answer:We need to calculate the number of trailing zeros in n! and (3n)! such that the zeros in (3n)! are four times those in n!. The formula for the number of trailing zeros in n! is: [ text{Number of zeros in } n! = leftlfloor frac{n}{5} rightrfloor + leftlfloor frac{n}{25} rightrfloor + leftlfloor frac{n}{125} rightrfloor + cdots ] Similarly, for (3n)!: [ text{Number of zeros in } (3n)! = leftlfloor frac{3n}{5} rightrfloor + leftlfloor frac{3n}{25} rightrfloor + leftlfloor frac{3n}{125} rightrfloor + cdots ] Setting up the given condition: [ 4 left( leftlfloor frac{n}{5} rightrfloor + leftlfloor frac{n}{25} rightrfloor + leftlfloor frac{n}{125} rightrfloor + cdots right) = leftlfloor frac{3n}{5} rightlfloor + leftlfloor frac{3n}{25} rightlfloor + leftlfloor frac{3n}{125} rightlfloor + cdots ] Checking for small values of n, we find that n = 10, 11, 17, 18 are the smallest values that satisfy the condition. Calculating the sum: [ s = 10 + 11 + 17 + 18 = 56 ] The sum of the digits of s is: [ 5 + 6 = 11 ] Thus, the sum of the digits of s is 11. The final answer is boxed{11}.

answer:To solve the problem, we start by differentiating the given curve y=e^{x+a} with respect to x to find its slope at any point. The derivative, denoted as y', is obtained using the chain rule: [y' = frac{d}{dx}e^{x+a} = e^{x+a} cdot frac{d}{dx}(x+a) = e^{x+a}] Next, we identify the point of tangency between the line and the curve as (x_{0}, y_{0}). At this point, the slope of the curve must equal the slope of the tangent line, and the curve and the line must intersect. Given the tangent line y=x-1, we can set up the following system of equations to represent these conditions: 1. The equation of the tangent line at (x_{0}, y_{0}): y_{0}=x_{0}-1 2. The curve's equation at (x_{0}, y_{0}): y_{0}=e^{x_{0}+a} 3. The slope of the curve at (x_{0}, y_{0}) equals the slope of the tangent line, which is 1: e^{x_{0}+a}=1 From equations (1) and (2), we can equate the expressions for y_{0} to get: [x_{0}-1 = e^{x_{0}+a} quad (4)] Equation (3) tells us that e^{x_{0}+a} = 1. Since the exponential function e^{x}=1 when x=0, we have: [x_{0}+a = 0 quad (5)] Substituting e^{x_{0}+a} = 1 into equation (4) gives us: [x_{0}-1 = 1] Solving for x_{0}, we find: [x_{0} = 2] Using equation (5) and substituting x_{0} = 2, we solve for a: [2 + a = 0] [a = -2] Therefore, the value of the real number a that satisfies the given conditions is -2. So, the correct answer is: [boxed{text{A}}]

answer:From the given information, we know that f₀(x) = sin(x), f₁(x) = f₀'(x), f₂(x) = f₁'(x), ..., f_{n+1}(x) = fₙ'(x), n ∈ N. So, according to the given information, we have: f₀(x) = sin(x), f₁(x) = cos(x), f₂(x) = -sin(x), f₃(x) = -cos(x), f₄(x) = sin(x). We can see that fₙ(x) has a period of 4. Therefore, 2021 ÷ 4 = 505 remainder 1. Hence, f₂₀₂₀(x) = f₀(x) = sin(x). So the answer is: boxed{text{A}}. From the given information, we can list out the terms and discover that the period is 4, which leads us to the answer. This question tests our understanding of derivative formulas and the periodicity of functions and can be considered a simple question.

answer:Cross-multiplication yields: [ x^2 + 3x + 4 = (x + 5)(x + 6) ] Expanding the right-hand side: [ x^2 + 3x + 4 = x^2 + 11x + 30 ] Subtracting (x^2) and rearranging terms: [ 0 = 8x + 26 ] Solving for (x): [ 8x = -26 x = -frac{26}{8} = -frac{13}{4} ] Thus, the solution is (boxed{-frac{13}{4}}).