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11. On a circle, there is 1 red point and 2009 blue points. Xiao Dan calculates the number of convex polygons where all vertices are blue points, while Xiao Dong calculates the number of convex polygons where one vertex is the red point. The difference between the two numbers they get is $\qquad$ | 11. 2017036.
For every convex $n$-sided polygon $(n \in \mathbf{N}_{+})$ calculated by Xiao Dan, there is a corresponding convex $(n+1)$-sided polygon calculated by Xiao Dong. Xiao Dong's convex $(n+1)$-sided polygon has one more red vertex than Xiao Dan's convex $n$-sided polygon. This correspondence does not hold wh... | 2017036 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. In an international chess tournament with 10 participants, each player must play exactly one game against every other player. After several games, it is found that among any three players, at least two have not yet played against each other. How many games have been played at most by this point? | 2. Suppose there are 5 men and 5 women among the 10 players, and all matches so far have been between men and women, thus satisfying the condition, i.e., 25 matches have been played.
Next, we prove: 25 is indeed the maximum value.
Let $k$ be the number of matches played by the player who has played the most matches, de... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Find the smallest positive integer that can be expressed as the sum of the squares of four positive integers and can divide some integer of the form $2^{n}+15\left(n \in \mathbf{N}_{+}\right)$. | 3. The smallest five positive integers that can be expressed as the sum of four positive integer squares are
$$
\begin{array}{l}
4=1+1+1+1, \\
7=4+1+1+1, \\
10=4+4+1+1, \\
12=9+1+1+1, \\
13=4+4+4+1 .
\end{array}
$$
Obviously, since $2^{n}+15$ is odd, the smallest positive integer cannot be 4, 10, or 12.
Also, $2^{n} ... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. (40 points) In each game of bridge, four players play together. It is known that the number of games played is exactly equal to the number of players, and any two players have played at least one game together. Try to find the maximum number of players.
| 9. There are $n$ players, and each game can produce 6 pairs of players.
From $\mathrm{C}_{n}^{2} \leqslant 6 n$ we get
$$
n \leqslant 13 \text {. }
$$
Next, we prove that $n=13$ is feasible.
As shown in Figure 13, number the 13
players from $0 \sim 12$.
The players in each game are as follows:
$$
\begin{array}{l}
(0,2... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9.1. A simplest fraction is equal to the sum of two simplest fractions with denominators 600 and 700, respectively. Find the smallest possible value of the denominator of such a simplest fraction. | 9. 1. $2^{3} \times 3 \times 7=168$.
Let the two simplest fractions be $\frac{a}{600}$ and $\frac{b}{700}$. Then $(a, 6)=(b, 7)=1$.
Thus, the sum $\frac{7 a+6 b}{4200}$ has a numerator that is coprime with 6 and 7.
Since $4200=2^{3} \times 3 \times 7 \times 5^{2}$, after canceling out the common factors, the denominat... | 168 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9.4. In a regular 100-gon, each vertex is covered by a cloth. It is known that exactly one of the cloths hides a coin. The following action is called an operation: choose any four cloths to check if there is a coin underneath, and after each operation, the cloths are returned to their original positions, while the coin... | 9.4.33 times.
Assume that a regular 100-gon is placed on a rotatable horizontal circular table, and the vector from the center to one vertex points due north. The initial positions of the vertices of the polygon are defined from due north in a counterclockwise direction as $0, 1, \cdots, 99$. Each operation and coin t... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
9.7. A diagonal of eight squares on a chessboard is called a "fence". A rook starts from a square outside the fence on the chessboard and moves, satisfying the following conditions:
(1) It stays on any square of the chessboard at most once;
(2) It never stays on a square of the fence.
Find the maximum number of times ... | 9.7.47 times.
Let the square at the $i$-th row and $j$-th column be denoted as $(i, j)$. Suppose the eight squares occupied by the fence are $(i, i) (i=1,2, \cdots, 8)$. The non-fence squares are divided into four categories $A, B, C, D$:
$$
\begin{aligned}
A= & \{(i, j) \mid 2 \leqslant j+1 \leqslant i \leqslant 4\} ... | 47 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 1, given $A(-3,0)$, $B(0,-4)$, and $P$ is any point on the hyperbola $y=\frac{12}{x} (x > 0)$. A perpendicular line from point $P$ to the x-axis meets at point $C$, and a perpendicular line from point $P$ to the y-axis meets at point $D$. Then the minimum value of the area $S$ of quadrilateral $A ... | 2. C.
Let $P\left(x, \frac{12}{x}\right)$. Then $C(x, 0)$ and $D\left(0, \frac{12}{x}\right)$.
It is easy to see that $C A=x+3, D B=\frac{12}{x}+4$, so
$$
S=\frac{1}{2} C A \cdot D B=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \text {. }
$$
Simplifying, we get $S=2\left(x+\frac{9}{x}\right)+12$.
Since $x>0, \frac{9}{... | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. As shown in Figure 2, the first polygon is "expanded" from an equilateral triangle, with the number of sides denoted as $a_{3}$, the second polygon is "expanded" from a square, with the number of sides denoted as $a_{4}, \cdots \cdots$ and so on. The polygon "expanded" from a regular $n(n \geqslant 3)$-sided polygon... | 2. 2009 .
From the extended definition, we know
$$
\begin{array}{l}
a_{3}=12=3 \times 4, \\
a_{4}=20=4 \times 5, \\
\cdots \cdots \\
a_{n}=n(n+1) . \\
\text { Also, } \frac{1}{a_{n}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \\
\Rightarrow \frac{1}{a_{3}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{n}} \\
=\left(\frac{1}{3}-\... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Among the natural numbers from 1 to 144, the number of ways to pick three numbers that form an increasing geometric progression with an integer common ratio is $\qquad$ . | 3. 78.
Let the three numbers $a$, $a q$, and $a q^{2}$ form an increasing geometric sequence. Then
$$
1 \leqslant a < a q < a q^{2} \leqslant 144 \text{.}
$$
From this, we have $2 \leqslant q \leqslant 12$.
When $q$ is fixed, the number of integers $a$ such that the three numbers $a$, $a q$, and $a q^{2}$ are integer... | 78 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of integer points within the square (including the four sides) formed by the four lines $y=x+10, y=-x+10, y=$ $x-10, y=-x-10$ in the Cartesian coordinate system is $\qquad$ . | (Tip: By plotting, it is known that the number of integer points in each quadrant is 45. The number of integer points on the coordinate axes is 41, so, the number of integer points is $(4 \times 45+41=) 221$. | 221 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Question Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$
$(2007$, National High School Mathematics League Jiangxi Province Preliminary Contest) | Another solution: Considering that $a_{n}$ is not continuous in decimal, and the digits of $a_{n}$ are all no greater than 3 (i.e., $0, 1, 2, 3$), then the form of $a_{n}$ is a sequence of consecutive integers in quaternary, thus simplifying the problem.
Let the set $\left\{\left(a_{n}\right)_{4}\right\}=\{$ consecuti... | 133113 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, the five-digit numbers formed by the digits $1, 2, 3, 4, 5$ with repetition, arranged in ascending order. Ask:
(1) What are the positions of 22435 and 43512?
(2) What is the 200th number? ${ }^{[1]}$ | Analysis: Let the $n$-th number after sorting be denoted as $a_{n}$. Since the numbers do not contain 0, we cannot directly use the quinary system. Therefore, subtract 11111 from all $a_{n}$, and denote the result as $b_{n}$. Thus, we can draw the following conclusions:
(i) Each number in $b_{n}$ consists of some of th... | 12355 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given in rectangle $A B C D$, $A B=72, A D=$ 56. If $A B$ is divided into 72 equal parts, and parallel lines to $A D$ are drawn through each division point; and if $A D$ is divided into 56 equal parts, and parallel lines to $A B$ are drawn through each division point, then these parallel lines divide the entire rect... | 4. D.
According to the problem, establish a Cartesian coordinate system such that $A(0,0)$, $B(72,0)$, and $D(0,56)$. Then, $C(72,56)$.
Since $AC$ intersects with every horizontal line (including $AB$ and $DC$) and every vertical line (including $AD$ and $BC$), there are $57 + 73 = 130$ intersection points (including... | 120 | Geometry | MCQ | Yes | Yes | cn_contest | false |
7. Calculate $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}}(n \geqslant 2$, $n \in \mathbf{N})$ 的值为
The value of $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}... | 7. 100 .
$$
\begin{array}{l}
\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}} \\
=\sqrt[n]{\left(10^{n}-1\right)^{2}+2 \times 10^{n}-1} \\
=\sqrt[n]{10^{2 n}}=100 .
\end{array}
$$ | 100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. As shown in Figure 3, fill in the 10 spaces in the annulus with the numbers $1,2, \cdots, 10$. Add up the absolute values of the differences between all adjacent cells (cells that share a common edge). If this sum is to be maximized, then the maximum value is $\qquad$ | 10.50.
Let the two numbers in adjacent cells be $a$ and $b$ ($a > b$). Then $|a-b| = a-b$, and there are 10 differences. To maximize the sum of these 10 differences, the 10 minuends $a$ should be as large as possible, and the 10 subtrahends $b$ should be as small as possible. Since each number is adjacent to two other... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The smallest positive integer $a$ that makes the inequality
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}<a-2007 \frac{1}{3}
$$
hold for all positive integers $n$ is $\qquad$ | 4. 2009 .
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$.
Obviously, $f(n)$ is monotonically decreasing. From the maximum value of $f(n)$, $f(1)<a-2007 \frac{1}{3}$, we get $a=2009$. | 2009 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. A station has exactly one bus arriving between $8:00 \sim 9:00$ and $9:00 \sim 10:00$ every morning, but the arrival times are random, and the two arrival times are independent of each other, as shown in Table 1. A passenger arrives at the station at $8:20$. What is the expected waiting time for the passenger (round... | 8. 27 .
The distribution of passenger waiting times is shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
Waiting Time \\
(min)
\end{tabular} & 10 & 30 & 50 & 70 & 90 \\
\hline Probability & $\frac{1}{2}$ & $\frac{1}{3}$ & $\frac{1}{6} \times \frac{1}{6}$ & $\frac{1}{2} \times \frac{1}... | 27 | Other | math-word-problem | Yes | Yes | cn_contest | false |
10.3. The function $f(x)=\prod_{i=1}^{2000} \cos \frac{x}{i}$ changes sign how many times in the interval $\left[0, \frac{2009 \pi}{2}\right]$? | 10.3. 75 times.
Let $n=2009$. Consider the function $\cos \frac{x}{k}$. It changes sign at $x=\frac{k(2 m+1) \pi}{2}$. This indicates that the zeros of $f(x)$ are $x_{i}=\frac{i \pi}{2}(1 \leqslant i \leqslant n)$.
We only need to consider the sign change of $f(x)$ at $x_{i}(i=1,2, \cdots, n-1)$.
$\cos \frac{x}{k}$ c... | 75 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
10. 4. In a regular 2009-gon, a non-negative integer not exceeding 100 is placed at each vertex. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k$ su... | 10. 4. $k_{\min }=100400$.
Let the numbers at each vertex be $a_{1}, a_{2}, \cdots, a_{2009}$. Let $N=100400$.
$$
\begin{array}{l}
\text { (1) Let } a_{2}=a_{4}=\cdots=a_{2008}=100, \\
a_{1}=a_{3}=\cdots=a_{2008}=0, \\
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right) .
\end{arr... | 100400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.5. Let $k$ be a positive integer. An infinite strictly increasing sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integer $n$, we have $a_{n+1}-a_{n} \leqslant k$, and $a_{n}$ is a multiple of 1005 or 1006, but not a multiple of 97. Find the smallest possible value of $k$. | 10.5. $k_{\text {min }}=2010$.
Given a positive integer $N$, such that
$$
a_{1}<1005 \times 1006 \times 97 \times N=D \text {. }
$$
Since $D$ is a multiple of 97, it is not any term in the sequence $a_{n}$, hence there exists a positive integer $n$, such that $a_{n}<D<a_{n+1}$.
The largest multiple of 1005 or 1006 l... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. The total score of an exam consists of six 6-point questions, six 9-point questions, and twelve 5-point questions. Therefore, the number of different scores that can be formed by this test paper is $\qquad$ | 8. 136.
Since there are scores, and the scores can only be generated from the 150 positive integers from $1 \sim 150$, and the sum of the scores and the deducted points is also 150, we only need to consider the number of positive integers from $1 \sim 75$ that can be represented by $6x + 9y + 5z$ (where $x, y, z$ are ... | 136 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. (14 points) Let $n$ be a positive integer greater than 16, and let $A(n, k)$ and $B(n, k) (k \geqslant 3)$ be the number of $k$-term monotonic arithmetic sequences and geometric sequences with integer common ratios, respectively, in the set $\{1,2, \cdots, n\}$.
(1) Find the value of $A(2009,50)$;
(2) Prove: $n-4 \s... | 1. (1) First, find the general $A(n, \dot{k})$.
Let the set $\{1,2, \cdots, n\}$ have a monotonically increasing arithmetic sequence $\left\{a_{i}\right\}$, with common difference $d$, and $a_{k}=a_{1}+(k-1) d$. Then
$$
\begin{array}{l}
1+(k-1) d \leqslant a_{1}+(k-1) d \leqslant n \\
\Rightarrow d \leqslant \frac{n-1}... | 80360 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Initially 261 the perimeter of an integer-sided triangle is 75, and squares are constructed on each side. The sum of the areas of the three squares is 2009. Find the difference between the longest and shortest sides of this triangle. | Solution: Given that the sum of three numbers is 75, their average is 25.
Let $2009=(25+a)^{2}+(25+b)^{2}+(25+c)^{2}$, where $a+b+c=0$. It is easy to see that
$a^{2}+b^{2}+c^{2}=134$.
Decomposing 134 into the sum of squares of three positive integers, we find the following four cases:
(1) $a^{2}+b^{2}+c^{2}=134=121+9+4... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Two, (50 points) The four-digit numbers $m$ and $n$ are reverse positive integers of each other, and $m+n=18 k+9\left(k \in \mathbf{N}_{+}\right), m$ and $n$ have 16 and 12 positive divisors (including 1 and themselves), respectively. The prime factors of $n$ are also prime factors of $m$, but $n$ has one fewer prime f... | Let $m=\overline{a b c d}, a d \neq 0$. Then $n=\overline{d c b a}$.
Given $m+n=9(2 k+1)$, then $9 \mid (m+n)$.
Thus, $9 \mid [(1000 a+100 b+10 c+d)+$
$$
\begin{array}{c}
(1000 d+100 c+10 b+a)], \\
9 \mid 2(a+b+c+d), 9 \mid (a+b+c+d).
\end{array}
$$
Therefore, $9 \mid m, 9 \mid n$.
Since $m+n$ is odd, it follows that ... | 1998 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Does there exist 4098 sets
\[
\begin{aligned}
B_{i} & =\left\{a_{i 1}, a_{i 2}, \cdots, a_{i n}\right\}(i=1,2, \cdots, 4098, \\
a_{i j} \in A_{j} & =\{3 j-2,3 j-1,3 j\}, j=1,2, \cdots, 12)
\end{aligned}
\]
satisfying the following three conditions:
(1) \( B_{i} \cap B_{i+1}=\varnothing(i=1,2, \cdots,... | Let $(s, t)$ denote a set circle with $s$ sets $A_{1}, A_{2}, \cdots, A_{s}$, and $t$ sets $B_{1}, B_{2}, \cdots, B_{t}$ that satisfy the given conditions. Use $\left(B_{i}, p\right)$ to denote a set formed by all elements of $B_{i}$ and the $p$-th element of $A_{i+1}$.
If $B_{1}, B_{2}, \cdots, B_{t}$ satisfy the give... | 4098 | Combinatorics | proof | Yes | Yes | cn_contest | false |
1. Calculate $\frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9$ The value equals $\qquad$ . | $\begin{array}{l} \text { II.1.3 481. } \\ \frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9 \\ = \frac{(45.1-13.9)\left(45.1^{2}+45.1 \times 13.9+13.9^{2}\right)}{45.1-13.9}+ \\ 45.1 \times 13.9 \\ = 45.1^{2}+2 \times 45.1 \times 13.9+13.9^{2} \\ =(45.1+13.9)^{2}=59^{2}=3481 .\end{array}$ | 3481 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 1, in $\square A B C D$, $A D=a, C D=b$, draw the heights $h_{a}, h_{b}$ from point $B$ to sides $A D$ and $C D$ respectively. Given $h_{a} \geqslant a, h_{b} \geqslant b$, and the diagonal $A C=20$. Then the area of $\square A B C D$ is $\qquad$ | 3. 200 .
Given $h_{a} \geqslant a, h_{b} \geqslant b$, in addition, $a \geqslant h_{b}, b \geqslant h_{a}$, then $h_{a} \geqslant a \geqslant h_{b} \geqslant b \geqslant h_{a}$, i.e., $h_{a}=a=h_{b}=b$, which means that quadrilateral ABCD is a square.
Since the diagonal of this square $AC=20$, the area of $\square ABC... | 200 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. There are $n (n>12)$ people participating in a mathematics invitational competition. The test consists of fifteen fill-in-the-blank questions, with each correct answer worth 1 point and no answer or a wrong answer worth 0 points. Analyzing every possible score situation, it is found that as long as the sum of the sc... | 7. The minimum possible value of $n$ is 911.
(1) First, prove: 911 satisfies the condition.
If each student answers at least three questions correctly, since the number of different ways a student can answer three questions correctly is $\mathrm{C}_{15}^{3}=455$, then if there are 911 students participating, by the pig... | 911 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Place 3 identical white balls, 4 identical red balls, and 5 identical yellow balls into three different boxes, allowing some boxes to contain balls of different colors. The total number of different ways to do this is (answer in numbers).
允许有的盒子中球的颜色不全的不同放法共有种 (要求用数字做答).
Allowing some boxes to contain balls of diff... | $\begin{array}{l}\text { 9. } 3150 \text {. } \\ \mathrm{C}_{5}^{2} \cdot \mathrm{C}_{6}^{2} \cdot \mathrm{C}_{7}^{2}=3150 \text {. }\end{array}$ | 3150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11.6. A $10 \times 10$ chessboard has $k$ rooks. A square on the board that can be attacked by a rook is called "dangerous" (the square occupied by the rook itself is also considered dangerous). If removing any rook results in at least one dangerous square becoming safe, find the maximum possible value of $k$. | 11. 6. $k_{\max }=16$.
Consider a chessboard with $k$ rooks satisfying the problem's conditions. There are two cases to consider.
(1) Each row (column) has a rook.
In this case, all squares are dangerous. If there is a row (column) with at least two rooks, removing one of these rooks will still leave all squares dange... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If a positive integer is written on each face of a cube, and then a number is written at each vertex, which is equal to the product of the two integers on the faces passing through that vertex, then, when the sum of the numbers at the vertices of the cube is 290, the sum of the numbers on the faces of the cube is | 3. 36 .
Let the numbers on each face of the cube be $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, $x_{5}$, $x_{6}$, and the numbers written at the vertices be
$$
\begin{array}{l}
x_{1} x_{2} x_{5}, x_{2} x_{3} x_{5}, x_{3} x_{4} x_{5}, x_{4} x_{1} x_{5}, x_{1} x_{2} x_{6}, \\
x_{2} x_{3} x_{6}, x_{3} x_{4} x_{6}, x_{4} x_{1} x_... | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Given an isosceles triangle with a vertex angle less than $60^{\circ}$, the lengths of its three sides are all positive integers. Construct a square outward on each side, such that the sum of the areas of the three squares is 2009. Find the perimeter of this isosceles triangle. | Let the length of the legs of the isosceles triangle be $a$, and the length of the base be $b$. According to the problem, we have
$$
2 a^{2}+b^{2}=2009 \text{. }
$$
Therefore, $a^{2}=\frac{2009-b^{2}}{2}\frac{2009}{3}>669$.
Thus, $669<a^{2}<1005 \Rightarrow 25<a<32$.
Since $a$ and $b$ are positive integers, it is veri... | 77 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $[x]$ represents the greatest integer not exceeding the real number $x$. If $P=\sum_{i=1}^{2010}\left[\frac{6^{i}}{7}\right]$, then the remainder when $P$ is divided by 35 is | 6. 20 .
First, consider $S=\frac{6}{7}+\frac{6^{2}}{7}+\cdots+\frac{6^{2010}}{7}$.
In equation (1), no term is an integer, but the sum of any two adjacent terms is an integer (since $\frac{6^{k}}{7}+\frac{6^{k+1}}{7}=6^{k}$ $(k \in \mathbf{Z})$ is an integer).
If the sum of two non-integer numbers is an integer, then... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (15 points) Given seven different points on a circle, vectors are drawn from any one point to another (for points $A$ and $B$, if vector $\overrightarrow{A B}$ is drawn, then vector $\overrightarrow{B A}$ is not drawn). If the four sides of a convex quadrilateral determined by any four points are four consecutive v... | 11. Let the seven points on the circumference be $P_{1}, P_{2}, \cdots, P_{7}$. The number of vectors starting from point $P_{i} (i=1,2, \cdots, 7)$ is $x_{i} (i=1,2, \cdots, 7)$, then $0 \leqslant x_{i} \leqslant 6$, and
$$
\sum_{i=1}^{7} x_{i}=\mathrm{C}_{7}^{2}=21 \text {. }
$$
First, find the minimum number of "no... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Try to find the smallest positive integer $M$, such that the sum of all positive divisors of $M$ is 4896. | Let $M=\prod_{i=1}^{n} p_{i}^{\alpha_{i}}\left(p_{i}\right.$ be a prime, $\alpha_{i} \in \mathbf{N}_{+}, i=1,2, \cdots, n)$. Denote
$$
f(p, \alpha)=\sum_{k=0}^{\alpha} p^{k}\left(p\right.$ be a prime, $\left.\alpha \in \mathbf{N}_{+}\right)$.
From the problem, we know
$$
\prod_{i=1}^{n} f\left(p_{i}, \alpha_{i}\right)=... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $n$ be a positive integer, and let $S_{n}$ be a subset of the set $A_{n}=\left\{m \mid m \in \mathbf{N}_{+}\right.$, and $\left.m \leqslant n\right\}$. Moreover, the difference between any two numbers in $S_{n}$ is not equal to 4 or 7. If the maximum number of elements in $S_{n}$ is denoted by $M_... | It is known that the difference between any two numbers among $1,4,6,7,9$ is not 4 or 7. Adding 11 to each of these numbers gives $12, 15, 17, 18, 20$, which clearly also have the same property, and the difference between any of these numbers and any of the first five numbers is also not 4 or 7. By this reasoning, for ... | 922503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $a b c<0$, let
$$
P=\frac{a}{|a|}+\frac{|b|}{b}+\frac{c}{|c|}+\frac{|a b|}{a b}+\frac{a c}{|a c|}+\frac{|b c|}{b c} \text {. }
$$
Find the value of $a P^{3}+b P^{2}+c P+2009$. | Given $a b c<0$, we get $\frac{a b c}{|a b c|}=-1$, and at least one of $a, b, c$ is negative. According to the corollary, we can change the position of the absolute value symbol, then
$$
\begin{aligned}
P & =\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b}{|a b|}+\frac{a c}{|a c|}+\frac{b c}{|b c|} \\
& =\left(1+\... | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that the altitude to the hypotenuse of right $\triangle A B C$ is 4. Then the minimum value of the area of $\triangle A B C$ is $\qquad$ . | 2. 16 .
Let the lengths of the two legs of the right triangle $\triangle ABC$ be $a$ and $b$, and the length of the hypotenuse be $c$. From the area relationship, we have $ab = 4c$.
By the Pythagorean theorem, we know $c^2 = a^2 + b^2 \geq 2ab = 8c$.
Thus, $c \geq 8$. Therefore, $S_{\triangle ABC} = 2c \geq 16$.
When ... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In rectangle $A B C D$, $A B=12, A D=3, E$ and $F$ are points on $A B$ and $D C$ respectively. Then the minimum length of the broken line $A F E C$ is $\qquad$ . | 3. 15 .
As shown in Figure 5, construct the symmetric points $A_{1}$ and $C_{1}$ of $A$ and $C$ with respect to $DC$ and $AB$, respectively. Connect $A_{1}C_{1}$ to intersect $AB$ and $DC$ at points $E_{1}$ and $F_{1}$, respectively, and connect $A_{1}F$ and $C_{1}E$. Draw a perpendicular from $A_{1}$ to the extension... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given a regular 200-gon $A_{1} A_{2} \cdots A_{200}$, connect the diagonals $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, where $A_{i+200}=$ $A_{i}(i=1,2, \cdots, 9)$. Then these 200 diagonals have $\qquad$ different intersection points inside the regular 200-gon. | 8. 1600.
For each diagonal $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, $A_{i+1}, A_{i+2}, \cdots, A_{i+8}$ each draw two diagonals intersecting with $A_{i} A_{i+9}$. And each such intersection point is counted twice, so the number of different intersection points of these 200 diagonals inside the regular 200-gon does not exc... | 1600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the sum:
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{60}\right)+\left(\frac{2}{3}+\frac{2}{4}+\cdots+\frac{2}{60}\right)+ \\
\left(\frac{3}{4}+\frac{3}{5}+\cdots+\frac{3}{60}\right)+\cdots+\left(\frac{58}{59}+\frac{59}{60}\right)
\end{array}
$$ | $$
\begin{aligned}
= & \frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\cdots+ \\
& \left(\frac{1}{60}+\frac{2}{60}+\cdots+\frac{59}{60}\right) \\
= & \frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\cdots+\frac{59}{2} \\
= & \frac{1}{2}(1+2+\cdots+59) \\
= & \frac{1}{2} \times \fr... | 885 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the set
$$
M=\left\{x|5-| 2 x-3 \mid \in \mathbf{N}_{+}\right\} \text {. }
$$
Then the number of all non-empty proper subsets of $M$ is ( ).
(A) 254
(B) 255
(C) 510
(D) 511 | - 1. C.
Given $5-|2 x-3| \in \mathbf{N}_{+}$, i.e., $|2 x-3|=0,1, 2,3,4$, we have
$$
M=\left\{-\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{7}{2}\right\} .
$$
Since $M$ contains 9 elements, the number of all non-empty proper subsets of $M$ is $2^{9}-2=510$. | 510 | Algebra | MCQ | Yes | Yes | cn_contest | false |
II. (40 points) Let the number of ways to choose $k$ pairwise coprime numbers from the set $\{1,2, \cdots, 28\}$ be $T(k)$. Find $T(2)+T(3)+\cdots+T(12)$. | Obviously, the prime numbers in $\{1,2, \cdots, 28\}$ are 2, 3, $5,7,11,13,17,19,23$, a total of 9.
When $k \geqslant 11$, the $k$ numbers taken out, except for 1, must include at least 10 numbers. By the pigeonhole principle, there must be two numbers divisible by the same prime number, which means these two numbers ... | 15459 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Distribute 2010 red cards and 2010 white cards arbitrarily to 2010 players participating in the game, with each person receiving two cards. All players sit in a circle facing down. The game rule is that each operation requires each player to simultaneously follow the following principle: if a player h... | In 2009.
Let $n=2$ 010. A wheel (divided into $n$ sections) can be placed in a circle of $n$ players. Each player, according to the game rules, when the section they face contains a white card and they hold a red card, they place the red card in the section and take back the white card; in all other cases, it is assume... | 2009 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
The initial 267 four-digit number $w_{1}$ and the sum of its four digits is the four-digit number $w_{2}, w_{2}$ and the sum of its four digits is the four-digit number $w_{3}, w_{3}$ and the sum of its four digits is the four-digit number $w_{4}, w_{4}$ and the sum of its four digits is the four-digit number $w_{5}, w... | Let $w_{5-i}=a_{i} b_{i} c_{i} d_{i}(i=0,1, \cdots, 4)$. Then $1001 a_{0}+101 b_{0}+11 c_{0}+2 d_{0}=2009$.
Obviously, $a_{0}=1$ (otherwise, if $a_{0}=2$, then $101 b_{0}+11 c_{0}+2 d_{0}=7$, which only has the solution $b_{0}=c_{0}=0, d_{0}=\frac{7}{2}$, which is not an integer).
Thus, $101 b_{0}+11 c_{0}+2 d_{0}=1008... | 1921 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 1, it is known that rectangle $A B C D$ can be exactly divided into seven small rectangles of the same shape and size. If the area of the small rectangle is 3, then the perimeter of rectangle $A B C D$ is | $-1.19$.
Let the length of the small rectangle be $a$ and the width be $b$. Then $3a = 4b \Rightarrow a = \frac{4}{3}b \Rightarrow ab = \frac{4}{3}b^2 = 3$. Solving this, we get $b = \frac{3}{2}, a = 2$.
Therefore, the perimeter of rectangle $ABCD$ is $2(4a + b) = 8a + 2b = 16 + 3 = 19$. | 19 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange the numbers $1,2, \cdots, 10$ in a row in some order, such that the sum of any three consecutive numbers does not exceed $n$. Answer the following:
(1) When $n=10$, can it be arranged? Please explain your reasoning;
(2) When it can be arranged, what is the minimum value of $n$? | 8. (1) Suppose $n=10$ is already arranged. Then the sum of the last nine numbers is less than or equal to 30. Thus, the first number is not less than 25, which is a contradiction. Therefore, it cannot be arranged.
(2) With the same consideration as (1).
When $n=11,12,13,14$, none can be arranged.
When $n=15$, the sum o... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $x_{1}, x_{2}, \cdots, x_{n}$ take values 7 or -7, and satisfy
(1) $x_{1}+x_{2}+\cdots+x_{n}=0$;
(2) $x_{1}+2 x_{2}+\cdots+n x_{n}=2009$.
Determine the minimum value of $n$. | 9. First, when $n=34$,
$$
\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{16}=x_{18}=-7, \\
x_{17}=x_{19}=x_{20}=\cdots=x_{34}=7
\end{array}
$$
satisfies the conditions of the problem.
Second, divide both sides of condition (2) by 7, and let $y_{i}=\frac{x_{i}}{7}(i=1,2, \cdots, n)$. Then
$$
\begin{array}{l}
y_{i}= \pm 1, \\
y... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. (15 points) Fill seven different perfect squares into seven consecutive boxes so that the sum of any three adjacent boxes is greater than 100. Find the minimum possible value of the sum of these seven perfect squares.
| 3. Since the three perfect squares at positions $1, 4$, and $7$ are all different, considering the smallest possible values, one of these numbers must be at least 9. After positioning 9, the remaining six positions can be divided into two groups, each with three consecutive numbers whose sum is at least 101. Therefore,... | 211 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If the three-digit decimal number $n=abc$ satisfies that $a$, $b$, $c$ form an arithmetic sequence, then the maximum possible value of a prime factor of $n$ is $\qquad$ | 3. 317 .
$$
\begin{array}{l}
31 n, 999=27 \times 37, 987=3 \times 7 \times 47, \\
963=9 \times 107, 951=3 \times 317 .
\end{array}
$$ | 317 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. The sum of the radii of all circles passing through point $A(1505,1008)$ and tangent to the lines $l_{1}: y=0$ and $l_{2}: y=\frac{4}{3} x$ is $\qquad$ | 5. 2009 .
From $\tan 2 \theta=\frac{4}{3} \Rightarrow \tan \theta=\frac{1}{2}$, the center of the circle lies on the line $y=\frac{x}{2}$, and the center is $(2 r, r)$ (where $r$ is the radius of the circle).
$$
\begin{array}{l}
\text { Hence }(1505-2 r)^{2}+(1008-r)^{2}=r^{2} \\
\Rightarrow 4 r^{2}-8036 r+1505^{2}+10... | 2009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the lines $l_{1} / / l_{2}$, and take 10 points $A_{1}, A_{2}, \cdots, A_{10}$ and $B_{1}, B_{2}, \cdots, B_{10}$ on $l_{1}$ and $l_{2}$ respectively. Then the line segments $A_{1} B_{1}, A_{2} B_{2}, \cdots, A_{10} B_{10}$ can divide the strip region between $l_{1}$ and $l_{2}$ into at most $\qquad$ non-overlap... | 8. 56 .
A line segment divides the original region into two parts. The $k$-th line segment can be divided into at most $k$ segments by the previous $k-1$ line segments, and it can add at most $k$ parts. Therefore, $k$ line segments can divide the region into at most $2+2+3+\cdots+k=\frac{k^{2}+k+2}{2}$ parts. | 56 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n+2}}$. Then $\sum_{n=1}^{2010} a_{n}=$ $\qquad$ . | 4.5 829 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$;
When $n \equiv 2,4,8(\bmod 10)$,
$$
\begin{array}{l}
(n+1)^{n+2} \equiv 1(\bmod 4) \\
\Rightarrow a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10) ;
\end{array}
$$
When $n \equiv 3,7,9(\bmod 10)$,
$$
\begin{array}... | 5829 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the function $f:\{0,1, \cdots, 2010\} \rightarrow \mathbf{N}$. If for all possible integers $x$, we have
$$
\begin{array}{l}
f(4 x+2)=f(4 x+1), \\
f(5 x+3)=f(5 x+2), \\
f(7 x+5)=f(7 x+4),
\end{array}
$$
then $f(x)$ can take at most $\qquad$ different values. | 8. 1033 .
For the function $f$, construct the function $g$, defined as:
$$
\begin{array}{l}
g(0)=1, \\
g(i+1)=\left\{\begin{array}{l}
1, f(i+1) \neq f(i) ; \\
0, f(i+1)=f(i),
\end{array}\right.
\end{array}
$$
where $i=0,1, \cdots, 2009$.
Then $g$ is a mapping from $\{0,1, \cdots, 2010\}$ to $\{0,1\}$, and for all pos... | 1033 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $\left(2+x-2 x^{2}\right)^{1005}=\sum_{k=0}^{2010} a_{k} x^{k}$. Then
$$
\begin{array}{l}
a_{1}+3 a_{3}+5 a_{5}+\cdots+2009 a_{2009} \\
=
\end{array}
$$ | 3. 1005.
Let $f(x)=\left(2+x-2 x^{2}\right)^{1005}$. Then $f^{\prime}(x)=1005\left(2+x-2 x^{2}\right)^{1004}(1-4 x)$. Therefore, $f^{\prime}(1)=1005 \times(-3)$, $f^{\prime}(-1)=1005 \times 5$.
Given $f(x)=\sum_{k=0}^{2010} a_{k} x^{k}$, we have
$$
\begin{array}{l}
f^{\prime}(x)=\sum_{k=1}^{2010} k a_{k} x^{k-1} . \\
... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Try to find $\left[a_{2009}^{2}\right]$. | 11. From $a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}$, we get $a_{n+1}^{2}=\frac{a_{n}^{2}}{n^{2}}+\frac{n^{2}}{a_{n}^{2}}+2$. Given $a_{1}=1$, we have $a_{2}^{2}=4, a_{3}^{2}=4, a_{4}^{2}=4+\frac{25}{36}$.
Next, we use mathematical induction to prove:
When $n \geqslant 4$, $n+\frac{2}{n}\frac{n+1}{n^{2}}+\frac{n^{2}}{n+1... | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The positive numbers $b_{1}, b_{2}, \cdots, b_{60}$ are arranged in sequence and satisfy $\frac{b_{2}}{b_{1}}=\frac{b_{3}}{b_{2}}=\cdots=\frac{b_{60}}{b_{59}}$.
Determine the value of $\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right)$. | Given that $\left\{b_{k}\right\}$ is a geometric sequence with a common ratio of $q$, then $b_{k}=b_{1} q^{k-1}$. Therefore, $b_{k} b_{61-k}=b_{1} b_{60}$. Hence,
$$
\begin{array}{l}
\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right) \\
=\log _{b_{1} b_{60}}\left(b_{1} b_{60}\right)^{30}=30 .
\end{array}
$$ | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. $[x]$ represents the greatest integer not exceeding the real number $x$. If
$$
\left[\log _{3} 6\right]+\left[\log _{3} 7\right]+\cdots+\left[\log _{3} n\right]=2009 \text {, }
$$
determine the value of the positive integer $n$. | 4. For $3^{k} \leqslant i \leqslant 3^{k+1}-1$, we have $\left[\log _{3} i\right]=k$. Therefore, $\sum_{i=3^{k}}^{3 k+1}\left[\log _{3} i\right]=2 \times 3^{k} k$.
Also, $3+\sum_{k=2}^{4} 2 \times 3^{k} k<2009<3+\sum_{k=2}^{5} 2 \times 3^{k} k$, hence $3^{5}<n<3^{6}-1$.
From $3+\sum_{k=2}^{4} 2 \times 3^{k} k+5\left(n... | 474 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the 6027-digit number $\frac{a b c a b c \cdots a b c}{2000 \uparrow a b c}$ is a multiple of 91. Find the sum of the minimum and maximum values of the three-digit number $\overline{a b c}$. | 7. From $91=7 \times 13, 1001=7 \times 11 \times 13$, we know 91 | 1001.
And $\overline{a b c a b c}=1001 \times \overline{a b c}$, thus, $91 \mid \overline{a b c a b c}$. $2009 \uparrow a b c$
$$
\overline{a_{2009 \uparrow a b c}^{a b c a b c}}=\underset{2 \times 1004 \uparrow a b c}{\overline{a b c a b c \cdots a b c... | 1092 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A three-digit number has 3 digits, none of which are 0, and its square is a six-digit number that has exactly 3 digits as 0. Write down one such three-digit number: $\qquad$ | 8.448 or 548 or 949 | 448 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For a positive integer $n$, let $t_{n}=\frac{n(n+1)}{2}$. Writing down the last digits of $t_{1}=1, t_{2}=3, t_{3}=6, t_{4}=10, t_{5}=15 \cdots \cdots$ can form an infinite repeating decimal: $0.13605 \cdots$. Find the length of the repeating cycle of this decimal. | $$
\begin{array}{l}
t_{n+20}-t_{n}=\frac{(n+20)(n+20+1)}{2}-\frac{n(n+1)}{2} \\
=20 n+210=10(2 n+21)
\end{array}
$$
That is, the last digit of $t_{n+20}$ is the same as that of $t_{n}$.
Therefore, 20 is the length of the repeating cycle of this repeating decimal. This repeating decimal is
$$
0.13605186556815063100 .
$... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0
\end{array}
$$
$\left(a 、 b \in \mathbf{N}_{+}\right)$ have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^... | Given the known equations, we have $a \neq 1, b \neq 1$. Therefore, $a > 1, b > 1$, and $a \neq b$. Let $x_{0}$ be the common root of the two equations. It is easy to see that $x_{0} \neq 1$. By the definition of the roots of the equation, $a$ and $b$ are the two distinct real roots of the equation
$$
\left(1-x_{0}\rig... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given that there are only three positive integers between the fractions $\frac{112}{19}$ and $\frac{112+x}{19+x}$. Find the sum of all possible integer values of $x$. | Notice that $50$ or $x<-19$.
Therefore, the three positive integers between $\frac{112+x}{19+x}$ and $\frac{112}{19}$ are $3,4,5$.
From $2<\frac{112+x}{19+x}<3$, we get $\frac{55}{2}<x<74$.
Since $x$ is an integer greater than 0, thus, $x=28,29, \cdots, 73$.
Therefore, the sum of these numbers is
$$
\frac{(28+73) \time... | 2310 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
At first 272, from the eight points consisting of the vertices and midpoints of the sides of a square, how many isosceles triangles can be formed by selecting three points?
Will the above text be translated into English, please retain the original text's line breaks and format, and output the translation result direct... | Considering that there are no equilateral triangles that meet the requirements in this problem, we classify the isosceles triangles by their vertices.
(1) Using the vertices of the square as the vertices of the isosceles triangle. As shown in Figure 3, if $A$ is the vertex,
then $\triangle A G H$, $\triangle A B D$, an... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. How many pairs $(n, r)$ are there in the array satisfying $0 \leqslant r \leqslant n \leqslant 63$ for which the binomial coefficient $\mathrm{C}_{n}^{r}$ is even (assuming $\left.\mathrm{C}_{0}^{0}=1\right) ?$ | From Example 5, we know that the number of odd numbers in $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ is $2^{S(n)},$ where $S(n)$ is the sum of the binary digits of $n$.
Since $63=2^{6}-1=(111111)_{2}$, when $0 \leqslant n \leqslant 63$, we have $0 \leqslant S(n) \leqslant 6$.
Classifying and... | 1351 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Divide the numbers $1,2, \cdots, 200$ into two groups arbitrarily, each containing 100 numbers. Arrange one group in ascending order (denoted as $a_{1}<a_{2}<\cdots<a_{100}$) and the other in descending order (denoted as $b_{1}>b_{2}>\cdots>b_{100}$). Try to find
$$
\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\... | First, prove: For any term in the algebraic expression
$$
\left|a_{k}-b_{k}\right|(k=1,2, \cdots, 100)
$$
the larger number among $a_{k}$ and $b_{k}$ must be greater than 100, and the smaller number must not exceed 100.
(1) If $a_{k} \leqslant 100$ and $b_{k} \leqslant 100$, then by
$$
a_{1}b_{k+1}>\cdots>b_{100}
$$
w... | 10000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) For $n$ distinct positive integers, among any six numbers, there are at least two numbers such that one can divide the other. Find the minimum value of $n$ such that among these $n$ numbers, there must exist six numbers where one can be divided by the other five.
| The smallest positive integer $n=26$.
The proof is divided into two steps.
【Step 1】When $n \leqslant 25$, the condition is not satisfied.
Construct the following 25 positive integers:
(1) $2^{5}, 2^{4}, 2^{3}, 2^{2}, 2^{1}$;
(2) $3^{5}, 3^{4}, 3^{3}, 3^{2}, 3^{1}$;
(3) $5^{5}, 5^{4}, 5^{3}, 5^{2}, 5^{1}$;
(4) $7^{5}, 7... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Divide the 100 natural numbers $1, 2, \cdots, 100$ into 50 groups, each containing two numbers. Now, substitute the two numbers in each group (denoted as $a$ and $b$) into $\frac{1}{2}(|a-b|+a+b)$ for calculation, and obtain 50 values. Find the maximum value of the sum of these 50 values. | Since the 100 numbers from $1 \sim 100$ are all different, in each pair of numbers, there must be a larger number. Therefore, the result of the calculation is the larger of the two numbers. Hence, the maximum sum of these 50 values is
$$
51+52+\cdots+100=3775
$$ | 3775 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Let $a, b, x, y \in \mathbf{R}$, satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$
Find the value of $a x^{5}+b y^{5}$.
$(1990$, American Mathematical Invitational) | Solution 1:
From
$a x^{3}+b y^{3}$
we get
$$
=\left(a x^{2}+b y^{2}\right)(x+y)-(a x+b y) x y,
$$
thus
$$
\begin{array}{l}
16=7(x+y)-3 x y . \\
\text { From } a x^{4}+b y^{4} \\
=\left(a x^{3}+b y^{3}\right)(x+y)-\left(a x^{2}+b y^{2}\right) x y,
\end{array}
$$
From $a x^{4}+b y^{4}$
we get
$$
42=16(x+y)-7 x y \text {... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Mother's Day is here, and Xiao Hong, Xiao Li, and Xiao Ying go to a flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the above flowers. Then she... | 10. 20 .
Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $z$ yuan, respectively. Then
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text {. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text {. }
... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b$ be the roots of the equation $x^{2}+68 x+1=0$, and $c, d$ be the roots of the equation $x^{2}-86 x+1=0$. Then
$$
(a+c)(b+c)(a-d)(b-d)
$$
the value is $\qquad$. | 6. 2772 .
$$
\begin{array}{l}
\text { Given } a b=c d=1, a+b=-68, \\
c^{2}-8 b c+1=d^{2}-86 d+1=0, \\
\text { then }(a+c)(b+c)(a-d)(b-d) \\
=\left[a b+(a+b) c+c^{2}\right]\left[a b-(a+b) d+d^{2}\right] \\
=\left(1-68 c+c^{2}\right)\left(1+68 d+d^{2}\right) \\
=18 c \cdot 154 d=2772 .
\end{array}
$$ | 2772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The number of all integer solutions $(x, y, z)$ for the equation $x y z=2009$ is $\qquad$ groups. | 8. 72 .
First consider the integer solutions for $00$, it is easy to know that there are
$$
2 \mathrm{~A}_{3}^{1}+2 \mathrm{~A}_{3}^{3}=18
$$
sets of positive integer solutions.
For each set of positive integer solutions, adding two negative signs can yield 3 sets of integer solutions with two negatives and one posit... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (15 points) Given a positive integer $n$ that satisfies the following condition: among any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$.
| Because $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor no greater than 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17,19,23$,
$29,31,37,41,43$.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive integers greater than 1 and not exce... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. If the odd function $y=f(x)$ defined on $\mathbf{R}$ is symmetric about the line $x=1$, and when $0<x \leqslant 1$, $f(x)=\log _{3} x$, then the sum of all real roots of the equation $f(x)=-\frac{1}{3}+f(0)$ in the interval $(0,10)$ is . $\qquad$ | 5. 30 .
Given that the graph of the function $y=f(x)$ is symmetric about the line $x=1$, and $f(x)$ is an odd function, we have
$$
f(x+2)=f(-x)=-f(x) .
$$
Therefore, $f(x+4)=-f(x+2)=f(x)$, which means $f(x)$ is a periodic function with 4 as one of its periods.
Since $f(x)$ is an odd function defined on $\mathbf{R}$, ... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Satisfy $0 \leqslant k_{i} \leqslant 20(i=1,2,3,4)$, and $k_{1}+k_{3}=k_{2}+k_{4}$ of the ordered integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ the number is $\qquad$ . | 10. 6181.
For $0 \leqslant m \leqslant 20$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(0 \leqslant j \leqslant m),
$$
there are $m+1$ solutions;
When $20<m \leqslant 40$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \... | 6181 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\}(i=1,2, \cdots, 2010)$. Let $S=x_{1} x_{2}+x_{3} x_{4}+\cdots+x_{2009} x_{2010}$.
(1) Can $S$ be equal to 2010? Prove your conclusion;
(2) How many different integer values can $S$ take? | 14. (1) Since $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
\begin{array}{l}
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, \\
(\sqrt{2}-1)(\sqrt{2}+1)=1
\end{array}
$$
Therefore, $x_{2 i-1} x_{2 i} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ terms of $3+2 \sqrt{2}$, $b$ terms of $3-2 \sqrt{2}$, and $c$ terms of 1. Then ... | 503 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a$, $b$, $x$, $y$ satisfy $x y=2008^{2008}$,
$$
\frac{1}{1+2008^{a} x}+\frac{1}{1+2008^{b-2009} y}=1 \text {. }
$$
Then the value of $2008^{a+b}$ is | Ni, 1.2008.
Given the equation, after removing the denominator, we get
$$
\begin{array}{l}
1+2008^{b-2009} y+1+2008^{a} x \\
=\left(1+2008^{a} x\right)\left(1+2008^{b-2009} y\right) .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
2008^{a+b-2009} x y=1 \\
\Rightarrow 2008^{a+b-2009} \times 2008^{2008}=1 \\
\Ri... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Function
$$
f(x)=27^{x}-3^{x+3}+1
$$
The minimum value of the function on the interval $[0,3]$ is $\qquad$ | - 1. -53.
Let $t=3^{x}(x \in[0,3])$. Then $f(x)=g(t)=t^{3}-27 t+1(t \in[0,27])$. And $g^{\prime}(t)=3 t^{2}-27=3(t-3)(t+3)$, so when $t \in[1,3]$, $g^{\prime}(t)0, g(t)$ is monotonically increasing.
Therefore, when $t=3$, $g(t)$ reaches its minimum value
$$
g(t)_{\min }=g(3)=-53,
$$
which means when $x=1$, $f(x)$ tak... | -53 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right] \text {, }
$$
where, $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
a_{2009}+a_{2010}=
$$ | 2. 2009.
Given $a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}$.
We will prove by mathematical induction:
$$
a_{n+2}-a_{n}=1, a_{n}+a_{n+1}=n \text {. }
$$
Obviously, when $n=1$, the conclusion holds.
Assume when $n=k$, the conclusion holds, i.e.,
$$
a_{k+2}-a_{k}=1, a_{k}+a_{k+1}=k \text {. }
$$
Then when ... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If the set $A=\{x \mid x=6 n-1, n \in \mathbf{N}\}$,
$$
B=\{x \mid x=8 n+3, n \in \mathbf{N}\},
$$
then the number of elements in $A \cap B$ that are less than 2010 is $\qquad$ | 3. 84 .
According to the problem, if $x \in A$, then $x \equiv 5(\bmod 6)$; if $x \in B$, then $x \equiv 3(\bmod 8)$. Therefore, if $x \in A \cap B$, then $x \equiv 11(\bmod 24)$, i.e., $x=24 k+11(k \in \mathbf{N})$.
Thus, there are 84 elements that satisfy the condition. | 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given a positive integer $n$ such that the last two digits of $3^{n}$ form a two-digit prime number. Then the sum of all $n$ that satisfy this condition and do not exceed 2010 is $\qquad$ . | $-, 1.909128$.
Considering $3^{k}(k=1,2, \cdots)$ modulo 100, the remainders are sequentially $3,9,27,81,43,29,87,61,83,49,47,41$, $23,69,7,21,63,89,67,1 ; 3,9, \cdots$,
where $43,29,61,83,47,41,23,89,67$ are two-digit primes, corresponding to
$$
k \equiv 5,6,8,9,11,12,13,18,19(\bmod 20) .
$$
Thus, the sum of all $n$ ... | 909128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
[\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}]
$$
is equal to
(there are a total of 2009 square roots). | 2. 45 .
Let $A=\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}$.
Then $A>\sqrt{2010+\sqrt{2009}}$
$$
\begin{array}{l}
>\sqrt{2010+44}>45, \\
A<\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{2009}}}} \\
<\sqrt{2010+46}<46 .
\end{array}
$$
Thus, $[A]=45$. | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Select 1005 numbers $a_{1}, a_{2}, \cdots, a_{1005}$ from 1 to 2010 such that their total sum is 1006779, and the sum of any two of these 1005 numbers is not equal to 2011.
(1) Prove that $\sum_{i=1}^{1005} a_{i}^{2}-4022 \sum_{i=1}^{1005} a_{i}^{3}+\sum_{i=1}^{1005} a_{i}^{4}$ is a constant;
(2) When... | Divide $\{1,2, \cdots, 2010\}$ into 1005 groups:
$$
A_{i}=\{i, 2011-i\}(i=1,2, \cdots, 1005) .
$$
Since the sum of any two numbers in $\left\{a_{i}\right\}$ is not equal to 2011, exactly one number is taken from each group. First, take the even numbers from each group to form
$\left\{b_{i}\right\}$, where $b_{i} \in A... | 44253 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its divisors 1, $2,17,34,59,118,1003,2006$ is 3240. Find the smallest good number. ${ }^{[3]}$
(... | \left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)=8.
Therefore, when $k=1$, $\alpha_{1}=7$;
when $k=2$, $\alpha_{1}=1, \alpha_{2}=3$ or $\alpha_{1}=3, \alpha_{2}=1$;
when $k=3$, $\alpha_{1}=\alpha_{2}=\alpha_{3}=1$;
when $k \geqslant 4$, there is no solution.
(1) If $n=p^{7}$ ( $p$ is a... | 1614 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. There is a bus, a truck, and a car each traveling in the same direction at a constant speed on a straight road. At a certain moment, the bus is in front, the car is at the back, and the truck is exactly in the middle between the bus and the car. After $10 \mathrm{~min}$, the car catches up with the truck; after anot... | 7. 15 .
Suppose at a certain moment, the distances between the truck and the bus, and the truck and the car are both $S \mathrm{~km}$. The speeds of the car, truck, and bus are $a, b, c(\mathrm{~km} / \mathrm{min})$, respectively, and it takes the truck $x \mathrm{~min}$ to catch up with the bus.
From the problem, we ... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. B. Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$
Find the minimum value of $u=9 a^{2}+72 b+2$. | 11. B. From
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {, }
$$
we get $(a-2 b)(3 a-4 b+5)=0$.
Therefore, $a-2 b=0$ or $3 a-4 b+5=0$.
(1) When $a-2 b=0$,
$$
\begin{array}{l}
u=9 a^{2}+72 b+2=36 b^{2}+72 b+2 \\
=36(b+1)^{2}-34 .
\end{array}
$$
Thus, when $b=-1$, the minimum value of $u$ is -34.
(2) When $3 a-4 b+5=0$,
... | -34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A. From the 2010 positive integers $1,2, \cdots, 2010$, what is the maximum number of integers that can be selected such that the sum of any three selected numbers is divisible by 33? | 14. A. First, the following 61 numbers: $11, 11+33, 11+2 \times 33, \cdots, 11+60 \times 33$ (i.e., 1991) satisfy the conditions of the problem.
On the other hand, let $a_{1}<a_{2}<\cdots<a_{n}$ be the numbers selected from 1, 2, $\cdots, 2010$ that satisfy the conditions of the problem. For any four numbers $a_{i}, a... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If the sum of $k$ consecutive positive integers is 2010, then the maximum value of $k$ is | Ni. 1.60 .
Let $2010=(n+1)+(n+2)+\cdots+(n+k)$.
Then $k(2 n+k+1)=4020$.
Notice that $k<2 n+k+1$, and
$$
4020=2^{2} \times 3 \times 5 \times 67 \text {, }
$$
To maximize the value of $k$, 4020 should be expressed as the product of the closest pair of factors, which is $4020=60 \times 67$.
Thus, $k_{\max }=60$. | 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. For the cyclic quadrilateral $ABCD$, the lengths of the four sides in sequence are $AB=2, BC=7, CD=6, DA=9$. Then the area of the quadrilateral is $\qquad$ . | 3. 30 .
Since $7^{2}+6^{2}=85=9^{2}+2^{2}$, that is,
$$
B C^{2}+C D^{2}=D A^{2}+A B^{2} \text {, }
$$
thus, $\triangle B C D$ and $\triangle D A B$ are both right triangles.
Therefore, the area of the quadrilateral is
$$
S_{\triangle B C D}+S_{\triangle D A B}=\frac{1}{2}(7 \times 6+9 \times 2)=30 \text {. }
$$ | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In $\pm 1 \pm 2 \pm 3 \pm 5 \pm 20$, by appropriately choosing + or -, different algebraic sums can be obtained $\qquad$.
| 4.24.
Among $1,2,3,5,20$, there are three odd numbers, so their algebraic sum must be odd.
Observing, we see that from $1,2,3,5$ we can obtain all odd numbers with absolute values not exceeding 11.
According to the problem, the expression must include 1, 2, 3, 5, and 20. Therefore, the integers that can be obtained ... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a_{1}, a_{2}, \cdots, a_{10} \in(1,+\infty)$. Then
$$
\frac{\log _{a_{1}} 2009+\log _{a_{2}} 2009+\cdots+\log _{a_{10}} 2009}{\log _{a_{1,2} \cdots a_{10}} 2009}
$$
the minimum value is | -1.100 .
$$
\begin{array}{l}
\text { Original expression }=\left(\sum_{i=1}^{10} \frac{\lg 2009}{\lg a_{i}}\right) \frac{\lg \left(\prod_{i=1}^{10} a_{i}\right)}{\lg 2009} \\
=\left(\sum_{i=1}^{10} \frac{1}{\lg a_{i}}\right)\left(\sum_{i=1}^{10} \lg a_{i}\right) \\
\geqslant 10^{2},
\end{array}
$$
Equality holds if an... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $A$ be the set of all positive integers not exceeding 2009, i.e., $A=\{1,2, \cdots, 2009\}$, and let $L \subseteq A$, where the difference between any two distinct elements of $L$ is not equal to 4. Then the maximum possible number of elements in the set $L$ is | 7. 1005.
Divide the set $A$ into the following 1005 subsets:
$$
\begin{array}{l}
A_{4 k+i}=\{8 k+i, 8 k+i+4\}(i=1,2,3,4 ; \\
k=0,1, \cdots, 250), \\
A_{1005}=\{2009\} .
\end{array}
$$
If the number of elements in $L$ is greater than 1005, then at least one of the first 1004 subsets is a subset of $L$, meaning there ... | 1005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. In a plane, given a convex decagon and all its diagonals, in such a graph, the number of triangles that have at least two vertices as vertices of the convex decagon is $\qquad$ (answer with a number). | 8. 960 .
The number of triangles with all three vertices being vertices of a convex decagon is $\mathrm{C}_{10}^{3}$.
The number of triangles with only two vertices being vertices of the convex decagon, and the other vertex being the intersection of two diagonals, can be determined as follows: Two diagonals define fo... | 960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (16 points) Let $A$ and $B$ be two different subsets of the set $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$, such that $A$ is not a subset of $B$, and $B$ is not a subset of $A$. Find the number of different ordered pairs $(A, B)$. | 11. The set $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ has $2^{5}$ subsets, and the number of different ordered pairs $(A, B)$ is $2^{5}\left(2^{5}-1\right)$.
If $A \subset B$, and suppose $B$ contains $k(1 \leqslant k \leqslant 5)$ elements. Then the number of ordered pairs $(A, B)$ satisfying $A \subset B$ i... | 570 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The body diagonal of a rectangular prism is 10, and the projection of this diagonal on one of the surfaces of the prism is 8. Then the maximum volume of this rectangular prism is $\qquad$ . | 4. 192 .
According to the problem, the height of the cuboid is $\sqrt{10^{2}-8^{2}}=6$. Let the side lengths of the base of the cuboid be $a$ and $b$. Then $a^{2}+b^{2}=64$.
Thus, the volume of the cuboid is $V=6 a b \leqslant 3\left(a^{2}+b^{2}\right)=192$.
The equality holds if and only if $a=b=4 \sqrt{2}$. Therefor... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $x, y$ are positive integers, and $x y+x+y=23, x^{2} y+x y^{2}=120$. Then $x^{2}+y^{2}=$ $\qquad$ | Solve: From $x y+x+y=23$, we can set $x y=\frac{23}{2}+t, x+y=\frac{23}{2}-t$.
Multiplying the two equations gives
$$
x^{2} y+x y^{2}=\left(\frac{23}{2}\right)^{2}-t^{2}=120 \text{. }
$$
Solving for $t$ yields $t= \pm \frac{7}{2}$.
Since $x$ and $y$ are both positive integers, when $t=\frac{7}{2}$, we have $x+y=8, x y... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, $c$, and $d$ are prime numbers, and $a b c d$ is the sum of 77 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ | Given 77 consecutive positive integers, with the middle one being $x$. Then $a b c d=77 x=7 \times 11 x$, and $x \geqslant 39$.
Since $a, b, c, d$ are prime numbers, thus, $x$ can be decomposed into the product of two prime numbers, and the sum of these two prime numbers is minimized.
It is easy to verify that when $x... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) From the natural numbers $1, 2, \cdots, 2010$, take $n$ numbers such that the sum of any three of the taken numbers is divisible by 21. Find the maximum value of $n$.
| $$
\begin{array}{l}
\Rightarrow\left(x^{2}+16+\frac{64}{x^{2}}\right)-\left(8 x+\frac{64}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}\right)^{2}-8\left(x+\frac{8}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}-\frac{17}{3}\right)\left(x+\frac{8}{x}-\frac{7}{3}\right)=0 \\
\Rightarrow x=\fra... | 96 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $A \cup B=\{1,2, \cdots, 10\},|A|=|B|$. Then the number of all possible ordered pairs of sets $(A, B)$ is | - 1. 9953.
If $|A|=k(k=5,6, \cdots, 10)$, then $A$ has $\mathrm{C}_{10}^{k}$ possibilities.
For each possibility of $A$, since $A \cup B=\{1,2, \cdots, 10\}$, we know that $C_{A \cup B} A \subseteq B$.
Thus, $B$ already has $10-k$ elements.
To ensure $|A|=|B|$, it is necessary to select $2 k-10$ elements from $A$ to ... | 8953 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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