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5 Several students take an exam, with a total of 4 multiple-choice questions, each with 3 options. It is known: any 3 students have one question where their answers are all different. Find the maximum number of students. (29th
| 5. Suppose there are $n$ candidates. It is easy to see that each candidate's answer sheet is a sequence of length 4, consisting of the letters $A$, $B$, and $C$. Thus, this problem is equivalent to 3-coloring an $n \times 4$ grid such that "any three rows have a column with 3 cells of different colors." Let the total n... | 9 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,846 |
Given 25 people, where every 5 people can form a committee, and any two committees have at most one common member. Prove: the number of committees is no more than 30. (5th All-Russian Mathematical Olympiad) | 6. If we understand that each pair of committees has at most one common member as $\left|A_{i} \cap A_{j}\right| \leqslant 1$, then we should denote $A_{i}$ as the set of people in the $i$-th committee. Let $X$ be the set of the given 25 people, and $F=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$. We will calculate the ... | 30 | Combinatorics | proof | Yes | Yes | inequalities | false | 737,847 |
Example 2: There are 1988 unit cubes, and they (all or part) are used to form three right square prisms $A$, $B$, and $C$ with heights of 1 and base side lengths of $a$, $b$, and $c$ ($a<b<c$). Now, place $A$, $B$, and $C$ in the first quadrant, with each base parallel to the coordinate axes, one vertex of $C$ at the o... | Analysis and Solution: From "boundary misalignment", we have $a \leqslant b-2 \leqslant c-4$. Thus, placing $A$ on $B$ has $(b-a-1)^{2}$ ways, and placing $B$ on $C$ has $(c-b-1)^{2}$ ways. Therefore, there are $P=(b-a-1)^{2}(c-b-1)^{2}$ different three-story buildings. Thus, the problem is equivalent to: for all posit... | 345^2 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,848 |
Example 1 In an $n \times n$ chessboard $C$, a number is filled in each cell, and the numbers filled in the edge cells of the board are all -1. For any other empty cell, fill it with the product of the two closest filled numbers on its left and right in the same row or above and below in the same column. Find the maxim... | Analysis and Solution First, it is clear that $f(3)=g(3)=1$.
When $n>3$, we hope that 1 appears as much as possible. Can they all be 1? Through attempts, we find that there must be at least one -1. In fact, consider the -1 cells tightly surrounding the outermost layer, among which at least one cell is filled with -1. O... | f(n)=\left\{\begin{array}{ll}
1 & (n=3) ; \\
(n-2)^{2}-1 & (n>3) .
\end{array}\right. \quad g(n)=n-2} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,849 |
Example 3: On a lottery ticket, there are 50 spaces arranged in sequence. Each participant fills in the integers from 1 to 50 on their ticket (each number appears exactly once on the same ticket). The host also fills out a ticket as the reference. If a participant's number sequence has a position where the number match... | Analysis and Solution If filling $k$ lottery tickets can result in a win, then $k$ is called a winning number. Arrange the $k$ filled lottery tickets into a $k \times 50$ number table:
\begin{tabular}{|c|c|c|c|c|c|}
\hline & $a_{1}$, & $a_{2}$, & $a_{3}$, & $\cdots$, & $a_{50}$ \\
\hline & $b_{1}$, & $b_{2}$, & $b_{3}$... | 26 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,851 |
Example 4 A dance and song troupe has $n(n>3)$ actors, who have arranged some performances, with each performance featuring 3 actors performing on stage together. During one performance, they found: it is possible to appropriately arrange several performances so that every 2 actors in the troupe perform on stage togeth... | To solve: Use $n$ points to represent $n$ actors. If two actors have performed together on stage, connect the corresponding points with an edge. Thus, the condition of the problem is equivalent to: the complete graph $K_{n}$ of order $n$ can be partitioned into several complete graphs $K_{3}$ of order 3, such that each... | 7 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,852 |
1 Let $A$ be a $k$-element subset of the set $\{1,2,3, \cdots, 16\}$, and any two subsets of $A$ have different sums of elements. For any $(k+1)$-element subset $B$ of the set $\{1,2,3, \cdots, 16\}$ that contains $A$, there exist two subsets of $B$ whose sums of elements are equal. (1) Prove: $k \leqslant 5$; (2) Find... | 1. (1) If $k \geqslant 7$, then the number of non-empty subsets of $A$ is $2^{k}-1$, and the sum of elements in each subset does not exceed $17k$. However, $2^{k}-1 > 17k$, so there must be two subsets with the same sum, which is a contradiction. If $k=6$, consider the one-, two-, three-, and four-element subsets of $A... | 66 | Combinatorics | proof | Yes | Yes | inequalities | false | 737,854 |
2 In the parliament, there are 2000 deputies who decide to review the budget, which consists of 200 expenditures. Each deputy prepares a budget draft, listing the amounts for each expenditure. The total amount listed by each does not exceed S. During the parliamentary review, each amount is set to a number agreed upon ... | 2. Obviously, the expenses filled by each member of parliament form a sequence of length 200, and the 2000 sequences filled by the members of parliament constitute a $2000 \times 200$ number table. The sum of each row in this table does not exceed $S$. Now, a row of numbers needs to be filled at the bottom of the table... | 1991 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,855 |
3 There are $n(n \geqslant 5)$ football teams participating in a league, where each pair of teams plays exactly one match. It is stipulated that a win earns 3 points, a draw earns 1 point, and a loss earns 0 points. After the league ends, some teams may be disqualified, and thus their match results will also be cancele... | 3. If all teams except team $i$ are disqualified, then team $i$ naturally becomes the champion, so $f_{i} \leqslant n-1$. Therefore, $F \leqslant n(n-1)$. Furthermore, when all teams draw, for all $i=1,2, \cdots, n$, we have $f_{i}=n-1$. At this point, $F=n(n-1)$. Hence, $F_{\text {max }}=n(n-1)$. When $n \geqslant 5$,... | F_{\min }=n, F_{\max }=n(n-1) | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,856 |
Example 1 A city has $n$ high schools, the $i$-th high school sends $c_{i}$ students $\left(1 \leqslant c_{i} \leqslant 39\right)$ to watch a ball game at the gymnasium, where $\sum_{i=1}^{n} c_{i}=1990$. Each row in the stand has 199 seats, and it is required that students from the same school sit in the same row. Que... | Analysis and Solution First, consider what the worst-case scenario is. The worst-case scenario refers to the situation where the number of empty seats in each row is maximized. Clearly, if the number of participants from each school varies, it is easier to arrange, as the remaining seats can be filled by students from ... | 12 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,857 |
Example 3 Given a positive integer $k$ and a positive number $a$, and $k_{1}+k_{2}+\cdots+k_{r}=k\left(k_{i}\right.$ are positive integers, $1 \leqslant$ $r \leqslant k$ ), find the maximum value of $F=a^{k_{1}}+a^{k_{2}}+\cdots+a^{k_{r}}$. (8th China Mathematical Olympiad Problem) | Take $k=6, r=3$, then $k_{1}+k_{2}+k_{3}=6, F=a^{k_{1}}+a^{k_{2}}+a^{k_{3}}$.
(1) If $\left(k_{1}, k_{2}, k_{3}\right)=(2,2,2)$, then $F_{1}=a^{2}+a^{2}+a^{2}=3 a^{2}$;
(2) If $\left(k_{1}, k_{2}, k_{3}\right)=(1,2,3)$, then $F_{2}=a+a^{2}+a^{3}$;
(3) If $\left(k_{1}, k_{2}, k_{3}\right)=(1,1,4)$, then $F_{3}=a+a+a^{4}... | \max \left\{a^{k}, k a\right\} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,859 |
Example 3 In a $19 \times 89$ chessboard, what is the maximum number of chess pieces that can be placed so that no $2 \times 2$ rectangle contains more than 2 pieces? | Analysis and Solution Note that 19 and 89 are both odd numbers, so we can consider a general $(2m-1) \times (2n-1)$ chessboard (where $m$ and $n$ are at least one greater than 1). We can consider induction on $m$. Let the maximum number of pieces that can be placed on the chessboard be $r_m$.
(1) When $m=1$, each cell... | 890 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,860 |
Example 4 A $9 \times 9$ chessboard is colored in black and white such that for each white square, the number of black squares adjacent to it is greater than the number of white squares adjacent to it, and for each black square, the number of white squares adjacent to it is greater than the number of black squares adja... | Analysis and Solution: To satisfy the coloring condition, each square can have at most one adjacent square of the same color, thus certain special cases cannot occur: (1) 3 squares in an L-shape are the same color. (2) 3 squares in a $1 \times 3$ rectangle are the same color.
If any two adjacent squares on the chessbo... | 3 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,861 |
Example 5 Let $n \geqslant 3$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}$ be any $n$ distinct real numbers with a positive sum. If a permutation $b_{1}, b_{2}, \cdots, b_{n}$ of these numbers satisfies: for any $k=1,2, \cdots, n$, $b_{1}+b_{2}+\cdots+b_{k}>0$, then this permutation is called good. Find the... | Analysis and Solution: Consider the worst-case scenario: for $k=1,2, \cdots, n, b_{1}+b_{2}+\cdots+b_{k}>0$ is difficult to satisfy. This only requires that the number of negative numbers in $a_{1}, a_{2}, \cdots, a_{n}$ is as many as possible. Let $a_{2}, \cdots, a_{n}$ all be negative, and $a_{1}=-a_{2}-\cdots-a_{n}+... | (n-1)! | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,862 |
Example 7 Let $m, n$ be positive integers, $m<2001, n<2002$. There are $2001 \times 2002$ different real numbers, which are filled into the squares of a $2001 \times 2002$ chessboard, so that each square contains exactly one number. If the number in a certain square is less than at least $m$ numbers in its column and a... | Analysis and Solution: Consider a special case: Fill the $2001 \times 2002$ chessboard with the numbers $1, 2, 3, \cdots, 2001 \times 2002$ in natural order (as shown in the table below), at this time the number of bad cells $S = (2001 - m)(2002 - n)$.
\begin{tabular}{|c|c|c|c|}
\hline 1 & 2 & $\cdots$ & 2002 \\
\hline... | (2001 - m)(2002 - n) | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,864 |
Example 8 A diagonal of a 2006-gon $P$ is called good if its endpoints divide the boundary of $P$ into two parts, each containing an odd number of sides of $P$. In particular, a side of $P$ is also considered good.
Suppose $P$ is triangulated by 2003 diagonals that do not intersect in the interior of $P$. Find the max... | Define an isosceles triangle with two good sides as a good triangle. First, consider special cases.
For a square, there is essentially only one type of division, and the number of good triangles is 2;
For a regular hexagon, there are essentially only 3 types of divisions, and the maximum number of good triangles is 3. ... | 1003 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,865 |
Example 9 Given a positive integer $a$, let $X=\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{n}\right\}$ be a set of positive integers, where $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant \cdots \leqslant a_{n}$. If for any integer $p(1 \leqslant p \leqslant a)$, there exists a subset $A$ of $X$ such that $S(A)=p$, where $... | To study special cases, for $a=1,2,3,4$, we conduct experiments and obtain the corresponding minimum values $n=1,2,2,3$, and from this, we discover that the set $X=\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{n}\right\}$, which makes $n$ reach its minimum, has the following property: for any $i=1,2, \cdots, n$, we have $a_{i... | r+1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,866 |
1 A certain exam consists of 5 multiple-choice questions, each with 4 different options, and each person selects exactly one option per question. Among 2000 answer sheets, there exists a number $n$ such that in any $n$ answer sheets, there are 4 sheets where any two of them have at most 3 answers in common. Find the mi... | 1. The minimum possible value of $n$ is 25. Denote the 4 possible answers to each question as $1, 2, 3, 4$, and each test paper's answers as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. For all 2000 answer sets $(g, h, i, j, k)$, group those with the last 4 components identical into one class, resulting in... | 25 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,867 |
2 Let $a_{1}, a_{2}, \cdots, a_{k}$ be a finite sequence of positive integers not exceeding $n$, where any term has two different adjacent terms, and there do not exist any four indices $p<q<r<s$, such that $a_{p}=$ $a_{1} \neq a_{q}=a_{s}$. Find the maximum value of $k$.
| 2. The sequence $n, n, n-1, n-1, \cdots, 2,2,1,1,2,2, \cdots, n-1, n-1, n, n$ meets the requirement, because if $a_{i}=a_{j}$, then $i$ and $j$ must be adjacent or symmetric about the middle two terms 1,1. Therefore, the maximum value of $k$ is no less than $4 n-2$. Furthermore, in any sequence that meets the condition... | 4n-2 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,868 |
Example 4 In a cyclic quadrilateral $ABCD$, the lengths of the four sides $AB, BC, CD, DA$ are all positive integers, $DA=2005, \angle ABC=\angle ADC=90^{\circ}$, and $\max \{AB, BC, CD\}<2005$. Find the maximum and minimum values of the perimeter of quadrilateral $ABCD$. (2005 China National Training Team Test Questio... | Let $A B=a, B C=b, C D=c$, then $a^{2}+b^{2}=A C^{2}=c^{2}+2005^{2}$, so $2005^{2}-a^{2}=b^{2}-c^{2}=(b+c)(b-c)$, where $a, b, c \in\{1,2, \cdots, 2004\}$.
Assume $a \geqslant b$, first fix $a$, let $a_{1}=2005-a$, then
$$(b+c)(b-c)=2005^{2}-a^{2}=a_{1}\left(4010-a_{1}\right)$$
From $a^{2}+b^{2}>2005^{2}$, we get $a>... | 4160 \text{ and } 7772 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,870 |
4 Place $r$ chess pieces on an $m \times n (m>1, n>1)$ chessboard, with at most one piece per square. If the $r$ pieces have the following property $p$: each row and each column contains at least one piece. However, removing any one of the pieces will cause them to no longer have the above property $p$. Find the maximu... | 4. $r(m, n)=m+n-2$. On the one hand, by construction, $r=m+n-2$ is possible (all cells except the first cell in the first row and the first column are placed with chess pieces). Below is the proof: if $r2$ and $n>2$, then since there are at least $m+n-1>m$ chess pieces on the board, there must be a row with at least tw... | r(m, n)=m+n-2 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,871 |
5ـ Let $F=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$ be a family of subsets of $X=\{1,2, \cdots, n\}$, and satisfy: (1) $\left|A_{i}\right|=3$, (2) $\left|A_{i} \cap A_{j}\right| \leqslant 1$. Denote the maximum value of $|F|$ by $f(n)$, prove that: $\frac{n^{2}-4 n}{6} \leqslant f(n) \leqslant \frac{n^{2}-n}{6}$. (6t... | 5. Since $\left|A_{j} \cap B_{j}\right| \leqslant 1$, let's calculate the number of binary subsets $S$. On one hand, $\left|A_{i}\right|=3$, so each $A_{i}$ contains $\mathrm{C}_{3}^{2}=3$ binary sets. Thus, the $k$ sets in $F$ can generate $3k$ binary sets. Since $\left|A_{i} \cap B_{j}\right| \leqslant 1$, these $3k$... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 737,872 |
6 In an $m \times n (m>1, n>1)$ chessboard $C$, each cell is filled with a number such that for any positive integers $p, q$ and any $p \times q$ rectangle, the sum of the numbers in the cells at the opposite corners is equal. If after filling in numbers in an appropriate $r$ cells, the numbers in the remaining cells a... | 6. After filling the first row and the first column, the number table is uniquely determined, at this point, the table has only been filled with $m+n-1$ numbers. That is, when $r=m+n-1$, there exists a corresponding filling method. Below, we prove that for all valid filling methods, $r \geqslant m+n-1$. We use proof by... | m+n-1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,873 |
Let $x, y, z$ be non-negative real numbers, $x+y+z=a(a \geqslant 1)$, find the maximum value of $F=2 x^{2}+y+3 z^{2$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
(Note: The note itself is not part of the translation, it's ... | 1. From $x+y+z=a$, we get $\frac{x}{a}+\frac{y}{a}+\frac{z}{a}=1$. Let $x=a u, y=a v, z=a w$, then $0 \leqslant u, v, w \leqslant 1, u+v+w=1, F=2 x^{2}+y+3 z^{2}=2 a^{2} u^{2}+a v+3 a^{2} w^{2}$. Since $a \geqslant 1$, we have $a^{2} \geqslant a \geqslant 1$, thus, $F=2 a^{2} u^{2}+a v+3 a^{2} w^{2} \leqslant 2 a^{2} u... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,874 |
3 Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and denote $H=\frac{x_{1}}{\left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{2}}+$ $\frac{x_{2}}{\left(1+x_{2}+x_{3}+\cdots+x_{n}\right)^{2}}+\cdots+\frac{x_{n}}{\left(1+x_{n}\right)^{2}}$ with the maximum value as $a_{n+1}$. When do $x_{1}$, $\dot{x}_{2}, \cdots... | 3. First prove the lemma: Let $g(x)=\frac{a}{x+b}+\frac{x}{(x+b)^{2}}$, where $a \geqslant 0, b \geqslant 1$. Using the discriminant method, it can be shown that when $x=\frac{b(1-a)}{1+a}$, the maximum value of $g(x)$ is $\frac{(1+a)^{2}}{4 b}$. Original problem solution: Fix $x_{2}, x_{3}, \cdots, x_{n}$, then $H$ is... | 1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,876 |
4. Let $n$ be a given integer $(n>1)$, and positive integers $a, b, c, d$ satisfy $\frac{b}{a}+\frac{d}{c}<1, b+d \leqslant n$. Find the maximum value of $\frac{b}{a}+\frac{d}{c}$. | 4. Let the maximum value of $\frac{b}{a}+\frac{d}{c}$ be $f(n)$, and assume $a \leqslant c$. If $a \geqslant n+1$, then $\frac{b}{a}+\frac{d}{c} \leqslant$ $\frac{b}{a}+\frac{d}{a}=\frac{b+d}{a} \leqslant \frac{n}{n+1}$. If $a \leqslant n$, then fix $a$, and let $x=a(n-a+1)+1$. If $c \leqslant x$, then by $\frac{b}{a}+... | 1-\frac{1}{a[a(n-a+1)+1]} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,877 |
Example 3 Let $P$ be a point inside (including the boundary) a regular tetrahedron $T$ with volume 1. Draw 4 planes through $P$ parallel to the 4 faces of $T$, dividing $T$ into 14 pieces. Let $f(P)$ be the sum of the volumes of those pieces which are neither tetrahedra nor parallelepipeds. Find the range of $f(P)$. (3... | Let the distances from point $P$ to the four faces of the regular tetrahedron $ABCD$ be $d_{1}, d_{2}, d_{3}, d_{4}$.
Let $x_{i}=\frac{d_{i}}{h}$, where $h$ is the height of the regular tetrahedron. Then $\sum_{i=1}^{4} x_{i}=1$.
Since the 14 pieces divided by $T$ include 4 tetrahedra with volumes $x_{i}^{3}$ each. Add... | 0 \leqslant f(P) \leqslant \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,878 |
Example 1 The sum of several positive integers is 1976, find the maximum value of their product. | First, consider the simple cases where the sum of several numbers is $4$, $5$, $6$, $7$, and $8$. The partitions that maximize the product are: $\square$
$$4=2+2,5=2+3,6=3+3,7=2+2+3,8=2+3+3.$$
From this, we conjecture: To maximize the product, the sum in the partition should only contain 2 and 3, and there should be a... | 3^{658} \times 2 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,879 |
Example 2 There are 1989 points in space, with no 3 points collinear, and they are divided into 30 groups with different numbers of points. From any 3 different groups, take one point from each, and form a triangle with these 3 points. Question: To maximize the total number of such triangles, how many points should eac... | Analysis Intuition tells us that when the number of points in each group is equal, the total number of triangles is maximized. However, upon careful reading of the problem, we find that the requirement is for the number of points in each group to be distinct, leading us to think that the number of points in each group ... | 51, 52, \cdots, 56, 58, 59, \cdots, 81 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,880 |
Example 3 Suppose point $P$ starts from lattice point $A(1,1)$ and moves to point $B(m, n)\left(m, n \in \mathbf{N}^{*}\right)$ along the grid with the shortest path, i.e., each time it moves to another lattice point, the x-coordinate or y-coordinate increases by 1. Find the maximum value of the sum $S$ of the products... | Let the points passed by $P$ be $P_{1}=A(1,1), P_{2}, \cdots, P_{m+n-1}=B(m, n)$, and the coordinates of $P_{i}$ be $\left(x_{i}, y_{i}\right)$, then $S=\sum_{i=1}^{m+n-1} x_{i} y_{i}$. To maximize $S$, intuitively, $x_{i}$ and $y_{i}$ should be as close as possible. However, it is not required that for any $i$, $x_{i}... | \frac{1}{6} n\left(3 m^{2}+n^{2}+3 m-1\right) \text{ or } \frac{1}{6} m\left(3 n^{2}+m^{2}+3 n-1\right) | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,881 |
Example 4 The IMO space station consists of 99 space stations, any two of which are connected by a tubular passage. It is stipulated that 99 of these passages are bidirectional main roads, while the rest are strictly one-way. If any 4 space stations can reach any other station among them through the passages, the set o... | Analysis and Solution: Directly finding the number of "four-way groups" is relatively difficult due to the stringent conditions. However, the conditions for a non-interconnected four-station group are relatively easier to satisfy. For example, imagine a space station with "no loops." This leads us to consider all one-w... | 2052072 | Combinatorics | proof | Yes | Yes | inequalities | false | 737,882 |
Example 5 Express 2006 as the sum of 5 positive integers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, and let $S=$ $\sum_{1 \leqslant i<j \leq 5} x_{i} x_{j}$. Questions:
(1) For what values of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, does $S$ reach its maximum value;
(2) Furthermore, if for any $1 \leqslant i, j \leqslant 5$, $\l... | (1) First, the set of such $S$ values is bounded, so there must exist a maximum and a minimum value.
If $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=2006$, and $S=\sum_{1 \leq i < j \leq 5} x_{i} x_{j}$ is maximized.
Rewrite $S$ as
$$S=\sum_{1 \leq i < j \leq 5} x_{i} x_{j}.$$
If there exist $x_{i}, x_{j}$ such that $|x_{i}-x_{j}| ... | 402, 402, 402, 400, 400 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,883 |
I divide 1989 into the sum of 10 positive integers, to maximize their product. | 1. When 1989 is decomposed into: $199+199+\cdots+199+198$, the corresponding product is $199^{9} \times$ 198. Below is the proof that it is the largest. Since the number of decompositions is finite, the maximum value must exist, so we can assume the decomposition $\left(x_{1}, x_{2}, \cdots, x_{10}\right)$, where $x_{1... | 199^{9} \times 198 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,884 |
2 In a non-decreasing sequence of positive integers $a_{1}, a_{2}, \cdots, a_{m}, \cdots$, for any positive integer $m$, define $b_{m}=$ $\min \left\{n \mid a_{n} \geqslant m\right\}$. It is known that $a_{19}=85$. Find the maximum value of $S=a_{1}+a_{2}+\cdots+a_{19}+b_{1}+b_{2}+\cdots+$ $b_{85}$. (1985 USA Mathemati... | 2. First, the largest number must exist. From the condition, we have $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{19}=85$. We conjecture that the extremum point is when each $a_{i}$ is as large as possible and all $a_{i}$ are equal, and all $b_{j}$ are equal. In fact, if there is $a_{i}a_{i}$, so $a_{i+1} \geqs... | 1700 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,885 |
4 Given real numbers $P_{1} \leqslant P_{2} \leqslant P_{3} \leqslant \cdots \leqslant P_{n}$, find real numbers $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, such that $d=\left(P_{1}-x_{1}\right)^{2}+\left(P_{2}-x_{2}\right)^{2}+\cdots+\left(P_{n}-x_{n}\right)^{2}$ is minimized. | 4. Intuitively, we conjecture that the point of extremum $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is uniform, i.e., $x_{1}=x_{2}=\cdots=x_{n}=x$ (to be determined). At this point, $d=\sum_{i=1}^{n}\left(P_{i}-x_{i}\right)^{2}=\sum_{i=1}^{n}\left(P_{i}-x\right)^{2}=n x^{2}-2 \sum_{i=1}^{n}\left(P_{i}\right) x + \sum_{... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,887 |
5】 Given a set of points on a plane $P=\left\{p_{1}, p_{2}, \cdots, p_{194}\right\},$ where no three points in $P$ are collinear. Divide the points in $P$ into 83 groups, with each group containing at least 3 points. Connect every pair of points within the same group, but do not connect points from different groups, to... | 5. (1) Since the number of grouping methods is finite, there must exist a grouping method that results in the minimum number of triangles. Note that $1994=83 \times 24+2=81 \times 24+2 \times 25$, thus, we can divide 1994 points into 83 groups, where 81 groups each have 24 points, and 2 groups each have 25 points. We p... | 168544 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,888 |
Example 4 Let $a_{1}, a_{2}, \cdots, a_{6}; b_{1}, b_{2}, \cdots, b_{6}$ and $c_{1}, c_{2}, \cdots, c_{6}$ all be permutations of $1,2, \cdots, 6$, find the minimum value of $\sum_{i=1}^{6} a_{i} b_{i} c_{i}$. | Let $S=\sum_{i=1}^{6} a_{i} b_{i} c_{i}$, by the AM-GM inequality we have
$$S \geqslant 6 \sqrt[6]{\prod_{i=1}^{6} a_{i} b_{i} c_{i}}=6 \sqrt[6]{(6!)^{3}}=6 \sqrt{6!}=72 \sqrt{5}>160$$
Next, we prove $S>161$.
Since the geometric mean of $a_{1} b_{1} c_{1}, a_{2} b_{2} c_{2}, \cdots, a_{6} b_{6} c_{6}$ is $12 \sqrt{5}$... | 162 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,889 |
Example 1 Let $0<p \leqslant a, b, c, d, e \leqslant q$, find the maximum value of $F=(a+b+c+d+$
e) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right)$. | Since $F$ is continuous on a closed domain, it must have a maximum value. Fix $a, b, c, d$, and let $a + b + c + d = A$, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = B$, then $A, B$ are constants, and $F = (A + e)(B + \frac{1}{e}) = 1 + AB + eB + \frac{A}{e}$.
Consider $f(e) = eB + \frac{A}{e}$. It is easy... | 25 + 6\left(\sqrt{\frac{p}{q}} - \sqrt{\frac{q}{p}}\right)^2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,891 |
Example 2 Let $0 \leqslant x_{i} \leqslant 1(1 \leqslant i \leqslant n)$, find the maximum value of $F=\sum_{1 \leqslant i<j \leq n}\left|x_{i}-x_{j}\right|$. | Since $F$ is continuous on a closed domain, there must exist a maximum value. Fixing $x_{2}, x_{3}, \cdots, x_{n}$, then $F\left(x_{1}\right)$ is a function of $x_{1}$:
$$F\left(x_{1}\right)=\left|x_{1}-x_{2}\right|+\left|x_{1}-x_{3}\right|+\cdots+\left|x_{1}-x_{n}\right|+\sum_{2 \leqslant i<j \leqslant n}\left|x_{i}-x... | \left[\frac{n^{2}}{4}\right] | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,892 |
Example 3 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy the following two conditions:
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.
Try to find the maximum value of $x_{1}^{12}+x_{2}^{12}+\cdots+x_{197}^{12}$, and explain... | Since $x_{1}+x_{2}+\cdots+x_{1997}$ is a continuous function on a closed domain, $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$ must have a maximum value. Fixing $x_{3}, x_{4}, \cdots, x_{1997}$, then $x_{1}+x_{2}=c$ (a constant), let $x_{1}=x$, then $x_{2}=c-x$, thus
$$x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}=x^{12}+(... | 189548 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,893 |
Example 4 Let $x_{1}, x_{2}, \cdots, x_{n}$ take values in some interval of length 1, denote $x=\frac{1}{n} \sum_{j=1}^{n} x_{j}$, $y=\frac{1}{n} \sum_{j=1}^{n} x_{j}^{2}$, find the maximum value of $f=y-x^{2}$. (High School Mathematics League Simulation) | Let $x_{1}, x_{2}, \cdots, x_{n} \in[a, a+1](a \in \mathbf{R})$, when $n=1$, $f=0$, at this time $f_{\max }=0$.
When $n>1$, if we fix $x_{2}, x_{3}, \cdots, x_{n}$, then
$$\begin{aligned}
f & =y-x^{2}=\frac{1}{n} \sum_{j=1}^{n} x_{j}^{2}-\left(\frac{1}{n} \sum_{j=1}^{n} x_{j}\right)^{2} \\
& =\frac{n-1}{n^{2}} x_{1}^{... | f_{\max }=\left\{\begin{array}{cl}
\frac{n^{2}-1}{4 n^{2}} & \text { ( } n \text { is odd }) ; \\
\frac{1}{4} & \text { ( } n \text { is even }) .
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,894 |
Given $0<p \leqslant a_{1}, a_{2}, \cdots, a_{n} \leqslant q$, find the maximum value of $F=\left(a_{1}+a_{2}+\cdots+\right.$ $\left.a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)$. | 1. The maximum value of $F$ is: $n^{2}+\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2}$. Since $F$ is continuous on a closed domain, it must have a maximum value. Fix $a_{1}, a_{2}, \cdots, a_{n-1}$, and let $a_{1}+a_{2}+\cdots+a_{n-1}=A$, $\frac{1}{a_{1}}+\frac{... | n^{2}+\left[\frac{n}{2}\right]\left[\frac{n+1}{2}\right]\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,895 |
2. Let $x_{i} \in \mathbf{R},\left|x_{i}\right| \leqslant 1(1 \leqslant i \leqslant n)$, find the minimum value of $F=\sum_{1 \leq i<j \leqslant n} x_{i} x_{j}$. | 2. Since $F$ is continuous on a closed domain, there must exist a minimum value. Fixing $x_{2}, x_{3}, \cdots, x_{n}$, then $F\left(x_{1}\right)$ is a linear function of $x_{1}$. Given $-1 \leqslant x_{1} \leqslant 1$, when $F\left(x_{1}\right)$ takes its minimum value, it must be that $x_{1} \in\{-1,1\}$. By symmetry,... | -\left[\frac{n}{2}\right] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,896 |
3 Given a natural number $n>2, \lambda$ is a given constant. Determine the maximum and minimum values of the function: $F=x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{n}^{2}+\lambda x_{1} x_{2} \cdots x_{n}$, where $x_{1}, x_{2}, \cdots, x_{n}$ are non-negative real numbers, and $x_{1}+x_{2}+\cdots+x_{n}=1$. (1994 China Mathematic... | 3. Since $F$ is continuous on the closed domain $0 \leqslant x_{i} \leqslant 1$, $F$ must have a maximum and a minimum value. Suppose $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is a point where $F$ attains its extremum. We will prove that for any $i \neq j, x_{i} x_{j}=0$, or $x_{i}=x_{j}$. In fact, by symmetry, we onl... | F_{\min }=\min \left\{\frac{1}{n-1}, \frac{\lambda+n^{n-1}}{n^{n}}\right\}, F_{\max }=\max \left\{1, \frac{\lambda+n^{n-1}}{n^{n}}\right\} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,897 |
Example 1 Given $2 n$ real numbers: $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots \leqslant b_{n}$. Let
$$F=a_{1} b_{i_{1}}+a_{2} b_{i_{2}}+\cdots+a_{n} b_{i_{n}}$$
where $b_{i_{1}}, b_{i_{2}}, \cdots, b_{i_{n}}$ is a permutation of $b_{1}, b_{2}, \cdots, b_{n}$. Find ... | If the expression of $F$ does not contain the term $a_{1} b_{1}$, then consider the two terms containing $a_{1}$ and $b_{1}$ respectively: $a_{1} b_{i_{1}}$ and $a_{j} b_{1}$. Adjust them to $a_{1} b_{1}$ and $a_{j} b_{i_{1}}$, respectively. Then,
$a_{1} b_{1} + a_{j} b_{i_{1}} - (a_{1} b_{i_{1}} + a_{j} b_{1}) = (a_{1... | a_{1} b_{n} + a_{2} b_{n-1} + \cdots + a_{n} b_{1} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,898 |
Example 2 Let $A, B, C$ be the three interior angles of a triangle, find the maximum value of $\sin A+\sin B+\sin C$.
untranslated text remains the same as requested. However, if you need the entire text to be translated, please let me know! | Assuming $A \leqslant B \leqslant C$, then $A \leqslant 60^{\circ} \leqslant C$, so by the lemma, we have
$$\sin A+\sin B+\sin C \leqslant \sin 60^{\circ}+\sin \left(A+C-60^{\circ}\right)+\sin B$$
Assuming again that $A+C-60^{\circ} \leqslant B$, then because $\left(A+C-60^{\circ}\right)+B=A+B+C-$ $60^{\circ}=120^{\ci... | \frac{3 \sqrt{3}}{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 737,899 |
Example 5 Given an integer $n \geqslant 3$, real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\min _{1 \leqslant i<j \leqslant n}\left|a_{i}-a_{j}\right|=1$, find the minimum value of $\sum_{k=1}^{n}\left|a_{k}\right|^{3}$. (2009 China Mathematical Olympiad Problem) | Let's assume $a_{1}<a_{2}<\cdots<a_{n}$, then for $1 \leqslant k \leqslant n$, we have
$$\left|a_{k}\right|+\left|a_{n-k+1}\right| \geqslant\left|a_{n-k+1}-a_{k}\right| \geqslant|n+1-2 k|$$
Therefore, $\sum_{k=1}^{n}\left|a_{k}\right|^{3}=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\left|a_{n+1-k}\right|^{3... | \frac{1}{32}\left(n^{2}-1\right)^{2} \text{ (for odd } n\text{)}, \frac{1}{32} n^{2}\left(n^{2}-2\right) \text{ (for even } n\text{)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,900 |
Example 3 Let $x_{i} \geqslant 0(1 \leqslant i \leqslant n), \sum_{i=1}^{n} x_{i}=1, n \geqslant 2$. Find the maximum value of $F=\sum_{1 \leq i<j \leqslant n} x_{i} x_{j}\left(x_{i}+\right.$ $x_{j}$ ). (32nd IMO Shortlist Problem) | Analysis and Solution: Since $F$ is continuous on a closed region, the maximum value of $F$ must exist, which suggests using local adjustments. Without loss of generality, let's try it. If we fix $x_{2}, x_{3}, \cdots, x_{n}$, then $x_{1}$ is also fixed, so we can only fix $n-2$ numbers. Let's fix $x_{3}, x_{4}, \cdots... | \frac{1}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,901 |
Example 4 Let $f(x)=a x^{2}+b x+c$ have all coefficients positive, and $a+b+c=1$. For all positive arrays $x_{1}, x_{2}, \cdots, x_{n}$ satisfying: $x_{1} x_{2} \cdots x_{n}=1$, find the minimum value of $f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right)$. (All-Russian Mathematical Olympiad problem) | Solve $f(1)=a+b+c=1$. If $x_{1}=x_{2}=\cdots=x_{n}=1$, then $f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right)=1$.
If $x_{1}, x_{2}, \cdots, x_{n}$ are not all 1, then from $x_{1} x_{2} \cdots x_{n}=1$ we know that there must be one less than 1 and one greater than 1. Without loss of generality, let $... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,902 |
Example 5 For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $S=\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$. (40th IMO Chinese National Team Selection Exam) | Let $x_{1}, x_{2}, \cdots, x_{n}$ be the number of non-zero elements as $k$, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{k}>0, x_{k+1}=x_{k+2}=\cdots=x_{n}=0$. If $k \geqslant 3$, then let $x_{i}^{\prime}=x_{i}(i=1,2, \cdots, k-2), x_{k-1}^{\prime}=x_{k-1}+x_{k}, x_{k}^{\prim... | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,903 |
Example 6 Let $x, y, z$ be non-negative real numbers, satisfying: $x+y+z=1$, find the minimum value of $Q=\sqrt{2-x}+$ $\sqrt{2-y}+\sqrt{2-z}$. | By symmetry, without loss of generality, let $x \leqslant y \leqslant z$, and set $x^{\prime}=0, y^{\prime}=y, z^{\prime}=z+x-x^{\prime}$, then $x^{\prime} \geqslant 0, y^{\prime} \geqslant 0, z^{\prime} \geqslant 0$, and $z^{\prime}-z=x-x^{\prime}, y^{\prime}+z^{\prime}=x+y+z-x^{\prime}=$ $x+y+z=1$, thus
$$\begin{alig... | 1+2\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,904 |
1 Let $x_{i} \geqslant 0(1 \leqslant i \leqslant n), \sum_{i=1}^{n} x_{i} \leqslant \frac{1}{2}, n \geqslant 2$. Find the minimum value of $F=\left(1-x_{1}\right)(1-$ $\left.x_{2}\right) \cdots\left(1-x_{n}\right)$. | 1. When $n=2$, $x_{1}+x_{2} \leqslant \frac{1}{2},\left(1-x_{1}\right)\left(1-x_{2}\right)=1+x_{1} x_{2}-\left(x_{1}+x_{2}\right) \geqslant 1-x_{1}-x_{2} \geqslant \frac{1}{2}$. The equality holds when $x_{1}+x_{2}=\frac{1}{2}, x_{1} x_{2}=0$, i.e., $x_{1}=\frac{1}{2}, x_{2}=0$. When $n=3$, $x_{1}+x_{2}+x_{3} \leqslant... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,905 |
2 Let $x_{i} \geqslant 0(1 \leqslant i \leqslant n, n \geqslant 4), \sum_{i=1}^{n} x_{i}=1$, find the maximum value of $F=\sum_{i=1}^{n} x_{i} x_{i+1}$. | 2. By the method similar to the above, the maximum value of $F$ can be obtained as $F\left(\frac{1}{2}, \frac{1}{2}, 0,0, \cdots, 0\right)=\frac{1}{4}$. | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,906 |
3. Let $x_{i} \geqslant 0(1 \leqslant i \leqslant n), \sum_{i=1}^{n} x_{i}=\pi, n \geqslant 2$. Find the maximum value of $F=\sum_{i=1}^{n} \sin ^{2} x_{i}$. | 3. When $n=2$, $x_{1}+x_{2}=\pi, F=\sin ^{2} x_{1}+\sin ^{2} x_{2}=2 \sin ^{2} x_{1} \leqslant 2$. The equality holds when $x_{1}=x_{2}=\frac{\pi}{2}$. When $n \geqslant 3$, let $x_{3}, x_{4}, \cdots, x_{n}$ be constants, then $x_{1}+x_{2}$ is also a constant. Consider $A=\sin ^{2} x_{1}+\sin ^{2} x_{2}, 2-2 A=1-2 \sin... | \frac{9}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,907 |
4. Let $0<a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}<\pi, a_{1}+a_{2}+\cdots+a_{n}=A$, find the maximum value of $\sin a_{1}+$ $\sin a_{2}+\cdots+\sin a_{n}$. | 4. By the method similar to the above, the maximum value of $\sin a_{1}+\sin a_{2}+\cdots+\sin a_{n}$ can be found to be $n \sin \frac{A}{n}$. | n \sin \frac{A}{n} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,908 |
5 Let $x_{1}, x_{2}, x_{3}, x_{4}$ all be positive real numbers, and $x_{1}+x_{2}+x_{3}+x_{4}=\pi$. Find the minimum value of
$A=\left(2 \sin ^{2} x_{1}+\frac{1}{\sin ^{2} x_{1}}\right)\left(2 \sin ^{2} x_{2}+\frac{1}{\sin ^{2} x_{2}}\right)\left(2 \sin ^{2} x_{3}+\frac{1}{\sin ^{2} x_{3}}\right)\left(2 \sin ^{2} x_{4... | 5. First note, if $x_{1}+x_{2}=a$ (a constant), then from $2 \sin x_{1} \sin x_{2}=\cos \left(x_{1}-x_{2}\right)-\cos a$, and $\left|x_{1}-x_{2}\right|\frac{\pi}{4}>x_{2}$, then fixing $x_{3}, x_{4}$, perform a smoothing transformation on $x_{1}, x_{2}$: that is, let $x_{1}^{\prime}=\frac{\pi}{4}$, $x_{2}^{\prime}=x_{1... | 81 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,909 |
Example 1 Let $M=\{1,2, \cdots, 2005\}, A$ be a subset of $M$. If for any $a_{i}, a_{j} \in A$, $a_{i} \neq a_{j}$, an isosceles triangle can be uniquely determined with $a_{i}$ and $a_{j}$ as side lengths, find the maximum value of $|A|$. | When $a<b$, $a, b, b$ must form an isosceles triangle. Therefore, two numbers $a, b$ uniquely determine an isosceles triangle with $a, b$ as its two sides, which is equivalent to $a, a, b$ not forming an isosceles triangle, i.e., $2a \leqslant b$.
Let $A=\left\{a_{1}<a_{2}<\cdots<a_{n}\right\}$ be a subset of $M$ that... | 11 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,910 |
For a positive integer $M$, if there exist integers $a, b, c, d$ such that $M \leqslant a < b \leqslant c < d \leqslant M+49, ad = bc$, then $M$ is called a good number; otherwise, $M$ is called a bad number. Find the largest good number and the smallest bad number. (2006 National Team Selection Examination Question) | The largest good number is 576, and the smallest bad number is 443.
$M$ is a good number if and only if there exist positive integers $u, v$ such that $uv \geqslant M, (u+1) \cdot (v+1) \leqslant M+49(*)$.
Sufficiency: Suppose $u, v$ exist, without loss of generality, assume $u \leqslant v$, then $M \leqslant uv < u(v... | 576, 443 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,911 |
Example 2 Let $A$ be a subset of the set of positive integers $\mathbf{N}^{*}$. For any $x, y \in A, x \neq y$, we have $\mid x-1$ $y \left\lvert\, \geqslant \frac{x y}{25}\right.$. Find the maximum value of $|A|$. (26th IMO Shortlist) | Let's assume $A=\left\{a_{1}\frac{n-1}{25}\right\}$. Therefore, $n-1>\frac{n-2}{25}$, so $2 \leqslant a_{2}<\frac{25}{n-2}, n<\frac{29}{2}, n \leqslant 14$. After testing, further refinement of the estimate is still needed.
Similarly, we have $3 \leqslant a_{3}<\frac{25}{n-3}, n \leqslant 11 ; 4 \leqslant a_{4}<\frac{... | 9 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,912 |
Example 3 Let $X=\{1,2, \cdots, n\}, A_{i}=\left\{a_{i}, b_{i}, c_{i}\right\}\left(a_{i}<b_{i}<c_{i}, i=1\right.$, $2, \cdots, m)$ be 3-element subsets of $X$, such that for any $A_{i}, A_{j}(1 \leqslant i<j \leqslant m), a_{i}=a_{j}, b_{i}=$ $b_{j}, c_{i}=c_{j}$ holds for at most one. Find the maximum value of $m$. | For $2 \leqslant k \leqslant n-1$, consider the 3-element subset $\left\{a_{i}, k, c_{i}\right\}$ with $k$ as the middle element, where $a_{i}<k<c_{i}$. Let the number of such sets be $f(k)$. Since $a_{i}$ can take values from $1,2, \cdots, k-1$, we have $f(k) \leqslant k-1$. Also, $c_{i}$ can take values from $n, n-1,... | m_{\max }=\left\{\begin{array}{ll}
\frac{n(n-2)}{4} & (n \text { is even }), \\
\left(\frac{n-1}{2}\right)^{2} & (n \text { is odd }) .
\end{array}\right.} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,913 |
Example 4 Let $a_{1}<a_{2}<\cdots<a_{n}=100$, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive integers. If for any $i \geqslant 2$, there exist $1 \leqslant p \leqslant q \leqslant r \leqslant i-1$, such that $a_{i}=a_{p}+a_{q}+a_{r}$, find the maximum and minimum values of $n$. (Original problem) $\quad$ | Solution (1) Clearly, $n \neq 1,2$, so $n \geqslant 3$.
Also, when $n=3$, take $a_{1}=20, a_{2}=60, a_{3}=100$, then $a_{2}=a_{1}+a_{1}+a_{1}$, $a_{3}=a_{1}+a_{1}+a_{2}$, so $n=3$ meets the condition, hence the minimum value of $n$ is 3.
(2) If $a_{1} \equiv 1(\bmod 2)$, then $a_{2}=3 a_{1} \equiv 3 \equiv 1(\bmod 2)$.... | 3, 13 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,914 |
I Let $X$ be a subset of $\mathbf{N}^{*}$, the smallest element of $X$ is 1, and the largest element is 100. For any number in $X$ that is greater than 1, it can be expressed as the sum of two numbers (which can be the same) in $X$. Find the minimum value of $|X|$. | 1. $|X|_{\min }=9$. Let $X=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, and $1=a_{1}<a_{2}<\cdots<a_{n}=100$. For any $a_{k} (2 \leq k \leq n-1)$, there exist $a_{i}, a_{j}$ such that $a_{k}=a_{i}+a_{j}$, where it must be that $a_{i}<a_{k}, a_{j}<a_{k}$, i.e., $a_{i} \leqslant a_{k-1}, a_{j} \leqslant a_{k-1}$. Therefo... | 9 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,915 |
3 Suppose 2005 line segments are connected end-to-end, forming a closed polyline, and no two segments of the polyline lie on the same straight line. Then, what is the maximum number of intersection points where the polyline intersects itself?
| 3. Consider a general problem: Let $f(n)$ be the maximum number of intersection points of a broken line with $n$ segments (where $n$ is odd). Starting from a fixed point and moving along the broken line, the segments encountered in sequence are called the 1st, 2nd, ..., $n$-th segments. When drawing the first two segme... | 2007005 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,917 |
Example 2 Let $A$ be a subset of the set of positive integers $\mathbf{N}^{*}$. For any $x, y \in A, x \neq y$, we have $\mid x- y \left\lvert\, \geqslant \frac{x y}{25}\right.$. Find the maximum value of $|A|$. (26th IMO Shortlist) | $$\begin{array}{l}
\text { Let } X_{1}=\{1\}, X_{2}=\{2\}, X_{3}=\{3\}, X_{4}=\{4\}, X_{5}=\{5,6\}, \\
X_{6}=\{7,8,9\}, X_{7}=\{10,11, \cdots, 16\}, X_{8}=\{17,18, \cdots, 53\}, X_{9}=\{54, \\
55, \cdots\}=\mathbf{N}^{*} \backslash\{1,2, \cdots, 53\} .
\end{array}$$
For $X_{9}$, when $x, y \in X_{9}$, $x>25$, so $y-x<... | 9 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,918 |
Example 3 Let $A \subseteq\{0,1,2, \cdots, 29\}$, satisfying: for any integer $k$ and any numbers $a$, $b$ in $A$ (where $a$, $b$ can be the same), $a+b+30k$ is not the product of two consecutive integers. Try to find all $A$ with the maximum number of elements. (2003 China National Training Team Selection Exam Problem... | Solve for $A=\{3 t+2 \mid 0 \leqslant t \leqslant 9, t \in \mathbf{Z}\}$.
Let $A$ satisfy the conditions in the problem and $|A|$ be maximized. Since for two consecutive integers $a, a+1$, we have $a(a+1) \equiv 0,2,6,12,20,26(\bmod 30)$. Therefore, for any $a \in A$, taking $b=a, k=0$, we know that $2 a \neq 0,2,6,12,... | A=\{2,5,8,11,14,17,20,23,26,29\} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,919 |
1 Let $a$, $b$, $c$, $a+b-c$, $b+c-a$, $c+a-b$, $a+b+c$ be 7 distinct prime numbers, and the sum of two of $a$, $b$, $c$ is 800. Let $d$ be the difference between the largest and smallest of these 7 prime numbers. Find the maximum possible value of $d$. (2001 China Mathematical Olympiad Problem) | 1. Let's assume $a0$, so $c<a+$ $b<a+c<b+c$, but one of $a+b$, $a+c$, $b+c$ is 800, so $c<800$. Also, $799=17 \times 47$, $798$ are not prime numbers, so $c \leqslant 797, d=2c \leqslant 1594$. Let $c=797$, $a+b=800$, noting that $b<c=797$, the smallest prime solution for $(a, b)=(13,787)$ (since $795,793=13 \times 61,... | 1594 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,920 |
Example 4 Let $A$ be a subset of $X=\{1,2,3, \cdots, 1989\}$, such that for any $x, y \in A$, $|x-y| \neq 4$ and 7. Find the maximum value of $|A|$. (7th United States of America Mathematical Olympiad Problem) | Let $A$ be a subset that meets the requirements. For $P=\{1,2, \cdots, 11\}$, we prove that $|A \cap P| \leqslant 5$.
Actually, divide $P$ into 6 subsets: $\{1,5\}, \{2,9\}, \{3,7\}, \{4,8\}, \{6,10\}, \{11\}$. For each of these subsets, $A$ can contain at most one element, so $|A \cap P| \leqslant 6$. If $|A \cap P|=... | 905 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,921 |
Example 5 Let $p$ be a given positive integer, $A$ is a subset of $X=\left\{1,2,3,4, \cdots, 2^{p}\right\}$, and has the property: for any $x \in A$, $2 x \notin A$. Find the maximum value of $|A|$. (1991 French Mathematical Olympiad) | Analysis and Solution: Divide $X$ into blocks and use induction on $p$.
When $p=1$, $X=\{1,2\}$, take $A=\{1\}$, then $f(1)=1$;
When $p=2$, $X=\{1,2,3,4\}$, divide $X$ into 3 subsets: $\{1,2\}$, $\{3\}$, $\{4\}$. Then $A$ can contain at most one number from each subset, so $|A| \leqslant 3$. Take $A=\{1,3,4\}$, then ... | \frac{2^{p+1}+(-1)^{p}}{3} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,922 |
Example 6 Let $X=\{1,2, \cdots, 2001\}$, find the smallest positive integer $m$, such that for any $m$-element subset $W$ of $X$, there exist $u, v \in W$ ($u$ and $v$ can be the same), such that $u+v$ is a power of 2. (2001 China Mathematical Olympiad Problem) | Analysis and Solution: For the convenience of describing the problem, if $u+v$ is a power of 2, then $u, v$ is called a pair. We consider the problem from the opposite side. If the subset $W$ of $X$ does not contain any pairs, then what is the maximum number of elements in $W$? Clearly, if we can divide $X$ into severa... | 999 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,923 |
Example 7 Let $n$ be a fixed positive even number, and consider an $n \times n$ square chessboard. If two squares share at least one common edge, they are called adjacent.
Now, mark $N$ squares on the chessboard so that every square on the chessboard (marked and unmarked) is adjacent to at least one marked square.
De... | Solve: Color an $n \times n$ chessboard as follows: If $n$ is not divisible by 4, color it as shown in Figure 7-1; otherwise, color it as shown in Figure 7-2. Consider all the black squares. If $n=4k$, after coloring according to Figure 7-2, there are $4 \times 3 + 4 \times 7 + \cdots + 4 \times (4k-1) = 2k(4k+2)$ blac... | \frac{1}{4} n(n+2) | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,924 |
Example 8 On the $x O y$ plane, there are 2002 points forming a point set $S$. It is known that the line connecting any two points in $S$ is not parallel to the coordinate axes. For any two points $P$ and $Q$ in $S$, consider the rectangle $M_{P Q}$ with $P Q$ as its diagonal, and its sides parallel to the coordinate a... | Solve $N_{\max }=400$. First, prove that there must be $P, Q$ such that $W_{P Q} \geqslant 400$. In fact, let $A$ be the point in $S$ with the largest y-coordinate, $B$ be the point in $S$ with the smallest y-coordinate, $C$ be the point in $S$ with the largest x-coordinate, and $D$ be the point in $S$ with the smalles... | 400 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,925 |
Example 9 In a $7 \times 8$ grid chessboard, each cell contains a chess piece. If two chess pieces are in cells that share a vertex, then these two pieces are said to be connected. Now, $r$ pieces are removed from these pieces, so that among the remaining pieces, there are no 5 pieces in a straight line (horizontal, ve... | We call the squares removed from the chessboard "holes," and let the chessboard have $k$ holes in total.
As shown in Figure 7-6, the chessboard is divided into 9 regions $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, and $O$ by 4 straight lines.
By the conditions, $A \cup G$ has at least 2 holes, $E \cup O$ has at least 3 ho... | 11 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,926 |
Example 10 Let the set $S=\{1,2, \cdots, 50\}, X$ be any subset of $S$, $|X|=n$. Find the smallest positive integer $n$, such that the set $X$ must contain three numbers that are the lengths of the three sides of a right-angled triangle. | Let the three sides of a right-angled triangle be $x, y, z$, with $x^{2}+y^{2}=z^{2}$. The positive integer solutions can be expressed as:
$$x=k\left(a^{2}-b^{2}\right), y=2 k a b, z=k\left(a^{2}+b^{2}\right)$$
where $k, a, b \in \mathbf{N}^{*}$ and $(a, b)=1, a>b$.
First, one of $x, y, z$ must be a multiple of 5. Oth... | 42 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,927 |
1 Let $X=\{1,2,3, \cdots, 1993\}, A$ be a subset of $X$, and satisfy: (1) For any two numbers $x \neq y$ in $A$, 93 does not divide $x \pm y$. (2) $S(A)=1993$. Find the maximum value of $|A|$. | 1. Divide $X$ into 47 subsets: $A_{i}=\{x \mid x \equiv i$, or $x \equiv 93-i(\bmod 93)\}$, $i=0,1,2, \cdots, 46$. For any $x, y \in A_{i}$, we have $x-y \equiv 0(\bmod 93)$, or $x+y \equiv$ $0(\bmod 93)$, so $A$ can contain at most one number from $A_{i}$. Therefore, $|A| \leqslant 47$. Let $A=\{10$, $11, \cdots, 46\}... | 47 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,928 |
2 Let $X=\{1,2,3, \cdots, 10\}, A$ be a subset of $X$, and for any $x<y<z, x, y, z \in A$, there exists a triangle with side lengths $x, y, z$. Find the maximum value of $|A|$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 2. First, let $A_{1}=\{1,2,3,5,8\}, A_{2}=\{4,6,10\}, A_{3}=\{7,9\}$, then any 3 numbers in $A_{i}$ do not form a triangle, so $A$ can contain at most 2 numbers from $A_{i}$, thus $|A| \leqslant 2 \times 3=6$. Next, let $A=\{5,6,7,8,9,10\}$, then $A$ meets the requirement. Therefore, the maximum value of $|A|$ is 6. | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,929 |
3 Let $X=\{1,2,3, \cdots, 20\}, A$ be a subset of $X$, and for any $x<y<z, x, y, z \in A$, there exists a triangle with side lengths $x, y, z$. Find the maximum value of $|A|$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. First, let $A_{1}=\{1,2,3,5,8,13\}, A_{2}=\{4,6,10,16\}, A_{3}=\{7$, $12,19\}, A_{4}=\{9,11,20\}, A_{5}=\{14,15\}, A_{6}=\{17,18\}$, then any 3 numbers in $A_{i}$ do not form a triangle, so $A$ can contain at most 2 numbers from $A_{i}$, thus $|A| \leqslant 2 \times$ $6=12$. But if $|A|=12$, then $14,15,17,18 \in A$... | 11 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,930 |
2. Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ satisfy the condition $\sum_{i=1}^{2 n-1}\left(a_{i+1}-a_{i}\right)^{2}=1$, find the maximum value of $\left(a_{n+1}+\right.$ $\left.a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$. (2003 Western Mathematical Olympiad) | 2. When $n=1$, $\left(a_{2}-a_{1}\right)^{2}=1$, so $a_{2}-a_{1}= \pm 1$. It is easy to see that the maximum value sought is 1. When $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i}, i=1,2, \cdots, 2 n-1$, then $\sum_{i=2}^{2 n} x_{i}^{2}=1$, and $a_{k}=x_{1}+x_{2}+\cdots+x_{k}, k=1,2, \cdots, 2 n$. Therefore, ... | \sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,931 |
4 Let $X=\{00,01, \cdots, 98,99\}$ be the set of 100 two-digit numbers, and $A$ be a subset of $X$ such that: in any infinite sequence of digits from 0 to 9, there are two adjacent digits that form a two-digit number in $A$. Find the minimum value of $|A|$. (52nd Moscow Mathematical Olympiad) | 4. Let $A_{ij}=\{\overline{ij}, \overline{ji}\}, i, j \in\{0,1,2, \cdots, 9\}$, then $A$ must contain at least one element from $A_{ij}$, otherwise, the infinite sequence ijijijij $\cdots$ would have no adjacent digits belonging to $A$, leading to a contradiction. Clearly, the set $A_{ij}$ has a total of $10+\mathrm{C}... | 55 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,932 |
6. The natural number $k$ satisfies the following property: in $1,2, \cdots, 1988$, $k$ different numbers can be selected such that the sum of any two of these numbers is not divisible by their difference. Find the maximum value of $k$. (26th Mo
将上面的文本翻译成英文，保留了源文本的换行和格式。 | 6. The maximum value of $k$ is 663. First, we prove that $k \leqslant 663$. We note the following fact: when $x - y = 1$ or 2, we have $x - y \mid x + y$. This implies that in any sequence of 3 consecutive natural numbers, if we take any two numbers $x$ and $y$, then $x - y \mid x + y$. If $k > 663$, let $A_{i} = \{3i ... | 663 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,934 |
7 In the subset $S$ of the set $X=\{1,2, \cdots, 50\}$, the sum of the squares of any two elements is not a multiple of 7. Find the maximum value of $|S|$. | 7. Divide $X$ into 2 subsets: $A_{1}=\{x \mid x \equiv 0(\bmod 7), x \in X\}, A_{2}=$ $X \backslash A_{1}$. Then $\left|A_{1}\right|=7,\left|A_{2}\right|=43$. Clearly, $S$ can contain at most 1 number from $A_{1}$, so $|S| \leqslant 43+1=44$. On the other hand, for any integer $x$, if $x \equiv 0, \pm 1, \pm 2, \pm 3(\... | 44 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,935 |
8 Let $X=\{1,2, \cdots, 1995\}, A$ be a subset of $X$, such that when $x \in A$, $19 x \notin A$. Find the maximum value of $|A|$.
When $x \in A$, $19 x \notin A$. | 8. Let $A_{k}=\{k, 19 k\}, k=6,7, \cdots, 105$, then $A$ contains at most 1 number from each $A_{k}$, so $|A| \leqslant 1995-100=1895$. On the other hand, let $A=\{1,2,3,4,5\} \cup$ $\{106,107, \cdots, 1995\}$, then $A$ meets the requirement, and in this case $|A|=1895$. Therefore, the maximum value of $|A|$ is 1895. | 1895 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,936 |
Example 1 If a set does not contain three numbers $x, y, z$ that satisfy $x+y=z$, then it is called simple. Let $M=\{1,2, \cdots, 2 n+1\}, A$ be a simple subset of $M$, find the maximum value of $|A|$. (1982 | Let $A=\{1,3,5, \cdots, 2 n+1\}$, then $A$ is simple, and at this point $|A|=n+1$. We will prove that for any simple subset $A$, $|A| \leqslant n+1$. We use proof by contradiction. Assume $|A|>n+1$, then $A$ contains at least $n+2$ elements, let them be: $a_{1} < a_{2} < \cdots < a_{n+2}$. Consider the $2n+2$ numbers: ... | n+1 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,937 |
Example 2 Let $X=\{1,2, \cdots, 100\}, A$ be a subset of $X$, if for any two elements $x$ 、 $y(x<y)$ in $A$, we have $y \neq 3 x$, find the maximum value of $|A|$ . | Let $A=\{3 k+1 \mid k=0,1,2, \cdots, 33\} \cup\{3 k+2 \mid k=0,1,2, \cdots, 32\} \bigcup\{9,18,36,45,63,72,81,90,99\}$, then $A$ clearly meets the conditions, at this point $|A|=76$.
On the other hand, consider 24 sets $A_{k}=\{k, 3 k\}(k=1,2,12,13, \cdots, 33)$, they contain 48 distinct numbers. Excluding these numbe... | 76 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,938 |
Example 3 For a set of numbers $M$, define the sum of $M$ as the sum of all numbers in $M$, denoted as $S(M)$. Let $M$ be a set composed of several positive integers not greater than 15, and any two disjoint subsets of $M$ have different sums. Find the maximum value of $S(M)$. | Let $M=\{15,14,13,11,8\}$, at this point, $S(M)=61$. Below, we prove: for any set $M$ that meets the problem's conditions, $S(M) \leqslant 61$. For this, we first prove: $|M| \leqslant 5$. (1)
By contradiction, assume $|M| \geqslant 6$, and consider all subsets $A$ of $M$ with no more than 4 elements, then $S(A) \leqs... | 61 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,939 |
Example 4 Given that in a mathematics competition, the number of competition problems is $n(n \geqslant 4)$, each problem is solved by exactly 4 people, and for any two problems, there is exactly one person who solves both problems.
If the number of participants is no less than $4 n$, find the minimum value of $n$ suc... | Analysis and Solution First, when $4 \leqslant n \leqslant 13$, counterexamples can be constructed. In fact, because the number of participants $\geqslant 4 n \geqslant 16$, consider the following 13 sets: $M_{1}=\{1,2,3,4\}$, $M_{2}=\{1,5,6,7\}$, $M_{3}=\{1,8,9,10\}$, $M_{4}=\{1,11,12,13\}$, $M_{5}=\{2,5,8,11\}$, $M_{... | 14 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,940 |
Example 5 There are 18 teams participating in a single round-robin tournament, meaning each round the 18 teams are divided into 9 groups, with each group's 2 teams playing one match. In the next round, the teams are regrouped to play, for a total of 17 rounds, ensuring that each team plays one match against each of the... | ```
Solution: Consider the following match program:
1. (1, 2) (3, 4) (5, 6) (7, 8) (9, 18) (10, 11) (12, 13) (14, 15)
(16, 17);
2. (1, 3) (2, 4) (5, 7) (6, 9) (8, 17) (10, 12) (11, 13) (14, 16)
(15, 18);
3. (1, 4) (2, 5) (3, 6) (8, 9) (7, 16) (10, 13) (11, 14) (12, 15)
(17, 18);
... | 7 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,941 |
3 Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of $1,2, \cdots, n$, find the maximum value of $S_{n}=\left|a_{1}-1\right|+\mid a_{2}-$ $2|+\cdots+| a_{n}-n \mid$. | 3. The key to solving the problem is to remove the absolute value symbol. Note that $\left|a_{i}-i\right|$ equals $a_{i}-i$ or $i-a_{i}$, so after removing the absolute value symbol, the number of negative signs in the sum remains unchanged. That is, regardless of how $a_{1}, a_{2}, \cdots, a_{n}$ are arranged, after r... | \left[\frac{n^{2}}{2}\right] | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,942 |
Example 6 Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two different elements $a, b$ in $S$, there exists an element $c$ in $S$ such that the greatest common divisor of $a$ and $c$ is 1, and the gre... | Analysis and Solution: Represent the positive integer $n \leq 100$ as $n=2^{\alpha_{1}} 3^{\alpha_{2}} 5^{\alpha_{3}} 7^{\alpha_{4}} 11^{\alpha_{5}} \cdot m$, where $m$ is a positive integer not divisible by $2, 3, 5, 7, 11$, and $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are natural numbers. Select t... | 72 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,943 |
Example 7 Let $a_{i}, b_{i} (i=1,2, \cdots, n)$ be rational numbers such that for any real number $x$ we have $x^{2}+x+4=\sum_{i=1}^{n}\left(a_{i} x+b_{i}\right)^{2}$. Find the minimum possible value of $n$. (2006 National Training Team Problem) | It is easy to find that $n=5$ works. In fact, $x^{2}+x+4=\left(x+\frac{1}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}+1^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}$.
From this, we can conjecture that the minimum possible value of $n$ is 5, which only requires proving that $n \neq 4$.
By contradiction,... | 5 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,944 |
Example 8 If a set contains an even number of elements, it is called an even set. Let $M=\{1,2, \cdots, 2011\}$, if there exist $k$ even subsets of $M$: $A_{1}, A_{2}, \cdots, A_{k}$, such that for any $1 \leqslant i<j \leqslant k$, $A_{i} \cap A_{j}$ is not an even set, find the maximum value of $k$. (Original problem... | First, it is easy to see that $k=2010$ meets the conditions. In fact, let $A_{i}=\{i, 2011\}(i=1,2, \cdots, 2010)$, then for any $1 \leqslant i<j \leqslant 2010$, we have $A_{i} \cap A_{j}=\{2011\}$, which is not an even set, so $k=2010$ meets the conditions.
From this, we can conjecture that $k \leqslant 2010$, and w... | 2010 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,945 |
1 Positive integer $n$ satisfies the following property: when $n$ different odd numbers are chosen from $1,2, \cdots, 100$, there must be two whose sum is 102. Find the minimum value of $n$. | 1. The minimum value of $n$ is 27. First, prove that $n \geqslant 27$. If $n \leqslant 26$, then let $A=\{1,3,5, \cdots, 2 n-1\}$, then $|A|=n$, but the sum of any two numbers in $A$: $(2 i-1)+(2 j-1)=2(i+j)-2 \leqslant 2(26+25)-2=100<102$ (where $0<i<j \leqslant n \leqslant 26$), which is a contradiction. Next, when $... | 27 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,946 |
2 Let $X=\{1,2, \cdots, 1995\}, A$ be a subset of $X$, if for any two elements $x$ 、 $y(x<y)$ in $A$, we have $y \neq 15 x$, find the maximum value of $|A|$.
untranslated text remains unchanged. | 2. Let $A=\{1,2, \cdots, 8\} \cup\{134,135, \cdots, 1995\}$, then $A$ clearly satisfies the condition, and at this point $|A|=1870$. On the other hand, consider 125 sets $A_{k}=\{k, 15 k\}(k=9,10, \cdots, 133)$, which contain 250 distinct numbers. Removing these numbers, there are still 1745 numbers in $X$. Form a sing... | 1870 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,947 |
3 Let $X=\{0,1,2, \cdots, 9\}, F=\left\{A_{1}, A_{2}, \cdots, A_{k}\right\}$ where each element $A_{i}$ is a non-empty subset of $X$, and for any $1 \leqslant i<j \leqslant k$, we have $\left|A_{i} \cap A_{j}\right| \leqslant 2$, find the maximum value of $k$. (No. | 3. Let $F$ satisfy the conditions. To maximize $|F|$, note that $\left|A_{i} \cap A_{j}\right| \leqslant 2$. We should include some sets with fewer elements in $F$. Clearly, all sets satisfying $\left|A_{i}\right| \leqslant 2$ can be included in $F$. Next, all sets satisfying $\left|A_{i}\right| \leqslant 3$ can also b... | 175 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,948 |
4. Let $A_{i}=\{i, i+1, i+2, \cdots, i+59\}(i=1,2, \cdots, 11), A_{11+j}=\{11+j, 12+j, \cdots, 70,1,2, \cdots, j\}(j=1,2, \cdots, 59)$. Among these 70 sets, there exist $k$ sets such that the intersection of any 7 of these sets is non-empty. Find the maximum value of $k$.
In these 70 sets, there exist $k$ sets, where ... | 4. Let $A_{i}=\{i, i+1, i+2, \cdots, i+59\}(i=1,2, \cdots, 70)$, where the elements in the set are understood modulo 70, i.e., if $x>70$, replace $x$ with $x-70$. Clearly, $60 \in A_{1}, A_{2}, \cdots, A_{60}$, so $k_{\text {max }} \geqslant 60$. We conjecture that $k_{\text {max }}=60$. This only requires proving that... | 60 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,949 |
5 Let $S$ be a non-empty subset of the set $\{1,2, \cdots, 108\}$, satisfying: (i) for any numbers $a, b$ in $S$, there exists a number $c$ in $S$ such that $(a, c)=(b, c)=1$; (ii) for any numbers $a, b$ in $S$, there exists a number $c'$ in $S$ such that $\left(a, c'\right)>1,\left(b, c'\right)>1$. Find the maximum po... | 5. The maximum possible number of elements in $S$ is 76. Let $|S| \geqslant 3, p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}} \in S, p_{1}, p_{2}, p_{3}$ be three different primes, $p_{1}1, (c_{3}, c_{2})>1$. From $(c_{1}, c_{2})=1$ we know the product of the smallest prime factors of $c_{1}$ and $c_{2}$ is $... | 76 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 737,950 |
6. Find the smallest positive integer $n$ such that: If each vertex of a regular $n$-gon is arbitrarily colored with one of the three colors red, yellow, or blue, then there must exist four vertices of the same color that are the vertices of an isosceles trapezoid. (2008 China Mathematical Olympiad) | 6. First construct a coloring method that does not meet the problem's requirements for $n \leqslant 16$.
Let $A_{1}, A_{2}, \cdots, A_{n}$ represent the vertices of a regular $n$-sided polygon (in clockwise order), and $M_{1}, M_{2}, M_{3}$ represent the sets of vertices of three different colors.
When $n=16$, let $M... | 17 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,951 |
Example 1 Find the largest positive integer $A$, such that in any permutation of $1,2, \cdots, 100$, there are 10 consecutive terms whose sum is not less than $A$. (22nd Polish Mathematical Olympiad Problem) | Let $a_{1}, a_{2}, \cdots, a_{100}$ be any permutation of $1,2, \cdots, 100$, and let $A_{i}=a_{i}+a_{i+1}+\cdots+a_{i+9}(i=1,2, \cdots, 91)$. Then $A_{1}=a_{1}+a_{2}+\cdots+a_{10}, A_{11}=a_{11}+a_{12}+\cdots+a_{20}, \cdots, A_{91}=a_{91}+a_{92}+\cdots+a_{100}$.
Notice that $A_{1}+A_{11}+\cdots+A_{91}=a_{1}+a_{2}+\cd... | 505 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,952 |
4 Let $x_{k}(k=1,2, \cdots, 1991)$ satisfy $\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\mid x_{1990}-$ $x_{1991} \mid=1991$. Let $y_{k}=\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}(k=1,2, \cdots, 1991)$. Find the maximum value of $F=$ $\left|y_{1}-y_{2}\right|+\left|y_{2}-y_{3}\right|+\cdots+\left|y_{1990}-y_{1991}... | 4. For each $k(1 \leqslant k \leqslant 1990)$, we have $\left|y_{k}-y_{k+1}\right|=\left\lvert\, \frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\right.$ $\frac{x_{1}+x_{2}+\cdots+x_{k+1}}{k+1}|=| \frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\left|=\frac{1}{k(k+1)}\right|\left(x_{1}-\right.$
$\left.x_{2}\right) \left.+2\left(x_... | 1990 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,953 |
Example 2 Let $A_{i}(i=1,2, \cdots, 30)$ be subsets of $M=\{1,2,3, \cdots, 1990\}$, with $\left|A_{i}\right| \geqslant 660$. Prove that there exist $i, j(1 \leqslant i<j \leqslant 30)$, such that $\left|A_{i} \cap A_{j}\right| \geqslant 200$. | Proof: Without loss of generality, assume all $\left|A_{i}\right|=660$. Otherwise, remove some elements from $A_{i}$ to get $A_{i}^{\prime}$. If we can prove that $\left|A_{i}^{\prime} \cap A_{j}^{\prime}\right| \geqslant 200$, then adding back the elements that were originally removed, it is clear that $\left|A_{i} \c... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 737,954 |
Example 3: 10 people go to a bookstore to buy books. It is known that each person bought three types of books, and any two people have at least one book in common. Question: What is the maximum number of people who bought the most purchased book, at a minimum? (No. 8
| Let the total number of books sold be $n$, and the set of books bought by the $i$-th person be $A_{i}(i=1,2, \cdots, 10)$. Construct a table of set element relationships, where the $i$-th row has $m_{i}$ ones.
Estimate the total number of times each element appears, we have
$$\sum_{i=1}^{n} m_{i}=S=\sum_{i=1}^{10}\lef... | 5 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 737,955 |
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