answer:According to the problem, we will discuss 2 cases: ① If 1 person joins the "Speech Team," there are C_{6}^{1} = 6 ways to choose 1 person from the 6 to join the "Speech Team." The remaining 5 people will join the remaining 4 clubs. There are two ways to arrange the number of people: 1, 1, 1, 2 or 1, 2, 2. For 1, 1, 1, 2 arrangement: There are frac{C_{5}^{1}C_{4}^{1}C_{3}^{1}C_{2}^{2}}{A_{3}^{3}} times A_{4}^{4} ways to assign the people to the clubs. For 1, 2, 2 arrangement: There are frac{C_{5}^{1}C_{4}^{2}C_{2}^{2}}{A_{2}^{2}} times A_{4}^{3} ways to assign the people to the clubs. So, there are 6 times (frac{C_{5}^{1}C_{4}^{1}C_{3}^{1}C_{2}^{2}}{A_{3}^{3}} times A_{4}^{4} + frac{C_{5}^{1}C_{4}^{2}C_{2}^{2}}{A_{2}^{2}} times A_{4}^{3}) = 6 times 600 = 3600 ways in this case. ② If no one joins the "Speech Team," 6 people will join 4 clubs. There are two ways to arrange the number of people: 1, 1, 2, 2 or 2, 2, 2. For 1, 1, 2, 2 arrangement: There are frac{C_{6}^{2}C_{4}^{2}C_{2}^{1}C_{1}^{1}}{A_{2}^{2}A_{2}^{2}} times A_{4}^{4} ways to assign the people to the clubs. For 2, 2, 2 arrangement: There are frac{C_{6}^{2}C_{4}^{2}C_{2}^{2}}{A_{3}^{3}} times A_{4}^{3} ways to assign the people to the clubs. So, there are frac{C_{6}^{2}C_{4}^{2}C_{2}^{1}C_{1}^{1}}{A_{2}^{2}A_{2}^{2}} times A_{4}^{4} + frac{C_{6}^{2}C_{4}^{2}C_{2}^{2}}{A_{3}^{3}} times A_{4}^{3} = 1440 ways in this case. Thus, there are 3600 + 1440 = boxed{5040} ways in total. The answer is C. According to the problem, the 6 people have at most 1 person joining the "Speech Team," so we discuss 2 cases: ① when 1 person joins the "Speech Team" and ② when no one joins the "Speech Team." We calculate the number of ways for each case and add them up to get the final answer. This problem tests the application of permutations and combinations, and it's essential to pay attention to the club's people limit and use the grouping formula correctly.

answer:1. **Identify the sequences**: - Odd numbers from 1 to 2023. - Even numbers from 2 to 2020. 2. **Sum of odd numbers**: - The number of terms in the odd sequence is: [ n = frac{2023 - 1}{2} + 1 = 1012 ] - The sum of the first n odd numbers is n^2: [ 1012^2 = 1024144 ] 3. **Sum of even numbers**: - The number of terms in the even sequence is: [ n = frac{2020 - 2}{2} + 1 = 1010 ] - The sum of the first n even numbers is n(n+1): [ 1010 times 1011 = 1021110 ] 4. **Calculate the difference**: - Subtract the sum of the even numbers from the sum of the odd numbers: [ 1024144 - 1021110 = 3034 ] The value of the expression is 3034. Conclusion: The new problem maintains a consistent arithmetic sequence structure and can be solved using given formulas, ensuring a valid solution. The final answer is boxed{textbf{(D)} 3034}

answer:Let the four-digit integer be ab25, where a and b denote digits. Since a number is divisible by 75 if and only if it is divisible by both 25 and 3, the last three digits of the number must form a number divisible by 25, and the entire number must be divisible by 3. The last three digits form the number "225," which is divisible by 25, so this condition is satisfied. Now, we need to check divisibility by 3. The sum of digits of ab25 is a + b + 2 + 5 = a + b + 7. This sum needs to be divisible by 3. We find the values of a and b such that a+b+7 is divisible by 3. We can express this as a+b equiv -7 pmod{3} equiv 2 pmod{3}. The possible pairs (a, b) such that a+b equiv 2 pmod{3} are: - a = 1, b = 1 - a = 1, b = 4 - a = 1, b = 7 - a = 1, b = 0 - a = 2, b = 0 - a = 2, b = 3 - a = 2, b = 6 - a = 2, b = 9 - a = 3, b = 2 - a = 3, b = 5 - a = 3, b = 8 - a = 4, b = 1 - a = 4, b = 4 - a = 4, b = 7 - a = 4, b = 0 - a = 5, b = 0 - a = 5, b = 3 - a = 5, b = 6 - a = 5, b = 9 - a = 6, b = 2 - a = 6, b = 5 - a = 6, b = 8 - a = 7, b = 1 - a = 7, b = 4 - a = 7, b = 7 - a = 7, b = 0 - a = 8, b = 0 - a = 8, b = 3 - a = 8, b = 6 - a = 8, b = 9 - a = 9, b = 2 - a = 9, b = 5 - a = 9, b = 8 Hence, there are boxed{33} such four-digit numbers.

answer:1. For the hyperbola x^{2}- frac{y^{2}}{b^{2}}=1 (b > 0), the left and right foci are F_{1} and F_{2}, respectively. Let a=1 and c^{2} = 1 + b^{2}. Line l passes through F_{2}, intersects the hyperbola at points A and B, and has an angle of inclination of frac{pi}{2}. Also, triangle F_{1}AB is an equilateral triangle. We can deduce that point A is (c, b^2). Using the properties of an equilateral triangle, we get frac{sqrt{3}}{2} cdot 2b^{2} = 2c. This simplifies to 3b^4 = 4(a^2 + b^2) = 4(1 + b^2), which gives us the equation 3b^4 - 4b^2 - 4 = 0. Since b > 0, we find that b^2 = 2. The equation of the hyperbola is x^{2} - frac{y^{2}}{2} = 1, and the equations of its asymptotes are y = pm sqrt{2}x. 2. Let b = sqrt{3}. The hyperbola is now described by the equation x^{2} - frac{y^{2}}{3} = 1, with foci F_{1}(-2, 0) and F_{2}(2, 0). Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}). The slope of line l is given by k = frac{y_{2} - y_{1}}{x_{2} - x_{1}}. The equation of line l is y = k(x - 2). Substituting the value of y from the equation of line l into the equation of the hyperbola and eliminating y, we get (3 - k^{2})x^{2} + 4k^{2}x - 4k^{2} - 3 = 0. The discriminant triangle = 36(1 + k^{2}) > 0, so x_{1} + x_{2} = frac{4k^{2}}{k^{2} - 3}. Then, y_{1} + y_{2} = k(x_{1} + x_{2} - 4) = kleft(frac{4k^{2}}{k^{2} - 3} - 4right) = frac{12k}{k^{2} - 3}. Since M is the midpoint of AB and overrightarrow{FM} cdot overrightarrow{AB} = 0, we have (x_{1} + x_{2} + 4, y_{1} + y_{2}) cdot (x_{1} - x_{2}, y_{1} - y_{2}) = 0. This simplifies to x_{1} + x_{2} + 4 + (y_{1} + y_{2})k = 0, so frac{4k^{2}}{k^{2} - 3} + 4 + frac{12k}{k^{2} - 3}k = 0. Solving this equation for k, we find that k^{2} = frac{3}{5}, so k = pm frac{sqrt{15}}{5}. Therefore, the slope of line l is boxed{k = pm frac{sqrt{15}}{5}}.