🧠:[<think>Okay, let's see. So there are four vehicles starting from point A at the same time and arriving at point B one after another with equal time intervals between their arrivals. The fastest vehicle has speed v₁ and the slowest has speed v₄. We need to find the speeds of the other two vehicles, which I guess would be v₂ and v₃. Hmm.First, let me try to visualize the problem. All four vehicles start together from A, but they have different speeds. The fastest one arrives first at B, then the next one arrives after some time interval, then the third, and finally the slowest one arrives. The key here is that the time intervals between their arrivals are equal. So if the first vehicle arrives at time t₁, the next arrives at t₁ + Δt, then t₁ + 2Δt, and the last one at t₁ + 3Δt. But since they all start at the same time, the actual arrival times depend on their speeds. The distance between A and B is the same for all vehicles, so the time each vehicle takes is distance divided by speed. Let me denote the distance as D.So, the time taken by the fastest vehicle is D / v₁, and the time taken by the slowest is D / v₄. The other two vehicles take times D / v₂ and D / v₃. Since they arrive successively with equal time intervals, the difference in their arrival times should be the same. So, (D / v₂) - (D / v₁) = Δt, (D / v₃) - (D / v₂) = Δt, and (D / v₄) - (D / v₃) = Δt. So, all these differences are equal. Therefore, the differences between the reciprocals of the speeds multiplied by D are equal. Let me write that out:(D / v₂) - (D / v₁) = (D / v₃) - (D / v₂) = (D / v₄) - (D / v₃)If I factor out D, it becomes:D(1/v₂ - 1/v₁) = D(1/v₃ - 1/v₂) = D(1/v₄ - 1/v₃)Since D is the same for all, we can divide through by D and get:1/v₂ - 1/v₁ = 1/v₃ - 1/v₂ = 1/v₄ - 1/v₃So the reciprocals of the speeds form an arithmetic sequence. Let me check that. If we let the reciprocals be terms in an arithmetic progression, then the differences between consecutive terms are equal, which matches the equations above. So, if we denote 1/v₁, 1/v₂, 1/v₃, 1/v₄ as four terms in an arithmetic sequence, that would satisfy the given condition. Let me confirm.Suppose 1/v₁, 1/v₂, 1/v₃, 1/v₄ are in arithmetic progression. Then, the common difference d would be equal to 1/v₂ - 1/v₁ = 1/v₃ - 1/v₂ = 1/v₄ - 1/v₃. Which is exactly what we have here. So, that's a good approach. Therefore, the reciprocals of the speeds form an arithmetic progression. Therefore, we can find v₂ and v₃ by finding the terms of this arithmetic sequence.Given that the first term is 1/v₁ and the fourth term is 1/v₄, since there are four vehicles. Wait, but arithmetic progression with four terms: term 1 is 1/v₁, term 2 is 1/v₂, term 3 is 1/v₃, term 4 is 1/v₄.So, in an arithmetic progression, the nth term is given by a_n = a₁ + (n - 1)d, where d is the common difference. Here, a₁ = 1/v₁, a₄ = 1/v₄. Therefore:a₄ = a₁ + 3dTherefore, 1/v₄ = 1/v₁ + 3dSo, solving for d gives:d = (1/v₄ - 1/v₁)/3Therefore, the common difference d is (1/v₄ - 1/v₁)/3. Then, the other terms can be found by adding d successively.So, a₂ = a₁ + d = 1/v₁ + (1/v₄ - 1/v₁)/3 = (3(1/v₁) + 1/v₄ - 1/v₁)/3 = (2/v₁ + 1/v₄)/3Similarly, a₃ = a₂ + d = (2/v₁ + 1/v₄)/3 + (1/v₄ - 1/v₁)/3 = [2/v₁ + 1/v₄ + 1/v₄ - 1/v₁]/3 = (1/v₁ + 2/v₄)/3Therefore, the reciprocals of the speeds are:1/v₂ = (2/v₁ + 1/v₄)/31/v₃ = (1/v₁ + 2/v₄)/3Therefore, to get v₂ and v₃, we can take reciprocals:v₂ = 1 / [(2/v₁ + 1/v₄)/3] = 3 / (2/v₁ + 1/v₄)Similarly,v₃ = 1 / [(1/v₁ + 2/v₄)/3] = 3 / (1/v₁ + 2/v₄)Alternatively, these can be written as:v₂ = 3v₁v₄ / (2v₄ + v₁)v₃ = 3v₁v₄ / (v₄ + 2v₁)Let me check this. Let's verify the steps.Starting with the reciprocals forming an arithmetic progression. Yes, since the time intervals between arrivals are equal, the differences in arrival times are equal, which translate to differences in the reciprocals of speeds multiplied by distance. Therefore, the reciprocals must be in AP. Then, using the formula for the nth term of an AP, we found the common difference and then the middle terms.Therefore, the speeds v₂ and v₃ are given by those expressions. Let me test with some numbers to see if this makes sense.Suppose the distance D is 60 units. Let’s say the fastest speed v₁ is 60 units per hour, so the time taken by the fastest is 1 hour. Let’s say the slowest speed v₄ is 20 units per hour, so time taken is 3 hours. Then, the arrival times are 1 hour, then something, then something, then 3 hours. The intervals between arrivals should be equal. So, if four vehicles arrive at t=1, t=1+Δt, t=1+2Δt, t=1+3Δt=3. Therefore, 1+3Δt=3 ⇒ 3Δt=2 ⇒ Δt=2/3 hours. Therefore, arrival times are 1, 1+2/3=5/3, 5/3+2/3=7/3, 7/3+2/3=9/3=3. So, 1, 1.666..., 2.333..., 3. So the times taken by the vehicles are 1, 5/3, 7/3, 3. Therefore, their speeds are D divided by these times. So:v₁ = 60 /1 =60v₂ =60 / (5/3)=36v₃=60 / (7/3)≈25.714v₄=60 /3=20But according to our formula:v₂=3v₁v₄/(2v₄ +v₁)=3*60*20/(2*20 +60)=3600/(40 +60)=3600/100=36. Correct.v₃=3v₁v₄/(v₄ +2v₁)=3*60*20/(20 +120)=3600/140≈25.714. Correct.So in this case, the formula gives the correct results. Therefore, the answer seems to be correct.Therefore, the speeds of the other two vehicles are 3v₁v₄/(2v₄ +v₁) and 3v₁v₄/(v₄ +2v₁).Final AnswerThe speeds of the other two vehicles are boxed{v_2 = dfrac{3v_1v_4}{2v_4 + v_1}} and boxed{v_3 = dfrac{3v_1v_4}{v_4 + 2v_1}}.</think>Given that four vehicles start from point ( A ) at the same time and arrive at point ( B ) with equal time intervals between their arrivals, we need to determine the speeds of the two intermediate vehicles, ( v_2 ) and ( v_3 ), given the fastest speed ( v_1 ) and the slowest speed ( v_4 ).The key observation is that the reciprocals of the speeds form an arithmetic progression. This is because the time intervals between arrivals are equal, leading to equal differences in the reciprocals of the speeds when multiplied by the distance ( D ).1. Let the reciprocals of the speeds ( frac{1}{v_1}, frac{1}{v_2}, frac{1}{v_3}, frac{1}{v_4} ) form an arithmetic progression.2. The common difference ( d ) of this arithmetic progression can be found using the first term ( frac{1}{v_1} ) and the fourth term ( frac{1}{v_4} ): [ frac{1}{v_4} = frac{1}{v_1} + 3d ] Solving for ( d ): [ d = frac{frac{1}{v_4} - frac{1}{v_1}}{3} ]3. The intermediate terms ( frac{1}{v_2} ) and ( frac{1}{v_3} ) are calculated by adding the common difference ( d ) successively: [ frac{1}{v_2} = frac{1}{v_1} + d = frac{1}{v_1} + frac{frac{1}{v_4} - frac{1}{v_1}}{3} = frac{2}{3v_1} + frac{1}{3v_4} ] [ frac{1}{v_3} = frac{1}{v_2} + d = frac{2}{3v_1} + frac{1}{3v_4} + frac{frac{1}{v_4} - frac{1}{v_1}}{3} = frac{1}{3v_1} + frac{2}{3v_4} ]4. Taking the reciprocals of these intermediate terms gives the speeds ( v_2 ) and ( v_3 ): [ v_2 = frac{3}{frac{2}{v_1} + frac{1}{v_4}} = frac{3v_1v_4}{2v_4 + v_1} ] [ v_3 = frac{3}{frac{1}{v_1} + frac{2}{v_4}} = frac{3v_1v_4}{v_4 + 2v_1} ]Thus, the speeds of the other two vehicles are boxed{v_2 = dfrac{3v_1v_4}{2v_4 + v_1}} and boxed{v_3 = dfrac{3v_1v_4}{v_4 + 2v_1}}.,<think>Okay, let's see. So there are four vehicles starting from point A at the same time and arriving at point B one after another with equal time intervals. The fastest has speed v1 and the slowest has speed v4. I need to find the speeds of the other two vehicles. Hmm.First, let me try to visualize the problem. All four vehicles leave A simultaneously. They all reach B, but not at the same time. The arrivals are successive with equal time intervals. So, if the first vehicle arrives at B at time t, the next one arrives at t + Δt, then t + 2Δt, and the last one at t + 3Δt, where Δt is the equal time interval between arrivals. The fastest vehicle must arrive first, right? Because it's the fastest, so it takes the least time. Then the next one, which is slower, arrives Δt later, and so on. So the order of arrival is from fastest to slowest: v1, then the next two vehicles with speeds v2 and v3, and finally v4.So, each vehicle takes a different amount of time to cover the distance from A to B. Let's denote the distance between A and B as D. Then, the time taken by each vehicle would be D divided by their respective speeds. Since they start at the same time, the arrival times are D/v1, D/v2, D/v3, D/v4. These arrival times should be equally spaced. So, the difference between consecutive arrival times is Δt.Therefore, we have:D/v2 - D/v1 = ΔtD/v3 - D/v2 = ΔtD/v4 - D/v3 = ΔtSo, three equations here. Each time difference is equal. So, if I can write these equations, maybe I can solve for v2 and v3 in terms of v1 and v4.Let me write them out more clearly:1. (D/v2) - (D/v1) = Δt2. (D/v3) - (D/v2) = Δt3. (D/v4) - (D/v3) = ΔtSince D is the same for all, maybe we can set D to 1 for simplicity, as it will cancel out later. Wait, but let me check. If we set D to a constant, then the equations become differences in reciprocal speeds. Let's see.Alternatively, maybe it's better to keep D as a variable. Let me proceed.From the first equation:D/v2 - D/v1 = Δt => D(1/v2 - 1/v1) = ΔtSimilarly, second equation:D(1/v3 - 1/v2) = ΔtThird equation:D(1/v4 - 1/v3) = ΔtSince all three right-hand sides are equal to Δt, the left-hand sides must be equal to each other. Therefore:1/v2 - 1/v1 = 1/v3 - 1/v2 = 1/v4 - 1/v3So, the differences between reciprocals of consecutive speeds are equal. Let me denote 1/v1, 1/v2, 1/v3, 1/v4 as t1, t2, t3, t4, respectively. Then, the differences t2 - t1, t3 - t2, t4 - t3 are equal. So, the reciprocals of the speeds form an arithmetic sequence. Wait, that's interesting. So, 1/v1, 1/v2, 1/v3, 1/v4 is an arithmetic progression?If that's the case, then the reciprocals of the speeds are in arithmetic progression. So, the difference between consecutive terms is constant. Let me verify. If 1/v2 - 1/v1 = 1/v3 - 1/v2 = 1/v4 - 1/v3 = Δt/D, then yes, it's an arithmetic progression with common difference Δt/D. So, if that's the case, then 1/v2, 1/v3 can be expressed in terms of 1/v1 and 1/v4.Given four terms in arithmetic progression: t1, t2, t3, t4. Then, t2 = t1 + d, t3 = t1 + 2d, t4 = t1 + 3d. So, the common difference d is (t4 - t1)/3. Therefore, t2 = t1 + (t4 - t1)/3 = (2t1 + t4)/3, and t3 = t1 + 2(t4 - t1)/3 = (t1 + 2t4)/3.Therefore, since t1 = 1/v1 and t4 = 1/v4, then t2 = (2(1/v1) + (1/v4))/3, and t3 = ((1/v1) + 2(1/v4))/3. Therefore, the reciprocals of the speeds v2 and v3 are the averages as such. Therefore, to get v2 and v3, we take the reciprocals:v2 = 1 / [(2/(3v1) + 1/(3v4))] = 3 / (2/v1 + 1/v4) = 3v1v4 / (2v4 + v1)Similarly, v3 = 1 / [(1/(3v1) + 2/(3v4))] = 3 / (1/v1 + 2/v4) = 3v1v4 / (v4 + 2v1)Wait, that seems plausible. Let me check. If the reciprocals of the speeds form an arithmetic progression, then the reciprocals of v2 and v3 would be the averages as per the AP. So, for four terms, the common difference is (1/v4 - 1/v1)/3. Then, 1/v2 = 1/v1 + (1/v4 - 1/v1)/3 = (3(1/v1) + 1/v4 - 1/v1)/3 = (2/v1 + 1/v4)/3, so yes, that's correct. Similarly for 1/v3 = 1/v1 + 2*(1/v4 - 1/v1)/3 = (1/v1 + 2/v4)/3. Therefore, inverting those gives v2 and v3 as above.Alternatively, maybe we can write it as harmonic mean. Since the reciprocals are in AP, the speeds are in harmonic progression. So, the intermediate terms can be expressed as harmonic means. Hmm.Alternatively, let me verify with an example. Suppose the distance D is, say, 60 units. Let v1 be 60 units per hour, so time taken is 1 hour. Let v4 be 20 units per hour, so time taken is 3 hours. Then, the time intervals between arrivals are equal. So, four vehicles arrive at 1 hour, 1 + Δt, 1 + 2Δt, 1 + 3Δt. The last one arrives at 3 hours, so 1 + 3Δt = 3 => Δt = (3 - 1)/3 = 2/3 hours. So, the arrival times are 1, 1 + 2/3 = 5/3, 5/3 + 2/3 = 7/3, 7/3 + 2/3 = 3. So, the times are 1, 1.666..., 2.333..., 3.So, the times taken by each vehicle are 1, 5/3, 7/3, 3. Therefore, the speeds would be D divided by time: 60/1 = 60, 60/(5/3) = 36, 60/(7/3) ≈ 25.714, 60/3 = 20. So, the speeds are 60, 36, 25.714..., 20. Let's check the reciprocals. 1/60, 1/36, 1/25.714..., 1/20. Let's compute the differences.1/36 - 1/60 = (5 - 3)/180 = 2/180 = 1/901/25.714... ≈ 1/(180/7) = 7/180 ≈ 0.0389. Then, 7/180 - 1/36 = 7/180 - 5/180 = 2/180 = 1/90Similarly, 1/20 - 7/180 = 9/180 - 7/180 = 2/180 = 1/90. So yes, the reciprocals have equal differences of 1/90. Therefore, reciprocals form an AP with common difference 1/90. So, that example confirms that the reciprocals of the speeds are in AP.Therefore, in general, if the arrival times are equally spaced, then the reciprocals of the speeds are in arithmetic progression. Therefore, given v1 and v4, the other two speeds can be found by taking the harmonic mean appropriately.So, to find v2 and v3:First, compute the reciprocals: t1 = 1/v1, t4 = 1/v4. The reciprocals t2 and t3 are spaced equally between t1 and t4. Since there are three intervals between four terms, the common difference d is (t4 - t1)/3. Therefore:t2 = t1 + d = t1 + (t4 - t1)/3 = (2t1 + t4)/3t3 = t1 + 2d = t1 + 2(t4 - t1)/3 = (t1 + 2t4)/3Therefore, converting back to speeds:v2 = 1/t2 = 3/(2t1 + t4) = 3/(2/v1 + 1/v4) = 3v1v4/(2v4 + v1)Similarly, v3 = 1/t3 = 3/(t1 + 2t4) = 3/(1/v1 + 2/v4) = 3v1v4/(v4 + 2v1)So, these are the expressions for v2 and v3 in terms of v1 and v4. Therefore, the speeds of the other two vehicles are 3v1v4/(2v4 + v1) and 3v1v4/(v4 + 2v1). Let me check if this matches the example I took earlier.In the example, v1 was 60 and v4 was 20.So, v2 = 3*60*20/(2*20 + 60) = 3600/(40 + 60) = 3600/100 = 36. Which matches.v3 = 3*60*20/(20 + 2*60) = 3600/(20 + 120) = 3600/140 ≈ 25.714..., which also matches. So, correct.Therefore, the answer is v2 = 3v1v4/(v1 + 2v4) and v3 = 3v1v4/(2v1 + v4). Wait, wait, hold on. Wait in the first case, denominator was (2v4 + v1). Wait, in the first calculation:v2 = 3v1v4/(2v4 + v1)Wait, 2v4 + v1 is equivalent to v1 + 2v4, but ordered differently. Similarly, v3's denominator is v4 + 2v1. So, written as:v2 = (3v1v4)/(v1 + 2v4)v3 = (3v1v4)/(2v1 + v4)Yes, that's equivalent. So, depending on how you write it. So, the two speeds are 3v1v4 divided by (v1 + 2v4) and 3v1v4 divided by (2v1 + v4). Therefore, the answer should be expressed as:v2 = dfrac{3v_1v_4}{v_1 + 2v_4}andv3 = dfrac{3v_1v_4}{2v_1 + v_4}Hence, that's the solution.Final AnswerThe speeds of the other two vehicles are boxed{v_2 = dfrac{3v_1v_4}{v_1 + 2v_4}} and boxed{v_3 = dfrac{3v_1v_4}{2v_1 + v_4}}.</think>Given four vehicles starting from point ( A ) at the same time and arriving at point ( B ) with equal time intervals, we need to determine the speeds of the two intermediate vehicles, ( v_2 ) and ( v_3 ), given the fastest speed ( v_1 ) and the slowest speed ( v_4 ).1. The arrival times of the vehicles are equally spaced. Let ( D ) be the distance between ( A ) and ( B ). The time taken by each vehicle is ( frac{D}{v_i} ) for ( i = 1, 2, 3, 4 ).2. The differences between the arrival times of consecutive vehicles are equal, leading to the following equations: [ frac{D}{v_2} - frac{D}{v_1} = Delta t ] [ frac{D}{v_3} - frac{D}{v_2} = Delta t ] [ frac{D}{v_4} - frac{D}{v_3} = Delta t ]3. Dividing through by ( D ), we find that the reciprocals of the speeds form an arithmetic progression: [ frac{1}{v_2} - frac{1}{v_1} = frac{1}{v_3} - frac{1}{v_2} = frac{1}{v_4} - frac{1}{v_3} ]4. Let ( t_1 = frac{1}{v_1} ), ( t_2 = frac{1}{v_2} ), ( t_3 = frac{1}{v_3} ), and ( t_4 = frac{1}{v_4} ). The common difference ( d ) in the arithmetic progression is: [ d = frac{t_4 - t_1}{3} ]5. Solving for ( t_2 ) and ( t_3 ): [ t_2 = t_1 + frac{t_4 - t_1}{3} = frac{2t_1 + t_4}{3} ] [ t_3 = t_1 + 2 cdot frac{t_4 - t_1}{3} = frac{t_1 + 2t_4}{3} ]6. Converting back to speeds: [ v_2 = frac{1}{t_2} = frac{3}{frac{2}{v_1} + frac{1}{v_4}} = frac{3v_1v_4}{2v_4 + v_1} ] [ v_3 = frac{1}{t_3} = frac{3}{frac{1}{v_1} + frac{2}{v_4}} = frac{3v_1v_4}{v_4 + 2v_1} ]The speeds of the other two vehicles are (boxed{v_2 = dfrac{3v_1v_4}{v_1 + 2v_4}}) and (boxed{v_3 = dfrac{3v_1v_4}{2v_1 + v_4}}).]

🔑:If the fastest vehicle covers the distance AB = s in time t_{1}, then t_{1} = frac{s}{v_{1}}. The slowest vehicle covers this distance in time t_{4}, t_{4} = frac{s}{v_{4}}.Since the vehicles arrive at B at equal intervals, the time interval t_{4} - t_{1} must be divided into 3 equal parts to find the time between the arrivals of the individual vehicles.frac{t_{4} - t_{1}}{3} = frac{frac{s}{v_{4}} - frac{s}{v_{1}}}{3} = Delta tThe travel time of the 2nd vehicle is thereforet_{2} = t_{1} + Delta t = frac{s}{v_{1}} + frac{frac{s}{v_{4}} - frac{s}{v_{1}}}{3} = frac{1}{3} left( frac{2s}{v_{1}} + frac{s}{v_{4}} right)and its speed isv_{2} = frac{s}{t_{2}} = frac{s}{frac{1}{3} left( frac{2s}{v_{1}} + frac{s}{v_{4}} right)} = frac{3 v_{1} v_{4}}{2 v_{4} + v_{1}}The travel time of the 3rd vehicle is similarlyt_{3} = t_{4} - Delta t = frac{s}{3} left( frac{2}{v_{4}} + frac{1}{v_{1}} right)and its speed isv_{3} = frac{s}{t_{3}} = frac{3 v_{1} v_{4}}{v_{4} + 2 v_{1}}Remark. The problem can be generalized to n vehicles. In this case, the size of the intervals is Delta t = frac{t_{n} - t_{1}}{n-1} = frac{frac{s}{v_{n}} - frac{s}{v_{1}}}{n-1}, and the speed of the k-th vehicle isv_{k} = frac{s}{t_{k}} = frac{(n-1) v_{1} v_{n}}{(n-1) v_{n} + (k-1) v_{1} - (k-1) v_{k}} = frac{(n-1) v_{1} v_{n}}{(n-k) v_{n} + (k-1) v_{1}}Béla Gajdos (Beregszász, Bethlen G. Gymnasium, 12th grade)

🧠:[<think>Okay, so I need to find all pairs of positive integers (a, b) that satisfy the inequalities 11ab ≤ a³ - b³ ≤ 12ab. Hmm, let's start by understanding what this means. Both a and b are positive integers, so they must be at least 1. The inequalities involve cubic terms and a product term. Maybe I can manipulate these inequalities to find relationships between a and b.First, let me rewrite the inequalities:11ab ≤ a³ - b³ ≤ 12abI can factor the middle term a³ - b³. Remember that a³ - b³ factors into (a - b)(a² + ab + b²). Let's write that down:(a - b)(a² + ab + b²) ≥ 11aband(a - b)(a² + ab + b²) ≤ 12abHmm, so both inequalities involve the same expression (a - b)(a² + ab + b²). Maybe I can find bounds for this expression in terms of ab. Let's think about possible values of a and b.Since a and b are positive integers, and we have a³ - b³ on the left and right sides, which suggests that a must be larger than b. Because if a were equal to or smaller than b, then a³ - b³ would be non-positive, but the right side is 12ab, which is positive. Therefore, a must be greater than b. So (a - b) is a positive integer, at least 1.Let me denote k = a - b, where k is a positive integer. Then a = b + k. Maybe substituting this into the inequalities will help. Let's try that.Set a = b + k. Then substitute into the inequalities:First inequality: 11ab ≤ (a³ - b³)Second inequality: (a³ - b³) ≤ 12abLet's compute a³ - b³ with a = b + k:a³ - b³ = (b + k)³ - b³ = 3b²k + 3bk² + k³So substituting into the inequalities:First inequality: 11ab ≤ 3b²k + 3bk² + k³Second inequality: 3b²k + 3bk² + k³ ≤ 12abNow, substitute a = b + k into ab:ab = (b + k) * b = b² + bkTherefore, the first inequality becomes:11(b² + bk) ≤ 3b²k + 3bk² + k³Similarly, the second inequality becomes:3b²k + 3bk² + k³ ≤ 12(b² + bk)Let me write these out:First inequality:11b² + 11bk ≤ 3b²k + 3bk² + k³Second inequality:3b²k + 3bk² + k³ ≤ 12b² + 12bkHmm, these look complicated, but perhaps we can rearrange terms.Starting with the first inequality:11b² + 11bk - 3b²k - 3bk² - k³ ≤ 0Similarly, the second inequality:3b²k + 3bk² + k³ - 12b² - 12bk ≤ 0Let's factor these expressions if possible. Maybe factor out common terms.First inequality:11b² - 3b²k + 11bk - 3bk² - k³ ≤ 0Factor b² terms:b²(11 - 3k) + bk(11 - 3k) - k³ ≤ 0Wait, 11b² - 3b²k = b²(11 - 3k), and 11bk - 3bk² = bk(11 - 3k). So:b²(11 - 3k) + bk(11 - 3k) - k³ ≤ 0Factor out (11 - 3k):(11 - 3k)(b² + bk) - k³ ≤ 0But (b² + bk) is ab, which is positive. Hmm, so the first term is (11 - 3k)ab - k³ ≤ 0Alternatively, maybe that's not helpful. Let's see.If we factor (11 - 3k) from the first two terms:(11 - 3k)(b² + bk) - k³ ≤ 0But since ab = b² + bk, so:(11 - 3k)ab - k³ ≤ 0But we have:11ab - 3kab - k³ ≤ 0But the original first inequality is 11ab ≤ 3b²k + 3bk² + k³, which is equivalent to 11ab ≤ (a³ - b³). So another way to write this is:3b²k + 3bk² + k³ - 11ab ≥ 0Which is the same as:3b²k + 3bk² + k³ - 11ab ≥ 0But perhaps this isn't helpful. Maybe I need to approach this differently.Alternatively, let's divide both sides of the inequalities by b³ to normalize them. Since b is a positive integer, this is allowed.Let me set t = a/b, which is a rational number greater than 1 since a > b. Then t = (b + k)/b = 1 + k/b. Since a and b are positive integers, t is a rational number greater than 1.Expressing the inequalities in terms of t:First inequality: 11ab ≤ a³ - b³Divide both sides by b³:11*(a/b)*(b/b) ≤ (a³/b³ - 1)11t ≤ t³ - 1Similarly, second inequality:a³ - b³ ≤ 12abDivide by b³:t³ - 1 ≤ 12tSo now, we have:11t ≤ t³ - 1 ≤ 12tWhich simplifies to:11t + 1 ≤ t³ ≤ 12t + 1Therefore, solving for t in the real numbers:11t + 1 ≤ t³ and t³ ≤ 12t + 1Let me solve these inequalities for t.First, 11t + 1 ≤ t³Let me rearrange this as t³ - 11t - 1 ≥ 0Similarly, t³ - 12t - 1 ≤ 0So t must satisfy both t³ - 11t - 1 ≥ 0 and t³ - 12t - 1 ≤ 0Let's find the real roots of these equations.First equation: t³ - 11t - 1 = 0Let me check approximate solutions. Let's test t=3: 27 - 33 -1 = -7 < 0t=4: 64 - 44 -1=19>0So there is a root between 3 and 4.Similarly, t³ - 12t -1=0At t=3: 27 - 36 -1= -10 <0t=4: 64 - 48 -1=15>0So another root between 3 and 4.Therefore, the real solutions for t lie in intervals where t³ -11t -1 ≥0 and t³ -12t -1 ≤0But since t must be greater than 1, and we have the approximate roots between 3 and 4, maybe the valid t is between approx 3.5 and 4?Wait, but let's check exact roots.Alternatively, maybe we can use calculus to find where t³ -11t -1 is increasing or decreasing.The derivative of t³ -11t -1 is 3t² -11. Setting to zero: t= sqrt(11/3) ≈ 1.915. So the function t³ -11t -1 has a local maximum at t ≈1.915 and then increases after that. At t=3, it's negative, and at t=4, positive. So only one real root after t≈1.915.Similarly for t³ -12t -1, derivative 3t² -12. Critical points at t=2 and t=-2. So increasing for t>2. At t=3, 27 -36 -1= -10; t=4, 64 -48 -1=15. So root between 3 and4.Thus, the interval of t where 11t +1 ≤ t³ ≤12t +1 is between the roots of t³ -11t -1=0 and t³ -12t -1=0. Since both roots are between 3 and4, maybe t is in (approx) 3.5 to 4.But since t is rational (because t = a/b, a and b integers), and t >1, perhaps possible values of t are fractions with small denominators. Maybe t=4, 3.5, 3.333..., etc.But let's check possible t values.But t must be a rational number, as t = a/b where a and b are positive integers. So t is a positive rational number greater than 1.Alternatively, maybe I can set t = n/m where n and m are positive integers with n > m and gcd(n, m)=1. Then a = n, b = m. Wait, but not exactly. Because t = a/b, so a = tb. If t is rational, then a must be a multiple of the denominator when t is expressed in lowest terms.But maybe instead of dealing with fractions, since a and b are integers, perhaps we can look for small integer values of k = a - b. Since k is a positive integer, and a = b +k.Given that t is around 3 to 4, maybe k is 1, 2, 3, or something like that. Let's try small values of k and see if we can find integer solutions.Let me start with k=1. So a = b +1.Then substitute into the original inequalities:First inequality: 11ab ≤ a³ - b³Second inequality: a³ - b³ ≤12abWith a = b +1, compute a³ - b³ = (b+1)^3 - b^3 = 3b² + 3b +1ab = b(b +1) = b² + bSo first inequality: 11(b² + b) ≤ 3b² + 3b +1Compute left side: 11b² +11bRight side: 3b² +3b +1Subtract left side from right side: (3b² +3b +1) - (11b² +11b) = -8b² -8b +1 ≥0So -8b² -8b +1 ≥0 ⇒ 8b² +8b -1 ≤0Solve 8b² +8b -1 ≤0Quadratic equation 8b² +8b -1 =0. Discriminant: 64 +32=96. sqrt(96)=4*sqrt(6)≈9.798Solutions: (-8 ±9.798)/16. Positive solution: (1.798)/16≈0.112. So b must be ≤0.112, but b is a positive integer. Therefore, no solutions for k=1.So k=1 is impossible.Next, k=2. Then a = b +2.Compute a³ -b³ = (b+2)^3 -b^3 = 6b² +12b +8ab = b(b +2) =b² +2bFirst inequality: 11(b² +2b) ≤6b² +12b +8Left side:11b² +22bRight side:6b² +12b +8Subtract left side from right side:6b² +12b +8 -11b² -22b = -5b² -10b +8 ≥0So -5b² -10b +8 ≥0 ⇒5b² +10b -8 ≤0Solve 5b² +10b -8=0. Discriminant:100 +160=260. sqrt(260)≈16.12Solutions: (-10 ±16.12)/10. Positive solution: (6.12)/10≈0.612. So b must be ≤0.612, but b is positive integer. No solutions for k=2.Next, k=3. Then a = b +3.Compute a³ -b³ = (b+3)^3 -b^3 =9b² +27b +27ab =b(b +3)=b² +3bFirst inequality:11(b² +3b) ≤9b² +27b +27Left side:11b² +33bRight side:9b² +27b +27Subtract left from right:9b² +27b +27 -11b² -33b = -2b² -6b +27 ≥0So -2b² -6b +27 ≥0 ⇒2b² +6b -27 ≤0Solve 2b² +6b -27=0. Discriminant:36 +216=252. sqrt(252)=6*sqrt(7)≈15.874Solutions: (-6 ±15.874)/4. Positive solution: (9.874)/4≈2.468. So b ≤2.468. Since b is a positive integer, possible b=1,2.Check b=1:a=1+3=4Check original inequalities:First inequality:11*1*4=44. a³ -b³=64 -1=63. 44 ≤63, which is true.Second inequality:63 ≤12*1*4=48? 63 ≤48? No. So fails second inequality.Check b=2:a=2+3=5First inequality:11*2*5=110. a³ -b³=125 -8=117. 110 ≤117, true.Second inequality:117 ≤12*2*5=120. 117 ≤120, true. So (5,2) is a solution.So with k=3, b=2 gives a valid solution. Let's check if there are other possible b.Wait, the quadratic inequality allows b up to approximately 2.468, so only b=1 and 2. But b=1 failed the second inequality, but b=2 worked. So for k=3, only b=2 works.Next, check k=4. a =b +4.Compute a³ -b³ = (b +4)^3 -b³ =12b² +48b +64ab =b(b +4)=b² +4bFirst inequality:11(b² +4b) ≤12b² +48b +64Left side:11b² +44bRight side:12b² +48b +64Subtract left from right:12b² +48b +64 -11b² -44b =b² +4b +64 ≥0. Which is always true. So first inequality holds for all b.Second inequality:12b² +48b +64 ≤12(b² +4b)Compute right side:12b² +48bLeft side:12b² +48b +64 ≤12b² +48bSubtract left from right:12b² +48b - (12b² +48b +64)= -64 ≥0? Which is -64 ≥0, which is false. Therefore, no solutions for k=4.So for k=4, the second inequality cannot be satisfied.Next, k=5. a =b +5.Compute a³ -b³= (b +5)^3 -b³=15b² +75b +125ab =b(b +5)=b² +5bFirst inequality:11(b² +5b) ≤15b² +75b +125Left side:11b² +55bRight side:15b² +75b +125Subtract left from right:4b² +20b +125 ≥0, which is always true.Second inequality:15b² +75b +125 ≤12(b² +5b)Right side:12b² +60bLeft side:15b² +75b +125 ≤12b² +60bSubtract left from right:12b² +60b -15b² -75b -125 = -3b² -15b -125 ≥0?Which is -3b² -15b -125 ≥0, which is impossible since all terms are negative. So no solutions for k=5.Similarly, higher k will only make the left side (a³ -b³) even larger compared to the right side 12ab, so inequalities won't hold. So k=3 is the only possible value so far with solution (5,2).Wait, but maybe there are other k values with b larger? Wait, for k=3, we found b=2. What about k=3 and b=3? Wait, when k=3, we substituted and found that the quadratic inequality allows b up to approximately 2.468, so only b=1,2. But b=1 didn't work, and b=2 did. So k=3, b=2 is the only solution in that case.Wait, let's check k=3 for b=3. Wait, no, according to our substitution, a = b +k. If k=3 and b=3, then a=6. Let's check if that works.Check (a,b)=(6,3):First inequality:11*6*3=198. a³ -b³=216 -27=189. 198 ≤189? No, so it doesn't satisfy the first inequality. Therefore, even though k=3 and b=3 would have a=6, it's invalid.Wait, but according to the earlier substitution for k=3, when we set a =b +3, we found that only b=2 works because the quadratic inequality restricted b to ≤2.468. Therefore, for k=3, only b=2 is valid. So (5,2) is the only solution for k=3.What about k=6? Probably similar to k=4 and k=5, where second inequality fails.But let's check k=0. Wait, but k=a -b must be at least 1, so k=0 is invalid.Wait, perhaps there are other possibilities where k is larger but with larger b? Let me think. Suppose k is fixed, but b is large. Let's see if for any k, there are solutions with large b.But as b increases, let's see the behavior of the inequalities. Let's take a = b +k.First inequality:11ab ≤a³ -b³Which is 11b(b +k) ≤ (b +k)^3 -b³ =3b²k +3bk² +k³Divide both sides by b² (assuming b is large):11(1 + k/b) ≤3k + 3k²/b +k³/b²As b approaches infinity, the left side approaches 11, and the right side approaches 3k. Therefore, for the inequality to hold for large b, we need 11 ≤3k. Therefore, k ≥4. But earlier when k=4, the second inequality failed. Let's check for k=4 and large b.Take k=4, so a =b +4. Then compute the second inequality:a³ -b³ =12b² +48b +64 ≤12ab=12b(b +4)=12b² +48bThus, 12b² +48b +64 ≤12b² +48bWhich simplifies to 64 ≤0, which is impossible. So even for large b, k=4 is invalid.Similarly, for k=5, a³ -b³=15b² +75b +125. 12ab=12b² +60b. Then 15b² +75b +125 ≤12b² +60b ⇒3b² +15b +125 ≤0, which is impossible.Therefore, for k≥4, even with large b, the second inequality is not satisfied. Hence, no solutions for k≥4.Therefore, only k=3 gives a solution with b=2, which gives a=5.Wait, but let's check if there are any other possible k values. Maybe k=3 and higher b? Wait, when k=3 and b=2, we have a=5. If b increases, but k=3, then a =b +3.Wait, let's check for k=3 and b=3. As before, a=6, but 11ab=11*6*3=198. a³ -b³=216-27=189. 198 ≤189 is false. So b=3 with k=3 doesn't work.Similarly, b=1 with k=3: a=4. Check first inequality:11*4*1=44. a³ -b³=64 -1=63. 44 ≤63 is true. Second inequality:63 ≤12*4*1=48? No. So fails.Therefore, only b=2 with k=3 works.But wait, perhaps there are other k values where b is larger? Let's try k=2. Wait, for k=2, we saw that there are no solutions. Similarly, k=1, no solutions.Alternatively, maybe there are solutions where k is not an integer? Wait, no, because a and b are integers, so k=a -b must be a positive integer.Alternatively, maybe a and b are such that a/b is exactly 4? For example, a=4, b=1. Let's check.(a,b)=(4,1):First inequality:11*4*1=44. a³ -b³=64 -1=63. 44 ≤63, true.Second inequality:63 ≤12*4*1=48? 63 ≤48? No. So fails.Another example, a=8, b=2. Then a/b=4. Check:First inequality:11*8*2=176. a³ -b³=512 -8=504. 176 ≤504, true.Second inequality:504 ≤12*8*2=192? No. 504 ≤192? No. So fails.Similarly, a=12, b=3: a³ -b³=1728 -27=1701. 12ab=12*12*3=432. 1701 ≤432? No. So even with higher multiples, the second inequality fails.So seems like the only solution is (5,2).Wait, let's check if there are other solutions with different k. Wait, what if k=3 but b=4? Wait, no, that would make a=7, let's check:a=7, b=4.First inequality:11*7*4=308. a³ -b³=343 -64=279. 308 ≤279? No. So fails.Similarly, a=6, b=3. As before, fails.Alternatively, a=5, b=2 is the only solution? Let's see.Wait, another approach: Let's consider the original inequalities:11ab ≤a³ -b³ ≤12abWe can divide all terms by b³ (since b is positive):11*(a/b)*(1/b²) ≤ (a/b)^3 -1 ≤12*(a/b)*(1/b²)But this complicates things. Alternatively, think of t = a/b as before. Then:11t ≤ t³ -1 ≤12tSo t³ -1 is between 11t and12t.Therefore, 11t ≤t³ -1 ≤12tSo first inequality: t³ -11t -1 ≥0Second inequality: t³ -12t -1 ≤0We can look for rational solutions t = p/q, p and q positive integers, p>q, such that t³ -11t -1 ≥0 and t³ -12t -1 ≤0.But solving for rational t is challenging. Alternatively, think of possible t values where this holds.Earlier, approximate roots between 3 and4. Let's check t=3.5. 3.5³=42.875, 11*3.5=38.5, 42.875 -1=41.875. 41.875 ≥38.5, which is true. 12*3.5=42. So 42.875 -1=41.875 ≤42? Yes. So t=3.5 would satisfy the inequalities.But t must be rational. 3.5 is 7/2. So a/b=7/2. Therefore, a=7, b=2. Let's check if that works.Check (7,2):First inequality:11*7*2=154. a³ -b³=343 -8=335. 154 ≤335, true.Second inequality:335 ≤12*7*2=168? 335 ≤168? No. So fails.Hmm, so even though t=3.5 is a solution to the real inequalities, when converted to integers, the second inequality fails. So this suggests that rational t's that are solutions in the real case might not correspond to integer solutions.Alternatively, maybe t=5/2=2.5. Let's check:t³ -11t -1 =15.625 -27.5 -1= -12.875 <0. Doesn't satisfy first inequality.t=4: 64 -44 -1=19 ≥0. 64 -48 -1=15 ≤0? No, 15>0. So t=4 doesn't satisfy the second inequality.t=3:27 -33 -1=-7 <0, so first inequality fails.t=3. Let's check another value.Wait, maybe t= (some fraction). For example, t= 5/1=5. Then t³ -11t -1=125 -55 -1=69 ≥0. t³ -12t -1=125 -60 -1=64 ≥0. So doesn't satisfy second inequality.t= 3.2, which is 16/5.16/5 cubed is (4096)/125 ≈32.768. 11*(16/5)=35.2. 32.768 -1=31.768 <35.2. Fails first inequality.t=3.4=17/5.17/5 cubed=4913/125≈39.304. 11*3.4=37.4. 39.304 -1=38.304. 38.304 ≥37.4, so first inequality holds. Second inequality:39.304 -1=38.304 ≤12*3.4=40.8. Yes. So t≈3.4 satisfies both inequalities. Then t=17/5. So a=17, b=5.Check (17,5):First inequality:11*17*5=935. a³ -b³=4913 -125=4788. 935 ≤4788, true.Second inequality:4788 ≤12*17*5=1020? 4788 ≤1020? No. So fails.Thus, even if t is a rational number that satisfies the real inequalities, the corresponding integer pair may not satisfy the original inequalities because multiplying by b³ can make the left side much larger.Therefore, this approach might not be helpful. Let's return to the earlier approach where we set a =b +k and found that only k=3, b=2 gives a solution.Alternatively, let's check for other possible k and b combinations. Maybe k=3 and higher b. But as we saw with b=3, it doesn't work. Let's check another k.Wait, maybe k=6 and b= something? Wait, let's check for k=3 and b=2, we have (5,2). What about k=4 and b=1?a=5, b=1.First inequality:11*5*1=55. a³ -b³=125 -1=124. 55 ≤124, true.Second inequality:124 ≤12*5*1=60? 124 ≤60? No. Fails.Alternatively, k=4, b=2. Then a=6.First inequality:11*6*2=132. a³ -b³=216 -8=208. 132 ≤208, true.Second inequality:208 ≤12*6*2=144? 208 ≤144? No. Fails.Another try: k=3, b=4. a=7.First inequality:11*7*4=308. a³ -b³=343 -64=279. 308 ≤279? No. Fails.Alternatively, k=2, b=3. a=5.First inequality:11*5*3=165. a³ -b³=125 -27=98. 165 ≤98? No. Fails.Alternatively, k=3, b=1. a=4. Which we checked before, fails second inequality.So it seems like (5,2) is the only solution.Wait, let's check (6,1):a=6, b=1.First inequality:11*6*1=66. a³ -b³=216 -1=215. 66 ≤215, true.Second inequality:215 ≤12*6*1=72? No.Same with a=7, b=1: First inequality:77 ≤342, true. Second:342 ≤84? No.Alternatively, a=4, b=1:44 ≤63, true. 63 ≤48? No.So none of these work.Another thought: Maybe there's a solution with b=2 and a=5.Wait, we already have that. Any other b=2?Check a=6, b=2:11*6*2=132. a³ -b³=216 -8=208. 132 ≤208. Second inequality:208 ≤12*6*2=144? No.a=4, b=2:11*4*2=88. 64 -8=56. 88 ≤56? No.So only a=5, b=2 works.Check b=3, a=6:11*6*3=198. a³ -b³=216 -27=189. 198 ≤189? No.Alternatively, b=4, a=7:11*7*4=308. 343 -64=279. 308 ≤279? No.Therefore, all other b with higher a do not work.Wait, perhaps b=2 and a=5 is the only solution. Let me check if there are other possibilities where k=3 but with larger b. For example, if k=3 and b=4, then a=7. As checked before, 11*7*4=308. 343-64=279. 308 ≤279? No.Alternatively, if k=3 and b=0, but b must be positive. So no.Thus, after checking all possible k values up to 6 and various b values, the only pair that satisfies both inequalities is (5,2).Therefore, the solution is boxed{(5, 2)}.Final AnswerThe only solution is boxed{(5, 2)}.</think>To find all pairs ((a, b)) of positive integers such that (11ab leq a^3 - b^3 leq 12ab), we start by factoring (a^3 - b^3) as ((a - b)(a^2 + ab + b^2)). We then consider the inequalities:1. (11ab leq (a - b)(a^2 + ab + b^2))2. ((a - b)(a^2 + ab + b^2) leq 12ab)We introduce (k = a - b) (where (k) is a positive integer) and set (a = b + k). Substituting (a = b + k) into the inequalities, we get:1. (11b(b + k) leq (b + k)^3 - b^3)2. ((b + k)^3 - b^3 leq 12b(b + k))Expanding and simplifying these inequalities, we test small values of (k) and corresponding (b) values:- For (k = 1, 2, 4, 5, 6), the inequalities fail for all positive integers (b).- For (k = 3), we substitute (a = b + 3) and simplify the inequalities: - First inequality: (11(b^2 + 3b) leq 9b^2 + 27b + 27) - Second inequality: (9b^2 + 27b + 27 leq 12(b^2 + 3b))Solving these inequalities, we find that (b = 2) works, leading to (a = 5). Checking the pair ((5, 2)):- First inequality: (11 cdot 5 cdot 2 = 110 leq 125 - 8 = 117)- Second inequality: (117 leq 12 cdot 5 cdot 2 = 120)Both inequalities are satisfied. Testing other values of (b) for (k = 3) and other values of (k) shows no additional solutions.Thus, the only solution is (boxed{(5, 2)}).,<think>Okay, let's try to tackle this problem. We need to find all pairs of positive integers (a, b) such that 11ab ≤ a³ - b³ ≤ 12ab. Hmm, so both a and b are positive integers, which means they are at least 1. Let me start by understanding the inequalities given.First, let's write down the inequalities:1. a³ - b³ ≥ 11ab2. a³ - b³ ≤ 12abSo, the difference between a³ and b³ is sandwiched between 11ab and 12ab. Since a and b are positive integers, a must be greater than b because a³ - b³ is positive (since 11ab is positive). Therefore, a > b. That makes sense because if a were equal to or less than b, then a³ - b³ would be non-positive, which can't satisfy the inequalities given that 11ab and 12ab are positive.So, a > b. Let's denote k = a - b, where k is a positive integer. Maybe substituting a = b + k could help. Let me try that.Let a = b + k. Then, we can express a³ - b³ in terms of b and k. The formula for a³ - b³ is (a - b)(a² + ab + b²). So substituting a = b + k:a³ - b³ = (b + k - b)[(b + k)² + (b + k)b + b²] = k[(b + k)² + b(b + k) + b²]Let me expand this:First, expand (b + k)²: b² + 2bk + k²Then, expand b(b + k): b² + bkThen, add the third term b².So, adding all these together:(b² + 2bk + k²) + (b² + bk) + b² = 3b² + 3bk + k²Therefore, a³ - b³ = k(3b² + 3bk + k²)So, substituting back into the original inequalities:11ab ≤ k(3b² + 3bk + k²) ≤ 12abBut since a = b + k, we can substitute a in terms of b and k in the left and right sides.First, let's write ab as (b + k)b = b² + bkSo, the inequalities become:11(b² + bk) ≤ k(3b² + 3bk + k²) ≤ 12(b² + bk)Let me divide all parts by b² (since b is positive, we can do this without changing the inequality direction). Let me set t = k/b. Wait, but since k and b are positive integers, t would be a positive rational number. Maybe this substitution complicates things. Alternatively, maybe we can divide through by b².Let me define x = k/b. Then, since k and b are positive integers, x can be a rational number such that x = m/n where m and n are positive integers with n dividing b. But perhaps this is not the easiest path. Alternatively, let's work with variables as they are.Let me rewrite the inequalities:Left inequality: 11(b² + bk) ≤ k(3b² + 3bk + k²)Right inequality: k(3b² + 3bk + k²) ≤ 12(b² + bk)Let me first work on the left inequality:11(b² + bk) ≤ k(3b² + 3bk + k²)Expanding the right side:3k b² + 3k² b + k³So, left inequality becomes:11b² + 11bk ≤ 3k b² + 3k² b + k³Bring all terms to the right side:0 ≤ 3k b² + 3k² b + k³ - 11b² - 11bkSimplify:0 ≤ (3k - 11)b² + (3k² - 11k)b + k³Similarly, for the right inequality:3k b² + 3k² b + k³ ≤ 12b² + 12bkBring all terms to the left:3k b² + 3k² b + k³ - 12b² - 12bk ≤ 0Simplify:(3k - 12)b² + (3k² - 12k)b + k³ ≤ 0So, now we have two inequalities:1. (3k - 11)b² + (3k² - 11k)b + k³ ≥ 02. (3k - 12)b² + (3k² - 12k)b + k³ ≤ 0These are quadratic in terms of b (if we treat k as a constant). Hmm, this seems a bit complicated, but maybe we can analyze for fixed k and find possible b. Since k and b are positive integers, and k = a - b, which is at least 1, perhaps we can check for small k values. Since k is a positive integer, starting from 1 upwards, and for each k, find the possible b.Let me try this approach. Let's let k = 1, 2, 3, etc., and see if there are possible b that satisfy both inequalities.Starting with k = 1:For k = 1, let's plug into the inequalities.First inequality:(3*1 - 11)b² + (3*1² - 11*1)b + 1³ ≥ 0(3 - 11)b² + (3 - 11)b + 1 ≥ 0(-8)b² + (-8)b + 1 ≥ 0This is a quadratic in b: -8b² -8b + 1 ≥ 0. Multiply both sides by -1 (which reverses the inequality):8b² + 8b -1 ≤ 0Solve 8b² +8b -1 ≤0. Find the roots of 8b² +8b -1 =0:Discriminant D = 64 + 32 = 96sqrt(96) = 4*sqrt(6) ≈ 9.798Thus, roots at (-8 ± 9.798)/16Positive root: (1.798)/16 ≈ 0.112, negative root: (-17.798)/16 ≈ -1.112So, the quadratic is ≤0 between the roots. Since b is a positive integer, the only possible b is when b is between ≈ -1.112 and 0.112, but b is positive. Therefore, no solution for k=1.Similarly, check the right inequality for k=1:(3*1 -12)b² + (3*1² -12*1)b +1³ ≤0(-9)b² + (-9)b +1 ≤0Again, quadratic in b: -9b² -9b +1 ≤0, multiply by -1:9b² +9b -1 ≥0Solve 9b² +9b -1 =0:Discriminant D=81 +36=117sqrt(117)≈10.816Roots: (-9 ±10.816)/18Positive root: (1.816)/18 ≈0.101, negative root: (-19.816)/18≈-1.101Thus, the quadratic 9b² +9b -1 is ≥0 when b ≤ -1.101 or b ≥0.101. Since b is positive integer, all b ≥1 satisfy this. But since the original inequality was -9b² -9b +1 ≤0, which becomes 9b² +9b -1 ≥0, so all b ≥1 satisfy this. But the left inequality had no solutions. Therefore, overall no solutions for k=1.Moving on to k=2:Left inequality:(3*2 -11)b² + (3*4 - 22)b +8 ≥0(6 -11)b² + (12 -22)b +8 ≥0(-5)b² + (-10)b +8 ≥0Multiply by -1 (reverse inequality):5b² +10b -8 ≤0Solve 5b² +10b -8 =0:Discriminant D=100 +160=260sqrt(260)=~16.124Roots: (-10 ±16.124)/10Positive root: (6.124)/10≈0.612, negative root: (-26.124)/10≈-2.612Thus, quadratic ≤0 between -2.612 and 0.612. Since b is positive integer, no solutions here. Therefore, left inequality not satisfied for k=2.Right inequality for k=2:(3*2 -12)b² + (3*4 -24)b +8 ≤0(6 -12)b² + (12 -24)b +8 ≤0(-6)b² + (-12)b +8 ≤0Multiply by -1 (reverse inequality):6b² +12b -8 ≥0Solve 6b² +12b -8=0:Discriminant D=144 +192=336sqrt(336)=~18.330Roots: (-12 ±18.330)/12Positive root: (6.330)/12≈0.5275, negative root: (-30.330)/12≈-2.5275So quadratic ≥0 when b ≤ -2.5275 or b ≥0.5275. Since b is positive integer, all b ≥1 satisfy this. Therefore, original inequality (right side) is satisfied for all b ≥1. But left inequality wasn't satisfied, so no solutions for k=2.Next, k=3:Left inequality:(9 -11)b² + (27 -33)b +27 ≥0(-2)b² + (-6)b +27 ≥0Which is -2b² -6b +27 ≥0. Multiply by -1:2b² +6b -27 ≤0Solve 2b² +6b -27 =0:Discriminant D=36 +216=252sqrt(252)=~15.874Roots: (-6 ±15.874)/4Positive root: (9.874)/4≈2.468, negative root≈-5.468Thus, quadratic ≤0 between -5.468 and 2.468. Since b is positive integer, possible b=1,2.Check if the original left inequality holds for b=1:-2(1)^2 -6(1) +27 = -2 -6 +27=19 ≥0, yes.For b=2:-2(4) -12 +27= -8 -12 +27=7 ≥0, yes.For b=3:-2(9) -18 +27= -18 -18 +27= -9 <0, which doesn't satisfy. So left inequality is satisfied for b=1,2.Now check the right inequality for k=3:(9 -12)b² + (27 -36)b +27 ≤0(-3)b² + (-9)b +27 ≤0Multiply by -1:3b² +9b -27 ≥0Divide both sides by 3:b² +3b -9 ≥0Solve b² +3b -9=0:Discriminant=9 +36=45sqrt(45)=6.708Roots: (-3 ±6.708)/2Positive root: (3.708)/2≈1.854, negative root≈-4.854Thus, quadratic ≥0 when b ≤ -4.854 or b ≥1.854. Since b is positive integer, b ≥2.Therefore, original right inequality is satisfied when b ≥2.So, for k=3:Left inequality holds for b=1,2.Right inequality holds for b ≥2.Therefore, overlapping values are b=2.Thus, check if b=2 satisfies both inequalities.Compute original inequalities for k=3, a = b +3=5:First inequality: 11ab =11*5*2=110a³ -b³=125 -8=117117 ≥110, which holds.Second inequality: a³ -b³=117 ≤12ab=12*5*2=120, which holds. 117 ≤120, yes.So, (a,b)=(5,2) is a solution.Check if b=1 with k=3 also satisfies both inequalities:a=4, b=1.First inequality:11ab=11*4*1=44a³ -b³=64 -1=6363 ≥44, which holds.Second inequality:63 ≤12ab=12*4*1=48. 63 ≤48? No. So, right inequality fails. Therefore, b=1 is not a solution.Thus, only b=2 for k=3 gives a solution.So, (5,2) is a solution.Now, move to k=4.Left inequality for k=4:(12 -11)b² + (48 -44)b +64 ≥0Wait, let me compute again.Wait, left inequality expression is (3k -11)b² + (3k² -11k)b +k³ ≥0For k=4:3*4=12; 12-11=13k²=3*16=48; 48 -44=4k³=64Therefore, left inequality becomes:1*b² +4*b +64 ≥0Which is b² +4b +64 ≥0. Since all terms are positive, this is always true. So left inequality is satisfied for all b ≥1.Right inequality:(3k -12)b² + (3k² -12k)b +k³ ≤0For k=4:3*4 -12=03*16 -48=48 -48=0k³=64Therefore, right inequality becomes:0*b² +0*b +64 ≤0Which is 64 ≤0, which is false. Therefore, no solutions for k=4.Hmm, interesting. So for k=4, right inequality is never satisfied, so no solutions.k=5:Left inequality:(15 -11)b² + (75 -55)b +125 ≥04b² +20b +125 ≥0, which is always true.Right inequality:(15 -12)b² + (75 -60)b +125 ≤03b² +15b +125 ≤0But 3b² +15b +125 is always positive for positive b, so no solutions.Similarly for higher k:For k ≥4, left inequality is (3k -11)b² + ... Since 3k -11 for k≥4:k=4:3*4-11=1, positive.k=5:3*5-11=4, positive.So the left inequality becomes positive coefficients for b², so as b increases, the left side is positive, hence inequality holds for all b.Right inequality for k ≥4: (3k -12)b² + ... For k=4, 3k-12=0, but as we saw, leads to 64 ≤0. For k ≥5, 3k -12 ≥3*5-12=3, positive. So the right inequality becomes (positive)b² + (positive)b + positive ≤0, which is impossible. Therefore, for k ≥4, there are no solutions.Thus, only k=3 gives a solution when b=2. Let's check k=3 and other possible b values. Wait, we already checked for k=3, b=1 and 2. Only b=2 works.Wait, maybe there are other k values. Wait, we tried k=1,2,3,4,5. k=1,2: no solutions. k=3: one solution. k≥4: no solutions. So maybe that's the only solution? Wait, but maybe there's another approach. Let me verify.Alternatively, let's consider that a³ - b³ = (a - b)(a² + ab + b²). The given inequalities are 11ab ≤ (a - b)(a² + ab + b²) ≤12ab. Let me denote d = a - b, so d is a positive integer. Then, the inequality becomes 11ab ≤ d(a² + ab + b²) ≤12ab.But since d = a - b, we can write a = b + d. Then substitute back into the expression:d[(b + d)² + (b + d)b + b²] = d[ b² + 2bd + d² + b² + bd + b² ] = d[3b² + 3bd + d²]Which is the same as before. So the same expression. Therefore, we can consider that expression.So, the inequality is 11ab ≤ d(3b² +3bd + d²) ≤12ab.But since a = b + d, ab = (b + d)b = b² + bd.So, substituting ab into the inequality:11(b² + bd) ≤ d(3b² + 3bd + d²) ≤12(b² + bd)Which is the same as before. So, perhaps, instead of fixing k, we can let x = d/b, assuming that b divides d, but that might not be necessary. Alternatively, set t = b/d. Hmm, maybe ratios.Alternatively, let's divide both sides of the inequality by b³. Let me set t = a/b, so t >1 since a >b. Then, the inequality becomes:11ab ≤ a³ -b³ ≤12abDivide all terms by b³:11 (a/b) (1) ≤ (a³/b³ -1) ≤12 (a/b)(1)Let t = a/b, so t >1, and t is rational since a and b are integers.Thus, the inequalities become:11t ≤ t³ -1 ≤12tSo now, we have to solve 11t ≤ t³ -1 ≤12t for t >1, and t is rational (since a and b are positive integers, t must be a rational number greater than 1).Therefore, we can solve the inequalities:First inequality: t³ -1 ≥11t ⇒ t³ -11t -1 ≥0Second inequality: t³ -1 ≤12t ⇒ t³ -12t -1 ≤0So, we need t such that t³ -11t -1 ≥0 and t³ -12t -1 ≤0.Let me find the real numbers t satisfying these inequalities, then check for rational t (which can be expressed as a fraction a/b in lowest terms, leading to possible integer solutions for a and b).First, solve t³ -11t -1 =0. Let's find approximate roots.Let f(t) = t³ -11t -1.Compute f(3)=27 -33 -1= -7f(4)=64 -44 -1=19So, there is a root between 3 and 4.Similarly, f(3.5)=42.875 -38.5 -1=3.375>0Wait, f(3.5)=3.5³ -11*3.5 -1=42.875 -38.5 -1=3.375>0Wait, but f(3)= -7, f(3.5)=3.375. So root between 3 and 3.5.Using the Intermediate Value Theorem.Similarly, f(3.2)= (3.2)^3 -11*3.2 -1=32.768 -35.2 -1= -3.432f(3.3)=35.937 -36.3 -1= -1.363f(3.4)=39.304 -37.4 -1=0.904Thus, root between 3.3 and 3.4.Similarly, solving t³ -12t -1=0.Let g(t)=t³ -12t -1.Compute g(3)=27 -36 -1= -10g(4)=64 -48 -1=15So, root between 3 and 4.g(3.5)=42.875 -42 -1= -0.125g(3.6)=46.656 -43.2 -1=2.456So, root between 3.5 and 3.6.Thus, the interval for t is between the root of t³ -11t -1=0 (≈3.3 to3.4) and the root of t³ -12t -1=0 (≈3.5 to3.6). So, t must be in this interval approximately 3.3 to 3.5.But t must be a rational number greater than1. Since t = a/b where a and b are positive integers, t must be a rational number. Let's see if there are any rational numbers in this interval.The approximate interval is between ~3.3 and ~3.5. So possible fractions could be 10/3 ≈3.333, 17/5=3.4, 7/2=3.5. Wait, 3.5 is 7/2, but the upper bound is around 3.5, so maybe 7/2 is too high?Wait, the upper equation is t³ -12t -1=0. Let's check t=3.5 (7/2):(7/2)^3 -12*(7/2) -1 = 343/8 -84/2 -1 = 42.875 -42 -1= -0.125 <0Wait, at t=3.5, g(t)= -0.125. So, the root is slightly above 3.5. Let's check t=3.52:3.52³≈3.52*3.52=12.3904*3.52≈12.3904*3 +12.3904*0.52≈37.1712 +6.442≈43.613212t≈12*3.52=42.24So, 43.6132 -42.24 -1≈0.3732>0. Therefore, root between 3.5 and3.52.Similarly, lower bound t≈3.35.So, t is approximately between3.35 and3.52.Possible rational numbers in this interval are 17/5=3.4, 24/7≈3.428, 31/9≈3.444, 34/10=3.4, 35/10=3.5, etc.But t must be a reduced fraction a/b where a and b are positive integers. Let's check possible fractions:First, 17/5=3.4. Is there a pair (a,b) where a/b=17/5? So a=17, b=5. Let's check if this satisfies the original inequalities.Compute a³ -b³=17³ -5³=4913 -125=478811ab=11*17*5=93512ab=12*17*5=1020But 4788 is way larger than 1020, so this does not satisfy the inequalities. So, even though 17/5 is in the interval, the actual a and b may not satisfy because t is not necessarily in lowest terms. Wait, but t is a rational number, which can be expressed as a fraction in lowest terms. For example, 5/2=2.5, which is not in our interval. Hmm, but we need t ≈3.3 to3.5.Wait, maybe the only t that works is t=5/2=2.5, but that's not in the interval. Wait, the solution we found earlier was (a,b)=(5,2). So t=5/2=2.5. Wait, but 2.5 is less than the lower bound of the interval (approx3.3). Wait, this seems contradictory.Wait, this approach via t=a/b and converting the inequality into t³ -11t -1 ≥0 and t³ -12t -1 ≤0 is leading to t ≈3.3 to3.5. However, the solution we found earlier was (5,2) which is t=2.5. But 2.5 is not in the interval. So something is wrong here.Wait, let's check the previous steps. We had:Original inequalities:11ab ≤ a³ -b³ ≤12abDivided by b³:11 (a/b) ≤ (a/b)^3 -1 ≤12 (a/b)So, 11t ≤ t³ -1 ≤12t, where t=a/b.But for the solution (5,2), t=5/2=2.5.Check the inequalities:11t=27.5t³ -1=15.625 -1=14.62514.625 ≥27.5? No, that's not true. Wait, this is a contradiction. But we know that (5,2) satisfies the original inequality.Wait, so maybe my approach to divide by b³ is flawed?Wait, let me check again.Original inequalities:11ab ≤ a³ -b³ ≤12abFor (a,b)=(5,2):Left inequality:11*5*2=110 ≤125 -8=117 ✔️Right inequality:117 ≤12*5*2=120 ✔️But when I set t=a/b=5/2, and then compute 11t=11*(5/2)=55/2=27.5t³ -1=(125/8) -1=117/8≈14.625So, 27.5 ≤14.625? Which is not true.Therefore, this approach must be wrong.Wait, why is there a discrepancy?Because when I divided the original inequality by b³, I assumed that the direction of inequalities would stay the same, but actually, if we divide by b³, which is positive, the inequalities should preserve direction. But let's check:Original inequalities:11ab ≤ a³ -b³ ≤12abDivide by b³:11ab / b³ ≤ (a³ -b³)/b³ ≤12ab /b³Simplify:11a /b² ≤ (a/b)^3 -1 ≤12a /b²But a/b = t, so:11 t /b ≤ t³ -1 ≤12 t /bBut since t = a/b, which is a rational number, but unless we can express this in terms of t and b, it's not straightforward. Therefore, my previous approach was incorrect.I made a mistake here. Dividing by b³ gives:11ab / b³ =11a / b² =11 t /bSimilarly, 12ab /b³=12 t /bTherefore, the inequalities become:11 t /b ≤ t³ -1 ≤12 t /bBut this introduces a term with both t and b, so we cannot analyze it purely in terms of t. Therefore, my previous method was flawed.This means that my initial idea to set t = a/b and analyze the inequalities is not helpful because of the division by b. Therefore, we have to approach it differently.Therefore, going back to the original substitution: a = b +k, leading to the inequalities in terms of k and b.We found that only k=3 gives a solution when b=2. But wait, when we checked (a,b)=(5,2), that works. Let's see if there are other solutions. Perhaps if we check k=3 with larger b?Wait, for k=3:Left inequality: -2b² -6b +27 ≥0. We saw that this holds for b=1,2. For b=3, it's -18 -18 +27= -9 <0, so no.Right inequality for k=3:3b² +9b -27 ≥0 ⇒ b² +3b -9 ≥0. This holds for b ≥2. But left inequality only holds for b=1,2. So for k=3, possible b=2. For b=2, works, for b=1, right inequality fails. So only b=2.Is there another k where we can have solutions? Wait, maybe k=6? Wait, but earlier analysis for k≥4 shows no solution. Let's see.Wait, perhaps we made a mistake in analysis for k=6? Let me check.Take k=6:Left inequality: (18 -11)b² + (108 -66)b +216 ≥0 ⇒7b² +42b +216 ≥0. Always true.Right inequality: (18 -12)b² + (108 -72)b +216 ≤0 ⇒6b² +36b +216 ≤0. Which is impossible. So no solutions.Hmm. So seems only k=3 gives a solution.Wait, but let's check another possible k. Let me take k=2, but maybe with higher b? Wait, for k=2, left inequality was -5b² -10b +8 ≥0. Which is 5b² +10b -8 ≤0. Solutions between ≈-2.612 and 0.612. Since b is positive, no solution. So, no.k=3: only b=2.Wait, let's check other possible pairs. For example, a=6, b=3. Then:11ab=11*6*3=198a³ -b³=216 -27=18912ab=12*6*3=216Check if 198 ≤189 ≤216. No, 189 <198. So, doesn't satisfy.Another pair: a=4, b=1:11ab=44a³ -b³=64 -1=6312ab=4863 ≥44, but 63 >48. Doesn't satisfy right inequality.Wait, but if a=6, b=3, then a=6, b=3, k=3. Then:Check left inequality: -2b² -6b +27= -2*9 -18 +27= -18 -18 +27= -9 <0. So left inequality fails. Thus, no.Alternatively, try a=7, b=4. k=3.Left inequality: -2*16 -24 +27= -32 -24 +27= -29 <0. Fails.So, no. Therefore, only (5,2) works.Wait, perhaps a=8, b=5. Let's check:11ab=11*40=440a³ -b³=512 -125=387 <440. Doesn't satisfy left inequality.Alternatively, maybe a=3, b=1:11ab=33a³ -b³=27 -1=26 <33. No.Another pair: a=4, b=2:11ab=88a³ -b³=64 -8=56 <88. No.a=6, b=4:11ab=264a³ -b³=216 -64=152 <264. No.a=7, b=5:11*35=385343 -125=218 <385. No.Wait, what about a=2, b=1:11*2=228 -1=7 <22. No.a=3, b=2:11*3*2=6627 -8=19 <66. No.a=4, b=3:11*12=13264 -27=37 <132. No.a=5, b=3:11*15=165125 -27=98 <165. No.a=5, b=4:11*20=220125 -64=61 <220. No.a=6, b=5:11*30=330216 -125=91 <330. No.So, seems like only (5,2) is a solution.Wait, but let me check a=7, b=3. k=4.Left inequality for k=4 is always satisfied, right inequality is 64 ≤0? No. Doesn't work.Another check: a=5, b=2.Yes, already verified. 11*5*2=110, 125-8=117, 12*5*2=120. 110 ≤117 ≤120. Yes.Is there another possible pair? Let's see. Suppose a=9, b=6. k=3.Left inequality: -2*(6)^2 -6*6 +27= -72 -36 +27= -81 <0. Fails.Another idea: Maybe multiple k's? But we saw that only k=3 gives solution for b=2.Alternatively, check for k=3 and b=2 gives (5,2). Let me see if scaling this pair would help. Suppose (a,b)=(5n,2n). Let's check if this works for some integer n>1.Let n=2: a=10, b=4.11ab=11*10*4=440a³ -b³=1000 -64=93612ab=12*10*4=480Check if 440 ≤936 ≤480. No, 936 >480. Doesn't satisfy.Similarly, n=1: works, n=2: doesn't. So scaling doesn't help.Alternatively, maybe other ratios. But from previous checks, it seems only (5,2) works.Wait, another possible pair: a=6, b=2. k=4.Left inequality: -2*(2)^2 -6*2 +27= -8 -12 +27=7 ≥0Right inequality:3*(2)^2 +9*2 -27=12 +18 -27=3 ≥0. So, right inequality was 3b² +9b -27 ≥0. For b=2, 12 +18 -27=3 ≥0, which implies that the original inequality was -3b² -9b +27 ≤0. Wait, original right inequality for k=3 is (3k -12)b² + (3k² -12k)b +k³ ≤0. For k=3, this is (-3)b² -9b +27 ≤0. For b=2: -12 -18 +27= -3 ≤0. Wait, no: -3*4 -9*2 +27= -12 -18 +27= -3 ≤0. So satisfies. Wait, then why did we get conflicting results?Wait, when k=3 and b=2:Original right inequality: (3*3 -12)b² + (3*9 -12*3)b +27 ≤0 ⇒(9-12)b² + (27-36)b +27 ≤0 ⇒-3b² -9b +27 ≤0.For b=2: -3*4 -9*2 +27= -12 -18 +27= -3 ≤0. Yes, which holds. So why when I computed with a=6, b=2, which would be k=4, but actually a=6, b=2 gives k=4. Wait, no. a=6, b=2, so k=4. Then, the right inequality for k=4 is 64 ≤0, which is false. But a=6, b=2: check original inequalities:11ab=11*6*2=132a³ -b³=216 -8=20812ab=12*6*2=144So, 132 ≤208 ≤144? 208 >144, so no. So even though when k=3, b=2 works, but if we take k=4, b=2, it's invalid.Wait, this is confusing. Let me check with a=5, b=2:k=3, b=2. So, works.If we take a=8, b=5, k=3. Check:11ab=11*8*5=440a³ -b³=512 -125=387 <440. Doesn't satisfy left inequality.Wait, so only when k=3 and b=2, which gives a=5, b=2, works.Therefore, the only solution is (5,2). Let me check once more to be sure.a=5, b=2.Left side:11*5*2=110Middle:125 -8=117Right side:12*5*2=120110 ≤117 ≤120. Yes. So, works.Another possible check: Maybe a=10, b=7. k=3.Left inequality: -2*49 -6*7 +27= -98 -42 +27= -113 <0. Fails.Therefore, only solution is (5,2).Final AnswerThe only solution is boxed{(5, 2)}.</think>Given the problem of finding all pairs ((a, b)) of positive integers such that (11ab leq a^3 - b^3 leq 12ab), we start by noting that (a > b) since (a^3 - b^3) must be positive. We substitute (a = b + k) where (k) is a positive integer, and rewrite the inequalities using the identity (a^3 - b^3 = (a - b)(a^2 + ab + b^2)).Substituting (a = b + k) into the inequalities, we get:[11(b^2 + bk) leq k(3b^2 + 3bk + k^2) leq 12(b^2 + bk)]We analyze these inequalities for different values of (k):1. For (k = 1, 2): - Both inequalities do not hold for any positive integer (b).2. For (k = 3): - The left inequality becomes (-2b^2 - 6b + 27 geq 0), which holds for (b = 1, 2). - The right inequality becomes (3b^2 + 9b - 27 geq 0), which holds for (b geq 2). - Overlapping values: (b = 2). - Check (a = b + 3 = 5) and (b = 2): - (11ab = 110), (a^3 - b^3 = 117), (12ab = 120) - (110 leq 117 leq 120) holds true.3. For (k geq 4): - The right inequality fails as it results in a positive expression being less than or equal to zero.After verifying all possible values, the only solution found is ((a, b) = (5, 2)).[boxed{(5, 2)}]]

🔑:Solution 1. Let a-b=t. Due to a^{3}-b^{3} geq 11 a b we conclude that a>b so t is a positive integer and the condition can be written as11 b(b+t) leq tleft[b^{2}+b(b+t)+(b+t)^{2}right] leq 12 b(b+t)Sincetleft[b^{2}+b(b+t)+(b+t)^{2}right]=tleft(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}right)=3 t b(b+t)+t^{3}the condition can be rewritten as(11-3 t) b(b+t) leq t^{3} leq(12-3 t) b(b+t)We can not have t geq 4 since in that case t^{3} leq(12-3 t) b(b+t) is not satisfied as the right hand side is not positive. Therefore it remains to check the cases when t in{1,2,3}. If t=3, the above condition becomes2 b(b+3) leq 27 leq 3 b(b+3)If b geq 3, the left hand side is greater than 27 and if b=1 the right hand side is smaller than 27 so there are no solutions in these cases. If b=2, we get a solution (a, b)=(5,2).If t leq 2, we have(11-3 t) b(b+t) geq(11-6) cdot 1 cdot(1+1)=10>t^{3}so there are no solutions in this case.In summary, the only solution is (a, b)=(5,2).

🧠:[<think>Okay, so I need to figure out the value of n and the coefficient of the term with the maximum binomial coefficient in the expansion of (2x + 1/√x)^n, given that the 5th term is a constant term. Hmm, let's start by recalling some binomial theorem basics.First, the binomial expansion of (a + b)^n has terms of the form C(n, k) * a^(n-k) * b^k, where C(n, k) is the combination of n choose k. The term number is k + 1, right? So the first term corresponds to k=0, the second to k=1, and so on. Therefore, the 5th term would be when k=4. That's important because they're saying the 5th term is a constant term, so when k=4, the term is constant.Given the expression (2x + 1/√x)^n, let's identify a and b here. a is 2x and b is 1/√x. So each term in the expansion will be C(n, k) * (2x)^(n - k) * (1/√x)^k. Let's write that out:Term(k) = C(n, k) * (2x)^(n - k) * (1/x^(1/2))^kSimplifying the exponents on x: (2x)^(n - k) gives us 2^(n - k) * x^(n - k), and (1/x^(1/2))^k is x^(-k/2). Multiplying those together gives x^(n - k - k/2) = x^(n - (3k)/2). So the exponent of x in the term when k is chosen is n - (3k)/2. For the term to be constant, the exponent must be zero. Since the 5th term is the constant term, that corresponds to k=4. Therefore:n - (3*4)/2 = 0Let me compute that:n - 12/2 = 0n - 6 = 0 => n = 6.Wait, so n is 6? Let me verify that. If n=6, then the 5th term (k=4) would have exponent x^(6 - (3*4)/2) = 6 - 6 = 0. Yep, that's a constant. So n=6. That seems straightforward. Now, the second part is to find the coefficient of the term with the maximum binomial coefficient. The binomial coefficients in the expansion are C(n, k). For n=6, the binomial coefficients C(6, k) are maximum at k=3, since for even n, the middle term(s) have the maximum coefficients. So the maximum binomial coefficient is C(6, 3) = 20. But wait, the question says "the coefficient of the term with the maximum binomial coefficient". Wait, so the term's coefficient isn't just the binomial coefficient, but also multiplied by the coefficients from the terms (2x) and (1/√x). So even though C(6, 3) is the maximum binomial coefficient, the actual coefficient of that term would be C(6, 3) * (2x)^(6 - 3) * (1/√x)^3. But we need to find the coefficient, so the numerical part.Let me compute that. The term with maximum binomial coefficient is when k=3 (since n=6, the middle term). So the coefficient is:C(6, 3) * (2)^(6 - 3) * (1)^3 * (x)^(6 - 3) * (x)^(-3/2)Wait, but the x exponent here is 6 - 3 - 3/2 = 3 - 1.5 = 1.5, so x^(3/2). But the question is about the coefficient of that term, regardless of the x part. So the coefficient is C(6,3)*2^(3)*1 = 20 * 8 = 160.Wait, but hold on. The problem says "the coefficient of the term with the maximum binomial coefficient". So the binomial coefficient itself is C(n, k), and the term with maximum binomial coefficient is when k=3. Then, the coefficient of that term (which includes the 2^3 and 1^3) would be 160. So is the answer n=6 and coefficient 160?Wait, let me just verify again. n=6, 5th term is k=4. For the term with maximum binomial coefficient, which is C(6,3)=20. Then the coefficient is 20*(2)^3*(1)^3 = 20*8=160. That seems correct.But let me check if the term with maximum coefficient (numerically) is indeed the same as the term with maximum binomial coefficient. Because sometimes, the coefficients from the terms can affect which term has the maximum numerical coefficient. For example, even if the binomial coefficient is maximum at k=3, the coefficients from (2x)^{6 - k} and (1/√x)^k might make another term have a larger numerical coefficient.Wait, the problem specifically says "the coefficient of the term with the maximum binomial coefficient". So maybe it's not asking for the term with the maximum coefficient in the expansion, but the coefficient of the term which has the maximum binomial coefficient. In that case, since the binomial coefficient C(n, k) is maximum at k=3, we just compute the coefficient of that term, which is 160.But just to be thorough, maybe I should check if the maximum numerical coefficient is different. Let's compute coefficients for different k and see.For n=6, k from 0 to 6:Term 0 (k=0): C(6,0)*(2x)^6*(1/√x)^0 = 1*64x^6*1 = 64x^6. Coefficient=64.Term 1 (k=1): C(6,1)*(2x)^5*(1/√x)^1 = 6*32x^5*x^(-0.5)=6*32x^4.5. Coefficient=192.Term 2 (k=2): C(6,2)*(2x)^4*(1/√x)^2 = 15*16x^4*x^(-1) =15*16x^3. Coefficient=240.Term 3 (k=3): C(6,3)*(2x)^3*(1/√x)^3 =20*8x^3*x^(-1.5)=20*8x^1.5. Coefficient=160.Term 4 (k=4): C(6,4)*(2x)^2*(1/√x)^4 =15*4x^2*x^(-2)=15*4x^0=60. Coefficient=60.Term 5 (k=5): C(6,5)*(2x)^1*(1/√x)^5=6*2x*x^(-2.5)=6*2x^(-1.5). Coefficient=12.Term 6 (k=6): C(6,6)*(2x)^0*(1/√x)^6=1*1*x^(-3). Coefficient=1.So the coefficients are 64, 192, 240, 160, 60, 12, 1. So the maximum numerical coefficient is 240 at k=2 (third term). But the problem says "the coefficient of the term with the maximum binomial coefficient". The maximum binomial coefficient is C(6,3)=20, which is at k=3. The coefficient of that term is 160. So even though the term at k=2 has a higher numerical coefficient, the problem is specifically asking for the coefficient of the term that has the maximum binomial coefficient. Therefore, the answer is 160.Therefore, the value of n is 6, and the coefficient is 160.Final AnswerThe value of ( n ) is (boxed{6}) and the coefficient is (boxed{160}).</think>Given the binomial expansion of ((2x + frac{1}{sqrt{x}})^n), the 5th term is a constant term. We need to find the value of (n) and the coefficient of the term with the maximum binomial coefficient.1. Finding the value of (n): - The general term in the expansion is given by (T_{k+1} = binom{n}{k} (2x)^{n-k} left(frac{1}{sqrt{x}}right)^k). - Simplifying the exponent of (x) in the term: (x^{n - k - frac{k}{2}} = x^{n - frac{3k}{2}}). - The 5th term corresponds to (k = 4). For this term to be constant, the exponent of (x) must be zero: [ n - frac{3 cdot 4}{2} = 0 implies n - 6 = 0 implies n = 6 ]2. Finding the coefficient of the term with the maximum binomial coefficient: - For (n = 6), the binomial coefficients (binom{6}{k}) are maximum at (k = 3). - The term corresponding to (k = 3) is: [ binom{6}{3} (2x)^{6-3} left(frac{1}{sqrt{x}}right)^3 = binom{6}{3} cdot 2^3 cdot x^3 cdot x^{-1.5} ] - The coefficient is: [ binom{6}{3} cdot 2^3 = 20 cdot 8 = 160 ]Thus, the value of (n) is (boxed{6}) and the coefficient of the term with the maximum binomial coefficient is (boxed{160}).,<think>Okay, let's try to solve this problem step by step. The problem says that in the binomial expansion of (2x + 1/√x)^n, the 5th term is a constant term. We need to find the value of n and the coefficient of the term with the maximum binomial coefficient.First, I need to recall the binomial expansion formula. The binomial expansion of (a + b)^n is given by the sum from k=0 to n of C(n, k) * a^(n-k) * b^k, where C(n, k) is the combination of n choose k. Each term in the expansion is C(n, k) * a^(n-k) * b^k.In this case, our a is 2x and b is 1/√x. So each term would be C(n, k) * (2x)^(n - k) * (1/√x)^k. We are told that the 5th term is a constant term. Remember that in binomial expansions, the terms are indexed starting from k=0, so the first term corresponds to k=0, the second to k=1, and so on. Therefore, the 5th term is when k=4.So, let's write out the general term for k=4:Term(k=4) = C(n, 4) * (2x)^(n - 4) * (1/√x)^4.Since this term is a constant term, the exponent of x in this term must be zero. Let's compute the exponent of x here.First, in (2x)^(n - 4), the x is raised to the (n - 4) power. Then, in (1/√x)^4, which is x^(-1/2 * 4) = x^(-2). So the total exponent of x in the term is (n - 4) + (-2) = n - 6.Since it's a constant term, the exponent must be zero:n - 6 = 0 → n = 6.Wait, is that correct? Let me verify. If n - 6 = 0, then n=6. So that's the value of n? Hmm. Let me check again.The term for k=4 is:C(n, 4) * (2x)^(n - 4) * (x^(-1/2))^4.Expanding the exponents:(2x)^(n - 4) gives 2^(n - 4) * x^(n - 4)(x^(-1/2))^4 gives x^(-2)Multiply those together: 2^(n - 4) * x^(n - 4) * x^(-2) = 2^(n - 4) * x^(n - 6)Set the exponent of x to 0:n - 6 = 0 → n = 6. So that's correct. So n=6.Okay, so that's the first part. Now, the second part is to find the coefficient of the term with the maximum binomial coefficient.Wait, the term with the maximum binomial coefficient. In the expansion of (a + b)^n, the binomial coefficients C(n, k) are maximum at k = floor(n/2) or k = ceil(n/2). So, since n=6 here, which is even, the maximum binomial coefficient occurs at k=3. Because 6/2=3, so the middle term.But the problem says "the coefficient of the term with the maximum binomial coefficient". So, the binomial coefficient C(n, k) is maximum at k=3, so the term is C(6, 3)*(2x)^{6 - 3}*(1/√x)^3. But we need the coefficient of that term. So compute the coefficient part, which is C(6, 3)*2^{3}*(1)^{3} (since (1/√x)^3 is x^{-3/2}, but the coefficient is 1^3). Wait, no. Wait, (1/√x)^3 is x^{-3/2}, but the coefficient part would be 1^3=1. So the coefficient is C(6,3)*2^3*1.Compute that. C(6,3)=20, 2^3=8, so 20*8=160. So the coefficient is 160.But let me verify again. First, the term with maximum binomial coefficient. For n=6, the binomial coefficients C(6, k) are 1,6,15,20,15,6,1. So indeed the maximum is at k=3, which is 20. So the term is k=3, which is the 4th term. Wait, the 4th term? Wait, k starts at 0, so term 1: k=0, term 2: k=1, term3: k=2, term4:k=3, term5:k=4, etc. So yes, the 4th term is the middle term with maximum coefficient.But in our problem, the 5th term (k=4) is the constant term. So the term with maximum binomial coefficient is the 4th term (k=3). So the coefficient of that term is C(6,3)*2^{6-3}*(1/√x)^3's coefficient. Wait, the coefficient would be C(6,3)*2^3*1^3. Since (1/√x)^3 is x^{-3/2}, but the coefficient part is 1^3. So 20*8*1=160. Therefore, the coefficient is 160.Wait, but let's check the term again. The term is C(6,3)*(2x)^3*(1/√x)^3. So expanding that:(2x)^3 = 8x^3(1/√x)^3 = x^{-3/2}Multiply them together: 8x^3 * x^{-3/2} = 8x^{3 - 3/2} = 8x^{3/2}So the term is 160x^{3/2} (since C(6,3)=20, 20*8=160). So the coefficient is 160.But the problem says "the coefficient of the term with the maximum binomial coefficient". So even though the term is not a constant term, but the coefficient is 160.Wait, but the binomial coefficient C(n,k) is maximum at k=3, which is 20, and the coefficient of the term is 160. So yes, the answer should be n=6 and coefficient 160.But let me make sure there isn't a trick here. The problem says "the coefficient of the term with the maximum binomial coefficient". The binomial coefficient is C(n, k), which for n=6 is maximum at k=3. The coefficient of the term includes both the binomial coefficient and the coefficients from the terms (2x)^{n - k} and (1/√x)^k. So yes, we need to multiply C(n, k) by 2^{n - k} and 1^k (since (1/√x)^k has coefficient 1^k). Therefore, the coefficient of the term is C(n, k)*2^{n - k}*1^k. So for k=3, n=6, that's 20*2^{3}*1=160.Therefore, the value of n is 6 and the coefficient is 160.So, to recap:1. The 5th term is when k=4. Setting the exponent of x to zero gives n=6.2. For n=6, the maximum binomial coefficient is at k=3. Compute the coefficient of that term: C(6,3)*2^{3}=20*8=160.Hence, the answers are n=6 and coefficient=160.But wait, just to make sure, let's check the 5th term. For n=6, k=4:Term(k=4)=C(6,4)*(2x)^{2}*(1/√x)^4.Compute exponent of x: 2 - 4*(1/2)=2 -2=0. So indeed, exponent zero. Coefficient is C(6,4)*2^2*1^4=15*4=60. So the constant term is 60. So the fifth term is 60, which is a constant. That checks out.And the term with maximum binomial coefficient is k=3: C(6,3)=20. The coefficient is 20*2^3=160. Correct.Therefore, the answers are n=6 and the coefficient is 160.Final AnswerThe value of ( n ) is (boxed{6}) and the coefficient is (boxed{160}).</think>Given the binomial expansion of ((2x + frac{1}{sqrt{x}})^n), the 5th term is a constant term. We need to find the value of (n) and the coefficient of the term with the maximum binomial coefficient.1. Finding the value of (n): - The general term in the binomial expansion is given by (C(n, k) cdot (2x)^{n-k} cdot left(frac{1}{sqrt{x}}right)^k). - The 5th term corresponds to (k=4). This term is: [ C(n, 4) cdot (2x)^{n-4} cdot left(frac{1}{sqrt{x}}right)^4 ] - Simplifying the exponent of (x): [ (2x)^{n-4} cdot left(x^{-1/2}right)^4 = 2^{n-4} cdot x^{n-4} cdot x^{-2} = 2^{n-4} cdot x^{n-6} ] - For this term to be constant, the exponent of (x) must be zero: [ n - 6 = 0 implies n = 6 ]2. Finding the coefficient of the term with the maximum binomial coefficient: - For (n = 6), the binomial coefficients (C(6, k)) are maximum at (k = 3) (since the maximum occurs at (k = lfloor n/2 rfloor) or (k = lceil n/2 rceil)). - The term with (k = 3) is: [ C(6, 3) cdot (2x)^{6-3} cdot left(frac{1}{sqrt{x}}right)^3 ] - The coefficient of this term is: [ C(6, 3) cdot 2^{3} cdot 1 = 20 cdot 8 = 160 ]Thus, the value of (n) is (boxed{6}) and the coefficient of the term with the maximum binomial coefficient is (boxed{160}).]

🔑:The general term Tr+1 of the binomial expansion (2x + frac {1}{ sqrt {x}})n is given by Tr+1= C_{ n }^{ r }•2n-r•x^{n- frac {3r}{2}}.Since the 5th term is a constant term, we have n - frac {3}{2} times 4 = 0, which gives n = 6.The binomial coefficient C_{ 6 }^{ r } is maximum when r = 3. Therefore, the coefficient of the term with the maximum binomial coefficient is C_{ 6 }^{ 3 }•23 = 160.Hence, the answer is boxed{6} and boxed{160}.The solution uses the general term formula of the binomial expansion to find the value of n and the properties of binomial coefficients to find the coefficient of the term with the maximum binomial coefficient. This problem primarily tests the application of the binomial theorem, properties of binomial coefficients, and the general term formula of binomial expansion. It is a basic problem.

🧠:[<think>Okay, let's tackle this problem step by step. So, Xiaoming has a total of 350 pens, which are a mix of pencils and ballpoint pens. Pencils sell for 4 yuan each, and ballpoint pens are 7 yuan each. He didn't sell all of them, but his sales revenue was 2013 yuan. We need to find out the minimum number of ballpoint pens he could have sold. Hmm, okay.First, let's parse the problem. He started with 350 pens total—pencils and ballpoint pens. Let me note down:Total pens initially: 350 (pencils + ballpoint pens)Pencil price: 4 yuan eachBallpoint pen price: 7 yuan eachTotal sales revenue: 2013 yuanHe didn't sell all the pens. So, the number of pens sold is less than 350. But we don't know exactly how many he sold. We need to figure out the least number of ballpoint pens he could have sold to reach 2013 yuan.Let me think. Let’s denote:Let x be the number of pencils sold.Let y be the number of ballpoint pens sold.Then, total revenue is 4x + 7y = 2013.But since he didn't sell all the pens, x + y < 350.We need to find the minimum value of y such that 4x + 7y = 2013 and x + y < 350, with x and y being non-negative integers.So, the problem reduces to solving the equation 4x + 7y = 2013, subject to x + y < 350, and x, y ≥ 0 integers. We need to find the minimal possible y.So, our goal is to minimize y while satisfying the above conditions.First, perhaps we can express x in terms of y from the equation:4x = 2013 - 7ySo, x = (2013 - 7y)/4Since x must be a non-negative integer, (2013 - 7y) must be divisible by 4 and non-negative.Similarly, x + y < 350.So, (2013 - 7y)/4 + y < 350.Let me compute that inequality.First, multiply both sides by 4 to eliminate the denominator:2013 - 7y + 4y < 1400Simplify left side:2013 - 3y < 1400Subtract 2013 from both sides:-3y < 1400 - 2013Which is:-3y < -613Multiply both sides by (-1), which reverses the inequality:3y > 613Hence, y > 613/3 ≈ 204.333...Since y must be an integer, y ≥ 205.But wait, this is the lower bound from the constraint x + y < 350. So, y must be at least 205. But we also need (2013 - 7y) divisible by 4 and non-negative. Let's check that.So, to recap:We need to find the smallest y ≥ 205 such that (2013 - 7y) is divisible by 4 and (2013 - 7y) ≥ 0.First, let's check the divisibility by 4.2013 divided by 4 is 503.25, so 2013 ≡ 1 mod 4 (since 4*503=2012, so 2013=2012+1).Similarly, 7y mod 4. Since 7 ≡ 3 mod 4, 7y ≡ 3y mod 4.Therefore, 2013 -7y ≡ 1 - 3y mod 4. We need this to be congruent to 0 mod 4.So,1 - 3y ≡ 0 mod 4=> -3y ≡ -1 mod 4Multiply both sides by -1:3y ≡ 1 mod 4Now, 3y ≡ 1 mod 4. Let's solve for y.Multiplicative inverse of 3 mod 4 is 3, since 3*3=9 ≡1 mod4.So, multiply both sides by 3:y ≡ 3*1 ≡3 mod4Therefore, y ≡3 mod4. So, y must be congruent to 3 modulo4.Therefore, possible y are 3,7,11,... etc. But since y ≥205, the smallest possible y is the smallest number ≥205 congruent to 3 mod4.Let me compute 205 divided by4. 4*51=204, so 205=4*51 +1, so 205 ≡1 mod4. Then, next numbers:205 ≡1, 206≡2, 207≡3 mod4. So, 207 is the first y ≥205 congruent to3 mod4.Therefore, possible y are 207,211,215,... up to the maximum y such that 7y ≤2013.Compute maximum possible y: 2013/7≈287.571. So, y ≤287.But since x + y <350, and x = (2013 -7y)/4, we need to check that x + y <350. Wait, but earlier we derived that y must be ≥205. But let's check for y=207.So, let's start with y=207.Compute x=(2013 -7*207)/4.7*207=1449. 2013-1449=564. Then, 564/4=141. So, x=141.Then x+y=141+207=348 <350. So, that's okay.So, y=207, x=141. Then total pens sold=348, which is less than 350. Revenue=4*141 +7*207=564 +1449=2013. Correct.Now, check if there is a smaller y ≥205 that satisfies the conditions. But since y must be congruent to3 mod4, the next lower y would be 207-4=203. But 203 is less than 205. But 203 is ≡3 mod4 (203=4*50 +3). So, y=203.Wait, but earlier we found that y must be ≥205. Wait, maybe our earlier inequality was incorrect? Let's check.Wait, we had:x + y <350.We substituted x=(2013 -7y)/4.So, (2013 -7y)/4 + y <350.Multiply by4:2013 -7y +4y <14002013 -3y <1400So, -3y <1400-2013= -613Multiply by -1 (reverse inequality):3y >613So, y >613/3≈204.333. So, y≥205.Hence, y must be at least205. Therefore, y=203 is invalid because it is less than205. Therefore, the minimal possible y is207.Wait, but let me check y=207.x=141, y=207. Total sold=348 <350. That works.Is there a y=206? Wait, y=206. Check if 206 satisfies the congruence.Earlier, we had y≡3 mod4. 206≡2 mod4, so it won't satisfy. Similarly, 205≡1, 204≡0. So, next is207≡3.Thus, the minimal y is207.Wait, but let me confirm. Let me check if there's any possible y between205 and207, but since y must be ≡3 mod4, the next y after205 is207. So, indeed, the minimal y is207. Therefore, the answer is207.But wait, let me check if perhaps even with a higher y, we can get a lower number? Wait, no, because we need the minimal y. But according to our constraints, the minimal y is207. However, let me think again.Is there any error here?Wait, perhaps. Let me re-examine the steps.We started with 4x +7y=2013, x+y <350, x,y non-negative integers.We need to find the minimal y such that x=(2013 -7y)/4 is a non-negative integer, and x+y <350.So, y must satisfy:1. 2013 -7y must be divisible by4.2. 2013 -7y ≥0 ⇒7y ≤2013 ⇒y ≤287.571 ⇒y ≤287.3. (2013 -7y)/4 +y <350 ⇒2013 -3y <1400 ⇒3y >613 ⇒y>204.333 ⇒y≥205.Also, from divisibility condition, y≡3 mod4.So, possible y are 207,211,215,...,287.Therefore, the minimal y is207.So, the answer is207.But just to be thorough, let's check y=207.Compute x=(2013 -7*207)/4=2013 -1449=564; 564/4=141.x=141, y=207. Total sold=141+207=348 <350. Correct.Revenue=4*141 +7*207=564 +1449=2013. Correct.Is there a smaller y? Let's check y=203, which is less than205, so violates the x+y <350 constraint. Let's see:If y=203, then x=(2013 -7*203)/4=(2013 -1421)/4=592/4=148.x=148, y=203. Total sold=148+203=351. But 351 is not less than350. So, that's invalid.Similarly, y=204. Let's check y=204.x=(2013 -7*204)/4=2013 -1428=585. 585/4=146.25. Not integer. So, invalid.y=205: 2013 -7*205=2013 -1435=578. 578/4=144.5. Not integer.y=206: 2013 -7*206=2013 -1442=571. 571/4=142.75. Not integer.y=207: as above, valid.Therefore, 207 is indeed the minimal y.So, the answer is boxed{207}Final Answerboxed{207}</think>Xiaoming has a total of 350 pens, consisting of pencils and ballpoint pens. Pencils are sold at 4 yuan each, and ballpoint pens at 7 yuan each. The total sales revenue was 2013 yuan, and not all pens were sold. We need to find the minimum number of ballpoint pens sold.Given:- Total pens initially: 350- Pencil price: 4 yuan each- Ballpoint pen price: 7 yuan each- Total sales revenue: 2013 yuan- Number of pens sold: less than 350Let ( x ) be the number of pencils sold and ( y ) be the number of ballpoint pens sold. We have the equations:[ 4x + 7y = 2013 ][ x + y < 350 ]We need to find the minimum ( y ) such that ( x ) and ( y ) are non-negative integers, ( 4x + 7y = 2013 ), and ( x + y < 350 ).1. Express ( x ) in terms of ( y ):[ x = frac{2013 - 7y}{4} ]2. ( x ) must be a non-negative integer, so ( 2013 - 7y ) must be divisible by 4 and non-negative.3. Check the congruence condition:[ 2013 equiv 1 mod 4 ][ 7y equiv 3y mod 4 ][ 1 - 3y equiv 0 mod 4 ][ 3y equiv 1 mod 4 ]The multiplicative inverse of 3 modulo 4 is 3, so:[ y equiv 3 mod 4 ]4. Solve for ( y ) such that ( y geq 205 ) (since ( x + y < 350 ) implies ( y > 204.333 )) and ( y equiv 3 mod 4 ).The smallest ( y geq 205 ) that satisfies ( y equiv 3 mod 4 ) is 207.5. Verify ( y = 207 ):[ x = frac{2013 - 7 times 207}{4} = frac{2013 - 1449}{4} = frac{564}{4} = 141 ][ x + y = 141 + 207 = 348 < 350 ]Thus, the minimum number of ballpoint pens sold is (boxed{207}).,<think>Alright, let's tackle this problem step by step. So, Xiaoming has a total of 350 pens, which are a mix of pencils and ballpoint pens. Pencils cost 4 yuan each, and ballpoint pens cost 7 yuan each. He didn't sell all of them, but his sales revenue was 2013 yuan. We need to find out the minimum number of ballpoint pens he must have sold. Hmm, okay.First, let's make sure I understand the problem correctly. He started with 350 pens total, some pencils and some ballpoint pens. He sold some of each, but not all, and made 2013 yuan. We need to find the least number of ballpoint pens he could have sold. So, the question is asking for the minimal number of ballpoint pens sold given these constraints. Got it.Let me break this down. Let's denote the number of pencils he sold as x and the number of ballpoint pens he sold as y. The total revenue from these sales is 4x + 7y = 2013. Additionally, since he didn't sell all the pens, the number of pens he sold, x + y, must be less than 350. Also, since he can't sell more pens than he had initially, x must be less than or equal to the initial number of pencils, and y must be less than or equal to the initial number of ballpoint pens. However, we don't know the exact initial numbers of each type, only that the total is 350. Hmm, so the initial counts of pencils and ballpoint pens are variables here too. That complicates things.Wait a minute, maybe we can model this as follows. Let me denote the initial number of pencils as P and the initial number of ballpoint pens as B. So, P + B = 350. Then, the number of pencils sold, x, must satisfy x ≤ P, and the number of ballpoint pens sold, y, must satisfy y ≤ B. The revenue equation is 4x + 7y = 2013. Our goal is to find the minimal y such that these constraints are satisfied.But we have four variables here: P, B, x, y. However, we know that P + B = 350, so we can express B as 350 - P. Therefore, we can write y ≤ 350 - P. Similarly, x ≤ P. So, substituting, since x ≤ P, then P must be at least x. Similarly, since y ≤ 350 - P, then 350 - P must be at least y, so P ≤ 350 - y. Therefore, combining these, x ≤ P ≤ 350 - y. Hmm, that's a bit abstract. Maybe we can approach this by focusing on the revenue equation.We have 4x + 7y = 2013. Let's see if we can express x in terms of y. Let's rearrange the equation: 4x = 2013 - 7y, so x = (2013 - 7y)/4. Since x must be an integer (you can't sell a fraction of a pencil), (2013 - 7y) must be divisible by 4. Similarly, since y must also be an integer, we can look for integer solutions (x, y) such that 4x + 7y = 2013 and x + y < 350, with x ≤ P and y ≤ B where P + B = 350.But we don't know P and B. However, since x ≤ P and y ≤ B, and P + B = 350, we can say that x + y ≤ P + B = 350. But wait, he didn't sell all the pens, so x + y < 350. Therefore, x + y ≤ 349. So, we have two equations:1. 4x + 7y = 20132. x + y ≤ 349But we can combine these. From the first equation, x = (2013 - 7y)/4. Then substituting into the second equation:(2013 - 7y)/4 + y ≤ 349Let me compute that:Multiply both sides by 4 to eliminate the denominator:2013 - 7y + 4y ≤ 4 * 349Simplify left side: 2013 - 3y ≤ 1396Subtract 2013 from both sides: -3y ≤ 1396 - 2013Compute 1396 - 2013: that's -617So, -3y ≤ -617Multiply both sides by (-1) reverses inequality:3y ≥ 617Therefore, y ≥ 617 / 3 ≈ 205.666...Since y must be an integer, y ≥ 206. Hmm, so the minimal possible y is 206. But we need to check if this is possible. Let's verify.If y = 206, then x = (2013 - 7*206)/4Compute 7*206: 7*200 = 1400, 7*6 = 42, so total 14422013 - 1442 = 571Then x = 571 / 4 = 142.75. Wait, that's not an integer. So y = 206 is invalid because x would have to be a non-integer. Therefore, y must be such that (2013 - 7y) is divisible by 4.So, we need to find the smallest y ≥ 206 such that (2013 - 7y) ≡ 0 mod 4.Let me compute 2013 mod 4. 2012 is divisible by 4, so 2013 mod 4 = 1.Similarly, 7y mod 4. 7 mod 4 = 3, so 7y mod 4 = (3y) mod 4. Therefore, 2013 - 7y ≡ 1 - 3y ≡ 0 mod 4. So, 1 - 3y ≡ 0 mod 4 ⇒ -3y ≡ -1 mod 4 ⇒ 3y ≡ 1 mod 4.Multiply both sides by the inverse of 3 mod 4. The inverse of 3 mod 4 is 3, since 3*3=9≡1 mod4. So, y ≡ 3*1 ≡3 mod4. Therefore, y ≡3 mod4. So y must be congruent to 3 modulo4.Therefore, the smallest y ≥206 that is ≡3 mod4.Compute 206 divided by 4: 4*51=204, so 206=204+2, so 206≡2 mod4. Then next number ≡3 mod4 would be 207. 207≡3 mod4. Then 207 +4=211, etc.So possible y are 207,211,215,... Let's check y=207.For y=207, x=(2013 -7*207)/4.7*200=1400, 7*7=49, so total 1449. 2013 -1449=564. 564/4=141. So x=141. Then check if x + y =141 +207=348. 348 <350, so that's okay. Then check whether the initial numbers P and B can accommodate x=141 and y=207. So we need P ≥141 and B ≥207. Since total P + B=350, then if B=207, P=350 -207=143. So P=143 ≥141, which is okay. So he had 143 pencils and 207 ballpoint pens initially, sold 141 pencils and 207 ballpoint pens. That works. So y=207 is possible.Wait, but the question is asking for the minimal number of ballpoint pens sold. So 207. But let's check if y=207 is indeed the minimal possible. Wait, earlier we thought y must be at least 206, but since y=206 gives x=142.75 which is not integer, the next possible y is 207. Therefore, 207 is the minimal y. But let's check if that works with the constraints.Wait, but another way to check: perhaps even if y=207 gives x=141, which is less than P=143, which is acceptable. So yes, that works. So the answer would be 207? But let's confirm.Alternatively, maybe there is a mistake here. Let me go through the steps again.First, we set up the equations:4x + 7y = 2013x + y < 350We need to find the minimal y such that these hold, with x and y non-negative integers.Then we derived that y >= 206 (from the inequality x + y <=349). But when we plugged in y=206, we found x was not an integer. Then, since y must be congruent to 3 mod4, the next possible y is 207, which works. So y=207 is the minimal possible.Therefore, the answer is 207. However, let me check if there's a possible lower y that's congruent to 3 mod4 and satisfies all constraints. Wait, 207 is the first after 206. Let's check y=203. Wait, 203 mod4 is 203/4=50*4 + 3, so 203≡3 mod4. But 203 is less than 206. But earlier we found y >=206. So how does that work?Wait, the inequality gave us y >= 617/3≈205.666, so y >=206. So 203 is too low. So even though 203 is congruent to3mod4, it is below the lower bound. Therefore, the minimal y is 207.But let's check y=207. Then x=141. Total pens sold:141+207=348, which is less than 350. So that's okay. The initial numbers would have to be P=143 and B=207, so x=141 <=143 and y=207<=207. So that works. Therefore, 207 is possible.Wait, but is there a possible y less than 207 that also satisfies 4x +7y=2013 and x + y <=349? For example, y=203. Let's check.If y=203, then x=(2013 -7*203)/4. 7*200=1400, 7*3=21, total 1421. 2013-1421=592. 592/4=148. So x=148. Then x + y=148+203=351, which exceeds 349. So that's invalid. So x + y=351, which is more than 349, so violates the constraint. Therefore, y=203 is invalid.Similarly, try y=207-4=203, which we saw. What about y=207-8=199? y=199, which is 199≡3 mod4. Compute x=(2013-7*199)/4. 7*200=1400, minus7=1393. 2013-1393=620. 620/4=155. x=155. x + y=155+199=354>349. Still too high.Similarly, y=211: x=(2013-7*211)/4. 7*200=1400, 7*11=77, so 1477. 2013-1477=536. 536/4=134. x=134. x+y=134+211=345<350. So that's acceptable. But y=211 is higher than 207, so 207 is better.Similarly, trying lower y with congruence 3 mod4, but above 206. So y=207 is the first one. Therefore, the answer is 207.But let me check another way. Suppose we have 4x +7y=2013, and x + y <=349. Let's express y in terms of x: y=(2013 -4x)/7. Then x + (2013 -4x)/7 <=349.Multiply both sides by7:7x +2013 -4x <=2443. So 3x +2013 <=2443. Then 3x <=2443 -2013=430. Therefore, x <=430/3≈143.333. So x<=143. Since x must be integer, x<=143.But also, y=(2013 -4x)/7 must be integer and y>=0. So 2013 -4x must be divisible by7 and non-negative. Let me compute 2013 mod7. 7*287=2009, so 2013-2009=4. So 2013≡4 mod7. Therefore, 2013 -4x ≡4 -4x mod7. We need 4 -4x ≡0 mod7. Therefore, 4x≡4 mod7. Divide both sides by4: x≡1 mod7. So x must be congruent to1 mod7. So x=1,8,15,..., up to x=141, since x<=143.Wait, x=1 +7k, where k=0,1,2,..., such that x<=143. The maximum k is when x=141: (141 -1)/7=20. So k=0,...,20. So x can be 1,8,15,...,141. Let's compute the corresponding y for x=141. x=141, y=(2013 -4*141)/7=(2013 -564)/7=1449/7=207. So y=207. Similarly, if x=141-7=134, then y=(2013 -4*134)/7=(2013 -536)/7=1477/7=211. So y=211. As x decreases by7, y increases by4. So the sequence is x=141, y=207; x=134,y=211; x=127,y=215; etc., up to x=1,y=(2013 -4)/7=2009/7=287.But in all these cases, x + y would be:For x=141, y=207: 141+207=348<350.For x=134,y=211:134+211=345<350.Similarly, x=127,y=215:127+215=342<350.So all these are valid. But we need the minimal y. So among these solutions, the minimal y is when x is maximum. Because as x increases by7, y decreases by4. Therefore, the maximum x is 141, leading to minimal y=207. So 207 is indeed the minimal possible y.Therefore, the answer is 207. But let's check if x=141 and y=207 can be achieved given the initial counts. The initial counts must satisfy P >=x=141 and B >=y=207. Since P + B=350, then if B=207, P=350 -207=143. So 143 >=141, which is true. Therefore, it's possible that he had 143 pencils and 207 ballpoint pens, and sold 141 pencils and all 207 ballpoint pens. Wait, but he didn't sell all the pens. Wait, he sold 141 pencils and 207 ballpoint pens, total sold is 141 +207=348, which is less than 350, so there are 2 pens left. So he didn't sell all. So that's acceptable.Alternatively, if he had more pencils and fewer ballpoint pens initially, but in that case, he couldn't have sold 207 ballpoint pens if he had fewer. So since he sold 207 ballpoint pens, he must have had at least 207 initially, which implies P=350 -B <=350 -207=143. Therefore, he could have had 143 pencils and 207 ballpoint pens. Therefore, selling 141 pencils and 207 ballpoint pens is feasible.Therefore, the minimal number of ballpoint pens sold is 207.Final AnswerThe minimum number of ballpoint pens Xiaoming sold is boxed{207}.</think>Xiaoming has a total of 350 pens, consisting of pencils and ballpoint pens. Pencils are sold at 4 yuan each, and ballpoint pens at 7 yuan each. The sales revenue was 2013 yuan, and he did not sell all the pens. We need to find the minimum number of ballpoint pens sold.1. Let ( x ) be the number of pencils sold and ( y ) be the number of ballpoint pens sold. The revenue equation is: [ 4x + 7y = 2013 ]2. The total number of pens sold ( x + y ) must be less than 350: [ x + y < 350 ]3. Express ( x ) in terms of ( y ): [ x = frac{2013 - 7y}{4} ] ( x ) must be an integer, so ( 2013 - 7y ) must be divisible by 4.4. Check the divisibility condition: - ( 2013 equiv 1 mod 4 ) - ( 7y equiv 3y mod 4 ) - Therefore, ( 1 - 3y equiv 0 mod 4 ) implies ( 3y equiv 1 mod 4 ) - The solution to this congruence is ( y equiv 3 mod 4 )5. Determine the minimum ( y ) such that ( x + y < 350 ): - From the inequality ( x + y leq 349 ): [ frac{2013 - 7y}{4} + y leq 349 ] - Solving this gives ( y geq 206 ). The smallest ( y geq 206 ) that satisfies ( y equiv 3 mod 4 ) is 207.6. Verify ( y = 207 ): - Calculate ( x ): [ x = frac{2013 - 7 times 207}{4} = frac{2013 - 1449}{4} = frac{564}{4} = 141 ] - Check ( x + y = 141 + 207 = 348 ), which is less than 350.7. Confirm initial counts: If there were 207 ballpoint pens initially, then pencils were ( 350 - 207 = 143 ). Sold 141 pencils and 207 ballpoint pens, which is feasible.Thus, the minimum number of ballpoint pens sold is (boxed{207}).,<think>Okay, so Xiaoming has a total of 350 pens, which are a combination of pencils and ballpoint pens. The pencils sell for 4 yuan each, and the ballpoint pens sell for 7 yuan each. He didn't sell all of them, but his sales revenue was 2013 yuan. We need to find the minimum number of ballpoint pens he could have sold. Hmm, let's break this down step by step.First, let me denote the number of pencils he sold as x and the number of ballpoint pens he sold as y. So, according to the problem, the total revenue from selling these pens would be 4x + 7y = 2013 yuan. Also, he started with a total of 350 pens, but he didn't sell all of them. That means the number of pens he sold, which is x + y, must be less than 350. So, we have x + y < 350. Our goal is to find the minimum value of y, which is the number of ballpoint pens sold.But wait, since we're supposed to find the minimum number of ballpoint pens sold, we need to maximize the number of pencils sold, right? Because if he sold more pencils, he could have sold fewer ballpoint pens to reach the same total revenue. However, there's a constraint here: he didn't sell all the pens. So, the total pens sold (x + y) has to be less than 350. Therefore, even if he tried to sell as many pencils as possible, he still couldn't exceed 349 pens. Let me formalize this.We need to solve the equation 4x + 7y = 2013 with the constraints that x + y < 350 and x, y are non-negative integers. Our objective is to minimize y. To minimize y, we need to maximize x. So, let's express x in terms of y: x = (2013 - 7y)/4. Since x must be an integer, (2013 - 7y) must be divisible by 4. So, 2013 - 7y ≡ 0 mod 4. Let's compute 2013 mod 4 first. 2012 is divisible by 4 (since 2012 ÷ 4 = 503), so 2013 ≡ 1 mod 4. Therefore, 1 - 7y ≡ 0 mod 4 → -7y ≡ -1 mod 4 → 7y ≡ 1 mod 4. Since 7 ≡ 3 mod 4, this becomes 3y ≡ 1 mod 4. Multiplying both sides by the modular inverse of 3 mod 4, which is 3 (since 3×3=9≡1 mod4), we get y ≡ 3×1 ≡ 3 mod4. Therefore, y must be congruent to 3 modulo 4. So, y can be written as y = 4k + 3, where k is a non-negative integer.Substituting back into x = (2013 - 7y)/4, we have:x = (2013 - 7(4k + 3))/4= (2013 - 28k - 21)/4= (1992 - 28k)/4= 498 - 7k.So, x = 498 - 7k and y = 4k + 3. Now, we have the constraints:1. x + y < 3502. x ≥ 03. y ≥ 0Let's plug x and y into the first constraint:498 - 7k + 4k + 3 < 350501 - 3k < 350-3k < -151Multiply both sides by -1 (remember to reverse the inequality sign):3k > 151k > 151/3k > 50.333...Since k must be an integer, the smallest integer greater than 50.333... is 51. Therefore, k ≥ 51.Now, check the second constraint x ≥ 0:x = 498 - 7k ≥ 0498 - 7k ≥ 07k ≤ 498k ≤ 498/7k ≤ 71.142...So, k ≤ 71. Therefore, k can be from 51 to 71 inclusive.Now, the third constraint y ≥ 0 is automatically satisfied because y = 4k + 3, and since k ≥ 51, y is definitely positive.Now, we need to find the minimum y. Since y = 4k + 3, the minimum y occurs when k is minimized. The smallest k is 51, so plugging that in:y = 4*51 + 3 = 204 + 3 = 207.But we need to check if with k=51, the total pens sold x + y is indeed less than 350. Let's compute x and y:x = 498 - 7*51 = 498 - 357 = 141y = 207x + y = 141 + 207 = 348 < 350. Okay, that's valid.Wait, so the minimum number of ballpoint pens sold is 207? Let me verify this because 207 seems quite high. Let's check the math again.Wait, the problem says he had a total of 350 pens (pencils and ballpoint pens) initially. He didn't sell all of them, so the number sold is less than 350. If he sold 348 pens, that's 2 pens left. But 348 pens sold with 141 pencils and 207 ballpoint pens. The revenue would be 141*4 + 207*7.Calculating that:141*4: 140*4=560, 1*4=4, total 564.207*7: 200*7=1400, 7*7=49, total 1449.Total revenue: 564 + 1449 = 2013. Correct.So, the numbers check out. But 207 seems like a lot of ballpoint pens. But perhaps that's correct? Let me see if there's a smaller y possible.Wait, we concluded that y must be congruent to 3 mod4. So, possible y values are 3,7,11,... up to a maximum. But with the constraints of x + y <350 and x,y non-negative.But according to the equations, k starts at 51, giving y=207. If we try k=50, which would give y=4*50 +3=203. Let's check if that's possible. Then x=498 -7*50=498-350=148. Then x+y=148+203=351. But 351 is not less than 350. So, that's invalid. Hence, k=50 is invalid. So, the next possible k is 51, giving x+y=348, which is valid.Therefore, 207 is indeed the minimum number of ballpoint pens he could have sold.But wait, let me check if there's another solution where y is smaller, maybe with different x and y that still satisfy 4x +7y=2013 and x+y <350. Suppose we consider different residues or something. Wait, we already considered that y must be congruent to 3 mod4. So, the next lower y would be 207 -4=203, but we saw that with y=203, x+y=351 which is over 350. Similarly, y=199 would require k=49, y=4*49+3=199. Then x=498 -7*49=498-343=155. Then x+y=155+199=354>350. Still invalid. Similarly, each decrement of y by 4 increases x by 7 (since 7y decreases by 28, so x increases by 28/4=7). Therefore, each time we reduce y by 4, x increases by 7, which adds 7-4=3 to x+y. Wait, so x+y increases by 3 each time we decrease y by 4. But since we need x+y to be less than 350, and when y=207, x+y=348, which is under 350. If we decrease y by 4 (y=203), x becomes 141+7=148, so x+y=148+203=351. Which is over 350. So, that's invalid. So, any lower y would result in x+y ≥351, which is not allowed. Therefore, 207 is indeed the minimal possible y.Therefore, the answer should be 207. Wait, but 207 seems really high. Let me check once more.Total pens sold: 348. Total revenue: 2013. If he sold all 350 pens, maximum revenue would be if all were ballpoint pens: 350*7=2450, which is way higher than 2013. So, 2013 is less than that. If he sold all pencils, 350*4=1400, so 2013 is between 1400 and 2450. The average revenue per pen is 2013/348 ≈ 5.78 yuan per pen. Which is between 4 and 7, so that makes sense. If he sold more pencils, the average would be lower, but he has to reach 2013. Since 348 pens sold gives average ~5.78, which is reasonable. Let me see, maybe there's a way to have a lower y with x+y even less than 348? Wait, but if x+y is less, that would mean more pens unsold. But since we need to reach 2013 revenue, perhaps having less pens sold would require higher-priced items, i.e., more ballpoint pens. Wait, no. Wait, if you sell fewer pens, but need the same revenue, you need more expensive pens. So, if we can find a combination where x + y is smaller (i.e., he sold fewer pens) but with more ballpoint pens, perhaps the number of ballpoint pens could be lower? Wait, that seems contradictory. Wait, no. If he sells fewer pens but needs the same revenue, he needs each pen to contribute more to the revenue. So, he needs more ballpoint pens. Therefore, maybe there's a solution with a higher proportion of ballpoint pens but fewer total pens sold. But in that case, the number of ballpoint pens would be higher, not lower. Therefore, the minimal number of ballpoint pens is achieved when he sells as many pencils as possible without exceeding the total pens. Therefore, the initial approach is correct. So, 207 is indeed the minimal number.Therefore, the answer is 207. But just to ensure, let me check with k=51:y=4*51+3=207x=498 -7*51=498 -357=141x+y=141+207=348 <350 ✔️Revenue: 141*4 +207*7=564 +1449=2013 ✔️Therefore, correct.Is there a possibility with a smaller y? Let's say y=206. Then, 4x +7*206=4x +1442=2013 →4x=571. 571 is not divisible by 4. So x would not be integer. Similarly, y=205: 7*205=1435. 2013-1435=578. 578/4=144.5. Not integer. y=204:7*204=1428. 2013-1428=585. 585/4=146.25. Not integer. y=203:7*203=1421. 2013-1421=592. 592/4=148. Then x=148. But x+y=148+203=351>350, which is invalid. So, even if y=203 gives integer x, it's invalid because total pens sold exceed 349. Therefore, the next possible y with x+y<350 is y=207. So, 207 is indeed the minimal.Final Answerboxed{207}</think>Xiaoming volunteered to sell pens at a stationery store, where pencils were sold at 4 yuan each and ballpoint pens at 7 yuan each. He started with a total of 350 pens and had a sales revenue of 2013 yuan without selling all the pens. We need to find the minimum number of ballpoint pens he sold.1. Let ( x ) be the number of pencils sold and ( y ) be the number of ballpoint pens sold. The revenue equation is ( 4x + 7y = 2013 ).2. The total number of pens sold ( x + y ) must be less than 350.3. To minimize ( y ), we need to maximize ( x ). Express ( x ) in terms of ( y ): ( x = frac{2013 - 7y}{4} ).4. For ( x ) to be an integer, ( 2013 - 7y ) must be divisible by 4. Solving ( 2013 equiv 1 mod 4 ) and ( 7y equiv 1 mod 4 ), we find ( y equiv 3 mod 4 ). Thus, ( y = 4k + 3 ) for some integer ( k ).5. Substituting ( y = 4k + 3 ) into the revenue equation, we get ( x = 498 - 7k ).6. The constraint ( x + y < 350 ) gives ( 498 - 7k + 4k + 3 < 350 ), leading to ( 501 - 3k < 350 ) or ( k > 50.333 ). The smallest integer ( k ) is 51.7. Checking ( k = 51 ), we find ( y = 4(51) + 3 = 207 ) and ( x = 498 - 7(51) = 141 ). The total pens sold ( x + y = 141 + 207 = 348 ), which is less than 350.Thus, the minimum number of ballpoint pens sold is (boxed{207}).,<think>Okay, let me try to figure out this problem. So, Xiaoming is selling pens—both pencils and ballpoint pens. The pencils are 4 yuan each, and the ballpoint pens are 7 yuan each. He started with a total of 350 pencils and ballpoint pens. He didn't sell all of them, but his sales revenue was 2013 yuan. The question is asking how many ballpoint pens he sold at least. So, we need to find the minimum number of ballpoint pens he could have sold to reach that revenue. Hmm.Let me break it down. Let's let x be the number of pencils he sold, and y be the number of ballpoint pens he sold. So, his total revenue would be 4x + 7y = 2013 yuan. Also, he didn't sell all of his pens, so the total number he sold, x + y, is less than 350. But he started with 350 pens total. Wait, does that mean that the total number of pens he had was 350, so the number he sold plus the number remaining is 350? Or does it mean that he had 350 pens in total, both pencils and ballpoint pens combined? The problem says "he had a total of 350 pencils and ballpoint pens." So, initial stock is 350 pens total, but he didn't sell them all. So, x + y < 350. But the problem doesn't specify how many he had of each initially, just the total. So maybe we don't need to know the initial numbers, just that he sold some number x and y such that x + y < 350, and 4x + 7y = 2013. We need to find the minimal possible y.So, the problem reduces to solving 4x + 7y = 2013 with x and y being non-negative integers and x + y < 350. We need to find the smallest possible y that satisfies these conditions.Okay, let's think about how to approach this. Since we need the minimal number of ballpoint pens (y), which are more expensive, we need to maximize the number of pencils sold (x). But we have the constraint that x + y < 350. So, even if we try to maximize x, we have to keep y as low as possible while still meeting the revenue and the total pens sold constraint.First, let's try to solve the equation 4x + 7y = 2013 for integer solutions. Let me rearrange this equation to express x in terms of y: 4x = 2013 - 7y, so x = (2013 - 7y)/4. Since x has to be an integer, (2013 - 7y) must be divisible by 4. So, 2013 - 7y ≡ 0 mod 4. Let's compute 2013 mod 4. 2012 is divisible by 4 (since 2012/4 = 503), so 2013 mod 4 is 1. Therefore, 1 - 7y ≡ 0 mod 4. Which simplifies to -7y ≡ -1 mod 4, which is equivalent to 7y ≡ 1 mod 4. Since 7 ≡ 3 mod 4, this becomes 3y ≡ 1 mod 4. To solve for y, we can multiply both sides by the inverse of 3 mod 4. The inverse of 3 mod 4 is 3, since 3*3=9≡1 mod4. Therefore, y ≡ 3*1 ≡ 3 mod4. So, y must be congruent to 3 modulo4. That means possible values of y are 3,7,11,15,... etc.So, the general solution for y is y=4k+3, where k is a non-negative integer. Then, substituting back into x=(2013 -7y)/4, we get x=(2013 -7*(4k+3))/4=(2013 -28k -21)/4=(1992 -28k)/4=498 -7k. So, x=498-7k and y=4k+3. Since both x and y must be non-negative integers, we have the constraints:1. x=498 -7k ≥0 => 498 -7k ≥0 => 7k ≤498 => k ≤498/7≈71.14. So k ≤71.2. y=4k+3 ≥0. Since k is a non-negative integer, this is automatically satisfied for all k≥0.Additionally, we have the constraint that x + y <350. Let's substitute x and y in terms of k:x + y = (498 -7k) + (4k +3) = 501 -3k <350.So, 501 -3k <350 => -3k < -151 => 3k >151 =>k >151/3≈50.333. Since k must be an integer, k≥51.So, combining the constraints on k: k must satisfy 51 ≤k ≤71, and k must be integer. Since k must be an integer, k=51,52,...,71.But we need to find the minimal y. Since y=4k+3, the minimal y occurs when k is minimal. So the minimal k is 51. Then, y=4*51+3=204+3=207. But we need to check if x=498 -7*51=498 -357=141. Then, x + y=141+207=348, which is indeed less than 350. So that works. So y=207 is possible.But wait, the problem says "he did not sell them all", so x + y <350. 348 is less than 350, so that's okay. So the minimal number of ballpoint pens he could have sold is 207? Wait, but let's check if this is the case.Wait, but let me verify again. The general solution is y=4k+3, x=498-7k. For k=51, y=207, x=141. Then, x+y=348. So, that's valid. If we take k=50, which would give y=4*50+3=203, x=498-7*50=498-350=148. Then x + y=203+148=351, which is more than 350, so that's invalid. Therefore, k cannot be 50. So k must be at least 51. Therefore, the minimal y is 207. But the answer is 207? Let me check if there's another way to have a smaller y. Wait, perhaps not. Let's see.Alternatively, maybe we can approach it differently. Let's set up the problem. We need to find the minimal y such that 4x +7y=2013 and x+y <350. So, perhaps we can express x in terms of y: x=(2013-7y)/4. Then, x must be a non-negative integer, so (2013 -7y) must be divisible by 4 and non-negative. So, 7y ≤2013 => y ≤2013/7≈287.571, so y ≤287. Also, (2013 -7y) must be divisible by 4, so as before, y≡3 mod4. So y=4k+3, k from 0 to 71.But then x+y=(2013 -7y)/4 + y=(2013 -7y +4y)/4=(2013 -3y)/4. So x+y=(2013 -3y)/4. This must be less than 350. So:(2013 -3y)/4 <350 => 2013 -3y <1400 => -3y < -613 => 3y >613 => y>613/3≈204.333. Therefore, y>204.333. Since y must be an integer, y≥205. But also, y must be ≡3 mod4, so the smallest y≥205 that is congruent to 3 mod4. Let's compute 205 divided by4: 4*51=204, so 204 +1=205, which is 1 mod4. Then 205,206,207. 207 is 3 mod4. So the smallest y≥205 and y≡3 mod4 is y=207. Therefore, the minimal y is 207. Which matches our previous result. Therefore, the answer is 207.But let me check if this is correct. Let's verify with y=207. Then x=(2013 -7*207)/4=(2013 -1449)/4=(564)/4=141. Then x+y=141+207=348<350. Revenue is 4*141 +7*207=564 +1449=2013. Correct. So that works. If we try y=203, which is 203≡3 mod4 (since 203=4*50 +3). Then x=(2013 -7*203)/4=(2013 -1421)/4=592/4=148. Then x+y=148+203=351≥350, which is invalid. So y=203 is invalid. Next possible y is 207. So y=207 is the minimal. Therefore, the answer is 207.Wait, but the problem says "he sold at least how many ballpoint pens". So maybe the answer is 207. Let me check again. Let me confirm that there's no smaller y with y≡3 mod4 and y≥205. The next y would be 207, since 205≡1,206≡2,207≡3 mod4. So indeed, 207 is the smallest y meeting y≥205 and y≡3 mod4.So yes, the answer should be 207. Therefore, he sold at least 207 ballpoint pens.Final Answerboxed{207}</think>Xiaoming sold pencils and ballpoint pens at a stationery store. Pencils are sold for 4 yuan each, and ballpoint pens for 7 yuan each. The total number of pencils and ballpoint pens he initially had was 350, but he did not sell all of them. His sales revenue was 2013 yuan. We need to find the minimum number of ballpoint pens he sold.We start with the equations:[ 4x + 7y = 2013 ][ x + y < 350 ]where ( x ) is the number of pencils sold and ( y ) is the number of ballpoint pens sold.First, we express ( x ) in terms of ( y ):[ x = frac{2013 - 7y}{4} ]For ( x ) to be an integer, ( 2013 - 7y ) must be divisible by 4. We find that ( 2013 equiv 1 mod 4 ), leading to:[ 1 - 7y equiv 0 mod 4 ][ 7y equiv 1 mod 4 ]Since ( 7 equiv 3 mod 4 ), we solve:[ 3y equiv 1 mod 4 ]The multiplicative inverse of 3 modulo 4 is 3, so:[ y equiv 3 mod 4 ]Thus, ( y = 4k + 3 ) for non-negative integers ( k ).Substituting ( y = 4k + 3 ) into the revenue equation, we get:[ x = 498 - 7k ]We also need ( x + y < 350 ):[ (498 - 7k) + (4k + 3) < 350 ][ 501 - 3k < 350 ][ 3k > 151 ][ k > frac{151}{3} approx 50.33 ]Thus, ( k geq 51 ).The minimum value of ( y ) occurs when ( k = 51 ):[ y = 4(51) + 3 = 207 ]Verification:- ( x = 498 - 7(51) = 141 )- ( x + y = 141 + 207 = 348 ) (which is less than 350)- Revenue: ( 4(141) + 7(207) = 564 + 1449 = 2013 )Thus, the minimum number of ballpoint pens sold is (boxed{207}).]

🔑:10. A. 207.Let x and y represent the number of pencils and ballpoint pens sold, respectively. Then[begin{array}{l}left{begin{array}{l}4 x+7 y=2013 ; x+y=204end{array}right.end{array}]Thus, y_{text {min }}=207, at which point, x=141.

🧠:[<think>Okay, so I need to express the number 3,300,000 in scientific notation. Hmm, scientific notation is a way to write really big or really small numbers more compactly, right? From what I remember, it's usually in the form of a number between 1 and 10 multiplied by 10 raised to some power. Let me think. For example, 5,000 would be 5 times 10^3 because you move the decimal point three places to the left. So, starting with 3,300,000. Let me count the digits. It's 3.3 million, which is 3,300,000. If I want to write this in scientific notation, I need to get a number between 1 and 10. So, I have to move the decimal point in 3,300,000 to the left until there's only one non-zero digit before the decimal. The original number is 3,300,000. The decimal point is at the end, right? So, 3,300,000. Starting from the end, moving the decimal to the left one place at a time. First move: 330,000.0. That's 3,300,000 divided by 10, which is 330,000. But that's still not between 1 and 10. Second move: 33,000.00. Still too big. Third move: 3,300.000. Hmm, still not there. Fourth move: 330.0000. Fifth move: 33.00000. Sixth move: 3.300000. Okay, now it's 3.3, which is between 1 and 10. So, I moved the decimal point 6 places to the left. That means I need to multiply by 10^6 to get back to the original number. Therefore, 3.3 multiplied by 10^6 should equal 3,300,000. Let me check that. 3.3 times 10^6 is 3.3 times 1,000,000, which is 3,300,000. Yep, that works. Wait, but just to make sure I didn't make a mistake. Let me count the number of zeros in 3,300,000. The number is 3,300,000. Breaking it down, the number is 3.3 million. 1 million is 10^6, so 3.3 million would be 3.3 times 10^6. That seems right. Alternatively, if I write out 3,300,000, there are 5 zeros after the 3 and the comma. Wait, commas are used as thousand separators. So, 3,300,000 is 3 million 300 thousand. So, 3,000,000 is 3x10^6, and 300,000 is 3x10^5. So, adding them together, 3x10^6 + 3x10^5 = 3.3x10^6. Because 3x10^5 is 0.3x10^6. So, yes, 3.3x10^6. Another way to think about it: the original number is 3,300,000. To convert to scientific notation, place the decimal after the first non-zero digit. The first non-zero digit is 3, so we put the decimal after that, making it 3.300000. Then, count how many places we moved the decimal from its original position. The original number is 3,300,000., so moving the decimal from after the last zero to between the 3 and the first comma, which is 6 places. So, that's 10^6. Therefore, 3.3 x 10^6. Let me make sure there's no confusion with the number of zeros. 10^6 is 1,000,000, which is a 1 followed by six zeros. So, 3.3 times that would be 3,300,000, which is correct. I think that's solid. So the answer should be 3.3 times 10 to the sixth power. In scientific notation, that's written as 3.3 × 10^6. So putting that into the required format, inside a box and all. Just to recap the steps: 1. Start with the number 3,300,000.2. Move the decimal point left until you have a number between 1 and 10. In this case, moving 6 places gives 3.3.3. Multiply that number by 10 raised to the number of places you moved the decimal. So, 3.3 × 10^6.Yep, that all checks out. No mistakes in my reasoning. I think that's the correct scientific notation for 3,300,000.Final AnswerThe number 3300000 expressed in scientific notation is boxed{3.3 times 10^6}.</think>To express the number 3,300,000 in scientific notation, we follow these steps:1. Identify the number: Start with 3,300,000.2. Place the decimal point: Move the decimal point to the left until we have a number between 1 and 10. For 3,300,000, moving the decimal point 6 places to the left results in 3.3.3. Determine the exponent: The decimal point was moved 6 places to the left, so we multiply by (10^6).Thus, the number 3,300,000 in scientific notation is (3.3 times 10^6).[boxed{3.3 times 10^6}],<think>Okay, so I need to express the number 3,300,000 in scientific notation. Hmm, let me think about how scientific notation works. From what I remember, scientific notation is a way to write really big or really small numbers more concisely by using powers of ten. The general form is a number between 1 and 10 multiplied by 10 raised to some exponent. Like, for example, 5,000 would be 5 × 10³ because you move the decimal point three places to the left.Alright, so applying that to 3,300,000. First, I need to convert this number into a number between 1 and 10 multiplied by 10 to the power of something. Let me start by placing the decimal point in the right spot. The original number is 3,300,000. If I put a decimal point after the first digit, that would be 3.300000, right? Because the first digit is 3, and then the rest are zeros or trailing numbers. But wait, there are two 3s here. Let me check again. The number is 3,300,000, which is 3 million plus 300 thousand. So written out in digits, it's 3 3 0 0 0 0 0. So starting from the left, the first non-zero digit is 3, then another 3, followed by five zeros. Hmm, so maybe the decimal should be after the first digit? But in that case, if the number is 3.3 million, right?Wait, let me think. Let's count the number of places we move the decimal point. The original number is 3,300,000. If I write it as 3.3 × 10^6, that makes sense because 3.3 multiplied by 10^6 is 3,300,000. Let me verify: 10^6 is 1,000,000, so 3.3 × 1,000,000 is 3,300,000. Yep, that works. So the decimal starts after the first 3, then we move it six places to the right to get back to the original number. So that means the exponent is 6. So the scientific notation would be 3.3 × 10^6.Wait, but I want to make sure I didn't make a mistake here. Let me step through it again. Starting with 3,300,000. To get a number between 1 and 10, I need to move the decimal point so that there's only one non-zero digit before the decimal. In this case, the first digit is 3, so the decimal should be after the 3, making it 3.300000. But since the number is 3,300,000, which is 3.3 million, that's 3.3 × 10^6. Alternatively, if I move the decimal six places to the left from the original number, I get 3.3, and since I moved six places, the exponent is 6. So that seems correct.Let me check another example to confirm. Take 45,000. In scientific notation, that's 4.5 × 10^4. Right, because moving the decimal four places to the left gives 4.5 and multiplying by 10^4 gets back to 45,000. So similarly, 3,300,000 is 3.3 × 10^6. That seems consistent.Another way to think about it is to count the number of digits after the first digit. In 3,300,000, after the first digit 3, there are six digits: 3, 0, 0, 0, 0, 0. Wait, actually, no. Let's write the number out: 3,300,000 is "3" comma "300,000". Breaking it down, 3 million is 3,000,000, and then 300,000. So the full number is 3,300,000. How many digits is that? Let's count: 3, 3, 0, 0, 0, 0, 0. So seven digits in total. Wait, that's seven digits. But in scientific notation, the exponent is one less than the number of digits if the number is greater than 1. Wait, is that a rule? Let me recall: For numbers greater than or equal to 10, the exponent is one less than the number of digits. For example, 10 has two digits, exponent is 1 (10^1). 100 has three digits, exponent 2 (10^2). So following that, if 3,300,000 has seven digits, the exponent would be 6, which matches what I had before. So 3.3 × 10^6.But just to be thorough, let's break down the number. 3,300,000 is 3.3 × 1,000,000. And 1,000,000 is 10^6. So indeed, 3.3 × 10^6. That makes sense.Alternatively, if I start at the decimal point in the original number, which is implicitly after the last zero, 3,300,000., and move it to the left until I have a number between 1 and 10. Each move to the left increases the exponent by 1. Starting from 3,300,000.0, moving the decimal to the left once: 330,000.0 which is ×10^1. Wait, no, maybe I'm confusing the direction. If I move the decimal to the left, the exponent increases. Let me do this step by step.Original number: 3,300,000.0Move decimal one place to the left: 330,000.00 (exponent 10^1)Move decimal two places: 33,000.000 (exponent 10^2)Three places: 3,300.0000 (exponent 10^3)Four places: 330.00000 (exponent 10^4)Five places: 33.000000 (exponent 10^5)Six places: 3.3000000 (exponent 10^6)So after moving six places to the left, we get 3.3, and the exponent is 10^6. Therefore, the scientific notation is 3.3 × 10^6.That seems consistent. Another check: if I take 3.3 and multiply by 10^6, that's 3.3 × 1,000,000 = 3,300,000. Exactly the number we started with. So that must be correct.I think I've verified it multiple ways. So the answer is 3.3 × 10^6.Final AnswerThe number 3300000 in scientific notation is boxed{3.3 times 10^{6}}.</think>To express the number 3,300,000 in scientific notation, we need to convert it into a number between 1 and 10 multiplied by a power of 10. 1. Start with the number 3,300,000.2. Place the decimal point after the first non-zero digit, which is 3, resulting in 3.300000.3. Count the number of places the decimal point has moved from its original position to the new position. The decimal point has moved 6 places to the left.4. Therefore, the exponent for the power of 10 is 6.Thus, the number 3,300,000 in scientific notation is (3.3 times 10^6).[boxed{3.3 times 10^{6}}]]

🔑:To express 3300000 in scientific notation, we follow the steps below:1. Identify the significant digits in the number, which are 3 and 3.2. Count the number of places we move the decimal point to the left until it is after the first significant digit. In this case, we move it 6 places to the left.3. Write the number as a product of the significant digits and 10 raised to the power of the number of places the decimal was moved.Therefore, 3300000 can be expressed as 3.3 times 10^{6}.So, the final answer in scientific notation is boxed{3.3 times 10^{6}}.

🧠:[<think>Okay, let's tackle this problem step by step. It's divided into two parts, so I'll handle them one at a time.Part (1): Find the Cartesian equation of curve C.The polar equation given is ρ = 4 sin(θ + π/3). I need to convert this into Cartesian coordinates. Remember that polar and Cartesian coordinates are related by the equations:- x = ρ cos θ- y = ρ sin θ- ρ² = x² + y²First, let's expand the sine term using the angle addition formula. The identity for sin(A + B) is sin A cos B + cos A sin B. Applying that here:sin(θ + π/3) = sin θ cos(π/3) + cos θ sin(π/3)We know that cos(π/3) = 1/2 and sin(π/3) = √3/2. Substituting these in:sin(θ + π/3) = sin θ*(1/2) + cos θ*(√3/2) = (1/2) sin θ + (√3/2) cos θSo, the polar equation becomes:ρ = 4 [ (1/2) sin θ + (√3/2) cos θ ] = 2 sin θ + 2√3 cos θNow, multiply both sides by ρ to make it easier to convert to Cartesian coordinates:ρ² = 2 ρ sin θ + 2√3 ρ cos θSubstitute ρ² with x² + y², ρ sin θ with y, and ρ cos θ with x:x² + y² = 2y + 2√3 xNow, let's rearrange the equation to bring all terms to one side:x² - 2√3 x + y² - 2y = 0To complete the squares for both x and y terms.For the x-terms: x² - 2√3 x. Take half of the coefficient of x, which is -2√3 / 2 = -√3, square it: (√3)² = 3. So, add and subtract 3.For the y-terms: y² - 2y. Half of -2 is -1, square is 1. Add and subtract 1.Putting it all together:(x² - 2√3 x + 3) - 3 + (y² - 2y + 1) - 1 = 0Simplify:(x - √3)² + (y - 1)² - 4 = 0Therefore,(x - √3)² + (y - 1)² = 4So, curve C is a circle with center at (√3, 1) and radius 2. That's the Cartesian equation.Part (2): Calculate |MA| + |MB| where M is (3, π/2) and A, B are intersections of line l and curve C.First, let's convert point M's polar coordinates to Cartesian coordinates. Polar coordinates (r, θ) convert to Cartesian as (r cos θ, r sin θ). For M(3, π/2):x = 3 cos(π/2) = 0y = 3 sin(π/2) = 3*1 = 3So, M is at (0, 3) in Cartesian coordinates.Next, we need to find points A and B where line l intersects curve C. Line l is given by parametric equations:x = -1/2 ty = 3 + (√3/2) tWe can substitute these into the Cartesian equation of curve C, which is (x - √3)² + (y - 1)² = 4.Substituting x and y:[ (-1/2 t - √3 )² + ( (3 + (√3/2 t) - 1 )² ] = 4Simplify each part:First term: (- (1/2)t - √3 )² = [ -√3 - (t/2) ]² = ( √3 + (t/2) )² = (t/2 + √3 )² = (t²)/4 + √3 t + 3Second term: (2 + (√3/2) t )² = [ (√3/2 t ) + 2 ]² = ( (√3/2 t )² + 2*(√3/2 t)*2 + 2² ) = ( (3/4)t² + 2√3 t + 4 )Wait, let me check:Wait, the second term is (3 + (√3/2)t - 1) = 2 + (√3/2)t. Then squaring that:(2 + (√3/2)t)^2 = 2² + 2*2*(√3/2)t + (√3/2 t)^2 = 4 + 2√3 t + (3/4)t²So, the entire equation becomes:[ (t²)/4 + √3 t + 3 ] + [ (3/4)t² + 2√3 t + 4 ] = 4Combine like terms:First, t² terms: (1/4)t² + (3/4)t² = (4/4)t² = t²Next, √3 t terms: √3 t + 2√3 t = 3√3 tConstants: 3 + 4 = 7So the equation is:t² + 3√3 t + 7 = 4Subtract 4 from both sides:t² + 3√3 t + 3 = 0Now, solve for t:This is a quadratic equation: t² + 3√3 t + 3 = 0Using the quadratic formula:t = [ -3√3 ± sqrt( (3√3)^2 - 4*1*3 ) ] / 2*1Calculate discriminant:(3√3)^2 - 12 = 27 - 12 = 15So,t = [ -3√3 ± sqrt(15) ] / 2Therefore, the two t parameters are t1 = [ -3√3 + sqrt(15) ] / 2 and t2 = [ -3√3 - sqrt(15) ] / 2Now, plug these t values into the parametric equations of line l to get points A and B.For point A (t1):x1 = -1/2 * t1 = -1/2 * [ (-3√3 + sqrt(15)) / 2 ] = (3√3 - sqrt(15)) / 4Wait, hold on:Wait, t1 is [ -3√3 + sqrt(15) ] / 2So x1 = -1/2 * t1 = -1/2 * [ (-3√3 + sqrt(15)) / 2 ] = (3√3 - sqrt(15)) / 4Similarly, y1 = 3 + (√3/2) t1 = 3 + (√3/2) * [ (-3√3 + sqrt(15)) / 2 ] = 3 + [ (-3*3 + √3*sqrt(15)) / 4 ]Wait, compute (√3/2)*( -3√3 + sqrt(15) )/2:Multiply out:√3/2 * (-3√3)/2 + √3/2 * sqrt(15)/2First term: -3*(√3*√3)/4 = -3*3/4 = -9/4Second term: (√3*sqrt(15))/4 = sqrt(45)/4 = 3*sqrt(5)/4Therefore, y1 = 3 - 9/4 + 3√5/4 = (12/4 - 9/4) + 3√5/4 = 3/4 + 3√5/4 = (3 + 3√5)/4 = 3(1 + √5)/4Similarly, for point B (t2):t2 = [ -3√3 - sqrt(15) ] / 2x2 = -1/2 * t2 = -1/2 * [ (-3√3 - sqrt(15)) / 2 ] = (3√3 + sqrt(15)) / 4y2 = 3 + (√3/2)*t2 = 3 + (√3/2)*[ (-3√3 - sqrt(15))/2 ] = 3 + [ (-3*3 - √3*sqrt(15))/4 ]Compute:First term: √3/2 * (-3√3)/2 = -9/4Second term: √3/2 * (-sqrt(15))/2 = -sqrt(45)/4 = -3√5/4Therefore, y2 = 3 - 9/4 - 3√5/4 = (12/4 - 9/4) - 3√5/4 = 3/4 - 3√5/4 = (3 - 3√5)/4 = 3(1 - √5)/4So points A and B have coordinates:A: ( (3√3 - sqrt(15))/4 , (3 + 3√5)/4 )B: ( (3√3 + sqrt(15))/4 , (3 - 3√5)/4 )Now, we need to find |MA| + |MB| where M is (0,3).So, compute the distances from M(0,3) to A and B, then sum them up.First, let's compute |MA|:Coordinates of A: ( (3√3 - sqrt(15))/4 , (3 + 3√5)/4 )Coordinates of M: (0, 3) = (0, 12/4 )So, the difference in x-coordinates: (3√3 - sqrt(15))/4 - 0 = (3√3 - sqrt(15))/4Difference in y-coordinates: (3 + 3√5)/4 - 12/4 = (3 + 3√5 - 12)/4 = (-9 + 3√5)/4So,|MA| = sqrt[ ( (3√3 - sqrt(15))/4 )² + ( (-9 + 3√5)/4 )² ]Similarly for |MB|:Coordinates of B: ( (3√3 + sqrt(15))/4 , (3 - 3√5)/4 )Difference in x-coordinates: (3√3 + sqrt(15))/4 - 0 = (3√3 + sqrt(15))/4Difference in y-coordinates: (3 - 3√5)/4 - 12/4 = (3 - 3√5 - 12)/4 = (-9 - 3√5)/4Thus,|MB| = sqrt[ ( (3√3 + sqrt(15))/4 )² + ( (-9 - 3√5)/4 )² ]This seems complicated. Maybe there's a smarter way. Let's see.Alternatively, maybe instead of computing coordinates, since A and B lie on line l, and M is a point, perhaps using properties of lines and circles?Wait, another thought: If points A and B are intersections of line l with circle C, then line l is a secant of circle C. The value |MA| + |MB| is the sum of distances from M to A and M to B. This might be related to properties of circles or ellipses, but I'm not sure. Alternatively, perhaps using power of a point?Power of a point M with respect to circle C is |MO|² - r², where O is the center of the circle. Then, for a point outside, the power is equal to MA * MB (if M is outside, which it might be here). Let's check.First, compute the power of M with respect to circle C.Center of circle C is (√3, 1), radius 2.Coordinates of M: (0,3). Compute distance from M to center:Distance MO = sqrt( (√3 - 0)^2 + (1 - 3)^2 ) = sqrt( 3 + 4 ) = sqrt(7)Power of M: MO² - r² = 7 - 4 = 3Therefore, MA * MB = 3But the problem asks for |MA| + |MB|. Hmm. If we know MA * MB and we could find MA + MB, but with just that, it's not directly possible unless we have more info. Alternatively, maybe we can parametrize the line and use some algebraic approach.Alternatively, perhaps parametrize the points A and B as t1 and t2, and use the parametrization of line l to compute |MA| + |MB|.Wait, given that in parametric terms, points A and B correspond to parameters t1 and t2. Since line l is parametrized with parameter t, and we have expressions for x(t) and y(t), we can write |MA| and |MB| in terms of t.But maybe even better, since M is on line l?Wait, check if M is on line l. Line l's parametric equations are x = -1/2 t, y = 3 + (√3/2) t. If M is (0,3), then set x=0: 0 = -1/2 t => t=0. Then y=3 + 0=3. So yes, M is on line l when t=0. Therefore, line l passes through M. So points A and B are the intersections of line l with circle C, and M is on line l. Therefore, A and B are points on line l, and M is also on line l.Therefore, MA and MB are distances along line l from M to A and M to B. But in that case, |MA| + |MB| is just the distance between A and B? Wait, no. Because M is between A and B, then |MA| + |MB| would be |AB|. But if M is not between them, then it's the sum of distances from M to each. Wait, but since the line passes through M and intersects the circle at A and B, M could be inside or outside the circle.Wait, let's check if M is inside the circle C. Circle C has center (√3, 1) and radius 2. Distance from M(0,3) to center (√3,1):sqrt( (√3 - 0)^2 + (1 - 3)^2 ) = sqrt(3 + 4) = sqrt(7) ≈ 2.645, which is greater than radius 2. So M is outside the circle. Therefore, line l passes through M and intersects the circle at two points A and B. Then, the power of point M with respect to circle C is MA * MB = 3 as calculated earlier.But how can we relate MA + MB? Since we know MA * MB = 3, but need MA + MB. If we can find MA + MB, perhaps by expressing it in terms of t1 and t2?Alternatively, let's recall that for points on a line parametrized by t, the distance from a fixed point (which is also on the line) can be expressed in terms of the parameter difference.Wait, since M corresponds to t=0, as we saw earlier (x=0, y=3). The parameters for A and B are t1 and t2. Since the parametric equations are linear in t, the distance from M to A would be proportional to |t1 - 0| = |t1|, and similarly for MB.But wait, in parametric terms, the parameter t isn't necessarily an arc length parameter. The speed of the parametrization needs to be considered. Let's check the speed of the parametrization.Given x(t) = -1/2 t, y(t) = 3 + (√3/2) t. Then, the velocity vector is (dx/dt, dy/dt) = (-1/2, √3/2). The speed is sqrt( (-1/2)^2 + (√3/2)^2 ) = sqrt(1/4 + 3/4) = sqrt(1) = 1. So the parameter t is actually the arc length parameter. Therefore, the distance from M (t=0) to A (t = t1) is |t1 - 0| = |t1|, and similarly for B, it's |t2|.Therefore, |MA| + |MB| = |t1| + |t2|But t1 and t2 are the roots of the quadratic equation t² + 3√3 t + 3 = 0.From quadratic equation, sum of roots t1 + t2 = -3√3, product t1 t2 = 3.But since the quadratic is t² + 3√3 t + 3 = 0, the roots are t = [ -3√3 ± sqrt(27 - 12) ] / 2 = [ -3√3 ± sqrt(15) ] / 2So t1 = [ -3√3 + sqrt(15) ] / 2 ≈ [ -5.196 + 3.872 ] / 2 ≈ [ -1.324 ] / 2 ≈ -0.662t2 = [ -3√3 - sqrt(15) ] / 2 ≈ [ -5.196 - 3.872 ] / 2 ≈ [ -9.068 ] / 2 ≈ -4.534Therefore, both t1 and t2 are negative. Therefore, |t1| + |t2| = (-t1) + (-t2) = - (t1 + t2 ) = - ( -3√3 ) = 3√3Because from the quadratic, t1 + t2 = -3√3. Therefore, the sum |MA| + |MB| = 3√3.Wait, that's a neat result. Because if the parametrization has unit speed, then distances correspond to |t|. Since both points A and B are on the line, which is parametrized with t as the arc length parameter, and since M corresponds to t=0, the distances MA and MB are |t1| and |t2|. Since t1 and t2 are both negative (as computed numerically above), their absolute values sum up to -(t1 + t2) = - (sum of roots). But sum of roots is -3√3, so -( -3√3 ) = 3√3.Therefore, the answer is 3√3.But let's verify this with the earlier calculation where I found MA * MB = 3 (from power of a point) and now MA + MB = 3√3. Let's check if (MA + MB)^2 = MA^2 + MB^2 + 2 MA * MB.Compute (3√3)^2 = 27MA^2 + MB^2 + 2*3 = ?But we need MA^2 + MB^2. Alternatively, from coordinates:But maybe it's too tedious, but let's trust the parametric method. Since the parametrization has unit speed, then the distance from M (t=0) to any point on the line is indeed |t|. Therefore, |MA| = |t1|, |MB| = |t2|, and since t1 and t2 are both negative, |MA| + |MB| = -t1 - t2 = -(t1 + t2) = -(-3√3) = 3√3. Therefore, the answer is 3√3.Final Answer(1) boxed{(x - sqrt{3})^2 + (y - 1)^2 = 4}(2) boxed{3sqrt{3}}</think>(1) To find the Cartesian equation of curve ( C ) given by the polar equation ( rho = 4 sin left( theta + frac{pi}{3} right) ):First, expand the sine term using the angle addition formula:[sin left( theta + frac{pi}{3} right) = sin theta cos frac{pi}{3} + cos theta sin frac{pi}{3} = frac{1}{2} sin theta + frac{sqrt{3}}{2} cos theta]Substituting this back into the polar equation:[rho = 4 left( frac{1}{2} sin theta + frac{sqrt{3}}{2} cos theta right) = 2 sin theta + 2sqrt{3} cos theta]Multiply both sides by ( rho ):[rho^2 = 2 rho sin theta + 2sqrt{3} rho cos theta]Convert to Cartesian coordinates using ( rho^2 = x^2 + y^2 ), ( rho sin theta = y ), and ( rho cos theta = x ):[x^2 + y^2 = 2y + 2sqrt{3}x]Rearrange and complete the square:[x^2 - 2sqrt{3}x + y^2 - 2y = 0][(x - sqrt{3})^2 + (y - 1)^2 = 4]Thus, the Cartesian equation of curve ( C ) is:[boxed{(x - sqrt{3})^2 + (y - 1)^2 = 4}](2) To calculate ( |MA| + |MB| ) where ( M ) has polar coordinates ( left( 3, frac{pi}{2} right) ) and ( A ), ( B ) are the intersections of line ( l ) and curve ( C ):Convert ( M ) to Cartesian coordinates:[M = (0, 3)]The parametric equations of line ( l ) are:[begin{cases}x = -frac{1}{2}t y = 3 + frac{sqrt{3}}{2}tend{cases}]Substitute these into the Cartesian equation of curve ( C ):[left( -frac{1}{2}t - sqrt{3} right)^2 + left( 3 + frac{sqrt{3}}{2}t - 1 right)^2 = 4]Simplify and solve the quadratic equation:[t^2 + 3sqrt{3}t + 3 = 0]The roots ( t_1 ) and ( t_2 ) are:[t = frac{-3sqrt{3} pm sqrt{15}}{2}]Since the parametrization has unit speed, the distances ( |MA| ) and ( |MB| ) are the absolute values of the parameters ( t_1 ) and ( t_2 ). The sum of the distances is:[|MA| + |MB| = |t_1| + |t_2| = -left( t_1 + t_2 right) = 3sqrt{3}]Thus, the value of ( |MA| + |MB| ) is:[boxed{3sqrt{3}}],<think>Alright, let's tackle this problem step by step. It's divided into two parts, so I'll start with part (1).Problem (1): Find the Cartesian equation of curve C.The polar equation given is ρ = 4 sin(θ + π/3). I need to convert this into Cartesian coordinates. Remember, polar coordinates relate to Cartesian coordinates through the equations:- x = ρ cos θ- y = ρ sin θ- ρ² = x² + y²So, first, let's expand the sine term using the angle addition formula. The formula for sin(A + B) is sin A cos B + cos A sin B. Applying that here:sin(θ + π/3) = sin θ cos(π/3) + cos θ sin(π/3)I know that cos(π/3) = 1/2 and sin(π/3) = √3/2. Substituting these values in:sin(θ + π/3) = sin θ*(1/2) + cos θ*(√3/2) = (sin θ)/2 + (√3 cos θ)/2So the original equation becomes:ρ = 4 [ (sin θ)/2 + (√3 cos θ)/2 ] = 2 sin θ + 2√3 cos θNow, multiply both sides by ρ to make it easier to substitute Cartesian coordinates:ρ² = 2ρ sin θ + 2√3 ρ cos θBut ρ² is x² + y², ρ sin θ is y, and ρ cos θ is x. So substituting:x² + y² = 2y + 2√3 xLet me rearrange this equation to standard form. Bring all terms to the left side:x² + y² - 2√3 x - 2y = 0Now, to write this as a circle equation, complete the squares for x and y.For the x terms: x² - 2√3 x. Take half of -2√3, which is -√3, square it to get (√3)² = 3. So add and subtract 3.For the y terms: y² - 2y. Take half of -2, which is -1, square it to get 1. Add and subtract 1.So:(x² - 2√3 x + 3) - 3 + (y² - 2y + 1) - 1 = 0Which simplifies to:(x - √3)² + (y - 1)² - 4 = 0Then,(x - √3)² + (y - 1)² = 4So, curve C is a circle with center at (√3, 1) and radius 2. That should be the Cartesian equation. Let me just check the steps again to make sure I didn't make an arithmetic mistake.Starting from ρ = 4 sin(θ + π/3), expanded correctly. Then multiplied by ρ, substituted terms, completed the squares. Looks okay. So part (1) is done.Problem (2): Calculate |MA| + |MB| where M is (3, π/2) and A, B are intersections of line l and curve C.First, let me note that M has polar coordinates (3, π/2). Converting this to Cartesian coordinates: x = ρ cos θ = 3 cos(π/2) = 0, y = 3 sin(π/2) = 3*1 = 3. So M is (0, 3) in Cartesian.Wait, but the parametric equations for line l are given as:x = -1/2 ty = 3 + (√3)/2 tSo, line l is parametrized by t, and we need to find where this line intersects curve C (which we now know is the circle (x - √3)^2 + (y - 1)^2 = 4). Then, once we find points A and B, we need to compute the sum of the distances from M to A and M to B.Alright, step by step:1. Find the intersections A and B between line l and curve C.First, substitute the parametric equations of line l into the Cartesian equation of curve C.From line l:x = - (1/2)ty = 3 + (√3/2)tSubstitute x and y into the circle equation:[ (-(1/2)t - √3 )² + (3 + (√3/2 t - 1) )² ] = 4Wait, let me check the substitution:The circle equation is (x - √3)^2 + (y - 1)^2 = 4So substituting x and y:[ (- (1/2)t - √3 ) ]² + [ (3 + (√3/2)t - 1 ) ]² = 4Simplify each bracket:First bracket: x - √3 = - (1/2)t - √3Second bracket: y - 1 = 3 + (√3/2)t - 1 = 2 + (√3/2)tSo the equation becomes:[ - (1/2)t - √3 ]² + [ 2 + (√3/2)t ]² = 4Let me compute each term:First term: [ - (1/2)t - √3 ]² = [ - (t/2 + √3) ]² = (t/2 + √3)² = (t²)/4 + √3 t + 3Second term: [ 2 + (√3/2)t ]² = 4 + 2*(√3/2)t*2 + ( (√3/2)t )² = Wait, no, expanding (a + b)^2:= 2² + 2*2*(√3/2)t + (√3/2 t)^2= 4 + 2*2*(√3/2)t + (3/4)t²Simplify each part:First term: 4Second term: 2*2*(√3/2) = 2*√3, so 2√3 tThird term: (3/4)t²So altogether, the second term is 4 + 2√3 t + (3/4)t²Now, combine both terms:First term: (t²)/4 + √3 t + 3Second term: (3/4)t² + 2√3 t + 4Adding them:(t²/4 + 3t²/4) + (√3 t + 2√3 t) + (3 + 4) = (t²) + 3√3 t + 7Wait, let me check:First term: (1/4 + 3/4)t² = (4/4)t² = t²Second term: √3 t + 2√3 t = 3√3 tThird term: 3 + 4 = 7So overall:t² + 3√3 t + 7 = 4So subtract 4:t² + 3√3 t + 3 = 0Now, this is a quadratic equation in t: t² + 3√3 t + 3 = 0Let me solve for t using quadratic formula.Discriminant D = (3√3)^2 - 4*1*3 = 27 - 12 = 15So roots:t = [ -3√3 ± √15 ] / 2So the two values of t are [ -3√3 + √15 ] / 2 and [ -3√3 - √15 ] / 2Therefore, the points A and B correspond to these t-values.Now, let's find coordinates for A and B.Using the parametric equations:x = -1/2 ty = 3 + (√3/2)tSo for each t, compute x and y.Let me denote t1 = [ -3√3 + √15 ] / 2 and t2 = [ -3√3 - √15 ] / 2Compute coordinates:For t1:x1 = -1/2 * t1 = -1/2 * [ (-3√3 + √15)/2 ] = (3√3 - √15)/4y1 = 3 + (√3/2)*t1 = 3 + (√3/2)*[ (-3√3 + √15)/2 ] = 3 + [ (-3*3 + √3*√15 ) /4 ] = 3 + [ (-9 + √45 ) /4 ]Simplify √45 = 3√5, so:y1 = 3 + ( -9 + 3√5 ) /4 = (12/4) + (-9 + 3√5)/4 = (12 -9 + 3√5)/4 = (3 + 3√5)/4 = 3(1 + √5)/4Similarly for t2:x2 = -1/2 * t2 = -1/2 * [ (-3√3 - √15 ) /2 ] = (3√3 + √15)/4y2 = 3 + (√3/2)*t2 = 3 + (√3/2)*[ (-3√3 - √15 ) /2 ] = 3 + [ (-3*3 - √3*√15 ) /4 ] = 3 + [ (-9 - √45 ) /4 ]Again, √45 = 3√5:y2 = 3 + (-9 - 3√5)/4 = (12/4) + (-9 -3√5)/4 = (12 -9 -3√5)/4 = (3 -3√5)/4 = 3(1 - √5)/4Therefore, points A and B have coordinates:A: ( (3√3 - √15)/4 , 3(1 + √5)/4 )B: ( (3√3 + √15)/4 , 3(1 - √5)/4 )Now, point M is (0, 3). Need to find |MA| + |MB|.This seems complicated because the coordinates of A and B are quite messy. Maybe there's a smarter way than computing each distance directly.Alternatively, since points A and B lie on the line l, and M is a fixed point, maybe we can parameterize the distances.Alternatively, maybe there's a geometric interpretation. Let's recall that in some cases, for example, with an ellipse, the sum of distances from two foci is constant, but this is a circle and a line, so perhaps not directly applicable.Alternatively, perhaps use properties of power of a point or inversion, but that might be overcomplicating.Alternatively, since MA + MB is the sum of distances from M to A and M to B, and A and B are points of intersection of line l with curve C. Maybe there is a way to express this sum without computing each coordinate.Alternatively, note that the problem gives us point M with polar coordinates (3, π/2), which in Cartesian is (0,3). Now, check if point M lies on line l. Let's see:Line l is parametrized by x = -1/2 t, y = 3 + (√3/2)t. If x=0, then -1/2 t=0 => t=0. Then y=3 + 0=3. So yes, when t=0, the point on line l is (0,3), which is M. So M is actually on line l.Therefore, line l passes through point M, and intersects curve C at points A and B. Therefore, MA and MB are the distances from M to A and M to B along line l. Since M is on line l, and A and B are two points on l, then MA and MB are just the distances from M to A and M to B along the line. So the sum |MA| + |MB| is equal to the distance between A and B.Wait, is that correct? If M is between A and B, then |MA| + |MB| = |AB|. If M is outside the segment AB, then |MA| + |MB| would be greater than |AB|.But in our case, since M is on line l, which intersects curve C at A and B, so depending on where M is with respect to A and B.Wait, but let's check. Since M is (0,3). Let's see where M is in relation to the circle.The circle C is centered at (√3, 1) with radius 2. The distance from M (0,3) to the center (√3,1) is sqrt( (√3 - 0)^2 + (1 - 3)^2 ) = sqrt( 3 + 4 ) = sqrt(7) ≈ 2.645. The radius is 2, so since sqrt(7) > 2, M is outside the circle. Therefore, the line l passes through M (which is outside the circle) and intersects the circle at A and B. Therefore, M is outside the circle, and line l passes through M and intersects the circle at A and B. Therefore, the points A and B are on line l, on either side relative to M? Or both on one side?Wait, but if the line passes through M and intersects the circle, which is located at (√3, 1), so let's think about the position.Alternatively, perhaps compute the parameter t for points A, B, and M.Given that M is at t=0 (since when t=0, x=0, y=3). So points A and B correspond to t1 and t2, which are [ -3√3 ± √15 ] / 2. Let's approximate these values numerically to see their positions.Compute t1 and t2:First, approximate √3 ≈ 1.732, √15 ≈ 3.872.t1 = [ -3*1.732 + 3.872 ] / 2 ≈ [ -5.196 + 3.872 ] / 2 ≈ (-1.324)/2 ≈ -0.662t2 = [ -3*1.732 - 3.872 ] /2 ≈ [ -5.196 -3.872 ] /2 ≈ (-9.068)/2 ≈ -4.534So t1 ≈ -0.662 and t2 ≈ -4.534. Since M is at t=0, which is greater than t1 and t2 (since both are negative), so on the line l, as t increases from -∞ to +∞, the point moves from (positive x, decreasing y) to (negative x, increasing y). Wait, original parametric equations:x = -1/2 t, so as t increases, x decreases.y = 3 + (√3/2)t, so as t increases, y increases.So when t is negative, x is positive, and y is less than 3.Since t1 ≈ -0.662 and t2 ≈ -4.534, both points A and B are at negative t parameters. So on the line l, starting from M (t=0), moving towards negative t, we reach A first (t ≈ -0.662), then further on to B (t ≈ -4.534). So M is outside the circle, and the line intersects the circle at A and B, both lying on the side of the line l where t is negative. Therefore, from M's perspective, both points A and B are in the same direction along line l, but since the line passes through the circle twice, there are two points.But in this case, since both t1 and t2 are negative, both points A and B are on the same side relative to M (since M is at t=0). Therefore, the distances MA and MB are the distances from M to A and M to B along the line l, which would be |t1 - 0| * speed along the line? Wait, parametric equations are x = -1/2 t, y = 3 + (√3/2)t. So the direction vector of the line is (-1/2, √3/2). The parameter t is not necessarily the arc length, but the distance between two points on the line can be computed using the difference in t multiplied by the magnitude of the direction vector.The direction vector is (-1/2, √3/2). Its magnitude is sqrt( ( (-1/2)^2 + (√3/2)^2 )) = sqrt( 1/4 + 3/4 ) = sqrt(1) = 1. So the parameter t is actually the arc length parameter because the direction vector has magnitude 1. Therefore, the distance between two points on the line with parameters t1 and t2 is |t2 - t1|.Wait, is that right? If the direction vector has magnitude 1, then each unit of t corresponds to moving 1 unit along the line. So yes, the distance between M (t=0) and A (t=t1) is |t1 - 0| = |t1|, and similarly for B.But in our case, t1 ≈ -0.662 and t2 ≈ -4.534. So the distances MA and MB would be | -0.662 - 0 | = 0.662 and | -4.534 - 0 | = 4.534. Then |MA| + |MB| ≈ 0.662 + 4.534 ≈ 5.196. Wait, 5.196 is approximately 3√3, which is about 5.196. That seems familiar. Let me check:3√3 ≈ 5.196, which matches. Let's see if the exact value is 3√3.So perhaps there's a way to compute |MA| + |MB| without going through all the coordinates.Since t is the parameter and the direction vector has magnitude 1, as we established, the distance from M (t=0) to A (t = t1) is |t1|, and to B (t = t2) is |t2|. Therefore, |MA| + |MB| = |t1| + |t2|. But t1 and t2 are the roots of the quadratic equation t² + 3√3 t + 3 = 0. Since both roots are negative (sum is -3√3 and product is 3), so |t1| + |t2| = (-t1) + (-t2) = - (t1 + t2) = - ( -3√3 ) = 3√3.Therefore, |MA| + |MB| = 3√3.Wow, that's a much simpler way! So instead of computing coordinates and then using distance formula, since the parameter t corresponds to arc length (because the direction vector is unit length), the sum |MA| + |MB| is simply the sum of the absolute values of the parameters t1 and t2, which is equal to the sum of the roots' magnitudes. Since the quadratic equation is t² + 3√3 t + 3 = 0, the sum of roots t1 + t2 = -3√3, so |t1| + |t2| = 3√3.Therefore, the answer is 3√3.Let me verify this with the approximate values. Earlier, t1 ≈ -0.662, t2 ≈ -4.534, so |t1| + |t2| ≈ 0.662 + 4.534 ≈ 5.196, which is approximately 3√3 ≈ 5.196. So that checks out.Therefore, part (2) answer is 3√3.Final Answer(1) The Cartesian equation of curve ( C ) is (boxed{(x - sqrt{3})^2 + (y - 1)^2 = 4}).(2) The value of ( |MA| + |MB| ) is (boxed{3sqrt{3}}).</think>(1) To find the Cartesian equation of the curve ( C ) given by the polar equation (rho = 4 sin left( theta + frac{pi}{3} right)):1. Expand (sin left( theta + frac{pi}{3} right)) using the angle addition formula: [ sin left( theta + frac{pi}{3} right) = sin theta cos frac{pi}{3} + cos theta sin frac{pi}{3} = frac{1}{2} sin theta + frac{sqrt{3}}{2} cos theta ]2. Substitute this back into the polar equation: [ rho = 4 left( frac{1}{2} sin theta + frac{sqrt{3}}{2} cos theta right) = 2 sin theta + 2sqrt{3} cos theta ]3. Multiply both sides by (rho) and convert to Cartesian coordinates: [ rho^2 = 2rho sin theta + 2sqrt{3} rho cos theta implies x^2 + y^2 = 2y + 2sqrt{3}x ]4. Rearrange and complete the square: [ x^2 - 2sqrt{3}x + y^2 - 2y = 0 implies (x - sqrt{3})^2 + (y - 1)^2 = 4 ] The Cartesian equation of curve ( C ) is (boxed{(x - sqrt{3})^2 + (y - 1)^2 = 4}).(2) To calculate the value of ( |MA| + |MB| ):1. Convert the polar coordinates of point ( M left( 3, frac{pi}{2} right) ) to Cartesian coordinates: [ M = (0, 3) ]2. Substitute the parametric equations of line ( l ) into the Cartesian equation of curve ( C ): [ x = -frac{1}{2}t, quad y = 3 + frac{sqrt{3}}{2}t ] Substituting these into the circle equation: [ left( -frac{1}{2}t - sqrt{3} right)^2 + left( 3 + frac{sqrt{3}}{2}t - 1 right)^2 = 4 ]3. Simplify and solve the quadratic equation in ( t ): [ t^2 + 3sqrt{3}t + 3 = 0 ] The roots ( t_1 ) and ( t_2 ) are: [ t_1 = frac{-3sqrt{3} + sqrt{15}}{2}, quad t_2 = frac{-3sqrt{3} - sqrt{15}}{2} ]4. Calculate the sum of distances using the fact that the parameter ( t ) corresponds to arc length (since the direction vector has magnitude 1): [ |MA| + |MB| = |t_1| + |t_2| = - (t_1 + t_2) = 3sqrt{3} ] The value of ( |MA| + |MB| ) is (boxed{3sqrt{3}}).]

🔑:Solution: (1) Expanding rho=4sin left(theta+ frac{pi}{3}right) yields rho =2sin theta +2sqrt{3}cos theta, Multiplying both sides by rho gives rho^2=2rho sin theta +2sqrt{3}rho cos theta, Since rho^2=x^2+y^2, rhosin theta=y, rhocos theta=x, therefore The Cartesian equation of curve C is x^2+y^2-2sqrt{3}x-2y=0; (2) Substituting begin{cases} x=-frac{1}{2}t, y=3+frac{sqrt{3}}{2}t end{cases} into x^2+y^2-2sqrt{3}x-2y=0, We get t^2+3sqrt{3}t+3=0, It is easy to know that the Cartesian coordinates of point M are (0,3). Let the two real roots of this equation be t_1 and t_2, therefore t_1+t_2=-3sqrt{3}, t_1t_2=3, Then, by the geometric meaning of parameter t, we have |MA|+|MB|=|t_1|+|t_2|=|t_1+t_2|=3sqrt{3}.Thus, the final answers are: (1) The Cartesian equation of curve C is boxed{x^2+y^2-2sqrt{3}x-2y=0}. (2) The value of |MA|+|MB| is boxed{3sqrt{3}}.