🧠:[<think>Okay, let's see. The problem says: If z is the conjugate of 3 + 4i, then the magnitude of the vector OA corresponding to z is which of the options given. The options are A: 1, B: sqrt(7), C: sqrt(13), D: 5. Hmm, right. So I need to recall what the conjugate of a complex number is and how to find the magnitude of a complex number. Let's start from the beginning.First, complex numbers have a real part and an imaginary part. The conjugate of a complex number is obtained by changing the sign of the imaginary part. For example, the conjugate of a complex number a + bi is a - bi. So if z is the conjugate of 3 + 4i, that means z = 3 - 4i. Is that right? Let me check. Yes, because the original complex number is 3 + 4i, so flipping the sign of the imaginary part gives 3 - 4i. So z = 3 - 4i.Next, the problem mentions the magnitude of the vector OA corresponding to z. I think in the context of complex numbers, the magnitude of the vector corresponding to a complex number is just the modulus or absolute value of the complex number. The modulus of a complex number a + bi is calculated as sqrt(a² + b²). So regardless of the sign of the imaginary part, the modulus will be the same because squaring a negative number gives a positive result. Wait, so does that mean the modulus of 3 + 4i and its conjugate 3 - 4i are the same? Let me confirm.Calculating modulus for 3 + 4i: sqrt(3² + 4²) = sqrt(9 + 16) = sqrt(25) = 5. Similarly, modulus for 3 - 4i: sqrt(3² + (-4)²) = sqrt(9 + 16) = sqrt(25) = 5. Oh, so they are the same. That makes sense because the modulus is the distance from the origin to the point (a, b) in the complex plane, and flipping the sign of the imaginary part reflects the point over the real axis, but the distance remains the same. Therefore, the magnitude of the vector OA for z, which is the conjugate, is also 5. So the answer should be D: 5.Wait, but let me make sure I didn't skip any steps. The problem says "the magnitude of the vector OA corresponding to z". So OA is probably the vector from the origin (O) to the point A, which is represented by the complex number z. In the complex plane, complex numbers can be represented as points with coordinates (real part, imaginary part). So z = 3 - 4i corresponds to the point (3, -4). The magnitude of the vector OA would then be the distance from (0, 0) to (3, -4), which is indeed sqrt((3 - 0)^2 + (-4 - 0)^2) = sqrt(9 + 16) = sqrt(25) = 5. So yes, that's correct.Let me check the answer options again. Option D is 5, so that's the one. The other options: A: 1 is way too small, B: sqrt(7) is approximately 2.645, which doesn't match. C: sqrt(13) is about 3.605, which is not 5. So D is the only one that fits. Alternatively, just to be thorough, if someone made a mistake and thought that the conjugate somehow affects the modulus, but no, as we saw, modulus is based on the sum of squares, which is the same whether the imaginary part is positive or negative. So even if z is the conjugate, its modulus is the same as the original. Therefore, since the original complex number 3 + 4i has modulus 5, the conjugate has modulus 5 as well. So the answer is D: 5.Final Answerboxed{D}</think>If ( z ) is the conjugate of ( 3 + 4i ), then ( z = 3 - 4i ). The magnitude of the vector ( overrightarrow{OA} ) corresponding to ( z ) is the modulus of the complex number ( z ). The modulus of a complex number ( a + bi ) is calculated as ( sqrt{a^2 + b^2} ). For ( z = 3 - 4i ), the modulus is:[sqrt{3^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5]Thus, the magnitude of the vector ( overrightarrow{OA} ) is ( 5 ).[boxed{D}],<think>Okay, let me see. The problem says that z is the conjugate of 3 + 4i, and then we need to find the magnitude of the vector OA corresponding to z. The options are A through D, with numbers 1, sqrt7, sqrt13, and 5. Hmm. Alright, step by step.First, I need to recall what the conjugate of a complex number is. If we have a complex number in the form a + bi, its conjugate is a - bi. So, the conjugate changes the sign of the imaginary part. So, if the original number is 3 + 4i, then its conjugate z should be 3 - 4i. That seems straightforward. Let me double-check. Yes, conjugating 3 + 4i would flip the sign of the 4i term, making it 3 - 4i. Got that.Next, the problem mentions the magnitude of the vector OA corresponding to z. I think in the context of complex numbers, the vector OA would just be the vector from the origin (0,0) to the point (a, b) where z = a + bi. So, if z is 3 - 4i, then the vector OA would be (3, -4). The magnitude of a vector (x, y) is given by the formula sqrt(x² + y²). So, the magnitude here would be sqrt(3² + (-4)²) = sqrt(9 + 16) = sqrt(25) = 5. Therefore, the magnitude is 5, which is option D.Wait, but let me confirm. The original complex number is 3 + 4i, its conjugate is 3 - 4i. The magnitude of the original is sqrt(3² + 4²) = 5. But does taking the conjugate affect the magnitude? Let me think. The magnitude of a complex number is sqrt(a² + b²), regardless of the sign of b. Because when you square both a and b, the negative sign goes away. So, the magnitude of 3 + 4i is the same as the magnitude of 3 - 4i. So, even if we take the conjugate, the magnitude remains the same. Therefore, the magnitude of z is also 5, which is option D.Therefore, the answer should be D:5. Let me check if that's one of the options. Yes, option D is 5. So, that's the correct answer.But just to make sure I didn't make any mistakes. Let's recapitulate:1. Given z is the conjugate of 3 + 4i. So z = 3 - 4i.2. The vector OA corresponding to z is (3, -4).3. The magnitude is sqrt(3² + (-4)²) = sqrt(9 + 16) = sqrt(25) = 5.Yes, that all checks out. The key point here is that conjugation doesn't change the magnitude of a complex number because squaring the imaginary part removes any negative sign. So both the original number and its conjugate have the same magnitude.Alternatively, if I think about the complex plane, the conjugate reflects the point over the real axis. So, 3 + 4i is in the first quadrant, and its conjugate 3 - 4i is in the fourth quadrant. But the distance from the origin (the magnitude) remains the same because reflection doesn't change the distance. So, whether you are at (3,4) or (3, -4), the distance from (0,0) is still 5. That's a good way to visualize it.Therefore, I feel confident that the answer is D:5.Final Answerboxed{D}</think>Given that ( z ) is the conjugate of ( 3 + 4i ), we need to find the magnitude of the vector (overrightarrow{OA}) corresponding to ( z ).1. Find the conjugate of ( 3 + 4i ): - The conjugate of a complex number ( a + bi ) is ( a - bi ). - Therefore, the conjugate of ( 3 + 4i ) is ( 3 - 4i ).2. Determine the vector corresponding to ( z ): - The complex number ( 3 - 4i ) corresponds to the vector ( (3, -4) ) in the complex plane.3. Calculate the magnitude of the vector: - The magnitude of a vector ( (a, b) ) is given by ( sqrt{a^2 + b^2} ). - For the vector ( (3, -4) ), the magnitude is ( sqrt{3^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5 ).The magnitude of the vector (overrightarrow{OA}) corresponding to ( z ) is ( 5 ).[boxed{D}]]

🔑:Given that z=3-4i, thus, the coordinates of the vector overrightarrow{OA} corresponding to z are (3,-4). Its magnitude is sqrt {3^{2}+(-4)^{2}}=5. Therefore, the correct choice is boxed{D}. From the given information, we find z and then obtain the coordinates of the vector overrightarrow{OA}. We calculate the magnitude of the vector using the formula for the magnitude of a vector. This question tests the algebraic representation of complex numbers and their geometric meaning, as well as the method of finding the magnitude of a vector. It is a basic question.

🧠:[<think>Okay, so I need to find the length of the arc for the polar curve given by ρ = 3φ, where φ ranges from 0 to 4/3. Hmm, let's see. I remember that the formula for the arc length in polar coordinates is different from Cartesian coordinates. Let me recall the formula. I think it's an integral that involves the square root of [ρ² + (dρ/dφ)²] dφ. Let me confirm that. Yeah, right, the general formula for the arc length L in polar coordinates is:L = ∫√[ρ² + (dρ/dφ)²] dφ from φ = a to φ = b.So in this case, ρ is given as 3φ, and the limits for φ are from 0 to 4/3. So first, I need to compute dρ/dφ. Since ρ = 3φ, the derivative dρ/dφ is just 3. That seems straightforward.Now plugging into the formula, the integrand becomes √[(3φ)² + (3)²] = √[9φ² + 9]. I can factor out a 9 from inside the square root, which gives √[9(φ² + 1)] = 3√(φ² + 1). So the integral simplifies to 3 times the integral of √(φ² + 1) dφ from 0 to 4/3.Alright, now I need to compute the integral of √(φ² + 1) dφ. I think this is a standard integral. Let me recall how to integrate √(x² + a²) dx. I remember that the integral is (x/2)√(x² + a²) + (a²/2) ln(x + √(x² + a²)) ) + C, or something like that. Let me check. Let me set a substitution here. Let me let φ = sinh t, since the integral of √(φ² + 1) dφ might be handled via hyperbolic substitution. Alternatively, maybe integration by parts. Let me try integration by parts.Let me set u = √(φ² + 1) and dv = dφ. Then du would be (1/2)(φ² + 1)^(-1/2) * 2φ dφ = φ / √(φ² + 1) dφ. And v = φ. So integration by parts gives uv - ∫v du = φ√(φ² + 1) - ∫ φ * [φ / √(φ² + 1)] dφ.Simplify the integral term: ∫ [φ² / √(φ² + 1)] dφ. Hmm, maybe rewrite this as ∫ [ (φ² + 1 - 1) / √(φ² + 1) ] dφ = ∫ √(φ² + 1) dφ - ∫ 1 / √(φ² + 1) dφ.So putting this back into the integration by parts equation:Original integral = φ√(φ² + 1) - [ Integral of √(φ² +1 ) dφ - Integral of 1 / √(φ² +1 ) dφ ]Let me denote the original integral as I. Then:I = φ√(φ² +1 ) - (I - ∫ 1 / √(φ² +1 ) dφ )So bring I to the left side:I + I = φ√(φ² +1 ) + ∫ 1 / √(φ² +1 ) dφThus, 2I = φ√(φ² +1 ) + ∫ 1 / √(φ² +1 ) dφNow, the remaining integral is ∫ 1 / √(φ² +1 ) dφ. I know that integral is sinh^{-1}(φ) + C, which is also ln(φ + √(φ² +1 )) + C. So putting it all together:2I = φ√(φ² +1 ) + ln(φ + √(φ² +1 )) + CTherefore, dividing both sides by 2:I = (1/2)[φ√(φ² +1 ) + ln(φ + √(φ² +1 ))] + CSo the integral of √(φ² +1 ) dφ is (1/2)(φ√(φ² +1 ) + ln(φ + √(φ² +1 )) ) + C. Let me verify this by differentiation. Let's compute d/dφ of (1/2)(φ√(φ² +1 ) + ln(φ + √(φ² +1 )) ).First term derivative: (1/2)[ √(φ² +1 ) + φ*(1/(2√(φ² +1 )))*2φ ] = (1/2)[ √(φ² +1 ) + (φ²)/√(φ² +1 ) ]Second term derivative: (1/2)*(1/(φ + √(φ² +1 ))) * [1 + (1/(2√(φ² +1 )))*2φ ] = (1/2)*(1/(φ + √(φ² +1 ))) * [1 + φ/√(φ² +1 ) ]Hmm, this seems a bit complicated, but let me see. Let me compute each part step by step.First term's derivative: Let's take the first part: (1/2)√(φ² +1 ). Its derivative is (1/2)*( (2φ)/(2√(φ² +1 )) ) = (1/2)*(φ/√(φ² +1 )).Wait, no. Wait, actually, derivative of √(φ² +1 ) is (1/(2√(φ² +1 ))) * 2φ = φ / √(φ² +1 ). So the first term in the first part is (1/2)*φ / √(φ² +1 ).Wait, no. Wait, the first term is (1/2) multiplied by the derivative of φ√(φ² +1 ). Let me compute that derivative properly. The derivative of φ√(φ² +1 ) is √(φ² +1 ) + φ*( derivative of √(φ² +1 ) ). Which is √(φ² +1 ) + φ*( (φ)/√(φ² +1 ) ) = √(φ² +1 ) + φ² / √(φ² +1 ). Then, multiplying by (1/2), we have (1/2)[ √(φ² +1 ) + φ² / √(φ² +1 ) ].The derivative of the second part, (1/2)ln(φ + √(φ² +1 )), is (1/2)*(1/(φ + √(φ² +1 )))*(1 + (φ)/√(φ² +1 )). Let's simplify this.Let me denote S = √(φ² +1 ). Then the derivative is (1/2)*(1/(φ + S))*(1 + φ/S ) = (1/2)*( (S + φ ) / (S(φ + S)) ) ) = (1/2)*(1/S ). Therefore, the derivative of the second term is (1/2)*(1/√(φ² +1 )).Therefore, combining both derivatives:First part derivative: (1/2)(√(φ² +1 ) + φ² / √(φ² +1 )) = (1/2)( (φ² +1 ) + φ² ) / √(φ² +1 ) = (1/2)(2φ² +1 ) / √(φ² +1 )Second part derivative: (1/2)(1 / √(φ² +1 ))So total derivative is (2φ² +1 )/(2√(φ² +1 )) + 1/(2√(φ² +1 )) = (2φ² +1 +1 )/(2√(φ² +1 )) = (2φ² +2 )/(2√(φ² +1 )) = (φ² +1 ) / √(φ² +1 ) = √(φ² +1 ). Which is correct, because that's the integrand. So yes, the integration is correct. Great.Therefore, going back, the integral of √(φ² +1 ) dφ is (1/2)(φ√(φ² +1 ) + ln(φ + √(φ² +1 )) ) + C.Therefore, returning to the arc length problem. The arc length L is 3 times the integral from 0 to 4/3 of √(φ² +1 ) dφ. So plugging in the antiderivative:L = 3 * [ (1/2)(φ√(φ² +1 ) + ln(φ + √(φ² +1 )) ) ] evaluated from 0 to 4/3.Compute this at 4/3 and subtract the value at 0. Let's compute each term.First, evaluate at φ = 4/3.Term 1: (4/3)√( (4/3)^2 +1 ) = (4/3)√(16/9 +1 ) = (4/3)√(25/9 ) = (4/3)*(5/3) = 20/9.Term 2: ln(4/3 + √( (4/3)^2 +1 )) = ln(4/3 + 5/3 ) = ln(9/3 ) = ln(3).So the first part at φ=4/3 is (20/9 + ln 3).Now evaluate at φ=0.Term 1: 0 * √(0 +1 ) = 0.Term 2: ln(0 + √(0 +1 )) = ln(1) = 0.Therefore, the value at 0 is 0.Thus, the integral from 0 to 4/3 is (1/2)(20/9 + ln3 - 0) = (1/2)(20/9 + ln3).Therefore, the arc length L is 3 * (1/2)(20/9 + ln3 ) = (3/2)(20/9 + ln3 ) = (3/2)*(20/9) + (3/2)*ln3 = (60/18) + (3/2)ln3 = (10/3) + (3/2)ln3.Simplify 10/3 is approximately 3.333..., and (3/2)ln3 is approximately (1.5)(1.0986) ≈ 1.6479, but since the problem probably wants an exact answer, we can leave it in terms of fractions and ln3.Thus, the final answer is 10/3 + (3/2) ln3. To write this as a single expression, factor out 1/2: (20/6 + 9/6 ln3 ) = (20 + 9 ln3 ) / 6. Wait, no. Wait, 10/3 is equal to 20/6, and (3/2) ln3 is (9/6) ln3. Therefore, combining them over a common denominator of 6: (20 + 9 ln3 ) / 6. Alternatively, leave it as 10/3 + (3/2) ln3. Both are correct, but perhaps the former is neater. Let me check.Alternatively, multiply numerator and denominator:10/3 + (3/2) ln3 = (20 + 9 ln3 ) / 6. Yes, because 10/3 = 20/6 and 3/2 = 9/6, so 20/6 + 9/6 ln3 = (20 + 9 ln3)/6. So either form is acceptable, but perhaps the problem expects the answer in terms of fractions combined. Let me check the problem statement. It just says "Calculate the lengths of the arcs...", and doesn't specify the form, so both are acceptable, but perhaps the combined fraction is better.Therefore, the final answer is (20 + 9 ln3 ) / 6. Alternatively, simplifying further, 10/3 + (3/2) ln3. Either form is correct, but let me check units and if there's any simplification. Since ln3 is just a constant, there's no further simplification. Let me also verify the calculations again to make sure I didn't make an arithmetic mistake.So the integral was 3 * [ (1/2)(φ√(φ² +1 ) + ln(φ + √(φ² +1 )) ) ] from 0 to 4/3.At φ=4/3: (4/3)*sqrt(16/9 +9/9)= (4/3)*sqrt(25/9)= (4/3)*(5/3)=20/9. ln(4/3 +5/3)=ln(9/3)=ln3. So 20/9 + ln3. Multiply by 1/2: (20/9 + ln3)/2. Then multiply by 3: (20/9 + ln3 )*(3/2)= (20/9)*(3/2) + (ln3)*(3/2)= (60/18) + (3/2)ln3=10/3 + (3/2)ln3. Correct. And 10/3 is 20/6, 3/2 is 9/6, so combined over 6: (20 +9 ln3)/6.Yes, that's correct. So the answer can be written as either (20 + 9 ln3)/6 or 10/3 + (3/2) ln3. Both are equivalent. Let me check if the problem expects a particular form. Since it's a calculus problem, often they prefer simplified fractions with combined terms. So perhaps (20 + 9 ln3)/6. Alternatively, if factored as 10/3 + (3/2) ln3, that's also acceptable.But let me check if I can simplify 20 +9 ln3 over 6. Since 20 and 9 are coprime with 6? 20/6 reduces to 10/3, and 9/6 is 3/2. So yeah, it's the same. So both forms are correct. So either is fine, but maybe the first form is preferable as a single fraction. Alternatively, the problem may prefer the non-combined version. Since the problem says "put your final answer within boxed{}", probably either is acceptable, but perhaps the second form is better because it's split into terms with rational coefficients. Let me check with standard integral answers. For example, in calculus textbooks, such integrals are often presented with separate terms rather than combined over a common denominator unless necessary. So 10/3 + (3/2) ln3 is probably the standard form. Therefore, I think that's the preferable answer.So, final answer: 10/3 + (3/2) ln3. Therefore, boxed, it's boxed{dfrac{10}{3} + dfrac{3}{2} ln 3}.Final AnswerThe length of the arc is boxed{dfrac{10}{3} + dfrac{3}{2} ln 3}.</think>To calculate the length of the arc for the polar curve given by (rho = 3varphi) where (0 leq varphi leq frac{4}{3}), we use the formula for arc length in polar coordinates:[L = int_{a}^{b} sqrt{rho^2 + left(frac{drho}{dvarphi}right)^2} , dvarphi]Given (rho = 3varphi), we find (frac{drho}{dvarphi} = 3). Substituting into the arc length formula, we get:[L = int_{0}^{frac{4}{3}} sqrt{(3varphi)^2 + 3^2} , dvarphi = int_{0}^{frac{4}{3}} sqrt{9varphi^2 + 9} , dvarphi = 3 int_{0}^{frac{4}{3}} sqrt{varphi^2 + 1} , dvarphi]The integral (int sqrt{varphi^2 + 1} , dvarphi) can be solved using integration by parts, resulting in:[int sqrt{varphi^2 + 1} , dvarphi = frac{1}{2} left( varphi sqrt{varphi^2 + 1} + ln left( varphi + sqrt{varphi^2 + 1} right) right) + C]Evaluating this from (0) to (frac{4}{3}):[int_{0}^{frac{4}{3}} sqrt{varphi^2 + 1} , dvarphi = left[ frac{1}{2} left( varphi sqrt{varphi^2 + 1} + ln left( varphi + sqrt{varphi^2 + 1} right) right) right]_{0}^{frac{4}{3}}]At (varphi = frac{4}{3}):[frac{1}{2} left( frac{4}{3} sqrt{left( frac{4}{3} right)^2 + 1} + ln left( frac{4}{3} + sqrt{left( frac{4}{3} right)^2 + 1} right) right) = frac{1}{2} left( frac{20}{9} + ln 3 right)]At (varphi = 0):[frac{1}{2} left( 0 + ln 1 right) = 0]Subtracting these results:[int_{0}^{frac{4}{3}} sqrt{varphi^2 + 1} , dvarphi = frac{1}{2} left( frac{20}{9} + ln 3 right)]Multiplying by 3:[L = 3 times frac{1}{2} left( frac{20}{9} + ln 3 right) = frac{3}{2} left( frac{20}{9} + ln 3 right) = frac{10}{3} + frac{3}{2} ln 3]Thus, the length of the arc is (boxed{dfrac{10}{3} + dfrac{3}{2} ln 3}).,<think>Okay, so I need to find the length of the arc for the polar curve given by ρ = 3φ, where φ ranges from 0 to 4/3. Hmm, let me recall how arc length is calculated in polar coordinates. I think the formula involves integrating some expression involving ρ and its derivative with respect to φ. Let me check my notes or maybe derive it.Right, in Cartesian coordinates, the arc length of a parametric curve x(φ), y(φ) from φ=a to φ=b is the integral from a to b of sqrt[(dx/dφ)^2 + (dy/dφ)^2] dφ. But since this is a polar coordinate system, where x = ρ cos φ and y = ρ sin φ. So maybe I can substitute these into the Cartesian formula. Let's try that.So, x = ρ cos φ = 3φ cos φ, and y = 3φ sin φ. Then, dx/dφ would be the derivative of 3φ cos φ with respect to φ. Using the product rule: 3 cos φ - 3φ sin φ. Similarly, dy/dφ would be derivative of 3φ sin φ, which is 3 sin φ + 3φ cos φ.Then, (dx/dφ)^2 + (dy/dφ)^2 would be [3 cos φ - 3φ sin φ]^2 + [3 sin φ + 3φ cos φ]^2. Let me expand these terms.First term: [3 cos φ - 3φ sin φ]^2 = 9 cos² φ - 18φ cos φ sin φ + 9φ² sin² φ.Second term: [3 sin φ + 3φ cos φ]^2 = 9 sin² φ + 18φ sin φ cos φ + 9φ² cos² φ.Adding these two together, the cross terms (-18φ cos φ sin φ and +18φ sin φ cos φ) cancel each other out. So we have:9 cos² φ + 9φ² sin² φ + 9 sin² φ + 9φ² cos² φ.Factor out the 9:9[cos² φ + sin² φ + φ² sin² φ + φ² cos² φ].But cos² φ + sin² φ = 1, and φ² sin² φ + φ² cos² φ = φ² (sin² φ + cos² φ) = φ². So this simplifies to:9[1 + φ²].Therefore, the integrand for the arc length is sqrt[9(1 + φ²)] = 3 sqrt(1 + φ²).So the arc length S is the integral from φ=0 to φ=4/3 of 3 sqrt(1 + φ²) dφ. So S = 3 ∫₀^(4/3) sqrt(1 + φ²) dφ.Now, I need to compute this integral. The integral of sqrt(1 + x²) dx is a standard one. Let me recall the formula. I think it can be integrated using substitution or hyperbolic substitution, but maybe trigonometric substitution. Alternatively, integration by parts.Let me try integration by parts. Let u = sqrt(1 + φ²), dv = dφ. Then du = (φ / sqrt(1 + φ²)) dφ, and v = φ. So integration by parts gives:uv - ∫ v du = φ sqrt(1 + φ²) - ∫ φ * (φ / sqrt(1 + φ²)) dφ.Simplify the integral: ∫ φ² / sqrt(1 + φ²) dφ. Let me write this as ∫ (φ² + 1 - 1)/sqrt(1 + φ²) dφ = ∫ sqrt(1 + φ²) dφ - ∫ 1/sqrt(1 + φ²) dφ.Therefore, the original integral, let's call it I, is:I = φ sqrt(1 + φ²) - [I - ∫ 1/sqrt(1 + φ²) dφ].So bringing I to the left side:I + I = φ sqrt(1 + φ²) + ∫ 1/sqrt(1 + φ²) dφ.Thus, 2I = φ sqrt(1 + φ²) + ∫ 1/sqrt(1 + φ²) dφ.The integral ∫ 1/sqrt(1 + φ²) dφ is a standard one, which is ln(φ + sqrt(φ² + 1)) + C.Therefore, 2I = φ sqrt(1 + φ²) + ln(φ + sqrt(φ² + 1)) + C.Hence, I = (1/2)[φ sqrt(1 + φ²) + ln(φ + sqrt(φ² + 1))] + C.Therefore, the arc length S is 3 times the evaluated integral from 0 to 4/3:S = 3 * [ (1/2)(φ sqrt(1 + φ²) + ln(φ + sqrt(φ² + 1))) ] from 0 to 4/3.Let's compute this. First, plug in φ = 4/3:Term1 = (4/3) sqrt(1 + (16/9)) = (4/3) sqrt(25/9) = (4/3)(5/3) = 20/9.Term2 = ln(4/3 + sqrt(1 + (16/9))) = ln(4/3 + sqrt(25/9)) = ln(4/3 + 5/3) = ln(9/3) = ln(3).Now, when φ = 0:Term1 = 0 * sqrt(1 + 0) = 0.Term2 = ln(0 + sqrt(0 + 1)) = ln(1) = 0.Therefore, the integral from 0 to 4/3 is (1/2)[20/9 + ln(3)] - (1/2)(0 + 0) = (1/2)(20/9 + ln 3).Thus, S = 3 * (1/2)(20/9 + ln 3) = (3/2)(20/9 + ln 3).Simplify 3/2 * 20/9: (3/2)*(20/9) = (60)/18 = 10/3.And 3/2 * ln 3 = (3/2) ln 3.Therefore, S = 10/3 + (3/2) ln 3.Alternatively, this can be written as (10/3) + (3/2) ln 3.Let me check if the integration steps are correct. Let me verify the integral of sqrt(1 + φ²) dφ. Let me look it up to confirm. Yes, the integral is indeed (1/2)(φ sqrt(1 + φ²) + sinh^{-1}(φ)) ) + C, but sinh^{-1}(φ) is equivalent to ln(φ + sqrt(φ² +1)). So that's correct. So the integral evaluation is correct.Thus, the final expression is 10/3 plus three-halves the natural logarithm of 3. Let me compute this numerically to check if it's reasonable. Let's see:10/3 ≈ 3.333..., ln 3 ≈ 1.0986, so (3/2)*1.0986 ≈ 1.6479. So total S ≈ 3.333 + 1.6479 ≈ 4.9809. So approximately 5 units. That seems plausible.Alternatively, if we keep it in exact terms, it's 10/3 + (3/2) ln 3, which is the exact value. So unless the problem wants a decimal approximation, we can leave it in terms of logarithms.Therefore, the arc length is (10/3) + (3/2) ln 3. Let me write that as the final answer.Final AnswerThe length of the arc is boxed{dfrac{10}{3} + dfrac{3}{2} ln 3}.</think>To calculate the length of the arc of the curve given by the equation (rho = 3 varphi) in polar coordinates, where (0 leq varphi leq frac{4}{3}), we use the formula for arc length in polar coordinates. The formula is:[S = int_{a}^{b} sqrt{left(frac{dx}{dvarphi}right)^2 + left(frac{dy}{dvarphi}right)^2} , dvarphi]For the given polar coordinates (x = 3varphi cos varphi) and (y = 3varphi sin varphi), we compute the derivatives:[frac{dx}{dvarphi} = 3 cos varphi - 3varphi sin varphi][frac{dy}{dvarphi} = 3 sin varphi + 3varphi cos varphi]Squaring and adding these derivatives:[left(frac{dx}{dvarphi}right)^2 + left(frac{dy}{dvarphi}right)^2 = 9 cos^2 varphi - 18varphi cos varphi sin varphi + 9varphi^2 sin^2 varphi + 9 sin^2 varphi + 18varphi sin varphi cos varphi + 9varphi^2 cos^2 varphi]Simplifying the expression, we get:[9 (cos^2 varphi + sin^2 varphi + varphi^2 (cos^2 varphi + sin^2 varphi)) = 9 (1 + varphi^2)]Thus, the integrand becomes:[sqrt{9 (1 + varphi^2)} = 3 sqrt{1 + varphi^2}]The arc length (S) is:[S = 3 int_{0}^{frac{4}{3}} sqrt{1 + varphi^2} , dvarphi]Using the integral result for (int sqrt{1 + varphi^2} , dvarphi):[int sqrt{1 + varphi^2} , dvarphi = frac{1}{2} left( varphi sqrt{1 + varphi^2} + ln left( varphi + sqrt{1 + varphi^2} right) right)]Evaluating this from (0) to (frac{4}{3}):[int_{0}^{frac{4}{3}} sqrt{1 + varphi^2} , dvarphi = frac{1}{2} left( frac{4}{3} sqrt{1 + left( frac{4}{3} right)^2} + ln left( frac{4}{3} + sqrt{1 + left( frac{4}{3} right)^2} right) right) - frac{1}{2} left( 0 + ln(1) right)]Simplifying the terms:[= frac{1}{2} left( frac{4}{3} cdot frac{5}{3} + ln(3) right) = frac{1}{2} left( frac{20}{9} + ln(3) right)]Multiplying by 3:[S = 3 cdot frac{1}{2} left( frac{20}{9} + ln(3) right) = frac{3}{2} left( frac{20}{9} + ln(3) right) = frac{10}{3} + frac{3}{2} ln(3)]The length of the arc is (boxed{dfrac{10}{3} + dfrac{3}{2} ln 3}).]

🔑:## SolutionThe length of the arc of a curve given by an equation in polar coordinates is determined by the formulaL=int_{varphi_{0}}^{varphi_{1}} sqrt{(rho(varphi))^{2}+left(rho^{prime}(varphi)right)^{2}} d varphiFor the curve given by the equation rho=3 varphi, we find: rho^{prime}=3We obtain:begin{aligned}L & =int_{0}^{4 / 3} sqrt{(3 varphi)^{2}+3^{2}} d varphi= & =3 int_{0}^{4 / 3} sqrt{varphi^{2}+1} d varphi={ }^{(1)} & =3left(left.frac{varphi}{2} sqrt{varphi^{2}+1}right|_{0} ^{4 / 3}+left.frac{1}{2} ln left|varphi+sqrt{varphi^{2}+1}right|right|_{0} ^{4 / 3}right)= & =3left(frac{1}{2} cdot frac{4}{3} sqrt{frac{16}{9}+1}-0right)+3left(frac{1}{2} ln left|frac{4}{3}+sqrt{frac{16}{9}+1}right|-ln |0+sqrt{0+1}|right)= & =2 cdot sqrt{frac{25}{9}}+3left(frac{1}{2} ln left|frac{4}{3}+sqrt{frac{25}{9}}right|-0right)= & =2 cdot frac{5}{3}+frac{3}{2} ln left|frac{4}{3}+frac{5}{3}right|=frac{10}{3}+frac{3}{2} ln 3end{aligned}In (1) we used the formula: int sqrt{x^{2}+a^{2}} d x=frac{x}{2} sqrt{x^{2}+a^{2}}+frac{a^{2}}{2} ln left|x+sqrt{x^{2}+a^{2}}right|Source — «http://pluspi.org/wiki/index.php/������������ %D0%9A%D1%83%D0%B7%D0%BD%D0%B5%D1%86%D0%BE%D0%B2_%D0%98%D0%BD % mathrm{D} 1 % 82 % mathrm{D} 0 % mathrm{~B} 5 % mathrm{D} 0 % mathrm{~B} 3 % mathrm{D} 1 % 80 % mathrm{D} 0 % mathrm{~B} 0 % mathrm{D} 0 % mathrm{BB} % mathrm{D} 1 % 8 mathrm{~B}+19-24 »Categories: Kuznetsov's Problem Book Integrals Problem 19 mid IntegralsUkrainian Banner Network- Last modified: 08: 15,28 May 2010.- Content is available under CC-BY-SA 3.0.Created by GeeTeatoo## Problem Kuznetsov Integrals 19-25## Material from PlusPi

🧠:[<think>Okay, let's see. I have two problems here related to a function f(x) = (1/2)a x² + b x + 1, where a can be either 2 or 4, and b can be 1 or 3. I need to find two probabilities: first, the probability that f(x) is decreasing on the interval (-∞, -1], and second, the probability that if I pick two such functions randomly, their tangent lines at the point (1, f(1)) are parallel.Starting with problem 1. The function is a quadratic, so its graph is a parabola. For a quadratic function ax² + bx + c, the derivative is 2ax + b. But in this case, the function is given as (1/2)a x² + b x + 1, so the derivative would be a x + b. Right? Because derivative of (1/2)a x² is a x, derivative of b x is b, and derivative of 1 is 0. So f'(x) = a x + b.A function is decreasing where its derivative is negative. So we need f'(x) ≤ 0 for all x in (-∞, -1]. Wait, but since it's a quadratic, the function is either concave up or concave down depending on the coefficient of x². Since a is either 2 or 4, which are positive, so the parabola opens upwards. Therefore, the function will have a minimum point. The derivative f'(x) = a x + b. The critical point is at x = -b/a. Since a is positive, the function is decreasing to the left of the critical point and increasing to the right. So, to have f(x) decreasing on (-∞, -1], we need that the critical point (the vertex) is at x ≥ -1. Because then, to the left of x = -1, the function is still decreasing. Wait, no. If the critical point is at x = -b/a, then for the function to be decreasing on (-∞, -1], the critical point must be at x ≥ -1. Because the function is decreasing from (-∞, critical point], then increasing. So if the critical point is at x ≥ -1, then on (-∞, -1], which is a subset of (-∞, critical point], the function is decreasing. If the critical point is less than -1, then on (-∞, -1], the function would first be decreasing until the critical point, and then increasing from critical point to -1. Wait, but that can't be. Wait, the function is decreasing on (-∞, critical point] and increasing on [critical point, ∞). So if the critical point is at x = -b/a, then if -b/a ≥ -1, then (-∞, -1] is entirely within the decreasing interval. If -b/a < -1, then on (-∞, -1], the function is decreasing only up to x = -b/a, and increasing afterward. Therefore, for f(x) to be decreasing on the entire interval (-∞, -1], we need that the critical point -b/a ≥ -1. Which is equivalent to -b/a ≥ -1 → multiplying both sides by a (positive, so inequality remains the same) → -b ≥ -a → a ≥ b.Therefore, the condition is a ≥ b. Since a can be 2 or 4, and b can be 1 or 3. So let's list all possible combinations. The possible (a, b) pairs are (2,1), (2,3), (4,1), (4,3). So total of 4 possible functions.Now, for each of these, check if a ≥ b:1. a=2, b=1: 2 ≥ 1 → yes2. a=2, b=3: 2 ≥ 3 → no3. a=4, b=1: 4 ≥ 1 → yes4. a=4, b=3: 4 ≥ 3 → yesSo out of 4 possible functions, 3 satisfy a ≥ b. Therefore, the probability is 3/4.Wait, but let's verify that. Let's compute -b/a for each case:1. a=2, b=1: critical point x = -1/2 ≈ -0.5. So -0.5 is greater than -1. So on (-∞, -1], the function is decreasing because it's left of the critical point. Wait, but if the critical point is at x=-0.5, then the function is decreasing on (-∞, -0.5] and increasing on [-0.5, ∞). So the interval (-∞, -1] is entirely within (-∞, -0.5], so the function is decreasing there. So that's okay.2. a=2, b=3: critical point x = -3/2 = -1.5. So the critical point is at -1.5. Then the function is decreasing on (-∞, -1.5] and increasing on [-1.5, ∞). So on (-∞, -1], since -1 is greater than -1.5, the interval (-∞, -1] includes points from (-∞, -1.5] where it's decreasing and from [-1.5, -1] where it's increasing. Therefore, the function is not decreasing on the entire interval.3. a=4, b=1: critical point x = -1/4 = -0.25. So similar to first case, the interval (-∞, -1] is entirely within (-∞, -0.25], so decreasing there.4. a=4, b=3: critical point x = -3/4 = -0.75. So again, (-∞, -1] is within (-∞, -0.75], so decreasing there.Therefore, yes, only the second case fails. So 3 out of 4 functions satisfy the condition. So probability is 3/4.Okay, that seems correct. So the answer to problem 1 is 3/4.Now moving to problem 2. We need to select two functions randomly from the set of possible functions (which are 4 in total) and find the probability that their tangent lines at the point (1, f(1)) are parallel.First, tangent lines at a point are parallel if their slopes at that point are equal. The slope of the tangent line at x=1 is given by f'(1). Since f'(x) = a x + b, so f'(1) = a*1 + b = a + b.Therefore, two functions will have parallel tangent lines at (1, f(1)) if their a + b are equal. So we need to compute a + b for each function and see how many pairs have the same a + b.First, let's list all possible functions with their a and b values:1. a=2, b=1: a + b = 32. a=2, b=3: a + b = 53. a=4, b=1: a + b = 54. a=4, b=3: a + b = 7So the a + b values are 3, 5, 5, 7.Now, how many pairs of functions have the same a + b? Let's list all possible pairs. Since we are selecting two functions randomly from the set, the number of possible pairs is C(4,2) = 6.The pairs are:1. Function 1 and Function 2: 3 vs 5 → different2. Function 1 and Function 3: 3 vs 5 → different3. Function 1 and Function 4: 3 vs 7 → different4. Function 2 and Function 3: 5 vs 5 → same5. Function 2 and Function 4: 5 vs 7 → different6. Function 3 and Function 4: 5 vs 7 → differentSo out of 6 pairs, only 1 pair (Function 2 and Function 3) have the same slope. Therefore, the probability is 1/6.Wait, but hold on. Wait, when we select two functions randomly, is the selection with replacement or without replacement? The problem says "two functions are randomly selected from the set of possible functions", so it's without replacement. Therefore, the number of possible selections is C(4,2)=6, as above.But let me confirm: Function 2 is (a=2, b=3) and Function 3 is (a=4, b=1). Both have a + b =5, so their slopes at x=1 are 5, which are equal. So this is the only pair.Hence, the probability is 1/6.But wait, maybe I made a mistake here. Let me check again.Wait, another way: total number of possible pairs is 6. The pairs that have equal a + b are those where a + b is equal. From the list:- a + b =3: only Function 1- a + b=5: Function 2 and 3- a + b=7: Function4Therefore, the number of pairs with same a + b is the pairs within the same group. So for a + b=5, there is C(2,2)=1 pair. The other groups have only one function each, so no pairs. Therefore, total favorable pairs is 1. Hence, probability 1/6.Therefore, answer is 1/6.But wait, hold on. Wait, but if we consider ordered pairs instead of unordered? No, the problem says "two functions are randomly selected", which is typically considered unordered unless specified otherwise. So it's 6 possible pairs, 1 favorable. So 1/6.Alternatively, if we thought of it as ordered pairs, then the number of ordered pairs is 4*3=12, and the number of ordered pairs where the slopes are equal would be 2 (since Function2 and Function3 can be ordered in two ways). Then the probability would be 2/12=1/6. So same answer. So regardless, the probability is 1/6.Therefore, the answer to the second problem is 1/6.But wait, hold on. Let me check again the a + b values:Function1: a=2, b=1: 3Function2: a=2, b=3:5Function3: a=4, b=1:5Function4: a=4, b=3:7So yes, only Function2 and Function3 have the same slope. So only that pair.Hence, 1/6.But I just want to make sure there isn't any other pair. For example, is there a case where two different (a, b) pairs give same a + b?Looking at the numbers:3,5,5,7. So only two functions have a + b=5. So yes, only one pair.Therefore, the answer is 1/6.Hmm. So problem 1 answer is 3/4, problem 2 answer is 1/6. Let me confirm once more.Problem1: 3 out of 4 functions satisfy a ≥ b. Correct.Problem2: Only one pair out of six possible pairs has equal a + b. Correct.Yes, that seems right. Therefore, the final answers are 3/4 and 1/6.Final Answer1. boxed{dfrac{3}{4}}2. boxed{dfrac{1}{6}}</think>Given (a in {2, 4}) and (b in {1, 3}), consider the function (f(x) = frac{1}{2}ax^2 + bx + 1).1. To find the probability that (f(x)) is a decreasing function on the interval ((-infty, -1]), we first compute the derivative (f'(x) = ax + b). A quadratic function is decreasing where its derivative is negative. The critical point is (x = -frac{b}{a}). For (f(x)) to be decreasing on ((-infty, -1]), the critical point must be at (x geq -1), which translates to (-frac{b}{a} geq -1) or (a geq b). Evaluating the possible pairs ((a, b)): - ((2, 1)): (2 geq 1) (satisfies) - ((2, 3)): (2 notgeq 3) (does not satisfy) - ((4, 1)): (4 geq 1) (satisfies) - ((4, 3)): (4 geq 3) (satisfies) Three out of four pairs satisfy the condition, so the probability is (frac{3}{4}).2. To find the probability that the tangent lines at the point ((1, f(1))) are parallel for two randomly selected functions, we note that the slope of the tangent line at (x = 1) is (f'(1) = a + b). We need pairs of functions with the same (a + b) value. Calculating (a + b) for each function: - ((2, 1)): (3) - ((2, 3)): (5) - ((4, 1)): (5) - ((4, 3)): (7) The pairs with the same slope are ((2, 3)) and ((4, 1)). There are (binom{4}{2} = 6) possible pairs, and only one pair has the same slope. Thus, the probability is (frac{1}{6}). Final Answer1. boxed{dfrac{3}{4}}2. boxed{dfrac{1}{6}},<think>Okay, so I have this problem here where I need to find probabilities related to a function f(x) = (1/2)a x² + b x + 1. The coefficients a and b are from the sets {2,4} and {1,3} respectively. There are two parts to the problem. Let me tackle them one by one.Starting with part 1: Find the probability that f(x) is a decreasing function on the interval (-∞, -1].Hmm. First, I remember that for a function to be decreasing on an interval, its derivative should be non-positive throughout that interval. Since f(x) is a quadratic function, its derivative f’(x) will be linear. Let me confirm that.The function is f(x) = (1/2)a x² + b x + 1. Taking the derivative with respect to x, we get f’(x) = a x + b. Right, so the derivative is linear in x. For f(x) to be decreasing on (-∞, -1], the derivative must be less than or equal to zero for all x in that interval. However, since the derivative is a linear function, its maximum or minimum occurs at the endpoints if we consider a closed interval. But here the interval is (-∞, -1], which is unbounded. So, since the derivative is linear, it's going to be decreasing if the coefficient of x in the derivative is negative, or increasing if the coefficient is positive. Wait, but the derivative here is f’(x) = a x + b. So the coefficient of x is a, which is either 2 or 4, both positive numbers. Therefore, the derivative is a linear function with a positive slope. That means as x increases, f’(x) increases. Therefore, the derivative is increasing. So the minimum value of f’(x) on (-∞, -1] would be as x approaches -∞, which goes to negative infinity, but since a is positive, f’(x) approaches negative infinity as x approaches negative infinity. Wait, that can't be. Wait, if x is approaching negative infinity, and a is positive, then a x is approaching negative infinity, so f’(x) = a x + b approaches negative infinity. But since the derivative is linear with a positive slope, it is increasing. Therefore, on the interval (-∞, -1], the maximum value of f’(x) would be at x = -1, because as x increases towards -1 from the left, the derivative increases. Therefore, the maximum derivative on that interval is at x = -1. So, if the maximum derivative at x = -1 is less than or equal to zero, then the entire interval would have the derivative less than or equal to zero. Therefore, f(x) is decreasing on (-∞, -1] if and only if f’(-1) ≤ 0.Therefore, the condition is f’(-1) ≤ 0. Let's compute f’(-1). f’(x) = a x + b, so f’(-1) = -a + b. So, we need -a + b ≤ 0. Which simplifies to b ≤ a.Therefore, for each possible a and b, we need to check whether b ≤ a. Since a can be 2 or 4, and b can be 1 or 3. Let's list all possible pairs (a, b):When a = 2:- b = 1: 1 ≤ 2, which is true.- b = 3: 3 ≤ 2, which is false.When a = 4:- b = 1: 1 ≤ 4, which is true.- b = 3: 3 ≤ 4, which is true.So total possible pairs are 2 (for a) * 2 (for b) = 4. Out of these 4, how many satisfy b ≤ a?For a=2: 1 satisfies (b=1), 1 doesn't (b=3)For a=4: both b=1 and b=3 satisfy (1 ≤4 and 3 ≤4)So total favorable cases: 1 (from a=2) + 2 (from a=4) = 3.Therefore, the probability is 3/4.Wait, but let me confirm once again. The possible functions are:1. a=2, b=1: f'(x)=2x +1. Then f’(-1) = -2 +1 = -1 ≤0. So decreasing on (-∞, -1]. Correct.2. a=2, b=3: f'(x)=2x +3. f’(-1)= -2 +3 =1 >0. So not decreasing.3. a=4, b=1: f’(x)=4x +1. f’(-1)= -4 +1 = -3 ≤0. So decreasing.4. a=4, b=3: f’(x)=4x +3. f’(-1)= -4 +3 = -1 ≤0. So decreasing.Thus, out of 4 possible functions, 3 of them satisfy the condition. Therefore, the probability is 3/4. So part 1 answer is 3/4.Moving on to part 2: If two functions are randomly selected from the set of possible functions, find the probability that their tangent lines at the point (1, f(1)) are parallel.Hmm. Tangent lines at a specific point being parallel. The tangent line at a point (x0, f(x0)) for a function f has a slope equal to the derivative at that point, f’(x0). So, for two tangent lines to be parallel at the same point (1, f(1)), their slopes must be equal. Therefore, the derivatives of the two functions at x=1 must be equal.So, first, we need to compute f’(1) for each function. The derivative of f(x) is a x + b, so f’(1) = a*1 + b = a + b. Therefore, the slope of the tangent line at x=1 is a + b. Therefore, two functions will have parallel tangent lines at (1, f(1)) if and only if their a + b values are equal.Therefore, the problem reduces to: when selecting two functions at random (without replacement, I assume, since it's a probability question), what is the probability that their (a + b) values are equal.First, let's list all possible functions with their a and b values and compute a + b for each.Possible functions:1. (a=2, b=1): a + b = 32. (a=2, b=3): a + b = 53. (a=4, b=1): a + b = 54. (a=4, b=3): a + b = 7So the a + b values are 3,5,5,7.So the possible pairs of functions. Since we are selecting two functions randomly, how many total possible pairs are there? There are 4 functions, so the number of ways to choose two is C(4,2) = 6.Now, how many of these pairs have the same a + b?Looking at the a + b values: 3,5,5,7.So the pairs with the same a + b would be the two functions with a + b =5, which are function 2 and function 3. So the pair (2,3) is the only pair with the same a + b. Wait, but let's check:Function 1: 3Function 2:5Function3:5Function4:7So functions 2 and 3 both have a + b=5. So the pair (2,3) is one such pair. Any others?Function1 and any other: 3 vs 5,5,7. No match.Function4 and any other: 7 vs 3,5,5. No match.So only one pair: function2 and function3.Therefore, number of favorable pairs:1.Total pairs:6.Therefore, probability is 1/6.Wait, but hold on. Let me confirm.Wait, the functions are distinct, right? Each function is determined by a and b. So the four functions are:1. (2,1)2. (2,3)3. (4,1)4. (4,3)Therefore, the pairs are:1 &2: (2,1) and (2,3): a + b =3 vs5 → different.1&3: (2,1) and (4,1):3 vs5→ different.1&4: (2,1) and (4,3):3 vs7→ different.2&3: (2,3) and (4,1):5 vs5→ same.2&4: (2,3) and (4,3):5 vs7→ different.3&4: (4,1) and (4,3):5 vs7→ different.So only the pair (2,3) is the one with same a + b. Hence, only 1 pair.Thus, probability is 1/6.But wait, maybe I made a mistake here. Let me check again. The a + b values:Function1: 2+1=3Function2:2+3=5Function3:4+1=5Function4:4+3=7So, pairs with same a + b:Function2 and Function3: both 5.That's the only pair. So yes, only 1 pair out of 6. So 1/6.Wait, but in the problem statement, it says "two functions are randomly selected from the set of possible functions". The set of possible functions has 4 elements. So the number of possible pairs is C(4,2)=6. And only one pair has the same slope. Therefore, 1/6.But wait, another thought: perhaps the tangent lines are considered as parallel even if they are the same line? Wait, no, the tangent lines at (1,f(1)) would only be the same line if both the slopes and the points (1,f(1)) are the same. However, even if two functions have the same slope at x=1, unless they also pass through the same point (1,f(1)), their tangent lines would not be the same, but they would be parallel. However, in this problem, the question is about the tangent lines being parallel, regardless of being the same line or not. So even if two different functions have the same slope at x=1, their tangent lines would be parallel but distinct.But in our case, we just need the slopes to be equal, which is equivalent to f’(1) being equal for both functions. So the key is whether a1 + b1 = a2 + b2.Therefore, the number of pairs where a1 + b1 = a2 + b2. In our case, only the pair (2,3) and (4,1) have a + b equal to 5. So the pair consisting of function2 and function3. So that is one pair.But wait, another thought: is the selection of functions with replacement or without replacement? The problem says "two functions are randomly selected from the set of possible functions". Typically, "randomly selected" without further specification implies without replacement. So it's combinations, not permutations. So 6 possible pairs, as we have.But let me check: is there another pair with the same a + b?Looking at the functions:Function2: a=2, b=3: sum 5Function3: a=4, b=1: sum 5So those two are the only ones with sum 5. So only one pair.Thus, the probability is 1/6. Therefore, the answer is 1/6.Wait, but hold on. Let me re-examine the functions and their a + b.Wait, Function1: a=2, b=1 → sum 3Function2: a=2, b=3 → sum5Function3: a=4, b=1 → sum5Function4: a=4, b=3 → sum7So indeed, only functions 2 and 3 have the same sum. So only one pair. Therefore, 1/6.But let me check if there are any other possible pairs. For example, is there a case where two different (a,b) pairs result in the same a + b? Let's see:Possible (a,b):(2,1):3(2,3):5(4,1):5(4,3):7So sums are 3,5,5,7. So duplicates only at sum=5. So only the pair (2,3) and (4,1). So yes, one pair. So the answer is 1/6.But wait, wait. Wait, hold on. Wait, the problem says "two functions are randomly selected from the set of possible functions". The set of possible functions has 4 elements. So the possible pairs are 6. But if we count the pairs, as I did before, only one pair has equal a + b.Alternatively, maybe I miscounted. Let me list all the pairs:1. Function1 and Function2: sums 3 and5 → different2. Function1 and Function3: sums3 and5→ different3. Function1 and Function4: sums3 and7→ different4. Function2 and Function3: sums5 and5→ same5. Function2 and Function4: sums5 and7→ different6. Function3 and Function4: sums5 and7→ differentSo only pair 4 is the one with same sums. So 1 out of 6. Therefore, the probability is 1/6.Therefore, part 2 answer is 1/6.But hold on, let me think again. Is there another way the slopes could be equal? For example, if two different (a,b) pairs have the same a + b. But in our case, only (2,3) and (4,1) have a + b=5. So, yes, only one pair.Alternatively, maybe the problem allows selecting the same function twice? But the problem says "two functions are randomly selected", which usually implies distinct functions, so without replacement. If it were with replacement, the number of possible pairs would be 16, but the problem doesn't specify that. So probably it's without replacement, so 6 pairs. So answer is 1/6.Therefore, the final answers are 3/4 and 1/6.But let me check once more for part 1. When a=4 and b=3, f’(-1) = -4 +3 = -1 ≤0, so that's decreasing. When a=4 and b=1, f’(-1)= -4 +1=-3 ≤0. When a=2 and b=1, f’(-1)=-2 +1=-1 ≤0. The only case where it's not decreasing is a=2 and b=3. So 3 out of 4. Correct. So 3/4.For part 2, only the pair (2,3) and (4,1) have same slope at x=1. So 1 pair out of 6. So 1/6. That seems correct.Final Answer1. boxed{dfrac{3}{4}} 2. boxed{dfrac{1}{6}}</think>Given (a in {2, 4}) and (b in {1, 3}), we consider the function (f(x) = frac{1}{2}ax^2 + bx + 1).1. To determine the probability that (f(x)) is a decreasing function on the interval ((-infty, -1]), we need the derivative (f'(x) = ax + b) to be non-positive on this interval. The maximum value of the derivative on ((-infty, -1]) occurs at (x = -1), so we check if (f'(-1) leq 0). This simplifies to (-a + b leq 0) or (b leq a). - For (a = 2): (b = 1) satisfies (1 leq 2), but (b = 3) does not satisfy (3 leq 2). - For (a = 4): Both (b = 1) and (b = 3) satisfy (1 leq 4) and (3 leq 4). There are 4 possible functions in total, and 3 of them satisfy the condition. Therefore, the probability is (frac{3}{4}).2. To find the probability that the tangent lines at ((1, f(1))) are parallel for two randomly selected functions, we need their derivatives at (x = 1) to be equal. The derivative at (x = 1) is (a + b). We check the values of (a + b) for all functions: - (a = 2, b = 1): (a + b = 3) - (a = 2, b = 3): (a + b = 5) - (a = 4, b = 1): (a + b = 5) - (a = 4, b = 3): (a + b = 7) The pairs with equal (a + b) are ((2, 3)) and ((4, 1)). There are (binom{4}{2} = 6) possible pairs of functions, and only 1 pair meets the condition. Therefore, the probability is (frac{1}{6}).Final Answers:1. (boxed{dfrac{3}{4}})2. (boxed{dfrac{1}{6}})]

🔑:1. For f(x) to be a decreasing function on the interval (-infty,-1], the inequality -frac{b}{2cdotfrac{1}{2}a}geq -1 must hold, which simplifies to bleq a.There are a total of C_{2}^1cdot C_{2}^1=4 possible pairs of (a,b). Among these, there are 3 pairs that satisfy bleq a. Therefore, the probability that f(x) is a decreasing function on the interval (-infty,-1] is frac{3}{4}.2. From part 1, we know that there are 4 possible functions f(x). If two functions are randomly selected, there are C_4^2=6 possible combinations.The slope of the tangent line to the function f(x) at the point (1,f(1)) is given by f'(1)=a+b. For two functions to have parallel tangent lines at this point, the sum of their respective a and b values must be equal.Only the pair (2,3) and (4,1) satisfy this condition, which corresponds to boxed{frac{1}{6}} of the possible combinations.

🧠:[<think>Okay, let's try to figure out this problem step by step. So, we need to find two years between 1700 and 1900 in France's history where two important victories happened on April 22nd. The conditions are that exactly 4382 days passed between the first and second victory, and the sum of the digits in the first victory's year is 23. Hmm, okay.First, let's understand the time difference. 4382 days. Let me convert that into years to get a rough idea. Since a year is approximately 365 days, dividing 4382 by 365 gives about 12 years. Let me check: 365 * 12 = 4380, so 4382 days is 12 years and 2 days. Wait, but leap years might affect this. So between two dates, if there are leap years involved, the actual number of days could vary.But since both events are on April 22nd, maybe the 4382 days is an exact count. Let me verify. If we have two dates, same month and day, April 22nd, separated by 12 years, how many days is that? Well, each regular year is 365 days, but every 4 years is a leap year, adding an extra day. So in 12 years, there would be 3 leap years (since 12 divided by 4 is 3). Therefore, 12 years would be 12*365 + 3 = 4380 + 3 = 4383 days. But the problem states 4382 days. Hmm, that's one day less than 12 years with 3 leap years. Interesting.Wait, maybe there's a leap year involved between the two dates? Let's think. Suppose the first victory is in year Y, and the second is in year Y + 12. If between Y and Y + 12 there are 3 leap years, then the total days would be 12*365 + 3 = 4383. But the problem says 4382 days. So perhaps there are only 2 leap years in between? But why would that happen?Ah, maybe the period between the two April 22nds doesn't include all the leap years. For example, if the first year Y is a leap year, and the period from April 22, Y to April 22, Y + 12 includes the leap day of Y only if February 29 is between those dates. Wait, April 22 is after February, so if Y is a leap year, the leap day is in February, which is before April 22. So when you go from April 22, Y to April 22, Y + 1, you don't include the leap day of Y unless the period crosses February 29. But since both dates are April 22, the period from April 22 to April 22 next year would include February 28 (or 29) of the next year, right? Wait, maybe I'm getting confused.Let me recall how date differences work. Suppose we have two dates: April 22, Y and April 22, Y + N years. The number of days between them depends on how many leap years are in between. Each leap year adds an extra day. However, if the start date is before March 1, then a leap day in year Y is included, but if it's after February 29, it's not. Similarly, if the end date is before March 1, then the leap day in the final year isn't included. So since both dates are April 22, which is after February 29, any leap year Y or Y + N would have their leap days not counted in the interval. Wait, so if we go from April 22, Y to April 22, Y + N, the leap days would be those in years Y + 1 to Y + N - 1, because the leap day in Y is before April 22, and the leap day in Y + N is also before April 22, but we are ending on April 22, Y + N, so we don't include that leap day either.Wait, that might not be right. Let me think. For example, take Y = 2020 (a leap year). The period from April 22, 2020 to April 22, 2021. How many days is that? Since 2020 is a leap year, but the leap day is February 29, 2020, which is before April 22, 2020. So from April 22, 2020 to April 22, 2021 is 366 days? Wait, no. Wait, April 22, 2020 to April 22, 2021 is 365 days because after April 22, 2020, the next year's April 22 is 365 days later, regardless of whether 2020 was a leap year or not. Because the leap day is already past by April. So, in general, the number of days between April 22, Y and April 22, Y + 1 is 365 days if Y is not a leap year, and 365 days if Y is a leap year? Wait, no, that can't be. Wait, wait, if you start on April 22, 2020 (a leap year), the next April 22, 2021 would be 365 days later because the leap day in 2020 is already before April 22. Similarly, starting from April 22, 2021 (not a leap year) to April 22, 2022 is 365 days. So in any case, moving from April 22 to April 22 next year is 365 days. Therefore, the number of days between April 22, Y and April 22, Y + N is 365*N + number of leap days in between. Wait, but where are those leap days?Wait, according to this, the number of leap days between two dates same month/day is equal to the number of leap years in between, where leap years are years divisible by 4, but not by 100 unless by 400. Since we are dealing with 1700-1900, so in that period, leap years are years divisible by 4, but not 1700, 1800, 1900 because those are divisible by 100 but not 400. So between 1700 and 1900, leap years are every 4 years except 1700, 1800, 1900.But getting back, the number of days between April 22, Y and April 22, Y + N is 365*N + number of leap years from Y + 1 to Y + N - 1, inclusive. Wait, why? Because each leap year adds a leap day, which is February 29. But since we are starting and ending on April 22, the leap days in years Y and Y + N are not counted because they are before April 22. So only the leap days in the years strictly between Y and Y + N are counted. Wait, if N is 12, then the leap years between Y + 1 and Y + 11. Wait, no, if N is 12, the period is Y to Y + 12, so the leap years in between would be Y + 1 to Y + 11. Hmm, but maybe I need to check.Wait, let's take an example. Suppose Y = 1796. April 22, 1796 to April 22, 1808. How many leap years are in between? Leap years are every 4 years: 1796, 1800, 1804, 1808. But since we're going from April 22, 1796 to April 22, 1808, the leap days in 1796 and 1808 are not counted because they are before April 22. So the leap days in between would be 1800 and 1804. But 1800 is a year divisible by 100, so not a leap year in the Gregorian calendar. Wait, France was using the Gregorian calendar by this time, right? France adopted the Gregorian calendar in 1582, but some regions might have transitioned later. But assuming it's the Gregorian calendar, 1800 is not a leap year. So from 1796 to 1808, the leap years would be 1804 only? So between 1796 and 1808, excluding 1796 and 1808, the leap years are 1800 (not leap) and 1804 (leap). So only 1 leap day. Therefore, days between April 22, 1796 and April 22, 1808 would be 12*365 + 1 = 4380 + 1 = 4381 days. But the problem states 4382 days. Hmm.Wait, maybe I miscalculated. Let me check another example. Suppose the two years are 1812 and 1824. From April 22, 1812 to April 22, 1824. Leap years between 1812 and 1824 (excluding 1812 and 1824) are 1816, 1820. Both are leap years (divisible by 4, not excluded). So two leap days. Therefore, days would be 12*365 + 2 = 4380 + 2 = 4382. That matches the problem's 4382 days. So in this case, the number of leap years between Y and Y + 12 is two. So the difference is 12 years with 2 leap days, totaling 4382 days. Therefore, this would mean that between Y and Y + 12, there are two leap years. But how does that happen?So to get two leap years in 12 years, we need that among the years Y + 1 to Y + 11, there are two leap years. Since leap years occur every 4 years, in 12 years, normally there would be 3 leap years (12 / 4 = 3). However, if one of those years is a century year not divisible by 400, like 1800, which is not a leap year, then in a 12-year span that includes such a year, you would have one less leap year. For example, between 1796 and 1808, we have 1800, which is not a leap year, so only two leap years: 1804 and 1808 (but 1808 is excluded as it's the end year). Wait, no, earlier we saw between 1796 and 1808, excluding the start and end, the leap years would be 1800 (not leap) and 1804 (leap). So only one. Hmm. Maybe another example.Take Y = 1800. Then from April 22, 1800 to April 22, 1812. The leap years between 1800 and 1812 (excluding 1800 and 1812) are 1804 and 1808. Both are leap years. So two leap days. Thus, days would be 12*365 + 2 = 4382. So that works. So in this case, Y = 1800, but 1800 is not a leap year. So the leap years in between are 1804 and 1808. So that gives two leap days. Therefore, if the 12-year period includes a century year which is not a leap year, then the number of leap days is two instead of three.Therefore, in order for the difference between two April 22nds to be 4382 days, the 12-year period must include a century year (like 1700, 1800, 1900) which is not a leap year, thereby reducing the number of leap days by one. Therefore, the two victories are 12 years apart, but spanning a century year, so that only two leap days are added instead of three.Therefore, the second victory is 12 years after the first, but during those 12 years, there was a century year (which is not a leap year), so only two leap days. Hence, the total days are 12*365 + 2 = 4382.Therefore, the first year (Y) and the second year (Y + 12) must be such that between Y and Y + 12, there is a year divisible by 100 (i.e., 1700, 1800, 1900). Since we're looking between 1700 and 1900, the possible century years are 1700, 1800, 1900. However, 1900 is the end of our range, so if Y + 12 is 1900, then Y would be 1888, but 1888 + 12 is 1900. However, victories have to be between 1700 and 1900, so 1900 is allowed as the end year. But let's see.So possible century years within the span are 1700, 1800, 1900. So let's consider each case:Case 1: The 12-year period includes 1700. So Y + 12 = 1700, meaning Y = 1688. But 1688 is before 1700, which is outside our range. Therefore, not possible.Case 2: The 12-year period includes 1800. So Y must be less than 1800, and Y + 12 must be greater than 1800. So Y + 12 > 1800 => Y > 1788. But Y must be less than 1800. So Y is between 1789 and 1799. Then Y + 12 would be between 1801 and 1811. Therefore, between Y and Y + 12, the year 1800 is included. Since 1800 is not a leap year, so the leap years in between would be Y + 4 and Y + 8 (assuming Y is a leap year or not). Wait, let's take Y = 1792. Then Y + 12 = 1804. Between 1792 and 1804, excluding the start and end, the years are 1793-1803. The leap years here would be 1796 and 1800. But 1800 is not a leap year, so only 1796 is a leap year. Wait, so only one leap year. Then days would be 12*365 + 1 = 4381, which is less than 4382. Hmm, maybe another Y.Wait, perhaps Y is 1788. Then Y + 12 = 1800. But 1800 is the excluded year. Wait, but if Y is 1788, then the period is April 22, 1788 to April 22, 1800. The leap years between 1789 and 1799. Leap years would be 1792, 1796. Both are leap years. So two leap days. Therefore, days would be 12*365 + 2 = 4382. So that works. But Y = 1788 is within the 1700-1900 range. However, does 1788 + 12 = 1800, which is allowed as the end is 1900. But 1788 is the first victory, and 1800 is the second. But wait, the problem states "between 1700 and 1900". So does 1700 count as the start? If "between 1700 and 1900" includes 1700 and 1900, then 1788 is after 1700, so that's okay. But 1800 is the end year here.Wait, let me confirm the date calculation. From April 22, 1788 to April 22, 1800. How many leap days are in between? The leap years after 1788 and before 1800 are 1792 and 1796. Both are leap years (since 1700 is a leap year in the Julian calendar but not Gregorian, but France was using Gregorian by then). So 1792 and 1796 are leap years. So two leap days. Therefore, 12*365 + 2 = 4380 + 2 = 4382 days. Perfect. So that works. So Y = 1788 and Y + 12 = 1800. Then check if the sum of the digits in 1788 is 23. Let's see: 1 + 7 + 8 + 8 = 1 + 7 + 8 + 8 = 24. Oh, that's 24, not 23. So that doesn't satisfy condition b). So even though the days are correct, the digit sum is 24, not 23. So that's out.Hmm. So perhaps another Y where the period includes 1800. Let's try Y = 1789. Then Y + 12 = 1801. The leap years between 1789 and 1801 are 1792, 1796, 1800. But 1800 is not a leap year, so only 1792 and 1796. So two leap days. Therefore, days = 12*365 + 2 = 4382. So that works. Now check the digit sum of 1789: 1 + 7 + 8 + 9 = 25. Still not 23.Next, Y = 1790. Y + 12 = 1802. Leap years between 1790 and 1802: 1792, 1796, 1800. Again, 1800 not a leap year. So two leap days. Days = 4382. Digit sum: 1 + 7 + 9 + 0 = 17. Not 23.Y = 1791. Sum: 1 + 7 + 9 + 1 = 18. Nope.Y = 1792. Sum: 1 + 7 + 9 + 2 = 19. No.Y = 1793. Sum: 1 + 7 + 9 + 3 = 20. No.Y = 1794. Sum: 1 + 7 + 9 + 4 = 21. No.Y = 1795. Sum: 1 + 7 + 9 + 5 = 22. Close, but not 23.Y = 1796. Sum: 1 + 7 + 9 + 6 = 23. Oh! That's 23. Okay, so Y = 1796. Then Y + 12 = 1808. Wait, but between 1796 and 1808, how many leap days? Let's check. The years between 1796 and 1808 (excluding 1796 and 1808) are 1797-1807. Leap years in this period: 1800 (not leap), 1804 (leap). So only 1804 is a leap year. Therefore, only one leap day. Wait, so days would be 12*365 + 1 = 4381. But the problem requires 4382 days. So conflict here.Wait, if Y = 1796, then Y + 12 = 1808. The period is April 22, 1796 to April 22, 1808. The leap years in between are 1796, 1800, 1804, 1808. But since we start after April 22, 1796, the leap day of 1796 is already passed. Similarly, the leap day of 1808 is April 22, 1808, so we don't count that. So the leap days in between are 1800 (not a leap year) and 1804 (leap year). So only one leap day. Hence, days = 12*365 + 1 = 4381. Not 4382. Therefore, even though the digit sum of 1796 is 23, the days don't add up. So this is a problem.Wait, but maybe my earlier assumption is wrong. Let me check again. If Y = 1796, then the period is April 22, 1796 to April 22, 1808. The leap years between those dates (excluding Y and Y + 12) would be 1800 and 1804. But 1800 is not a leap year, so only 1804. Therefore, one leap day. Hence, 4381 days. But the problem requires 4382. Therefore, this doesn't work. So even though the digit sum is 23, the days don't match. Therefore, Y = 1796 is invalid.Hmm. So what's going on. Maybe another case where the 12-year span includes 1900? Let's check.Case 3: The 12-year period includes 1900. So Y + 12 = 1900, which gives Y = 1888. Check if 1888 is within 1700-1900. Yes. Then, between 1888 and 1900, the leap years would be 1892, 1896. 1900 is not a leap year. So two leap days. Therefore, days = 12*365 + 2 = 4382. That works. Now check the digit sum of 1888: 1 + 8 + 8 + 8 = 25. Not 23. So no good.Alternatively, if Y is 1889, Y + 12 = 1901, which is outside the range. So not allowed.Alternatively, Y = 1876. Then Y + 12 = 1888. Wait, but 1900 is not in between. Wait, no, 1876 to 1888: leap years 1880, 1884, 1888. But 1888 is the end year, so excluded. So 1880, 1884. Two leap days. So days = 12*365 + 2 = 4382. Digit sum of 1876: 1 + 8 + 7 + 6 = 22. Not 23.Wait, maybe we need to look elsewhere.Wait, perhaps the two victories are not exactly 12 years apart but some other number of years with the right number of leap days. Wait, 4382 days. Let me check how many years that is more precisely.Since 365 * 12 = 4380, so 4382 is 2 days more. But depending on leap years, it could be 12 years with 2 leap days, as we saw. Alternatively, maybe 11 years with some leap days. Let me check.Wait, 11 years would be 365*11 = 4015 days. Plus number of leap days. If in 11 years there are 3 leap days, then 4015 + 3 = 4018. Not 4382. So no. 12 years is more plausible.Alternatively, 12 years and 2 days. Wait, but the problem states that exactly 4382 days elapsed between the first and second victory. If they are both on April 22, then the difference should be exact years, right? Because same month and day. So the difference should be a whole number of years, with leap days accounted for. Therefore, the difference is 12 years with 2 leap days, totaling 4382 days. So that must be the case.Therefore, the only possible cases are when the two victories are 12 years apart, with two leap days in between. Which happens when the 12-year span includes a non-leap year century year (1700, 1800, 1900). So we need to check all such possibilities within 1700-1900.We checked Y = 1788 (sum 24), Y = 1789 (25), Y = 1790 (17)... up to Y = 1796 (23, but days don't match). Y = 1888 (sum 25). Hmm.Wait, let's check another span. Suppose the two victories are 1800 and 1812. Then Y = 1800, sum of digits: 1 + 8 + 0 + 0 = 9. Not 23. Also, 1800 is not a leap year, so between 1800 and 1812, leap years are 1804, 1808. Two leap days. So 12*365 + 2 = 4382. But the digit sum is 9. Not 23.Alternatively, maybe another century year. Wait, 1700 is a century year. Let's see. If Y = 1688, then Y + 12 = 1700. But 1688 is before 1700. So not allowed. If Y = 1700, then Y + 12 = 1712. Digit sum of 1700 is 1 + 7 + 0 + 0 = 8. Not 23.Alternatively, victories in 1808 and 1820. Then Y = 1808, sum 1 + 8 + 0 + 8 = 17. Not 23. Days between 1808 and 1820: leap years 1812, 1816. Two leap days. So 4382 days. But sum is 17.Wait, maybe I need to check other 12-year intervals that include a non-leap century year but start earlier. Wait, for example, Y = 1796 to 1808, which we checked earlier. But the digit sum for Y = 1796 is 23, but the days are 4381. Hmm. Conflict. What's the issue here?Wait, maybe my calculation of leap days is wrong. Let's check the exact days between April 22, 1796 and April 22, 1808.First, note that 1796 is a leap year (divisible by 4), but since we start on April 22, the leap day (February 29, 1796) is already past. Then, from April 22, 1796 to April 22, 1797: 365 days.From April 22, 1797 to April 22, 1800: 3 years. Let's count leap days in between. 1797-1799: no leap years. 1800 is not a leap year. So 3*365 = 1095 days.From April 22, 1800 to April 22, 1804: 4 years. Leap years in 1800 (no), 1804 (yes, but leap day is February 29, 1804). Since we go from April 22, 1800 to April 22, 1804, that includes the leap day in 1804. Wait, but 1804 is a leap year, so from April 22, 1803 to April 22, 1804 is 366 days? Wait, no. Wait, each year from April 22 to April 22 is 365 days, regardless of leap years, unless the period includes February 29. Wait, confusion arises.Wait, perhaps the best way is to use an online date calculator, but since I can't do that, let's think differently.The total number of days between April 22, 1796 and April 22, 1808.Each year has 365 days, except for leap years, which have 366. However, since we're starting and ending on April 22, the leap day only affects the total if the leap day is within the period.So from April 22, 1796 to April 22, 1797: 365 days (no leap day included).Same for each subsequent year, except when there's a leap year. So between April 22, 1796 and April 22, 1808, the leap days occur in 1796, 1800, 1804, 1808. But since we start after February 29, 1796, and end before February 29, 1808, we don't count the leap days of 1796 and 1808. But what about 1800 and 1804?Wait, 1800 is not a leap year. 1804 is a leap year, so February 29, 1804 is included in the period from April 22, 1803 to April 22, 1804. Wait, no. From April 22, 1803 to April 22, 1804: this period includes February 29, 1804. Therefore, that year would have 366 days. Therefore, the total days would be 366 instead of 365 for that year.Similarly, any leap year where February 29 falls within the period from April 22 of the previous year to April 22 of the leap year. Wait, this is getting complicated. Maybe the proper way is:From April 22, Y to April 22, Y + 1 is 365 days if Y is not a leap year, and 366 days if Y is a leap year? No, because if Y is a leap year, but the start date is after February 29, then the next April 22 would still be 365 days later. Wait, no, the number of days between April 22, Y and April 22, Y + 1 is always 365 days, regardless of whether Y is a leap year or not, because the leap day, if any, is before April 22. So even if Y is a leap year, from April 22, Y to April 22, Y + 1 is 365 days because you skip over the leap day.Wait, let me verify with an example. Take Y = 2020 (leap year). April 22, 2020 to April 22, 2021: how many days? 366 days (because 2020 is a leap year) minus the days from January 1 to April 22, 2020, plus the days from January 1 to April 22, 2021? Wait, no. Alternatively, use a date calculator. From April 22, 2020 to April 22, 2021 is exactly 365 days because even though 2020 is a leap year, the leap day is February 29, 2020, which is before April 22. So the period from April 22, 2020 to April 22, 2021 is 365 days. Similarly, from April 22, 2021 to April 22, 2022 is 365 days. If we take a leap year like 2024, then from April 22, 2023 to April 22, 2024 is 366 days because it includes February 29, 2024. Wait, no. Wait, April 22, 2023 to April 22, 2024: this period does include February 29, 2024. Therefore, it's 366 days. So the key is whether the period from April 22, Y to April 22, Y + 1 includes a February 29. If Y + 1 is a leap year, then yes, the period includes February 29, Y + 1, making it 366 days. Wait, but April 22, Y to April 22, Y + 1: if Y + 1 is a leap year, then between those dates, there is February 29, Y + 1. So the length is 366 days. If Y + 1 is not a leap year, it's 365 days.Wait, that seems correct. Therefore, the number of days between April 22, Y and April 22, Y + 1 is 366 if Y + 1 is a leap year, otherwise 365. Therefore, the total days between April 22, Y and April 22, Y + N is the sum over each year Y + 1 to Y + N of 366 if that year is a leap year, else 365.So in that case, to compute the days between April 22, 1796 and April 22, 1808, we need to check each year from 1797 to 1808 and see if they are leap years.Wait, no. Wait, the period is from April 22, Y to April 22, Y + N. So for each year Y + 1 to Y + N, check if it is a leap year. If yes, add 366, else 365.Wait, no. Wait, actually, each year Y + i to Y + i + 1, the number of days is 366 if Y + i + 1 is a leap year, else 365. Therefore, to compute the total days from April 22, 1796 to April 22, 1808:Break it down year by year:- 1796 to 1797: 1797 is not a leap year (1797 not divisible by 4). So 365 days.- 1797 to 1798: 365- 1798 to 1799: 365- 1799 to 1800: 1800 is not a leap year. 365- 1800 to 1801: 365- 1801 to 1802: 365- 1802 to 1803: 365- 1803 to 1804: 1804 is a leap year. 366- 1804 to 1805: 365- 1805 to 1806: 365- 1806 to 1807: 365- 1807 to 1808: 1808 is a leap year. 366Wait, but we stop at April 22, 1808. So the last segment is from April 22, 1807 to April 22, 1808. Since 1808 is a leap year, but we end on April 22, 1808, which is after February 29, 1808. Therefore, does that count? Wait, no. Because the period from April 22, 1807 to April 22, 1808 includes February 29, 1808. Therefore, it should be 366 days. So total days:Let's count:From 1796-1797: 3651797-1798: 3651798-1799: 3651799-1800: 3651800-1801: 3651801-1802: 3651802-1803: 3651803-1804: 3661804-1805: 3651805-1806: 3651806-1807: 3651807-1808: 366Total days: 365*10 + 366*2 = 3650 + 732 = 4382 days. Oh! Wait, that's exactly 4382 days. But earlier, I thought that from 1796 to 1808 would have only one leap day, but according to this count, there are two leap days: 1804 and 1808. But 1808 is the end year, but since the period includes up to April 22, 1808, which is after February 29, 1808, so that leap day is included. Wait, but according to the earlier logic, if you go from April 22, Y to April 22, Y + N, the leap days are in the years Y + 1 to Y + N. Wait, maybe my previous assumption was wrong.Wait, according to this breakdown, the period from April 22, 1796 to April 22, 1808 includes the leap days of 1804 and 1808. But 1808 is the end year. So does the leap day of 1808 count?Yes, because the period from April 22, 1807 to April 22, 1808 includes February 29, 1808. Therefore, that segment is 366 days. Therefore, the total number of days is 4382. Therefore, Y = 1796 and Y + 12 = 1808 gives 4382 days. And the sum of digits in 1796 is 1 + 7 + 9 + 6 = 23. Perfect! So this satisfies both conditions.But earlier, I thought that leap day in 1808 wouldn't count because it's the end year, but actually, since the period from April 22, 1807 to April 22, 1808 includes the leap day, it does count. Therefore, my initial mistake was not considering that the leap day in the final year (1808) is included because the period spans from April to April, thus including February 29 of the final year if it's a leap year.Therefore, the correct answer is 1796 and 1808. Let's confirm once more:Sum of digits in 1796: 1 + 7 + 9 + 6 = 23. Check.Days between April 22, 1796 and April 22, 1808: As calculated above, 4382 days. Check.Therefore, the two years are 1796 and 1808.But wait, let me check historical events. Is there a famous victory in France on April 22, 1796 or 1808?Well, April 22, 1796, during the French Revolutionary Wars, the Battle of Mondovì was fought, but that was on April 21. Hmm. Alternatively, the Battle of Millesimo on April 13-14, 1796. Not sure. Similarly, in 1808, Napoleonic Wars. The Battle of Baylen was in July 1808. Not April. Maybe the problem is fictional, given the lack of specific historical events on those dates. But the question doesn't require the actual historical events, just the years based on the mathematical conditions. So as per the conditions given, 1796 and 1808 are the correct answers.Final AnswerThe two years are boxed{1796} and boxed{1808}.</think>To determine the two years between 1700 and 1900 in which important victories were recorded on April 22, given that exactly 4382 days elapsed between them and the sum of the digits in the first year is 23, we proceed as follows:1. Convert Days to Years: - 4382 days is approximately 12 years. - Considering leap years, 12 years typically include 3 leap years, adding 3 days. However, 4382 days is 2 days more than 12*365 = 4380 days, indicating 2 leap years instead of 3.2. Identify Leap Years: - A 12-year period spanning a century year (1700, 1800, 1900) which are not leap years reduces the number of leap days by one. - Therefore, the 12-year period must include a century year, resulting in 2 leap days.3. Check Possible Years: - The first year (Y) and the second year (Y + 12) must span a century year. - Calculate the sum of the digits of potential years to find 23.4. Verify Conditions: - For the year 1796: - Sum of digits: 1 + 7 + 9 + 6 = 23. - Check days between April 22, 1796, and April 22, 1808: - Includes leap years 1804 and 1808, resulting in 2 leap days. - Total days: 12*365 + 2 = 4382.Thus, the two years that satisfy the conditions are 1796 and 1808.boxed{1796} and boxed{1808},<think>Okay, let's try to figure out this problem step by step. So, we need to find two years between 1700 and 1900 in France's history where there were victories on April 22nd. The conditions are that exactly 4382 days passed between the first and second victory, and the sum of the digits in the first year is 23. Hmm, let's break this down.First, let's understand the time elapsed between the two victories: 4382 days. Since we're dealing with years, I should convert days into years to get a rough estimate. But wait, converting days to years isn't straightforward because of leap years. Each regular year has 365 days, and a leap year has 366. So, depending on how many leap years are in between, the number of years could vary. Let me see... 4382 divided by 365 is approximately 12 years (365*12=4380), so almost 12 years. But since it's 4382 days, that's two days more than 12 years. But wait, if those 12 years include three leap years, then the total days would be 365*9 + 366*3 = 3285 + 1098 = 4383 days. Hmm, that's actually one day more than 4382. So maybe the period includes three leap years, but the total is 4383, which is one day over. Alternatively, maybe there's a different number of leap years?Alternatively, maybe the time between the two dates is exactly 12 years, but considering the exact days. Let's check if April 22nd, Y + 12 years is the same date. But wait, 12 years can have 3 or 4 leap years, depending on the starting year. For example, if the first year is a leap year, then in 12 years, there could be 3 leap years if the starting year is not divisible by 4, or 4 if it is? Wait, no. Leap years are every 4 years, so in a 12-year span, there should be 3 leap years, right? Unless the period includes a year divisible by 100 but not by 400, but since we're between 1700 and 1900, let's check. The leap years are every year divisible by 4, except for years divisible by 100 unless they're also divisible by 400. So between 1700 and 1900, 1700 is not a leap year, 1800 is not, 1900 is not. So, in the 18th and 19th centuries, leap years are all years divisible by 4 except 1700, 1800, 1900. So, if the period between the two victories crosses one of these years, it might affect the count.But let's start by assuming that the time difference is approximately 12 years. 12 years times 365 days is 4380 days, so 4382 days is 2 days more. Alternatively, if there are 3 leap years in that period, adding 3 days, making it 4383, which is one day over. Hmm. Maybe the actual dates cross a leap day, so depending on whether the period includes February 29th.Wait, let's think differently. If the two victories are both on April 22nd, then the number of days between them can be calculated by the number of years times 365 plus the number of leap days in between. For example, from April 22, Y1 to April 22, Y2 is (Y2 - Y1) years, which is 365*(Y2 - Y1) + number of leap days in between. So if we can model that, then 365*(Y2 - Y1) + L = 4382, where L is the number of leap days. So we need to find Y2 - Y1 such that 365*N + L = 4382, where N = Y2 - Y1, and L is the number of leap years between Y1 and Y2 (exclusive? Or inclusive? Let me check.)Wait, the leap days between two dates would be the number of February 29ths that occur between April 22, Y1 and April 22, Y2. Since April 22 is after February, if Y1 is a leap year, then the leap day in Y1 has already passed by April 22, so the next leap day would be in Y1 + 4, etc. Similarly, if Y2 is a leap year, the leap day in Y2 would be February 29, which is before April 22, so it would be included if Y2 is a leap year. So to count the number of leap years between Y1 and Y2, exclusive of Y1 if we're starting after February, but inclusive of Y2 if we're ending after February. Wait, actually, from April 22, Y1 to April 22, Y2: any leap year between Y1 and Y2 (exclusive) would have their February 29th in that interval. Additionally, if Y1 itself is a leap year, but since we start on April 22, the leap day of Y1 is already past. Similarly, Y2 is a leap year, then the leap day of Y2 is before April 22, so included. Therefore, the number of leap days between April 22, Y1 and April 22, Y2 is equal to the number of leap years in (Y1, Y2], i.e., exclusive of Y1, inclusive of Y2.Therefore, if Y2 - Y1 = N years, then the number of leap years L is the number of years in (Y1, Y2] divisible by 4, excluding those divisible by 100 unless by 400. Since we're between 1700-1900, we can ignore the 400-year exception because 1700, 1800, 1900 are not leap years.So, the formula is 4382 = 365*N + L, where L is the number of leap years in (Y1, Y2]. Let's try possible N. Let's start with N=12. Then 365*12=4380, so 4382-4380=2. So L=2. Therefore, between Y1 and Y2, there must be 2 leap years. Let's check if that's possible.Alternatively, N=11. 365*11=4015, 4382-4015=367. But 367 is more than 1 year, which is impossible. Similarly, N=13: 365*13=4745, which is more than 4382. So N=12 seems the only possible. Then L=2.Therefore, Y2 - Y1 = 12 years, and in those 12 years, there are 2 leap years. Let's check if that's feasible.Wait, in 12 years, normally there are 3 leap years (since 12/4=3). But if the period includes a year like 1700, 1800, or 1900, which are not leap years, then maybe there's one less. So if Y1 is, say, 1788, then Y2 would be 1800. From 1788 to 1800 is 12 years. The leap years in between would be 1792, 1796, and 1800. But 1800 is not a leap year, so only 1792 and 1796. So L=2. Then 365*12 + 2 = 4380 + 2 = 4382. Perfect! So that works.Wait, so if Y1 is 1788 and Y2 is 1800, then from April 22, 1788 to April 22, 1800 is exactly 12 years with 2 leap days (1792 and 1796), since 1800 is not a leap year. Therefore, 365*12 + 2 = 4382 days. That seems to fit.Now, check if the sum of the digits in Y1 (1788) is 23. Let's compute: 1 + 7 + 8 + 8 = 1+7=8, 8+8=16, total 8+16=24. Wait, that's 24, not 23. Hmm, so that's not good.Wait, maybe another pair of years. Let's think. If Y2 - Y1 =12, and in those 12 years, there are 2 leap years. So possible if the 12-year period skips a leap year because of a century year. For example, if Y1 is 1796, Y2 would be 1808. Wait, 1800 is not a leap year. Let's see, between 1796 and 1808, exclusive of 1796, inclusive of 1808. The leap years would be 1804, 1808. But 1800 is not a leap year. So from 1797 to 1808: leap years are 1800 (no), 1804, 1808. So L=2. Then 365*12 +2=4382. So if Y1 is 1796, Y2 is 1808. Then check sum of digits in 1796: 1+7+9+6=23. Yes! 1+7=8, 9+6=15, 8+15=23. So that works.Wait, but wait. From April 22, 1796 to April 22, 1808. Let's check how many leap days are in between. Since we start in 1796, which is a leap year, but since we start on April 22, 1796, the leap day in 1796 (February 29) has already passed. Then the next leap years would be 1800 (not a leap year), 1804, 1808. So between 1796 and 1808, the leap days are 1804 and 1808. But 1808 is the end year, and since we're ending on April 22, 1808, the leap day of 1808 (February 29) is included because it's before April 22. Wait, 1808 is a leap year, and February 29 is before April 22, so that leap day is included in the period. Therefore, between April 22, 1796 and April 22, 1808, there are two leap days: 1804 and 1808. So L=2. Therefore, days: 365*12 +2=4382. Perfect. And the sum of the digits in 1796 is 1+7+9+6=23. That matches condition b). So that seems to fit.But wait, the problem says "between 1700 and 1900". So the first victory is in 1796, the second in 1808, both within 1700-1900. That works. But are there other possibilities?Let me check another possible case. Suppose Y1 is 1789. Then Y2 would be 1801. But 1789 to 1801 is 12 years. Leap years in between: 1792, 1796, 1800. But 1800 is not a leap year. So L=2 (1792 and 1796). Then 365*12 +2=4382. Then sum of digits in 1789: 1+7+8+9=25. Not 23. Doesn't work.Another example: Y1=1798, Y2=1810. 1798 to 1810 is 12 years. Leap years: 1800 (no), 1804, 1808. So L=2. Days: 4382. Sum of digits in 1798: 1+7+9+8=25. Not 23.Y1=1780, Y2=1792. 12 years. Leap years: 1784, 1788, 1792. 1792 is included. So L=3. Then days: 365*12+3=4383. Not 4382. Doesn't work.Y1=1776, Y2=1788. 12 years. Leap years: 1780, 1784, 1788. So L=3. 4383 days. Not 4382.Y1=1801, Y2=1813. 12 years. Leap years: 1804, 1808, 1812. L=3. 4383 days. Not 4382.Wait, so seems like the only possible pair where L=2 is when the 12-year span includes a non-leap year century year (like 1800), thereby reducing the number of leap years by one. So that case seems unique to around the turn of the century. So 1796 to 1808 gives L=2. Another example could be around 1700: if Y1=1688, Y2=1700. But 1700 is excluded because we start from 1700. Wait, the problem states between 1700 and 1900, so 1700 is included as the start. So if Y1=1696, Y2=1708, but 1696 is before 1700. So not allowed.Similarly, near 1900: Y1=1896, Y2=1908. But 1908 is outside the upper limit of 1900. So no.So the only valid pair within 1700-1900 is 1796 and 1808. Let me verify:April 22, 1796 to April 22, 1808. Let's count the leap years in between. Starting from 1796, the next leap years after 1796 would be 1804 and 1808 (since 1800 is not a leap year). So in between 1796 and 1808, exclusive of 1796 but inclusive of 1808, we have 1804 and 1808 as leap years. Therefore, two leap days. So 12*365 + 2 = 4382. Correct.Sum of digits in 1796: 1 + 7 + 9 + 6 = 23. Perfect. So that must be the answer.But wait, let me check if there are any other years where the sum of digits is 23. Let's think of other years between 1700-1900 where the digits sum to 23. Let's consider a year ABCD, where A+B+C+D=23. Since A is 1 or 2 (because 1700-1900). Let's see:If A=1, then B can be 7,8,9. Let's see:For 17CD: 1+7+C+D=23 ⇒ C+D=15. Possible C and D (digits 0-9):C+D=15. Possible pairs: (6,9), (7,8), (8,7), (9,6). So years could be:1769: 1+7+6+9=231778: 1+7+7+8=231787:1+7+8+7=231796:1+7+9+6=23Similarly, if A=1 and B=8: 1+8+C+D=23 ⇒ C+D=14. But B=8 would make the year 18CD. Wait, A=1 and B=8 would be 18CD, which is 1800-1899. So 1+8+C+D=23 ⇒ C+D=14. Possible pairs: (5,9), (6,8), (7,7), (8,6), (9,5). So years like 1859, 1868, 1877, 1886, 1895.Similarly, A=1 and B=9: 1+9+C+D=23 ⇒ C+D=13. But 19CD is 1900-1999, but our upper limit is 1900. So 1900-1900 is allowed? The problem says between 1700 and 1900, so up to 1900 inclusive. 1900 sum is 1+9+0+0=10, which is not 23. So if A=1 and B=9, C+D=13, but the years would be 19CD, which would be 1903 (1+9+0+3=13) but sum is 13, not 23. Wait, no. Wait, if A=1 and B=9, then 1+9+C+D=23 ⇒ C+D=13. So possible years like 1949 (1+9+4+9=23), but 1949 is outside 1900. So not allowed. Hence, the only possible years with sum 23 in 1700-1900 are:1769, 1778, 1787, 1796, 1859, 1868, 1877, 1886, 1895.Now, from our earlier analysis, only 1796 paired with 1808 gives the 4382 days. Let's check if any other of these years could pair with another year 12 years later (or another interval) to give 4382 days.Take 1769: If Y1=1769, then Y2=1769 +12=1781. Check if between 1769 and 1781, there are 2 leap years. Leap years between 1769 and 1781: 1772, 1776, 1780. So three leap years. So days would be 365*12 +3=4383. Not 4382. So no.Next, 1778: Y1=1778, Y2=1790. Leap years between 1778 and 1790: 1780, 1784, 1788. Three leap years. 4383 days. Doesn't work.1787: Y1=1787, Y2=1799. Leap years between 1787 and 1799: 1788, 1792, 1796. Three leap years. 4383 days. No.1796: Y2=1808 as before. That works.1859: Y1=1859, Y2=1871. Leap years between 1859 and 1871: 1860, 1864, 1868. Three leap years. 4383 days.1868: Y1=1868, Y2=1880. Leap years between 1868 and 1880: 1872, 1876, 1880. Wait, 1880 is included. So 1872, 1876, 1880. Three leap years. 4383 days.1877: Y2=1889. Leap years: 1880, 1884, 1888. Three. 4383.1886: Y2=1898. Leap years: 1888, 1892, 1896. Three. 4383.1895: Y2=1907. But 1907 is beyond 1900. So invalid.So the only possible pair is Y1=1796 and Y2=1808.Therefore, the two years are 1796 and 1808. Let me check historical events. April 22nd victories in France... Hmm, the Battle of Millesimo in 1796? Wait, actually, Napoleon's victory at Mondovi on April 22, 1796, during the Italian campaign. Then, another victory on April 22, 1808? Not sure about the specific event, but according to the problem, we just need to find the years based on the mathematical conditions, which point to 1796 and 1808. So the answer should be these two years.</think>The two years are determined by calculating the 4382-day interval and verifying the sum of the digits in the first year. 1. Convert days to years: 4382 days ≈ 12 years (365×12=4380 days). The extra 2 days suggest two leap years within the period.2. Leap year adjustment: Between 1700–1900, century years (1700, 1800, 1900) are not leap years. A 12-year span crossing 1800 would reduce the leap years from 3 to 2.3. First year condition: The sum of the digits must be 23. Testing years, 1796 (1+7+9+6=23) fits.4. Second year calculation: Adding 12 years to 1796 gives 1808. The period includes leap years 1804 and 1808 (excluding 1800), totaling 2 leap days: ( 365 times 12 + 2 = 4382 ) days.Answer: The years are boxed{1796} and boxed{1808}.]

🔑:The time between the two victories is 4382=12 cdot 365+2 days, which is 12 years, including two leap years.According to the Gregorian calendar, which has been in use since 1582, years divisible by 4 are leap years, except for those divisible by 100, unless they are also divisible by 400.The 12 years can only include two leap years if this period contains a century year. The years 1700 and 1900 are clearly not applicable due to the conditions, so this century year must be 1800. Therefore, the first victory must have occurred between 1788 and 1800.In this interval, only one number, 1796, satisfies the second condition. Indeed:On April 22, 1796, Napoleon defeated the Piedmontese forces at Mondovi during the Italian campaign, and on April 22, 1808, he captured Madrid.

🧠:[<think>Okay, so I need to figure out the total number of digits when both 2^2000 and 5^2000 are written out in sequence. Hmm, let's start by understanding what the question is asking. When they say "written in sequence," I think it just means writing one number after the other, like concatenating them. So first write all the digits of 2^2000, then immediately write all the digits of 5^2000, and count how many digits there are in total. That would be the sum of the number of digits in each number.So the key here is to find the number of digits in 2^2000 and the number of digits in 5^2000 separately, and then add them together. I remember that the number of digits in a number N in base 10 is given by the floor of log10(N) plus 1. So, digits = floor(log10(N)) + 1. Let me verify that. For example, take N = 100. log10(100) is 2, floor(2) +1 = 3, which is correct because 100 has three digits. Another example: N=999. log10(999) is approximately 2.999565, floor of that is 2, 2 +1=3 digits. N=1000, log10 is 3, floor(3)+1=4, which is right. So yes, that formula works.Therefore, for both 2^2000 and 5^2000, I can compute log10 of each, take the floor, add 1, and then sum those two results.But wait, 2^2000 multiplied by 5^2000 is (2*5)^2000 = 10^2000. So 2^2000 * 5^2000 = 10^2000. That means that 2^2000 and 5^2000 are multiplicative inverses in terms of magnitude relative to 10^2000. So, 2^2000 = 10^2000 / 5^2000, and 5^2000 = 10^2000 / 2^2000. So, if I take log10(2^2000) + log10(5^2000) that would be log10(10^2000) = 2000. So that's 2000. But the individual logs would be log10(2^2000) = 2000*log10(2) and log10(5^2000)=2000*log10(5). Their sum is 2000*(log10(2) + log10(5)) = 2000*log10(10) = 2000*1=2000. So that's correct.But how does that help? Well, if I can find log10(2^2000) and log10(5^2000), then add 1 to each (after taking floor) to get the digits. But maybe there's a relationship between the two? Since their product is 10^2000, which is a 1 followed by 2000 zeros. Therefore, 2^2000 and 5^2000 are numbers that when multiplied together give a 2001-digit number (since 10^1 is 10 which has 2 digits, 10^2 is 100 which has 3 digits, so 10^n has n+1 digits). So 10^2000 has 2001 digits. But how does that help with the individual digits of 2^2000 and 5^2000? Let me think. Let’s denote N = 2^2000. Then 5^2000 = 10^2000 / N. So if N is a number with d digits, then 10^2000 / N is a number with (2001 - d + 1) digits? Wait, not exactly. Let's think. Suppose N is a d-digit number, then 10^2000 / N would be a number between 10^{2000 - d} and 10^{2000 - d + 1}, right? Because N is between 10^{d-1} and 10^d, so 10^2000 / N is between 10^{2000}/10^d = 10^{2000 - d} and 10^{2000}/10^{d-1} = 10^{2001 - d}. Therefore, the number of digits of 10^2000 / N would be floor(log10(10^2000 / N)) +1 = floor(2000 - log10(N)) +1. But log10(N) is approximately (d -1) + something. Wait, maybe there's a relation here.Alternatively, let's note that if N has d digits, then 10^{d-1} ≤ N < 10^d. Then 10^2000 / N would satisfy 10^{2000}/10^d < 10^2000 / N ≤ 10^{2000}/10^{d-1}, which is 10^{2000 - d} < 10^2000 / N ≤ 10^{2001 - d}. Taking log10, we get 2000 - d < log10(10^2000 / N) ≤ 2001 - d. Therefore, the floor of log10(10^2000 / N) would be either 2000 - d or 2001 - d -1? Wait, maybe not. Let me write:Let’s denote M = 10^2000 / N. Then log10(M) = 2000 - log10(N). Since N has d digits, log10(N) is between d -1 and d. Therefore, log10(M) is between 2000 - d and 2000 - (d -1) = 2001 - d. So, log10(M) ∈ [2000 - d, 2001 - d). Therefore, floor(log10(M)) is 2000 - d if the fractional part of log10(N) is non-zero? Wait, maybe.Wait, if log10(N) is in [d -1, d), then 2000 - log10(N) is in (2000 - d, 2001 - (d -1)] = (2000 - d, 2001 - d +1] = (2000 - d, 2001 - d +1]. Wait, maybe not. Let's make it precise.If d -1 ≤ log10(N) < d, then 2000 - d < 2000 - log10(N) ≤ 2001 - d. Therefore, log10(M) is in (2000 - d, 2001 - d]. Therefore, the floor of log10(M) is either 2000 - d or 2000 - d. Wait, if log10(M) is in (2000 - d, 2001 - d], then the integer part is either 2000 - d if log10(M) is in [2000 - d, 2000 - d +1), but since log10(M) is greater than 2000 - d, but less than or equal to 2001 - d. So if 2000 - d is an integer, then floor(log10(M)) is 2000 - d when log10(M) is in [2000 - d, 2000 - d +1), and 2000 - d when log10(M) is in [2000 - d, 2001 - d). Wait, no, if the upper limit is 2001 - d, which is (2000 - d) +1. So floor(log10(M)) would be 2000 - d if log10(M) is in [2000 - d, 2000 - d +1), which is the case here because log10(M) is in (2000 - d, 2001 - d]. Therefore, if log10(M) is in (2000 - d, 2001 - d], then the floor is 2000 - d if log10(M) < 2000 - d +1, which is always true because 2001 - d = (2000 - d) +1. So the entire interval is (2000 - d, 2000 - d +1], so the floor is 2000 - d. Wait, but the interval is (2000 - d, 2001 - d], which is the same as (2000 - d, 2000 - d +1). So actually, floor(log10(M)) = floor(2000 - log10(N)) = 2000 - d. Wait, not sure. Maybe an example would help.Suppose N is a 3-digit number. Then log10(N) is between 2 and 3. So log10(M) = 2000 - log10(N) is between 1997 and 1998. So floor(log10(M)) is 1997, which is 2000 -3. Therefore, the number of digits of M would be floor(log10(M)) +1 = 1997 +1 = 1998. But 1998 digits? That seems a lot, but considering M = 10^2000 / N, and N is a 3-digit number, so M is roughly 10^2000 / 100 = 10^1998, which has 1999 digits. Wait, but 10^1998 is 1 followed by 1998 zeros, which has 1999 digits. Wait, so my previous logic is flawed here.Wait, in the example: If N is a 3-digit number, then M = 10^2000 / N. The minimum N is 100, so M is 10^2000 / 100 = 10^1998, which has 1999 digits. The maximum N is 999, so M is approximately 10^2000 / 10^3 = 10^1997, but actually a bit more, so 10^1997.??? which would have 1998 digits. Wait, so depending on N, M could have 1998 or 1999 digits? But how?Wait, let's check with N=100. Then M=10^2000 /100=10^1998, which is 1 followed by 1998 zeros. That's 1999 digits. If N=999, then M=10^2000 /999 ≈ 1.001... x 10^1997, which is a number just over 10^1997, so it has 1998 digits. Therefore, depending on the value of N, M can have either 1998 or 1999 digits. So, in this case, if N is a 3-digit number, M can have 1998 or 1999 digits. Therefore, the number of digits in M is either (2000 - d) or (2000 - d +1). Wait, in this case, d=3, so 2000 - d=1997, but 1997 +1=1998 digits. Wait, but M is 10^1998 which has 1999 digits. So something's wrong here.Wait, maybe my initial approach was wrong. Let's step back.The number of digits of a number x is floor(log10(x)) +1. So if I have x = 2^2000, then log10(x) = 2000*log10(2). Similarly, log10(5^2000) = 2000*log10(5). Then, the number of digits of 2^2000 is floor(2000*log10(2)) +1, and the number of digits of 5^2000 is floor(2000*log10(5)) +1. Then total digits is the sum.But since 2^2000 *5^2000 =10^2000, which has 2001 digits, maybe there's a relationship between the number of digits of each. For example, if x has d digits, and y has e digits, and x*y has k digits, is there a relationship between d, e, and k? Not directly obvious. But in this specific case, x=2^2000, y=5^2000, and x*y=10^2000. So, perhaps the number of digits of x and y can be related through their product.But perhaps it's easier to compute each separately. Let me compute log10(2) ≈ 0.3010, so log10(2^2000)=2000*0.3010=602. So, floor(602) +1=602 +1=603 digits. Similarly, log10(5)=0.69897, so log10(5^2000)=2000*0.69897≈1397.94, so floor(1397.94)=1397, so digits=1397 +1=1398. Then total digits is 603 +1398=2001. Wait, that's the same as the number of digits in 10^2000. Interesting.Wait, but 2^2000 *5^2000=10^2000, which is a 1 followed by 2000 zeros, so it has 2001 digits. But if 2^2000 has 603 digits and 5^2000 has 1398 digits, then 603 +1398=2001. So, the total digits when writing them in sequence is the same as the number of digits in their product. Is this a coincidence? Probably not. Let me check with smaller exponents.Take 2^1 and 5^1. 2*5=10. 2 has 1 digit, 5 has 1 digit, total 2 digits. 10 has 2 digits. Hmm, same total. 2^2=4, 5^2=25. 4 has 1 digit, 25 has 2 digits. Total 3 digits. 4*25=100, which has 3 digits. Same. 2^3=8, 5^3=125. 8 has 1, 125 has 3. Total 4. 8*125=1000, which has 4 digits. So, seems like the total number of digits is equal to the number of digits in the product. Which in this case, the product is 10^n, which has n+1 digits. So for 2^k and 5^k, the total digits when written in sequence is k+1. Wait, in the previous examples, k=1: 1+1=2, correct. k=2: 2+1=3, correct. k=3:3+1=4, correct. So for k=2000, would the total digits be 2000 +1=2001? But according to the earlier calculation, 2^2000 has 603 digits, 5^2000 has 1398, 603+1398=2001. So that holds.But why does this happen? Is there a general formula here? Let me think. If we have two numbers, a and b, such that a*b=10^n, then the total number of digits of a and b when written together is equal to the number of digits of 10^n, which is n+1. But why?Because if a has d digits and b has e digits, then d + e = n +1. But why? Let's see. If a*b=10^n, then log10(a) + log10(b)=n. Let’s denote log10(a)=x, so log10(b)=n -x. The number of digits of a is floor(x) +1, and the number of digits of b is floor(n -x) +1. Therefore, the total digits is floor(x) + floor(n -x) +2. But x is a real number between 0 and n. Let's see. Let x = k + f, where k is integer and 0 ≤ f <1. Then floor(x) =k, floor(n -x)=floor(n -k -f)=n -k -1 if f >0, because n -k -f would be (n -k) -f, which is between (n -k -1, n -k). So floor(n -x)=n -k -1 if f>0, and floor(n -x)=n -k if f=0.Therefore, total digits is k + (n -k -1) +2 =n +1 if f>0, or k + (n -k) +2= n +2 if f=0. Wait, but f=0 implies that x is integer, so a=10^k, so b=10^{n -k}, which are numbers with k+1 digits and n -k +1 digits. Then total digits would be (k+1)+(n -k +1)=n +2. But in the case when a and b are powers of 10, their product is 10^n, which has n+1 digits, but the total digits written in sequence would be n+2. But in our previous examples, when a and b were not powers of 10, the total digits was n+1, but when they are powers of 10, it's n+2. So in general, the total digits is either n+1 or n+2 depending on whether a and b are exact powers of 10 or not.But in the problem here, 2^2000 and 5^2000. Their product is 10^2000, which has 2001 digits. If 2^2000 is a power of 10, then it must be that 2^2000=10^k, which would imply 5^2000=10^{2000 -k}. But 2^2000 is not a power of 10 unless 2000=0, which it's not. Therefore, in our case, neither 2^2000 nor 5^2000 is a power of 10. Therefore, in the formula above, f>0, so total digits would be n +1. Wait, n here is 2000, so total digits is 2000 +1=2001. Which matches our previous calculation. So that's the case.Therefore, the general rule seems to be that if a*b=10^n, and neither a nor b is a power of 10, then the total number of digits in a and b written in sequence is n+1. If one of them is a power of 10, then it would be n+2. But since in this problem, neither 2^2000 nor 5^2000 is a power of 10, the total number of digits should be 2000 +1=2001.But let's confirm with another example. Take a=2^1=2, b=5^1=5. Their product is 10. Neither a nor b is a power of 10. The digits of a is 1, digits of b is 1, total 2, which is 1+1=2. But n=1, so n+1=2. Correct.Another example: a=4, b=25. a*b=100=10^2. n=2. Number of digits: 1 +2=3. Which is n+1=3. Correct. If a=8, b=125. Product=1000=10^3. Digits:1 +3=4=n+1=4. Correct.If a=10, b=10^{n-1}, then a has 2 digits, b has n digits, total digits 2 +n= n+2. Which is the case when one of them is a power of 10. So in such cases, total digits is n+2. So the general formula is: if a and b are positive integers such that a*b=10^n, then the total number of digits in a and b is n+1 if neither a nor b is a power of 10, and n+2 otherwise.Therefore, in our problem, since 2^2000 and 5^2000 are not powers of 10, the total digits should be 2000 +1=2001. Therefore, without calculating each individually, we can just state the answer as 2001. But wait, in our initial calculation, we found 603 +1398=2001. So that's consistent. So maybe there's a general principle here that can shortcut the calculation. But since the problem is about 2^2000 and 5^2000, which multiply to 10^2000, and neither is a power of 10, the total number of digits when written together is 2000 +1=2001.But I should verify this with the direct calculation as well, to make sure. Let's compute the number of digits of 2^2000. log10(2^2000)=2000*log10(2). log10(2)=0.30102999566 approximately. So 2000*0.30102999566≈2000*0.30103=602.06. So floor(602.06)=602, so digits=602 +1=603.Similarly, log10(5^2000)=2000*log10(5). log10(5)=0.69897000434. So 2000*0.69897000434≈2000*0.69897=1397.94. Floor(1397.94)=1397, so digits=1397 +1=1398. Adding 603 +1398=2001. So yes, same answer.Therefore, the total number of digits written in total is 2001. So the answer is 2001.But let me check another exponent to ensure. Let's take exponent 4. 2^4=16 (2 digits), 5^4=625 (3 digits). Total digits 5. 2+3=5. 10^4=10000 (5 digits). So 4+1=5. Correct.Another example: exponent 5. 2^5=32 (2 digits), 5^5=3125 (4 digits). Total digits 6. 5+1=6. Correct.So, this seems to hold. Therefore, the answer is 2001.Final Answerboxed{2001}</think>To determine the total number of digits when (2^{2000}) and (5^{2000}) are written in sequence, we use the formula for the number of digits of a number (N), which is (lfloor log_{10} N rfloor + 1).1. Number of digits in (2^{2000}): [ log_{10}(2^{2000}) = 2000 cdot log_{10}(2) approx 2000 cdot 0.30102999566 approx 602.06 ] The number of digits is: [ lfloor 602.06 rfloor + 1 = 602 + 1 = 603 ]2. Number of digits in (5^{2000}): [ log_{10}(5^{2000}) = 2000 cdot log_{10}(5) approx 2000 cdot 0.69897000434 approx 1397.94 ] The number of digits is: [ lfloor 1397.94 rfloor + 1 = 1397 + 1 = 1398 ]3. Total number of digits: [ 603 + 1398 = 2001 ]Additionally, recognizing that (2^{2000} cdot 5^{2000} = 10^{2000}), which has 2001 digits, we use the property that the total number of digits of two numbers whose product is a power of 10 (and neither is a power of 10) is equal to the number of digits of their product. Thus, the total number of digits is confirmed to be (2001).[boxed{2001}],<think>Okay, so I need to figure out how many digits are written in total when I write down the numbers 2^2000 and 5^2000 one after the other. Hmm, let me think about this step by step.First, I remember that the number of digits in a number can be found using logarithms. Specifically, the formula for the number of digits of a number N in base 10 is given by floor(log10(N)) + 1. Yeah, that makes sense because log10(N) gives the exponent to which 10 must be raised to get N. Taking the floor of that gives the highest power of 10 that's less than N, and adding 1 gives the total number of digits. For example, if N is 100, log10(100) is 2, floor(2) is 2, and 2 + 1 = 3 digits, which is correct.So, applying this formula to both 2^2000 and 5^2000, I can find the number of digits each has and then add them together. Let me write that down:Number of digits in 2^2000: floor(log10(2^2000)) + 1Number of digits in 5^2000: floor(log10(5^2000)) + 1Then total digits = digits in 2^2000 + digits in 5^2000Let me compute log10(2^2000) first. Using logarithm properties, log10(2^2000) = 2000 * log10(2). Similarly, log10(5^2000) = 2000 * log10(5). Wait a second, if I add these two logarithms together, log10(2^2000) + log10(5^2000) = 2000*(log10(2) + log10(5)) = 2000*log10(10) = 2000*1 = 2000. Because log10(2) + log10(5) = log10(2*5) = log10(10) = 1. So that's interesting. So the sum of the two logarithms is 2000, but how does that relate to the number of digits?Wait, the number of digits for each is floor(log10(N)) +1. So for 2^2000, the number of digits is floor(2000*log10(2)) +1, and for 5^2000, it's floor(2000*log10(5)) +1. So total digits would be floor(2000*log10(2)) + floor(2000*log10(5)) + 2.But since 2000*log10(2) + 2000*log10(5) = 2000, then floor(a) + floor(b) where a + b = 2000. But floor(a) + floor(b) can be either equal to floor(a + b) -1 or floor(a + b). Because if a and b are both integers, then floor(a) + floor(b) = a + b. But if they have fractional parts, the sum of the floors is less than the sum of the numbers.Since a = 2000*log10(2) and b = 2000*log10(5), their sum is exactly 2000. But since 2^2000 * 5^2000 = (2*5)^2000 = 10^2000, which is a 1 followed by 2000 zeros. So 2^2000 and 5^2000 are factors of 10^2000, and they multiply to 10^2000. Therefore, 2^2000 = 10^2000 / 5^2000, so they are reciprocals in a sense scaled by 10^2000.But how does this help me? Hmm. Maybe I can recall that if two numbers multiply to a power of 10, then their number of digits might relate in a specific way.Alternatively, let's consider that 2^2000 has a certain number of digits, D1, and 5^2000 has D2 digits. Then D1 + D2 = (floor(2000 log10 2) +1) + (floor(2000 log10 5) +1) = floor(2000 log10 2) + floor(2000 log10 5) + 2.But since log10 2 + log10 5 = 1, then 2000 log10 2 + 2000 log10 5 = 2000. Let me denote a = 2000 log10 2 and b = 2000 log10 5. Then a + b = 2000. So floor(a) + floor(b) = floor(a + b) - k, where k is 1 if there is a fractional part in either a or b, otherwise 0. Since a and b are not integers (because log10 2 and log10 5 are irrational?), their decimal representations will have fractional parts. Therefore, floor(a) + floor(b) = a + b - fractional(a) - fractional(b) = 2000 - (fractional(a) + fractional(b)).But since a + b is exactly 2000, which is an integer, then fractional(a) + fractional(b) must be an integer. Since fractional parts are between 0 and 1, the sum can be either 0 or 1. But since a and b are not integers, their fractional parts are each in (0,1). Therefore, fractional(a) + fractional(b) must be in (0,2). The only integer in that interval is 1. Therefore, fractional(a) + fractional(b) = 1, so floor(a) + floor(b) = 2000 -1 = 1999.Therefore, the total number of digits D1 + D2 = 1999 + 2 = 2001.Wait, that seems too straightforward. Let me verify this with a smaller exponent to see if this holds.Let's take exponent 1: 2^1 = 2, 5^1 = 5. Written in sequence, that's "25", which has 2 digits. According to the formula: floor(1*log10 2) +1 + floor(1*log10 5) +1 = floor(0.3010) +1 + floor(0.69897) +1 = 0 +1 + 0 +1 = 2. Correct. And according to the previous logic, a + b =1, floor(a) + floor(b) =0 +0=0, total digits 0 +2=2. Which matches.Another example: exponent 3. 2^3=8, 5^3=125. Writing them together: "8125", which has 4 digits. Let's check the formula:Digits for 8: floor(3*log10 2) +1 = floor(0.9030) +1 =0 +1=1Digits for 125: floor(3*log10 5) +1 = floor(2.0969) +1=2 +1=3Total digits:1 +3=4. Correct.According to the logic: floor(a) + floor(b) where a=3 log10 2≈0.9030, b≈2.0969, floor(a)=0, floor(b)=2, sum=2, total digits=2 +2=4. Also correct. And here, fractional(a)=0.9030, fractional(b)=0.0969, sum≈1.0. So fractional parts sum to 1, hence floor(a)+floor(b)=a + b -1=3 -1=2. Hence, total digits 2 +2=4. So this works.Another example: exponent 4. 2^4=16, 5^4=625. Combined: "16625", which is 5 digits.Digits for 16: floor(4 log10 2) +1≈floor(1.2041) +1=1 +1=2Digits for 625: floor(4 log10 5) +1≈floor(2.7959) +1=2 +1=3Total digits:2 +3=5. Correct.According to the formula: floor(a) + floor(b) +2=1 +2 +2=5. Correct. The fractional parts of a and b are 0.2041 and 0.7959, which sum to 1. So floor(a) + floor(b)=1 +2=3, total digits=3 +2=5. Correct.Therefore, this seems to hold. Therefore, in the original problem, the total number of digits would be floor(a) + floor(b) +2= (a - fractional(a)) + (b - fractional(b)) +2= a + b - (fractional(a) + fractional(b)) +2= 2000 -1 +2=2001.Therefore, the answer is 2001 digits in total.But let me check another case where perhaps the fractional parts don't sum to 1. Wait, but in the previous examples, they did. Because a + b is integer, and each a and b is non-integer (since log10 2 and log10 5 are irrational), so their fractional parts must sum to 1. Let me confirm that.Since a = 2000 log10 2, which is not an integer because log10 2 is irrational. Similarly, b = 2000 log10 5, also irrational. Since a + b =2000, an integer. Therefore, the fractional part of a plus the fractional part of b must be equal to 1. Because when you add two numbers, their integer parts add up and their fractional parts add up. If the total sum is integer, then the sum of fractional parts must be integer. Since fractional parts are in [0,1), the only possible integer is 0 or 1. But since a and b are not integers, each fractional part is in (0,1), so their sum must be in (0,2). Since the total sum is an integer, the only possibility is 1. Therefore, fractional(a) + fractional(b) =1. Hence, floor(a) + floor(b) = (a - fractional(a)) + (b - fractional(b)) = a + b - (fractional(a) + fractional(b)) =2000 -1=1999. Therefore, total digits is 1999 + 2=2001.Therefore, regardless of the exponent, as long as the two numbers multiply to 10^n, their total number of digits when written together is n +1. Wait, in our case, n=2000, so total digits 2001. So in general, if you have two numbers a and b such that a*b=10^n, then the total number of digits of a and b when written together is n +1. That seems to be a general formula.So, in this problem, since 2^2000 *5^2000=10^2000, the total number of digits is 2000 +1=2001. Therefore, the answer is 2001.But just to ensure there isn't a case where this might not hold, let me think. Suppose I take a=10^(n - 0.5) and b=10^0.5, so that a*b=10^n. Then the number of digits of a is floor(n -0.5) +1= n -0 (since n -0.5 is not an integer), wait, floor(n -0.5) would be n -1 if n is an integer. For example, n=3: a=10^(2.5)=316.227..., which has floor(2.5) +1=2 +1=3 digits. b=10^0.5≈3.16227..., which has floor(0.5) +1=0 +1=1 digit. Total digits=3+1=4. But n=3, so 3 +1=4, which matches. So seems to hold. Even in this case, the fractional parts would sum to 1, so the formula works.Therefore, the conclusion is that when two numbers multiply to 10^n, their total number of digits is n +1. Therefore, in this problem, the answer is 2000 +1=2001.Final AnswerThe total number of digits written is boxed{2001}.</think>To determine the total number of digits written when (2^{2000}) and (5^{2000}) are written in sequence, we use the formula for the number of digits of a number (N) in base 10, which is (lfloor log_{10} N rfloor + 1).1. Calculate the number of digits in (2^{2000}): [ text{Number of digits} = lfloor 2000 log_{10} 2 rfloor + 1 ]2. Calculate the number of digits in (5^{2000}): [ text{Number of digits} = lfloor 2000 log_{10} 5 rfloor + 1 ]3. Recognize that (2^{2000} times 5^{2000} = 10^{2000}), which implies: [ 2000 log_{10} 2 + 2000 log_{10} 5 = 2000 ]4. Let (a = 2000 log_{10} 2) and (b = 2000 log_{10} 5). Since (a + b = 2000) and both (a) and (b) are not integers, their fractional parts sum to 1. Therefore: [ lfloor a rfloor + lfloor b rfloor = 2000 - 1 = 1999 ]5. The total number of digits is the sum of the digits of (2^{2000}) and (5^{2000}): [ text{Total digits} = (lfloor 2000 log_{10} 2 rfloor + 1) + (lfloor 2000 log_{10} 5 rfloor + 1) = 1999 + 2 = 2001 ]Thus, the total number of digits written is (boxed{2001}).]

🔑:If the number 2^{2000} contains m digits, then 10^{mathrm{m}-1}<2^{2000}<10^{mathrm{m}}.## SolutionLet the number 2^{2000} contain mathrm{m} digits, and the number 5^{2000} contain mathrm{n} digits. Then the following inequalities hold: 10^{mathrm{m}-1}<2^{2000}<10^{mathrm{m}}, 10^{mathrm{n}-1}<5^{2000}<10^{mathrm{n}} (the inequalities are strict because the power of two or five is not equal to the power of ten). Multiplying these inequalities, we get: 10^{mathrm{m}+mathrm{n}-2}<10^{2000}<10^{mathrm{m}+mathrm{n}}. From this, it follows that the exponent 2000 is between mathrm{m}+mathrm{n}-2 and mathrm{m}+mathrm{n}, so 2000=mathrm{m}+mathrm{n}-1 and mathrm{m}+mathrm{n}=2001. This means that a total of 2001 digits are written.## Answer2001.00

🧠:[<think>Okay, let's tackle this problem. So, we need to find the total number of four-digit numbers made from the digits 0, 1, 2, and 5 without repeating any digits, and then determine how many of those numbers are not divisible by 5. The options are A: 18, B: 12, C: 10, D: 8. Hmm, let me think through this step by step.First, let's understand the problem. We have four digits: 0, 1, 2, 5. We need to form four-digit numbers without repeating any digits. Then, among those numbers, we need to count how many are not divisible by 5. The answer choices are given, so maybe after calculating, we can check which one matches.Let me recall that a number is divisible by 5 if its last digit is either 0 or 5. Therefore, for a number to not be divisible by 5, its last digit must be something other than 0 or 5. In our case, the digits we can use are 0,1,2,5, so the last digit has to be either 1 or 2. That seems like a crucial point. So maybe we can approach this by calculating the total number of four-digit numbers possible with those digits, then subtract the number of four-digit numbers that are divisible by 5. Alternatively, we can directly calculate the numbers that end with 1 or 2. Let me see which method is better.First, let's compute the total number of four-digit numbers we can form with digits 0,1,2,5 without repetition. Since it's a four-digit number, the first digit cannot be 0. So, the thousands place has to be 1, 2, or 5. Then, the remaining three digits can be any of the remaining three digits, including 0, but without repetition.So, total numbers:First digit (thousands place): 3 choices (1,2,5).Then, the next three digits can be any permutation of the remaining three digits. Since we can't repeat digits, after choosing the first digit, we have 3 remaining digits, so:For the hundreds place: 3 choices,Tens place: 2 choices,Units place: 1 choice.Therefore, total four-digit numbers: 3 (for first digit) * 3 * 2 * 1 = 18.Wait, so total numbers are 18. Then, the answer options have 18 as an option. But we need the numbers that cannot be divided by 5, so subtract those that can be divided by 5. Alternatively, maybe calculate the non-divisible ones directly.But let's confirm total numbers first. Let me think again. The first digit: 1,2,5. Then, after choosing the first digit, the remaining three digits can be arranged in 3! ways. So, total numbers: 3 * 3! = 3*6=18. Yes, that's correct. So total four-digit numbers possible are 18.Now, numbers divisible by 5 must end with 0 or 5. So, we can calculate how many numbers end with 0 or 5, then subtract that from 18 to get the answer.Let me compute numbers ending with 0:If the number ends with 0, then the last digit is fixed as 0. The first digit can be 1,2,5 (since 0 is already used in the last digit). Then, the remaining two middle digits can be any of the remaining two digits. So:First digit: 3 choices (1,2,5).Then, hundreds and tens places: permutations of the remaining two digits. So, 2! = 2.Therefore, numbers ending with 0: 3 * 2 = 6.Similarly, numbers ending with 5:If the last digit is 5, then first digit cannot be 0 or 5 (since 5 is already used). So, first digit can be 1 or 2. Then, the remaining two middle digits can be any of the remaining two digits (including 0).So:First digit: 2 choices (1,2).Then, hundreds and tens places: permutations of the remaining two digits. So, 2! = 2.Therefore, numbers ending with 5: 2 * 2 = 4.Therefore, total numbers divisible by 5: 6 (ending with 0) + 4 (ending with 5) = 10.Therefore, numbers not divisible by 5: total numbers (18) - divisible by 5 (10) = 8. So, the answer is D:8.Wait, but let me check this again. Let's verify by calculating directly the numbers not divisible by 5. These numbers must end with 1 or 2.Numbers ending with 1:Last digit is 1. Then, first digit can be 2,5 (since 0 is allowed in other positions, but first digit can't be 0). Wait, no, first digit can be 0? No, it's a four-digit number, so first digit must be non-zero. Since digits allowed are 0,1,2,5, but 1 is already used in the last digit. So, remaining digits are 0,2,5. So, first digit can be 2,5 (since 0 can't be first). So:First digit: 2 choices (2,5).Then, the hundreds and tens places can be filled with the remaining two digits (including 0). So, permutations of 2 digits: 2! = 2.Therefore, numbers ending with 1: 2 * 2 = 4.Similarly, numbers ending with 2:Last digit is 2. Then, first digit can be 1,5 (since 0 is not allowed as first digit, and 2 is already used in the last digit). Remaining digits are 0,1,5 minus first digit. Wait, let's see:Digits available: 0,1,2,5. Last digit is 2, so remaining digits are 0,1,5. First digit can't be 0, so choices are 1,5.Therefore, first digit: 2 choices (1,5).Then, hundreds and tens places: permutations of the remaining two digits (including 0). So, 2! = 2.Therefore, numbers ending with 2: 2 * 2 = 4.Therefore, total numbers not divisible by 5: 4 + 4 = 8. Same result as before. So, 8 numbers, which is option D. So that seems consistent.But let me make sure there are no mistakes here.Another way to think about it: For numbers not divisible by 5, last digit is 1 or 2.Case 1: Last digit is 1.Then, first digit: cannot be 0 or 1. So, possible first digits: 2,5. So, 2 choices.Then, the hundreds and tens digits can be any of the remaining two digits (since 0 is allowed here). So, for hundreds place: 2 choices (since two digits left), then tens place: 1 choice. So, 2*1=2 permutations.Therefore, numbers ending with 1: 2*2=4. Correct.Case 2: Last digit is 2.Similarly, first digit: cannot be 0 or 2. So, possible first digits: 1,5. 2 choices.Then, hundreds and tens digits: remaining two digits (including 0). So, 2 permutations. Hence, 2*2=4 numbers.Total: 4 +4=8. Correct.Therefore, the answer is D:8. So, even if I compute it both ways, subtracting the divisible numbers from total, or directly calculating the non-divisible ones, I get 8. Therefore, the answer is D.But just to be thorough, let's enumerate all possible numbers and check. Since there are only 18 numbers, maybe listing them would help.Digits available: 0,1,2,5.Four-digit numbers with no repetition:First digit: 1,2,5.Let me list them by first digit.First digit 1:Possible numbers:1 0 2 51 0 5 21 2 0 51 2 5 01 5 0 21 5 2 0Similarly for first digit 2:2 0 1 52 0 5 12 1 0 52 1 5 02 5 0 12 5 1 0First digit 5:5 0 1 25 0 2 15 1 0 25 1 2 05 2 0 15 2 1 0So, total 18 numbers. Now, let's count how many are divisible by 5. Divisible by 5 means last digit 0 or 5.Looking at the numbers:First digit 1:1025: ends with 5, divisible.1052: ends with 2, not.1205: ends with 5, divisible.1250: ends with 0, divisible.1502: ends with 2, not.1520: ends with 0, divisible.So, first digit 1: 4 numbers divisible by 5 (1025,1205,1250,1520). Wait, 1025,1205,1250,1520. That's four. Then two not:1052,1502.First digit 2:2015: ends with 5, divisible.2051: ends with 1, not.2105: ends with 5, divisible.2150: ends with 0, divisible.2501: ends with 1, not.2510: ends with 0, divisible.So, divisible numbers:2015,2105,2150,2510. Four numbers. Two not:2051,2501.First digit 5:5012: ends with 2, not.5021: ends with1, not.5102: ends with 2, not.5120: ends with0, divisible.5201: ends with1, not.5210: ends with0, divisible.So, divisible numbers:5120,5210. Two numbers. Four not:5012,5021,5102,5201.Total divisible by 5:First digit 1:4First digit 2:4First digit 5:2Total:4+4+2=10. So, 10 divisible by 5, 18 total. Therefore, non-divisible:18-10=8. So, same result. Therefore, the answer is D:8.But let's check the non-divisible numbers as per the list.First digit 1:1052,1502.First digit 2:2051,2501.First digit 5:5012,5021,5102,5201.So, total non-divisible numbers:2+2+4=8. Correct. So, the answer is D. Thus, I can confirm that the correct answer is D:8.Final Answerboxed{D}</think>To determine the total number of four-digit numbers composed of the digits 0, 1, 2, and 5 without any repetition that cannot be divided by 5, we start by calculating the total number of such four-digit numbers. A four-digit number cannot start with 0, so the thousands place has 3 choices (1, 2, or 5). The remaining three digits can be any permutation of the remaining three digits. Therefore, the total number of four-digit numbers is:[3 times 3 times 2 times 1 = 18]Next, we calculate the number of four-digit numbers divisible by 5. A number is divisible by 5 if its last digit is 0 or 5. 1. Numbers ending with 0: - The thousands place has 3 choices (1, 2, 5). - The remaining two middle digits are permutations of the remaining two digits. - Total: (3 times 2 = 6)2. Numbers ending with 5: - The thousands place has 2 choices (1, 2). - The remaining two middle digits are permutations of the remaining two digits. - Total: (2 times 2 = 4)Total numbers divisible by 5:[6 + 4 = 10]Subtracting the numbers divisible by 5 from the total gives the numbers not divisible by 5:[18 - 10 = 8]To verify, we directly calculate numbers ending with 1 or 2:- Ending with 1: 2 choices for the thousands place (2, 5) and permutations of the remaining two digits: (2 times 2 = 4)- Ending with 2: 2 choices for the thousands place (1, 5) and permutations of the remaining two digits: (2 times 2 = 4)Total numbers not divisible by 5:[4 + 4 = 8]Thus, the total number of numbers that cannot be divided by 5 is (boxed{D}).,<think>Okay, let's tackle this problem step by step. So, we need to find out how many four-digit numbers made from the digits 0, 1, 2, and 5 without repeating any digits are not divisible by 5. The options given are A: 18, B: 12, C: 10, D: 8. Hmm, let's break this down.First, let me recall that a number is divisible by 5 if its last digit is either 0 or 5. So, to find numbers that cannot be divided by 5, we need those numbers that don't end with 0 or 5. Therefore, the last digit has to be either 1 or 2. Got it.But before jumping into that, maybe I should first figure out how many total four-digit numbers can be formed with the digits 0, 1, 2, 5 without repetition. Then, subtract the number of those divisible by 5. That might be a more straightforward approach. Let me see.Total four-digit numbers: Since it's a four-digit number, the first digit can't be 0. So, the thousands place has 3 options: 1, 2, or 5. Then, the remaining three digits can be arranged from the remaining three digits (including 0). So, total permutations would be 3 (for first digit) * 3 * 2 * 1 = 3*3*2*1 = 18? Wait, no. Wait, let's do it properly.For the thousands place: 3 choices (1, 2, 5). Then, hundreds place: 3 remaining digits (since one is used and 0 is allowed here). Then tens place: 2 remaining digits. Then units place: 1 remaining digit. So total numbers: 3 * 3 * 2 * 1 = 18. Wait, that's 18 total four-digit numbers. But wait, that seems low. Let me check again.Digits available: 0,1,2,5. We need four-digit numbers without repetition. So, first digit: 3 choices (can't be 0). Then the next three digits: permutations of the remaining three digits. So, permutations of 3 digits is 3! = 6. Therefore, total numbers: 3 * 3! = 3*6=18. Yes, that's correct. So, total 18 four-digit numbers.Now, out of these 18 numbers, how many are divisible by 5? As we know, numbers ending with 0 or 5 are divisible by 5. So, let's calculate numbers ending with 0 and numbers ending with 5, then add them together.First, numbers ending with 0: If the last digit is fixed as 0, then the first digit has to be 1, 2, or 5 (3 choices). The middle two digits are permutations of the remaining two digits. So, for example, if first digit is 1, then the middle digits are permutations of 2 and 5, which is 2! = 2. Similarly for each first digit. So, total numbers ending with 0: 3 (first digits) * 2! (middle digits) = 3*2=6.Now, numbers ending with 5: Similarly, if the last digit is 5, then the first digit can't be 0 or 5 (since no repetition). So, first digit options: 1 or 2 (2 choices). The middle two digits are permutations of the remaining two digits (since 0 is allowed here). For example, if first digit is 1, the remaining digits are 0 and 2, so permutations of 0 and 2: 2! = 2. Similarly, if first digit is 2, remaining digits are 0 and 1, permutations 2. So, total numbers ending with 5: 2 (first digits) * 2! (middle digits) = 2*2=4.Therefore, total numbers divisible by 5: 6 (ending with 0) + 4 (ending with 5) = 10.Hence, numbers not divisible by 5 = total numbers - numbers divisible by 5 = 18 - 10 = 8. Hmm, so the answer is D: 8. Wait, but let me confirm again to be sure.Wait, but let me check if there's another way to calculate the numbers not divisible by 5 directly. Since numbers not divisible by 5 must end with 1 or 2. Let's see.Numbers ending with 1: Last digit is 1. Then first digit can't be 0 or 1. So first digit options: 2 or 5. Then the middle two digits are permutations of the remaining two digits (including 0). For example, if first digit is 2, remaining digits are 0 and 5, so permutations: 2! = 2. Similarly, if first digit is 5, remaining digits are 0 and 2, permutations: 2. So total numbers ending with 1: 2 (first digits) * 2! = 4.Similarly, numbers ending with 2: Last digit is 2. First digit can't be 0 or 2. So first digit options: 1 or 5. Middle two digits: permutations of remaining two digits. For example, first digit 1: remaining digits 0 and 5, permutations 2. First digit 5: remaining digits 0 and 1, permutations 2. So total numbers ending with 2: 2 * 2 = 4.Therefore, total numbers not divisible by 5: 4 + 4 = 8. Which matches the previous result. So, 8 numbers. Therefore, answer D: 8.But wait, let me verify this with actual enumeration to ensure we haven't missed anything. Let's list all possible four-digit numbers with digits 0,1,2,5 without repetition.Possible four-digit numbers:First, thousands digit can be 1, 2, or 5.Let's start with thousands digit 1:1 _ _ _The remaining digits are 0,2,5.Possible numbers:1025, 1052, 1205, 1250, 1502, 1520That's 6 numbers. Now, check which are divisible by 5:End with 0 or 5. So:1025 ends with 5: divisible.1052 ends with 2: not.1205 ends with 5: divisible.1250 ends with 0: divisible.1502 ends with 2: not.1520 ends with 0: divisible.So, among these 6 numbers, 4 are divisible by 5 (1025, 1205, 1250, 1520), 2 not (1052, 1502).Next, thousands digit 2:2 _ _ _Remaining digits: 0,1,5.Possible numbers:2015, 2051, 2105, 2150, 2501, 2510Again, 6 numbers. Check divisibility:2015 ends with 5: divisible.2051 ends with 1: not.2105 ends with 5: divisible.2150 ends with 0: divisible.2501 ends with 1: not.2510 ends with 0: divisible.So, divisible by 5: 2015, 2105, 2150, 2510 (4 numbers). Not divisible: 2051, 2501 (2 numbers).Next, thousands digit 5:5 _ _ _Remaining digits: 0,1,2.Possible numbers:5012, 5021, 5102, 5120, 5201, 5210Check divisibility:5012 ends with 2: not.5021 ends with 1: not.5102 ends with 2: not.5120 ends with 0: divisible.5201 ends with 1: not.5210 ends with 0: divisible.So, divisible by 5: 5120, 5210 (2 numbers). Not divisible: 5012, 5021, 5102, 5201 (4 numbers).Now, total numbers:From thousands digit 1: 6 numbers.From thousands digit 2: 6 numbers.From thousands digit 5: 6 numbers.Total: 18, which matches our initial calculation.Now, numbers divisible by 5:From 1xxx: 4From 2xxx: 4From 5xxx: 2Total divisible: 4 + 4 + 2 = 10. So numbers not divisible by 5: 18 - 10 = 8.Looking at the non-divisible numbers:From 1xxx: 1052, 1502From 2xxx: 2051, 2501From 5xxx: 5012, 5021, 5102, 5201Total: 2 + 2 + 4 = 8. Yes, that's correct. So, the answer is D:8.But let me check once more if all numbers are correctly listed and counted. For thousands digit 1, non-divisible numbers are 1052 and 1502. Correct. For thousands digit 2: 2051 and 2501. Correct. For thousands digit 5: 5012, 5021, 5102, 5201. Correct. So 8 numbers in total. Hence, the answer is D.Final Answerboxed{D}</think>To determine the total number of four-digit numbers composed of the digits 0, 1, 2, and 5 without any repetition that cannot be divided by 5, we first calculate the total number of such four-digit numbers. A four-digit number cannot start with 0, so the thousands place has 3 choices (1, 2, or 5). The remaining three digits can be any permutation of the remaining three digits. This gives us:[3 times 3 times 2 times 1 = 18]total four-digit numbers.Next, we calculate the number of these four-digit numbers that are divisible by 5. A number is divisible by 5 if its last digit is 0 or 5.1. Numbers ending with 0: - The thousands place has 3 choices (1, 2, 5), and the remaining two digits can be any permutation of the remaining two digits. - Total: (3 times 2 = 6).2. Numbers ending with 5: - The thousands place has 2 choices (1, 2), and the remaining two digits can be any permutation of the remaining two digits (including 0). - Total: (2 times 2 = 4).Adding these, the total number of numbers divisible by 5 is:[6 + 4 = 10]Subtracting the numbers divisible by 5 from the total gives us the numbers not divisible by 5:[18 - 10 = 8]Thus, the total number of four-digit numbers that cannot be divided by 5 is (boxed{D}).]

🔑:According to the problem, the numbers that cannot be divided by 5 are those whose last digit is not 0 or 5. Therefore, we can exclude the numbers whose last digit is 0 or 5 from all the eligible four-digit numbers. The total number of four-digit numbers is A_3^1 times A_3^3 = 18. When the last digit is 0, there are A_3^3 = 6 numbers. When the last digit is 5, there are A_2^1 times A_2^2 = 4 numbers. Therefore, the total number of numbers that cannot be divided by 5 is 18 - 6 - 4 = 8. Hence, the correct answer is boxed{D}.