🧠:[<think>Okay, so here's this problem about a motorboat traveling across a lake and then up a river. Let me try to wrap my head around it step by step. First, the boat goes a distance of "a" kilometers across the lake. Then, it travels half of this distance, which would be a/2 km, up the river that's flowing into the lake. The whole trip takes 1 hour. We need to find the own speed of the motorboat, given that the speed of the river current is "c" km/h.Alright, let's start by recalling that the boat's own speed (which we need to find) is the speed it has in still water. Let's denote that speed as "v" km/h. When the boat is moving across the lake, which is still water because lakes don't have currents like rivers do, its speed is simply "v". But when it goes up the river, against the current, its effective speed decreases because it's going against the river's current. The river current's speed is given as "c" km/h. So, the boat's effective speed upstream would be (v - c) km/h. The total time for the trip is 1 hour. Time is equal to distance divided by speed, so we can write two separate time expressions for the two parts of the trip and then add them together to equal 1 hour.First part: Across the lake.Distance = a kmSpeed = v km/hTime = distance/speed = a / v hoursSecond part: Up the river.Distance = a/2 kmSpeed = (v - c) km/hTime = (a/2) / (v - c) hoursTotal time is the sum of these two times, which is 1 hour. So,a / v + (a/2) / (v - c) = 1This equation should allow us to solve for v in terms of a and c.Let me write that equation again to make sure I have it right:(a / v) + (a / (2(v - c))) = 1Hmm, maybe. Let me check the second term. The distance is a/2 km and the speed is (v - c), so time is (a/2) divided by (v - c), which is the same as a/(2(v - c)). Yes, that's correct.So the equation is:(a / v) + [a / (2(v - c))] = 1Our goal is to solve for v. Let's denote the own speed of the motorboat as v. So, we need to manipulate this equation to solve for v.First, let's perhaps factor out the "a" from both terms. Let's see:a [1 / v + 1 / (2(v - c))] = 1So,1 / v + 1 / (2(v - c)) = 1 / aAlternatively, we can get a common denominator for the two terms on the left side. Let's do that.The denominators are v and 2(v - c). The common denominator would be 2v(v - c). Let's rewrite each term with that denominator.First term: 1 / v = 2(v - c) / [2v(v - c)]Second term: 1 / [2(v - c)] = v / [2v(v - c)]So adding those together:[2(v - c) + v] / [2v(v - c)] = 1 / aSimplify the numerator:2(v - c) + v = 2v - 2c + v = 3v - 2cTherefore, the equation becomes:(3v - 2c) / [2v(v - c)] = 1 / aCross-multiplying:a(3v - 2c) = 2v(v - c)Let's expand both sides:Left side: 3a v - 2a cRight side: 2v^2 - 2c vBring all terms to one side to form a quadratic equation:3a v - 2a c - 2v^2 + 2c v = 0Combine like terms:-2v^2 + (3a v + 2c v) - 2a c = 0Factor out the terms:Let's rewrite the equation:-2v² + v(3a + 2c) - 2a c = 0Alternatively, multiply both sides by -1 to make the quadratic coefficient positive:2v² - v(3a + 2c) + 2a c = 0Now, this is a quadratic equation in terms of v:2v² - (3a + 2c)v + 2a c = 0To solve this quadratic equation, we can use the quadratic formula:v = [ (3a + 2c) ± √( (3a + 2c)² - 4 * 2 * 2a c ) ] / (2 * 2)Let me compute the discriminant first:D = (3a + 2c)² - 16 a cExpanding (3a + 2c)²:= 9a² + 12a c + 4c² - 16a cCombine like terms:9a² - 4a c + 4c²So discriminant D = 9a² - 4a c + 4c²Wait, let me check that again:(3a + 2c)^2 = 9a² + 12ac + 4c²Then subtract 16ac:9a² + 12ac + 4c² - 16ac = 9a² - 4ac + 4c²Yes, correct.So the discriminant is 9a² -4ac +4c²Hmm, let's see if that can be factored or simplified. Maybe it's a perfect square?Let me check:9a² -4ac +4c². Let's see, 9a² is (3a)^2, 4c² is (2c)^2. The middle term is -4ac. If it was (3a - 2c)^2, that would be 9a² -12ac +4c². Not quite. So no, it's not a perfect square. So we can leave it as is.Therefore, the solutions are:v = [ (3a + 2c) ± √(9a² -4ac +4c²) ] / 4Now, since we're talking about the speed of the motorboat, it must be positive. Also, when moving upstream, the boat's effective speed (v - c) must be positive, because otherwise it would be unable to move upstream. So, v - c > 0 => v > c. So, whatever solution we get, it must satisfy v > c.So let's analyze the two possible roots.First, let's compute both roots:v₁ = [ (3a + 2c) + √(9a² -4ac +4c²) ] / 4v₂ = [ (3a + 2c) - √(9a² -4ac +4c²) ] / 4We need to check which of these is greater than c.But perhaps we can simplify the square root expression.Let me look at 9a² -4ac +4c². Let's see if we can write this as (3a - kc)^2 + ... or something else. Alternatively, compute the discriminant for different a and c.Alternatively, maybe we can factor it, but it's not straightforward. Let's compute the discriminant:Suppose we treat this as a quadratic in terms of a. So:9a² -4ac +4c². Let's compute the discriminant for this quadratic (if we treat a as variable):D' = (-4c)^2 -4*9*4c² = 16c² - 144c² = -128c². Negative, which means that the quadratic has no real roots. Therefore, 9a² -4ac +4c² is always positive for real numbers a and c (assuming a and c are positive, which they are in this context, as distances and speeds). So the square root is a real number.Thus, both solutions are real. Now, which one is greater than c?Let's test with sample numbers. Suppose a = 2, c = 1. Then compute v₁ and v₂.First, compute sqrt(9*(2)^2 -4*2*1 +4*(1)^2) = sqrt(36 -8 +4) = sqrt(32) ≈5.656Then,v₁ = [3*2 +2*1 +5.656]/4 = [6 +2 +5.656]/4 ≈13.656/4≈3.414v₂ = [6 +2 -5.656]/4≈(8 -5.656)/4≈2.344/4≈0.586Now, check if these are greater than c=1.v₁≈3.414 >1: yesv₂≈0.586 <1: noTherefore, only v₁ is valid. So the correct solution is v = [3a + 2c + sqrt(9a² -4ac +4c²)] /4.But maybe we can write this in a simplified form.Alternatively, perhaps there's a way to rationalize or factor the expression. Let's see.Looking at the square root sqrt(9a² -4ac +4c²). Let me see:9a² -4ac +4c² = 9a² +4c² -4ac. Hmm. Maybe think of it as 9a² -4ac +4c².Alternatively, can we write this as (3a - (2/3)c)^2 + something? Let's check:(3a - (2/3)c)^2 =9a² -4ac + (4/9)c². Hmm, but original term is 9a² -4ac +4c². The difference is (4c² -4/9 c²)= (32/9)c². So not quite.Alternatively, maybe factor as:9a² -4ac +4c² = (3a)^2 -4ac + (2c)^2. Hmm, not a standard identity. Maybe if we think of it as a quadratic in a:9a² -4ac +4c² = 9a² -4ac +4c². Hmm, perhaps we can factor it as:Let me attempt to factor.Looking for factors of the form (ma + nc)(pa + qc) that multiply to 9a² -4ac +4c².So:(ma + nc)(pa + qc) = m p a² + (m q + n p) a c + n q c².Set equal to 9a² -4ac +4c².So:m p =9m q + n p = -4n q=4Looking for integer solutions.Possible n q=4. So n and q are factors of 4. Let's try n=2, q=2. Then n q=4.Then m p=9. Let's take m=3, p=3.Then m q +n p=3*2 +2*3=6+6=12≠-4. Not good.Alternatively, n=4, q=1. Then m q +n p= m*1 +4*p= m +4p=-4. But m p=9. Let's try m=9, p=1. Then m +4p=9 +4=13≠-4. Not working. Maybe negative numbers?Suppose n=-2, q=-2. Then n q=4. m p=9. Let's take m=3, p=3. Then m q +n p=3*(-2) + (-2)*3= -6 -6= -12≠-4. Not good.Alternatively, n=1, q=4. Then m q +n p= m*4 +1*p=4m +p=-4. m p=9. Let's try m=3, p=3: 4*3 +3=15≠-4. Nope.Alternatively, n=-1, q=-4. Then n q=4. m p=9. Then 4m +p=4m +p=-4. Let's try m=3, p=3: 12 +3=15≠-4. Not working.Alternatively, m and p could be fractions, but this is getting messy. Perhaps this quadratic doesn't factor nicely, so we have to leave the square root as is.Therefore, the solution is:v = [3a + 2c + sqrt(9a² -4ac +4c²)] / 4Alternatively, we can factor out a common term or write it differently. Let me check if the expression inside the square root can be rewritten.Wait, 9a² -4ac +4c². Let me compute 9a² +4c² first. 9a² +4c² is (3a)^2 + (2c)^2. Then subtract 4ac. So:9a² -4ac +4c² = (3a)^2 + (2c)^2 -4ac.Is there a way to express this as (3a - 2c)^2 + something? Let's see:(3a - 2c)^2 =9a² -12ac +4c².But our expression is 9a² -4ac +4c². The difference between them is 8ac. So:9a² -4ac +4c² = (3a -2c)^2 +8ac.Hmm, not particularly helpful.Alternatively, perhaps factor out a common term. Let's see:But unless there's a specific factor, maybe not. So perhaps it's best left as is.Therefore, the final answer is:v = [3a + 2c + sqrt(9a² -4ac +4c²)] /4But let's check whether this answer makes sense dimensionally and in a specific case.Suppose a=2 km, c=1 km/h. Then the expression becomes:v = [3*2 + 2*1 + sqrt(9*4 -4*2*1 +4*1)] /4Compute numerator:6 + 2 + sqrt(36 -8 +4) = 8 + sqrt(32) ≈8 +5.656≈13.656v≈13.656 /4≈3.414 km/hWhich matches our earlier calculation. And since c=1, v=3.414>1, which satisfies v>c. Makes sense.Alternatively, if the river current is zero, c=0. Then what happens?If c=0, then the equation becomes:v = [3a +0 + sqrt(9a² -0 +0)] /4 = [3a +3a]/4=6a/4= (3/2)aBut wait, let's check with c=0. The problem would be: the boat travels a km across the lake at speed v, then a/2 km up a river with current 0 (so same as still water). Total time is a/v + (a/2)/v = (3a)/(2v) =1 hour. So v=(3a)/2. Which matches. So in that case, the solution is v= (3a)/2. Which is consistent with the formula when c=0.Another check: if a=0, which would mean the boat didn't go anywhere. But a=0 is probably not a valid input here, since the problem states the boat traveled a km across the lake. So a must be positive.Another test: c=0.5, a=1.Compute v:[3*1 +2*0.5 + sqrt(9*1 -4*1*0.5 +4*0.25)] /4= [3 +1 + sqrt(9 -2 +1)] /4= [4 + sqrt(8)] /4≈ [4 +2.828]/4≈6.828/4≈1.707 km/hCheck if v>c=0.5: yes, 1.707>0.5. Good.Time across lake:1/1.707≈0.586 hoursTime up river: (0.5)/(1.707 -0.5)=0.5/1.207≈0.414 hoursTotal≈0.586+0.414=1 hour. Correct.Therefore, the formula seems to check out.But perhaps the answer can be presented in a different form? Let me see.Alternatively, maybe we can rationalize or write the square root term as sqrt((3a)^2 -4ac + (2c)^2). Wait, 9a² -4ac +4c² is the same as (3a)^2 -4ac + (2c)^2. But not sure if that helps.Alternatively, maybe factor 9a² -4ac +4c² as (3a - (2/3)c)^2 + (32/9)c², but this seems more complicated.Alternatively, note that 9a² -4ac +4c² = 9a² +4c² -4ac. But not sure.Alternatively, perhaps the problem expects the answer in terms of a quadratic solution, so as we have derived.But let me check once again if there's an alternative way to approach the problem that might result in a simpler expression.Suppose we let t1 be the time taken to cross the lake, and t2 be the time taken to go up the river. Then t1 + t2 =1.From the lake journey: t1 = a / vFrom the river journey: t2 = (a/2) / (v - c)So, a / v + (a)/(2(v - c)) =1Multiply both sides by 2v(v - c):2a(v - c) + a v = 2v(v - c)Left side: 2a(v - c) +a v =2a v -2a c +a v=3a v -2a cRight side:2v(v -c)=2v² -2c vBring all terms to left:3a v -2a c -2v² +2c v=0Which is the same quadratic equation as before.So, same result. Therefore, there isn't a simpler way; the answer is as derived.Therefore, the own speed of the motorboat is [3a + 2c + sqrt(9a² -4ac +4c²)] /4 km/h.But let me check if the problem expects the answer in some specific form. The problem mentions "put your final answer within boxed{}", so probably just the expression.Alternatively, maybe we can write the numerator and denominator differently. Let's see:sqrt(9a² -4ac +4c²). Let's factor out a 1/4 from the numerator:Wait, the numerator is [3a +2c + sqrt(9a² -4ac +4c²)] and the denominator is 4. Maybe we can factor out something from the numerator.Alternatively, perhaps write the entire expression as:v = [3a + 2c + sqrt{(3a - 2c)^2 + 8ac}]/4Wait, since earlier we saw that 9a² -4ac +4c² = (3a -2c)^2 +8ac. So:sqrt{(3a -2c)^2 +8ac}But not sure if that's helpful. Maybe not. Alternatively, leave it as is.Alternatively, if we can write it as:sqrt{(3a)^2 + (2c)^2 -4ac}But that's the same as original.So, probably the answer is best left in the form we derived. Therefore, the own speed of the motorboat is [3a + 2c + sqrt(9a² -4ac +4c²)] divided by 4.Thus, boxed:boxed{dfrac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}}Final AnswerThe own speed of the motorboat is boxed{dfrac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}} km/h.</think>The motorboat traveled (a) km across the lake and then half of this distance up the river. The entire trip lasted 1 hour. We need to find the own speed of the motorboat if the speed of the river current is (c) km/h.1. Let (v) be the own speed of the motorboat.2. The time taken to travel across the lake is (frac{a}{v}) hours.3. The effective speed of the motorboat upstream is (v - c) km/h, so the time taken to travel (frac{a}{2}) km up the river is (frac{a/2}{v - c}) hours.4. The total time for the trip is given by the equation: [ frac{a}{v} + frac{a}{2(v - c)} = 1 ]5. To solve for (v), we first combine the terms: [ frac{a}{v} + frac{a}{2(v - c)} = 1 ]6. Multiply through by the common denominator (2v(v - c)): [ 2a(v - c) + av = 2v(v - c) ]7. Simplify and combine like terms: [ 3av - 2ac = 2v^2 - 2cv ]8. Rearrange to form a quadratic equation: [ 2v^2 - (3a + 2c)v + 2ac = 0 ]9. Solve the quadratic equation using the quadratic formula (v = frac{-b pm sqrt{b^2 - 4ac}}{2a}): [ v = frac{(3a + 2c) pm sqrt{(3a + 2c)^2 - 16ac}}{4} ]10. Simplify the discriminant: [ sqrt{9a^2 - 4ac + 4c^2} ]11. The valid solution (since speed must be positive and greater than the current) is: [ v = frac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4} ]The own speed of the motorboat is boxed{dfrac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}} km/h.,<think>Okay, so I need to solve this problem about a motorboat traveling across a lake and then up a river. The whole trip took one hour, and I have to find the boat's own speed. The river's current speed is given as c km/h. Let me try to break this down step by step.First, let me visualize the scenario. The motorboat goes across a lake for a distance of 'a' kilometers. Then it turns around and travels half of that distance, which would be a/2 km, up a river that's flowing into the lake. The entire trip, both going across the lake and up the river, took exactly one hour. I need to find the boat's own speed, which I'll denote as 'v' km/h. The river current's speed is c km/h. Since the boat is moving across the lake, which I assume is still water (no current), its speed there would just be its own speed, v. But when it goes up the river, it's going against the current, so its effective speed would be reduced by the current's speed. Therefore, the upstream speed would be v - c km/h.The problem involves distances, speeds, and time, so I should use the formula: time = distance / speed. The total time for the trip is the sum of the time taken to cross the lake and the time taken to go up the river. Since the total time is 1 hour, I can set up an equation where the sum of these two times equals 1.Let me write down the given information:1. Distance across the lake: a km2. Distance up the river: (a/2) km3. River current speed: c km/h4. Total trip time: 1 hour5. Boat's own speed: v km/h (this is what we need to find)So, time taken to cross the lake is distance divided by speed, which is a / v.Time taken to go up the river is (a/2) divided by (v - c), because going upstream against the current.Therefore, the total time is:a / v + (a/2) / (v - c) = 1This equation should let me solve for v. Let me write that more clearly:(a / v) + (a / (2(v - c))) = 1Now, I need to solve this equation for v. Let's work through the algebra step by step.First, let's factor out the common term 'a' from both terms:a [ 1 / v + 1 / (2(v - c)) ] = 1Then, divide both sides by 'a':1 / v + 1 / (2(v - c)) = 1 / aNow, the equation is:1/v + 1/(2(v - c)) = 1/aTo solve for v, I need to combine these two fractions on the left side. The common denominator would be 2v(v - c). Let's rewrite each term with that denominator:[ 2(v - c) + v ] / [2v(v - c)] = 1/aWait, let me check that. To combine 1/v and 1/(2(v - c)), we need a common denominator. The denominators are v and 2(v - c), so the common denominator is 2v(v - c). Therefore:1/v = 2(v - c) / [2v(v - c)]and1/(2(v - c)) = v / [2v(v - c)]Adding them together:[2(v - c) + v] / [2v(v - c)] = 1/aSimplify the numerator:2(v - c) + v = 2v - 2c + v = 3v - 2cSo the equation becomes:(3v - 2c) / [2v(v - c)] = 1/aNow, cross-multiplying:a(3v - 2c) = 2v(v - c)Expanding both sides:3a v - 2a c = 2v² - 2c vBring all terms to one side to form a quadratic equation:3a v - 2a c - 2v² + 2c v = 0Let me rearrange terms:-2v² + (3a v + 2c v) - 2a c = 0Factor out the v from the middle terms:-2v² + v(3a + 2c) - 2a c = 0It's usually better to have the quadratic term positive, so let's multiply both sides by -1:2v² - v(3a + 2c) + 2a c = 0Now, the quadratic equation is:2v² - (3a + 2c)v + 2a c = 0Now, we can apply the quadratic formula to solve for v. For a quadratic equation of the form ax² + bx + c = 0, the solutions are x = [-b ± sqrt(b² - 4ac)] / (2a). Here, in our equation, the coefficients are:a = 2b = -(3a + 2c) (Wait, here a is a coefficient in the quadratic equation, but the problem uses a as a distance. Hmm, maybe I should have used different letters to avoid confusion. Let's check: in the quadratic equation, the coefficients are:Quadratic term: 2 (coefficient of v²)Linear term: -(3a + 2c) (coefficient of v)Constant term: 2a cTherefore, using the quadratic formula:v = [ (3a + 2c) ± sqrt( (3a + 2c)² - 4*2*2a c ) ] / (2*2)Simplify inside the square root:(3a + 2c)² - 16a cLet me compute that:(3a + 2c)^2 = 9a² + 12a c + 4c²Subtract 16a c:9a² + 12a c + 4c² - 16a c = 9a² - 4a c + 4c²Therefore, the discriminant becomes 9a² - 4a c + 4c²So, plugging back into the quadratic formula:v = [3a + 2c ± sqrt(9a² - 4a c + 4c²)] / 4Now, we have two possible solutions. But since speed cannot be negative, we need to check which of the solutions makes sense in the context.Let me compute sqrt(9a² -4a c +4c²). Let's see if this can be simplified. Let's check if the discriminant is a perfect square.Looking at 9a² -4a c +4c². Let's see if this factors.Let me consider this as a quadratic in a:9a² -4c a +4c²Alternatively, perhaps factor it:Is there a way to write 9a² -4a c +4c² as (3a - something)^2?(3a - k c)^2 = 9a² -6k a c +k² c²Comparing with 9a² -4a c +4c²:-6k a c = -4a c => 6k = 4 => k = 4/6 = 2/3Then, k² c² = (4/9)c², but in our original expression, the constant term is 4c². So that doesn't match. So maybe it's not a perfect square.Alternatively, perhaps factor as (something a + something c)^2. But it might not be necessary. Let's compute the square root as is.sqrt(9a² -4a c +4c²)Alternatively, maybe completing the square?9a² -4a c +4c² = 9a² -4a c + (4c²)Let me factor out the coefficient of a²:= 9(a² - (4c)/(9) a) +4c²Then complete the square inside the brackets:a² - (4c/9)a = a² - (4c/9)a + ( (4c)/(2*9) )² - ( (4c)/(2*9) )²= [a - (2c/9)]² - ( (2c/9)^2 )Therefore,9(a² - (4c/9)a) +4c² = 9[ (a - 2c/9)^2 - (4c²)/(81) ] +4c²= 9(a - 2c/9)^2 - (9*4c²)/81 +4c²= 9(a - 2c/9)^2 - (4c²)/9 +4c²= 9(a - 2c/9)^2 + ( -4c² +36c² ) /9= 9(a - 2c/9)^2 +32c²/9Hmm, that seems more complicated. Maybe it's not necessary. Let's just leave it as sqrt(9a² -4a c +4c²).Therefore, the solutions are:v = [3a + 2c ± sqrt(9a² -4a c +4c²)] /4Now, we need to consider both solutions. Let me compute both possibilities.First, the positive root:v = [3a + 2c + sqrt(9a² -4a c +4c²)] /4Second, the negative root:v = [3a + 2c - sqrt(9a² -4a c +4c²)] /4We need to check which of these solutions is physically meaningful. Since speed must be positive, and greater than the current's speed (since when going upstream, the boat's speed relative to the water must be greater than the current's speed; otherwise, it couldn't make headway). Therefore, v must be greater than c.So, let's check both solutions.First, let's compute the positive root:v1 = [3a + 2c + sqrt(9a² -4a c +4c²)] /4Second, the negative root:v2 = [3a + 2c - sqrt(9a² -4a c +4c²)] /4We need to verify whether these are positive and whether v > c.But perhaps we can analyze the discriminant expression first.sqrt(9a² -4a c +4c²) = sqrt(9a² +4c² -4a c)Note that 9a² +4c² -4a c can be written as 9a² -4a c +4c².Alternatively, perhaps factor 9a² -4a c +4c² as follows:Wait, 9a² -4a c +4c². Let's check if this is equal to (3a - 2c/3)^2 + (some positive term). Let me compute:(3a - (2c)/3)^2 = 9a² - 2*3a*(2c)/3 + (2c/3)^2 = 9a² -4a c +4c²/9So, 9a² -4a c +4c²/9. That's less than 9a² -4a c +4c². So, the discriminant is bigger than this squared term. Therefore, sqrt(9a² -4a c +4c²) is greater than 3a - (2c)/3. Not sure if that helps.Alternatively, maybe compare 9a² -4a c +4c² with (3a - 2c)^2.Compute (3a - 2c)^2 = 9a² -12a c +4c².Compare with our discriminant: 9a² -4a c +4c².So, the discriminant is (3a - 2c)^2 +8a c.Wait, because 9a² -4a c +4c² = (9a² -12a c +4c²) +8a c = (3a -2c)^2 +8a c.Therefore, sqrt( (3a -2c)^2 +8a c ). Since 8a c is positive (assuming a and c are positive, which they are, as distances and speeds), then sqrt( (3a -2c)^2 +8a c ) is greater than |3a -2c|. But since all terms are positive, this square root is a positive real number.Therefore, in any case, sqrt(9a² -4a c +4c²) is a real number, so both solutions are real.Now, check if both solutions are positive.First, consider v1:[3a + 2c + sqrt(9a² -4a c +4c²)] /4All terms in the numerator are positive (since a and c are positive), so v1 is definitely positive.Now, v2:[3a + 2c - sqrt(9a² -4a c +4c²)] /4We need to check if the numerator is positive. Let's see:3a + 2c - sqrt(9a² -4a c +4c²) > 0?Let me denote S = sqrt(9a² -4a c +4c²). Then, we need 3a + 2c - S > 0.Let me square both sides to see if this holds, but we have to be careful with inequalities when squaring. Alternatively, let's compute (3a + 2c)^2 and compare it to S^2.(3a + 2c)^2 = 9a² +12a c +4c²Compare to S^2 = 9a² -4a c +4c²So, (3a + 2c)^2 - S^2 = (9a² +12a c +4c²) - (9a² -4a c +4c²) = 16a cSince 16a c is positive, (3a + 2c)^2 > S^2. Therefore, 3a + 2c > S, since both sides are positive. Therefore, 3a + 2c - S >0. Therefore, v2 is also positive.So both solutions are positive. Now, we need to check which one is physically meaningful. Remember that when going upstream, the boat's speed relative to the ground is v - c, which must be positive, so v > c. Therefore, the boat's own speed must be greater than the current speed.Therefore, we need to check which of the solutions v1 and v2 is greater than c.Let me compute both v1 and v2.First, v1 = [3a + 2c + sqrt(9a² -4a c +4c²)] /4Second, v2 = [3a + 2c - sqrt(9a² -4a c +4c²)] /4Let me compute v1 - c and v2 - c to see which is positive.v1 - c = [3a + 2c + sqrt(9a² -4a c +4c²)] /4 - c= [3a + 2c -4c + sqrt(9a² -4a c +4c²)] /4= [3a -2c + sqrt(9a² -4a c +4c²)] /4Similarly, v2 - c = [3a + 2c - sqrt(9a² -4a c +4c²)] /4 -c= [3a + 2c -4c - sqrt(9a² -4a c +4c²)] /4= [3a -2c - sqrt(9a² -4a c +4c²)] /4Now, for v1 - c: numerator is 3a -2c + sqrt(9a² -4a c +4c²). Since sqrt(...) is positive, and 3a -2c could be positive or negative depending on a and c, but given that a and c are positive, if a is large enough, 3a -2c is positive, but maybe not always. However, let's see:But in the problem statement, the boat travels a km across the lake and half of that up the river. So, if the boat can go up the river, then the time taken to go up the river must be positive and finite. So, the denominator in the upstream time, which is (v - c), must be positive. Therefore, v > c is required. Therefore, both solutions v1 and v2 must satisfy v > c? Wait, but maybe not. Let's check.Wait, even if one solution gives v > c and the other gives v <= c, the latter would be invalid. So, we need to check if both solutions satisfy v > c or only one does.Let's take an example. Let me plug in some numbers for a and c to see.Suppose a = 2 km, c = 1 km/h.Then compute v1 and v2.First, compute sqrt(9a² -4a c +4c²):sqrt(9*(4) -4*2*1 +4*1) = sqrt(36 -8 +4) = sqrt(32) ≈5.656Then, v1 = [3*2 +2*1 +5.656]/4 = [6 +2 +5.656]/4 ≈13.656/4 ≈3.414 km/hv2 = [6 +2 -5.656]/4 ≈(8 -5.656)/4 ≈2.344/4 ≈0.586 km/hNow, check if v1 and v2 are greater than c=1 km/h.v1 ≈3.414 >1, so valid.v2 ≈0.586 <1, which would mean the boat can't go upstream, so invalid.Therefore, in this case, only v1 is valid.Similarly, take another example. Let a =1 km, c = 3 km/h.Compute sqrt(9*1 -4*1*3 +4*9) = sqrt(9 -12 +36) = sqrt(33) ≈5.744v1 = [3*1 +2*3 +5.744]/4 = [3 +6 +5.744]/4 ≈14.744/4 ≈3.686 km/hv2 = [3 +6 -5.744]/4 ≈(9 -5.744)/4≈3.256/4≈0.814 km/hAgain, v1 ≈3.686 >3, so valid.v2≈0.814 <3, invalid.Another example: a=1, c=1.sqrt(9*1 -4*1*1 +4*1)=sqrt(9-4+4)=sqrt(9)=3v1=(3+2+3)/4=8/4=2>1, valid.v2=(3+2-3)/4=2/4=0.5<1, invalid.Hence, in these examples, only v1 is valid. Therefore, likely the correct solution is v1.Therefore, the boat's own speed is [3a + 2c + sqrt(9a² -4a c +4c²)] /4But let me check algebraically whether v1 > c and v2 < c.We can consider the expression for v1 - c:v1 - c = [3a -2c + sqrt(9a² -4a c +4c²)] /4Since sqrt(9a² -4a c +4c²) = sqrt( (3a)^2 -4a c + (2c)^2 )Wait, perhaps factor this expression as follows:9a² -4a c +4c². Let me check discriminant:It's a quadratic in a: 9a² -4c a +4c². The discriminant would be ( -4c )² -4*9*4c² = 16c² - 144c² = -128c² <0. Wait, but this is a quadratic in a, but since we are taking sqrt(9a² -4a c +4c²), which is a real number because 9a² -4a c +4c² is always positive?Wait, for real numbers, 9a² -4a c +4c². Let me check if this is always positive.Since 9a² -4a c +4c² can be considered as 9a² -4a c +4c² = 9a² -4a c + (4/9)c² + (32/9)c² = ?Wait, maybe not. Alternatively, check the discriminant as a quadratic in a. For 9a² -4a c +4c², the discriminant is (-4c)^2 -4*9*4c² = 16c² - 144c² = -128c² <0. Therefore, as a quadratic in a, this expression has no real roots, which means it is always positive (since the coefficient of a² is positive). Therefore, sqrt(9a² -4a c +4c²) is always real, and the expression under the square root is always positive.Therefore, going back to v1 - c:[3a -2c + sqrt(9a² -4a c +4c²)] /4Since sqrt(9a² -4a c +4c²) > sqrt(9a² -12a c +4c²) = |3a - 2c|Because 9a² -4a c +4c² = 9a² -12a c +4c² +8a c = (3a -2c)^2 +8a cTherefore, sqrt( (3a -2c)^2 +8a c ) > |3a -2c|Therefore, if 3a -2c is positive, then sqrt(...) >3a -2c, so 3a -2c + sqrt(...) >3a -2c +3a -2c=6a -4c. But since sqrt(...) is more than |3a -2c|, so if 3a -2c is positive, then sqrt(...) >3a -2c, so numerator becomes 3a -2c + (something bigger than 3a -2c) = something bigger than 6a -4c. Divided by 4, which may or may not be positive. Wait, maybe not the right approach.Alternatively, since sqrt(9a² -4a c +4c²) > |3a -2c|, then:If 3a -2c is positive, then sqrt(...) >3a -2c, so 3a -2c + sqrt(...) >3a -2c +3a -2c=6a -4c. If 6a -4c >0, which is 3a >2c. If a and c are such that 3a >2c, then yes.But even if 3a -2c is negative, sqrt(...) > -(3a -2c), so sqrt(...) +3a -2c >0. Let's check:If 3a -2c is negative, let 3a -2c = -k, where k>0.Then sqrt(...) >k, so sqrt(...) + (-k) >0. Therefore, numerator is positive. Therefore, in any case, 3a -2c + sqrt(...) is positive. Therefore, v1 -c is positive. Therefore, v1 >c.For v2 -c:[3a -2c - sqrt(9a² -4a c +4c²)] /4Again, sqrt(...) > |3a -2c|If 3a -2c is positive, sqrt(...) >3a -2c, so 3a -2c - sqrt(...) <0.If 3a -2c is negative, sqrt(...) > -(3a -2c), so 3a -2c - sqrt(...) = -k - sqrt(...) < -k -k = -2k <0.Therefore, in both cases, v2 -c is negative, so v2 <c. Therefore, v2 is invalid, as the boat cannot go upstream. Hence, only v1 is the valid solution.Therefore, the boat's own speed is:v = [3a + 2c + sqrt(9a² -4a c +4c²)] /4But let me see if this expression can be simplified further. Let's check the sqrt(9a² -4a c +4c²) term again.Earlier, we saw that 9a² -4a c +4c² = (3a)^2 -4a c + (2c)^2. Hmm, but maybe factor it differently. Wait, 9a² -4a c +4c².Alternatively, factor it as follows:Let me check if 9a² -4a c +4c² can be expressed as (3a - kc)^2 + m c².Compute (3a - kc)^2 = 9a² -6k a c +k² c²Comparing to 9a² -4a c +4c²:-6k a c = -4a c => k=4/6=2/3k² c²= (4/9)c²Therefore, (3a - (2/3)c)^2 +4c² - (4/9)c²= (3a - (2/3)c)^2 + (32/9)c²But I don't know if that helps. Alternatively, since it's not a perfect square, maybe leave it as is.Therefore, the final answer is:v = [3a + 2c + sqrt(9a² -4a c +4c²)] /4Alternatively, factor out a 9 from the square root:sqrt(9a² -4a c +4c²) = sqrt(9a² +4c² -4a c) = sqrt(9a² -4a c +4c²)Alternatively, maybe factor out 1/4 from the expression:But perhaps not necessary.Alternatively, note that 9a² -4a c +4c² can be written as (3a)^2 - 4a c + (2c)^2. However, it doesn't resemble a standard algebraic identity. Therefore, perhaps this is as simplified as it gets.Alternatively, maybe we can write the solution in another form. Let me try:Multiply numerator and denominator by something? Let's see.Alternatively, rationalize if possible. But I don't think so.Alternatively, note that the expression under the square root is 9a² -4a c +4c². Maybe think of it as a quadratic in c:4c² -4a c +9a² =4c² -4a c +9a². The discriminant here would be ( -4a )² -4*4*9a² =16a² -144a² = -128a² <0, so no real roots, so it's always positive as a quadratic in c.But that doesn't help.Alternatively, perhaps use substitution. Let me set t = c/a, which is the ratio of current speed to distance. Then, the expression becomes:sqrt(9a² -4a c +4c²) =a*sqrt(9 -4(c/a) +4(c/a)^2)=a*sqrt(9 -4t +4t²)Then, the speed v is:[3a + 2c +a*sqrt(9 -4t +4t²)] /4 =a[3 + 2t + sqrt(9 -4t +4t²)] /4But this might not necessarily be simpler. It depends on the context. Since the problem doesn't specify to express in terms of t, perhaps it's better to leave it in terms of a and c.Therefore, the final answer is:v = [3a + 2c + sqrt(9a² -4a c +4c²)] /4Alternatively, factor out a 1/4:v = (3a + 2c)/4 + sqrt(9a² -4a c +4c²)/4But this doesn't particularly simplify things.Alternatively, note that 9a² -4a c +4c² can be rewritten as 4c² -4a c +9a², which is a quadratic in c. But again, not helpful.Alternatively, to check if this expression can be written as (something)^2:But as we saw earlier, it's not a perfect square, so probably not.Therefore, the answer is as simplified as it can be. So, boxed, it would be:boxed{dfrac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}}Final AnswerThe own speed of the motorboat is boxed{dfrac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}}.</think>The motorboat traveled (a) km across the lake and then half of this distance up the river. The entire trip lasted 1 hour. We need to find the own speed of the motorboat, (v), given the river current speed is (c) km/h.1. Time taken to cross the lake: (frac{a}{v})2. Time taken to go up the river: (frac{a/2}{v - c})The total time for the trip is given by:[frac{a}{v} + frac{a}{2(v - c)} = 1]Factoring out (a):[a left( frac{1}{v} + frac{1}{2(v - c)} right) = 1]Dividing both sides by (a):[frac{1}{v} + frac{1}{2(v - c)} = frac{1}{a}]Finding a common denominator for the left side:[frac{2(v - c) + v}{2v(v - c)} = frac{1}{a}]Simplifying the numerator:[frac{3v - 2c}{2v(v - c)} = frac{1}{a}]Cross-multiplying:[a(3v - 2c) = 2v(v - c)]Expanding and rearranging terms:[3av - 2ac = 2v^2 - 2cv][2v^2 - (3a + 2c)v + 2ac = 0]Solving the quadratic equation using the quadratic formula (v = frac{-b pm sqrt{b^2 - 4ac}}{2a}) where (a = 2), (b = -(3a + 2c)), and (c = 2ac):[v = frac{(3a + 2c) pm sqrt{(3a + 2c)^2 - 16ac}}{4}]Simplifying the discriminant:[(3a + 2c)^2 - 16ac = 9a^2 + 12ac + 4c^2 - 16ac = 9a^2 - 4ac + 4c^2]Thus, the solutions are:[v = frac{3a + 2c pm sqrt{9a^2 - 4ac + 4c^2}}{4}]Since the speed must be positive and greater than the current speed (c), the valid solution is:[v = frac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}]The own speed of the motorboat is:[boxed{dfrac{3a + 2c + sqrt{9a^2 - 4ac + 4c^2}}{4}}]]

🔑:Solution. If x is the speed of the motorboat (km/h), then frac{a}{x} is the time it takes to travel on the lake, and frac{a}{2(x-c)} is the time it takes to travel on the river.From this,frac{a}{x}+frac{a}{2(x-c)}=1 Rightarrow 2 x^{2}-x(2 c+3 a)+2 a c=0 Rightarrow x=frac{3 a+2 c pm sqrt{9 a^{2}-4 a c+4 c^{2}}}{4}.One root is discarded because x>c.Answer: frac{3 a+2 c+sqrt{9 a^{2}-4 a c+4 c^{2}}}{4} text{km}/ h.

🧠:[<think>Okay, so I need to figure out the smallest positive integer N where the sum of the first N numbers in this list is an odd integer greater than 2021. Let me try to break this down step by step.First, let's understand how the list is generated. The first two numbers are given: 3 and 4. Then each subsequent number is calculated by taking the previous number, adding 1, and then dividing by the number before that. The example given is that the third number is (4 + 1)/3 = 5/3. So, the recursion formula is c = (b + 1)/a, where a, b, c are three consecutive numbers.Let me write out the first few terms to see if there's a pattern or cycle. Starting with:Term 1: 3Term 2: 4Term 3: (4 + 1)/3 = 5/3 ≈ 1.666...Term 4: (5/3 + 1)/4 = (8/3)/4 = (8/3)*(1/4) = 2/3 ≈ 0.666...Term 5: (2/3 + 1)/(5/3) = (5/3)/(5/3) = 1Term 6: (1 + 1)/(2/3) = 2/(2/3) = 3Term 7: (3 + 1)/1 = 4/1 = 4Term 8: (4 + 1)/3 = 5/3Wait a second, term 7 is 4 and term 8 is 5/3, which is the same as term 3. So, starting from term 3, the sequence seems to repeat every 5 terms: 5/3, 2/3, 1, 3, 4, then back to 5/3, 2/3, etc. Let me verify this.Term 9: (5/3 + 1)/4 = same as term 4: (8/3)/4 = 2/3. Yep, that's term 4 again.Term 10: (2/3 + 1)/(5/3) = same as term 5: 1.Term 11: (1 + 1)/(2/3) = same as term 6: 3.Term 12: (3 + 1)/1 = same as term 7: 4.Term 13: (4 + 1)/3 = same as term 8: 5/3.So, yes, from term 3 onwards, there's a cycle of 5 terms: 5/3, 2/3, 1, 3, 4, and then repeats. So, the sequence is:1: 32: 43: 5/34: 2/35: 16: 37: 48: 5/39: 2/310: 111: 312: 4...and so on.Therefore, the cycle length is 5 starting from term 3. So, terms 3-7: 5/3, 2/3, 1, 3, 4; terms 8-12: same as terms 3-7, etc.So, to find the sum of the first N terms, we need to consider the first two terms (3 and 4), and then starting from term 3, every 5 terms repeat. So, the sum can be broken down as:Sum = 3 + 4 + sum of cycles.Each cycle (5 terms) has the sum: 5/3 + 2/3 + 1 + 3 + 4.Calculating that:5/3 + 2/3 = 7/37/3 + 1 = 10/310/3 + 3 = 19/319/3 + 4 = 19/3 + 12/3 = 31/3 ≈ 10.333...Wait, 31/3 is approximately 10.333... So each cycle of 5 terms adds 31/3 to the sum. But the problem requires the sum to be an odd integer greater than 2021. Hmm. So, we need to find N such that the total sum is an odd integer >2021.But first, let me check if the cycle sum is indeed 31/3 every 5 terms. Let me compute the sum from term 3 to term 7:Term3: 5/3Term4: 2/3Term5: 1Term6: 3Term7: 4Sum: 5/3 + 2/3 = 7/3; 7/3 + 1 = 10/3; 10/3 + 3 = 19/3; 19/3 + 4 = 31/3. Yes, that's correct. Each cycle of 5 terms adds 31/3. Then, starting from term 8, the cycle repeats.So, the total sum for N terms is 3 + 4 + k*(31/3) + partial cycle sum, where k is the number of full cycles after the first two terms, and the partial cycle sum is the sum of the first m terms of the cycle, if N is not a multiple of 5 after the first two terms.But since we need the total sum to be an integer, and 31/3 is not an integer, this complicates things. However, perhaps over multiple cycles, the fractions might cancel out? Let's check.Each cycle contributes 31/3, so two cycles would contribute 62/3, three cycles 93/3 = 31, which is an integer. Wait, 31/3 * 3 = 31. So every three cycles, the fractional parts add up to an integer. Therefore, if we have k cycles, where k is a multiple of 3, then the total sum contributed by the cycles is an integer.But the total sum is 3 + 4 + cycles. 3 + 4 = 7. So, 7 + integer (from cycles) would be an integer. So, to get the total sum as an integer, we need the cycles part to be an integer, which requires that k is a multiple of 3. Because each cycle contributes 31/3, so 3 cycles contribute 31, 6 cycles contribute 62, etc.Therefore, the total sum can be written as 7 + 31/3 * k + partial sum. If k is a multiple of 3, then 31/3 * k is integer. For example, if k = 3, then 31/3 *3 = 31. So total sum is 7 +31 = 38. Then if k=6, sum is 7 + 62 = 69, etc.But wait, if k is not a multiple of 3, then the sum would have a fractional part, so the total sum would not be an integer. Therefore, to have the sum be an integer, the number of cycles k must be a multiple of 3. Because each cycle contributes 31/3, which is 10 + 1/3. So three cycles would contribute 31, which is integer.Therefore, the total sum is 7 + 31 * (k / 3). Wait, no. If k is the number of cycles, each cycle is 5 terms. So total terms contributed by k cycles is 5k. So total N is 2 + 5k + m, where m is the number of terms in the partial cycle (0 ≤ m <5). But the sum would be 7 + 31/3 * k + sum of first m terms of the cycle.But if we need the total sum to be integer, then 31/3 *k must be integer, which implies that k must be a multiple of 3, as 31 is prime. So k = 3t for some integer t ≥0. Then, 31/3 *3t =31t. Therefore, total sum is 7 +31t + sum of first m terms.But sum of first m terms of the cycle (the cycle is 5 terms: 5/3, 2/3, 1, 3, 4). Let's note the cumulative sum of the cycle:Term3: 5/3 ≈1.666...Term3-4: 5/3 +2/3 =7/3≈2.333...Term3-5: 7/3 +1=10/3≈3.333...Term3-6:10/3 +3=19/3≈6.333...Term3-7:19/3 +4=31/3≈10.333...So, the cumulative sums for the cycle:m=1: 5/3m=2:7/3m=3:10/3m=4:19/3m=5:31/3Therefore, if we have k=3t cycles (i.e., 5*3t=15t terms), then the sum from cycles is 31t. Then, adding the initial 7, the total sum is 7 +31t. Then, if we have a partial cycle of m terms, the sum would be 7 +31t + sum(m). So to have the total sum integer, 31t must be integer, which it is if t is integer, and sum(m) must be such that sum(m) is a fraction that when added to 7 +31t gives integer. But sum(m) is either 5/3,7/3,10/3,19/3,31/3. So, unless sum(m) is also a multiple of 1/3 with denominator dividing 3, but adding 7 (which is 21/3) +31t (which is 93t/3) + sum(m) (which is 5/3,7/3, etc.), so total would be (21 +93t + sum(m)*3)/3. For this to be integer, the numerator must be divisible by 3.Wait, 21 +93t is divisible by 3, since 21 is 7*3, 93t is 31t*3. Then sum(m)*3 must also be divisible by 3. Sum(m)*3 is either 5,7,10,19,31. These numbers: 5,7,10,19,31. None of these are divisible by 3. For example, 5: 5 mod3=2, 7 mod3=1, 10 mod3=1, 19 mod3=1, 31 mod3=1. Therefore, adding 21 +93t + sum(m)*3 would be 21 +93t +5, which is 26 +93t. 26 mod3=2, 93t mod3=0, so total mod3=2. Not divisible by3. Similarly for other sum(m)*3:If sum(m)*3=7: 21+93t+7=28+93t. 28 mod3=1, so total mod3=1. Not divisible by3.sum(m)*3=10:21+93t+10=31+93t. 31 mod3=1. Total mod3=1.sum(m)*3=19:21+93t+19=40+93t. 40 mod3=1. Total mod3=1.sum(m)*3=31:21+93t+31=52+93t. 52 mod3=1. Total mod3=1.So regardless of the partial sum, the total numerator is not divisible by3, which would mean that the total sum is not an integer unless m=0. Because if m=0, there is no partial sum, then total sum is 7 +31t, which is an integer. So if m=0, i.e., N=2 +5k, where k=3t, so N=2 +15t. Then, the sum is 7 +31t. So in that case, the sum is 7 +31t, which is an integer.But if m>0, even if k=3t, the total sum is 7 +31t + sum(m), which is not an integer. Therefore, the only way the sum is integer is if there are no partial cycles, i.e., N=2 +15t, and the sum is 7 +31t. So, to have the sum be an odd integer greater than 2021, we need 7 +31t >2021, and 7 +31t is odd.So let's solve 7 +31t >2021Subtract 7: 31t >2014Divide by31: t>2014/31Calculate 2014 ÷31. 31*64= 2014- let's check: 31*60=1860, 31*64=1860+31*4=1860+124=1984. Wait, 31*64=1984. Then 2014-1984=30. So 2014=31*64 +30. Therefore, 2014/31=64 +30/31≈64.967. So t must be greater than 64.967, so t≥65.Therefore, the minimal t is 65. Then, the sum is 7 +31*65. Let's compute 31*65: 30*65=1950, 1*65=65, so 1950+65=2015. Then 2015 +7=2022. So sum=2022. Check if 2022 is odd? 2022 is even, since it ends with 2. So that's a problem. We need the sum to be odd. So 7 +31t must be odd. 7 is odd, 31t is odd if t is odd, even if t is even. Because 31 is odd, so odd*odd=odd, odd*even=even. So 7 +odd=even, 7 +even=odd.Therefore, to have 7 +31t be odd, 31t must be even, so t must be even. So t has to be even. Therefore, the minimal t is 65, but 65 is odd, so the next t is 66. Let's check t=66.31*66=2046. Then 2046 +7=2053. 2053 is odd, and 2053>2021. So that's a candidate. Then, the corresponding N is 2 +15t=2 +15*66=2 +990=992. So N=992.But wait, but maybe there's a smaller t where t is even and 31t +7>2021. Wait, t=65 gives sum=2022 which is even, so not acceptable. Next t=66 gives sum=2053 which is odd. Let me confirm whether there are any even t between 65 and 66? No, t must be integer. So t=66 is the next even integer after 65. So that's the minimal t. Therefore, N=992.But hold on. Is there a possibility that with partial cycles, even if the sum is fractional, but when added to the integer parts, it might result in an integer? But earlier reasoning suggested that unless m=0, the sum is not integer. Let me check an example.Suppose N=2 +15t +m, where m=1,2,3,4,5. Then, sum=7 +31t + sum(m). For sum(m)=5/3,7/3, etc. So 7 +31t +5/3= (7*3 +31t*3 +5)/3= (21 +93t +5)/3=(26 +93t)/3. Since 93t is divisible by3, 26 is 2 mod3, so total is (2 +0)/3=2/3, not integer. Similarly, if m=2, sum(m)=7/3. Then total sum=(21 +93t +7)/3=(28 +93t)/3. 28 mod3=1, so 1/3, not integer. Same for m=3: sum(m)=10/3. Total sum=(21 +93t +10)/3=(31 +93t)/3. 31 mod3=1, so 1/3, not integer. For m=4: sum(m)=19/3. Total sum=(21 +93t +19)/3=(40 +93t)/3. 40 mod3=1, so same. m=5: sum(m)=31/3. Total sum=(21 +93t +31)/3=(52 +93t)/3. 52 mod3=1. So none of these give integer sums. Therefore, only when m=0, meaning N=2 +15t, the sum is integer. Therefore, the only possible sums that are integers are 7 +31t where t is integer, and N=2 +15t.But for the sum to be odd, 7 +31t must be odd. Since 7 is odd, 31t must be even. Since 31 is odd, t must be even. Therefore, t=2s for some integer s. Therefore, the sum becomes 7 +31*(2s)=7 +62s. We need 7 +62s >2021. So 62s >2014. s>2014/62=32.483... So s≥33. Then, t=2s=66, which is the same as before. Therefore, the minimal t is 66, leading to N=2 +15*66=992. The sum is 7 +31*66=7 +2046=2053, which is odd and greater than 2021.But wait, is there a smaller N where, even if we include partial cycles, the total sum is an integer? The previous analysis says no, because any partial cycle introduces a fractional part that can't be canceled out. Therefore, the only way to have an integer sum is with N=2 +15t, leading to sum=7 +31t. So in order to have that sum be odd, t must be even, so t=66 is minimal. Hence, N=992.But let me verify this by checking smaller values. Suppose t=64 (even). Then sum=7 +31*64=7 +1984=1991. 1991 is odd, but 1991 <2021, so it's not greater. Then t=66 gives 2053>2021. So yes, that's the minimal t.Therefore, the answer is N=992.But just to be thorough, let's check N=992.Sum=3 +4 +31/3*(66 cycles). Wait, each cycle is 5 terms, so 66 cycles would be 330 terms. Then total N=2 +330=332? Wait, wait, wait, no. Wait, the number of cycles k is 3t, but earlier we had k=3t? Wait, no. Wait, let's clarify.Earlier, I thought that k cycles correspond to 5k terms, starting from term3. So N=2 +5k. But then, in order for the sum from the cycles to be integer, k must be multiple of3, i.e., k=3t. Therefore, N=2 +5*(3t)=2 +15t. So yes, N=2 +15t. Therefore, t=66 gives N=2 +15*66=2 +990=992.So the total sum is 7 +31*66=7 +2046=2053, which is indeed odd and greater than 2021. So this seems correct.But just to make sure, let me compute the sum for t=65 (which is odd), N=2 +15*65=2 +975=977. Sum=7 +31*65=7 +2015=2022, which is even, so not acceptable. Then t=66 gives 2053. Correct.Therefore, the answer is N=992.Final Answerboxed{992}</think>The problem requires finding the smallest positive integer ( N ) such that the sum of the first ( N ) numbers in a specific list is an odd integer greater than 2021. The list is generated with the first two numbers being 3 and 4, and each subsequent number ( c ) is calculated as ( c = frac{b + 1}{a} ) where ( a ) and ( b ) are the two preceding numbers.Key steps to solve the problem:1. Identify the Sequence Pattern: - The first few terms are calculated as follows: - Term 1: 3 - Term 2: 4 - Term 3: ( frac{5}{3} ) - Term 4: ( frac{2}{3} ) - Term 5: 1 - Term 6: 3 - Term 7: 4 - From term 3 onwards, the sequence repeats every 5 terms: ( frac{5}{3}, frac{2}{3}, 1, 3, 4 ).2. Sum of the Sequence: - The sum of the first two terms (3 and 4) is 7. - Each cycle of 5 terms starting from term 3 sums to ( frac{31}{3} ). - For the total sum to be an integer, the number of cycles must be a multiple of 3 (since ( frac{31}{3} times 3 = 31 ), an integer).3. Determine ( N ) for Odd Integer Sum: - The total sum after ( t ) cycles (each cycle being 3 sets of 5 terms) is ( 7 + 31t ). - We need ( 7 + 31t ) to be an odd integer greater than 2021. - Solving ( 7 + 31t > 2021 ): - ( 31t > 2014 ) - ( t > 64.967 ), so ( t geq 65 ) - ( t ) must be even to ensure ( 31t ) is even, making ( 7 + 31t ) odd. The smallest even ( t ) greater than 64.967 is 66.4. Calculate ( N ): - Each ( t ) corresponds to ( 15t ) terms (3 cycles of 5 terms each). - ( N = 2 + 15t ) - For ( t = 66 ): - ( N = 2 + 15 times 66 = 2 + 990 = 992 )Thus, the smallest positive integer ( N ) for which the sum of the first ( N ) numbers is an odd integer greater than 2021 is (boxed{992}).,<think>Okay, let's see. The problem is about a sequence where the first two numbers are 3 and 4. Each subsequent number is calculated by taking the previous number, adding 1, and then dividing by the number before that. So, for any three consecutive numbers a, b, c, we have c = (b + 1)/a. The example given is that the third number is (4 + 1)/3 = 5/3. The question is asking for the smallest positive integer N such that the sum of the first N numbers is an odd integer greater than 2021. First, I need to understand the sequence better. Let me write down the first few terms. The first term is 3, the second is 4. The third term is (4 + 1)/3 = 5/3. Then the fourth term would be (5/3 + 1)/4. Let's compute that: 5/3 + 1 is 8/3, divided by 4 is (8/3)/4 = 2/3. So the fourth term is 2/3. The fifth term would then be (2/3 + 1)/ (5/3). Let's compute that: 2/3 + 1 is 5/3, divided by 5/3 is 1. So the fifth term is 1. The sixth term would be (1 + 1)/(2/3) = 2 divided by (2/3) = 3. Then the seventh term is (3 + 1)/1 = 4. Wait a minute, that's interesting. The sixth term is 3 and the seventh term is 4. Then the eighth term would be (4 + 1)/3 = 5/3, which is the same as the third term. So it looks like the sequence is starting to repeat here. Let me check:Term 1: 3Term 2: 4Term 3: 5/3Term 4: 2/3Term 5: 1Term 6: 3Term 7: 4Term 8: 5/3Term 9: 2/3Term 10: 1Term 11: 3Term 12: 4Yes, so from term 6 onwards, the sequence repeats 3, 4, 5/3, 2/3, 1, and then again 3, 4, etc. So the period here is 5 terms: 3, 4, 5/3, 2/3, 1, and then repeats. Wait, but starting from term 6, which is 3, and term 7 is 4, which is the same as terms 1 and 2. So the cycle length is 5. Let's verify:Term 6: 3Term 7: 4Term 8: (4 + 1)/3 = 5/3Term 9: (5/3 + 1)/4 = (8/3)/4 = 2/3Term 10: (2/3 + 1)/(5/3) = (5/3)/(5/3) = 1Term 11: (1 + 1)/(2/3) = 2/(2/3) = 3Yes, so term 11 is 3 again, same as term 6. So the cycle is 5 terms: 3, 4, 5/3, 2/3, 1. Then it repeats. So the sequence has a period of 5 starting from term 6. So the initial terms 1 and 2 are 3 and 4, then term 3, 4, 5 follow the pattern, and term 6 starts the cycle again.Therefore, the sequence is:3, 4, 5/3, 2/3, 1, 3, 4, 5/3, 2/3, 1, 3, 4, 5/3, 2/3, 1, ...So the cycle is 3, 4, 5/3, 2/3, 1, repeating every 5 terms. Therefore, starting from term 1: the first two terms are 3 and 4, then the next three terms start forming the cycle? Wait, no. Let's count:Term 1: 3Term 2: 4Term 3: 5/3Term 4: 2/3Term 5: 1Term 6: 3Term 7: 4So starting from term 1, the first five terms are 3, 4, 5/3, 2/3, 1. Then term 6 is 3, which is the same as term 1. So actually, the cycle is of length 5 starting from term 1. Wait, that can't be, because term 6 is 3, term 1 is also 3. Term 7 is 4, term 2 is 4. So the cycle is the same as the first five terms. But term 1 is 3, term 6 is 3; term 2 is 4, term 7 is 4, etc. Therefore, the entire sequence is periodic with period 5 starting from term 1. So the cycle is 3, 4, 5/3, 2/3, 1, and repeats every 5 terms. Therefore, every group of five terms is the same: 3, 4, 5/3, 2/3, 1. Wait, but term 6 is 3, which is the same as term 1. Term 7 is 4, same as term 2. So the cycle is indeed 5 terms. Therefore, the sequence is periodic with period 5, starting from the first term. That's interesting. So the sequence cycles every five terms. Therefore, the sum of the sequence can be broken down into the sum of the first few terms plus the sum of complete cycles. Given that the sum of each cycle (5 terms) is 3 + 4 + 5/3 + 2/3 + 1. Let's compute that. 3 + 4 is 7. 5/3 + 2/3 is 7/3. Then 1 is 1. So total sum per cycle is 7 + 7/3 + 1. Let's compute 7 + 1 is 8, and 8 + 7/3 is 8 + 2 + 1/3 = 10 + 1/3. Wait, 7/3 is 2 and 1/3. So 8 + 2 + 1/3 = 10 + 1/3. So each cycle of 5 terms sums to 10 + 1/3. But we need the sum to be an odd integer greater than 2021. So 10 + 1/3 is 10.333... per cycle. Since each cycle is 5 terms, and each cycle adds approximately 10.333 to the sum. But wait, the first cycle (terms 1-5) sum to 10 + 1/3. The next cycle (terms 6-10) also sum to 10 + 1/3, and so on. So the total sum after N terms is equal to the number of complete cycles times 10 + 1/3 plus the sum of the remaining terms if N is not a multiple of 5. But the problem is that the sum must be an odd integer. However, each cycle contributes 10 + 1/3, which is not an integer. Therefore, the sum after a multiple of 5 terms is a multiple of 10 + 1/3, which is not an integer. Therefore, to have the sum be an integer, we need that the fractional parts from the cycles cancel out with the remaining terms. Alternatively, maybe the sum of the remaining terms after some number of cycles plus the cycles can result in an integer. Hmm. Let's think. Alternatively, perhaps we need to look for when the sum is an integer. Since the total sum after N terms is S_N. Since the sequence is periodic with period 5, we can model the sum as k cycles (each contributing 10 + 1/3) plus the sum of the first m terms of the cycle (where m is from 1 to 4). So, if N = 5k + m, where k is the number of complete cycles and m is the number of additional terms (0 <= m <5), then S_N = k*(10 + 1/3) + sum_{i=1}^m a_i, where a_i are the terms in the cycle: a1=3, a2=4, a3=5/3, a4=2/3, a5=1.But we need S_N to be an odd integer. So we need S_N = integer, odd, >2021.First, let's note that the sum S_N is a sum of fractions. For it to be an integer, the fractional parts must cancel out. Since each cycle adds 1/3 to the sum (because 10 + 1/3 per cycle), and each additional term might add a fractional part. So, for example, if we have k cycles, then the total sum from cycles is k*(10 + 1/3) = 10k + k/3. Then, if we add m terms from the cycle (each term can be integer or fraction), we need the total sum to be integer. Therefore, the total sum would be 10k + k/3 + sum_{i=1}^m a_i. For this to be integer, the fractional parts from k/3 and from sum_{i=1}^m a_i must add up to an integer. Let me compute the fractional contributions:First, for each cycle, the fractional part is 1/3. So k cycles contribute k*(1/3). Then, the additional m terms can contribute fractions depending on which terms they are. Let's compute the fractional parts for each possible m:If m=0: no additional terms. Then sum is 10k + k/3. For this to be integer, k/3 must be integer. So k must be divisible by 3. Then S_N = 10k + k/3 = (10 + 1/3)k = (31/3)k. If k is divisible by 3, say k=3t, then S_N = 31t, which is an integer. Then 31t must be odd and greater than 2021. Since 31 is odd, t must be odd. So possible if t is odd and 31t >2021. Let's compute t_min: 2021 /31 ≈ 65.1935. So t_min is 66, but t must be odd. The next odd integer after 65.1935 is 67. Therefore, 31*67=2077. Then N would be 5k + 0 =5*(3*67)=5*201=1005. So N=1005. But wait, need to check if this is the minimal N. Because maybe there are smaller N where m>0 but the sum is integer and odd.But let's first check m=0. For m=0, N=5*3t=15t. The sum is 31t. To have 31t >2021 and odd. t must be odd and >=67. So N=15*67=1005. But perhaps there are smaller N with m>0. Let's check for other m values.Let's compute for m=1,2,3,4. For each m, we need to find k and m such that S_N =10k +k/3 + sum_{i=1}^m a_i is integer, odd, and greater than 2021. Let's consider m from 1 to4:First, compute sum_{i=1}^m a_i for each m:m=1: sum=3m=2: 3+4=7m=3: 3+4+5/3= 7 +5/3=26/3 ≈8.666...m=4: 3+4+5/3+2/3=7 +7/3=28/3≈9.333...These sums have fractional parts:m=1: 3 (integer)m=2:7 (integer)m=3:26/3 (8 + 2/3)m=4:28/3 (9 + 1/3)Therefore, when adding these to k*(10 +1/3), which is 10k +k/3, the total fractional part would be k/3 + fractional part from sum_{i=1}^m a_i. For the total sum to be integer, the sum of fractional parts must be integer. Let's analyze each case.Case m=1: sum=3 (integer). So total sum is 10k +k/3 +3. For this to be integer, k/3 must be integer. So k divisible by 3. Let k=3t. Then sum=10*3t + t +3=30t +t +3=31t +3. This must be odd and >2021. So 31t +3 is odd: 31t is even, so t must be even. So t even, 31t +3 >2021. 31t >2018. t>2018/31≈65.096. So t >=66 (even). The smallest t is 66. Then N=5*3t +1=15*66 +1=990 +1=991. Sum is 31*66 +3=2046 +3=2049, which is odd and greater than 2021. So N=991.Case m=2: sum=7 (integer). Similarly, total sum is 10k +k/3 +7. For this to be integer, k/3 must be integer. So k=3t. Sum=31t +7. Which must be odd and >2021. 31t +7 odd implies 31t is even, so t even. 31t +7 >2021 => 31t >2014 => t>2014/31≈64.967. So t>=65, but t must be even, so t=66. Then sum=31*66 +7=2046 +7=2053, which is odd. N=5*3*66 +2=990 +2=992. So N=992.Case m=3: sum=26/3. So total sum is 10k +k/3 +26/3. Let's combine the fractional parts: k/3 +26/3=(k +26)/3. For the total sum to be integer, (k +26)/3 must be integer. Therefore, k +26 ≡0 mod3 =>k ≡1 mod3. So k=3t +1. Then total sum=10*(3t +1) + (3t +1)/3 +26/3=30t +10 + t +1/3 +26/3=31t +10 + (1 +26)/3=31t +10 +27/3=31t +10 +9=31t +19. So 31t +19 must be odd and >2021. 31t +19 is odd =>31t is even =>t even. 31t +19 >2021 =>31t >2002 =>t>2002/31≈64.58. So t>=65. Since t must be even, next even is 66. Then sum=31*66 +19=2046 +19=2065. N=5*(3*66 +1) +3=5*(199) +3=995 +3=998. Case m=4: sum=28/3. Total sum=10k +k/3 +28/3. Combine fractions: k/3 +28/3=(k +28)/3. For integer sum, (k +28)/3 must be integer =>k +28≡0 mod3 =>k≡2 mod3. So k=3t +2. Then total sum=10*(3t +2) + (3t +2)/3 +28/3=30t +20 + t +2/3 +28/3=31t +20 + (2 +28)/3=31t +20 +30/3=31t +20 +10=31t +30. So sum=31t +30. This must be odd and >2021. 31t +30 is odd =>31t must be odd =>t odd. 31t +30 >2021 =>31t >1991 =>t>1991/31≈64.225. So t>=65, which is odd. Then t=65. Sum=31*65 +30=2015 +30=2045. N=5*(3*65 +2) +4=5*(197) +4=985 +4=989.So for each m=0,1,2,3,4, we have possible N:- m=0: N=1005, sum=2077- m=1: N=991, sum=2049- m=2: N=992, sum=2053- m=3: N=998, sum=2065- m=4: N=989, sum=2045Wait, so the minimal N here is 989, which gives sum 2045. Is 2045 an odd integer greater than 2021? Yes, 2045 >2021, and it's odd. So N=989. But need to verify this calculation, because if N=989 is the minimal, then that's the answer.Wait, but need to check if for m=4, t=65. Let's confirm:k=3t +2=3*65 +2=197 +2=197? Wait, 3*65=195, +2=197. Then N=5k +4=5*197 +4=985 +4=989. Then sum=31t +30 where t=65. 31*65=2015, 2015 +30=2045. Correct. So 2045 is indeed an odd integer greater than 2021, and N=989.Compare with the other cases:For m=1, N=991 gives sum=2049.For m=4, N=989 gives sum=2045. Since 989 is smaller than 991, N=989 is better. Similarly, m=2 gives N=992, which is larger than 989. m=3 gives N=998, which is also larger. m=0 gives N=1005, which is the largest. So the minimal N is 989.But wait, is there a possibility that for some smaller k and m, the sum is an odd integer greater than 2021? Let me check the calculations again for m=4.Wait, when m=4, we have k=3t +2. Then sum=31t +30. For t=65, sum=2045. If t=63, which is odd and less than 65, let's see: t=63, sum=31*63 +30=1953 +30=1983, which is less than 2021. t=64 is even, but we need t odd. So the first t is 65. So yes, t=65 gives sum=2045. N=989. Similarly, for m=1, t=66 gives sum=2049. But 989 is smaller than 991. So N=989 is the answer.But let me check the cycle again. Wait, does the sequence really cycle every 5 terms starting from term1?Wait, term1:3, term2:4, term3:5/3, term4:2/3, term5:1, term6:3, term7:4, etc. So term6=3=term1, term7=4=term2, term8=5/3=term3, etc. Therefore, the cycle is indeed 5 terms starting from term1. Therefore, the sum of each cycle (5 terms) is 3 +4 +5/3 +2/3 +1=10 +1/3. So that part is correct.Then, for N=989, how many cycles is that? Let's compute N=989. 989 divided by5 is 197.8. So 197 complete cycles (each 5 terms) and 4 additional terms. 197*5=985, so 989-985=4, so m=4. Therefore, the sum is 197*(10 +1/3) + sum of first4 terms of the cycle. The first4 terms are3,4,5/3,2/3. Sum is3 +4 +5/3 +2/3=7 +7/3=28/3≈9.333... Then total sum=197*(31/3) +28/3=(197*31 +28)/3. Let's compute 197*31:197*30=5910, 197*1=197, so total 5910+197=6107. Then +28=6107 +28=6135. So sum=6135/3=2045. Which is correct. So 2045 is integer, odd, and greater than2021. Now, check if there's a smaller N. For example, let's see for m=4, if t=65 gives N=989. If there's a smaller t, but t must be >=65. Since t=65 is the minimal t for m=4. So N=989 is the minimal. But wait, another way: let's compute the sum progression. Since each cycle adds 10 +1/3, which is about10.333, and we need the total sum to cross 2021. Let's estimate how many cycles we need. 2021 /10.333≈195.6 cycles. So approximately 196 cycles would give around 196*10.333≈2025. So perhaps around there. But since we have cycles and partial terms, the exact calculation is needed. But we already found that with N=989 (197 cycles and 4 terms), the sum is 2045. The next smaller candidate would be 197-1=196 cycles plus some m terms. Let's check.Wait, for example, if we take 196 cycles (which is196*5=980 terms) and then add m terms. Let's see:If m=4, sum=196*(31/3) +28/3=(196*31 +28)/3=(6076 +28)/3=6104/3≈2034.666..., which is not integer. So not valid.If m=1: sum=196*(31/3)+3=(196*31 +9)/3=(6076 +9)/3=6085/3≈2028.333..., not integer.If m=2: sum=196*(31/3)+7=(6076 +21)/3=6097/3≈2032.333..., not integer.m=3: sum=196*(31/3)+26/3=(6076 +26)/3=6102/3=2034, which is integer. 2034 is even. So we need an odd integer. So 2034 is not acceptable. Therefore, with 196 cycles and m=3 terms, sum=2034 even. So not valid. So next is 196 cycles and m=4 terms: sum≈2034.666..., not integer. So no. Then 197 cycles (197*5=985 terms) plus m=4 terms (total N=989) gives sum=2045. Which is odd. So indeed, 2045 is the first odd integer greater than2021 achieved at N=989. But let's check if there is any N between, say, 980 and 989 where the sum is an odd integer greater than2021. For example, N=985 (197 cycles) sum=197*(10 +1/3)=197*(31/3)=6107/3≈2035.666..., which is not integer. So fractional. Then N=986: 197 cycles +1 term. Sum=6107/3 +3=(6107 +9)/3=6116/3≈2038.666..., not integer. N=987: +2 terms: sum=6107/3 +7=(6107 +21)/3=6128/3≈2042.666..., not integer. N=988: +3 terms: sum=6107/3 +26/3=6133/3≈2044.333..., not integer. N=989: +4 terms: sum=6107/3 +28/3=6135/3=2045, which is integer. So yes, 2045 is the first sum that is an odd integer greater than2021, achieved at N=989. Therefore, the answer is N=989.But let me just check with m=4 and t=65. Wait, earlier we said for m=4, k=3t +2. So k=3*65 +2=197. Then N=5k +4=5*197 +4=989. So that aligns. Therefore, the calculations are consistent.Therefore, the smallest positive integer N is 989.Final Answerboxed{989}</think>The sequence starts with 3 and 4, and each subsequent term is calculated by adding 1 to the previous term and dividing by the term before that. The sequence is periodic with a cycle of 5 terms: 3, 4, (frac{5}{3}), (frac{2}{3}), 1. The sum of one cycle (5 terms) is (10 + frac{1}{3}).To find the smallest (N) such that the sum of the first (N) terms is an odd integer greater than 2021, we consider the sum of complete cycles and additional terms. The sum of each cycle contributes (10 + frac{1}{3}), which is approximately 10.333. We need to determine the number of cycles (k) and additional terms (m) such that the total sum is an odd integer greater than 2021. After detailed calculations, we find that:- For (m = 4) (additional terms), (k = 3t + 2) where (t) must be odd.- The smallest (t) such that the sum (31t + 30) is greater than 2021 and odd is (t = 65).This results in (k = 3 times 65 + 2 = 197) cycles, leading to (N = 5 times 197 + 4 = 989). The sum of the first 989 terms is 2045, which is an odd integer greater than 2021.Thus, the smallest positive integer (N) is (boxed{989}).]

🔑:5. Using the given rules, the first 7 numbers in the list aremathbf{3}, mathbf{4}, frac{5}{3}, frac{(5 / 3)+1}{4}=frac{8 / 3}{4}=frac{2}{3}, frac{(2 / 3)+1}{5 / 3}=frac{5 / 3}{5 / 3}=mathbf{1}, frac{1+1}{2 / 3}=frac{2}{2 / 3}=mathbf{3}, frac{3+1}{1}=mathbf{4}Since the 6 th number equals the 1st number, the 7 th number equals the 2 nd number, and each number in the list depends on the previous two numbers, then the 8 th number equals the 3rd number, which will mean that the 9 th number will equal the 4 th number, and so on.In other words, the list will repeat every 5 numbers since each group of 5 numbers is produced in exactly the same way as the previous 5 numbers.The sum of the first 5 numbers is 3+4+frac{5}{3}+frac{2}{3}+1=8+frac{7}{3}=10 frac{1}{3}.Since 2021 div 10 frac{1}{3} approx 195.6, then 195 groups of 5 numbers will still have a sum less than 2021 .In fact, since 195 times 10 frac{1}{3}=1950+65=2015 and 195 times 5=975, then the sum of the first 975 numbers in the list is 2015 .We make a chart of the next terms and the sum to that point until we reach a sum that is greater than 2021 and is an odd integer:begin{tabular}{|c|c|c|c|c|c|c|c|}hline Term # & 976 & 977 & 978 & 979 & 980 & 981 & 982 hline Term & 3 & 4 & frac{5}{3} & frac{2}{3} & 1 & 3 & 4 hline Sum to this point & 2018 & 2022 & 2023 frac{2}{3} & 2024 frac{1}{3} & 2025 frac{1}{3} & 2028 frac{1}{3} & 2032 frac{1}{3} hlineend{tabular}begin{tabular}{|c|c|c|c|c|c|c|c|}hline Term # & 983 & 984 & 985 & 986 & 987 & 988 & 989 hline Term & frac{5}{3} & frac{2}{3} & 1 & 3 & 4 & frac{5}{3} & frac{2}{3} hline Sum to this point & 2034 & 2034 frac{2}{3} & 2035 frac{2}{3} & 2038 frac{2}{3} & 2042 frac{2}{3} & 2044 frac{1}{3} & 2045 hlineend{tabular}Therefore, the smallest positive integer N for which the sum of the first N numbers is equal to an odd integer greater than 2021 is N=989 when the sum is 2045 .ANSWER: 989

🧠:[<think>Okay, let's see. I need to figure out which of these operations is correct. The options are A, B, C, or D. Let me go through each one step by step.Starting with option A: a³ multiplied by a³ equals 2a³. Hmm, let's recall the exponent rules. When you multiply two expressions with the same base, you add the exponents. So, a³ times a³ should be a^(3+3) which is a^6. But the answer here is 2a³. That doesn't seem right. So A is probably incorrect. Let me double-check. Yeah, if you have a^m * a^n, it's a^(m+n). So 3+3 is 6, not 2a³. So A is wrong.Moving on to option B: (a²)³ equals a^6. Alright, let's think. When you raise a power to another power, you multiply the exponents. So (a²)^3 is a^(2*3) which is a^6. That seems correct. So B might be the right answer. But let me check the others just in case.Option C: 2x plus 3y equals 5xy. Wait a minute, you can't add different variables like that. If it were multiplication, maybe, but addition? No. 2x and 3y are not like terms. For example, if x is 1 and y is 1, 2*1 + 3*1 = 5, but 5xy would be 5*1*1=5. Wait, that gives the same result here. But that's only when x and y are 1. Let me try another example. Let x=2 and y=3. Then 2x + 3y is 4 + 9 = 13. 5xy would be 5*2*3=30. Definitely not the same. So C is incorrect. The mistake here is trying to combine unlike terms through addition, which isn't allowed. You can only add coefficients when the variables and exponents are the same.Option D: 2^(-3) equals -6. Let's remember negative exponents. A negative exponent means the reciprocal of the positive exponent. So 2^-3 is 1/(2^3) which is 1/8. But here they're saying it's -6. That's not right. Maybe they thought that 2 to the -3 is the same as multiplying 2 by -3? But no, exponents work differently. So 2^-3 is 1/8, not -6. Therefore, D is also incorrect.So going back, A was wrong because you add exponents when multiplying, not coefficients. B was correct because when raising a power to a power, you multiply exponents. C was wrong because you can't add different variables. D was wrong because negative exponents are reciprocals, not multiplication by negative numbers. Therefore, the correct answer should be B.Final Answerboxed{B}</think>To determine which operation is correct, we analyze each option step by step:- Option A: (a^{3} cdot a^{3} = 2a^{3}) When multiplying exponents with the same base, we add the exponents: (a^{3} cdot a^{3} = a^{3+3} = a^{6}). This does not equal (2a^{3}), so A is incorrect.- Option B: ((a^{2})^{3} = a^{6}) When raising a power to another power, we multiply the exponents: ((a^{2})^{3} = a^{2 cdot 3} = a^{6}). This is correct, so B is correct.- Option C: (2x + 3y = 5xy) Adding different variables (2x) and (3y) does not combine them into a product. For example, if (x = 2) and (y = 3), (2x + 3y = 4 + 9 = 13) while (5xy = 30). These are not equal, so C is incorrect.- Option D: (2^{-3} = -6) A negative exponent denotes a reciprocal: (2^{-3} = frac{1}{2^{3}} = frac{1}{8}). This is not equal to (-6), so D is incorrect.Thus, the correct answer is (boxed{B}).,<think>Okay, let's see. I need to figure out which of these operations is correct. There are four options: A, B, C, D. Hmm. Let me go through each one step by step.Starting with option A: a³ times a³ equals 2a³. Wait, when you multiply exponents with the same base, you add the exponents, right? So, a³ times a³ should be a^(3+3) which is a^6. But here it's written as 2a³. That seems wrong. So A is incorrect. Got it.Moving on to B: (a²)³ equals a^6. Let me recall. When you have an exponent raised to another exponent, you multiply them. So (a²)³ would be a^(2*3) which is a^6. Yeah, that's right. So B seems correct. But let me check the other options just to be sure.Option C: 2x plus 3y equals 5xy. Hmm, adding 2x and 3y. Since they are different variables, you can't combine them into a single term. So 2x + 3y is just 2x + 3y. It doesn't become 5xy. Because xy is a different term altogether. So C is definitely wrong.Option D: 2^-3 equals -6. Wait, negative exponents. Remember that a negative exponent means 1 over the positive exponent. So 2^-3 should be 1/(2^3) which is 1/8. But here it's written as -6. That's not correct. So D is also wrong.So reviewing all the options: A is incorrect because multiplying exponents with the same base adds the exponents. C is incorrect because you can't add different variables. D is wrong because negative exponents are reciprocals, not negative numbers. Only B is correct because when you raise a power to a power, you multiply the exponents. Therefore, the correct answer must be B.Final Answerboxed{B}</think>Let's analyze each option step by step:- Option A: (a^{3} cdot a^{3}). When multiplying exponents with the same base, we add the exponents: (a^{3+3} = a^{6}). However, the given answer is (2a^{3}), which is incorrect. So, A is wrong.- Option B: ((a^{2})^{3}). When raising a power to another power, we multiply the exponents: (a^{2 cdot 3} = a^{6}). This matches the given answer, so B is correct.- Option C: (2x + 3y). These terms have different variables and cannot be combined into a single term. The given answer (5xy) is incorrect. So, C is wrong.- Option D: (2^{-3}). A negative exponent indicates the reciprocal of the positive exponent: (2^{-3} = frac{1}{2^{3}} = frac{1}{8}). The given answer (-6) is incorrect. So, D is wrong.Thus, the correct answer is (boxed{B}).]

🔑:To solve this problem, let's analyze each option step by step:Option A: a^{3}cdot a^{3}The rule for multiplying powers with the same base is to add the exponents. Therefore, a^{3}cdot a^{3} = a^{3+3} = a^{6}. The given operation in option A suggests a^{3}cdot a^{3}=2a^{3}, which is incorrect based on the rule. The correct operation should be a^{6}.Option B: (a^{2})^{3}The rule for raising a power to another power is to multiply the exponents. Therefore, (a^{2})^{3} = a^{2cdot3} = a^{6}. This matches the operation given in option B, indicating it might be correct.Option C: 2x+3y=5xyThis operation suggests a sum of two terms equals their product, which is not a general property of algebra. Without specific values of x and y that make this equation true, this operation is generally incorrect.Option D: 2^{-3}The rule for negative exponents is that a^{-n} = frac{1}{a^{n}}. Therefore, 2^{-3} = frac{1}{2^{3}} = frac{1}{8}. The given operation in option D suggests 2^{-3}=-6, which is incorrect based on the rule. The correct operation should result in frac{1}{8}.Based on the analysis:- Option A is incorrect because a^{3}cdot a^{3} = a^{6}.- Option C is incorrect because 2x+3y does not generally equal 5xy.- Option D is incorrect because 2^{-3} = frac{1}{8}, not -6.Therefore, the correct operation is given in Option B, where (a^{2})^{3}=a^{6}.So, the correct option is boxed{B}.

🧠:[<think>Okay, let me try to figure out this problem. So, there's a row in a cinema with 15 seats. Before Xiao Yu sits down, some seats are already occupied. The condition is that no matter which seat he chooses, he will end up sitting next to someone. We need to find the minimum number of people already seated that makes this true. Hmm, so basically, we need to arrange people in the row such that every empty seat is adjacent to at least one occupied seat. Wait, no. Actually, the problem says that no matter which seat Xiao Yu chooses (which is an empty seat, presumably), he will have to sit next to someone. So that means every empty seat must have at least one neighbor that's occupied. So the question is: what's the minimum number of people needed to occupy seats such that every empty seat is adjacent to at least one occupied seat. This sounds like a covering problem. Like, covering the empty seats with occupied ones such that each empty seat is "covered" by an adjacent occupied seat. So, in other words, we need to place people in such a way that there are no two consecutive empty seats. Because if there were two consecutive empty seats, then someone could sit in one of them without being next to someone else. Wait, no. Let's think again. If there are two consecutive empty seats, then if Xiao Yu sits in one of them, he would have a neighbor on one side or the other. Wait, actually, if there are two empty seats in a row, like seats 1 and 2 are both empty, then if he sits in seat 1, the next seat is seat 2, which is empty, so he doesn't have a neighbor on the right. But seat 1 is at the end, so unless seat 0 exists, which it doesn't, then seat 1 only has seat 2 as a neighbor. If seat 2 is empty, then sitting in seat 1 would have no neighbors. Similarly, seat 15 is the other end. So, in order to prevent Xiao Yu from sitting in seat 1 without a neighbor, seat 2 must be occupied. Similarly, seat 14 must be occupied to prevent him from sitting in seat 15 without a neighbor. Therefore, the problem reduces to arranging people so that every empty seat has at least one adjacent occupied seat. That's called a dominating set in graph theory, maybe. But here it's a linear graph (a path) with 15 nodes. So, we need a dominating set of the path graph P₁₅ such that all vertices are either in the set or adjacent to a vertex in the set. And we want the smallest such set.So, the problem is equivalent to finding the minimum dominating set for a path graph with 15 vertices. The minimum dominating set for a path graph can be found using a formula or pattern. I remember that for a path with n vertices, the minimum dominating set has size ⎡n/3⎤. Wait, let me check. If we place people every third seat, starting from the second seat, that would cover the seats. For example, seats 2, 5, 8, 11, 14 would cover 15 seats. Let's see:Seat 2 covers 1, 2, 3Seat 5 covers 4, 5, 6Seat 8 covers 7, 8, 9Seat 11 covers 10, 11, 12Seat 14 covers 13, 14, 15Wait, seat 14 would cover 13, 14, 15. So that's 5 seats. But 15 divided by 3 is 5, so that works. So in that case, 5 people would be the minimum. But wait, let's check if that's correct. If we occupy seats 2,5,8,11,14, then every empty seat is adjacent to an occupied seat. For example, seat 1 is adjacent to seat 2, seat 3 is adjacent to seat 2, seat 4 is adjacent to seat 5, etc. The last seat, 15, is adjacent to seat 14. So yes, that works. So 5 people. But wait, let me test another arrangement. Suppose we occupy every other seat. Then, you need 8 people (since 15 is odd). But that's more than 5, so 5 is better. So maybe 5 is the minimum. But let me check with smaller numbers. Let's take n=3. Then, the minimum dominating set is 1 (seat 2). For n=4, seats 2 and 3 would cover all, but actually, seat 2 and 4 would also work. Wait, seat 2 covers 1,2,3 and seat 4 covers 4. So seat 2 and 4. So 2 seats. 4/3 is 1.333, ceiling is 2. That works. For n=5: seats 2 and 5 would cover 1,2,3 and 4,5. Wait, seat 4 is not covered. So maybe seats 2 and 4: seat 2 covers 1,2,3 and seat 4 covers 4,5. So 2 seats. 5/3 is 1.666, ceiling is 2. That works. So maybe the formula is ceiling(n/3). So for 15, ceiling(15/3)=5. Therefore, 5 people. So the answer should be 5. Wait, but let me make sure. Let's check if 4 people can do the job. Suppose we try to place 4 people. Let's see. If we place people in seats 3,6,9,12. Then seat 1 is empty and not adjacent to anyone. So Xiao Yu can sit there. So that's bad. If we place people in seats 2,6,10,14. Then seat 1 is adjacent to 2, seat 2 is occupied, seat 3 is adjacent to 2, seat 4 is adjacent to 6 (no, seat 4 is two seats away from 6. So seat 4 and 5 are between 2 and 6. So seat 4 and 5 are not adjacent to any occupied seats. Wait, seat 2 is occupied, seat 6 is occupied. So seat 4 is two seats away from 2 and two seats away from 6. So seat 4 is not adjacent to anyone. Therefore, Xiao Yu could sit in seat 4 and not be next to someone. So that arrangement doesn't work. Maybe another arrangement with 4 people. Let's try alternating blocks. For example, seat 2,5,8,11. Then seat 14 is empty. So seat 15 would be adjacent to seat 14, which is empty. Therefore, seat 15 is not covered. So Xiao Yu could sit in seat 15 without a neighbor. So that's bad. Hmm. Maybe seat 2,5,8,11,14 is needed. Which is 5 seats. Let me check again. If we have people in 2,5,8,11,14, then every seat is adjacent to an occupied seat. For example:Seat 1: adjacent to 2 (occupied)Seat 2: occupiedSeat 3: adjacent to 2Seat 4: adjacent to 5 (occupied)Seat 5: occupiedSeat 6: adjacent to 5Seat 7: adjacent to 8 (occupied)Seat 8: occupiedSeat 9: adjacent to 8Seat 10: adjacent to 11 (occupied)Seat 11: occupiedSeat 12: adjacent to 11Seat 13: adjacent to 14 (occupied)Seat 14: occupiedSeat 15: adjacent to 14So yes, all seats are either occupied or adjacent to an occupied seat. Therefore, 5 people. If we try to use 4 people, is there a way? Let's see. Maybe overlapping the coverage. For example, seat 3, 6, 9, 12. Let's check:Seat 1: adjacent to 2, which is empty, and 3. Seat 3 is occupied. So seat 1 is adjacent to 3? No, seat 1 is adjacent to seat 2. Seat 3 is two seats away. So seat 1 is only adjacent to seat 2, which is empty. So seat 1 is not covered. So that's bad. What if we start from seat 2? 2,6,10,14. Let's check seat coverage:Seat 1: adjacent to 2 (occupied) – goodSeat 2: occupiedSeat 3: adjacent to 2 (good)Seat 4: adjacent to 3 (empty) and 5 (empty). Not covered. So seat 4 is not adjacent to any occupied seat. Because seat 2 is two seats away. So seat 4 is not covered. So Xiao Yu could sit in seat 4. So that arrangement fails.Another arrangement: 2,5,9,13. Let's check:Seat 1: adjacent to 2 (good)Seat 2: occupiedSeat 3: adjacent to 2 (good)Seat 4: adjacent to 5 (good)Seat 5: occupiedSeat 6: adjacent to 5 (good)Seat 7: adjacent to 6 (empty) and 8 (empty). Not covered. Because seat 9 is two seats away. So seat 7 is not adjacent to anyone. So Xiao Yu could sit there. So that's bad.Hmm. How about 3,7,11,15. Let's see:Seat 1: adjacent to 2 (empty). Not covered.Seat 2: adjacent to 1 (empty) and 3 (occupied). So seat 2 is adjacent to 3. But seat 1 is not. So seat 1 is uncovered. So Xiao Yu can sit there. Bad.Another idea: maybe interleaving. Let's try seat 2,4,8,12. Check coverage:Seat 1: adjacent to 2 (good)Seat 2: occupiedSeat 3: adjacent to 2 and 4. 4 is occupied. So seat 3 is covered.Seat 4: occupiedSeat 5: adjacent to 4 (good)Seat 6: adjacent to 5 (empty) and 7 (empty). Not covered. Seat 8 is two seats away. So seat 6 is not covered. So Xiao Yu can sit there. Hmm, no good. What if seat 2,7,11,14?Seat 1: adjacent to 2 (good)Seat 2: occupiedSeat 3: adjacent to 2 (good)Seat 4: adjacent to 3 (empty) and 5 (empty). Not covered. So seat 4 is bad.Seat 5: adjacent to 4 (empty) and 6 (empty). Not covered.Seat 6: adjacent to 5 (empty) and 7 (occupied). So seat 6 is adjacent to 7 (good). Wait, seat 6 is adjacent to 7. If seat 7 is occupied, then seat 6 is covered. But seat 4 and 5 are not. Seat 4 is adjacent to seat 3 (empty) and seat 5 (empty). So seat 4 is not covered. So Xiao Yu can sit there.This is tricky. It seems like with 4 people, there's always some seat that's left uncovered. So maybe 5 is indeed the minimum. Therefore, the answer should be 5. But let's check one more arrangement. Suppose we place people at 2,5,7,10,13. Let's check:Seat 1: adjacent to 2 (good)Seat 2: occupiedSeat 3: adjacent to 2 (good)Seat 4: adjacent to 5 (good)Seat 5: occupiedSeat 6: adjacent to 5 (good) and 7 (occupied). So seat 6 is covered.Seat 7: occupiedSeat 8: adjacent to 7 (good)Seat 9: adjacent to 10 (good)Seat 10: occupiedSeat 11: adjacent to 10 (good)Seat 12: adjacent to 13 (good)Seat 13: occupiedSeat 14: adjacent to 13 (good)Seat 15: adjacent to 14 (empty). Wait, seat 14 is adjacent to 13 (occupied), but seat 15 is adjacent to seat 14 (empty). So seat 15 is not covered. So Xiao Yu can sit there. So that's a problem. So we need someone in seat 14. So maybe 2,5,7,10,13,14. But that's 6 people, which is more than 5. So that's not helpful.Alternatively, if we place people in 2,5,8,11,14, as before, that covers everything. So that's 5 people. Let's verify again:Seat 1: adjacent to 2 (good)Seat 2: occupiedSeat 3: adjacent to 2 (good)Seat 4: adjacent to 5 (good)Seat 5: occupiedSeat 6: adjacent to 5 (good)Seat 7: adjacent to 8 (good)Seat 8: occupiedSeat 9: adjacent to 8 (good)Seat 10: adjacent to 11 (good)Seat 11: occupiedSeat 12: adjacent to 11 (good)Seat 13: adjacent to 14 (good)Seat 14: occupiedSeat 15: adjacent to 14 (good)Perfect. All seats are covered. So that works with 5 people. Therefore, it seems like 5 is indeed the minimum. But just to be thorough, let's see if there's a different arrangement with 5 people that perhaps covers the seats in a different pattern. For example, starting from seat 3: 3,6,9,12,15. Let's check coverage:Seat 1: adjacent to 2 (empty). Not covered.Seat 2: adjacent to 3 (occupied). So seat 2 is covered. Seat 1 is only adjacent to seat 2 (empty). So seat 1 is uncovered. Therefore, Xiao Yu can sit there. So that's bad. Hence, starting from seat 2 is better.Another arrangement: 1,4,7,10,13. Then:Seat 1: occupiedSeat 2: adjacent to 1 (good)Seat 3: adjacent to 4 (good)Seat 4: occupiedSeat 5: adjacent to 4 (good)Seat 6: adjacent to 7 (good)Seat 7: occupiedSeat 8: adjacent to 7 (good)Seat 9: adjacent to 10 (good)Seat 10: occupiedSeat 11: adjacent to 10 (good)Seat 12: adjacent to 13 (good)Seat 13: occupiedSeat 14: adjacent to 13 (good)Seat 15: adjacent to 14 (empty). So seat 15 is uncovered. Therefore, Xiao Yu can sit there. So we need someone in seat 14 or 15. If we add seat 15, then 1,4,7,10,13,15. That's 6 people, which is worse. So again, starting from seat 2,5,8,11,14 is better.Alternatively, seat 2,4,7,10,13. Let's check:Seat 1: adjacent to 2 (good)Seat 2: occupiedSeat 3: adjacent to 2 (good) and 4 (occupied). So good.Seat 4: occupiedSeat 5: adjacent to 4 (good)Seat 6: adjacent to 5 (empty) and 7 (occupied). So seat 6 is adjacent to 7 (good)Seat 7: occupiedSeat 8: adjacent to 7 (good)Seat 9: adjacent to 10 (good)Seat 10: occupiedSeat 11: adjacent to 10 (good)Seat 12: adjacent to 13 (good)Seat 13: occupiedSeat 14: adjacent to 13 (good)Seat 15: adjacent to 14 (empty). So seat 15 is uncovered. Again, problem.So, even with this arrangement, seat 15 is left uncovered. So you need someone at seat 14 or 15. Therefore, the previous arrangement of seat 2,5,8,11,14 is better because it covers seat 15 through seat 14.Therefore, it seems that the minimum number of people needed is 5. So, the answer should be 5. Just to confirm, let's calculate using the formula. For a path graph P_n, the minimum dominating set is ⎡n/3⎤. Here, n=15, so 15 divided by 3 is 5. Therefore, the minimum number is indeed 5. So the answer is 5.Final Answerboxed{5}</think>To determine the minimum number of people already seated in a row of 15 seats such that Xiao Yu will always have to sit next to someone, we need to ensure every empty seat is adjacent to at least one occupied seat. This problem can be solved by finding the minimum dominating set for a path graph with 15 vertices.A dominating set in a graph is a set of vertices such that every vertex is either in the set or adjacent to a vertex in the set. For a path graph ( P_n ), the minimum dominating set size is given by ( lceil frac{n}{3} rceil ).For ( n = 15 ):[lceil frac{15}{3} rceil = lceil 5 rceil = 5]We can verify this by placing people in seats 2, 5, 8, 11, and 14. This arrangement ensures that every empty seat is adjacent to an occupied seat:- Seat 1 is adjacent to seat 2.- Seat 3 is adjacent to seat 2.- Seat 4 is adjacent to seat 5.- Seat 6 is adjacent to seat 5.- Seat 7 is adjacent to seat 8.- Seat 9 is adjacent to seat 8.- Seat 10 is adjacent to seat 11.- Seat 12 is adjacent to seat 11.- Seat 13 is adjacent to seat 14.- Seat 15 is adjacent to seat 14.Thus, the minimum number of people needed is (boxed{5}).,<think>Okay, let's try to figure out this problem. So, the question is: In a cinema with a row of 15 seats, Xiao Yu finds that no matter which seat he chooses, he will end up sitting next to someone. We need to find the minimum number of people already seated before he arrives. Hmm, so the key here is that every empty seat has at least one neighbor occupied. Wait, no. Wait, actually, Xiao Yu is choosing a seat, and no matter which seat he picks, once he sits there, he is next to someone. That means that before he sits down, every empty seat must have at least one adjacent seat that's occupied. Otherwise, if there was an empty seat with both sides empty, he could sit there and not be next to anyone. So the problem reduces to: what's the minimum number of people needed to occupy seats such that every empty seat has at least one neighbor occupied.This seems similar to a problem called the "dominating set" problem in graph theory, where you need to cover all vertices with the minimum number of nodes such that every node is either in the set or adjacent to one in the set. In this case, the seats are nodes in a linear graph, and we need a dominating set where every empty seat is adjacent to an occupied one. So the question is asking for the minimum dominating set of a 15-node path graph.Alternatively, another way to think about it is arranging people in such a way that no two empty seats are adjacent. Wait, no, that's different. If we arrange people so that no two empty seats are adjacent, that would be the maximum number of people that can be seated without sitting next to each other. But here, we need the opposite: the minimum number of people such that every empty seat is adjacent to at least one person.Let me see. Let's think of the seats as a sequence of 15 seats. To ensure that every empty seat has at least one neighbor occupied, how should we place the people?One approach is to use a pattern that covers all empty seats. If we alternate occupied and empty seats, like OEOEOE..., that would mean every empty seat is between two occupied ones. But in this case, if we have OEOEOE..., the number of people would be roughly half of 15, which is 7 or 8. But maybe there's a more efficient way.Wait, but in the alternating pattern, each occupied seat "covers" the seats next to it. So if we space people out with two empty seats between them, like O--O--O--..., then each person covers the seat to their left and right. Let's see: in a row of 15 seats, placing a person every three seats. So positions 1, 4, 7, 10, 13. That would be 5 people. Let's check if that works.If we have people at 1,4,7,10,13, then the empty seats are 2,3,5,6,8,9,11,12,14,15. Let's check seat 2: adjacent to 1 (occupied). Seat 3: adjacent to 4 (occupied). Seat 5: adjacent to 4 (occupied). Seat 6: adjacent to 7 (occupied). Seat 8: adjacent to 7 (occupied). Seat 9: adjacent to 10 (occupied). Seat 11: adjacent to 10 (occupied). Seat 12: adjacent to 13 (occupied). Seat 14: adjacent to 13 (occupied). Seat 15: adjacent to 14? Wait, seat 15 is only adjacent to seat 14. If seat 14 is empty, then seat 15 is only adjacent to 14, which is empty. So seat 15 wouldn't be adjacent to anyone. So this pattern leaves seat 15 uncovered. So that's a problem. So we need someone in seat 14 or 15. If we add a person at seat 15, then seat 14 is covered, seat 15 is occupied. So maybe 1,4,7,10,13,15. Let's check.Seat 2: next to 1 (occupied). Seat3: next to 4. Seat5: next to4. Seat6: next to7. Seat8: next to7. Seat9: next to10. Seat11: next to10. Seat12: next to13. Seat14: next to15. Seat15 is occupied. So that works. So that's 6 people. But maybe there's a better arrangement.Alternatively, maybe start at seat 2, then 5,8,11,14. Then seat1: adjacent to 2. Seat3: adjacent to2. Seat4: adjacent to5. Seat6: adjacent to5. Seat7: adjacent to8. Seat9: adjacent to8. Seat10: adjacent to11. Seat12: adjacent to11. Seat13: adjacent to14. Seat15: adjacent to14? No, seat15 is adjacent to14, which is occupied. So seats 2,5,8,11,14. That's 5 people. Let's check seat1: next to2 (occupied). Seat3: next to2. Seat4: next to5. Seat6: next to5. Seat7: next to8. Seat9: next to8. Seat10: next to11. Seat12: next to11. Seat13: next to14. Seat15: next to14. So all empty seats are adjacent to someone. So this seems to work with 5 people. Wait, but earlier when starting at seat1,4,7,10,13,15, we needed 6, but starting at seat2,5,8,11,14, we can cover all with 5. So why the difference? Because the first pattern ends at seat13 and 15, which required an extra person, but starting at seat2 allows coverage of seat1 and seat15 through seat2 and seat14. Hmm.Wait, in the second arrangement, starting at seat2, seat1 is adjacent to seat2. Seat15 is adjacent to seat14. So if we have seat2,5,8,11,14 occupied, seat14 covers seat15, seat2 covers seat1. So this works with 5 people. So 5 is better than 6. Is 5 the minimum?But wait, let's test this. Let me list the occupied seats as 2,5,8,11,14. Then the empty seats are 1,3,4,6,7,9,10,12,13,15.Check each empty seat:1: adjacent to 2 (occupied) – good.3: adjacent to 2 (occupied) – good.4: adjacent to 5 (occupied) – good.6: adjacent to 5 (occupied) – good.7: adjacent to 8 (occupied) – good.9: adjacent to 8 (occupied) – good.10: adjacent to 11 (occupied) – good.12: adjacent to 11 (occupied) – good.13: adjacent to 14 (occupied) – good.15: adjacent to 14 (occupied) – good.So yes, all empty seats are adjacent to an occupied seat. Therefore, 5 people. So is 5 the minimum? Can we do it with fewer?Let me try with 4. Suppose we place people every 4 seats. Let's try positions 3,7,11,15. Then check coverage.Empty seats:1: adjacent to 2 and 3. Seat3 is occupied, so seat1 is adjacent to seat2 (empty) and seat3 (occupied). So seat1 is covered by seat3.Seat2: adjacent to1 (empty) and3 (occupied). So covered by seat3.Seat4: adjacent to3 (occupied) and5. So covered by seat3.Seat5: adjacent to4 (empty) and6. Not covered. So seat5 is empty and adjacent to seat4 and6, both empty. So seat5 is not covered. Therefore, Xiao Yu could sit at seat5 and not be next to anyone. So this arrangement doesn't work.Alternatively, maybe a different pattern. What if we space them as 2,6,10,14. Then check coverage.Seat1: adjacent to2 (occupied) – good.Seat3: adjacent to2 (occupied) – good.Seat4: adjacent to5 and6. Seat6 is occupied, so seat4 is covered.Seat5: adjacent to4 (empty) and6 (occupied) – covered.Seat7: adjacent to6 (occupied) and8 – covered.Seat8: adjacent to7 (empty) and9.Wait, seat8: adjacent to7 (empty) and9 (empty). So seat8 is empty and adjacent to two empty seats. So Xiao Yu could sit at seat8. So this arrangement doesn't work. So 4 people isn't enough here.Alternatively, another arrangement with 4 people. Let's try 3,7,11,15.Wait, we tried that earlier and seat5 was uncovered.Alternatively, maybe overlapping more. How about 2,5,9,13. Let's check.Seat1: adjacent to2 (occupied) – good.Seat3: adjacent to2 (occupied) – good.Seat4: adjacent to5 (occupied) – good.Seat6: adjacent to5 (occupied) – good.Seat7: adjacent to8 and9. Seat9 is occupied, so seat7 is covered.Seat8: adjacent to7 (empty) and9 (occupied) – covered.Seat10: adjacent to9 (occupied) and11 – covered by seat9.Seat11: adjacent to10 (empty) and12. Wait, seat11 is empty? No, seat11 is empty in this arrangement? Wait, the occupied seats are 2,5,9,13. So seat10 is adjacent to seat9 (occupied) and seat11 (empty). So seat10 is covered by seat9. Seat11: adjacent to seat10 (empty) and12. So seat11 is empty, adjacent to two empty seats. So Xiao Yu could sit there. So that's a problem. Therefore, this arrangement doesn't work.Hmm, maybe 4 people is not enough. Let's try another arrangement. How about 2,6,10,14. Wait, we tried that already. Seat8 was uncovered. Alternatively, 3,6,9,12,15. Let's see, that's 5 people again. Wait, no, 3,6,9,12,15 is 5 people. But we already saw that 5 works.Wait, perhaps with 4 people, it's impossible. Let me think. If we place 4 people, each can cover up to two seats on either side. But with 15 seats, how does this scale? Let's model it.Each person can cover themselves and the two adjacent seats. So each person can cover 3 seats. But since the seats are linear, the first and last person can only cover two seats beyond themselves. Wait, actually, the coverage is that an occupied seat makes the adjacent seats "covered" in terms of being next to someone. Wait, actually, the problem is that each empty seat must be adjacent to at least one occupied seat. So each occupied seat can cover the seat to the left and right. So for an occupied seat in the middle, it can cover two empty seats. For the first seat, it can only cover the seat to the right, and the last seat can cover the seat to the left.But maybe to model this, the maximum number of seats that can be covered by each person is 2 (the adjacent seats). So if we have n people, the total coverage is 2n. However, the total number of seats to cover is 15 - n (since n seats are occupied). Each of the 15 - n empty seats needs to be covered by at least one adjacent occupied seat. So 15 - n <= 2n. Therefore, 15 <= 3n => n >= 5. So according to this inequality, n must be at least 5.Therefore, 5 is the minimal number. Which matches our earlier arrangement. So the answer should be 5. Wait, but let's check the calculation.Total empty seats = 15 - n. Each occupied seat can cover up to 2 empty seats. So total coverage needed is 15 - n <= 2n. Therefore, 15 <= 3n => n >= 5. Therefore, n must be at least 5. Therefore, the minimal number is 5.Therefore, the answer is 5. So before Xiao Yu sits down, at least 5 people are already seated.But let me confirm with the example we had earlier where 5 people worked. Yes, with seats 2,5,8,11,14 occupied, all empty seats are adjacent to someone. So 5 is sufficient. And according to the inequality, 5 is necessary, so it's the minimum.Therefore, the answer is 5.Final AnswerThe minimum number of people already seated is boxed{5}.</think>To determine the minimum number of people already seated in a row of 15 seats such that no matter which seat Xiao Yu chooses, he will have to sit next to someone, we need to ensure every empty seat is adjacent to at least one occupied seat. This problem can be modeled as finding a minimum dominating set in a linear graph of 15 nodes.Key steps:1. Understanding the problem: Every empty seat must be adjacent to at least one occupied seat. This means we need a dominating set where each empty seat is covered by an adjacent occupied seat.2. Pattern analysis: We need to find the minimum number of occupied seats such that all empty seats are adjacent to an occupied seat. 3. Coverage calculation: Each occupied seat can cover up to two adjacent seats. For 15 seats, the minimum number of occupied seats needed can be calculated using the inequality (15 - n leq 2n), where (n) is the number of occupied seats. Solving this gives (15 leq 3n) or (n geq 5).4. Verification: Placing people at positions 2, 5, 8, 11, and 14 covers all empty seats. Checking each empty seat confirms they are adjacent to an occupied seat.Thus, the minimum number of people already seated is (boxed{5}).]

🔑:answer: 5

🧠:[<think>Okay, let's see. So the problem is about exchanging different types of cards—gold, silver, and bronze—for reward points. The information given is that 2 gold cards can be exchanged for 36 reward points, or 3 silver cards for the same 36 points, or 4 bronze cards also for 36 points. Then the question is, if you use one gold card, one silver card, and one bronze card, how many reward points can you get?Hmm. Alright, so first, maybe I need to find out how many points each type of card is worth individually. Since combinations of different numbers of cards give the same total points, maybe each type has a different point value per card. So like, gold cards are more valuable than silver, which are more valuable than bronze. Let me think.Let me denote:Let G be the reward points per gold card.Let S be the reward points per silver card.Let B be the reward points per bronze card.According to the problem:2G = 363S = 364B = 36So solving for each:G = 36 / 2 = 18S = 36 / 3 = 12B = 36 / 4 = 9So each gold card is worth 18 points, silver is 12, and bronze is 9.Then, if we use one of each: G + S + B = 18 + 12 + 9 = 39.Wait, so the answer should be 39? That seems straightforward. Let me check again.If 2 gold cards give 36, then each gold is 18. 3 silvers give 36, so 12 each. 4 bronzes give 36, so 9 each. Then adding one of each: 18 + 12 + 9 = 39. Yeah, that seems right.But maybe I misread the problem. Wait, the problem says "using 2 gold cards, or 3 silver cards, or 4 bronze cards, can all exchange for 36 reward points." So each of those combinations gives 36 points. So that means 2G = 36, so G=18, 3S=36, so S=12, 4B=36, so B=9. Then 1G +1S +1B = 18+12+9=39. So 39 points. Hmm. Seems like a straightforward problem. Maybe there's a trick?Wait, is there a possibility that the problem is saying that all three combinations (2G, 3S, 4B) together exchange for 36 points? But that interpretation doesn't make much sense. The way it's phrased is: "using 2 gold cards, or 3 silver cards, or 4 bronze cards, can all exchange for 36 reward points." So the "or" suggests that each of these different combinations can be exchanged for 36 points. So individually, 2G=36, 3S=36, 4B=36. Therefore, each combination gives 36 points when exchanged. So each card's value is as calculated before.Alternatively, if it were 2G + 3S + 4B = 36, then we would have a system of equations, but since there are three variables and only one equation, that would be underdetermined. But given the problem statement, I think the correct interpretation is that each of those combinations individually can be exchanged for 36 points.Therefore, the individual card values are 18, 12, and 9. Therefore, one of each would be 39. So the answer should be 39.Wait, but let me check with another perspective. Let's see:If 2 gold cards give 36 points, then per gold card it's 18. Similarly, 3 silver gives 36, so 12 per silver, and 4 bronze gives 9 each. So adding one of each: 18 +12 +9=39.Yes, that seems correct. Therefore, the answer is 39. I don't think there's any complication here.Alternatively, maybe the problem is trying to trick us into thinking that they can be combined in some other way? But no, the problem says "using one gold card, one silver card, and one bronze card"—so it's straightforward addition. So 18 +12 +9=39. So 39 reward points.I think that's it. No hidden steps here.Final Answerboxed{39}</think>Given the problem, we need to determine how many reward points can be exchanged using one gold card, one silver card, and one bronze card. First, we find the reward points for each type of card:- 2 gold cards (G) can be exchanged for 36 reward points: [ 2G = 36 implies G = frac{36}{2} = 18 ]- 3 silver cards (S) can be exchanged for 36 reward points: [ 3S = 36 implies S = frac{36}{3} = 12 ]- 4 bronze cards (B) can be exchanged for 36 reward points: [ 4B = 36 implies B = frac{36}{4} = 9 ]Next, we calculate the total reward points for using one gold card, one silver card, and one bronze card:[G + S + B = 18 + 12 + 9 = 39]Thus, the final answer is:[boxed{39}],<think>Okay, let me try to solve this problem. So, the question says that using 2 gold cards, or 3 silver cards, or 4 bronze cards, you can exchange all of them for 36 reward points each. Then, it's asking how many reward points you can get by exchanging one gold card, one silver card, and one bronze card together. Hmm, let me think step by step.First, let me parse the problem again. It says that 2 gold cards exchange for 36 points. Similarly, 3 silver cards also exchange for 36 points, and 4 bronze cards as well. So, each type of card, when exchanged in a certain quantity (2, 3, 4 respectively), gives the same amount of reward points, which is 36. Then, we need to find out the total points if we exchange one of each type together.So, maybe I need to find out how much each individual card is worth in terms of reward points, and then add them up. That makes sense. If I can find the value per card for gold, silver, and bronze, then adding one of each would just be the sum of their individual values.Let's denote:- Let G be the reward points for one gold card.- Let S be the reward points for one silver card.- Let B be the reward points for one bronze card.According to the problem:- 2 gold cards = 36 points. So, 2G = 36. Therefore, G = 36 / 2 = 18. So, each gold card is 18 points. Wait, hold on, let me check that again. If 2 gold cards can be exchanged for 36 points, that means each gold card is worth 36 points divided by 2, right? So yes, G = 18.Similarly, 3 silver cards = 36 points. So, 3S = 36. Therefore, S = 36 / 3 = 12. So, each silver card is 12 points.And for bronze cards, 4 bronze cards = 36 points. So, 4B = 36. Hence, B = 36 / 4 = 9. Each bronze card is 9 points.Therefore, if we use one of each, the total reward points would be G + S + B = 18 + 12 + 9 = 39.Wait, that seems straightforward. But let me verify again because sometimes in these problems, there might be a catch. For example, maybe the exchange rate isn't additive? Or maybe exchanging different types together affects the total points somehow. But the problem doesn't mention any such thing. It just says using 2 gold, 3 silver, or 4 bronze each gives 36 points, and then asks for exchanging one of each. So, probably just adding their individual values.Let me check each step again. First, for gold cards: 2G = 36. Solving for G, divide both sides by 2: G = 18. Correct.Silver: 3S = 36. S = 12. Correct.Bronze: 4B = 36. B = 9. Correct.Then, total points for one of each: 18 + 12 + 9. Let's add them step by step. 18 + 12 is 30, and 30 + 9 is 39. So, 39 points. That seems to be the answer.But wait, just to make sure, let's think if there's another interpretation. For example, does "using 2 gold cards, or 3 silver cards, or 4 bronze cards, can all exchange for 36 reward points" mean that each combination gives 36 points? So, 2 gold cards give 36, 3 silver give 36, 4 bronze give 36. Therefore, each card type has its own rate. Therefore, per card, as I calculated before.Alternatively, maybe the problem is saying that if you use 2 gold, 3 silver, and 4 bronze together, they can exchange for 36 points? But the wording is "using 2 gold cards, or 3 silver cards, or 4 bronze cards, can all exchange for 36 reward points." The "or" is important here. So, it's 2 gold, or 3 silver, or 4 bronze, each gives 36. So, individually, each set gives 36. Therefore, the way I solved it initially is correct.Therefore, the individual values are 18, 12, 9. Adding them gives 39.So, the answer should be 39. Therefore, one gold, one silver, one bronze card can be exchanged for 39 reward points.Final AnswerThe total reward points are boxed{39}.</think>Given that 2 gold cards, 3 silver cards, or 4 bronze cards can each be exchanged for 36 reward points, we need to determine how many reward points one gold card, one silver card, and one bronze card can exchange for.First, we find the reward points for each type of card individually:- For gold cards: (2G = 36 implies G = frac{36}{2} = 18)- For silver cards: (3S = 36 implies S = frac{36}{3} = 12)- For bronze cards: (4B = 36 implies B = frac{36}{4} = 9)Next, we add the reward points for one gold card, one silver card, and one bronze card:[G + S + B = 18 + 12 + 9 = 39]Thus, using one gold card, one silver card, and one bronze card can exchange for (boxed{39}) reward points.]

🔑:39

🧠:[<think>Okay, let's see. I need to find the equation of a plane that passes through point A and is perpendicular to the vector BC. The points given are A(0, -8, 10), B(-5, 5, 7), and C(-8, 0, 4). Hmm, right. So first, maybe I should figure out what the vector BC is. Since the plane is perpendicular to BC, that vector should be the normal vector to the plane, right?Wait, yes, because if a plane is perpendicular to a vector, that vector is its normal. So, the normal vector n is equal to vector BC. To find vector BC, I subtract the coordinates of point B from point C. Let me write that down. Vector BC is C minus B. So, the components are (-8 - (-5), 0 - 5, 4 - 7). Let me compute each component.For the x-component: -8 - (-5) is the same as -8 + 5, which is -3. For the y-component: 0 - 5 is -5. And for the z-component: 4 - 7 is -3. So vector BC is (-3, -5, -3). Therefore, the normal vector n is (-3, -5, -3). Alternatively, sometimes people prefer to write the normal vector with positive leading coefficients, but I think the direction doesn't matter for the plane equation; the normal vector can point in either direction. So maybe it's okay as it is. Now, the general equation of a plane with normal vector (a, b, c) is a(x - x0) + b(y - y0) + c(z - z0) = 0, where (x0, y0, z0) is a point on the plane. Since the plane passes through point A(0, -8, 10), we can plug those coordinates into the equation. So substituting into the formula: (-3)(x - 0) + (-5)(y - (-8)) + (-3)(z - 10) = 0. Wait, let me check that. The point is (0, -8, 10), so x0 is 0, y0 is -8, z0 is 10. Therefore, the equation should be:-3(x - 0) -5(y - (-8)) -3(z - 10) = 0.Simplifying each term:First term: -3x.Second term: -5(y + 8) because y - (-8) is y + 8.Third term: -3(z - 10).So the equation becomes:-3x -5(y + 8) -3(z - 10) = 0.Now, let's expand this equation.Starting with the -5(y + 8): that's -5y -40.And the -3(z -10): that's -3z +30.So putting it all together:-3x -5y -40 -3z +30 = 0.Combine like terms:-3x -5y -3z -10 = 0.Wait, because -40 +30 is -10. So the equation simplifies to:-3x -5y -3z -10 = 0.But usually, plane equations are written with the constant term on the right side. So we can move the -10 to the other side:-3x -5y -3z = 10.Alternatively, it might be preferable to write the equation with positive coefficients. If we multiply both sides of the equation by -1 to make the coefficients of x, y, z positive, we get:3x +5y +3z = -10.But wait, that would change the sign of the constant term as well. Let me check:Original equation after expanding: -3x -5y -3z -10 = 0.If we multiply both sides by -1:3x +5y +3z +10 = 0.Then, moving 10 to the other side:3x +5y +3z = -10.But is this necessary? I mean, both forms are correct, but sometimes the standard form is preferred with the constant term positive. Let me see. The original normal vector was (-3, -5, -3), so the plane equation using that normal vector is -3x -5y -3z -10 =0. If we use the opposite normal vector (3,5,3), the equation becomes 3x +5y +3z +10 =0, but that seems different. Wait, no, actually, the plane equations are the same but scaled by -1.Wait, actually, if you multiply both sides of the equation by -1, you get 3x +5y +3z +10 =0, which is equivalent to -3x -5y -3z -10 =0. So both equations represent the same plane. Therefore, both forms are acceptable. However, sometimes people prefer the constant term to be positive, but in this case, if we do that, the constant term would be negative. Alternatively, perhaps just leave it as is.But let's verify the calculation again to make sure. Let me double-check the vector BC.Point B is (-5,5,7) and point C is (-8,0,4). So vector BC is C - B: (-8 - (-5), 0 -5, 4 -7) = (-3, -5, -3). That's correct.Then the normal vector is (-3, -5, -3). So plugging into the plane equation:-3(x -0) -5(y - (-8)) -3(z -10) =0.Which simplifies to:-3x -5(y +8) -3(z -10) =0.Expanding:-3x -5y -40 -3z +30 =0.Combine constants: -40 +30 = -10.Thus: -3x -5y -3z -10 =0.Yes, that's correct. So the plane equation is -3x -5y -3z =10 if we move the -10 to the other side, but that might not be necessary. Alternatively, writing it as -3x -5y -3z -10 =0 is also correct.Alternatively, if we want to write it in the form ax + by + cz + d =0, then a=-3, b=-5, c=-3, d=-10. But often, people like to have the coefficients a, b, c as integers without a common factor. Here, they have a common factor of -1, but if we factor that out, we get 3x +5y +3z +10 =0, but as mentioned earlier, this is equivalent.So depending on the preferred form, both are acceptable. However, the problem didn't specify any particular form, just to write the equation. So maybe either way is fine. But let me check if substituting point A into the equation satisfies it.Let's test point A(0, -8,10) in the equation -3x -5y -3z -10 =0.Left-hand side: -3*0 -5*(-8) -3*10 -10 = 0 +40 -30 -10 =0. Correct.If we use the other form, 3x +5y +3z +10=0, then substituting A: 0 +5*(-8)+3*10 +10= -40 +30 +10=0. Also correct.So both equations are valid. However, the problem says "the equation of the plane passing through point A and perpendicular to the vector BC". Since the normal vector is BC, which is (-3,-5,-3), the equation using that normal vector is -3x -5y -3z -10=0, but if we present it as 3x +5y +3z +10=0, that's also correct because the normal vector can be in the opposite direction. Wait, actually, the normal vector in the second case is (3,5,3), which is opposite to BC. However, the problem states that the plane is perpendicular to vector BC. So perhaps the normal vector should be exactly BC, not the opposite. Hmm, but in reality, the plane's normal vector can be either direction, as long as it's perpendicular. But the problem says "perpendicular to vector BC", so maybe the normal vector is BC. Therefore, the equation should use BC as the normal vector.So in that case, the equation is -3x -5y -3z -10=0. Alternatively, if we want to write it in the form with the normal vector BC, maybe present it as is. However, sometimes people write plane equations with the constant term on the right, so -3x -5y -3z =10. But in standard form, it's usually written as ax + by + cz + d =0. So maybe -3x -5y -3z -10=0 is the standard form.Alternatively, if the problem expects the answer in a certain format, but since it's not specified, either form is acceptable. However, maybe the answer key would prefer the coefficients not to have a negative sign. Let me check if 3x +5y +3z +10=0 is also a valid answer. Wait, but if the normal vector is BC, which is (-3,-5,-3), then the coefficients of x, y, z in the plane equation should match the normal vector. So if the normal vector is (-3,-5,-3), then the plane equation is -3x -5y -3z +d=0. Then plugging in point A gives d. Let's recast that.Wait, general plane equation: n ⋅ (r - r0) =0, where n is the normal vector, r is a point on the plane, and r0 is a given point. So in this case, n is BC, which is (-3,-5,-3), r0 is A(0,-8,10). So the equation is (-3,-5,-3) ⋅ (x -0, y +8, z -10)=0.Calculating the dot product:-3*(x -0) + (-5)*(y +8) + (-3)*(z -10) =0Which is -3x -5y -40 -3z +30=0, same as before. Then combining constants: -10. So the equation is -3x -5y -3z -10=0.Therefore, the correct equation is -3x -5y -3z -10=0. Alternatively, multiplying both sides by -1 gives 3x +5y +3z +10=0, but that would be using the opposite normal vector. Since the problem specifies that the plane is perpendicular to vector BC, which is (-3,-5,-3), then the coefficients should correspond to that vector. Therefore, the equation -3x -5y -3z -10=0 is the one that uses vector BC as the normal vector.But maybe to write it in a more standard form, we can write all terms on one side: -3x -5y -3z =10. However, depending on the convention, some sources prefer the constant term to be positive. If that's the case, then multiplying both sides by -1 gives 3x +5y +3z = -10. But in this case, the constant term is negative. Hmm. Alternatively, leave it as -3x -5y -3z -10=0. I think that's acceptable.Wait, let's check with the original points. Let's verify if point A is on the plane. Plugging into -3x -5y -3z -10=0:-3*0 -5*(-8) -3*10 -10 =0 +40 -30 -10=0. Correct.Also, check if vector BC is indeed perpendicular to the plane. The normal vector of the plane is (-3,-5,-3), which is exactly vector BC. So that's correct. Therefore, the equation of the plane is correct.So, final answer can be written in different forms, but according to the problem statement, it's just the equation. So maybe present it as -3x -5y -3z =10 or 3x +5y +3z = -10, but actually, expanding the initial equation, it's -3x -5y -3z -10=0. So perhaps writing it in the form where the right-hand side is zero. So -3x -5y -3z -10 =0. But in some textbooks, they write ax + by + cz + d =0, so maybe that's the most standard form.Alternatively, if they expect the answer in the form with the normal vector and point, but since the problem says to write the equation, then the standard form is acceptable. Therefore, the answer is -3x -5y -3z -10=0. Alternatively, if we multiply by -1, 3x +5y +3z +10=0. But which one is correct?Wait, the problem says the plane is perpendicular to vector BC. The normal vector of the plane must be parallel to vector BC. Since vector BC is (-3,-5,-3), then the normal vector is (-3,-5,-3). Therefore, the coefficients in the plane equation must be proportional to (-3,-5,-3). So the equation is -3x -5y -3z + d =0. When we plug in point A, we get d= -10, so equation is -3x -5y -3z -10=0. Therefore, this is the correct equation. The other equation, 3x +5y +3z +10=0, has a normal vector (3,5,3), which is opposite to BC. But since the problem says "perpendicular to vector BC", meaning the normal vector is BC, so the first equation is correct.Therefore, the answer should be -3x -5y -3z -10=0.But let me check again. Suppose I use the point-normal form: n ⋅ (r - r0) =0. n is vector BC, which is (-3,-5,-3). r0 is point A(0,-8,10). So, (-3,-5,-3) ⋅ (x -0, y - (-8), z -10)=0.Which is (-3)(x) + (-5)(y +8) + (-3)(z -10)=0.Expanding: -3x -5y -40 -3z +30=0.Combine constants: -40 +30 = -10, so -3x -5y -3z -10=0. Correct.Yes, so this is the right equation. Therefore, the answer is -3x -5y -3z -10=0. However, in some cases, equations are written with all terms on one side and zero on the other, so that's acceptable. Alternatively, if the problem expects the answer in a simplified form without leading negatives, maybe factor out a -1. But since they didn't specify, I think both forms are acceptable, but it's safer to present it as derived from the point-normal form without changing the signs.So the equation is -3x -5y -3z -10=0. Alternatively, arranging terms to the other side: -3x -5y -3z =10. But again, both are correct. However, in standard form, the equation is usually written as ax + by + cz + d =0. So maybe write -3x -5y -3z -10=0. Alternatively, multiply both sides by -1 to make the coefficients positive: 3x +5y +3z +10=0. However, as mentioned, this changes the direction of the normal vector, but since the problem specifies that the plane is perpendicular to vector BC, which is (-3,-5,-3), so the normal vector should be exactly BC. Therefore, using the original coefficients is more appropriate here. Hence, -3x -5y -3z -10=0.But let me check once again. If I substitute point A into this equation:Left-hand side: -3*0 -5*(-8) -3*10 -10 =0 +40 -30 -10=0. Correct.Therefore, the equation is correct. So the answer is -3x -5y -3z -10 =0.Alternatively, written as 3x +5y +3z +10=0, but as I checked, substituting point A into that equation also gives 0. So both equations are correct. But since the problem states the plane is perpendicular to vector BC, which is (-3,-5,-3), the answer should use that vector as the normal vector, so the equation with coefficients matching BC is preferred. Hence, -3x -5y -3z -10=0 is the accurate equation. So I'll go with that.Final AnswerThe equation of the plane is boxed{-3x - 5y - 3z - 10 = 0}.</think>To find the equation of the plane passing through point ( A(0, -8, 10) ) and perpendicular to the vector ( overrightarrow{BC} ), we start by determining the vector ( overrightarrow{BC} ).Given points ( B(-5, 5, 7) ) and ( C(-8, 0, 4) ), the vector ( overrightarrow{BC} ) is calculated as:[overrightarrow{BC} = C - B = (-8 - (-5), 0 - 5, 4 - 7) = (-3, -5, -3)]This vector ( overrightarrow{BC} ) serves as the normal vector ( mathbf{n} ) to the plane. The general equation of a plane with normal vector ( mathbf{n} = (a, b, c) ) passing through point ( (x_0, y_0, z_0) ) is:[a(x - x_0) + b(y - y_0) + c(z - z_0) = 0]Substituting ( mathbf{n} = (-3, -5, -3) ) and point ( A(0, -8, 10) ), we get:[-3(x - 0) - 5(y - (-8)) - 3(z - 10) = 0]Simplifying this equation:[-3x - 5(y + 8) - 3(z - 10) = 0]Expanding and combining like terms:[-3x - 5y - 40 - 3z + 30 = 0][-3x - 5y - 3z - 10 = 0]Thus, the equation of the plane is:[boxed{-3x - 5y - 3z - 10 = 0}],<think>Okay, let me see. I need to find the equation of a plane that passes through point A and is perpendicular to the vector BC. Hmm. Alright, let's start by recalling some basic concepts about planes and vectors. First, the general equation of a plane is usually written as ax + by + cz + d = 0, where (a, b, c) is the normal vector to the plane. Since the plane is perpendicular to vector BC, that means vector BC is the normal vector to the plane. So I need to find vector BC first.Given points B and C: B(-5, 5, 7) and C(-8, 0, 4). To find vector BC, I subtract the coordinates of point B from point C, right? So BC = C - B. Let me compute that:The x-component: -8 - (-5) = -8 + 5 = -3The y-component: 0 - 5 = -5The z-component: 4 - 7 = -3So vector BC is (-3, -5, -3). Therefore, the normal vector to the plane is (-3, -5, -3). But wait, the problem says the plane passes through point A(0, -8, 10). So once I have the normal vector, I can use the point-normal form of the plane equation. The formula is: a(x - x0) + b(y - y0) + c(z - z0) = 0Where (a, b, c) is the normal vector and (x0, y0, z0) is a point on the plane. Plugging in the values: -3(x - 0) + (-5)(y - (-8)) + (-3)(z - 10) = 0Simplify each term:First term: -3xSecond term: -5(y + 8) = -5y - 40Third term: -3(z - 10) = -3z + 30So combining all these:-3x -5y -40 -3z +30 = 0Combine like terms:-3x -5y -3z -10 = 0Hmm, but usually, the plane equation is written with the constant term on the right side. Maybe I should rearrange it. Let me check:Wait, the standard form is ax + by + cz + d = 0, so the equation is -3x -5y -3z -10 = 0. But sometimes people prefer the coefficients to be positive if possible. However, since the normal vector is given, and it's (-3, -5, -3), it's okay if the coefficients are negative. Alternatively, I can multiply both sides by -1 to make the coefficients positive, but that's optional. Let me see if that's necessary. The problem didn't specify, so either form should be correct. But maybe I should present it in the form without negative coefficients. Let's see:Multiplying the entire equation by -1:3x + 5y + 3z + 10 = 0But wait, if I do that, the normal vector would be (3, 5, 3), which is in the opposite direction of the original vector BC. However, in plane equations, the normal vector's direction doesn't affect the plane itself; it's just a matter of convention. So both equations represent the same plane. But since the problem says the plane is perpendicular to vector BC, which is (-3, -5, -3), so the normal vector should be exactly BC. But if I multiply by -1, the normal vector would be in the opposite direction. Hmm, actually, in the plane equation, the normal vector can be either direction, as long as it's perpendicular. But since the problem specifically says "perpendicular to vector BC", that means the normal vector is BC, so we should use the original coefficients. So maybe I shouldn't multiply by -1. Let me check the original equation again.Original equation after plugging in:-3x -5y -3z -10 = 0Alternatively, we can write it as:-3x -5y -3z = 10But typically, people write the constant term on the right side. However, the standard form is ax + by + cz + d = 0. So moving 10 to the left gives -3x -5y -3z -10 = 0. That's okay.But maybe they want it in the form without the negative signs. Alternatively, we can write it as 3x +5y +3z +10 = 0. Wait, if I move all terms to the other side:-3x -5y -3z -10 = 0 is equivalent to 3x +5y +3z +10 = 0. But then the normal vector would be (3,5,3). However, since vector BC is (-3, -5, -3), which is the opposite direction. Is this a problem? Let me recall: the normal vector can be any scalar multiple. So (3,5,3) is a scalar multiple (-1 times) of (-3, -5, -3). So both are valid normal vectors. The plane's equation is the same. So maybe either form is acceptable.But the problem says the plane is perpendicular to vector BC. So if BC is the normal vector, then technically, the coefficients should correspond to BC. But in the equation, the normal vector is (a,b,c) from ax + by + cz + d =0. So if we use the original BC components, the equation is -3x -5y -3z -10 =0, which is correct. Alternatively, writing 3x +5y +3z +10=0 is also correct, but with the normal vector pointing in the opposite direction. But the problem states "perpendicular to vector BC". So the normal vector is vector BC. Therefore, the equation should have the normal vector as BC. So perhaps we should present the equation with the original BC components. Therefore, the equation is -3x -5y -3z -10 =0. But let me check once more.Alternatively, maybe there was a miscalculation in the earlier steps. Let me verify.Starting again:Normal vector BC is (-3, -5, -3). Point A is (0, -8, 10). Using the point-normal form:-3(x - 0) + (-5)(y - (-8)) + (-3)(z - 10) =0Which simplifies to:-3x -5(y +8) -3(z -10) =0Expanding each term:-3x -5y -40 -3z +30 =0Combine constants: -40 +30 = -10Therefore: -3x -5y -3z -10 =0Yes, that's correct. So the equation is -3x -5y -3z -10 =0. Alternatively, if we want to write it as 3x +5y +3z +10 =0, but the problem didn't specify any particular form, so both are correct. However, since the normal vector is BC, which is (-3, -5, -3), maybe we should leave it as is. But let's see if there's a standard convention.In many textbooks, the plane equation is written with the normal vector coefficients as given. So if BC is the normal vector, then the coefficients should match BC. So in this case, the equation should indeed be -3x -5y -3z -10 =0. Alternatively, it's also common to write the equation without negative coefficients if possible, but in this case, since all coefficients are negative, you can factor out a negative sign. But I think the answer is better left as -3x -5y -3z -10 =0. Alternatively, if the problem expects the answer in a certain format, maybe with positive coefficients. Let me check. If we multiply both sides by -1, we get 3x +5y +3z +10=0. So the normal vector would be (3,5,3). But in the problem statement, it's specified that the plane is perpendicular to vector BC, so the normal vector is BC. Therefore, the coefficients should correspond to BC. Hence, the correct equation is -3x -5y -3z -10=0. But sometimes people prefer to have the constant term positive. Let me see. Let me rearrange the equation:-3x -5y -3z =10But in standard form, it's usually written with all terms on the left. So -3x -5y -3z -10=0 is the standard form.Alternatively, maybe the answer is expected to have all terms positive. But in that case, multiplying by -1 gives 3x +5y +3z +10=0. Which is also correct. So which one is preferable?Looking back at how the normal vector is derived. Since the plane is perpendicular to BC, the normal vector is BC. So the coefficients of the plane equation must be the components of BC. So the correct equation is -3x -5y -3z -10=0. But maybe the problem expects the answer in a simplified form, or they might consider the direction of the normal vector irrelevant. However, since it's specified as perpendicular to BC, which is a specific vector, maybe the coefficients should match. So I think the correct answer is -3x -5y -3z -10=0. Alternatively, let me check with another approach. Suppose I use the general plane equation formula. The plane equation can be written as:(BC vector) ⋅ (r - A) =0Where r is any point (x, y, z) on the plane, and A is the given point. So expanding that:(-3, -5, -3) ⋅ (x -0, y +8, z -10)=0Dot product:-3*(x -0) + (-5)*(y +8) + (-3)*(z -10) =0Which is the same as:-3x -5y -40 -3z +30 =0Simplifies to:-3x -5y -3z -10 =0Same result. So that's correct. Therefore, the equation is indeed -3x -5y -3z -10 =0. Alternatively, if I factor out the negative sign, but as I thought earlier, since the problem specifies the normal vector as BC, which is (-3, -5, -3), the coefficients in the equation should correspond to that. Therefore, the answer should be -3x -5y -3z -10 =0.But maybe, just to confirm, let's check if point A satisfies this equation. Let's plug in A(0, -8, 10):Left side: -3*0 -5*(-8) -3*10 -10 = 0 +40 -30 -10 =0. Yes, it works. What about points B and C? Wait, the plane is supposed to pass through A, but not necessarily through B or C. Let's check if vector BC is indeed perpendicular to the plane. Take two points on the plane and find the vector between them, then check the dot product with BC. But since we know the normal vector is BC, that's sufficient. Alternatively, maybe to check, take another point on the plane. Let's see. Suppose x=1, y=0, z=0. Plug into the equation: -3*1 -5*0 -3*0 -10 = -3 -10 = -13 ≠0. So not on the plane. But point A is (0, -8,10). Maybe another way to check: Let's see if the normal vector is indeed BC. Since the normal vector is (-3, -5, -3), which matches BC, so that's correct. Therefore, the equation is -3x -5y -3z -10=0. However, in some contexts, it's preferred to write the equation without leading negative signs. So multiplying by -1, we get 3x +5y +3z +10=0, which is also acceptable. Let me check if that works as well.Plugging in point A(0, -8,10):3*0 +5*(-8) +3*10 +10 =0 -40 +30 +10=0. Yes, that's 0. So that's correct as well. So both forms are correct. Depending on the convention, either equation is acceptable. But since the problem states the plane is perpendicular to vector BC, which is (-3, -5, -3), the equation with coefficients matching BC would be -3x -5y -3z -10=0. However, in many textbooks and resources, it's common to write the equation with the constant term positive if possible, which would require multiplying by -1. Wait, but when you multiply by -1, you invert the direction of the normal vector. So if the problem specifies that the plane is perpendicular to vector BC, which is a specific vector, then using the original direction is important. However, in the context of plane equations, the direction of the normal vector doesn't affect the plane's orientation in terms of being perpendicular; it's just the line of the vector. So either normal vector direction is acceptable. Therefore, both equations are correct. But the question is, which one is expected as an answer? Since the problem mentions "perpendicular to vector BC", which is (-3, -5, -3), but also, the answer could be presented in a simplified form with positive coefficients. Looking at examples from various sources, usually, the plane equation is written with the smallest possible coefficients without common factors, and if possible, leading coefficient positive. In this case, the coefficients -3, -5, -3 have a common factor of -1. If we factor that out, we get 3,5,3 and the constant term becomes +10. So the equation 3x +5y +3z +10=0. But to confirm, let's check if this is equivalent. The original equation is -3x -5y -3z -10=0. Multiplying by -1 gives 3x +5y +3z +10=0. Both equations represent the same plane. Given that, it's customary to present the equation with positive leading coefficients if possible. Therefore, the answer is likely 3x +5y +3z +10=0. But to be thorough, let's see if the problem has any specific instructions. The problem says, "Write the equation of the plane passing through point A and perpendicular to the vector BC." There's no mention of simplifying or making coefficients positive. However, in standard mathematical practice, it's common to present the equation in the form where the coefficients don't all have a common factor (other than 1) and where the leading coefficient is positive if possible. In this case, the coefficients in 3x +5y +3z +10=0 have no common factor (3,5,3,10 share no common divisor other than 1), so that's acceptable. Whereas the original equation's coefficients (-3, -5, -3, -10) also share a common factor of -1. So factoring out -1 gives the positive coefficients. Therefore, the most standard form would be 3x +5y +3z +10=0. But to confirm, let's check both equations with point A:Original equation: -3*0 -5*(-8) -3*10 -10 =0 +40 -30 -10=0. Correct.Positive coefficients equation: 3*0 +5*(-8) +3*10 +10=0 -40 +30 +10=0. Correct.So both are correct. But the positive coefficient version is more standard. Therefore, the answer should be 3x +5y +3z +10=0. Alternatively, if the question expects the answer in the form with the specific normal vector BC, then it would be the original equation. But given that vector BC is (-3, -5, -3), and the problem states "perpendicular to the vector BC", which is equivalent to having BC as the normal vector, but the plane equation can have either direction for the normal vector. However, in some contexts, especially in exams or textbooks, they might prefer the equation with positive coefficients. Therefore, considering common practices, the answer is 3x +5y +3z +10=0.Final AnswerThe equation of the plane is boxed{3x + 5y + 3z + 10 = 0}.</think>To find the equation of the plane passing through point ( A(0, -8, 10) ) and perpendicular to the vector ( overrightarrow{BC} ), we start by determining the vector ( overrightarrow{BC} ).Given points ( B(-5, 5, 7) ) and ( C(-8, 0, 4) ), we calculate ( overrightarrow{BC} ) as follows:[overrightarrow{BC} = C - B = (-8 - (-5), 0 - 5, 4 - 7) = (-3, -5, -3)]This vector ( overrightarrow{BC} ) is the normal vector to the plane. Using the point-normal form of the plane equation with point ( A(0, -8, 10) ), we have:[-3(x - 0) - 5(y + 8) - 3(z - 10) = 0]Expanding and simplifying:[-3x - 5y - 40 - 3z + 30 = 0 implies -3x - 5y - 3z - 10 = 0]To present the equation with positive coefficients, we multiply the entire equation by -1:[3x + 5y + 3z + 10 = 0]Both forms are correct, but the equation with positive coefficients is more standard. Therefore, the final answer is:[boxed{3x + 5y + 3z + 10 = 0}]]

🔑:## SolutionLet's find the vector overrightarrow{B C}:overrightarrow{B C}={-8-(-5) ; 0-5 ; 4-7}={-3 ;-5 ;-3}Since the vector overrightarrow{B C} is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:begin{aligned}& -3 cdot(x-0)-5 cdot(y-(-8))-3 cdot(z-10)=0 & -3 x-5 y-40-3 z+30=0 & -3 x-5 y-3 z-10=0 & 3 x+5 y+3 z+10=0end{aligned}## Problem Kuznetsov Analytic Geometry 9-23