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7. (40 points) A cat caught 81 mice and arranged them in a circle, numbering them from $1 \sim 81$ in a clockwise direction. The cat starts counting from a certain mouse in a clockwise direction, continuously counting “$1, 2, 3$” and eating all the mice that are counted as 3. As the cat continues to count, the circle gets smaller and smaller until only two mice are left. It is known that one of the remaining mice with a higher number is 40. Question: From which numbered mouse did the cat start counting?
|
7. First, arrange the numbers of all the mice in 9 columns as follows.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\
19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 \\
28 & 29 & 30 & 31 & 32 & 33 & 34 & 35 & 36 \\
37 & 38 & 39 & 40 & 41 & 42 & 43 & 44 & 45 \\
46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 & 54 \\
55 & 56 & 57 & 58 & 59 & 60 & 61 & 62 & 63 \\
64 & 65 & 66 & 67 & 68 & 69 & 70 & 71 & 72 \\
73 & 74 & 75 & 76 & 77 & 78 & 79 & 80 & 81
\end{tabular}
Assuming the cat starts counting from mouse number 1. In the first round, the mice in columns 3, 6, and 9 are eaten. In the second round, the mice in columns 4 and 8 are eaten. The remaining mice's numbers are rearranged into 9 columns as follows:
\begin{tabular}{ccccccccc}
1 & 2 & 5 & 7 & 10 & 11 & 14 & 16 & 19 \\
20 & 23 & 25 & 28 & 29 & 32 & 34 & 37 & 38 \\
41 & 43 & 46 & 47 & 50 & 52 & 55 & 56 & 59 \\
61 & 64 & 65 & 68 & 70 & 73 & 74 & 77 & 79
\end{tabular}
After the third and fourth rounds, the 3rd, 4th, 6th, 8th, and 9th columns in the rearranged 9 columns are eaten. At this point, the 16 surviving mice's numbers are:
\begin{tabular}{cccccccc}
1 & 2 & 10 & 14 & 20 & 23 & 29 & 34 \\
41 & 43 & 50 & 55 & 61 & 64 & 70 & 74
\end{tabular}
Continuing until only two mice remain, their numbers are 1 and 34. Given that the larger number is 40, the cat must have started counting from \(1 + 40 - 34 = 7\).
|
7
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (40 points) Given positive integers $a$, $b$, and $c$ to 甲, 乙, and 丙 respectively, each person only knows their own number. They are told that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, and each is asked the following two questions:
(1) Do you know the value of $a+b+c$?
(2) Do you know the values of $a$, $b$, and $c$ respectively?
甲 answers "no" to both questions, and 乙, after hearing 甲's answers, answers "yes" to the first question and "no" to the second question. After hearing both of their answers, how should 丙 answer these two questions?
|
10. First solve the equation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.
Assume $a \geqslant b \geqslant c$. If $c=3$, then $1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{3}{c}$.
Thus, $a=b=c=3$. Otherwise, $c=2$.
Then, $\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$.
Solving this, we get $a=b=4$ or $b=3, a=6$.
The sums corresponding to these three solutions are
$$
3+3+3=9,4+4+2=10,6+3+2=11 \text {. }
$$
Since Jia's answer to question (1) is negative, therefore, $a \neq 6$ or 4.
Similarly, $b=6$ or 4.
If $b=4$, since $a \neq 4$, then $a=2, c=4$.
Since Yi's answer to question (2) is negative, hence $b=6$.
Therefore, Bing knows $b=6$ and $a+b+c=11$.
Since Bing also knows the value of $c$, he must also know the value of $a$.
Thus, Bing answers "know" to both questions.
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a, b$ be positive real numbers, and
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}, \\
(a-b)^{2}=4(a b)^{3} .
\end{array}
$$
Then $\log _{a} b=$ $\qquad$
|
3. -1 .
From $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}$, we get $a+b \leqslant 2 \sqrt{2} a b$.
$$
\begin{array}{l}
\text { Also, }(a+b)^{2}=4 a b+(a-b)^{2} \\
=4 a b+4(a b)^{3} \\
\geqslant 4 \times 2 \sqrt{a b(a b)^{3}}=8(a b)^{2},
\end{array}
$$
which means $a+b \geqslant 2 \sqrt{2} a b$.
Thus, $a+b=2 \sqrt{2} a b$.
From the condition for equality in equation (1), we get $a b=1$.
Solving this together with equation (2), we have
Therefore, $\log _{a} b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Now arrange for 7 students to participate in 5 sports events, requiring that students A and B cannot participate in the same event, each event must have participants, and each person can only participate in one event. The number of different arrangements that meet the above requirements is $\qquad$ (answer in numbers).
安排 7 名同学去参加 5 个运动项目 -> Arrange for 7 students to participate in 5 sports events
要求甲、乙两同学不能参加同一个项目 -> Requiring that students A and B cannot participate in the same event
每个项目都有人参加 -> Each event must have participants
每人只参加一个项目 -> Each person can only participate in one event
则满足上述要求的不同安排方案数为 $\qquad$ (用数字作答) -> The number of different arrangements that meet the above requirements is $\qquad$ (answer in numbers)
|
5. 15000 .
According to the problem, there are two scenarios that meet the conditions:
(1) One project has 3 participants, with a total of
$$
C_{7}^{3} \times 5! - C_{5}^{1} \times 5! = 3600
$$
schemes;
(2) Two projects each have 2 participants, with a total of
$$
\frac{1}{2}\left(\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2}\right) \times 5! - \mathrm{C}_{5}^{2} \times 5! = 11400
$$
schemes.
Therefore, the total number of schemes is
$$
3600 + 11400 = 15000 .
$$
|
15000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given
$$
a_{n}=\mathrm{C}_{200}^{n}(\sqrt[3]{6})^{200-n}\left(\frac{1}{\sqrt{2}}\right)^{n}(n=1,2, \cdots, 95) \text {. }
$$
The number of integer terms in the sequence $\left\{a_{n}\right\}$ is $\qquad$
|
8. 15 .
Notice that $a_{n}=\mathrm{C}_{200}^{n} \times 3^{\frac{200-n}{3}} \times 2^{\frac{200-5 n}{6}}$.
To make $a_{n}(1 \leqslant n \leqslant 95)$ an integer, it must be that $\frac{200-n}{3}$ and $\frac{400-5 n}{6}$ are both integers, i.e., $61(n+4)$.
When $n=6 k+2(k=0,1, \cdots, 13)$, $\frac{200-n}{3}, \frac{400-5 n}{6}$ are both non-negative integers. Therefore, $a_{n}$ is an integer, with a total of 14.
When $n=86$, $a_{86}=\mathrm{C}_{200}^{86} \times 3^{38} \times 2^{-5}$, in $\mathrm{C}_{200}^{86}=\frac{200!}{86!\times 114!}$, the number of factor 2 in $200!$ is $\left[\frac{200}{2}\right]+\left[\frac{200}{2^{2}}\right]+\cdots+\left[\frac{200}{2^{7}}\right]=197$.
Similarly, the number of factor 2 in 86! is 82, and the number of factor 2 in 114! is 110.
Thus, the number of factor 2 in $\mathrm{C}_{200}^{86}$ is
$197-82-110=5$.
Therefore, $a_{86}$ is an integer.
When $n=92$, $a_{92}=\mathrm{C}_{200}^{92} \times 3^{36} \times 2^{-10}$.
Similarly, the number of factor 2 in $\mathrm{C}_{200}^{92}$ is less than 10.
Therefore, $a_{92}$ is not an integer.
Thus, the number of integer terms is $14+1=15$.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Let $A$ be a $3 \times 9$ grid, with each small cell filled with a positive integer. If the sum of all numbers in an $m \times n (1 \leqslant m \leqslant 3, 1 \leqslant n \leqslant 9)$ subgrid of $A$ is a multiple of 10, then it is called a "good rectangle"; if a $1 \times 1$ cell in $A$ is not contained in any good rectangle, then it is called a "bad cell". Find the maximum number of bad cells in $A$.
|
First, we prove by contradiction that there are no more than 25 bad cells in $A$.
Assume the conclusion is not true. Then, in the grid $A$, there is at most 1 cell that is not a bad cell. By the symmetry of the grid, we can assume that all cells in the first row are bad cells.
Let the numbers filled in the $i$-th column from top to bottom be $a_{i}, b_{i}, c_{i} (i=1,2, \cdots, 9)$. Define
$$
\begin{array}{l}
S_{k}=\sum_{i=1}^{k} a_{i}, \\
T_{k}=\sum_{i=1}^{k}\left(b_{i}+c_{i}\right)(k=0,1, \cdots, 9),
\end{array}
$$
where $S_{0}=T_{0}=0$.
We will prove that the three sets of numbers $S_{0}, S_{1}, \cdots, S_{9}; T_{0}, T_{1}, \cdots, T_{9}$, and $S_{0}+T_{0}, S_{1}+T_{1}, \cdots, S_{9}+T_{9}$ are all complete residue systems modulo 10.
In fact, suppose there exist $m, n (0 \leqslant m < n \leqslant 9)$ such that $S_{m} \equiv S_{n} (\bmod 10)$. Then
$$
\sum_{i=m+1}^{n} a_{i} = S_{n} - S_{m} \equiv 0 (\bmod 10),
$$
which means the cells from the $(m+1)$-th column to the $n$-th column in the first row form a good rectangle, contradicting the assumption that all cells in the first row are bad cells.
Similarly, suppose there exist $m, n (0 \leqslant m < n \leqslant 9)$ such that
$$
\begin{array}{l}
T_{m} \equiv T_{n} (\bmod 10). \\
\text{Then } \sum_{i=m+1}^{n}\left(b_{i}+c_{i}\right) = T_{n} - T_{m} \equiv 0 (\bmod 10),
\end{array}
$$
which means the cells from the $(m+1)$-th column to the $n$-th column in the second and third rows form a good rectangle.
Thus, there are at least 2 cells that are not bad cells, which is a contradiction.
Similarly, there do not exist $m, n (0 \leqslant m < n \leqslant 9)$ such that
$$
\begin{array}{l}
S_{m} + T_{m} \equiv S_{n} + T_{n} (\bmod 10). \\
\text{Hence } \sum_{k=0}^{9} S_{k} \equiv \sum_{k=0}^{9} T_{k} \equiv \sum_{k=0}^{2}\left(S_{k} + T_{k}\right) \\
\equiv 0 + 1 + \cdots + 9 \equiv 5 (\bmod 10). \\
\text{Then } \sum_{k=0}^{2}\left(S_{k} + T_{k}\right) \equiv \sum_{k=0}^{2} S_{k} + \sum_{k=0}^{2} T_{k} \\
\equiv 5 + 5 \equiv 0 (\bmod 10),
\end{array}
$$
which is a contradiction.
Therefore, the assumption is false, meaning that the number of bad cells cannot exceed 25.
Next, we construct a $3 \times 9$ grid (Table 1), and it can be verified that each cell not filled with 10 is a bad cell. In this case, there are 25 bad cells.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline 1 & 1 & 1 & 2 & 1 & 1 & 1 & 1 & 10 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 1 & 1 & 1 & 10 & 1 & 1 & 1 & 1 & 2 \\
\hline
\end{tabular}
In conclusion, the maximum number of bad cells is 25.
(Ding Longyun provided)
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given a sequence of numbers
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \cdots, \frac{k}{1}, \frac{k-1}{2}, \cdots, \frac{1}{k} \text {. }
$$
In this sequence, the index of the 40th term that equals 1 is ( ).
(A) 3120
(B) 3121
(C) 3200
(D) 3201
|
- 1. B.
For the terms where the sum of the numerator and denominator is $k+1$, we denote them as the $k$-th group. According to the arrangement rule, the 40th term with a value of 1 should be the 40th number in the $2 \times 40-1=79$ group, with the sequence number being
$$
\begin{array}{l}
(1+2+\cdots+78)+40 \\
=\frac{(1+78) \times 78}{2}+40=3121 .
\end{array}
$$
|
3121
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{199}-\frac{1}{200}}{\frac{1}{201^{2}-1^{2}}+\frac{1}{202^{2}-2^{2}}+\cdots+\frac{1}{300^{2}-100^{2}}} \\
= \\
\end{array}
$$
|
3. 400 .
Original expression
$$
\begin{array}{l}
=\frac{\left(1+\frac{1}{2}+\cdots+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{200}\right)}{\frac{1}{202 \times 200}+\frac{1}{204 \times 200}+\cdots+\frac{1}{400 \times 200}} \\
=\frac{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}{\frac{1}{400}\left(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}\right)}=400 .
\end{array}
$$
|
400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A real-coefficient polynomial $P(x)$ of degree not exceeding 2011 takes integer values for any integer $x$, and the remainders when $P(x)$ is divided by $x-1, x-2, \cdots, x-2011$ are $1, 2, \cdots, 2011$ respectively. Then $\max _{x \in \{-1, -2, \cdots, -2011\}}|P(x)|$ has the minimum value of $\qquad$
|
$-1.2011$
First, $P(x)-x$. can be divided by $x-1, x-2, \cdots$, $x-2011$ respectively, so we can assume
$$
P(x)-x=q(x-1)(x-2) \cdots(x-2011) \text {, }
$$
where, $q \in \mathbf{Q}$.
If $q \neq 0$, then
$$
\begin{array}{l}
|P(-2011)|=\left|-2011-q \cdot \frac{4022!}{2011!}\right| \\
\geqslant\left|q \cdot \frac{4022!}{2011!}\right|-2011 \\
>4024|q \cdot 2012!|-2011 \\
\geqslant 4024-2011=2013 .
\end{array}
$$
If $q=0$, then
$$
\begin{array}{l}
P(x)=x, \\
\max _{x \in 1-1,-2, \cdots,-20111}|P(x)|=2011 .
\end{array}
$$
In summary, $\max _{x \in 1-1,-2, \cdots,-20111}|P(x)|$'s minimum value is 2011 .
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $x$ is a positive integer, and $2011-x$ is a perfect cube. Then the minimum value of $x$ is $\qquad$ .
|
1. 283.
From $12^{3}=1728<2011<13^{3}=2197$, we know the minimum value of $x$ is
$$
2011-1728=283 \text {. }
$$
|
283
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Divide a cube with an edge length of a positive integer into 99 smaller cubes, among which, 98 smaller cubes are unit cubes. Find the surface area of the original cube.
|
Let the side length of the original cube be $x$, and the side lengths of the 99 smaller cubes be 1 and $y(x>y>1)$. Then
$$
\begin{array}{l}
x^{3}-y^{3}=98 \\
\Rightarrow(x-y)\left[(x-y)^{2}+3 x y\right]=98 \\
\Rightarrow(x-y) \mid 98=7^{2} \times 2 \\
\Rightarrow x-y=1,2,7,14,49,98 . \\
\text { By }(x-y)^{3}<x^{3}-y^{3}=98 \\
\Rightarrow x-y=1,2 .
\end{array}
$$
When $x-y=1$, we have $3 x y=97$, which is a contradiction;
When $x-y=2$, we have $x y=15$, solving gives $(x, y)=(5,3)$.
Therefore, the surface area of the original cube is $6 \times 5^{2}=150$.
|
150
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, chord $CD$ intersects $AB$ at point $E$. A tangent to $\odot O$ is drawn from point $A$ and intersects the extension of $CD$ at point $F$. Given $AC=8$, $CE:ED=6:5$, and $AE:EB=2:3$. Find the length of $AB$ and the value of $\tan \angle ECB$.
---
Translate the text into English, preserving the original text's line breaks and format. Directly output the translation result.
|
Let $C E=6 x, E D=5 x, A E=2 y$, $E B=3 y, D F=z$.
By the intersecting chords theorem, we have
$$
A E \cdot B E=C E \cdot E D \Rightarrow y=\sqrt{5} x \text {. }
$$
By the secant-tangent theorem, we have
$$
A F^{2}=D F \cdot C F=z(z+11 x) \text {. }
$$
By the Pythagorean theorem, we have
$$
\begin{array}{l}
A F^{2}=E F^{2}-A E^{2} \\
\Rightarrow z(z+11 x)=(5 x+z)^{2}-(2 y)^{2} .
\end{array}
$$
Substituting $y=\sqrt{5} x$ yields $z=5 x$, meaning $D$ is the midpoint of $E F$.
Therefore, $A D=D E=5 x$.
It is easy to see that $B C=B E=3 \sqrt{5} x, A B=5 \sqrt{5} x$.
$$
\begin{array}{l}
\text { Also, } A B^{2}-B C^{2}=A C^{2} \\
\Rightarrow(5 \sqrt{5} x)^{2}-(3 \sqrt{5} x)^{2}=8^{2} \\
\Rightarrow x=\frac{2 \sqrt{5}}{5} \Rightarrow A B=10 .
\end{array}
$$
Since $\angle E C B=\angle E A D=\angle D E A$, we have
$$
\tan \angle E C B=\tan \angle D E A=\frac{A F}{A E}=2 .
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let the function
$$
f(x)=\frac{1+\ln (x+1)}{x} \text {, }
$$
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
untranslated part:
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
(Note: The last part was already in English, so it remains unchanged.)
|
11. Since $x>0$, we have
$$
k0)
$$
always holds.
$$
\text{Let } g(x)=\frac{(x+1)[1+\ln (x+1)]}{x}(x>0) \text{.}
$$
Then $g^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
Let $h(x)=x-1-\ln (x+1)$.
Obviously, $h(2)0$.
If $h\left(x_{0}\right)=0$, then $20$ (since $x>0$), then $h(x)$ is monotonically increasing in $(0,+\infty)$.
When $x \in\left(0, x_{0}\right)$, $h(x)0$, i.e., $g^{\prime}(x) >0$, then $g(x)$ is increasing in the interval $\left(x_{0},+\infty\right)$.
Thus, when $x>0$,
$$
\begin{array}{l}
g(x)_{\text{min}}=g\left(x_{0}\right) \\
=\frac{\left(x_{0}+1\right)\left[1+\ln \left(x_{0}+1\right)\right]}{x_{0}}.
\end{array}
$$
From equation (1) and the above equation, we get $g(x)_{\min }=x_{0}+1$.
Given that $k<g(x)_{\text{min}}$, i.e., $k<x_{0}+1$.
Since $3<x_{0}+1<4$ and $k$ is a positive integer, the maximum value of $k$ is 3.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the number of ordered integer pairs $(a, b)$ such that
$$
x^{2}+a x+b=167 y
$$
has integer solutions $(x, y)$, where $1 \leqslant a, b \leqslant 2004 .^{[4]}$ (2004, Singapore Mathematical Olympiad)
|
If $x^{2}+a x+b \equiv 0(\bmod 167)$, then completing the square gives
$$
a^{2}-4 b \equiv(2 x+a)^{2}(\bmod 167) .
$$
Therefore, for a fixed $a$, $a^{2}-4 b$ is a quadratic residue modulo 167.
Hence, by the lemma, $b$ can take $\frac{167-1}{2}+1=84$ values modulo 167.
And $\frac{2004}{167}=12$, so each $a$ corresponds to $84 \times 12$ values of $b$, totaling
$$
2004 \times 84 \times 12=2020032
$$
ordered integer solutions.
The lemma can lead to the important Euler's criterion.
Theorem 3 (Euler's criterion) Let $p$ be an odd prime. Then
$$
a^{\frac{p-1}{2}} \equiv\left(\frac{a}{p}\right)(\bmod p) \text {. }
$$
Proof: Take a primitive root of $p$ to be $g$, so the residue system $\{1,2, \cdots, p-1\}$ can be represented as
$$
\left\{g, \cdots, g^{p-2}, g^{p-1}=1\right\} \text {. }
$$
By the lemma, there are exactly $\frac{p-1}{2}$ quadratic residues.
Notice that, $g^{2 k}\left(1 \leqslant k \leqslant \frac{p-1}{2}\right)$ has an even exponent and must be a quadratic residue. Therefore, exactly the even powers of $g$ are quadratic residues, and the odd powers are quadratic non-residues.
Thus, when $\left(\frac{a}{p}\right)=1$,
$a^{\frac{p-1}{2}} \equiv\left(g^{2 k}\right)^{\frac{p-1}{2}} \equiv 1(\bmod p) ;$
when $\left(\frac{a}{p}\right)=-1$,
$$
a^{\frac{p-1}{2}} \equiv\left(g^{2 k+1}\right)^{\frac{p-1}{2}} \equiv g^{\frac{p-1}{2}} \equiv-1(\bmod p) \text {. }
$$
The last equation is due to
$$
g^{p-1} \equiv 1(\bmod p) \Rightarrow g^{\frac{p-1}{2}} \equiv \pm 1(\bmod p),
$$
and since $g$ is a primitive root with order $p-1$, it must be
$$
g^{\frac{p-1}{2}} \equiv -1(\bmod p) \text {. }
$$
Since each quadratic residue modulo $p$ can be characterized by a power of a primitive root of $p$, it is easy to see that the Legendre symbol is multiplicative.
|
2020032
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a=\sqrt{3}-1$. Then $a^{2012}+2 a^{2011}-2 a^{2010}=$
|
2. 0 .
Notice that $a^{2}=(\sqrt{3}-1)^{2}=4-2 \sqrt{3}$. Then $a^{2}+2 a-2=0$.
Therefore, $a^{2012}+2 a^{2011}-2 a^{2010}$ $=a^{2010}\left(a^{2}+2 a-2\right)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Figure 2 is a rectangle composed of 6 squares. If the area of the smallest square is 1, then the area of this rectangle is $\qquad$ .
|
3. 143 .
Let the side lengths of the six squares, from smallest to largest, be
$$
1, x, x, x+1, x+2, x+3 \text{.}
$$
Then, by the equality of the top and bottom sides of the rectangle, we have
$$
\begin{array}{l}
x+x+(x+1)=(x+2)+(x+3) \\
\Rightarrow x=4 .
\end{array}
$$
Thus, the length and width of the rectangle are
$$
x+x+(x+1)=13, x+(x+3)=11 \text{.}
$$
Therefore, the area of the rectangle is $13 \times 11=143$.
|
143
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $\triangle A B C$
the lengths of the three medians are 3,
4, 5. Then $S_{\triangle M B C}$ is $\qquad$
|
4. 8 .
As shown in Figure 4, extend GD to point $D^{\prime}$, making it twice as long.
Then the side lengths of $\triangle G D^{\prime} C$ are $\frac{2}{3}$ times the lengths of the three medians of $\triangle A B C$.
Therefore, it is a right triangle, and its area is $\frac{8}{3}$.
Additionally, the area of $\triangle G D^{\prime} C$ is equal to the area of $\triangle B G C$, and is $\frac{1}{3}$ of the area of $\triangle A B C$, hence
$$
S_{\triangle A B C}=8 .
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) There are $m$ regular $n$-sided polygons, and the sum of the interior angles of these $m$ regular polygons can be divided by 8. Find the minimum value of $m+n$.
|
Three, from the problem, we know that the total sum of the interior angles of these $m$ regular polygons is $m(n-2) \times 180$.
From $81[180 m(n-2)]$
$\Rightarrow 21 \mathrm{~m}(n-2) \Rightarrow 21 \mathrm{mn}$.
Thus, at least one of $m$ and $n$ is even.
Also, $m \geqslant 1, n \geqslant 3$, and both are integers.
To make $m+n$ the smallest, then
$(m, n)=(1,4),(2,3)$.
Therefore, the minimum value of $m+n$ is 5.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (25 points) There are several (more than enough) socks in red, yellow, blue, and white. If any two socks of the same color can make 1 pair, the question is: What is the minimum number of socks needed to ensure that 10 pairs of socks can be formed?
|
Solution 1 Since there are 4 colors, among 5 socks, there must be 1 pair.
After taking out 1 pair, 3 socks remain. By adding 2 more socks, another pair can be formed.
Following this logic, the number of pairs of socks $(x)$ and the number of socks needed $(y)$ have the following relationship:
$$
y=2 x+3 \text {. }
$$
Thus, to form 10 pairs of socks, only 23 socks are needed.
If 22 socks are taken out, 9 pairs can certainly be formed. If the remaining 4 socks are all different colors, then 22 socks cannot form 10 pairs.
Therefore, at least 23 socks are needed to ensure 10 pairs can be formed.
Solution 2 Since the maximum number of single-color socks that can remain is 4, 24 socks can certainly form 10 pairs.
When 23 socks are taken out, 9 pairs can certainly be formed, leaving 5 socks;
Among the 5 socks, 1 pair can be formed, so 23 socks can also form 10 pairs.
The rest is the same as Solution 1.
|
23
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 As shown in Figure 5, in $\triangle A B C$, $\angle C=90^{\circ}$, $I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$
(2004, Beijing Middle School Mathematics Competition (Grade 8))
|
Solve As shown in Figure 5, construct the symmetric points $F, G$ of points $E, D$ with respect to $AD, BE$ respectively. Then $F, G$ lie on $AB$. Connect $IF, IG$. It is easy to know that
$$
\angle AIB=90^{\circ}+\frac{1}{2} \angle C=135^{\circ}.
$$
By the properties of axial symmetry, we have
$$
\begin{array}{l}
IF=IE, ID=IG, \\
\angle AIE=\angle AIF=\angle BID=\angle BIG=45^{\circ}. \\
\text{Therefore, } \angle FIG=\angle AIB-\angle AIF-\angle BIG \\
=135^{\circ}-45^{\circ}-45^{\circ}=45^{\circ}=\angle BID.
\end{array}
$$
Construct $DH \perp BE$ at point $H$, and $GK \perp IF$ at point $K$.
It is easy to prove $\triangle IDH \cong \triangle IGK$.
Thus, $GK=DH$.
Therefore, $\frac{1}{2} IE \cdot DH=\frac{1}{2} IF \cdot GK$, which means
$S_{\triangle DDE}=S_{\triangle ICF}$.
Hence, $S_{\text{quadrilateral } ABDE}=2 S_{\triangle AIB}=24$.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the set $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can be found two elements $a, b$, such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$ (where $|M|$ denotes the number of elements in the set $M$). (Supplied by Feng Zhigang)
|
2. When
$$
M=\left\{1,2,2^{2}, \cdots, 2^{10}, 3,3 \times 2,3 \times 2^{2}, \cdots, 3 \times 2^{9}\right\}
$$
it satisfies the condition, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011,
\end{array}
$$
contradiction.
In summary, the maximum value of $|M|$ is 21.
|
21
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given an integer $n \geqslant 2$.
(1) Prove: The set $\{1,2, \cdots, n\}$ can be properly arranged into subsets $A_{1}, A_{2}, \cdots, A_{2^{n}}$, such that the number of elements in $A_{i}$ and $A_{i+1}\left(i=1,2, \cdots, 2^{n}\right.$, and $\left.A_{2^{n+1}}=A_{1}\right)$ differ by exactly 1.
(2) For subsets $A_{1}, A_{2}$,
$$
\begin{array}{l}
\cdots, A_{2^{n}}, \text { find all possible values of } \sum_{i=1}^{2 n}(-1)^{i} S\left(A_{i}\right) \text {, where } \\
\text { } S\left(A_{i}\right)=\sum_{x \in \lambda_{i}} x, S(\varnothing)=0 .
\end{array}
$$
(Supplied by Liang Yingde)
|
3. (1) Prove by mathematical induction: there exists a subsequence $A_{1}, A_{2}, \cdots, A_{2^{n}}$ that meets the requirements, and $A_{1}=\{1\}$, $A_{2^{n}}=\varnothing$.
When $n=2$, the sequence $\{1\}, \{1,2\}, \{2\}, \varnothing$ satisfies the requirements.
Assume that when $n=k$, there exists a subsequence $B_{1}, B_{2}, \cdots, B_{2^{k}}$ that meets the requirements.
For $n=k+1$, construct the sequence as follows:
$A_{1}=B_{1}=\{1\}$,
$A_{i}=B_{i-1} \cup\{k+1\}\left(i=2,3, \cdots, 2^{k}+1\right)$,
$A_{j}=B_{j-2^{k}}\left(j=2^{k}+2,2^{k}+3, \cdots, 2^{k+1}\right)$.
It is easy to verify that the sequence $A_{1}, A_{2}, \cdots, A_{2^{k+1}}$ meets the requirements.
In summary, for any positive integer $n(n \geqslant 2)$, there exists a subsequence that meets the requirements.
(2) Without loss of generality, let $A_{1}=\{1\}$. Since the number of elements in adjacent two subsets differs by 1, one must be odd and the other even. Therefore, the number of elements in a subset is the same as the parity of its index.
$$
\text { Hence } \sum_{i=1}^{2 n}(-1)^{i} S\left(A_{i}\right)=\sum_{i \in P} S(A)-\sum_{A \in Q} S(A) \text {, }
$$
where $P$ is the set of all even-element subsets of $\{1,2, \cdots, n\}$, and $Q$ is the set of all odd-element subsets of $\{1,2, \cdots, n\}$.
For any $x \in\{1,2, \cdots, n\}$, since it appears $\mathrm{C}_{n-1}^{k-1}$ times in all $k$-element subsets, its contribution to $\sum_{i \in P} S(A)-\sum_{i \in Q} S(A)$ is
$$
\sum_{k=1}^{n}(-1)^{k} \mathrm{C}_{n-1}^{k-1}=-(1-1)^{n-1}=0 .
$$
Therefore, $\sum_{i=1}^{2 n}(-1)^{i} S\left(A_{i}\right)=0$.
|
0
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. Nine positive real numbers $a_{1}, a_{2}, \cdots, a_{9}$ form a geometric sequence, and
$$
a_{1}+a_{2}=\frac{3}{4}, a_{3}+a_{4}+a_{5}+a_{6}=15 .
$$
Then $a_{7}+a_{8}+a_{9}=$ . $\qquad$
|
1. 112.
Let the common ratio be $q$. Then, from the given conditions, we have
$$
\begin{array}{l}
a_{1}(1+q)=\frac{3}{4} \\
a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15
\end{array}
$$
Dividing the above two equations yields $q^{2}\left(1+q^{2}\right)=20$.
Thus, $q=2, a_{1}=\frac{1}{4}$.
Therefore, $a_{7}+a_{8}+a_{9}=a_{1} q^{6}\left(1+q+q^{2}\right)=112$.
|
112
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The function $f(x)$ is defined on $(0, \infty)$, and satisfies
$$
f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0 \text {. }
$$
Then the minimum value of $f(x)$ is $\qquad$
|
5.3.
From $f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0$, we get $f\left(\frac{1}{x}\right)-\frac{2}{x} f(x)+\frac{3}{x^{2}}=0$.
By solving the above two equations simultaneously, we obtain $f(x)=x^{2}+\frac{2}{x}$.
By the AM-GM inequality,
$$
x^{2}+\frac{2}{x}=x^{2}+\frac{1}{x}+\frac{1}{x} \geqslant 3\left(x^{2} \cdot \frac{1}{x} \cdot \frac{1}{x}\right)^{\frac{1}{3}}=3 \text {, }
$$
and when $x=1$, the equality holds.
Therefore, the minimum value of $f(x)$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The complex number $z$ satisfies
$$
|z|(3 z+2 \mathrm{i})=2(\mathrm{i} z-6) \text {. }
$$
Then $|z|$ equals $\qquad$ .
|
6. 2 .
Solution 1 Direct calculation shows that
$$
|3 z+2 \mathrm{i}|^{2}-|\mathrm{i} z-6|^{2}=8\left(|z|^{2}-4\right) \text {. }
$$
From this, if $|z|>2$, then
$$
\begin{array}{l}
|3 z+2 \mathrm{i}|>|\mathrm{i} z-6| \\
\Rightarrow|| z|(3 z+2 \mathrm{i})|>|2(\mathrm{i} z-6)|,
\end{array}
$$
which contradicts the given condition;
if $|z|<2$, then
$$
|| z \mid(3 z+2 \text { i) }|<| 2(z+6 \text { i) } \mid,
$$
which is also a contradiction.
Therefore, $|z|=2$.
Solution 2 Let $|z|=r$. Substituting into the condition, we get
$$
z=\frac{-(12+2 r \mathrm{i})}{3 r-2 \mathrm{i}} \text {. }
$$
Then $r^{2}=|z|^{2}=\frac{12^{2}+(2 r)^{2}}{(3 r)^{2}+(-2)^{2}}$.
Solving this, we get $r=2$, i.e., $|z|=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+\cdots+a_{20} x^{20}$. Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=$ $\qquad$
|
7. -9 .
Let $x=0$, we get $a_{0}=1$.
Differentiating both sides of the given equation, we get
$$
\begin{array}{l}
10\left(1+x-x^{2}\right)^{9}(1-2 x) \\
=a_{1}+2 a_{2} x+\cdots+20 a_{20} x^{19} . \\
\text { Let } x=1 \text {, we get. } \\
a_{1}+2 a_{2}+\cdots+20 a_{20}=-10 .
\end{array}
$$
Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=-9$.
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 10, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition)
|
As shown in Figure 10, construct the symmetric point $B^{\prime}$ of point $B$ with respect to line $A C$, and let $B B^{\prime}$ intersect $A C$ at point $E$. Draw $B^{\prime} N \perp A B$ at point $N$, and let $B^{\prime} N$ intersect $A C$ at point $M$. Then, $M$ and $N$ are the required points. The minimum value sought is $B^{\prime} N$.
From $S_{\triangle A B C}=\frac{1}{2} A B \cdot B C=\frac{1}{2} A C \cdot B E$, we get $B E=4 \sqrt{5}$.
Therefore, $B B^{\prime}=2 B E=8 \sqrt{5}$.
From $\triangle B^{\prime} N B \sim \triangle A B C$
$\Rightarrow \frac{B^{\prime} N}{A B}=\frac{B^{\prime} B}{A C} \Rightarrow B^{\prime} N=16$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the set of positive real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$, and the set $S=\{(a, b) \mid a \in A, b \in A, a-b \in A\}$. Then the set $S$ can have at most $\qquad$ elements.
|
9.4950 .
The number of ordered pairs of real numbers $(a, b)$ formed by the elements of set $A$ is $100^{2}=10000$.
Since $a_{i}-a_{i}=0 \notin A$, we have $\left(a_{i}, a_{i}\right) \notin S(i=1,2, \cdots, 100)$.
When $\left(a_{i}, a_{j}\right) \in S$, then $\left(a_{j}, a_{i}\right) \notin S$.
Therefore, the maximum number of elements in set $S$ is
$$
\frac{1}{2}(10000-100)=4950 \text {. }
$$
When $a_{i}=i(i=1,2, \cdots, 100)$, the number of elements in $S$ is 4950 .
|
4950
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Each vertex of the convex pentagon $A B C D E$ is colored with one of five colors, such that the two endpoints of each diagonal have different colors. The number of such coloring methods is $\qquad$ (answer with a number),
|
12. 1020 .
The number of coloring ways where all vertices have different colors is $A_{5}^{5}=$ 120. The number of coloring ways where two adjacent vertices have the same color and the rest of the vertices have different colors is
$$
A_{5}^{1} A_{5}^{4}=5(5 \times 4 \times 3 \times 2)=600 \text { ways. }
$$
The number of coloring ways where two pairs of adjacent vertices have the same color and the remaining vertex has a different color is
$$
A_{5}^{1} A_{5}^{3}=5(5 \times 4 \times 3)=300 \text { ways. }
$$
Therefore, there are a total of 1020 coloring ways.
|
1020
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 1, in the tetrahedron $D-ABC$, it is known that $DA \perp$ plane $ABC$, and $\triangle ABC$ is an equilateral triangle with a side length of 2. Then, when the tangent value of the dihedral angle $A-BD-C$ is 2, $V_{D-ABC}=$ $\qquad$
|
4.2.
Given $D A \perp$ plane $A B C$, we know
$D A \perp A B$, plane $D A B \cdot \perp$ plane $A B C$.
As shown in Figure 3, take the midpoint $O$ of $A B$.
Then, by the problem statement, $C O \perp A B, C O$
is on plane $D A B$, and
$$
C O=\sqrt{3} \text {. }
$$
Draw $O E \perp D B$ at point $E$, and connect $C E$. Then
$$
C E \perp D B \text {. }
$$
Thus, $\angle O E C$ is the plane angle of the dihedral angle $A-B D-C$.
Let $D A=x$. Then
$$
\begin{array}{l}
O E=\frac{1}{2} \times \frac{D A \cdot A B}{D B} \\
=\frac{1}{2} \times \frac{2 x}{\sqrt{x^{2}+4}}=\frac{x}{\sqrt{x^{2}+4}} .
\end{array}
$$
Given $\tan \angle O E C=\frac{O C}{O E}=2$, we have
$$
\sqrt{3} \times \sqrt{x^{2}+4}=2 x \Rightarrow x=2 \sqrt{3} \text {. }
$$
Therefore, $V_{D \rightarrow A B C}=\frac{1}{3} \times 2 \sqrt{3}\left(\frac{\sqrt{3}}{4} \times 2^{2}\right)=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given positive integers $x, y, z$ satisfying $x y z=(14-x)(14-y)(14-z)$, and $x+y+z<28$. Then the maximum value of $x^{2}+y^{2}+z^{2}$ is . $\qquad$
|
7. 219 .
From the problem, we know that $x$, $y$, and $z$ are all positive integers less than 14. On the other hand, expanding the given equation, we get
$$
2 x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x) \text {. }
$$
Thus, $71 x y z$.
Since $x$, $y$, and $z$ are all less than 14, at least one of $x$, $y$, or $z$ must be 7.
Without loss of generality, let $z=7$. Then the given equation simplifies to
$$
x y=(14-x)(14-y) \Rightarrow x+y=14 \text {. }
$$
In this case, the maximum value of $x^{2}+y^{2}$ is $1^{2}+13^{2}$.
Therefore, the maximum value of $x^{2}+y^{2}+z^{2}$ is
$$
1^{2}+13^{2}+7^{2}=219 \text {. }
$$
|
219
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. For a positive integer $n$, let $x_{n}$ be the real root of the equation
$$
n x^{3}+2 x-n=0
$$
with respect to $x$, and let
$$
a_{n}=\left[(n+1) x_{n}\right](n=2,3, \cdots),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
\frac{1}{1005}\left(a_{2}+a_{3}+\cdots+a_{2011}\right)=
$$
$\qquad$
|
9.2013.
Let $f(x)=n x^{3}+2 x-n$.
It is easy to see that when $n$ is a positive integer, $f(x)$ is an increasing function.
When $n \geqslant 2$,
$$
\begin{array}{l}
f\left(\frac{n}{n+1}\right)=n\left(\frac{n}{n+1}\right)^{3}+2 \times \frac{n}{n+1}-n \\
=\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$.
Therefore, when $n \geqslant 2$, the equation $n x^{3}+2 x-n=0$ has a unique real root $x_{n}$, and $x_{n} \in\left(\frac{n}{n+1}, 1\right)$.
Thus, $n<(n+1) x_{n}<n+1$,
$$
a_{n}=\left[(n+1) x_{n}\right]=n \text {. }
$$
Then $\frac{1}{1005} \sum_{n=2}^{2011} a_{n}=\frac{1}{1005} \sum_{n=2}^{2011} n=2013$ :
|
2013
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. In the Cartesian coordinate system, given the point set $I=\{(x, y) \mid x, y$ are integers, and $0 \leqslant x, y \leqslant 5\}$. Then the number of different squares with vertices in the set $I$ is . $\qquad$
|
10. 105.
It is easy to know that there are only two types of squares that meet the conditions: squares whose sides lie on lines perpendicular to the coordinate axes, called "standard squares," and squares whose sides lie on lines not perpendicular to the coordinate axes, called "oblique squares."
(1) In standard squares, the number of squares with side length $k(k=1,2$, $\cdots, 5)$ is $(6-k)^{2}$.
(2) Since the oblique squares with vertices in the point set $I$ are all inscribed squares of some standard square, we only need to consider the number of inscribed squares of standard squares.
Obviously, a square with side length $k(k=1,2, \cdots, 5)$ has $k-1$ inscribed squares.
In summary, the number of squares that meet the conditions is
$$
\sum_{k=1}^{5}\left[(6-k)^{2}+(6-k)^{2}(k-1)\right]=105(\text { squares }) .
$$
|
105
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ .
|
$$
\begin{array}{l}
\left(x^{4}+2 x^{3}+x^{2}\right)+\left[(2+k) x^{2}+(2+k) x\right]+2 k=0 \\
\Rightarrow\left(x^{2}+x\right)^{2}+(2+k)\left(x^{2}+x\right)+2 k=0 \\
\Rightarrow\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 .
\end{array}
$$
Since the original equation has real roots, and $x^{2}+x+2=0$ has no real roots, the equation $x^{2}+x+k=0$ has two real roots $x_{1}$ and $x_{2}$.
According to the problem, $x_{1} x_{2}=k=-2$.
Also, $x_{1}+x_{2}=-1$, so
$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=5 \text{. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. There are 4 colors of light bulbs (with enough of each color), and we need to install a light bulb at each vertex of the triangular prism $A B C-A_{1} B_{1} C_{1}$. The requirement is that the light bulbs at the two endpoints of the same edge must be of different colors, and each color of light bulb must be used at least once. The number of installation methods is $\qquad$ kinds.
|
9.216 .
We can first install $A$, $B$, and $C$, which has $\mathrm{A}_{4}^{3}$ ways; then select one vertex from $A_{1}$, $B_{1}$, and $C_{1}$ to install the fourth color of the light bulb, which has $\mathrm{C}_{3}^{1}$ ways; finally, there are 3 ways to install the remaining two vertices.
Therefore, there are 216 different installation methods.
|
216
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers is $\qquad$ .
|
10. 495 .
$$
\begin{array}{l}
\text { Let } t=l+(l+1)+\cdots+(l+8) \\
=m+(m+1)+\cdots+(m+9) \\
=n+(n+1)+\cdots+(n+10)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $l=n+2+\frac{2 n+1}{9}$,
$$
m=\frac{n}{10}+n+1 .
$$
Therefore, $2 n+1 \equiv 0(\bmod 9)$,
$$
n \equiv 0(\bmod 10) \text {. }
$$
Thus, the smallest positive integer $n$ that satisfies the condition is $n=40$, i.e.,
$$
t_{\min }=40+41+\cdots+50=495 .
$$
|
495
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $A$ and $B$ be the common vertices of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
and the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
Let $P$ and $M$ be two moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A M}+\overrightarrow{B M}),
$$
where $\lambda \in \mathbf{R},|\lambda|>1$. Let the slopes of the lines $A P, B P, A M, B M$ be $k_{1}, k_{2}, k_{3}, k_{4}$, respectively, and $k_{1}+k_{2}=5$. Then $k_{3}+k_{4}=$ $\qquad$
|
11. -5 .
Let $A(-a, 0), B(a, 0), P\left(x_{1}, y_{1}\right), M\left(x_{2}, y_{2}\right)$.
From $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A M}+\overrightarrow{B M})$, we know that points $O, P, M$ are collinear, and we can find that
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a} \\
=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}} .
\end{array}
$$
Similarly, $k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}$.
Therefore, $k_{3}+k_{4}=-\left(k_{1}+k_{2}\right)=-5$.
|
-5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 5, given $\triangle A B C$ is inscribed in $\odot O$, chord $A F \perp B C$ at point $H, G$ is the midpoint of $B F$. Then $\frac{A C}{O G}=$
|
4. 2 .
Connect $F O$. Then
$$
\begin{array}{l}
\text { Rt } \triangle A B H \backsim \text { Rt } \triangle O \\
\Rightarrow \frac{A H}{B H}=\frac{O G}{F G} . \\
\text { Also } \triangle A C H \backsim \triangle B F H \\
\Rightarrow \frac{A H}{B H}=\frac{A C}{B F} .
\end{array}
$$
Therefore $A C=2 O G \Rightarrow \frac{A C}{O G}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of real numbers $a$ that make the equation
$$
x^{2}+a x+8 a=0
$$
have only integer solutions is $\qquad$ .
|
2. 8 .
Let the equation (1) have integer solutions $m, n (m \leqslant n)$. Then $m+n=-a, mn=8a$.
Thus, $(m+8)(n+8)=64$.
Solving for $(m, n)$
$$
\begin{array}{c}
=(-72,-9),(-40,-10),(-24,-12), \\
(-16,-16),(-7,56),(-6,24), \\
(-4,8),(0,0) .
\end{array}
$$
Correspondingly,
$$
\begin{array}{l}
a=-(m+n) \\
=81,50,36,32,-49,-18,-4,0,
\end{array}
$$
a total of 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. All prime numbers $p$ that make $2 p^{4}-p^{2}+36$ a perfect square are $\qquad$ .
|
5.2.
When $p=2$,
$$
2 p^{4}-p^{2}+36=64
$$
is a perfect square, thus, $p=2$ is the solution.
When $p=3$,
$$
2 p^{4}-p^{2}+36=189
$$
is not a perfect square.
When $p$ is an odd prime greater than 3, let
$2 p^{4}-p^{2}+36=k^{2}$ ( $k$ is an odd positive integer).
Then $p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6)$
$\Rightarrow p \mid(k-6)$ or $p \mid(k+6)$.
However, $p$ cannot divide both $k-6$ and $k+6$. Therefore, one and only one of $k-6$ and $k+6$ is divisible by $p^{2}$.
If $p^{2} \mid(k-6)$, let $k-6=s p^{2}$ ( $s$ is an odd positive integer).
$$
\begin{array}{l}
\text { Then } p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6) \\
=s p^{2}\left(s p^{2}+12\right) \\
\Rightarrow s\left(s p^{2}+12\right)=2 p^{2}-1 \\
\Rightarrow s^{2}<2 \Rightarrow s=1 \Rightarrow p^{2}=13,
\end{array}
$$
which is a contradiction.
If $p^{2} \mid(k+6)$, let $k+6=s p^{2}$ ( $s$ is an odd positive integer).
Similarly, there does not exist a prime $p$ greater than 3.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and it satisfies
$$
S_{n+1}-5 S_{n}-4 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the last four digits of $a_{2 \text { on }}$ are $\qquad$
|
7.8124.
Given $S_{n+1}-5 S_{n}-4 n-4=0$, and
$$
S_{n}-5 S_{n+1}-4 n=0 \text {, }
$$
subtracting the two equations yields
$$
\begin{array}{l}
a_{n+1}-5 a_{n}-4=0 \\
\Rightarrow a_{n+1}+1=5\left(a_{n}+1\right) \\
\Rightarrow a_{n}=5^{n}-1 .
\end{array}
$$
Notice that, when $k \in \mathbf{N}_{+}$,
$$
\begin{aligned}
5^{4 k+1} & \equiv 3125\left(\bmod 10^{4}\right), \\
5^{4 k+2} & \equiv 5625\left(\bmod 10^{4}\right), \\
5^{4 k+3} & \equiv 8125\left(\bmod 10^{4}\right), \\
5^{4 k+4} & \equiv 625\left(\bmod 10^{4}\right) .
\end{aligned}
$$
Thus, $a_{2011} \equiv 5^{4 \times 502+3}-1 \equiv 8124\left(\bmod 10^{4}\right)$.
|
8124
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 㷵 On a board, there is a convex 2011-gon, and Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? [3]
|
To prove by induction: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. The following $2n-6$ diagonals can be drawn sequentially:
$A_{2} A_{4}, A_{3} A_{5}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.
When $n=3$, the conclusion is obviously true.
Assume the proposition holds for all values less than $n$.
For a convex $n$-sided polygon, let the last diagonal drawn be $A_{1} A_{k}$, which intersects at most one of the previously drawn diagonals (if it exists, let it be $l$) at an interior point. All the diagonals drawn, except for $A_{1} A_{k}$ and $l$, lie entirely within the $k$-sided polygon $A_{1} A_{2} \cdots A_{k}$ or the $(n+2-k)$-sided polygon $A_{k} A_{k+1} \cdots A_{n} A_{1}$.
By the induction hypothesis, there are at most
$$
(2k-6) + [2(n+2-k)-6] = 2n-8
$$
diagonals; adding $A_{1} A_{k}$ and $l$ results in at most $2n-6$ diagonals.
Therefore, for a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Thus, Betja can draw at most 4016 diagonals.
|
4016
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given a moving point $O$ inside $\triangle A B C$, rays $A O, B O, C O$ intersect the opposite sides at points $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Let $B^{\prime} C^{\prime}$ intersect $A O$ at $D$, $C^{\prime} A^{\prime}$ intersect $B O$ at $E$, and $A^{\prime} B^{\prime}$ intersect $C O$ at $F$. Prove: Regardless of the position of point $O$, $\frac{O D}{A D}+\frac{O E}{B E}+\frac{O F}{C F}$ is always a constant.
|
Prove that, as shown in Figure 3, there exist three complete quadrilaterals.
From the complete quadrilateral \( A C^{\prime} O B^{\prime} B C \), the complete quadrilateral \( B A^{\prime} O C^{\prime} C A \), and the complete quadrilateral \( C B^{\prime} O A^{\prime} A B \), we get
\[
\begin{array}{l}
\frac{O D}{A D}=\frac{O A^{\prime}}{A A^{\prime}}=\frac{S_{\triangle O B C}}{S_{\triangle A B C}}, \\
\frac{O E}{B E}=\frac{O B^{\prime}}{B B^{\prime}}=\frac{S_{\triangle O C A}}{S_{\triangle A B C}}, \\
\frac{O F}{C F}=\frac{O C^{\prime}}{C C^{\prime}}=\frac{S_{\triangle O A B}}{S_{\triangle A B C}} . \\
\text { Therefore, } \frac{O D}{A D}+\frac{O E}{B E}+\frac{O F}{C F} \\
=\frac{S_{\triangle O B C}+S_{\triangle O C A}+S_{\triangle O A B}}{S_{\triangle A B C}} \\
=\frac{S_{\triangle A B C}}{S_{\triangle A B C}}=1 \text { (a constant). }
\end{array}
\]
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { 11. If } a+b-2 \sqrt{a-1}-4 \sqrt{b-2} \\
=3 \sqrt{c-3}-\frac{1}{2} c-5 \text {, }
\end{array}
$$
then $a+b+c=$ . $\qquad$
|
Ni, 11.20.
Notice,
$$
(\sqrt{a-1}-1)^{2}+(\sqrt{b-2}-2)^{2}+\frac{1}{2}(\sqrt{c-3}-3)^{2}=0 \text {. }
$$
Therefore, $a=2, b=6, c=12$.
So $a+b+c=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given real numbers $a, b$ satisfy
$$
6^{a}=2010,335^{b}=2010 \text {. }
$$
Then the value of $\frac{1}{a}+\frac{1}{b}$ is $\qquad$
|
12. 1 .
Notice, $6^{a b}=2010^{b}, 335^{a b}=2010^{a}$.
Then $(6 \times 335)^{a b}=2010^{a+b}$.
Thus $a b=a+b \Rightarrow \frac{1}{a}+\frac{1}{b}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the function $y=f(x)$ have the domain $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
2 f\left(x^{2}+x\right)+f\left(x^{2}-3 x+2\right) \\
=9 x^{2}-3 x-6 .
\end{array}
$$
Then the value of $f(60)$ is . $\qquad$
|
2. 176.
Substituting $1-x$ for $x$ in equation (1) yields
$$
\begin{array}{l}
2 f\left(x^{2}-3 x+2\right)+f\left(x^{2}+x\right) \\
=9 x^{2}-15 x .
\end{array}
$$
Combining equations (1) and (2) gives
$$
\begin{array}{l}
f\left(x^{2}+x\right)=3 x^{2}+3 x-4 \\
=3\left(x^{2}+x\right)-4 .
\end{array}
$$
Therefore, $f(60)=3 \times 60-4=176$.
|
176
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $[x]$ represents the greatest integer not exceeding the real number $x$. Then, in the Cartesian coordinate system $x O y$, the area of the figure formed by all points $(x, y)$ that satisfy $[x][y]=$ 2011 is $\qquad$
|
3.4.
Let $[x]=a,[y]=b$, meaning all such points $(x, y)$ form the region
$$
a \leqslant x<a+1, b \leqslant y<b+1
$$
bounded by these inequalities, with an area of 1.
Since 2011 is a prime number, the points $(x, y)$ that satisfy
$$
[x][y]=2011
$$
form 4 regions, each with an area of 1, making the total area 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If the two quadratic equations
$$
x^{2}+x+m=0 \text { and } m x^{2}+x+1=0
$$
each have two distinct real roots, but one of them is a common real root $\alpha$, then the range of the real root $\alpha$ is $\qquad$.
|
Given that the common real root of the two equations is $\alpha$, and $m \neq 1$. Then
$$
\begin{array}{l}
\alpha^{2}+\alpha+m=0, \\
m \alpha^{2}+\alpha+1=0 .
\end{array}
$$
From equation (1), we get $m=-\alpha^{2}-\alpha$.
Substituting into equation (2), we get
$$
\begin{array}{l}
\alpha^{4}+\alpha^{3}-\alpha-1=0 \\
\Rightarrow\left(\alpha^{2}-1\right)\left(\alpha^{2}+\alpha+1\right)=0 .
\end{array}
$$
Since $\alpha^{2}+\alpha+1=\left(\alpha+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$, then
$$
\alpha^{2}-1=0 \Rightarrow \alpha= \pm 1 \text {. }
$$
When $\alpha=1$, from equation (1) we get $m=-2$, at this time, the given two equations become
$$
x^{2}+x-2=0 \text { and } 2 x^{2}-x-1=0,
$$
with distinct roots -2 and $\frac{1}{2}$.
When $\alpha=-1$, from equation (1) we get $m=0$, at this time, the equation $m x^{2}+x+1=0$ becomes a linear equation $x+1=0$, which contradicts the problem statement.
In summary, the value of the real root $\alpha$ is $\alpha=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
20. Given the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, a line is drawn through its left focus $F_{1}$ intersecting the ellipse at points $A$ and $B$. Point $D(a, 0)$ is a point to the right of $F_{1}$. Connecting $A D$ and $B D$ intersects the left directrix of the ellipse at points $M$ and $N$. If the circle with diameter $M N$ passes exactly through point $F_{1}$, find the value of $a$.
|
20. It is known that $F_{1}(-3,0)$, the equation of the left directrix is $x=-\frac{25}{3}$, and $l_{A B}: y=k(x+3)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=k(x+3), \\ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1\end{array}\right.$
$$
\Rightarrow\left(16+25 k^{2}\right) x^{2}+150 k^{2} x+225 k^{2}-400=0 .
$$
Then $x_{1}+x_{2}=-\frac{150 k^{2}}{16+25 k^{2}}, x_{1} x_{2}=\frac{225 k^{2}-400}{16+25 k^{2}}$
$$
\Rightarrow y_{1} y_{2}=k^{2}\left(x_{1}+3\right)\left(x_{2}+3\right)=-\frac{256 k^{2}}{16+25 k^{2}} \text {. }
$$
Let $M\left(-\frac{25}{3}, y_{3}\right), N\left(-\frac{25}{3}, y_{4}\right)$.
From the collinearity of $M, A, D$ we get
$$
y_{3}=\frac{(3 a+25) y_{1}}{3\left(a-x_{1}\right)} \text {. }
$$
Similarly, $y_{4}=\frac{(3 a+25) y_{2}}{3\left(a-x_{2}\right)}$.
$$
\text { Also, } \overrightarrow{F_{1} M}=\left(-\frac{16}{3}, y_{3}\right), \overrightarrow{F_{1} N}=\left(-\frac{16}{3}, y_{4}\right) \text {. }
$$
From the given information,
$$
\begin{array}{l}
\overrightarrow{F_{1} M} \perp \overrightarrow{F_{1} N} \Rightarrow \overrightarrow{F_{1} M} \cdot \overrightarrow{F_{1} N}=0 \\
\Rightarrow y_{3} y_{4}=-\frac{256}{9} \text {. } \\
\end{array}
$$
$$
\begin{array}{l}
\text { And } y_{3} y_{4}=\frac{(3 a+25)^{2} y_{1} y_{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)} \\
\Rightarrow-\frac{256 k^{2}}{16+25 k^{2}} \cdot \frac{(3 a+25)^{2}}{9\left(a-x_{1}\right)\left(a-x_{2}\right)}=-\frac{256}{9} \\
\Rightarrow\left(1+k^{2}\right)\left(16 a^{2}-400\right)=0
\end{array}
$$
$\Rightarrow a= \pm 5$ (negative value is discarded).
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let positive real numbers $a, b, c, d, e$ satisfy $a<b<c<d$ $<e$, and the smallest three of the 10 products of any two numbers are $28, 32, 56$, and the largest two are 128, 240. Then $e=$ $\qquad$
|
2.16.
From the problem, we know
$$
a b=28, a c=32, c e=128, d e=240 \text {. }
$$
Then $c=\frac{8}{7} b, d=\frac{15}{8} c=\frac{15}{7} b$.
Thus, $a d=\frac{15}{7} a b=60>56$.
Therefore, $b c=56 \Rightarrow \frac{8}{7} b^{2}=56 \Rightarrow b=7$.
Hence, $a=4, c=8, d=15, e=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) If placing the positive integer $N$ to the left of the positive integer $n$ results in a new number that is divisible by 7, then $N$ is called a "magic number" of $n$. $M$ is a set of positive integers such that for any positive integer $n$, there exists a positive integer in set $M$ that is a magic number of $n$. When $|M|$ is minimized, find the minimum value of the sum of all elements in set $M$.
|
For $n=1,2, \cdots, 7$.
If $|M| \leqslant 6$, then by the pigeonhole principle, there must be a positive integer $N$ in set $M$ that is a common magic number of $i, j(1 \leqslant i<j \leqslant 7)$, i.e.,
$71(10 N+i), 7 I(10 N+j)$.
Then $7!(j-i)$, but $0<j-i \leqslant 6$, a contradiction.
Therefore, $|M| \geqslant 7$.
Thus, the sum of all elements in set $M$ is no less than
$$
1+2+\cdots+7=28 \text {. }
$$
When $M=\{1,2, \cdots, 7\}$, for any positive integer $n$, let it be a $k\left(k \in \mathbf{N}_{+}\right)$-digit number. Then $10^{k} i+n$ $(i=1,2, \cdots, 7)$ have distinct remainders when divided by 7.
Otherwise, there exist positive integers $i, j(1 \leqslant i<j \leqslant 7)$, such that
$$
\begin{array}{l}
7 \mid\left[\left(10^{k} j+n\right)-\left(10^{k} i+n\right)\right] \\
\Rightarrow 7\left|10^{k}(j-i) \Rightarrow 7\right|(j-i),
\end{array}
$$
a contradiction.
Therefore, there must exist a positive integer $i(1 \leqslant i \leqslant 7)$, such that $71\left(10^{k} i+n\right)$, i.e., $i$ is a magic number of $n$.
Thus, when $|M|$ is minimal, the minimum value of the sum of all elements in set $M$ is 28.
|
28
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=4, A A_{1}=A D=2$, points $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}, B C$ respectively. Then the volume of the tetrahedron $B_{1}-E F G$ is $\qquad$
|
2.3.
Take point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{2}$. Then $H E / / B_{1} G$. Therefore, $H E / /$ plane $B_{1} F G$.
Thus, $V_{B_{1}-E F G}=V_{E-B_{1} F C}=V_{H-B_{1} F G}=V_{G-B_{1} F H}$.
And $S_{\triangle B_{1} F H}=\frac{9}{2}$, the distance from point $G$ to plane $B_{1} F H$ is 2. Hence $V_{B_{1}-E F G}=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the three sides of $\triangle A B C$ are $a, b, c$. If $a+b+c=16$, then
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
$=$
|
6. 64 .
Let the circumradius of $\triangle ABC$ be $R$. Then
$$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\right. \\
\left.2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\frac{1-\cos B}{2}+\right. \\
\left.\frac{1-\cos C}{2}+\frac{\cos A+\cos B+\cos C-1}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{A}{2} \cdot \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2} \\
= R^{2}(\sin A+\sin B+\sin C)^{2} \\
= \frac{1}{4}(a+b+c)^{2}=64 .
\end{array}
$$
|
64
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. An $8 \times 8$ chessboard is colored in the usual way, with 32 black squares and 32 white squares. A "path" consists of 8 white squares, one in each row, and adjacent white squares share a common vertex. The number of such paths is $\qquad$.
|
8. 296
We can use the number labeling method:
\begin{tabular}{cccccccc}
1 & Black & 1 & Black & 1 & Black & 1 & Black \\
Black & 2 & Black & 2 & Black & 2 & Black & 1 \\
2 & Black & 4 & Black & 4 & Black & 3 & Black \\
Black & 6 & Black & 8 & Black & 7 & Black & 3 \\
6 & Black & 14 & Black & 15 & Black & 10 & Black \\
Black & 20 & Black & 29 & Black & 25 & Black & 10 \\
20 & Black & 49 & Black & 54 & Black & 35 & Black \\
Black & 69 & Black & 103 & Black & 89 & Black & 35
\end{tabular}
Therefore, there are $69+103+89+35=296$ (ways).
|
296
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) In the geometric sequence $\left\{a_{n}\right\}$ where all terms are positive, what is the maximum number of terms that can be integers between $100 \sim 1000$?
|
9. Let the geometric sequence $\left\{a q^{n-1}\right\}$ satisfy
$$
100 \leqslant a q^{n-1} \leqslant 1000 \quad (a, q > 1) \text{ are integers.}
$$
Clearly, $q$ must be a rational number.
Let $q=\frac{t}{s}(t>s \geqslant 1, (t, s)=1)$.
Since $a q^{n-1}=a\left(\frac{t}{s}\right)^{n-1}$ is an integer, $a$ must be a multiple of $s^{n-1}$.
Let $t=s+1$. Thus, the sequence satisfies
$100 \leqslant a < a \cdot \frac{s+1}{s} < \cdots < a\left(\frac{s+1}{s}\right)^{n-1} \leqslant 1000$.
If $s \geqslant 3$, then
$1000 \geqslant a\left(\frac{s+1}{s}\right)^{n-1} \geqslant (s+1)^{n-1} \geqslant 4^{n-1}$.
Thus, $n \leqslant 5$.
If $s=1$, then
$1000 \geqslant a \cdot 2^{n-1} \geqslant 100 \times 2^{n-1}$.
Thus, $n \leqslant 4$.
If $s=2$, then
$1000 \geqslant a\left(\frac{3}{2}\right)^{n-1} \geqslant 100\left(\frac{3}{2}\right)^{n-1}$.
Thus, $n \leqslant 6$.
On the other hand, the sequence
$128, 192, 288, 432, 648, 972$
satisfies the given conditions.
Therefore, the maximum number of terms between 100 and 1000 is 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) There are 12 points on a plane, and no three points are collinear. Using any one of these points as the starting point and another point as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of zero triangles that can be formed with these 12 points as vertices.
|
Let the 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $C_{12}^{3}$. Let the number of vectors starting from $P_{i}(i=1,2, \cdots, 12)$ be $x_{i}$ $\left(0 \leqslant x_{i} \leqslant 11\right)$.
If a triangle with three points as vertices is a non-zero triangle, then there is and only one point that is the starting point of the two sides vectors of this triangle. Therefore, the number of non-zero triangles with $P_{i}$ as one of the vertices and as the starting point of two sides is $\mathrm{C}_{x_{i}}^{2}$. Thus, the total number of non-zero triangles with these points as vertices is $\sum_{i=1}^{12} c_{x_{i}}^{2}$.
Therefore, the number of zero triangles is $\mathrm{C}_{12}^{3}-\sum_{i=1}^{12} \mathrm{C}_{\mathrm{x}_{i}}^{2}$.
First, find the minimum value of $\sum_{i=1}^{12} C_{x_{i}}^{2}$.
Since $\sum_{i=1}^{12} x_{i}=\mathrm{C}_{12}^{2}=66$, we have
$$
\begin{array}{l}
\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}=\sum_{i=1}^{12} \frac{x_{i}\left(x_{i}-1\right)}{2} \\
=\frac{1}{2} \sum_{i=1}^{12}\left(x_{i}^{2}-x_{i}\right)=\frac{1}{2} \sum_{i=1}^{12} x_{i}^{2}-33 .
\end{array}
$$
And the non-negative integer $x_{i}$ does not exceed 11, so $\sum_{i=1}^{12} x_{i}^{2}$ has a minimum value.
By the adjustment method, when
$$
\left|x_{i}-x_{j}\right|=0,1(1 \leqslant i<j \leqslant 12),
$$
that is, when
$$
\begin{array}{l}
\left\{x_{1}, x_{2}, \cdots, x_{12}\right\} \\
=\{5,5,5,5,5,5,6,6,6,6,6,6\}
\end{array}
$$
$\sum_{i=1}^{12} x_{i}^{2}$ takes the minimum value 366.
Thus, the minimum value of $\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$ is
$$
\frac{1}{2} \times 366-33=150 \text {. }
$$
Therefore, the maximum number of zero triangles with these 12 points as vertices is 70.
|
70
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given 315, find
$$
S=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\cdots+\sqrt[2011]{2011}}}}
$$
the integer part of \( S \).
|
Let $n=2011$, and set
$$
\begin{array}{l}
a_{1}=\sqrt[n]{n}, \\
a_{k}=\sqrt[n-k+1]{n-k+1+a_{k-1}}(2 \leqslant k \leqslant n-1) .
\end{array}
$$
Then $S=a_{n-1}$.
We will prove by induction that:
$$
1<a_{m}<2(m=1,2, \cdots, n-1) \text {. }
$$
(1) When $m=1$, since $1<n<2^{n}$, we have
$$
1<\sqrt[n]{n}<2 \text {. }
$$
(2) Assume that when $m=k(1 \leqslant k \leqslant n-2)$, the conclusion holds. Then when $m=k+1$, since
$$
\begin{array}{l}
1<n-k+1<n-k+a_{k}<n-k+2 \\
\leqslant(1+1)^{n-k}=1+(n-k)+\cdots+1,
\end{array}
$$
we have $1<a_{k+1}=\sqrt[n-k]{n-k+a_{k}}<2$.
Therefore, when $m=k+1$, the conclusion holds.
In summary, $1<a_{m}<2(m=1,2, \cdots, n-1)$.
Thus, the integer part of $S$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
316 Given that the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ is a monotonically increasing function. If $f(f(n))=3 n$, find $f(2011)$.
|
Let $f(1)=m$. Then $f(m)=f(f(1))=3$.
Thus, $m>1 \Rightarrow f(m)>f(1) \Rightarrow 3>m$.
Therefore, $f(1)=m=2 \Rightarrow f(2)=f(f(1))=3$.
Similarly, $f(3)=6, f(6)=9$.
From $f(3)<f(4)<f(5)<f(6)$, we get
$6<f(4)<f(5)<9$.
Hence, $f(4)=7, f(5)=8$.
Furthermore, $f(7)=f(f(4))=12$,
$f(8)=f(f(5))=15$,
$f(9)=f(f(6))=18$,
$f(12)=f(f(7))=21$,
$f(10)=19, f(11)=20$.
We will prove by mathematical induction:
$$
f(n)=\left\{\begin{array}{ll}
n+3^{m}, & 3^{m} \leqslant n<2 \times 3^{m} ; \\
3 n-3^{m+1}, & 2 \times 3^{m} \leqslant n<3^{m+1}
\end{array}\left(m \in N_{+}\right) .\right.
$$
When $m=0,1$, the conclusion has been verified.
Assume the conclusion holds for $m=k-1$.
When $2 \times 3^{k-1} \leqslant n<3^{k}$,
$f(n)=3 n-3^{k} \Rightarrow f\left(3 n-3^{k}\right)=3 n$.
Let $3 n-3^{k}=p \in\left[3^{k}, 2 \times 3^{k}\right)$. Then
$$
f(p)=p+3^{k} \text {. }
$$
By the monotonicity of $f$, when $3^{k} \leqslant n<2 \times 3^{k}$, we always have $f(n)=n+3^{k}$.
Then $f\left(n+3^{k}\right)=f(f(n))=3 n$.
Let $n+3^{k}=q \in\left[2 \times 3^{k}, 3^{k+1}\right)$. Then
$$
f(q)=3 q-3^{k+1} \text {. }
$$
Therefore, the conclusion holds for $n=k$.
By mathematical induction, the conclusion holds for $n \in \mathbf{N}_{+}$.
Since $2 \times 3^{6}<2011<3^{7}$, we have
$$
f(2011)=3 \times 2011-3^{7}=3846 .
$$
|
3846
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given $a, b \in \mathbf{R}$, the equation about $x$
$$
x^{4}+a x^{3}+2 x^{2}+b x+1=0
$$
has one real root. Find the minimum value of $a^{2}+b^{2}$.
|
【Analysis】This is a quartic equation, and parameters $a, b$ are not easy to handle. We might as well regard $a, b$ as the main variables and $x$ as a parameter. Since $a^{2}+b^{2}$ represents the square of the distance from the moving point $P(a, b)$ to the origin, and $P(a, b)$ lies on the line
$$
x^{3} a + x b + x^{4} + 2 x^{2} + 1 = 0,
$$
then
$$
\begin{array}{l}
\sqrt{a^{2}+b^{2}} \geqslant \frac{x^{4}+2 x^{2}+1}{\sqrt{x^{6}+x^{2}}} = \frac{x^{4}+1+2 x^{2}}{|x| \sqrt{x^{4}+1}} \\
= \frac{\sqrt{x^{4}+1}}{|x|} + \frac{2|x|}{\sqrt{x^{4}+1}} \geqslant 2 \sqrt{2}.
\end{array}
$$
The equality holds if and only if $x = \pm 1$.
Therefore, $\left(a^{2}+b^{2}\right)_{\text {min }} = 8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If the two roots of the equation
$$
x^{2}-3 x-1=0
$$
are also roots of the equation
$$
x^{4}+a x^{2}+b x+c=0
$$
then the value of $a+b-2 c$ is ( ).
(A) -13
(B) -9
(C) 6
(D) 0
|
$$
\begin{array}{l}
\text { From the problem, we know that } x^{4}+a x^{2}+b x+c \text { can definitely be divided by } x^{2}-3 x-1. \\
=\left(x^{2}-3 x-1\right)\left(x^{2}+3 x+a+10\right)+ \\
{[(3 a+b+33) x+(a+c+10)], } \\
\text { then }\left\{\begin{array}{l}
3 a+b+33=0, \\
a+c+10=0
\end{array}\right. \\
\Rightarrow a+b-2 c=-13 .
\end{array}
$$
Therefore, the answer is A.
|
-13
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a$, $b$, and $c$ in $S$ such that $a+b$, $b+c$, and $c+a$ are all in the set $A$.
|
6. Let $a<b<c$. Let
$x=a+b, y=a+c, z=b+c$.
Then $x\langle y\langle z, x+y\rangle z$, and $x+y+z$ is even. (1)
Conversely, if there exist $x, y, z \in A$ satisfying property (1), then take
$$
a=\frac{x+y-z}{2}, b=\frac{x+z-y}{2}, c=\frac{y+z-x}{2},
$$
we have $a, b, c \in \mathbf{Z}, 1 \leqslant a<b<c \leqslant 2012$, and
$$
x=a+b, y=a+c, z=b+c.
$$
Thus, the condition of the problem is equivalent to the fact that for any $k$-element subset $A$, there exist $x, y, z \in A$ satisfying property (1).
If $A=\{1,2,3,5,7, \cdots, 2011\}$, then $|A|=$ 1007, and the set $A$ does not contain three elements satisfying property (1). Therefore, $k \geqslant 1008$.
Next, we prove that any 1008-element subset contains three elements satisfying property (1).
We will prove a more general conclusion:
For any integer $n(n \geqslant 4)$, any $(n+2)$-element subset of the set $\{1,2, \cdots, 2 n\}$ contains three elements satisfying property (1).
We use induction on $n$.
When $n=4$, let $A$ be a six-element subset of $\{1,2, \cdots, 8\}$. Then $A \cap\{3,4, \cdots, 8\}$ has at least 4 elements.
If $A \cap\{3,4, \cdots, 8\}$ contains three even numbers, then $4, 6, 8 \in A$ and they satisfy property (1);
If $A \cap\{3,4, \cdots, 8\}$ contains exactly two even numbers, then it also contains at least two odd numbers. Taking these two odd numbers, at least two even numbers from $4, 6, 8$ can form a three-element set satisfying property (1), since there are at least two even numbers, there exist three numbers satisfying property (1);
If $A \cap\{3,4, \cdots, 8\}$ contains exactly one even number, then it contains all three odd numbers, and this even number with 5, 7 forms a three-element set satisfying property (1).
Therefore, the conclusion holds when $n=4$.
Assume the conclusion holds for $n(n \geqslant 4)$, consider the case for $n+1$.
Let $A$ be an $(n+3)$-element subset of $\{1,2, \cdots, 2 n+2\}$. If $|A \cap\{1,2, \cdots, 2 n\}| \geqslant n+2$, then by the induction hypothesis, the conclusion holds. Thus, we only need to consider the case
$$
|A \cap\{1,2, \cdots, 2 n\}| = n+1,
$$
and $2 n+1, 2 n+2 \in A$.
In this case, if $\{1,2, \cdots, 2 n\}$ contains an odd number $x$ greater than 1 in the set $A$, then $x, 2 n+1, 2 n+2$ form a three-element set satisfying property (1);
If $\{1,2, \cdots, 2 n\}$ contains no odd numbers greater than 1 in the set $A$, then
$$
A \subset\{1,2,4,6, \cdots, 2 n, 2 n+1,2 n+2\},
$$
and the latter has exactly $n+3$ elements, hence
$$
A=\{1,2,4,6, \cdots, 2 n, 2 n+1,2 n+2\},
$$
in this case, $4, 6, 8 \in A$ satisfy property (1).
In conclusion, the smallest $k$ is 1008.
|
1008
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The graph of the quadratic function $y=x^{2}-a x+2$ is symmetric about the line $x=1$. Then the minimum value of $y$ is $\qquad$ .
|
$$
=, 1.1 .
$$
From the condition, we know that $a=2$. Then $y=(x-1)^{2}+1$ has a minimum value of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a=\sqrt{3}-1$. Then the value of $a^{2012}+2 a^{2011}-$ $2 a^{2010}$ is $\qquad$.
|
2. 0 .
From the condition we know
$$
\begin{array}{l}
(a+1)^{2}=3 \Rightarrow a^{2}+2 a-2=0 \text {. } \\
\text { Then } a^{2012}+2 a^{2011}-2 a^{2010} \\
=a^{2010}\left(a^{2}+2 a-2\right)=0 .
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In a chess tournament, there are $n$ female players and $9 n$ male players. Each player plays one game against each of the other $10 n-1$ players. The scoring system is as follows: the winner gets 2 points, the loser gets 0 points, and in the case of a draw, each player gets 1 point. After the tournament, it was found that the total score of all male players is 4 times the total score of all female players. Then, all possible values of $n$ are $\qquad$
|
4.1.
Let the total score of the girls be $m$. Then the total score of the boys is $4 m$. According to the problem,
$$
\begin{array}{l}
\frac{10 n(10 n-1)}{2} \times 2=5 m \\
\Rightarrow m=2 n(10 n-1) .
\end{array}
$$
Since each player competes in $10 n-1$ matches, the maximum score is $2(10 n-1)$, so the total score of the girls
$$
m \leqslant 2 n(10 n-1) \text {. }
$$
Therefore, each girl must win all her matches.
Thus, $n=1$.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Two quadratic equations with unequal leading coefficients $(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0$, (1) and $(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0$ ( $a, b$ are positive integers) have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$.
|
Given the conditions $a>1, b>1, a \neq b$.
Assume the common root of equations (1) and (2) is $x_{0}$. Then
$$
\begin{array}{l}
(a-1) x_{0}^{2}-\left(a^{2}+2\right) x_{0}+\left(a^{2}+2 a\right)=0, \\
(b-1) x_{0}^{2}-\left(b^{2}+2\right) x_{0}+\left(b^{2}+2 b\right)=0 .
\end{array}
$$
(3) $\times(b-1)$ - (4) $\times(a-1)$ and simplifying, we get
$$
(a-b)(a b-a-b-2)\left(x_{0}-1\right)=0 \text {. }
$$
Since $a \neq b$, we have
$x_{0}=1$ or $a b-a-b-2=0$.
If $x_{0}=1$, substituting into equation (3) yields $a=1$, which is a contradiction.
Therefore, $x_{0} \neq 1$.
Hence, $a b-a-b-2=0$
$\Rightarrow(a-1)(b-1)=3$
$\Rightarrow a=2, b=4$ or $a=4, b=2$.
Substituting into the required algebraic expression, we get
$$
\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=\frac{2^{4}+4^{2}}{2^{-4}+4^{-2}}=256 .
$$
|
256
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. When $s$ and $t$ take all real values,
$$
(s+7-|\cos t|)^{2}+(s-2|\sin t|)^{2}
$$
the minimum value is $\qquad$
|
6. 18
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "6. 18" is already in a numerical form that is universal, it does not require translation. Here is the retained format:
6. 18
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) It is known that $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 9 and common difference 7.
(1) Prove: The sequence $\left\{a_{n}\right\}$ contains infinitely many perfect squares;
(2) The 100th perfect square in the sequence $\left\{a_{n}\right\}$ is the nth term?
|
(1) The general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=9+7(n-1) \text {. }
$$
When $n=7 k^{2} \pm 6 k+1(k=0,1, \cdots)$, we have
$$
\begin{array}{l}
a_{n}=9+7(n-1) \\
=9+7\left(7 k^{2} \pm 6 k\right)=(7 k \pm 3)^{2} .
\end{array}
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has infinitely many terms that are perfect squares.
(2) Let $a_{n}=9+7(n-1)=m^{2}(m$ is a positive integer). Then
$$
7(n-1)=(m-3)(m+3) \text {. }
$$
Thus, $7 \mid (m-3)(m+3)$.
Therefore, $7 \mid (m-3)$ or $7 \mid (m+3)$.
Furthermore, when $7 \mid (m-3)$ or $7 \mid (m+3)$, let $m \pm 3 = 7 k$. Then
$$
n=7 k^{2} \mp 6 k+1 \text {. }
$$
From (1), we know that at this time, $a_{n}$ is a perfect square. Hence, $a_{n}$ is a perfect square if and only if $n=7 k^{2} \mp 6 k+1(k=0,1, \cdots)$.
Since $7(k+1)^{2}-6(k+1)+1$
$$
>7 k^{2}+6 k+1 \text {, }
$$
the 100th perfect square in the sequence $\left\{a_{n}\right\}$ is the 17201st term of the sequence $\left\{a_{n}\right\}$, which is $a_{17201}=347^{2}$.
|
17201
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the equations
$$
x^{2}+b x+1=0 \text { and } x^{2}-x-b=0
$$
have a common root, find the value of $b$.
|
Prompt: Example 1. Use the substitution method to find the common root \( x_{0}=-1 \).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The foci of the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ are $F_{1}$ and $F_{2}$. If a point $P$ on the ellipse makes $P F_{1} \perp P F_{2}$, then the area of $\triangle P F_{1} F_{2}$ is $\qquad$
|
2. 9 .
It is known that $F_{1} F_{2}=8, P F_{1}+P F_{2}=10$.
Then $\left(P F_{1}+P F_{2}\right)^{2}=10^{2}$.
In the right triangle $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}^{2}+P F_{2}^{2}=8^{2} \text{. }
$$
From equations (1) and (2), we get
$$
S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=9 .
$$
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $4n+1$ and $6n+1$ are both perfect squares, then the smallest positive integer $n$ is $\qquad$
|
4. 20 .
Obviously, $4 n+1, 6 n+1$ are both odd square numbers.
Let $6 n+1=(2 m+1)^{2}=4 m(m+1)+1$.
Then $3 n=2 m(m+1)$.
Since $m(m+1)$ is even, $4 \mid n$.
Let $n=4 k$. Then
$4 n+1=16 k+1, 6 n+1=24 k+1$.
When $k=1,2,3,4$, $4 n+1, 6 n+1$ are not both square numbers, but when $k=5$, i.e., $n=20$, $4 n+1=81$, $6 n+1=121$ are both square numbers.
Therefore, the smallest positive integer $n$ is 20.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If the four digits of the four-digit number $\overline{a b c d}$ satisfy $a+b=c+d$, then it is called a "good number" (for example, 2011 is a good number). Then, the number of good numbers is $\qquad$
|
8. 615 .
Let $k=a+b=c+d$.
Since $1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant 9$, then
$1 \leqslant k \leqslant 18$.
When $1 \leqslant k \leqslant 9$, in the above equation, $a$ can take any value in $\{1,2$, $\cdots, k\}$, $c$ can take any value in $\{0,1, \cdots, k\}$, and once $a 、 c$ are determined, $b 、 d$ are also determined. Therefore, the number of four-digit numbers $\overline{a b c d}$ that satisfy the condition is $k(k+1)$.
When $10 \leqslant k \leqslant 18$, $a 、 b 、 c 、 d$ cannot be 0.
$$
\begin{array}{l}
\text { Let } a_{1}=10-a, b_{1}=10-b, \\
c_{1}=10-c, d_{1}=10-d .
\end{array}
$$
Then $1 \leqslant a_{1} 、 b_{1} 、 c_{1} 、 d_{1} \leqslant 9$.
Let $k_{1}=a_{1}+b_{1}=c_{1}+d_{1}$. Then $2 \leqslant k_{1} \leqslant 10$, and
the four-digit number $\overline{a b c d}$ corresponds one-to-one with the four-digit number $\overline{a_{1} b_{1} c_{1} d_{1}}$.
In the above equation, $a_{1}$ and $c_{1}$ can take any value in $\left\{1,2, \cdots, k_{1}-1\right\}$, and once $a_{1} 、 c_{1}$ are determined, $b_{1} 、 d_{1}$ are also determined.
Therefore, the number of four-digit numbers $\overline{a_{1} b_{1} c_{1} d_{1}}$ that satisfy $k_{1}=a_{1}+b_{1}=c_{1}+d_{1}$ is $\left(k_{1}-1\right)^{2}$.
Thus, the total number of good numbers is
$$
\begin{array}{l}
\sum_{k=1}^{9} k(k+1)+\sum_{k_{1}=2}^{10}\left(k_{1}-1\right)^{2} \\
=330+285=615 .
\end{array}
$$
|
615
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given an arithmetic sequence $\left\{a_{n}\right\}$, the sum of the first 15 terms $S_{15}=30$. Then $a_{1}+a_{8}+a_{15}=$ $\qquad$
|
12.6.
From $S_{15}=30 \Rightarrow a_{1}+7 d=2$.
Therefore, $a_{1}+a_{8}+a_{15}=3\left(a_{1}+7 d\right)=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let $x, y$ be real numbers. Then
$$
\max _{S x^{2}+4 y^{2}=10 x}\left(x^{2}+y^{2}\right)=
$$
$\qquad$
|
15. 4 .
$$
\begin{array}{l}
\text { Given } 5 x^{2}+4 y^{2}=10 x \\
\Rightarrow 4 y^{2}=10 x-5 x^{2} \geqslant 0 \\
\Rightarrow 0 \leqslant x \leqslant 2 .
\end{array}
$$
Then $4\left(x^{2}+y^{2}\right)=10 x-x^{2}$
$$
\begin{array}{l}
=25-(5-x)^{2} \leqslant 25-3^{2} \\
\Rightarrow x^{2}+y^{2} \leqslant 4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: There are $n$ people registered to participate in four sports events: A, B, C, and D. It is stipulated that each person must participate in at least one event and at most two events, but events B and C cannot be registered for simultaneously. If in all different registration methods, there must be at least one method where at least 20 people register, then the minimum value of $n$ is ( ).
(A) 171
(B) 172
(C) 180
(D) 181
|
Let the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{T}}\right)$ represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is 1 (for example, if participating in event 甲, then $a_{\text {甲 }}=1$), otherwise, the corresponding number is 0.
Thus, each person has 9 possible ways to register for the events:
$$
\begin{array}{l}
(1,0,0,0),(0,1,0,0),(0,0,1,0), \\
(0,0,0,1),(1,1,0,0),(1,0,1,0), \\
(1,0,0,1),(0,1,0,1),(0,0,1,1) .
\end{array}
$$
Therefore, for $n$ people, there are 9 possible ways to register for the events, which can be considered as 9 drawers.
By the pigeonhole principle, when
$$
n=19 \times 9+r(r \geqslant 1)
$$
there must be at least one way in which at least 20 people have registered.
So, when $r=1$, $n$ takes the minimum value
$$
19 \times 9+1=172 \text {. }
$$
Hence, the answer is B.
|
172
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $M=\{1,2,3,4,5\}$. Then the number of mappings $f: M \rightarrow M$ such that
$$
f(f(x))=f(x)
$$
is $\qquad$
|
7. 196.
For the case where $M$ contains $n$ elements, from
$$
f(f(x))=f(x) \text {, }
$$
we know that for any $a \in M$ (let $f(c)=a$), we have
$$
f(f(c))=f(a)=f(c)=a .
$$
If the range of $f$ contains $k$ elements, then the $n-k$ elements of set $M$ that are not in the range have $k^{n-k}$ possible mappings. Therefore, the number of such mappings $f$ is $\mathrm{C}_{n}^{k} k^{n-k}$.
Thus, the number of mappings $f$ that satisfy the condition is
$$
\sum_{k=1}^{n} \mathrm{C}_{n}^{k} k^{n-k} \text {. }
$$
Taking $n=5$, the number of mappings $f: M \rightarrow M$ is
196.
|
196
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Find
$$
f(x)=|x-1|+2|x-2|+\cdots+2011|x-2011|
$$
the minimum value.
|
10. Obviously, when $x=2011$, $f(x)$ has no minimum value.
The following assumes $x \in[1,2011]$.
When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2010)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2011} i(i-x) \\
=\left(k^{2}+k-2011 \times 1006\right) x+ \\
\quad \frac{2011 \times 2012 \times 4023}{6}- \\
\quad \frac{k(k+1)(2 k+1)}{3}
\end{array}
$$
is a linear function, and its minimum value is attained at the endpoints of the interval $[k, k+1]$.
Therefore, the minimum term of the sequence $\{f(k)\}(1 \leqslant k \leqslant 2010)$ is the desired value.
Notice that,
$$
\begin{array}{l}
f(k+1)>f(k) \\
\Leftrightarrow k^{2}+k-2011 \times 1006>0 \\
\Leftrightarrow k \geqslant 1422 . \\
\text { Hence } f(x)_{\min }=f(1422) \\
=1422\left(1422^{2}+1422-2011 \times 1006\right)+ \\
\frac{2011 \times 2012 \times 4023}{6}-\frac{1422 \times 1423 \times 2845}{3} \\
=794598996 .
\end{array}
$$
|
794598996
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\},
$$
where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedron. Question: What is the maximum number of triangles in graph $G$?
|
Four, first prove a lemma.
Lemma If in a space graph with $n$ points there is no triangle, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$ ( $[x]$ represents the largest integer not exceeding the real number $x$).
Proof Let the $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, among which, the number of edges drawn from point $A_{1}$ is the most (let's say there are $k$ edges) as $A_{1} A_{n}, A_{1} A_{n-1}, \cdots, A_{1} A_{n-k+1}$.
Since there is no triangle, there are no edges between the points $A_{n}, A_{n-1}, \cdots, A_{n-k+1}$.
Thus, each edge in the space graph has at least one endpoint among the points $A_{1}, A_{2}, \cdots, A_{n-k}$, and each $A_{i}(1 \leqslant i \leqslant n-k)$ can draw at most $k$ edges.
Therefore, the total number of edges is less than or equal to $k(n-k) \leqslant\left[\frac{n^{2}}{4}\right]$.
Back to the original problem.
Next, prove: If graph $G$ already has (at least) 28 triangles, then there is at least one tetrahedron.
Otherwise, in the 9-point set $M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\}$, by the pigeonhole principle, there must be a point that is at least
$$
\left[\frac{28 \times 3}{9}\right]+1=10
$$
vertices of triangles. Thus, at least 5 edges are drawn from this point (let this point be $A_{1}$).
(1) If 5 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{6}$ are drawn from point $A_{1}$, according to the problem, since there is no tetrahedron, the subgraph formed by the 5 points $A_{2}, A_{3}, \cdots, A_{6}$ has no triangle. By the lemma, this subgraph has at most $\left[\frac{5^{2}}{4}\right]=6$ edges. Thus, the number of triangles with $A_{1}$ as a vertex is at most 6, a contradiction.
(2) If 6 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{7}$ are drawn from point $A_{1}$, similarly to (1), the number of triangles with $A_{1}$ as a vertex is at most $\left[\frac{6^{2}}{4}\right]=9$, a contradiction.
(3) If 7 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{8}$ are drawn from point $A_{1}$, since there is no tetrahedron, the subgraph formed by the 7 points $A_{2}, A_{3}, \cdots, A_{8}$ has no triangle. This subgraph has at most $\left[\frac{7^{2}}{4}\right]=12$ edges. Thus, the number of triangles with $A_{1}$ as a vertex is at most 12, and the triangles not having $A_{1}$ as a vertex must have point $A_{9}$ as a vertex. Similarly, there are at most 12 such triangles. Therefore, the total number of triangles is less than or equal to $12 \times 2=24<28$, a contradiction.
(4) If 8 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{9}$ are drawn from point $A_{1}$, the subgraph formed by the 8 points $A_{2}, A_{3}, \cdots, A_{9}$ has no triangle. This subgraph has at most $\left[\frac{8^{2}}{4}\right]=16$ edges. Hence, the graph $G$ has at most 16 triangles, a contradiction.
Thus, the number of triangles satisfying the condition is at most 27.
Divide the 9-point set $M$ into three groups
$$
\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}, A_{6}\right\},\left\{A_{7}, A_{8}, A_{9}\right\},
$$
such that no two points in the same group are connected, while any two points in different groups are connected, thus, there are $C_{3}^{1} C_{3}^{1} C_{3}^{1}=27$ triangles, of course, there is no tetrahedron.
|
27
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given a set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$
Then the number of complex numbers in $D$ whose real and imaginary parts are both integers is $\qquad$ [s]
|
Solve: Establish a complex plane, let
$$
z=x+y \mathrm{i}(x, y \in \mathbf{R}) \text {. }
$$
From $\left|z_{1}\right|=1$, we know $z_{1}^{2}$ is also on the unit circle $\odot 0$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \\
=\left|z_{1}^{2}-1\right|^{2}
\end{array}
$$
This represents the square of the distance from a point on the unit circle $\odot O$ to the point $(1,0)$ (as shown in Figure 7), with a value range of $[0,4]$.
$$
\begin{array}{l}
\text { Therefore }|z-2005-2006 \mathrm{i}| \leqslant 4 \\
\Rightarrow(x-2005)^{2}+(y-2006)^{2} \leqslant 16,
\end{array}
$$
The number of integer points (both coordinates are integers) is the same as the number of integer points in
$$
x^{2}+y^{2} \leqslant 16
$$
which is 49.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given that $x, y, z$ are real numbers, satisfying
$$
\begin{array}{c}
x=\sqrt{y^{2}-\frac{1}{16}}+\sqrt{z^{2}-\frac{1}{16}}, \\
y=\sqrt{z^{2}-\frac{1}{25}}+\sqrt{x^{2}-\frac{1}{25}}, \\
z=\sqrt{x^{2}-\frac{1}{36}}+\sqrt{y^{2}-\frac{1}{36}},
\end{array}
$$
and $x+y+z=\frac{m}{\sqrt{n}}\left(m, n \in \mathbf{Z}_{+}, \sqrt{n}\right.$ is the simplest radical form). Find $m+n$.
|
Solve as shown in Figure 11, construct an acute triangle $\triangle ABC$, such that
$$
\begin{array}{l}
BC=x, \\
CA=y, \\
AB=z .
\end{array}
$$
Draw perpendiculars from $A$, $B$, and $C$ to $BC$, $CA$, and $AB$ respectively, with the feet of the perpendiculars being $D$, $E$, and $F$. Let $AD=u$, $BE=v$, and $CF=w$. Then
$$
\begin{array}{l}
x=\sqrt{y^{2}-u^{2}}+\sqrt{z^{2}-u^{2}}, \\
y=\sqrt{z^{2}-v^{2}}+\sqrt{x^{2}-v^{2}}, \\
z=\sqrt{x^{2}-w^{2}}+\sqrt{y^{2}-w^{2}} .
\end{array}
$$
Consider the function
$$
f(u)=\sqrt{y^{2}-u^{2}}+\sqrt{z^{2}-u^{2}}.
$$
It is easy to see that $f(u)$ is strictly decreasing on $(0,+\infty)$. Hence $u=\frac{1}{4}$.
Similarly, $v=\frac{1}{5}$, $w=\frac{1}{6}$.
Let the area of $\triangle ABC$ be $S$. Then
$$
S=\frac{1}{2} AB \cdot CF=\frac{1}{2} BC \cdot AD=\frac{1}{2} CA \cdot BE,
$$
which means
$$
\begin{array}{l}
2 S=z w=x u=y v \\
\Rightarrow x=8 S, y=10 S, z=12 S \\
\Rightarrow x+y+z=30 S .
\end{array}
$$
By Heron's formula,
$$
\begin{array}{l}
S^{2}=15 S(15 S-8 S)(15 S-10 S)(15 S-12 S) \\
\Rightarrow S=\frac{1}{15 \sqrt{7}} \Rightarrow x+y+z=\frac{2}{\sqrt{7}} \\
\Rightarrow m+n=9 .
\end{array}
$$
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, 32006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find
$$
a_{1}+a_{2}+\cdots+a_{2000}
$$
the minimum value. ${ }^{[1]}$
|
First, we prove: For any 119 positive integers \( b_{1}, b_{2}, \cdots, b_{119} \), there must exist a subset (at least one, or all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
\[ b_{1}, b_{1}+b_{2}, b_{1}+b_{2}+b_{3}, \cdots, b_{1}+b_{2}+\cdots+b_{119}. \]
If one of them is a multiple of 119, then the conclusion holds.
If none of them is a multiple of 119, then the remainders when they are divided by 119 can only be \( 1, 2, \cdots, 118 \), which are 118 different cases.
By the pigeonhole principle, we know that there must be two of them that have the same remainder when divided by 119. Let's assume these two are
\[ b_{1}+b_{2}+\cdots+b_{i} \]
and \( b_{1}+b_{2}+\cdots+b_{j} (1 \leqslant i < j \leqslant 119) \).
Then, \( 119 \mid \left(b_{i+1}+b_{i+2}+\cdots+b_{j}\right) \).
Next, for any 119 numbers among \( a_{1}, a_{2}, \cdots, a_{2006} \), by the previous proof, we know that there must be a subset of these numbers whose sum (let's denote it as \( x \)) is a multiple of 119.
By the problem's condition, we know that \( x \) is not equal to 119, so \( x \) is at least \( 2 \times 119 \).
Since \( 2006 = 16 \times 119 + 102 \), we have
\[ \begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2006} \\
\geqslant 16 \times 238 + 102 = 3910.
\end{array} \]
If we set \( a_{119}=a_{238}=\cdots=a_{1904}=120 \), and the rest of the numbers are 1, then the equality holds.
Therefore, the minimum value of \( a_{1}+a_{2}+\cdots+a_{2006} \) is 3910.
|
3910
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 In the positive integers not exceeding 2012, take $n$ numbers arbitrarily, so that there are definitely two numbers whose ratio is within the interval $\left[\frac{2}{3} ; \frac{3}{2}\right]$. Try to find the minimum value of $n$.
|
Solve for
$$
\begin{array}{l}
\{1,2,4,7,11,17,26,40,61,92,139, \\
209,314,472,709,1064,1597\}
\end{array}
$$
When any two numbers are taken, their ratio is either greater than $\frac{3}{2}$ or less than $\frac{2}{3}$.
Therefore, $n \geqslant 18$.
Next, we prove: When 18 numbers are taken, there exist two numbers whose ratio lies within the interval $\left[\frac{2}{3}, \frac{3}{2}\right]$.
Divide $1,2, \cdots, 2012$ into the following 17 sets:
$\{1\},\{2,3\},\{4,5,6\},\{7,8,9,10\}$,
$\{11,12, \cdots, 16\},\{17,18, \cdots, 25\}$,
$\{26,27, \cdots, 39\},\{40,41, \cdots, 60\}$,
$\{61,62, \cdots, 91\},\{92,93, \cdots, 138\}$,
$\{139,140, \cdots, 208\},\{209,210, \cdots, 313\}$,
$\{314,315, \cdots, 471\},\{472,473, \cdots, 708\}$,
$\{709,710, \cdots, 1063\}$,
$\{1064,1065, \cdots, 1596\}$,
$\{1597,1598, \cdots, 2012\}$.
In each of these 17 sets, the ratio $k$ of any two numbers satisfies
$$
\frac{2}{3} \leqslant k \leqslant \frac{3}{2} \text {. }
$$
By the pigeonhole principle, we know that from $1,2, \cdots, 2012$, if 18 numbers are arbitrarily selected, there must be two numbers from the same set, and their ratio lies within $\left[\frac{2}{3}, \frac{3}{2}\right]$.
In summary, the minimum value of $n$ is 18.
|
18
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given two equations about $x$
$$
x^{2}-x+3 m=0, x^{2}+x+m=0(m \neq 0) \text {. }
$$
If one root of the first equation is three times a root of the second equation, then the value of the real number $m$ is $\qquad$
|
- 1. -2 .
Let one root of the latter equation be $\alpha$. And the former equation has a root $3 \alpha$, then
$$
\alpha^{2}+\alpha+m=0,
$$
and
$$
\begin{array}{l}
9 \alpha^{2}-3 \alpha+3 m=0 \\
\Rightarrow 3 \alpha^{2}-\alpha+m=0 .
\end{array}
$$
(2) - (1) gives
$$
2 \alpha^{2}-2 \alpha=0 \Rightarrow \alpha=0 \text { or } 1 \text {. }
$$
If $\alpha=0$, then $m=0$, which contradicts the assumption that $m \neq 0$. Therefore, $\alpha=1$. Consequently, $m=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange the eight numbers $-7, -5, -3, -2, 2, 4, 6, 13$ as $a, b, c, d, e, f, g, h$, such that
$$
(a+b+c+d)^{2}+(e+f+g+h)^{2}
$$
is minimized. Then this minimum value is $\qquad$
|
4.34.
Let $x=a+b+c+d$. Then
$$
\begin{array}{l}
e+f+g+h=8-x, \\
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
=x^{2}+(8-x)^{2}=2(x-4)^{2}+32 .
\end{array}
$$
From the known eight numbers, the sum of any four numbers $x$ is an integer and cannot be 4, so we get
$$
\begin{array}{l}
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
\geqslant 2+32=34,
\end{array}
$$
and when $x=(-3)+(-2)+4+6=5$, the equality holds.
Therefore, the minimum value sought is 34.
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. There are 10 chess players participating in a round-robin tournament (i.e., each pair of players competes in one match). The rules state that a win earns 2 points, a draw earns 1 point for each player, and a loss earns 0 points. After the tournament, it is found that each player's score is unique, and the second-place player's score is $\frac{4}{5}$ of the sum of the scores of the last five players. What is the score of the second-place player? $\qquad$
|
7. 16 .
The last five contestants have to play 10 matches, so the total score of the last five contestants is greater than or equal to $2 \times 10=20$.
Thus, the score of the second-place contestant is greater than or equal to
$$
\frac{4}{5} \times 20=16 \text {. }
$$
The score of the first-place contestant is less than or equal to $2 \times 9=18$.
If the second-place contestant scores 17 points (i.e., 8 wins and 1 draw), then the match between the first and second-place contestants must be a draw. This way, both the first and second-place contestants would score 17 points, which contradicts the rule that each contestant has a unique score.
Therefore, the second-place contestant scores 16 points.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $a b c d$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ .
|
8. 22 .
$$
\begin{array}{l}
\text { Let } a b c d=k+(k+1)+\cdots+(k+34) \\
=35(k+17),
\end{array}
$$
where $k$ is a positive integer. Then the prime numbers $a, b, c, d$ must include one 5 and one 7.
Assume $a=5, b=7$. Then $c d=k+17$.
When $k \geqslant 8$,
$$
c+d \geqslant 2 \sqrt{c d}=2 \sqrt{k+17} \geqslant 2 \sqrt{8+17}=10 \text {. }
$$
When $1 \leqslant k \leqslant 7$, there are only two cases where $k+17$ can be written as the product of two primes:
$k=4$ when, $k+17=21=3 \times 7$. In this case,
$c+d=10$;
$k=5$ when, $k+17=22=2 \times 11$. In this case,
$$
c+d=13 \text {. }
$$
In summary, the minimum value of $a+b+c+d$ is
$$
5+7+3+7=22
$$
|
22
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given a positive integer $n$ that satisfies the following condition: In any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$. ${ }^{[2]}$
|
Since $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor not exceeding 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17$,
$19,23,29,31,37,41,43$.
On one hand,
$2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}$,
$19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$
are 14 positive integers greater than 1 and not exceeding 2009 that are pairwise coprime, and none of them are prime.
Therefore, $n \geqslant 15$.
On the other hand, let $a_{1}, a_{2}, \cdots, a_{15}$ be 15 positive integers greater than 1 and not exceeding 2009 that are pairwise coprime.
If they are all composite, let $p_{i}$ be the prime factor of $a_{i}$ not exceeding 44 $(i=1,2, \cdots, 15)$.
By the pigeonhole principle, there exist $i, j (1 \leqslant i<j \leqslant 15)$ such that $p_{i}=p_{j}$.
Thus, $a_{i}$ and $a_{j}$ are not coprime, which is a contradiction.
Therefore, among $a_{1}, a_{2}, \cdots, a_{15}$, there exists a prime number.
In conclusion, the minimum value of $n$ is 15.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The quadratic trinomial
$$
x^{2}+a x+b(a 、 b \in \mathbf{N})
$$
has real roots, and $a b=2^{2011}$. Then the number of such quadratic trinomials is $\qquad$.
|
$-1.1341$.
It is known that the numbers $a$ and $b$ are powers of 2 with non-negative integer exponents, i.e., $a=2^{k}, b=2^{2011-k}$. Therefore, we have
$$
\begin{array}{l}
\Delta=a^{2}-4 b \geqslant 0 \\
\Rightarrow 2 k \geqslant 2013-k \Rightarrow k \geqslant \frac{2013}{3}=671 .
\end{array}
$$
But $k \leqslant 2011$, so $k$ can take
$$
2011-671+1=1341
$$
different integer values.
Each $k$ corresponds exactly to one of the required quadratic trinomials, hence there are 1341 such quadratic trinomials.
|
1341
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (15 points) On a plane, $n$ points are called a "standard $n$-point set" if among any three of these points, there are always two points whose distance is no more than 1. To ensure that a circular paper with a radius of 1 can cover at least 25 points of any standard $n$-point set, find the minimum value of $n$.
|
First, prove: $n_{\text {min }}>48$.
Draw a line segment $AB$ of length 5 on the plane, and construct two circles with radii of 0.5 centered at $A$ and $B$, respectively. Take 24 points in each circle. Then there are 48 points on the plane that satisfy the problem's condition (any three points must have at least two points with a distance no greater than 1).
Obviously, it is impossible to construct a circle with a radius of 1 that contains 25 of the selected points.
Therefore, $n_{\text {min }}>48$.
Next, prove: $n_{\text {min }}=49$.
If $n=49$, let $A$ be one of the points. Construct a circle $\odot A$ with a radius of 1. If all the points are within $\odot A$, then the condition of the problem is satisfied.
Otherwise, there is at least one point $B$ not in $\odot A$. Construct another circle $\odot B$ with a radius of 1. Then the distance between points $A$ and $B$ is greater than 1 (as shown in Figure 4).
Thus, for the remaining 47 points, each point $P$ forms a triplet with $A$ and $B$, and it must be true that $PA \leqslant 1$ or $PB \leqslant 1$, meaning point $P$ is either in $\odot A$ or in $\odot B$.
According to the pigeonhole principle, one of the circles must contain at least 24 of these 47 points (let's assume it is $\odot A$). Adding the center point $A$, there are at least 25 points in the circle $\odot A$ with a radius of 1 (inside or on the circumference).
Therefore, the minimum value of $n$ is 49.
|
49
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (15 points) Given a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$ we have
$$
\begin{array}{l}
\frac{1}{2} f(x y)+\frac{1}{2} f(x z)-f(x) f(y z) \geqslant \frac{1}{4} . \\
\text { Find }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)]
\end{array}
$$
where $[a]$ denotes the greatest integer not exceeding the real number $a$.
|
$$
f(1)=\frac{1}{2} \text {. }
$$
Let $y=z=0$. Then
$$
-\frac{1}{2} f(0)+\frac{1}{2} f(0)-f(x) f(0) \geqslant \frac{1}{4} \text {. }
$$
Substituting $f(0)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \leqslant \frac{1}{2} \text {. }
$$
Now let $y=z=1$. Then
$$
\frac{1}{2} f(x)+\frac{1}{2} f(x)-f(x) f(1) \geqslant \frac{1}{4} \text {. }
$$
Substituting $f(1)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \geqslant \frac{1}{2} \text {. }
$$
Combining (1) and (2), for any real number $x$,
$$
f(x)=\frac{1}{2} \text {. }
$$
Upon verification, $f(x)=\frac{1}{2}$ satisfies the given conditions.
$$
\begin{array}{l}
\text { Therefore, }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)] \\
=\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\cdots+\left[\frac{2011}{2}\right] \\
=0+1+1+\cdots+1005+1005 \\
=2(1+2+\cdots+1005) \\
=(1+1005) \times 1005 \\
=1011030 .
\end{array}
$$
Five, let $x=y=z=0$. Then
$$
\begin{array}{l}
\frac{1}{2} f(0)+\frac{1}{2} f(0)-f^{2}(0) \geqslant \frac{1}{4} \\
\Rightarrow\left(f(0)-\frac{1}{2}\right)^{2} \leqslant 0 \\
\Rightarrow f(0)=\frac{1}{2} .
\end{array}
$$
Now let $x=y=z=1$. Then similarly we have
|
1011030
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $A$ and $B$ be two sets, and call $(A, B)$ a "pair". When $A \neq B$, consider $(A, B)$ and $(B, A)$ as different pairs. Then the number of different pairs $(A, B)$ that satisfy the condition
$$
A \cup B=\{1,2,3,4\}
$$
is $\qquad$
|
2. 81.
When set $A$ has no elements, i.e., $A=\varnothing$, set $B$ has 4 elements, there is 1 case; when set $A$ contains $k(k=1,2,3,4)$ elements, set $B$ contains the other $4-k$ elements besides these $k$ elements, the elements in set $A$ may or may not be in set $B$, there are $\mathrm{C}_{4}^{k} \times 2^{k}$ cases.
In summary, the total number of different pairs is
$$
1+\sum_{k=1}^{4} \mathrm{C}_{4}^{k} \times 2^{k}=81
$$
|
81
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation
$$
a_{n+1}=2 a_{n}+2^{n}-1\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
and $\left\{\frac{a_{n}+\lambda}{2^{n}}\right\}$ is an arithmetic sequence. Then the value of $\lambda$ is $\qquad$
|
6. -1 .
Notice,
$$
\begin{array}{l}
\frac{a_{n+1}+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2 a_{n}+2^{n}-1+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2^{n}-1-\lambda}{2^{n+1}} . \\
\text { From } \frac{2^{n}-1-\lambda}{2^{n+1}} \text { being a constant, we know } \lambda=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given three points $A, B, C$ in a plane satisfying $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=4,|\overrightarrow{C A}|=5$.
Then $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}=$ $\qquad$
|
8. -25 .
Given that $\overrightarrow{A B} \perp \overrightarrow{B C}$. Then
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B} \\
=\overrightarrow{C A} \cdot(\overrightarrow{A B}+\overrightarrow{B C})=-\overrightarrow{C A}^{2}=-25
\end{array}
$$
|
-25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Fill in each small square of a $9 \times 9$ grid with a number, with no more than four different numbers in each row and each column. What is the maximum number of different numbers that can be in this grid? ${ }^{[4]}$
|
If there are 29 different numbers in this grid, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is the first row).
The remaining 25 numbers are in rows $2 \sim 9$. Similarly, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is the second row).
Next, observe the numbers in each column of the grid.
Each column already has two different numbers at the top. Therefore, each column below the second row can have at most two different numbers. This way, the maximum number of different numbers is $8 + 9 \times 2 = 26$, which is a contradiction.
Therefore, the number of different numbers is less than 29.
Table 1 constructs a $9 \times 9$ grid containing 28 different numbers, with each row and column having exactly four different numbers.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. There are 8 red, 8 white, and 8 yellow chopsticks. Without looking, how many chopsticks must be taken out to ensure that at least two pairs of chopsticks are of different colors? $\qquad$
|
10. 11.
Since among 11 chopsticks there must be a pair of the same color (let's say yellow), the number of black or white chopsticks must be at least 3, among which there must be a pair of the same color, i.e., both black or both white. Therefore, 11 chopsticks ensure success. However, if only 10 chopsticks are taken, it is possible to have 8 yellow, 1 black, and 1 white, which does not meet the requirement.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ (1)
|
Solve As shown in Figure 1, the figure determined by inequality (1) is the quadrilateral $\square A B C D$ and its interior, enclosed by four straight lines, where,
$$
\begin{array}{l}
A(-1,4), B(1,1), \\
C(-1,-2), D(-3,1) .
\end{array}
$$
Consider the family of lines $2 x-3 y=k$, i.e.,
$$
y=\frac{2}{3} x-\frac{k}{3} \text {, }
$$
which has the smallest intercept $-\frac{4}{3}$ at the lowest point $C$ of $\triangle A B C D$.
Therefore, when $x=-1, y=-2$, $2 x-3 y$ attains its maximum value of 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the sum of the first $n$ terms of the real geometric sequence $\left\{a_{n}\right\}$ be
$S_{n}$. If $S_{10}=10, S_{30}=70$, then $S_{40}=$ $\qquad$ .
|
9. 150 .
$$
\begin{array}{l}
\text { Let } b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, \\
b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30} .
\end{array}
$$
Let $q$ be the common ratio of $\left\{a_{n}\right\}$. Then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with common ratio $r=q^{10}$. Therefore,
$$
\begin{array}{l}
70=S_{30}=b_{1}+b_{2}+b_{3} \\
=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right) .
\end{array}
$$
Thus, $r^{2}+r-6=0$.
Solving, we get $r=2$ or $r=-3$ (discard).
Hence, $S_{40}=10\left(1+r+r^{2}+r^{3}\right)=150$.
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $x$ be a real number, and define $\lceil x\rceil$ as the smallest integer not less than the real number $x$ (for example, $\lceil 3.2 \rceil = 4, \lceil -\pi \rceil = -3$). Then, the sum of all real roots of the equation
$$
\lceil 3 x+1\rceil=2 x-\frac{1}{2}
$$
is equal to
|
10. -4 .
Let $2 x-\frac{1}{2}=k \in \mathbf{Z}$. Then
$$
x=\frac{2 k+1}{4}, 3 x+1=k+1+\frac{2 k+3}{4} \text {. }
$$
Thus, the original equation is equivalent to
$$
\begin{array}{l}
{\left[\frac{2 k+3}{4}\right]=-1 \Rightarrow-2<\frac{2 k+3}{4} \leqslant-1} \\
\Rightarrow-\frac{11}{2}<k \leqslant-\frac{7}{2} \\
\Rightarrow k=-5 \text { or }-4 .
\end{array}
$$
The corresponding $x$ values are $-\frac{9}{4},-\frac{7}{4}$.
Therefore, the sum of all real roots is -4.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$.
|
Three, 13. Let $t=\sqrt{100-m x}$. Then $x=\frac{100-t^{2}}{m}$.
Hence $y=\frac{100-t^{2}}{m}+t$
$$
=-\frac{1}{m}\left(t-\frac{m}{2}\right)^{2}+\frac{100}{m}+\frac{m}{4} \text {. }
$$
When $t=\frac{m}{2}$, $y$ has a maximum value $\frac{100}{m}+\frac{m}{4}$, that is,
$$
g(m)=\frac{100}{m}+\frac{m}{4} \geqslant 2 \sqrt{\frac{100}{m} \times \frac{m}{4}}=10 .
$$
Therefore, when and only when $m=20$, $g(m)$ has a minimum value of 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given the function
$$
f(x)=2\left(\sin ^{4} x+\cos ^{4} x\right)+m(\sin x+\cos x)
$$
has a maximum value of 5 for $x \in\left[0, \frac{\pi}{2}\right]$. Find the value of the real number $m$.
|
14. Notice,
$$
f(x)=2-(2 \sin x \cdot \cos x)^{2}+m(\sin x+\cos x)^{4} \text {. }
$$
Let $t=\sin x+\cos x$
$$
=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \in[1, \sqrt{2}] \text {. }
$$
Thus, $2 \sin x \cdot \cos x=t^{2}-1$.
Therefore, $f(x)=2-\left(t^{2}-1\right)^{2}+m t^{4}$
$=(m-1) t^{4}+2 t^{2}+1$.
Let $u=t^{2} \in[1,2]$.
By the problem, $g(u)=(m-1) u^{2}+2 u+1$ has a maximum value of 5 for $u \in[1,2]$.
When $m-1=0$,
$$
g(u)=2 u+1 \leqslant g(2)=5,
$$
Thus, $m=1$ satisfies the condition;
When $m-1>0$,
$$
g(u)_{\max } \geqslant g(2)>2 \times 2+1=5 \text {, }
$$
which is a contradiction;
When $m-1<0$,
$$
g(u)<2 u+1 \leqslant 5 \text {, }
$$
which is a contradiction.
In conclusion, the real number $m=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{aligned}
M= & |2012 x-1|+|2012 x-2|+\cdots+ \\
& |2012 x-2012|
\end{aligned}
$$
The minimum value of the algebraic expression is . $\qquad$
|
2. 1012036 .
By the geometric meaning of absolute value, we know that when
$$
1006 \leqslant 2012 x \leqslant 1007
$$
$M$ has the minimum value.
$$
\begin{array}{l}
\text { Then } M_{\text {min }}=(-1-2-\cdots-1006)+ \\
(1007+1008+\cdots+2012) \\
=1006 \times 1006=1012036 .
\end{array}
$$
|
1012036
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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