answer:Let's denote the ages of A, B, and C as A, B, and C respectively. According to the information given: 1. The total age of A and B is some years more than the total age of B and C. This can be written as: A + B > B + C 2. C is 12 years younger than A. This can be written as: C = A - 12 We want to find out how many years more the total age of A and B is compared to the total age of B and C. Let's denote this difference as D. So we have: D = (A + B) - (B + C) Substituting the second equation into the first, we get: D = (A + B) - (B + (A - 12)) Now, let's simplify the equation: D = A + B - B - A + 12 D = 12 Therefore, the total age of A and B is boxed{12} years more than the total age of B and C.

answer:Let's define the variables to clearly express the problem. 1. Let x be the number of 甲 tickets sold. 2. Let y be the number of 乙 tickets sold. 3. Let z be the number of 丙 tickets sold. From the problem, we have the following system of equations: 1. The total number of tickets sold is 85: [ x + y + z = 85 ] 2. The total revenue generated from selling the tickets is 500: [ 7x + 4y + 2z = 500 ] 3. The number of 甲 tickets sold is 31 more than the number of 乙 tickets sold: [ x = y + 31 ] We now have a system of linear equations. Let's substitute x in the first and second equations using x = y + 31: Substituting x in x + y + z = 85: [ (y + 31) + y + z = 85 ] Simplifying, [ 2y + z + 31 = 85 ] [ 2y + z = 54 quad text{(1)} ] Next, substitute x in 7x + 4y + 2z = 500: [ 7(y + 31) + 4y + 2z = 500 ] Expanding and simplifying, [ 7y + 217 + 4y + 2z = 500 ] [ 11y + 2z + 217 = 500 ] [ 11y + 2z = 283 quad text{(2)} ] Substitute equation (1) into equation (2): Solve for z from (1): [ z = 54 - 2y quad text{(3)} ] Now substitute (3) into (2): [ 11y + 2(54 - 2y) = 283 ] [ 11y + 108 - 4y = 283 ] Combine like terms, [ 7y + 108 = 283 ] Subtract 108 from both sides, [ 7y = 175 ] Divide by 7, [ y = 25 ] Substitute y = 25 back into (3): [ z = 54 - 2(25) ] [ z = 54 - 50 ] [ z = 4 ] Finally, use x = y + 31 to find x: [ x = 25 + 31 ] [ x = 56 ] Thus, the number of 甲 (A) tickets sold is boxed{56}.

answer:(I) Since vectors overrightarrow{m} = (ln x, 1-a ln x) and overrightarrow{n} = (x, f(x)) are parallel, we have: ln x cdot f(x) = x (1-a ln x) quad (x > 0). Hence, the function can be expressed as: f(x) = frac{x(1-a ln x)}{ln x} = frac{x}{ln x} - ax. The derivative of f(x) is given by: f'(x) = frac{ln x - 1}{(ln x)^2} - a. Since the function f(x) is monotonically decreasing on (1, +infty), we require that: f'(x) leq 0 quad text{for } x > 1, or equivalently: a geq frac{ln x - 1}{(ln x)^2} quad text{for } x > 1. Let g(x) = frac{ln x - 1}{(ln x)^2} = frac{1}{ln x} - left( frac{1}{ln x} right)^2. Substitute frac{1}{ln x} = t, noting that x > 1 implies ln x > 0 and t in (0, +infty). The function g(t) = -t^2 + t is a downward-opening parabola with axis of symmetry at t = frac{1}{2}. The maximum value of g(t) occurs at t = frac{1}{2}, hence g(t) leq frac{1}{4}. Therefore, the minimum value of a is: boxed{a geq frac{1}{4}}. (II) The proposition "There exist x_1, x_2 in [e, e^2] such that f(x_1) leq f'(x_2) + a" is equivalent to: "For x in [e, e^2], f(x)_{text{min}} leq f'(x)_{text{max}} + a." From part (I), for x in [e, e^2], we have ln x in [1, 2], and thus frac{1}{ln x} in [frac{1}{2}, 1]. The derivative f'(x) can be rewritten as: f'(x) = -left( frac{1}{ln x} - frac{1}{2} right)^2 + frac{1}{4} - a leq frac{1}{4} - a. Therefore, f'(x)_{text{max}} + a = frac{1}{4}, and we must have: f(x)_{text{min}} leq frac{1}{4}. 1. When a leq -frac{1}{4}, f(x) is decreasing on [e, e^2], thus: f(x)_{text{min}} = f(e^2) = frac{e^2}{2} - ae^2 leq frac{1}{4}, implying: a geq frac{1}{2} - frac{1}{4e^2}, which is contradictory to a leq -frac{1}{4}. 2. When a > -frac{1}{4}, as x lies in [e, e^2], ln x lies in [1, 2]. Since f'(x) = frac{ln x - 1}{(ln x)^2} - a is increasing on [e, e^2], there exists a unique x_0 in (e, e^2) such that f'(x_0) = 0, and consequently: f(x)_{text{min}} = f(x_0) = frac{x_0}{ln x_0} - ax_0. To satisfy f(x)_{text{min}} leq frac{1}{4}, we must have: a geq frac{1}{ln x_0} - frac{1}{4x_0} < frac{1}{4}. Thus, for this case, a geq frac{1}{4}. Combining both cases, the range of values for the real number a is: boxed{a in [frac{1}{4}, +infty)}.

answer:1. textbf{Definition:} Consider a prime number p and the sum S = 1^i + 2^i + cdots + (p-1)^i. We need to show that: [ S equiv 0 pmod{p} ] 2. textbf{Primitive Root:} Let g be a primitive root modulo p. This means (g) is a generator of the multiplicative group of integers modulo (p), that is (g^1, g^2, ldots, g^{p-1}) are all distinct modulo (p). 3. textbf{PGCD Definition:} Let d = gcd(i, p-1). According to number theory, the sum can be expressed as: [ S equiv dleft(sum_{k=0}^{frac{p-1}{d}-1} g^{kd}right) pmod{p} ] 4. textbf{Geometric Series:} The inner sum is a geometric series: [ sum_{k=0}^{frac{p-1}{d}-1} g^{kd} ] 5. textbf{Sum of Geometric Series:} The sum of a geometric series is given by: [ sum_{k=0}^{n-1} r^k = frac{1-r^n}{1-r} ] Applying this to our case: [ sum_{k=0}^{frac{p-1}{d}-1} g^{kd} = frac{1-g^{(p-1)}}{1-g^d} ] 6. textbf{Properties of Primitive Root:} By the properties of a primitive root: [ g^{p-1} equiv 1 pmod{p} ] Thus, [ frac{1-g^{p-1}}{1-g^d} = frac{1-1}{1-g^d} = 0 pmod{p} ] 7. textbf{Final Calculation:} So, we have: [ left(g^{d} - 1right) S equiv g^{p-1} - 1 equiv 0 pmod{p} ] Since ( g^{p-1} equiv 1 pmod{p} ), therefore: [ left(g^{d} - 1right) S equiv 0 pmod{p} ] And since ( g^{d} notequiv 1 pmod{p} ) as (d < p-1), it follows that: [ S equiv 0 pmod{p} ] # Conclusion: [ boxed{S equiv 0 pmod{p}} ]