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1. Find the smallest positive integer $n$, such that
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2002}
$$
has integer solutions.
(Uzbekistan provided) | Solution: Since $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), 2002$ $=667 \times 3+1$, therefore,
$$
2002^{2002} \equiv 4^{2002} \equiv 4(\bmod 9) \text {. }
$$
Also, $x^{3} \equiv 0, \pm 1(\bmod 9)$, where $x$ is an integer, thus,
$$
x_{1}^{3}, x_{1}^{3}+x_{2}^{3}, x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \equiv 4(\bmod 9) ... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that there are exactly 600 integer-sided triangles with unequal sides and the longest side exactly $n$. Find the value of $n$. | Explanation: The key to solving this problem is to establish an equation about $n$. Note that one side of the triangle, $n$, is fixed, and the three sides $x, y, n$ are all integers. Therefore, the number of such triangles is equal to the number of lattice points $(x, y)$ in the coordinate plane.
Let the lengths of th... | 51 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. For all non-negative integers $x, y$, find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, satisfying $f(3 x+2 y)=f(x) f(y)$, where $\mathbf{N}$ is the set of non-negative integers.
(53rd Romanian Mathematical Olympiad (Final)) | Solution: Let $x=y=0$, we get $f(0)=f(0)^{2}$. Therefore, $f(0)=0$ or $f(0)=1$.
If $f(0)=0$, for $x=0$ or $y=0$, we get $f(2 y)=$ $f(3 x)=0$ for all $x, y \in \mathrm{N}$. Let $f(1)=a$, then
$$
\begin{array}{l}
f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\
f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} .
\end{array... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Among all triangles with side lengths as consecutive positive integers and a perimeter not exceeding 100, the number of acute triangles is | (Tip: Establish an inequality relationship, determine the range of side lengths. Answer: 29.) | 29 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Consider a square on the complex plane, whose 4 vertices correspond to the 4 roots of a certain monic quartic equation with integer coefficients $x^{4}+p x^{3}+q x^{2}+r x+s=0$. Find the minimum value of the area of such a square.
| 3. According to the problem, the 4 roots of the equation can only be in two scenarios: 2 real roots and 1 pair of conjugate complex roots; 2 pairs of conjugate complex roots.
(1) If the 4 roots of the equation are 2 real roots and 1 pair of conjugate complex roots, then we can set these 4 roots as $a \pm b, a \pm b \ma... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find a point $P$ on the plane of equilateral $\triangle A B C$ such that $\triangle P A B$, $\triangle P B C$, and $\triangle P C A$ are all isosceles triangles. How many points $P$ with this property are there? | Solution: Let any side of the known triangle be denoted as $a$. When $a$ is the base of the sought isosceles triangle, point $P$ lies on the perpendicular bisector of $a$; when $a$ is one of the legs of the sought isosceles triangle, point $P$ lies on the circumference of the circle with the vertex of $\triangle ABC$ a... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given an $n \times n$ ($n$ is an odd number) chessboard where each unit square is colored in a checkerboard pattern, and the 4 corner unit squares are colored black. A figure formed by 3 connected unit squares in an L-shape is called a "domino". For what value of $n$ can all the black squares be covered by non-overl... | Solution: Let $n=2m+1$, consider the odd rows, then each row has $m+1$ black cells, with a total of $(m+1)^2$ black cells. Any two black cells cannot be covered by a single "domino", therefore, at least $(m+1)^2$ "dominoes" are needed to cover all the black cells on the chessboard. Since when $n=1,3,5$, we have $3(m+1)... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 As shown in Figure 4, in a $4 \times 4$ grid square,
construct a grid point $\triangle A B C$
$(A B=\sqrt{5}, B C=$
$\sqrt{13}, C A=\sqrt{10}$ ). How many
grid point triangles congruent to
$\triangle A B C$ (including
$\triangle A B C)$ can be
constructed in Figure 4? | (1) As shown in Figure 5, $\triangle ABC$ is an inscribed lattice triangle in a $3 \times 3$ square, and there are 4 such $3 \times 3$ squares in Figure 4.
(2) As shown in Figure 5, one vertex of the triangle is at a vertex of the $3 \times 3$ square (point $C$, which has 4 possible positions), and the other two vertic... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $T$ be a set of ordered triples $(x, y, z)$, where $x, y, z$ are integers, and $0 \leqslant x, y, z \leqslant 9$. Two players, A and B, play the following game: A selects a triple $(x, y, z)$ from $T$, and B has to guess A's chosen triple using several "moves". One "move" consists of: B giving A a triple $(a, b,... | Solution: Two "movements" are not enough. Because each answer is an even number between 0 and 54, i.e., there are 28 possible values for each answer. The maximum number of possible results from two "movements" is $28^{2}$, which is less than the 1000 possible values for $(x, y, z)$.
Below, we prove that 3 "movements" ... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
7. There are 120 people. Any two people are either friends or not friends. A set of four people that contains exactly one pair of friends is called a "weak quartet." Find the maximum number of "weak quartets." (Provided by New Zealand)
| Solution: Let the 120 people be 120 points in graph $G$. If two people know each other, then connect the points corresponding to these two people with a line segment. Let $Q(G)$ be the number of "weak quartets" in graph $G$. If $x, y$ are two points in $G$, and there is a line segment between $x, y$, let $G^{\prime}$ s... | 4769280 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
33. The function $f: \mathbf{N} \rightarrow \mathbf{N} (\mathbf{N}$ being the set of non-negative integers), satisfies:
(1) For any non-negative integer $n$,
$$
f(n+1)>f(n) \text{; }
$$
(2) For any $m, n \in \mathbf{N}$,
$$
f(n+f(m))=f(n)+m+1 \text{. }
$$
Find the value of $f(2001)$.
(2001, British Mathematical Olym... | Solution: Let $f(0)=k$, where $k$ is a non-negative integer. From (2), we have
$$
f(n+k)=f(n)+1 \text {. }
$$
If $k=0$, then $f(n)=f(n)+1$, which is a contradiction. Therefore, $k \neq 0$.
According to (1), we have
$$
f(n+k-1) \cdot 1 \text{, if } k>1, \text{ then } n+k-1 \geqslant n+1. \text{ Thus, by (1), we get }
$... | 2002 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Real numbers $x_{1}, x_{2}, \cdots, x_{2} 0001$ satisfy
$$
\begin{array}{c}
\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001, \\
\text { let } y_{k}=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2, \cdots,
\end{array}
$$
2001 . Find the maximum possible value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\r... | Explanation: Since the relationship between $x_{k}$ and $x_{k+1}$ is unknown, the difference $x_{k}-x_{k+1}$ can be treated as a whole, and the condition $\sum_{k=1}^{2000} |x_{k}-x_{k+1}|=2001$ can be viewed as a constraint on $x_{k}-x_{k+1}$. By making the substitution $a_{0}=x_{1}, a_{k}=x_{k+1}-x_{k}, k=1,2, \cdots... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 For a quadrilateral, there can be 2 ways to dissect it into triangles, $a_{4}=2$ (Figure 15); for a pentagon, there can be 5 ways to dissect it into triangles, $a_{5}=$ 5 (Figure 16). How many ways are there to dissect a hexagon into triangles? | Solution: Consider the triangles containing the edge $A_{1} A_{2}$, using $\triangle A_{1} A_{2} A_{k}(k=3,4,5,6)$ as a standard, the figure can be divided into 4 categories.
(1) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{3}$, we can take out $\triangle A_{1} A_{2} A_{3}$, leaving the pentagon ... | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}$. Then the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{200 B}+1}$ is $\qquad$ | 4. 1 .
From the problem, we know that $\left\{x_{n}\right\}$ is an increasing sequence, and $x_{3}>1$. From $\frac{1}{x_{k+1}}=\frac{1}{x_{k}\left(x_{k}+1\right)}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, we get $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$.
Then $S=\sum_{k=1}^{2008} \frac{1}{x_{k}+1}=\sum_{k=1}^{20... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) There are 800 points on a circle, labeled $1, 2, \cdots, 800$ in a clockwise direction. They divide the circumference into 800 gaps. Now, choose one point and color it red, then proceed to color other points red according to the following rule: if the $k$-th point has been colored red, then move $k$ ... | Consider a circle with $2n$ points.
There are two cases:
(1) On the circle with $2n$ points, if the first red point is an even-numbered point, for example, the $2k$-th point, then according to the coloring rule, each red point that is subsequently colored will also be an even-numbered point. At this time, if the 2nd, 4... | 25 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Given an infinite sequence $\left\{a_{n}\right\}$ where all terms are positive integers, and the sum of any consecutive terms is not equal to 100. Find the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$. | Solution: The minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
First, we prove: If all terms of the sequence $\left\{a_{n}\right\}$ are 1 or 2, then there must exist a continuous subsequence whose sum equals 100.
Consider the first 100 terms of the sequence $\left\{a_{n}\right\}$, and let
$$
S_{0}=0... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
131 Given that the function $f(n)$ is a strictly increasing function defined on $\mathbf{N}_{+}$, with its range also in $\mathbf{N}_{+}$, and satisfies $f(f(n))=3n$. Find $f(2003)$. | Solution: (1) First, $f(1) \neq 1$. Otherwise, on one hand, $f(f(1))=f(1)=1$, and on the other hand, according to the given condition, $f(f(1))=3 \times 1=3$. This is a contradiction.
Second, $f(1) \neq 3$. This contradicts the known condition $f(f(1))=3 \times 1$.
In summary, $f(1)=2$ (since $f(n) \in \mathbf{N}_{+}$... | 3822 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
The second question: Let the three sides of a triangle be integers $l$, $m$, and $n$, and $l>m>n$. It is known that $\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=$ $\left\{\frac{3^{n}}{10^{4}}\right\}$, where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding $x$. Find the... | Proof: Since $l, m, n$ are the lengths of the sides of a triangle, hence $m+n>l$.
From the given information,
$$
3^{l-n} \equiv 3^{m-n} \equiv 3^{l-m} \equiv 1\left(\bmod 10^{4}\right).
$$
Without loss of generality, let $\left\{\begin{array}{l}l-m=s, \\ m-n=t\end{array}\left(s, t \in \mathbf{N}_{+}\right)\right.$, th... | 3003 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 As shown in Figure 22, use 6 different colors to color the 4 different regions $A, B, C, D$, requiring that adjacent regions be colored with different colors. How many different coloring methods are there? | Solution 1: Complete in four steps according to the order of $A-B-C-D$.
(1) Color region $A$ with no restrictions, 6 colors can be used, giving 6 ways to color.
(2) Color region $B$, which must be a different color from $A$, only 5 remaining colors can be used, giving 5 ways to color.
(3) Color region $C$, which must b... | 600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.8 Try to find the largest positive integer $N$, such that no matter how the positive integers from 1 to 400 are filled into the cells of a $20 \times 20$ grid, one can always find two numbers in the same row or the same column whose difference is not less than $N$. | 10.8209.
First, we illustrate that $N \leqslant 209$. Divide the grid into two $20 \times 10$ grids using a vertical line through the center. Fill the numbers from 1 to 200 into the left grid in increasing order row by row, and fill the numbers from 201 to 400 into the right grid in the same manner. This way, the maxi... | 209 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the unit disk be $D=\left\{(x, y) \in \mathbf{R}^{2} \mid x^{2}+y^{2} \leqslant\right.$ 1\}, and define the width of a strip formed by two parallel lines as the distance between these lines. If the unit disk can be covered by some strips, prove: the sum of the widths of these strips is at least 2.
(19th Iranian ... | Solution: Consider the unit sphere that contains this unit circle. Let the unit circle belong to plane II. For each strip, replace the two parallel lines that form this strip with two planes that pass through these lines and are perpendicular to plane $I$, then for each strip, we can obtain two parallel planes, which w... | 2 | Geometry | proof | Yes | Yes | cn_contest | false |
8. As shown in Figure $3, A C=$ $B C, A C \perp B C$ at point $C, A B=A D=B D$, $C D=C E=D E$. If $A B=\sqrt{2}$, then $B E=$ | 8. 1 .
In $\triangle A D C$ and $\triangle B D C$, given $A D=D B, D C=D C$, $A C=B C$, we can conclude that
$\triangle A D C \cong \triangle B D C, \angle A D C=\angle B D C$.
Since $\angle A D B=60^{\circ}$, it follows that,
$\angle A D C=\angle B D C=\angle E D B=30^{\circ}$.
Therefore, $D B \perp C E, B C=B E$.
Mo... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that the real number $a$ satisfies $a^{2}-a-1=0$. Then the value of $a^{8}+7 a^{-4}$ is - $\qquad$ | 10.48 .
From $a^{2}-a-1=0$, we get $a-a^{-1}=1$, then, $a^{2}+a^{-2}$
$$
\begin{aligned}
=3, & a^{4}+a^{-4}=7 . \\
& a^{8}+7 a^{-4}=a^{4}\left(a^{4}+a^{-4}\right)+7 a^{-4}-1 \\
& =7\left(a^{4}+a^{-4}\right)-1=7 \times 7-1=48 .
\end{aligned}
$$ | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Observe the array: $(1),(3,5),(7,9,11),(13,15,17$,
19), $\cdots \cdots$. Then 2003 is in the group. | 11.45 .
Using the trial method.
Assume 2003 is in the 50th group, then the first 50 groups have
$$
1+2+3+\cdots+50=51 \times 25=1275
$$
numbers, the last number in this group is 2549, and the first number is 2451, so 2003 is not in the 50th group;
Assume 2003 is in the 40th group, then the first 40 groups have
$$
1+2+... | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 2, with the sides $AB$ and $AD$ of the square $ABCD$ as diameters, semicircles are drawn outside the square. A line is drawn through $A$ intersecting the two semicircles at $E$ and $F$. If the area of the square $ABCD$ is $1997$, and $AE$ and $AF$ are integers, then the length of $EF$ is ( ).
(A) ... | 5.A.
Connect $B E, D F$, then $\angle B E A=\angle A F D=90^{\circ}$.
Also, $\angle E B A+\angle E A B=90^{\circ}, \angle E A B+\angle F A D=90^{\circ}$, hence $\angle E B A=\angle F A D$.
Since $A B=A D$, thus,
$\triangle B E A \cong \triangle A F D, D F=A E$.
Let $A E=x, A F=y$, then
$x^{2}+y^{2}=A D^{2}=1997$ (wher... | 63 | Geometry | MCQ | Yes | Yes | cn_contest | false |
2. As shown in Figure 3, $A_{0} A_{1}$ is the diameter of a semicircle, $A_{0} A_{1}=2$, $A_{2}, A_{3}, \cdots, A_{k}$, $A_{k+1}, \cdots$, are points on the semicircle, $\angle A_{0} A_{1} A_{2}=1^{\circ}, \angle A_{1} A_{2} A_{3}=2^{\circ}$, $\angle A_{2} A_{3} A_{4}=3^{\circ}, \cdots, \angle A_{k-1} A_{k} A_{k+1}=k^{... | 2. 11 .
As shown in Figure 6, connect
$$
A_{k} O, A_{k+1} O, A_{k} A_{1} \text {, }
$$
then
$$
\begin{array}{l}
\angle O A_{k} A_{k+1}= \\
\angle A_{1} A_{k} A_{k+1}+\angle A_{1} A_{k} O \\
=\angle A_{1} A_{k} A_{k+1}+\angle A_{k} A_{1} O=\left(\frac{k(k+1)}{2}\right)^{\circ} .
\end{array}
$$
Since $A_{k} A_{k+1}60^... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $a, b, c$ are real numbers. The functions are $y_{1} = a x^{2} + b x + c, y_{2} = a x + b (a > 0)$. When $-1 \leqslant x \leqslant 1$, it is given that $-1 \leqslant y_{1} \leqslant 1$ and $y_{2}$ has a maximum value of 2. Try to find the area of the figure formed by connecting in sequence... | Three, from $a>0$, we know that $y_{2}$ increases as $x$ increases.
Therefore, $a+b=2$.
When $x=0,1$, $-1 \leqslant c \leqslant 1, -1 \leqslant a+b+c \leqslant 1$, so $-1 \leqslant c=(a+b+c)-2 \leqslant 1-2 \leqslant-1$.
Thus, $c=-1$. Therefore, when $x=0$, $y_{1}=-1$ is the minimum value of $y_{1}=$ $a x^{2}+b x+c$ in... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $\lg a<0, \lg b<0, \lg c<0$, and $\lg (a+b+c)=0$. Then the maximum value of $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. 0 .
Given that $a, b, c$ are positive numbers less than 1, and $a+b+c=1$. We have
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$$
$\geqslant 9$ (Cauchy-Schwarz inequality).
The equality holds when $a=b=c=\frac{1}{3}$.
Transforming the inequality (1), we get
$$
\begi... | 0 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
6. The sum of $n$ consecutive natural numbers starting from the positive integer $m$ is 2004, and $(m, n)>1$ (not coprime). Then the greatest common divisor $(m, n, 2004)=$
| 6. 12 .
According to the problem, $(m, n)>1$ and $m+(m+1)+\cdots+(m+n-1)=2004$, that is,
$$
\frac{(2 m+n-1) n}{2}=2004 \text {. }
$$
And $(2 m+n-1) n$
$$
=1 \times 4008=3 \times 1336=167 \times 24=501 \times 8 \text {, }
$$
where $2 m+n-1$ and $n$ are one odd and one even, and $2 m+n-1$ is greater than $n$. Therefor... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Try to find a positive integer $k$ other than 1, such that $k$ and $k^{4}$ can both be expressed as the sum of squares of two consecutive integers, and prove that such a $k$ is unique.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation ... | $$
\begin{array}{l}
\text { Solution: Let } k=n^{2}+(n+1)^{2} . \\
\text { By }\left(a^{2}+b^{2}\right)^{2}=\left(a^{2}-b^{2}\right)^{2}+(2 a b)^{2}, \text { we have } \\
k^{2}=(2 n+1)^{2}+[2 n(n+1)]^{2}, \\
k^{4}=\left(4 n^{4}+8 n^{3}-4 n-1\right)^{2}+ \\
\quad\left(8 n^{3}+12 n^{2}+4 n\right)^{2} . \\
\text { Let }\l... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, let $M$ be a set of $n$ points in the plane, satisfying:
(1) There exist 7 points in $M$ that are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it.
Find the minimum... | Three, prove $n \geqslant 11$.
Consider a convex heptagon $A_{1} A_{2} \cdots A_{7}$ with vertices in $M$, and connect $A_{1} A_{5}$. By condition (2), there is at least one point in $M$ within the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, denoted as $P_{1}$. Connect $P_{1} A_{1}$ and $P_{1} A_{5}$. Then, in the... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. A class has 47 students, and the classroom has 6 rows, with 8 seats in each row. The seat located in the $i$th row and the $j$th column is denoted as $(i, j)$. For the new semester, the seats are to be rearranged. If a student's original seat is $(i, j)$ and the new seat is $(m, n)$, then the student's movement is d... | 2. Let the empty seat last semester be $\left(i_{0}, j_{0}\right)$, and the empty seat this semester be $\left(i_{1}, j_{1}\right)$.
Then
$$
\begin{aligned}
S= & {\left[\sum_{i=1}^{6} \sum_{j=1}^{8}(i+j)-\left(i_{0}+j_{0}\right)\right] } \\
& -\left[\sum_{i=1}^{6} \sum_{j=1}^{8}(i+j)-\left(i_{1}+j_{1}\right)\right] \\... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Does there exist a natural number $n$, such that the sum of the digits of $n^{2}$ equals: (1) 2003; (2) 2002. | Solution: (1) Since the remainder of any natural number divided by 9 is equal to the remainder of the sum of its digits divided by 9, and $n^{2} \equiv$ $0,1,4,7(\bmod 9)$, while $2003 \equiv 5(\bmod 9)$, there does not exist such a natural number $n$ that the sum of the digits of $n^{2}$ equals 2003.
$$
\begin{array}{... | 2002 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Place the numbers $1,2,3,4,5,6,7,8$ on the eight vertices of a cube such that the sum of any three numbers on each face is not less than 10. Find the minimum value of the sum of the four numbers on each face. | 1. Let the four numbers on a certain face be $a_{1}, a_{2}, a_{3}, a_{4}$, whose sum reaches the minimum value, and $a_{1}<a_{2}<a_{3}<a_{4}$. Since the sum of three different positive integers less than 5 is at most 9, it follows that $a_{4} \geqslant 6$. Therefore,
$$
a_{1}+a_{2}+a_{3}+a_{4} \geqslant 16 \text {. }
$... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8.1650 students are arranged in 22 rows and 75 columns. It is known that among any two columns in the same row, the number of pairs of students of the same gender does not exceed 11. Prove: the number of boys does not exceed 928. | 8. Let the number of boys in the $i$-th row be $a_{i}$, then the number of girls is $75-a_{i}$. According to the problem, we have
$$
\sum_{i=1}^{2}\left(C_{a_{i}}^{2}+C_{n s-a_{i}}^{2}\right) \leqslant 11 \times C_{n s}^{2} .
$$
This is because, for any two columns, the number of pairs of students in the same row who ... | 928 | Combinatorics | proof | Yes | Yes | cn_contest | false |
7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{2003}+2002(x-1)=-1, \\
(y-2)^{2008}+2002(y-2)=1 .
\end{array}\right.
$$
Then $x+y=$ | 7. 3
Construct the function $f(t)=t^{2008}+2002 t$. It is easy to see that $f(t)$ is an odd function on $\mathbf{R}$, and it is also a monotonically increasing function. From this, we can get $f(x-1)=-f(y-2)$, which means $f(x-1)=f(2-y)$. Therefore, $x-1=2-y, x+y=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $x, y, z$ be positive real numbers, and $x+y+z=1$. Find the minimum value of the function
$$
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
$$
and provide a proof. | 14. Consider the function $g(t)=\frac{t}{1+t^{2}}$, we know that $g(t)$ is an odd function. Since when $t>0$, $\frac{1}{t}+t$ is decreasing in $(0,1)$, it is easy to see that $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in $(0,1)$. For $t_{1} \backslash t_{2} \in(0,1)$ and $t_{1}<t_{2}$, we have
$$
\left(t_{1}-t_{2}\ri... | 0 | Algebra | proof | Yes | Yes | cn_contest | false |
8. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a, b, c$ form an arithmetic sequence, and $c=10, a \cos A=$ $b \cos B, A \neq B$, then the inradius of $\triangle A B C$ is $\qquad$ | 8. 2 .
Let the inradius of $\triangle ABC$ be $r$. Since $a \cos A = b \cos B$, by the Law of Sines, we get $b \sin A = a \sin B$. Therefore, $\sin 2A = \sin 2B$.
Since $A \neq B$, then $A + B = 90^{\circ}$. Thus, $\triangle ABC$ is a right triangle, $\angle C = 90^{\circ}, a^2 + b^2 = c^2$.
Also, since $c = 10$, and ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
12. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with all terms being positive integers, an odd number of terms, a non-zero common difference, and the sum of all terms equal to 2004. Then the value of the second term $a_{2}$ is $\qquad$ | 12. 668 .
Let the first term and common difference of an arithmetic sequence be $a$ and $d$, and the number of terms be $n$. Then, $n a+\frac{1}{2} n(n-1) d=2004$, which means
$$
[2 a+(n-1) d] n=2004 \times 2=2^{3} \times 3 \times 167 \text {. }
$$
Since $n$ is odd and $d \neq 0$,
when $n=501$, $a+250 d=4$, which doe... | 668 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. A store purchased a certain number of student-specific scientific calculators at the beginning of the school year for 880 yuan, and sold them at 50 yuan each, which were quickly sold out. It was found that students still needed 3 times as many calculators, so the store purchased the required calculators for 2580 yua... | 2.B.
Let the first purchase of calculators be $x$ units, then the second purchase of calculators is $3x$ units. According to the problem, we have
$$
\frac{880}{x}=\frac{2580}{3 x}+1 \text {. }
$$
Solving for $x$ gives $x=20$. Therefore, the total profit is
$$
(50 \times 76+4 \times 50 \times 0.9)-(880+2580)=520 \text ... | 520 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Real numbers $x, y, z$ satisfy $x=y+\sqrt{2}, 2 x y+$ $2 \sqrt{2} z^{2}+1=0$. Then the value of $x+y+z$ is $\qquad$ | $=1.0$.
Since $x=y+\sqrt{2}$, we have,
$$
\begin{array}{l}
(x-y)^{2}=2, (x+y)^{2}-4 x y=2, \\
2 x y=\frac{1}{2}(x+y)^{2}-1 .
\end{array}
$$
Substituting this into $2 x y+2 \sqrt{2} z^{2}+1=0$ yields
$$
\frac{1}{2}(x+y)^{2}+2 \sqrt{2} z^{2}=0 \text {. }
$$ | 1.0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Points $A(-4,0)$ and $B(2,0)$ are two fixed points on the $x O y$ plane, and $C$ is a moving point on the graph of $y=-\frac{1}{2} x+2$. How many right triangles $\triangle A B C$ can be drawn that satisfy the above conditions? | 3. 4 .
As shown in Figure 6, draw perpendiculars from $A$ and $B$ to the $x$-axis, intersecting the line $y=-\frac{1}{2} x + 2$ at points $C_{1}$ and $C_{2}$, respectively; with $AB$ as the diameter, draw a semicircle intersecting the line
$$
y=-\frac{1}{2} x + 2 \text { at }
$$
points $C_{3}$ and $C_{4}$. Then,
$\t... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $a$, $b$, $c$, and $d$ are all prime numbers, and satisfy 10 $<c<d<20$, also $c$ and $a$ differ by a larger prime number, $d^{2}-c^{2}=a^{3} b(a+b)$. Then the value of $a b(c+d)$ is $\qquad$. | 4. 180.
Since $a, b, c, d$ are all prime numbers, and $10<c<d<20$, $c, d$ can only be 11, 13, 17, or 19, and $c \neq 19$. Also, $c-a$ is a larger prime number (not even), so $a=2$. By taking $c=11, 13, 17$, $c-a$ are 9, 11, 15, respectively. Only when $c=13, c-a=11$ does it meet the condition. Substituting $c=13, a=2$... | 180 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) A woodworking factory produces two types of chairs, A and B. Each chair must go through two processes, woodworking and painting, to be completed. The daily working hours of the woodworkers and painters and the time required to make (paint) one chair are shown in Table 1. It is known that the factory ea... | Let's do $x$ chairs of type A and $y$ chairs of type B each day, with daily profit being $M$, then we have
$$
\left\{\begin{array}{l}
x+2 y \leqslant 8, \\
3 x+y \leqslant 9, \\
M=2 x+3 y, \\
x \geqslant 0, y \geqslant 0 .
\end{array}\right.
$$
From Figure 7, it is clear that,
$$
\text { when } x=2, y=3
$$
$M$ is max... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $n$ be a positive integer not exceeding 2003. If there is an angle $\theta$ such that $(\sin \theta+\mathrm{i} \cos \theta)^{n}=\sin n \theta+\mathrm{i} \cos n \theta$ holds. Then the total number of such $n$ is $\qquad$. | 9. 501 .
$$
\begin{array}{l}
(\sin \theta + i \cos \theta)^{n} = [i(\cos \theta - i \sin \theta)]^{n} \\
= i^{n}[\cos (-\theta) + i \sin (-\theta)]^{n} \\
= i^{n}(\cos n \theta - i \sin n \theta) = i^{n-1}(\sin n \theta + i \cos n \theta).
\end{array}
$$
If \( i^{n-1}(\sin n \theta + i \cos n \theta) = \sin n \theta +... | 501 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Among three-digit numbers, if the digit in the tens place is smaller than the digits in the hundreds and units places, the number is called a concave number, such as 504, 746, etc., which are all concave numbers. Therefore, among three-digit numbers with no repeated digits, the number of concave numbers is $\qquad$... | 10. 240 .
When the tens digit is 0, there are $9 \times 8$ concave numbers that meet the condition, when the tens digit is 1, there are $8 \times 7$ concave numbers that meet the condition, .....
When the tens digit is 7, there are $2 \times 1$ concave numbers that meet the condition, in total there are $72+56+42+30+2... | 240 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. Given that $x$, $y$, and $z$ are all positive integers. Then the equation $x+y+z=$ 15 has solutions. | 12. 91 .
Express 15 as 15 ones, i.e., $1+1+\cdots+1=15$, where any 2 of the 14 plus signs are chosen, and the 1s separated by these two plus signs are combined into a single number to form a solution of the equation. Therefore, the number of solutions is $\mathrm{C}_{14}^{2}=91$. | 91 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. City $A$ has 4 suburban counties $(B, C, D, E)$, as shown in Figure 2. There are 5 colors available. How many different coloring methods are there such that no two adjacent areas share the same color, and each area is painted with only one color? | 15. The coloring method that meets the requirements must use at least three colors, so it can be divided into three categories:
(1) Using five colors, there are $P_{5}^{s}=120$ methods.
(2) Using four colors. The number of ways to choose four colors is $\mathrm{C}_{5}^{4}$. Among them, choosing one color to paint $A$ h... | 420 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $p>0, q>0$, and satisfy $2 p+\sqrt{p q}-$ $q+\sqrt{p}+\sqrt{q}=0$. Then $(2 \sqrt{p}-\sqrt{q}+2)^{3}=$ $\qquad$ | 2. 1 .
The original expression can be transformed into
$$
2(\sqrt{p})^{2}+\sqrt{p q}-(\sqrt{q})^{2}+(\sqrt{p}+\sqrt{q})=0 \text {, }
$$
which is $(\sqrt{p}+\sqrt{q})(2 \sqrt{p}-\sqrt{q}+1)=0$.
Since $p>0, q>0$, then $\sqrt{p}+\sqrt{q}>0$. Therefore, $2 \sqrt{p}-\sqrt{q}+1=0$, which means $2 \sqrt{p}-\sqrt{q}+2=1$.
Th... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) A research study group from Class 1, Grade 3 of Yuhong Middle School conducted a survey on students' lunch time at the school canteen. It was found that within a unit of time, the number of people buying lunch at each window and the number of people choosing to eat outside due to unwillingness to wait ... | Let each window sell lunch to $x$ people per minute, and $y$ people go out to eat per minute, with the total number of students being $z$ people, and assume that at least $n$ windows need to be open simultaneously. According to the problem, we have
$$
\left\{\begin{array}{l}
45 x=z-45 y, \\
2 \times 30 x=z-30 y, \\
20 ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In a tetrahedron $ABCD$ with volume 12, points $E$, $F$, and $G$ are on edges $AB$, $BC$, and $AD$ respectively, such that $AE = 2EB$, $BF = FC$, and $AG = 2GD$. A plane through points $E$, $F$, and $G$ intersects the tetrahedron in a section $EFGH$, and the distance from point $C$ to this section is 1. The area of ... | 2. 7 .
As shown in Figure 3, it is easy to know that $G E / /$ $D B$. Since $D B / /$ plane $E F H G$, then $H F$ // BD. Therefore, $H$ is also the midpoint of $D C$.
Let the distances from points $A, B, C, D$ to the section $E F H G$ be $h_{a}, h_{b}, h_{c}, h_{d}$, respectively. We have
$$
\begin{array}{l}
h_{b}=h_... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $f(x)=a x^{2}+b x+c$ be a quadratic trinomial with integer coefficients. If integers $m, n$ satisfy $f(m)-f(n)=1$. Then $|m-n|=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4. 1 .
$$
\begin{array}{l}
f(m)-f(n)=a\left(m^{2}-n^{2}\right)+b(m-n) \\
=(m-n)[a(m+n)+b],
\end{array}
$$
then $(m-n)(a m+a n+b)=1$.
Since $m-n$ and $a m+a n+b$ are both integers, it must be that
$$
|m-n|=1 \text{. }
$$ | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Given the set $M=|A|$ where $A$ is a ten-digit positive integer with all distinct digits, and $11111|A|$. Find $|M|$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: Since the digits of $A$ are all different, we have
$A \equiv 0+1+\cdots+9 \equiv 0(\bmod 9)$, i.e., $9 \mid A$.
Also, $11111 \mid A$, and since $(9,11111)=1$, it follows that 99 999. .
Let $A=99999 A_{0}, A_{0} \in \mathbf{Z}_{+}$.
Since $10^{9}\frac{10^{9}}{10^{5}}=10^{4}, 10^{5}+10, a_{i} \in\{0,1, \cdots, ... | 3456 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Someone wants to go upstairs, and this person can go up 1 step or 2 steps at a time. If one floor has 18 steps, how many different ways can he go up one floor? | Solution 1: This person can climb one floor in a maximum of 18 steps and a minimum of 9 steps. Here, $a_{18}, a_{17}, a_{16}, \cdots, a_{9}$ represent the number of ways this person can climb the stairs in 18 steps, 17 steps, 16 steps, $\cdots, 9$ steps, respectively (for any two consecutive step counts, the latter tak... | 4181 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Before the World Cup, a coach of a certain country, in order to examine the players $A_{1}, A_{2}, \cdots, A_{7}$, plans to have them all play in three training matches (each 90 minutes). Assuming that at any moment during the matches, only one of these players is on the field, and the total playing time (in ... | Let $A_{i}(i=1,2, \cdots, 7)$ be the total time in minutes that each player is on the field. According to the problem, we can set
$$
a_{i}=7 k_{i}(i=1,2,3,4), a_{i}=13 k_{i}(i=5,6,7),
$$
where $k_{i}(i=1,2, \cdots, 7) \in \mathbf{Z}_{+}$.
$$
\text { Let } \sum_{i=1}^{4} k_{i}=m, \sum_{i=5}^{7} k_{i}=n \text {, where }... | 42244 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5: A total of 240 students participating in a large group performance stand in a row facing the coach, and they report numbers in sequence from left to right as 1, 2, 3, 4, 5, ... The coach asks all students to remember their reported numbers and perform the following actions: First, all students who reported n... | Solution: (1) Let $U=\{1,2, \cdots, 240\}, A_{i}$ represent the set of all multiples of $i$ in $U$. Then
$|U|=240 ;\left|A_{3}\right|=\left[\frac{240}{3}\right]=80$,
$\left|A_{5}\right|=\left[\frac{240}{5}\right]=48,\left|A_{7}\right|=\left[\frac{240}{7}\right]=34$;
$\left|A_{15}\right|=\left[\frac{240}{15}\right]=16,\... | 118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a \neq b$, and satisfy
$$
(a+1)^{2}=3-3(a+1), 3(b+1)=3-(b+1)^{2} \text {. }
$$
Then the value of $b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}}$ is ( ).
(A) 23
(B) -23
(C) -2
(D) -13 | $-、 1 . B$.
Since $a, b$ are the roots of the equation with respect to $x$
$$
(x+1)^{2}+3(x+1)-3=0
$$
Rearranging this equation, we get
$$
x^{2}+5 x+1=0 \text {. }
$$
Since $\Delta=25-4>0$, we have
$$
a+b=-5, a b=1 \text {. }
$$
Therefore, $a, b$ are both negative. Thus,
$$
\begin{array}{l}
b \sqrt{\frac{b}{a}}+a \s... | -23 | Algebra | MCQ | Yes | Yes | cn_contest | false |
8. Given real numbers $a, b, x, y$ satisfy $a+b=x+y=2$, $a x+b y=5$. Then the value of $\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right)$ is $\qquad$ . | 8. -5 .
Given $a+b=x+y=2$, we have
$$
(a+b)(x+y)=a x+b y+a y+b x=4 \text{. }
$$
Since $a x+b y=5$, then,
$$
\begin{array}{l}
a y+b x=-1 . \\
\text{ Thus, }\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right) \\
=(a y+b x)(a x+b y)=-5 .
\end{array}
$$ | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. $n$ is a positive integer, $f(n)=\sin \frac{n \pi}{2}$. Then
$$
f(1991)+f(1992)+\cdots+f(2003)=
$$
$\qquad$ | 7. -1 .
It is easy to know that $f(1991)=-1, f(1992)=0, f(1993)=1$, $f(1994)=0, \cdots, f(2003)=-1$. Therefore, $f(1991)+f(1992)+\cdots+f(2003)=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $x$ and $y$ are positive integers, and satisfy $x y + x + y = 71$, $x^{2} y + x y^{2} = 880$. Then $x^{2} + y^{2} =$ $\qquad$ . | $-、 1.146$.
Let $a=x+y, b=x y, x y+x+y=a+b=71$, $x^{2} y+x y^{2}=x y(x+y)=a b=880$.
Therefore, $a, b$ are the two roots of $t^{2}-71 t+880=0$.
Solving, we get $a=x+y, b=x y$ equal to 16 and 55, respectively.
If $x+y=55, x y=16$, there are clearly no positive integer solutions. Therefore, only $x+y=16, x y=55$.
Thus, $x... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The function defined on the set of positive integers is
$$
f(x)=\left\{\begin{array}{ll}
3 x-1, & x \text { is odd, } \\
\frac{x}{2}, & x \text { is even. }
\end{array}\right.
$$
Let $x_{1}=12, x_{n+1}=f\left(x_{n}\right), n \in \mathbf{N}$, then the number of elements in the set $\left\{x \mid x=x_{n}\right.$, $n ... | 3. 7 .
From $x_{1}=12$ we get $x_{2}=6$, then $x_{3}=3, x_{4}=8, x_{5}=4$, $x_{6}=2, x_{7}=1, x_{8}=2, x_{9}=1, \cdots$, and thereafter it is a cycle of $2,1,2,1$, $\cdots$. Therefore, $x_{n}$ takes a total of 7 different values, i.e., the set $|x| x$ $\left.=x_{n}, n \in \mathbf{N}\right\}$ contains 7 elements. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given a geometric sequence $\left\{b_{n}\right\}$ with all terms being positive and an arithmetic sequence $\left\{a_{n}\right\}$, satisfying $b_{3}-b_{1}=9, b_{5}-b_{3}=36$, and $b_{1}=$ $a_{1}, b_{2}=a_{3}$. Let the sum of the first 6 terms of the geometric sequence be $G_{6}$, and the sum of the first 12 terms of... | 4. 324.
Let the common ratio be $q(q>0)$, then
$$
\left\{\begin{array}{l}
b_{1} q^{2}-b_{1}=9, \\
b_{1} q^{4}-b_{1} q^{2}=36,
\end{array}\right.
$$
which is equivalent to $\left\{\begin{array}{l}b_{1} q^{2}-b_{1}=9, \\ q^{2}\left(b_{1} q^{2}-b_{1}\right)=36 .\end{array}\right.$
Substituting (1) into (2) gives $q^{2}=... | 324 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. $M=\{-2,0,1\}, N=\{1,2,3,4,5\}$. Mapping $f: M$ $\rightarrow N$, such that for any $x \in M$, $x+f(x)+x f(x)$ is an odd number. Then the number of different mappings is $\qquad$ . | 5. 45 .
Given the mapping $f: M \rightarrow N$ where the independent variable is taken from $M=\{-2,0, 1\}$ and the values of $f(x)$ are taken from $N=\{1,2,3,4,5\}$. The problem is to find the number of different mappings such that for any $x \in M$, $x+f(x)+x f(x)$ is an odd number.
And $x+f(x)+x f(x)=x+x f(x)+1+f(x... | 45 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. In the expansion of $\left(4 x^{2}-2 x-5\right)\left(1+\frac{1}{x^{2}}\right)^{5}$, the constant term is $\qquad$ | 8. 15 .
$$
\text { Original expression }=\left(4 x^{2}-2 x-5\right)\left(1+\frac{5}{x^{2}}+\frac{10}{x^{4}}+\cdots\right) \text {. }
$$
Therefore, the constant term in the expanded form is $(-5) \times 1+4 \times 5=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given
$$
\left(x+\sqrt{x^{2}+2000}\right)\left(y+\sqrt{y^{2}+2002}\right)=2002 \text {. }
$$
Then $x^{2}-3 x y-4 y^{2}-6 x-6 y+58=$ $\qquad$
(2002, Junior High School Mathematics Competition, Jiangsu Province) | Explain: Construct the rationalizing factors $x-\sqrt{x^{2}+2002}$ and $y-\sqrt{y^{2}+2002}$ for $x+\sqrt{x^{2}+2002}$ and $y+\sqrt{y^{2}+2002}$. Because
$$
\begin{array}{l}
\left(x+\sqrt{x^{2}+2002}\right)\left(x-\sqrt{x^{2}+2002}\right) . \\
\quad\left(y+\sqrt{y^{2}+2002}\right)\left(y-\sqrt{y^{2}+2002}\right) \\
=(-... | 58 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Set $R$ represents the set of points $(x, y)$ in the plane, where $x^{2}+6 x+1+y^{2}+6 y+1 \leqslant 0$, and $x^{2}+6 x+1-\left(y^{2}+\right.$ $6 y+1) \leqslant 0$. Then, the area of the plane represented by set $R$ is closest to ( ).
(A) 25
(B) 24
(C) 23
(D) 22 | 6.A.
From the given, we have $(x+3)^{2}+(y+3)^{2} \leqslant 16$,
$$
(x-y)(x+y+6) \leqslant 0 \text {. }
$$
The first inequality represents the interior and boundary of a circle with center $(-3,-3)$ and radius 4; the second inequality can be written as $x-y \geqslant 0$ and $x+y+6 \leqslant 0$, or $x-y \leqslant 0$ a... | 25 | Geometry | MCQ | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{49}-\frac{1}{50}\right) \div\left(\frac{1}{20}+\frac{1}{77}+\cdots+\frac{1}{50}\right) \\
= \\
\end{array}
$$ | $=、 1.1$
$$
\begin{array}{l}
\text { Original expression }=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{49}+\frac{1}{50}\right)-\right. \\
\left.\quad\left(1+\frac{1}{2}+\cdots+\frac{1}{25}\right)\right] \div\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right) \\
=\left(\frac{1}{26}+\frac{1}{... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of real roots of the equation $x^{2}|x|-5 x|x|+2 x=0$ is $\qquad$.
The equation $x^{2}|x|-5 x|x|+2 x=0$ has $\qquad$ real roots. | 3.4 .
From the given, we have $x(x|x|-5|x|+2)=0$.
Thus, $x=0$ or $x|x|-5|x|+2=0$.
(1) When $x>0$, $x^{2}-5 x+2=0$, which has two distinct positive real roots;
(2) When $x<0$, $-x^{2}+5 x+2=0$, which has one negative real root.
Therefore, the original equation has 4 real roots. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For each positive integer $n$, let $f(n)$ represent the last digit of $1+2+\cdots+n$. For example, $f(1)=$
$$
\begin{aligned}
1, f(2)=3, f(5) & =5 \text {. Find } \\
f(1)+f(2) & +f(3)+\cdots+f(2004) .
\end{aligned}
$$ | $$
\begin{array}{l}
\text { Given } f(1)=1, f(2)=3, f(3)=6, f(4)=0, \\
f(5)=5, f(6)=1, f(7)=8, f(8)=6, f(9)=5, \\
f(10)=5, f(11)=6, f(12)=8, f(13)=1, f(14)=5, \\
f(15)=0, f(16)=6, f(17)=3, f(18)=1, f(19)=0, \\
f(20)=0, f(21)=1, f(22)=3, f(23)=6, \cdots
\end{array}
$$
From the above, we can see that the period of $f(n)... | 7010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) The integers $x_{0}, x_{1}, \cdots, x_{2000}$ satisfy the conditions: $x_{0}=0,\left|x_{1}\right|=\left|x_{0}+1\right|,\left|x_{2}\right|=\left|x_{1}+1\right|$, $\cdots,\left|x_{2004}\right|=\left|x_{2003}+1\right|$. Find the minimum value of $\mid x_{1}+x_{2}+\cdots+$ $x_{2004} \mid$. | Three, from the known we can get
$$
\left\{\begin{array}{l}
x_{1}^{2}=x_{0}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots . \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$
Thus, $x_{2004}^{2}=x_{0}^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$.
Given $x_{0}=0$, then
$$
\begi... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In the Cartesian coordinate system, color the point set
$$
\{(m, n) \mid m, n \in \mathbf{N}, 1 \leqslant m, n \leqslant 6\}
$$
with red or blue. Then, the number of different coloring schemes where each unit square has exactly two red vertices is | 4.126.
First, color the first row (points with a y-coordinate of 6), which has $2^{6}$ ways of coloring. We can divide these into two cases:
(1) No two adjacent points are the same color (i.e., red and blue alternate), which gives 2 ways of coloring. In this case, there are only 2 ways to color the second row. Similar... | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. The four sides of the quadrilateral pyramid are isosceles triangles with a leg length of $\sqrt{7}$ and a base length of 2. Then the maximum possible volume of this quadrilateral pyramid is | 5. 3 .
There are three scenarios for the quadrilateral pyramid that meet the requirements.
(1) As shown in Figure 4, all four lateral edges are $\sqrt{7}$, and the volume is calculated as $V_{1}=\frac{4}{3} \sqrt{5}$.
(2) As shown in Figure 5, two lateral edges are $\sqrt{7}$.
Construct $O E \perp$ plane $A B C D$, a... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Given a height of 137, find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two distinct elements $a$ and $b$ in $S$, there exists another element $c$ in $S$ such that the greatest common divisor (gcd) of $... | Solution: Construct 50 arrays:
$$
(1,100),(2,99) \cdots,(50,51) \text {, }
$$
The sum of the two numbers in each array is 101.
Since 101 is a prime number, there does not exist an element $c$ in $S$ such that the greatest common divisor of 101 and $c$ is greater than 1. Therefore, it is impossible for $S$ to contain b... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3: In front of each number in $1, 2, 3, \cdots, 1989$, add a “+” or “-” sign to make their algebraic sum the smallest non-negative number, and write out the equation.
(1989, All-Russian Mathematical Olympiad) | Proof: First, we prove that the algebraic sum is an odd number.
Consider the simplest case:
If all are filled with “+”, then at this moment
$$
1+2+\cdots+1989=995 \times 1989
$$
is an odd number.
For the general case, it only requires adjusting some “+” to “-”.
Since $a+b$ and $a-b$ have the same parity, the parity o... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given that $\alpha, \beta$ are the roots of the equation $x^{2}-x-1=0$. Then the value of $\alpha^{4}+3 \beta$ is $\qquad$
(2003, National Junior High School Mathematics Competition, Tianjin Preliminary Contest) | Explanation: $\alpha^{4}+3 \beta$ is not a symmetric expression of the two roots of the equation, and it is obviously impossible to directly substitute it using Vieta's formulas. We can construct the dual expression $\beta^{4}+3 \alpha$ of $\alpha^{4}+3 \beta$, and calculate the sum and difference of the two expression... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $a b c \neq 0$, and $a+b+c=0$. Then the value of the algebraic expression $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | $\begin{array}{l}\text { I. 1.A. } \\ \text { Original expression }=\frac{-(b+c) a}{b c}+\frac{-(a+c) b}{a c}+\frac{-(a+b) c}{a b} \\ =-\left(\frac{a}{b}+\frac{a}{c}\right)-\left(\frac{b}{a}+\frac{b}{c}\right)-\left(\frac{c}{a}+\frac{c}{b}\right) \\ =\frac{a}{a}+\frac{b}{b}+\frac{c}{c}=3 .\end{array}$ | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 4 Real numbers $a, b, x, y$ satisfy $a x + b y = 3$, $a x^2 + b y^2 = 7$, $a x^3 + b y^3 = 16$, $a x^4 + b y^4 = 42$. Find the value of $a x^5 + b y^5$.
(8th American High School Mathematics Examination) | Explanation: Let $S_{n}=a x^{n}+b y^{n}$. Then
$$
S_{n}=(x+y) S_{n-1}-x y S_{n-2}, n=3,4, \cdots \text {. }
$$
Substituting the known conditions into the above equation, we get
$$
\begin{array}{l}
7(x+y)-3 x y=16,16(x+y)-7 x y=42 . \\
\text { Solving, we get } x+y=-14, x y=-38 . \\
\text { Therefore, } S_{n}=-14 S_{n-... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. As shown in Figure 1, on a $2 \times 3$ rectangular grid paper, the vertices of each small square are called lattice points. Then the number of isosceles right triangles with lattice points as vertices is ( ) .
(A) 24
(B) 38
(C) 46
(D) 50 | 6.D.
The length of line segments with grid points as vertices can take 8 values: $1, \sqrt{2}$, $2, \sqrt{5}, 2 \sqrt{2}, 3, \sqrt{10}, \sqrt{13}$. The isosceles right triangles formed by these line segments can be classified into 4 cases based on their side lengths:
$$
1,1, \sqrt{2} ; \sqrt{2}, \sqrt{2}, 2 ; 2,2,2 \s... | 50 | Geometry | MCQ | Yes | Yes | cn_contest | false |
3. Let $m$ be the largest integer that cannot be expressed as the sum of three distinct composite numbers. Then $m=$ $\qquad$
Set $m$ is the largest integer that cannot be represented as the sum of three distinct composite numbers. Then $m=$ $\qquad$ | 3.17 .
The smallest three composite numbers are 4, 6, and 8, and $4+6+8=18$. Therefore, 17 is an integer that cannot be expressed as the sum of three distinct composite numbers.
When $m>18$, if $m=2k>18$, then $m=4+6+2 \times (k-5)$; if $m=2k-1>18$, then $m=4+9+2(k-7)$.
Therefore, any integer greater than 18 can be ... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The minimum value of the function $f(x)=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$ is $\qquad$ . | 2. 5 .
Obviously, if $xf(-x)$.
Therefore, when $f(x)$ takes its minimum value, there must be $x \geqslant 0$.
As shown in Figure 6, draw a line segment $A B=4$, $A C \perp A B, D B \perp A B$, and $A C=$ $1, B D=2$. For any point $O$ on $A B$, let $O A=x$, then
$$
\begin{array}{l}
O C=\sqrt{x^{2}+1}, \\
O D=\sqrt{(4-x... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$2 . a 、 b 、 c$ are non-zero real numbers, and $a+b+c \neq 0$. If $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$, then $\frac{(a+b)(b+c)(c+a)}{a b c}$ equals ( ).
(A) 8
(B) 4
(C) 2
(D) 1 | 2.A.
Let $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$, then $a+b+c=k(a+b+c)$.
Since $a+b+c \neq 0$, we have $k=1$.
Therefore, the original expression $=\frac{2 c \cdot 2 a \cdot 2 b}{a b c}=8$. | 8 | Algebra | MCQ | Yes | Yes | cn_contest | false |
7. Given $x=\frac{\sqrt{3}-1}{\sqrt{3}+1}, y=\frac{\sqrt{3}+1}{\sqrt{3}-1}$. Then $x^{4}+y^{4}$ equals | $=.7 .194$.
Given $x=2-\sqrt{3}, y=2+\sqrt{3}$. Then
$$
\begin{array}{l}
x+y=4, x y=1, \\
x^{2}+y^{2}=(x+y)^{2}-2 x y=14, \\
x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2 x^{2} y^{2}=194 .
\end{array}
$$ | 194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. On the beach, there is a pile of apples that belongs to 3 monkeys. The first monkey comes, divides the apples into 3 equal piles with 1 apple left over, then it throws the extra apple into the sea and takes one pile; the second monkey comes, divides the remaining apples into 3 equal piles, again with 1 apple left ov... | 9. 25 .
Let there be $n$ apples initially, and the remaining number of apples each time be $y_{1}, y_{2}, y_{3}$. Then
$$
\begin{array}{l}
y_{1}=\frac{2}{3}(n-1)=\frac{2}{3}(n+2)-2, \\
y_{2}=\left(\frac{2}{3}\right)^{2}(n+2)-2, \\
y_{3}=\left(\frac{2}{3}\right)^{3}(n+2)-2 .
\end{array}
$$
To make $y_{3}$ a positive i... | 25 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If $(2 x-1)^{5}=a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+$ $a_{1} x+a_{0}$, then $a_{2}+a_{4}=$ $\qquad$ | ,$- 1 .-120$.
Let $x=0$, we get $a_{0}=-1$.
Let $x=1$, we get $a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=1$;
Let $x=-1$, we get
$-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0}=-243$.
Adding the last two equations gives $a_{4}+a_{2}+a_{0}=-121$.
Therefore, $a_{2}+a_{4}=-120$. | -120 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (18 points) A student, in order to plot the graph of the function $y=a x^{2}+$ $b x+c(a \neq 0)$, took 7 values of the independent variable: $x_{1}<x_{2}<$ $\cdots<x_{7}$, and $x_{2}-x_{1}=x_{3}-x_{2}=\cdots=x_{7}-x_{6}$, and calculated the corresponding $y$ values, listing them in Table 1.
Table 1
\begin{tabular... | Let $x_{2}-x_{1}=x_{3}-x_{2}=\cdots=x_{7}-x_{6}=d$, and the function value corresponding to $x_{i}$ be $y_{i}$. Then
$$
\begin{array}{l}
\Delta_{k}=y_{k+1}-y_{k} \\
=\left(a x_{k+1}^{2}+b x_{k+1}+c\right)-\left(a x_{k}^{2}+b x_{k}+c\right) \\
=a\left[\left(x_{k}+d\right)^{2}-x_{k}^{2}\right]+b\left[\left(x_{k}+d\right)... | 551 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let $x, y$ be non-negative integers, $x+2y$ is a multiple of 5, $x+y$ is a multiple of 3, and $2x+y \geqslant 99$. Then the minimum value of $7x+5y$ is $\qquad$
(15th "Five Sheep Forest" Junior High School Mathematics Competition (Initial $\exists)$) | Explanation: Let $x+2y=5A, x+y=3B, A, B$ be integers.
Since $x, y \geqslant 0$, hence $A, B \geqslant 0$, solving we get
$$
\begin{array}{l}
x=6B-5A, y=5A-3B, \\
2x+y=9B-5A \geqslant 99, \\
S=7x+5y=27B-10A .
\end{array}
$$
The problem is transformed to: integers $A, B \geqslant 0, 6B \geqslant 5A \geqslant$ $3B, 9B \g... | 366 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (16 points) Let the four-digit number $\overline{a b c d}$ be a perfect square, and $\overline{a b}=2 \overline{c d}+1$. Find this four-digit number. | Let $\overline{a b c d}=m^{2}$, then $32 \leqslant m \leqslant 99$.
Suppose $\overline{c d}=x$, then $\overline{a b}=2 x+1$. Therefore,
$$
100(2 x+1)+x=m^{2} \text {, }
$$
which simplifies to $67 \times 3 x=(m+10)(m-10)$.
Since 67 is a prime number, at least one of $m+10$ and $m-10$ must be a multiple of 67.
(1) If $m... | 5929 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$ is
$\qquad$ .
(11th "Hope Cup" National Mathematics Invitational Competition (Grade 8)) | (Tip: Construct a symmetrical figure. Answer: 13) | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$\begin{array}{l}\text { 1. If } n \text { satisfies }(n-2003)^{2}+(2004-n)^{2} \\ =1 \text {, then }(2004-n)(n-2003)=\end{array}$ | $\begin{array}{l} \text { II.1.0. } \\ \text { From }(n-2003)^{2}+2(n-2003)(2004-n)+ \\ (2004-n)^{2}-2(n-2003)(2004-n) \\ =(n-2003+2004-n)^{2}-2(n-2003)(2004-n) \\ =1-2(n-2003)(2004-n)=1, \\ \text { we get }(n-2003)(2004-n)=0 .\end{array}$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $n$ is a natural number, $11 n+10$ and $10 n+3$ are both multiples of some natural number $d$ greater than 1. Then $d=$ $\qquad$ | 2.67.
From $11 n+10$ and $10 n+3$ both being multiples of $d$, we know
$$
10(11 n+10)-11(10 n+3)=67
$$
is still a multiple of $d$. But 67 is a prime number, $d \neq 1$, so $d=67$. | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. If real numbers $x, y$ satisfy $x \geqslant 0$, and
$$
\max \{1-x, x-1\} \leqslant y \leqslant x+2 \text {, }
$$
then the minimum value of the bivariate function $u(x, y)=2 x+y$ is
$\qquad$ . | 6.1.
From the given information, we have
$$
\begin{array}{l}
u(x, y)=2 x+y \geqslant 2 x+\max \{1-x, x-1\} \\
=\max \{2 x+(1-x), 2 x+(x-1)\} \\
=\max \{x+1,3 x-1\} \\
\geqslant \max \{1,-1\}=1,
\end{array}
$$
i.e., $u(x, y) \geqslant 1$, with equality holding if and only if $x=0, y=1$. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Given the set $\{1,2,3,4,5,6,7,8,9,10\}$. Find the number of subsets of this set that have the following property: each subset contains at least 2 elements, and the absolute difference between any two elements in each subset is greater than 1.
(1996, Shanghai High School Mathematics Competition) | Let $a_{n}$ be the number of subsets of the set $\{1,2, \cdots, n\}$ that have the given property. Then, the subsets of $\{1,2, \cdots, k, k+1, k+2\}$ that have the given property can be divided into two categories: the first category does not contain $k+2$, and there are $a_{k+1}$ such subsets; the second category con... | 133 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $M=\{1,2, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A$, then $15 x \notin A$. The maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition) | Solution: From the given, we know that at least one of the numbers $k$ and $15k$ does not belong to $A$.
Since $\left[\frac{1995}{15}\right]=133$, when $k=134,135$, $\cdots, 1995$, $15k$ definitely does not belong to $A$.
Similarly, $\left[\frac{133}{15}\right]=8$, when $k=9,10, \cdots, 133$, $k$ and $15k$ cannot bot... | 1870 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let the set $M=\{1,2, \cdots, 1000\}$, and for any non-empty subset $X$ of $M$, let $a_{x}$ denote the sum of the largest and smallest numbers in $X$. Then, the arithmetic mean of all such $a_{x}$ is $\qquad$
(1991, National High School Mathematics Competition) | Analysis: For any subset $X=\left\{b_{1}, b_{2}, \cdots, b_{k}\right\}$ of set $M$, assume $b_{1}<b_{2}<\cdots<b_{k}$, then there must exist another subset $X^{\prime}=\left\{1001-b_{1}, 1001-\right.$ $\left.b_{2}, \cdots, 1001-b_{k}\right\}$, at this time
$$
\begin{array}{l}
a_{x}=b_{1}+b_{k}, \\
a_{x^{\prime}}=\left(... | 1001 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Given $a>0$, and
$$
\sqrt{b^{2}-4 a c}=b-2 a c \text {. }
$$
Find the minimum value of $b^{2}-4 a c$.
(2004, "TRULY ${ }^{\circledR}$ Xinli Cup" National Junior High School Mathematics Competition) | Let $y=a x^{2}+b x+c$.
From $a<0$, we know $\Delta=b^{2}-4 a c>0$.
Therefore, the graph of this quadratic function is a parabola opening downwards, and it intersects the $x$-axis at two distinct points $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$.
Since $x_{1} x_{2}=\frac{c}{a}<0$, let's assume $x_{1}<x_{2}$.
The... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let $S$ be a subset of the set $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements of $S$ cannot be divisible by 7. Then, what is the maximum number of elements that $S$ can have?
(43rd American High School Mathematics Examination) | Solution: For two different natural numbers $a$ and $b$, if $7 \times (a+b)$, then the sum of the remainders when they are divided by 7 is not 0. Therefore, the set $\{1,2, \cdots, 50\}$ can be divided into 7 subsets based on the remainders when divided by 7. In $A_{i}$, each element has a remainder of $i$ when divided... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Let the set $A=\{1,2, \cdots, 366\}$. If a binary subset $B=\{a, b\}$ of $A$ satisfies 17 | $(a+$ $b)$, then $B$ is said to have property $P$.
(1) Find the number of binary subsets of $A$ that have property $P$;
(2) Find the number of a set of binary subsets of $A$, which are pairwise disjoint and have prope... | Solution: (1) Divide $1,2, \cdots, 366$ into 17 classes based on the remainder when divided by 17: $[0],[1], \cdots,[16]$.
Since $366=17 \times 21+9$, the classes $[1],[2], \cdots$, [9] each contain 22 numbers; [10], [11], $\cdots$, [16] and [0] each contain 21 numbers.
(i) When $a, b \in[0]$, the number of subsets wi... | 3928 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$ be a collection of non-empty subsets of $A$, and when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Then the maximum value of $k$ is $\qquad$
(1999, National High School Mathematics League Guangxi Preliminary Contest (High) Three)... | (It is easy to see that the family of all subsets of $A$ containing at most three elements meets the requirements of the problem, where the number of subsets is $\mathrm{C}_{10}^{\mathrm{l}}+$ $\mathrm{C}_{10}^{2}+\mathrm{C}_{10}^{3}=175$. It remains to prove that this is the maximum value.) | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $S=\{1,2,3,4\}$, and the sequence $a_{1}, a_{2}, \cdots, a_{n}$ has the following property: for any non-empty subset $B$ of $S$, there are consecutive $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shanghai High School Mathematics Competition) | (Since a binary subset containing a fixed element of $S$ has 3 elements, any element of $S$ appears at least twice in the sequence. Therefore, the minimum value of $n$ is estimated to be 8. On the other hand, an 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition, so the minimum value of $n$ is 8.) | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Let $x_{1}, x_{2}, \cdots, x_{19}$ all be positive integers, and satisfy $x_{1}+x_{2}+\cdots+x_{19}=95$. Find the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{19}^{2}$.
(1995, Hebei Province Junior High School Mathematics Competition) | Explanation: $\bar{x}=\frac{1}{19}\left(x_{1}+x_{2}+\cdots+x_{19}\right)=5$.
From the variance formula $S^{2}=\frac{1}{19}\left(\sum_{i=1}^{19} x_{i}^{2}-19 \bar{x}^{2}\right)$, we know that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{19}^{2}$ and $S^{2}$ reach their maximum values simultaneously.
For the positive integers $x_{1},... | 5947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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