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question:Let m and n be two natural numbers. A school has n students eating in the cafeteria. Over the course of the year, the cafeteria has offered 100 different dishes. Each student has tasted 10 different dishes. Furthermore, if we take any two dishes out of the 100, we know that exactly m students have tasted each of the two dishes. Calculate the ratio: frac{n}{m}.
answer:Let E denote the set of students, M the set of menus, and A the set of triplets ({a, b}, e) where a, b in M, a neq b and e in E such that e has tasted menu a and menu b. Then, if e_{0} in E, we denote A_{e_{0}} as the set of triplets ({a, b}, e) in A such that e=e_{0}, and A_{left{a_{0}, b_{0}right}} as the set of triplets ({a, b}, e) in A such that left{a_{0}, b_{0}right}={a, b}. If B is a set, we denote |B| as the number of elements in B. According to the statement, for all a, b in M such that a neq b,left|A_{{a, b}}right|=m and, since each student has tasted 10 menus, for all e in E,left|A_{e}right|=binom{10}{2}=45. Finally, we have:begin{aligned}|A| & =sum_{e in E}left|A_{e}right| & =n times 45 |A| & =sum_{a, b in M}left|A_{{a, b}}right| & =left(begin{array}{c}100 2 2end{array}right) m & =4950 m 45 n & =4950 m frac{n}{m} & =110end{aligned}
question:7. In a chess tournament, 8 players participated. They scored 7, 6, 4, 4, 3, 2, 1, 5 and 0.5 points respectively. How many points did the players who took the first four places lose in matches against the others?
answer:7. From the condition of the problem, it follows that the tournament winner and the participant who took second place did not lose a single point in their matches against the others. Further, the participants who shared third and fourth places could have scored 8 points against the rest, but scored 7 (they played one point in their match against each other). Thus, the players who took the first four places lost 1 point in their matches against the others.73 Tournaments
question:Transform the equation x^{2}-15y^{2}=15 into its standard form. The correct answer is ( ).A: dfrac {x^{2}}{15}-y^{2}=1B: dfrac {y^{2}}{15}-x^{2}=1C: x^{2}- dfrac {y^{2}}{15}=1D: dfrac {x^{2}}{15}+y^{2}=1
answer:Given x^{2}-15y^{2}=15, we get dfrac {x^{2}}{15}- dfrac {15y^{2}}{15}=1, which simplifies to dfrac {x^{2}}{15}-y^{2}=1. Therefore, the correct choice is boxed{A}. This solution is obtained by dividing both sides of the given equation by 15. This question tests the understanding of the standard equation of a hyperbola, and it is a basic problem.
question:2. 10 people go to the bookstore to buy books, it is known that: (1) each person bought three types of books; (2) any two people have at least one book in common.How many people at most could have bought the book that was purchased by the fewest people?
answer:2. Suppose among them, A bought three books, and since A has at least one book in common with each of the other 9 people, the book that the most people bought among A's three books is purchased by no less than 4 people. If the book that the most people bought is purchased by 4 people, then all three books bought by A are each purchased by 4 people, and each book bought by the other 9 people is also purchased by 4 people. Therefore, the total number of books bought by the 10 people is a multiple of 4, i.e., 4 mid 30, which is a contradiction. Thus, the book that the most people bought is purchased by at least 5 people. Consider the following purchasing method:begin{array}{l}quadleft{B_{1}, B_{2}, B_{3}right},left{B_{1}, B_{2}, B_{4}right},left{B_{2}, B_{3},right. left.B_{5}right},left{B_{1}, B_{3}, B_{6}right},left{B_{1}, B_{4}, B_{5}right},left{B_{2}, B_{4},right. left.B_{6}right},left{B_{3}, B_{4}, B_{5}right},left{B_{1}, B_{5}, B_{6}right},left{B_{2}, B_{5},right. left.B_{6}right},left{B_{3}, B_{4}, B_{6}right}end{array}
question:5. Given the equation a x^{2}+b x+c=0(a neq 0), the sum of the roots is s_{1}, the sum of the squares of the roots is s_{2}, and the sum of the cubes of the roots is s_{3}. Then the value of a s_{3}+ leftlangle s_{2}right. +c s_{1} is . qquad
answer:(Tip: Let the two roots of the equation be x_{1}, x_{2}.Then by definition, we have a x_{1}^{2}+b x_{1}+c=0, a x_{2}^{2}+b x_{2}+c=0. The original expression =0.)
question:7. The solution set of the inequality log _{14}(sqrt{x}+sqrt[3]{x}+sqrt[6]{x})^{6} geqslant log _{2} x is
answer:Let x=64^{t}, then the original inequality becomes log _{14}left(8^{t}+4^{t}+2^{t}right) geqslant t Rightarrow 8^{t}+4^{t}+2^{t} geqslant 14^{t} Rightarrowleft(frac{8}{14}right)^{t}+left(frac{4}{14}right)^{t}+left(frac{2}{14}right)^{t} geqslant 1. Since the function f(t)=left(frac{8}{14}right)^{t}+left(frac{4}{14}right)^{t}+left(frac{2}{14}right)^{t} is monotonically decreasing and f(1)=1, we have f(t) geqslant 1 Rightarrow t leqslant 1 Rightarrow 0<x leqslant 64.Therefore, the solution set of the inequality log _{14}(sqrt{x}+sqrt[3]{x}+sqrt[6]{x})^{6} geqslant log _{2} x is (0,64].