answer:(I) The derivative of the function f(x) = frac{ mx}{ln x} is f'(x) = frac{m(ln x - 1)}{(ln x)^2}. Based on the given condition, the slope of the tangent line at x = e^2 is the negative reciprocal of the slope of the line 2x + y = 0, which is -frac{1}{2}. Therefore, we have: f'(e^2) = -frac{1}{2} = frac{m(ln e^2 - 1)}{(ln e^2)^2} Rightarrow frac{2m}{4} = -frac{1}{2} Rightarrow m = -1. Thus, the explicit expression for f(x) becomes f(x) = frac{-x}{ln x}. For the function f(x) to be monotonically decreasing, we need f'(x) leq 0. With our value of m = -1, f'(x) = frac{-(ln x - 1)}{(ln x)^2} leq 0 implies 0 < x leq 1 or e leq x < infty. Therefore, the intervals where f(x) is monotonically decreasing are (0,1] and [e, infty). (II) We have g(x) = f(x) - frac{kx^2}{x-1} Rightarrow g(x) = xleft(frac{-1}{ln x} - frac{kx}{x-1}right), with the domain (0,1) cup (1, infty). For g(x) to have no zeros, the equation frac{-1}{ln x} = frac{kx}{x-1} must have no solution for x in (0,1) cup (1, infty). Consider the function h(x) = kln x - frac{2(x-1)}{x} Rightarrow h'(x) = frac{kx-2}{x^2}. ① When k leq 0, h'(x) < 0 for all x in (0,1) cup (1, infty), so h(x) is monotonically decreasing in both intervals. Since h(1) = 0, there are no zeros in these intervals, satisfying the condition. ② When k > 0, we examine h'(x) again: - If 0 < k < 2, h(x) is monotonically decreasing in (0,1), also decreasing in left(1, frac{2}{k}right), and increasing in left(frac{2}{k}, infty right). With h(1)=0, there are no zeros in (0,1). Since hleft(frac{2}{k}right) < 0 and hleft(e^{frac{2}{k}}right) = frac{2}{k} - 2 + frac{2}{e^{frac{2}{k}}} > 0, h(x) has one zero in left(frac{2}{k}, infty right), which does not satisfy the condition. - If k = 2, h(x) is monotonically decreasing in (0,1) and increasing in (1, infty). Since h(1)=0, h(x) > 0 for all x in (0,1) cup (1, infty), which satisfies the condition. - If k > 2, h(x) is decreasing in left(0, frac{2}{k}right), increasing in left(frac{2}{k}, 1right), and increasing in (1, infty). Since h(1) = 0, there are no zeros in left(frac{2}{k}, 1right) and (1, infty). Knowing hleft(frac{2}{k}right) < 0 and h(e^{-k}) = 2e^k - k^2 - 2 > 0 when k > 2, h(x) has one zero in left(0, frac{2}{k}right), which does not satisfy the condition. Combining these results, the range of values for k is k leq 0 or k = 2. boxed{k leq 0 text{ or } k = 2}

answer:1. **Choose (10^{2002}) arbitrary primes.** Let these primes be (p_1, p_2, ldots, p_{10^{2002}}). 2. **Calculate the product (P) and the sum (S) of these primes.** [ P = p_1 cdot p_2 cdot ldots cdot p_{10^{2002}} ] [ S = p_1 + p_2 + ldots + p_{10^{2002}} ] 3. **Consider the difference (P - S).** Since (P) is a product of primes and (S) is a sum of primes, (P - S) is an integer. 4. **Find the smallest prime factor (q) of (P - S).** Let (q) be the smallest prime factor of (P - S). 5. **Express (P - S) in terms of (q).** Since (q) is a prime factor of (P - S), we can write: [ P - S = kq quad text{for some integer } k ] 6. **Add (q) to the list of primes (N) times such that the new sum (S') equals the new product (P').** Let (N) be the number of times we need to add (q) to the list of primes to make the new sum (S') equal to the new product (P'). We need: [ S' = S + Nq = P ] [ P' = q^N cdot P ] 7. **Determine (N).** From the equation (S' = S + Nq = P), we solve for (N): [ Nq = P - S ] [ N = frac{P - S}{q} ] Since (q) is a prime factor of (P - S), (N) is an integer. 8. **Verify that (P') is divisible by (S').** [ P' = q^N cdot P ] [ S' = S + Nq = P ] Since (P' = q^N cdot P) and (S' = P), it follows that (P') is divisible by (S'). Thus, we have constructed a number (P') with at least (10^{2002}) distinct prime factors that is divisible by the sum of its prime factors, making it a "wonderful" number. (blacksquare)

answer:Dave picked up 14 sticks, and there were 4 left. The difference between the number of sticks Dave picked up and the number left is: 14 (picked up) - 4 (left) = 10 sticks So, the difference is boxed{10} sticks.

answer:x^2 - 2xy + y^2 - 1 = (x^2 - 2xy + y^2) - 1 = (x - y)^2 - 1 = (x - y + 1)(x - y - 1). Thus, the factorized form is boxed{(x - y + 1)(x - y - 1)}.