+1} binom{2n+1}{k} = 2^{2n+1} quad text{and} quad binom{2n+1}{1} + binom{2n+1}{3} + cdots binom{2n+1}{2n+1} = 2^{2n} quad (text{sum of odd terms}) ]7. Thus, the remaining part for even indices: [ sum_{k=0, 2, 4, ldots, 2n}^{2(2n+1)} binom{2n+1}{k} = 2^{2n} ] Taking half without the zero term: [ sum_{k=2, 4, 6, ldots, 2n} binom{2n+1}{k} = frac{1}{2} (2^{2n+1} - 2) ]8. Substituting back, we have: [ (-30) + 2 left(frac{1}{2} cdot (2^{2n+1} - 2) right) = 0 ]9. Simplifying the equation: [ (-30) + (2^{2n+1} - 2) = 0 ] [ 2^{2n+1} - 2 - 30 = 0 ] [ 2^{2n+1} = 32 ]10. Solving ( 2^{2n+1} = 32 ): [ 32 = 2^5 ] Therefore, we equate exponents: [ 2n+1 = 5 ] Solving for ( n ): [ 2n = 4 quad Rightarrow quad n = 2 ] Conclusion:[boxed{2}]

🧠:Okay, so I need to figure out for which integers ( n geq 1 ) the expression ( 3^{2n} - 2^n ) is a prime number. Hmm, let's start by understanding the problem. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself. So, I need to check for each ( n geq 1 ), compute ( 3^{2n} - 2^n ), and see if the result is prime.First, let's rewrite the expression ( 3^{2n} ). Since ( 3^{2n} = (3^2)^n = 9^n ), so the expression simplifies to ( 9^n - 2^n ). That might make it a bit easier to handle. So now, we have ( 9^n - 2^n ).Maybe I should start by checking small values of ( n ) first. Let's compute the expression for ( n = 1, 2, 3, ) and so on, and see if the result is prime.Starting with ( n = 1 ):( 9^1 - 2^1 = 9 - 2 = 7 ). 7 is a prime number. So, n=1 works.Next, ( n = 2 ):( 9^2 - 2^2 = 81 - 4 = 77 ). 77 is 7 times 11, which is not prime. So, n=2 doesn't work.Then, ( n = 3 ):( 9^3 - 2^3 = 729 - 8 = 721 ). Let me check if 721 is prime. Hmm, 721 divided by 7 is 103 (since 7*103=721). So, 721 is composite.Moving on to ( n = 4 ):( 9^4 - 2^4 = 6561 - 16 = 6545 ). Is 6545 prime? Let's check divisibility. It ends with a 5, so it's divisible by 5. 6545 /5 = 1309. So, composite.n=5: ( 9^5 - 2^5 = 59049 - 32 = 59017 ). Check if 59017 is prime. Not sure. Let's see. Let's check divisibility by small primes. Divided by 2? No, it's odd. Divided by 3: 5+9+0+1+7=22, 22 isn't divisible by 3. Divided by 5? Ends with 7, so no. 7? Let's do 59017 divided by 7. 7*8431=59017? 7*8000=56000, 7*400=2800, 7*31=217. So, 56000+2800+217=59017. Wow, so 7*8431=59017. Therefore, composite.n=6: ( 9^6 - 2^6 = 531441 - 64 = 531377 ). Checking for primality. Again, it's even? No. Sum of digits: 5+3+1+3+7+7=26, not divisible by 3. Ends with 7, not divisible by 5. Let's check 7: 531377 divided by 7. 7*75911=531377? 7*70000=490000, 7*5000=35000, 7*591=4137. So, 490000 + 35000 = 525000, plus 4137 = 529137. Hmm, that's less than 531377. Wait, maybe my calculation is off. Alternatively, use a calculator approach: 531377 ÷7. 7*75000=525000. 531377 - 525000 = 6377. 6377 ÷7=911. So, 7*75000 +7*911=7*(75000+911)=7*75911=531377. So yes, divisible by 7. Therefore, composite.n=7: ( 9^7 - 2^7 = 4782969 - 128 = 4782841 ). Check if this is prime. Let's try dividing by small primes. It's odd, not divisible by 2. Sum of digits: 4+7+8+2+8+4+1=34, not divisible by 3. Doesn't end with 5. Check divisibility by 7: 4782841 ÷7. 7*680000=4760000. 4782841 -4760000=22841. 22841 ÷7≈3263, since 7*3000=21000, 22841-21000=1841. 1841 ÷7=263. So, total is 680000 + 3000 +263=683263. 7*683263=4782841. Therefore, composite.Hmm, so up to n=7, only n=1 gives a prime. Let's check n=8 just in case.n=8: ( 9^8 - 2^8 = 43046721 - 256 = 43046465 ). Check if prime. Ends with 5, so divisible by 5. Therefore, composite.n=9: ( 9^9 - 2^9 = 387420489 - 512 = 387419977 ). Checking primality. This is a large number. Let's see if it's even? No. Sum of digits: 3+8+7+4+1+9+9+7+7= 3+8=11, +7=18, +4=22, +1=23, +9=32, +9=41, +7=48, +7=55. 55 isn't divisible by 3. Not ending with 5. Check divisibility by 7: 387419977 ÷7. Let's compute 7*55,000,000=385,000,000. 387,419,977 -385,000,000=2,419,977. 2,419,977 ÷7. 7*345,000=2,415,000. 2,419,977 -2,415,000=4,977. 4,977 ÷7=711. So total is 55,000,000 +345,000 +711=55,345,711. 7*55,345,711=387,419,977. So, divisible by 7, hence composite.n=10: ( 9^{10} - 2^{10} = 3486784401 - 1024 = 3486783377 ). Check if prime. Again, large number. Let's check divisibility by small primes. It's odd. Sum digits: 3+4+8+6+7+8+3+3+7+7= Let's compute: 3+4=7, +8=15, +6=21, +7=28, +8=36, +3=39, +3=42, +7=49, +7=56. 56 isn't divisible by 3. Ends with 7, not 5. Check 7: 3486783377 ÷7. 7*498,111,900=3,486,783,300. 3,486,783,377 -3,486,783,300=77. 77 ÷7=11. So total is 498,111,900 +11=498,111,911. Therefore, 7*498,111,911=3,486,783,377. Hence, divisible by 7. Composite.Hmm, so far only n=1 gives a prime. Let's think if there's a mathematical reason why for n>1, ( 9^n - 2^n ) is composite. Maybe using factorization.Recall that ( a^n - b^n ) can be factored if n is composite, but even if n is prime, there is a factorization:( a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + dots + ab^{n-2} + b^{n-1}) ).So, for any n, ( 9^n - 2^n ) factors as ( (9 - 2)(9^{n-1} + 9^{n-2}*2 + dots + 9*2^{n-2} + 2^{n-1}) ).Wait, that's true. So, ( 9^n - 2^n = 7 * text{[some integer]} ). Therefore, unless that "some integer" is 1, the number is divisible by 7 and hence composite. So, when is ( frac{9^n - 2^n}{7} = 1 )? That would mean ( 9^n - 2^n =7 ), which only holds when n=1, since for n=1: 9-2=7. For n>1, 9^n - 2^n is much larger than 7, and divided by 7 gives a number greater than 1. Hence, for n>1, ( 9^n - 2^n =7 * k ), where k>1, so it's composite.Therefore, the only integer ( n geq1 ) for which ( 3^{2n} -2^n ) is prime is n=1.Wait, let me confirm this logic. The factorization ( a^n -b^n = (a -b)(a^{n-1} + ... + b^{n-1}) ). Since a=9, b=2, then ( 9 -2=7 ), so 7 is a factor. Therefore, unless the other factor is 1, which would require that ( 9^{n} -2^{n} =7*1=7 ), which only happens when n=1. For n≥2, the other factor is larger than 1, hence the number is composite.Therefore, the conclusion is that the only n≥1 where ( 3^{2n} -2^n ) is prime is n=1.But wait, let me check n=1 again. 3^(2*1) -2^1=9-2=7, which is prime. Correct.So, this seems like a solid argument. Since for n≥2, the expression factors into 7 times another integer greater than 1, hence composite. Therefore, only n=1.But wait, just to be thorough, suppose someone argues that maybe for some n, the other factor is 1, but that's only when 9^n -2^n=7. Let's solve 9^n -2^n=7.We can check for n=1: 9 -2=7. For n=2: 81 -4=77≠7. For n≥2, 9^n grows exponentially, while 2^n also grows but slower. So, 9^n -2^n will be way larger than 7 for n≥2. Hence, the equation 9^n -2^n=7 has only solution n=1.Therefore, the only n is 1.Thus, the answer is n=1.

✅:The expression ( 3^{2n} - 2^n ) simplifies to ( 9^n - 2^n ). To determine for which integers ( n geq 1 ) this value is prime, we analyze its factorization and evaluate specific cases:1. Factorization Insight: For any ( n geq 1 ), ( 9^n - 2^n ) can be factored as: [ 9^n - 2^n = (9 - 2)(9^{n-1} + 9^{n-2} cdot 2 + cdots + 9 cdot 2^{n-2} + 2^{n-1}) = 7 cdot k, ] where ( k = 9^{n-1} + 9^{n-2} cdot 2 + cdots + 2^{n-1} ). Since ( 7 ) is a factor, ( 9^n - 2^n ) is composite unless ( k = 1 ). This occurs only if ( 9^n - 2^n = 7 ), which uniquely holds for ( n = 1 ).2. Case Analysis: - ( n = 1 ): ( 9^1 - 2^1 = 7 ), which is prime. - ( n geq 2 ): For ( n = 2, 3, 4, ldots ), direct computation shows ( 9^n - 2^n ) results in composite numbers (e.g., ( 77, 721, 6545 )), all divisible by 7 or other primes. The factorization confirms these values cannot be prime.Conclusion: The only integer ( n geq 1 ) for which ( 3^{2n} - 2^n ) is prime is ( boxed{1} ).

🔑:1. We start with the given expression: [ 3^{2n} - 2^n = 9^n - 2^n ]2. To determine if this expression can be prime for any integer ( n geq 1 ), we need to check if ( 9^n - 2^n ) can be a prime number.3. First, we note a property of the expression modulo 7: [ 9 equiv 2 (text{mod} 7) ] Hence, [ 9^n equiv 2^n (text{mod} 7) ] Therefore, [ 9^n - 2^n equiv 2^n - 2^n equiv 0 (text{mod} 7) ]4. This shows that ( 9^n - 2^n ) is always divisible by 7.5. Next, we investigate if ( 9^n - 2^n ) can be exactly 7 when ( n geq 2 ). For this, we need to verify if ( 9^n - 2^n > 7 ) when ( n > 1 ). We approach this by mathematical induction: - Base Case: Let ( n = 2 ): [ 9^2 - 2^2 = 81 - 4 = 77 ] Clearly, ( 77 > 7 ). - Induction Hypothesis: Assume for some ( n geq 2 ): [ 9^n - 2^n > 7 ] - Induction Step: We need to show: [ 9^{n+1} - 2^{n+1} > 7 ] Starting from: [ 9^{n+1} = 9 cdot 9^n ] Using the induction hypothesis ( 9^n > 7 + 2^n ), we have: [ 9^{n+1} = 9 cdot 9^n geq 9 cdot (7 + 2^n) = 9 cdot 7 + 9 cdot 2^n = 63 + 9 cdot 2^n ] So, [ 9^{n+1} - 2^{n+1} geq 63 + 9 cdot 2^n - 2 cdot 2^n = 63 + 7 cdot 2^n > 7 ] Therefore, the inequality holds for ( n + 1 ).6. By induction, we conclude that ( 9^n - 2^n > 7 ) for all ( n geq 2 ).7. Given that ( 9^n - 2^n ) is always greater than 7 and always divisible by 7, it cannot be prime for any ( n > 1 ).8. Finally, for ( n = 1 ): [ 9^1 - 2^1 = 9 - 2 = 7 ] Here, 7 is a prime number. Conclusion:The only integer ( n geq 1 ) for which ( 9^n - 2^n ) is a prime number is:[boxed{n = 1}]

🧠:Okay, so I need to find the limit as h approaches 0 of [f(x + h) - f(h)] divided by x, evaluated at x = π. The function f(x) is sin(sin x). Let me start by writing down exactly what the problem is asking for. First, let's substitute x with π in the expression. That would make the expression [f(π + h) - f(h)] / π, and then we take the limit as h approaches 0. Wait, hold on, the original expression is [f(x + h) - f(h)] / x. So when x is π, it's [f(π + h) - f(h)] divided by π, and then the limit as h goes to 0. But wait a second, usually when I see a limit involving h approaching 0, it's often related to the derivative. The standard derivative definition is the limit as h approaches 0 of [f(a + h) - f(a)] / h. But here, the expression is a bit different. Instead of f(x + h) - f(x), it's f(x + h) - f(h), and then divided by x instead of h. Also, we're evaluating this at x = π. Hmm. Let me parse this carefully.So the limit is as h approaches 0, and the expression is [f(π + h) - f(h)] / π. Since π is a constant here (because we're evaluating at x = π), the denominator is just π, not h. So this is different from a derivative. The numerator is f(π + h) - f(h). So as h approaches 0, what happens to the numerator?Let me write this out step by step. Let's substitute f(x) = sin(sin x) into the expression. So f(π + h) is sin(sin(π + h)), and f(h) is sin(sin h). Therefore, the numerator becomes sin(sin(π + h)) - sin(sin h). Then, we divide that by π and take the limit as h approaches 0.So the question is: what is the limit of [sin(sin(π + h)) - sin(sin h)] / π as h approaches 0. Let's see if I can simplify sin(π + h). Remember that sin(π + θ) = -sin θ. So sin(π + h) = -sin h. Therefore, sin(sin(π + h)) = sin(-sin h) = -sin(sin h), since sin is an odd function. So substituting back into the numerator: sin(sin(π + h)) - sin(sin h) becomes -sin(sin h) - sin(sin h) = -2 sin(sin h). Therefore, the entire expression becomes [-2 sin(sin h)] / π. Then we take the limit as h approaches 0. So the limit becomes lim_{h→0} [-2 sin(sin h) / π]. Since π is a constant, we can factor it out of the limit. So this is -2/π times lim_{h→0} sin(sin h). Now, as h approaches 0, sin h approaches 0, so sin(sin h) approaches sin(0) which is 0. Therefore, the limit is -2/π * 0 = 0. Wait, but that seems straightforward, but let me check if I made any mistakes here. Let's verify each step again. First, we have f(x) = sin(sin x). At x = π, the expression is [f(π + h) - f(h)] / π. Substituting f into this gives [sin(sin(π + h)) - sin(sin h)] / π. Then, using the identity sin(π + h) = -sin h, so sin(sin(π + h)) = sin(-sin h) = -sin(sin h). Therefore, the numerator is -sin(sin h) - sin(sin h) = -2 sin(sin h). Then, dividing by π gives -2 sin(sin h) / π. Taking the limit as h approaches 0, sin h approaches 0, so sin(sin h) approaches sin 0 = 0. Therefore, the entire limit is 0. Is there any error here? Let me think. Perhaps in the application of the trigonometric identity? Let me check sin(π + h) = -sin h. Yes, that's correct. Because sin(π + θ) = -sin θ. So that's right. Then, sin(-sin h) = -sin(sin h), which is correct because sin is an odd function. So that step is correct. Then combining the terms gives -sin(sin h) - sin(sin h) = -2 sin(sin h). That's straightforward. Then dividing by π, which is a constant, so that's fine. Then the limit as h approaches 0 of sin(sin h). Since h approaches 0, sin h approaches 0, so sin(sin h) approaches sin 0 = 0. Therefore, the entire expression approaches -2/π * 0 = 0. Therefore, the final answer should be 0. But wait, let me think again. Is there another way this limit could be interpreted? For example, maybe there's a typo in the problem statement, but assuming it's as written: [f(x + h) - f(h)] / x evaluated at x = π, so the denominator is x, not h. If it were h, it would be the derivative, but here it's x. Since x is fixed at π, the denominator is just π, so we're looking at [f(π + h) - f(h)] / π as h approaches 0. Alternatively, could the problem have meant [f(x + h) - f(x)] / h? Then evaluated at x = π, which would be the derivative. But the problem states [f(x + h) - f(h)] / x. Maybe that's correct. Alternatively, perhaps the problem was miswritten, but given the problem as stated, we have to work with that. Alternatively, maybe the user intended to write [f(x + h) - f(x)] / h, but mistyped it as [f(x + h) - f(h)] / x. If that were the case, the answer would be different. Let me check both interpretations just to be thorough. First, as written: [f(π + h) - f(h)] / π, limit as h→0. As above, that gives 0. Alternatively, if it's supposed to be [f(π + h) - f(π)] / h, which would be the derivative at π. Then, f’(π) = cos(sin π) * cos π. Because the derivative of sin(sin x) is cos(sin x) * cos x. Then, sin π = 0, so cos(0) = 1, and cos π = -1. So f’(π) = 1 * (-1) = -1. Therefore, if the problem had been the derivative, the answer would be -1. But according to the problem statement, it's [f(x + h) - f(h)] / x, so the answer would be 0. Therefore, assuming the problem is written correctly, the answer is 0. Wait, but just to make sure, maybe I need to consider expanding f(π + h) and f(h) using Taylor series or something? Let's try that. First, f(π + h) = sin(sin(π + h)) = sin(-sin h) = -sin(sin h). Then, f(h) = sin(sin h). So the numerator is -sin(sin h) - sin(sin h) = -2 sin(sin h). Then, as h approaches 0, sin h ~ h - h^3/6 + ..., so sin(sin h) ~ sin(h - h^3/6 + ...) ~ (h - h^3/6) - (h - h^3/6)^3 / 6 + ... ≈ h - h^3/6 - h^3/6 + higher order terms ≈ h - h^3/3. Therefore, sin(sin h) ≈ h - h^3/3. Therefore, the numerator is -2(h - h^3/3), so dividing by π gives -2(h - h^3/3)/π. Then, taking the limit as h approaches 0 would be -2(0)/π = 0. So even using the Taylor expansion, the limit is 0. Alternatively, maybe I can use L’Hospital’s Rule? But in this case, since the numerator approaches 0 and the denominator is a constant π, it's not a 0/0 or infinity/infinity form. The numerator approaches -2*0 = 0, and the denominator is π, so the overall expression approaches 0. So L’Hospital’s Rule isn't applicable here. Alternatively, if the problem had been [f(x + h) - f(h)] / h, then as h approaches 0, that would be different. Let's compute that for curiosity's sake. Then, the expression would be [sin(sin(π + h)) - sin(sin h)] / h. Then, as h approaches 0, sin(sin(π + h)) = sin(-sin h) = -sin(sin h), so the numerator is -sin(sin h) - sin(sin h) = -2 sin(sin h). Then, the expression is -2 sin(sin h)/h. Then, we can compute the limit as h approaches 0: -2 * [sin(sin h)/h]. Let’s set t = sin h, so as h approaches 0, t approaches 0, and sin t / t approaches 1. But here, sin(sin h)/h = [sin(sin h)/sin h] * [sin h / h]. As h approaches 0, sin h ~ h, so sin h / h ~ 1, and sin(sin h)/sin h ~ 1. Therefore, the limit would be -2 * 1 * 1 = -2. But that's only if the denominator was h instead of x. But in our problem, the denominator is x, which is π. So the answer remains 0. Alternatively, perhaps the problem had a typo and wanted x approaching 0? But no, the limit is as h approaches 0. Wait, just to double-check, let's plug in h = 0. If h approaches 0, plugging h = 0 into the expression gives [f(π + 0) - f(0)] / π = [f(π) - f(0)] / π. f(π) = sin(sin π) = sin(0) = 0. f(0) = sin(sin 0) = sin(0) = 0. So [0 - 0]/π = 0. But that's just the value at h = 0, which is 0, but the limit as h approaches 0 might not necessarily be the same as the value at h = 0. Wait, no, actually, in this case, the expression when h approaches 0 is being evaluated. But when h approaches 0, the numerator approaches f(π) - f(0) = 0 - 0 = 0, so the limit would be 0/π = 0. But that seems too straightforward. Wait, but actually, this is only if the numerator is approaching f(π) - f(0). However, when h is approaching 0, f(π + h) approaches f(π) and f(h) approaches f(0). Therefore, the numerator approaches f(π) - f(0) = 0 - 0 = 0. So the limit is 0. But wait, is that always the case? If the function f is continuous at π and 0, then yes. Since f(x) = sin(sin x) is continuous everywhere, so f(π + h) approaches f(π) and f(h) approaches f(0). Therefore, the numerator's limit is f(π) - f(0) = 0 - 0 = 0, so the entire expression's limit is 0. But in our earlier step-by-step approach, we actually computed the numerator as -2 sin(sin h), which for small h is approximately -2 sin h, which is approximately -2h. Therefore, the numerator is -2h, so the expression is -2h / π. Then, as h approaches 0, that limit is 0. So even if we consider the linear term, it still goes to 0. Therefore, regardless of the method, the limit is 0. Therefore, I think the answer is 0. But to be absolutely thorough, let's consider another approach. Suppose we expand f(π + h) and f(h) in Taylor series around h = 0. First, f(π + h) = sin(sin(π + h)) = sin(-sin h) = -sin(sin h). Let's expand sin h around h = 0: sin h = h - h^3/6 + h^5/120 - ... Then, sin(sin h) = sin(h - h^3/6 + ...) = (h - h^3/6) - (h - h^3/6)^3 / 6 + ... Let's compute up to the first few terms. First term: h - h^3/6. Second term: - [h^3 - 3h^4/6 + ... ] / 6 ≈ -h^3/6 + ... So sin(sin h) ≈ h - h^3/6 - h^3/6 = h - h^3/3. Therefore, sin(sin h) ≈ h - h^3/3. Similarly, f(π + h) = -sin(sin h) ≈ -h + h^3/3. Then, f(h) = sin(sin h) ≈ h - h^3/3. So the numerator is f(π + h) - f(h) ≈ (-h + h^3/3) - (h - h^3/3) = -h + h^3/3 - h + h^3/3 = -2h + 2h^3/3. Therefore, the entire expression [f(π + h) - f(h)] / π ≈ (-2h + 2h^3/3) / π. Then, taking the limit as h approaches 0, the dominant term is -2h / π, which approaches 0. The next term is 2h^3/(3π), which also approaches 0. So even with the Taylor expansion, the limit is 0. Another way: Let's consider the Mean Value Theorem. If we have [f(π + h) - f(h)] / π, perhaps we can consider this as [f(π + h) - f(h)] / π = [f(π + h) - f(h)] / (π + h - h) = [f(π + h) - f(h)] / (π). But I don't know if the Mean Value Theorem applies here. The MVT states that there exists a c between h and π + h such that [f(π + h) - f(h)] / π = f’(c). But since h is approaching 0, the interval between h and π + h becomes (0, π). But then as h approaches 0, c would approach some point in (0, π). But f’(c) = cos(sin c) * cos c. So the limit as h approaches 0 would be f’(c) as c approaches some point? But this seems not directly helpful. Wait, but actually, the MVT would require that for each h, there's a c_h between h and π + h such that [f(π + h) - f(h)] / π = f’(c_h). Then, as h approaches 0, c_h is between h and π + h, so as h approaches 0, c_h approaches some value between 0 and π. However, since h is approaching 0, the lower bound of the interval (h) approaches 0, and the upper bound (π + h) approaches π. Therefore, c_h approaches some limit point in [0, π]. But unless we can find the exact limit of c_h, this approach might not yield the answer. However, if we suppose that c_h approaches some c in [0, π], then the limit would be f’(c). But this is speculative. However, even if we accept this, unless f’(c) is the same for all c in [0, π], which it's not. For example, f’(0) = cos(sin 0) * cos 0 = cos(0) * 1 = 1. f’(π) = cos(sin π) * cos π = cos(0) * (-1) = -1. So depending on where c_h approaches, the limit could be different. Therefore, this approach is inconclusive. But in reality, since we already computed via substitution and via Taylor series that the limit is 0, the MVT approach must reconcile with that. Therefore, perhaps the issue is that the MVT gives us a value c_h such that [f(π + h) - f(h)] = f’(c_h) * π. But as h approaches 0, c_h approaches somewhere between 0 and π, so f’(c_h) approaches f’(c) where c is in [0, π]. But unless f’(c) is 0, but we know f’(c) = cos(sin c) * cos c. For example, at c = π/2, f’(π/2) = cos(sin(π/2)) * cos(π/2) = cos(1) * 0 = 0. But c_h could approach different points. However, since we know the limit is 0 via other methods, this suggests that the average rate of change [f(π + h) - f(h)] / π approaches 0, which would imply that f’(c_h) approaches 0. But how?Wait, perhaps because as h approaches 0, the interval [h, π + h] becomes [0, π], and the average rate of change over [0, π] is [f(π) - f(0)] / π = (0 - 0)/π = 0. So in the limit as h approaches 0, the average rate of change approaches 0, which is consistent with our previous result. Therefore, by the MVT, for each h, there exists c_h in (h, π + h) such that [f(π + h) - f(h)] / π = f’(c_h). Then, as h approaches 0, c_h is in (0, π). But the problem is that c_h could approach any point in [0, π], and unless f’(c) is continuous and we know it approaches 0, but f’(c) varies. However, the limit of [f(π + h) - f(h)] / π as h approaches 0 is 0, so even if f’(c_h) is varying, the product f’(c_h) * π = [f(π + h) - f(h)], which tends to 0. Hence, f’(c_h) tends to 0 as h approaches 0. But how does f’(c_h) approach 0? For example, if c_h approaches π/2, then f’(π/2) = 0 as we saw. If c_h approaches some other point where f’(c) = 0, then that works. Alternatively, even if c_h approaches a point where f’(c) is non-zero, but the average rate still tends to 0. However, given that we already know the answer is 0, perhaps this approach is not necessary. In any case, all methods point to the limit being 0. Therefore, I can be confident that the answer is 0. But just to recap:1. Substitute x = π into the expression: [f(π + h) - f(h)] / π.2. Expand f(π + h) using sin(π + h) = -sin h, leading to sin(sin(π + h)) = -sin(sin h).3. Subtract f(h) = sin(sin h), giving -sin(sin h) - sin(sin h) = -2 sin(sin h).4. Divide by π: -2 sin(sin h) / π.5. Take the limit as h approaches 0: sin(sin h) approaches sin(0) = 0.6. Therefore, the limit is 0.Yes, that all checks out. I think this is solid. No mistakes detected in the reasoning.Final Answerboxed{0}

✅:To evaluate the limit (lim_{h rightarrow 0} frac{f(x+h) - f(h)}{x}) at (x = pi) for the function (f(x) = sin(sin x)), we proceed as follows:1. Substitute (x = pi) into the expression: [ frac{f(pi + h) - f(h)}{pi} ]2. Expand (f(pi + h)) and (f(h)): - Using the trigonometric identity (sin(pi + h) = -sin h), we get: [ f(pi + h) = sin(sin(pi + h)) = sin(-sin h) = -sin(sin h) ] - For (f(h)), we have: [ f(h) = sin(sin h) ]3. Simplify the numerator: [ f(pi + h) - f(h) = -sin(sin h) - sin(sin h) = -2sin(sin h) ]4. Form the expression: [ frac{-2sin(sin h)}{pi} ]5. Evaluate the limit as (h rightarrow 0): - As (h rightarrow 0), (sin h rightarrow 0), so (sin(sin h) rightarrow sin(0) = 0). - Therefore, the limit becomes: [ lim_{h rightarrow 0} frac{-2 cdot 0}{pi} = 0 ]Thus, the final answer is (boxed{0}).

🔑:1. Define the function ( f(x) ): [ f(x) = sin (sin x) ]2. Express and analyze the limit: We want to evaluate: [ lim _{h rightarrow 0} frac{f(pi+h) - f(h)}{pi} ]3. Assess the function behavior around ( h = 0 ): We observe that ( f(x) ) is composed of sinusoidal functions, and the limit involves differences of these functions.4. Simplify the given expression: Notice that the expression ( frac{f(pi+h) - f(h)}{pi} ) can be approximated for small ( h ) if ( f(x) ) is sufficiently smooth.5. Approximation and continuity: Given the smoothness and continuity of trigonometric functions, ( f(x) ) can be continuous around ( h = 0 ). Consequently: [ lim_{h rightarrow 0} (f(x+h) - f(h)) = f(x) - f(0) ] 6. Evaluate the simplified expression directly: This means the limit reduces to examining this simplified expression for ( x = pi ): [ frac{f(pi) - f(0)}{pi} ]7. Compute specific function values: We substitute ( x = pi ): [ f(pi) = sin (sin pi) = sin (0) = 0 ] Similarly for: [ f(0) = sin (sin 0) = sin (0) = 0 ]8. Calculate the result: Therefore, the expression becomes: [ frac{0 - 0}{pi} = 0 ] Concluding, we find: [ lim _{h rightarrow 0} frac{f(pi+h) - f(h)}{pi} = 0 ] Hence the solution is: [ boxed{0} ]

🧠:Okay, let me try to figure out how to prove the area ratio between the new triangle A'B'C' and the original triangle ABC. The formula given is S'/S = 1 + λ1 + λ2 + λ3 + λ1λ2 + λ2λ3 + λ3λ1. Hmm, so the problem is about extending the sides of triangle ABC with certain ratios and then forming a new triangle from those extensions. I need to relate the area of this new triangle to the original one.First, I need to visualize the problem. Let me sketch triangle ABC. The sides AB, BC, and CA are being extended to points B', C', and A' respectively. The extensions are such that BB' = λ1 AB, CC' = λ2 BC, and AA' = λ3 CA. So, for example, BB' is an extension of AB beyond B, and its length is λ1 times AB. Similarly for the others. Then connecting these points A', B', C' forms the new triangle. The goal is to find the ratio of their areas.Maybe coordinate geometry can help here. Assign coordinates to the original triangle ABC and then compute the coordinates of A', B', C'. Once I have those, I can use the shoelace formula or determinant method to compute the areas S and S', then take the ratio. Let me try this approach.Let me assign coordinates. Let's place point A at (0, 0) for simplicity. Let me let AB lie along the x-axis. Let’s say point B is at (c, 0). Then point C can be at some coordinates (d, e), which will determine the shape and area of the original triangle. The area S of triangle ABC can be calculated using the determinant formula: (1/2)| (B - A) × (C - A) |. But maybe assigning specific coordinates will make the calculations easier. Let me choose coordinates such that A is (0,0), B is (1,0), and C is (0,1). Then triangle ABC is a right-angled triangle with area 1/2. Maybe this simplifies the calculation, and since the formula is general, the ratio should hold regardless of the original triangle's shape. Wait, but I need to make sure that the result is independent of the original triangle's configuration. So perhaps using a specific coordinate system where ABC is a right-angled triangle is okay, as affine transformations preserve ratios of areas. So if I can prove it for a specific triangle, it should hold generally.Alternatively, maybe using barycentric coordinates or vectors. But coordinate geometry seems straightforward.Let me proceed with A(0,0), B(1,0), and C(0,1). Then area S = 1/2. Let's compute points A', B', C'.First, BB' = λ1 AB. AB is from A(0,0) to B(1,0), so AB vector is (1,0). BB' is an extension beyond B, so the direction from B to B' is the same as AB. Since BB' = λ1 AB, which is λ1 times the length of AB. The length of AB is 1, so BB' has length λ1*1 = λ1. Therefore, the coordinates of B' would be B + λ1*AB vector. The AB vector is (1,0), so moving from B(1,0) in the direction of AB (which is already along AB) would be adding λ1*(1,0) to B. Wait, but AB is from A to B, so if we are extending AB beyond B to B', then the vector from B to B' should be λ1 times AB. AB vector is (1,0), so BB' vector is λ1*(1,0). Therefore, B' is (1 + λ1*1, 0 + λ1*0) = (1 + λ1, 0). Got it.Similarly, CC' = λ2 BC. BC is from B(1,0) to C(0,1). The vector BC is (-1,1). CC' is an extension beyond C, so the vector CC' is λ2 times BC. Wait, the problem says CC' = λ2 BC. So the length of CC' is λ2 times the length of BC. But BC has length sqrt( (-1)^2 + 1^2 ) = sqrt(2). So CC' length is λ2*sqrt(2). But the direction is the same as BC. Wait, but the problem states "extend the sides AB, CB, CA to B', C', A' respectively". Wait, for CC', it's extending CB. Wait, the original sides are AB, CB, and CA. Wait, CB is from C to B. So extending CB beyond C? Wait, the problem says: "extend the sides AB, CB, CA to B', C', A' respectively". So for each side, AB is extended to B', CB is extended to C', and CA is extended to A'.So for CB: the side CB is from C to B. So extending CB beyond C would be in the direction opposite to B. Wait, but the problem says "extend CB to C' such that CC' = λ2 BC". Hmm. Wait, BC is the vector from B to C, but CB is the side from C to B. Wait, maybe there's a confusion here. Let me clarify.The problem states: "extend the sides AB, CB, CA to B', C', A' respectively such that BB' = λ1 AB, CC' = λ2 BC, and AA' = λ3 CA".So for each side:- AB is extended to B', with BB' = λ1 AB. Since AB is from A to B, extending AB beyond B would give B'.- CB is extended to C', with CC' = λ2 BC. CB is from C to B, so extending CB beyond C would give C'.- CA is extended to A', with AA' = λ3 CA. CA is from C to A, so extending CA beyond A would give A'.Wait, let's confirm:For AB: extending AB beyond B to B', such that BB' = λ1 AB. AB is the vector from A to B. So BB' is λ1 times AB. So the direction from B to B' is same as AB. So B' is located at B + λ1*AB vector.For CB: extending CB beyond C to C', such that CC' = λ2 BC. CB is the side from C to B. So BC is the vector from B to C. Wait, the problem says CC' = λ2 BC. So CC' is a vector equal to λ2 times BC. BC is from B to C, so BC vector is (C - B). So CC' = λ2*(C - B). Therefore, C' is located at C + CC' = C + λ2*(C - B). Wait, but extending CB beyond C would mean the direction from C to B is CB, so extending beyond C would be opposite direction. Wait, maybe the problem is written in terms of lengths. Let me parse the problem again.The problem says: "extend the sides AB, CB, CA to B', C', A' respectively such that BB' = λ1 AB, CC' = λ2 BC, and AA' = λ3 CA".So BB' is a segment such that its length is λ1 times AB. Similarly for CC' and AA'. So BB' is an extension of AB beyond B, with length λ1 AB. Similarly, CC' is extension of CB beyond C, with length λ2 BC. AA' is extension of CA beyond A, with length λ3 CA.Therefore, in terms of vectors:For B', since AB is from A to B, the direction from B to B' is same as AB. The length BB' = λ1 AB. Since AB has length |AB|, then BB' has length λ1 |AB|. Therefore, the vector BB' is λ1 times the vector AB. Therefore, B' = B + λ1*(AB vector). Since AB vector is (B - A), which in coordinates is (1,0) - (0,0) = (1,0). So BB' vector is λ1*(1,0), so B' is (1,0) + λ1*(1,0) = (1 + λ1, 0).Similarly for CC': CC' is an extension of CB beyond C, with length λ2 BC. CB is from C to B, so BC is the reverse vector. Let me be careful. The problem states CC' = λ2 BC. So BC is the vector from B to C. But CC' is an extension of CB beyond C, so the direction is from C to beyond, opposite to B. So perhaps CC' is λ2 times BC vector. Wait, BC is from B to C, which is vector (C - B). If we extend CB beyond C, then CC' would be in the direction opposite to B, which is the direction of (C - B). Wait, no. Wait, CB is the side from C to B, so the direction from C to B is CB. Extending CB beyond C would mean going from C in the direction opposite to B, which is the direction of (C - B). Wait, no, if CB is from C to B, then extending beyond C would be going away from B. Wait, maybe better to think in terms of coordinates.Original points:A(0,0), B(1,0), C(0,1).CB is the side from C(0,1) to B(1,0). So the vector CB is B - C = (1,0) - (0,1) = (1,-1). So extending CB beyond C would be in the direction away from B, which is the direction of CB vector. Wait, CB vector is (1,-1), so from C to B. If we extend CB beyond C, we need to go in the opposite direction, which is (-1,1). Wait, no. Wait, if you have a line CB going from C to B, then beyond C, the line continues in the direction away from B. So the direction vector is still CB, which is (1,-1). Wait, no. If you are at point C and you go along the line CB beyond C, then the direction is the same as CB but starting at C. Wait, actually, the line CB is parametrized as C + t*(B - C) where t >= 0 gives the segment CB, and t < 0 extends beyond C. Wait, no. If you parametrize the line CB, starting at C when t=0 and going to B when t=1. To extend beyond C, you would have t negative. So the direction vector is (B - C) = (1,-1). So CC' is such that the length CC' = λ2 BC. BC is the length from B to C, which is sqrt( (1-0)^2 + (0 -1)^2 ) = sqrt(2). So CC' has length λ2*sqrt(2). But since we are extending beyond C, the vector CC' is in the direction opposite to BC, which would be (C - B) = (-1,1). So CC' vector is (C - B) scaled by λ2. Wait, but the problem says CC' = λ2 BC. BC is the vector from B to C. If CC' is λ2 times BC, then CC' vector is λ2*(C - B) = λ2*(-1,1). Therefore, the coordinates of C' would be C + CC' = (0,1) + λ2*(-1,1) = (0 - λ2, 1 + λ2).Wait, let's verify. The problem states CC' = λ2 BC. If BC is the vector from B to C, then BC is (0,1) - (1,0) = (-1,1). So CC' is λ2 times BC vector. So CC' = λ2*(-1,1). Therefore, C' = C + CC' = (0,1) + (-λ2, λ2) = (-λ2, 1 + λ2). Okay, that makes sense.Similarly, AA' is an extension of CA beyond A. CA is from C to A, which is vector (0,0) - (0,1) = (0,-1). The problem states AA' = λ3 CA. So AA' is λ3 times CA vector. CA vector is (A - C) = (-0, -1). So AA' vector is λ3*(-0, -1) = (0, -λ3). Therefore, A' = A + AA' = (0,0) + (0, -λ3) = (0, -λ3). Wait, but CA is the side from C to A. Extending CA beyond A would be in the direction from A away from C, which is the direction of (A - C) = (0,-1). So adding that vector scaled by λ3 gives A'(0, -λ3).Wait, let me check again. The problem says "extend the sides AB, CB, CA to B', C', A' respectively such that BB' = λ1 AB, CC' = λ2 BC, and AA' = λ3 CA". So:- BB' is the extension of AB beyond B, with BB' = λ1 AB. AB is from A to B, so BB' is λ1 times AB vector. So vector BB' = λ1*(B - A) = λ1*(1,0). Therefore, B' = B + BB' = (1,0) + (λ1,0) = (1 + λ1, 0).- CC' is the extension of CB beyond C, with CC' = λ2 BC. BC is from B to C, so vector BC = C - B = (-1,1). Therefore, CC' = λ2*(BC) = λ2*(-1,1). So C' = C + CC' = (0,1) + (-λ2, λ2) = (-λ2, 1 + λ2).- AA' is the extension of CA beyond A, with AA' = λ3 CA. CA is from C to A, so vector CA = A - C = (0, -1). Therefore, AA' = λ3*(CA) = λ3*(0, -1). So A' = A + AA' = (0,0) + (0, -λ3) = (0, -λ3).Okay, so coordinates are:A'(0, -λ3)B'(1 + λ1, 0)C'(-λ2, 1 + λ2)Now, we need to compute the area of triangle A'B'C' and compare it to the original area S = 1/2.Using the shoelace formula for the area of a triangle given coordinates.The shoelace formula states that the area is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|.Let me assign:A'(x1, y1) = (0, -λ3)B'(x2, y2) = (1 + λ1, 0)C'(x3, y3) = (-λ2, 1 + λ2)Plugging into the shoelace formula:Area S' = (1/2)| 0*(0 - (1 + λ2)) + (1 + λ1)*((1 + λ2) - (-λ3)) + (-λ2)*(-λ3 - 0) |Simplify term by term:First term: 0*(0 - (1 + λ2)) = 0Second term: (1 + λ1)*( (1 + λ2) + λ3 ) = (1 + λ1)(1 + λ2 + λ3)Third term: (-λ2)*(-λ3 - 0) = (-λ2)*(-λ3) = λ2λ3So overall:S' = (1/2)| 0 + (1 + λ1)(1 + λ2 + λ3) + λ2λ3 |Since all terms are positive if λ1, λ2, λ3 are positive (assuming they are extension factors, so positive), we can drop the absolute value:S' = (1/2)[ (1 + λ1)(1 + λ2 + λ3) + λ2λ3 ]Let me expand (1 + λ1)(1 + λ2 + λ3):= 1*(1 + λ2 + λ3) + λ1*(1 + λ2 + λ3)= 1 + λ2 + λ3 + λ1 + λ1λ2 + λ1λ3Then add λ2λ3:Total inside the brackets:1 + λ2 + λ3 + λ1 + λ1λ2 + λ1λ3 + λ2λ3So S' = (1/2)(1 + λ1 + λ2 + λ3 + λ1λ2 + λ1λ3 + λ2λ3 )Original area S = 1/2.Therefore, S'/S = [ (1/2)(1 + λ1 + λ2 + λ3 + λ1λ2 + λ1λ3 + λ2λ3 ) ] / (1/2) = 1 + λ1 + λ2 + λ3 + λ1λ2 + λ1λ3 + λ2λ3.Which matches the given formula. Therefore, proved.Wait, but the given formula in the problem is S'/S = 1 + λ1 + λ2 + λ3 + λ1λ2 + λ2λ3 + λ3λ1. The terms λ1λ3 and λ3λ1 are the same, so yes, the expression I obtained is the same as the given one. Therefore, the proof is complete.But wait, let me check if this is dependent on the coordinate system. I chose a specific coordinate system where A is at (0,0), B at (1,0), and C at (0,1). The original area S is 1/2. But the problem states a general triangle with area S. However, area ratios are preserved under affine transformations, so the ratio S'/S should be the same regardless of the original triangle. Therefore, by proving it for this specific triangle, we have proven it in general. Alternatively, if I had used a general triangle, the algebra might be more complicated, but the ratio would still hold. Since the problem is stated for any triangle ABC, the result is general. So this coordinate approach suffices.Alternatively, another approach could be using mass point geometry or vectors, but coordinate geometry worked here.Wait, let me just verify my steps once again to be sure.1. Assigned coordinates to ABC as right-angled triangle for simplicity.2. Calculated coordinates of A', B', C' based on extension ratios.3. Applied shoelace formula to find area of A'B'C'.4. Expanded the expression and found that the ratio S'/S gives the desired expression.Yes, each step seems correct. The key was to model the extensions properly, especially CC', which was the trickiest since it's an extension of CB beyond C. But by breaking down the vectors, it worked out.Another check: when all λ1, λ2, λ3 are zero, the new triangle A'B'C' should coincide with the original triangle ABC, so S'/S should be 1. Plugging λ1=λ2=λ3=0 into the formula gives 1 + 0 + 0 + 0 + 0 + 0 + 0 = 1, which is correct.Another test case: take λ1=1, λ2=0, λ3=0. Then S'/S should be 1 + 1 + 0 + 0 + 0 + 0 + 0 = 2. Let's check with coordinates.B' becomes (1 + 1, 0) = (2,0), C' remains (0,1), A' remains (0,0). Wait, no: λ2=0, so C' is (-0, 1 + 0) = (0,1). A' with λ3=0 is (0,0). So triangle A'(0,0), B'(2,0), C'(0,1). The area is (1/2)*2*1 = 1. Original area was 1/2, so ratio is 2, which matches the formula. Good.Another test case: λ1=1, λ2=1, λ3=0. Then formula gives 1 +1 +1 +0 +1*1 +1*0 +0*1 = 1+1+1+0+1+0+0=4. Let's compute coordinates.B'(2,0), C'(-1, 2), A'(0,0). Area via shoelace:Coordinates: A'(0,0), B'(2,0), C'(-1,2).Area = (1/2)|0*(0-2) + 2*(2 - 0) + (-1)*(0 - 0)| = (1/2)|0 + 4 + 0| = 2. Original area 1/2, ratio 4. Correct. So formula holds.Therefore, confident that the proof is correct.

✅:To prove the area ratio (frac{S'}{S} = 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_2 lambda_3 + lambda_3 lambda_1) for the extended triangle ( triangle A'B'C' ), follow these steps:1. Assign Coordinates: Place ( triangle ABC ) in a coordinate system with ( A(0,0) ), ( B(1,0) ), and ( C(0,1) ). The area ( S = frac{1}{2} ).2. Determine Coordinates of Extended Points: - B': Extend ( AB ) beyond ( B ). The vector ( overrightarrow{AB} = (1,0) ), so ( B' = B + lambda_1 overrightarrow{AB} = (1 + lambda_1, 0) ). - C': Extend ( CB ) beyond ( C ). The vector ( overrightarrow{BC} = (-1,1) ), so ( C' = C + lambda_2 overrightarrow{BC} = (-lambda_2, 1 + lambda_2) ). - A': Extend ( CA ) beyond ( A ). The vector ( overrightarrow{CA} = (0,-1) ), so ( A' = A + lambda_3 overrightarrow{CA} = (0, -lambda_3) ).3. Compute Area of ( triangle A'B'C' ): Apply the shoelace formula to ( A'(0, -lambda_3) ), ( B'(1 + lambda_1, 0) ), ( C'(-lambda_2, 1 + lambda_2) ): [ S' = frac{1}{2} left| 0(0 - (1 + lambda_2)) + (1 + lambda_1)(1 + lambda_2 + lambda_3) + (-lambda_2)(-lambda_3) right| ] Simplify: [ S' = frac{1}{2} left[ (1 + lambda_1)(1 + lambda_2 + lambda_3) + lambda_2 lambda_3 right] ] Expand the product: [ S' = frac{1}{2} left[ 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_1 lambda_3 + lambda_2 lambda_3 right] ]4. Calculate the Ratio: Since ( S = frac{1}{2} ), the ratio is: [ frac{S'}{S} = frac{frac{1}{2} left[ 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_1 lambda_3 + lambda_2 lambda_3 right]}{frac{1}{2}} = 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_2 lambda_3 + lambda_3 lambda_1 ]Conclusion: The area ratio (frac{S'}{S}) is indeed (1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_2 lambda_3 + lambda_3 lambda_1). (boxed{frac{S'}{S} = 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_2 lambda_3 + lambda_3 lambda_1})

🔑:1. Consider the triangle Delta ABC with area S. Extend sides AB, BC, CA to points B', C', A' respectively such that [ BB' = lambda_1 AB, quad CC' = lambda_2 BC, quad AA' = lambda_{3} CA. ]2. We need to find the area S' of the new triangle A'B'C' and show that: [ frac{S'}{S} = 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_2 lambda_3 + lambda_3 lambda_1. ]3. From the given conditions and the properties of similar triangles, in the triangle A'B'C', consider the sub-triangle Delta B'CB and Delta BCA. Notice that the base B'B is lambda_1 AB and the height remains the same, thus the area of Delta B'CB relative to Delta BCA is: [ frac{S_{Delta B'CB}}{S} = lambda_1. ]4. Similarly, in Delta CB'C' and Delta BB'C, observe that C'C is extended by lambda_2 BC: [ frac{S_{Delta CB'C'}}{S_{Delta BB'C}} = lambda_2. ] Combining these, we get: [ frac{S_{Delta CB'C'}}{S} = lambda_1 lambda_2. ]5. For the next sub-triangle Delta B B'C' relative to Delta BCA, since B'B = lambda_1 AB and B'C extends proportionally: [ frac{S_{Delta BB'C'}}{S} = lambda_1 + lambda_1 lambda_2. ]6. Applying similar logic to other segments: - In Delta CCA' and Delta BCA: [ frac{S_{Delta CCA'}}{S} = lambda_2 + lambda_2 lambda_3, ] - In Delta ABA' relative to Delta BCA: [ frac{S_{Delta ABA'}}{S} = lambda_3 + lambda_3 lambda_1. ]7. Summing up all these relative areas, the total area relation becomes: [ frac{S'}{S} = 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_2 lambda_3 + lambda_3 lambda_1. ] Conclusion:[boxed{frac{S'}{S} = 1 + lambda_1 + lambda_2 + lambda_3 + lambda_1 lambda_2 + lambda_2 lambda_3 + lambda_3 lambda_1}]