answer:: To solve the problem where a four-digit number overline{abcd} is divisible by 3, and a, b, c are three consecutive integers in some permutation, we proceed as follows: 1. Assume that a, b, and c are three consecutive integers. Let: [ a = k, quad b = k+1, quad c = k+2 ] where k is an integer. 2. Since the number overline{abcd} is divisible by 3, the sum of its digits must be divisible by 3. Therefore, we have: [ 3 mid (a + b + c + d) ] 3. Substitute a, b, and c with k, k+1, and k+2 respectively: [ a + b + c = k + (k+1) + (k+2) = 3k + 3 ] Hence: [ 3 mid (3k + 3 + d) ] 4. Simplify the expression: [ 3 mid (3(k+1) + d) ] This means d must be such that the expression 3(k+1) + d is divisible by 3. Since 3(k+1) is always divisible by 3, d must also be divisible by 3. Thus, possible values of d are: [ d = 0, 3, 6, 9 ] 5. Consider different values of k and the permutations of a, b, c: - For k = 0, the consecutive integers are 0, 1, 2. Permutations of (a, b, c) are: [ (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0) ] These give us 6 permutations. - For other values of k = 1, 2, 3, 4, 5, 6, 7, there are 3 consecutive integers (k, k+1, k+2) each. The number of permutations of three numbers is 3! = 6. So, each k gives 6 permutations. 6. Summarize the number of valid digits for a, b, c: - For k = 0, we have 4 valid permutations. - For k = 1,2,...,7, we have 7 sets of 3 consecutive integers, each with 6 permutations. Therefore, 7 times 6 = 42 permutations in total. Thus, the total number of four-digit numbers overline{abcd} is: [ 4 + 42 = 46 ] 7. Considering the 4 possible values of d, the total number of valid four-digit numbers (each unique quadruple overline{abcd}): [ 46 times 4 = 184 ] # Conclusion: The total number of such four-digit numbers is: [ boxed{184} ]

answer:Let's denote the amount of money Jethro has as J. According to the information given: 1. Carmen needs 7 more to have twice the amount of money that Jethro has. So, the amount of money Carmen has is 2J - 7. 2. Patricia has 60, which is a certain multiple of the amount of money Jethro has. So, we can write this as 60 = kJ, where k is the multiple. 3. The sum of all their money is 113. So, we can write this as J + (2J - 7) + 60 = 113. Now, let's solve for J: J + 2J - 7 + 60 = 113 3J + 53 = 113 3J = 113 - 53 3J = 60 J = 60 / 3 J = 20 Now that we know Jethro has 20, we can find the ratio of the amount of money Patricia has to the amount of money Jethro has. Since Patricia has 60, we can write the ratio as: Patricia's money : Jethro's money = 60 : 20 To simplify the ratio, we divide both numbers by the greatest common divisor, which is 20: 60 / 20 : 20 / 20 = 3 : 1 So, the ratio of the amount of money Patricia has to the amount of money Jethro has is boxed{3:1} .

answer:(1) From the given conditions, we have the system of equations: begin{cases} f(1) = 1 + b + c = 0 f(3) = 9 + 3b + c = 0 end{cases} Solving this system, we get: boxed{begin{cases} b = -4 c = 3 end{cases}} (6 points) (2) From ①, we know f(x) = x^2 - 4x + 3. Let x_1, x_2 in (2, +infty), but x_1 < x_2. Then, f(x_1) - f(x_2) = x_1^2 - 4x_1 - x_2^2 + 4x_2 = (x_1 + x_2)(x_1 - x_2) - 4(x_1 - x_2) = (x_1 - x_2)[(x_1 + x_2) - 4] Since x_1 < x_2, we have x_1 - x_2 < 0. And since x_1 > 2 and x_2 > 2, we have (x_1 + x_2) - 4 > 0. Therefore, f(x_1) - f(x_2) < 0, which implies f(x_1) < f(x_2). Thus, f(x) is an increasing function on (2, +infty). boxed{text{Proved}} (12 points)

answer:**Analysis** This problem mainly examines the use of derivatives to study the extreme values of functions. **Solution** Since the function f(x) has a minimum value in the interval (-1,0), it means that f(x) has a local minimum in the interval (-1,0). Because f'(x) = e^x + 2x + 3a + 2 has a zero in the interval (-1,0), it follows that -3a-2 = e^x + 2x has a zero, Since y = e^x + 2x is an increasing function in the interval (-1,0), we have dfrac{1}{e}-2 < -3a-2 < 1, solving this yields -1 < a < -dfrac{1}{3e}, Therefore, the correct choice is boxed{D}.