answer:1. **Roots of Unity**: Since omega^3 = 1 and omega neq 1, it implies omega is a primitive third root of unity. Hence, omega^2 + omega + 1 = 0. By the quadratic formula, omega = -frac{1}{2} pm frac{sqrt{3}}{2} i. 2. **Conjugate and Sum**: The conjugate of the given sum is [ frac{1}{b_1 + overline{omega}} + frac{1}{b_2 + overline{omega}} + dots + frac{1}{b_m + overline{omega}} = 3 - 3i. ] For any real b, [ frac{1}{b + omega} + frac{1}{b + overline{omega}} = frac{b + omega + b + overline{omega}}{(b + omega)(b + overline{omega})} = frac{2b - 1}{b^2 - b + 1}. ] Adjusting for the new coefficients, [ frac{3b - 2}{b^2 - b + 1} = frac{1}{b + omega} + frac{1}{b + overline{omega}}. ] 3. **Final Sum**: Hence, [ sum_{k = 1}^m frac{3b_k - 2}{b_k^2 - b_k + 1} = sum_{k = 1}^m left( frac{1}{b_k + omega} + frac{1}{b_k + overline{omega}} right) = 3 + 3i + 3 - 3i = boxed{6}. ]

answer:** 1. **Identify possible digits**: The prime digits to use are 2, 3, 5, and 7. 2. **Find combinations where the sum is even**: We need an even number of digits that are odd (3, 5, 7) for the total sum to be even after including the even digit 2. - Possibilities with two identical odd digits: - (3, 3, 2), (5, 5, 2), (7, 7, 2) - Possibilities with two different odd digits: - (3, 5, 2), (3, 7, 2), (5, 7, 2) - Each of these groups can be permuted (Position of '2' can change), so the arrangements would be: - (3, 3, 2) - three permutations: (3, 3, 2), (3, 2, 3), (2, 3, 3) and similarly for others. 3. **Calculate total options**: - Three options (identical odd digits) with three permutations each, total = 3 times 3 = 9. - Three options (different odd digits) with three permutations each, total = 3 times 3 = 9. - **Total count**: 9 (identical) + 9 (different) = boxed{18} Conclusion: There are 18 such three-digit integers where each digit is prime and the sum of the digits is even.

answer:1. **Introduction and Assumption**: Assume that the radii of the circumscribed circles of the triangles triangle ACE and triangle BDF are both greater than 1. Let O be the center of the circumscribed circle of triangle triangle ACE. 2. **Angles Comparison**: We need to compare the exterior angles of the hexagon ABCDEF with the angles subtended by the sides of triangles triangle ACE and triangle BDF at the respective circumcenters. 3. **Circumcenter Angles**: For the triangle triangle ACE with circumcenter O: - The angle angle AOC subtends the arc AC of the circumscribed circle. However, angle ABC subtends the same arc in the hexagon itself. Since the circumradius is greater than 1, angle ABC > angle AOC because the closer point B accentuates the external angle due to smaller internal angle. - Similarly, angle COE and angle EOA behave the same way compared to angles angle CDE and angle EFA respectively: [ angle ABC > angle AOC ] [ angle CDE > angle COE ] [ angle EFA > angle EOA ] 4. **Sum of Angles in Hexagon**: Since the sum of angles in any polygon is (n-2)pi for an n-sided polygon (where n = 6 for a hexagon): - The total sum of the exterior angles of a hexagon is 4pi. Hence: [ angle A + angle B + angle C + angle D + angle E + angle F = 4pi ] 5. **Contradiction**: Given the condition of radii being more than 1: - Summarizing our angle comparisons, for both triangles triangle ACE and triangle BDF, leads to: [ (angle A + angle C + angle E) > 2pi ] [ (angle B + angle D + angle F) > 2pi ] By adding these inequalities, we find that the total exceeds 4pi: [ (angle A + angle C + angle E) + (angle B + angle D + angle F) > 4pi ] This implies: [ (angle A + angle B + angle C + angle D + angle E + angle F) > 4pi ] 6. **Conclusion**: The sum of interior angles of hexagon ABCDEF is exactly 4pi, leading to a contradiction if one assumes the radii of the circumscribed circles of either triangle ACE or triangle BDF are greater than 1. Therefore, it must be true that for at least one of these triangles, the circumradius is not greater than 1. [ boxed{8}

answer:To solve the system of inequalities left{begin{array}{l}{frac{3x+1}{2} > x}{4(x-2) leq x-5}end{array}right., we proceed as follows: 1. For the first inequality frac{3x+1}{2} > x, we multiply both sides by 2 to eliminate the denominator and simplify: begin{align*} frac{3x+1}{2} &> x 3x + 1 &> 2x 3x - 2x &> -1 x &> -1 end{align*} 2. For the second inequality 4(x-2) leq x-5, we distribute the 4 and then simplify: begin{align*} 4(x-2) &leq x-5 4x - 8 &leq x - 5 4x - x &leq -5 + 8 3x &leq 3 x &leq 1 end{align*} Combining the results from both inequalities, we find that the solution set for the system is -1 < x leq 1. Therefore, the solution set of the inequality system is boxed{-1 < x leq 1}.