🧠:[<think>Okay, so I need to find the limit as x approaches -1 of [(x³ - 2x -1)(x +1)] divided by (x⁴ +4x² -5). Hmm, let me start by understanding what's going on here. Limits often involve substituting the value into the function, but if that leads to an indeterminate form like 0/0, I might need to simplify the expression somehow. Let me check if substituting x = -1 gives an indeterminate form.First, plugging x = -1 into the numerator: ( (-1)^3 -2*(-1) -1 )*( -1 +1 ). Let's compute each part. (-1)^3 is -1, then 2*(-1) is -2, so subtracting that would be -(-2) which is +2. So -1 +2 -1 is 0. Then multiply by (-1 +1) which is 0. So numerator is 0.Denominator: (-1)^4 +4*(-1)^2 -5. (-1)^4 is 1, 4*(-1)^2 is 4*1 =4. So 1 +4 -5 =0. So yes, it's 0/0, which is indeterminate. That means I need to manipulate the expression to resolve the indeterminate form. Probably factor both numerator and denominator and cancel out the common terms.Let me start with the numerator: (x³ -2x -1)(x +1). Maybe I can factor x³ -2x -1. Let's see if x = -1 is a root here. Plugging x = -1 into x³ -2x -1: (-1)^3 -2*(-1) -1 = -1 +2 -1 = 0. Oh, so x = -1 is a root of the cubic polynomial in the numerator. That means (x +1) is a factor of x³ -2x -1. So maybe I can factor the cubic into (x +1)(quadratic). Let's perform polynomial division or use synthetic division.Let me divide x³ -2x -1 by (x +1). Using synthetic division for root x = -1:Coefficients of x³ +0x² -2x -1. So coefficients are 1, 0, -2, -1.Set up synthetic division:-1 | 1   0   -2   -1    |     -1   1   1    -----------------      1  -1   -1   0So the cubic factors into (x +1)(x² -x -1). Therefore, the numerator is (x +1)(x² -x -1)*(x +1) = (x +1)^2 (x² -x -1).So numerator simplifies to (x +1)^2 (x² -x -1).Now the denominator is x⁴ +4x² -5. Let's try to factor this quartic polynomial. Maybe substitute y =x², so it becomes y² +4y -5. Let's factor this quadratic in y: y² +4y -5. Find two numbers that multiply to -5 and add to 4. Hmm, 5 and -1? Yes: (y +5)(y -1). So x⁴ +4x² -5 = (x² +5)(x² -1). Then, x² -1 factors further into (x -1)(x +1). So denominator is (x² +5)(x -1)(x +1).So denominator factors into (x +1)(x -1)(x² +5).So now, the original expression becomes:[(x +1)^2 (x² -x -1)] / [ (x +1)(x -1)(x² +5) ]We can cancel out one (x +1) from numerator and denominator, leaving:[(x +1)(x² -x -1)] / [ (x -1)(x² +5) ]Now, since we are taking the limit as x approaches -1, and after canceling, the remaining expression is continuous at x = -1 (since denominator at x=-1 would be (-1 -1)( (-1)^2 +5 ) = (-2)(1 +5) = (-2)(6) = -12, which is not zero). So we can substitute x = -1 into the simplified expression.Compute the simplified function at x = -1:Numerator: (-1 +1)( (-1)^2 - (-1) -1 ) = (0)(1 +1 -1) = 0*1 =0Wait, but the denominator was -12. So the simplified expression is 0 / (-12) =0. But wait, hold on. Did I make a mistake here?Wait, no. Let me check again. Wait, after canceling, the numerator is (x +1)(x² -x -1). So when x approaches -1, the numerator becomes (0)( (1 - (-1) -1 ) = (0)(1 +1 -1) =0*(1)=0. The denominator is (x -1)(x² +5) which is (-2)(1 +5)= -12. So the limit is 0 / (-12) =0. But that contradicts the initial thought that after factoring, we have a 0/0 form. Wait, but after canceling, the expression is no longer 0/0. So perhaps the limit is 0.Wait, but let me double-check my factoring steps because I might have made a mistake.Original numerator: (x³ -2x -1)(x +1). Factored the cubic as (x +1)(x² -x -1). So numerator becomes (x +1)^2 (x² -x -1). Denominator factors as (x² +5)(x² -1) = (x² +5)(x -1)(x +1). Therefore, canceling one (x +1) gives (x +1)(x² -x -1) / [ (x -1)(x² +5) ]. So yes, that's correct.Then plugging x = -1 into numerator: ( -1 +1 )*( ( (-1)^2 - (-1) -1 ) = 0*(1 +1 -1) =0. Denominator: ( -1 -1 )*( ( (-1)^2 +5 ) = (-2)*(1 +5)= -12. So 0 / -12 =0. So the limit is 0. But wait, let me verify this because sometimes after canceling, you might still have some hidden factors, but in this case, no. The remaining (x +1) in the numerator becomes 0 when x approaches -1, and the denominator is non-zero. So the limit should be 0.But let me check another way, maybe using L'Hospital's Rule. Since the original limit is 0/0, maybe differentiate numerator and denominator. But since we have multiple variables, maybe need to apply L'Hospital's Rule multiple times. Wait, but the original expression is [(x³ -2x -1)(x +1)] / (x⁴ +4x² -5). Let's call the numerator N(x) and denominator D(x). Then, since N(-1) =0 and D(-1)=0, so apply L'Hospital's Rule once: take derivatives of numerator and denominator.First, compute N'(x):N(x) = (x³ -2x -1)(x +1). Use product rule: N’(x) = (3x² -2)(x +1) + (x³ -2x -1)(1). Evaluate at x = -1:First term: (3*(-1)^2 -2)(-1 +1) = (3*1 -2)(0) = (1)(0)=0Second term: ( (-1)^3 -2*(-1) -1 )*1 = (-1 +2 -1)=0. So N’(-1)=0+0=0.Similarly, compute D’(x): derivative of x⁴ +4x² -5 is 4x³ +8x. Evaluate at x=-1: 4*(-1)^3 +8*(-1) = -4 -8= -12.So applying L’Hospital’s Rule gives 0 / (-12)=0. So same result. Therefore, the limit is 0.But wait, since after first derivative, it's still 0/-12, so the limit is 0.Alternatively, let's check if maybe I made a mistake in the factoring. Wait, when I factor the denominator x⁴ +4x² -5, I set y =x², so y² +4y -5 factors into (y +5)(y -1), which is correct. So x⁴ +4x² -5 = (x² +5)(x² -1) = (x² +5)(x -1)(x +1). Correct.In the numerator, factoring x³ -2x -1 by (x +1). We found that using synthetic division, the cubic factors into (x +1)(x² -x -1). Let me verify that:(x +1)(x² -x -1) = x³ -x² -x +x² -x -1 =x³ -2x -1. Yes, that's correct.Therefore, the numerator is (x +1)^2 (x² -x -1). Correct. Then denominator is (x +1)(x -1)(x² +5). So after canceling, we have (x +1)(x² -x -1)/[(x -1)(x² +5)]. Then when x approaches -1, substituting gives 0 in numerator and -12 in denominator. Therefore, the limit is 0. So both factoring and L’Hospital’s Rule confirm that the limit is 0.But wait, just to make sure, let's try plugging in a value close to -1, say x = -1.1, and see what the function approaches.Compute numerator: ( (-1.1)^3 -2*(-1.1) -1 )*( -1.1 +1 )First compute (-1.1)^3: -1.331Then -2*(-1.1)=2.2So -1.331 +2.2 -1= (-1.331 -1) +2.2= -2.331 +2.2= -0.131Multiply by (-1.1 +1)= -0.1So numerator: -0.131*(-0.1)=0.0131Denominator: (-1.1)^4 +4*(-1.1)^2 -5Compute (-1.1)^4= (1.1)^4=1.46414*(1.21)=4.84So 1.4641 +4.84 -5= 6.3041 -5=1.3041So the value is approx 0.0131 /1.3041≈0.01, which is close to 0. Similarly, trying x = -0.9:Numerator: ( (-0.9)^3 -2*(-0.9) -1 )*( -0.9 +1 )First, (-0.9)^3= -0.729-2*(-0.9)=1.8So -0.729 +1.8 -1=0.071Multiply by (0.1): 0.071*0.1=0.0071Denominator: (-0.9)^4 +4*(-0.9)^2 -5=0.6561 +4*(0.81)=0.6561 +3.24=3.8961 -5= -1.1039Wait, but denominator is x⁴ +4x² -5. If x=-0.9, then x⁴ is positive, 4x² is positive, so 0.6561 +3.24=3.8961 -5= -1.1039. So denominator is negative. So numerator is positive (0.0071) and denominator is negative, so overall value is approx -0.0064. Wait, but approaching from the right side (x approaches -1 from above, like -0.9 is to the right of -1 on the number line). So the limit from the right is approaching -0.0064, and from the left (like x=-1.1) was positive 0.01. But the actual limit should be 0. Maybe these approximations aren't precise enough? Wait, but when x approaches -1, the simplified expression after factoring is (x +1)(x² -x -1)/[(x -1)(x² +5)]. So as x approaches -1, (x +1) approaches 0, and the rest approaches [ ( (-1)^2 - (-1) -1 ) / ( (-1 -1)( (-1)^2 +5 ) ) ] = [ (1 +1 -1)/(-2*(1 +5)) ] = (1)/(-12). So the limit is 0 * (1/-12) =0. So regardless of the side, it's approaching 0. The negative sign in the approximation when approaching from the right might be because (x +1) is positive when x > -1, but (x -1) is negative, and (x² -x -1) when x is near -1: let's compute x² -x -1 at x near -1. At x=-1, it's 1 - (-1) -1=1. So near x=-1, x² -x -1 is approximately 1. Therefore, (x +1) is small, so the whole expression is (x +1)*(approx 1)/(-2*6)= (x +1)*(-1/12). So as x approaches -1, (x +1) approaches 0, hence the entire expression approaches 0. The sign difference from left and right would be due to (x +1) being negative when approaching from the left (x < -1) and positive from the right. But since we multiply by (x +1), the overall limit from both sides is 0. So even if one side is positive approaching 0 and the other negative approaching 0, the limit is still 0.Therefore, after all these checks, I think the limit is indeed 0.Final AnswerThe limit is boxed{0}.</think>To calculate the limit:[lim _{x rightarrow-1} frac{left(x^{3}-2 x-1right)(x+1)}{x^{4}+4 x^{2}-5}]we start by checking if substituting ( x = -1 ) results in an indeterminate form. Both the numerator and denominator evaluate to 0, confirming a 0/0 indeterminate form. Next, we factor both the numerator and the denominator. The numerator ((x^3 - 2x - 1)(x + 1)) is factored by recognizing ( x = -1 ) as a root of the cubic polynomial ( x^3 - 2x - 1 ). Using synthetic division, we factor the cubic polynomial as ((x + 1)(x^2 - x - 1)). Thus, the numerator becomes ((x + 1)^2(x^2 - x - 1)).For the denominator ( x^4 + 4x^2 - 5 ), we substitute ( y = x^2 ) and factor the quadratic ( y^2 + 4y - 5 ) into ((y + 5)(y - 1)). This gives the denominator as ((x^2 + 5)(x^2 - 1)), which further factors into ((x + 1)(x - 1)(x^2 + 5)).Canceling the common factor ((x + 1)) from the numerator and denominator, we get:[frac{(x + 1)(x^2 - x - 1)}{(x - 1)(x^2 + 5)}]Substituting ( x = -1 ) into the simplified expression, the numerator evaluates to 0 (due to the factor ( x + 1 )) and the denominator evaluates to (-12). Thus, the limit is:[frac{0}{-12} = 0]Verification using L'Hospital's Rule confirms the same result. The limit is:[boxed{0}],<think>Okay, so I need to find the limit as x approaches -1 of the function [(x³ - 2x - 1)(x + 1)] divided by (x⁴ + 4x² - 5). Hmm, let's start by plugging in x = -1 into the function to see what happens. If I substitute, both the numerator and denominator might become zero, which would mean we have an indeterminate form 0/0. In that case, I can use L'Hôpital's Rule or factor both the numerator and denominator to cancel out the common terms. Let me check.First, let's plug in x = -1 into the numerator: ( (-1)^3 - 2*(-1) - 1 ) * (-1 + 1). Calculating the first part: (-1) + 2 - 1 = 0. Then, multiplied by (0) gives 0. So numerator is 0.Now the denominator: (-1)^4 + 4*(-1)^2 - 5. That's 1 + 4*1 -5 = 1 + 4 -5 = 0. So yes, it's 0/0. Indeterminate form. So I need to factor both numerator and denominator.Let's start with the numerator: (x³ - 2x -1)(x + 1). Maybe I can factor x³ -2x -1. Let's see if x = -1 is a root here. Plugging into x³ -2x -1: (-1)^3 -2*(-1) -1 = -1 +2 -1 = 0. Oh, so x = -1 is a root of x³ -2x -1. That means (x + 1) is a factor. So we can factor x³ -2x -1 as (x +1)(something). Let's perform polynomial division or use synthetic division.Using synthetic division for polynomial x³ -2x -1 divided by (x +1). The coefficients are 1 (x³), 0 (x²), -2 (x), -1 (constant term). So coefficients: 1, 0, -2, -1.Divide by x = -1. Bring down the 1. Multiply by -1: -1. Add to next term: 0 + (-1) = -1. Multiply by -1: 1. Add to next term: -2 +1 = -1. Multiply by -1: 1. Add to last term: -1 +1 = 0. So the quotient is x² -x -1. Therefore, x³ -2x -1 = (x +1)(x² -x -1).So the numerator becomes (x +1)(x² -x -1)(x +1) = (x +1)²(x² -x -1).Now the denominator: x⁴ +4x² -5. Let me try to factor this. Let’s set y = x², so the equation becomes y² +4y -5. Factor that: looking for two numbers that multiply to -5 and add to 4. Hmm, 5 and -1. So y² +4y -5 = (y +5)(y -1). Substitute back y = x²: (x² +5)(x² -1). Then, x² -1 factors further as (x -1)(x +1). So denominator is (x² +5)(x -1)(x +1).Therefore, the denominator is (x +1)(x -1)(x² +5).Now, the original function after factoring becomes:Numerator: (x +1)²(x² -x -1)Denominator: (x +1)(x -1)(x² +5)We can cancel out one (x +1) term from numerator and denominator, so we have:[(x +1)(x² -x -1)] / [(x -1)(x² +5)]Now, since we're taking the limit as x approaches -1, we can substitute x = -1 into this simplified expression, since the (x +1) term has been canceled and the remaining denominator isn't zero (let's check):Denominator after substitution: (-1 -1)( (-1)^2 +5 ) = (-2)(1 +5) = (-2)(6) = -12Numerator after substitution: ( (-1) +1 )( (-1)^2 - (-1) -1 ) = (0)(1 +1 -1) = 0 *1 = 0Wait, so numerator becomes 0 and denominator becomes -12. So 0 / (-12) = 0. Therefore, the limit is 0. But wait, that seems too straightforward. Let me verify.Wait, hold on. When I canceled the (x +1) term, the remaining expression is (x +1)(x² -x -1) in the numerator and (x -1)(x² +5) in the denominator. So when substituting x = -1 into the simplified expression:Numerator: (-1 +1)( (-1)^2 - (-1) -1 ) = (0)(1 +1 -1) = 0*(1) = 0Denominator: (-1 -1)( (-1)^2 +5 ) = (-2)(1 +5) = -12So the simplified expression approaches 0/-12 as x approaches -1, which is 0. Therefore, the limit is 0. That makes sense. So the answer is 0.But let me double-check the factoring steps to make sure I didn't make a mistake.First, factoring the numerator: original numerator (x³ -2x -1)(x +1). We found that (x +1) is a factor of x³ -2x -1, which gave us (x +1)(x² -x -1). Then multiplied by (x +1) gives (x +1)^2 (x² -x -1). That seems correct.Denominator: x⁴ +4x² -5. Letting y = x², so y² +4y -5. Factoring into (y +5)(y -1), which is correct because 5*-1 = -5 and 5 + (-1) =4. Then substituting back, (x² +5)(x² -1) which factors into (x² +5)(x -1)(x +1). Correct.So after canceling, we have (x +1) in the numerator and denominator cancels, leaving (x +1)(x² -x -1)/[(x -1)(x² +5)]. Then substituting x = -1 gives 0 in numerator and -12 in denominator. Therefore, limit is 0. So the answer is 0.Alternatively, I could use L'Hôpital's Rule since it was 0/0. Let's try that to confirm.Original function: [(x³ -2x -1)(x +1)] / [x⁴ +4x² -5]. Let’s compute derivatives of numerator and denominator.First, derivative of numerator: d/dx [ (x³ -2x -1)(x +1) ]Use product rule: ( derivative of first term * second term ) + ( first term * derivative of second term )First term: x³ -2x -1, derivative: 3x² -2Second term: x +1, derivative:1So numerator derivative: (3x² -2)(x +1) + (x³ -2x -1)(1)At x = -1:First part: (3*(-1)^2 -2)(-1 +1) = (3*1 -2)(0) = (1)(0) = 0Second part: ( (-1)^3 -2*(-1) -1 ) = (-1 +2 -1) = 0So numerator derivative is 0 +0 =0. Hmm, still 0.Denominator derivative: d/dx [x⁴ +4x² -5] =4x³ +8xAt x =-1: 4*(-1)^3 +8*(-1) = -4 -8 = -12So using L’Hospital’s Rule once gives 0/-12, which is 0. But since after first derivative, numerator is 0 and denominator is -12, the limit is 0/-12 =0. So same result. Therefore, confirms the limit is 0.Alternatively, if after first derivative, we still had 0/0, we would have to differentiate again. But here it's 0/non-zero, so we can stop.Therefore, both methods confirm the limit is 0.Final AnswerThe limit is boxed{0}.</think>To calculate the limit:[lim _{x rightarrow-1} frac{left(x^{3}-2 x-1right)(x+1)}{x^{4}+4 x^{2}-5}]we start by substituting ( x = -1 ) into the function. Both the numerator and the denominator evaluate to 0, resulting in an indeterminate form ( frac{0}{0} ). We factor both the numerator and the denominator to simplify the expression.Factoring the numerator:The numerator is ( (x^3 - 2x - 1)(x + 1) ). We factor ( x^3 - 2x - 1 ) using synthetic division, which gives:[x^3 - 2x - 1 = (x + 1)(x^2 - x - 1)]Thus, the numerator becomes:[(x + 1)(x^2 - x - 1)(x + 1) = (x + 1)^2(x^2 - x - 1)]Factoring the denominator:The denominator is ( x^4 + 4x^2 - 5 ). Substituting ( y = x^2 ), we get:[y^2 + 4y - 5 = (y + 5)(y - 1)]Substituting back ( y = x^2 ), we factor:[x^4 + 4x^2 - 5 = (x^2 + 5)(x^2 - 1) = (x^2 + 5)(x - 1)(x + 1)]Simplifying the expression:After factoring, the function becomes:[frac{(x + 1)^2(x^2 - x - 1)}{(x + 1)(x - 1)(x^2 + 5)}]We cancel the common ( (x + 1) ) term:[frac{(x + 1)(x^2 - x - 1)}{(x - 1)(x^2 + 5)}]Substituting ( x = -1 ):The simplified expression is evaluated at ( x = -1 ):- Numerator: ( (-1 + 1)((-1)^2 - (-1) - 1) = 0 times 1 = 0 )- Denominator: ( (-1 - 1)((-1)^2 + 5) = (-2)(6) = -12 )Thus, the limit evaluates to:[frac{0}{-12} = 0]Using L'Hôpital's Rule confirms the result, as differentiating the numerator and denominator once each and substituting ( x = -1 ) also yields 0 in the numerator and -12 in the denominator.The limit is:[boxed{0}]]

🔑:## Solutionbegin{aligned}& lim _{x rightarrow-1} frac{left(x^{3}-2 x-1right)(x+1)}{x^{4}+4 x^{2}-5}=left{frac{0}{0}right}=lim _{x rightarrow-1} frac{left(x^{3}-2 x-1right)(x+1)}{left(x^{3}-x^{2}+5 x-5right)(x+1)}= & =lim _{x rightarrow-1} frac{x^{3}-2 x-1}{x^{3}-x^{2}+5 x-5}=frac{(-1)^{3}-2(-1)-1}{(-1)^{3}-(-1)^{2}+5(-1)-5}= & =frac{-1+2-1}{-1-1-5-5}=frac{0}{-12}=0end{aligned}## Problem Kuznetsov Limits 10-1

🧠:[<think>Okay, let's try to figure out this problem. So, there's a rectangle with a square cut out, and the area of the resulting shape is 369 cm². The shortened sides are 21 cm and 9 cm. We need to find the area of the cut-out square. Hmm, okay. Let me visualize this. If there's a rectangle and a square is cut out from it, the resulting shape would have some missing part. The area of the remaining shape is given, so maybe the original area of the rectangle minus the area of the square equals 369 cm²? That makes sense. So, if I can find the original area of the rectangle and subtract the square's area, it should equal 369. But how do I find the original rectangle's dimensions?The problem mentions "shortened sides" of 21 cm and 9 cm. I think that means after cutting out the square, the remaining sides of the rectangle are 21 cm and 9 cm. But wait, when you cut out a square from a rectangle, you're removing a square from one of the corners, right? So, the original rectangle would have two sides that are longer by the length of the square's side. Let me draw this in my mind. Suppose the original rectangle has length L and width W. Then, we cut out a square of side length x from one corner. The remaining shape would have sides: the original length minus x, and the original width minus x? Wait, no. If you cut out a square from a corner, the remaining sides would be (L - x) and (W - x), but depending on which corner you cut from. But actually, maybe both sides get reduced by x? Wait, no. Let me think again.Imagine the rectangle with length L and width W. If you cut out a square of side x from, say, the top right corner. Then, the remaining figure would have one side that's L - x (the original length minus the square's side) and the width would be W - x (the original width minus the square's side). But then, the problem says the shortened sides are 21 cm and 9 cm. So, maybe after cutting out the square, the two adjacent sides of the remaining shape are 21 cm and 9 cm. That is, L - x = 21 and W - x = 9? If that's the case, then the original rectangle's length and width would be 21 + x and 9 + x, respectively.Then, the original area of the rectangle is (21 + x)(9 + x), and the area of the square is x². So, according to the problem, the area of the resulting shape is 369 cm². Therefore, the original area minus the square's area equals 369. So:(21 + x)(9 + x) - x² = 369Let me compute that. Expanding the original area:(21*9) + 21x + 9x + x² - x² = 369Simplify:189 + 30x + x² - x² = 369The x² terms cancel out:189 + 30x = 369Then, subtract 189 from both sides:30x = 369 - 18930x = 180Divide both sides by 30:x = 180 / 30 = 6So, the side length of the square is 6 cm. Therefore, the area of the square is 6² = 36 cm².Wait, let me check if that makes sense. Let's verify.Original rectangle dimensions: 21 + 6 = 27 cm and 9 + 6 = 15 cm. So, original area is 27*15 = 405 cm². Subtract the square's area 36 cm², 405 - 36 = 369 cm², which matches the given area. So that works.But wait, let me make sure I interpreted the problem correctly. The image shows a rectangle with a square cut out, and the lengths of the shortened sides are 21 cm and 9 cm. Maybe the shortened sides refer to the sides of the remaining figure adjacent to where the square was cut out. So, when you remove the square, the remaining figure has two sides that are shorter than the original by x. That interpretation leads us to the correct answer, and the verification checks out.Alternatively, is there another way to interpret the problem? For example, maybe the original rectangle has sides 21 cm and 9 cm, and after cutting out the square, the area becomes 369 cm². But wait, if the original sides were 21 and 9, then the original area would be 189, which is less than 369, so that can't be. So, the original rectangle must be larger, and after cutting out the square, the remaining area is 369. So, my first interpretation seems correct.Therefore, the area of the cut-out square is 36 cm².Final AnswerThe area of the cut-out square is boxed{36} mathrm{cm}^{2}.</think>Given a rectangle with a square cut out, the area of the resulting shape is 369 cm². The lengths of the shortened sides are 21 cm and 9 cm. We need to find the area of the cut-out square.1. Let the side length of the square be ( x ) cm.2. The original dimensions of the rectangle are ( 21 + x ) cm and ( 9 + x ) cm.3. The area of the original rectangle is ( (21 + x)(9 + x) ) cm².4. The area of the cut-out square is ( x^2 ) cm².5. The area of the resulting shape is given by the equation: [ (21 + x)(9 + x) - x^2 = 369 ]6. Expanding and simplifying the equation: [ 189 + 21x + 9x + x^2 - x^2 = 369 ] [ 189 + 30x = 369 ]7. Solving for ( x ): [ 30x = 180 ] [ x = 6 ]8. The area of the cut-out square is ( 6^2 = 36 ) cm².Verification:- Original rectangle dimensions: ( 21 + 6 = 27 ) cm and ( 9 + 6 = 15 ) cm.- Original area: ( 27 times 15 = 405 ) cm².- Remaining area: ( 405 - 36 = 369 ) cm², which matches the given area.The area of the cut-out square is boxed{36} (mathrm{cm}^{2}).,<think>Okay, let's see. There's a rectangle with a square cut out, and the area of the resulting shape is 369 cm². The lengths of the shortened sides are 21 cm and 9 cm. I need to find the area of the cut-out square. Hmm, let me try to visualize this. So, imagine a rectangle. If you cut out a square from one of its corners, the resulting shape would have two sides that are shorter than the original rectangle. The problem mentions the shortened sides are 21 cm and 9 cm. That probably means that after cutting out the square, the remaining parts of the original rectangle's length and width are 21 cm and 9 cm. Wait, but how exactly is the square cut out? Let me think. If you remove a square from a corner, both the length and the width of the original rectangle get reduced by the side length of the square. Let me denote the side length of the square as 's'. Then the original rectangle's length would be 21 cm plus 's', and the original width would be 9 cm plus 's'. Because when you cut out a square of side 's', the remaining length is 21 cm and the remaining width is 9 cm. So original dimensions: (21 + s) cm and (9 + s) cm. The area of the original rectangle would be (21 + s)(9 + s). Then, subtracting the area of the square, which is s², should give the area of the resulting shape, which is 369 cm². Wait, but is that right? Wait, no. If you have a rectangle and you cut out a square from one corner, the area of the resulting shape is just the area of the rectangle minus the area of the square. So the formula would be:Area of resulting shape = Area of original rectangle - Area of squareSo, (21 + s)(9 + s) - s² = 369Let me compute that. Let's expand (21 + s)(9 + s):21*9 + 21s + 9s + s² = 189 + 30s + s²Then subtract s²: 189 + 30s + s² - s² = 189 + 30sSo, 189 + 30s = 369Solving for s:30s = 369 - 189 = 180Therefore, s = 180 / 30 = 6So the side length of the square is 6 cm, which makes the area of the square 6² = 36 cm².Wait, let me check that again. Let's plug back into the original equation. Original rectangle dimensions would be 21 + 6 = 27 cm and 9 + 6 = 15 cm. Area of original rectangle is 27*15 = 405 cm². Subtract the square area, which is 36 cm², gives 405 - 36 = 369 cm². Yes, that matches the given area. So the answer should be 36 cm².But let me make sure there's no other interpretation. The problem says "the lengths of the shortened sides are 21 cm and 9 cm". So after cutting the square, the remaining sides are 21 and 9. So the original sides would be 21 + s and 9 + s. Then the original area minus the square area equals 369. Yes, that seems correct. Another way could be that the figure after cutting out the square has some sides of 21 and 9. Maybe it's a composite shape? Let me think.Alternatively, maybe the resulting figure is an L-shape, and the two legs of the L-shape have lengths 21 and 9? But then how does that relate to the area? Let me consider that possibility. Suppose we have a rectangle, and from one corner, we cut out a square. The remaining shape would have two sides: one of length original length minus s, and the original width; and another side of original width minus s and the original length? Hmm, not sure. Wait, maybe the sides that are shortened are adjacent sides? Like, the original rectangle had length L and width W. After cutting out a square of side s, the new lengths are L - s and W - s? Then the area would be (L)(W) - s² = 369. But then the shortened sides are 21 and 9. So L - s = 21 and W - s = 9. Then L = 21 + s and W = 9 + s. Then the area is (21 + s)(9 + s) - s² = 369. Which is exactly the same equation as before. So expanding that gives 21*9 + 21s + 9s + s² - s² = 189 + 30s = 369. So 30s = 180, s = 6. Therefore, area of the square is 36. So that's the same answer. So regardless of the interpretation, it seems the answer is 36. Is there another way to approach it? Maybe by considering the remaining area. The L-shape area can also be calculated as the sum of two rectangles. One rectangle is 21 cm by (9 + s) cm, and the other is 9 cm by s cm. Wait, no, maybe not. Let me visualize again. If you have a big rectangle, and you cut out a square from the corner, the remaining shape would consist of two rectangles: one that's (original length - s) by original width, and another that's s by (original width - s). Wait, no. Let me think. Suppose the original rectangle is length L and width W. We cut out a square of side s from the corner. Then the remaining area is L*W - s². But if we break down the remaining shape, it can be considered as a larger rectangle (L - s) by W plus a smaller rectangle s by (W - s). Wait, but that would be overlapping. Alternatively, it's two rectangles: one with dimensions (L - s) by W, and another with dimensions s by (W - s). But actually, no, that's not correct because cutting out the square would leave an L-shape which is (L - s) by W plus s by (W - s). Wait, maybe. Let me check:Total area of L-shape = (L - s)*W + s*(W - s) - but wait, that would be overlapping. Wait, no. Actually, the L-shape can be divided into two rectangles: one with length L and width (W - s), and the other with length (L - s) and width s. So total area would be L*(W - s) + (L - s)*s. Let's compute that:L*(W - s) + s*(L - s) = LW - Ls + Ls - s² = LW - s². Which is the same as the original area minus the square. So that checks out. Alternatively, if the problem states that the lengths of the shortened sides are 21 cm and 9 cm, perhaps those are the dimensions of the two rectangles in the L-shape. So one part is 21 cm long and the other is 9 cm long. Wait, but the problem says "the lengths of the shortened sides are 21 cm and 9 cm". So maybe the sides that were cut are now shorter, so the original sides were longer by 's', and now they're 21 and 9. So original length was 21 + s, original width was 9 + s. Then the area is (21 + s)(9 + s) - s² = 369. Which again gives 189 + 30s = 369, so s = 6. Therefore, area of square is 36. Therefore, the answer should be 36. Wait, is there a way this could be a trick question? For example, if the square is cut from the middle, but the problem says "a square cut out", and the image shows a rectangle with a square cut out. Typically, cutting out a square from a corner would lead to the remaining sides being shortened by the square's side length. So the original sides are longer by 's'. So I think the first approach is correct. Let me just verify once more. If the square has side 6, original rectangle is 27 by 15. Area 405. Subtract 36, get 369. Correct. The shortened sides are 27 - 6 = 21 and 15 - 6 = 9, which matches the problem statement. So yes, that's consistent. Therefore, the area of the cut-out square is 36 cm².Final AnswerThe area of the cut-out square is boxed{36} mathrm{cm}^{2}.</think>Given a rectangle with a square cut out, the area of the resulting shape is 369 cm². The lengths of the shortened sides are 21 cm and 9 cm. We need to find the area of the cut-out square.1. Let the side length of the square be ( s ).2. The original dimensions of the rectangle are ( (21 + s) ) cm and ( (9 + s) ) cm.3. The area of the original rectangle is ( (21 + s)(9 + s) ).4. The area of the resulting shape is the area of the original rectangle minus the area of the square, which is ( (21 + s)(9 + s) - s^2 ).5. This area is given as 369 cm², so we set up the equation: [ (21 + s)(9 + s) - s^2 = 369 ]6. Expanding and simplifying the equation: [ 189 + 21s + 9s + s^2 - s^2 = 369 ] [ 189 + 30s = 369 ]7. Solving for ( s ): [ 30s = 180 ] [ s = 6 ]8. The area of the square is ( s^2 = 6^2 = 36 ) cm².Thus, the area of the cut-out square is (boxed{36}) cm².]

🔑:5. Let the side of the square be denoted by a, and the area of the shape by P.We divide the obtained shape into three parts whose areas are P_{1}, P_{2}, and P_{3}.[^0][^0]: The area P_{1} is P_{1}=21 mathrm{~cm} cdot 9 mathrm{~cm}=189 mathrm{~cm}^{2}. 2 POINTS Then, P_{2}+P_{3}=P-P_{1}=369 mathrm{~cm}^{2}-189 mathrm{~cm}^{2}=180 mathrm{~cm}^{2}. 2 POINTS It holds that: P_{2}=21 a, P_{3}=9 a, so P_{2}+P_{3}=21 a+9 a=30 a. 2 POINTS From 30 a=180 we calculate a=180: 30=6 mathrm{~cm}. 2 POINTS The area of the square is 36 mathrm{~cm}^{2}. 2 POINTS TOTAL 10 POINTS

🧠:[<think>Okay, so there's this problem about lotuses in a lake. Let me read it again to make sure I understand. The problem says that each lotus splits in half every day, so one becomes two, and this happens every day. After 30 days, the lake is completely covered with lotuses. The question is asking how long it took for the lake to be half full. Hmm.Let me think. If the lotuses double each day, then the number of lotuses on day n is double the number from day n-1. So, for example, if on day 1 there's 1 lotus, on day 2 there are 2, day 3 there are 4, and so on. So it's exponential growth, right? Each day it's multiplied by 2.Now, if the lake is full on day 30, that means the number of lotuses on day 30 is exactly enough to cover the lake. Then, the day before, which is day 29, the number of lotuses would have been half of that, because they double each day. So, if day 30 is full, day 29 must be half full. Wait, let me check that logic. Let's say the lake's capacity is C lotuses. On day 30, the number of lotuses is C. Then on day 29, since they double each day, the number must have been C/2. Therefore, the lake was half full on day 29. But maybe I should model it with equations to be sure. Let's let the number of lotuses on day n be N(n). According to the problem, N(n) = 2 * N(n-1). So this is a recurrence relation where each day's population is twice the previous day's. Assuming that on day 0, there was 1 lotus. Then on day 1, it becomes 2, day 2, 4, and so on. So, in general, N(n) = 2^n. If the lake is full on day 30, that means N(30) = 2^30. So the capacity of the lake is 2^30 lotuses. Therefore, the lake is half full when N(n) = 2^30 / 2 = 2^29. But N(n) = 2^n, so we set 2^n = 2^29. Solving for n gives n = 29. So, it took 29 days for the lake to be half full. But wait, is there a different starting point? The problem doesn't specify how many lotuses there were initially. It just says "each lotus splits in half" each day. So maybe the initial number is 1? But perhaps it doesn't matter. Because regardless of the initial number, the doubling each day would mean that the day before it's full, it's half. For example, suppose there was more than one lotus initially. Let's say there were k lotuses on day 0. Then on day 1, there would be 2k, day 2: 4k, etc. So on day n, there are k * 2^n lotuses. The lake is full on day 30, so k * 2^30 is the total capacity. Then, the lake is half full when the number of lotuses is (k * 2^30)/2 = k * 2^29. Which occurs on day 29, since k * 2^29 is exactly the number on day 29. So regardless of the initial number k, the day before it's full, it's half. Therefore, the answer should be 29 days. But let me check with another example. Suppose we start with 1 lotus. On day 1: 2 lotuses, day 2: 4, day 3: 8... So each day, it's doubling. If day 30 is full, then day 29 is half. If day 30 has 2^30 lotuses, day 29 has 2^29, which is half. So yeah, exactly. Alternatively, if you think about it in reverse. If the lake is full on day 30, then each day prior, it's half. So to go backward, each day the number of lotuses is halved. So day 30: 100% full, day 29: 50%, day 28: 25%, etc. So yes, the lake was half full on day 29. Therefore, the answer should be 29 days.</think>The lotuses double in number each day, meaning the lake's coverage progresses exponentially. If the lake is fully covered on day 30, it must have been half covered the day before, since doubling half the lake's capacity results in full coverage the next day. Answer: boxed{29},<think>Okay, so there's this problem about lotuses in a lake. Let me try to understand it step by step. The problem says that each lotus splits in half every day, meaning each one becomes two after a day. Then, the next day, those two split again, and so on. After 30 days, the lake is completely covered with lotuses. The question is, how long did it take for the lake to be half full?Hmm. Let me think. So, the lotuses are doubling every day. That sounds like exponential growth. If they double each day, then the number of lotuses on day n is 2^n times the original number. Wait, but we don't know the original number. Maybe it starts with one lotus? Or maybe some other number? Hmm, the problem doesn't specify. Let me check the problem again."In the lake, lotuses grow. Over the course of a day, each lotus splits in half, and instead of one lotus, two appear. After another day, each of the resulting lotuses splits in half, and so on. After 30 days, the lake is completely covered with lotuses. How long did it take for the lake to be half full?"So, it doesn't mention the initial number of lotuses. But maybe we don't need it. Because the key here is that the growth is doubling every day. So, if on day 30 the lake is full, then on day 29, it must have been half full. Because the next day, it doubles to become full. Similarly, day 28 would be a quarter, and so on. So, this is a classic problem where the lake is half covered the day before it's fully covered. So, the answer would be 29 days. Let me verify that.Suppose on day 1, there is 1 lotus. Then day 2: 2 lotuses, day 3: 4, day 4: 8, etc. So, on day n, there are 2^(n-1) lotuses. Wait, if day 1 is 1, day 2 is 2, then day 30 would be 2^29 lotuses. So, if the lake is full on day 30, then on day 29, there would be half that number, which is 2^28. So, day 29 is half full. Therefore, the answer is 29 days.But let me think again. Maybe the problem starts with more than one lotus? Wait, the problem says "each lotus splits in half," so maybe the initial number is one? Because otherwise, if there were, say, x lotuses initially, then after n days, the number would be x*2^n. So, if after 30 days it's full, then x*2^30 = full. Then the day before, x*2^29 would be half full. So regardless of the initial number x, the day before it's half full. So, whether the initial number is 1 or some other number, the time to go from half to full is one day. So, since it took 30 days to be full, it was half full at 29 days.Therefore, the answer should be 29 days.Wait, but let me check with a smaller number of days to see if my logic holds. Suppose the lake is full on day 3. Then on day 2, it's half full. Let's see: start with 1 lotus. Day 1: 1 becomes 2. Day 2: 2 becomes 4. Day 3: 4 becomes 8. If the lake is full on day 3, then on day 2, it's half full. So, the answer would be day 2. That seems to hold. So, if full on day n, half on day n-1.Therefore, in the original problem, since it's full on day 30, half full is day 29. So the answer is 29 days.</think>The lotuses double in number each day. If the lake is full on day 30, then the day before (day 29), it must have been half full because doubling from half results in the full coverage the next day. Answer: boxed{29}]

🔑:Note that the number of lotus flowers doubles in one day.## SolutionIf you read the condition of the problem carefully, you will understand that the lake was half full after 29 days. The day before the lake is completely filled, it will be exactly half full.## AnswerAfter 29 days

🧠:[<think>Okay, let me try to solve this problem step by step. The function f is defined on natural numbers with two conditions:1. f(1) = 12. For n ≥ 1, - f(2n) = f(n) - f(2n + 1) = f(2n) + 1We need to find the maximum value u of f(n) for 1 ≤ n ≤ 1989 and how many numbers n satisfy f(n) = u.First, let me understand how the function f works. It seems recursive. Let's try to compute some values manually to see the pattern.Starting with n=1:- f(1) = 1 (given)n=2:- Since 2 is even, f(2) = f(2/2) = f(1) = 1n=3:- 3 is odd, so f(3) = f(2*1 + 1) = f(2*1) + 1 = f(2) + 1 = 1 + 1 = 2n=4:- Even, so f(4) = f(4/2) = f(2) = 1n=5:- Odd, f(5) = f(4) + 1 = 1 + 1 = 2n=6:- Even, f(6) = f(3) = 2n=7:- Odd, f(7) = f(6) + 1 = 2 + 1 = 3n=8:- Even, f(8) = f(4) = 1n=9:- Odd, f(9) = f(8) + 1 = 1 + 1 = 2n=10:- Even, f(10) = f(5) = 2n=11:- Odd, f(11) = f(10) + 1 = 2 + 1 = 3n=12:- Even, f(12) = f(6) = 2n=13:- Odd, f(13) = f(12) + 1 = 2 + 1 = 3n=14:- Even, f(14) = f(7) = 3n=15:- Odd, f(15) = f(14) + 1 = 3 + 1 = 4n=16:- Even, f(16) = f(8) = 1Hmm, interesting. The function seems to reset to 1 at powers of 2. For example, n=2,4,8,16 all have f(n)=1. Then, when n is odd, it's f(previous even) +1, which seems to be counting something. Let me check what this represents.Looking at the values, f(n) might be the number of 1s in the binary representation of n. Let's verify:n=1: 1 in binary is 1 → one 1. f(1)=1 ✔️n=2: 10 → one 1. f(2)=1 ✔️n=3: 11 → two 1s. f(3)=2 ✔️n=4: 100 → one 1. f(4)=1 ✔️n=5: 101 → two 1s. f(5)=2 ✔️n=6: 110 → two 1s. f(6)=2 ✔️n=7: 111 → three 1s. f(7)=3 ✔️n=8: 1000 → one 1. f(8)=1 ✔️n=9: 1001 → two 1s. f(9)=2 ✔️n=10: 1010 → two 1s. f(10)=2 ✔️n=11: 1011 → three 1s. f(11)=3 ✔️n=12: 1100 → two 1s. f(12)=2 ✔️n=13: 1101 → three 1s. f(13)=3 ✔️n=14: 1110 → three 1s. f(14)=3 ✔️n=15: 1111 → four 1s. f(15)=4 ✔️n=16: 10000 → one 1. f(16)=1 ✔️Yes! So f(n) is indeed the number of 1s in the binary representation of n. That makes sense because:- When n is even (i.e., ends with 0 in binary), removing the trailing 0 (which is dividing by 2) doesn't change the number of 1s. So f(2n) = f(n).- When n is odd (i.e., ends with 1 in binary), writing it as 2n + 1, then f(2n +1) = f(2n) +1. Since 2n is even, f(2n) is f(n), and adding 1 to the count (the trailing 1). So yes, this counts the number of 1s.Therefore, the problem reduces to finding the maximum number of 1s in the binary representations of numbers from 1 to 1989, and how many numbers have that many 1s.So, first step: find the maximum number of 1s in the binary representation of numbers up to 1989.Let me find the binary representation of 1989.Divide 1989 by 2 repeatedly:1989 ÷2 = 994 rem 1994 ÷2 = 497 rem 0497 ÷2 = 248 rem 1248 ÷2 = 124 rem 0124 ÷2 = 62 rem 062 ÷2 = 31 rem 031 ÷2 = 15 rem 115 ÷2 = 7 rem 17 ÷2 = 3 rem 13 ÷2 = 1 rem 11 ÷2 = 0 rem 1So writing the remainders from last to first: 11111000101Let's check: 11111000101.Compute its decimal value:From right to left (positions 0 to 10):Positions with 1s:Position 0: 1 (2^0 = 1)Position 2: 1 (2^2 = 4)Position 6: 1 (2^6 = 64)Positions 7: 1 (2^7 = 128)Position 8: 1 (2^8 = 256)Position 9: 1 (2^9 = 512)Position 10: 1 (2^10 = 1024)Sum: 1 + 4 + 64 + 128 + 256 + 512 + 1024 = Let's compute step by step:1 + 4 = 55 + 64 = 6969 + 128 = 197197 + 256 = 453453 + 512 = 965965 + 1024 = 1989. Perfect.So binary is 11111000101, which is 11 bits. Let's count the number of 1s: 1,1,1,1,1,0,0,0,1,0,1 → total 7 ones.So the binary representation of 1989 has 7 ones. Therefore, the maximum number of 1s possible in numbers up to 1989 is at least 7. But is it possible to have more?Wait, but 1989 is 11111000101. To have more than 7 ones, the number must be less than or equal to 1989. Let's check the next number after 1989. For example, 1989 +1 = 1990 is 11111000110. That has 6 ones. Hmm.Wait, actually, numbers with higher number of ones would have to be smaller numbers, because adding more ones in binary would make the number larger. Wait, that's not necessarily true. For example, 1023 is 1111111111 (10 ones), but 1023 is less than 1989.Wait, 1023 is 10 ones in binary. So if 1023 is within 1 to 1989, then the maximum number of ones would be 10. Wait, but 1023 is 1023, which is less than 1989. So why does 1989 only have 7 ones?Wait, so maybe my initial thought is wrong. The maximum number of 1s is not necessarily in the largest number. For example, 1023 (which is 2^10 -1) has 10 ones. 511 is 2^9 -1 has 9 ones. So numbers of the form 2^k -1 have k ones. So 1023 is 1111111111 (10 ones), which is 1023. Since 1023 ≤1989, then the maximum number of ones would be 10. But wait, 1023 is 10 ones, but 1989 is 7. So maybe the maximum is 10?But wait, 1023 is less than 1989, so it's included in the range. Therefore, the maximum number of ones is 10? But that contradicts with 1989's count. Wait, maybe I miscalculated.Wait, 2^10 is 1024, so 2^10 -1 is 1023. So 1023 is 10 ones. So yes, 1023 is included in 1 to 1989. Therefore, f(1023)=10. Then, numbers larger than 1023 would have more bits, but 1024 is 10000000000 (1 followed by ten 0s), which has only 1 one. Numbers between 1024 and 1989 have 11 bits. Let's check how many 1s can they have.Wait, 1024 to 1989: the first bit is 1 (position 10). Then bits 9 to 0.The maximum number of ones would be 1 (from position 10) plus the maximum number of ones in the remaining 10 bits. But the remaining bits can go up to 1989 - 1024 = 965. Let's see what's the maximum number of 1s in numbers from 0 to 965.Wait, 965 in binary is, let's compute:965 ÷2=482 rem1482 ÷2=241 rem0241 ÷2=120 rem1120 ÷2=60 rem060 ÷2=30 rem030 ÷2=15 rem015 ÷2=7 rem17 ÷2=3 rem13 ÷2=1 rem11 ÷2=0 rem1So binary: 1111000101 (positions 0 to 9). Count the ones: 1,1,1,1,0,0,0,1,0,1 → 7 ones. So 965 in binary has 7 ones. Therefore, the maximum number of ones in numbers from 0 to 965 is 7 (for 965). Then, numbers from 1024 to 1989 have binary 1 followed by 10 bits, so the maximum number of 1s would be 1 + 7 = 8. But 1023 has 10 ones. Therefore, the maximum number of ones in 1 to 1989 is 10, achieved at 1023.Wait, but 1023 is included in 1 to 1989. So why does the problem mention 1989? Wait, maybe the maximum is 10, but then why does 1989 have 7?But 1023 is less than 1989. So 1023 is within the range, so the maximum u should be 10. But maybe there are numbers larger than 1023 with more 1s? For example, 1023 + 1 = 1024, which has 1. 1025 is 10000000001, which has 2. 1023 is the maximum.Wait, 1535 is 10111111111 which is 10 ones as well. Wait, 1535 is 1024 + 511 = 1535. 511 is 111111111 (9 ones). So 1535 is 10111111111 (10 ones). So 1535 is less than 1989. So 1535 is another number with 10 ones.Similarly, 1791 is 11011111111 (10 ones). 1791 = 1024 + 512 + 255 = 1791. 255 is 11111111 (8 ones). Wait, maybe 1791 is 11011111111, which is 10 ones. So 1791 is also within 1-1989.Similarly, 1983 is 11110111111 (10 ones). Let's check:1983 in binary: 1983 ÷2=991 rem1; 991÷2=495 rem1; 495÷2=247 rem1; 247÷2=123 rem1; 123÷2=61 rem1; 61÷2=30 rem1; 30÷2=15 rem0; 15÷2=7 rem1; 7÷2=3 rem1; 3÷2=1 rem1; 1÷2=0 rem1.So binary is 11110111111. Let's count the ones: 1,1,1,1,0,1,1,1,1,1,1. That's 10 ones. So 1983 is 11110111111 (10 ones). 1983 is less than 1989, so included. Therefore, numbers up to 1989 can have 10 ones. So the maximum u is 10.Wait, but how many numbers between 1 and 1989 have 10 ones in their binary representations?Wait, numbers with 10 ones must be numbers where the binary representation has exactly 10 ones. Since the maximum number of bits needed for numbers up to 1989 is 11 bits (since 1989 is 11111000101, 11 bits). So numbers with 10 ones in 11 bits would be numbers where one bit is 0 and the rest are 1s. But wait, how many such numbers are there?Wait, the total number of 11-bit numbers with exactly 10 ones is C(11,1) = 11. But these numbers must be less than or equal to 1989. However, some of these numbers may exceed 1989. So we need to count how many 11-bit numbers with exactly 10 ones are ≤1989.Alternatively, maybe numbers with 10 ones can also be in 10 bits. For example, 1023 is a 10-bit number with all 10 bits set to 1. So 1023 is 10 ones. So numbers with 10 ones can be either 10-bit numbers (only 1023) or 11-bit numbers where 10 bits are 1s and 1 bit is 0.Wait, let's clarify. If a number has 10 ones in its binary representation, it can be represented in:- 10 bits: all ones, which is 1023 (as 2^10 -1).- 11 bits: one 0 and ten 1s. How many such numbers are there? There are 11 positions where the 0 can be, so 11 numbers. Each of these numbers is 2^11 -1 - 2^k, where k is the position of the 0 bit (counting from 0). For example, if the 0 is in the least significant bit (position 0), the number is 11111111110 (binary) which is 2046 -1 = 2045. Wait, 2^11 is 2048, so 2^11 -1 is 2047. Subtracting 2^k gives 2047 - 1 = 2046, 2047 -2=2045, etc.Wait, but 2^11 -1 -2^k = 2047 - 2^k. For k from 0 to 10. These numbers are all 11-bit numbers with exactly 10 ones. However, we need to check which of these are ≤1989.So let's compute 2047 - 2^k ≤1989 → 2047 -1989 ≥2^k → 58 ≥2^k. So 2^k ≤58. The values of k where 2^k ≤58 are k=0 (1),1(2),2(4),3(8),4(16),5(32),5(32). 2^5=32, 2^6=64. 64 >58. So k can be 0,1,2,3,4,5. Therefore, there are 6 such positions. Hence, there are 6 numbers in 11-bit numbers with 10 ones that are ≤1989. Additionally, the 10-bit number 1023 (which is 1111111111) is also ≤1989. Therefore, total numbers with 10 ones are 6 +1=7.Wait, let's check:First, the 10-bit number: 1023 (10 ones).Then, the 11-bit numbers with 10 ones where the 0 is in positions 5,4,3,2,1,0. Wait, 2^k ≤58, so k can be up to 5, since 2^5=32, and 2^6=64 which is more than 58. Therefore, the positions k=0 to 5.But the positions in 11-bit numbers: the bits are from position 10 (MSB) down to 0 (LSB). So setting a 0 in position k (0 to 10) gives numbers 2047 -2^k. We need 2047 -2^k ≤1989.So solving 2047 -2^k ≤1989:2047 -1989 = 58 ≤2^k → k ≥ log2(58) ≈5.85. Therefore, k must be ≥6. Wait, but wait, 2047 -2^k ≤1989 ⇒ 2^k ≥58 ⇒ k ≥6 (since 2^6=64). Wait, this contradicts my previous conclusion.Wait, let me re-examine:2047 -2^k ≤1989→ -2^k ≤1989 -2047→ -2^k ≤ -58Multiply both sides by -1 (reverse inequality):2^k ≥58Thus, k must satisfy 2^k ≥58.Compute 2^5=32 <58, 2^6=64 ≥58. Therefore, k ≥6.Therefore, the positions k where 0 is placed must be from 6 to 10. Because in 11-bit numbers, k ranges from 0 (LSB) to 10 (MSB). Wait, but position 10 is the MSB. If we set the 0 in position 10, then the number is 0 followed by ten 1s, which is 01111111111, but that's actually a 11-bit number with MSB 0, which is invalid because numbers are represented without leading zeros. Wait, actually, in binary representations, numbers don't have leading zeros, so setting a 0 in position 10 (the MSB) would effectively make it a 10-bit number. But 2047 -2^10 = 2047 -1024 =1023, which is the 10-bit number with all ones. So actually, the 11-bit numbers with 0 in position 10 is equivalent to the 10-bit number 1023. So we need to be careful here.Therefore, when we compute 2047 -2^k for k=6 to 10:k=6: 2047 -64=1983k=7:2047 -128=1919k=8:2047 -256=1791k=9:2047 -512=1535k=10:2047 -1024=1023Thus, the numbers are 1983,1919,1791,1535,1023. Wait, k=6 gives 1983, which is 11110111111 (binary). Let's check 1983: 1983 ÷2=991 rem1; 991÷2=495 rem1; 495÷2=247 rem1; 247÷2=123 rem1; 123÷2=61 rem1; 61÷2=30 rem1; 30÷2=15 rem0; 15÷2=7 rem1; 7÷2=3 rem1; 3÷2=1 rem1; 1÷2=0 rem1. So binary: 11110111111, which is 11 bits with a 0 at position 6 (counting from 0). So 10 ones.Similarly, k=7: 2047 -128=1919. Binary: Let's divide by 2:1919 ÷2=959 rem1959 ÷2=479 rem1479 ÷2=239 rem1239 ÷2=119 rem1119 ÷2=59 rem159 ÷2=29 rem129 ÷2=14 rem114 ÷2=7 rem07 ÷2=3 rem13 ÷2=1 rem11 ÷2=0 rem1So binary:11101111111. Here, the 0 is at position 7. So 10 ones.Similarly, k=8: 1791 is 11011111111 (0 at position8)k=9:1535 is 10111111111 (0 at position9)k=10:1023 is 01111111111 (0 at position10, which is actually 10-bit 1111111111)So these are all valid numbers. So numbers with 10 ones in 11-bit numbers where the 0 is in positions 6 to 10. That gives 5 numbers (k=6 to k=10). But also, 1023 (k=10) is a 10-bit number. So total numbers with 10 ones are 5 (from 11-bit) +1 (1023) =6? Wait, but 1023 is included when k=10. So 2047 -2^10 =1023, which is the 10-bit number.Therefore, the total numbers with 10 ones are 5 (from k=6 to k=10) +1 (the 10-bit 1023). Wait, but 1023 is already counted when k=10. Wait, no. When k=10, the 0 is in the 10th position (MSB of 11-bit number), which effectively removes the leading 1, resulting in 10-bit 1111111111. So actually, there are 5 numbers with 11 bits (1983,1919,1791,1535,1023), but 1023 is also a 10-bit number. Wait, but in 11-bit numbers, 1023 is 01111111111, which is the same as the 10-bit number. So in the range 1 to1989, 1023 is counted once. So the numbers with 10 ones are:- 1023 (10-bit)- 1983,1919,1791,1535 (11-bit with 0 in positions6,7,8,9)Wait, when k=10, we get 1023, which is the same as the 10-bit number. So perhaps the 11-bit numbers with 10 ones that are ≤1989 are 1983,1919,1791,1535, and 1023 (but 1023 is already included as 10-bit). However, 1023 as an 11-bit number would be 01111111111, which is equal to 1023. So perhaps there's overlap.But how does this affect the count? Since 1023 is only one number, whether we count it as a 10-bit or an 11-bit number, it's still just one instance. Therefore, maybe the total numbers with 10 ones are:1 (1023) +4 (1983,1919,1791,1535) =5? Wait, but when k=6 to10, that's 5 numbers. However, 1023 (k=10) is 1023. But in the 11-bit representation, k=10 is the MSB. Setting that to 0 gives 01111111111=1023. So in total, there are 5 numbers with 10 ones: 1983,1919,1791,1535,1023. But 1023 is the same as the 10-bit number. Therefore, total is 5 numbers. Wait, but 1023 is already covered as a 10-bit number. So perhaps it's 5 numbers. Wait, but earlier I thought that k=6 to10 gives 5 numbers, which are 1983,1919,1791,1535,1023. So 5 numbers. So 1023 is included here. So the total number of numbers with 10 ones is 5.Wait, but 1023 is 10 ones (as a 10-bit number). So when we count 11-bit numbers with 10 ones, we have 5 numbers (including 1023). So does that mean the total is 5 numbers?Wait, perhaps we need to check each of these numbers:- 1023: 01111111111 (10 ones)- 1535:10111111111 (10 ones)- 1791:11011111111 (10 ones)- 1919:11101111111 (10 ones)- 1983:11110111111 (10 ones)So those are 5 numbers. But 1023 is also a 10-bit number. So in the range 1-1989, how many numbers have exactly 10 ones?It's these 5 numbers. So the count is 5.But wait, 1023 is a 10-bit number, but when represented in 11 bits, it's 01111111111. So it's considered as an 11-bit number with a leading zero, which is not standard. Therefore, perhaps 1023 is a 10-bit number, and the other four numbers (1535,1791,1919,1983) are 11-bit numbers with 10 ones. Therefore, total numbers with 10 ones are 1 (1023) +4=5.Wait, let's verify:1535 is 10111111111 (10 ones)1791 is 11011111111 (10 ones)1919 is 11101111111 (10 ones)1983 is 11110111111 (10 ones)1023 is 1111111111 (10 ones)So these are 5 distinct numbers. Therefore, there are 5 numbers in 1-1989 with 10 ones. But wait, 1023 is 1023, 1535 is 1535, 1791 is 1791, 1919 is 1919, 1983 is 1983. That's 5 numbers. So the answer would be u=10, count=5.But wait, earlier I thought that numbers like 2047-2^5=2047-32=2015, but 2015 is greater than 1989, so not included. So numbers with 0 in positions k=0 to5 would give numbers greater than 1989. For example, k=5: 2047 -32=2015>1989, so excluded. Therefore, only k=6 to10 are included, giving 5 numbers.Therefore, the maximum u=10, and the count is 5. Wait, but in my initial calculation, when I thought about numbers with 10 ones, I considered 1023 (10-bit) and 4 others (11-bit). But in reality, when k=10, the number is 1023, which is the same as the 10-bit number. So the total numbers with 10 ones are 5.Wait, but let's confirm with another example. For instance, 511 is 111111111 (9 ones). 1023 is the next. 1535 is 10111111111. So in binary, each of these numbers has 10 ones. So total 5 numbers. Therefore, the answer should be u=10, count=5.But wait, let me check another number. For example, 1023 +512=1535, which is 10111111111 (10 ones). Similarly, 1535+256=1791 (11011111111). 1791+128=1919 (11101111111). 1919+64=1983 (11110111111). 1983+64=2047, which is beyond 1989. So these are the 5 numbers:1023,1535,1791,1919,1983.Thus, each of these numbers has 10 ones, and they are all within 1-1989. Therefore, the maximum value u is 10, and there are 5 numbers that achieve this.Wait, but wait. 1023 is 1111111111 (10 ones). 1535 is 10111111111 (10 ones). 1791 is 11011111111 (10 ones). 1919 is 11101111111 (10 ones). 1983 is 11110111111 (10 ones). So that's five numbers. So the count is 5.But I recall that in the problem statement, the maximum value u is to be found for 1 ≤n ≤1989. Wait, but 1983 is 1983, which is less than 1989. So these five numbers are all within the range.But let me check if there are any other numbers with 10 ones. For example, 1279. Let's check 1279 in binary:1279 ÷2=639 rem1639 ÷2=319 rem1319 ÷2=159 rem1159 ÷2=79 rem179 ÷2=39 rem139 ÷2=19 rem119 ÷2=9 rem19 ÷2=4 rem14 ÷2=2 rem02 ÷2=1 rem01 ÷2=0 rem1So binary:100111111111. Wait, that's 12 bits? Wait, 1279 is less than 1989. Let's count the ones: 1,0,0,1,1,1,1,1,1,1,1,1. So 9 ones. Hmm. So no, 1279 has 9 ones.Wait, maybe 1023,1535,1791,1919,1983 are the only ones. Let me check another number, say 1407. Let's compute 1407 in binary:1407 ÷2=703 rem1703 ÷2=351 rem1351 ÷2=175 rem1175 ÷2=87 rem187 ÷2=43 rem143 ÷2=21 rem121 ÷2=10 rem110 ÷2=5 rem05 ÷2=2 rem12 ÷2=1 rem01 ÷2=0 rem1So binary:10101111111. That's 1,0,1,0,1,1,1,1,1,1,1. So 8 ones. Not 10.Hmm. So seems like only those five numbers have 10 ones.Therefore, the answer should be u=10, count=5.But wait, let me verify one more time with the original function f(n). For example, take 1983: f(1983)=number of ones=10.Similarly, f(1535)=10, etc.Yes, so the maximum u is 10, achieved by 5 numbers.Wait, but I just need to confirm if there's any other number with 10 ones. For example, 2047 is 11111111111 (11 ones), but 2047>1989, so not included. The next is 1983=11110111111 (10 ones). So yes, 1983 is the largest number in the range with 10 ones. So that's 5 numbers. So answer is u=10, count=5.Wait, but why does the problem say 1989? Maybe there's a different approach. Let's think again.Alternatively, since f(n) is the number of 1s in binary, the maximum number of 1s is floor(log2(n))+1 only for numbers like 2^k -1, but maybe not. Wait, for example, 1989 is 11 bits, so the maximum number of 1s is 11, but 1989 is only 7. The maximum number of 1s is achieved by numbers that are as close as possible to 2^11 -1=2047, but less than or equal to 1989. So the closest is 1983=2047 -64=11110111111 (10 ones). So 10 is the maximum.Alternatively, maybe considering the binary representation of 1989:11111000101, which has 7 ones. So if we try to find the maximum number of 1s in numbers up to 1989, we need to see the maximum possible. Since 1989 is less than 2048 (2^11), the maximum number of 1s possible is 10, as in 1023 (10 ones), and numbers formed by 11 bits with a single 0. Since 1023 is 10 ones, which is less than 11, but in 11-bit numbers, we can have 10 ones by having one 0.So the maximum number of ones is 10, achieved by numbers:- 1023 (10-bit)- 1535,1791,1919,1983 (11-bit)Total of 5 numbers. Therefore, the answer is u=10, count=5.But let me check another approach. If the function f(n) counts the number of 1s, then perhaps we can model it as the sum of digits in base 2. So the problem is equivalent to finding the maximum digit sum in base 2 for numbers up to 1989, and how many numbers have that sum.Yes, exactly. So, digit sum in base 2 is called the Hamming weight or population count. So the maximum Hamming weight in numbers from 1 to1989 is 10, occurring 5 times.But let's make sure there's no other numbers with 10 ones. For example, let's take 1023 (10 ones), then adding 512 gives 1535 (10 ones), adding 256 to 1535 gives 1791 (10 ones), adding 128 gives 1919 (10 ones), adding 64 gives 1983 (10 ones). Adding next 32 would give 2015, which is over 1989. So those are the four additions after 1023. So total 5 numbers:1023,1535,1791,1919,1983.Thus, the count is indeed 5.Therefore, the maximum value u is 10, and the number of n is 5.But wait, I just need to confirm the count again. Let's list all numbers:1. 1023: 1111111111 (10 ones)2. 1535:10111111111 (10 ones)3. 1791:11011111111 (10 ones)4. 1919:11101111111 (10 ones)5. 1983:11110111111 (10 ones)So yes, five numbers. Therefore, the answer is u=10, count=5.But wait, I just need to check if there are any other numbers in between with 10 ones. For example, 1279 is 100111111111 in binary. Wait, how many ones is that? Let's count:1,0,0,1,1,1,1,1,1,1,1,1. That's 9 ones. So no. Similarly, 1407 is 101011111111, which is 8 ones. No. So no other numbers. Therefore, the answer should be u=10 and count=5.Therefore, the maximum value u is 10, and there are 5 numbers n satisfying f(n)=u.Final AnswerThe maximum value ( u ) is (boxed{10}) and the number of such ( n ) is (boxed{5}).</think>The function ( f ) is defined on the set of natural numbers with the following conditions:1. ( f(1) = 1 )2. ( f(2n) = f(n) ) and ( f(2n+1) = f(2n) + 1 ) for ( n geq 1 ).We need to find the maximum value ( u ) of ( f(n) ) for ( 1 leq n leq 1989 ) and determine how many ( n ) satisfy ( f(n) = u ).First, we recognize that ( f(n) ) counts the number of 1s in the binary representation of ( n ). This is because:- For even ( n ), the binary representation ends in 0, so removing the last 0 (dividing by 2) does not change the count of 1s.- For odd ( n ), the binary representation ends in 1, so removing the last 1 (subtracting 1 and then dividing by 2) reduces the count by 1.To find the maximum number of 1s in the binary representations of numbers from 1 to 1989, we consider the binary representation of 1989, which is ( 11111000101 ) (7 ones). However, numbers like 1023 (binary ( 1111111111 )) have 10 ones, which is the highest possible count within the range.We identify the numbers with 10 ones in their binary representations:1. ( 1023 ) (binary ( 1111111111 ))2. ( 1535 ) (binary ( 10111111111 ))3. ( 1791 ) (binary ( 11011111111 ))4. ( 1919 ) (binary ( 11101111111 ))5. ( 1983 ) (binary ( 11110111111 ))These numbers are derived by setting one bit to 0 in an 11-bit number (except for 1023, which is a 10-bit number). Each of these numbers has exactly 10 ones in their binary representation.Thus, the maximum value ( u ) is 10, and there are 5 such numbers.[boxed{10} quad text{and} quad boxed{5}],<think>Okay, so I have this problem here about a function defined on natural numbers with some recursive conditions. Let me try to parse it step by step. The problem says:The function f is defined on the set of natural numbers and satisfies:1. f(1) = 1;2. For n ≥ 1, f(2n) = f(n) and f(2n+1) = f(2n) + 1.We need to find the maximum value u of f(n) for 1 ≤ n ≤ 1989, and determine how many numbers n in that range satisfy f(n) = u.Hmm. Let's start by understanding what this function f does. It seems recursive, where even indices just take the value of half the index, and odd indices take the value of the previous even index plus 1. Let me try computing some small values to see if there's a pattern.Starting with n=1: f(1) = 1.n=2: since 2 is even, f(2) = f(2/2) = f(1) = 1.n=3: 3 is odd, so f(3) = f(2*1 +1) = f(2) +1 = 1 +1 = 2.n=4: even, so f(4) = f(4/2) = f(2) =1.n=5: odd, so f(5) = f(4) +1 =1 +1=2.n=6: even, f(6)=f(3)=2.n=7: odd, f(7)=f(6)+1=2 +1=3.n=8: even, f(8)=f(4)=1.n=9: odd, f(9)=f(8)+1=1+1=2.n=10: even, f(10)=f(5)=2.n=11: odd, f(11)=f(10)+1=2+1=3.n=12: even, f(12)=f(6)=2.n=13: odd, f(13)=f(12)+1=2+1=3.n=14: even, f(14)=f(7)=3.n=15: odd, f(15)=f(14)+1=3+1=4.n=16: even, f(16)=f(8)=1.Hmm, so the function seems to be counting something. Let me see. For even numbers, it's just the function evaluated at half the number. For odd numbers, it's the function at the previous even number (which is half the odd number rounded down) plus 1. Wait a minute, this seems familiar. If we write each number in binary, maybe f(n) is the number of 1s in the binary representation? Let me check:n=1: 1 -> 1 -> 1 one. f(1)=1. Correct.n=2: 10 -> 1 one. f(2)=1. Correct.n=3: 11 -> 2 ones. f(3)=2. Correct.n=4: 100 ->1 one. f(4)=1. Correct.n=5:101 ->2 ones. f(5)=2. Correct.n=6:110 ->2 ones. f(6)=2. Correct.n=7:111->3 ones. f(7)=3. Correct.n=8:1000->1 one. f(8)=1. Correct.n=9:1001->2 ones. f(9)=2. Correct.n=10:1010->2 ones. Correct.n=11:1011->3 ones. Correct.n=12:1100->2 ones. Correct.n=13:1101->3 ones. Correct.n=14:1110->3 ones. Correct.n=15:1111->4 ones. Correct.n=16:10000->1 one. Correct.So indeed, it looks like f(n) counts the number of 1s in the binary representation of n. So the function f(n) is the Hamming weight or the binary weight of n. Therefore, the problem reduces to: For numbers from 1 to 1989 inclusive, what is the maximum number of 1s in their binary representations, and how many numbers have that maximum number of 1s?So now, the problem becomes a binary number problem. We need to find the maximum number of 1s in the binary representations of numbers from 1 up to 1989, and count how many numbers have that many 1s.First, let's figure out the binary representation of 1989 to understand the maximum possible number of 1s.Let me convert 1989 to binary. Let's divide by 2 repeatedly:1989 ÷2 = 994, remainder 1994 ÷2=497, rem 0497 ÷2=248, rem 1248 ÷2=124, rem 0124 ÷2=62, rem 062 ÷2=31, rem 031 ÷2=15, rem 115 ÷2=7, rem 17 ÷2=3, rem 13 ÷2=1, rem 11 ÷2=0, rem 1So writing the remainders from last to first: 1 1 1 1 0 1 1 1 0 0 0 1 0 1Wait, let's count how many times we divided:1989: 1st remainder 1994: 0497:1248:0124:062:031:115:17:13:11:1So starting from the last division, the remainders are 1 (from 1), then 1 (from 3), 1 (from 7), 1 (from 15), 1 (from 31), 0 (from 62), 0 (from 124), 0 (from 248), 1 (from 497), 0 (from 994), 1 (from 1989). Wait, maybe I missed a step. Let's do it step by step:1989 ÷2: quotient 994, remainder 1994 ÷2: quotient 497, remainder 0497 ÷2: quotient 248, remainder 1248 ÷2: quotient 124, remainder 0124 ÷2: quotient 62, remainder 062 ÷2: quotient 31, remainder 031 ÷2: quotient 15, remainder 115 ÷2: quotient 7, remainder 17 ÷2: quotient 3, remainder 13 ÷2: quotient 1, remainder 11 ÷2: quotient 0, remainder 1So writing remainders from last to first: 1 (from last division), then 1,1,1,1,0,1,0,0,0,1,0,1. Wait, that seems 12 bits? Let me check:Let me count the number of divisions:Starting with 1989:1. 1989 /2: rem 12. 994 /2: rem 03. 497 /2: rem 14. 248 /2: rem 05. 124 /2: rem 06. 62 /2: rem 07. 31 /2: rem18.15 /2: rem19.7 /2: rem110.3 /2: rem111.1 /2: rem1So total 11 divisions, so 11 bits. Wait, but when you divide until quotient is 0, the number of bits is the number of divisions. Wait, no. Actually, the number of bits is the number of divisions. For example, 1 is 1 bit, 2 is 2 bits (10), etc. Wait, 1989 is between 2^10=1024 and 2^11=2048, so it should be 11 bits. Let me confirm:2^10 = 10242^11 = 2048So 1989 is less than 2048, so 11 bits. So 1989 in binary is 11111000101.Let me verify:128*15 = 1920, 1989-1920=69. 64 is 64, so 64+5=69. So 11111000101. Let's see:From the remainders above:The remainders collected from last to first (LSB to MSB) would be:1 (from 1), 1 (from 3), 1 (from 7), 1 (from 15), 1 (from 31), 0 (from 62), 0 (from 124), 0 (from 248), 1 (from 497), 0 (from 994), 1 (from 1989). Wait, but that's 11 bits. Wait, writing from last division to first:Starting from the last division (which gave quotient 0), the remainders are:1 (from 1), then 1 (from 3), 1 (from7),1 (from15),1 (from31),0(from62),0(from124),0(from248),1(from497),0(from994),1(from1989). So the binary is 1 1 1 1 1 0 0 0 1 0 1. Let me write this as 11111000101. Let's check this value:Let's compute the value of 11111000101:From right to left (LSB to MSB):Position 0: 1 (2^0 =1)Position 2: 0Position 3:1 (2^3=8)Position 4:0Position 5:0Position 6:0Position 7:1 (2^7=128)Position 8:1 (256)Position 9:1 (512)Position 10:1 (1024)Wait, actually, positions start from 0 at the right. Let me write the binary number as:1 1 1 1 1 0 0 0 1 0 1Starting from the left, the first bit is 2^10:2^10 (1024) + 2^9 (512) + 2^8 (256) + 2^7 (128) + 2^6 (64) + 0 + 0 + 0 + 2^2 (4) + 0 + 2^0 (1)Wait, but hold on, the positions:Bit 10:1 (1024)Bit 9:1 (512)Bit8:1 (256)Bit7:1 (128)Bit6:1 (64)Bit5:0 (32)Bit4:0 (16)Bit3:0 (8)Bit2:1 (4)Bit1:0 (2)Bit0:1 (1)Adding these up: 1024+512+256+128+64+4+1 = Let's compute:1024 + 512 = 15361536 +256=17921792 +128=19201920 +64=19841984 +4=19881988 +1=1989Yes, that's correct. So 1989 in binary is 11111000101, which has 1+1+1+1+1+0+0+0+1+0+1 = 7 ones. Wait, let me count the 1s:Positions 10,9,8,7,6,2,0: that's 7 ones. So the binary representation of 1989 has 7 ones. Hmm. So the maximum number of 1s in numbers up to 1989 would be at least 7.But wait, maybe numbers less than 1989 can have more? For example, 2047 is 11111111111 (11 ones), but 2047 is larger than 1989, so not included. What's the maximum possible number of ones in numbers up to 1989?Since 1989 is 11111000101 (11 bits), the maximum number of ones would be 10? Wait, but 1989 is less than 2048, which is 2^11. So the numbers up to 1989 are 11-bit numbers, but not all 11-bit numbers are allowed. The maximum number of ones would be when the number is as many ones as possible within 11 bits but less than or equal to 1989.Alternatively, since 1989 is 11111000101, which is 11 bits with 7 ones. To get more ones, we need numbers with more 1s but still less than or equal to 1989. For example, if a number has 8 ones. Let's see if such a number exists below 1989.For example, let's take 11111111111 is 2047, too big.What about 11111111101 (2045), still too big.Wait, perhaps numbers with leading bits as 11111, then the rest. Wait, 1989 is 11111000101. So in the first five bits (from the left), it's 11111. Then bits 6-10 are 00010, but actually, in the 11-bit structure, the first five bits are 11111, followed by 000101.Wait, maybe another approach. Let me think. The maximum number of 1s possible in an 11-bit number is 11, but since numbers go up to 1989, which is less than 2048, the maximum possible number of 1s would be 10? Wait, but to have 10 ones, the number would be 11111111110 (2046), which is still larger than 1989. Similarly, 11111111101 (2045), etc., all above 1989. So maybe the maximum number of 1s is 7? But wait, 1989 itself has 7 ones. But maybe there's a number less than 1989 with more ones?Wait, let's check 1535. 1535 in binary is 10111111111, which is 10 ones. Wait, 1535 is less than 1989. Wait, 1535 is 1535. Let me check:1535 ÷2=767 rem1767 ÷2=383 rem1383 ÷2=191 rem1191 ÷2=95 rem195 ÷2=47 rem147 ÷2=23 rem123 ÷2=11 rem111 ÷2=5 rem15 ÷2=2 rem12 ÷2=1 rem01 ÷2=0 rem1So binary is 10111111111. Let's count the ones: starting from the left, 1,0,1,1,1,1,1,1,1,1,1. That's 10 ones. So 1535 is 10111111111 with 10 ones, and 1535 is less than 1989. So the maximum number of ones is at least 10. Wait, that contradicts the previous thought. Wait, 1535 is indeed less than 1989. So how come?Wait, 1535 is 1535, which is 1989 - 1535 = 454 less than 1989. So it's way below. So 1535 is a number with 10 ones. So then the maximum number of ones must be 10. Wait, let me check another number. For example, 1023 is 1111111111 (10 ones), which is 1023, less than 1989. So 1023 has 10 ones. Wait, 1023 is 10 bits, but in 11 bits, it's 01111111111. But leading zeros don't count, so 1023 is 10 ones. So numbers like 1023, 1535, etc., have 10 ones. So maybe the maximum number of ones is 10? But 1535 is 10111111111 which is 10 ones. Let me check 2047 is 11 ones, but too big.So, numbers up to 1989 can have up to 10 ones. Wait, but 1023 is 10 ones. 1023 is 1111111111 (10 bits). Then, 1535 is 10111111111 (11 bits, 10 ones). Similarly, 1989 is 11111000101 (7 ones). So why does 1535 have 10 ones? Because it's 10111111111. So that's 10 ones. So if we can have 10 ones, then the maximum u is 10. But wait, let's check 2047 is 11 ones, but it's over. So 10 is the max. But how?Wait, hold on. If 1023 is 10 ones, and 1535 is also 10 ones. So numbers with 10 ones do exist below 1989. So the maximum u is 10.Wait, but 1989 is 11111000101 (7 ones). So how can there be numbers with 10 ones? Because 10 ones require that in the binary representation, 10 bits are 1. But the numbers with 10 ones would be numbers like 1111111111 (1023), which is 10 bits. But in 11 bits, numbers can have up to 11 ones. But since we are limited to numbers up to 1989, which is 11111000101, maybe numbers with 10 ones are possible. Let me check 1989 - 1 = 1988. What's 1988 in binary? 1988 is 11111000100. That's 6 ones. Hmm. So 1989 and 1988 have 7 and 6 ones respectively.Wait, but 1535 is 10111111111. Let me check 1535 is indeed less than 1989. Yes, 1535 is 1.5k, which is less than 1989. So 1535 has 10 ones. Therefore, the maximum u is 10. So the answer for the maximum value u is 10. Then, how many numbers between 1 and 1989 have exactly 10 ones in their binary representations?So now the problem becomes: count the numbers from 1 to 1989 inclusive that have exactly 10 ones in their binary representations. To do this, we need to consider the numbers in binary with 10 ones and ensure they are ≤1989.First, note that numbers with 10 ones can be represented in 11 bits or fewer. However, numbers with 10 ones in 10 bits (i.e., 1111111111) is 1023, which is 10 ones. Numbers with 10 ones in 11 bits would have a 1 followed by 10 ones, but that would be 2047, which is over 1989. Wait, actually, in 11 bits, the positions are from 0 to 10. To have 10 ones in 11 bits, the number would be 2^11 -1 -2^k for some k? Wait, not exactly.Wait, numbers with exactly 10 ones in 11 bits would have one 0 and ten 1s. There are 11 such numbers (each with a 0 in one of the 11 positions). But these numbers range from 2^11 -1 -2^0 = 2047 -1 = 2046 (11111111110) down to 2047 -2^10 = 2047 - 1024 = 1023 (01111111111). But 1023 is 10 ones in 10 bits (since leading zero doesn't count), but in 11 bits, it's 01111111111. However, 1023 is less than 1989, so numbers like 01111111111 (1023) and 10111111111 (1535), etc., up to 11111111110 (2046). But 2046 is over 1989, so we need to find how many 11-bit numbers with 10 ones are less than or equal to 1989.Alternatively, since 1989 is 11111000101, we can think of numbers with 10 ones in their binary representations as follows:Either:1. 10 ones in 10 bits: which is 1111111111 (1023). But this is only one number.2. 10 ones in 11 bits, where the 11th bit is 1, and the remaining 10 bits have 9 ones. Wait, no. Wait, to have 10 ones in 11 bits, one bit is 0, the rest are 1s.But the numbers with 10 ones in 11 bits are all numbers of the form 2^11 -1 - 2^k for k from 0 to 10. Each of these has exactly 10 ones. But these numbers range from 1023 (k=10) to 2046 (k=0). However, numbers greater than 1989 (which is 11111000101) must be excluded.So first, let's count all 11-bit numbers with 10 ones (there are 11 of them) and subtract those that are greater than 1989. Then add the 10-bit number with 10 ones (which is 1023).But wait, 1023 is included in the 11-bit numbers with 10 ones (as 01111111111), but in reality, when written as an 11-bit number, it's 01111111111, but as a number, it's 1023. So when considering numbers from 1 to 1989, the 11-bit numbers with 10 ones that are ≤1989 are those from 1023 up to 1989. The 11-bit numbers with 10 ones above 1989 are from 1989+1=1990 up to 2046. So we need to find how many of the 11-bit numbers with 10 ones are ≤1989.So total 11-bit numbers with 10 ones: 11 numbers (each missing one bit from 11 bits). Out of these, how many are ≤1989?To find that, we need to check which of the 11 numbers (each with one 0 in different positions) are ≤1989.Given that 1989 is 11111000101.First, let's represent 1989 in binary as 11111000101. Let's index the bits from left (MSB) to right (LSB) as positions 10 to 0.So:Position 10: 1Position 9:1Position8:1Position7:1Position6:1Position5:0Position4:0Position3:0Position2:1Position1:0Position0:1So, in positions 10-0: 1 1 1 1 1 0 0 0 1 0 1.Now, numbers with 10 ones are numbers where exactly one bit is 0. So each such number has a 0 in exactly one position. We need to count how many such numbers are ≤1989.So, for each position k (from 0 to 10), if the number with a 0 in position k is ≤1989, then count it.But note that 1989 has certain bits set. So if we set a 0 in a position where 1989 has a 0, that might not affect the number being less than or equal. Wait, let's think.Alternatively, when constructing a number with 10 ones (i.e., one 0), if the 0 is in a position that, when compared to 1989's binary, makes the number less than or equal.To do this systematically, perhaps we can compare bit by bit.Let me consider each possible position where the 0 can be, and check if the resulting number is ≤1989.Starting from the highest position (position 10):Case 1: 0 in position 10.This gives the number 01111111111, which is 1023. Since 1023 ≤1989, this is valid.Case 2: 0 in position 9.The number would be 10111111111. Let's compute its value:Position 10:1, 9:0, 8:1, 7:1, 6:1, 5:1, 4:1, 3:1, 2:1, 1:1, 0:1.Wait, actually, no. Wait, if we have a 0 in position 9, then the number is:1 0 1 1 1 1 1 1 1 1 1.Wait, but in an 11-bit number, positions 10 to 0.So 0 in position 9: bits are:Position10:1, Position9:0, Positions8-0: all 1s.So the value is 2^10 + (2^8 + 2^7 + ... +2^0) = 1024 + (256 + 128 + 64 + 32 +16 +8 +4 +2 +1) = 1024 + 511 = 1535. 1535 ≤1989, so valid.Case3: 0 in position8.Number: 11011111111.Compute value:2^10 + 2^9 + (2^7 + ... +2^0) = 1024 + 512 + 255 = 1024 +512=1536 +255=1791. 1791 ≤1989. Valid.Case4: 0 in position7.Number: 11101111111.Value: 2^10 +2^9 +2^8 + (2^6 +...+2^0) = 1024+512+256 + 127= 1792 +127=1919 ≤1989. Valid.Case5: 0 in position6.Number: 11110111111.Value: 2^10 +2^9 +2^8 +2^7 + (2^5 +...+2^0)= 1024+512+256+128 + 63= 1920 +63=1983 ≤1989. Valid.Case6: 0 in position5.Number: 11111011111.Wait, position5 is the 6th bit from the left (positions 10-0). Wait, position5 corresponds to 2^5=32.But in the original number 1989, position5 is 0. So if we set position5 to 0 in the 10-ones number, how does that compare?Wait, the number would be 11111011111. Let's compute this:2^10 +2^9 +2^8 +2^7 +2^6 + (2^4 +2^3 +2^2 +2^1 +2^0)= 1024 +512 +256 +128 +64 + (16+8+4+2+1)= 1024+512=1536; 1536+256=1792; 1792+128=1920; 1920+64=1984; 1984 +31=2015.But 2015 is greater than 1989. So this number is 2015, which exceeds 1989. Therefore, setting position5 to 0 gives a number greater than 1989. Therefore, invalid.But wait, let me check again:The number with 0 in position5 is 11111011111.Breaking down:Bits from position10 to0: 1 1 1 1 1 0 1 1 1 1 1.Compute the value:2^10 (1024) + 2^9 (512) +2^8 (256)+2^7 (128)+2^6 (64) +0*32 +2^4 (16)+2^3 (8)+2^2 (4)+2^1 (2)+2^0 (1)So sum: 1024+512+256+128+64+16+8+4+2+1.Let me compute this:1024+512=15361536+256=17921792+128=19201920+64=19841984+16=20002000+8=20082008+4=20122012+2=20142014+1=2015.Yes, 2015. Which is greater than 1989. So this number is invalid.Case7: 0 in position4.Number: 11111101111.Similarly, this would be:1 1 1 1 1 1 0 1 1 1 1.Compute value:2^10 +2^9 +2^8 +2^7 +2^6 +2^5 +0*16 +2^3 +2^2 +2^1 +2^0.=1024+512+256+128+64+32 +0 +8 +4 +2 +1.Sum:1024+512=15361536+256=17921792+128=19201920+64=19841984+32=20162016+8=20242024+4=20282028+2=20302030+1=2031.2031 >1989. So invalid.Case8: 0 in position3.Number:11111110111.Value:1 1 1 1 1 1 1 0 1 1 1.Calculating:2^10+2^9+2^8+2^7+2^6+2^5+2^4 +0*8 +2^2 +2^1 +2^0=1024+512+256+128+64+32+16 +0 +4 +2 +1=1024+512=15361536+256=17921792+128=19201920+64=19841984+32=20162016+16=20322032+4=20362036+2=20382038+1=2039>1989. Invalid.Case9: 0 in position2.Number:11111111011.Value:1 1 1 1 1 1 1 1 0 1 1.Computing:2^10 + ... +2^3 +0*4 +2^1 +2^0=1024+512+256+128+64+32+16+8 +0 +2 +1=1024+512=15361536+256=17921792+128=19201920+64=19841984+32=20162016+16=20322032+8=20402040+0=20402040+2=20422042+1=2043>1989. Invalid.Case10: 0 in position1.Number:11111111101.Value:1 1 1 1 1 1 1 1 1 0 1.Compute:2^10 +...+2^2 +0*2 +2^0.=1024+512+256+128+64+32+16+8+4 +0 +1=1024+512=15361536+256=17921792+128=19201920+64=19841984+32=20162016+16=20322032+8=20402040+4=20442044+0=20442044+1=2045>1989. Invalid.Case11: 0 in position0.Number:11111111110.Value:1 1 1 1 1 1 1 1 1 1 0.Compute:2^10 +...+2^1 +0*1.=1024+512+256+128+64+32+16+8+4+2+0=1024+512=15361536+256=17921792+128=19201920+64=19841984+32=20162016+16=20322032+8=20402040+4=20442044+2=2046>1989. Invalid.So among the 11 numbers with 10 ones (each missing one bit), the first five cases (positions10,9,8,7,6) give numbers ≤1989, and the rest (positions5 to0) give numbers >1989.Wait, let me recap:Positions where 0 is placed:- Position10: 01111111111 (1023) – valid- Position9: 10111111111 (1535) – valid- Position8:11011111111 (1791) – valid- Position7:11101111111 (1919) – valid- Position6:11110111111 (1983) – valid- Position5:11111011111 (2015) – invalid- Position4:11111101111 (2031) – invalid- Position3:11111110111 (2039) – invalid- Position2:11111111011 (2043) – invalid- Position1:11111111101 (2045) – invalid- Position0:11111111110 (2046) – invalidSo only the first five positions (10,9,8,7,6) when zeroed result in numbers ≤1989. Thus, there are 5 numbers with 10 ones in their binary representations that are ≤1989.Additionally, the number 1023 (01111111111) is considered in the 11-bit numbers, but as a 10-bit number, it's 1111111111. But since leading zeros don't count, 1023 is a 10-bit number with 10 ones. However, in our count above, when we considered 0 in position10, that gives 1023, which is the same as 10-bit number. So the total numbers with 10 ones are:- The 5 numbers from 11-bit numbers (with 0 in positions10,9,8,7,6)- The 10-bit number 1023 (which is 1111111111)Wait, but 1023 was already counted in the 11-bit numbers when we put a 0 in position10. So in our earlier count of 11 numbers, 1023 is included as one of them. So actually, the 11 numbers with 10 ones include all numbers from 1023 (01111111111) up to 2046 (11111111110). Therefore, in our previous analysis, the 5 numbers are:1023 (0 in position10),1535 (0 in position9),1791 (0 in position8),1919 (0 in position7),1983 (0 in position6).These are all the 11-bit numbers with 10 ones that are ≤1989. So total of 5 numbers.But wait, 1023 is 01111111111, which as a number is 1023, and when considered in 11 bits, it's allowed. So the total numbers with 10 ones are these 5 numbers. The 10-bit number 1111111111 (1023) is the same as 01111111111 in 11 bits, so it's already included.Therefore, the total number of numbers from 1 to 1989 with exactly 10 ones is 5.But wait, wait. Let me confirm. For example, 1535 is 10111111111. If we write this in binary, how many bits does it have? It's 11 bits. Similarly, 1791 is 11011111111, also 11 bits. 1919 is 11101111111, 1983 is 11110111111, and 1023 is 01111111111. So these are 5 numbers with exactly 10 ones. But is there another number with 10 ones?Wait, for example, 1023 is 10 ones (in 10 bits). 1535 is 10 ones (11 bits). So these are separate numbers. But how do we count them? Since in the 11-bit numbers, each number with 10 ones has exactly one 0. So the numbers are distinct. So 5 numbers in total.But let's check another example. Let's take 2047 - 2^5 = 2047 -32=2015, which is the first invalid number. So indeed, only 5 numbers with 10 ones are valid.Thus, according to this, the maximum u is 10, and the count is 5. But wait, but earlier, we thought that 1023 is 10 ones, and the other numbers 1535,1791,1919,1983. So total 5 numbers.But wait, maybe there are other numbers with 10 ones that are not of the form 11-bit numbers with one 0. For example, can we have a number like 10111111110 (which is 1534). But this number has 9 ones. Wait, no. Wait, 10111111110 is 1 followed by 0, then nine 1s and a 0. So that's two zeros. So it would have 9 ones. So that's less.Alternatively, numbers with multiple zeros but overlapping with the zeros in 1989. Wait, but if we are looking for numbers with exactly 10 ones, they must have exactly one 0. So all numbers with exactly 10 ones in their binary representations within 11 bits must have exactly one 0. Therefore, the count is indeed 5.But let me check n=1983. 1983 is 11110111111. Let me confirm that 1983 is ≤1989. Yes, 1983 is 6 less than 1989.But wait, how do we know there aren't other numbers with 10 ones that are not in the form of 11-bit numbers with one 0? For example, suppose there's a number where the 0 is not in a single position but spread out, but still, the total number of ones is 10. But in an 11-bit number, to have 10 ones, you must have exactly one 0. So there's no other way. So all numbers with exactly 10 ones in 11 bits have exactly one 0.Therefore, the total numbers with 10 ones ≤1989 are the 5 numbers where the 0 is in positions10,9,8,7,6. Therefore, the count is 5.But wait, 1023 is included here as 0 in position10. So 1023 is 01111111111 (with 10 ones). Then, 1535 is 10111111111 (0 in position9), 1791 is 11011111111 (0 in position8), 1919 is 11101111111 (0 in position7), and 1983 is 11110111111 (0 in position6). So these 5 numbers.But wait, what about numbers with 10 ones in fewer bits? For example, 511 is 111111111 (9 ones). But that's 9 bits. So numbers with 10 ones must be in 10 bits or more. The only 10-bit number with 10 ones is 1023. But as we saw, 1023 is also counted as an 11-bit number with a leading zero. So when considering numbers from 1 to 1989, 1023 is included once. So perhaps the answer is 5 numbers.But let's verify by checking the numbers:1. 1023: binary 1111111111 (10 ones) or 01111111111 (10 ones in 11 bits). Either way, it's one number.2. 1535: 10111111111 (10 ones). Check.3. 1791:11011111111 (10 ones). Check.4. 1919:11101111111 (10 ones). Check.5. 1983:11110111111 (10 ones). Check.These are five numbers. Each has exactly 10 ones. All are less than or equal to 1989. Therefore, the maximum value u is 10, and the number of such n is 5.Wait, but let's check another number. For example, 511 in binary is 111111111 (9 ones). 767 is 1011111111 (9 ones). Wait, but 1023 is 10 ones. Then 1535,1791,1919,1983.But how about a number like 2047 - 2^11 - 2^10? No, that's not correct. Wait, perhaps there's another number with 10 ones. Let me think. For example, 1989 is 11111000101 (7 ones). But we're looking for numbers with 10 ones. Since 11-bit numbers with 10 ones are only those with exactly one 0, and we found 5 such numbers ≤1989, there are no other numbers with 10 ones.Therefore, the conclusion is u=10 and the number of such n is 5.But wait, this seems conflicting with the initial thought that 1023 is a separate number. But 1023 is included in the 11-bit numbers as 01111111111, which is indeed 1023. So the five numbers are:1. 01111111111 (1023)2. 10111111111 (1535)3. 11011111111 (1791)4. 11101111111 (1919)5. 11110111111 (1983)Yes, five numbers. All of these have exactly 10 ones. Therefore, the answer is maximum value u=10, and there are 5 numbers.But let me confirm with another approach. Let's compute how many numbers less than or equal to 1989 have exactly 10 ones in their binary representation.The number of such numbers is equal to the number of ways to choose 10 positions out of the 11 bits (since numbers can be up to 11 bits) such that the number formed is ≤1989.But as we saw, not all combinations are allowed because some choices would result in numbers exceeding 1989.Alternatively, we can model this as a combinatorial problem where we need to count the number of binary numbers with exactly 10 ones, of length up to 11 bits, and ≤1989.But perhaps the way we did before is correct.Alternatively, think recursively: the number of 1s in the binary representation relates to the function f(n). Since f(n) = number of 1s, then the maximum number of 1s up to n is the maximum Hamming weight, and the count is the number of numbers with that Hamming weight.Given that 1989 in binary is 11111000101, which has 7 ones.But numbers below 1989 can have higher numbers of ones. For example, 1983 is 11110111111, which has 10 ones.So the maximum is indeed 10, achieved by 5 numbers.Therefore, the answer should be u=10 and the count is 5.Wait, but let's check an online encyclopedia or something for the function. Wait, since f(n) is defined as f(1)=1, f(2n)=f(n), f(2n+1)=f(n)+1. So yes, it's the number of ones in binary. Therefore, the problem reduces to finding the maximum Hamming weight in numbers up to 1989 and the count.According to our analysis, the maximum Hamming weight is 10, achieved by 5 numbers. Therefore, the answer is u=10 and the number is 5.But just to confirm, let me check one more example. Take 1983: 11110111111. Let's count the ones: 1,1,1,1,0,1,1,1,1,1,1. That's 10 ones. Yes. 1983 is less than 1989. Similarly, 1919:11101111111. Count the ones: 1,1,1,0,1,1,1,1,1,1,1. That's 10 ones. Yes. 1791:11011111111. 10 ones. 1535:10111111111. 10 ones. 1023:01111111111. 10 ones. So total 5 numbers. Therefore, the answer is 5.Thus, the final answer is u=10 and the count is 5.Final AnswerThe maximum value ( u ) is (boxed{10}) and the number of ( n ) such that ( f(n) = u ) is (boxed{5}).</think>The function ( f ) is defined on the set of natural numbers with the following conditions:1. ( f(1) = 1 )2. ( f(2n) = f(n) ) and ( f(2n+1) = f(2n) + 1 ) for ( n geq 1 ).We need to find the maximum value ( u ) of ( f(n) ) for ( 1 leq n leq 1989 ) and determine how many ( n ) satisfy ( f(n) = u ).By analyzing the function ( f ), we recognize that ( f(n) ) counts the number of 1s in the binary representation of ( n ). This is because:- For even ( n ), ( f(n) = f(n/2) ), which corresponds to right-shifting the binary representation.- For odd ( n ), ( f(n) = f((n-1)/2) + 1 ), which corresponds to right-shifting and adding 1.To find the maximum number of 1s in the binary representations of numbers from 1 to 1989, we convert 1989 to binary:- ( 1989 ) in binary is ( 11111000101 ), which has 7 ones.However, we need to check if there are numbers less than 1989 with more ones. The number ( 1535 ) in binary is ( 10111111111 ), which has 10 ones. Similarly, other numbers like ( 1791 ), ( 1919 ), ( 1983 ), and ( 1023 ) also have 10 ones in their binary representations.These numbers are:1. ( 1023 ) (binary: ( 01111111111 ))2. ( 1535 ) (binary: ( 10111111111 ))3. ( 1791 ) (binary: ( 11011111111 ))4. ( 1919 ) (binary: ( 11101111111 ))5. ( 1983 ) (binary: ( 11110111111 ))Each of these numbers has exactly 10 ones in their binary representation and is less than or equal to 1989. Therefore, the maximum value ( u ) is 10, and there are 5 such numbers. Final AnswerThe maximum value ( u ) is (boxed{10}) and the number of ( n ) such that ( f(n) = u ) is (boxed{5}).]

🔑:[Solution] We will prove: If a is a number represented in binary, then f(a) is the sum of the digits of a. Indeed, when a is a single-digit number, f(1)=1, the proposition holds; assuming that when a is an n-digit number, the proposition holds, i.e., when a=a_{0} 2^{n-1}+a_{1} 2^{n-2}+cdots+a_{n-1}, f(a)=a_{0}+a_{1}+cdots+a_{n-1}.A binary number with n+1 digits has two cases:(1) 2 a=a_{0} 2^{n}+a_{1} 2^{n-1}+cdots+a_{n-1} cdot 2, the sum of its digits is a_{0}+a_{1}+cdots+a_{n-1}=f(a), and f(2 a)=f(a), i.e., the sum of the digits of f(2 cdot a) is the sum of the digits of 2 a.(2) 2 a+1=a_{0} 2^{n}+a_{1} 2^{2-1}+cdots+a_{n-1} cdot 2+1 the sum of its digits isa_{0}+a_{1}+cdots+a_{n-1}+1=f(a)+1 .Also, f(2 a+1)=f(2 a)+1=f(a)+1, i.e., f(2 a+1) is also the sum of the digits of 2 a+1.Thus, we have proved by mathematical induction that when a is a number represented in binary, f(a) is the sum of the digits of a.Since 2^{10}+2^{9}+cdots+2+1=2^{11}-1=2047>1989, and 1989=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{2}+1=(11111000101)_{2},so M_{text {max }}=10.The binary representation of a number not exceeding 1989 in decimal is a ten-digit number, (1111111111)_{2}= 1023 and four eleven-digit numbers (10111111111)_{2}=1535,begin{array}{l}(11011111111)_{2}=1791,(11101111111)_{2}=1919, (11110111111)_{2}=1983 .end{array}Therefore, M_{text {max }}=10, when 1 leqslant n leqslant 1989, there are 5 n such that f(n)=10.

🧠:[<think>嗯，我现在得解决这个问题：内接于球体的圆锥的最大体积与球体体积的比值是多少。好，我得仔细想一下怎么解这个问题。首先，这个问题应该是关于求圆锥的最大体积，然后跟球体的体积相比对吧？那首先，我需要回忆一下圆锥和球体体积的公式。圆锥体积是1/3πr²h，而球体的体积是4/3πR³，其中R是球的半径。不过这里的问题里，圆锥是内接在球里的，所以圆锥的底面半径和高度应该和球的半径有关系吧？首先，我需要建立一个几何模型。假设球体的半径是R，圆锥内接于这个球体内。那么圆锥的顶点应该在球的表面上吗？或者说，圆锥可能完全位于球体内？这个时候可能圆锥的底面是一个圆，位于球内的某个位置，顶点可能位于球的另一个端点？比如，假设圆锥的顶点在球的北极点，而底面则是在球内的某个纬度的圆上。这样的话，圆锥的高度h就是从顶点到底面圆心的距离，而底面圆的半径r则和这个纬度有关。那这个时候，圆锥的高度和底面半径应该满足某种几何关系，因为它们都在球体内。可能我需要用球体的方程来关联这些变量。比如，假设球的中心在原点，顶点在点(0,0,R)处，而底面圆的圆心在(0,0,R - h)的位置，这里h是圆锥的高度。那么底面圆的半径r应该满足，在球体表面上的点，即底面圆周上的每个点都必须位于球体上。例如，底面圆的半径r和位置z坐标之间有什么关系呢？根据球体的方程x² + y² + z² = R²。当底面圆的圆心在(0,0,a)的位置，其中a = R - h，那么底面圆周上的任意一点坐标应该是(x, y, a)，其中x² + y² = r²。这个点必须满足球体的方程，所以带入进去，得到x² + y² + a² = R²，即r² + a² = R²。所以，r² = R² - a²。这里a就是底面圆心的z坐标，而圆锥的高度h是顶点到底面圆心的距离，也就是顶点在(0,0,R)处，底面圆心在(0,0,a)处，那么h = R - a。所以，a = R - h，代入上式，得到r² = R² - (R - h)² = R² - (R² - 2Rh + h²) = 2Rh - h²。因此，r² = 2Rh - h²。这样，圆锥的体积就可以表示为V = (1/3)πr²h = (1/3)π(2Rh - h²)h = (1/3)π(2R h² - h³)。现在问题就转化为，对于h的取值范围，找到这个函数V(h)的最大值。然后，将这个最大值除以球体的体积，得到比值。接下来，我需要用微积分来求这个体积的最大值。首先，确定h的取值范围。因为圆锥必须内接于球体内，所以底面圆的位置不能超过球体的底部，即当a = R - h ≥ -R，因为球体的底部在z = -R处。所以，当底面圆心在(0,0,a)时，a必须大于等于 -R。因此，R - h ≥ -R → -h ≥ -2R → h ≤ 2R。同时，h必须大于0，因为高度不能为负或零。所以h ∈ (0, 2R]。接下来，对V(h) = (1/3)π(2R h² - h³)求导，找到临界点。导数V’(h) = (1/3)π(4R h - 3h²). 令导数为零，解得临界点：4R h - 3h² = 0 → h(4R - 3h) = 0 → h = 0 或 h = 4R/3。但是h=0的时候体积为零，所以极大值可能在h=4R/3处。不过需要验证这个临界点是否在定义域内。之前定义的h范围是0 < h ≤ 2R。4R/3 ≈1.333R，确实小于2R，所以这个临界点在定义域内。接下来，需要验证这是极大值。可以用二阶导数或者观察函数的变化情况。这里计算二阶导数：V''(h) = (1/3)π(4R - 6h)。当h=4R/3时，V''(h) = (1/3)π[4R - 6*(4R/3)] = (1/3)π[4R - 8R] = (1/3)π*(-4R) < 0。因此，在h=4R/3处，函数取得极大值。接下来，计算最大体积V_max：V_max = (1/3)π[2R*(4R/3)^2 - (4R/3)^3]先计算括号里的部分：2R*(16R²/9) - (64R³/27) = 32R³/9 - 64R³/27 = 把分母统一成27，得到：32R³/9 = 96R³/27，所以96R³/27 - 64R³/27 = 32R³/27。所以V_max = (1/3)π*(32R³/27) = (32R³ π)/(81)。然后，球体的体积是V_sphere = (4/3)πR³。所以比值是V_max / V_sphere = (32R³π/81) / (4πR³/3) ) = (32/81) * (3/4) = (32*3)/(81*4) = 96/324 = 约分的话，分子分母同除以12，得到8/27。所以，最大体积的圆锥体积与球体体积的比值是8/27。不过，我需要确认一下这个结果是否正确，因为在之前的步骤中，可能哪里出错了。比如，在确定底面圆的位置时是否正确？比如，当圆锥的高度是h=4R/3，这个时候底面圆心的位置是a = R - h = R - 4R/3 = -R/3。也就是说，底面圆心位于球心下方R/3的位置，这确实是在球体内的，因为球体的半径是R，所以从球心到该位置的距离是R/3 < R，所以这个位置在球体内。那么底面圆的半径r根据之前的公式r² = 2Rh - h²，代入h=4R/3：r² = 2R*(4R/3) - (4R/3)^2 = 8R²/3 - 16R²/9 = 24R²/9 -16R²/9 = 8R²/9，所以r = 2R√2/3。这个结果是否合理呢？因为底面圆的位置在z=-R/3，所以这个圆的半径是否在球体内？根据球体的方程，在z=-R/3处的截面的半径应该是sqrt(R² - ( (-R/3)^2 )) = sqrt(R² - R²/9) = sqrt(8R²/9) = 2R√2/3，刚好和r相同，这说明底面圆确实位于球体的内部，正确。所以计算是正确的，得到的比值是8/27。也就是说，内接圆锥的最大体积是球体体积的8/27倍。因此，答案应该是这个比值，即8/27。</think>内接于球体的圆锥的最大体积与球体体积的比值为 (boxed{dfrac{8}{27}})。步骤解析：1. 设定变量： - 球体半径 ( R )，圆锥高度 ( h )，底面半径 ( r )。2. 几何关系： - 圆锥顶点在球顶 ((0, 0, R))，底面圆心在 ((0, 0, R - h))。 - 球体方程 ( x^2 + y^2 + z^2 = R^2 )，代入底面圆周点 ((x, y, R - h))： [ r^2 + (R - h)^2 = R^2 implies r^2 = 2Rh - h^2 ]3. 体积函数： - 圆锥体积 ( V = dfrac{1}{3}pi r^2 h = dfrac{1}{3}pi (2Rh^2 - h^3) )。4. 求导找极值： - ( dfrac{dV}{dh} = dfrac{1}{3}pi (4Rh - 3h^2) )，令导数为零得临界点 ( h = dfrac{4R}{3} )。5. 验证极大值： - 二阶导数 ( dfrac{d^2V}{dh^2} = dfrac{1}{3}pi (4R - 6h) )，代入 ( h = dfrac{4R}{3} ) 为负，确认极大值。6. 计算最大体积： - 代入 ( h = dfrac{4R}{3} ): [ V_{text{max}} = dfrac{1}{3}pi left( 2R left(dfrac{4R}{3}right)^2 - left(dfrac{4R}{3}right)^3 right) = dfrac{32pi R^3}{81} ]7. 球体体积及比值： - 球体体积 ( V_{text{sphere}} = dfrac{4}{3}pi R^3 )。 - 比值： [ dfrac{V_{text{max}}}{V_{text{sphere}}} = dfrac{dfrac{32pi R^3}{81}}{dfrac{4pi R^3}{3}} = dfrac{8}{27} ]最终答案为 (boxed{dfrac{8}{27}})。,<think>嗯，好的，我现在要解决的问题是，内接于球体的圆锥的最大体积与球体体积的比值是多少。这个问题看起来像是一个优化问题，需要用到微积分来找最大值对吧。那我得先理清楚题目的条件，然后一步步推导。首先，题目说的是一个圆锥内接于一个球体，也就是说这个圆锥的底面是一个圆，顶点在球面上，而底面圆周上的所有点也在球面上吗？或者说，圆锥完全包含在球体内部，顶点和底面圆周都在球面上？应该是后者，对吧。这样的话，整个圆锥都位于球体内，顶点和底面的圆周都在球的表面上。这样的话，圆锥的高和底面半径都会和球的半径有关系。假设球的半径为R，我需要找到这样的圆锥，使得它的体积最大。然后比较这个最大体积和球体积的比值。球体积的公式是已知的，4/3πR³，所以问题转化为求圆锥的最大体积，再除以球的体积即可。首先，我需要建立圆锥体积的表达式。圆锥的体积公式是V = (1/3)πr²h，其中r是底面半径，h是高。不过这里的r和h都要用球的半径R来表示，并且满足内接的条件。接下来，如何把r和h用R表示呢？可能需要用到几何关系。考虑圆锥内接于球体的情况，想象球心到圆锥底面的距离，然后圆锥的高度应该是从顶点到底面的垂直距离。假设球心到圆锥底面的距离为d，那么球心到圆锥顶点的距离就是h - d，对吗？不过可能这里需要更仔细的分析。或者，可以建立坐标系，把球心放在原点，圆锥的顶点位于球的一个点，比如在坐标系的(0,0,a)，然后底面的圆心在球心下方某个位置，比如说(0,0,b)，这样底面圆的半径r满足球面方程。这时候底面圆周上的任何一点(x,y,b)都应该满足x² + y² + b² = R²，对吗？而底面圆的半径r满足x² + y² = r²，所以r² + b² = R²，即r² = R² - b²。然后圆锥的高h应该是顶点到底面的距离，也就是a到b的距离，即h = |a - b|。不过顶点(0,0,a)也应该在球面上，所以a² = R²，即a = ±R。如果顶点在球的上顶点，可能设为(0,0,R)，那么底面圆心在(0,0,b)，而b应该小于R，这样圆锥的高度h = R - b。这时候底面半径r由r² = R² - b²。这时候圆锥体积V可以表示为(1/3)πr²h = (1/3)π(R² - b²)(R - b)。这个时候，变量是b，我需要找到b的值使得V最大。这里可能需要将V关于b求导，然后找临界点。不过这个表达式可能有点麻烦，或许换一个变量更方便。比如，设h = R - b，那么b = R - h，代入r² = R² - (R - h)^2 = R² - (R² - 2Rh + h²) = 2Rh - h²。这样，体积V可以表示为(1/3)π(2Rh - h²)h = (1/3)π(2Rh² - h³)。这时候V是关于h的函数，求导的话可能更容易。接下来对V关于h求导：dV/dh = (1/3)π(4Rh - 3h²)。令导数等于零，解得4Rh - 3h² = 0 → h(4R - 3h) = 0。所以h=0或者h=4R/3。显然h=0的话体积为零，所以最大值可能在h=4R/3处。不过这时候需要检查这个h是否合理，因为原来的h是R - b，而原来的b是底面圆心到球心的距离，所以当h=4R/3时，b = R - 4R/3 = -R/3，也就是底面圆心在球心下方R/3的位置，这时候底面半径r² = R² - (-R/3)^2 = R² - R²/9 = 8R²/9，所以r = (2√2 R)/3。这时候体积V就是(1/3)π*(8R²/9)*(4R/3) = (1/3)*(8*4)/(9*3) * πR³ = 32/81 πR³。对吗？不过这里有个问题，因为如果h=4R/3，那么圆锥的高度超过了球的直径2R吗？球的直径是2R，所以如果h=4R/3，这显然超过了球的直径，因为当h=4R/3时，圆锥的高度是4R/3，而球的直径是2R，所以显然不可能，这时候可能哪里出错了。这说明我在变量替换或者几何分析的时候有问题。可能我的坐标设定或者变量替换有误。让我再仔细检查一下。首先，假设球心在原点，圆锥顶点在球面上，坐标可能是(0,0,R)或者(0,0,-R)，假设顶点在(0,0,R)，底面圆心在(0,0,b)，其中b < R，这样圆锥的高度h就是R - b。底面圆的半径r满足x² + y² + z² = R²，对于底面圆上的点，z = b，所以x² + y² + b² = R² → r² = R² - b²，所以r = √(R² - b²)。因此体积V = (1/3)πr²h = (1/3)π(R² - b²)(R - b)。这时候，变量是b，取值范围应该是从某个最小值到R。因为当b接近R时，底面半径趋近于0，体积趋近于0；当b趋近于负无穷时，显然不可能，因为圆锥必须在球内。但这里的b必须满足底面圆在球内，也就是底面圆心在球内，所以b的取值范围应该是使得底面圆的所有点都在球内。但根据之前的设定，底面圆周在球面上，所以底面圆上的点必须满足z = b，所以x² + y² + b² ≤ R²，所以当底面圆周上的点刚好在球面上时，x² + y² + b² = R²，也就是r² + b² = R²，这时候底面圆确实位于球面上。因此，此时b的取值范围应该是满足这个条件的任何实数，只要r是实数，即R² - b² ≥ 0，即|b| ≤ R。但是当b < 0时，底面圆心位于球心的另一侧，这时候圆锥的高度h = R - b就会超过R，当b为-R时，h = R - (-R) = 2R，即圆锥的高度等于球的直径，这时候底面半径r = 0，体积为0。所以可能当b为负数时，圆锥的高度更大，但底面半径也更大，但需要找到平衡点使得体积最大。所以之前的分析中，当h = 4R/3时，对应的b = R - h = R - 4R/3 = -R/3，此时底面半径r = √(R² - (-R/3)^2) = √(R² - R²/9) = √(8R²/9) = (2√2 R)/3，这确实在球内，因为底面圆所在的平面距离球心-R/3，而底面半径是(2√2 R)/3，所以这个圆的边缘到球心的距离是√[(-R/3)^2 + (2√2 R/3)^2] = √[R²/9 + 8R²/9] = √[9R²/9] = R，所以确实在球面上，所以这样的圆锥是存在的，尽管它的高度超过了R，但球本身的半径是R，所以高度h=4R/3是可能的，只要圆锥的顶点在(0,0,R)，底面圆心在(0,0,-R/3)，这样整个圆锥都包含在球体内，对吗？例如，当h=4R/3，顶点在(0,0,R)，底面圆心在(0,0,-R/3)，这时底面半径是(2√2 R)/3，而球的半径是R，所以从球心到底面圆周的距离是√[ (-R/3)^2 + (2√2 R/3)^2 ] = R，符合球面方程，没问题。所以这时候体积是(1/3)π*(8R²/9)*(4R/3) = (32/81)πR³，而球体积是(4/3)πR³，所以比值是 (32/81)/(4/3) = (32/81)*(3/4) = 8/27，对吗？所以最大圆锥体积是球体积的8/27倍，对吗？不过让我再检查一下导数的过程是否正确。我之前把V表示为关于h的函数，即V = (1/3)π(2Rh - h²)h，也就是V = (1/3)π(2Rh² - h³)。然后对h求导，得到dV/dh = (1/3)π(4Rh - 3h²) = 0 → 4Rh - 3h² = 0 → h(4R - 3h) = 0，解得h=0或h=4R/3。这样看来，当h=4R/3时，体积取极大值，对吗？这时候确实当h=4R/3时，b=R - h = R - 4R/3 = -R/3，这在球的内部，所以这个解是可行的。所以结果正确，对吗？那么最大体积是32/81 πR³，球体积是4/3 πR³，比值是32/81 ÷ 4/3 = 32/81 * 3/4 = 8/27。所以这个比值是8/27，对吗？不过，我记得以前好像有个类似的题目，可能结果不一样，但可能我记错了。或者可能我的推导中有哪里出错了？再检查一下圆锥体积的表达式。当用b作为变量时，V = (1/3)π(R² - b²)(R - b)。这时候我们可以展开这个表达式，或者把它转化为关于b的函数，然后求导。假设以b为变量，求导的话：V(b) = (1/3)π(R² - b²)(R - b)展开一下：V(b) = (1/3)π [ (R²)(R - b) - b²(R - b) ] = (1/3)π [ R³ - R²b - R b² + b³ ]对b求导：dV/db = (1/3)π [ -R² - 2Rb + 3b² ]令导数等于零：-R² - 2Rb + 3b² = 0 → 3b² - 2Rb - R² = 0解这个二次方程，判别式D = ( -2R )² - 4*3*(-R²) = 4R² + 12R² = 16R² → 根为 [2R ± √(16R²)]/(2*3) = [2R ±4R]/6即b = (2R +4R)/6 = 6R/6 = R 或者 b=(2R -4R)/6 = (-2R)/6 = -R/3所以临界点是b=R和b=-R/3。当b=R时，体积为0，因为R - R = 0，h=0，所以体积为0。另一个临界点是b=-R/3，这时候h=R - (-R/3)=4R/3，此时体积如前所述，为32/81 πR³，对吗？所以无论是用h作为变量还是用b作为变量，结果都是正确的，得到最大体积时h=4R/3，此时体积是32/81 πR³，与球体积的比值为8/27。看来这个结果是对的。不过为了确保，可以再检查一下代数运算：V = (1/3)π*(R² - b²)*(R - b) 当b=-R/3时，R² - (-R/3)^2 = R² - R²/9 = 8R²/9，R - (-R/3) = R + R/3 = 4R/3，所以体积V= (1/3)π*(8R²/9)*(4R/3) = (1/3)*(8*4)/(9*3) πR³ = 32/81 πR³，没错。球体积是4/3 πR³，所以比值是32/81 ÷ 4/3 = (32/81)*(3/4)= (32*3)/(81*4)= 96/324= 8/27，化简正确。所以结论是，内接于球体的圆锥的最大体积是球体积的8/27倍，即它们的比值是8:27。不过，我是否应该考虑其他可能的圆锥位置呢？比如顶点不在球体的顶端，而是其他地方？或者是否存在另一种内接方式能得到更大的体积？比如，如果顶点在球体的某个位置，而底面圆的位置不同，但根据对称性，最大的体积应该出现在顶点在球体的一个极点，底面圆在另一侧的某个位置，这样的对称结构，所以之前的分析应该是正确的，无需考虑其他位置，因为对称性可以保证最大值出现在这种轴对称的情况下。因此，最终的结论是，内接于球体的圆锥的最大体积与球体积的比值为8/27。</think>内接于球体的圆锥的最大体积与球体积的比值为 (boxed{dfrac{8}{27}})。推导过程：1. 设定变量： 设球半径为 (R)，圆锥顶点位于球面 ((0,0,R))，底面圆心在 ((0,0,b))（(b < R)）。圆锥高度为 (h = R - b)，底面半径 (r) 满足球面方程 (r^2 + b^2 = R^2)，即 (r = sqrt{R^2 - b^2})。2. 体积表达式： 圆锥体积 (V = dfrac{1}{3}pi r^2 h = dfrac{1}{3}pi (R^2 - b^2)(R - b))。3. 变量替换与求导： 以 (h = R - b) 为变量，则 (b = R - h)，代入得 (r^2 = 2Rh - h^2)，体积变为： [ V = dfrac{1}{3}pi (2Rh^2 - h^3). ] 对 (h) 求导并令导数为零： [ dfrac{dV}{dh} = dfrac{1}{3}pi (4Rh - 3h^2) = 0 implies h(4R - 3h) = 0. ] 解得 (h = dfrac{4R}{3})（舍去 (h = 0)）。4. 验证几何合理性： 当 (h = dfrac{4R}{3}) 时，底面圆心 (b = R - dfrac{4R}{3} = -dfrac{R}{3})，底面半径 (r = sqrt{R^2 - left(-dfrac{R}{3}right)^2} = dfrac{2sqrt{2}R}{3})。此时底面圆周位于球面上，符合内接条件。5. 最大体积计算： 代入 (h = dfrac{4R}{3}) 和 (r = dfrac{2sqrt{2}R}{3})： [ V_{text{max}} = dfrac{1}{3}pi left(dfrac{8R^2}{9}right)left(dfrac{4R}{3}right) = dfrac{32}{81}pi R^3. ] 球体积为 (dfrac{4}{3}pi R^3)，比值为： [ dfrac{V_{text{max}}}{V_{text{球}}} = dfrac{dfrac{32}{81}pi R^3}{dfrac{4}{3}pi R^3} = dfrac{8}{27}. ]最终答案为 (boxed{dfrac{8}{27}})。]

🔑:2. Let the radius of the sphere be R, the height of the inscribed cone be h, and the base radius be r.Using any plane passing through the center of the sphere and the vertex of the inscribed cone to cut, the resulting section is as shown in the figure.It is easy to know that r^{2}=h(2 R-h).begin{array}{c}therefore V_{text {cone}}=frac{1}{3} pi r^{2} h =frac{1}{3} pi h^{2}(2 R-h) =frac{1}{3} pi cdot frac{1}{2}[h cdot h cdot(4 R-2 h)] .end{array}because h+h+(4 R-2 h)=4 R is a constant, # h+1+h^{-1}text { ( } left.V_{text {cone}}right)_{max }=frac{1}{3} pi cdotleft(frac{4}{3} Rright)^{2} cdotleft(2 R-frac{4}{3} Rright)=frac{32}{81} pi R^{3} text {, }Also,V_{text {sphere}}=frac{4}{3} pi R^{3},therefore quadleft(V_{text {cone}}right)_{max }: V_{text {sphere}}=frac{32}{81}: frac{4}{3}=8: 27

🧠:[<think>Okay, so I need to figure out the smallest number n such that rotating Vasya's triangle around vertex 3 n times (each time by the angle at that vertex) brings it back to its original position. The problem also mentions that for vertices 1 and 2, rotating 12 and 6 times respectively does the same. Let me start by understanding what's going on here.First, when you rotate a shape around one of its vertices by the angle at that vertex, the number of rotations needed to return to the original position relates to the angle itself. Specifically, if the angle is θ, then the rotation is essentially a rotation by θ degrees (or radians) each time. The number of times you need to rotate to get back to the original position is the smallest integer k such that kθ is a multiple of 360 degrees (or 2π radians). That is, kθ = 360°m for some integer m, so the order of the rotation is k when θ = 360°m/k. But since θ is an angle of a triangle, it must be between 0° and 180°, right?Wait, actually, in a triangle, each angle is between 0° and 180°, and the sum is exactly 180°. So each angle θ_i (i=1,2,3) must satisfy 0° < θ_i < 180°, and θ1 + θ2 + θ3 = 180°.Now, for vertex 1, rotating 12 times by θ1 brings the triangle back. So, 12θ1 must be a multiple of 360°, which means θ1 = 360° / 12 = 30°, or θ1 could be 360° * m / 12, where m is an integer such that θ1 is less than 180°. But since θ1 is an angle in a triangle, m must be 1, because if m=2, θ1 would be 60°, which is still okay, but wait, the problem says that rotating 12 times brings it back to original position. So the angle must be 360° / 12 = 30°, because each rotation is θ1, so 12θ1 = 360°, hence θ1 = 30°. Similarly for vertex 2, rotating 6 times gives 6θ2 = 360°, so θ2 = 60°. Therefore, the angles at vertices 1 and 2 are 30° and 60°, respectively. Then, the angle at vertex 3 must be 180° - 30° - 60° = 90°, right?Wait, so θ3 = 90°. Then, rotating around vertex 3 by 90° each time. How many such rotations are needed to get back to the original position? Well, similar to the other vertices, we need nθ3 = 360°m for some integer m. So n * 90° = 360°m. Simplifying, n = 4m. The smallest positive integer n here is 4 when m=1. So, rotating 4 times by 90° would give a full 360° rotation, bringing it back. Therefore, the minimal n is 4. But wait, the problem says "What is the smallest n that Vasya could have named so that his claim is true for at least one cardboard triangle?" So maybe there's a possibility that θ3 could be different? But hold on, the angles at vertices 1 and 2 are fixed as 30° and 60°, so θ3 must be 90°, right? Then, n must be 4. Hmm, but maybe there's another interpretation?Wait, let me check again. The problem states that when rotating around vertex 1 (which has angle θ1) 12 times, the triangle returns to its original position. So, each rotation is θ1 degrees. Then, 12θ1 must be a multiple of 360°, so θ1 = 360°k / 12 = 30°k, where k is an integer such that θ1 < 180°, because it's a triangle angle. Similarly, for vertex 2, 6θ2 must be a multiple of 360°, so θ2 = 360°m / 6 = 60°m, with θ2 < 180°. Since θ1 and θ2 must satisfy θ1 + θ2 < 180° (because the third angle θ3 is also positive), let's see:If θ1 = 30°k, then k can be 1, 2, 3, 4, or 5 (since 30°*6=180°, which is invalid). But θ1 must be less than 180°, and θ1 + θ2 < 180°, so let's check possible k and m.Wait, maybe k and m can't be too large. Let's take the minimal case first. If k=1, then θ1=30°, and m=1, θ2=60°, so θ3=90°, as before. Then, n=4. But maybe if we take different k and m?Suppose k=2, then θ1=60°. Then, for vertex 2, θ2=60°m. Let's take m=1, then θ2=60°, so θ3=180° - 60° - 60° = 60°, but then θ3=60°, so rotating around vertex 3 would require n such that n*60°=360°m, so n=6. But in this case, the original problem's statements would be different: rotating around vertex 1 (60°) would require 6 times (since 60°*6=360°). But the problem says that rotating around vertex 1 12 times. So if θ1=60°, then 60°*12=720°, which is 2 full rotations, so yes, it would return to original position. But the problem states that 12 rotations around vertex 1 bring it back. Similarly, for vertex 2, if θ2=60°, 6 rotations give 360°, as in the problem. So in this case, θ3 would be 60°, and then rotating 6 times around vertex 3 would bring it back. But the problem says "Vasya claims that if his triangle is rotated n times around the vertex numbered 3...", so in this case, minimal n would be 6.But wait, but the problem says "for at least one cardboard triangle". So maybe there exists a triangle where θ1=30°, θ2=60°, θ3=90°, leading to n=4, and another triangle where θ1=60°, θ2=60°, θ3=60°, leading to n=6. So the minimal n is 4. Wait, but the question is asking for the smallest n that Vasya could have named so that his claim is true for at least one cardboard triangle. So if there exists a triangle where n=4 works, then 4 is the answer. Even if other triangles require higher n. So in the first case, θ3=90°, n=4. So 4 is possible, so the answer is 4. But wait, let me check again.Wait, the problem says that "it turned out that if Vasya's triangle is rotated 12 times... around vertex 1... returns to original position. Similarly, 6 times around vertex 2... returns. Vasya claims that if his triangle is rotated n times around vertex 3... returns. What is the smallest n that Vasya could have named so that his claim is true for at least one cardboard triangle."So Vasya is talking about his triangle. But the question is, given that Vasya's triangle has the properties that rotating around 1 12 times and around 2 6 times works, what's the minimal n such that there exists at least one triangle with these properties where rotating around 3 n times works. So Vasya's triangle could be any triangle with angles θ1, θ2, θ3 such that 12θ1 ≡ 0 mod 360°, 6θ2 ≡ 0 mod 360°, and we need to find the minimal n where nθ3 ≡ 0 mod 360°, given θ1 + θ2 + θ3 = 180°. So θ1 must be 360° / 12 = 30°, or a divisor, but θ1 must be such that 12θ1 is a multiple of 360°, so θ1 = 360° * k / 12 = 30°k, where k is an integer. Similarly, θ2 = 360° * m /6 = 60°m, with m integer. But θ1 and θ2 must satisfy θ1 + θ2 < 180°, since θ3 is positive.So θ1 = 30°k, θ2 = 60°m. Then, θ1 + θ2 = 30k + 60m < 180. Therefore, 30k + 60m < 180 ⇒ k + 2m < 6. Also, k and m are positive integers (since angles must be positive). So possible (k, m):k=1, m=1: 1 + 2 = 3 < 6 ⇒ θ1=30°, θ2=60°, θ3=90°k=1, m=2: 1 + 4 =5 <6 ⇒ θ1=30°, θ2=120°, θ3=30°k=2, m=1: 2 +2=4 <6 ⇒ θ1=60°, θ2=60°, θ3=60°k=2, m=2: 2 +4=6 not <6 ⇒ invalidk=3, m=1: 3 +2=5 <6 ⇒ θ1=90°, θ2=60°, θ3=30°k=1, m=3: 1 +6=7 invalidSimilarly, other combinations may not work. So possible angles:Case 1: θ1=30°, θ2=60°, θ3=90°Case 2: θ1=30°, θ2=120°, θ3=30°Case 3: θ1=60°, θ2=60°, θ3=60°Case 4: θ1=90°, θ2=60°, θ3=30°So for each case, compute θ3 and find minimal n where nθ3 is multiple of 360°.Case 1: θ3=90°, then n=4 (90*4=360)Case 2: θ3=30°, n=12 (30*12=360)Case 3: θ3=60°, n=6 (60*6=360)Case 4: θ3=30°, n=12So among these, the minimal n is 4. Therefore, since there exists at least one triangle (Case 1) where n=4 works, the answer is 4.But wait, hold on. Wait, in Case 4, θ1=90°, θ2=60°, θ3=30°. Then rotating around vertex 1 (90°) would require 4 rotations to get back (90*4=360). But the problem states that rotating around vertex 1 twelve times brings it back. Wait, in this case, θ1=90°, so 12 rotations would be 12*90=1080°, which is 3 full rotations (3*360=1080). So, indeed, after 12 rotations, it returns to original position. Similarly, rotating around vertex 2 (θ2=60°), 6*60=360°, so that works. And θ3=30°, so n=12. But in this case, n=12 for vertex 3. But in Case 1, θ3=90°, n=4. So since the problem asks for the minimal n across all possible triangles that satisfy the given conditions, the minimal n is 4. Therefore, the answer should be 4.But let me check if there are other possibilities. For example, θ1 and θ2 could be different. Wait, but we considered all possible (k, m) pairs. Let me verify.θ1=30k, θ2=60m, k + 2m <6.Possible (k, m):(1,1): 30°, 60°, sum 90°, θ3=90°(1,2): 30°, 120°, sum 150°, θ3=30°(2,1): 60°, 60°, sum 120°, θ3=60°(3,1): 90°, 60°, sum 150°, θ3=30°Other combinations would make k +2m >=6. So these are all possible.So the possible θ3 are 90°, 30°, 60°, and 30° again. So minimal n is 4, as 4*90=360. So the answer is 4. Therefore, the minimal n is 4.But wait, another thought: the rotation is clockwise. Does the direction affect the result? For example, rotating clockwise vs counterclockwise. But since we're rotating multiple times in the same direction, the total rotation is additive. So direction shouldn't affect the total angle required to reach a multiple of 360°. So whether you rotate clockwise or counterclockwise, the total angle after n rotations is nθ. To get back to original, you need nθ ≡0 mod 360°, regardless of direction. So direction doesn't matter here. Therefore, the minimal n is 4. So the answer is 4.Final AnswerThe smallest ( n ) is boxed{4}.</think>Vasya's triangle has vertices numbered 1, 2, and 3. The problem states that rotating the triangle 12 times clockwise around vertex 1 (with angle ( theta_1 )) returns it to its original position, and rotating it 6 times clockwise around vertex 2 (with angle ( theta_2 )) also returns it to its original position. We need to find the smallest ( n ) such that rotating the triangle ( n ) times around vertex 3 (with angle ( theta_3 )) returns it to its original position.1. Determine angles at vertices 1 and 2: - For vertex 1: ( 12 theta_1 = 360^circ implies theta_1 = 30^circ ). - For vertex 2: ( 6 theta_2 = 360^circ implies theta_2 = 60^circ ).2. Calculate the angle at vertex 3: - The sum of angles in a triangle is ( 180^circ ). Therefore, ( theta_3 = 180^circ - 30^circ - 60^circ = 90^circ ).3. Determine the smallest ( n ) for vertex 3: - Rotating by ( theta_3 = 90^circ ) each time, we need ( n times 90^circ = 360^circ implies n = 4 ).4. Verify other possible configurations: - Considering other possible configurations of ( theta_1 ) and ( theta_2 ) that satisfy the given rotation conditions, we find that the minimal ( n ) for any valid triangle configuration is still 4.Thus, the smallest ( n ) that Vasya could have named is (boxed{4}).,<think>Okay, so Vasya has this triangle with vertices labeled 1, 2, 3. The problem is about rotational symmetry, I think. Let me try to parse the information step by step.First, when they say Vasya rotated the triangle 12 times clockwise around vertex 1 and it returned to its original position. So, each rotation is by the angle at vertex 1. Then, after 12 such rotations, the triangle is back to where it started. That must mean that 12 times the angle at vertex 1 is a multiple of 360 degrees. So, 12 * angle1 = 360 * k, where k is some integer. Since angle1 is an internal angle of a triangle, it's between 0 and 180 degrees. So, 12 * angle1 = 360 * k. Let's solve for angle1: angle1 = (360 * k)/12 = 30 * k degrees. Since angle1 has to be less than 180, 30k < 180 => k < 6. Also, since the sum of angles in a triangle is 180, angle1 must be such that angle2 and angle3 add up to 180 - angle1. Similarly, for the second condition: rotating around vertex 2 six times gives the original position. So, 6 * angle2 = 360 * m => angle2 = 60 * m degrees, where m is an integer. Again, angle2 must be less than 180, so 60m < 180 => m < 3. So m can be 1 or 2, leading to angle2 = 60 or 120 degrees.Now, we need to find possible angles at vertex 1 and 2 such that these equations hold, and then find the minimal n such that n * angle3 is a multiple of 360 degrees. Because then, rotating around vertex 3 n times would bring it back. The question is, what's the minimal such n given the constraints from angles 1 and 2.Let me note that angle1 must be 30k degrees where k is 1 to 5, and angle2 must be 60m degrees where m is 1 or 2. Then angle3 is 180 - angle1 - angle2. Then we need n such that n * angle3 is a multiple of 360. So, n must be the minimal integer where angle3 divides 360/n. Wait, actually, n * angle3 must be a multiple of 360, so angle3 must be equal to 360/n * t, where t is an integer. But since angle3 is fixed once angle1 and angle2 are fixed, n would be the minimal integer such that 360 is divisible by angle3. But angle3 might not divide 360, so n would be the least common multiple denominator? Wait, maybe n is the order of the rotation, which is 360 divided by the greatest common divisor of 360 and angle3. Wait, no. If you have a rotation by angle θ, then the minimal n such that rotating n times gives a full rotation (i.e., nθ ≡ 0 mod 360) is n = 360 / gcd(θ, 360). Wait, is that right?Wait, let's think. Suppose θ is 60 degrees. Then gcd(60, 360) is 60, so n = 360 / 60 = 6. Which matches the second condition: 6 rotations around vertex 2 (angle2=60) brings it back. Similarly, if angle1 is 30 degrees, then gcd(30, 360)=30, so n=360/30=12. Which matches the first condition. So yes, that formula seems to hold. Therefore, for angle3, the minimal n is 360 / gcd(angle3, 360). Therefore, we need to compute angle3 for possible angle1 and angle2, compute gcd(angle3, 360), then compute n=360/gcd(...), and find the minimal such n over all possible angle1 and angle2 that satisfy the given conditions.So the problem reduces to finding possible angle1 and angle2 such that angle1 = 30k, angle2 = 60m, with k from 1 to 5 and m from 1 to 2, and angle1 + angle2 < 180 (since angle3 = 180 - angle1 - angle2 must be positive). Then for each possible angle3, compute n=360/gcd(angle3,360), and find the minimal n.So let's list all possible angle1 and angle2:Possible angle1: 30, 60, 90, 120, 150 degrees (k=1 to 5)Possible angle2: 60, 120 degrees (m=1 to 2)Now, we need to check which combinations of angle1 and angle2 satisfy angle1 + angle2 < 180.Let's list all possible combinations:1. angle1=30, angle2=60: sum=90 < 180 → angle3=902. angle1=30, angle2=120: sum=150 < 180 → angle3=303. angle1=60, angle2=60: sum=120 < 180 → angle3=604. angle1=60, angle2=120: sum=180 → invalid (angle3=0)5. angle1=90, angle2=60: sum=150 < 180 → angle3=306. angle1=90, angle2=120: sum=210 → invalid7. angle1=120, angle2=60: sum=180 → invalid8. angle1=120, angle2=120: sum=240 → invalid9. angle1=150, angle2=60: sum=210 → invalid10. angle1=150, angle2=120: sum=270 → invalidSo valid combinations are 1,2,3,5.Therefore, possible angle3 values:1. angle3=902. angle3=303. angle3=605. angle3=30So angle3 can be 30, 60, or 90 degrees.Now compute n for each:1. angle3=90: gcd(90,360)=90. So n=360/90=42. angle3=30: gcd(30,360)=30. So n=360/30=123. angle3=60: gcd(60,360)=60. So n=360/60=65. angle3=30: same as case 2, n=12Thus, possible n values are 4, 12, 6. The minimal n is 4.But wait, the problem says "the smallest n that Vasya could have named so that his claim is true for at least one cardboard triangle". So even if other triangles require higher n, as long as there exists a triangle where n=4 is sufficient, then the answer is 4.But let's check if angle3=90 is possible. For angle3=90, we need angle1=30 and angle2=60. Let's confirm that:angle1=30, angle2=60, angle3=90. Sum is 180. That's a valid triangle. So such a triangle exists. Therefore, n=4 is possible. So the minimal n is 4.Wait, but before finalizing, let's check again.Wait, in the problem statement, it says "rotated around vertex 3 by an angle equal to the angle at this vertex". So the angle of rotation is angle3. Therefore, each rotation is angle3 degrees. Then, the total rotation after n steps is n*angle3. To get back to the original position, n*angle3 must be a multiple of 360 degrees. So n*angle3 ≡ 0 mod 360. So n must be such that angle3 divides 360/n? Wait, no. Let me think again. If you rotate by angle3 each time, then after n rotations, the total rotation is n*angle3. For the triangle to return to its original position, the total rotation must be an integer multiple of 360 degrees, i.e., n*angle3 = 360*k for some integer k. The minimal n is the minimal positive integer such that 360 divides n*angle3. So n must be the minimal integer such that n*angle3 is a multiple of 360. Which is equivalent to n = 360 / gcd(angle3, 360). As per the earlier formula. So for angle3=90: gcd(90, 360)=90, so n=360/90=4. For angle3=60: gcd(60,360)=60, n=6. For angle3=30: gcd(30,360)=30, n=12.Yes, so the minimal n is 4. Therefore, the answer should be 4. But wait, let's check if in the case when angle3=90, the triangle with angles 30,60,90 is possible. Yes, it's a 30-60-90 triangle, which is a standard right-angled triangle.Therefore, such a triangle exists, so n=4 is indeed possible, and since 4 is the minimal possible n, that's the answer.Final Answerboxed{4}</think>Vasya's triangle has vertices numbered 1, 2, and 3. The problem states that rotating the triangle 12 times clockwise around vertex 1 (with the angle at vertex 1) returns it to its original position, and rotating it 6 times clockwise around vertex 2 (with the angle at vertex 2) also returns it to its original position. We need to find the smallest ( n ) such that rotating the triangle ( n ) times around vertex 3 (with the angle at vertex 3) returns it to its original position.1. Angle at Vertex 1: Since rotating 12 times around vertex 1 returns the triangle to its original position, the angle at vertex 1 must be ( 30^circ ) (because ( 12 times 30^circ = 360^circ )).2. Angle at Vertex 2: Since rotating 6 times around vertex 2 returns the triangle to its original position, the angle at vertex 2 must be ( 60^circ ) (because ( 6 times 60^circ = 360^circ )).3. Angle at Vertex 3: The remaining angle at vertex 3 is calculated as ( 180^circ - 30^circ - 60^circ = 90^circ ).To find the smallest ( n ) such that rotating ( n ) times around vertex 3 (with angle ( 90^circ )) returns the triangle to its original position, we need ( n times 90^circ ) to be a multiple of ( 360^circ ). The smallest ( n ) is ( frac{360^circ}{gcd(90^circ, 360^circ)} ). The greatest common divisor (gcd) of ( 90^circ ) and ( 360^circ ) is ( 90^circ ), so ( n = frac{360^circ}{90^circ} = 4 ).Thus, the smallest ( n ) is (boxed{4}).]

🔑:Answer: 4.Solution: Since after 12 turns around the first vertex of the triangle, it returns to its original position, it makes one or several full rotations of 360^{circ}. Therefore, the angle at the first vertex is not less than 30^{circ}. Reasoning similarly, we get that the angle at the second vertex is not less than 60^{circ}. Thus, the angle at the third vertex, by the sum of the angles in a triangle, is not more than 90^{circ} and n geq 4.Four turns are sufficient: a triangle with angles 30^{circ}, 60^{circ}, and 90^{circ} will work.## Variant 2Vasya cut out a triangle from cardboard and numbered its vertices with the digits 1,2,3. It turned out that if Vasya's triangle is rotated 24 times clockwise around its vertex numbered 1 by an angle equal to the angle at this vertex, it will return to its original position. If Vasya's triangle is rotated 8 times clockwise around its vertex numbered 2 by an angle equal to the angle at this vertex, it will return to its original position. Vasya claims that if his triangle is rotated n times around the vertex numbered 3 by an angle equal to the angle at this vertex, it will return to its original position. What is the minimum n that Vasya could have named so that his claim is true for at least some cardboard triangle?Answer: 3.## Variant 3Vasya cut out a triangle from cardboard and numbered its vertices with the digits 1,2,3. It turned out that if Vasya's triangle is rotated 10 times clockwise around its vertex numbered 1 by an angle equal to the angle at this vertex, it will return to its original position. If Vasya's triangle is rotated 5 times clockwise around its vertex numbered 2 by an angle equal to the angle at this vertex, it will return to its original position. Vasya claims that if his triangle is rotated n times around the vertex numbered 3 by an angle equal to the angle at this vertex, it will return to its original position. What is the minimum n that Vasya could have named so that his claim is true for at least some cardboard triangle?Answer: 5.## Variant 4Vasya cut out a triangle from cardboard and numbered its vertices with the digits 1,2,3. It turned out that if Vasya's triangle is rotated 18 times clockwise around its vertex numbered 1 by an angle equal to the angle at this vertex, it will return to its original position. If Vasya's triangle is rotated 3 times clockwise around its vertex numbered 2 by an angle equal to the angle at this vertex, it will return to its original position. Vasya claims that if his triangle is rotated n times around the vertex numbered 3 by an angle equal to the angle at this vertex, it will return to its original position. What is the minimum n that Vasya could have named so that his claim is true for at least some cardboard triangle?Answer: 9.## Variant 5Vasya cut out a triangle from cardboard and numbered its vertices with the digits 1,2,3. It turned out that if Vasya's triangle is rotated 15 times clockwise around its vertex numbered 1 by an angle equal to the angle at this vertex, it will return to its original position. If Vasya's triangle is rotated 3 times clockwise around its vertex numbered 2 by an angle equal to the angle at this vertex, it will return to its original position. Vasya claims that if his triangle is rotated n times around the vertex numbered 3 by an angle equal to the angle at this vertex, it will return to its original position. What is the minimum n that Vasya could have named so that his claim is true for at least some cardboard triangle?Answer: 10.