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7. In $\triangle A B C$, it is known that
$$
\begin{array}{l}
|\overrightarrow{A B}|=\sqrt{3},|\overrightarrow{B C}|=1, \\
|\overrightarrow{A C}| \cos B=|\overrightarrow{B C}| \cos A \text {. } \\
\text { Then } \overrightarrow{A C} \cdot \overrightarrow{A B}=
\end{array}
$$
|
7.2.
Let the three sides of $\triangle A B C$ be $a, b, c$. From the given condition and using the Law of Sines, we have
$$
\begin{array}{l}
\sin B \cdot \cos B=\sin A \cdot \cos A \\
\Rightarrow \sin 2 B=\sin 2 A \\
\Rightarrow \angle B=\angle A \text { or } \angle B+\angle A=90^{\circ} .
\end{array}
$$
We will discuss two cases.
(1) If $\angle B=\angle A$, then
$$
b=a=1, c=\sqrt{3}, \angle A=30^{\circ} \text {. }
$$
Thus, $\overrightarrow{A C} \cdot \overrightarrow{A B}=b c \cos A=1 \times \sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}$.
(2) If $\angle B+\angle A=90^{\circ}$, then
$$
b=\sqrt{2}, c=\sqrt{3}, \cos A=\frac{\sqrt{2}}{\sqrt{3}} \text {. }
$$
Thus, $\overrightarrow{A C} \cdot \overrightarrow{A B}=b c \cos A=\sqrt{2} \times \sqrt{3} \times \frac{\sqrt{2}}{\sqrt{3}}=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Ed and Ann have lunch together, Ed and Ann order medium and large lemonade drinks, respectively. It is known that the large size has a capacity 50% more than the medium size. After both have drunk $\frac{3}{4}$ of their respective drinks, Ann gives the remaining $\frac{1}{3}$ plus 2 ounces to Ed. After lunch, they find that they have drunk the same amount of lemonade. Then the total amount of lemonade they drank is ( ) ounces.
(A) 30
(B) 32
(C) 36
(D) 40
(E) 50
|
6. D.
Let the size of the medium lemon drink be $x$ ounces. According to the problem, we have
$$
x+\frac{3}{2} \times \frac{1}{4} \times \frac{1}{3} x+2=\frac{3}{2} x-\frac{3}{2} \times \frac{1}{4} \times \frac{1}{3} x-2 \text {. }
$$
Solving for $x$, we get $x=16$.
Therefore, the total amount they drank is $16+\frac{3}{2} \times 16=40$ ounces.
|
40
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
24. Let pentagon $A B C D E$ be a cyclic pentagon, $A B=$ $C D=3, B C=D E=10, A E=14$, the sum of the lengths of all diagonals is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$. Then $m+n=$ ( ).
(A) 129
(B) 247
(C) 353
(D) 391
(E) 421
|
24. D.
As shown in Figure 3, from the given conditions, we know that quadrilaterals $ABCD$ and $BCDE$ are both isosceles trapezoids.
Thus,
$$
\begin{array}{c}
AC=BD=CE. \\
\text{Let } AC=BD=CE \\
=x, AD=y, BE=z.
\end{array}
$$
By Ptolemy's theorem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}=10 y+9, \\
x^{2}=3 z+100, \\
y z=14 x+30
\end{array}\right. \\
\Rightarrow \frac{x^{2}-9}{10} \times \frac{x^{2}-100}{3}=14 x+30 \\
\Rightarrow x^{3}-109 x-420=0 \\
\Rightarrow(x-12)(x+5)(x+7)=0 \\
\Rightarrow x_{1}=12, x_{2}=-5, x_{3}=-7.
\end{array}
$$
Since $x>0$, then $x=12$.
Thus, $y=\frac{27}{2}, z=\frac{44}{3}, 3 x+y+z=\frac{385}{6}$.
Therefore, $m+n=385+6=391$.
|
391
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given integers $a, b, c, d$. If the roots of the equation
$$
z^{4}+a z^{3}+b z^{2}+c z+d=0
$$
correspond to four points $A, B, C, D$ forming the vertices of a square in the complex plane, then the minimum value of the area of square $A B C D$ is $\qquad$
|
7. 2 .
Let the complex number corresponding to the center of the square be $m$. Then, after translating the origin of the complex plane to $m$, the vertices of the square are distributed on a circle, i.e., they are the solutions to the equation $(z-m)^{4}=n$ (where $n$ is a complex number).
From $z^{4}+a z^{3}+b z^{2}+c z+d=(z-m)^{4}-n$, comparing coefficients we know that $m=-\frac{a}{4}$ is a rational number.
Furthermore, from $-4 m^{3}=c$, we know that $m$ is an integer.
Then, from $d=m^{4}-n$, we know that $n$ is an integer.
Thus, the roots of the equation $(z-m)^{4}=n$ are
$$
z_{k}=m+\sqrt[4]{|n|}\left(\cos \frac{k \pi}{2}+\mathrm{i} \sin \frac{k \pi}{2}\right) .
$$
Therefore, the length of the diagonal of the square is $2 \sqrt[4]{|n|}$, and its area is
$$
2 \sqrt{|n|} \geqslant 2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Let $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{Z}_{+}\right)$. Find the smallest positive real number $\lambda$, such that for any $n \geqslant 2$, we have
$$
a_{n}^{2}<\lambda \sum_{k=1}^{n} \frac{a_{k}}{k} .
$$
|
Sure, here is the translated text:
```
9. Notice that,
\[
\begin{array}{l}
a_{k}^{2}-a_{k-1}^{2}=\left(a_{k}-a_{k-1}\right)\left(a_{k}+a_{k-1}\right) \\
=\frac{1}{k}\left(2 a_{k}-\frac{1}{k}\right)(k \geqslant 2) .
\end{array}
\]
Then \(a_{n}^{2}-a_{1}^{2}=\sum_{k=2}^{n}\left(a_{k}^{2}-a_{k-1}^{2}\right)=2 \sum_{k=2}^{n} \frac{a_{k}}{k}-\sum_{k=2}^{n} \frac{1}{k^{2}}\)
\[
\begin{array}{l}
\Rightarrow a_{n}^{2}=2 \sum_{k=1}^{n} \frac{a_{k}}{k}-\sum_{k=1}^{n} \frac{1}{k^{2}} \\
\Rightarrow a_{n}^{2}\sum_{k=1}^{n} \frac{1}{k}.
\]
Taking \(n=2^{m}\), thus, when \(m\) is sufficiently large, we have
\[
\begin{array}{l}
\sum_{k=1}^{n} \frac{1}{k}=1+\sum_{i=0}^{m-1} \sum_{i=1}^{2^{i}} \frac{1}{2^{t}+s} \\
>1+\sum_{i=0}^{m-1} 2^{t} \times \frac{1}{2^{t+1}} \\
=1+\frac{m}{2}>\frac{2}{2-\lambda} .
\end{array}
\]
Therefore, equation (1) does not hold.
Thus, \(\lambda \geqslant 2\).
In conclusion, \(\lambda_{\min }=2\).
```
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given that $AB$ is the major axis of the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and $CD$ is a chord of the ellipse $\Gamma$. The tangents at points $C$ and $D$ intersect at point $P$, the extension of $AD$ intersects the extension of $CB$ at point $E$, and the extension of $AC$ intersects the extension of $DB$ at point $F$. If $E$, $P$, and $F$ are collinear, find $\frac{EP}{PF}$.
|
10. As shown in Figure 3, let the center of the ellipse be $O$. Connect $O P$, intersecting $C D$ at point $M$.
Let $C\left(x_{1}, y_{1}\right), D\left(x_{2}, y_{2}\right), P\left(x_{0}, y_{0}\right)$.
Then the equation of line $C D$ is $\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1$.
From $\left\{\begin{array}{l}\frac{x_{0} x}{a^{2}}+\frac{y_{0} y}{b^{2}}=1, \\ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\end{array}\right.$ we get
$$
\left(\frac{x_{0}^{2}}{a^{4}}+\frac{y_{0}^{2}}{a^{2} b^{2}}\right) x^{2}-\frac{2 x_{0}}{a^{2}} x+1-\frac{y_{0}^{2}}{b^{2}}=0 \text {. }
$$
By Vieta's formulas, the coordinates of the midpoint $N$ of $C D$ are
$$
x_{N}=\frac{x_{1}+x_{2}}{2}=\frac{x_{0}}{\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}}, y_{N}=\frac{y_{0}}{\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}} .
$$
Thus, $k_{O N}=\frac{y_{0}}{x_{0}}=k_{O P}$.
Therefore, points $O, N, P$ are collinear.
Hence, point $N$ coincides with $M$.
By Newton's theorem, $\frac{E P}{P F}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Natural numbers $a, b$ make $4a + 7b$ and $5a + 6b$ both multiples of 11, and $a + 21b \geq 792$. Find the minimum value of $T = 21a + b$.
|
11. Let $4a + 7b = 11x$, $5a + 6b = 11y$. Then
$$
\begin{array}{l}
a = 7y - 6x \geqslant 0, \\
b = 5x - 4y \geqslant 0.
\end{array}
$$
Thus, $T = 11(13y - 11x)$.
Let $13y - 11x = r$.
From $a + 21b \geqslant 792$, we get
$$
9x - 7y \geqslant 72.
$$
Substituting equation (3) into equation (1) to eliminate $y$ gives $x \leqslant 7r$.
(1) + (4) gives $x \geqslant 24$.
Thus, $7r \geqslant 24 \Rightarrow r \geqslant 4$.
Therefore, $T \geqslant 44$.
On the other hand, when $x = 28$, $y = 24$, we have $a = 0$, $b = 44$, $a + 21b = 924 > 792$, and $T = 44$. In summary, $T_{\min} = 44$.
|
44
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1.2014 Arrange chairs in a circle, with $n$ people sitting on the chairs, such that when one more person sits down, they will always sit next to one of the original $n$ people. Then the minimum value of $n$ is $\qquad$
|
-、1. 672 .
From the problem, we know that after $n$ people sit down, there are at most two empty chairs between any two people.
If we can arrange for there to be exactly two empty chairs between any two people, then $n$ is minimized.
Thus, if we number the chairs where people are sitting, we easily get the arithmetic sequence: $1,4,7, \cdots, 2014$.
Therefore, $2014=1+3(n-1) \Rightarrow n=672$.
|
672
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. It is known that 99 wise men are seated around a large round table, each wearing a hat of one of two different colors. Among them, 50 people's hats are of the same color, and the remaining 49 people's hats are of the other color. However, they do not know in advance which 50 people have the same color and which 49 people have the other color. They can only see the colors of the hats on others' heads, but not their own. Now they are required to simultaneously write down the color of their own hat on the paper in front of them. Question: Can they pre-arrange a strategy to ensure that at least 74 of them write the correct color?
|
8. Sure.
Suppose there are 50 white hats and 49 black hats. Those who see 50 white hats and 48 black hats are all wearing black hats. Thus, they can all write down the correct color, and there are 49 such people.
Next, let those who see 49 white hats and 49 black hats follow this strategy:
Observe which color of hat is more prevalent among the next 49 people in the clockwise direction, and write down that color.
Assume $A$ is such a person. And $B$ is the 25th person wearing a white hat after $A$ in the clockwise direction. If there are no more than 48 people between them, $A$ will write “white” on his paper. Otherwise, $B$ will write “white” on his paper.
Since the 50 people wearing white hats can be divided into 25 such $(A, B)$ pairs, in each pair, one person will write the correct color. Therefore, adding the 49 people wearing black hats, a total of $25 + 49 = 74$ people can write the correct color.
|
74
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Each point on a circle is colored one of three colors: red, yellow, or blue, and all three colors appear. Now, $n$ points are chosen from the circle. If among these points, there always exist three points that form a triangle with vertices of the same color and an obtuse angle, then the minimum possible value of $n$ is $\qquad$ .
|
6.13.
First, we state that \( n \geqslant 13 \).
If \( n \geqslant 13 \), then by the pigeonhole principle, among these \( n \) points, there must be \(\left[\frac{13-1}{3}\right]+1=5\) points of the same color (let's assume they are red).
Take a diameter, whose endpoints do not belong to these five points.
By the pigeonhole principle, among these five red points, there are always three on the same side of this diameter, and the triangle formed by these three points must be an acute triangle.
When \( n=12 \), consider the following scenario: select three inscribed squares \( A_{1} A_{2} A_{3} A_{4} \), square \( B_{1} B_{2} B_{3} B_{4} \), and square \( C_{1} C_{2} C_{3} C_{4} \). The vertices of these three squares are respectively red, yellow, and blue. In this case, all triangles with three vertices of the same color are right triangles. Therefore, \( n=12 \) does not meet the requirement.
In summary, the smallest possible value of \( n \) is 13.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For $I$ being the excenter of $\triangle ABC$ with respect to $\angle A$, prove:
$$
\frac{A I^{2}}{C A \cdot A B}-\frac{B I^{2}}{A B \cdot B C}-\frac{C I^{2}}{B C \cdot C A}=1 .
$$
|
Proof As shown in Figure 5, let $\odot I$ be tangent to $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$ respectively.
$$
\begin{array}{l}
\text { Let } \angle A I F=\alpha, \\
\angle B I D=\beta, \\
\angle C I E=\gamma .
\end{array}
$$
Then $\alpha=\beta+\gamma$.
Assume $I D=I E=I F=1$. Then $A I^{2}=\frac{1}{\cos ^{2} \alpha}$,
$$
\begin{array}{l}
A B=A F-B F=\tan \alpha-\tan \beta, \\
A C=A E-C E=\tan \alpha-\tan \gamma . \\
\text { Therefore, } \frac{A I^{2}}{C A \cdot A B}=\frac{1}{\cos ^{2} \alpha(\tan \alpha-\tan \beta)(\tan \alpha-\tan \gamma)} \\
=\cos ^{2} \alpha \cdot \frac{\sin (\alpha-\beta)}{\cos \alpha \cdot \cos \beta} \cdot \frac{\sin (\alpha-\gamma)}{\cos \alpha \cdot \cos \gamma} \\
=\frac{1}{\tan \beta \cdot \tan \gamma} .
\end{array}
$$
Similarly, $\frac{B I^{2}}{A B \cdot B C}=\frac{1}{\tan \gamma \cdot \tan \alpha}$,
$$
\begin{array}{l}
\frac{C I^{2}}{B C \cdot C A}=\frac{1}{\tan \alpha \cdot \tan \beta} . \\
\text { Therefore, } \frac{A I^{2}}{C A \cdot A B}-\frac{B I^{2}}{A B \cdot B C}-\frac{C I^{2}}{B C \cdot C A} \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{1}{\tan \gamma \cdot \tan \alpha}-\frac{1}{\tan \alpha \cdot \tan \beta} \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{1}{\tan \alpha} \cdot \frac{\tan \beta+\tan \gamma}{\tan \beta \cdot \tan \gamma} \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{\tan (\beta+\gamma)}{\tan \alpha \cdot \tan \beta \cdot \tan \gamma}(1-\tan \beta \cdot \tan \gamma) \\
=\frac{1}{\tan \beta \cdot \tan \gamma}-\frac{1}{\tan \beta \cdot \tan \gamma}(1-\tan \beta \cdot \tan \gamma) \\
=1 .
\end{array}
$$
|
1
|
Geometry
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the largest positive integer $x$, such that for every positive integer $y$, we have $x \mid\left(7^{y}+12 y-1\right)$.
|
When $y=1$, $7^{y}+12 y-1=18$.
Let 18 I $\left(7^{y}+12 y-1\right)$. Notice that,
$$
\begin{array}{l}
7^{y+1}+12(y+1)-1 \\
=6 \times\left(7^{y}+2\right)+\left(7^{y}+12 y-1\right) .
\end{array}
$$
Since $7^{y}+2 \equiv 1+2 \equiv 0(\bmod 3)$, therefore,
$$
18 \mid\left[7^{y+1}+12(y+1)-1\right] \text {. }
$$
Thus, for any positive integer $y, 18 \mid\left(7^{y}+12 y-1\right)$.
Hence, the required maximum positive integer $x=18$.
|
18
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Find the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{2012}\right]$.
(Adapted from the 25th IMO Preliminary Question)
|
Let $a_{n}=(\sqrt{29}+\sqrt{21})^{2 n}+(\sqrt{29}-\sqrt{21})^{2 n}$
$$
=(50+2 \sqrt{609})^{n}+(50-2 \sqrt{609})^{n} \text {. }
$$
Then the characteristic equation of the sequence $\left\{a_{n}\right\}$ is
$$
x^{2}-100 x+64=0 \text {. }
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has the recurrence relation
$$
a_{n+2}=100 a_{n+1}-64 a_{n}(n=0,1, \cdots) \text {, }
$$
and $a_{0}=2, a_{1}=100$.
By mathematical induction, we know that for any $(n=0,1, \cdots), a_{n}$ is an integer.
$$
\begin{array}{l}
\text { Also, } 0<(\sqrt{29}-\sqrt{21})^{2 n}<1, \text { thus, } \\
{\left[(\sqrt{29}+\sqrt{21})^{2 n}\right]=a_{n}-1 .} \\
\text { Hence } a_{n} \equiv-64 a_{n-2} \equiv 6^{2} a_{n-2} \equiv 6^{4} a_{n-4} \\
\equiv \cdots \equiv 6^{n} a_{0}=2 \times 6^{n}(\bmod 100) .
\end{array}
$$
Since $6^{4}=1296 \equiv-4(\bmod 100)$, we have
$$
\begin{array}{l}
a_{1006} \equiv 2(-4)^{252} \equiv 2^{505} \\
\equiv\left(2^{24}\right)^{21} \times 2(\bmod 100) . \\
\text { Also, } 2^{12}=4096 \equiv-4(\bmod 100), \\
2^{24} \equiv 16 \equiv 2^{4}(\bmod 100), \\
\text { thus } a_{1006} \equiv\left(2^{4}\right)^{21} \times 2=2^{85}=\left(2^{24}\right)^{3} \times 2^{13} \\
\equiv 2^{25} \equiv 2^{5} \equiv 32(\bmod 100),
\end{array}
$$
which means the last two digits of $a_{1000}$ are $32$, and the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{2 n}\right]=a_{1006}-1$ are 31.
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 3 Given $x, y \in(0,1)$, and $3x+7y$, $5x+y$ are both integers. Then there are $\qquad$ pairs of $(x, y)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
【Analysis】Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v .\end{array}\right.$
We can first determine the integer values of $u$ and $v$ based on $x, y \in(0,1)$.
Solution Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v,\end{array}\right.$ where the possible values of $u$ and $v$ are $u \in\{1,2, \cdots, 9\}, v \in\{1,2, \cdots, 5\}$.
Thus, $\left\{\begin{array}{l}x=\frac{7 v-u}{32}, \\ y=\frac{5 u-3 v}{32} .\end{array}\right.$
Since $x, y \in(0,1)$, we have
$$
\left\{\begin{array}{l}
0<7 v-u<32, \\
0<5 u-3 v<32 .
\end{array}\right.
$$
Upon verification, when $v=1$, $u=1,2,3,4,5,6$;
when $v=2$, $u=2,3,4,5,6,7$;
when $v=3$, $u=2,3,4,5,6,7,8$;
when $v=4$, $u=3,4,5,6,7,8$;
when $v=5$, $u=4,5,6,7,8,9$.
In summary, there are 31 pairs of $(u, v)$ that satisfy the conditions, which means there are 31 pairs of $(x, y)$ that satisfy the conditions.
|
31
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Below is the inscription by the famous Chinese mathematician, Academician Wang Yuan:
数棈妻好
If different Chinese characters represent different digits from $0 \sim 9$, and assuming “数学竞赛好” represents the largest five-digit number that is a perfect square, formed by different digits. Then this five-digit number is $\qquad$
|
Notice that, $300^{2}=90000,310^{2}=96100$,
$$
320^{2}=102400>100000 \text {. }
$$
Thus, this five-digit number with distinct digits might be among the nine square numbers between $310^{2} \sim$ $320^{2}$.
$$
\begin{array}{l}
\text { Also, } 311^{2}=96721,312^{2}=97344,313^{2}=97969, \\
314^{2}=98596,315^{2}=99225,316^{2}=99856, \\
317^{2}=100489>100000,
\end{array}
$$
Therefore, the largest square number among the five-digit numbers with distinct digits is 96721.
|
96721
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Try to determine the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$
|
4. 3 .
Notice that,
$$
\begin{array}{l}
\frac{\sqrt{14}+2}{\sqrt{14}-2}=\frac{(\sqrt{14}+2)^{2}}{(\sqrt{14}-2)(\sqrt{14}+2)} \\
=\frac{14+4+4 \sqrt{14}}{14-4}=\frac{18+4 \sqrt{14}}{10} .
\end{array}
$$
And $3<\sqrt{14}<4$, thus, $30<18+4 \sqrt{14}<34$.
Therefore, $3<\frac{18+4 \sqrt{14}}{10}<3.4$.
Hence, the largest integer not exceeding $\frac{\sqrt{14}+2}{\sqrt{14}-2}$ is $=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. On each face of a cube, write a positive integer, and at each vertex, write the product of the positive integers on the three faces meeting at that vertex. If the sum of the numbers written at the eight vertices is 2014, then the sum of the numbers written on the six faces is $\qquad$ .
|
6. 74 .
As shown in Figure 4, let the pairs of positive integers written on the opposite faces of the cube be $(a, b),(c, d),(e, f)$.
Then the sum of the numbers written at each vertex is
$$
\begin{array}{l}
a c f+a d f+a c e+a d e+b c f+b c e+b d f+b d e \\
=(a+b)(c+d)(e+f) \\
=2014=2 \times 19 \times 53 .
\end{array}
$$
Thus, the sum of the numbers written on each face is
$$
\begin{array}{l}
(a+b)+(c+d)+(e+f) \\
=2+19+53=74 .
\end{array}
$$
|
74
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $A, B$ are digits in the set $\{0,1, \cdots, 9\}$, $r$ is a two-digit integer $\overline{A B}$, $s$ is a two-digit integer $\overline{B A}$, $r, s \in\{00,01$, $\cdots, 99\}$. When $|r-s|=k^{2}$ ( $k$ is an integer), the number of ordered pairs $(A, B)$ is $\qquad$.
|
7. 42 .
Notice,
$$
|(10 A+B)-(10 B+A)|=9|A-B|=k^{2} \text {. }
$$
Then $|A-B|$ is a perfect square.
When $|A-B|=0$, there are 10 integer pairs:
$$
(A, B)=(0,0),(1,1), \cdots,(9,9) \text {; }
$$
When $|A-B|=1$, there are 18 integer pairs:
$$
(A, B)=(0,1),(1,2), \cdots,(8,9) \text {, }
$$
and their reverse numbers;
When $|A-B|=4$, there are 12 integer pairs:
$$
(A, B)=(0,4),(1,5), \cdots,(5,9) \text {, }
$$
and their reverse numbers;
When $|A-B|=9$, there are 2 integer pairs:
$$
(A, B)=(0,9) \text {, }
$$
and its reverse number.
Thus, the number of ordered pairs that satisfy the condition is
$$
10+18+12+2=42 \text {. }
$$
|
42
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $a, b$ be real numbers, for any real number $x$ satisfying $0 \leqslant x \leqslant 1$ we have $|a x+b| \leqslant 1$. Then the maximum value of $|20 a+14 b|+|20 a-14 b|$ is . $\qquad$
|
9.80 .
Let $x=0$, we know $|b| \leqslant 1$;
Let $x=1$, we know $|a+b| \leqslant 1$.
Therefore, $|a|=|a+b-b| \leqslant|a+b|+|b| \leqslant 2$, when $a=2, b=-1$, the equality can be achieved.
If $|20 a| \geqslant|14 b|$, then
$$
\begin{array}{l}
|20 a+14 b|+|20 a-14 b| \\
=2|20 a|=40|a| \leqslant 80 ;
\end{array}
$$
If $|20 a| \leqslant|14 b|$, then
$$
\begin{array}{l}
|20 a+14 b|+|20 a-14 b| \\
=2|14 b|=28|b| \leqslant 28 .
\end{array}
$$
In summary, the maximum value sought is 80, which can be achieved when $a=2, b=-1$.
|
80
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given that vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ are two mutually perpendicular unit vectors in a plane, and
$$
(3 \alpha-\gamma) \cdot(4 \beta-\gamma)=0 .
$$
Then the maximum value of $|\boldsymbol{\gamma}|$ is . $\qquad$
|
11.5.
As shown in Figure 2, let
$$
\begin{array}{l}
\overrightarrow{O A}=3 \alpha, \\
\overrightarrow{O B}=4 \beta, \\
\overrightarrow{O C}=\gamma .
\end{array}
$$
From the given information, $\overrightarrow{A C} \perp \overrightarrow{B C}$. Therefore, point $C$ lies on the circle with $A B$ as its diameter, and this circle passes through the origin.
Thus, the maximum value of $|\overrightarrow{O C}|$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given $k$ as a positive integer, the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=3, a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
Let $b_{n}=\frac{1}{n} \log _{3} a_{1} a_{2} \cdots a_{n}\left(n \in \mathbf{Z}_{+}\right)$, and define
$$
T_{k}=\sum_{i=1}^{2 k}\left|b_{i}-\frac{3}{2}\right| \text {. }
$$
If $T_{k} \in \mathbf{Z}_{+}$, find all possible values of $k$.
|
15. From the problem, we know
Also, $a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3$,
$$
a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n-1}+3(n \geqslant 2) \text {, }
$$
Thus, $a_{n+1}-a_{n}=\left(3^{\frac{2}{2 k-1}}-1\right) a_{n}$
$\Rightarrow a_{n+1}=3^{\frac{2}{2 k-1}} a_{n}$
$$
\Rightarrow a_{n}=a_{2}\left(3^{\frac{2}{2 k-1}}\right)^{n-2}=3^{\frac{2 n+2 k-3}{2 k-1}}(n \geqslant 2) \text {. }
$$
Clearly, $n=1$ also fits.
Therefore, the general term formula for the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=3^{\frac{2 n+2 k-3}{2 k-1}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Thus, $b_{n}=\frac{1}{n} \log _{3} a_{1} a_{2} \cdots a_{n}$
$$
\begin{array}{l}
=\frac{1}{n} \cdot \frac{1}{2 k-1} \sum_{i=1}^{n}(2 i+2 k-3) \\
=1+\frac{n-1}{2 k-1} .
\end{array}
$$
Therefore, $b_{n}-\frac{3}{2}=\frac{n-\left(k+\frac{1}{2}\right)}{2 k-1}$.
Thus, when $n \leqslant k$, $b_{n}-\frac{3}{2}<0$; when $n \geqslant k+1$, $b_{n}-\frac{3}{2}>0$.
$$
\begin{array}{l}
\text { Then } T_{k}=\sum_{i=1}^{2 k}\left|b_{i}-\frac{3}{2}\right| \\
=\sum_{i=1}^{k}\left(\frac{3}{2}-b_{i}\right)+\sum_{i=k+1}^{2 k}\left(b_{i}-\frac{3}{2}\right) \\
=\frac{k^{2}}{2 k-1} .
\end{array}
$$
Since $T_{k} \in \mathbf{Z}_{+}$, i.e., $(2 k-1) \mid k^{2}$, we have
$$
(2 k-1) \mid\left(4 k^{2}-1+1\right) \text {. }
$$
Thus, $(2 k-1) \mid 1$.
Hence, $2 k-1=1 \Rightarrow k=1$.
Therefore, the only possible value of $k$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The value of the complex number $\left(\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{6 n}\left(n \in \mathbf{Z}_{+}\right)$ is
|
3. 1 .
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{6 n}=\left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{3}\right]^{2 n} \\
=\left(\frac{1}{8}+\frac{3}{4} \times \frac{\sqrt{3}}{2} i-\frac{3}{2} \times \frac{3}{4}-\frac{3 \sqrt{3}}{8} i\right) \\
=(-1)^{2 n}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $(\sqrt{2}+1)^{21}=a+b \sqrt{2}$, where $a, b$ are positive integers. Then $(b, 27)=$ $\qquad$
|
8. 1 .
Notice,
$$
\begin{array}{l}
(\sqrt{2}+1)^{21} \\
=(\sqrt{2})^{21}+\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}+\cdots+ \\
\mathrm{C}_{21}^{20} \sqrt{2}+\mathrm{C}_{21}^{21}, \\
(\sqrt{2}-1)^{21} \\
=(\sqrt{2})^{21}-\mathrm{C}_{21}^{1}(\sqrt{2})^{20}+\mathrm{C}_{21}^{2}(\sqrt{2})^{19}-\cdots+ \\
\quad \mathrm{C}_{2 \mathrm{~N}}^{20}-\mathrm{C}_{21}^{21} .
\end{array}
$$
Therefore, $(\sqrt{2}-1)^{21}=-a+b \sqrt{2}$.
Thus, $2 b^{2}-a^{2}=1$.
If $b \equiv 0(\bmod 3)$, then $a^{2} \equiv-1(\bmod 3)$.
This contradicts $a^{2} \equiv 0,1(\bmod 3)$.
Therefore, $(27, b)=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1.200 people stand in a circle, some of whom are honest people, and some are liars. Liars always tell lies, while honest people tell the truth depending on the situation. If both of his neighbors are honest people, he will definitely tell the truth; if at least one of his neighbors is a liar, he may sometimes tell the truth and sometimes lie. It is known that among these 200 people, 100 people say they are liars, and 100 people say they are honest. How many honest people can there be at most in these 200 people?
|
1. There can be at most 150 honest people.
Since liars do not say they are liars, the 100 people who say they are liars are all honest people, and what they say is false. This indicates that they are all adjacent to liars.
Since each liar is adjacent to at most two honest people, there must be at least 50 liars. This indicates that the number of honest people will not exceed 150.
Next, let's provide an example.
150 honest people are divided into 50 groups, with 3 honest people standing together in each group, and one liar between each group.
The honest person in the middle of each group tells the truth, saying they are honest, while the honest people at the ends of each group lie, saying they are liars, and the 50 liars all say they are honest.
|
150
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A $3 \times 3$ grid with the following properties is called a "T-grid":
(1) Five cells are filled with 1, and four cells are filled with 0;
(2) Among the three rows, three columns, and two diagonals, at most one of these eight lines has three numbers that are pairwise equal.
Then the number of different T-grids is $\qquad$.
|
8. 68.
First, the number of all ways to fill a $3 \times 3$ grid with five 1s and four 0s is $\mathrm{C}_{9}^{4}=126$.
Next, consider the number of ways that do not satisfy property (2), i.e., methods that result in at least two lines of three equal numbers (hereafter referred to as good lines).
We will count these by categorizing them. Good lines can be rows, columns, or diagonals.
If the two good lines are both diagonals, then they are all 1s, with 1 method;
If the two good lines are both rows (or columns), then one row is filled with 0s and one row is filled with 1s, there are $\mathrm{C}_{3}^{2} \times 2 \times \mathrm{C}_{3}^{1}=18$ methods, by the symmetry of rows and columns, there are $2 \times 18=36$ methods;
If the two good lines are one row and one column, in this case, the good lines are all 1s, there are $3 \times 3=9$ methods;
If the two good lines are one row and one diagonal, in this case, the good lines are all 1s, there are $3 \times 2=6$ methods;
If the two good lines are one column and one diagonal, in this case, the good lines are all 1s, there are $3 \times 2=6$ methods.
Therefore, the number of methods that satisfy the property is
$$
126-1-36-9-6-6=68 \text {. }
$$
|
68
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 3, two equilateral triangles overlap to form a six-pointed star. Now, the positive integers 1, 2, $\cdots, 12$ are to be filled in the 12 nodes of the figure, such that the sum of the four numbers on each straight line is equal.
(1) Try to find the minimum value of the sum of the numbers at the six vertices $a_{1}, a_{2}, \cdots, a_{6}$ of the six-pointed star;
(2) Prove: The number of different filling schemes that meet the conditions is even (if filling scheme $K$ can coincide with filling scheme $T$ after rotation or reflection, they are considered the same scheme).
|
(1) Solution: For any arrangement that satisfies the conditions, the number filled at point $a_i$ $(i=1,2, \cdots, 12)$ is still denoted as $a_i$. If the sum of the four numbers on each line is $s$, then
$6 s=2(1+2+\cdots+12) \Rightarrow s=26$.
Therefore, in $\triangle a_{1} a_{3} a_{5}$ and $\triangle a_{2} a_{4} a_{6}$, the sums of the numbers on the three sides are respectively
$$
\begin{array}{l}
\left(a_{1}+a_{8}+a_{9}+a_{3}\right)+ \\
\left(a_{3}+a_{10}+a_{11}+a_{5}\right)+ \\
\left(a_{5}+a_{12}+a_{7}+a_{1}\right) \\
=3 s, \\
\left(a_{2}+a_{9}+a_{10}+a_{4}\right)+ \\
\left(a_{4}+a_{11}+a_{12}+a_{6}\right)+ \\
\left(a_{6}+a_{7}+a_{8}+a_{2}\right) \\
=3 s .
\end{array}
$$
Subtracting the two equations gives $a_{1}+a_{3}+a_{5}=a_{2}+a_{4}+a_{6}$.
Thus, the sum of the numbers at the vertices of the two triangles is equal.
If we denote $m=\left(a_{1}+a_{3}+a_{5}\right)+\left(a_{2}+a_{4}+a_{6}\right)$, then $m$ is an even number.
Because in $1,2, \cdots, 12$, the sum of the smallest six numbers is
$1+2+\cdots+6=21$ (an odd number),
so, $m \geqslant 22$.
If $m=22$, since in $1,2, \cdots, 12$, the only six numbers that sum to 22 are $1,2,3,4,5,7$, and dividing them into two groups with equal sums, there is only one way: $\{1,3,7\}$ and $\{2,4,5\}$.
At this point, if we directly attempt to construct or exclude, it will be too many cases to classify and exhaust, often leading to no good end.
Therefore, we reverse the problem (i.e., "reverse engineering") and use the "clustering" method.
Now, ignoring the specific numbers on the outer circle, and changing $\{1,3,7\}$ and $\{2,4,5\}$ to $\{$ odd, odd, odd $\}$ and $\{$ even, even, odd $\}$, as shown in Figure 4. Thus, under rotation, the way to fill the numbers at the six vertices is unique.
Next, consider the way to fill the remaining six numbers, using the notation $a \sim b$ to indicate that integers $a$ and $b$ have the same parity.
Since the sum of the four numbers on each line $s$ is even, then
$a_{9} \sim a_{8} \sim a_{7} \sim a_{12}, a_{10} \sim a_{11}$.
Among the six numbers in the inner circle $6,8,9,10,11,12$, there are exactly four even numbers and two odd numbers, so $\left\{a_{10}, a_{11}\right\}=\{9,11\}$.
Since $a_{10}$ and $a_{11}$ are on one side of the triangle with odd vertices, the sum of the numbers at the two vertices of that side should be $s-(9+11)=6$. However, there are no two numbers in $\{1,3,7\}$ that sum to 6, which is a contradiction. Therefore, $m \neq 22$, and thus, the even number $m \geqslant 24$.
When $m=24$, as shown in Figure 5, a valid filling method can be given.
Therefore, the minimum value sought is 24.
(2) Proof: For any filling scheme $K$ that satisfies the conditions, let the numbers filled at each point be $a_{1}, a_{2}, \cdots, a_{12}$.
Let $b_{i}=13-a_{i}(i=1,2, \cdots, 12)$.
Then when $a_{i} \in\{1,2, \cdots, 12\}$, $b_{i} \in\{1,2, \cdots, 12\}$.
In the filling scheme $K$, replace $a_{1}, a_{2}, \cdots, a_{12}$ with $b_{1}, b_{2}, \cdots, b_{12}$, to get the filling scheme $T$. Then $T$ is also a filling scheme that satisfies the conditions.
Obviously, $\sum_{i=1}^{6} a_{i}+\sum_{i=1}^{6} b_{i}=\sum_{i=1}^{6}\left(a_{i}+b_{i}\right)=78$.
And $\sum_{i=1}^{6} a_{i}$ and $\sum_{i=1}^{6} b_{i}$ are both even numbers, so
$\sum_{i=1}^{6} a_{i} \neq 39, \sum_{i=1}^{6} b_{i} \neq 39$.
Let the set of all filling schemes that satisfy the conditions be $M$, and divide $M$ into two disjoint subsets: $M=M_{1} \cup M_{2}$, where,
$M_{1}$ contains each filling scheme $K_{1}=\left(a_{1}, a_{2}, \cdots, a_{12}\right)$, the sum of the numbers at the six vertex points $\sum_{i=1}^{6} a_{i}39$.
Make the correspondence between the filling scheme $K_{1}=\left(a_{1}, a_{2}, \cdots, a_{12}\right)$ in set $M_{1}$ and the filling scheme $K_{2}=\left(b_{1}, b_{2}, \cdots, b_{12}\right)$ in set $M_{2}$, such that
$$
b_{i}=13-a_{i}(i=1,2, \cdots, 12) \text {, }
$$
This correspondence is one-to-one.
Therefore, $\left|M_{1}\right|=\left|M_{2}\right|, M_{1} \cap M_{2}=\varnothing$.
And $|M|=\left|M_{1} \cup M_{2}\right|=2\left|M_{1}\right|$, i.e., the number of elements in set $M$ is even, which means there are an even number of different filling schemes that satisfy the conditions.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Solve the equation
$$
x^{2}-x-1000 \sqrt{1+8000 x}=1000 .
$$
|
Solution: Clearly, $x \neq 0$. The original equation can be transformed into
$$
\begin{array}{l}
x^{2}-x=1000(1+\sqrt{1+8000 x}) \\
=\frac{8 \times 10^{6} x}{\sqrt{1+8000 x}-1} .
\end{array}
$$
Then $x-1=\frac{8 \times 10^{6}}{\sqrt{1+8000 x}-1}$.
Let $a=10^{3}, t=\sqrt{1+8 a x}-1$.
Thus, $x=\frac{t^{2}+2 t}{8 a}$, and we have $\frac{t^{2}+2 t}{8 a}-1=\frac{8 a^{2}}{t}$.
Rearranging gives $t^{3}+2 t^{2}-8 a t-64 a^{3}=0$.
Factoring yields
$$
(t-4 a)\left[t^{2}+(4 a+2) t+16 a^{2}\right]=0 \text {. }
$$
Clearly, $t^{2}+(4 a+2) t+16 a^{2}>0$.
Therefore, $t=4 a, x=\frac{t^{2}+2 t}{8 a}=2 a+1=2001$.
|
2001
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $a+b=\sqrt{5}$, then
$$
\frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b=(\quad) \text {. }
$$
(A) 5
(B) $\frac{3 \sqrt{5}}{2}$
(C) $2 \sqrt{5}$
(D) $\frac{5 \sqrt{5}}{2}$
|
$\begin{array}{l}\text {-1. A. } \\ \frac{a^{4}+a^{2} b^{2}+b^{4}}{a^{2}+a b+b^{2}}+3 a b \\ =\frac{\left(a^{2}+a b+b^{2}\right)\left(a^{2}-a b+b^{2}\right)}{a^{2}+a b+b^{2}}+3 a b \\ =\left(a^{2}-a b+b^{2}\right)+3 a b \\ =(a+b)^{2}=5 .\end{array}$
|
5
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Among the natural numbers from $1 \sim 10000$, the integers that are neither perfect squares nor perfect cubes are $\qquad$ in number.
|
In the natural numbers from $1 \sim 10000$, there are 100 perfect squares.
$$
\begin{array}{l}
\text { Because } 22^{3}=10648>10000, \\
21^{3}=9261<10000,
\end{array}
$$
Therefore, there are 21 perfect cubes.
Next, consider the number of natural numbers from $1 \sim 10000$ that are both perfect squares and perfect cubes (i.e., perfect sixth powers). Since $4^{6}=4096,5^{6}=15625$, there are 4 perfect sixth powers.
By the principle of inclusion-exclusion, the number of natural numbers from $1 \sim 10000$ that are neither perfect squares nor perfect cubes is
$$
10000-100-21+4=9883 .
$$
|
9883
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $M$ is the least common multiple of 15 consecutive natural numbers $1,2, \cdots, 15$. If a divisor of $M$ is divisible by exactly 14 of these 15 natural numbers, it is called a "good number" of $M$. Then the number of good numbers of $M$ is $\qquad$.
|
4.4.
It is known that $M=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$.
Since $2 \times 11, 2 \times 13$ are both greater than 15, therefore,
$$
\begin{array}{l}
\frac{M}{11}=2^{3} \times 3^{2} \times 5 \times 7 \times 13, \\
\frac{M}{13}=2^{3} \times 3^{2} \times 5 \times 7 \times 11,
\end{array}
$$
each is a divisor of $M$, and is exactly divisible by the other 14 numbers.
If the power of 2 in $M$ is reduced by 1, then
$$
\frac{M}{2}=2^{2} \times 3^{2} \times 5 \times 7 \times 11 \times 13
$$
is a divisor of $M$, and is exactly divisible by the 14 numbers except 8.
Similarly, if the power of 3 is reduced by 1, then
$$
\frac{M}{3}=2^{3} \times 3 \times 5 \times 7 \times 11 \times 13
$$
is a divisor of $M$, and is exactly divisible by the 14 numbers except 9;
if the power of 5 in $M$ is reduced by 1, then $\frac{M}{5}$ is a divisor of $M$ and is exactly divisible by the 12 numbers except 5, 10, and 15; if 7 is removed, then $\frac{M}{7}$ is a divisor of $M$ and is exactly divisible by the 13 numbers except 7 and 14.
In summary, there are four divisors of $M$ that are exactly divisible by 14 of the 15 natural numbers, namely $\frac{M}{2}, \frac{M}{3}, \frac{M}{11}, \frac{M}{13}$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the sequence of natural numbers from $1 \sim 8$ be $a_{1}, a_{2}$, $\cdots, a_{8}$. Then
$$
\begin{array}{l}
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+\left|a_{4}-a_{5}\right|^{\prime}+ \\
\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+\left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1}\right|^{\prime}
\end{array}
$$
The maximum value is $\qquad$
|
5. 32 .
From the problem, we have
$$
\begin{aligned}
S= & \left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\left|a_{3}-a_{4}\right|+ \\
& \left|a_{4}-a_{5}\right|+\left|a_{5}-a_{6}\right|+\left|a_{6}-a_{7}\right|+ \\
& \left|a_{7}-a_{8}\right|+\left|a_{8}-a_{1}\right| .
\end{aligned}
$$
Removing the absolute value signs in any term of this sum, we get one positive and one negative natural number. To achieve the maximum possible value of the sum, we need to take the numbers from 1 to 4 as negative and the numbers from 5 to 8 as positive. Thus,
$$
\begin{array}{l}
S=2[(8+7+6+5)-(4+3+2+1)]=32 \text {. } \\
\text { For example, }|8-4|+|4-7|+|7-1|+|1-5|+ \\
|5-2|+|2-6|+|6-3|+|3-8| \\
=32 .
\end{array}
$$
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Find all positive integers $n$, such that there exists an integer-coefficient polynomial $P(x)$, satisfying $P(d)=\left(\frac{n}{d}\right)^{2}$, where, $d$ is each divisor of $n$. [3]
|
For a polynomial $P(x)$ with integer coefficients, and positive integers $a, b (a \neq b)$, we always have
$$
(a-b) \mid (P(a)-P(b)).
$$
This is an invariant.
If $n=1$, then $P(1)=1$. Hence, we can take the polynomial $P(x)=x$.
If $n$ is a prime number, then $n$ has only two factors, 1 and $n$. The polynomial $P(x)$ must satisfy
$$
P(1)=n^{2}, P(n)=1.
$$
Let $P(x)=a x+b$.
From $P(1)=n^{2}, P(n)=1$, we get
$$
P(x)=-(n+1) x+n^{2}+n+1.
$$
If $n=k l(1<k \leqslant l)$ is a composite number, then
$$
\begin{array}{l}
P(1)=n^{2}, P(l)=k^{2}, P(k)=l^{2}, P(n)=1. \\
\text { By }(n-k) \mid (P(n)-P(k)) \\
\Rightarrow k(l-1) \mid (1-l)(1+l) \\
\Rightarrow k \mid (1+l).
\end{array}
$$
Similarly, by
$$
(n-l) \mid (P(n)-P(l)) \Rightarrow l \mid (1+k).
$$
Thus, $k l \mid (1+k)(1+l) \Rightarrow k l \mid (1+k+l)$
$$
\Rightarrow k l \leqslant 1+k+l \Rightarrow (k-1)(l-1) \leqslant 2.
$$
Solving this, we get $k=2, l=3$, i.e., $n=6$.
Then $P(1)=36, P(2)=9, P(3)=4, P(6)=1$.
Let $Q(x)=P(x)-1$.
Then $Q(1)=35, Q(2)=8, Q(3)=3, Q(6)=0$.
Let $Q(x)=(x-6) R(x)$. Then
$$
R(1)=-7, R(2)=-2, R(3)=-1.
$$
Let $R(x)=-1+(x-3) S(x)$. Then
$$
S(1)=3, S(2)=1.
$$
Thus, we find $S(x)=5-2 x$.
Therefore, $P(x)=1+(x-6)[-1+(x-3)(5-2 x)]$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, (15 points) A school assigns numbers to the contestants participating in a math competition, with the smallest number being 0001 and the largest number being 2014. No matter which contestant steps forward to calculate the average of the numbers of all other contestants in the school, the average is always an integer. How many contestants can the school have at most?
Will the above text be translated into English, preserving the original text's line breaks and format? Please output the translation directly.
|
Let the school have a total of $n$ participants, whose admission numbers are
$$
1=x_{1}<x_{2}<\cdots<x_{n-1}<x_{n}=2014 .
$$
According to the problem, we have
$$
S_{k}=\frac{x_{1}+x_{2}+\cdots+x_{n}-x_{k}}{n-1}(k=1,2, \cdots, n) \in \mathbf{Z}_{+} .
$$
For any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
S_{i}-S_{j}=\frac{x_{j}-x_{i}}{n-1} \in \mathbf{Z}_{+} \text {. }
$$
Thus, $x_{j}-x_{i} \geqslant n-1$.
$$
\begin{array}{l}
\text { Hence } x_{n}-x_{1} \\
=\left(x_{n}-x_{n-1}\right)+\left(x_{n-1}-x_{n-2}\right)+\cdots+\left(x_{2}-x_{1}\right) \\
\geqslant(n-1)^{2} \\
\Rightarrow(n-1)^{2} \leqslant x_{n}-x_{1}=2013 \Rightarrow n \leqslant 45 .
\end{array}
$$
Since $\frac{2014-1}{n-1}$ is an integer, it follows that $n-1$ is a divisor of 2013.
Noting that $2013=3 \times 11 \times 61$, the largest divisor not exceeding 45 is 33. Therefore, the maximum value of $n$ is 34, meaning the maximum number of participants is 34.
Such 34 participants' numbers can be realized. For example,
$$
x_{i}=33 i-32(i=1,2, \cdots, 33), x_{34}=2014 \text {. }
$$
Therefore, the maximum number of participants in the competition is 34.
|
34
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If positive real numbers $a, b$ satisfy
$$
\log _{8} a+\log _{4} b^{2}=5, \log _{8} b+\log _{4} a^{2}=7 \text {, }
$$
then $\log _{4} a+\log _{8} b=$ $\qquad$
|
$=1.4$.
Let $a=2^{x}, b=2^{y}$. Then $\frac{x}{3}+y=5, \frac{y}{3}+x=7$.
Thus, $x=6, y=3$.
Therefore, $\log _{4} a+\log _{8} b=\frac{x}{2}+\frac{y}{3}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}+a_{n-1}(n \geqslant 2) \text {. }
$$
If $a_{7}=8$, then $a_{1}+a_{2}+\cdots+a_{10}=$ $\qquad$
|
3.88.
From the problem, we know that $a_{7}=8 a_{2}+5 a_{1}$.
Therefore, $a_{1}+a_{2}+\cdots+a_{10}$
$$
=88 a_{2}+55 a_{1}=11 a_{7}=88 .
$$
|
88
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Arrange the numbers in the set $\left\{2^{x}+2^{y}+2^{z} \mid x 、 y 、 z \in \mathbf{N}, x<y<z\right\}$ in ascending order. The 100th number is $\qquad$ (answer with a number).
|
5.577.
Notice that the number of combinations $(x, y, z)$ such that $0 \leqslant x<y<z \leqslant n$ is $\mathrm{C}_{n+1}^{3}$.
Since $\mathrm{C}_{9}^{3}=84<100<120=\mathrm{C}_{10}^{3}$, the 100th number must satisfy $z=9$.
Also notice that the number of combinations $(x, y)$ such that $0 \leqslant x<y \leqslant m$ is $\mathrm{C}_{m+1}^{2}$.
Since $\mathrm{C}_{9}^{3}+\mathrm{C}_{6}^{2}=99$, the 100th number must satisfy $y=6, x=0$, which means the 100th number is
$$
2^{0}+2^{6}+2^{9}=577 .
$$
|
577
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the function $f(x)$ satisfies
$$
f(x)=\left\{\begin{array}{ll}
x-3, & x \geqslant 1000 ; \\
f(f(x+5)), & x<1000 .
\end{array}\right.
$$
Then $f(84)=$ . $\qquad$
|
6.997.
Let $f^{(n)}(x)=\underbrace{f(f(\cdots f(x)))}_{n \uparrow}$. Then
$$
\begin{aligned}
& f(84)=f(f(89))=\cdots=f^{(184)}(999) \\
= & f^{(185)}(1004)=f^{(184)}(1001)=f^{(183)}(998) \\
= & f^{(184)}(1003)=f^{(183)}(1000)=f^{(182)}(997) \\
= & f^{(183)}(1002)=f^{(182)}(999)=f^{(183)}(1004) \\
= & f^{(182)}(1001)=f^{(181)}(998)=f^{(182)}(1003) \\
= & f^{(181)}(1000)=\cdots=f(1000)=997 .
\end{aligned}
$$
Therefore, $f(84)=997$.
|
997
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the unit digit of $(2+\sqrt{3})^{2013}$.
|
Notice,
$$
\begin{array}{l}
{\left[(2+\sqrt{3})^{2013}\right]=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}-1 .} \\
\text { Let } a_{n}=(2+\sqrt{3})^{2013}+(2-\sqrt{3})^{2013}=4 a_{n-1}-a_{n-2}, \\
a_{0}=2, a_{1}=4 .
\end{array}
$$
Then $2 \mid a_{n}$,
$$
\begin{array}{l}
a_{n} \equiv 2,4,4,2,4,4,2, \cdots(\bmod 5) \\
\Rightarrow a_{2013} \equiv 2(\bmod 5) .
\end{array}
$$
Therefore, the last digit is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $a$ and $b$ are two different positive integers. Ask:
$$
\begin{array}{l}
a(a+2), a b, a(b+2), (a+2) b, \\
(a+2)(b+2), b(b+2)
\end{array}
$$
Among these six numbers, what is the maximum number of perfect squares?
|
2. At most two numbers are perfect squares (such as $a=2, b=16$). Notice that,
$$
\begin{array}{l}
a(a+2)=(a+1)^{2}-1, \\
b(b+2)=(b+1)^{2}-1
\end{array}
$$
cannot be perfect squares;
$$
a b \cdot a(b+2)=a^{2}\left(b^{2}+2 b\right)
$$
is not a perfect square, so at most one of $a b$ and $a(b+2)$ is a perfect square.
Similarly, at most one of $(a+2) b$ and $(a+2)(b+2)$ is a perfect square.
Thus, at most two numbers are perfect squares.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $N>1$ be a positive integer, and $m$ denote the largest divisor of $N$ that is less than $N$. If $N+m$ is a power of 10, find $N$.
|
5. $N=75$.
Let $N=m p$. Then $p$ is the smallest prime factor of $N$.
By the problem, we know $m(p+1)=10^{k}$.
Since $10^{k}$ is not a multiple of 3, therefore, $p>2$.
Hence, $N$ and $m$ are both odd.
Thus, $m=5^{*}$.
If $s=0, N=p=10^{k}-1$ is a multiple of 9, which is a contradiction.
Then $s \geqslant 1,5 \mid N$.
Therefore, $p \leqslant 5$.
If $p=3$, then $4 \times 5^{s}=10^{k}$.
So $k=2, m=25, N=75$.
If $p=5$, we get $6110^{k}$, which is a contradiction.
|
75
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 There are three piles of stones. Each time, A moves one stone from one pile to another, and A can receive a reward from B for each move, which is equal to the difference between the number of stones in the pile to which A moves the stone and the number of stones in the pile from which A moves the stone. If this difference is negative, A should return this amount of reward to B (if unable to pay, A can owe it temporarily). It is known that at a certain moment, all the stones are in their initial piles. Try to find the maximum reward A can earn at this moment.
---
The translation maintains the original text's line breaks and format.
|
【Analysis】Due to the uncertainty of A's operations, it is necessary to start from the whole and establish a "substitute" for the increase or decrease of A's reward each time, which must be simple to calculate.
Solution A's reward is 0.
In fact, the three piles of stones can be imagined as three complete graphs (each stone in the same pile is connected by a line segment, and stones in different piles are not connected).
Each time A operates, a vertex $u$ is moved out from a complete graph $A_{1}$ and into another complete graph $A_{2}$, forming two new complete graphs $A_{1}^{\prime}$ and $A_{2}^{\prime}$. The reward is the difference in the number of vertices between $A_{2}^{\prime}$ and $A_{1}$, which is the difference in the sum of the number of edges between $A_{1}^{\prime}$ and $A_{2}^{\prime}$ and the sum of the number of edges between $A_{1}$ and $A_{2}$.
Thus, A's reward for each operation is the change in the number of edges in the graph.
Since it eventually returns to the initial state, A's reward is 0.
【Comment】By establishing the above substitute, the conclusion becomes clear at a glance.
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $n(n \geqslant 4)$ be a positive integer. $n$ players each play a table tennis match against every other player (each match has a winner and a loser). Find the minimum value of $n$ such that after all the matches, there always exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$, satisfying that when $1 \leqslant i<j \leqslant 4$, player $A_{i}$ defeats player $A_{j}$.
(Supplied by He Yixie)
|
2. First, prove: when $n=8$, there always exists an ordered quartet that satisfies the problem's conditions.
Since 8 players have played a total of $\mathrm{C}_{8}^{2}=28$ matches, there must be at least one player who has won at least $\left\lceil\frac{28}{8}\right\rceil=4$ matches (where $\left\lceil x \right\rceil$ denotes the smallest integer not less than the real number $x$).
Assume player $A_{1}$ has defeated $a_{1}, a_{2}, a_{3}, a_{4}$.
Among players $a_{1}, a_{2}, a_{3}, a_{4}$, a total of six matches have been played, so at least one player must have won at least $\left\lceil\frac{6}{4}\right\rceil=2$ matches. Assume player $a_{1}$ has defeated players $a_{2}$ and $a_{3}$, and assume player $a_{2}$ has defeated player $a_{3}$. Then, set players $A_{2}, A_{3}, A_{4}$ to be $a_{1}, a_{2}, a_{3}$, respectively. Therefore, the ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$ satisfies the conditions.
Next, prove: when $n \leqslant 7$, an ordered quartet that satisfies the conditions does not necessarily exist.
It suffices to disprove the case for $n=7$.
Label the 7 players as $b_{1}, b_{2}, \cdots, b_{7}$, and agree that $b_{7+k}=b_{k}$. Construct the following scenario:
For $i=1,2, \cdots, 7$, let player $b_{i}$ defeat players $b_{i+1}, b_{i+2}, b_{i+4}$, but lose to players $b_{i+3}, b_{i+5}, b_{i+6}$.
This precisely determines the outcome of every match.
Assume there exists an ordered quartet $\left(A_{1}, A_{2}, A_{3}, A_{4}\right)$ that meets the conditions.
Since player $A_{1}$ must be one of the $b_{1}$, players $A_{2}, A_{3}, A_{4}$ can only be a permutation of $b_{i+1}, b_{i+2}, b_{i+4}$. However, because player $b_{i+1}$ defeats player $b_{i+2}$, player $b_{i+2}$ defeats player $b_{i+4}$, and player $b_{i+4}$ defeats player $b_{i+1}$, none of the players $b_{i+1}, b_{i+2}, b_{i+4}$ can serve as player $A_{4}$, leading to a contradiction.
In conclusion, the smallest value of $n$ that satisfies the conditions is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. In the figure shown in Figure 3, on both sides of square $P$, there are $a$ and $b$ squares to the left and right, and $c$ and $d$ squares above and below, where $a$, $b$, $c$, and $d$ are positive integers, satisfying
$$
(a-b)(c-d)=0 \text {. }
$$
The shape formed by these squares is called a "cross star".
There is a grid table consisting of 2014 squares, forming a $38 \times 53$ grid. Find the number of cross stars in this grid table.
(Tao Pingsheng, provided)
|
8. For a cross, the cell $P$ referred to in the problem is called the "center block" of the cross.
When $a=b$, the cross is called "standing"; when $c=d$, it is called "lying" (some crosses are both standing and lying).
If the union of a row and a column of a rectangle $R$ is exactly a cross $S$, then $R$ is called the "source rectangle" of $S$. In the source rectangle $R$, the cross $S$ is uniquely determined by the position of the center block, and the crosses corresponding to different source rectangles must be different. Moreover, by the definition of the cross, the number of rows and columns of the source rectangle is at least 3; a source rectangle with an even number of rows does not contain a lying cross; a source rectangle with an even number of columns does not contain a standing cross.
Now consider the $m \times n$ grid $P_{m, n}$. The number of source rectangles of size $(2k+1) \times l$ is $(m-2k)(n-l+1)$, where $k, l$ satisfy
$$
3 \leqslant 2k+1 \leqslant m, 3 \leqslant l \leqslant n.
$$
For each $(2k+1) \times l$ source rectangle, the center block of a lying cross must be in the $(k+1)$-th row and in the 2nd to $(l-1)$-th columns, so this source rectangle corresponds to $l-2$ lying crosses.
$$
\begin{array}{l}
\text { Hence } A=\sum_{k=1}^{\left[\frac{m-1}{2}\right]} \sum_{l=3}^{n}(l-2)(m-2k)(n-l+1) \\
=\left[\sum_{k=1}^{\left[\frac{m-1}{2}\right]}(m-2k)\right]\left[\sum_{i=1}^{n-2} l(n-l-1)\right], \\
\text { where, } \sum_{k=1}^{\left[\frac{m-1}{2}\right]}(m-2k) \\
=\left[\frac{m-1}{2}\right] \frac{(m-2)+\left(m-2\left[\frac{m-1}{2}\right]\right)}{2} \\
\quad=\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right), \\
\sum_{i=1}^{n-2} l(n-l-1)=\left(\sum_{l=1}^{n-2} l\right)(n-1)-\sum_{l=1}^{n-2} l^{2} \\
=\frac{(n-2)(n-1)^{2}}{2}-\frac{(n-2)(n-1)(2n-3)}{6} \\
=\frac{n(n-1)(n-2)}{6} .
\end{array}
$$
Thus, $A=\left[\frac{m-1}{2}\right]\left(n-\left[\frac{m+1}{2}\right]\right) \frac{n(n-1)(n-2)}{6}$.
Similarly, the number of source rectangles of size $k \times (2l+1)$ in the grid $P_{m, n}$ is $(m-k+1)(n-2l)$, and each $k \times (2l+1)$ source rectangle corresponds to $k-2$ standing crosses. Therefore,
$$
\begin{aligned}
B & =\sum_{k=3}^{m}\left[\sum_{k=1}^{\left.n-\frac{n-1}{2}\right]}(k-2)(m-k+1)(n-2l)\right. \\
& =\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right) \frac{m(m-1)(m-2)}{6} .
\end{aligned}
$$
Each $(2k+1) \times (2l+1)$ source rectangle corresponds to exactly one cross that is both standing and lying (the center block is at the center of the source rectangle), and the grid $P_{m, n}$ contains $(m-2k)(n-2l)$ such source rectangles, so similarly we get
$$
\begin{aligned}
C & \left.=\sum_{k=1}^{\left[\frac{m-1}{2}\right]}\right]\left[\frac{n-1}{2}\right] \\
& =\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right)\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right), \text { (3) }
\end{aligned}
$$
which is the number of crosses that are counted twice in $A+B$.
In particular, for the case $m=38, n=53$, we have
$$
\begin{array}{l}
{\left[\frac{m-1}{2}\right]\left(m-\left[\frac{m+1}{2}\right]\right)=18 \times 19=342,} \\
\frac{m(m-1)(m-2)}{6}=8436, \\
{\left[\frac{n-1}{2}\right]\left(n-\left[\frac{n+1}{2}\right]\right)=26 \times 26=676,} \\
\frac{n(n-1)(n-2)}{6}=23426 .
\end{array}
$$
Substituting into equations (1), (2), and (3), we get
$$
\begin{array}{l}
A=342 \times 23426=8011692, \\
B=676 \times 8436=5702736, \\
C=342 \times 676=231192 .
\end{array}
$$
Thus, the number of crosses in the grid $P_{38,53}$ is
$$
A+B-C=13483236
$$
|
13483236
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given distinct complex numbers $a$ and $b$ satisfying $a b \neq 0$, the set $\{a, b\}=\left\{a^{2}, b^{2}\right\}$. Then $a+b=$ $\qquad$ .
|
$-1 .-1$.
If $a=a^{2}, b=b^{2}$, by $a b \neq 0$, we get $a=b=1$, which is a contradiction. If $a=b^{2}, b=a^{2}$, by $a b \neq 0$, we get $a^{3}=1$.
Clearly, $a \neq 1$.
Thus, $a^{2}+a+1=0 \Rightarrow a=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
Similarly, $b=-\frac{1}{2} \mp \frac{\sqrt{3}}{2} \mathrm{i}$.
Therefore, $a+b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a positive integer $a$ such that the function
$$
f(x)=x+\sqrt{13-2 a x}
$$
has a maximum value that is also a positive integer. Then the maximum value of the function is $\qquad$ .
|
2.7.
Let $t=\sqrt{13-2 a x} \geqslant 0$. Then
$$
\begin{aligned}
y= & f(x)=\frac{13-t^{2}}{2 a}+t \\
& =-\frac{1}{2 a}(t-a)^{2}+\frac{1}{2}\left(a+\frac{13}{a}\right) .
\end{aligned}
$$
Since $a$ is a positive integer, $y_{\max }=\frac{1}{2}\left(a+\frac{13}{a}\right)$ is also a positive integer, so, $y_{\max }=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the positive geometric sequence $\left\{a_{n}\right\}$,
$$
a_{5}=\frac{1}{2}, a_{6}+a_{7}=3 \text {. }
$$
Then the maximum positive integer $n$ that satisfies $a_{1}+a_{2}+\cdots+a_{n}>a_{1} a_{2} \cdots a_{n}$ is $\qquad$
|
3. 12 .
According to the problem, $\frac{a_{6}+a_{7}}{a_{5}}=q+q^{2}=6$.
Since $a_{n}>0$, we have $q=2, a_{n}=2^{n-6}$.
Thus, $2^{-5}\left(2^{n}-1\right)>2^{\frac{n(n-11)}{2}} \Rightarrow 2^{n}-1>2^{\frac{n(n-11)}{2}+5}$.
Estimating $n>\frac{n(n-11)}{2}+5$, we get $n_{\max }=12$.
Upon verification, $n=12$ meets the requirement.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. As shown in Figure 1, given a regular tetrahedron $P-A B C$ with all edge lengths equal to 4, points $D, E, F$ are on edges $P A, P B, P C$ respectively. Then the number of $\triangle D E F$ that satisfy $D E = E F = 3, D F = 2$ is $\qquad$.
|
7.3.
Let $P D=x, P E=y, P F=z$. Then
$$
\left\{\begin{array}{l}
x^{2}+y^{2}-x y=9, \\
y^{2}+z^{2}-y z=9, \\
z^{2}+x^{2}-z x=4 .
\end{array}\right.
$$
(1) - (2) gives $x=z$ or $x+z=y$.
When $x=z$, we get $x=z=2, y=1+\sqrt{6}$;
When $x+z=y$, $x z=\frac{5}{2}, x^{2}+z^{2}=\frac{13}{2}$, there are two sets of solutions.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $1 \leqslant x, y, z \leqslant 6$.
The number of cases where the product of the positive integers $x, y, z$ is divisible by 10 is
$\qquad$ kinds.
|
8. 72 .
(1) The number of ways to choose $x, y, z$ is $6^{3}$;
(2) The number of ways to choose $x, y, z$ without taking $2, 4, 6$ is $3^{3}$; (3) The number of ways to choose $x, y, z$ without taking 5 is $5^{3}$;
(4) The number of ways to choose $x, y, z$ without taking $2, 4, 5, 6$ is $2^{3}$. Therefore, the number of ways for the product of $x, y, z$ to be divisible by 10 is $6^{3}-3^{3}-5^{3}+2^{3}=72$.
|
72
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given the sequence $\left\{a_{n}\right\}_{n \geqslant 0}$ satisfies $a_{0}=0$, $a_{1}=1$, and for all positive integers $n$,
$$
a_{n+1}=2 a_{n}+2013 a_{n-1} \text {. }
$$
Find the smallest positive integer $n$ such that $2014 \mid a_{n}$.
|
10. Below are $a_{n}$ modulo 2014.
Then $a_{n+1} \equiv 2 a_{n}-a_{n-1} \Rightarrow a_{n+1}-a_{n} \equiv a_{n}-a_{n-1}$.
Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2014.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, the smallest positive integer $n$ such that $2014 \mid a_{n}$ is 2014.
|
2014
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the function
$$
f(x)=a \sin x+b \cos x \quad(a, b \in \mathbf{Z}),
$$
and it satisfies
$$
\{x \mid f(x)=0\}=\{x \mid f(f(x))=0\} .
$$
Then the maximum value of $a$ is . $\qquad$
|
2.3.
Let $A=\{x \mid f(x)=0\}, B=\{x \mid f(f(x))=0\}$.
Obviously, set $A$ is non-empty.
Take $x_{0} \in A$, i.e., $x_{0} \in B$, hence
$$
b=f(0)=f\left(f\left(x_{0}\right)\right)=0 \text {. }
$$
Thus, $f(x)=a \sin x(a \in \mathbf{Z})$.
When $a=0$, obviously, $A=B$.
Now assume $a \neq 0$, in this case,
$$
\begin{array}{l}
A=\{x \mid a \sin x=0\}, \\
B=\{x \mid a \sin (a \sin x)=0\} \\
=\{x \mid a \sin x=k \pi, k \in \mathbf{Z}\} .
\end{array}
$$
It is easy to see that $A=B$ if and only if for any $x \in \mathbf{R}$, $a \sin x \neq k \pi(k \in \mathbf{Z}, k \neq 0)$,
i.e., $|a|<\pi$.
Therefore, the maximum value of the integer $a$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let any real numbers $a>b>c>d>0$. To make
$$
\begin{array}{l}
\log _{\frac{b}{a}} 2014+\log _{\frac{d}{b}} 2014+\log _{\frac{d}{c}} 2014 \\
\geqslant m \log _{\frac{d}{a}} 2014
\end{array}
$$
always hold, then the minimum value of $m$ is $\qquad$.
|
5.9.
$$
\begin{array}{l}
\text { Let } x_{1}=-\log _{2014} \frac{b}{a}, x_{2}=-\log _{2014} \frac{c}{b}, \\
x_{3}=-\log _{2014} \frac{d}{c} .
\end{array}
$$
Since $a>b>c>d>0$, we have
$$
x_{1}>0, x_{2}>0, x_{3}>0 \text {. }
$$
Thus, the given inequality can be transformed into
$$
\begin{array}{l}
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}} \leqslant m \cdot \frac{1}{x_{1}+x_{2}+x_{3}} \\
\Rightarrow m \geqslant\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right) \geqslant 9 .
\end{array}
$$
When $x_{1}=x_{2}=x_{3}$, i.e., $a, b, c, d$ form a geometric sequence, the equality holds.
Therefore, the minimum value of $m$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$. If $f(4)=2014$, then
$$
f(2014)=
$$
$\qquad$
|
6.4024.
Let $g(x)=f(x)-x$, then we have
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Also, from $f(x+3) \leqslant f(x)+3$,
$$
f(x+2) \geqslant f(x)+2 \text {, }
$$
we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
From equation (1), we have
$$
\begin{array}{l}
g(x+4) \geqslant f(x+2)-x-2 \\
\geqslant f(x)+2-x-2=f(x)-x .
\end{array}
$$
Thus, $g(x+6) \geqslant f(x+2)-x-2 \geqslant f(x)-x$.
From equation (2), we have
$$
g(x+6) \leqslant f(x+3)-x-3 \leqslant f(x)-x \text {. }
$$
Therefore, $g(x+6)=f(x)-x=g(x)$.
Hence, $g(x)$ is a periodic function with a period of 6.
Notice that,
$$
\begin{array}{l}
g(2014)=g(335 \times 6+4)=g(4) \\
=f(4)-4=2014-4=2010 .
\end{array}
$$
Thus, $f(2014)=g(2014)+2014=4024$.
|
4024
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If non-negative integers $m, n$ add up with exactly one carry (in decimal), then the ordered pair $(m, n)$ is called "good". The number of all good ordered pairs whose sum is 2014 is $\qquad$ .
|
7. 195 .
If the carry is in the units place, then the combination of units and tens is $5+9$, $6+8$, $7+7$, $8+6$, $9+5$, a total of 5 kinds; the hundreds place can only be $0+0$, a total of 1 kind; the thousands place is $0+2$, $1+1$, $2+0$, a total of 3 kinds. In this case, there are $5 \times 1 \times 3=15$ pairs.
(1) If the carry is in the tens place, this would cause the hundreds place to also carry, which does not meet the requirement;
(2) If the carry is in the hundreds place, then the units place is $0+4$, $1+3$, $2+2$, $3+1$, $4+0$, a total of 5 kinds; the tens place is $0+1$, $1+0$, a total of 2 kinds; the combination of hundreds and thousands place is $1+19$, $2+18$, ..., $9+11$, $11+9$, $12+8$, ..., $19+1$, a total of 18 kinds. In this case, there are $5 \times 2 \times 18=180$ pairs.
The thousands place cannot carry.
Therefore, the number of ordered pairs is $180+15=195$.
|
195
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let \( x, y, z \) all be positive real numbers, and
\[
x+y+z=1 \text{. }
\]
Find the minimum value of the function
\[
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}
\]
and provide a proof.
|
11. Notice that, $\frac{3 x^{2}-x}{1+x^{2}}=\frac{x(3 x-1)}{1+x^{2}}$.
Consider the function $g(t)=\frac{t}{1+t^{2}}$.
It is easy to see that $g(t)$ is an odd function.
Since when $t>0$, $\frac{1}{t}+t$ is decreasing in the interval $(0,1)$, hence $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in the interval $(0,1)$.
Therefore, for $t_{1}, t_{2} \in(0,1)$, we have
$$
\left(t_{1}-t_{2}\right)\left(g\left(t_{1}\right)-g\left(t_{2}\right)\right) \geqslant 0 \text {. }
$$
Thus, for any $x \in(0,1)$, we have
$$
\begin{array}{l}
\left(x-\frac{1}{3}\right)\left(\frac{x}{1+x^{2}}-\frac{3}{10}\right) \geqslant 0 \\
\Rightarrow \frac{3 x^{2}-x}{1+x^{2}} \geqslant \frac{3}{10}(3 x-1) .
\end{array}
$$
Similarly, $\frac{3 y^{2}-y}{1+y^{2}} \geqslant \frac{3}{10}(3 y-1)$,
$$
\frac{3 z^{2}-z}{1+z^{2}} \geqslant \frac{3}{10}(3 z-1) \text {. }
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}} \\
\geqslant \frac{3}{10}[3(x+y+z)-3]=0 .
\end{array}
$$
When $x=y=z=\frac{1}{3}$, $f(x, y, z)=0$.
Therefore, the minimum value sought is 0.
|
0
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (50 points) Prove: There exists a set $S$ consisting of 2014 positive integers, with the following property: if a subset $A$ of $S$ satisfies that for any $a, a' \in A, a \neq a'$, we have $a + a' \notin S$, then $|A| \leq 152$.
---
The translation maintains the original text's formatting and line breaks.
|
For $1<k<2014$, let
$2014=k q+r(0 \leqslant r<k)$.
For $i=1,2, \cdots, k$, let
$S_{i}=\left\{2^{i-1} m \mid q \leqslant m \leqslant 2 q-1\right\}$.
Then $\left|S_{i}\right|=q$, and for any
$1 \leqslant i<j \leqslant k, q \leqslant m_{1}, m_{2} \leqslant 2 q-1$,
we have $2^{i-1} m_{1}=2^{j-1} m_{2} \Leftrightarrow 2^{j-i} m_{2}=m_{1}$.
From $2 m_{2} \leqslant 2^{j-i} m_{2}=m_{1} \leqslant 2 q-1$, we get $m_{2}<q$, which is a contradiction.
Thus, $S_{i} \cap S_{j}=\varnothing$.
Let $S_{0}$ be a set composed of $r$ positive integers, such that for any $1 \leqslant i \leqslant k$, we have $S_{0} \cap S_{i}=\varnothing$.
Let $S=\bigcup_{i=0}^{k} S_{i}$. Then $|S|=k q+r=2014$.
Thus, the set $S$ has a subset $A$, satisfying for any $a, a^{\prime} \in A$, $a \neq a^{\prime}$, we have
$a+a^{\prime} \notin S$ (for example, $S_{1}=\{q, q+1, \cdots, 2 q-1\}$).
Therefore, for any subset $A$ and $1 \leqslant i \leqslant k-1$, we have $\left|S_{i} \cap A\right| \leqslant 2$.
If there exists $i(1 \leqslant i \leqslant k-1)$ such that $\left|S_{i} \cap A\right| \geqslant 3$, let $2^{i-1} m_{1}, 2^{i-1} m_{2}, 2^{i-1} m_{3}$ be three different numbers in $S_{i} \cap A$, where $q \leqslant m_{1}, m_{2}, m_{3} \leqslant 2 q-1$. Then among $m_{1}, m_{2}, m_{3}$, there are two with the same parity (let's say $m_{1}, m_{2}$), so,
$m_{1}+m_{2}=2 m\left(m \in \mathbf{Z}_{+}\right)$, and $q \leqslant m \leqslant 2 q-1$.
Thus, $2^{i-1} m_{1}+2^{i-1} m_{2}=2^{i} m \in S_{i+1}$.
But $S_{i+1} \subseteq S$, which means the sum of the numbers $2^{i-1} m_{1}, 2^{i-1} m_{2}$ in subset $A$ is in set $S$, a contradiction.
Therefore, $\left|S_{i} \cap A\right| \leqslant 2$ for $1 \leqslant i \leqslant k-1$.
From this and $r \leqslant k-1$, we get
$$
\begin{array}{l}
|A| \leqslant\left|S_{0}\right|+\left|S_{k}\right|+2(k-1) \\
=r+q+2(k-1) \\
\leqslant 3 k+\left[\frac{2014}{k}\right]-3
\end{array}
$$
$$
\leqslant 3 k+\frac{2014}{k}-3=3\left(k+\frac{\frac{2014}{3}}{k}\right)-3,
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Taking $k=\left[\sqrt{\frac{2014}{3}}\right]+1=26$, then
$$
|A| \leqslant 3 \times 26+\frac{2014}{26}-3 \text {. }
$$
Thus, $|A| \leqslant 3 \times 26+\left[\frac{2014}{26}\right]-3=152$.
(Wang Xin, Fuzhou No.1 High School, 350001, China)
|
152
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: Twelve acrobats numbered $1, 2, \cdots, 12$ are divided into two groups, $A$ and $B$, each with six people. Let the actors in group $A$ form a circle, and each actor in group $B$ stand on the shoulders of two adjacent actors in the $A$ circle. If the number of each actor in the $B$ circle is equal to the sum of the numbers of the two actors beneath him, then such a combination is called a "tower". How many structurally different towers can be built?
[Note] The six people in group $A$, $a_{1}, a_{2}, \cdots, a_{6}$, forming a circle in this order in a clockwise or counterclockwise direction can be considered the same structure. A circle is also considered the same structure after rotation. For convenience, draw a circle, fill in the numbers of the bottom actors inside the circle, and fill in the numbers of the top actors outside the circle. Figure 20 is a tower of 14 people.
|
Let the sum of the elements in groups $A$ and $B$ be denoted as $x$ and $y$ respectively.
Then $y = 2x$.
Therefore, $3x = x + y = 1 + 2 + \cdots + 12 = 78 \Rightarrow x = 26$.
Clearly, $1, 2 \in A, 11, 12 \in B$.
Let $A = \{1, 2, a, b, c, d\} (a < b < c < d)$.
If $d \leq 7$, then
$$
a + b + c + d \leq 4 + 5 + 6 + 7 = 22.
$$
Thus, $a + b + c + d = 23$, and $a \geq 3, 8 \leq d \leq 10$.
(1) $d = 8$.
Then $A = \{1, 2, a, b, c, 8\}, c \leq 7, a + b + c = 15$.
Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, i.e.,
$A = \{1, 2, 3, 5, 7, 8\}$ or $A = \{1, 2, 4, 5, 6, 8\}$.
(i) If $A = \{1, 2, 3, 5, 7, 8\}$, then
$B = \{4, 6, 9, 10, 11, 12\}$.
Since set $B$ contains $4, 6, 11, 12$, set $A$ must have 1 adjacent to 3, 1 adjacent to 5, 5 adjacent to 7, and 8 adjacent to 3.
In this case, there is only one arrangement, resulting in the tower in Figure 21.
Use 21
(ii) If $A = \{1, 2, 4, 5, 6, 8\}$, then
$$
B = \{3, 7, 9, 10, 11, 12\}.
$$
Similarly, set $A$ must have 1 adjacent to 2, 5 adjacent to 6, and 4 adjacent to 8. In this case, there are two arrangements, resulting in the two towers in Figure 22.
(2) $d = 9$.
Then $A = \{1, 2, a, b, c, 9\}, c \leq 8, a + b + c = 14$.
Thus, $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, i.e.,
$A = \{1, 2, 3, 5, 6, 9\}$ or $A = \{1, 2, 3, 4, 7, 9\}$.
(i) If $A = \{1, 2, 3, 5, 6, 9\}$, then
$B = \{4, 7, 8, 10, 11, 12\}$.
To get $4, 10, 12$ in set $B$, set $A$ must have 1, 3, 9 all adjacent, which is impossible.
(ii) If $A = \{1, 2, 3, 4, 7, 9\}$, then
$B = \{5, 6, 8, 10, 11, 12\}$.
To get $6, 8, 12$ in set $B$, set $A$ must have 2 adjacent to 4, 1 adjacent to 7, and 9 adjacent to 3. Thus, there are two arrangements, resulting in the two towers in Figure 23.
(3) $d = 10$.
Then $A = \{1, 2, a, b, c, 10\}, c \leq 9, a + b + c = 13$.
Thus, $(a, b, c) = (3, 4, 6)$, i.e.,
$$
\begin{array}{l}
A = \{1, 2, 3, 4, 6, 10\}, \\
B = \{5, 7, 8, 9, 11, 12\}.
\end{array}
$$
To get $8, 9, 11, 12$ in set $B$, set $A$ must have 6 adjacent to 2, 6 adjacent to 3, 10 adjacent to 1, and 10 adjacent to 2. Thus, there is only one arrangement, resulting in the tower in Figure 24.
Therefore, there are 6 structurally different towers, i.e.,
$T(6) = 6$.
As the value of $n$ increases, the number of such arrangements $T(n)$ increases rapidly, making the situation more complex.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying $f(0) \neq 0$, and for any $x, y \in \mathbf{R}$ we have
$$
f\left((x-y)^{2}\right)=f^{2}(x)-2 x f(y)+y^{2} .
$$
Then $f(2012)=$ $\qquad$
|
Let $x=y=0$.
Then $f(0)=f^{2}(0) \Rightarrow f(0)=1$ or 0 (discard 0).
Let $y=x$.
$$
\begin{array}{l}
\text { Then } f(0)=f^{2}(x)-2 x f(x)+x^{2}=(f(x)-x)^{2} \\
\Rightarrow f(x)=x \pm 1 .
\end{array}
$$
If there exists $x_{0}$ such that $f\left(x_{0}\right)=x_{0}-1$, let
$$
\begin{array}{l}
x=x_{0}, y=0 \text {. } \\
\text { Then } f\left(x_{0}^{2}\right)=f^{2}\left(x_{0}\right)-2 x_{0} \\
=\left(x_{0}-1\right)^{2}-2 x_{0}=x_{0}^{2}-4 x_{0}+1 \text {. }
\end{array}
$$
Now let $x_{0}=0, y=x_{0}$.
$$
\begin{array}{l}
\text { Then } f\left(x_{0}^{2}\right)=f^{2}(0)+x_{0}^{2}=1+x_{0}^{2} \\
\Rightarrow x_{0}^{2}-4 x_{0}+1=1+x_{0}^{2} \\
\Rightarrow x_{0}=0 \Rightarrow f(0)=-1 .
\end{array}
$$
This contradicts $f(0)=1$,
Therefore, for any $x \in \mathbf{R}$, we have
$$
f(x)=x+1 \text {. }
$$
Thus, $f(2012)=2013$.
|
2013
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
f(x)=x\left(\sqrt{36-x^{2}}+\sqrt{64-x^{2}}\right)
$$
Find the maximum value of the function.
|
Algebraic solution Using the Cauchy-Schwarz inequality, we get
$$
\begin{aligned}
f(x) & =x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
& \leqslant \sqrt{\left(x^{2}+36-x^{2}\right)\left(64-x^{2}+x^{2}\right)}=48 .
\end{aligned}
$$
Geometric solution Construct $\triangle A B C, A D \perp B C$, and let $A B=6$, $A C=8, A D=x$. Then
$$
\begin{array}{l}
2 S_{\triangle A B C}=2 S_{\triangle A D B}+2 S_{\triangle A D C} \\
=x \sqrt{36-x^{2}}+x \sqrt{64-x^{2}} \\
=6 \times 8 \sin \angle B A C .
\end{array}
$$
Obviously, when
$$
\angle B A C=90^{\circ} \Leftrightarrow A B \perp A C \Leftrightarrow B C=10, A D=\frac{24}{5}
$$
$S_{\triangle A B C}$ achieves its maximum value of 48.
|
48
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 The function $f$ is defined on the set of ordered pairs of positive integers, and satisfies
$$
\begin{array}{c}
f(x, x)=x, f(x, y)=f(y, x), \\
(x+y) f(x, y)=y f(x, x+y) .
\end{array}
$$
Calculate $f(14,52)$.
|
Since $f(x, x+y)=\frac{x+y}{y} f(x, y)$, we have,
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac{26}{19} f(14,14+24)=\frac{13}{6} f(14,24) \\
= & \frac{13}{6} f(14,14+10)=\frac{26}{5} f(14,10) \\
= & \frac{26}{5} f(10,14)=\frac{26}{5} f(10,10+4) \\
= & \frac{91}{5} f(10,4)=\frac{91}{5} f(4,10)=\frac{91}{3} f(4,6) \\
= & 91 f(4,2)=91 f(2,4)=182 f(2,2) \\
= & 182 \times 2=364 .
\end{aligned}
$$
|
364
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If positive numbers $a, b$ satisfy
$$
2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b) \text {, }
$$
then $\frac{1}{a}+\frac{1}{b}=$ $\qquad$ .
|
$-, 1.108$
Let $2+\log _{2} a=3+\log _{3} b=\log _{6}(a+b)=k$. Then
$$
\begin{array}{l}
a=2^{k-2}, b=3^{k-3}, a+b=6^{k} . \\
\text { Therefore } \frac{1}{a}+\frac{1}{b}=\frac{a+b}{a b}=\frac{6^{k}}{2^{k-2} \times 3^{k-3}} \\
=2^{2} \times 3^{3}=108 .
\end{array}
$$
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example: Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x, y \in \mathbf{R}$, we have
$$
f(x+3) \leqslant f(x)+3, f(x+2) \geqslant f(x)+2 .
$$
Let $g(x)=f(x)-x$.
(1) Prove: $g(x)$ is a periodic function;
(2) If $f(998)=1002$, find the value of $f(2000)$.
|
(1) Proof: From $g(x)=f(x)-x$, we get
$$
\begin{array}{l}
g(x+2)=f(x+2)-x-2, \\
g(x+3)=f(x+3)-x-3 .
\end{array}
$$
Substituting into the inequality in the problem, we get
$$
\begin{array}{l}
g(x+2) \geqslant f(x)+2-x-2=f(x)-x, \\
g(x+3) \leqslant f(x)+3-x-3=f(x)-x .
\end{array}
$$
From equation (1), we get
$$
\begin{array}{l}
g(x+4) \geqslant f(x+2)-x-2 \\
\geqslant f(x)+2-x-2=f(x)-x, \\
g(x+6) \geqslant f(x+2)-x-2 \geqslant f(x)-x .
\end{array}
$$
From equation (2), we get
$$
g(x+6) \leqslant f(x+3)-x-3 \leqslant f(x)-x \text {. }
$$
From equations (3) and (4), we know
$$
g(x+6)=f(x)-x=g(x) \text {. }
$$
Therefore, $g(x)$ is a periodic function (6 is one of its periods).
(2) Solution: Note that, $2000-998=1002$ is a multiple of 6, thus,
$$
\begin{array}{l}
g(2000)=g(998) \\
\Rightarrow f(2000)-2000=f(998)-998 \\
\Rightarrow f(2000)=f(998)+1002 \\
=1002+1002=2004 .
\end{array}
$$
|
2004
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{\pi}{6}, a_{n+1}=\arctan \left(\sec a_{n}\right)\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the positive integer $m$ such that
$$
\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}=\frac{1}{100} .
$$
|
10. From the problem, we know that for any positive integer $n$,
$$
a_{n+1} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),
$$
and $\tan a_{n+1}=\sec a_{n}$.
Since $\sec a_{n}>0$, then $a_{n+1} \in\left(0, \frac{\pi}{2}\right)$.
From equation (1), we get $\tan ^{2} a_{n+1}=\sec ^{2} a_{n}=1+\tan ^{2} a_{n}$.
Thus, $\tan ^{2} a_{n}=n-1+\tan ^{2} a_{1}$
$$
\begin{array}{l}
=n-1+\frac{1}{3}=\frac{3 n-2}{3} \\
\Rightarrow \tan a_{n}=\sqrt{\frac{3 n-2}{3}} .
\end{array}
$$
Therefore, $\sin a_{1} \cdot \sin a_{2} \cdots \cdot \sin a_{m}$
$=\frac{\tan a_{1}}{\sec a_{1}} \cdot \frac{\tan a_{2}}{\sec a_{2}} \cdots \cdots \cdot \frac{\tan a_{m}}{\sec a_{m}}$
$=\frac{\tan a_{1}}{\tan a_{2}} \cdot \frac{\tan a_{2}}{\tan a_{3}} \cdots \cdot \frac{\tan a_{m}}{\tan a_{m+1}}$ (using equation (1))
$=\frac{\tan a_{1}}{\tan a_{m+1}}=\sqrt{\frac{1}{3 m+1}}$.
From $\sqrt{\frac{1}{3 m+1}}=\frac{1}{100} \Rightarrow m=3333$.
|
3333
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x, y$ be positive real numbers. Find the minimum value of
$$
x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}
$$
(Gu Bin, Jin Aiguo)
|
1. Let $f(x, y)=x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$.
If $x \geqslant 1, y \geqslant 1$, then
$$
f(x, y) \geqslant x+y \geqslant 2 \text {; }
$$
If $0<x<1, 0<y<1$, then
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{1-y}{x} = \left(x + \frac{1-x}{y}\right) + \left(y + \frac{1-y}{x}\right) \geqslant 2.
$$
For $0<x<1, y \geqslant 1$, we have
$$
f(x, y) = x + y + \frac{1-x}{y} + \frac{y-1}{x} \geqslant x + y + \frac{1-x}{y} \geqslant 2.
$$
For $x \geqslant 1, 0<y<1$, we have
$$
f(x, y) = x + y + \frac{x-1}{y} + \frac{1-y}{x} \geqslant x + y + \frac{1-y}{x} \geqslant 2.
$$
For $x>0, y>0$, we have $f(x, y) \geqslant 2$. Also, $f(1,1)=2$, hence the minimum value is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, in an election, there are 12 candidates, and each member of the electoral committee casts 6 votes. It is known that any two members' votes have at most 2 candidates in common. Find the maximum number of members in the committee.
(Proposed by the Problem Committee)
|
The maximum number of committee members is 4.
Let the number of committee members be $k$, and the candidates be represented by $1,2, \cdots, 12$. Each person's vote is a set $A_{i}(1 \leqslant i \leqslant k)$, and the number of votes each candidate receives is $m_{i}(1 \leqslant i \leqslant 12)$. Then,
$$
\sum_{i=1}^{12} m_{i}=\sum_{i=1}^{k} 6=6 k .
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
2 \sum_{1 \leqslant i<j \leqslant k}\left|A_{i} \cap A_{j}\right|=2 \sum_{i=1}^{12} \mathrm{C}_{m_{i}}^{2}=\sum_{i=1}^{12} m_{i}^{2}-\sum_{i=1}^{12} m_{i} \\
\geqslant \frac{\left(\sum_{i=1}^{12} m_{i}\right)^{2}}{\sum_{i=1}^{12} 1^{2}}-\sum_{i=1}^{12} m_{i}=\frac{(6 k)^{2}}{12}-6 k \\
=3 k^{2}-6 k .
\end{array}
$$
Notice that,
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant k}\left|A_{i} \cap A_{j}\right| \leqslant \sum_{1 \leqslant i<j \leqslant k} 2=2 \mathrm{C}_{k}^{2}=k^{2}-k \\
\Rightarrow & 3 k^{2}-6 k \leqslant 2\left(k^{2}-k\right) \Rightarrow k \leqslant 4 .
\end{aligned}
$$
When $k=4$, a construction that satisfies the conditions is
$$
\begin{array}{l}
A_{1}=\{1,2,3,4,5,6\}, \\
A_{2}=\{1,2,7,8,9,10\}, \\
A_{3}=\{3,4,7,8,11,12\}, \\
A_{4}=\{5,6,9,10,11,12\} .
\end{array}
$$
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Rational numbers $a, b$ have decimal expansions that are repeating decimals with the smallest period of 30. It is known that the decimal expansions of $a-b$ and $a+kb$ have the smallest period of 15. Find the smallest possible value of the positive integer $k$.
|
3. $k_{\min }=6$.
Notice that, $a, b, a-b, a+k b$ can simultaneously become pure repeating decimals by multiplying them by a power of 10. Therefore, assume they are all pure repeating decimals.
Since the decimal part of a rational number is a pure repeating decimal with a period of $T$ if and only if it can be written in the form $\frac{m}{10^{T}-1}(m \in \mathbf{Z})$, we set
$$
a=\frac{m}{10^{30}-1}, b=\frac{n}{10^{30}-1}.
$$
By $a-b=\frac{m-n}{10^{30}-1}, a+k b=\frac{m+k n}{10^{30}-1}$, both are repeating decimals with a period of 15, subtracting them gives
$$
(k+1) b=\frac{(k+1) n}{10^{30}-1}
$$
is a repeating decimal with a period of 15, hence
$$
\left(10^{15}+1\right) \mid(k+1) n.
$$
Since $n$ is not a multiple of $10^{15}+1$ (otherwise, the period length of $b$ would be 15), $k+1$ must be a multiple of some prime factor of $10^{15}+1$.
Also, $10^{15}+1$ is not a multiple of $2,3,5$, thus,
$k+1 \geqslant 7 \Rightarrow k \geqslant 6$.
Next, provide an example for $k=6$.
Take $a=\frac{8}{7\left(10^{15}-1\right)}, b=\frac{1}{7\left(10^{15}-1\right)}$.
Notice that, $10^{3}+1=7 \times 143$, so the smallest period of $a, b$ is 30, $a-b=\frac{1}{10^{15}-1}, a+6 b=\frac{2}{10^{15}-1}$ has the smallest period of 15.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Let $f(n)$ be a function defined on $\mathbf{N}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}$ we have
$$
f(m+n)-f(m)-f(n)=0 \text{ or } 1 \text{, }
$$
and $f(2)=0, f(3)>0, f(6000)=2000$.
Find $f(5961)$.
|
Solve: From $0=f(2) \geqslant 2 f(1) \Rightarrow f(1)=0$;
From $f(3)-f(2)-f(1)=0$ or 1
$$
\Rightarrow 0 \leqslant f(3) \leqslant 1 \text {. }
$$
But $f(3)>0$, hence $f(3)=1$.
By the problem statement, we know
$$
\begin{array}{l}
f(3 n+3)=f(3 n)+3+0 \text { or } 1 \\
\Rightarrow f(3(n+1)) \geqslant f(3 n)+1 .
\end{array}
$$
In the above, let $n=1,2, \cdots, k$, respectively, we get
$$
\begin{array}{l}
f(3 \times 2) \geqslant f(3 \times 1)+1, \\
f(3 \times 3) \geqslant f(3 \times 2)+1, \\
\cdots \cdots \\
f(3 k) \geqslant f(3(k-1))+1 .
\end{array}
$$
Adding all the equations, we get
$$
f(3 k) \geqslant f(3)+(k-1)=k \text {. }
$$
Thus, for all natural numbers $k$,
$$
f(3 k) \geqslant k \text {. }
$$
When $k2000 .
\end{array}
$$
This contradicts the problem statement.
Therefore, $f(3 k)=k$.
Since $1987<2000$, we have
$$
f(5961)=f(3 \times 1987)=1987 .
$$
|
1987
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.
Let $p$ be a prime number, and $k$ be a positive integer. When the equation $x^{2}+p x+k p-1=0$ has at least one integer solution, find all possible values of $k$.
|
Three, let the equation $x^{2}+p x+k p-1=0$ have integer roots $x_{1}$ and another root $x_{2}$.
By the relationship between roots and coefficients, we know
$$
x_{1}+x_{2}=-p, x_{1} x_{2}=k p-1 \text {. }
$$
Thus, $x_{2}$ must also be an integer.
Assume $k>1$.
Notice,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right) \\
=x_{1} x_{2}+\left(x_{1}+x_{2}\right)+1=(k-1) p .
\end{array}
$$
Since $p$ is a prime number and $k-1>0$, it follows that $p \mid\left(x_{1}+1\right)$ or $p \mid\left(x_{2}+1\right)$.
Without loss of generality, assume $p \mid\left(x_{1}+1\right)$. Then
$$
x_{1}+1= \pm m p, x_{2}+1= \pm \frac{k-1}{m} \text {, }
$$
where $m$ is a positive integer, and $m \mid(k-1)$.
Adding the two equations, we get
$$
x_{1}+x_{2}+2= \pm\left(m p+\frac{k-1}{m}\right) \text {, }
$$
which means $-p+2= \pm\left(m p+\frac{k-1}{m}\right)$.
If the right side of equation (1) is positive, then
$$
(m+1) p+\frac{k-1}{m}=2 \text {. }
$$
Clearly, the left side of equation (1) is greater than 2, which is a contradiction.
If the right side of equation (1) is negative, then
$$
(m-1) p+2+\frac{k-1}{m}=0 \text {. }
$$
Clearly, the left side of equation (1) is greater than 0, which is a contradiction.
Therefore, the assumption is false.
Since $k$ is a positive integer, the only possible value for $k$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The function $f(x)(x \neq 1)$ defined on $\mathbf{R}$ satisfies $f(x)+2 f\left(\frac{x+2002}{x-1}\right)=4015-x$. Then $f(2004)=(\quad)$.
|
Let $x=2, x=2004$, we get
$$
f(2004)=2005 .
$$
|
2005
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. People numbered $1,2, \cdots, 2015$ are arranged in a line, and a position-swapping game is played among them, with the rule that each swap can only occur between adjacent individuals. Now, the person numbered 100 and the person numbered 1000 are to swap positions, with the minimum number of swaps required being $\qquad$ times.
|
$$
-, 1.1799 .
$$
Using the formula, the minimum number of swaps required is
$$
(1000-100) \times 2-1=1799
$$
times.
|
1799
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c=-1, a+b+c=4, \\
\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}=1 .
\end{array}
$$
Find the value of $a^{2}+b^{2}+c^{2}$.
|
From the conditions given in the problem, we have
$$
\frac{1}{a}=-b c, \quad a=4-b-c \text {. }
$$
Notice that,
$$
\begin{array}{l}
\frac{a}{a^{2}-3 a-1}=\frac{1}{a-3-\frac{1}{a}}=\frac{1}{b c-b-c+1} \\
=\frac{1}{(b-1)(c-1)} .
\end{array}
$$
Similarly, $\frac{b}{b^{2}-3 b-1}=\frac{1}{(c-1)(a-1)}$,
$\frac{c}{c^{2}-3 c-1}=\frac{1}{(a-1)(b-1)}$.
Therefore, $\frac{a}{a^{2}-3 a-1}+\frac{b}{b^{2}-3 b-1}+\frac{c}{c^{2}-3 c-1}$
$=\frac{1}{(b-1)(c-1)}+\frac{1}{(c-1)(a-1)}+\frac{1}{(b-1)(c-1)}$
$=\frac{a+b+c-3}{(a-1)(b-1)(c-1)}$
$=\frac{1}{(a-1)(b-1)(c-1)}=1$
$\Rightarrow (a-1)(b-1)(c-1)$
$=a b c-(a b+b c+c a)+(a+b+c)-1$
$=1$
$\Rightarrow a b+b c+c a=1$
$\Rightarrow a^{2}+b^{2}+c^{2}$
$=(a+b+c)^{2}-2(a b+b c+c a)=14$.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If positive numbers $a, b, c$ satisfy
$$
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}=3 \text {, }
$$
find the value of the algebraic expression
$$
\frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b}
$$
|
Notice,
$$
\begin{array}{l}
\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}-1 \\
=\left(\frac{b^{2}+c^{2}-a^{2}+2 b c}{2 b c}\right)\left(\frac{b^{2}+c^{2}-a^{2}-2 b c}{2 b c}\right) \\
=\frac{(b+c+a)(b+c-a)(b-c+a)(b-c-a)}{4 b^{2} c^{2}} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}-1 \\
=\frac{(c+a+b)(c+a-b)(c-a+b)(c-a-b)}{4 c^{2} a^{2}}, \\
\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}-1 \\
=\frac{(a+b+c)(a+b-c)(a-b+c)(a-b-c)}{4 a^{2} b^{2}} . \\
\text { Therefore, }\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)^{2}+\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)^{2}+\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}-3 \\
=-(a+b+c)(a+b-c)(b+c-a)(a+c-b) . \\
\left(\frac{1}{4 b^{2} c^{2}}+\frac{1}{4 c^{2} a^{2}}+\frac{1}{4 a^{2} b^{2}}\right) \\
=0 .
\end{array}
$$
Since \(a, b, c\) are positive numbers, therefore,
$$
\begin{array}{l}
a+b+c>0, \\
\frac{1}{4 b^{2} c^{2}}+\frac{1}{4 c^{2} a^{2}}+\frac{1}{4 a^{2} b^{2}}>0 .
\end{array}
$$
Then \((a+b-c)(b+c-a)(a+c-b)=0\).
Hence, among \(a, b, c\), there must be two numbers whose sum equals the third number.
Without loss of generality, let \(a+b=c\). Then
$$
\begin{array}{l}
b^{2}+c^{2}-a^{2}=b^{2}+c^{2}-(c-b)^{2}=2 b c, \\
c^{2}+a^{2}-b^{2}=c^{2}+a^{2}-(c-a)^{2}=2 a c, \\
a^{2}+b^{2}-c^{2}=a^{2}+b^{2}-(a+b)^{2}=-2 a b . \\
\text { Therefore, } \frac{b^{2}+c^{2}-a^{2}}{2 b c}+\frac{c^{2}+a^{2}-b^{2}}{2 c a}+\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
=1+1-1=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given real numbers $x, y, z$ satisfy
$$
x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+\frac{1}{x}=\frac{7}{3} \text {. }
$$
Find the value of $x y z$.
|
Multiplying the three conditional expressions yields
$$
\frac{28}{3}=x y z+\frac{1}{x y z}+\frac{22}{3} \Rightarrow x y z=1 .
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $x, y, z$ satisfy
$$
\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y}=0 \text {. }
$$
Find the value of the algebraic expression $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}$.
|
Notice,
$$
\begin{array}{l}
x+y+z \\
=\left(\frac{x^{2}}{y+z}+x\right)+\left(\frac{y^{2}}{z+x}+y\right)+\left(\frac{z^{2}}{x+y}+z\right) \\
=(x+y+z)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right) .
\end{array}
$$
Thus, $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=-3$ or 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Find the maximum value of the positive integer $r$ that satisfies the following condition: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. ${ }^{\text {[2] }}$
(2013, Romanian National Team Selection Exam)
|
【Analysis】Similarly, map the five subsets of 500 elements each to five vectors in a 1000-dimensional linear space. Since the requirement is the number of elements rather than their parity, we can consider the Euclidean space.
Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be the five vectors after transformation.
Notice,
$$
\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}=\sum_{i=1}^{5}\left|v_{i}\right|^{2}+2 \sum_{1<i<j \in s} v_{i} \cdot v_{j} \text {, (1) }
$$
and $\left|v_{i}\right|^{2}=500$.
Therefore, to minimize the maximum value of $v_{i} \cdot v_{j}$, we need $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}$ to be as small as possible, and the values of $v_{i} \cdot v_{j}$ to be as evenly distributed as possible.
First, find the minimum value of $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}$.
Let $v_{1}+v_{2}+\cdots+v_{5}=v=\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000}\right)$, where $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000} \in \mathbf{N}$, and their sum is 2500.
Since $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}=\sum_{i=1}^{1000} \alpha_{i}^{2}$, the minimum value of the right-hand side is conjectured to be when 500 of $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000}$ are 2 and the other 500 are 3.
This can be proven using the adjustment method, or simply by
$$
\left(\alpha_{i}-2\right)\left(\alpha_{i}-3\right) \geqslant 0 \Rightarrow \alpha_{i}^{2} \geqslant 5 \alpha_{i}-6 \text {. }
$$
Summing over $i=1,2, \cdots, 1000$ gives
$$
\sum_{i=1}^{1000} \alpha_{i}^{2} \geqslant 5 \times 2500-6000=6500 \text {. }
$$
Combining equation (1) and $\left|v_{i}\right|^{2}=500$, we get
$$
\sum_{1 \leqslant i<j \leqslant 5} v_{i} \cdot v_{j} \geqslant 2000 \Rightarrow \max _{1 \leqslant i<j \leqslant 5} v_{i} \cdot v_{j} \geqslant 200 \text {. }
$$
Next, consider the construction part. We only need to construct five vectors in a 10-dimensional space, each with five components being 1 and the other five components being 0, and the inner product of any two vectors is 2. Then, by replicating each of these five vectors 100 times, we obtain the required vectors in the 1000-dimensional space.
Using the cyclic construction method (the first five components cycle, and the last five components cycle), it is not difficult to construct the five vectors as:
$$
\begin{array}{l}
\boldsymbol{w}_{1}=(1,1,1,0,0,1,0,1,0,0), \\
w_{2}=(0,1,1,1,0,0,1,0,1,0), \\
w_{3}=(0,0,1,1,1,0,0,1,0,1), \\
w_{4}=(1,0,0,1,1,1,0,0,1,0), \\
w_{5}=(1,1,0,0,1,0,1,0,0,1) .
\end{array}
$$
In summary, the maximum value of $r$ is 200.
|
200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2$. Find the value of the algebraic expression $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}$.
|
Solve: Regarding $a$ and $b$ as the main elements, then
$$
\begin{array}{l}
\left\{\begin{array}{l}
a+b=2-c, \\
a^{2}+b^{2}=2-c^{2} .
\end{array}\right. \\
\text { Hence } a b=\frac{(a+b)^{2}-\left(a^{2}+b^{2}\right)}{2} \\
=c^{2}-2 c+1=(c-1)^{2} \\
\Rightarrow \frac{(1-c)^{2}}{a b}=1 .
\end{array}
$$
Similarly, $\frac{(1-a)^{2}}{b c}=1, \frac{(1-b)^{2}}{c a}=1$.
Thus, $\frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{3}-x y z=-5, \\
y^{3}-x y z=2, \\
z^{3}-x y z=21 .
\end{array}\right.
$$
Find the value of $x+y+z$.
|
Let $x y z=k$. Then
$$
\left\{\begin{array}{l}
x^{3}=k-5, \\
y^{3}=k+2, \\
z^{3}=k+21 .
\end{array}\right.
$$
Multiplying the above three equations, we get
$$
\begin{array}{l}
k^{3}=(x y z)^{3}=(k-5)(k+2)(k+21) \\
\Rightarrow 18 k^{2}-73 k-210=0 \\
\Rightarrow k_{1}=6, k_{2}=-\frac{35}{18} \text { (not suitable, discard). }
\end{array}
$$
Substituting $k=6$ into the system of equations (1), we solve to get
$$
x=1, y=2, z=3 \text {. }
$$
Therefore, $x+y+z=6$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Function
$$
y=\tan 2013 x-\tan 2014 x+\tan 2015 x
$$
The number of zeros of the function in the interval $[0, \pi]$ is $\qquad$.
|
2. 2014 .
$$
\begin{aligned}
y & =\tan 2013 x-\tan 2014 x+\tan 2015 x \\
& =\frac{\sin (2013 x+2015 x)}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 4028 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{2 \sin 2014 x \cdot \cos 2014 x}{\cos 2013 x \cdot \cos 2015 x}-\frac{\sin 2014 x}{\cos 2014 x} \\
& =\frac{\sin 2014 x \cdot(2 \cos 2014 x-\cos 2013 x \cdot \cos 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} \\
& =\frac{\sin 2014 x \cdot(1+\cos 4028 x-\cos 2013 x \cdot \cos 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} \\
& =\frac{\sin 2014 x \cdot(1-\sin 2013 x \cdot \sin 2015 x)}{\cos 2013 x \cdot \cos 2014 x \cdot \cos 2015 x} .
\end{aligned}
$$
From the domain of $y$, we know that
$$
\sin 2013 x \cdot \sin 2015 x \neq 1 \text{.}
$$
Therefore, the zeros of $y$ are
$$
x=\frac{k \pi}{2014}(k=0,1, \cdots, 2014) \text{.}
$$
However, when $k=1007$, the function is undefined.
Thus, $y$ has 2014 zeros in the interval $[0, \pi]$.
|
2014
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $P_{1}$ and $P_{2}$ be two points on a plane, $P_{2 k+1}\left(k \in \mathbf{Z}_{+}\right)$ be the symmetric point of $P_{2 k}$ with respect to $P_{1}$, and $P_{2 k+2}$ be the symmetric point of $P_{2 k+1}$ with respect to $P_{2}$. If $\left|P_{1} P_{2}\right|=1$, then $\left|P_{2013} P_{2014}\right|=$ $\qquad$ .
|
4.4024.
From the problem, we know
$$
\begin{array}{l}
\left\{\begin{array}{l}
P_{2 k+1}=2 P_{1}-P_{2 k}, \\
P_{2 k+2}=2 P_{2}-P_{2 k+1}
\end{array}\right. \\
\Rightarrow P_{2 k+2}=2\left(P_{2}-P_{1}\right)+P_{2 k} \\
\Rightarrow\left\{\begin{array}{l}
P_{2 k+2}=2 k\left(P_{2}-P_{1}\right)+P_{2}, \\
P_{2 k+1}=2 k\left(P_{1}-P_{2}\right)+P_{2}
\end{array}\right. \\
\Rightarrow \overrightarrow{P_{2 k+1} P_{2 k+2}}=4 k \overrightarrow{P_{1} P_{2}} \text {. } \\
\end{array}
$$
In particular, $\left|P_{2013} P_{2014}\right|=4024$.
|
4024
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=0, a_{2}=1$, and for all $n \geqslant 3, a_{n}$ is the smallest positive integer greater than $a_{n-1}$ such that there is no subsequence of $a_{1}, a_{2}, \cdots, a_{n}$ that forms an arithmetic sequence. Find $a_{2014}$.
|
4. First, prove a lemma using mathematical induction.
Lemma A non-negative integer appears in the sequence if and only if its ternary expansion contains only 0 and 1.
Proof It is obvious that the proposition holds for 0.
Assume the proposition holds for all non-negative integers less than \( N \), and consider \( N \).
If the ternary expansion of \( N \) contains the digit 2, replace all occurrences of 2 with 0 to get the number \( N_{0} \); replace all occurrences of 2 with 1 to get the number \( N_{1} \). Thus, the ternary expansions of \( N_{0} \) and \( N_{1} \) do not contain the digit 2.
By the induction hypothesis, the numbers \( N_{0} \) and \( N_{1} \) are in the sequence.
Since \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence, \( N \) does not appear in the sequence. If the ternary expansion of \( N \) does not contain the digit 2, then we need to prove that \( N \) must appear in the sequence. If not, then there exist terms \( N_{0} \) and \( N_{1} \) in the sequence such that \( N_{0} \), \( N_{1} \), and \( N \) form an arithmetic sequence. Let the common difference be \( d \), and \( 3^{k} \| d \). Then the ternary expansions of these three numbers have the same lowest \( k-1 \) digits, and the \( k \)-th digit is different for each pair, so one of the numbers must have a 2 in the \( k \)-th digit, which contradicts the assumption.
Returning to the original problem.
We only need to find the 2014th non-negative integer whose ternary expansion does not contain the digit 2.
Notice that under this restriction, the ternary carry-over method is equivalent to binary, so we just need to write 2013 (note \( a_{1}=0 \)) in binary \((11111011101)_{2}\) and then convert it to ternary \( a_{2014}=88327 \).
|
88327
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Prove: In the prime factorization of the product of any 10 consecutive three-digit numbers, there are at most 23 distinct prime factors.
|
First, in the prime factorization of each three-digit number, at most two prime factors greater than 10 can appear; otherwise, their product would exceed 1000, which is impossible.
Second, in any sequence of 10 consecutive three-digit numbers, there is one that is a multiple of 10, and in its prime factorization, at most one prime factor greater than 10 can appear. Therefore, at most 19 prime factors greater than 10 can appear. Including $2, 3, 5, 7$, at most 23 distinct prime numbers can appear.
|
23
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the 100th year of Besmiki's tenure as the President of the Currency Authority, he decided to issue new gold coins. In this year, he put into circulation an unlimited number of gold coins with a face value of $2^{100}-1$ yuan. In the following year, he put into circulation an unlimited number of gold coins with a face value of $2^{101}-1$ yuan. This continued every year until the day when the face value of a newly issued gold coin equaled the sum of the face values of some gold coins issued in previous years, at which point he was dismissed. When did this situation occur in Besmiki's tenure as President?
|
3. It happens in the 200th year of Besmiki's presidency.
Assume that the described scenario occurs in the $k$-th year of Besmiki's presidency. Then,
$$
2^{k}-1=a_{1}+a_{2}+\cdots+a_{n}=N-n,
$$
where $N$ is the sum of some powers of 2, all of which are divisible by $2^{100}$.
Since $2^{k}$ is also divisible by $2^{100}$, it follows that $n-1$ is also divisible by $2^{100}$.
Clearly, $n>1$.
Thus, $n \geqslant 2^{100}+1$.
This implies
$$
2^{k}-1 \geqslant\left(2^{100}-1\right)\left(2^{100}+1\right) \geqslant 2^{200}-1 \text {. }
$$
This indicates that $k \geqslant 200$.
Therefore, Besmiki cannot be dismissed from the presidency before the 200th year.
However, from $\left(2^{100}-1\right)\left(2^{100}+1\right)=2^{200}-1$, we know that Besmiki is dismissed in the 200th year.
|
200
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let the decimal representation of the positive integer $N$ consist only of the digits 1 and 2. By deleting digits from $N$, one can obtain all 10000 different positive integers formed by 9999 digits 1 and 1 digit 2. Find the minimum possible number of digits in $N$.
|
7. The minimum possible number of digits in $N$ is 10198.
For example, $N=\underbrace{1 \cdots 1}_{99 \uparrow} \underbrace{1 \cdots}_{100 \uparrow} 12 \underbrace{1 \cdots}_{100 \uparrow} \underbrace{2}_{98 \uparrow} \underbrace{1 \cdots 1}_{99 \uparrow}$.
For a number formed by 9999 digits 1 and 1 digit 2, if there are $100 m+n (0 \leqslant m, n \leqslant 99)$ digits 1 before the digit 2, then delete the other 2, leaving the $(m+1)$-th digit 2 from the left, and then delete the $99-n$ digits 1 before it and the $n$ digits 1 after it.
The following explains: The number of digits in $N$ cannot be less than 10198.
Obviously, in $N$, there do not need to be two digits 2 next to each other, otherwise one of them can be deleted.
Assume there are $k$ digits 2 in $N$, with $a_{0}$ digits 1 before the first 2, $a_{1}$ digits 1 between the first 2 and the second 2, and so on, with $a_{k}$ digits 1 after the $k$-th 2.
Let $s=a_{0}+a_{1}+\cdots+a_{k}$.
To ensure that the number obtained has only one 1 before the 2, at least $a_{0}-1$ digits 1 need to be deleted, thus, $s-\left(a_{0}-1\right)$ should be no less than 9999, i.e., $s-a_{0} \geqslant 9998$.
To ensure that the number obtained has $a_{0}+1$ digits 1 before the 2, the first 2 must be deleted and at least $a_{1}-1$ digits 1 must be deleted, thus, $s-a_{1} \geqslant 9998$.
To ensure that the number obtained has $a_{0}+a_{1}+1$ digits 1 before the 2, the first two 2s must be deleted and at least $a_{2}-1$ digits 1 must be deleted, thus, $s-a_{2} \geqslant 9998$.
By similar reasoning, for $i=0,1, \cdots, k-1$, we have $s-a_{i} \geqslant 9998$.
And to ensure that the 2 appears in the last position in the number obtained, we need $s-a_{k} \geqslant 9999$.
Adding these inequalities, we get
$$
\begin{array}{l}
(k+1) s-s \geqslant 9998(k+1)+1 \\
\Rightarrow k s>9998(k+1) \\
\Rightarrow s>9998+\frac{9998}{k} .
\end{array}
$$
Since there are also $k$ digits 2 in $N$, the number of digits in $N$ is more than
$$
9998+\frac{9998}{k}+k \geqslant 9998+2 \sqrt{9998}>10197 .
$$
|
10198
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The arithmetic sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}+a_{2}+\cdots+a_{14}=77 \text {, and } a_{1} 、 a_{11} \in \mathbf{Z}_{+} \text {. }
$$
Then $a_{18}=$ $\qquad$
|
2. -5 .
From the formula for the sum of an arithmetic sequence, we get
$$
\begin{array}{l}
a_{1}+a_{14}=11 \Rightarrow 2 a_{1}+13 \times \frac{a_{11}-a_{1}}{10}=11 \\
\Rightarrow 7 a_{1}+13 a_{11}=110 \\
\Rightarrow a_{1}=12(\bmod 13), a_{11}=2(\bmod 7) \\
\Rightarrow\left(a_{1}, a_{11}\right)=(12,2) \\
\Rightarrow a_{18}=a_{11}+7 \times \frac{a_{11}-a_{1}}{10}=-5 .
\end{array}
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the function
$$
f(x)=x \log _{2} x+(a-x) \log _{2}(a-x)
$$
be symmetric about the line $x=\frac{1}{2}$. Then for any real numbers $x_{i} \in(0,1)(1 \leqslant i \leqslant 4)$ satisfying $\sum_{i=1}^{4} x_{i}=1$, the minimum value of $s=\sum_{i=1}^{4} x_{i} \log _{2} x_{i}$ is . $\qquad$
|
5. -2 .
From the problem, we know that the midpoint of the interval $(0, a)$ is $\frac{1}{2}$.
Thus, $a=1$.
Then, $f(x)=x \log _{2} x+(1-x) \log _{2}(1-x)$
$\Rightarrow f^{\prime}(x)=\log _{2} \frac{x}{1-x}$.
Let $f^{\prime}(x)=0$, we get $x=\frac{1}{2}$.
For any $x \in\left(0, \frac{1}{2}\right)$, $f^{\prime}(x) < 0$, and for any $x \in\left(\frac{1}{2}, 1\right)$, $f^{\prime}(x) > 0$, so
$$
f(x)_{\text {min }}=f\left(\frac{1}{2}\right)=-1 .
$$
Let $x_{1}+x_{2}=x \in(0,1)$. Then
$$
x_{3}+x_{4}=1-x \in(0,1) \text {. }
$$
From $\frac{x_{1}}{x}+\frac{x_{2}}{x}=1$, we get
$$
\frac{x_{1}}{x} \log _{2} \frac{x_{1}}{x}+\frac{x_{2}}{x} \log _{2} \frac{x_{2}}{x}=f\left(\frac{x_{1}}{x}\right) \geqslant-1,
$$
which means $x_{1} \log _{2} x_{1}+x_{2} \log _{2} x_{2} \geqslant-x+x \log _{2} x$.
Similarly, from the above, we get
$$
\begin{array}{l}
x_{3} \log _{2} x_{3}+x_{4} \log _{2} x_{4} \\
\geqslant-(1-x)+(1-x) \log _{2}(1-x) .
\end{array}
$$
Adding the two inequalities, we get
$$
\sum_{i=1}^{4} x_{i} \log _{2} x_{i} \geqslant-1+f(x) \geqslant-2 \text {. }
$$
When $x_{1}=x_{2}=x_{3}=x_{4}=\frac{1}{4}$, the equality holds.
Thus, $s_{\min }=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ is $=$.
|
6. -2015 .
Notice that for any $x \in(-1,+\infty)$ we have
$$
\frac{x}{1+x} \leqslant \ln (1+x) \leqslant x \text {. }
$$
Then for $f(x)=\left(1+\frac{1}{x}\right)^{x+1}(x>0)$ and
$$
g(x)=\left(1+\frac{1}{x}\right)^{x}(x>0)
$$
the derivatives are respectively
$$
\begin{array}{l}
f^{\prime}(x)=\left(1+\frac{1}{x}\right)^{x+1}\left[\ln \left(1+\frac{1}{x}\right)-\frac{1}{x}\right]0 .
\end{array}
$$
Thus, $f(x)$ is decreasing on the interval $(0,+\infty)$, $g(x)$ is increasing on the interval $(0,+\infty)$, and for any $x \in(0,+\infty)$ we have $f(x)>\mathrm{e}>g(x)$.
Therefore, for any $m 、 n \in \mathbf{Z}_{+}$ we have
$$
\left(1+\frac{1}{n}\right)^{n+1}>\mathrm{e}>\left(1+\frac{1}{m}\right)^{m} \text {. }
$$
Thus, the integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ must be negative.
Let $n=-k\left(k \in \mathbf{Z}_{+}\right)$, substituting into the given equation we get
$$
\left(1+\frac{1}{2014}\right)^{2014}=\left(1-\frac{1}{k}\right)^{-k+1}=\left(1+\frac{1}{k-1}\right)^{k-1} \text {. }
$$
Hence $k-1=2014, n=-k=-2015$.
|
-2015
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If $x, y, z > 0$ satisfy
$$
\left\{\begin{array}{l}
\frac{2}{5} \leqslant z \leqslant \min \{x, y\}, \\
x z \geqslant \frac{4}{15}, \\
y z \geqslant \frac{1}{5},
\end{array}\right.
$$
|
7.13.
From the problem, we have
$$
\frac{1}{\sqrt{x}} \leqslant \frac{\sqrt{15 z}}{2}, \frac{1}{z} \leqslant \frac{5}{2}, \frac{1}{\sqrt{y}} \leqslant \sqrt{5 z} \text {. }
$$
Then
$$
f=\frac{2}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}+\frac{1}{z}\left(1-\frac{z}{x}\right)+
$$
$$
\begin{aligned}
& 2\left[\frac{2}{\sqrt{y}} \cdot \frac{1}{\sqrt{y}}+\frac{1}{z}\left(1-\frac{z}{y}\right)\right] \\
\leqslant & \frac{2}{\sqrt{x}} \cdot \frac{\sqrt{15 z}}{2}+\frac{5}{2}\left(1-\frac{z}{x}\right)+ \\
& 2\left[\frac{2}{\sqrt{y}} \cdot \sqrt{5 z}+\frac{5}{2}\left(1-\frac{z}{y}\right)\right] \\
= & \frac{5}{2}+\sqrt{15} \cdot \sqrt{\frac{z}{x}}-\frac{5}{2} \cdot \frac{z}{x}+ \\
& 2\left[\frac{5}{2}+2 \sqrt{5} \cdot \sqrt{\frac{z}{y}}-\frac{5}{2} \cdot \frac{z}{y}\right] \\
= & 13-\frac{5}{2}\left(\sqrt{\frac{z}{x}}-\sqrt{\frac{3}{5}}\right)^{2}-5\left(\sqrt{\frac{z}{y}}-\frac{2}{\sqrt{5}}\right)^{2}
\end{aligned}
$$
$$
\leqslant 13 \text {. }
$$
When $(x, y, z)=\left(\frac{2}{3}, \frac{1}{2}, \frac{2}{5}\right)$, the equality holds. Therefore, $f_{\max }=13$.
|
13
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Rearrange the six-element array $(1,2,3,4,5,6)$ to
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\right), \\
\text { and } \quad B=\left(b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}\right) .
\end{array}
$$
Then the minimum value of $P=\sum_{i=1}^{6} i a_{i} b_{i}$ is
|
8. 162 .
From the geometric mean
$$
G=\sqrt[6]{\prod_{i=1}^{6} i a_{i} b_{i}}=\sqrt[6]{(6!)^{3}}=12 \sqrt{5} \in(26,27),
$$
we know that there exists at least one term not less than 27, and at least one term not greater than 25.
Let $i_{1} a_{i_{1}} b_{i_{1}} \leqslant 25, i_{2} a_{i_{2}} b_{i_{2}} \geqslant 27$.
$$
\begin{array}{l}
\text { Then } P \geqslant\left(\sqrt{i_{2} a_{i_{2}} b_{i_{2}}}-\sqrt{i_{1} a_{i_{1}} b_{i_{1}}}\right)^{2}+ \\
2\left(\sqrt{i_{1} a_{i_{1}} b_{i_{1}} i_{2} a_{i_{2}} b_{i_{2}}}+\sqrt{i_{3} a_{i_{3}} b_{i_{3}} i_{4} a_{i_{4}} b_{i_{4}}}+\right. \\
\left.\sqrt{i_{5} a_{i 5} b_{i 5} i_{6} a_{i_{6}} b_{i_{6}}}\right) \\
\geqslant(3 \sqrt{3}-5)^{2}+6 \sqrt[6]{(6!)^{3}} \\
=0.19^{2}+72 \sqrt{5}>161 \text {. } \\
\end{array}
$$
Thus, $P \geqslant 162$, and the equality holds when
$$
A=(4,6,3,2,5,1) \text { and } B=(6,2,3,4,1,5) \text {. }
$$
Therefore, $P_{\min }=162$.
|
162
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given
$$
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2}, \frac{1}{y}+\frac{1}{z+x}=\frac{1}{3}, \frac{1}{z}+\frac{1}{x+y}=\frac{1}{4} \text {. }
$$
Find the value of $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}$.
|
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{x+y+z}{x(y+z)}=\frac{1}{2} \\
\Rightarrow \frac{2}{x}=\frac{y+z}{x+y+z} .
\end{array}
$$
Similarly, $\frac{3}{y}=\frac{z+x}{x+y+z}, \frac{4}{z}=\frac{x+y}{x+y+z}$.
Adding the three equations yields $\frac{2}{x}+\frac{3}{y}+\frac{4}{z}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=1, \\
\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}=1 .
\end{array}
$$
Then $a b c=$
|
$$
\begin{aligned}
& \frac{1}{1-2 c}+\frac{1}{1-2 a}+\frac{1}{1-2 b}=1 \\
\Rightarrow & (1-2 a)(1-2 b)+(1-2 b)(1-2 c)+(1-2 a)(1-2 c) \\
& =(1-2 a)(1-2 b)(1-2 c) \\
\Rightarrow & 2-2(a+b+c)=8 a b c \\
\Rightarrow & a b c=0 .
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The largest positive integer $n$ for which the inequality $\frac{9}{17}<\frac{n}{n+k}<\frac{8}{15}$ holds for a unique integer $k$ is $\qquad$
|
2. 144 .
From the problem, we know that $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$.
By the uniqueness of $k$, we have
$$
\begin{array}{l}
\frac{k-1}{n} \leqslant \frac{7}{8}, \text { and } \frac{k+1}{n} \geqslant \frac{8}{9} . \\
\text { Therefore, } \frac{2}{n}=\frac{k+1}{n}-\frac{k-1}{n} \geqslant \frac{8}{9}-\frac{7}{8}=\frac{1}{72} \\
\Rightarrow n \leqslant 144 .
\end{array}
$$
When $n=144$, from $\frac{7}{8}<\frac{k}{n}<\frac{8}{9}$, we get
$$
126<k<128 \text {. }
$$
Thus, $k$ can take the unique integer value 127. Therefore, the maximum value of the positive integer $n$ that satisfies the condition is 144.
|
144
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given positive integers $a, b, c$ satisfy
$$
1<a<b<c, a+b+c=111, b^{2}=a c \text {. }
$$
then $b=$ $\qquad$
|
4. 36.
Let $(a, c)=d, a=a_{1} d, c=c_{1} d, a_{1}, c_{1}$ be positive integers, and $\left(a_{1}, c_{1}\right)=1, a_{1}<c_{1}$.
Then $b^{2}=a c=d^{2} a_{1} c_{1} \Rightarrow d^{2}\left|b^{2} \Rightarrow d\right| b$.
Let $b=b_{1} d\left(b_{1} \in \mathbf{Z}_{+}\right)$. Then $b_{1}^{2}=a_{1} c_{1}$.
Since $\left(a_{1}, c_{1}\right)=1$, $a_{1}, c_{1}$ are both perfect squares.
Let $a_{1}=m^{2}, c_{1}=n^{2}$. Then
$$
b_{1}=m n\left(m, n \in \mathbf{Z}_{+} \text {, and }(m, n)=1, m<n\right) \text {. }
$$
Also, $a+b+c=111$, so
$$
\begin{array}{l}
d\left(a_{1}+b_{1}+c_{1}\right)=111 \\
\Rightarrow d\left(m^{2}+n^{2}+m n\right)=111 .
\end{array}
$$
Notice that,
$$
m^{2}+n^{2}+m n \geqslant 1^{2}+2^{2}+1 \times 2=7 \text {. }
$$
Therefore, $d=1$ or 3.
If $d=1$, then $m^{2}+n^{2}+m n=111$.
By trial, only $m=1, n=10$ satisfy the equation, but in this case, $a=1$ (discard).
If $d=3$, then $m^{2}+n^{2}+m n=37$.
By trial, only $m=3, n=4$ satisfy the equation, in this case, $a=27, b=36, c=48$, which meets the requirements.
Therefore, $b=36$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $n$ be an integer. If there exist integers $x, y, z$ satisfying
$$
n=x^{3}+y^{3}+z^{3}-3 x y z \text {, }
$$
then $n$ is said to have property $P$.
(1) Determine whether $1, 2, 3$ have property $P$;
(2) Among the 2014 consecutive integers $1, 2, \cdots, 2014$, how many do not have property $P$?
|
(1) Let $x=1, y=z=0$, we get
$$
1=1^{3}+0^{3}+0^{3}-3 \times 1 \times 0 \times 0 \text {. }
$$
Thus, 1 has property $P$.
Let $x=y=1, z=0$, we get
$$
2=1^{3}+1^{3}+0^{3}-3 \times 1 \times 1 \times 0 \text {. }
$$
Thus, 2 has property $P$.
If 3 has property $P$, then there exist integers $x, y, z$ such that
$$
\begin{array}{l}
3=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) \\
\Rightarrow 3\left|(x+y+z)^{3} \Rightarrow 3\right|(x+y+z) \\
\Rightarrow 9 \mid\left[(x+y+z)^{3}-3(x+y+z)(xy+yz+zx)\right] \\
\Rightarrow 9 \mid 3,
\end{array}
$$
which is impossible.
Therefore, 3 does not have property $P$.
(2) Let $f(x, y, z)=x^{3}+y^{3}+z^{3}-3xyz$. Then
$$
\begin{array}{l}
f(x, y, z)=(x+y)^{3}+z^{3}-3xy(x+y)-3xyz \\
=(x+y+z)^{3}-3z(x+y)(x+y+z)- \\
3xy(x+y+z) \\
=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) \\
= \frac{1}{2}(x+y+z)\left(x^{2}+y^{2}+z^{2}-xy-yz-zx\right) \\
= \frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right],
\end{array}
$$
i.e.,
$$
\begin{array}{l}
f(x, y, z) \\
=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] .
\end{array}
$$
Assume without loss of generality that $x \geqslant y \geqslant z$.
If $x-y=1, y-z=0, x-z=1$, i.e.,
$$
x=z+1, y=z \text {, }
$$
then $f(x, y, z)=3z+1$;
If $x-y=0, y-z=1, x-z=1$, i.e.,
$$
x=y=z+1 \text {, }
$$
then $f(x, y, z)=3z+2$;
If $x-y=1, y-z=1, x-z=2$, i.e., $x=z+2, y=z+1$,
then $f(x, y, z)=9(z+1)$.
Thus, numbers of the form $3k+1$ or $3k+2$ or $9k$ $(k \in \mathbf{Z})$ all have property $P$.
Note that,
$$
\begin{array}{l}
f(x, y, z) \\
=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) .
\end{array}
$$
If $3 \mid f(x, y, z)$, then
$3\left|(x+y+z)^{3} \Rightarrow 3\right|(x+y+z)$
$\Rightarrow 9 \mid f(x, y, z)$.
In summary, integers $n$ do not have property $P$ if and only if $n=9k+3$ or $n=9k+6$ $(k \in \mathbf{Z})$.
Since $2014=9 \times 223+7$, in the 2014 consecutive integers from 1 to 2014, the number of integers that do not have property $P$ is $224 \times 2=448$.
|
448
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If $n$ is a positive integer, then
$$
\sum_{n=1}^{2014}\left(\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]\right)=
$$
$\qquad$
|
7. 2027091.
Let $f(n)=\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]$. Then
$$
\begin{array}{l}
f(0)=0, f(1)=0, f(2)=1, \\
f(3)=2, f(4)=3, f(5)=3 .
\end{array}
$$
For a positive integer $k$, we have
$$
\begin{array}{l}
f(6 k)=\left[\frac{6 k}{2}\right]+\left[\frac{6 k}{3}\right]+\left[\frac{6 k}{6}\right] \\
=3 k+2 k+k=6 k \text {, } \\
f(6 k+1) \\
=\left[\frac{6 k+1}{2}\right]+\left[\frac{6 k+1}{3}\right]+\left[\frac{6 k+1}{6}\right] \\
=3 k+2 k+k=6 k \text {, } \\
f(6 k+2) \\
=\left[\frac{6 k+2}{2}\right]+\left[\frac{6 k+2}{3}\right]+\left[\frac{6 k+2}{6}\right] \\
=(3 k+1)+2 k+k=6 k+1 \text {, } \\
f(6 k+3) \\
=\left[\frac{6 k+3}{2}\right]+\left[\frac{6 k+3}{3}\right]+\left[\frac{6 k+3}{6}\right] \\
=(3 k+1)+(2 k+1)+k=6 k+2 \text {, } \\
f(6 k+4) \\
=\left[\frac{6 k+4}{2}\right]+\left[\frac{6 k+4}{3}\right]+\left[\frac{6 k+4}{6}\right] \\
=(3 k+2)+(2 k+1)+k=6 k+3 \text {, } \\
f(6 k+5) \\
=\left[\frac{6 k+5}{2}\right]+\left[\frac{6 k+5}{3}\right]+\left[\frac{6 k+5}{6}\right] \\
=(3 k+2)+(2 k+1)+k=6 k+3 \text {. } \\
\text { Hence } \sum_{n=1}^{2014}\left(\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]\right) \\
=\sum_{n=1}^{2014} f(n)=\sum_{n=0}^{2014} f(n) \\
=\sum_{k=0}^{335} \sum_{i=0}^{5} f(6 k+i)-f(2015) \\
=\sum_{k=0}^{335}(36 k+9)-f(6 \times 335+5) \\
=\frac{(9+12069) \times 336}{2}-(6 \times 335+3) \\
=2027091 . \\
\end{array}
$$
|
2027091
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
|
Let $N$ be the square of the desired natural number.
Below, we discuss different cases.
(1) When $n \leqslant 8$,
$$
N=2^{n}\left(2^{8-n}+2^{11-n}+1\right) \text {. }
$$
Since the result inside the parentheses is odd, for $N$ to be a square number, $n$ must be even.
By verifying $n=2,4,6,8$ one by one, we find that $N$ is not a square number in any of these cases.
(2) When $n \geqslant 9$, $N=2^{8}\left(9+2^{n-8}\right)$.
For $N$ to be a square number, $9+2^{n-8}$ must be the square of an odd number.
Assume $9+2^{n-8}=(2 k+1)^{2}$. Hence,
$$
2^{n-10}=(k-1)(k+2) \text {. }
$$
Since $k-1$ and $k+2$ are one odd and one even, the left side of the equation is a power of 2, so it must be that $k-1=1$, i.e., $k=2$.
From $2^{n-10}=2^{2}$, we know that $n=12$ is the solution.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given a regular tetrahedron $P-ABC$, points $P, A, B, C$ are all on a sphere with radius $\sqrt{3}$. If $PA, PB, PC$ are mutually perpendicular, then the distance from the center of the sphere to the plane $ABC$ is $\qquad$.
|
$=13 \cdot \frac{\sqrt{3}}{3}$.
Let $P A=P B=P C=x$. Then $A B=\sqrt{2} x$.
Let the centroid of $\triangle A B C$ be $M$, and the center of the circumscribed sphere of the regular tetrahedron be $O, O M=y$. Then
$$
\left.\begin{array}{l}
A M=\frac{2 \sqrt{3}}{3} \times \frac{\sqrt{2}}{2} x=\frac{\sqrt{6}}{3} x .
\end{array}\right\}
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. In the right trapezoid $A B C D$, it is known that $A D \perp A B$, $A B / / D C, A B=4, A D=D C=2$. Let $N$ be the midpoint of side $D C$, and $M$ be a moving point within or on the boundary of trapezoid $A B C D$. Then the maximum value of $\overrightarrow{A M} \cdot \overrightarrow{A N}$ is
|
8. 6 .
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. As shown in Figure 1, under the rules of Chinese chess, the "pawn" at point $A$ can reach point $B$ through a certain path (before crossing the river, the pawn can only move to the adjacent intersection directly in front of it each step; after crossing the river, it can move to the adjacent intersections in front, to the left, or to the right each step, but it cannot move backward. The "river" refers to the area between the 5th and 6th horizontal lines on the chessboard in Figure 1). During the pawn's movement, if each intersection on the chessboard is not visited more than once by the pawn, such a path is called a "non-repeating path." Therefore, the number of different non-repeating paths is $\qquad$.
|
7. 6561.
Assume the chessboard has 10 horizontal lines from bottom to top, sequentially labeled as the 1st, 2nd, ..., 10th rows, and 9 vertical lines from left to right, sequentially labeled as the 1st, 2nd, ..., 9th columns. For example, point $A$ is located at the 4th row and 5th column.
Note that, during the movement of the pawn from point $A$ to $B$, it cannot move downward. Therefore, the step from the $i$-th row to the $(i+1)$-th row is unique. If the starting and ending points of this step are in the $j$-th column, we denote $a_{i}=j(4 \leqslant i \leqslant 9,1 \leqslant j \leqslant 9)$.
According to the rules, it is easy to see that
$$
a_{4}=a_{5}=5, a_{6} 、 a_{7} 、 a_{8} 、 a_{9} \in\{1,2, \cdots, 9\} \text {. }
$$
There are $1^{2} \times 9^{4}=6561$ such ordered arrays $\left(a_{4}, a_{5}, \cdots, a_{9}\right)$, and each array corresponds one-to-one to a unique non-repeating path from point $A$ to $B$. This is because, after the pawn reaches the $i$-th row ($i=4,5, \cdots, 9$), there is exactly one way to move to the $a_{i}$-th column in the $i$-th row (otherwise, the path would have a repeat), and then move forward to the $(i+1)$-th row. Therefore, the total number of different non-repeating paths is 6561.
|
6561
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ and an inscribed parallelogram with one pair of opposite sides passing through the foci $F_{1}$ and $F_{2}$ of the ellipse. Find the maximum area of the parallelogram. ${ }^{[4]}$
(2013, National High School Mathematics League Shandong Province Preliminary Contest)
|
Solve the general ellipse:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
Corresponding problems.
With $F_{1}$ as the pole and $F_{1} x$ as the polar axis, the equation of the ellipse is transformed into
$$
\rho=\frac{e p}{1+e \cos \theta}=\frac{b^{2}}{a-c \cos \theta} .
$$
Therefore, the length of the chord passing through point $F_{1}$
$$
\begin{array}{l}
A B=\frac{b^{2}}{a-c \cos \theta}+\frac{b^{2}}{a+c \cos \theta} \\
=\frac{2 a b^{2}}{a^{2}-c^{2} \cos ^{2} \theta},
\end{array}
$$
The area of $\square A B C D$
$$
\begin{aligned}
S & =A B \cdot F_{1} F_{2} \sin \theta=\frac{4 a b^{2} c \sin \theta}{a^{2}-c^{2} \cos ^{2} \theta} \\
& =\frac{4 a b^{2} c \sin \theta}{b^{2}+c^{2} \sin ^{2} \theta}=\frac{4 a b^{2} c}{c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}} . \\
& \text { Let } f(\theta)=c^{2} \sin \theta+\frac{b^{2}}{\sin \theta}\left(0<\theta<\pi\right) . \text { Then } f^{\prime}(\theta)=c^{2} \cos \theta-\frac{b^{2} \cos \theta}{\sin ^{2} \theta}=\frac{\cos \theta\left(c^{2} \sin ^{2} \theta-b^{2}\right)}{\sin ^{2} \theta} .
\end{aligned}
$$
Let $f^{\prime}(\theta)=0$, we get $\sin \theta=\frac{b}{c}$. When $\theta \in\left(0, \frac{b}{c}\right)$, $f^{\prime}(\theta)<0$; when $\theta \in\left(\frac{b}{c}, \pi\right)$, $f^{\prime}(\theta)>0$. Therefore, $f(\theta)$ is monotonically decreasing in the interval $\left(0, \frac{b}{c}\right]$ and monotonically increasing in the interval $\left[\frac{b}{c}, \pi\right)$. Hence,
(1) When $\frac{b}{c} \geqslant 1$, i.e., $b\sqrt{2} b$, $g(t)$ is monotonically decreasing in the interval $\left(0, \frac{b}{c}\right]$ and monotonically increasing in the interval $\left[\frac{b}{c}, 1\right)$. Therefore, the minimum value of $g(t)$ is $g\left(\frac{b}{c}\right)=2 b c$. At this point, $S$ reaches its maximum value $\frac{4 a b^{2} c}{2 b c}=2 a b$.
In summary, when $b\sqrt{2} b$, $S$ reaches its maximum value $2 a b$.
For this problem, $a=2, b=\sqrt{3}, c=1, S$ reaches its maximum value $\frac{4 b^{2} c}{a}=6$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $n+2$ real numbers
$$
a_{1}, a_{2}, \cdots, a_{n}, 16, a_{n+2} \text {, }
$$
where the average of the first $n$ numbers is 8, the average of the first $n+1$ numbers is 9, and the average of these $n+2$ numbers is 10. Then the value of $a_{n+2}$ is $\qquad$
|
2. 18 .
From the condition, $\frac{8 n+16}{n+1}=9 \Rightarrow n=7$.
Also, $\frac{8 \times 7+16+a_{n+2}}{7+1+1}=10 \Rightarrow a_{n+2}=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If $a \in A$, and $a-1 \notin A, a+1 \notin A$, then $a$ is called an isolated element of set $A$. Therefore, the number of four-element subsets of set $M=\{1,2, \cdots, 9\}$ without isolated elements is $\qquad$ .
|
9. 21 .
Consider the smallest element $i$ and the largest element $j$ in a set that satisfies the condition.
Let this set be $A$. Then $i+1 \in A, j-1 \in A$ (otherwise, $i$ or $j$ would be an isolated element).
Thus, $A=\{i, i+1, j-1, j\}$.
And $2 \leqslant i+1 < j-1 \leqslant 8$, so the number of ways to choose $i+1, j-1$ is $\mathrm{C}_{7}^{2}=21$. Therefore, the number of four-element subsets of the set $M=\{1,2, \cdots, 9\}$ without isolated elements is 21.
|
21
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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