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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,
$$
x_{1}^{3}, x_{1}^{3}+x_{2}^{3}, x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \equiv 4(\bmod 9) \text {. }
$$
Since $2002=10^{3}+10^{3}+1^{3}+1^{3}$, then
$$
\begin{array}{l}
2002^{2002}=2002 \times\left(2002^{667}\right)^{3} \\
=\left(10 \times 2002^{667}\right)^{3}+\left(10 \times 2002^{667}\right)^{3}+ \\
\left(2002^{667}\right)^{3}+\left(2002^{667}\right)^{3} .
\end{array}
$$
Therefore, $n=4$.
|
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 the other two sides of the triangle be $x, y$. Thus, the number of positive integer solutions $(x, y)$ that satisfy the inequality system $\left\{\begin{array}{l}x+y>n, \\ x<y<n\end{array}\right.$ is the number of such triangles.
When $\dddot{n}$ is odd, the number of triangles is the number of lattice points inside $\triangle B C D$ in Figure 4, i.e.,
$$
\begin{aligned}
f(n) & =2+4+6+\cdots+(n-3) \\
& =\frac{1}{4}(n-1)(n-3) .
\end{aligned}
$$
When $n$ is even, the number of triangles is the number of lattice points inside $\triangle B C D$ in Figure 5, i.e.,
$$
\begin{array}{l}
f(n)=1+3+5+\cdots+(n-3) \\
=\frac{1}{4}(n-2)^{2} . \\
\text { If } \frac{1}{4}(n-1)(n-3)=600, \text { then } n=51 ; \\
\text { If } \frac{1}{4}(n-2)^{2}=600 \text {, then there is no integer solution. }
\end{array}
$$
Therefore, the required $n=51$.
Note: This problem can also be solved by examining the length of the shortest side one by one, establishing an equation about $n$ and solving it.
|
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
$$
\begin{array}{l}
f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\
f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} .
\end{array}
$$
Since $f(25)=f(2 \times 2+3 \times 7)=f(2) f(7)=0$, we have $a=0$.
For any odd number $k>4$, there exists $y \in \mathbb{N}$, such that $k=3+2 y$. Then $f(k)=0$, hence for all $x \in \mathbb{N}, f(x)=0$ satisfies the condition.
If $f(0)=1$, let $x=0$ or $y=0$, we get
$f(2 y)=f(y)$ or $f(3 x)=f(x)$.
Let $f(1)=a$, then
$$
\begin{array}{l}
f(2)=a, \\
f(5)=f(3 \times 1+2 \times 1)=f(1) f(1)=a^{2}, \\
f(25)=f(3 \times 5+2 \times 5)=f(5) f(5)=a^{4} .
\end{array}
$$
Since $f(25)=f(3 \times 3+2 \times 8)=f(3) f(8)=$ $f(1) f(4)=f(1) f(2)=a^{2}$, we have $a=0$ or $a=1$.
Similarly, we have
$$
f(x)=\left\{\begin{array}{ll}
1, & x=0, \\
0, & x>0
\end{array} \text { or } f(x)=1, x \in \mathbb{N} .\right.
$$
In summary, there are 3 functions that satisfy the conditions.
|
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 $a \pm b, a \pm b \mathrm{i}$. Thus, the original equation is
$$
(x-a)^{4}=b^{4},
$$
which is $x^{4}-4 a x^{3}+6 a^{2} x^{2}-4 a^{3} x+a^{4}-b^{4}=0$.
From $-4 a \in \mathbf{Z}$ and $4 a^{3} \in \mathbf{Z}$, we know $a \in \mathbf{Z}$. From $a^{4}-b^{4} \in \mathbf{Z}$, we know $b^{4} \in \mathbf{Z}$, so $b^{4} \geqslant 1, b^{2} \geqslant 1$. In this case, the area of the square is $2 b^{2} \geqslant 2$, with equality holding when $a \in \mathbf{Z}, b= \pm 1$.
(2) If the 4 roots of the equation are 2 pairs of conjugate complex roots, then we can set these 4 roots as $a \pm b \mathrm{i}, a+2 b \pm b \mathrm{i}$. These 4 roots are the roots of the equation $(x-(a+b))^{4}=-4 b^{4}$. Similarly, we know $4 b^{4} \in \mathbf{Z}$, thus $4 b^{4} \geqslant 1, b^{2} \geqslant \frac{1}{2}$. Therefore, the area of the square is $4 b^{2} \geqslant 2$. Equality holds when $b= \pm \frac{\sqrt{2}}{2}, a+b \in \mathbf{Z}$.
In summary, the area of such a square is greater than or equal to 2. The equation $x^{4}=1$ has 4 roots which are the 4 vertices of a square on the complex plane with an area of 2, so the minimum area of such a square is 2.
|
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 $\triangle ABC$ as the center and $a$ as the radius. Thus, the 3 perpendicular bisectors and 3 circles yield a total of 10 intersection points (as shown in Figure 1).
Therefore, there are 10 points $P$ that satisfy the conditions. These can be categorized into 4 types: the first type is the intersection of the 3 perpendicular bisectors $\left(P_{1}\right)$. The 2nd, 3rd, and 4th types are the intersections of the circles and the perpendicular bisectors, each with 3 points $\left(P_{2}, P_{3}\right.$, $\left.P_{4}\right),\left(P_{5}, P_{6}, P_{7}\right),\left(P_{8}, P_{9}, P_{10}\right)$.
|
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-overlapping "dominoes"? If it can be covered, what is the minimum number of "dominoes" needed?
|
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 $3(m+1)^2 > n^2$, thus, $n \geqslant 7$.
Below, we use mathematical induction to prove: when $n \geqslant 7$, $(m+1)^2$ "dominoes" can cover all the black cells on the chessboard.
When $n=7$, since two "dominoes" can form a $2 \times 3$ rectangle, and two $2 \times 3$ rectangles can form a $4 \times 3$ rectangle, then 4 such $4 \times 3$ rectangles can be placed on a $7 \times 7$ chessboard, such that all cells except the middle black cell are covered (as shown in Figure 1). Adjust the "domino" adjacent to the middle black cell, so that this "domino" covers the middle black cell and also covers the only black cell that the original "domino" covered. Thus, 16 "dominoes" cover all cells except one white cell on the chessboard (as shown in Figure 1).
Assume that when $n=2m-1$, on a $(2m-1) \times (2m-1)$ chessboard, $m^2$ "dominoes" can cover all the black cells. When $n=2m+1$, divide the $(2m+1) \times (2m+1)$ chessboard into three parts: $(2m-1) \times (2m-1)$, $(2m-1) \times 2$, and $(2m+1) \times 2$. Since the $(2m-1) \times 2$ rectangle can be divided into $m-2$ $2 \times 2$ squares and one $2 \times 3$ rectangle, the black cells in the $(2m-1) \times 2$ rectangle can be covered by $(m-2) + 2$ "dominoes". Similarly, the $(2m+1) \times 2$ rectangle can be covered by $(m-1) + 2$ "dominoes" (as shown in Figure 2). Therefore, the $(2m+1) \times (2m+1)$ chessboard can be covered by $m^2 + m + (m+1) = (m+1)^2$ "dominoes".
|
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 vertices $(A, B)$ are on the two adjacent sides of this vertex.
Solution 1: First, determine how many lattice $\triangle ABC$ can be formed in one $3 \times 3$ square. This can be done in two steps (Figure 6).
(1) First, find the side $AB$, whose endpoints are on the adjacent sides of the square (corresponding to 3 vertices of the square), with 2 possible choices, $AB$ and $A_1B_1$.
(2) Then, take the 4th vertex, which is not on the adjacent sides, as the 3rd vertex $C$ of the triangle, with only 1 possible choice.
Multiplying these, we get $2 \times 1 = 2$ lattice triangles.
Since a $3 \times 3$ square has 4 pairs of adjacent sides, there are a total of $2 \times 4 = 8$ lattice triangles.
Since there are 4 $3 \times 3$ squares in a $4 \times 4$ square, the number of lattice triangles congruent to $\triangle ABC$ is $8 \times 4 = 32$.
Analysis: From Solution 1, we know that once $AB$ is determined, $C$ is uniquely determined, so we only need to find the sides of length $\sqrt{5}$ in Figure 6.
Solution 2: In a $3 \times 3$ square, once $AB = \sqrt{5}$ is determined, $\triangle ABC$ is uniquely determined, so we only need to determine the number of inscribed squares with side length $\sqrt{5}$. Directly drawing the diagram yields 2 such squares (Figure 7).
Since there are 4 $3 \times 3$ squares in a $4 \times 4$ square, we get a total of 8 squares with side length $\sqrt{5}$. Each such square corresponds to 4 triangles, yielding a total of $4 \times 8 = 32$ triangles congruent to $\triangle ABC$.
Analysis: Here, a corresponding technique is used, converting the calculation of triangles to the calculation of squares with side length $\sqrt{5}$.
(To be continued)
|
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 $(a, b, c)$ from $T$, and A responding with the number $|x+y-a-b|+|y+z-b-c|+|z+x-c-a|$. Find the minimum number of "moves" required for B to determine A's chosen triple.
(Bulgaria provided)
|
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" are sufficient to determine the triplet chosen by A.
In the first "movement," B gives A the triplet $(0,0,0)$, and A's response is $2(x+y+z)$, so B can determine the value of $s=x+y+z$. Clearly, $0 \leqslant s \leqslant 27$. Without loss of generality, assume $s \leqslant 13$. In fact, if $s \geqslant 14$, then the $(a, b, c)$ given by B to A in the following process can be changed to $(9-a, 9-b, 9-c)$, and the triplet $(x, y, z)$ can still be determined.
If $s \leqslant 9$, in the second "movement," B gives A $(9,0,0)$, and A's response is $(9-x-y)+y+z+(9-x-z)=18-2x$, thus the value of $x$ can be determined. Similarly, in the third "movement," B gives A $(0,9,0)$, and the value of $y$ can be determined. Therefore, $z=s-x-y$ can be determined.
If $9 < s \leqslant 13$, in the second "movement," B gives A $(0, s, 0)$, and A's response is $|s-x-y|+|s-y-z|+|s-x-z|=3s-2(x+y+z)=3s-2s=s$, thus the value of $s$ can be determined. In the third "movement," B gives A $(9, s-9, 0)$, and A's response is
$$
\begin{array}{l}
|s-9-x-y|+|s-9-y-z|+|9-x-z| \\
=(9-x)+(s-9-y)+(9-x-z) \\
=18-2(x+z) .
\end{array}
$$
Thus, the value of $x+z$ can be determined. Similarly, the values of $x$ and $z$ can be determined. Therefore, the value of $y$ can be determined.
In summary, the minimum number of "movements" required is 3.
|
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 $G^{\prime}$ satisfy the following conditions:
$G^{\prime}$ is the "copy of $G$ from $y$ to $x$", for every point $z$ (different from $x, y$), if $yz$ is an edge in graph $G$, then add the edge $xz$ in graph $G^{\prime}$; if $yz$ is not an edge in graph $G$, then remove the edge $xz$ in graph $G^{\prime}$.
Similarly, define $G^{\prime \prime}$ to satisfy the following conditions:
$G^{\prime \prime}$ is the "copy of $G$ from $x$ to $y$", for every point $z$ (different from $x, y$), if $xz$ is an edge in graph $G$, then add the edge $yz$ in graph $G^{\prime \prime}$; if $xz$ is not an edge in graph $G$, then remove the edge $yz$ in graph $G^{\prime \prime}$.
Below, we prove that $Q(G) \leqslant \frac{1}{2}\left[Q\left(G^{\prime}\right)+Q\left(G^{\prime \prime}\right)\right]$.
If a "weak quartet" does not contain $x$ or $y$, then the number of "weak quartets" in $G$, $G^{\prime}$, and $G^{\prime \prime}$ is the same; the number of "weak quartets" containing $x$ and $y$ in $G^{\prime}$ and $G^{\prime \prime}$ is not less than the number of "weak quartets" containing $x$ and $y$ in $G$; the number of "weak quartets" containing $y$ but not $x$ or containing $x$ but not $y$ in $G^{\prime}$ is at least twice the number of "strong quartets" containing $y$ but not $x$ in $G$; the number of "weak quartets" containing $x$ but not $y$ or containing $y$ but not $x$ in $G^{\prime \prime}$ is at least twice the number of "weak quartets" containing $x$ but not $y$ in $G$. Therefore, we have
$$
Q(G) \leqslant \frac{1}{2}\left[Q\left(G^{\prime}\right)+Q\left(G^{\prime \prime}\right)\right].
$$
Now consider an extreme case, such that
$$
Q(G)=Q\left(G^{\prime}\right)=Q\left(G^{\prime \prime}\right)
$$
We repeatedly copy $G$ pair by pair. If there is an edge between $x$ and $y$, then perform the "copy from $y$ to $x$" and "copy from $x$ to $y$". In the "copy from $y$ to $x$", for any point $z$ adjacent to $x$, perform the "copy from $x$ to $z$". And so on. Finally, we get that $G$ is composed of several disconnected complete graphs. At this point, we have $Q(G) = Q\left(G^{\prime}\right) = Q\left(G^{\prime \prime}\right)$. Therefore, the maximum number of "weak quartets" is achieved in a graph $G$ composed of several disconnected complete graphs.
Suppose graph $G$ contains $n$ complete graphs, each with $a_{1}, a_{2}, \cdots, a_{n}$ points, and $a_{1} \geqslant 0, i=1,2, \cdots, n$, then
$$
Q(G)=\sum_{i=1}^{n} \mathrm{C}_{a_{i}}^{2} \sum_{\substack{1 \leq k \\ i, k \neq 1}} a_{k} a_{k},
$$
where $\mathrm{C}_{a}^{2}=\frac{1}{2} a(a-1)$.
For two numbers in $a_{1}$, say $a_{1}$ and $a_{2}$, when changing the values of $a_{1}$ and $a_{2}$ while keeping $a_{1}+a_{2}=s$ constant, we can write $Q(G)$ as an expression in terms of $a_{1}$ and $a_{2}$:
$$
\begin{array}{l}
Q(G)=C_{a_{1}}^{2}\left(a_{2} \sum_{j=3}^{n} a_{1}+\sum_{i, j=3}^{n} a_{1} a_{j}\right)+C_{a_{2}}^{2}\left(a_{1} \sum_{j=3}^{n} a_{j}+\sum_{i, j=3}^{n} a_{i} a_{j}\right)+ \\
\sum_{j=3}^{n}\left\{c_{j}^{2}\left(a_{1} a_{2}+a_{1} \sum_{\substack{k=3 \\
k=j}}^{n} a_{k}+a_{2} \sum_{\substack{k=3 \\
k=j}}^{n} a_{k}+\sum_{\substack{k=1 \\
k i=1 \\
k i=1}}^{n} a_{k} a_{i}\right)!\right. \\
=A\left(C_{a_{1}}^{2}+C_{a_{2}}^{2}\right)+B\left(a_{1}+a_{2}\right)+C\left(a_{1} C_{a_{2}}^{2}+a_{2} C_{a_{1}}^{2}\right)+D a_{1} a_{2}+E \text {. } \\
\end{array}
$$
where $A=\sum_{\substack{j=3 \\ i0$, then the values of any $a_{i}, a_{j}$ are either equal or differ by 1.
Assume all $a_{1}>0$, when $n$ is divisible by 120,
$$
\begin{array}{l}
Q(G)=n \mathrm{C}_{\frac{100}{2}}^{2} \mathrm{C}_{n-1}^{2}\left(\frac{120}{n}\right)^{2} \\
=\frac{120^{3}(120-n)(n-1)(n-2)}{4 n^{3}} .
\end{array}
$$
When $n \leqslant 6$, it is easy to see that $Q(G)$ achieves its maximum value when $n=5$, and the maximum value is 4769280.
Also, in (1), considering all $a_{1} \in R$, when all $a$ are equal, $Q(G)$ achieves its maximum value. Therefore, when $n \geqslant 6$,
$$
Q(G) \leqslant \frac{120^{3}(120-n)(n-1)(n-2)}{4 n^{3}}.
$$
And the function $\frac{120^{3}(120-n)(n-1)(n-2)}{4 n^{3}}$ is monotonically decreasing with respect to $n$.
In summary, the maximum value of $Q(G)$ is 4769280.
|
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 Olympiad (Round 1))
|
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
$$
f(n+k-1) \cdot 1 \text{, if } k>1, \text{ then } n+k-1 \geqslant n+1. \text{ Thus, by (1), we get }
$$
$$
f(n+k-1) \geqslant f(n+1) \geqslant f(n)+1 \text {. }
$$
(2) and (3) are contradictory, so $k=1$ is the only solution.
When $k=1$, equation (1) becomes $f(n+1)=f(n)+1$. This function satisfies conditions (1) and (2), hence the unique solution is $f(2001)=2002$.
|
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 $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$.
(2001, Shanghai High School Mathematics Competition)
|
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 $a_{0}=x_{1}, a_{k}=x_{k+1}-x_{k}, k=1,2, \cdots, 2000$, we have $x_{1}=a_{0}, x_{k}=\sum_{i=0}^{k-1} a_{i}, k=2,3, \cdots, 2001$, and the condition becomes $\sum_{k=1}^{2000} |a_{k}|=2001$. At this point,
$$
\begin{array}{l}
y_{k}=\frac{1}{k}\left[a_{0}+\left(a_{0}+a_{1}\right)+\cdots+\sum_{i=0}^{k-1} a_{i}\right] \\
=\frac{1}{k}\left[k a_{0}+(k-1) a_{1}+\cdots+a_{k-1}\right], \\
y_{k+1}=\frac{1}{k+1}\left[(k+1) a_{0}+k a_{1}+\cdots+2 a_{k-1}+a_{k}\right].
\end{array}
$$
Then $\left|y_{k}-y_{k+1}\right|$
$$
\begin{array}{l}
=\frac{1}{k(k+1)}\left|-a_{1}-2 a_{2}-\cdots-k a_{k}\right| \\
\leqslant \frac{1}{k(k+1)}\left(\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+k\left|a_{k}\right|\right).
\end{array}
$$
Let $A_{k}=\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+k\left|a_{k}\right|, k=1,2, \cdots, 2000, A_{0}=0$, then
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \leqslant \sum_{k=1}^{2000}\left(\frac{1}{k}-\frac{1}{k+1}\right) A_{k} \\
=\sum_{k=1}^{2000} \frac{1}{k}\left(A_{k}-A_{k-1}\right)-\frac{1}{2001} \cdot A_{2000} \\
=\sum_{k=1}^{2000}\left|a_{k}\right|-\frac{1}{2001} \cdot A_{2000}.
\end{array}
$$
Also, $A_{2000}=\left|a_{1}\right|+2\left|a_{2}\right|+\cdots+2000\left|a_{2000}\right| \geqslant \sum_{k=1}^{2000}\left|a_{k}\right|$, so
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right| \\
\leqslant \sum_{k=1}^{2000}\left|a_{k}\right|-\frac{1}{2001} \cdot \sum_{k=1}^{2000}\left|a_{k}\right| \\
=\frac{2000}{2001} \cdot \sum_{k=1}^{2000}\left|a_{k}\right|=2000.
\end{array}
$$
From the above process, we know that the equality holds if and only if $\left|a_{1}\right|=2001$, $a_{2}=a_{3}=\cdots=a_{2000}=0$. Therefore, the maximum value sought is 2000.
|
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 $A_{3} A_{4} A_{5} A_{6} A_{1}$, which has 5 triangulation methods (Figure 17).
(2) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{4}$, we can take out $\triangle A_{1} A_{2} A_{4}$, leaving the triangle $\triangle A_{2} A_{3} A_{4}$ and the quadrilateral $A_{1} A_{4} A_{5} A_{6}$, which have 2 triangulation methods (Figure 18).
(3) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{5}$, the situation is the same as (2), with 2 triangulation methods (Figure 19).
(4) When the triangulation graph must contain $\triangle A_{1} A_{2} A_{6}$, the situation is the same as (1), with 5 triangulation methods (Figure 20).
In total, there are $5+2+2+5=14$ triangulation methods.
Analysis: If we denote the number of triangulation methods for an $n$-sided polygon as $a_{n}$, and set $a_{2}=a_{3}=1$, then by the same method of taking out 1 $\triangle A_{1} A_{2} A_{k}(3 \leqslant k \leqslant n)$ (Figure 21), we get 1 $(k-1)$-sided polygon and 1 $(n-k+2)$-sided polygon, with $a_{k-1} a_{n-k+2}$ triangulation methods. Taking all $k=3,4, \cdots, n$, we get (the recursive formula)
$$
a_{n}=a_{2} a_{n-1}+a_{3} a_{n-2}+\cdots+a_{n-1} a_{2} .
$$
|
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 $S=\sum_{k=1}^{2008} \frac{1}{x_{k}+1}=\sum_{k=1}^{2008}\left(\frac{1}{x_{k}}-\frac{1}{x_{k+1}}\right)$ $=\frac{1}{x_{1}}-\frac{1}{x_{2004}}=2-\frac{1}{x_{2004}}$.
Since $x_{2} \cos >x_{3}>1$, then $1<S<2$, so $[S]=1$.
|
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 $k$ gaps in the clockwise direction and color the endpoint reached red. Continue this process. How many red points can be obtained at most on the circumference? Prove your conclusion.
|
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, 4th, 6th, ..., $2k$-th, ..., $2n$-th points are respectively renamed as $1^{*}, 2^{*}, 3^{*}, \cdots, k^{*}, \cdots, n^{*}$, it is easy to see that the number of red points that can be colored starting from the $2k$-th point on the circle with $2n$ points is the same as the number of red points that can be colored starting from the $k$-th point on the circle with $n$ points, and they form a one-to-one correspondence.
(2) On the circle with $2n$ points, if the first red point is an odd-numbered point, then the second red point must be an even-numbered point. From (1), we know that in this case, it can also correspond to a coloring method starting from the $k$-th point on the circle with $n$ points, and the number of red points colored is exactly one more than the number of red points obtained by this coloring method on the circle with $n$ points.
Let the maximum number of red points that can be colored on a circle with $n$ points be $f(n)$, and suppose the coloring method starting from the $k$-th point achieves this maximum value. Then, on the circle with $2n$ points, the coloring method starting from the $k$-th point should achieve the maximum value $f(n)+1$ (since the second red point is the $2k$-th point). Therefore, $f(2n)=f(n)+1$. Hence, we have
$$
\begin{array}{l}
f(800)=f(400)+1=f(200)+2 \\
=f(100)+3=f(50)+4=f(25)+5 .
\end{array}
$$
To calculate $f(25)$, note that on a circle with 25 points, if the number of the first red point is a multiple of 5, then the number of each red point colored afterward will also be a multiple of 5. Therefore, the number of red points that can be colored starting from such a point will not exceed 5.
If the number of the first red point is not a multiple of 5, then the number of each red point colored afterward will also not be a multiple of 5. Therefore, starting from such a point, at most 20 red points can be colored. On the other hand, starting from a certain such point, we can actually color 20 red points. For example, starting from the 1st point, we can sequentially color $1, 2, 4, 8, 16, 7, 14, 3, 6, 12, 24, 23, 21, 17, 9, 18, 11, 22, 19, 13$. Therefore, $f(25)=20$. Substituting into equation (1), we get
$$
f(800)=20+5=25 .
$$
Thus, the maximum number of red points that can be colored is 25.
|
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
$$
S_{0}=0, S_{1}=a_{1}, \cdots, S_{k}=\sum_{i=1}^{k} a_{i}, k \leqslant 100 .
$$
By the pigeonhole principle, there exist $0 \leqslant m<n \leqslant 100$ such that $S_{n}-S_{m}=100 t$, i.e.,
$$
\sum_{i=m+1}^{n} a_{i}=100 t .
$$
Since $a_{i}=1$ or 2, then $t=1$ or 2.
If $t=1$, the conclusion holds.
If $t=2$, by $a_{i} \leqslant 2, n \leqslant 100$, we get
$$
200 \geqslant \sum_{i=1}^{100} a_{i} \geqslant \sum_{i=m+1}^{n} a_{i}=200 .
$$
Then $a_{i}=2, i=1,2, \cdots, 100$. Hence $\sum_{i=1}^{50} a_{i}=100$.
In this case, the conclusion still holds. Therefore, $\max \left\{a_{n}, n \in \mathbf{N}\right\} \geqslant 3$.
Next, we construct an infinite periodic sequence such that the largest term is 3. The first 100 terms of the sequence $\left\{a_{n}\right\}$ are
$$
1,2, \cdots, 2,3,2, \cdots, 2, \cdots,
$$
where $a_{100 k+1}=1, a_{100 k+51}=3, k=0,1, \cdots$, and the other terms $a_{i}=2$.
It is easy to verify that the sum of any continuous subsequence of this sequence does not equal 100.
In summary, the minimum value of $\max \left\{a_{n}, n \in \mathbf{N}\right\}$ is 3.
(Zhang Yanwei, Suqian Education Bureau, Jiangsu Province, 223800)
|
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 $f(n) \in \mathbf{N}_{+}$).
(2) From $f(f(n))=3 n$, we get
$$
\begin{array}{l}
f(3 n)=f(f(f(n)))=3 f(n). \\
\text { Therefore, } f\left(3^{n}\right)=3 f\left(3^{n-1}\right)=3^{2} f\left(3^{n-2}\right)=\cdots \\
=3^{n} f(1)=2 \times 3^{n}(n \in \mathbf{N}) .
\end{array}
$$
(3) From (2), $f\left(2 \times 3^{n}\right)=f\left(f\left(3^{n}\right)\right)=3 \times 3^{n}$, so,
$$
f\left(2 \times 3^{n}\right)-f\left(3^{n}\right)=3 \times 3^{n}-2 \times 3^{n}=3^{n}.
$$
And $2 \times 3^{n}-3^{n}=3^{n}$, and $f(n+1) \geqslant f(n)+1$ (since $f(n)$ is a strictly increasing function on $\mathbf{N}_{+}$, $\left.f(n) \in \mathbf{N}_{+}\right)$. Therefore,
$$
\begin{array}{l}
f\left(3^{n}+l\right)=f\left(3^{n}\right)+l=2 \times 3^{n}+l\left(l=1,2, \cdots, 3^{n}\right). \\
\text { (4) } f\left(2 \times 3^{n}+l\right)=f\left(f\left(3^{n}+l\right)\right)=3 \times\left(3^{n}+l\right) \\
=3^{n+1}+3 l \quad\left(l=1,2, \cdots, 3^{n}\right).
\end{array}
$$
Thus, $f(2003)=f\left(2 \times 3^{6}+545\right)=3^{7}+3 \times 545=3822$.
(He Bin, Hubei Province Gucheng County Third Senior High School,
441700)
|
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 minimum perimeter of such a triangle.
|
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 $\left\{\begin{array}{l}l-m=s, \\ m-n=t\end{array}\left(s, t \in \mathbf{N}_{+}\right)\right.$, then $3^{s} \equiv 3^{t} \equiv 1\left(\bmod 10^{4}\right)$.
The perimeter of the triangle is
$$
c=n+n+t+n+s+t=3 n+2 t+s.
$$
Let the smallest positive integer $x$ that satisfies $3^{x} \equiv 1\left(\bmod 10^{4}\right)$ be $k$, so $3^{k} \equiv 1\left(\bmod 10^{4}\right)$. Therefore, $s \geqslant k, t \geqslant k$.
From (1) we get $n+n+t>n+s+t \Rightarrow n>s$.
Thus, $n \geqslant s+1 \geqslant k+1$,
$$
c \geqslant 3(k+1)+2 k+k=6 k+3.
$$
Now we analyze the minimum value of $k$.
According to Euler's theorem, since $\left(3,10^{4}\right)=1$, we have
$$
3^{\varphi\left(10^{4}\right)} \equiv 1\left(\bmod 10^{4}\right).
$$
Here, $\varphi\left(10^{4}\right)=10^{4}\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=4000$.
Thus, $k \mid 4000$.
It is easy to see that the last digit of $3^{k}$ cycles through 3, 9, 7, 1. Since $3^{k} \equiv 1\left(\bmod 10^{4}\right)$, hence $4 \mid k$.
Let $k=4 h\left(h \in \mathbf{N}_{+}\right)$, then
$$
\begin{aligned}
3^{k} & =9^{2 h}=(10-1)^{2 h} \\
= & 10^{2 h}-C_{2 h}^{2 h-1} 10^{2 h-1}+\cdots+C_{2 h}^{4} 10^{4}- \\
& C_{2 h}^{3} 10^{3}+C_{2 h}^{2} 10^{2}-C_{2 h}^{1} 10+1.
\end{aligned}
$$
Therefore, $3^{k} \equiv 1\left(\bmod 10^{4}\right)$
$$
\begin{array}{l}
\Leftrightarrow 3^{k} \equiv-\mathrm{C}_{2 h}^{3} 10^{3}+\mathrm{C}_{2 h}^{2} 10^{2}-\mathrm{C}_{2 h}^{1} 10+1 \\
\equiv 1\left(\bmod 10^{4}\right) \\
\Leftrightarrow-\mathrm{C}_{2 h}^{3} 10^{3}+\mathrm{C}_{2 h}^{2} 10^{2}-\mathrm{C}_{2 h}^{1} 10 \equiv 0\left(\bmod 10^{4}\right) \\
\Leftrightarrow \frac{2 h(2 h-1)}{2} \times 100 \\
\equiv 2 h \cdot 10+\frac{2 h(2 h-1)(2 h-2)}{3 \times 2} \times \\
1000\left(\bmod 10^{4}\right) \text {. } \\
\Leftrightarrow 20 h[50(2 h-1)(2 h-2)+3-15 h(2 h-1)] \\
\equiv 0\left(\bmod 10^{4}\right) \\
\Leftrightarrow 20 h\left(170 h^{2}-285 h+103\right) \equiv 0\left(\bmod 10^{4}\right) \\
\Leftrightarrow h\left(170 h^{2}-285 h+103\right) \equiv 0(\bmod 500) \text {. } \\
\end{array}
$$
Since 5 does not divide $170 h^{2}-285 h+103$, we have $125 \mid h$, i.e., $h=125 p, p \in \mathbf{N}_{+}$, and $p \equiv 1(\bmod 4)$ or $p \equiv 0(\bmod 4)$.
When $p=1$, $h_{\min }=125$. Therefore,
$$
k_{\text {min }}=4 h_{\text {min }}=500.
$$
Thus, the minimum perimeter $c_{\min }=3003$. At this time,
$$
\begin{array}{l}
\left\{\begin{array}{l}
l-m=s=k=500, \\
m-n=t=k=500, \\
n=k+1=501
\end{array}\right. \\
\Rightarrow l=1501, m=1001, n=501.
\end{array}
$$
(Dang Zhongliang, Chongqing Yucai High School, 400050)
|
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 be a different color from both $A$ and $B$, only 4 remaining colors can be used, giving 4 ways to color.
(4) Color region $D$, which only needs to be a different color from $C$, hence there are 5 ways to color.
Multiplying gives $6 \times 5 \times 4 \times 5=600$ different ways to color.
Solution 2: Complete in steps according to the order of $A-D-C-B$, coloring $A$ has 6 methods; coloring $D$ has 6 methods; coloring $C$ must be different from $A$ and $D$, hence there are 4 methods; coloring $B$ must be different from $A$ and $C$, hence there are 4 methods.
Multiplying gives $6 \times 6 \times 4 \times 4=576$ different ways to color.
Solution 3: Complete in steps according to the order of $A-D-B-C$, coloring $A$ has 6 methods; coloring $D$ has 6 methods; coloring $B$ must be different from $A$, hence there are 5 methods; coloring $C$ must be different from $A$, $B$, and $D$, hence there are 3 methods.
Multiplying gives $6 \times 6 \times 5 \times 3=540$ different ways to color.
Analysis: All three solutions use the step-by-step calculation (multiplication) method, but due to different orders, the answers are different, and there must be repetition or omission.
Reviewing Solution 1, no problems are found. Reflecting on Solution 2, we can see that regions $A$ and $D$ are not adjacent, allowing the same color (hence both have 6 ways to color). When $A$ and $D$ are the same color, $C$ is not 4 ways to color, but 5 ways to color, resulting in $6 \times 5 \times 4 = 120$ ways to color. However, adding 576 and 120 would result in repetition. The correct approach is to consider the cases where $A$ and $D$ are different colors and the same color, then add them, resulting in
$6 \times 5 \times 4 \times 4 + 6 \times 5 \times 4 = 600$.
Similarly, in Solution 3, there is also an omission of the coloring of region $C$ when $A$ and $D$ are the same color or $B$ and $D$ are the same color. The correct solution can be divided into 3 cases:
$A$ and $D$ are the same color and $B$ and $D$ are different colors, or $A$ and $D$ are different colors and $B$ and $D$ are the same color, which is equivalent to 3 regions each being colored 1 color, each having $6 \times 5 \times 4 = 120$ ways to color; $A$ and $D$ are different colors and $B$ and $D$ are different colors, which is equivalent to 4 regions each being colored 1 color, having $6 \times 5 \times 4 \times 3 = 360$ ways to color.
In total, $120 + 120 + 360 = 600$ different ways to color.
To reduce errors, the following two suggestions are proposed:
(1) Correspond regions to points, and adjacent relationships to lines between two points. This can simplify the graph and highlight the relationships.
(2) Each time coloring starts from the point with the most adjacent regions. This can better avoid repetition and omission.
Solution 4: Correspond regions $A$, $B$, $C$, $D$ to points $A$, $B$, $C$, $D$, and draw a line between two points when two regions are adjacent, resulting in a graph (Figure 23). The problem is transformed into point coloring, and coloring is done in steps according to the number of lines, in the order of $C-A-B-D$. Point $C$ has 6 ways to color, point $A$ has 5 ways to color, point $B$ has 4 ways to color, and point $D$ has 5 ways to color.
Multiplying gives $6 \times 5 \times 4 \times 5 = 600$ different ways to color.
Coloring in the order of $C-B-A-D$ or $C-A-D-B$ has the same effect. Additionally, the problem can also be solved by classifying based on the use of colors.
|
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 maximum difference between the numbers in each row does not exceed $210-1=209$; the maximum difference between the numbers in each column does not exceed $191-1=190$. Therefore, $N \leqslant 209$.
Next, we prove that $N$ cannot be less than 209. Consider the subsets
$$
M_{1}=\{1,2, \cdots, 91\} \text { and } M_{2}=\{300,301, \cdots, 400\} \text {. }
$$
Color the rows and columns that contain numbers from $M_{1}$ in red; color the rows and columns that contain numbers from $M_{2}$ in blue. We need to show that the number of red rows and columns is at least 20, and the number of blue rows and columns is at least 21. If this is true, then there must be a row or column that is both red and blue, meaning there are two numbers in it with a difference of at least $300-91=209$.
Suppose there are $i$ rows and $j$ columns colored red. Then, all elements of $M_{1}$ are located at the intersections of these rows and columns, so $i j \geqslant 91$.
Thus, $i+j \geqslant 2 \sqrt{\ddot{j}} \geqslant 2 \sqrt{91} \geqslant 19$.
Similarly, the sum of the number of blue rows and columns
$$
i^{\prime}+j^{\prime} \geqslant 2 \sqrt{i^{\prime} j^{\prime}} \geqslant 2 \sqrt{101}>20 .
$$
|
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 Mathematical Olympiad (Second Round))
|
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 we call a "super box". Each "super box" intersects the unit sphere and forms a region on the sphere's surface. Therefore, the sphere's surface can be covered by these spherical regions, meaning the sum of the areas of these spherical regions is greater than or equal to the area of the sphere's surface. Since the area of a spherical cap is $2 \pi R h$, we have $\sum 2 \pi R \cdot d_{t} \geqslant 4 \pi R^{2}$. Since $R=1$, we have $\sum d_{i} \geqslant 2$.
|
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, $D B \perp C E, B C=B E$.
Moreover, $\triangle A C B$ is an isosceles right triangle, and $A B=\sqrt{2}$, so, $B C=B E=1$.
|
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
$$
1+2+3+\cdots+40=41 \times 20=820
$$
numbers, the last number is 1639, so 2003 is not in the 40th group.
Therefore, 2003 is between the 40th and 50th groups.
Assume 2003 is in the 45th group, then the first 45 groups have
$$
1+2+3+\cdots+45=46 \times 22+23=1035
$$
numbers, the last number in this group is 2069, and the first number is 1981, so 2003 is in the 45th group.
|
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) 63
(B) 62
(C) 61
(D) 59
|
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
$x^{2}+y^{2}=A D^{2}=1997$ (where $x, y$ are positive integers).
Obviously, $x, y$ must be one odd and one even. Without loss of generality, let $x$ be odd and $y$ be even, then the last digit of $x$ can only be 1 or 9, and the last digit of $y$ can be 4 or 6.
Since $1997<45^{2}$, hence $x$ does not exceed 39, and $y$ does not exceed 44.
Upon inspection, only when $x=29, y=34$, the equation $x^{2}+y^{2}=1997$ holds. Therefore, $E F=x+y=63$.
|
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, $\angle A_{0} A_{1} A_{2}=1^{\circ}, \angle A_{1} A_{2} A_{3}=2^{\circ}$, $\angle A_{2} A_{3} A_{4}=3^{\circ}, \cdots, \angle A_{k-1} A_{k} A_{k+1}=k^{\circ}(k$ is a positive integer). If the length of the chord $A_{k} A_{k+1}$ is less than 1, then the minimum value of $k$ is $\qquad$.
|
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 $A_{k} A_{k+1}60^{\circ}$, that is
$$
\begin{array}{l}
\frac{k(k+1)}{2}>60, \\
\left(k+\frac{1+\sqrt{481}}{2}\right)\left(k-\frac{-1+\sqrt{481}}{2}\right)>0 .
\end{array}
$$
Solving gives $k>\frac{\sqrt{481}-1}{2}$.
Also, $10<\frac{\sqrt{481}-1}{2}<11, k$ is a positive integer, so the minimum value of $k$ is 11.
|
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 all the lattice points (points with integer coordinates) within and on the boundary of the closed region enclosed by the parabola $y_{1} = a x^{2} + b x + c$ and the line $y_{2} = a x + b$.
|
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 the interval $-1 \leqslant x \leqslant 1$.
Hence, $x=0$ is the axis of symmetry of the function $y_{1}=a x^{2}+b x+c$, i.e., $-\frac{b}{2 a}=0, b=0, a=2$.
Thus, $y_{1}=2 x^{2}-1, y_{2}=2 x$.
Solving the system of equations $\left\{\begin{array}{l}y=2 x^{2}-1, \\ y=2 x\end{array}\right.$, we get
$$
\left\{\begin{array}{l}
x_{1}=\frac{1+\sqrt{3}}{2}, \\
y_{1}=1+\sqrt{3};
\end{array} \left\{\begin{array}{l}
x_{2}=\frac{1-\sqrt{3}}{2}, \\
y_{2}=1-\sqrt{3}.
\end{array}\right.\right.
$$
Therefore, the coordinates of the intersection points of the two graphs are $A\left(\frac{1-\sqrt{3}}{2}, 1-\sqrt{3}\right)$ and $B\left(\frac{1+\sqrt{3}}{2}, 1+\sqrt{3}\right)$. The closed figure formed by the two graphs is shown in Figure 8. Since $\frac{1-\sqrt{3}}{2} \leqslant x \leqslant \frac{1+\sqrt{3}}{2}$ has integer solutions $x=0,1$, the corresponding lattice points are $O(0,0), M(0,-1), N(1,1), P(1,2)$. The quadrilateral $O M N P$ is exactly a parallelogram, with area $S_{\text {quadrilateral }}$ MNP $=$ $1 \times 1=1$.
|
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
$$
\begin{array}{l}
9 a b c \leqslant a b+b c+c a \\
=\frac{1}{2}\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right] \\
=\frac{1}{2}\left[1-\left(a^{2}+b^{2}+c^{2}\right)\right],
\end{array}
$$
which implies $a^{2}+b^{2}+c^{2}+18 a b c \leqslant 1$.
Therefore, $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right) \leqslant 0$.
When $a=b=c=\frac{1}{3}$, $\lg \left(a^{2}+b^{2}+c^{2}+18 a b c\right)$ can achieve its maximum value of 0.
|
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 $n$. Therefore, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 2 m + n - 1 = 4 0 0 8 , } \\
{ n = 1 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=2004, \\
n=1 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 2 m + n - 1 = 1 3 3 6 , } \\
{ n = 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=667, \\
n=3 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 2 m + n - 1 = 1 6 7 , } \\
{ n = 2 4 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=72, \\
n=24 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 2 m + n - 1 = 5 0 1 , } \\
{ n = 8 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=247, \\
n=8 .
\end{array}\right.\right.
\end{array}
$$
Given that $m$ and $n$ are not coprime, we can only take
$$
m=72=2^{3} \times 3^{2}, n=24=2^{3} \times 3 \text {. }
$$
Also, $2004=2^{2} \times 3 \times 167$. Therefore, $(m, n, 2004)=12$.
|
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 result directly.
|
$$
\begin{array}{l}
\text { Solution: Let } k=n^{2}+(n+1)^{2} . \\
\text { By }\left(a^{2}+b^{2}\right)^{2}=\left(a^{2}-b^{2}\right)^{2}+(2 a b)^{2}, \text { we have } \\
k^{2}=(2 n+1)^{2}+[2 n(n+1)]^{2}, \\
k^{4}=\left(4 n^{4}+8 n^{3}-4 n-1\right)^{2}+ \\
\quad\left(8 n^{3}+12 n^{2}+4 n\right)^{2} . \\
\text { Let }\left|\left(4 n^{4}+8 n^{3}-4 n-1\right)-\left(8 n^{3}+12 n^{2}+4 n\right)\right| \\
=1 \text {, i.e., }| 4 n^{4}-12 n^{2}-8 n-1 \mid=1 . \\
\text { If } 4 n^{4}-12 n^{2}-8 n-1=1 \text {, then } \\
4 n^{4}-12 n^{2}-8 n=2,
\end{array}
$$
The left side is divisible by 4, but the right side is not, which is a contradiction;
If $4 n^{4}-12 n^{2}-8 n-1=-1$, then
$$
4 n(n+1)^{2}(n-2)=0 \text {, }
$$
Thus, $n=-1,0,2$.
When $n=-1$ or 0, $k=1$, which contradicts the given condition;
When $n=2$, $k=13$, in this case we have
$$
13=2^{2}+3^{2}, 13^{4}=119^{2}+120^{2} \text {, }
$$
$k$ is unique.
This method uses the formula from $k$ to $k^{4}$,
$$
\left(a^{2}+b^{2}\right)^{2}=\left(a^{2}-b^{2}\right)^{2}+(2 a b)^{2} \text {, }
$$
We can also use the solution formula for Pythagorean equations to prove from $k^{4}$ to $k$, but the process is more complicated and will not be detailed here.
$$
|
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 value of $n$.
(Leng Gangsong, provided)
|
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 convex pentagon $A_{1} P_{1} A_{5} A_{6} A_{7}$, there is at least one point in $M$, denoted as $P_{2}$, and $P_{2}$ is different from $P_{1}$. Connect $P_{1} P_{2}$, then at least 5 of the vertices $A_{1}, A_{2}, \cdots, A_{7}$ are not on the line $P_{1} P_{2}$. By the pigeonhole principle, on one side of the line $P_{1} P_{2}$, there must be 3 vertices, and these 3 vertices, together with points $P_{1}$ and $P_{2}$, form a convex pentagon that contains at least one point $P_{3}$ in $M$.
Next, draw the lines $P_{1} P_{3}$ and $P_{2} P_{3}$. Let the region $\Pi_{3}$ be the half-plane defined by the line $P_{1} P_{2}$ and on the side opposite to $\triangle P_{1} P_{2} P_{3}$ (excluding the line $P_{1} P_{2}$). Similarly, define regions $\Pi_{1}$ and $\Pi_{2}$. Thus, the regions $\Pi_{1}$, $\Pi_{2}$, and $\Pi_{3}$ cover all points on the plane except for $\triangle P_{1} P_{2} P_{3}$. By the pigeonhole principle, among the 7 vertices $A_{1}, A_{2}, \cdots, A_{7}$, there must be $\left[\frac{7}{3}\right]+1=3$ vertices in the same region (let's say $\Pi_{3}$). These 3 vertices, together with $P_{1}$ and $P_{2}$, form a convex pentagon with vertices in $M$, and its interior contains at least one point $P_{4}$ in $M$. Therefore, $n \geqslant 11$.
Next, construct an example to show that $n=11$ is possible.
As shown in Figure 3, the heptagon $A_{1} A_{2} \cdots A_{7}$ is an integer-point heptagon, and the point set $M$ consists of the 7 vertices $A_{1}, A_{2}, \cdots, A_{7}$ and 4 integer points inside it. Clearly, this satisfies condition (1). This point set $M$ also satisfies condition (2). The proof is as follows.
Assume there exists an integer-point convex pentagon whose interior does not contain any integer points. Since the area of any integer-point polygon can be expressed as $\frac{n}{2}\left(n \in \mathbf{N}_{+}\right)$, by the principle of the smallest number, there must be a smallest area integer-point convex pentagon $A B C D E$ whose interior does not contain any integer points. Considering the parity of the coordinates of the vertices, there are only 4 cases: (odd, even), (even, odd), (odd, odd), (even, even). Therefore, among the vertices of the pentagon $A B C D E$, there must be two vertices with the same parity for both coordinates. Thus, the midpoint $P$ of the line segment connecting these two vertices is also an integer point. Since $P$ is not inside the convex pentagon $A B C D E$, $P$ must lie on one of the sides of the pentagon. Without loss of generality, assume $P$ lies on side $A B$, then $P$ is the midpoint of $A B$. Connect $P E$, then $P B C D E$ is a smaller area integer-point convex pentagon whose interior does not contain any integer points. This is a contradiction.
In conclusion, the minimum value of $n$ is 11.
|
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 denoted as $[a, b] = [i-m, j-n]$, and the sum $a+b$ is called the student's position number. The total sum of all students' position numbers is denoted as $S$. Find the difference between the maximum and minimum possible values of $S$.
|
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] \\
= & \left(i_{1}+j_{1}\right)-\left(i_{0}+j_{0}\right) .
\end{aligned}
$$
Therefore, $S_{\max }=(6+8)-(1+1)=12$,
$$
S_{\min }=(1+1)-(6+8)=-12 \text {. }
$$
Thus, the difference between the maximum and minimum values of $S$ is 24.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Does there exist a natural number $n$, such that the sum of the digits of $n^{2}$ equals: (1) 2003; (2) 2002.
|
Solution: (1) Since the remainder of any natural number divided by 9 is equal to the remainder of the sum of its digits divided by 9, and $n^{2} \equiv$ $0,1,4,7(\bmod 9)$, while $2003 \equiv 5(\bmod 9)$, there does not exist such a natural number $n$ that the sum of the digits of $n^{2}$ equals 2003.
$$
\begin{array}{l}
\text { (2) Let } n=10^{22}-3 \text {, then } \\
n^{2}=10^{44}-6 \times 10^{22}+9 \\
=\underbrace{99 \cdots 94}_{21} \underbrace{00 \cdots 09}_{21} \text {. } \\
\end{array}
$$
Thus, the sum of the digits is $9 \times 222+4=2002$.
The first part of the question uses the method of elimination; the second part uses the method of providing an example as mentioned in Example 1.
|
2002
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Place the numbers $1,2,3,4,5,6,7,8$ on the eight vertices of a cube such that the sum of any three numbers on each face is not less than 10. Find the minimum value of the sum of the four numbers on each face.
|
1. Let the four numbers on a certain face be $a_{1}, a_{2}, a_{3}, a_{4}$, whose sum reaches the minimum value, and $a_{1}<a_{2}<a_{3}<a_{4}$. Since the sum of three different positive integers less than 5 is at most 9, it follows that $a_{4} \geqslant 6$. Therefore,
$$
a_{1}+a_{2}+a_{3}+a_{4} \geqslant 16 \text {. }
$$
The example shown in Figure 1
demonstrates that 16 can be achieved.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8.1650 students are arranged in 22 rows and 75 columns. It is known that among any two columns in the same row, the number of pairs of students of the same gender does not exceed 11. Prove: the number of boys does not exceed 928.
|
8. Let the number of boys in the $i$-th row be $a_{i}$, then the number of girls is $75-a_{i}$. According to the problem, we have
$$
\sum_{i=1}^{2}\left(C_{a_{i}}^{2}+C_{n s-a_{i}}^{2}\right) \leqslant 11 \times C_{n s}^{2} .
$$
This is because, for any two columns, the number of pairs of students in the same row who have the same gender does not exceed 11, so the total number of pairs of students in the same row who have the same gender is no more than $11 \times \mathrm{C}_{75}^{2}$. Therefore, we have
$$
\sum_{i=1}^{2}\left(a_{i}^{2}-75 a_{i}\right) \leqslant-30525,
$$
which simplifies to $\sum_{i=1}^{2}\left(2 a_{i}-75\right)^{2} \leqslant 1650$.
Using the Cauchy-Schwarz inequality, we know
$$
\left[\sum_{i=1}^{n}\left(2 a_{i}-75\right)\right]^{2} \leqslant 22 \sum_{i=1}^{n}\left(2 a_{i}-75\right)^{2} \leqslant 36300 \text {. }
$$
Thus, $\sum_{i=1}^{2}\left(2 a_{i}-75\right)<191$.
Therefore, $\sum_{i=1}^{n} a_{i}<\frac{191+1650}{2}<921$.
Hence, the number of boys does not exceed 928.
(2003 Western China Mathematical Olympiad Problem Committee provided)
|
928
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $x, y \in \mathbf{R}$, and satisfy
$$
\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}\right)\left[g\left(t_{1}\right)-g\left(t_{2}\right)\right] \geqslant 0 .
$$
Therefore, for any $x \in(0,1)$, we have
$$
\begin{array}{l}
\left(x-\frac{1}{3}\right)\left(\frac{x}{1+x^{2}}-\frac{3}{10}\right) \geqslant 0, \\
\text { hence } \frac{3 x^{2}-x}{1+x^{2}} \geqslant \frac{3}{10}(3 x-1) .
\end{array}
$$
Similarly, $\frac{3 y^{2}-y}{1+y^{2}} \geqslant \frac{3}{10}(3 y-1)$;
$$
\frac{3 z^{2}-z}{1+z^{2}} \geqslant \frac{3}{10}(3 z-1) \text {. }
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
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}} \\
\geqslant \frac{3}{10}[3(x+y+z)-3]=0 .
\end{array}
$$
When $x=y=z=\frac{1}{3}$, $f(x, y, z)=0$, hence the minimum value is 0.
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $a, b, c$ form an arithmetic sequence, and $c=10, a \cos A=$ $b \cos B, A \neq B$, then the inradius of $\triangle A B C$ is $\qquad$
|
8. 2 .
Let the inradius of $\triangle ABC$ be $r$. Since $a \cos A = b \cos B$, by the Law of Sines, we get $b \sin A = a \sin B$. Therefore, $\sin 2A = \sin 2B$.
Since $A \neq B$, then $A + B = 90^{\circ}$. Thus, $\triangle ABC$ is a right triangle, $\angle C = 90^{\circ}, a^2 + b^2 = c^2$.
Also, since $c = 10$, and $a, b, c$ form an arithmetic sequence, we have $a = 6, b = 8, (8 - r) + (6 - r) = 10$.
Solving this, we get $r = 2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Let $\left\{a_{n}\right\}$ be an arithmetic sequence with all terms being positive integers, an odd number of terms, a non-zero common difference, and the sum of all terms equal to 2004. Then the value of the second term $a_{2}$ is $\qquad$
|
12. 668 .
Let the first term and common difference of an arithmetic sequence be $a$ and $d$, and the number of terms be $n$. Then, $n a+\frac{1}{2} n(n-1) d=2004$, which means
$$
[2 a+(n-1) d] n=2004 \times 2=2^{3} \times 3 \times 167 \text {. }
$$
Since $n$ is odd and $d \neq 0$,
when $n=501$, $a+250 d=4$, which does not meet the condition;
when $n=167$, $a+83 d=12$, which does not meet the condition;
when $n=3$, $a+d=4 \times 167=668$.
Therefore, $a_{2}=668$.
|
668
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A store purchased a certain number of student-specific scientific calculators at the beginning of the school year for 880 yuan, and sold them at 50 yuan each, which were quickly sold out. It was found that students still needed 3 times as many calculators, so the store purchased the required calculators for 2580 yuan, with each unit price being 1 yuan cheaper due to the larger quantity. The store still sold them at 50 yuan each, and finally sold the remaining 4 at a 10% discount. The total profit from this business is ( ) yuan.
(A) 508
(B) 520
(C) 528
(D) 560
|
2.B.
Let the first purchase of calculators be $x$ units, then the second purchase of calculators is $3x$ units. According to the problem, we have
$$
\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 { (yuan). }
$$
|
520
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $x, y, z$ satisfy $x=y+\sqrt{2}, 2 x y+$ $2 \sqrt{2} z^{2}+1=0$. Then the value of $x+y+z$ is $\qquad$
|
$=1.0$.
Since $x=y+\sqrt{2}$, we have,
$$
\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,
$\triangle A B C_{1} 、 \triangle A B C_{2} 、 \triangle A B C_{3} 、 \triangle A B C_{4}$ are the required triangles.
|
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$ into $d^{2}-c^{2}=a^{3} b(a+b)$, we get $d^{2}-13^{2}=8 b(2+b)$.
(1) If $d=17$, substituting gives $17^{2}-13^{2}=8 b(2+b)$, which simplifies to $b^{2}+2 b-15=0$. Solving this, we get $b=3$ or $b=-5$ (discard the latter);
(2) If $d=19$, substituting gives $19^{2}-13^{2}=8 b(2+b)$, which simplifies to $b^{2}+2 b-24=0$. Solving this, we get $b=4$ or $b=-6$, neither of which meet the requirements.
In summary, $a=2, b=3, c=13, d=17$.
Therefore, $a b(c+d)=6 \times(13+17)=180$.
|
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 earns a profit of 2 yuan for selling one chair of type A and 3 yuan for selling one chair of type B. If one woodworker and one painter form a group, how many chairs of types A and B should each group make daily to maximize profit?
Table 1
\begin{tabular}{|c|c|c|c|}
\hline \multirow{2}{*}{ Worker Type } & \multicolumn{2}{|c|}{ Time Required per Chair } & \multirow{2}{*}{ Daily Working Hours } \\
\cline { 1 - 3 } & A & B & \\
\hline Woodworker & 1 & 2 & 8 \\
\hline Painter & 3 & 1 & 9 \\
\hline
\end{tabular}
|
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
$$
$M$ is maximized, with its maximum value being 13.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $n$ be a positive integer not exceeding 2003. If there is an angle $\theta$ such that $(\sin \theta+\mathrm{i} \cos \theta)^{n}=\sin n \theta+\mathrm{i} \cos n \theta$ holds. Then the total number of such $n$ is $\qquad$.
|
9. 501 .
$$
\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 \( i^{n-1}(\sin n \theta + i \cos n \theta) = \sin n \theta + i \cos n \theta \), then \( i^{n-1} = 1 \). Therefore, \( n = 4k + 1 \leq 2003 \), where \( k \) (with \( k \geq 0 \)) is an integer.
Thus, the number of \( n \) that satisfy the condition is \(\left[\frac{2003-1}{4}\right]+1=501\).
|
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 $72+56+42+30+20+12+6+2=240$.
|
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 $A$ has $\mathrm{C}_{4}^{1}$ ways, and the remaining 4 blocks are painted with the remaining three colors, with one and only one pair of non-adjacent regions painted the same color. There are 2 ways to choose a pair of non-adjacent regions $(B, D$ or $C, E)$, and choosing one color from the remaining three colors to paint these non-adjacent regions has $\mathrm{C}_{3}^{1}$ ways. Finally, the number of ways to paint the remaining two regions with the remaining two colors is $\mathrm{P}_{2}^{2}$. According to the multiplication principle, there are $C_{5}^{4} \cdot C_{4}^{1} \cdot 2 \cdot C_{3}^{1} \cdot P_{2}^{2}=240$ methods.
(3) Using three colors. The number of ways to choose three colors is $\mathrm{C}_{5}^{3}$. Painting $A, B$ and $D, C$ and $E$ with one color each has $\mathrm{P}_{3}^{3}$ ways, so there are $\mathrm{C}_{5}^{3} \mathrm{P}_{5}^{3}=60$ methods.
According to the addition principle, there are a total of $120+240+60=420$ methods.
|
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 $2 \sqrt{p}-\sqrt{q}+2=1$.
Thus, $(2 \sqrt{p}-\sqrt{q}+2)^{3}=1$.
|
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 for a long time are both fixed numbers. Furthermore, it was found that if only 1 window is opened, it takes $45 \mathrm{~min}$ to ensure that all waiting students can buy lunch; if 2 windows are opened simultaneously, it takes $30 \mathrm{~min}$. It was also discovered that if all waiting students can buy lunch within $25 \mathrm{~min}$, the number of people eating outside can be reduced by $80 \%$ within a unit of time. Assuming the total number of students in the school remains constant and everyone needs to have lunch, to facilitate student dining, the research group suggests that the school canteen should sell out lunch within $20 \mathrm{~min}$. How many windows should be opened simultaneously at least?
|
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
$$
\left\{\begin{array}{l}
45 x=z-45 y, \\
2 \times 30 x=z-30 y, \\
20 n x \geqslant z-20(1-80 \%) y .
\end{array}\right.
$$
From (1) and (2), we get $y=x, z=90 x$. Substituting into (3) yields
$$
20 n x \geqslant 90 x-4 x, n \geqslant 4.3 \text{. }
$$
Therefore, at least 5 windows need to be open simultaneously.
|
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 this section is $\qquad$
|
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
$$
\begin{array}{l}
h_{b}=h_{d}, h_{r}=h_{b}, \\
h_{a}=2 h_{b} .
\end{array}
$$
Since $h_{c}=1$, then $h_{b}=h_{d}=1, h_{a}=2$.
Since $S_{i n k}=\frac{1}{6} S_{\text {Mec }}$, then $V_{D-B E F}=\frac{1}{6} V_{\text {Mec }}$.
Since $S_{\triangle A F:}=\frac{1}{2} S_{\triangle A B C}$, then $V_{H-\triangle F C}=-\frac{1}{4} V_{\text {Hexa } A B C C D}=3$.
Let the area of the section $E F H G$ be $x$, then $V_{A-E F H G}=\frac{2}{3} x, V_{D-E F H G}=\frac{1}{3} x$.
As shown in Figure 3, we have $12=-\frac{2}{3},-3+2+\frac{1}{3} x$.
Therefore, $x=7$.
|
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 $10^{9}\frac{10^{9}}{10^{5}}=10^{4}, 10^{5}+10, a_{i} \in\{0,1, \cdots, 9\},(i=0,1,2,3,4)$, then $A=A_{0}\left(10^{5}-1\right)$
$$
=\overline{a_{0} a_{1} a_{2} a_{3} a_{4} 00000}-\overline{a_{0} a_{1} a_{2} a_{3} a_{4}} \text {. }
$$
$$
\text { Let } A=\overline{c_{0} c_{1} c_{2} c_{3} c_{4} c_{0}^{\prime} c_{1}^{\prime} c_{2}^{\prime} c_{3}^{\prime} c_{4}^{\prime}} \quad\left(c_{0} \neq 0\right) \text {. }
$$
If $a_{4} \neq 0$, then $c_{0}=a_{0}, c_{1}=a_{1}, c_{2}=a_{2}, c_{3}=a_{3}$,
$$
\begin{array}{l}
c_{4}=a_{4}-1, c_{0}^{\prime}=9-a_{0}, c_{1}^{\prime}=9-a_{1}, c_{2}^{\prime}=9-a_{2}, c_{3}^{\prime}= \\
9-a_{3}, c_{4}^{\prime}=10-a_{4} ;
\end{array}
$$
If $a_{4}=0$, then $a_{3} \neq 0$ (otherwise $A$ would have two
$$
\begin{array}{l}
9) \text {. Thus, } c_{0}=a_{0}, c_{1}=a_{1}, c_{2}=a_{2}, c_{3}=a_{3}-1, c_{4}=9, \\
c_{0}^{\prime}=9-a_{0}, c_{1}^{\prime}=9-a_{1}, c_{2}^{\prime}=9-a_{2}, c_{3}^{\prime}=10-a_{3}, \\
c_{4}^{\prime}=0 .
\end{array}
$$
Therefore, $c_{i}+c_{i}^{\prime}=9(0 \leqslant i \leqslant 4$, and $i \in \mathbf{N})$.
$$
\begin{array}{l}
\text { Hence, }\left\{\left(c_{i}, c_{i}^{\prime}\right) \mid 0 \leqslant i \leqslant 4, i \in \mathbf{N}\right. \\
\quad=\{(0,9),(1,8),(2,7),(3,6),(4,5),(9,0), \\
(8,1),(7,2),(6,3),(5,4)\} .
\end{array}
$$
Since $c_{0} \neq 0$, the number of $A$ satisfying condition (1) is $5!\times 2^{5}-4!\times 2^{4}=3456$.
Conversely, any number satisfying condition (1) can be expressed as:
$$
\begin{array}{l}
\overline{c_{0} c_{1} c_{2} c_{3} c_{4} 00000}+\overline{c_{0}^{\prime} c_{1}^{\prime} c_{2}^{\prime} c_{3}^{\prime} c_{4}^{\prime}} \\
=\overline{c_{0} c_{1} c_{2} c_{3} c_{4}} \cdot\left(10^{5}-1\right)+99999 \\
\end{array}
$$
which is necessarily divisible by 11111.
Therefore, $|M|=3456$.
|
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 takes 2 fewer 1-step moves and 1 more 2-step move, resulting in 1 fewer step). Considering the 2-step moves in each step, we easily get
$$
a_{18}=\mathrm{C}_{18}^{1}, a_{17}=\mathrm{C}_{17}^{1}, a_{16}=\mathrm{C}_{16}^{2}, \cdots, a_{9}=\mathrm{C}_{9}^{9} .
$$
In summary, the number of ways he can climb one floor is
$$
\begin{array}{l}
\mathrm{C}_{18}^{0}+\mathrm{C}_{17}^{1}+\mathrm{C}_{16}^{2}+\cdots+\mathrm{C}_{9}^{9} \\
=1+17+120+\cdots+1=4181
\end{array}
$$
Solution 2: Let $F_{n}$ represent the number of ways to climb $n$ steps.
Obviously, $F_{1}=1, F_{2}=2$. For $F_{3}, F_{4}, \cdots$, there are only two different ways to start: climb 1 step or climb 2 steps.
Therefore, for $F_{3}$, if the first step is 1 step, there are 3-1 $=2$ steps left, which counts as $F_{2}$ different ways; if the first step is 2 steps, there are 3-2 =1 step left, which counts as $F_{1}$ different ways. In total, $F_{3}=F_{2}+F_{1}=2+1=3$.
$$
\begin{array}{l}
\text { Similarly, } F_{4}=F_{3}+F_{2}=3+2=5, \\
F_{5}=F_{4}+F_{3}=5+3=8, \\
\cdots \cdots \\
F_{18}=F_{17}+F_{16}=2584+1597=4181 .
\end{array}
$$
|
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 minutes) of $A_{1}, A_{2}, A_{3}, A_{4}$ can all be divided by 7, while the total playing time (in minutes) of $A_{5}, A_{6}, A_{7}$ can all be divided by 13. If there are no limits on the number of substitutions per match, then, based on the total playing time of each player, how many different scenarios are there?
|
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 $k_{i}(i=1,2, \cdots, 7) \in \mathbf{Z}_{+}$.
$$
\text { Let } \sum_{i=1}^{4} k_{i}=m, \sum_{i=5}^{7} k_{i}=n \text {, where } m \geqslant 4, n \geqslant
$$
3 , and $m, n \in \mathbf{Z}_{+}$, then $7 m+13 n=270$. It is easy to find an integer particular solution as $m=33, n=3$. Since $(7,13)=1$, the general integer solution is $m=33+13 t, n=3-7 t$.
Also, because $\left\{\begin{array}{l}33+13 t \geqslant 4, \\ 3-7 t \geqslant 3,\end{array}\right.$ we have $-\frac{29}{13} \leqslant t \leqslant 0$.
Thus, the integer $t=0,-1,-2$.
Therefore, all integer solutions that satisfy the conditions are
$$
(m, n)=(33,3),(20,10),(7,17) \text {. }
$$
For $\sum_{i=1}^{4} k_{i}=33$, the positive integer solutions can be written as a row of 33 ones, and in the 32 spaces between each pair of ones, choose 3 to insert “+” signs. Then, count the ones in the 4 parts separated by the 3 “+” signs to get one positive integer solution, so $\sum_{i=1}^{4} k_{i}=33$ has $\mathrm{C}_{32}^{3}$ positive integer solutions $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$;
Similarly, $\sum_{i=5}^{7} k_{i}=3$ has $\mathrm{C}_{2}^{2}$ positive integer solutions $\left(k_{5}, k_{6}, k_{7}\right)$.
Thus, the number of positive integer solutions $\left(k_{1}, k_{2}, k_{3}, k_{4}, k_{5}, k_{6}, k_{7}\right)$ that satisfy the conditions is $\mathrm{C}_{32}^{3} \cdot \mathrm{C}_{2}^{2}$...
Therefore, the total number of positive integer solutions $\left(k_{1}, k_{2}, k_{3}, k_{4}, k_{5}, k_{6}, k_{7}\right)$ that satisfy the conditions is
$$
\begin{array}{l}
\mathrm{C}_{32}^{3} \cdot \mathrm{C}_{2}^{2}+\mathrm{C}_{19}^{3} \cdot \mathrm{C}_{9}^{2}+\mathrm{C}_{6}^{3} \cdot \mathrm{C}_{16}^{2} \\
=4960+34884+2400=42244 \text { solutions. }
\end{array}
$$
Hence, based on the total time each player is on the field, there are 42244 different scenarios.
|
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 numbers that are multiples of 3 turn around; then, all students who reported numbers that are multiples of 5 turn around; finally, all students who reported numbers that are multiples of 7 turn around. Questions:
(1) How many students are still facing the coach?
(2) Among the students facing the coach, what is the number reported by the 66th student from left to right?
|
Solution: (1) Let $U=\{1,2, \cdots, 240\}, A_{i}$ represent the set of all multiples of $i$ in $U$. Then
$|U|=240 ;\left|A_{3}\right|=\left[\frac{240}{3}\right]=80$,
$\left|A_{5}\right|=\left[\frac{240}{5}\right]=48,\left|A_{7}\right|=\left[\frac{240}{7}\right]=34$;
$\left|A_{15}\right|=\left[\frac{240}{15}\right]=16,\left|A_{21}\right|=\left[\frac{240}{21}\right]=11$,
$\left|A_{35}\right|=\left[\frac{240}{35}\right]=6 ;\left|A_{105}\right|=\left[\frac{240}{105}\right]=2$.
Therefore, we have $|U|-\left(\left|A_{3}\right|+\left|A_{5}\right|+\left|A_{7}\right|\right)$
$+2\left(\left|A_{15}\right|+\left|A_{21}\right|+\left|A_{35}\right|\right)-4\left|A_{105}\right|=136$ students facing the coach.
If we use a Venn diagram for analysis, fill in the data obtained above into Figure 1, paying attention to the order from inside to outside.
As shown in Figure 1, the students facing the coach are clearly visible, with $109+14+4+9=136$ students.
(2) Using a similar method, we can calculate that among the first 105 students from left to right, 60 students are facing the coach.
Considering that students who report numbers that are not multiples of $3,5,7$ do not turn, they face the coach; students who report numbers that are multiples of two of $3,5,7$ turn twice, they still face the coach; the rest of the students turn once or three times, and they face away from the coach.
Starting from the 106th student, consider the numbers that are multiples of 3, 5, 7.
Multiples of 3 (starting from 105 and adding 3 each time):
$108,111,114,117,120, \cdots$
Multiples of 5 (starting from 105 and adding 5 each time):
$110,115,120, \cdots$
Multiples of 7 (starting from 105 and adding 7 each time):
$112,119, \cdots$
Therefore, starting from the 106th student, the numbers reported by students facing the coach from left to right are:
$$
106,107,109,113,116,118,120, \cdots \text {. }
$$
From this, we can see that the number reported by the 66th student facing the coach is 118.
|
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,
$$
\begin{array}{l}
b \sqrt{\frac{b}{a}}+a \sqrt{\frac{a}{b}} \\
=-\frac{(a+b)^{2}-2 a b}{\sqrt{a b}}=-23 .
\end{array}
$$
|
-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, $x^{2}+y^{2}=(x+y)^{2}-2 x y=146$.
|
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 $\left\{x \mid x=x_{n}\right.$, $n \in \mathbf{N}\}$ is $\qquad$.
|
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 the arithmetic sequence be $A_{12}$, then $G_{6}+A_{12}=$ $\qquad$ .
|
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 $q^{2}=4$. Hence, $q=2$ or $q=-2$. Since $q>0$, then $q=2$. At this point, $b_{1}=3$.
Therefore, $G_{6}=\frac{b_{1}\left(1-q^{6}\right)}{1-q}=\frac{3(1-64)}{1-2}=189$.
Let $d$ be the common difference of the arithmetic sequence $\left\{a_{n}\right\}$, then according to the given conditions, $\left\{\begin{array}{l}a_{1}=3, \\ a_{1}+2 d=6 .\end{array}\right.$ This yields $d=\frac{3}{2}$. Therefore,
$$
a_{12}=a_{1}+11 d=\frac{39}{2} \text {. }
$$
Thus, $A_{12}=\frac{\left(a_{1}+a_{12}\right) \times 12}{2}=\left(3+\frac{39}{2}\right) \times 6=135$.
Hence, $G_{6}+A_{12}=189+135=324$.
|
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 $x+f(x)+x f(x)=x+x f(x)+1+f(x)-1$
$=(x+1)[1+f(x)]-1$.
When $x=1$, $x+1$ is even, so regardless of which value $f(x)$ takes from $N$, $x+f(x)+x f(x)$ is always an odd number. Therefore, $f(x)$ has 5 possible values.
When $x=-2$, since $x+1$ is odd, for $x+f(x)+x f(x)$ to be odd, $1+f(x)$ must be even, which means $f(x)$ must be odd. Therefore, $f(x)$ has 3 possible values.
Similarly, when $x=0$, $f(x)$ also has 3 possible values.
Thus, the total number of different mappings is $3 \times 3 \times 5=45$.
|
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
$$
\begin{array}{l}
\left(x+\sqrt{x^{2}+2002}\right)\left(x-\sqrt{x^{2}+2002}\right) . \\
\quad\left(y+\sqrt{y^{2}+2002}\right)\left(y-\sqrt{y^{2}+2002}\right) \\
=(-2002) \times(-2002)=2002^{2}, \\
\left(x+\sqrt{x^{2}+2002}\right)\left(y+\sqrt{y^{2}+2002}\right) \\
=2002,
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }\left(x-\sqrt{x^{2}+2002}\right)\left(y-\sqrt{y^{2}+2002}\right) \\
=2002 \text {. }
\end{array}
$$
Thus, $x+\sqrt{x^{2}+2002}=\frac{2002}{y+\sqrt{y^{2}+2002}}$
$$
\begin{array}{l}
=-\left(y-\sqrt{y^{2}+2002}\right), \\
x-\sqrt{x^{2}+2002}=-\left(y+\sqrt{y^{2}+2002}\right) .
\end{array}
$$
Adding the two equations gives $2 x=-2 y$, i.e., $x+y=0$.
Therefore, $x^{2}-3 x y-4 y^{2}-6 x-6 y+58$
$$
=(x-4 y)(x+y)-6(x+y)+58=58 \text {. }
$$
|
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 $x-y \leqslant 0$ and $x+y+6 \geqslant 0$.
The area enclosed by the above figures is $8 \pi \approx 25.13$. Therefore, the area of the plane represented by the set $R$ is closest to 25.
|
25
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{49}-\frac{1}{50}\right) \div\left(\frac{1}{20}+\frac{1}{77}+\cdots+\frac{1}{50}\right) \\
= \\
\end{array}
$$
|
$=、 1.1$
$$
\begin{array}{l}
\text { Original expression }=\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{49}+\frac{1}{50}\right)-\right. \\
\left.\quad\left(1+\frac{1}{2}+\cdots+\frac{1}{25}\right)\right] \div\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right) \\
=\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right) \div\left(\frac{1}{26}+\frac{1}{27}+\cdots+\frac{1}{50}\right)=1 .
\end{array}
$$
|
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 $f(n)$ is 20, and
$$
f(1)+f(2)+\cdots+f(20)=70 \text {. }
$$
Also, $2004=20 \times 100+4$, so,
$$
\begin{array}{l}
f(1)+f(2)+\cdots+f(2004) \\
=70 \times 100+1+3+6+0=7010
\end{array}
$$
|
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
$$
\begin{array}{l}
2\left(x_{1}+x_{2}+\cdots+x_{2004}\right) \\
=x_{2004}^{2}+2 x_{2004}-2004=\left(x_{2004}+1\right)^{2}-2005,
\end{array}
$$
which means $\left.\left|x_{1}+x_{2}+\cdots+x_{2001}\right|=\frac{1}{2} \right\rvert\,\left(x_{2001}+1\right)^{2}-20051$.
Since $x_{1}+x_{2}+\cdots+x_{2004}$ is an integer, $x_{2004}+1$ must be an odd number. By comparing $143^{2}-20051$ and $145^{2}-20051$, we get
$$
\left|x_{1}+x_{2}+\cdots+x_{2004}\right| \geqslant \frac{1}{2}\left|45^{2}-2005\right|=10 \text {. }
$$
When $x_{0}=x_{2}=\cdots=x_{1900}=0, x_{1}=x_{3}=\cdots=x_{1999}=$ $-1, x_{11}=1, x_{1962}=2, x_{193}=3, \cdots, x_{2004}=44$, the equality holds.
Therefore, the minimum value of $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|$ is 10.
|
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. Similarly, each row has 2 ways of coloring, resulting in a total of $2^{6}$ ways.
(2) At least two adjacent points are the same color, which gives $\left(2^{6}-2\right)$ ways of coloring. In this case, when coloring the second row, it can be observed that the valid coloring is unique. Similarly, each row has a unique way of coloring. Therefore, there are $\left(2^{6}-2\right)$ ways of coloring.
Combining (1) and (2), the total number of valid coloring methods is
$$
2^{6}+\left(2^{6}-2\right)=2^{7}-2=126 \text { ways. }
$$
|
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 $A B C D$, and let $A E$ $=y, O E=x$. Thus,
$$
x^{2}+y^{2}=7 \text {, and } 4-x^{2}=7-y^{2} \text {. }
$$
Solving these equations, we get $x=\sqrt{2}, y=\sqrt{5}$. At this point,
$$
V_{2}=\frac{1}{3} \times \sqrt{2} \times 2 \sqrt{5} \times \sqrt{2}=\frac{4}{3} \sqrt{5}=V_{1} \text {. }
$$
(3) As shown in Figure 6, three lateral edges $(O B, O C, O D)$ are $\sqrt{7}$, and one lateral edge $O A=2$.
Let $A D$ and $B C$ intersect at point $E$, and let $B E=x$.
Since $A B=A C=\sqrt{7}, B D=D C=2$, we have
$$
D E^{2}=4-x^{2}, O E^{2}=7-x^{2}, A E^{2}=7-x^{2} \text {. }
$$
Because $\cos \angle O E D=-\cos \angle O E A$, we have
$$
\frac{O E^{2}+E D^{2}-O D^{2}}{20 E \cdot E D}=-\frac{O E^{2}+E A^{2}-O A^{2}}{20 E \cdot E A} \text {. }
$$
Substituting the previous results into the above equation, we solve to get $x=\sqrt{3}$ or $2 \sqrt{2}$. Clearly, $x \neq 2 \sqrt{2}$, so $x=\sqrt{3}$. At this point, $V_{3}=\frac{1}{3} \sqrt{3} \times 3 \sqrt{3}=3$.
In summary, $V_{3}>V_{2}=V_{1}$. Therefore, $V_{\text {max }}=3$.
|
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 $a+b$ and $c$ is 1;
(3) For any two distinct elements $a$ and $b$ in $S$, there exists another element $c$ in $S$ such that the gcd of $a+b$ and $c$ is greater than 1.
|
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 both numbers from the same array. By the pigeonhole principle, the number of elements in set $S$ is no more than 50.
On the other hand, we construct the set $A=\{2,1,3,5,7, \cdots, 95,97\}$. This set contains 2 and 49 odd numbers less than 98.
We now show that set $A$ satisfies the given conditions.
For any two elements $a$ and $b$ in set $A$:
(i) If $a=2$, then $b$ is an odd number.
If $b=1$, it is easy to see that there exists an element $c$ in $A$ that satisfies the given conditions;
If $3 \leqslant b \leqslant 95$, then the greatest common divisor of the element 1 in $A$ and $a+b$ is 1, and the greatest common divisor of the element $b+2$ in $A$ and $a+b$ is $b+2$, which is greater than 1;
If $b=97$, it is easy to see that there exists an element $c$ in $A$ that satisfies the given conditions.
(ii) If $a$ and $b$ are both not equal to 2, then $a$ and $b$ are both odd numbers, and $a+b$ is even. Thus, the greatest common divisor of $a+b$ and 2 is 2, which is greater than 1, and $a+b$ must be coprime with either 89 or 91.
Therefore, set $A$ satisfies the given conditions.
Thus, the maximum number of elements in set $S$ is 50.
(Zhang Yanwei, Bureau of Education, Suqian City, Jiangsu Province, 223800)
|
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 of the algebraic sum remains unchanged with each adjustment, meaning the total sum is always odd. Therefore,
$$
1+(2-3-4+5)+(6-7-8+9)+\cdots+
$$
$$
(1986-1987-1988+1989)=1 \text {, }
$$
Thus, the minimum value is 1.
|
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 expressions, then solve for the value through a system of equations. Because
$$
\begin{array}{l}
\alpha^{4}+3 \beta+\beta^{4}+3 \alpha \\
=\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]^{2}-2(\alpha \beta)^{2}+3(\alpha+\beta) \\
=10, \\
\left(\alpha^{4}+3 \beta\right)-\left(\beta^{4}+3 \alpha\right) \\
=(\alpha-\beta)\left\{(\alpha+\beta)\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]-3\right\} \\
=0,
\end{array}
$$
Therefore, $\alpha^{4}+3 \beta=\frac{10+0}{2}=5$.
|
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
$$
\begin{array}{l}
7(x+y)-3 x y=16,16(x+y)-7 x y=42 . \\
\text { Solving, we get } x+y=-14, x y=-38 . \\
\text { Therefore, } S_{n}=-14 S_{n-1}+38 S_{n-2} \quad(n \geqslant 3) . \\
\text { Hence, } S_{5}=a x^{5}+b y^{5}=-14 S_{4}+38 S_{3} \\
=-14 \times 42+38 \times 16=20 .
\end{array}
$$
|
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:
$$
1,1, \sqrt{2} ; \sqrt{2}, \sqrt{2}, 2 ; 2,2,2 \sqrt{2} ; \sqrt{5}, \sqrt{5}, \sqrt{10} \text {. }
$$
Below, we count the triangles by classifying them according to the length of the hypotenuse.
(1) When the hypotenuse is $\sqrt{2}$, the hypotenuse must be the diagonal of a small square. There are 12 such line segments, and each such line segment corresponds to two isosceles right triangles, totaling $2 \times 12=24$.
(2) When the hypotenuse is 2, there are 10 line segments of length 2 in the figure, 6 of which are on the perimeter of a $2 \times 3$ rectangle, each corresponding to one isosceles right triangle; the other 4 are inside the $2 \times 3$ rectangle, each corresponding to two isosceles right triangles. In total, there are $6+2 \times 4=14$.
(3) When the hypotenuse is $2 \sqrt{2}$, the hypotenuse must be the diagonal of a $2 \times 2$ square. There are 4 such line segments, each corresponding to two isosceles right triangles, totaling $2 \times 4=8$.
(4) When the hypotenuse is $\sqrt{10}$, there are 4 such line segments, each corresponding to one isosceles right triangle, totaling 4.
Therefore, the number of isosceles right triangles with grid points as vertices is $24+14+8+4=50$.
|
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 expressed as the sum of three distinct composite numbers. Hence, $m=17$.
|
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
$$
\begin{array}{l}
O C=\sqrt{x^{2}+1}, \\
O D=\sqrt{(4-x)^{2}+4} .
\end{array}
$$
Thus, the problem is to find a point $O$ on $A B$ such that $O C+O D$ is minimized.
Let the symmetric point of $C$ with respect to $A B$ be $E$, then the intersection of $D E$ and $A B$ is point $O$. At this point, $O C+O D=O E+O D=D E$. Draw $E F / / A B$ intersecting the extension of $D B$ at $F$.
In the right triangle $\triangle D E F$, it is easy to know that $E F=A B=4, D F=3$, so, $D E=5$.
Therefore, the function
$$
f(x)=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}
$$
has a minimum value of 5.
|
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 over, it also throws the extra apple into the sea and takes one pile; the third monkey does the same. Then, there were at least how many apples originally.
|
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 integer, $n+2$ must be a multiple of $3^{3}$, so the minimum value of $n$ is 25.
|
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}{|c|c|c|c|c|c|c|c|}
\hline$x$ & $x_{1}$ & $x_{2}$ & $x_{3}$ & $x_{4}$ & $x_{5}$ & $x_{6}$ & $x_{7}$ \\
\hline$y$ & 51 & 107 & 185 & 285 & 407 & 549 & 717 \\
\hline
\end{tabular}
However, due to carelessness, one of the $y$ values was calculated incorrectly. Please point out which value is incorrect, what the correct value should be, and explain your reasoning.
|
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
$$
\begin{array}{l}
\Delta_{k}=y_{k+1}-y_{k} \\
=\left(a x_{k+1}^{2}+b x_{k+1}+c\right)-\left(a x_{k}^{2}+b x_{k}+c\right) \\
=a\left[\left(x_{k}+d\right)^{2}-x_{k}^{2}\right]+b\left[\left(x_{k}+d\right)-x_{k}\right] \\
=2 a d x_{k}+\left(a d^{2}+b d\right) .
\end{array}
$$
Therefore, $\Delta_{k+1}-\Delta_{k}=2 a d\left(x_{k+1}-x_{k}\right)=2 a d^{2}$ (a constant).
From the given data
$$
\begin{array}{cllllllllll}
& y_{k} & : & 51 & 107 & 185 & 285 & 407 & 549 & 717 \\
\text { we get } & \Delta_{k} & : & & 56 & 78 & 100 & 122 & 142 & 168 \\
\Delta_{k+1}-\Delta_{k} & : & & 22 & 22 & 22 & 20 & 26
\end{array}
$$
From this, it is evident that 549 is the incorrect $y$ value, and its correct value should be 551.
|
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 $A, B \geqslant 0, 6B \geqslant 5A \geqslant$ $3B, 9B \geqslant 5A+99$, find the minimum value of $S$.
Since $15A \geqslant 9B \geqslant 99+5A$, hence
$$
10A \geqslant 99, A \geqslant 10.
$$
Thus, $9B \geqslant 50+99=149, B \geqslant 17$.
Then, $5A \geqslant 51, A \geqslant 11$.
Further, $9B \geqslant 55+99=154, B \geqslant 18$.
If $B \geqslant 19$, then
$$
\begin{array}{l}
S=9B+2(9B-5A) \\
\geqslant 9 \times 19+2 \times 99=369.
\end{array}
$$
If $B=18$, then
$$
5A \leqslant 9B-99=63, A \leqslant 12.
$$
Therefore, $A$ can only be 12 or 11.
When $A=12$, $S=27 \times 18-10 \times 12=366$ reaches the minimum value, and this is the only case, at this time $x=48, y=6$.
|
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 $m+10=67 k$ (where $k$ is a positive integer), because $32 \leqslant m \leqslant 99$, then $m+10=67$, i.e., $m=57$.
Upon verification, $57^{2}=3249$, which does not meet the condition, so it is discarded.
(2) If $m-10=67 k$ (where $k$ is a positive integer), then $m-10=67, m=77$.
Thus, $\overline{a b c d}=77^{2}=5929$.
|
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 contains $k+2$, and these subsets are either the union of a corresponding subset of $\{1,2, \cdots, k\}$ with $\{k+2\}$, or the union of a single-element subset of $\{1,2, \cdots, k\}$ with $\{k+2\}$, totaling
\[
\begin{array}{l}
a_{k}+k \text { subsets. } \\
\text { Therefore, } a_{k+2}=a_{k+1}+a_{k}+k . \\
\text { Clearly, } a_{3}=1, a_{4}=3 . \\
\text { Thus, } a_{5}=7, a_{6}=14, a_{7}=26, \\
a_{8}=46, a_{9}=79, a_{10}=133 .
\end{array}
\]
|
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 both belong to $A$. At this time, at least $133-$ $8=125$ numbers do not belong to $A$. Therefore,
$|A| \leqslant 1995-125=1870$.
Furthermore, we can take $A=\{1,2, \cdots, 8\} \cup\{134,135, \cdots$, $1995\}$, so the maximum value of $|A|$ is 1870.
|
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
$$
\begin{array}{l}
a_{x}=b_{1}+b_{k}, \\
a_{x^{\prime}}=\left(1001-b_{1}\right)+\left(1001-b_{n}\right) \\
=2002-b_{1}-b_{k},
\end{array}
$$
The arithmetic mean of $a_{x}$ and $a_{x^{\prime}}$ is 1001.
Solution: Pair the non-empty subsets of $M$. For each non-empty subset $X \subset M, X^{\prime}=\{1001-x \mid x \in X\}$, then $X^{\prime} \subset M$.
If $X \neq X^{\prime}$, then $a_{x}+a_{x^{\prime}}=2002$.
If $X=X^{\prime}$, then $a_{x}=1001$.
In summary, the arithmetic mean of all such $a_{x}$ is 1001.
|
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 $x_{1}<x_{2}$.
Then $x_{1}<0<x_{2}$, and the axis of symmetry is $x=-\frac{b}{2 a} \leqslant 0$.
Figure 4
The graph is shown in Figure 4.
Thus, $\left|x_{2}\right|=\left|\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}\right|$
$$
=\frac{b-\sqrt{b^{2}-4 a c}}{2 a}=c \text {. }
$$
Therefore, $\frac{4 a c-b^{2}}{4 a} \geqslant c=\frac{b-\sqrt{b^{2}-4 a c}}{2 a} \geqslant-\frac{\sqrt{b^{2}-4 a c}}{2 a}$.
So, $b^{2}-4 a c \geqslant 4$.
When $a=-1, b=0, c=1$, the equality holds.
Thus, the minimum value of $b^{2}-4 a c$ is 4.
|
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 by 7 ($i=1,2, \cdots, 6$), then
$$
\begin{array}{l}
A_{0}=\{7,14,21,28,35,42,49\}, \\
A_{1}=\{1,8,15,22,29,36,43,50\}, \\
A_{2}=\{2,9,16,23,30,37,44\}, \\
A_{3}=\{3,10,17,24,31,38,45\}, \\
A_{4}=\{4,11,18,25,32,39,46\}, \\
A_{5}=\{5,12,19,26,33,40,47\}, \\
A_{6}=\{6,13,20,27,34,41,48\},
\end{array}
$$
According to the problem:
(1) $S$ can contain at most one element from $A_{0}$;
(2) If $S$ contains one element from $A_{i}$, then it can contain all elements from this set, but it cannot contain any elements from $A_{7-i}$;
(3) $A_{1}$ contains 8 elements, while the other subsets contain only 7 elements, so the largest subset $S$ must contain all elements from $A_{1}$.
In summary, the largest subset $S$ has $1+8+7+7=23$ elements.
|
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 property $P$. (1994, Hebei Province High School Mathematics Competition)
|
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 with property $P$ is $\mathrm{C}_{21}^{2}=210$;
(ii) When $a \in[k], b \in[17-k], k=1,2$, $\cdots, 7$, the number of subsets with property $P$ is $\mathrm{C}_{22}^{1} \cdot \mathrm{C}_{21}^{1}=$ 462;
(iii) When $a \in[8], b \in[9]$, the number of subsets with property $P$ is $\mathrm{C}_{22}^{1} \cdot \mathrm{C}_{22}^{1}=484$.
Therefore, the number of subsets of $A$ with property $P$ is $210+462 \times 7+484=3928$.
(2) To ensure that the binary subsets do not intersect, when $a, b \in$ [0], 10 subsets can be paired;
When $a \in[k], b \in[17-k], k=1,2, \cdots, 7$, 21 subsets can be paired for each;
When $a \in[8], b \in$ [9], 22 subsets can be paired.
Therefore, the number of pairwise disjoint subsets with property $P$ is $10+21 \times 7+22=179$.
Note: Finding an appropriate criterion, i.e., using the remainder to partition the set, is the key to solving this problem.
|
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 $x_{1}, x_{2}, \cdots, x_{19}$, when only one of them is 77 and the other 18 are 1, it is clear that their fluctuation is the greatest, i.e., $S^{2}$ is the largest. In this case,
$$
\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{19}^{2}\right)_{\max }=18 \times 1+77^{2}=5947 .
$$
|
5947
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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