problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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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 disc... | 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 f... | 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... | 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
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}=10 y+9, \\
... | 391 | Geometry | MCQ | Yes | Yes | cn_contest | false |
7. Given integers $a, b, c, d$. If the roots of the equation
$$
z^{4}+a z^{3}+b z^{2}+c z+d=0
$$
correspond to four points $A, B, C, D$ forming the vertices of a square in the complex plane, then the minimum value of the area of square $A B C D$ is $\qquad$ | 7. 2 .
Let the complex number corresponding to the center of the square be $m$. Then, after translating the origin of the complex plane to $m$, the vertices of the square are distributed on a circle, i.e., they are the solutions to the equation $(z-m)^{4}=n$ (where $n$ is a complex number).
From $z^{4}+a z^{3}+b z^{2... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let $a_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}\left(n \in \mathbf{Z}_{+}\right)$. Find the smallest positive real number $\lambda$, such that for any $n \geqslant 2$, we have
$$
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}... | 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 $AC$ int... | 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... | 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 ... | 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 s... | 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 p... | 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... | 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 $n$... | 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 pig... | 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... | 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 ... | 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
$$
a... | 31 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Given $x, y \in(0,1)$, and $3x+7y$, $5x+y$ are both integers. Then there are $\qquad$ pairs of $(x, y)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 【Analysis】Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v .\end{array}\right.$
We can first determine the integer values of $u$ and $v$ based on $x, y \in(0,1)$.
Solution Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v,\end{array}\right.$ where the possible values of $u$ and $v$ are $u \in\{1,2, \cdots, 9\}, v \in\... | 31 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Below is the inscription by the famous Chinese mathematician, Academician Wang Yuan:
数棈妻好
If different Chinese characters represent different digits from $0 \sim 9$, and assuming “数学竞赛好” represents the largest five-digit number that is a perfect square, formed by different digits. Then this five-digit number is $\q... | Notice that, $300^{2}=90000,310^{2}=96100$,
$$
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... | 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 ... | 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
$$
\begin{array}{l}
a c f+a d f+a c e+a d e+b c f+b c e+b d f+b d e \\
=(a+b)(c+d)(e+f) \\
=2014=2 \times 19 \times 53 .
\end{arra... | 74 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $A, B$ are digits in the set $\{0,1, \cdots, 9\}$, $r$ is a two-digit integer $\overline{A B}$, $s$ is a two-digit integer $\overline{B A}$, $r, s \in\{00,01$, $\cdots, 99\}$. When $|r-s|=k^{2}$ ( $k$ is an integer), the number of ordered pairs $(A, B)$ is $\qquad$. | 7. 42 .
Notice,
$$
|(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 {, }
$$
and their reverse numbers;
W... | 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $a, b$ be real numbers, for any real number $x$ satisfying $0 \leqslant x \leqslant 1$ we have $|a x+b| \leqslant 1$. Then the maximum value of $|20 a+14 b|+|20 a-14 b|$ is . $\qquad$ | 9.80 .
Let $x=0$, we know $|b| \leqslant 1$;
Let $x=1$, we know $|a+b| \leqslant 1$.
Therefore, $|a|=|a+b-b| \leqslant|a+b|+|b| \leqslant 2$, when $a=2, b=-1$, the equality can be achieved.
If $|20 a| \geqslant|14 b|$, then
$$
\begin{array}{l}
|20 a+14 b|+|20 a-14 b| \\
=2|20 a|=40|a| \leqslant 80 ;
\end{array}
$$
If... | 80 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given that vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ are two mutually perpendicular unit vectors in a plane, and
$$
(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}
$$
From the given information, $\overrightarrow{A C} \perp \overrightarrow{B C}$. Therefore, point $C$ lies on the circle with $A B$ as its diameter, and thi... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Given $k$ as a positive integer, the sequence $\left\{a_{n}\right\}$ satisfies
$$
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 {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$.
Let $b_{n}=\frac{1}{n} \log _{3} a_{1}... | 15. From the problem, we know
Also, $a_{n+1}=\left(3^{\frac{2}{2 k-1}}-1\right) S_{n}+3$,
$$
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^... | 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}-\c... | 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 ... | 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 ... | 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 $... | 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 ... | 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... | (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... | 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}
$$
Then $x-1=\frac{8 \times 10^{6}}{\sqrt{1+8000 x}-1}$.
Let $a=10^{3}, t=\sqrt{1+8 a x}-1$.
Thus, $x=\frac{t^{2}+2 t}{8 a}$, an... | 2001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a+b=\sqrt{5}$, then
$$
\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}
$$
Therefore, there are 21 perfect cubes.
Next, consider the number of natural numbers from $1 \sim 10000$ that are both perfect squares and perfect cub... | 9883 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $M$ is the least common multiple of 15 consecutive natural numbers $1,2, \cdots, 15$. If a divisor of $M$ is divisible by exactly 14 of these 15 natural numbers, it is called a "good number" of $M$. Then the number of good numbers of $M$ is $\qquad$.
| 4.4.
It is known that $M=2^{3} \times 3^{2} \times 5 \times 7 \times 11 \times 13$.
Since $2 \times 11, 2 \times 13$ are both greater than 15, therefore,
$$
\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}
$$
each... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the sequence of natural numbers from $1 \sim 8$ be $a_{1}, a_{2}$, $\cdots, a_{8}$. Then
$$
\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... | 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 s... | 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 $P(x)$ ... | 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 inte... | 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
$$
S_{i}-S_{... | 34 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If positive real numbers $a, b$ satisfy
$$
\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 .
$$ | 88 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Arrange the numbers in the set $\left\{2^{x}+2^{y}+2^{z} \mid x 、 y 、 z \in \mathbf{N}, x<y<z\right\}$ in ascending order. The 100th number is $\qquad$ (answer with a number).
| 5.577.
Notice that the number of combinations $(x, y, z)$ such that $0 \leqslant x<y<z \leqslant n$ is $\mathrm{C}_{n+1}^{3}$.
Since $\mathrm{C}_{9}^{3}=84<100<120=\mathrm{C}_{10}^{3}$, the 100th number must satisfy $z=9$.
Also notice that the number of combinations $(x, y)$ such that $0 \leqslant x<y \leqslant m$ i... | 577 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function $f(x)$ satisfies
$$
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^... | 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 $2 \mid a_{n}$,
$$
\begin{array}{l}
a_{n} \equiv 2,4,4,2,4,4,2, \cdots(\bmod 5) \\
\Righ... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $a$ and $b$ are two different positive integers. Ask:
$$
\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.
... | 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$.
T... | 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 th... | 【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 st... | 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 w... | 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$... | 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 {. }
$$
The shape formed by these squares is called a "cross star".
There i... | 8. For a cross, the cell $P$ referred to in the problem is called the "center block" of the cross.
When $a=b$, the cross is called "standing"; when $c=d$, it is called "lying" (some crosses are both standing and lying).
If the union of a row and a column of a rectangle $R$ is exactly a cross $S$, then $R$ is called t... | 13483236 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given distinct complex numbers $a$ and $b$ satisfying $a b \neq 0$, the set $\{a, b\}=\left\{a^{2}, b^{2}\right\}$. Then $a+b=$ $\qquad$ . | $-1 .-1$.
If $a=a^{2}, b=b^{2}$, by $a b \neq 0$, we get $a=b=1$, which is a contradiction. If $a=b^{2}, b=a^{2}$, by $a b \neq 0$, we get $a^{3}=1$.
Clearly, $a \neq 1$.
Thus, $a^{2}+a+1=0 \Rightarrow a=-\frac{1}{2} \pm \frac{\sqrt{3}}{2} \mathrm{i}$.
Similarly, $b=-\frac{1}{2} \mp \frac{\sqrt{3}}{2} \mathrm{i}$.
Ther... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive integer $a$ such that the function
$$
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 {. }
$$
Then the maximum positive integer $n$ that satisfies $a_{1}+a_{2}+\cdots+a_{n}>a_{1} a_{2} \cdots a_{n}$ is $\qquad$ | 3. 12 .
According to the problem, $\frac{a_{6}+a_{7}}{a_{5}}=q+q^{2}=6$.
Since $a_{n}>0$, we have $q=2, a_{n}=2^{n-6}$.
Thus, $2^{-5}\left(2^{n}-1\right)>2^{\frac{n(n-11)}{2}} \Rightarrow 2^{n}-1>2^{\frac{n(n-11)}{2}+5}$.
Estimating $n>\frac{n(n-11)}{2}+5$, we get $n_{\max }=12$.
Upon verification, $n=12$ meets the re... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure 1, given a regular tetrahedron $P-A B C$ with all edge lengths equal to 4, points $D, E, F$ are on edges $P A, P B, P C$ respectively. Then the number of $\triangle D E F$ that satisfy $D E = E F = 3, D F = 2$ is $\qquad$. | 7.3.
Let $P D=x, P E=y, P F=z$. Then
$$
\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.
$$
(1) - (2) gives $x=z$ or $x+z=y$.
When $x=z$, we get $x=z=2, y=1+\sqrt{6}$;
When $x+z=y$, $x z=\frac{5}{2}, x^{2}+z^{2}=\frac{13}{2}$, there are two sets of solutions. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $1 \leqslant x, y, z \leqslant 6$.
The number of cases where the product of the positive integers $x, y, z$ is divisible by 10 is
$\qquad$ kinds. | 8. 72 .
(1) The number of ways to choose $x, y, z$ is $6^{3}$;
(2) The number of ways to choose $x, y, z$ without taking $2, 4, 6$ is $3^{3}$; (3) The number of ways to choose $x, y, z$ without taking 5 is $5^{3}$;
(4) The number of ways to choose $x, y, z$ without taking $2, 4, 5, 6$ is $2^{3}$. Therefore, the number ... | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}_{n \geqslant 0}$ satisfies $a_{0}=0$, $a_{1}=1$, and for all positive integers $n$,
$$
a_{n+1}=2 a_{n}+2013 a_{n-1} \text {. }
$$
Find the smallest positive integer $n$ such that $2014 \mid a_{n}$. | 10. Below are $a_{n}$ modulo 2014.
Then $a_{n+1} \equiv 2 a_{n}-a_{n-1} \Rightarrow a_{n+1}-a_{n} \equiv a_{n}-a_{n-1}$.
Therefore, the sequence $\left\{a_{n}\right\}$ has the characteristics of an arithmetic sequence modulo 2014.
Since $a_{0}=0, a_{1}=1$, we have $a_{n} \equiv n$.
Thus, the smallest positive integer ... | 2014 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function
$$
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... | 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
$$
\begin{array}{l}
\frac{1}{x_{1}}+\fra... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $f(x)$ be a function defined on $\mathbf{R}$, for any $x \in \mathbf{R}$, we have
$$
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}
$$
Fr... | 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.
... | 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 $(0,1)$.
T... | 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... | 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... | 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
$$
a + b + c + d \leq 4 + 5 + 6 + 7 = 2... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$, satisfying $f(0) \neq 0$, and for any $x, y \in \mathbf{R}$ we have
$$
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 {. } \\... | 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 $A B=6$, $A C=8, A D=x$... | 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,
$$
\begin{aligned}
& f(14,52)=f(14,14+38)=\frac{52}{38} f(14,38) \\
= & \frac{26}{19} f(14,14+24)=\frac{13}{6} f(14,24) \\
= & \frac{13}{6} f(14,14+10)=\frac{26}{5} f(14,10) \\
= & \frac{26}{5} f(10,14)=\frac{26}{5} f(10,10+4) \\
= & \frac{91}{5} f(10,4)=\frac{91}{5} f(... | 364 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If positive numbers $a, b$ satisfy
$$
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{... | 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, $\ta... | 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 \... | 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,
$$
\sum_{i=1}^{1... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Rational numbers $a, b$ have decimal expansions that are repeating decimals with the smallest period of 30. It is known that the decimal expansions of $a-b$ and $a+kb$ have the smallest period of 15. Find the smallest possible value of the positive integer $k$.
| 3. $k_{\min }=6$.
Notice that, $a, b, a-b, a+k b$ can simultaneously become pure repeating decimals by multiplying them by a power of 10. Therefore, assume they are all pure repeating decimals.
Since the decimal part of a rational number is a pure repeating decimal with a period of $T$ if and only if it can be writte... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let $f(n)$ be a function defined on $\mathbf{N}$ taking non-negative integer values, and for all $m, n \in \mathbf{N}$ we have
$$
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{arra... | 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 ... | 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,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)(x \neq 1)$ defined on $\mathbf{R}$ satisfies $f(x)+2 f\left(\frac{x+2002}{x-1}\right)=4015-x$. Then $f(2004)=(\quad)$. | Let $x=2, x=2004$, we get
$$
f(2004)=2005 .
$$ | 2005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. People numbered $1,2, \cdots, 2015$ are arranged in a line, and a position-swapping game is played among them, with the rule that each swap can only occur between adjacent individuals. Now, the person numbered 100 and the person numbered 1000 are to swap positions, with the minimum number of swaps required being $\q... | $$
-, 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}
$$
Similarly, $\frac{b}{b^{2}-3 b-1}=\frac{1}{(c-1)(a-1)}$,
$\frac{c}{c^{2}-3 c... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If positive numbers $a, b, c$ satisfy
$$
\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 ... | 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,
$$
\begin{array}{l}
\left(\frac{c^{2}+a^{2}-b^{2}}{2... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given real numbers $x, y, z$ satisfy
$$
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}
$$
Thus, $\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=-3$ or 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the maximum value of the positive integer $r$ that satisfies the following condition: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. ${ }^{\text {[2] }}$
(2013, Romanian National Team Selection Exam) | 【Analysis】Similarly, map the five subsets of 500 elements each to five vectors in a 1000-dimensional linear space. Since the requirement is the number of elements rather than their parity, we can consider the Euclidean space.
Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be the five vectors after transformation.
Notice,
$$
\... | 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, $\fra... | 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
$$
\begin{array}{l}
k^{3}=(x y z)^{3}=(k-5)(k+2)(k+21) \\
\Rightarrow 18 k^{2}-73 k-210=0 \\
\Rightarrow k_{1}=6, k_{2}=-\frac{35}{18} \text { (not suitable, disca... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Function
$$
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 \c... | 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(... | 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 \( N \)... | 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 m... | 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 ... | 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^{10... | 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 ... | 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
$$
\begin{array}{l}
a_{1}+a_{14}=11 \Rightarrow 2 a_{1}+13 \times \frac{a_{11}-a_{1}}{10}=11 \\
\Rightarrow 7 a_{1}+13 a_{11}=110 \\
\Rightarrow a_{1}=12(\bmod 13), a_{11}=2(\bmod 7) \\
\Rightarrow\left(a_{1}, a_{11}\right)=(12,2) \\
\Rightarrow a_... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the function
$$
f(x)=x \log _{2} x+(a-x) \log _{2}(a-x)
$$
be symmetric about the line $x=\frac{1}{2}$. Then for any real numbers $x_{i} \in(0,1)(1 \leqslant i \leqslant 4)$ satisfying $\sum_{i=1}^{4} x_{i}=1$, the minimum value of $s=\sum_{i=1}^{4} x_{i} \log _{2} x_{i}$ is . $\qquad$ | 5. -2 .
From the problem, we know that the midpoint of the interval $(0, a)$ is $\frac{1}{2}$.
Thus, $a=1$.
Then, $f(x)=x \log _{2} x+(1-x) \log _{2}(1-x)$
$\Rightarrow f^{\prime}(x)=\log _{2} \frac{x}{1-x}$.
Let $f^{\prime}(x)=0$, we get $x=\frac{1}{2}$.
For any $x \in\left(0, \frac{1}{2}\right)$, $f^{\prime}(x) < 0$... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The integer $n$ that satisfies $\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{2014}\right)^{2014}$ is $=$. | 6. -2015 .
Notice that for any $x \in(-1,+\infty)$ we have
$$
\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}\rig... | -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
$$
f=\frac{2}{\sqrt{x}} \cdot \frac{1}{\sqrt{x}}+\frac{1}{z}\left(1-\frac{z}{x}\right)+
$$
$$
\begin{aligned}
& 2\left[\frac{2}{\sqrt{y}... | 13 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Rearrange the six-element array $(1,2,3,4,5,6)$ to
$$
\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),
$$
we know that there exists at least one term not less than 27, and at least one term not greater than 25.
Let $i_{1} a_{i_{1}} b_{i_{1}} \leqslant 25, i_{2} a_{i_{2}} b_{i_{2}} \geqslant 27$.
$$
... | 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
$$
\begin{array}{l}
\frac{k-1}{n} \leqslant \frac{7}{8}, \text { and } \frac{k+1}{n} \geqslant \frac{8}{9} . \\
\text { Therefore, } \frac{2}{n}=\frac{k+1}{n}-\frac{k-1}{n} \geqslant \frac{8}{9}-\frac{7}{8}... | 144 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given positive integers $a, b, c$ satisfy
$$
1<a<b<c, a+b+c=111, b^{2}=a c \text {. }
$$
then $b=$ $\qquad$ | 4. 36.
Let $(a, c)=d, a=a_{1} d, c=c_{1} d, a_{1}, c_{1}$ be positive integers, and $\left(a_{1}, c_{1}\right)=1, a_{1}<c_{1}$.
Then $b^{2}=a c=d^{2} a_{1} c_{1} \Rightarrow d^{2}\left|b^{2} \Rightarrow d\right| b$.
Let $b=b_{1} d\left(b_{1} \in \mathbf{Z}_{+}\right)$. Then $b_{1}^{2}=a_{1} c_{1}$.
Since $\left(a_{1},... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $n$ be an integer. If there exist integers $x, y, z$ satisfying
$$
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{a... | 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
$$
\begin{array}{l}
f(6 k)=\left[\frac{6 k}{2}\right]+\left[\frac{6 k}{3}\right]+\left[\fra... | 2027091 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (16 points) Find all natural numbers $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square. | Let $N$ be the square of the desired natural number.
Below, we discuss different cases.
(1) When $n \leqslant 8$,
$$
N=2^{n}\left(2^{8-n}+2^{11-n}+1\right) \text {. }
$$
Since the result inside the parentheses is odd, for $N$ to be a square number, $n$ must be even.
By verifying $n=2,4,6,8$ one by one, we find that $... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. Given a regular tetrahedron $P-ABC$, points $P, A, B, C$ are all on a sphere with radius $\sqrt{3}$. If $PA, PB, PC$ are mutually perpendicular, then the distance from the center of the sphere to the plane $ABC$ is $\qquad$. | $=13 \cdot \frac{\sqrt{3}}{3}$.
Let $P A=P B=P C=x$. Then $A B=\sqrt{2} x$.
Let the centroid of $\triangle A B C$ be $M$, and the center of the circumscribed sphere of the regular tetrahedron be $O, O M=y$. Then
$$
\left.\begin{array}{l}
A M=\frac{2 \sqrt{3}}{3} \times \frac{\sqrt{2}}{2} x=\frac{\sqrt{6}}{3} x .
\end{a... | 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, t... | 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 movem... | 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... | 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 ... | 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+... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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