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Given $a, b, x, y$ are non-negative real numbers, and $a+b=27$. Try to find the maximum value of $\lambda$, such that the inequality
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant \lambda\left(a x^{2} y+b x y^{2}\right)^{2}
$$
always holds, and find the conditions for equality.
|
Let $a=0, b=27, x=27, y=2$, then the original inequality is $\lambda \leqslant 4$.
We will now prove that $\lambda=4$ makes the given inequality always true.
It is only necessary to prove under the original conditions that
$$
\left(a x^{2}+b y^{2}+4 x y\right)^{3} \geqslant 4(a x+b y)^{2} x^{2} y^{2}.
$$
When $x$ or $y$ is 0, inequality (1) is obviously true.
When $x$ and $y$ are both greater than 0, divide both sides of inequality (1) by $x^{3} y^{3}$ and let $t=\frac{x}{y}$, then inequality (1) becomes
$$
\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 4\left(a^{2} t+\frac{b^{2}}{t}+2 a b\right).
$$
Here $t \in \mathbf{R}_{+}$. Since:
$$
\begin{array}{l}
\left(a t+\frac{b}{t}\right)(a+b) \\
=a^{2} t+\frac{b^{2}}{t}+a b\left(t+\frac{1}{t}\right) \\
\geqslant a^{2} t+\frac{b^{2}}{t}+2 a b,
\end{array}
$$
it is only necessary to prove
$$
\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 4\left(a t+\frac{b}{t}\right) \times 27.
$$
Since $a t+\frac{b}{t}+4=a t+\frac{b}{t}+2+2$
$$
\geqslant 3 \sqrt[3]{4\left(a t+\frac{b}{t}\right)},
$$
we have $\left(a t+\frac{b}{t}+4\right)^{3} \geqslant 27 \times 4\left(a t+\frac{b}{t}\right)$.
Therefore, $\lambda_{\text {max }}=4$.
The equality in inequality (1) holds if and only if $a t+\frac{b}{t}=2, t=1$ or $x=0, b=0$ or $x=0, y=0$ or $y=0, a=0$.
Since $a+b=27>2$, the equality in inequality (1) holds if and only if $x=0, a=27, b=0$ or $y=0, a=0, b=27$. Or $x=0, y=0$.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 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.
But due to carelessness, one of the $y$ values was calculated incorrectly. Please point out which value is incorrect? What is the correct value? And explain the reason.
(2003, Shanghai Junior High School Mathematics Competition)
|
Explanation: Let $x_{2}-x_{1}=x_{3}-x_{2}=\cdots=x_{7}-x_{6}$ $=d$, and the function value corresponding to $x_{i}$ is $y_{i}$. 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, $\Delta_{k+1}-\Delta_{k}$
$$
=2 a d\left(x_{k+1}-x_{k}\right)=2 a d^{2} \text { (constant). }
$$
From the given data
$$
\begin{array}{llllllllll}
& y_{k} & : & 51 & 107 & 185 & 285 & 407 & 549 & 717 \\
\text { we get } & \Delta_{k} & : & & 50 & 78 & 100 & 122 & 142 & 168 \\
\Delta_{k+1}-\Delta_{k} & : & & 22 & 22 & 22 & 20 & 26
\end{array}
$$
From this, we can see that 549 is the incorrect $y$ value, and its correct value should be 551.
Note: A quadratic function is essentially a second-order arithmetic sequence with the independent variable taking real numbers. This problem is designed based on such a background. The key to solving the problem is to grasp the essential characteristics through the phenomena.
|
551
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $\sqrt{x^{2}+32}-\sqrt{65-x^{2}}=5$. Then $3 \sqrt{x^{2}+32}+2 \sqrt{65-x^{2}}=$ $\qquad$
|
3.35.
Observing the experiment, we know that $x^{2}=49$, so the original expression $=3 \times 9+2 \times 4=35$.
|
35
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. There are two positive integers $a$ and $b$, the sum of their squares is 585, and the sum of their greatest common divisor and least common multiple is 87. Then $a+b=$ $\qquad$ .
|
4.33.
Let the required positive integers be $a<b$, and let $(a, b)=d$, then we have $a=d x, b=d y$, and $(x, y)=1$.
Thus, $[a, b]=d x y$. According to the problem, we have $\left\{\begin{array}{l}d^{2} x^{2}+d^{2} y^{2}=585, \\ d+d x y=87,\end{array}\right.$ which simplifies to $\left\{\begin{array}{l}x^{2}+y^{2}=\frac{585}{d^{2}}, \\ 1+x y=\frac{87}{d} .\end{array}\right.$
Since $585=3^{2} \times 5 \times 13$, and $d^{2} \mid 585$, we have $d=3$. Thus, $\left\{\begin{array}{l}x^{2}+y^{2}=65, \\ 1+x y=29,\end{array}\right.$ which leads to $\left\{\begin{array}{l}x+y=11, \\ x y=28 .\end{array}\right.$
Solving this, we get $x=4, y=7$.
Therefore, $a=3 \times 4=12, b=3 \times 7=21$.
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (17 points) The quadratic trinomial $x^{2}-x-2 n$ can be factored into the product of two linear factors with integer coefficients.
(1) If $1 \leqslant n \leqslant 30$, and $n$ is an integer, how many such $n$ are there?
(2) When $n \leqslant 2005$, find the largest integer $n$.
|
(1) Notice that
$$
x^{2}-x-2 n=\left(x-\frac{1+\sqrt{1+8 n}}{2}\right)\left(x-\frac{1-\sqrt{1+8 n}}{2}\right),
$$
then we should have
$$
1+8 n=9,25,49,81,121,169,225,289, \cdots \text {. }
$$
The corresponding solutions for $n$ are $1,3,6,10,15,21,28,36$ (discard the last one).
Therefore, when $1 \leqslant n \leqslant 30$, there are 7 integers $n$ that satisfy the condition.
(2) Observing the sequence $1,3,6,10, \cdots$ and the relationship between its terms, we can deduce that
$$
1=1,3=1+2,6=1+2+3,10=1+2+3+4 \text {. }
$$
Thus, as long as $n=1+2+3+\cdots+k \leqslant 2005$, and $k$ is the largest integer satisfying this inequality, then the sum $n$ is what we are looking for.
From $n=\frac{(1+k) k}{2} \leqslant 2005$, we can solve by observation and trial:
When $k=61$, we get $n=1891$;
When $k=62$, we get $n=1953$;
When $k=63$, we get $n=2016>2005$.
Therefore, the largest integer $n$ is 1953.
|
1953
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (17 points) As shown in Figure 4, given the line
$$
l: y=k x+2-4 k \text{ ( } k \text{ is a real number). }
$$
- (1) Prove that regardless of the value of $k$, the line $l$ always passes through a fixed point $M$, and find the coordinates of point $M$;
(2) If the line $l$ intersects the positive $x$-axis and $y$-axis at points $A$ and $B$ respectively, find the minimum value of the area of $\triangle A O B$.
|
(1) Let $k=1$, we get $y=x-2$; let $k=2$, we get $y=2x-6$. Solving these simultaneously, we get $x=4, y=2$. Therefore, the fixed point is $M(4,2)$.
Substituting the coordinates of point $M(4,2)$ into the equation of line $l$, we get $2=2$, which is an identity independent of $k$. Hence, regardless of the value of $k$, line $l$ always passes through the fixed point $M(4,2)$.
(2) Taking $x=0$, we get $OB=2-4k$ ($k<0$).
Taking $y=0$, we get $OA=\frac{4k-2}{k}$ ($k<0$).
Thus, the area of $\triangle AOB$ is
$$
S=\frac{1}{2} OA \cdot OB=\frac{1}{2} \cdot \frac{4k-2}{k}(2-4k) \quad (k<0).
$$
Converting the above expression into a quadratic equation in $k$, we get
$$
8k^2 + (S-8)k + 2 = 0.
$$
Since $k$ is a real number, we have $\Delta = (S-8)^2 - 64 \geq 0$, which simplifies to $S^2 - 16S \geq 0$, hence $S \geq 16$.
Substituting $S=16$ into equation (1), we get $k=-\frac{1}{2}$.
Therefore, when $k=-\frac{1}{2}$, $S_{\text{min}}=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. On the coordinate plane, a point $(x, y)$ where both the x-coordinate and y-coordinate are integers is called an integer point. If the area enclosed by the quadratic function $y=-x^{2}+8 x-\frac{39}{4}$ and the x-axis is colored red, then the number of integer points inside this red region and on its boundary is ( ) points.
(A) 16
(B) 20
(C) 25
(D) 30
|
5.C.
$y=-\left(x-\frac{3}{2}\right)\left(x-\frac{13}{2}\right)$ intersects the $x$-axis at two points $M\left(\frac{3}{2}, 0\right)$ and $N\left(\frac{13}{2}, 0\right)$. Between $x=\frac{3}{2}$ and $x=\frac{13}{2}$, there are 5 integers: $2, 3, 4, 5, 6$.
Rewrite the function as
$$
y=-(x-4)^{2}+\frac{25}{4}.
$$
When $x=2$ or $6$, $y=\frac{9}{4}$, and the integers satisfying $0 \leqslant y \leqslant \frac{9}{4}$ are $0, 1, 2$, a total of 6 integers;
When $x=3$ or $5$, $y=\frac{21}{4}$, and the integers satisfying $0 \leqslant y \leqslant \frac{21}{4}$ are $0, 1, 2, 3, 4, 5$, a total of 12 integers;
When $x=4$, $y=\frac{25}{4}$, and the integers satisfying $0 \leqslant y \leqslant \frac{25}{4}$ are $0, 1, 2, 3, 4, 5, 6$, a total of 7 integers.
In total, there are 25 integer points.
|
25
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a, b$ are positive integers, $a=b-2005$. If the equation $x^{2}-a x+$ $b=0$ has positive integer solutions, then the minimum value of $a$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
2.95.
Let the two roots of the equation be $x_{1}$ and $x_{2}$, then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{1} x_{2}=b .
\end{array}\right.
$$
Since $x_{1}$ and $x_{2}$ include one positive integer, the other must also be a positive integer. Without loss of generality, let $x_{1} \leqslant x_{2}$. From equation (1), we have
$$
\begin{array}{l}
x_{1} x_{2}-\left(x_{1}+x_{2}\right)=b-a=2005 . \\
\text { Hence }\left(x_{1}-1\right)\left(x_{2}-1\right)=2006=2 \times 17 \times 59 .
\end{array}
$$
Since 59 is a prime number, one of $x_{1}-1$ and $x_{2}-1$ must be a multiple of 59.
In equation (2), if we take $x_{1}-1=34, x_{2}-1=59$, then $x_{1}+x_{2}=95$, i.e., $a=95$.
If we take $x_{1}-1x_{2} \geqslant 119$.
Therefore, the minimum value of $a$ is 95.
|
95
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. A person adds the page numbers of a book in the order $1,2,3, \cdots$, with one page number being added an extra time, resulting in an incorrect total sum of 2005. Then the page number that was added extra is $\qquad$ .
|
3.52 pages.
Let the total number of pages in the book be $n$, and the page number that was added extra be $x(1 \leqslant x \leqslant n)$. We have $\frac{n(n+1)}{2}+x=2005$.
And $\frac{n(n+1)}{2}+1 \leqslant 2005 \leqslant \frac{n(n+1)}{2}+n$, which means $n^{2}+n+2 \leqslant 4010 \leqslant n(n+3)$.
Since $\sqrt{4010} \approx 63$, verification shows that the only $n$ satisfying equation (2) is $n=62$.
Substituting into equation (1), we get $x=2005-\frac{62 \times 63}{2}=52$.
|
52
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $n$ be a natural number. If 2005 can be written as the sum of $n$ positive odd composite numbers, then $n$ is called a "good number". The number of such good numbers is $\qquad$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
4.111.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be odd composite numbers, and $a_{1}+a_{2}+\cdots+a_{n}=$ 2005, then $n$ is odd. Since 9 is the smallest odd composite number, and 2005 $<2007=9 \times 223$, hence $n<223$. Therefore, $n \leqslant 221$.
$$
2005=1980+25=\underbrace{9+9+\cdots}_{20 \uparrow}+9+25 \text {, }
$$
Thus, 221 is a good number.
Furthermore, when $(2 k-1) \times 9$ is an odd composite number,
$$
(2 k-1) \times 9+9+9=(2 k+1) \times 9
$$
is also an odd composite number. Therefore, the terms on the right side of equation (1) can be combined step by step, so $n$ can take 221, 219, ..., 5, 3.
Since 2005 itself is also an odd composite number, then $n$ can take 1.
Thus, 1, 3, 5, ..., 221 are all good numbers, totaling 111.
|
111
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, let the arithmetic mean of all elements in the set $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be denoted as $P(A)$ $\left(P(A)=\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)$. If $B$ is a non-empty subset of $A$ and $P(B)=P(A)$, then $B$ is called a "balanced subset" of $A$. Try to find the number of all "balanced subsets" of the set $M=\{11,2,3,4,5,6,7,8,9\}$.
|
$$
\begin{array}{l}
\text { Six, since } P(M)=5 \text {, let } \\
M^{\prime}=\{x-5 \mid x \in M\} \\
=\{-4,-3,-2,-1,0,1,2,3,4\},
\end{array}
$$
then $P\left(M^{\prime}\right)=0$. According to this translation relationship, the balanced subsets of $M$ and $M^{\prime}$ can be one-to-one correspondence. Let $f(k)$ denote the number of $k$-element balanced subsets of $M^{\prime}$. Clearly, $f(9)=1, f(1)=1$ (the 9-element balanced subset of $M^{\prime}$ is only $M^{\prime}$, and the 1-element balanced subset is only $\{0\}$).
The 2-element balanced subsets of $M^{\prime}$ are 4, which are $\left.B_{i}=\mid-i, i\right\}$, $i=1,2,3,4$.
Therefore, $f(2)=4$.
The 3-element balanced subsets of $M^{\prime}$ fall into two cases:
(1) Those containing the element 0 are $B_{i} \cup\{0\}=\{-i, 0, i\}, i=$ $1,2,3,4$. There are 4 in total;
(2) Those not containing the element 0, since the equations $3=1+2,4=1+3$ can be expressed as $-3+1+2=0,3-1-2=0$ and $-4+1+3$ $=0,4-1-3=0$, yielding 4 balanced subsets $\{-3,1,2\}, \{3, -1,-2\},\{-4,1,3\},\{4,-1,-3\}$.
Therefore, $f(3)=4+4=8$.
The 4-element balanced subsets of $M^{\prime}$ fall into three cases:
(1) The union of every two 2-element balanced subsets, $B_{i} \cup B_{j}, 1 \leqslant i<j$ $\leqslant 4$, there are 6 in total;
(2) The union of 3-element balanced subsets not containing the element 0 with $\{0\}$, there are 4 in total;
(3) Those not covered by the above two cases, since the equation $1+4=2+3$ can be expressed as $-1-4+2+3=0$ and $1+4-2-3=0$, yielding 2 balanced subsets $\{-1,-4,2,3\}$ and $\{1,4,-2,-3\}$.
Therefore, $f(4)=6+4+2=12$.
Also note that, except for $M^{\prime}$ itself, if $B^{\prime}$ is a balanced subset of $M^{\prime}$, then its complement $C_{M} \cdot B^{\prime}$ is also a balanced subset of $M^{\prime}$, and they are in one-to-one correspondence. Therefore, $f(9-k)=f(k), k=1,2,3,4$.
Thus, the number of balanced subsets of $M^{\prime}$ is
$$
\begin{array}{l}
\sum_{k=1}^{9} f(k)=f(9)+2 \sum_{k=1}^{4} f(k) \\
=1+2(1+4+8+12)=51 .
\end{array}
$$
Therefore, $M$ has 51 balanced subsets.
|
51
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(1) Discuss the number of roots of the equation with respect to $x$
$$
|x+1|+|x+2|+|x+3|=a
$$
(2) Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence, and
$$
\begin{array}{l}
\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right| \\
=\left|a_{1}+1\right|+\left|a_{2}+1\right|+\cdots+\left|a_{n}+1\right| \\
=\left|a_{1}-2\right|+\left|a_{2}-2\right|+\cdots+\left|a_{n}-2\right|=507 .
\end{array}
$$
Find the maximum value of the number of terms $n$. (Lin Chang)
|
(1) According to the graph of the function $y=|x+1|+|x+2|+|x+3|$ (as shown in Figure 3), we can see that:
When $a2$, the equation has two solutions.
(2) Since the equation $|x|=|x+1|=|x-2|$ has no solution, it follows that $n \geqslant 2$ and the common difference is not 0. Let's assume the terms of the sequence are $a-k d(1 \leqslant k \leqslant n, d>0)$. Construct the function:
$$
f(x)=\sum_{k=1}^{n}|x-k d| .
$$
Thus, the condition of the problem is equivalent to $f(x)=507$ having at least three different solutions $a, a+1, a-2$. This condition is also equivalent to the graph of $y=f(x)$ intersecting the horizontal line $y=507$ at least three different points.
Since the graph of $y=f(x)$ is a downward convex broken line with $n+1$ segments symmetric about the line $x=\frac{(n+1) d}{2}$, it intersects the horizontal line $l$ at three points if and only if the broken line has a horizontal segment on $l$, which occurs if and only if $n=2 m$ and $a, a+1, a-2 \in[m d, (m+1) d]$, $f(m d)=507$, i.e., $d \geqslant 3$ and $m^{2} d=507$.
This gives $m^{2} \leqslant \frac{507}{3}, m \leqslant 13$.
Clearly, when $m=13$, taking $d=3, a=41$ satisfies the conditions of the problem.
Therefore, the maximum value of $n$ is 26.
|
26
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A university has 10001 students, some of whom join and form several clubs (a student can belong to different clubs), and some clubs join together to form several associations (a club can belong to different associations). It is known that there are $k$ associations in total. Assume the following conditions are met:
(1) Every pair of students (i.e., any two students) belong to exactly one club;
(2) For each student and each association, the student belongs to exactly one club of that association;
(3) Each club has an odd number of students, and a club with $2m$ +1 students belongs to exactly $m$ associations, where $m$ is a positive integer.
Find all possible values of $k$.
|
1. Replace 10001 with $n$, and use two methods to calculate the number of ordered triples $(a, R, S)$, where $a, R, S$ represent a student, a club, and a society, respectively, and satisfy $a \in R, R \in S$. We call such triples "acceptable".
Fix a student $a$ and a society $S$. By condition (2), there is a unique club $R$ such that $(a, R, S)$ is an acceptable triple. Since there are $n k$ ways to choose the ordered pair $(a, S)$, there are $n k$ acceptable triples.
Fix a club $R$, and let the number of students in $R$ be $|R|$. By condition (3), $R$ belongs to exactly $\frac{|R|-1}{2}$ societies. Therefore, the number of acceptable triples containing club $R$ is $\frac{|R|(|R|-1)}{2}$. If $M$ is the set of all clubs, then the total number of acceptable triples is $\sum_{R \in M} \frac{|R|(|R|-1)}{2}$. Therefore,
$$
n k=\sum_{R \in M} \frac{|R|(|R|-1)}{2} .
$$
Since $\sum_{R \in M} \frac{|R|(|R|-1)}{2}=\sum_{R \in M} \mathrm{C}_{|R|}^{2}$, by condition (1), we have $\sum_{k \in M} \mathrm{C}_{|k|}^{2}=\mathrm{C}_{n}^{2}$. Hence, $n k=\frac{n(n-1)}{2}$, which implies $k=\frac{n-1}{2}$.
When $n=10001$, $k=5000$.
|
5000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $x=\frac{13}{4+\sqrt{3}}$, then $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+3}{x^{3}-7 x^{2}+5 x+15}=$ $\qquad$
|
$$
\text { II.1. }-5 \text {. }
$$
Since $x=\frac{13}{4+\sqrt{3}}=4-\sqrt{3}$, then $x^{2}-8 x+13=0$.
$$
\begin{array}{l}
\text { Hence } x^{4}-6 x^{3}-2 x^{2}+18 x+3 \\
=\left(x^{2}+2 x+1\right)\left(x^{2}-8 x+13\right)-10=-10, \\
x^{3}-7 x^{2}+5 x+15 \\
=(x+1)\left(x^{2}-8 x+13\right)+2=2 .
\end{array}
$$
Therefore, the original expression $=\frac{-10}{2}=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A natural number minus 69 is a perfect square, and this natural number plus 20 is still a perfect square. Then this natural number is $\qquad$ .
|
2.2005 .
Let this natural number be $x$. According to the problem, we have
$$
\left\{\begin{array}{l}
x-69=m^{2}, \\
x+20=n^{2},
\end{array}\right.
$$
where $m, n$ are both natural numbers.
Subtracting the two equations gives $n^{2}-m^{2}=89$, which is $(n-m)(n+m)=89$.
Since $n>m$, and 89 is a prime number, we have
$$
\left\{\begin{array}{l}
n-m=1, \\
n+m=89 .
\end{array}\right.
$$
Solving these, we get $n=45, x=2005$.
|
2005
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given real numbers $x_{1}, x_{2}, y_{1}, y_{2}$ satisfy
$$
\begin{array}{l}
x_{1}^{2}+25 x_{2}^{2}=10, \\
x_{2} y_{1}-x_{1} y_{2}=25, \\
x_{1} y_{1}+25 x_{2} y_{2}=9 \sqrt{55} .
\end{array}
$$
Then $y_{1}^{2}+25 y_{2}^{2}=$ $\qquad$
|
3.2008 .
Notice that
$$
\begin{array}{l}
\left(x_{1}^{2}+a x_{2}^{2}\right)\left(y_{1}^{2}+a y_{2}^{2}\right) \\
=\left(x_{1} y_{1}+a x_{2} y_{2}\right)^{2}+a\left(x_{2} y_{1}-x_{1} y_{2}\right)^{2},
\end{array}
$$
where $a$ is any real number.
For this problem, take $a=25$, and it is easy to find that $y_{1}^{2}+25 y_{2}^{2}=2008$.
|
2008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given that the sum of 10 natural numbers is 1001. What is the maximum possible value of their greatest common divisor?
untranslated text:
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Example 2 Given that the sum of 10 natural numbers is 1001. What is the maximum possible value of their greatest common divisor?
|
Solution: Let the greatest common divisor be $d$, and the 10 numbers be $b_{1} d, b_{2} d, \cdots, b_{10} d$. According to the problem, we have
$$
\left(b_{1}+b_{2}+\cdots+b_{10}\right) d=1001 .
$$
Since $b_{1}+b_{2}+\cdots+b_{10} \geqslant 10$, it follows that $d \leqslant$ 100 and is a divisor of 1001.
Given $1001=7 \times 11 \times 13$, the largest possible value of the divisor less than 100 is $d=7 \times 13=91$.
When the 10 numbers are $91,91, \cdots, 91,182$, their sum is 1001, so $d=91$ satisfies the condition.
|
91
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the smallest natural number $n$ such that $\frac{n-13}{5 n+6}$ is a non-zero reducible fraction.
(6th IMO)
|
Solution: Let the common divisor of the numerator and denominator be $d(d>1)$, and let
$$
\begin{array}{l}
n-13=k d, \\
5 n+6=l d .
\end{array}
$$
(2) $-5 \times$ (1) gives
$$
(l-5 k) d=71 \text{. }
$$
Since 71 is a prime number and $d>1$, we have
$$
d=71 \text{ and } l-5 k=1 \text{. }
$$
Thus, $n=13+k \times 71$.
When $k=0$, $n=13$, the numerator is 0, which does not meet the requirement.
Taking $k=1$, $n=84$ is the smallest natural number sought.
|
84
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given that $x, y$ are integers, and
$$
15 x^{2} y^{2}=35 x^{2}-3 y^{2}+412 \text {. }
$$
then $15 x^{2} y^{2}=$ $\qquad$ .
|
Solution: Transform the given equation into
$$
\left(5 x^{2}+1\right)\left(3 y^{2}-7\right)=405=3^{4} \times 5 \text {. }
$$
Since $5 x^{2}+1$ is not a multiple of 5, and $3 y^{2}-7$ is not a multiple of 3, it can only be
$$
5 x^{2}+1=3^{4}, 3 y^{2}-7=5 \text {. }
$$
Therefore, $5 x^{2}=80,3 y^{2}=12$.
Thus, $15 x^{2} y^{2}=80 \times 12=960$.
|
960
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3: There are $n$ people, and it is known that any two of them make at most one phone call. The total number of phone calls made among any $n-2$ people is equal, and it is $3^{k}$ times, where $k$ is a positive integer. Find all possible values of $n$.
(2000, National High School Mathematics Competition)
Analysis: Carefully savor the quantitative relationships given in the problem, and you can obtain its quantitative characteristics.
Assume the $n$ people are $A_{1}, A_{2}, \cdots, A_{n}$, and the number of calls made by $A_{i}$ is $m_{i}$. The number of calls between $A_{i}$ and $A_{j}$ is $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1. Thus, the quantitative characteristic is
$$
m_{i}+m_{j}-\lambda_{i j}=\frac{1}{2} \sum_{s=1}^{n} m_{s}-3^{k}=c,
$$
where $c$ is a constant. Consequently, we conjecture that $m_{i}(i=1,2, \cdots, n)$ is a constant.
|
Let $n$ people be denoted as $A_{1}, A_{2}, \cdots, A_{n}$. Let the number of calls made by $A_{i}$ be $m_{i}$, and the number of calls between $A_{i}$ and $A_{j}$ be $\lambda_{i j}(1 \leqslant i, j \leqslant n)$, where $\lambda_{i j}=0$ or 1.
Clearly, $n \geqslant 5$. Therefore,
$$
\begin{array}{l}
\left|m_{i}-m_{j}\right|=\left|\left(m_{i}+m_{s}\right)-\left(m_{j}+m_{s}\right)\right| \\
=\left|\lambda_{i s}-\lambda_{j}\right| \leqslant 1,1 \leqslant i, j, s \leqslant n .
\end{array}
$$
Let $m_{i}=\max \left\{m_{s}, 1 \leqslant s \leqslant n\right\}$,
$$
m_{j}=\min \left\{m_{s}, 1 \leqslant s \leqslant n\right\} .
$$
Thus, $m_{i}-m_{j} \leqslant 1$.
If $m_{i}-m_{j}=1$, then for any $s \neq i, j, 1 \leqslant s \leqslant n$, we have
$$
\begin{array}{l}
\left(m_{i}+m_{s}-\lambda_{i s}\right)-\left(m_{j}+m_{s}-\lambda_{j s}\right) \\
=1-\left(\lambda_{i s}-\lambda_{j s}\right)=0,
\end{array}
$$
which means $\lambda_{i s}-\lambda_{j s} \equiv 1,1 \leqslant i, j, s \leqslant n$.
Therefore, $\lambda_{i s} \equiv 1, \lambda_{j i} \equiv 0, s \neq i, j$, and $1 \leqslant s \leqslant n$.
Thus, $m_{i} \geqslant n-2, m_{j} \leqslant 1$.
Hence, $m_{i}-m_{j} \geqslant n-3 \geqslant 2$, which is a contradiction.
Therefore, $m_{i}-m_{j}=0$.
Then $\lambda_{i j} \equiv 0$ or $\lambda_{i j} \equiv 1$.
If $\lambda_{i j} \equiv 0$, then $m_{s} \equiv 0,1 \leqslant s \leqslant n$, which is a contradiction.
If $\lambda_{i j} \equiv 1$, then $m_{s}=n-1,1 \leqslant s \leqslant n$.
Thus, $(n-2)(n-3)=3^{k} \times 2$.
Let $n-2=2 \times 3^{k_{1}}, n-3=3^{k_{2}}, k_{1} \geqslant k_{2}$.
From $2 \times 3^{k_{1}}-3^{k_{2}}=1$, we get
$$
3^{k_{2}}=1 \text{. }
$$
Therefore, $k_{1}=k_{2}=0$. This contradicts $k \geqslant 1$.
Let $n-2=3^{k_{1}}, n-3=2 \times 3^{k_{2}}, k_{1} \geqslant k_{2}+1$.
From $3^{k_{1}}-2 \times 3^{k_{2}}=1$, we get
$$
k_{2}=0, k_{1}=1 \text{. }
$$
Therefore, $n=5$ is the solution.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 The polynomial $x^{4}+m x^{3}+n x-16$ contains the factors $x-1$ and $x-2$. Then $m n=$ $\qquad$
(1989, Sichuan Province Junior High School Mathematics League)
|
Solution: Let $f(x)=x^{4}+m x^{3}+n x-16$. By the Factor Theorem, we have
$$
\left\{\begin{array} { l }
{ f ( 1 ) = 0 , } \\
{ f ( 2 ) = 0 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
m+n=15, \\
8 m+2 n=0 .
\end{array}\right.\right.
$$
Solving, we get $m=-5, n=20$.
Therefore, $m n=-100$.
|
-100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $a 、 b 、 c$ are real numbers, and
$$
a^{2}+b^{2}+c^{2}+2 a b=1, a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{8} \text {, }
$$
The roots of the quadratic equation $(a+b) x^{2}-(2 a+c) x-(a+b)=0$ are $\alpha 、 \beta$. Find the value of $2 \alpha^{3}+\beta^{-5}-\beta^{-1}$.
|
Solution: From the given, we have
$$
\left\{\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)+2 a b=1, \\
2 a b\left(a^{2}+b^{2}+c^{2}\right)=\frac{1}{4} .
\end{array}\right.
$$
Thus, \(a^{2}+b^{2}+c^{2}\) and \(2 a b\) are the roots of the equation \(t^{2}-t+\frac{1}{4}=0\).
Since the roots of the equation \(t^{2}-t+\frac{1}{4}=0\) are \(t_{1}=t_{2}=\frac{1}{2}\), we have
$$
a^{2}+b^{2}+c^{2}=2 a b=\frac{1}{2} \text {. }
$$
Solving this, we get \(a=b= \pm \frac{1}{2}, c=0\).
Thus, the given equation can be transformed into
$$
x^{2}-x-1=0 \text {. }
$$
Since \(\alpha, \beta\) are the roots of equation (1), then \(\alpha+\beta=1\), and
$$
\left\{\begin{array}{l}
\alpha^{2}-\alpha-1=0, \\
\beta^{2}-\beta-1=0 .
\end{array}\right.
$$
From equation (2), we get \(\alpha^{2}=\alpha+1\). Therefore,
$$
\alpha^{3}=\alpha \cdot \alpha^{2}=\alpha(\alpha+1)=\alpha^{2}+\alpha=2 \alpha+1 \text {. }
$$
Clearly, \(\beta \neq 0\).
Dividing both sides of equation (3) by \(\beta\) and \(\beta^{2}\), and rearranging, we get
$$
\frac{1}{\beta}=\beta-1, \frac{1}{\beta^{2}}=1-\frac{1}{\beta}=2-\beta \text {. }
$$
$$
\begin{array}{l}
\text { and } \beta^{-3}=\beta^{-1} \cdot \beta^{-2}=(\beta-1)(2-\beta) \\
=3 \beta-\beta^{2}-2=2 \beta-3, \\
\beta^{-5}=\beta^{-2} \cdot \beta^{-3} \\
=(2-\beta)(2 \beta-3)=7 \beta-2 \beta^{2}-6 \\
=7 \beta-2(\beta+1)-6=5 \beta-8,
\end{array}
$$
Thus, \(2 \alpha^{3}+\beta^{-5}-\beta^{-1}=4(\alpha+\beta)-5=-1\).
(Zhang Zhecai, Shengzhou Middle School, Jiaojiang District, Taizhou City, Zhejiang Province, 318000)
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 When $m=$ $\qquad$, the polynomial
$$
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m
$$
can be factored into the product of two linear factors.
(1992, Zhengzhou City Junior High School Mathematics Competition)
|
Solution: First, factorize the quadratic term, we have
$$
12 x^{2}-10 x y+2 y^{2}=(3 x-y)(4 x-2 y) \text {. }
$$
Therefore, we can set
$$
\begin{array}{l}
12 x^{2}-10 x y+2 y^{2}+11 x-5 y+m \\
=(3 x-y+a)(4 x-2 y+b) \\
=(3 x-y)(4 x-2 y)+(4 a+3 b) x- \\
(2 a+b) y+a b .
\end{array}
$$
Comparing the coefficients of the linear terms, we get
$$
\left\{\begin{array}{l}
4 a+3 b=11, \\
2 a+b=5 .
\end{array}\right.
$$
Solving these, we find $a=2, b=1$.
Comparing the constant term, we get $m=a b=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The axial cross-section of a wine glass is part of a parabola, whose equation is $x^{2}=2 y(0 \leqslant y<15)$. If a glass ball with a radius of 3 is placed inside the cup, then the distance from the highest point of the ball to the bottom of the cup is $\qquad$
|
5.8 .
As shown in the figure, let the center of the sphere be at $(0, b)$.
Then we have
$$
x^{2}+(y-6)^{2}=9 \text {. }
$$
Therefore, the system of equations
$$
\left\{\begin{array}{l}
x^{2}=2 y \\
x^{2}+(y-b)^{2}=9
\end{array}\right.
$$
has two solutions. For $y$,
$$
y^{2}+2(1-b) y+b^{2}-9=0 \text {. }
$$
By $\Delta=0$, we solve to get $b=5$.
Thus, the distance from the highest point of the sphere to the bottom of the cup is
$$
d=b+r=5+3=8 .
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Fill the numbers $1,2, \cdots, 8$ into the 8 squares surrounding the four sides of a $3 \times 3$ chessboard, so that the sum of the absolute values of the differences between adjacent numbers in these 8 squares is maximized. Find this maximum value.
|
Solution: Let the sum of the absolute values of the differences between adjacent numbers in the 8 squares be $M$. Note that rotating the 8 numbers in the outer ring of the $3 \times 3$ chessboard does not change the value of $M$. As shown in Figure 6, let the numbers in each square be denoted as $a_{1}, a_{2}, \cdots, a_{8}$. It is known that $M$ reaches its maximum value when $a_{1}, a_{2}, \cdots, a_{8}$ are arranged in an alternating order of size.
Otherwise, assume that when $M$ is at its maximum, $a_{1}, a_{2}, a_{3}$ are in increasing order, i.e., $a_{1}<a_{2}<a_{3}$. Then, we have:
$$
\begin{array}{l}
M=\left(a_{1}-a_{2}\right)+\left(a_{3}-a_{2}\right)+\left(a_{3}-a_{4}\right)+ \\
\quad \cdots+\left(a_{7}-a_{8}\right)+\left(a_{1}-a_{8}\right) \\
=2\left(a_{1}+a_{3}+a_{5}+a_{7}\right)-2\left(a_{2}+a_{4}+a_{6}+a_{8}\right) .
\end{array}
$$
Among these 8 numbers, if we take $a_{1}, a_{3}, a_{5}, a_{7}$ as the 4 largest numbers and $a_{2}, a_{4}, a_{6}, a_{8}$ as the 4 smallest numbers, then $M$ reaches its maximum value, and $M \leqslant 2 \times(8+7+6+5)-2 \times (4+3+2+1)=32$. Therefore, the maximum value of $M$ is 32.
As shown in Figure 7, there are multiple valid ways to fill in the numbers.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Fill the numbers $1,2, \cdots, 9$ into a $3 \times 3$ chessboard, such that the sum of the absolute values of the differences between adjacent (sharing a common edge) cells is maximized. Find this maximum value.
|
Solution: Let the sum of the absolute values of the differences between adjacent cells be $M$. Note that $M$ can be divided into two parts:
(1) The sum of the absolute values of the differences between adjacent numbers in the 8 surrounding cells, denoted as $M_{1}$;
(2) The sum of the 4 absolute values of the differences between the numbers adjacent to the center cell and the number in the center cell, denoted as $M_{2}$.
Thus, we have $M=M_{1}+M_{2}$.
From Example 4, we know that when the numbers in the 8 surrounding cells are rotated, the value of $M_{1}$ remains unchanged, and at this time, $M_{2}$ has only two different values, hence, $M$ also has only two different values. Let the numbers in the cells of the chessboard be $a_{1}, a_{2}, \cdots, a_{9}$, with $a_{9}$ in the center cell, and the rest $a_{i}(i=1,2, \cdots, 8)$ in the 8 surrounding cells in sequence.
From Example 4, we also know that when $a_{9}$ is fixed, $a_{1}, a_{2}, \cdots, a_{8}$ are arranged in an alternating order of size, $M_{1}$ is maximized.
If $6 \leqslant a_{9} \leqslant 9$, when the 4 larger numbers are placed in the four corner cells of the $3 \times 3$ chessboard, $M_{2}$ is maximized;
If $1 \leqslant a_{9} \leqslant 5$, when the 4 smaller numbers are placed in the four corner cells of the $3 \times 3$ chessboard, $M_{2}$ is maximized.
Below, we use the enumeration method to analyze:
When $a_{9}=9$,
$M_{1} \leqslant 2 \times(8+7+6+5)-2 \times(4+3+2+1)=32$,
$M_{2} \leqslant(9-1)+(9-2)+(9-3)+(9-4)=26$,
Thus, $M=M_{1}+M_{2} \leqslant 32+26=58$;
When $a_{9}=1$,
$M_{1} \leqslant 2 \times(9+8+7+6)-2 \times(5+4+3+2)=32$,
$M_{2} \leqslant(9-1)+(8-1)+(7-1)+(6-1)=26$,
Thus, $M=M_{1}+M_{2} \leqslant 32+26=58$;
Similarly,
When $a_{9}=8$ or 2, $M_{1} \leqslant 34, M_{2} \leqslant 22$, then
$M \leqslant 56$;
When $a_{9}=7$ or 3, $M_{1} \leqslant 36, M_{2} \leqslant 18$, then
$M \leqslant 54$;
When $a_{9}=6$ or 4, $M_{1} \leqslant 38, M_{2} \leqslant 14$, then
$M \leqslant 52$;
When $a_{9}=5$, $M_{1} \leqslant 40, M_{2} \leqslant 10$, then
$M \leqslant 50$.
In summary, the $3 \times 3$ chessboard filled as shown in Figure 8, makes $M$ have the maximum value, and the maximum value is 58.
|
58
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 4 If real numbers $a, b, c$ satisfy
$$
\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1 \text{, }
$$
find the value of $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}$.
(1999, Changsha Junior High School Mathematics Competition)
|
Solution: Construct the identity
$$
\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1 \text {. }
$$
Subtract the above identity from the given equation to get
$$
\begin{array}{l}
\left(\frac{a}{b+c}-\frac{a}{a+b+c}\right)+\left(\frac{b}{a+c}-\frac{b}{a+b+c}\right)+ \\
\left(\frac{c}{a+b}-\frac{c}{a+b+c}\right)=0 .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
\frac{a^{2}}{(b+c)(a+b+c)}+\frac{b^{2}}{(a+c)(a+b+c)}+ \\
\frac{c^{2}}{(a+b)(a+c+b)}=0,
\end{array}
$$
which means $\square$
$$
\begin{array}{l}
\frac{1}{a+b+c}\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}\right)=0 . \\
\text { Hence } \frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find the smallest positive integer $n$ such that there exists a positive integer $k$ for which $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds.
|
Solution: For the inequality $\frac{8}{15}1$, i.e., $m>8$.
Therefore, $n>56$.
When $n=7 m+p\left(m \in \mathbf{N}_{+}, p \in \mathbf{N}\right.$, and $1 \leqslant p$ $\leqslant 6$), we have
$$
\begin{array}{l}
6 m+\frac{6}{7} p0$, i.e., $m>p$.
Since $1 \leqslant p \leqslant 6$, to find the smallest $m$, when $p=1$, $m_{\text {min }}=2$. Thus,
$$
n_{\text {min }}=7 m_{\min }+p=7 \times 2+1=15 .
$$
When $n=15$, we have $\frac{90}{7}<k<\frac{105}{8}$, i.e.,
$$
12 \frac{6}{7}<k<13 \frac{1}{8} \text {. }
$$
Therefore, $k=13$.
Hence, the smallest positive integer $n$ that satisfies the condition is $15$.
|
15
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 1, the diagonals $A C$ and $B D$ of quadrilateral $A B C D$ intersect at point $O$. $S_{\triangle A O B}=4, S_{\triangle C O D}=9$. Then the minimum value of $S_{\text {quadrilateral } A B C D}$ is ( ).
(A) 22
(B) 25
(C) 28
(D) 32
|
4. B.
As shown in Figure 1, let $S_{\triangle M O D}=x, S_{\triangle B O C}=y$. Then $S_{\text {quadrilateral } A B C D}=4+9+x+y \geqslant 13+2 \sqrt{x y}$. From $\frac{x}{9}=\frac{4}{y}$, we have $x y=36$. Therefore, $S_{\text {quadrilateral } A B C D} \geqslant 13+2 \sqrt{x y}=13+12=25$. Hence, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 25. At this point, $A B \parallel D C$, meaning quadrilateral $A B C D$ is a trapezoid.
|
25
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 2, two diameters $A C$ and $B D$ of the large circle intersect perpendicularly at point $O$. Four semicircles are drawn outside the large circle with $A B$, $B C$, $C D$, and $D A$ as diameters, respectively. The total area of the four "crescent" shaded regions in the figure is $2 \mathrm{~cm}^{2}$. Then the radius of the large circle is $\qquad$ $\mathrm{cm}$.
|
$=.1 .1$.
By the Pythagorean theorem, we know $A D^{2}+C D^{2}=A C^{2}$, so the area of the upper half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A D$ and $C D$. Similarly, the area of the lower half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A B$ and $B C$. Therefore, the area of the square $A B C D$ is equal to the total area of the four "lunar" shapes. It is easy to calculate that the radius of the large circle $O D$ is $1 \mathrm{~cm}$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $a^{2}+b c=14, b^{2}-2 b c=-6$. Then $3 a^{2}+4 b^{2}-5 b c=$ $\qquad$
|
3.18 .
$$
\begin{array}{l}
3 a^{2}+4 b^{2}-5 b c=3\left(a^{2}+b c\right)+4\left(b^{2}-2 b c\right) \\
=3 \times 14+4 \times(-6)=18
\end{array}
$$
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. There are 6 natural numbers that have the same remainder when divided by 12, and their product is 971425. Then the minimum value of the sum of these 6 natural numbers is $\qquad$ .
|
5. 150.
Since 971425 leaves a remainder of 1 when divided by 12, and
$$
971425=5 \times 5 \times 7 \times 7 \times 13 \times 61,
$$
among the prime factors, there are two each that leave a remainder of 5, 7, and 1 when divided by 12. Since the product of two numbers that leave a remainder of 5 (or 7) when divided by 12 also leaves a remainder of 1, and 971425 multiplied by several 1s is still 971425, and 1 leaves a remainder of 1 when divided by 12, it can only be that 971425 is the product of 6 numbers that leave a remainder of 1 when divided by 12. Calculation shows:
$$
971425=1 \times 1 \times 1 \times 1 \times 1 \times 971425,
$$
the sum of these 6 factors is
$$
\begin{array}{l}
1+1+1+1+1+971425=971430 ; \\
971425=1 \times 1 \times 1 \times 1 \times 13 \times 74725,
\end{array}
$$
the sum of these 6 factors is
$$
\begin{array}{l}
1+1+1+1+13+74725=74742 ; \\
971425=1 \times 1 \times 1 \times 13 \times 25 \times 2989,
\end{array}
$$
the sum of these 6 factors is
$$
1+1+1+13+25+2989=3030 \text {. }
$$
In fact, let $a$ and $b$ be natural numbers greater than 1 that leave a remainder of 1 when divided by 12, and $a \geqslant b$, then $a \geqslant b>2$. It is easy to see that
$$
a b>a \times 2=a+a>a+b .
$$
According to equation (1), we get
$$
\begin{array}{l}
971425=13 \times 74725>13+74725 \\
=13+25 \times 2989>13+25+2989 \\
=13+25+49 \times 61>13+25+49+61 .
\end{array}
$$
Since $971425=5^{2} \times 7^{2} \times 13 \times 61=1 \times 1 \times 13 \times 25$ $\times 49 \times 61$, 971425 can be expressed as the product of 6 natural numbers that leave a remainder of 1 when divided by 12, and the smallest sum of these numbers is
$$
1+1+13+25+49+61=150 \text {. }
$$
|
150
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (15 points) Given non-zero real numbers $a, b, c$ satisfy $a+b+c=0$. Prove:
(1) $a^{3}+b^{3}+c^{3}=3 a b c$;
(2) $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9$.
|
Three, (1) From $a+b+c=0$, we get $a+b=-c$. Therefore, $(a+b)^{3}=-c^{3}$.
Thus, $a^{3}+3 a^{2} b+3 a b^{2}+b^{3}=-c^{3}$.
Hence $a^{3}+b^{3}+c^{3}=-3 a b(a+b)=-3 a b(-c)=3 a b c$.
$$
\begin{array}{l}
\text { (2) }\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{c}{a-b} \\
=1+\left(\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{c}{a-b}=1+\frac{2 c^{2}}{a b} \text {. }
\end{array}
$$
Similarly, $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{a}{b-c}=1+\frac{2 a^{2}}{b c}$,
$$
\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right) \cdot \frac{b}{c-a}=1+\frac{2 b^{2}}{a c} \text {. }
$$
Therefore, $\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)$ $=1+\frac{2 c^{2}}{a b}+1+\frac{2 a^{2}}{b c}+1+\frac{2 b^{2}}{a c}$
$$
=3+\frac{2\left(a^{3}+b^{3}+c^{3}\right)}{a b c}=3+\frac{2 \times 3 a b c}{a b c}=9 \text {. }
$$
|
9
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
12. If the sum of the digits of a natural number $a$ equals 7, then $a$ is called a "lucky number". Arrange all lucky numbers in ascending order as $a_{1}, a_{2}, a_{3}, \cdots$, if $a_{n}=$ 2005, then $a_{5 n}=$ $\qquad$
|
12.52000 .
It is known that the number of integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{k}=m$ that satisfy $x_{1} \geqslant 1$, $x_{i} \geqslant 0(i \geqslant 2)$ is $\mathrm{C}_{m+k-2}^{m-1}$. Taking $m=7$, we know that the number of $k$-digit lucky numbers is $P(k)=\mathrm{C}_{k+5}^{6}$.
Since 2005 is the smallest lucky number in the form $\overline{2 a b c}$, and $P(1)=\mathrm{C}_{6}^{6}=1, P(2)=\mathrm{C}_{7}^{6}=7, P(3)=\mathrm{C}_{8}^{6}=28$, for the four-digit lucky number $\overline{1 a b c}$, the number of such numbers is the number of non-negative integer solutions to $a+b+c=6$, which is $\mathrm{C}_{6+3-1}^{6}=28$. Therefore, 2005 is the $1+7+28+28+1=65$-th lucky number. Hence, $a_{65}=2005$. Thus, $n=65,5 n=325$.
Furthermore, $P(4)=\mathrm{C}_{9}^{6}=84, P(5)=\mathrm{C}_{10}^{6}=210$, and
$$
\sum_{k=1}^{5} P(k)=330 \text {, }
$$
Therefore, the last six five-digit lucky numbers in descending order are
$$
70000,61000,60100,60010,60001,52000 \text {. }
$$
Thus, the 325th lucky number is 52000.
|
52000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $n(n \geqslant 3)$ be a positive integer. If there are $n$ lattice points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane such that: when $\left|P_{i} P_{j}\right|$ is a rational number, there exists $P_{k}$ such that $\left|P_{i} P_{k}\right|$ and $\left|P_{j} P_{k}\right|$ are both irrational; when $\left|P_{i} P_{j}\right|$ is an irrational number, there exists $P_{k}$ such that $\left|P_{i} P_{k}\right|$ and $\left|P_{j} P_{k}\right|$ are both rational, then $n$ is called a "good number".
(1) Find the smallest good number;
(2) Is 2005 a good number?
|
6. We assert that the smallest good number is 5, and 2005 is a good number.
In a triplet $\left(P_{i}, P_{j}, P_{k}\right)$, if $\left|P_{i} P_{j}\right|$ is a rational number (or irrational number), and $\left|P_{i} P_{k}\right|, \left|P_{j} P_{k}\right|$ are irrational numbers (or rational numbers), then $\left(P_{i}, P_{j}, P_{k}\right)$ is called a good triplet.
(1) $n=3$ is clearly not a good number.
$n=4$ is also not a good number.
If otherwise, assume $P_{1}, P_{2}, P_{3}, P_{4}$ satisfy the conditions. As shown in Figure 4, without loss of generality, assume $\left|P_{1} P_{2}\right|$ is a rational number and $\left(P_{1}, P_{2}, P_{3}\right)$ is a good triplet, then $\left(P_{2}, P_{3}, P_{4}\right)$ is a good triplet. Clearly, $\left(P_{2}, P_{4}, P_{1}\right)$ and $\left(P_{2}, P_{4}, P_{3}\right)$ are not good triplets. Therefore, $P_{1}, P_{2}, P_{3}, P_{4}$ do not satisfy the conditions. Contradiction.
$n=5$ is a good number.
The following five lattice points satisfy the conditions:
$$
A_{5}=\{(0,0),(1,0),(5,3),(8,7),(0,7)\} \text {. }
$$
As shown in Figure 5.
(2) Let $A=\{(1,0),(2,0), \cdots,(669,0)\}$,
$$
\begin{array}{l}
B=\{(1,1),(2,1), \cdots,(668,1)\}, \\
C=\{(1,2),(2,2), \cdots,(668,2)\}, \\
S_{2005}=A \cup B \cup C .
\end{array}
$$
For any positive integer $n$, it is easy to prove that $n^{2}+1$ and $n^{2}+4$ are not perfect squares. It is not difficult to prove that for any two points $P_{i}$ and $P_{j}$ in the set $S_{2005}, \left|P_{i} P_{j}\right|$ is a rational number if and only if $P_{i} P_{j}$ is parallel to one of the coordinate axes. Therefore, 2005 is a good number.
Note: When $n=6$, we have $A_{6}=A_{5} \cup\{(-24,0)\}$;
When $n=7$, we have $A_{7}=A_{6} \cup\{(-24,7)\}$.
It can be verified that $n=6,7$ are both good numbers.
When $n \geqslant 8$, they can be arranged in three rows like $n=2005$, indicating that for $n \geqslant 8$, all $n$ are good numbers.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Among the seven-digit numbers formed by the digits $0,1,2,3,4,5,6$, the number of permutations that do not contain "246" and "15" is ( ).
(A) 3606
(B) 3624
(C) 3642
(D) 4362
|
6.C.
Since the total number of permutations of the 7 digits $0,1, \cdots, 6$ into a seven-digit number with the first digit not being zero is $\mathrm{A}_{7}^{7}-\mathrm{A}_{6}^{6}=4320$, but among these, the number where “246” are together is $\mathrm{A}_{4}^{4} \mathrm{C}_{4}^{1}=96$, and the number where “15” are together is $\mathrm{A}_{5}^{5} \mathrm{C}_{5}^{1}=600$, and the number where both “246” and “15” are together is $A_{3}^{3} C_{3}^{1}=18$. Therefore, the number of permutations that do not contain “246” and “15” is $4320-96-600+18=3642$ (cases).
|
3642
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of non-negative integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{99}+2 x_{100}=3$ is $\qquad$ .
|
2. 166749 .
Classify by $x_{100}$:
(1) When $x_{100}=1$, there are 99 non-negative integer solutions;
(2) When $x_{100}=0$, there are
$$
\mathrm{C}_{99}^{1}+\mathrm{A}_{99}^{2}+\mathrm{C}_{99}^{3}=166 \text { 650 (solutions). }
$$
Combining (1) and (2), the total number of non-negative integer solutions is
$$
99+166650=166749 \text { (solutions). }
$$
|
166749
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given positive integers $a, b, c, d$ satisfying $b<a<d<c$, and the sums of each pair are $26, 27, 41, 101, 115, 116$. Find the value of $(100a + b) - (100d - c)$.
|
Three, from $b<a<d<c$, we can get
$$
\begin{array}{l}
a+b<b+d<b+c<a+c<d+c, \\
b+d<a+d<a+c .
\end{array}
$$
Therefore, $a+d$ and $b+c$ are both between $b+d$ and $a+c$.
It is easy to know that $a+b=26, b+d=27, a+c=115, d+c=116$.
If $a+d=101$, then $b+c=41$.
Solving this gives $b=-24$ (discard).
Thus, $a+d=41, b+c=101$.
Solving this gives $a=20, b=6, c=95, d=21$.
Therefore, $(100 a+b)-(100 d-c)=2006-2005=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the Cartesian coordinate system $x O y$, it is known that points $A(6,4)、B(4,0)、C(0,3)$, and the line $l: y=k x$ bisects the area of $\triangle A B C$, $k=\frac{-b+\sqrt{c}}{a}(a、b、c$ are all positive integers, and $c$ has no square factors). Then $a+b+c=$ $\qquad$ .
|
6.584.
As shown in Figure 5, take the midpoint $M\left(2, \frac{3}{2}\right)$ of side $BC$. Let the line $l$ intersect sides $BC$ and $CA$ at points $P$ and $Q$, respectively.
Since $AM$ and $PQ$ both bisect the area of $\triangle ABC$, we have $S_{\triangle PQC} = S_{\triangle MMC}$. Subtracting the common area $S_{\triangle CMQ}$, we get $S_{\triangle PMQ} = S_{\triangle \triangle MQ}$. Therefore, the distances from points $P$ and $A$ to the line $MQ$ are equal. Hence, $PA \parallel MQ$.
The equation of line $l_{BC}$ is $\frac{x}{4} + \frac{y}{3} = 1$, or $y = 3 - \frac{3}{4}x$;
The equation of line $l_{AC}$ is $\frac{y-3}{x-0} = \frac{4-3}{6-0}$, or $y = \frac{1}{6}x + 3$;
The equation of line $l$ is $y = kx$.
From equations (1) and (3), we get $P\left(\frac{12}{4k+3}, \frac{12k}{4k+3}\right)$,
From equations (2) and (3), we get $Q\left(\frac{18}{6k-1}, \frac{18k}{6k}-1\right)$.
Since $PA \parallel MQ$, we have $k_{PA} = k_{MQ}$, leading to $132k^2 + 77k - \frac{231}{2} = 0$.
Thus, $k = \frac{-7 + \sqrt{553}}{24}$.
Therefore, $a + b + c = 24 + 7 + 553 = 584$.
|
584
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For each positive integer $n$, define the function
$$
f(n)=\left\{\begin{array}{cl}
0, & \text { when } n \text { is a perfect square; } \\
{\left[\frac{1}{\{\sqrt{n}\}}\right],} & \text { when } n \text { is not a perfect square. }
\end{array}\right.
$$
where $[x]$ denotes the greatest integer not exceeding $x$, and $\{x\}=x-[x]$. Find: $\sum_{k=1}^{200} f(k)$.
|
Solution: If $k$ is not a perfect square, then there exists $a \in \mathbf{N}$, such that $a^{2}<k<(a+1)^{2}$, then $a<\sqrt{k}<a+1$, i.e.,
$$
\{\sqrt{k}\}=\sqrt{k}-a \text {. }
$$
Therefore, $\left[\frac{1}{\{\sqrt{k}\}}\right]=\left[\frac{1}{\sqrt{k}-a}\right]=\left[\frac{\sqrt{k}+a}{k-a^{2}}\right]$.
$$
\begin{aligned}
\text { Hence } & \sum_{k=1}^{2+0} f(k) \\
= & 15 \times 0+\sum_{k=2}^{3}\left[\frac{\sqrt{k}+1}{k-1}\right]+\sum_{k=5}^{8}\left[\frac{\sqrt{k}+2}{k-4}\right]+ \\
& \sum_{k=10}^{15}\left[\frac{\sqrt{k}+3}{k-9}\right]+\sum_{k=17}^{24}\left[\frac{\sqrt{k}+4}{k-16}\right]+\cdots+ \\
& \sum_{k=197}^{24}\left[\frac{\sqrt{k}+14}{k-196}\right]+\sum_{k=20}^{2+0}\left[\frac{\sqrt{k}+15}{k-225}\right] \\
= & 0+(2+1)+(4+2+1+1)+ \\
& (6+3+2+1 \times 3)+(8+4+2+2+1 \times 4)+ \\
& (10+5+3+2+2+1 \times 5)+ \\
& (12+6+4+3+2+2+1 \times 6)+ \\
& (14+7+4+3+2+2+2+1 \times 7)+ \\
& (16+8+5+4+3+2+2+2+1 \times 8)+ \\
& (18+9+6+4+3+3+2+2+2+1 \times 9)+ \\
& (20+10+6+5+4+3+2+2+2+2+1 \times 10)+ \\
& (22+11+7+5+4+3+3+2+2+2+ \\
& 2+1 \times 11)+(24+12+8+6+4+4+ \\
& 3+3+2+2+2+2+1 \times 12)+(26+13+8+6+5+ \\
& 4+3+3+2+2+2+2+2+1 \times 13)+ \\
& (28+14+9+7+5+4+4+3+3+2+ \\
& 2+2+2+2+1 \times 14)+(30+15+10+7+ \\
& 6+5+4+3+3+3+2+2+2+2+2) \\
= & 768
\end{aligned}
$$
|
768
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$6 \rightarrow$ In a regular $n$-sided polygon, the difference between any two adjacent interior angles is $18^{\circ}$, find the maximum value of $n$,
|
Hedgehog, this problem can be solved in three steps:
(1) Prove that $n$ is even
(2) Prove that $n<40$, and $n \leqslant 38$;
(3) Prove that $n=38$, i.e., the maximum value of $n$ is 38.
|
38
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The six faces of a unit cube are painted in 6 different colors, and a different number of roosters are drawn on each face. The colors and the number of roosters on each face correspond as shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Color on the face & Red & Yellow & Blue & Green & Purple & Green \\
\hline Number of roosters on the face & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\end{tabular}
Take 4 of the above unit cubes and arrange them to form a horizontally placed rectangular prism as shown in Figure 1. Then, the total number of roosters on the bottom face of this rectangular prism is $\qquad$.
Note: There is a minor error in the table where "Green" is listed twice. It should be "Green" and "Cyan" to match the original Chinese text. Here is the corrected version:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Color on the face & Red & Yellow & Blue & Cyan & Purple & Green \\
\hline Number of roosters on the face & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
\end{tabular}
|
5.17.
Since the 4 aforementioned unit cubes are identical, it can be observed that the colors of the 4 adjacent faces to the red face are blue, yellow, cyan, and purple, so the color of the face opposite the red face is green. The colors of the 4 adjacent faces to the yellow face are cyan, blue, red, and green, thus the color of the face opposite the yellow face is purple. The only remaining possibility is that the face opposite the cyan face is blue, which is the color of the top face of the rectangular prism. The corresponding bottom face color and the number of golden roosters on that face are:
$$
\begin{array}{l}
\text { Top }(\text { Yellow }) \longleftrightarrow \text { Bottom }(\text { Purple }) \longleftrightarrow 5 \text {; } \\
\text { Top }(\text { Purple }) \longleftrightarrow \text { Bottom }(\text { Yellow }) \longleftrightarrow 2 \text {; } \\
\text { Top }(\text { Red }) \longleftrightarrow \text { Bottom }(\text { Green }) \longleftrightarrow 6 \text {; } \\
\text { Top }(\text { Blue }) \longleftrightarrow \text { Bottom }(\text { Cyan }) \longleftrightarrow 4 \text { . }
\end{array}
$$
Therefore, the bottom face of this rectangular prism has a total of 17 golden roosters.
|
17
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. In the arithmetic sequences $3,10,17, \cdots, 2005$ and $3,8, 13, \cdots, 2003$, the number of terms that have the same value is $\qquad$.
|
9.58.
Subtract 3 from each term of the two sequences, transforming them into $0,7,14, \cdots$, 2002 and $0,5,10, \cdots, 2000$. The first sequence represents multiples of 7 not exceeding 2002, and the second sequence can be seen as multiples of 5 within the same range. Therefore, the common terms are multiples of 35. Hence $\left[\frac{2002}{35}\right]+1=58$.
|
58
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. If $2^{6}+2^{9}+2^{n}$ is a perfect square, then the positive integer $n=$ $\qquad$ .
|
11. 10 .
$$
2^{6}+2^{9}=2^{6}\left(1+2^{3}\right)=2^{6} \times 3^{2}=24^{2} \text {. }
$$
Let $24^{2}+2^{n}=a^{2}$, then
$$
(a+24)(a-24)=2^{n} \text {. }
$$
Thus, $a+24=2^{r}, a-24=2^{t}$,
$$
2^{r}-2^{t}=48=2^{4} \times 3,2^{t}\left(2^{r-t}-1\right)=2^{4} \times 3 \text {. }
$$
Then $t=4, r-t=2$.
So $r=6, n=t+r=10$.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Using weights labeled $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}, 15 \mathrm{~g}, 40 \mathrm{~g}$, each one of them, to weigh objects on a balance scale without graduations. If weights can be placed on both ends of the balance, then the maximum number of different gram weights (positive integer weights) that can be measured on this balance is.
|
12.55.
Using $1 \mathrm{~g}, 2 \mathrm{~g}, 3 \mathrm{~g}$ these three weights, all integer grams in the interval $A=$ $[1,6]$ can be measured; after adding a $15 \mathrm{~g}$ weight, the measurement range is expanded to the interval $B=[15-6,15+6]=[9,21]$. After adding a $40 \mathrm{~g}$ weight, the measurement range is further expanded to 3 intervals:
$$
\begin{array}{l}
C_{1}=[40-6,40+6]=[34,46], \\
C_{2}=[40-21,40-9]=[19,31], \\
C_{3}=[40+9,40+21]=[49,61] .
\end{array}
$$
However, interval $B$ overlaps with $C_{2}$ by 3 integers. Calculating the number of integers in each of the above intervals, the total number of different grams that can be measured is
$$
6+13+(13+13+13)-3=55 \text { (types). }
$$
|
55
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Find the smallest positive integer $n$, such that among any $n$ consecutive positive integers, there is at least one number whose sum of digits is a multiple of 7.
|
15. First, we can point out 12 consecutive positive integers, for example,
$$
994,995, \cdots, 999,1000,1001, \cdots, 1005 \text {, }
$$
where the sum of the digits of any number is not a multiple of 7. Therefore, $n \geqslant 13$.
Next, we prove that in any 13 consecutive positive integers, there must be one number whose sum of digits is a multiple of 7.
For each non-negative integer $a$, we call the set of 10 numbers
$$
A_{a}=\{10 a, 10 a+1, \cdots, 10 a+9\}
$$
a "basic segment." 13 consecutive positive integers either belong to two basic segments or belong to three basic segments.
When 13 numbers belong to two basic segments, by the pigeonhole principle, there must be 7 consecutive numbers belonging to the same basic segment;
When 13 consecutive numbers belong to three basic segments $A_{a-1} 、 A_{a} 、 A_{a+1}$, there must be 10 consecutive numbers all belonging to $A_{a}$. Now, let
be 7 numbers belonging to the same basic segment, and their sums of digits are
$$
\sum_{i=0}^{k} a_{i}, \sum_{i=0}^{k} a_{i}+1, \cdots, \sum_{i=0}^{k} a_{i}+6 .
$$
Clearly, these 7 sums have different remainders when divided by 7, and one of them must be a multiple of 7.
Therefore, the smallest value is $n=13$.
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.3. On an island, there live 100 people, some of whom always lie, while the rest always tell the truth. Each resident of the island worships one of three gods: the Sun God, the Moon God, and the Earth God. Each resident was asked three questions:
(1) Do you worship the Sun God?
(2) Do you worship the Moon God?
(3) Do you worship the Earth God?
To the first question, 60 people answered: “Yes”; to the second question, 40 people answered: “Yes”; to the third question, 30 people answered: “Yes”. How many of them were lying?
|
7.3. People who always tell the truth are called "honest people", and those who always lie are called "liars". Each honest person will only answer "yes" to one question, while each liar will answer "yes" to two questions. Let the number of honest people be $x$, and the number of liars be $y$. Therefore, $x+2y=130$. Since there are 100 people living on the island, we have $x+y=100$. Thus, it is known that $y=30$ people are telling lies.
|
30
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.3. In the guard of a grand duke, there are 1000 warriors. Any two warriors are either friends, enemies, or do not know each other. The warriors are all honest and only speak to their friends. However, the current situation makes every warrior unhappy, because for each warrior, any two of his friends are enemies, and any two of his enemies are friends. Prove: To ensure that all warriors know about a new decision of the grand duke, the grand duke needs to notify at least 200 warriors.
|
7.3. We point out that any samurai has no more than 2 friends. In fact, if a certain samurai has at least 3 friends $A, B, C$, then according to the problem, $A, B, C$ must be enemies with each other. This leads to a contradiction: since $B$ and $C$ are both enemies of $A$, they should be friends, not enemies. Therefore, each samurai has at most 2 friends. Thus, the samurais form several chains (which can be graphs): the first person is a friend of the second person, the second person is a friend of the third person, and so on.
Consider a continuous sequence of 5 people $A, B, C, D, E$ on a chain: since $A$ and $C$ are both friends of $B$, they are enemies; since $C$ and $E$ are both friends of $D$, they are also enemies; thus, $A$ and $E$ are both enemies of $C$, so they are friends. Therefore, each of these 5 people already has 2 friends, and thus they do not have any other friends. This means that the length of each such friend chain is at most 5.
The above facts indicate that each samurai can inform at most 5 people (including themselves). Therefore, if the Duke only informs fewer than 200 samurais, then fewer than $5 \times 200 = 1000$ samurais will know the new decision. Hence, he needs to inform at least 200 samurais.
|
200
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8.3. Let $p(n, k)$ denote the number of divisors of the positive integer $n$ that are not less than $k$. Try to find
$$
\begin{array}{l}
p(1001,1)+p(1002,2)+\cdots+ \\
p(2000,1000) .
\end{array}
$$
|
8.3. If we write down all the divisors of each positive integer $1000+k$ that are not less than $k$ $(k=1,2, \cdots, 1000)$, then the sum we are looking for is the sum of the number of all these divisors.
Below we prove that each positive integer $n$ from 1 to 2000 is written exactly once. If $n>1000$, then it is only written when it is a divisor of itself. If $n \leqslant 1000$, then for $k \leqslant n$, when $1000+k$ is a multiple of $n$, it is listed among the divisors of $1000+k$. Since there is exactly one number that is a multiple of $n$ between $1000+1$ and $1000+n$, $n$ is also written exactly once. Therefore, each positive integer $n$ from 1 to 2000 is written exactly once, and they total 2000.
|
2000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x=\frac{3-\sqrt{5}}{2}$. Then $x^{3}-3 x^{2}+3 x+$ $\frac{6}{x^{2}+1}=$ . $\qquad$
|
$$
=, 1.6 \text {. }
$$
Given $x=\frac{3-\sqrt{5}}{2}$, we know that
$$
\begin{array}{l}
x^{2}-3 x+1=0, x+\frac{1}{x}=3 . \\
\text { Therefore, } x^{3}-3 x^{2}+3 x+\frac{6}{x^{2}+1} \\
=x\left(x^{2}-3 x\right)+3 x+\frac{6}{x^{2}+1}=2 x+\frac{6}{3 x} \\
=2 x+\frac{2}{x}=2\left(x+\frac{1}{x}\right)=6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a$ and $b$ are real numbers, and $a \geqslant 1$. If the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ has real solutions, and satisfies $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, then $a^{2}+b^{2}=$
|
3.6 .
From the equation $x^{2}-2 b x-\left(a-2 b^{2}\right)=0$ having real solutions, we get $\Delta=4 b^{2}+4\left(a-2 b^{2}\right) \geqslant 0$, which means $a \geqslant b^{2}$.
From $2 a^{2}-a b^{2}-5 a+b^{2}+4=0$, we can derive $2 a^{2}-5 a+4=a b^{2}-b^{2}=b^{2}(a-1) \leqslant a(a-1)$. Therefore, $2 a^{2}-5 a+4 \leqslant a^{2}-a$, which simplifies to $(a-2)^{2} \leqslant 0$.
Solving this, we get $a=2, b^{2}=2$.
Thus, $a^{2}+b^{2}=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 A mall installs an escalator from the first floor to the second floor, which moves upward at a uniform speed. A boy and a girl start walking up the escalator at the same time (the escalator is also moving). If both of their movements are considered uniform, and the boy walks twice as many steps per minute as the girl, it is known that the boy walked 27 steps to reach the top of the escalator, while the girl walked 18 steps to reach the top (assuming the boy and the girl each step up one step at a time).
(1) How many steps are visible on the escalator?
(2) If there is a staircase near the escalator that leads from the second floor to the first floor, with the same number of steps as the escalator, and both individuals walk down the stairs at their original speeds, then reach the bottom and ride the escalator again (ignoring the distance between the escalator and the stairs), how many steps will the boy have walked when he catches up to the girl for the first time?
(9th Jiangsu Province Junior High School Mathematics Competition)
|
Explanation: (1) Let the girl's speed be $x$ steps/min, the escalator's speed be $y$ steps/min, the boy's speed be $2x$ steps/min, and the stairs have $s$ steps. Then
$$
\left\{\begin{array}{l}
\frac{27}{2 x}=\frac{s-27}{y}, \\
\frac{18}{x}=\frac{s-18}{y} .
\end{array}\right.
$$
(1)
$$
\frac{3}{4}=\frac{s-27}{s-18} \text {. }
$$
Solving for $s$ gives $s=54$.
Therefore, the part of the escalator exposed is 54 steps.
(2) From (1), we know $y=2x$.
Let the boy pass the girl for the first time after walking the escalator $m$ times and the stairs $n$ times, then the girl has walked the escalator $m-1$ times and the stairs $n-1$ times. The boy walks up the escalator at $4x$ steps/min, and the girl walks up the escalator at $3x$ steps/min. Then
$$
\frac{54 m}{4 x}+\frac{54 n}{2 x}=\frac{54(m-1)}{3 x}+\frac{54(n-1)}{x} \text {. }
$$
Thus, $\frac{m}{4}+\frac{n}{2}=\frac{m-1}{3}+\frac{n-1}{1}$, which simplifies to
$$
6 n+m=16 \text {. }
$$
Since $m$ and $n$ must be positive integers, and $0 \leqslant m-n \leqslant 1$. Therefore, we can get Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline$m$ & 1 & 2 & 3 & 4 & 5 \\
\hline$n=\frac{16-m}{6}$ & $2 \frac{1}{2}$ & $2 \frac{1}{3}$ & $2 \frac{1}{6}$ & 2 & $1 \frac{5}{6}$ \\
\hline$n$ & 1 & 2 & & & \\
\hline$m=16-6 n$ & 10 & 4 & & & \\
\hline
\end{tabular}
From Table 1, we can see that the only solution is $m=3, n=2 \frac{1}{6}$, meaning the boy walks up the escalator 3 times and down the stairs $2 \frac{1}{6}$ times.
Therefore, when the boy first catches up with the girl, he has walked $3 \times 27 + 2 \frac{1}{6} \times 54 = 198$ steps.
Note: Indeterminate equation problems in math competitions can often be solved using the tabulation method.
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 A pedestrian and a cyclist are traveling south simultaneously on a road parallel to a railway. The pedestrian's speed is $3.6 \mathrm{~km} / \mathrm{h}$, and the cyclist's speed is $10.8 \mathrm{~km} / \mathrm{h}$. If a train comes from behind and takes $22 \mathrm{~s}$ to pass the pedestrian and $26 \mathrm{~s}$ to pass the cyclist, then the length of the train is $\qquad$ $\mathrm{m}$.
(2002, Hubei Province Junior High School Mathematics Competition)
|
Explanation: First, convert the two speed units to: pedestrian $1 \mathrm{~m} / \mathrm{s}$, cyclist $3 \mathrm{~m} / \mathrm{s}$. Let the length of the train be $l \mathrm{~m}$, and the speed of the train be $v \mathrm{~m} / \mathrm{s}$. According to the problem, we have
$$
\left\{\begin{array}{l}
l=(v-1) \times 22, \\
l=(v-3) \times 26 .
\end{array}\right.
$$
Solving these equations gives $l=286, v=14$.
Therefore, the length of the train is $286 \mathrm{~m}$.
|
286
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the function $f(n)=\frac{20^{n}+3^{n}}{n!}, n \in \mathbf{N}$. Then, the value of $n$ that maximizes $f(n)$ is, $n=$
|
3.19.
$$
\begin{array}{l}
G(n)=f(n+1)-f(n) \\
=\frac{20^{n+1}+3^{n+1}}{(n+1)!}-\frac{20^{n}+3^{n}}{n!} \\
=\frac{(19-n) \times 20^{n}-(n-2) \times 3^{n}}{(n+1)!} \\
=\frac{3^{n}}{(n+1)!}\left[(19-n) \times\left(\frac{20}{3}\right)^{n}-(n-2)\right] .
\end{array}
$$
It is easy to see that when $n \geqslant 19$, $G(n) < 0$.
Therefore, $f(19)$ is the maximum value.
|
19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The maximum value of the algebraic expression $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$ is $\qquad$ .
|
5.2.
It is known that $|a| \leqslant \sqrt{2},|b| \leqslant \sqrt{2}$. Let
$$
a=\sqrt{2} \sin \alpha, b=\sqrt{2} \sin \beta, \alpha, \beta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \text {. }
$$
Then $a \sqrt{2-b^{2}}+b \sqrt{2-a^{2}}$
$$
\begin{array}{l}
=2(\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \alpha) \\
=2 \sin (\alpha+\beta) \leqslant 2 .
\end{array}
$$
Equality holds if and only if $\alpha+\beta=\frac{\pi}{2}$, i.e., $a^{2}+b^{2}=2, a, b \geqslant 0$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six. (15 points) A square is divided into 4 squares, the number of edges in the division diagram is 12. If a square is divided into 2005 convex polygons, try to find the maximum number of edges in the division diagram.
|
Six, according to Euler's theorem, the number of vertices $a$, the number of faces $b$, and the number of edges $e$ of a simple polyhedron have the following relationship: $a+b-e=2$.
From Euler's theorem, it is easy to see that if a convex polygon is divided into $n$ convex polygons, then the number of vertices $a$, the number of polygons $n$, and the number of edges $e$ in the division graph satisfy
$$
a+n-e=1 \text {. }
$$
Below, in the general case, that is, when a square is divided into $n$ convex polygons, find the maximum value of the number of edges in the division graph. Let the number of vertices in the division graph be $a$, the number of polygons be $n$, and the number of edges be $e$.
(1) First, find the upper bound of the number of edges.
Let the four vertices of the original square be $A, B, C, D$. If the vertex $V \notin\{A, B, C, D\}$ of the convex polygon, it is easy to know that
$d(V) \geqslant 3$ (here $d(V)$ represents the number of edges passing through vertex $V$). Hence, $d(V) \leqslant 3(d(V)-2)$.
Noting that there are $a-4$ such vertices $V$, we have $a-4$ inequalities of the above form. Adding them up and noting that each edge, except for the four sides of the square, is the edge of two convex polygons, we have
$$
\begin{array}{l}
2 e-[d(A)+d(B)+d(C)+d(D)] \\
\leqslant 3[2 e-(d(A)+d(B)+d(C)+d(D))]- \\
\quad 6(a-4),
\end{array}
$$
i.e., $4 e \geqslant 2(d(A)+d(B)+d(C)+d(D))+6(a-4)$.
Since $d(A) \geqslant 2, d(B) \geqslant 2, d(C) \geqslant 2, d(D) \geqslant 2$,
then $2 e \geqslant 8+3(a-4)=3 a-4$.
From equation (1), we have
$$
3 a+3 n-3 e=3 \text {. }
$$
Substituting equation (2) into equation (3) and rearranging, we get
$$
3 e+3=3 a+3 n \leqslant 2 e+4+3 n,
$$
i.e., $e \leqslant 3 n+1$.
(2) Construct an example where the number of edges $e=3 n+1$.
As shown in Figure 3, draw $n-1$ parallel lines to one side of the square successively, dividing the square into $n$ rectangles. It is easy to see that the number of edges
$$
\begin{array}{l}
e=4+3(n-1) \\
=3 n+1 .
\end{array}
$$
In summary, the maximum value of the number of edges in the division graph is $3 n+1$.
Therefore, when the square is divided into 2005 convex polygons, the maximum number of edges is 6016.
(Wang Xiaohui provided)
|
6016
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a regular $n$-sided polygon inscribed in a circle. If the line connecting a vertex to the center of the circle is exactly on the perpendicular bisector of a certain side, then $n \equiv$ $\qquad$ $(\bmod 2)$
|
2.1.
As shown in Figure 8, if the line connecting vertex $A_{1}$ and the center $O$ perpendicularly bisects side $A_{i} A_{j}$, then $\triangle A_{1} A_{i} A_{j}$ is an isosceles triangle. Since equal chords subtend equal arcs, the number of vertices in $\overparen{A_{1} A_{i}}$ is equal, and their sum is even. Adding the points $A_{1} 、 A_{i} 、 A_{j}$, $n$ must be odd. Therefore, $n \equiv 1(\bmod 2)$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given a sequence of numbers $a_{1}, a_{2}, \cdots, a_{2006}$, where $a_{1}$ $=1$, and the sum of each pair of consecutive terms is 3. Then $a_{1}-a_{2}+$ $a_{3}-a_{4}+\cdots+a_{2003}-a_{2004}+a_{2005}=$ $\qquad$
|
3. -1001 .
Given $a_{1}+a_{2}=a_{2}+a_{3}=\cdots=a_{2004}+a_{2000}=3$, and $a_{1}=1$, we can sequentially deduce that
$$
\begin{array}{l}
a_{1}=a_{3}=\cdots=a_{2005}=1, a_{2}=a_{4}=\cdots=a_{2004}=2 . \\
\text { Then } a_{1}-a_{2}+a_{3}-a_{4}+\cdots-a_{2004}+a_{2000} \\
=\left(a_{1}-a_{2}\right)+\left(a_{3}-a_{4}\right)+\cdots+\left(a_{2003}-a_{2004}\right)+a_{2000} \\
=-1002+1=-1001 .
\end{array}
$$
|
-1001
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Find the smallest positive integer $k$, such that there exists a positive integer $n$, satisfying $10^{n}=29 k+2$.
---
The translation maintains the original text's format and line breaks.
|
$$
\begin{array}{l}
\text { Given } \frac{1}{29}=0.0334482758620689655172413793 \text { i, } \\
\frac{2}{29}=0.0068965517241379310344827586 \text { 2e, }
\end{array}
$$
we know that when $n=11$, $k$ takes the minimum value 3448275862.
Explanation: By appending the 28-digit repeating cycle of $\frac{1}{29}$ to 3448275862, infinitely many solutions for $k$ can be obtained.
(Ro Zengru, Department of Mathematics, Shaanxi Normal University, 710062)
|
3448275862
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $A=\{1,2,3,4,5\}$. Then the number of mappings $f: A \rightarrow A$ that satisfy the condition $f(f(x))$ $=f(x)$ is $\qquad$ (answer with a number)
|
3.196 .
It can be generalized to the case where $A$ has $n$ elements. From the condition, if $a \in A$ and $a$ is in the range of $f$, then it must be that $f(a)=a$. Therefore, we can classify $f$ based on the number of elements in its range. The number of $f$ with $k\left(k \in \mathbf{N}_{+}\right)$ elements in its range is $\mathrm{C}_{n}^{k} k^{n-k}$. Hence, the number of $f$ that satisfy the condition is $\sum_{k=1}^{n} \mathrm{C}_{n}^{k} k^{n-k}$. Taking $n=5$ yields the result.
|
196
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the geometric sequence $\left\{a_{n}\right\}$, $a_{1}=\frac{1}{8}$, the geometric mean of the first $n$ terms is 8. If the geometric mean of the remaining terms after removing one term from the first $n$ terms is $4 \sqrt{2}$, then the removed term is the $\qquad$th term.
|
6.13.
Let the common ratio be $q$, the product of the first $n$ terms be $M$, and the $k$-th term be removed, then $M=8^{n}$, and $\frac{M}{a_{k}}=(4 \sqrt{2})^{n-1}$.
Thus, $a_{k}=(\sqrt{2})^{n+5}$, which means $\frac{1}{8} q^{k-1}=(\sqrt{2})^{n+5}$.
Therefore, $q^{k-1}=(\sqrt{2})^{n+11}$.
Also, $M=8^{n}$, i.e., $\left(\frac{1}{8}\right)^{n} q^{\frac{n(n-1)}{2}}=8^{n}$, which gives $q^{n-1}=(\sqrt{2})^{24}$.
Comparing (1) and (2), we have $k-1=\frac{(n-1)(n+11)}{24}$.
Given $k \leqslant n$, we solve to get $n \leqslant 13$, and
$k-1=\frac{(n-1)^{2}+12(n-1)}{24}$.
Thus, 12 divides $n-1$.
Therefore, $n=1$ (contradicts the given) or $n=13$. Hence, $k=13$.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) If a positive integer $n$ has the sum of its digits in base 3 divisible by 3, then $n$ is called a "proper number." Find the sum of all proper numbers in $S=\{1,2, \cdots, 2005\}$.
|
For $m \in \mathbf{N}$, in ternary (base 3) representation, there are $3^{m+1}$ non-negative integers with at most $m+1$ digits. Let the number of these integers whose digit sums are congruent to 0, 1, and 2 modulo 3 be $a_{m}$, $b_{m}$, and $c_{m}$, respectively. When $m \geqslant 1$, by classifying the numbers whose digit sums are congruent to 0 modulo 3 according to their leading digit (0, 1, 2) (for numbers with fewer than $m+1$ digits, prepend leading zeros to make them $m+1$ digits long, and similarly for the following), we easily get
$$
a_{m}=a_{m-1}+b_{m-1}+c_{m-1}.
$$
Similarly,
$$
b_{m}=a_{m-1}+b_{m-1}+c_{m-1},
$$
$$
c_{m}=a_{m-1}+b_{m-1}+c_{m-1}.
$$
Thus, $a_{m}=b_{m}=c_{m}=3^{m} (m \geqslant 1)$. Clearly, $a_{0}=b_{0}=c_{0}=1$, so for $m \in \mathbf{N}$, we always have $a_{m}=b_{m}=c_{m}=3^{m}$. Therefore, in ternary representation, there are $3^{m}-1$ appropriate numbers with at most $m+1$ digits.
For $m \in \mathbf{N}$, among the $3^{m+1}$ non-negative integers with at most $m+1$ digits in binary (base 2) representation, let the sums of the numbers whose digit sums are congruent to 0, 1, and 2 modulo 3 be $A_{m}$, $B_{m}$, and $C_{m}$, respectively. Then $A_{m}+B_{m}+C_{m}$ equals the sum of these $3^{m+1}$ numbers, i.e.,
$$
\begin{array}{l}
A_{m}+B_{m}+C_{m}=1+2+\cdots+\left(3^{m+1}-1\right) \\
=\frac{1}{2} \times 3^{m+1}\left(3^{m+1}-1\right).
\end{array}
$$
When $m \geqslant 1$, among the $a_{m}$ numbers whose digit sums are congruent to 0 modulo 3, the sum of the $a_{m-1}$ numbers with leading digit 0 is $A_{m-1}$, the sum of the $c_{m-1}$ numbers with leading digit 1 is $1 \times 3^{m} c_{m-1}+C_{m-1}=3^{2 m-1}+C_{m-1}$, and the sum of the $b_{m-1}$ numbers with leading digit 2 is $2 \times 3^{m} b_{m-1}+B_{m-1}=2 \times 3^{2 m-1}+B_{m-1}$. Thus,
$$
\begin{array}{l}
A_{m}=A_{m-1}+3^{2 m-1}+C_{m-1}+2 \times 3^{2 m-1}+B_{m-1} \\
=3^{2 m}+A_{m-1}+B_{m-1}+C_{m-1}=3^{2 m}+\frac{1}{2} \times 3^{m}\left(3^{m}-1\right) \\
=\frac{1}{2} \times 3^{m}\left(3^{m+1}-1\right)(m \geqslant 1).
\end{array}
$$
Therefore, the sum of the $3^{m}-1$ appropriate numbers with at most $m+1$ digits in ternary representation is $\frac{1}{2} \times 3^{m}\left(3^{m+1}-1\right)(m \geqslant 1)$.
Since $2005=(2202021)_{3}$ is a 7-digit number, and from the previous analysis, among the positive integers with at most 7 digits, there are $3^{6}-1$ appropriate numbers, and their sum is $\frac{1}{2} \times 3^{6}\left(3^{7}-1\right)=796797$. This sum includes 60 appropriate numbers that are greater than $(2202021)_{3}$ and less than or equal to $(2222222)_{3}$. Arranging these 60 numbers in ascending order, we get
$$
\begin{array}{l}
(2202102)_{3},(2202111)_{3},(2202120)_{3}, \\
(2202201)_{3},(2202210)_{3},(2202222)_{3} ; \\
(2210001)_{3},(2210010)_{3},(2210022)_{3}, \\
(2210100)_{3},(2210112)_{3},(2210121)_{3} ; \\
(2210202)_{3},(2210211)_{3},(2210220)_{3}, \\
(2211000)_{3},(2211012)_{3},(2211021)_{3} ; \\
(2211102)_{3},(2211111)_{3},(2211120)_{3}, \\
(2211201)_{3},(2211210)_{3},(2211222)_{3} ; \\
(2212002)_{3},(2212011)_{3},(2212020)_{3}, \\
(2212101)_{3},(2212110)_{3},(2212122)_{3}, \\
(2212200)_{3},(2212212)_{3},(2212221)_{3} ; \\
(2220000)_{3},(2220012)_{3},(2220021)_{3}, \\
(2220102)_{3},(2220111)_{3},(2220120)_{3}, \\
(2220201)_{3},(2220210)_{3},(2220222)_{3} ; \\
(2221002)_{3},(2221011)_{3},(2221020)_{3} \\
(2221101)_{3},(2221110)_{3},(2221122)_{3} ; \\
(2221200)_{3},(2221212)_{3},(2221221)_{3}, \\
(2222001)_{3},(2222010)_{3},(2222022)_{3} ; \\
(2222100)_{3},(2222112)_{3},(2222121)_{3} \\
(2222202)_{3},(2222211)_{3},(2222220)_{3} ;
\end{array}
$$
It is easy to calculate that the sum of these 60 numbers is
$$
\begin{array}{r}
120 \times\left(3^{6}+3^{5}\right)+81 \times 3^{4}+(18 \times 1+24 \times 2) \times \\
3^{3}+(21 \times 1+21 \times 2) \times 3^{2}+60 \times(3+1)=125790 .
\end{array}
$$
Therefore, the sum of all appropriate numbers in $S=\{1,2, \cdots, 2005\}$ is $796797-125790=671007$.
(Fang Tinggang, Chengdu No. 7 High School, Chengdu, Sichuan, 610041)
|
671007
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 A positive integer $n$ cannot be divisible by $2$ or $3$, and there do not exist non-negative integers $a$, $b$ such that $\left|2^{a}-3^{b}\right|=n$. Find the minimum value of $n$.
(2003, National Training Team Problem)
|
Explanation: When $n=1$, $\left|2^{1}-3^{1}\right|=1$;
When $n=5$, $12^{2}-3^{2} \mid=5$;
When $n=7$, $\left|2^{1}-3^{2}\right|=7$;
When $n=11$, $\left|2^{4}-3^{3}\right|=11$;
When $n=13$, $\left|2^{4}-3^{1}\right|=13$;
When $n=17$, $12^{6}-3^{4} \mid=17$;
When $n=19$, $12^{3}-3^{3} \mid=19$;
When $n=23$, $12^{5}-3^{2} \mid=23$;
When $n=25$, $12^{1}-3^{3} \mid=25$;
When $n=29$, $12^{5}-3^{1} \mid=29$;
When $n=31$, $\left|2^{5}-3^{0}\right|=31$.
Below, we use proof by contradiction to show that $n=35$ satisfies the requirement.
If not, there exist non-negative integers $a, b$ such that
$$
\left|2^{a}-3^{b}\right|=35.
$$
(1) If $2^{a}-3^{b}=35$, clearly, $a \neq 0,1,2$, hence $a \geqslant 3$. Taking modulo 8, we get $-3^{b} \equiv 3(\bmod 8)$, i.e., $3^{b} \equiv 5(\bmod 8)$, but $3^{b} \equiv 1,3(\bmod 8)$, which is impossible.
(2) If $3^{b}-2^{a}=35$, it is easy to see that $b \neq 0,1$. Taking modulo 9, we get $2^{a} \equiv 1(\bmod 9)$. And
$$
\begin{array}{l}
2^{6 k} \equiv 1(\bmod 9), 2^{6 k+1} \equiv 2(\bmod 9), \\
2^{6 k+2} \equiv 4(\bmod 9), 2^{6 k+3} \equiv 8(\bmod 9), \\
2^{6 k+4} \equiv 7(\bmod 9), 2^{6 k+5} \equiv 5(\bmod 9).
\end{array}
$$
Thus, $a=6 k(k \in \mathbf{N})$.
Therefore, $3^{b}-8^{2 k}=35$.
Taking modulo 7, we get $3^{b} \equiv 1(\bmod 7)$.
And $\left\{3^{k}(\bmod 7)\right\}: 3,2,6,4,5,1,3,2,6,4$, $5,1, \cdots$, hence $b=6 k^{\prime}\left(k^{\prime} \in \mathbf{N}_{+}\right)$. Therefore, $3^{6 k^{\prime}}-2^{6 k}=35$,
i.e., $\left(3^{3 k^{\prime}}-2^{3 k}\right)\left(3^{3 k^{\prime}}+2^{3 k}\right)=35$.
Thus, $\left\{\begin{array}{l}3^{3 k^{\prime}}-2^{3 k}=1, \text { or }\left\{\begin{array}{l}3^{3 k^{\prime}}-2^{3 k}=5, \\ 3^{3 k^{k}}+2^{3 k}=35\end{array} 2^{3 k^{k}}+2^{3 k}=7 .\right.\end{array}\right.$
Therefore, $3^{3 k^{\prime}}=18$ or 6, which is impossible.
In conclusion, the smallest value of $n$ is 35.
|
35
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Find the smallest positive integer $a$, such that there exists a positive odd integer $n$, satisfying $2001 \mid \left(55^{n}+a \times 32^{n}\right)$.
（14th Irish Mathematical Olympiad）
|
(Given $2001=87 \times 23$, we get $87 \mid\left(55^{n}+a \times\right.$ $\left.32^{n}\right)$, and $231\left(55^{n}+a \times 32^{n}\right)$. Then $0 \equiv 55^{n}+a \times 32^{n} \equiv$ $(-32)^{n}+a \times 32^{n}=32^{n}(a-1)(\bmod 87)$. Hence $a-1=$ $0(\bmod 87)$. Similarly, we get $a+1 \equiv 0(\bmod 23)$. Let $a=87 k+$ $1(k \in \mathbf{N})$, then 23 । $(87 k+2)$. Thus, 23 । $(-5 k+2)$, i.e., $k_{\min }=5$. Therefore, $a_{\min }=436$.)
|
436
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 As shown in Figure 1, quadrilateral $ABCD$ is a rectangle. Two people, A and B, start from points $A$ and $B$ respectively at the same time, and move counterclockwise along the rectangle. In which circle does B possibly catch up with A for the first time? In which circle does B certainly catch up with A at the latest? Please explain your reasoning.
|
Explanation: Let $A D=B C=a \mathrm{~m}, A B=C D=b . \mathrm{m}$, and suppose the first time Yi catches up with Jia is after
$$
\frac{2 a+b}{74-65}=\frac{2 a+b}{9}
$$
minutes. The distance Yi has run when he first catches up with Jia is
$$
\frac{2 a+b}{9} \times 74(\mathrm{~m}) \text {. }
$$
At this point, the number of laps Yi has run is $p$, then
$$
\begin{array}{l}
p=\frac{74(2 a+b)}{9} \div(2 a+2 b) \\
=\frac{37(2 a+b)}{9(a+b)}=4+\frac{38 a+b}{9(a+b)} \\
=9-\frac{7 a+44 b}{9(a+b)} .
\end{array}
$$
Thus, when $4\frac{35}{2} b$, we have $\frac{7 a+44 b}{9(a+b)}<1$, so Yi will certainly catch up with Jia by the time he completes the 9th lap at most.
Note: The key to solving this problem is the skillful use of partial fraction decomposition.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: Team A and Team B each send out 7 members to participate in a Go team tournament according to a pre-arranged order. Both sides start with the No. 1 member competing, the loser is eliminated; the winner then competes with the No. 2 member of the losing side, ... until all members of one side are eliminated, and the other side is declared the winner, forming a competition process. How many different possible competition processes are there?
|
Solution: Since the losing side is completely eliminated, the total number of eliminations is 7. Let's assume side A wins, and let the $i$-th member of side A eliminate $x_{i}$ members of side B. Then the problem is equivalent to finding the number of non-negative integer solutions to the equation
$$
x_{1}+x_{2}+\cdots+x_{7}=7
$$
which has $\mathrm{C}_{13}^{6}$ solutions.
Similarly, if side B wins, there are also $\mathrm{C}_{13}^{6}$ solutions.
Therefore, the total number of possible match processes is $2 \mathrm{C}_{13}^{6}=3432$.
Thought: If side A only used its first 5 members to defeat side B's 7 members, how many possible match processes are there?
|
3432
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 From the numbers $1,2, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways of selection are there that meet the conditions?
|
Solution: Notice that
$$
a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\left(14-a_{3}\right)=14 \text {, }
$$
where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-a_{3}$ $\geqslant 0$, i.e.,
$$
\begin{array}{l}
a_{1}-1+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+ \\
\left(14-a_{3}\right)=7 .
\end{array}
$$
$$
\text { Let } x_{1}=a_{1}-1, x_{2}=a_{2}-a_{1}-3, x_{3}=
$$
$a_{3}-a_{2}-3, x_{4}=14-a_{3}$, then
$$
x_{1}+x_{2}+x_{3}+x_{4}=7 \text {, }
$$
where $x_{1} \geqslant 0, x_{2} \geqslant 0, x_{3} \geqslant 0, x_{4} \geqslant 0$.
By the theorem, the equation has $\mathrm{C}_{10}^{3}=120$ solutions. Therefore, there are $\mathrm{C}_{10}^{3}=120$ ways that meet the conditions.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4: Arrange 5 white stones and 10 black stones in a horizontal row, such that the right neighbor of each white stone must be a black stone. How many arrangements are there?
(1996, Japan Mathematical Olympiad Preliminary)
|
Let the number of black stones between the $i$-th white stone and the $(i+1)$-th white stone be $x_{i+1}(i=1,2,3,4)$, with $x_{1}$ black stones at the far left and $x_{6}$ black stones at the far right, then
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=10,
$$
where $x_{1} \geqslant 0, x_{k} \geqslant 1, k=2,3,4,5,6$.
$$
\begin{array}{l}
\text { Let } t_{k}=x_{k}-1, k=2,3,4,5,6, \text { then } \\
x_{1}+t_{2}+t_{3}+t_{4}+t_{5}+t_{6}=10,
\end{array}
$$
where $x_{1} \geqslant 0, t_{k} \geqslant 0, k=2,3,4,5,6$.
By the theorem, the equation has $C_{10}^{5}=252$ solutions, meaning there are 252 arrangements.
|
252
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 In coin tossing, if Z represents heads and F represents tails, then the sequence of coin tosses is represented by a string composed of Z and F. We can count the number of occurrences of heads followed by tails (ZF), heads followed by heads (ZZ)...... For example, the sequence ZZFFZZZZFZZFFFF is the result of 15 coin tosses, which includes 5 ZZ, 3 ZF, 2 FZ, and 4 FF. How many sequences of 15 coin tosses have exactly 2 ZZ, 3 ZF, 4 FZ, and 5 FF?
(4th American Invitational Mathematics Examination (AIME))
|
Solution: The sequences that meet the requirements have the following two possible forms:
(1) Starting with F: $F \cdots F Z \cdots Z F \cdots F Z \cdots Z$;
(2) Starting with Z: $\mathrm{Z} \cdots \mathrm{ZF} \cdots \mathrm{FZ} \cdots \mathrm{ZF} \cdots \mathrm{F}$.
Since the sequence is required to have exactly 3 $\mathrm{ZF}$, if the sequence belongs to (2), it should have the form
$$
\text { Z } \cdots \underline{Z F} \cdots F Z \cdots Z F \cdots F Z \cdots Z F \cdots F \text {. }
$$
which contains only 2 $\mathrm{FZ}$, failing to meet the requirement of 4 $\mathrm{FZ}$. Therefore, the sequence that meets the requirements can only be of form (1).
Since the sequence has exactly 4 $\mathrm{FZ}$, when considering the sequence with exactly two ZZ, it can be divided into the following two categories: $\qquad$ ZZZ $\qquad$
$\qquad$
$\qquad$ Z, $\qquad$ ZZ $\qquad$ ZZ $\qquad$ Z $\qquad$ Z
and the different positions of $\mathrm{Z}$, with the spaces filled with $\mathrm{F}$.
Let the number of $\mathrm{F}$ in each space be $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, then $x_{1}+x_{2}+x_{3}+x_{4}=9$ has $\mathrm{C}_{8}^{3}=56$ positive integer solutions.
On the other hand, for (1), the position of ZZZ has 4 possibilities, and for (2), the arrangement of $\mathrm{ZZ}, \mathrm{ZZ}, \mathrm{Z}, \mathrm{Z}$ has 6 possibilities, so the arrangement of $\mathrm{Z}$ has 10 possibilities.
Therefore, the number of sequences that meet the requirements is
$$
10 \times 56=560 \text { (sequences) }
$$
|
560
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 In a live military exercise, the Red side has set up 20 posts along a straight line. To test 5 new types of weapons, it is planned to equip 5 posts with these new weapons, with the requirement that the first and last posts are not equipped with new weapons, and among every 5 adjacent posts, at least one post is equipped with a new weapon, and no two adjacent posts are equipped with new weapons at the same time. How many ways are there to equip the new weapons?
(2005, National High School League Zhejiang Province Preliminary Competition)
|
Let the 20 positions be ordered as $1,2, \cdots, 20$, and let the sequence number of the $k$-th new weapon be $a_{k}$,
$$
k=1,2,3,4,5 \text {. }
$$
$$
\begin{array}{l}
\text { Let } x_{1}=a_{1}, x_{2}=a_{2}-a_{1}, x_{3}=a_{3}-a_{2}, \\
x_{4}=a_{4}-a_{3}, x_{5}=a_{5}-a_{4}, x_{6}=20-a_{5} .
\end{array}
$$
Then we have
$$
x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=20,
$$
where $2 \leqslant x_{k} \leqslant 5(k=1,2,3,4,5), 1 \leqslant x_{6} \leqslant 4$.
Make the substitution
$$
y_{k}=x_{k}-1(k=1,2,3,4,5), y_{6}=x_{6} \text {. }
$$
Then $y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}=15$,
where $1 \leqslant y_{k} \leqslant 4(k=1,2,3,4,5,6)$.
Let $I$ be the set of all positive integer solutions to equation (1), and $A_{k}$ be the set of solutions in $I$ where $y_{k} > 4$. Then
$$
\begin{array}{l}
\left|\bigcap_{k=1}^{6} \overline{A_{k}}\right|=|I|-\left|\bigcap_{k=1}^{6} A_{k}\right| \\
=|I|-\sum_{k=1}^{6}\left|A_{k}\right|+\sum_{j4 、 y_{j}>4 、 y_{k}>4$ of the positive integer solutions to $\sum_{k=1}^{6} y_{k}=15$, so,
$$
I \bigcap_{k=1}^{6} \widehat{A_{k}} \mid=\mathrm{C}_{14}^{5}-6 \mathrm{C}_{10}^{5}+\mathrm{C}_{6}^{2} \mathrm{C}_{6}^{5}=580 .
$$
Since the 5 new weapons are all different, swapping their positions results in different arrangements, thus the number of ways to equip the new weapons is $580 \times 5!=69600$.
|
69600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 In a family photo album, there are 10 photos. Each photo has 3 men, the person standing on the left is the son of the person in the middle, and the person on the right is the middle person's brother. It is known that the 10 people in the middle of the 10 photos are all different. How many different people are there at least in these 10 photos?
|
Solution: As shown in Figure 1, there are 16 people, where the horizontal lines represent brothers, and the vertical and diagonal lines represent father-son relationships. The 10 photos taken by them are as follows:
$$
\begin{array}{l}
\quad\{3,1,2\},\{16, \\
2,1\},\{5,3,4\},\{15, \\
4,3\},\{7,5,6\},\{9, \\
6,5\},\{11,7,8\}, \\
\{12,8,7\},\{13,9, \\
10\},\{14,10,9\} .
\end{array}
$$
The 10 people in the middle are exactly $1,2, \cdots, 10$, and they are all different. Therefore, the minimum number of people required is no more than 16.
On the other hand, we use the "grouping method" to prove that it is impossible to take 10 photos that meet the requirements with only 15 different people.
If not, suppose there are 15 people who can meet the requirements. We group these 15 people by brothers, placing brothers in the same group and non-brothers in different groups, and those without brothers form a group by themselves.
(1) Since the 10 people in the middle of the 10 photos are all different, i.e., the 10 fathers in the middle are all different, the 10 sons on the left of the photos are in different groups, i.e., they belong to 10 different groups. In addition, the person with the highest seniority among the 15 people in the 10 photos must belong to another group, so the 15 people must be divided into at least 11 groups.
(2) We examine the following two types of groups:
(i) Each person in the group has appeared in the middle position of a photo;
(ii) At least one person in the group has never appeared in the middle position of a photo.
Since the person on the right of each photo is the brother of the person in the middle, the group of the person in the middle must have at least 2 people. Since there are only 10 people in the middle, there can be at most 5 groups of type (i).
Because there are 15 people in the 10 photos, and only 5 people have never appeared in the middle position, there can be at most 5 groups of type (ii).
Thus, the 15 people can be divided into at most 10 groups, which is a contradiction.
In conclusion, there must be at least 16 different people in the 10 photos.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 4 Let the function $f(x)=\frac{1}{k-x}$, and denote
$$
\underbrace{f(f(\cdots)}_{\text {j times }}
$$
as $f_{j}(x)$. Given that $f_{n}(x)=x$, and
$$
f_{i}(x) \neq f_{j}(x)(i \neq j, 1 \leqslant i, j \leqslant n) \text {. }
$$
Prove: $\prod_{i=1}^{n} f_{i}(x)=-1$.
|
Proof: Consider the recursive sequence $g_{j}=k g_{j-1}-g_{j-2}$, $g_{0}=0, g_{1}=1$. We have
$$
\begin{array}{l}
f_{1}(x)=\frac{1}{k-x}=\frac{x g_{0}-g_{1}}{x g_{1}-g_{2}}, \\
f_{2}(x)=\frac{x g_{1}-g_{2}}{x g_{2}-g_{3}}, \\
\cdots \cdots \\
f_{n}(x)=\frac{x g_{n-1}-g_{n}}{x g_{n}-g_{n+1}} .
\end{array}
$$
Multiplying the above equations, we get
$$
\begin{array}{l}
\prod_{i=1}^{n} f_{i}(x)=\frac{x g_{0}-g_{1}}{x g_{n}-g_{n+1}} . \\
\text { Also, } f_{n}(x)=x=\frac{x g_{n-1}-g_{n}}{x g_{n}-g_{n+1}} \\
=\frac{x g_{n-1}-g_{n}}{x g_{n}-k g_{n}+g_{n-1}},
\end{array}
$$
Then, $g_{n} x^{2}-g_{n} k x+g_{n}=0$ always holds, i.e.,
$$
\left\{\begin{array}{l}
g_{n}=0, \\
-g_{n} k=0, \\
g_{n}=0 .
\end{array}\right.
$$
Thus, $g_{n}=0$.
So, $g_{n}=0=k g_{n-1}-g_{n-2}$.
This gives $g_{n-1}=\frac{g_{n-2}}{g_{2}}=k g_{n-2}-g_{n-3}$.
Therefore, $g_{n-2}=\frac{g_{n-3}}{\frac{g_{3}}{g_{2}}}$, i.e., $\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}$.
Thus, $g_{n-1}=\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}=\cdots$.
If $n$ is even,
$$
g_{n-1}=\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}=\cdots=\frac{g_{\frac{n}{2}}}{g_{\frac{n}{2}}}=1 ;
$$
If $n$ is odd,
$$
\begin{array}{l}
g_{n-1}=\frac{g_{n-2}}{g_{2}}=\frac{g_{n-3}}{g_{3}}=\cdots=\frac{g_{\left[\frac{n}{2}\right]+1}}{g_{\left[\frac{n}{2}\right]}} \\
=\frac{g_{\left[\frac{n}{2}\right]}}{g_{\left[\frac{n}{2}\right]+1}}=\frac{g_{\left[\frac{n}{2}\right]}+g_{\left[\frac{n}{2}\right]+1}}{g_{\left[\frac{n}{2}\right]}+g_{\left[\frac{n}{2}\right]+1}}=1 .
\end{array}
$$
Hence, $g_{n+1}=k g_{n}-g_{n-1}=-1$.
Therefore, $\prod_{i=1}^{n} f_{i}(x)=\frac{x g_{0}-g_{1}}{x g_{n}-g_{n+1}}=-1$.
|
-1
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. For a positive integer $n$, if there exist positive integers $a$ and $b$ such that $n=a+b+a b$, then $n$ is called a "good number". For example, $3=1+1+1 \times 1$, so 3 is a good number. Then, among the 20 positive integers from $1 \sim 20$, the number of good numbers is $\qquad$ .
|
4.12.
$n+1=a+b+a b+1=(a+1)(b+1)$ is a composite number, so the required $n$ is the number obtained by subtracting 1 from the composite numbers between 2 and 21. The composite numbers between 2 and 21 are $4,6,8,9,10,12,14,15,16,18,20,21$, a total of 12. Therefore, the required $n$ has 12 values: $3,5,7,8,9,11,13,14,15,17,19,20$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Place five different files $A, B, C, D, E$ into a row of seven drawers numbered $1, 2, 3, 4, 5, 6, 7$, with each drawer containing at most one file. If files $A$ and $B$ must be placed in adjacent drawers, and files $C$ and $D$ must also be placed in adjacent drawers, then the number of different ways to place the files into the drawers that satisfy the conditions is ( ) ways.
(A) 60
(B) 120
(C) 240
(D) 480
|
5.C.
Place files $A$ and $B$ in adjacent drawers, denoted as “$AB$”, place files $C$ and $D$ in adjacent drawers, denoted as “$CD$”, and place file $E$ in a drawer, denoted as “$E$”. Thus, “$AB$”, “$CD$”, “$E$”, and the two empty drawers can be considered as five elements, and the number of all permutations of these five elements is $\mathrm{A}_{5}^{5}$. Since the permutations of files $A$ and $B$ and files $C$ and $D$ are both $\mathrm{A}_{2}^{2}$, and the two empty drawers are identical elements, the total number of different methods that satisfy the conditions is $\frac{A_{5}^{5} \cdot \mathrm{A}_{2}^{2} \cdot \mathrm{A}_{2}^{2}}{2}=240$.
|
240
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let the set $M=\left\{a \left\lvert\, a=\frac{x+y}{t}\right., 2^{x}+2^{y}=\right.$ $2^{t}$, where $x, y, t, a$ are all integers $\}$. Then the sum of all elements in the set $M$ is ( ).
(A) 1
(B) 4
(C) 7
(D) 8
|
6. D.
Assume $x \leqslant y$, then $2^{x}=2^{x}+2^{y} \leqslant 2^{y}+2^{y}=2^{y+1}$.
Thus, $t \leqslant y+1$.
From $2^{x}>0$, we get $2^{t}=2^{x}+2^{y}>2^{y}$. Therefore, $t>y$.
So, $y<t \leqslant y+1$.
Given that $x, y, t$ are all integers, then $t=y+1$. Hence, $2^{y+1}=$ $2^{x}+2^{y}$, which means $2^{x}=2^{y}$. Therefore, $x=y=t-1$.
Thus, $a=\frac{x+y}{t}=2-\frac{2}{t}$.
Here $a, t \in \mathbf{Z}$, so $t= \pm 1, \pm 2$.
Then $a=0,1,3,4$.
Therefore, the sum of all elements in set $M$ is $0+1+3+4=8$.
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given a fixed point $A(4, \sqrt{7})$. If a moving point $P$ is on the parabola $y^{2}=4 x$, and the projection of point $P$ on the $y$-axis is point $M$, then the maximum value of $|P A|-|P M|$ is $\qquad$.
|
$$
\begin{array}{l}
|P M|=|P N|-|M N|=|P F|-1 \downarrow \\
\text { Then }|P A|-|P M|=|P A|-(|P F|-1) \\
=(|P A|-|P F|)+1 \leqslant|A F|+1=4+1=5 .
\end{array}
$$
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. In the sequence $\left\{a_{n}\right\}$, it is known that
$$
a_{1}=2, a_{n}+a_{n+1}=1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
If $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, then the value of $S_{2003}-2 S_{2004}+S_{2000}$ is $\qquad$
|
9.3.
According to the problem, when $n$ is even, we have $a_{1}+a_{2}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n}{2}$ pairs, so $S_{n}=\frac{n}{2}$. Therefore, $S_{2004}=1002$.
When $n$ is odd, we have $a_{2}+a_{3}=1, a_{3}+a_{4}=1, \cdots, a_{n-1}+a_{n}=1$, a total of $\frac{n-1}{2}$ pairs, so
$$
S_{n}=a_{1}+\frac{n-1}{2}=2+\frac{n-1}{2}=\frac{n+3}{2} \text {. }
$$
Therefore, $S_{2003}=1003, S_{2005}=1004$.
Thus, $S_{2003}-2 S_{2004}+S_{2005}$
$$
=1003-2 \times 1002+1004=3 .
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. As shown in Figure 2, the side length of rhombus $A B C D$ is 1, and $\angle A B C=120^{\circ}$. If
$E$ is any point on the extension of $B C$,
$A E$
intersects $C D$ at point $F$, then
the angle between vector $B F$ and $E D$
is
$\qquad$
|
10.120.
As shown in Figure 5, establish a Cartesian coordinate system. Then
$$
\begin{array}{l}
A\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), \\
B(0,0), \\
C(1,0), \\
D\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) .
\end{array}
$$
Let $E(a, 0)(a>1)$. Then
$l_{c p}: y=-\sqrt{3}(x-1)$,
$$
l_{\text {AE }}: y=-\frac{\sqrt{3}}{2 a+1}(x-a) \text {. }
$$
By solving the two equations simultaneously, we get the intersection point of line $C D$ and line $A E$
$$
F\left(\frac{a+1}{2 a}, \frac{\sqrt{3}(a-1)}{2 a}\right) \text {. }
$$
Thus, $B F=\left(\frac{a+1}{2 a}, \frac{\sqrt{3}(a-1)}{2 a}\right)$,
$$
E D=\left(\frac{1-2 a}{2}, \frac{\sqrt{3}}{2}\right) \text {. }
$$
$B F \cdot E D=-\frac{a^{2}-a+1}{2 a},|B F|=\frac{\sqrt{a^{2}-a+1}}{a}$,
$$
|E D|=\sqrt{a^{2}-a+1} \text {. }
$$
Let $\theta$ be the angle between $\boldsymbol{B F}$ and $\boldsymbol{E D}$, then $\cos \theta=\frac{\boldsymbol{B F} \cdot \boldsymbol{E} \boldsymbol{D}}{|\boldsymbol{B F}| \cdot|E D|}=-\frac{1}{2}$, which means $\theta=120^{\circ}$.
|
120
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. As shown in Figure 3(a), the eight vertices of a cube are assigned values $a, b, c, d, e, f, g, h$, respectively. Then, the arithmetic mean of the values of the three adjacent vertices of each vertex, denoted as $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{f}, \bar{g}, \bar{h}$, is placed at the corresponding vertex of another cube, as shown in Figure 3(b).
If $\bar{a}=9, \bar{b}=8, \bar{c}=11, \bar{d}=10, \bar{e}=13, \bar{f}=$ $12, \bar{g}=15, \bar{h}=14$, then the value of $a+g$ is
|
11.20.
According to the problem, we have
$$
\begin{array}{l}
\bar{a}=\frac{b+d+e}{3}, \bar{b}=\frac{a+c+f}{3}, \bar{c}=\frac{b+d+g}{3}, \\
\bar{d}=\frac{a+c+h}{3}, e=\frac{a+f+h}{3}, \bar{f}=\frac{b+e+g}{3}, \\
\bar{g}=\frac{c+f+h}{3}, \bar{h}=\frac{d+e+g}{3} .
\end{array}
$$
Then, $a=(\bar{b}+\bar{d}+\bar{e})-2 \bar{g}, g=(\bar{c}+f+\bar{h})-2 \bar{a}$.
Thus, $a+g=(\bar{b}+\bar{c}+\bar{d}+\ddot{e}+f+\bar{h})-2(\vec{a}+\bar{g})$.
Given $\bar{a}=9, \bar{b}=8, \bar{c}=11, \bar{d}=10, e=13, \bar{f}=12, \bar{g}=$ $15, \bar{h}=14$, therefore, $a+g=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 A coach can travel from Nanjing to Shanghai on the Shanghai-Nanjing Expressway in $4 \mathrm{~h}$. It is known that a coach departs from Nanjing to Shanghai every half hour, and at the same time, a coach also departs from Shanghai to Nanjing. If a passenger takes a bus from Nanjing to Shanghai, how many coaches heading from Shanghai to Nanjing can he encounter during his entire journey? (Including the vehicles arriving at the Nanjing station and the vehicles leaving the Shanghai station.)
|
Explanation: This problem can be easily solved by just drawing a schematic diagram (Figure 6).
The traveler encounters 8 trains (including the one arriving at Nanjing station) in the first 2 hours (i.e., traveling half the distance). 2 hours after departure, they encounter exactly 1 train that departed from Shanghai to Nanjing at the same time. In the next 2 hours, they encounter another 8 trains (including the one leaving Shanghai station), totaling 17 trains, as shown by the number of intersection points in Figure 6.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let two vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ in a plane be perpendicular to each other, and $|\boldsymbol{a}|=2,|\boldsymbol{b}|=1$. Also, $k$ and $t(t \geqslant 0)$ are two real numbers that are not both zero. If the vectors
$$
\boldsymbol{x}=\boldsymbol{a}+(3-t) \boldsymbol{b} \text { and } \boldsymbol{y}=-k \boldsymbol{a}+t^{2} \boldsymbol{b}
$$
are perpendicular to each other, then the maximum value of $k$ is $\qquad$
|
13.1.
Given $\boldsymbol{a} \cdot \boldsymbol{b}=0$.
$$
\begin{array}{l}
\boldsymbol{x} \cdot \boldsymbol{y}=[\boldsymbol{a}+(3-t) \boldsymbol{b}] \cdot\left(-k \boldsymbol{a}+t^{2} \boldsymbol{b}\right) \\
=-k \boldsymbol{a}^{2}+\left[-k(3-t)+t^{2}\right] \boldsymbol{a} \cdot \boldsymbol{b}+t^{2}(3-t) \boldsymbol{b}^{2} \\
=-k|\boldsymbol{a}|^{2}+t^{2}(3-t)|\boldsymbol{b}|^{2} \\
=-4 k+t^{2}(3-t) .
\end{array}
$$
According to the problem, $-4 k+t^{2}(3-t)=0$.
Therefore, $k=-\frac{1}{4} t^{3}+\frac{3}{4} t^{2}$.
Let $k=f(t)$, then
$$
f^{\prime}(t)=-\frac{3}{4} t^{2}+\frac{3}{2} t=-\frac{3}{4} t(t-2) .
$$
Since $t \neq 0$ (otherwise $k=0$), when $t=2$, $f^{\prime}(t)=0$, and $00, t>2$ when $f^{\prime}(t)<0$. Therefore, when $t=2$, $k$ reaches its maximum value of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. (15 points) There are 800 points on a circle, labeled $1, 2, \cdots, 800$ in a clockwise direction. They divide the circle into 800 gaps. Arbitrarily choose one point and color it red, then color the remaining points red according to the following rule: if the $k$-th point has been colored red, then move clockwise through $k$ gaps and color the point reached red. Continue this process. How many red points can be obtained at most on the circle? Prove your conclusion.
|
19. Generally, for a circle with $n$ points, we denote the maximum number of points that can be colored red according to the given rules as $f(n)$.
If the circle has $2 n$ points, and the first point to be colored red is labeled $i$.
(1) If $i=2 k (k \geqslant 1)$ is an even number, then all the points that are colored red have even labels. This process is equivalent to coloring points on a circle with $n$ points, where the first point to be colored red is labeled $k$. Therefore, the number of red points on both circles is the same;
(2) If $i=2 k-1 (k \geqslant 1)$, the second point to be colored red has the label $2(2 k-1)$, which is even. Therefore, the number of red points is one more than the number of red points obtained by coloring points on a circle with $n$ points, where the first point to be colored red is labeled $2 k-1$.
In summary, we have $f(2 n)=f(n)+1$.
From this, we get
$$
\begin{array}{l}
f(800)=f(400)+1=f(200)+2 \\
=f(100)+3=f(50)+4=f(25)+5 .
\end{array}
$$
For a circle with 25 points, starting from point 1, we can sequentially obtain the labels of 20 red points: $1,2,4,8,16,7,14,3,6,12,24,23$, $21,17,9,18,11,22,19,13$. Therefore, we have:
$$
f(25) \geqslant 20.
$$
Conversely, if a red point has a label that is a multiple of 5, then all red points have labels that are multiples of 5. In this case, the number of red points is less than or equal to 5. Therefore, the process that achieves the maximum number of red points does not include points with labels that are multiples of 5. Thus, we have
$$
f(25) \leqslant 25-5=20, \text{ i.e., } f(25)=20.
$$
Therefore, $f(800)=20+5=25$.
|
25
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the set $S=\{1,2,3,4,5,6\}$, a one-to-one mapping $f: S \rightarrow S$ satisfies the condition: for any $x \in S$, $f(f(f(x)))=x$. Then the number of mappings $f$ that satisfy the condition is ( ).
(A) 81
(B) 80
(C) 40
(D) 27
|
6.A.
It is clear that there does not exist $x \in S$ such that $f(x) \neq x$ and $f(f(x)) = x$. Otherwise, $f(f(f(x))) = f(x) \neq x$, which contradicts the given condition. Therefore, for any $x \in S$, either $f(x) = x$, or $f(x) = x_{1}$, $f(x_{1}) = x_{2}$, $f(x_{2}) = x$, and $x, x_{1}, x_{2}$ are distinct. Hence, there are only the following three cases:
(1) For any $x \in S, f(x) = x$. There is only 1 such $f$;
(2) $f$ contains one cycle: $a \rightarrow b \rightarrow c \rightarrow a$, and the other 3 elements satisfy $f(x) = x$. There are $\mathrm{C}_{6}^{3} \cdot \mathrm{C}_{2}^{1} = 40$ such $f$;
(3) $f$ contains two cycles: $a \rightarrow b \rightarrow c \rightarrow a$ and $x \rightarrow y \rightarrow z \rightarrow x$. There are $\frac{\mathrm{C}_{6}^{3} \cdot \mathrm{C}_{3}^{3}}{2!} \times \mathrm{C}_{2}^{1} \cdot \mathrm{C}_{2}^{1} = 40$ such $f$.
In summary, there are a total of 81 $f$ that satisfy the conditions.
|
81
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. If there exists a permutation $a_{1}$, $a_{2}, \cdots, a_{n}$ of $1,2, \cdots, n$, such that $k+a_{k}(k=1,2, \cdots, n)$ are all perfect squares, then $n$ is called a "middle number". Then, in the set $\{15,17,2000\}$, the number of elements that are middle numbers is $\qquad$ .
|
5.3.
(1) 15 is a median number. Because in the arrangement of Table 1, $k+a_{k}$ $(k=1,2, \cdots, 15)$ are all perfect squares:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$k$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline$a_{k}$ & 15 & 14 & 13 & 12 & 11 & 10 & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\hline$k+a_{k}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ & $4^{2}$ \\
\hline
\end{tabular}
(2) 17 is a median number. Because in the arrangement of Table 2, $k+a_{k}$ $(k=1,2, \cdots, 17)$ are all perfect squares:
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$k$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\
\hline$a_{k}$ & 3 & 7 & 6 & 5 & 4 & 10 & 2 & 17 & 16 & 15 & 14 & 13 & 12 & 11 & 1 & 9 & 8 \\
\hline$k+a_{k}$ & $2^{2}$ & $3^{2}$ & $3^{2}$ & $3^{2}$ & $3^{2}$ & $4^{2}$ & $3^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $5^{2}$ & $4^{2}$ & $5^{2}$ & $5^{2}$ \\
\hline
\end{tabular}
(3) 2006 is a median number. Because it can be arranged in such a way that $k+a_{k}(k=1,2, \cdots, 2006)$ are all perfect squares:
For $k$ taking $1,2, \cdots, 17, a_{k}$ can correspond as in (2); for $k$ taking 18, $a_{k}=18$ can correspond; for $k$ taking $19,20, \cdots, 2006$, $a_{k}$ can sequentially take $2006,2005, \cdots, 19$ to correspond.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) A $98 \times$ 98 chessboard is displayed on a computer screen, colored in the usual way (i.e., with two colors alternating). A person can drag the mouse to select a rectangle with borders along the chessboard lines, then click the mouse, and all the colors within this frame will change (i.e., white to black, black to white). How many times at least must the mouse be clicked to turn the entire chessboard into a single color? Prove your conclusion.
|
Three, we prove that for an $n \times n$ chessboard, if $n$ is odd, then at least $n-1$ mouse clicks are required to turn the entire chessboard into one color; if $n$ is even, then at least $n$ mouse clicks are required to turn the entire chessboard into one color.
Consider the $4(n-1)$ small squares along the border lines. Since they alternate in black and white, there are a total of $4(n-1)$ pairs of adjacent squares of different colors. Each click can reduce at most 4 such pairs of different-colored squares, so at least $n-1$ mouse clicks are required to turn the entire chessboard into one color.
(1) $n$ is odd.
(i) If $n=1$, the chessboard is already one color, so no mouse clicks are needed;
(ii) If $n=2k+1$ ($k$ is a positive integer), then you can first click on rows $2, 4, \cdots, 2k$, and then click on columns $2, 4, \cdots, 2k$ (each click can reduce 4 pairs of different-colored squares), a total of $n-1$ mouse clicks, which will turn the entire chessboard into one color.
(2) $n$ is even.
Let $n=2k$ ($k$ is a positive integer). Since the four corners of this chessboard are different colors, there must be a rectangle containing these corners. Clicking such a rectangle once can reduce at most 2 pairs of different-colored squares, so at least $n$ mouse clicks are required to turn the entire chessboard into one color—first click on rows $2, 4, \cdots, 2k$, then click on columns $2, 4, \cdots, 2k$. The $k$-th and $2k$-th mouse clicks each reduce 2 pairs of different-colored squares, and the other clicks each reduce 4 pairs of different-colored squares, a total of $n$ clicks.
Therefore, at least 98 mouse clicks are required to turn a $98 \times 98$ chessboard into one color.
(Liu Kangning, Xi'an Tieyi Middle School, Shaanxi Province, 710054)
|
98
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Let $F$ be the set of real polynomials $f(x)$ satisfying
(1) the degree of $f(x)$ is less than or equal to 3;
(2) for any $x \in [0,1]$, $|f(x)| \leqslant 1$.
Find $\max _{f \in F} f(2)$.
|
Solution: Let $f(x)=a x^{3}+b x^{2}+c x+d$. Denote
$$
\left.\begin{array}{l}
A=f(1)=a+b+c+d, \\
B=f\left(\frac{3}{4}\right)=\frac{27}{64} a+\frac{9}{16} b+\frac{3}{4} c+d, \\
C=f\left(\frac{1}{4}\right)=\frac{1}{64} a+\frac{1}{16} b+\frac{1}{4} c+d, \\
D=f(0)=d .
\end{array}\right\}
$$
By condition (2), we know $A, B, C, D \in[-1,1]$.
Let $f(2)=8 a+4 b+2 c+d=x A+y B+z C+w D$.
Substituting (1) into the above equation and comparing the coefficients on both sides, we get
$$
\left\{\begin{array}{l}
x+\frac{27}{64} y+\frac{1}{64} z=8, \\
x+\frac{9}{16} y+\frac{1}{16} z=4, \\
x+\frac{3}{4} y+\frac{1}{4} z=2, \\
x+y+z+w=1 .
\end{array}\right.
$$
Solving, we get $(x, y, z, w)=\left(\frac{70}{3},-\frac{112}{3}, \frac{80}{3},-\frac{35}{3}\right)$.
Thus, $f(2)=\frac{70}{3} A-\frac{112}{3} B+\frac{80}{3} C-\frac{35}{3} D$
$$
\leqslant \frac{70}{3}+\frac{112}{3}+\frac{80}{3}+\frac{35}{3}=99 \text {. }
$$
On the other hand, let $f(x)=32 x^{3}-48 x^{2}+18 x-1$ (i.e., if (2) holds with equality, then $A=C=1, B=D=-1$, solving for $a, b, c, d$ gives $f(x))$, then $f(2)=99$.
Clearly, $f(x)$ satisfies condition (1). Next, we prove that $f(x)$ satisfies condition (2).
Since $f^{\prime}(x)=96 x^{2}-96 x+18=6(4 x-1)(4 x-3)$, the maximum and minimum values of $f(x)$ must be attained at $0, \frac{1}{4}, \frac{3}{4}, 1$.
And $f(0)=f\left(\frac{3}{4}\right)=-1, f\left(\frac{1}{4}\right)=f(1)=1$,
so, $f(x)$ satisfies condition (2) (or directly verify $f\left(\frac{1+\cos x}{2}\right)=\cos 3 x$ to show that $f(x)$ satisfies condition (2)).
Therefore, $f \in F$.
Hence $\max _{f \in F} f(2)=99$.
(Zhu Qing San, Peking University School of Mathematical Sciences, Class 2, Grade 04, 100871)
|
99
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Two positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (which can be the same) are such that $\frac{a_{1}}{a_{2}}, \frac{a_{2}}{a_{3}}, \cdots, \frac{a_{2005}}{a_{2006}}$ are all distinct. How many different numbers are there at least among $a_{1}$, $a_{2}, \cdots, a_{2006}$?
（Chen Yonggao, problem contributor）
|
Due to the fact that the pairwise ratios of 45 distinct positive integers are at most $45 \times 44 + 1 = 1981$, the number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2000}$ is greater than 45.
Below is an example to show that 46 can be achieved.
Let $p_{1}, p_{2}, \cdots, p_{46}$ be 46 distinct prime numbers, and construct $a_{1}, a_{2}, \cdots, a_{2000}$ as follows:
$$
\begin{array}{l}
p_{1}, p_{1}, \\
p_{2}, p_{1}, \\
p_{3}, p_{2}, p_{3}, p_{1}, \\
p_{4}, p_{3}, p_{4}, p_{2}, p_{4}, p_{1}, \\
\ldots \\
p_{k}, p_{k-1}, p_{k}, p_{k-2}, \cdots, p_{k}, p_{2}, p_{k}, p_{1}, \\
\ldots \\
p_{45}, p_{44}, p_{45}, p_{43}, \cdots, p_{45}, p_{2}, p_{45}, p_{1}, \\
p_{46}, p_{45}, p_{46}, p_{44}, \cdots, p_{46}, p_{34} .
\end{array}
$$
These 2006 positive integers meet the requirements.
Therefore, the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2000}$ is 46.
|
46
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, let $X$ be a 56-element set. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any 7 of them has at least $n$ elements, then there must exist 3 of these 15 subsets whose intersection is non-empty.
(Cold Gangsong provided the problem)
|
Six, the minimum value of $n$ is 41.
First, prove that $n=41$ meets the requirement. Use proof by contradiction.
Assume there exist 15 subsets of $X$ such that the union of any 7 of them contains at least 41 elements, and the intersection of any 3 of them is empty. Since each element belongs to at most 2 subsets, we can assume each element belongs to exactly 2 subsets (otherwise, add some elements to some subsets, and the conditions still hold). By the pigeonhole principle, there must be a subset, say $A$, containing at least $\left[\frac{2 \times 56}{15}\right]+1=8$ elements. Let the other 14 subsets be $A_{1}, A_{2}, \cdots, A_{14}$. Consider any 7 subsets that do not include $A$, each corresponding to 41 elements in $X$. All 7-subset groups that do not include $A$ together contain at least $41 C_{14}^{7}$ elements. Additionally, for an element $a$, if $a \notin A$, then 2 of $A_{1}, A_{2}, \cdots, A_{14}$ contain $a$. Hence, $a$ is counted $\mathrm{C}_{14}^{7}-\mathrm{C}_{12}^{7}$ times. If $a \in A$, then 1 of $A_{1}, A_{2}, \cdots, A_{14}$ contains $a$. Hence, $a$ is counted $\mathrm{C}_{14}^{7}-\mathrm{C}_{13}^{7}$ times. Therefore,
$$
\begin{array}{l}
41 C_{14}^{7} \leqslant(56-|A|)\left(C_{14}^{7}-C_{12}^{7}\right)+|A|\left(C_{14}^{7}-C_{13}^{7}\right) \\
=56\left(C_{14}^{7}-C_{12}^{7}\right)-|A|\left(C_{13}^{7}-C_{12}^{7}\right) \\
\leqslant 56\left(C_{14}^{7}-C_{12}^{7}\right)-8\left(C_{13}^{7}-C_{12}^{7}\right) .
\end{array}
$$
This leads to $196 \leqslant 195$, a contradiction.
Next, prove that $n \geqslant 41$. Use proof by contradiction.
Assume $n \leqslant 40$, let $X=\{1,2, \cdots, 56\}$, and define
$$
\begin{array}{l}
A_{i}=\{k \equiv i(\bmod 7), k \in X\}, i=1,2, \cdots, 7, \\
B_{i}=\{k \equiv j(\bmod 8), k \in X\}, j=1,2, \cdots, 8 .
\end{array}
$$
Clearly, $\left|A_{i}\right|=8(i=1,2, \cdots, 7),\left|A_{i} \cap A_{j}\right|=0$
$$
\begin{array}{l}
(1 \leqslant i<j \leqslant 7),\left|B_{j}\right|=7(j=1,2, \cdots, 8),\left|B_{i} \cap B_{j}\right|=0 \\
(1 \leqslant i<j \leqslant 8),\left|A_{i} \cap B_{j}\right|=1(1 \leqslant i \leqslant 7,1 \leqslant j \leqslant 8) . \text {. }
\end{array}
$$
Thus, for any 3 subsets, there must be 2 that are both $A_{i}$ or both $B_{j}$, and their intersection is empty.
For any 7 subsets $A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{s}}, B_{i_{1}}, B_{j_{2}}, \cdots, B_{j}(s+t=7)$, we have
$$
\begin{aligned}
\mid & \left|A_{i_{1}} \cup A_{i_{2}} \cup \cdots \cup A_{i_{s}} \cup B_{i_{1}} \cup B_{i_{2}} \cup \cdots \cup B_{i_{t}}\right| \\
= & \left|A_{i_{1}}\right|+\left|A_{i_{2}}\right|+\cdots+\left|A_{i_{s}}\right|+ \\
& \left|B_{j_{1}}\right|+\left|B_{j_{2}}\right|+\cdots+\left|B_{i_{1}}\right|-s t \\
= & 8 s+7 t-s t=8 s+7(7-s)-s(7-s) \\
= & (s-3)^{2}+40 \geqslant 40 .
\end{aligned}
$$
Therefore, $n \geqslant 41$.
In conclusion, the minimum value of $n$ is 41.
|
41
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let set $A=\{-2,0,1\}, B=\{1,2,3,4, 5\}$, and the mapping $f: A \rightarrow B$ such that for any $x \in A$, $x + f(x) + x f(x)$ is an odd number. Then the number of such mappings $f$ is ( ).
(A) 45
(B) 27
(C) 15
(D) 11
|
6.A.
When $x=-2$, $x+f(x)+x f(x)=-2-f(-2)$ is odd, then $f(-2)$ can take $1,3,5$, which gives 3 possible values;
When $x=0$, $x+f(x)+x f(x)=f(0)$ is odd, then $f(0)$ can take $1,3,5$, which gives 3 possible values;
When $x=1$, $x+f(x)+x f(x)=1+2 f(1)$ is odd, then $f(1)$ can take $1,2,3,4,5$, which gives 5 possible values.
By the multiplication principle, there are $3 \times 3 \times 5=45$ mappings.
|
45
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given the set $M=\left\{x \left\lvert\, x=\lim _{n \rightarrow \infty} \frac{2^{n+1}-2}{\lambda^{n}+2^{n}}\right.\right.$, $\lambda$ is a constant, and $\lambda+2 \neq 0\}$. Then the sum of all elements of $M$ is $\qquad$ .
|
10.3.
When $|\lambda|>2$, $x=\lim _{n \rightarrow \infty} \frac{2\left(\frac{2}{\lambda}\right)^{n}-2\left(\frac{1}{\lambda}\right)^{n}}{1+\left(\frac{2}{\lambda}\right)^{n}}=0$;
When $\lambda=2$, $x=\lim _{n \rightarrow \infty}\left(1-\frac{1}{2^{n}}\right)=1$;
When $|\lambda|<2$, $x=\lim _{n \rightarrow \infty} \frac{2-2\left(\frac{1}{2}\right)^{n}}{\left(\frac{\lambda}{2}\right)^{n}+1}=2$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=2, a_{n}=\frac{1+a_{n-1}}{1-a_{n-1}}(n \geqslant 2) \text {. }
$$
Then the value of $a_{2006}$ is $\qquad$ .
|
3.2.
Since $a_{1}=2, a_{2}=-3, a_{3}=-\frac{1}{2}, a_{4}=\frac{1}{3}, a_{5}=$ 2, we conjecture that $a_{n}$ is a periodic sequence with a period of 4. And
$$
a_{n+4}=\frac{1+a_{n+3}}{1-a_{n+3}}=\frac{1+\frac{1+a_{n+2}}{1-a_{n+2}}}{1-\frac{1+a_{n+2}}{1-a_{n+2}}}=-\frac{1}{a_{n+2}}=a_{n} .
$$
Therefore, $a_{2008}=a_{4 \times 501+1}=a_{1}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In the non-decreasing sequence of positive odd numbers $\{1,3,3,3,5,5,5, 5,5, \cdots\}$, each positive odd number $k$ appears $k$ times. It is known that there exist integers $b$, $c$, and $d$, such that for all integers $n$, $a_{n}=$ $b[\sqrt{n+c}]+d$, where $[x]$ denotes the greatest integer not exceeding $x$. Then $b+c+d$ equals
|
5.2.
Divide the known sequence into groups as follows:
$$
\begin{array}{l}
(1),(3,3,3),(5,5,5,5,5), \cdots, \\
(\underbrace{2 k-1,2 k-1, \cdots, 2 k-1}_{2 k-1 \uparrow}),
\end{array}
$$
Let $a_{n}$ be in the $k$-th group, where $a_{n}=2 k-1$. Then we have
$$
\begin{array}{l}
1+3+5+\cdots+2 k-3+1 \\
\leqslant n0$, solving this gives $\sqrt{n-1}<k \leqslant \sqrt{n-1}+1$.
Therefore, $k=[\sqrt{n-1}+1]=[\sqrt{n-1}]+1$.
Thus, $a_{n}=2[\sqrt{n-1}]+1$.
Hence, $b+c+d=2+(-1)+1=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6.4. A five-digit number is a multiple of 54, and none of its digits are 0. After deleting one of its digits, the resulting four-digit number is still a multiple of 54; after deleting one of the digits of this four-digit number, the resulting three-digit number is still a multiple of 54; after deleting one of the digits of this three-digit number, the resulting two-digit number is still a multiple of 54. Find the original five-digit number.
|
6.4. The original five-digit number is 59994.
Since the number faced after each deletion of the front or back is a multiple of 9, the number deleted each time is a multiple of 9. This means that only a 9 can be deleted each time, and the last remaining two-digit number can only be 54. Among the three-digit numbers 549, 594, 954, only 594 is a multiple of 54. Therefore, the three-digit number obtained after two deletions is 594. Among the four-digit numbers 9594, 5994, 5949, only 5994 is a multiple of 54, so the four-digit number obtained after the first deletion is 5994. Finally, among the five-digit numbers 95994, 59994, 59949, only 59994 is a multiple of 54.
|
59994
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.2. The sum of three positive integers (not necessarily distinct) is 100. By subtracting them pairwise (the larger minus the smaller), three difference numbers can be obtained. What is the maximum possible value of the sum of these three difference numbers?
|
7.2. The maximum possible value of the sum of these three differences is 194.
Let the three positive integers be $a$, $b$, and $c$, and assume without loss of generality that $a \geqslant b \geqslant c$. The sum of their pairwise differences is
$$
(a-b)+(a-c)+(b-c)=2(a-c).
$$
Since $b \geqslant 1$ and $c \geqslant 1$, and $a+b+c=100$, it follows that $a \leqslant 98$. Therefore,
$$
2(a-c) \leqslant 2(98-1)=194.
$$
The equality holds when $a=98$, $b=c=1$.
|
194
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7.4. Toma thought of a positive integer and found the remainders when it is divided by 3, 6, and 9. It is known that the sum of these three remainders is 15. Try to find the remainder when the number is divided by 18.
|
7.4. The remainder when this number is divided by 18 is 17.
Since the remainders when divided by $3$, $6$, and $9$ do not exceed $2$, $5$, and $8$ respectively, the sum of these three remainders is always no more than $2+5+8=15$. Since their sum is equal to 15, these three remainders must be $2$, $5$, and $8$. Furthermore, since the remainder when the number is divided by 9 is 8, the remainder when it is divided by 18 can only be 8 or 17. If the remainder were 8, then the number would be even. Thus, the remainder when it is divided by 6 could not be 5, leading to a contradiction.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6.4. Find all such four-digit numbers: they are all 83 times the sum of their digits.
|
6.4. There is only one such four-digit number: $1494=18 \times 83$.
Solution 1: Let the number we are looking for be $x$, and the sum of its digits be $a$. From the problem, we know that $x=83a$. Since the remainder when $x$ is divided by 9 is the same as the remainder when $a$ is divided by 9, $x-a=82a$ must be divisible by 9. This indicates that $a$ is a multiple of 9. Since the sum of the digits of any four-digit number does not exceed 36, $a$ can only be $9, 18, 27$, or 36.
(1) $9 \times 83 < 1000$, which is not a four-digit number;
(2) $18 \times 83 = 1494$, which meets the requirements of the problem;
(3) $27 \times 83 = 2241$, but the sum of its digits is not 27;
(4) $36 \times 83 \neq 9999$, and the sum of its digits cannot be 36.
Solution 2: The four-digit number we are looking for is a multiple of 83 and does not exceed $36 \times 83 = 2988$. There are 24 such numbers. After checking each one, we find that only 1494 meets the conditions of the problem.
|
1494
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The inequality about $x$ $\left\{\begin{array}{l}x-1>0, \\ x^{3}-x^{2}+x \leqslant k\end{array}\right.$ has only the positive integer solutions 2 and 3. Then the value of $| k-21|+| k-52 \mid+$ 1975 is ( ).
(A) 2004
(B) 2005
(C) 2006
(D) 2007
|
2.C.
From the problem, we know that $x>1$.
Since $x=2, 3$ are solutions to $x^{3}-x^{2}+x \leqslant k$, we have,
$$
\left\{\begin{array}{l}
2^{3}-2^{2}+2 \leqslant k, \\
3^{3}-3^{2}+3 \leqslant k
\end{array}\right.
$$
Solving this, we get $k \geqslant 21$.
From the given information, $x=4,5,6, \cdots$ are solutions to $x^{3}-x^{2}+x>k$, so,
$$
\left\{\begin{array}{l}
4^{3}-4^{2}+4>k \\
5^{3}-5^{2}+5>k, \\
6^{3}-6^{2}+6>k, \\
\ldots \ldots
\end{array}\right.
$$
Solving this, we get $k<52$.
Therefore, $21 \leqslant k<52$.
Thus, the original expression is $(k-21)+(52-k)+1975=2006$.
|
2006
|
Inequalities
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $x$, $y$, $z$ are positive real numbers, and $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is $\qquad$
|
\begin{array}{l}=1.2. \\ (x+y)(y+z)=y^{2}+(x+z) y+x z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 .\end{array}
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) In a certain competition, each player plays exactly one game against every other player. The winner of each game gets 1 point, the loser gets 0 points, and in the case of a draw, both get 0.5 points. After the competition, it is found that each player's score is exactly half from games played against the 10 lowest-scoring players (the 10 lowest-scoring players each have exactly half of their points from games played against each other). Find the number of participants in the competition.
|
Given $n$ players, the $n$ players scored a total of $\frac{n(n-1)}{2}$ points. The 10 "math players" scored a total of $\frac{10 \times 9}{2}=45$ points through their matches with each other, which is half of their total score, so they scored a total of 90 points. The remaining $n-10$ players scored a total of $\frac{(n-10)(n-11)}{2}$ points through their matches with each other, which is also half of their total score, so they scored a total of $(n-10)(n-11)$ points. Therefore,
$$
\frac{n(n-1)}{2}=90+(n-10)(n-11) \text {, }
$$
which simplifies to $(n-16)(n-25)=0$.
Solving this, we get $n_{1}=25, n_{2}=16$.
If there are 16 players, then there are only 6 "winners". Their total score is 30 points, averaging 5 points per person, which is clearly less than the average score of the losers, $\frac{90}{10}=9$ points. Therefore, $n=25$.
Thus, there are 25 players in the competition.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given the quadratic trinomial $a x^{2}+b x+c$ $(a>0)$.
(1) When $c<0$, find the maximum value of the function
$$
y=-2\left|a x^{2}+b x+c\right|-1
$$
(2) For any real number $k$, the line $y=k(x-$
1) $-\frac{k^{2}}{4}$ intersects the parabola $y=a x^{2}+b x+c$ at exactly one point, find the value of $a+b+c$.
|
(1) Given $a>0, c<0$, we know that $y^{\prime}=a x^{2}+b x+c$ intersects the $x$-axis, and $y_{\text {nuin }}^{\prime}<0$. Therefore,
$$
\left|y^{\prime}\right|=\left|a x^{2}+b x+c\right| \geqslant 0 \text {. }
$$
Thus, the minimum value of $\left|y^{\prime}\right|$ is 0.
At this point, $y_{\text {man }}=-2 \times 0-1=-1$.
(2) From $\left\{\begin{array}{l}y=a x^{2}+b x+c, \\ y=k(x-1)-\frac{k^{2}}{4} \text {, we get }\end{array}\right.$ $a x^{2}+(b-k) x+c+\frac{k^{2}}{4}+k=0$.
Let $\Delta=(b-k)^{2}-4 a\left(c+\frac{k^{2}}{4}+k\right)=0$.
Rearranging gives
$$
(1-a) k^{2}-(2 b+4 a) k+b^{2}-4 a c=0 \text {. }
$$
According to the problem,
$$
1-a=0,2 b+4 a=0, b^{2}-4 a c=0 \text {. }
$$
Solving these equations, we get $a=1, b=-2, c=1$.
Thus, $a+b+c=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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