problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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2. Let $A_{n}$ and $B_{n}$ be the sums of the first $n$ terms of the arithmetic sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively. If $\frac{A_{n}}{B_{n}}=\frac{5 n-3}{n+9}$, then $\frac{a_{8}}{b_{8}}=$ $\qquad$ | 2.3.
Let $\left\{a_{n}\right\}, \left\{b_{n}\right\}$ have common differences $d_{1}, d_{2}$, respectively. Then $\frac{A_{n}}{B_{n}}=\frac{a_{1}+\frac{1}{2}(n-1) d_{1}}{b_{1}+\frac{1}{2}(n-1) d_{2}}=\frac{5 n-3}{n+9}$.
Let $d_{2}=d$. Then $d_{1}=5 d$.
Thus, $a_{1}=d, b_{1}=5 d$.
Therefore, $\frac{a_{8}}{b_{8}}=\frac{d+7 \times 5 d}{5 d+7 d}=\frac{36 d}{12 d}=3$. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $O$ be the origin, $A$ be a moving point on the parabola $x=\frac{1}{4} y^{2}+1$, and $B$ be a moving point on the parabola $y=x^{2}+4$. Then the minimum value of the area of $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3.2.
Let $A\left(s^{2}+1,2 s\right), B\left(t, t^{2}+4\right)$. Then $l_{O B}:\left(t^{2}+4\right) x-t y=0$.
Let the distance from point $A$ to line $O B$ be $h$, we have
$$
h=\frac{\left|\left(t^{2}+4\right)\left(s^{2}+1\right)-t \cdot 2 s\right|}{\sqrt{\left(t^{2}+4\right)^{2}+t^{2}}} \text {. }
$$
Therefore, $S_{\triangle O A B}=\frac{1}{2} O B h$
$$
\begin{array}{l}
=\frac{1}{2}\left|\left(t^{2}+4\right)\left(s^{2}+1\right)-2 s t\right| \\
=\frac{1}{2}\left[\left(s^{2} t^{2}+3 s^{2}+(s-t)^{2}+4\right] \geqslant 2 .\right.
\end{array}
$$
When $s=t=0$, the equality holds. | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the complex number $z=\cos \frac{4 \pi}{7}+\mathrm{i} \sin \frac{4 \pi}{7}$. Then
$$
\left|\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}\right|
$$
is equal to $\qquad$ (answer with a number). | 5. 2 .
Given that $z$ satisfies the equation $z^{7}-1=0$.
From $z^{7}-1=(z-1) \sum_{i=0}^{6} z^{i}$, and $z \neq 1$, we get $z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=0$. After combining the fractions of $\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}$, the denominator is
$$
\begin{array}{l}
\left(1+z^{2}\right)\left(1+z^{4}\right)\left(1+z^{6}\right) \\
=1+z^{2}+z^{4}+z^{6}+z^{6}+z^{8}+z^{10}+z^{12} \\
=1+z^{2}+z^{4}+z^{6}+z^{6}+z+z^{3}+z^{5}=z^{6},
\end{array}
$$
The numerator is
$$
\begin{array}{l}
z\left(1+z^{4}\right)\left(1+z^{6}\right)+z^{2}\left(1+z^{2}\right)\left(1+z^{6}\right)+ \\
z^{3}\left(1+z^{2}\right)\left(1+z^{4}\right) \\
=\left(1+z^{4}\right)(z+1)+\left(1+z^{2}\right)\left(z^{2}+z\right)+ \\
\left(1+z^{2}\right)\left(z^{3}+1\right) \\
=1+z+z^{4}+z^{5}+z+z^{2}+z^{3}+z^{4}+1+z^{2}+z^{3}+z^{5} \\
=2\left(1+z+z^{2}+z^{3}+z^{4}+z^{5}\right)=-2 z^{6} \text {. } \\
\end{array}
$$
Therefore, $\left|\frac{z}{1+z^{2}}+\frac{z^{2}}{1+z^{4}}+\frac{z^{3}}{1+z^{6}}\right|=\left|\frac{-2 z^{6}}{z^{6}}\right|=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b, c, d$ be real numbers, satisfying
$$
a+2 b+3 c+4 d=\sqrt{10} \text {. }
$$
Then the minimum value of $a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}$ is $\qquad$ | 6. 1 .
From the given equation, we have
$$
\begin{array}{l}
(1-t) a+(2-t) b+(3-t) c+(4-t) d+ \\
t(a+b+c+d)=\sqrt{10} .
\end{array}
$$
By the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
{\left[(1-t)^{2}+(2-t)^{2}+(3-t)^{2}+(4-t)^{2}+t^{2}\right] .} \\
{\left[a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2}\right] \geqslant 10} \\
\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}+(a+b+c+d)^{2} \\
\quad \geqslant \frac{10}{5 t^{2}-20 t+30}=\frac{10}{5(t-2)^{2}+10}=1 .
\end{array}
$$
Equality holds if and only if
$$
t=2, a=-\frac{\sqrt{10}}{10}, b=0, c=\frac{\sqrt{10}}{10}, d=\frac{\sqrt{10}}{5}
$$
Therefore, the minimum value sought is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Assign five college students to three villages in a certain town. If each village must have at least one student, then the number of different assignment schemes is $\qquad$ . | 2. 150 .
According to the number of college students allocated to each village, the only two types that meet the requirements are $1,1,3$ and $1, 2, 2$. Therefore, the number of different allocation schemes is
$$
C_{3}^{1} C_{5}^{3} C_{2}^{1}+C_{3}^{1} C_{5}^{2} C_{3}^{2}=60+90=150
$$ | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. If $\left(x^{2}-x-2\right)^{3}=a_{0}+a_{1} x+\cdots+a_{6} x^{6}$, then $a_{1}+a_{3}+a_{5}=$ | 3. -4 .
Let $x=0, x=1$, we get
$$
\begin{array}{l}
a_{0}=-8, \\
a_{0}+a_{1}+\cdots+a_{6}=\left(1^{2}-1-2\right)^{3}=-8 .
\end{array}
$$
Thus, $a_{1}+a_{2}+\cdots+a_{6}=0$.
Let $x=-1$, we get
$$
\begin{array}{l}
a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6} \\
=\left[(-1)^{2}-(-1)-2\right]^{3}=0 .
\end{array}
$$
Thus, $-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}+a_{6}=8$.
Therefore, $a_{1}+a_{3}+a_{5}=-4$. | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given an isosceles triangle with a vertex angle of $20^{\circ}$ and a base length of $a$, the length of the legs is $b$. Then the value of $\frac{a^{3}+b^{3}}{a b^{2}}$ is $\qquad$ | 4.3.
Given $a=2 b \sin 10^{\circ}$.
Thus $a^{3}+b^{3}=8 b^{3} \sin ^{3} 10^{\circ}+b^{3}$
$$
\begin{array}{l}
=8 b^{3} \cdot \frac{1}{4}\left(3 \sin 10^{\circ}-\sin 30^{\circ}\right)+b^{3}=6 b^{3} \sin 10^{\circ} \\
\Rightarrow \frac{a^{3}+b^{3}}{a b^{2}}=\frac{6 b^{3} \sin 10^{\circ}}{2 b \sin 10^{\circ} \cdot b^{2}}=3 .
\end{array}
$$ | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a_{n}=2^{n}, b_{n}=5 n-1\left(n \in \mathbf{Z}_{+}\right)$,
$$
S=\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{a_{2015}}\right\} \text {. }
$$
Then the number of elements in the set $S$ is | 5.504.
Since the set $\left\{b_{1}, b_{2}, \cdots, b_{2015}\right\}$ contains $2^{2015}$ elements, forming an arithmetic sequence with a common difference of 5, and each term has a remainder of 4 when divided by 5, we only need to consider the number of terms in the set $\left\{a_{1}, a_{2}, \cdots, a_{2015}\right\}$ that have a remainder of 4 when divided by 5.
For $k \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{4 k}=2^{4 k} \equiv 1(\bmod 5), \\
a_{4 k-1}=2^{4(k-1)} \times 2^{3} \equiv 2^{3} \equiv 3(\bmod 5), \\
a_{4 k-2}=2^{4(k-1)} \times 2^{2} \equiv 2^{2} \equiv 4(\bmod 5), \\
a_{4 k-3}=2^{4(k-1)} \times 2 \equiv 2(\bmod 5) . \\
\text { Let } 4 k-2 \leqslant 2015 \Rightarrow k \leqslant 504+\frac{1}{4}
\end{array}
$$
Therefore, the number of elements in set $S$ is 504. | 504 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a regular tetrahedron $P-ABC$ with the side length of the base being 6 and the side length of the lateral edges being $\sqrt{21}$. Then the radius of the inscribed sphere of the tetrahedron is $\qquad$ | 7.1 .
Let $P O \perp$ plane $A B C$ at point $O$. Then $O$ is the center of the equilateral $\triangle A B C$. Connect $A O$ and extend it to intersect $B C$ at point $D$, and connect $P D$. Thus, $D$ is the midpoint of $B C$.
It is easy to find that $P D=2 \sqrt{3}, O D=\sqrt{3}, P O=3$.
Let the radius of the inscribed sphere of this tetrahedron be $r$. Then
$$
\begin{aligned}
r & =\frac{3 V_{\text {tetrahedron } P-A B C}}{S_{\triangle A B C}+3 S_{\triangle P A B}} \\
& =\frac{3 \times \frac{\sqrt{3}}{4} \times 6^{2}}{\frac{\sqrt{3}}{4} \times 6^{2}+3 \times \frac{1}{2} \times 6 \times 2 \sqrt{3}}=1 .
\end{aligned}
$$ | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. The largest prime $p$ such that $\frac{p+1}{2}$ and $\frac{p^{2}+1}{2}$ are both perfect squares is $\qquad$. | 10.7.
Let $\frac{p+1}{2}=x^{2} ， \frac{p^{2}+1}{2}=y^{2}\left(x, y \in \mathbf{Z}_{+}\right)$.
Obviously, $p>y>x, p>2$.
From $p+1=2 x^{2}, p^{2}+1=2 y^{2}$, subtracting the two equations gives $p(p-1)=2(y-x)(y+x)$.
Since $p(p>2)$ is a prime number and $p>y-x$, then $p \mid(y+x)$.
Because $2 p>y+x$, so $p=y+x$.
Thus, $p-1=2(y-x) \Rightarrow p+1=4 x$
$$
\Rightarrow 4 x=2 x^{2} \Rightarrow x=2 \Rightarrow p=7, y=5 \text {. }
$$
Therefore, the only $p$ that satisfies the condition is $p=7$. | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. On a plane, there is an $8 \times 8$ grid colored in a black and white checkerboard pattern. Basil arbitrarily selects one of the cells. Each turn, Peter draws a polygon (which can be concave but not self-intersecting) on the grid, and Basil will honestly inform Peter whether the selected cell is inside or outside the polygon. To ensure that he can determine whether the cell Basil chose is white or black, what is the minimum number of turns Peter needs to ask? | 2. If the polygon drawn by Peter includes only all the cells of a certain color, then this polygon must intersect itself. Therefore, Peter cannot determine the color of the cell chosen by Basil in just one round.
Next, two strategies are given that can determine the color of the selected cell in two rounds.
【Strategy 1】The two polygon drawing methods shown in Figure 1 can determine whether the selected cell is white or black in two rounds.
If the selected cell is white, the answers in both rounds will either be inside the drawn polygon or outside the drawn polygon. If the selected cell is black, then in these two rounds, it will appear once inside the polygon and once outside the polygon.
【Strategy 2】The two polygon drawing methods shown in Figure 2 can also determine the color of the selected cell in two rounds. In the first round, draw the polygon in Figure 2 (a). If the selected cell is outside the polygon, it must be in the 2nd, 4th, 6th, or 8th column from left to right. In the second round, draw the polygon in Figure 2 (b), and determine whether the selected cell is black or white based on whether it is inside or outside the polygon. If the answer in the first round is inside the polygon, then the cell must be in the 1st, 3rd, 5th, or 7th column from left to right, or in the last row from top to bottom. In the second round, use the polygon in Figure 2 (c).
If the answer is inside the polygon, the selected cell is white; if the answer is outside the polygon, the selected cell is black.
【Note】If the problem does not require drawing the polygon within the table (i.e., it can be understood as being able to draw outside the table), then Strategy 1 can be simplified as shown in Figure 3. | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the complex numbers be
$$
\begin{array}{l}
z_{1}=(6-a)+(4-b) \mathrm{i}, \\
z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, \\
z_{3}=(3-a)+(3-2 b) \mathrm{i},
\end{array}
$$
where, $a, b \in \mathbf{R}$.
When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ reaches its minimum value, $3 a+4 b$ $=$ | 2. 12 .
Notice that,
$$
\begin{array}{l}
\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right| \\
=|12+9 \mathrm{i}|=15 .
\end{array}
$$
Equality holds if and only if
$$
\frac{6-a}{4-b}=\frac{3+2 a}{2+3 b}=\frac{3-a}{3-2 b}=\frac{12}{9} \text {, }
$$
i.e., when $a=\frac{7}{3}, b=\frac{5}{4}$, the minimum value 15 is achieved. Therefore, $3 a+4 b=12$. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the line $l$ passing through the origin intersect the graph of the function $y=|\sin x|$ $(x \geqslant 0)$ at exactly three points, with $\alpha$ being the largest of the x-coordinates of these intersection points. Then
$$
\frac{\left(1+\alpha^{2}\right) \sin 2 \alpha}{2 \alpha}=
$$
$\qquad$ . | 3. 1 .
As shown in Figure 2, let the line $l$ be tangent to the function $y=|\sin x|(x \geqslant 0)$
at point $P(\alpha,-\sin \alpha)$, and $k_{l}=-\cos \alpha$.
Then the line $l: y+\sin \alpha=-(x-\alpha) \cos \alpha$.
Substituting $(0,0)$, we get $\alpha=\tan \alpha$.
Therefore, $\frac{\left(1+\alpha^{2}\right) \sin 2 \alpha}{2 \alpha}=\frac{\left(1+\tan ^{2} \alpha\right) \sin 2 \alpha}{2 \tan \alpha}$
$$
=\frac{\sin 2 \alpha}{2 \cos ^{2} \alpha \cdot \tan \alpha}=1 .
$$ | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P(1,4,5)$ is a fixed point in the rectangular coordinate system $O-x y z$, a plane is drawn through $P$ intersecting the positive half-axes of the three coordinate axes at points $A$, $B$, and $C$ respectively. Then the minimum value of the volume $V$ of all such tetrahedrons $O-A B C$ is $\qquad$ | 4. 90 .
Let the plane equation be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where the positive numbers $a$, $b$, and $c$ are the intercepts of the plane on the $x$-axis, $y$-axis, and $z$-axis, respectively.
Given that point $P$ lies within plane $ABC$, we have $\frac{1}{a}+\frac{4}{b}+\frac{5}{c}=1$.
From $1=\frac{1}{a}+\frac{4}{b}+\frac{5}{c} \geqslant 3 \sqrt[3]{\frac{1}{a} \times \frac{4}{b} \times \frac{5}{c}}$ $\Rightarrow \frac{1}{6} a b c \geqslant 90$.
When $(a, b, c)=(3,12,15)$, $V$ reaches its minimum value of 90. | 90 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the number of solutions to the equation
$$
|| \cdots|||x|-1|-2| \cdots|-2011|=2011
$$
The number of solutions. ${ }^{[4]}$ | 【Analysis】Remove the absolute value symbols from outside to inside, step by step.
From the original equation, we get
$$
|| \cdots|||x|-1|-2| \cdots|-2010|=0
$$
or ||$\cdots|||x|-1|-2| \cdots|-2010|=4002$.
For equation (1), we have
$$
\begin{array}{l}
|| \cdots|||x|-1|-2| \cdots|-2009|=2010 \\
\Rightarrow|| \cdots|||x|-1|-2| \cdots|-2008| \\
\quad=2009+2010 \\
\Rightarrow|x|=1+2+\cdots+2010=2011 \times 1005 .
\end{array}
$$
Similarly, for equation (2), we have
$$
|x|=1+2+\cdots+2010+4022=2011 \times 1007 .
$$
Therefore, $x= \pm 2011 \times 1005$ or
$$
x= \pm 2011 \times 1007 \text {, }
$$
there are four solutions in total. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let real numbers $x, y$ satisfy
$$
x^{2}+\sqrt{3} y=4, y^{2}+\sqrt{3} x=4, x \neq y \text {. }
$$
Then the value of $\frac{y}{x}+\frac{x}{y}$ is $\qquad$ | 3. -5 .
From the conditions, we have
$$
\left\{\begin{array}{l}
x^{2}-y^{2}+\sqrt{3} y-\sqrt{3} x=0, \\
x^{2}+y^{2}+\sqrt{3}(x+y)=8 .
\end{array}\right.
$$
From equation (1) and $x \neq y$, we know $x+y=\sqrt{3}$.
Substituting into equation (2) gives $x^{2}+y^{2}=5$.
$$
\begin{array}{l}
\text { Also, } 2 x y=(x+y)^{2}-\left(x^{2}+y^{2}\right)=-2 \\
\Rightarrow x y=-1 . \\
\text { Therefore, } \frac{y}{x}+\frac{x}{y}=\frac{x^{2}+y^{2}}{x y}=-5 .
\end{array}
$$ | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups.
The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups. | 7.6 .
Rewrite the given equation as
$$
\begin{array}{l}
x^{2}-x(2+y)+y^{2}-2 y=0 \\
\Rightarrow \Delta=(2+y)^{2}-4\left(y^{2}-2 y\right)=-3 y^{2}+12 y+4 \\
\quad=-3(y-2)^{2}+16 \geqslant 0 \\
\Rightarrow|y-2| \leqslant \frac{4}{\sqrt{3}}<3 .
\end{array}
$$
Since $y$ is an integer, thus, $y=0,1,2,3,4$.
Upon calculation, only when $y=0,2,4$, $x$ is an integer, and we obtain
$$
\begin{aligned}
(x, y)= & (0,0),(2,0),(0,2),(4,2), \\
& (2,4),(4,4) .
\end{aligned}
$$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) From the 2015 positive integers 1, 2, $\cdots$, 2015, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$.
untranslated part:
在 1,2 , $\cdots, 2015$ 这 2015 个正整数中选出 $k$ 个数,使得其中任意两个不同的数之和均不为 50 的倍数. 求 $k$ 的最大值.
translated part:
From the 2015 positive integers 1, 2, $\cdots$, 2015, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$. | 10. Classify $1 \sim 2015$ by their remainders when divided by 50.
Let $A_{i}$ represent the set of numbers from $1 \sim 2015$ that have a remainder of $i$ when divided by 50, where $i=0,1, \cdots, 49$. Then $A_{1}, A_{2}, \cdots, A_{15}$ each contain 41 numbers, and $A_{0}, A_{16}, A_{17}, \cdots, A_{49}$ each contain 40 numbers.
According to the problem, at most one number can be selected from $A_{0}$ and $A_{25}$; and no numbers can be selected simultaneously from $A_{i}$ and $A_{50-i}$ (for $i=1,2, \cdots, 24$).
Notice that, when $1 \leqslant i \leqslant 15$, $A_{i}$ contains more numbers than $A_{50-i}$.
Therefore, all numbers from $A_{1}, A_{2}, \cdots, A_{15}$ can be selected.
For $16 \leqslant i \leqslant 24$, all numbers from $A_{i}$ (or $A_{50-i}$) can be selected.
In summary, the maximum value of $k$ is
$$
15 \times 41 + 9 \times 40 + 1 + 1 = 977 \text{. }
$$ | 977 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) As shown in Figure 4, the three sides of $\triangle ABC$ are all positive integers, and the perimeter is 35. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find the length of side $AB$. | 11. Extend $G I$, intersecting $C B$ and $C A$ (or their extensions) at points $P$ and $Q$ respectively.
Since $C I$ is the angle bisector of $\angle C$ and $\angle G I C=90^{\circ}$, we know that $\triangle C P Q$ is an isosceles triangle.
Draw $G E \perp P C$ and $G F \perp C Q$ from point $G$, and draw $I R \perp Q C$ from point $I$.
Let the inradius of $\triangle A B C$ be $r$, then $I R=r$.
Since $G$ is a point on the base of the isosceles $\triangle C P Q$, and $I$ is the midpoint of the base, then
$$
G E+G F=2 I R=2 r .
$$
Since $G$ is the centroid of $\triangle A B C$, we have
$$
G E=\frac{1}{3} h_{a}, G F=\frac{1}{3} h_{b} \text {. }
$$
Thus, $S_{\triangle A B C}=\frac{1}{3}\left(h_{a}+h_{b}\right)=2 r$
$$
\begin{array}{l}
\Rightarrow \frac{1}{3}\left(\frac{S}{\frac{1}{2} a}+\frac{S}{\frac{1}{2} b}\right)=2 \times \frac{S}{\frac{1}{2}(a+b+c)} \\
\Rightarrow 6 a b=(a+b)(a+b+c) \\
\Rightarrow 6 a b=35(a+b) \Rightarrow 6 I(a+b) .
\end{array}
$$
Given $18 \leqslant a+b<35$, thus $a+b$ can take the values 18, 24, 30.
Upon verification, only when $(a+b, a b)=(24,140)$, $a$ and $b$ have positive integer solutions.
Therefore, $c=35-(a+b)=11$, i.e., $A B=11$. | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the function $f: \mathbf{N} \rightarrow \mathbf{N}$ defined as follows:
$$
f(x)=\left\{\begin{array}{ll}
\frac{x}{2}, & x \text { is even; } \\
\frac{x+7}{2}, & x \text { is odd. }
\end{array}\right.
$$
Then the number of elements in the set $A=\{x \in \mathbf{N} \mid f(f(f(x)))=x\}$ is | - 1.8.
On one hand, when $x \in \{0,1, \cdots, 7\}$, it is calculated that $f(x) \in \{0,1, \cdots, 7\}$,
and it can be verified that $\{0,1, \cdots, 7\} \subseteq A$.
On the other hand, when $x \geqslant 8$,
$$
\frac{x}{2}<x \text{, and } \frac{x+7}{2}<\frac{2 x}{2}<x \text{. }
$$
Then $f(f(f(x)))<8$, which does not hold.
Therefore, $A=\{0,1, \cdots, 7\}$, with a total of 8 elements. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Real numbers $x, y, a$ satisfy $x+y=a+1$ and $xy=a^{2}-7a+16$. Then the maximum value of $x^{2}+y^{2}$ is $\qquad$ | 5. 32 .
Notice that,
$$
\begin{array}{l}
x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(a+1)^{2}-2\left(a^{2}-7 a+16\right) \\
=-a^{2}+16 a-31=-(a-8)^{2}+33 . \\
\text { Also, }(x-y)^{2}=(a+1)^{2}-4\left(a^{2}-7 a+16\right) \geqslant 0 \\
\Rightarrow-3 a^{2}+30 a-63 \geqslant 0 \Rightarrow 3 \leqslant a \leqslant 7 .
\end{array}
$$
Thus, on the closed interval $[3,7]$, the function
$$
f(a)=-(a-8)^{2}+33
$$
is increasing.
Therefore, its maximum value is obtained at $a=7$, i.e., the maximum value of $x^{2}+y^{2}$ is 32. | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a triangle with sides as three consecutive natural numbers, the largest angle is twice the smallest angle. Then the perimeter of the triangle is $\qquad$ | II, 7.15.
Let the three sides of $\triangle ABC$ be $AB=n+1, BC=n, AC=n-1$, and the angle opposite to side $AC$ be $\theta$, and the angle opposite to side $AB$ be $2\theta$.
According to the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \frac{n+1}{n-1}=2 \cos \theta, \\
\cos \theta=\frac{(n+1)^{2}+n^{2}-(n-1)^{2}}{2 n(n+1)} \\
=\frac{n+4}{2(n+1)} .
\end{array}
$$
From equations (1) and (2), we get
$$
\frac{n+1}{n-1}=\frac{n+4}{n+1} \Rightarrow n=5 \text {. }
$$
Therefore, the perimeter of the triangle is 15. | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the minimum value of the function
$$
f(x)=|x-1|+|x-2|+\cdots+|x-10|
$$
. ${ }^{[1]}$ | 【Analysis】By the geometric meaning of absolute value, $\sum_{i=1}^{n}\left|x-a_{i}\right|$ represents the sum of distances from the point corresponding to $x$ on the number line to the points corresponding to $a_{i}(i=1,2, \cdots, n)$. It is easy to know that when the point corresponding to $x$ is in the middle of the points corresponding to $a_{i}$ (when $n=2 k+1$ $(k \in \mathbf{N})$) or between the middle two points (when $n=2 k$ $\left(k \in \mathbf{Z}_{+}\right)$), the sum of distances is minimized.
When $x \in[5,6]$, $f(x)$ achieves its minimum value of 25.
By generalizing Example 1, it is not difficult to obtain
Conclusion 1 Suppose $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. For the function
$$
f(x)=\sum_{i=1}^{n}\left|x-a_{i}\right| \text {, }
$$
(1) When $n=2 k+1(k \in \mathbf{N})$, the point where $f(x)$ achieves its minimum value is $a_{k+1}$, and
$$
f(x)_{\min }=\sum_{i=k+2}^{n} a_{i}-\sum_{i=1}^{k} a_{i} ;
$$
(2) When $n=2 k\left(k \in \mathbf{Z}_{+}\right)$, every point in the interval $\left[a_{k}, a_{k+1}\right]$
is a point where $f(x)$ achieves its minimum value, and
$$
f(x)_{\min }=\sum_{i=k+1}^{n} a_{i}-\sum_{i=1}^{k} a_{i} .
$$ | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (16 points) As shown in Figure 4, $A$ and $B$ are the common vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. $P$ and $Q$ are moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})(\lambda \in \mathbf{R},|\lambda|>1) \text {. }
$$
Let $k_{1} 、 k_{2} 、 k_{3} 、 k_{4}$, then $k_{1}+k_{2}+k_{3}+k_{4}$ is a constant. | Prove: (1) $O, P, Q$ are collinear;
(2) If the slopes of the lines $AP, BP, AQ, BQ$ are respectively
14. (1) Note that,
$\overrightarrow{A P}+\overrightarrow{B P}=2 \overrightarrow{O P}, \overrightarrow{A Q}+\overrightarrow{B Q}=2 \overrightarrow{O Q}$.
Also, $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A Q}+\overrightarrow{B Q})$, hence $\overrightarrow{O P}=\lambda \overrightarrow{O Q}$.
Therefore, $O, P, Q$ are collinear.
(2) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$. Then
$$
x_{1}^{2}-a^{2}=\frac{a^{2}}{b^{2}} y_{1}^{2} \text {. }
$$
Thus, $k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x_{1}-a}$
$$
=\frac{2 x_{1} y_{1}}{x_{1}^{2}-a^{2}}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}} \text {. }
$$
Similarly, $k_{3}+k_{4}=-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}}$.
From (1), we know that $O, P, Q$ are collinear, i.e., $\frac{x_{1}}{y_{1}}=\frac{x_{2}}{y_{2}}$.
Therefore, from equations (1) and (2), we get
$$
\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4}=\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{1}}{y_{1}}-\frac{2 b^{2}}{a^{2}} \cdot \frac{x_{2}}{y_{2}} \\
=\frac{2 b^{2}}{a^{2}}\left(\frac{x_{1}}{y_{1}}-\frac{x_{2}}{y_{2}}\right)=0 .
\end{array}
$$ | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\sqrt{a+b+c+d}+\sqrt{a^{2}-2 a+3-b}- \\
\sqrt{b-c^{2}+4 c-8}=3 .
\end{array}
$$
Then the value of $a-b+c-d$ is ( ).
(A) $-7 \quad \square \quad$ -
(B) -8
(C) -4
(D) -6 | 5. A.
$$
\begin{array}{l}
\text { Given } a^{2}-2 a+3-b \geqslant 0, \\
b-c^{2}+4 c-8 \geqslant 0,
\end{array}
$$
we know
$$
\begin{array}{l}
b \leqslant-(a+1)^{2}+4 \leqslant 4, \\
b \geqslant(c-2)^{2}+4 \geqslant 4 .
\end{array}
$$
Thus, $b=4, a=-1, c=2$.
Substituting these into the known equations gives $d=4$.
Therefore, $a-b+c-d=-7$. | -7 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$
then $x+\frac{1}{x}=$ | 3. 4 .
Transform the given equation into
$$
\begin{aligned}
& \left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow & \left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{aligned}
$$
Notice that, $x^{3}+x^{-3}+53>0$.
Thus, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
Then $x^{3}+x^{-3}=\left(x+x^{-1}\right)\left[\left(x+x^{-1}\right)^{2}-3\right]$
$$
\begin{array}{l}
\Rightarrow b\left(b^{2}-3\right)=52 \\
\Rightarrow(b-4)\left(b^{2}+4 b+13\right)=0 \\
\Rightarrow b=4 .
\end{array}
$$
Therefore, $x+\frac{1}{x}=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given a moving large circle $\odot O$ that is externally tangent to a fixed small circle $\odot O_{1}$ with radius 3 at point $P, AB$ is the external common tangent of the two circles, with $A, B$ being the points of tangency. A line $l$ parallel to $AB$ is tangent to $\odot O_{1}$ at point $C$ and intersects $\odot O$ at points $D, E$. Then $C D \cdot C E=$ | 4.36.
As shown in Figure 5, connect $AP$, $PB$, $O_{1}C$, $BO_{1}$, $PC$, and draw the common tangent line of the two circles through point $P$, intersecting $AB$ at point $Q$.
It is easy to prove that $\angle APB=90^{\circ}$, points $B$, $O_{1}$, and $C$ are collinear, $\angle BPC=90^{\circ}$, and points $A$, $P$, and $C$ are collinear.
By the secant theorem and the projection theorem, we get
$$
CD \cdot CE = AC \cdot PC = BC^2 = 36 .
$$ | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) There are 288 sets of cards, totaling 2016 cards, each set consisting of $1,2, \cdots, 7$ and stacked in the order $1,2, \cdots, 7$ from top to bottom. Now, these 288 sets of cards are stacked together from top to bottom. First, discard the top five cards, then place the top card at the bottom, and continue discarding the top five cards, then placing the top card at the bottom, and so on, until only one card remains.
(1) In the above process, when only 301 cards are left, how many cards have been discarded?
(2) What is the last remaining card, and which set does it belong to? | (1) For the first 42 cards (in 6 groups each), according to the operation rule, after discarding 5 cards, the cards placed at the bottom are $6,5, \cdots, 1,7$. Thus, each number discards 5 cards.
And $288 \div 6=48, 48 \times 5 \times 7=1680$ (cards), $48 \times 5=240$ (cards), $2016-1680=336$ (cards). That is, when 1680 cards are discarded, each number has discarded 240 cards, leaving 336 cards. The arrangement of these 336 cards is
$$
6,5, \cdots, 1,7,6, \cdots, 1,7,6, \cdots, 1,7,
$$
When 35 more cards are discarded, the number 7 discards 4 more cards, so the card 7 has discarded a total of 244 cards.
(2) Note that if there are only $6^{n}$ cards, the last card discarded is the card numbered $6^{n}$.
$$
\begin{array}{l}
\text { Also } 6^{4}=1296<2016<6^{5}, \\
2016-1296=720,
\end{array}
$$
After discarding 720 cards, the card placed at the bottom is the one we are looking for.
When 700 cards are discarded, each 6 groups discard 35 cards, at this point, it is already the $700 \div 35 \times 6=120$th group. When discarding up to the 720th card, it is already the 2nd card of the 124th group, and the 3rd card is placed at the bottom. Therefore, the last card is the 3rd card of the 124th group. | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let real numbers $x_{1}, x_{2}, \cdots, x_{1999}$ satisfy the condition $\sum_{i=1}^{1990}\left|x_{i}-x_{i+1}\right|=1991$.
And $y_{k}=\frac{1}{k} \sum_{i=1}^{k} x_{i}(k=1,2, \cdots, 1991)$. Try to find the maximum value of $\sum_{i=1}^{1990}\left|y_{i}-y_{i+1}\right|$. ${ }^{[3]}$ | 【Analysis】For $k=1,2, \cdots, 1990$, we have
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right|=\left|\frac{1}{k} \sum_{i=1}^{k} x_{i}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}\right| \\
=\left|\frac{1}{k(k+1)}\left(\sum_{i=1}^{k} x_{i}-k x_{k+1}\right)\right| \\
\leqslant \frac{1}{k(k+1)} \sum_{i=1}^{k} i\left|x_{i}-x_{i+1}\right| . \\
\text { Hence } \sum_{i=1}^{190}\left|y_{i}-y_{i+1}\right| \\
\leqslant \sum_{i=1}^{1990}\left[i\left|x_{i}-x_{i+1}\right| \sum_{k=i}^{1990} \frac{1}{k(k+1)}\right] \\
=\left(\sum_{i=1}^{1990}\left|x_{k}-x_{k+1}\right|\right)\left(1-\frac{k}{1991}\right) \\
\leqslant 1991\left(1-\frac{1}{1991}\right)=1990 .
\end{array}
$$
The maximum value 1990 is achieved when $x=1991, x_{2}=x_{3}=\cdots=x_{1991}=0$. | 1990 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}, \cdots, b_{31}$ satisfy
$$
\begin{array}{l}
\text { (1) } a_{1}<a_{2}<\cdots<a_{31} \leqslant 2015, \\
b_{1}<b_{2}<\cdots<b_{31} \leqslant 2015 ; \\
\text { (2) } a_{1}+a_{2}+\cdots+a_{31}=b_{1}+b_{2}+\cdots+b_{31} \text {. } \\
\text { Find } S=\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{31}-b_{31}\right|
\end{array}
$$
the maximum value.
(Supplied by He Yijie) | 1. Define the sets
$$
\begin{array}{l}
A=\left\{m \mid a_{m}>b_{m}, 1 \leqslant m \leqslant 31\right\}, \\
B=\left\{n \mid a_{n}<b_{n}, 1 \leqslant n \leqslant 31\right\} . \\
\text { Let } S_{1}=\sum_{m \in A}\left(a_{m}-b_{m}\right), S_{2}=\sum_{m \in B}\left(b_{n}-a_{n}\right) .
\end{array}
$$
Then $S=S_{1}+S_{2}$.
From condition (2), we know
$$
\begin{array}{l}
S_{1}-S_{2}=\sum_{m \in A \cup B}\left(a_{m}-b_{m}\right)=0 \\
\Rightarrow S_{1}=S_{2}=\frac{S}{2} . \\
\text { When } A=\varnothing, S=2 S_{1}=0 .
\end{array}
$$
Assume $A \neq \varnothing$, then $B \neq \varnothing$. In this case, $|A|, |B|$ are positive integers, and $|A|+|B| \leqslant 31$.
$$
\begin{array}{l}
\text { Let } u=a_{k}-b_{k}=\max _{m \in A}\left\{a_{m}-b_{m}\right\}, \\
v=b_{l}-a_{l}=\max _{n \in B}\left\{b_{n}-a_{n}\right\} .
\end{array}
$$
We will prove: $u+v \leqslant 1984$.
Without loss of generality, assume $1 \leqslant k<l \leqslant 31$.
Then $u+v=a_{k}-b_{k}+b_{l}-a_{l}$
$$
=b_{31}-\left(b_{31}-b_{l}\right)-b_{k}-\left(a_{l}-a_{k}\right) \text {. }
$$
From condition (1), we have
$$
\begin{array}{l}
b_{31} \leqslant 2015, b_{31}-b_{l} \geqslant 31-l, \\
b_{k} \geqslant k, a_{l}-a_{k} \geqslant l-k .
\end{array}
$$
Thus, $u+v \leqslant 2015-(31-l)-k-(l-k)$ $=1984$.
Clearly, $S_{1} \leqslant u|A|, S_{2} \leqslant v|B|$, hence,
$$
\begin{array}{l}
1984 \geqslant u+v \geqslant \frac{S_{1}}{|A|}+\frac{S_{2}}{|B|} \geqslant \frac{S_{1}}{|A|}+\frac{S_{2}}{31-|A|} \\
=\frac{S}{2} \cdot \frac{31}{|A|(31-|A|)} \geqslant \frac{31 S}{2 \times 15 \times 16} \\
\Rightarrow S \leqslant \frac{2 \times 15 \times 16}{31} \times 1984=30720 .
\end{array}
$$
On the other hand, if we take
$$
\begin{array}{l}
\left(a_{1}, a_{2}, \cdots, a_{16}, a_{17}, a_{18}, \cdots, a_{31}\right) \\
=(1,2, \cdots, 16,2001,2002, \cdots, 2015), \\
\left(b_{1}, b_{2}, \cdots, b_{31}\right)=(961,962, \cdots, 991),
\end{array}
$$
then conditions (1) and (2) are satisfied. In this case,
$$
S=2 S_{1}=2 \times 16 \times 960=30720 \text {. }
$$
In conclusion, the maximum value of $S$ is 30720. | 30720 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the set $A=\left\{x \left\lvert\, \frac{x+4}{x-3} \leqslant 0\right., x \in \mathbf{Z}\right\}$, and from set $A$ a random element $x$ is drawn, denoted by $\xi=x^{2}$. Then the mathematical expectation of the random variable $\xi$ is $\mathrm{E} \xi=$ $\qquad$ | -1.5 .
From the conditions, we know that
$$
A=\{-4,-3,-2,-1,0,1,2\},
$$
The values of the random variable $\xi$ are $0, 1, 4, 9, 16$.
It is easy to obtain that the probability distribution of $\xi$ is shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline$\xi$ & 0 & 1 & 4 & 9 & 16 \\
\hline$P$ & $\frac{1}{7}$ & $\frac{2}{7}$ & $\frac{2}{7}$ & $\frac{1}{7}$ & $\frac{1}{7}$ \\
\hline
\end{tabular}
Therefore, $\mathrm{E} \xi$
$$
\begin{array}{l}
=0 \times \frac{1}{7}+1 \times \frac{2}{7}+4 \times \frac{2}{7}+9 \times \frac{1}{7}+16 \times \frac{1}{7} \\
=5 .
\end{array}
$$ | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=x+g(x)$, where $g(x)$ is a function defined on $\mathbf{R}$ with the smallest positive period of 2. If the maximum value of $f(x)$ in the interval $[2,4)$ is 1, then the maximum value of $f(x)$ in the interval $[10,12)$ is $\qquad$
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 2.9.
According to the problem, we have
$$
\begin{array}{l}
f(x+2)=(x+2)+g(x+2) \\
=x+g(x)+2=f(x)+2 .
\end{array}
$$
Given that the maximum value of $f(x)$ in the interval $[2,4)$ is 1, we know that the maximum value of $f(x)$ in the interval $[4,6)$ is 3, ... and the maximum value of $f(x)$ in the interval $[10,12)$ is $3+2 \times 3=9$. | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y, z$ satisfy $x^{2}+2 y^{2}+3 z^{2}=24$.
Then the minimum value of $x+2 y+3 z$ is $\qquad$ . | 4. -12 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x+2 y+3 z)^{2} \\
=(1 \times x+\sqrt{2} \times \sqrt{2} y+\sqrt{3} \times \sqrt{3} z)^{2} \\
\leqslant\left[1^{2}+(\sqrt{2})^{2}+(\sqrt{3})^{2}\right]\left(x^{2}+2 y^{2}+3 z^{2}\right) \\
\quad=144 .
\end{array}
$$
Therefore, $x+2 y+3 z \geqslant-12$, with equality holding if and only if $x=y=z=-2$.
Hence, the minimum value of $x+2 y+3 z$ is -12. | -12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. If $\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\cdots+\sin \frac{n \pi}{9}=\frac{1}{2} \tan \frac{4 \pi}{9}$, then the smallest positive integer $n$ is $\qquad$. | 10.4.
Notice,
$$
\begin{array}{l}
2 \sin \frac{\pi}{18}\left(\sum_{k=1}^{n} \sin \frac{k \pi}{9}\right) \\
=\sum_{k=1}^{n}\left(\cos \frac{(2 k-1) \pi}{18}-\cos \frac{(2 k+1) \pi}{18}\right) \\
=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \tan \frac{4 \pi}{9} \cdot \sin \frac{\pi}{18}=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \cos \frac{\pi}{18}=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \cos \frac{(2 n+1) \pi}{18}=0 \\
\Rightarrow n=9 k+4(k \in \mathbf{Z}) .
\end{array}
$$
Therefore, the smallest positive integer $n$ is 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. In the coordinate plane, points with both coordinates as integers are called integer points, and triangles with all three vertices as integer points are called integer point triangles. Find the number of integer point right triangles $OAB$ with a right angle at the origin $O$ and with $I(2015,7 \times 2015)$ as the incenter. | 14. Let point $A$ be in the first quadrant.
Let $\angle x O I=\alpha$.
Then $\tan \alpha=7$,
$k_{O A}=\tan \left(\alpha-\frac{\pi}{4}\right)=\frac{\tan \alpha-1}{1+\tan \alpha}=\frac{3}{4}$
$\Rightarrow k_{O B}=-\frac{4}{3}$.
Since $A$ and $B$ are integer points, set
$$
A\left(4 t_{1}, 3 t_{1}\right), B\left(-3 t_{2}, 4 t_{2}\right)\left(t_{1} 、 t_{2} \in \mathbf{Z}_{+}\right) \text {. }
$$
Thus, $|O A|=5 t_{1},|O B|=5 t_{2}$.
Let the inradius of $\triangle O A B$ be $r$. Then
$r=\frac{\sqrt{2}}{2}|O I|=\frac{\sqrt{2}}{2} \times 5 \sqrt{2} \times 2015=5 \times 2015$.
Also, $r=\frac{|O A|+|O B|-|A B|}{2}$
$\Rightarrow|A B|=|O A|+|O B|-2 r$
$\Rightarrow|A B|^{2}=(|O A|+|O B|-2 r)^{2}=|O A|^{2}+|O B|^{2}$
$\Rightarrow\left(5 t_{1}+5 t_{2}-2 \times 5 \times 2015\right)^{2}=25 t_{1}^{2}+25 t_{2}^{2}$
$\Rightarrow\left(t_{1}+t_{2}-2 \times 2015\right)^{2}=t_{1}^{2}+t_{2}^{2}$.
Set $t_{1}=x+2015, t_{2}=y+2015$. Then
$(x+y)^{2}=(x+2015)^{2}+(y+2015)^{2}$
$\Rightarrow x y=2015 x+2015 y+2015^{2}$
$\Rightarrow(x-2015)(y-2015)=2 \times 2015^{2}$
$=2 \times 5^{2} \times 13^{2} \times 31^{2}$.
Since $|O A|>2 r,|O B|>2 r$, we know
$x-2015, y-2015 \in \mathbf{Z}_{+}$.
Note that, $2 \times 5^{2} \times 13^{2} \times 31^{2}$ has $2 \times 3 \times 3 \times 3=54$ positive divisors.
Thus, there are 54 pairs of $(x, y)$ that satisfy the conditions.
Therefore, there are 54 triangles that satisfy the conditions. | 54 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. For any positive integer $m$, the set
$$
\{m, m+1, m+2, \cdots, m+99\}
$$
any $n(n \geqslant 3)$-element subset of it, always contains three elements that are pairwise coprime. Find the minimum value of $n$. | 15. Consider the set $\{1,2, \cdots, 100\}$ with $m=1$, and its 67-element subset, whose elements are even numbers and odd numbers divisible by 3, i.e.,
$$
P=\{2,4, \cdots, 100,3,9, \cdots, 99\} .
$$
Clearly, there do not exist three pairwise coprime elements in set $P$.
Thus, $n \leqslant 67$ does not meet the requirement.
We now prove that $n=68$ meets the requirement.
First, we prove a lemma.
Lemma For any positive integer $m$, in any five-element subset of the set
$$
\{m, m+1, \cdots, m+5\}
$$
there are always three elements that are pairwise coprime.
Proof Let $A$ be a five-element subset of the set $\{m, m+1, \cdots, m+5\}$.
Notice that among $m, m+1, \cdots, m+5$, there are three odd and three even numbers, and exactly one of them is a multiple of 5.
Thus, if set $A$ contains three odd numbers, these three odd numbers must be pairwise coprime, and the conclusion holds.
If set $A$ contains two odd and three even numbers, since at most one of the three even numbers is a multiple of 3, and at most one is a multiple of 5, there must be one even number that is neither a multiple of 3 nor 5. This number is coprime with the two odd numbers, and the conclusion holds.
Returning to the original problem.
For any positive integer $m$, partition the set
$$
\{m, m+1, \cdots, m+99\}
$$
into the following 17 sets:
$$
A_{i}=\{m+6(i-1)+j \mid j=0,1, \cdots, 5\}
$$
$(i=1,2, \cdots, 16)$,
$$
A_{17}=\{m+96, m+97, m+98, m+99\} .
$$
Let $M$ be a 68-element subset of the set $\{m, m+1, \cdots, m+99\}$.
(1) If set $M$ has four elements from set $A_{17}$, since when $m$ is odd, $m+96, m+97, m+98$ are pairwise coprime; and when $m$ is even, $m+97, m+98, m+99$ are pairwise coprime, therefore, set $M$ contains at least three elements that are pairwise coprime.
(2) If set $M$ has at most three elements from set $A_{17}$, then set $M$ has at least 65 elements from sets $A_{1}, A_{2}, \cdots, A_{16}$.
By the pigeonhole principle, set $M$ must have at least five elements from one of these sets, say $A_{1}$. By the lemma, there exist three pairwise coprime elements among them. Therefore, set $M$ always contains three pairwise coprime elements.
Thus, $n=68$ meets the requirement, i.e., for any positive integer $m$, any 68-element subset of the set $\{m, m+1, \cdots, m+99\}$ contains three elements that are pairwise coprime.
In conclusion, the minimum value of $n$ is 68.
(Provided by Chen Deyan) | 68 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the function $f(x)=\frac{(x+1)^{2}+\sin x}{x^{2}+1}$ have the maximum value and minimum value as $M$ and $N$, respectively. Then $M+N=$ | 3. 2 .
From the given information, we have
$$
f(x)=\frac{(x+1)^{2}+\sin x}{x^{2}+1}=1+\frac{2 x+\sin x}{x^{2}+1} \text {. }
$$
Notice that the function $g(x)=\frac{2 x+\sin x}{x^{2}+1}$ is an odd function. Therefore, the maximum value $M_{0}$ and the minimum value $N_{0}$ of $g(x)$ satisfy
$$
\begin{array}{l}
M_{0}+N_{0}=0 . \\
\text { Also, } M=M_{0}+1, N=N_{0}+1 \Rightarrow M+N=2 .
\end{array}
$$ | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. After removing all perfect squares from the sequence of positive integers $\{1,2, \cdots\}$, the remaining numbers form a sequence $\left\{a_{n}\right\}$ in their original order. Then $a_{2015}=$ $\qquad$ . | 6.2060.
Let $k^{2}2014
\end{array}\right. \\
\Rightarrow k=45, a_{2015}=2060 .
\end{array}
$ | 2060 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and
$$
|b-c| \cos \frac{A}{2}=8,(b+c) \sin \frac{A}{2}=15 \text {. }
$$
Then $a=$ | 2. 17 .
From the cosine theorem, we get
$$
\begin{aligned}
a^{2}= & b^{2}+c^{2}-2 b c \cos A \\
= & \left(b^{2}+c^{2}\right)\left(\sin ^{2} \frac{A}{2}+\cos ^{2} \frac{A}{2}\right)- \\
& 2 b c\left(\cos ^{2} \frac{A}{2}-\sin ^{2} \frac{A}{2}\right) \\
= & (b+c)^{2} \sin ^{2} \frac{A}{2}+(b-c)^{2} \cos ^{2} \frac{A}{2} \\
= & 15^{2}+8^{2}=17^{2} .
\end{aligned}
$$
Therefore, $a=17$. | 17 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $\left\{a_{n}\right\}$ be a monotonically increasing sequence of positive integers, satisfying
$$
a_{n+2}=3 a_{n+1}-a_{n}, a_{6}=280 \text {. }
$$
Then $a_{7}=$ | 7.733.
From the problem, we have
$$
\begin{array}{l}
a_{3}=3 a_{2}-a_{1}, \\
a_{4}=3 a_{3}-a_{2}=8 a_{2}-3 a_{1}, \\
a_{5}=3 a_{4}-a_{3}=21 a_{2}-8 a_{1}, \\
a_{6}=3 a_{5}-a_{4}=55 a_{2}-21 a_{1}=280=5 \times 7 \times 8, \\
a_{7}=3 a_{6}-a_{5}=144 a_{2}-55 a_{1} .
\end{array}
$$
Since $(21,55)=1$, and $a_{2}>a_{1}$ are both positive integers, it follows that $5\left|a_{1}, 7\right| a_{2}$.
Let $a_{1}=5 b, a_{2}=7 c\left(b, c \in \mathbf{Z}_{+}\right)$, we get $11 c-3 b=8$.
It is easy to verify that $a_{1}=5, a_{2}=7$.
Then $a_{7}=144 a_{2}-55 a_{1}=733$. | 733 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Four numbers
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \cdot \sqrt{2-\sqrt{2-\sqrt{3}}} 、 \\
\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}}
\end{array}
$$
The product of these is ( ).
(A) $2+\sqrt{3}$
(B) 2
(C) 1
(D) $2-\sqrt{3}$ | $-1 . \mathrm{C}$.
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \times \\
\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \\
\quad \sqrt{2^{2}-\left(\sqrt{2-\sqrt{2-\sqrt{3}})^{2}}\right.} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \\
\quad \sqrt{2+\sqrt{2-\sqrt{3}}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2^{2}-(\sqrt{2-\sqrt{3}})^{2}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2+\sqrt{3}}=\sqrt{4-3}=1 .
\end{array}
$$ | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given positive integers $a$, $b$, $c$, $d$ satisfy $a^{2}=c(d+29)$, $b^{2}=c(d-29)$. Then the value of $d$ is $\qquad$. | $=, 1.421$
Let $(a, b)=d^{\prime}, a=d^{\prime} a_{1}, b=d^{\prime} b_{1},\left(a_{1}, b_{1}\right)=1$.
From the given conditions, dividing the two equations yields
$$
\begin{array}{l}
\frac{a^{2}}{b^{2}}=\frac{d+29}{d-29} \Rightarrow \frac{a_{1}^{2}}{b_{1}^{2}}=\frac{d+29}{d-29} \\
\Rightarrow\left\{\begin{array}{l}
d+29=a_{1}^{2} t, \\
d-29=b_{1}^{2} t
\end{array}\left(t \in \mathbf{Z}_{+}\right)\right. \\
\Rightarrow 2 \times 29=\left(a_{1}+b_{1}\right)\left(a_{1}-b_{1}\right) t .
\end{array}
$$
Since $a_{1}+b_{1}$ and $a_{1}-b_{1}$ have the same parity,
$$
t=2, a_{1}+b_{1}=29, a_{1}-b_{1}=1 \text {. }
$$
Thus, $a_{1}=15, b_{1}=14, d=421$. | 421 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Several different numbers are written on the blackboard, such that the sum of any three of them is a rational number, while the sum of any two is an irrational number. The maximum number of numbers that can be written on the blackboard is $\qquad$ | 4. 3 .
Assume that the numbers written on the blackboard are no less than four, denoted as $a, b, c, d$. Then, $a+b+c$ and $b+c+d$ are both rational numbers, which implies that their difference
$$
(b+c+d)-(a+b+c)=d-a
$$
is also a rational number.
Similarly, $b-a$ and $c-a$ are also rational numbers.
Therefore, $b=a+r_{1}, c=a+r_{2}, d=a+r_{3}$, where $r_{1}, r_{2}, r_{3}$ are rational numbers.
Furthermore, since $a+b+c=3a+r_{1}+r_{2}$ is a rational number, it follows that $a$ is also a rational number. This indicates that $a+b=2a+r_{1}$ is a rational number, which contradicts the condition that "the sum of any two is irrational."
Therefore, the numbers written on the blackboard do not exceed three, such as $\sqrt{2}$, $2\sqrt{2}$, and $-3\sqrt{2}$. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $a$ and $b$ are positive integers, the fraction $\frac{a}{b}$, when converted to a decimal, contains the consecutive digits $\overline{2015}$. Then the minimum value of $b$ is . $\qquad$ | 5. 129.
Given that multiplying a fraction by an appropriate power of 10 can result in the first four digits after the decimal point being $\overline{2015}$, we only need to find the minimum value of $b$ for the fraction $\frac{a}{b}=0.2015 \cdots$.
On one hand,
$\frac{a}{b}-\frac{1}{5}=0.2015 \cdots-0.2=0.0015$
$$
125 \text {. }
$$
We will test the values below.
When $b=126$,
$\frac{25}{126}=0.1984 \cdots \neq 0.2015 \cdots$;
When $b=127$, $\frac{26}{127}=0.2047 \cdots$;
When $b=128$, $\frac{26}{128}=0.2031 \cdots$;
When $b=129$, $\frac{26}{129}=0.2015 \cdots$.
Therefore, the minimum value of $b$ is 129. | 129 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the sequence $\left\{a_{n}\right\}$ :
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=5, \\
a_{n}=a_{n-1}-a_{n-2}+a_{n-3}(n=4,5, \cdots) .
\end{array}
$$
Then the sum of the first 2015 terms of this sequence $S_{2015}=$ | -、1.6045.
From the given information, we have $a_{4}=3$, and the sequence $\left\{a_{n}\right\}$ has a period of 4. Therefore, $S_{2015}$
$$
\begin{array}{l}
=503\left(a_{1}+a_{2}+a_{3}+a_{4}\right)+a_{1}+a_{2}+a_{3} \\
=503(1+3+5+3)+1+3+5=6045 .
\end{array}
$$ | 6045 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$1.2014^{2015}$ 的个位数为
The unit digit of $1.2014^{2015}$ is | -1.4 .
Let $g(n)$ denote the unit digit of a natural number $n$.
$$
\begin{array}{l}
\text { Then } g\left(2014^{2015}\right) \\
=g\left((201 \times 10+4)^{2015}\right) \\
=g\left(4^{2015}\right)=g\left(\left(4^{2}\right)^{1007} \times 4\right) \\
=g\left((10+6)^{1007} \times 4\right)=g\left(6^{1007} \times 4\right) \\
=g\left(g\left(6^{1007}\right) g(4)\right)=g(6 \times 4)=4 .
\end{array}
$$
Therefore, the unit digit of $2014^{2015}$ is 4. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $P$ be any point on the graph of the function $f(x)=x+\frac{2}{x}(x>0)$, and draw perpendiculars from point $P$ to the $x$-axis and $y$-axis, with the feet of the perpendiculars being $A$ and $B$ respectively. Then the minimum value of $|P A|+|P B|$ is $\qquad$ | 2. 4 .
Let $P(x, y)$. According to the problem,
$$
|P A|=|y|=y,|P B|=|x|=x,
$$
where, $y=x+\frac{2}{x}(x>0)$.
$$
\begin{array}{l}
\text { Hence }|P A|+|P B|=y+x=2\left(x+\frac{1}{x}\right) \\
\geqslant 2 \times 2 \sqrt{x \cdot \frac{1}{x}}=4,
\end{array}
$$
The equality holds if and only if $x=\frac{1}{x}$.
Since $x>0$, thus, when $x=1$, the above inequality takes the equality.
Therefore, when and only when $x=1$, $|P A|+|P B|$ has the minimum value 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)=a x^{5}+b x^{3}+c x+10$, and $f(3)$ $=3$. Then $f(-3)=$ $\qquad$ | 5. 17 .
Notice that, for any $x$, we have $f(x)+f(-x)=20$.
Therefore, $f(-3)=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Try to find the largest integer $k$, such that for each positive integer $n$, we have
$$
1980^{k} \left\lvert\, \frac{(1980 n)!}{(n!)^{1980}} .\right.
$$ | Notice,
$$
\begin{array}{l}
1980=2^{2} \times 3^{2} \times 5 \times 11, \\
v_{11}(1980!)=\sum_{i \geqslant 1}\left[\frac{1980}{11^{i}}\right]=197 . \\
\text { Since }\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right], \text { hence } \\
v_{5}(m!) \geqslant 2 v_{11}(m!) .
\end{array}
$$
Since $\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right]$, hence
similarly, $v_{3}(m!) \geqslant 3 v_{11}(m!)$,
$$
v_{2}(m!) \geqslant 5 v_{11}(m!) \text {. }
$$
Then the maximum $k$ satisfying $1980^{k} \mid 1980!$ is 197.
For any prime $p$,
$$
\begin{array}{l}
v_{p}\left(\frac{(1980 n)!}{(n!)^{1980}}\right) \\
=v_{p}((1980 n)!)-1980 v_{p}(n!) \\
=\frac{1980 S_{p}(n)-S_{p}(1980 n)}{p-1} .
\end{array}
$$
Using $S_{p}(m n) \leqslant S_{p}(m) S_{p}(n)$, we get
$$
\begin{array}{l}
v_{p}\left(\frac{(1980 n)!}{(n!)^{1980}}\right) \geqslant \frac{1980-S_{p}(1980)}{p-1} S_{p}(n) \\
\geqslant v_{p}(1980!) .
\end{array}
$$
Therefore, for any positive integer $n$, we have
$$
1980^{197} \left\lvert\, \frac{(1980 n)!}{(n!)^{1980}}\right. \text {. }
$$
In summary, the maximum $k$ is 197. | 197 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. As shown in Figure 3, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through point $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
then the value of $m+n$ is
$\qquad$ | 14. 2 .
Since $O$ is the midpoint of side $B C$, we have
$$
\overrightarrow{A O}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C})=\frac{m}{2} \overrightarrow{A M}+\frac{n}{2} \overrightarrow{A N} \text {. }
$$
Since points $M, O, N$ are collinear, it follows that,
$$
\frac{m}{2}+\frac{n}{2}=1 \Rightarrow m+n=2 \text {. }
$$ | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. Given $x, y \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right], a \in \mathbf{R}$, and $x^{3}+\sin x-2 a=0,4 y^{3}+\frac{1}{2} \sin 2 y+a=0$. Then the value of $\cos (x+2 y)$ is $\qquad$ | 15. 1 .
Solving the system of equations and eliminating $a$ yields
$$
x=-2 y \Rightarrow \cos (x+2 y)=1
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Polynomial
$$
p(x)=x^{3}-224 x^{2}+2016 x-d
$$
has three roots that form a geometric progression. Then the value of $d$ is $\qquad$ | 2. 729 .
Let the three roots of the polynomial be $a$, $b$, and $c$, and $b^{2}=a c$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
a+b+c=224, \\
a b+b c+c a=2016, \\
a b c=d .
\end{array}\right.
$$
Then $b=\frac{b(a+b+c)}{a+b+c}=\frac{a b+b c+a c}{a+b+c}=9$.
Thus, $d=a b c=b^{3}=729$. | 729 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $a, b, c, d$ are prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. [1] | Let
\[
\begin{array}{l}
a b c d=x+(x+1)+\cdots+(x+34) \\
=5 \times 7(x+17) .
\end{array}
\]
Assume \( a=5, b=7, c \leqslant d, c d=x+17 \).
Thus, \( d_{\text {min }}=5 \).
If \( d=5 \), then \( c_{\text {min }}=5 \), at this point,
\[
x=8, a+b+c+d=22 \text {; }
\]
If \( d=7 \), then \( c_{\text {min }}=3 \), at this point,
\[
x=4, a+b+c+d=22 \text {; }
\]
If \( d \geqslant 11 \), then
\[
a+b+c+d \geqslant 5+7+2+11=25 \text {. }
\]
In summary, the minimum value of \( a+b+c+d \) is 22. | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given positive integer $n=a b c<10000, a, b, c$ are all prime numbers, and $2 a+3 b=c, 4 a+c+1=4 b$. Find the value of $n$.
| From the given equations, we have
$$
b=6 a+1, c=20 a+3 \text{. }
$$
Then $a(6 a+1)(20 a+3)<10000$
$\Rightarrow 12 a^{3}<10000 \Rightarrow$ prime $a<5$.
Combining $b=6 a+1, c=20 a+3$ being primes, we can determine
$$
a=2, b=13, c=43, n=1118 \text{. }
$$ | 1118 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. For any $\alpha, \beta \in\left(0, \frac{2 \pi}{3}\right)$, we have
$$
\begin{array}{l}
4 \cos ^{2} \alpha+2 \cos \alpha \cdot \cos \beta+4 \cos ^{2} \beta- \\
3 \cos \alpha-3 \cos \beta-k<0 .
\end{array}
$$
Then the minimum value of $k$ is | 5.6.
Substitution $\left(x_{1}, x_{2}\right)=(2 \cos \alpha, 2 \cos \beta)$.
Then the given inequality becomes
$$
2 x_{1}^{2}+x_{1} x_{2}+2 x_{2}^{2}-3 x_{1}-3 x_{2}-2 k<0 \text {. }
$$
Since $x_{1}, x_{2} \in(-1,2)$, we have
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{1}-2\right)+\left(x_{2}+1\right)\left(x_{2}-2\right)+ \\
\frac{1}{2}\left(x_{1}+1\right)\left(x_{2}-2\right)+\frac{1}{2}\left(x_{1}-2\right)\left(x_{2}+1\right)<0 \\
\Rightarrow 2 x_{1}^{2}+x_{1} x_{2}+2 x_{2}^{2}-3 x_{1}-3 x_{2}-12<0 .
\end{array}
$$
When $x_{1}=x_{2}=x \rightarrow 2$ (or -1), the left side of the above equation
$$
\rightarrow 0 \text {. }
$$
Therefore, $k_{\min }=6$. | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let $a_{1} \in \mathbf{Z}_{+}$, and $a_{1} \leqslant 18$, define the sequence $\left\{a_{n}\right\}:$
$$
a_{n+1}=\left\{\begin{array}{ll}
2 a_{n}, & a_{n} \leqslant 18 ; \\
2 a_{n}-36, & a_{n}>18
\end{array}(n=1,2, \cdots) .\right.
$$
Find the maximum number of elements in the set $M=\left\{a_{n} \mid n \in \mathbf{Z}_{+}\right\}$. | 11. Given the positive integer $a_{1} \leqslant 18$ and the recursive formula for the sequence $\left\{a_{n}\right\}$, the following properties can be derived:
(1) $a_{n+1} \equiv 2 a_{n}(\bmod 36)$, i.e.,
$a_{n+1} \equiv 2 a_{n}(\bmod 4)$, and $a_{n+1} \equiv 2 a_{n}(\bmod 9)$;
(2) All terms are positive integers, and for any $n \in$ $\mathbf{Z}_{+}, a_{n} \leqslant 36$;
(3) From the third term onwards, all terms are multiples of 4, i.e., for any $n \geqslant 3,4 \mid a_{n}$;
(4) $3\left|a_{n} \Leftrightarrow 3\right| a_{1}$, i.e., if $a_{1}$ is a multiple of 3, then all terms are multiples of 3; otherwise, none of the terms are multiples of 3.
Consider the maximum value of $|M|$ in the following two scenarios.
【Scenario 1】If $3 \mid a_{1}$, then consider $a_{3}$.
(i) $\left\{\begin{array}{l}a_{3} \equiv 3(\bmod 9), \\ a_{3} \equiv 0(\bmod 4) .\end{array}\right.$
By the Chinese Remainder Theorem, we get
$$
\begin{array}{l}
a_{3} \equiv 12(\bmod 36) \Rightarrow a_{3} \equiv 12 \\
\Rightarrow\left(a_{3}, a_{4}, a_{5}, a_{6}, \cdots\right) \equiv(12,24,12,24, \cdots) .
\end{array}
$$
Considering the first two terms, we get $|M| \leqslant 4$.
(ii) $\left\{\begin{array}{l}a_{3} \equiv 6(\bmod 9) \\ a_{3} \equiv 0(\bmod 4)\end{array}\right.$.
By the Chinese Remainder Theorem, we get
$$
\begin{array}{l}
a_{3} \equiv 24(\bmod 36) \\
\Rightarrow\left(a_{3}, a_{4}, a_{5}, a_{6}, \cdots\right) \equiv(24,12,24,12, \cdots) .
\end{array}
$$
Considering the first two terms, we get $|M| \leqslant 4$.
(iii) $\left\{\begin{array}{l}a_{3} \equiv 0(\bmod 9), \\ a_{3} \equiv 0(\bmod 4)\end{array}\right.$.
Then $a_{3} \equiv 0(\bmod 36)$
$\Rightarrow\left(a_{3}, a_{4}, a_{5}, \cdots\right) \equiv(36,36,36, \cdots)$.
Considering the first two terms, we get $|M| \leqslant 3$.
【Scenario 2】If $a_{1}$ is not a multiple of 3, then none of the terms in the sequence $\left\{a_{n}\right\}$ are multiples of 3.
In this case, $a_{3} \equiv 1,2,4,5,7,8(\bmod 9)$.
Also, $a_{3} \equiv 0(\bmod 4)$, so by the Chinese Remainder Theorem, we get $a_{3} \equiv 28,20,4,32,16,8(\bmod 36)$.
Therefore, $a_{3} \in\{28,20,4,32,16,8\}$.
Hence, when $n \geqslant 3$, $a_{n} \in\{28,20,4,32,16,8\}$, and $\{28,20,4,32,16,8\} \subset M$.
Considering the first two terms, we get $|M| \leqslant 8$.
Taking $a_{1}=1$, we get the sequence $\left\{a_{n}\right\}$:
$1,2,4,8,16,32,28,20,4, \cdots$.
Thus, the maximum number of elements in the set $M$ is 8. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $S=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{980100}}$. Find the greatest positive integer $[S]$ that does not exceed the real number $S$. | Notice that, $\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}$.
Therefore, $\frac{1}{\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$,
$2(\sqrt{n+1}-\sqrt{n})=\frac{2}{\sqrt{n+1+\sqrt{n}}}<\frac{1}{\sqrt{n}}$.
Thus, $1977<2(\sqrt{980101}-\sqrt{2})$
$$
<S<2(\sqrt{980100}-1)=1978 \text {. }
$$
Hence, $[S]=1977$. | 1977 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 For integer pairs $(a, b)(0<a<b<1000)$, a set $S \subseteq\{1,2, \cdots, 2003\}$ is called a "jump set" of the pair $(a, b)$: if for any element pair $\left(s_{1}, s_{2}\right)$, $s_{1} 、 s_{2} \in$ $S,\left|s_{1}-s_{2}\right| \notin\{a, b\}$.
Let $f(a, b)$ be the maximum number of elements in a jump set of the pair $(a, b)$. Find the maximum and minimum values of $f(a, b)$. | 【Analysis】This method is quite typical, the minimum value is obtained using the greedy idea, while the maximum value is derived using the pigeonhole principle. Unfortunately, during the exam, most candidates only answered one part correctly.
For the minimum value, take $a=1, b=2$, then
$$
\begin{array}{l}
\{1,2,3\},\{4,5,6\}, \cdots, \\
\{1999,2000,2001\},\{2002,2003\}
\end{array}
$$
each set can contribute at most one element.
Thus, $f(1,2) \leqslant 668$.
Next, we prove that the minimum value is indeed 668, which requires proving that for any $(a, b), f(a, b) \geqslant 668$.
This proof process is similar to Example 4.
Since adding $x$ to the jump set $S$ prevents $x+a$ and $x+b$ from being added, after the first 1 is taken, 1, $1+a$, and $1+b$ are removed. Each time, "greedily" take the smallest number $x$ that can be taken, and remove $x$, $x+a$, and $x+b$. This means that taking one element can prevent at most three elements from being taken, thus,
$$
f(a, b) \geqslant\left\lceil\frac{2003}{3}\right\rceil=668 \text {, }
$$
where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.
In summary, $f_{\text {min }}=668$. | 668 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}$, $\cdots, b_{31}$ satisfy
$$
\begin{array}{l}
\text { (1) } a_{1}<a_{2}<\cdots<a_{31} \leqslant 2015, \\
b_{1}<b_{2}<\cdots<b_{31} \leqslant 2015 ;
\end{array}
$$
$$
\text { (2) } a_{1}+a_{2}+\cdots+a_{31}=b_{1}+b_{2}+\cdots+b_{31} \text {. }
$$
Find the maximum value of $S=\sum_{i=1}^{31}\left|a_{i}-b_{i}\right|$. | 【Detailed Estimation】It is known that, $a_{31} \leqslant 2015, a_{30} \leqslant 2014, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, we have $a_{i} \leqslant 1984+i$.
Similarly, $b_{i} \leqslant 1984+i$.
Also, $a_{1} \geqslant 1, a_{2} \geqslant 2, \cdots \cdots$
Thus, for all $1 \leqslant i \leqslant 31$, we have $a_{i} \geqslant i$.
Similarly, $b_{i} \geqslant i$.
Furthermore, $a_{i+1} \geqslant a_{i}+1$,
$a_{i+2} \geqslant a_{i+1}+1 \geqslant a_{i}+2, \cdots \cdots$.
Thus, for all $1 \leqslant i < j$. Then
$$
\begin{array}{l}
\left(a_{i}-b_{i}\right)+\left(b_{j}-a_{j}\right)=a_{i}-\left(b_{i}-b_{j}\right)-a_{j} \\
\leqslant(1984+i)-(i-j)-j=1984 . \\
\text { Hence } \frac{31 A}{240} \leqslant 1984 \\
\Rightarrow A \leqslant 64 \times 240=15360 \\
\Rightarrow S=2 A \leqslant 30720 .
\end{array}
$$
Finally, starting from the condition for equality, we can take
$$
p=15, q=16 \text {. }
$$
Also note that,
$$
15360=1024 \times 15=960 \times 16 \text {. }
$$
We can construct
$$
\begin{aligned}
& \left(a_{1}, a_{2}, \cdots, a_{31}\right) \\
& =(1,2, \cdots, 15,2000,2001, \cdots, 2015), \\
& \left(b_{1}, b_{2}, \cdots, b_{31}\right)=(1025,1026, \cdots, 1055), \\
& \text { at this time, } S=1024 \times 15+960 \times 16=30720, \\
\text { and } & a_{1}+a_{2}+\cdots+a_{31} \\
& =b_{1}+b_{2}+\cdots+b_{31}=32240,
\end{aligned}
$$
which meets the requirements.
In summary, the maximum value of $S$ is 30720. | 30720 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $S=\frac{1}{1^{3}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2011^{3}}$.
Then the integer part of $4 S$ is ( ). ${ }^{[2]}$
(A) 4
(B) 5
(C) 6
(D) 7
(2011, "Mathematics Weekly" Cup National Junior High School Mathematics Competition) | When $k=2,3, \cdots, 2011$,
$$
\frac{1}{k^{3}}<\frac{1}{k\left(k^{2}-1\right)}=\frac{1}{2}\left[\frac{1}{(k-1) k}-\frac{1}{k(k+1)}\right] \text {. }
$$
Then $1<S=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots+\frac{1}{2011^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2011 \times 2012}\right)<\frac{5}{4} \text {. }
$$
Thus, $4<4 S<5$.
Hence, the integer part of $4 S$ is 4. | 4 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
1. The smallest positive odd number that cannot be expressed as $7^{x}-3 \times 2^{y}\left(x 、 y \in \mathbf{Z}_{+}\right)$ is $\qquad$ | -1.3 .
Since $x, y \in \mathbf{Z}_{+}$, therefore, $7^{x}-3 \times 2^{y}$ is always an odd number, and $7^{1}-3 \times 2^{1}=1$.
If $7^{x}-3 \times 2^{y}=3$, then $317^{x}$.
And $7^{x}=(1+6)^{x}=1(\bmod 3)$, thus, there do not exist positive integers $x, y$ such that
$$
7^{x}-3 \times 2^{y}=3 \text {. }
$$
Therefore, the smallest positive odd number is 3. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $f(x)=\frac{\sin \pi x}{x^{2}}(x \in(0,1))$. Then
$$
g(x)=f(x)+f(1-x)
$$
the minimum value of $g(x)$ is . $\qquad$ | 7.8.
From the given, we have
$$
\begin{aligned}
f^{\prime}(x) & =\frac{\pi x \cos \pi x-2 \sin \pi x}{x^{3}}, \\
f^{\prime \prime}(x) & =\frac{\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x}{x^{3}} .
\end{aligned}
$$
Next, we need to prove
$$
\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x>0.
$$
Let $t=\pi x \in(0, \pi)$. Then
$\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x$
$$
=\left(6-t^{2}\right) \sin t-2 t \cos t=p(t).
$$
Thus, $p^{\prime}(t)=\left(4-t^{2}\right) \cos t$.
It is easy to see that $p^{\prime}(t)$ is increasing on the interval $\left(0, \frac{\pi}{2}\right)$, decreasing on the interval $\left(\frac{\pi}{2}, 2\right)$, and increasing on the interval $(2, \pi)$.
Also, $p(0)=0, p^{\prime}(0)>0$, so for $t \in(0, \pi)$, $p(t)>0$, which means that for $x \in(0,1)$, $f^{\prime \prime}(x)>0$.
Therefore, $f(x)$ is a convex function on the interval $(0,1)$. By Jensen's inequality, we know that for $x \in(0,1)$,
$$
\begin{array}{l}
\frac{f(x)+f(1-x)}{2} \\
\geqslant f\left(\frac{x+(1-x)}{2}\right)=f\left(\frac{1}{2}\right)=4.
\end{array}
$$
Thus, $g(x) \geqslant 8$, with equality holding if and only if $x=\frac{1}{2}$.
Therefore, the minimum value of $g(x)$ is 8. | 8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the set $M=\{1,2, \cdots, 12\}$, and the three-element set $A=$ $\{a, b, c\}$ satisfies $A \subset M$, and $a+b+c$ is a perfect square. Then the number of sets $A$ is $\qquad$. | ,- 1.26 .
From $6 \leqslant a+b+c \leqslant 33$, we know that the square numbers within this range are $9, 16, 25$. Let's assume $a<b<c$.
If $a+b+c=9$, then the possible values for $c$ are 6, 5, 4, in which case,
$$
A=\{1,2,6\},\{1,3,5\},\{2,3,4\} \text {; }
$$
If $a+b+c=16$, then $7 \leqslant c \leqslant 12$, we can have
$$
\begin{aligned}
A= & \{1,3,12\},\{1,4,11\},\{2,3,11\}, \\
& \{1,5,10\},\{2,4,10\},\{1,6,9\}, \\
& \{2,5,9\},\{3,4,9\},\{1,7,8\}, \\
& \{2,6,8\},\{3,5,8\},\{3,6,7\}, \\
& \{4,5,7\} ;
\end{aligned}
$$
If $a+b+c=25$, then $10 \leqslant c \leqslant 12$, in this case,
$$
\begin{aligned}
A= & \{2,11,12\},\{3,10,12\},\{4,9,12\}, \\
& \{4,10,11\},\{5,8,12\},\{5,9,11\}, \\
& \{6,7,12\},\{6,8,11\},\{6,9,10\}, \\
& \{7,8,10\} .
\end{aligned}
$$
In total, we have 26 sets $A$. | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The volume of a rectangular prism is 8 cubic centimeters, and the total surface area is 32 square centimeters. If the length, width, and height form a geometric sequence, then the sum of all the edges of this rectangular prism is $\qquad$ | 4.32 cm.
Let the common ratio be $q$, and the length, width, and height be $q a, a, \frac{a}{q}$, respectively. Then $a^{3}=8 \Rightarrow a=2$.
$$
\begin{array}{l}
\text { Also } 2\left(q a \cdot a+a \cdot \frac{a}{q}+q a \cdot \frac{a}{q}\right)=32 \\
\Rightarrow q+\frac{1}{q}+1=4 .
\end{array}
$$
Therefore, the sum of the 12 edges is
$$
4\left(q a+\frac{a}{q}+a\right)=8\left(q+\frac{1}{q}+1\right)=32 \text {. }
$$ | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) If the subset $A$ of the set $M=\{1,2, \cdots, 200\}$ consists of elements each of which can be expressed as the sum of the squares of two natural numbers (allowing the same number), find the maximum number of elements in the set $A$.
| 11. Notice that, the squares not exceeding 200 are $0^{2}, 1^{2}, 2^{2}, \cdots, 14^{2}$.
First, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form $k^{2}+0^{2}$, and there are 14 such numbers; while the sum of each pair of numbers in $1^{2}$, $2^{2}, \cdots, 10^{2}$ (allowing the same number) is in the set $M$, and there are $\mathrm{C}_{10}^{2}+10=55$ such numbers, among which, numbers of the form $x^{2}+x^{2}$ are 10, and numbers of the form $x^{2}+y^{2}(x \neq y)$ are $\mathrm{C}_{10}^{2}$.
Second, numbers of the form $11^{2}+x^{2}(x=1,2, \cdots, 8)$ are 8,
numbers of the form $12^{2}+x^{2}(x=1,2, \cdots, 7)$ are 7, numbers of the form $13^{2}+x^{2}(x=1,2, \cdots, 5)$ are 5, and numbers of the form $14^{2}+x^{2}(x=1,2)$ are 2, totaling 22.
Consider the repeated cases.
Notice that, if
$$
\begin{array}{l}
x=a^{2}+b^{2}, y=c^{2}+d^{2}(a \neq b, c \neq d), \\
\text { then } x y=(a c+b d)^{2}+(a d-b c)^{2} \\
=(a c-b d)^{2}+(a d+b c)^{2} .
\end{array}
$$
Numbers not exceeding 40 that can be expressed as the sum of squares of two different positive integers are
$$
5 、 10 、 13 、 17 、 20 、 25,26,29 、 34,37 、 40 \text {, }
$$
The product of each number in this group with 5, and $13^{2}$, are all in the set $M$, and can each be expressed as the sum of squares in two ways, hence each is counted twice, totaling 12 repeated counts (10, $13 、 17 、 20$ with 10 are already included in the above product group).
Therefore, the maximum number of elements in set $A$ is $14+55+22-12=79$. | 79 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If $2016+3^{n}$ is a perfect square, then the positive integer $n=$ . $\qquad$ | 4. 2 .
Obviously, $2016+3^{n}$ is an odd perfect square.
So $2016+3^{n} \equiv 1(\bmod 8)$
$$
\Rightarrow 3^{n} \equiv 1(\bmod 8)
$$
$\Rightarrow n$ must be even.
Let $n=2+2 k(k \in \mathbf{N})$. Then
$$
2016+3^{n}=9\left(224+3^{2 k}\right) \text {. }
$$
So $224+3^{2 k}=224+\left(3^{k}\right)^{2}$ is a perfect square.
Let $224+\left(3^{k}\right)^{2}=t^{2}\left(t \in \mathbf{Z}_{+}\right)$. Then
$$
\left(t+3^{k}\right)\left(t-3^{k}\right)=224=2^{5} \times 7 \text {, }
$$
and $t+3^{k}>t-3^{k}$, both are even.
So $\left(t+3^{k}, t-3^{k}\right)$
$$
=\left(2^{4} \times 7,2\right),\left(2^{3} \times 7,2^{2}\right),\left(2^{2} \times 7,2^{3}\right),\left(2^{4}, 2 \times 7\right) \text {. }
$$
Solving gives $k=0$, i.e., $n=2$. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}} \neq x \text {, }
$$
find all positive integers $a$ that meet the requirement. | $$
\text { Three, let } \sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}}=t \text {. }
$$
Then $t^{2} x+x+t^{2}-a=0$,
$$
(t+1) x^{2}+t-a=0 \text {. }
$$
$t \times$ (2) $-x \times$ (1) gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\frac{a-x^{2}}{1+x^{2}}=\frac{a-x}{1-x}$
$$
\Rightarrow 2 x^{3}-(1+a) x^{2}+(1-a) x=0 \text {. }
$$
Given that $x$ is a positive integer,
$$
2 x^{2}-(1+a) x+(1-a)=0 \text {. }
$$
Therefore, the discriminant of equation (1) must be a perfect square.
Let $(1+a)^{2}-8(1-a)=k^{2}(k \in \mathbf{N})$, i.e.,
$$
(a+5+k)(a+5-k)=32 \text {. }
$$
Since $a+5+k$ and $a+5-k$ are both even, and $a+5+k > a+5-k$, we have $(a+5+k, a+5-k)=(16,2)$ or $(8,4)$.
Solving gives $a=4$ or 1.
Substituting into equation (1) yields
$(a, x)=(4,3)$ or $(1,1)$.
Upon verification, only $(a, x)=(1,1)$ satisfies the given conditions, i.e., only $a=1$ meets the requirement. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In the plane, $m$ points have no three points collinear, and their convex hull is an $n$-sided polygon. By appropriately connecting lines, a grid region composed of triangles can be obtained. Let the number of non-overlapping triangles be $f(m, n)$. Then $f(2016,30)=$ $\qquad$ | One, 1.4000.
Since no three points are collinear, we have
$$
f(m, n)=(n-2)+2(m-n)=2 m-n-2 \text {. }
$$
Therefore, $f(2016,30)=2 \times 2016-30-2=4000$. | 4000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$ satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$ | 2. 2 .
According to the problem, for any $(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking $(x, y)=(1,0),(0,1)$, we get the fixed point $B(a, b)$ satisfying the necessary conditions $\left\{\begin{array}{l}a \leqslant 1, \\ b \leqslant 1,\end{array}\right.$, which implies $a+b \leqslant 2$.
Thus, $(a+b)_{\text {max }} \leqslant 2$.
For the fixed point $B(1,1)$, for any $A(x, y) \in D$, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=(1,1) \cdot(x, y)=x+y \leqslant 1 \text {, }
$$
with equality holding at point $A\left(\frac{1}{2}, \frac{1}{2}\right)$.
Therefore, the point $B(1,1)$ satisfies the problem's conditions, and in this case, $a+b=2$.
Hence, $(a+b)_{\text {max }}=2$. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the set $T=\{1,2, \cdots, 2010\}$, for each non-empty subset of $T$, calculate the reciprocal of the product of all its elements. Then the sum of all such reciprocals is $\qquad$ | 6. 2010 .
Notice that, in the expansion of the product
$$
(1+1)\left(1+\frac{1}{2}\right) \cdots\left(1+\frac{1}{2010}\right)
$$
the $2^{2010}$ terms are all the reciprocals of positive integers, and each is precisely the reciprocal of the product of the numbers in one of the $2^{2010}$ subsets of set $T$. Removing the term 1, which corresponds to the empty set, the sum of the reciprocals is
$$
2 \times \frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{2011}{2010}-1=2010 .
$$ | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the incenter $I(-1,7)$ of a right-angled triangle $\triangle O A B$ with all three vertices as integer points, and the origin $O$ as the right-angle vertex. The number of such right-angled triangles $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 8.2.
As shown in Figure 4.
Let $\angle x O A=\alpha, \angle x O I=\beta$.
Then $\alpha=\beta-\frac{\pi}{4}$.
Also, $\tan \beta=-7$, so
$$
\begin{array}{l}
\tan \alpha=\tan \left(\beta-\frac{\pi}{4}\right) \\
=\frac{\tan \beta-1}{1+\tan \beta}=\frac{4}{3}, \\
\tan \angle x O B=\tan \left(\frac{\pi}{2}+\alpha\right) \\
=-\cot \alpha=-\frac{1}{\tan \alpha}=-\frac{3}{4} .
\end{array}
$$
Then the point $A(3 t, 4 t), B(-4 k, 3 k)\left(k \backslash t \in \mathbf{Z}_{+}\right)$.
Thus, $O A=5 t, O B=5 k$, and the inradius of the right triangle $\triangle O A B$ is
$$
\begin{array}{l}
r=O I \sin 45^{\circ}=\frac{\sqrt{2}}{2} O I=5 . \\
\text { Also } r=\frac{1}{2}(O A+O B-A B) \\
\Rightarrow \frac{1}{2}\left[5 t+5 k-\sqrt{(5 t)^{2}+(5 k)^{2}}\right]=5 \\
\Rightarrow t+k-2=\sqrt{t^{2}+k^{2}} \\
\Rightarrow(t-2)(k-2)=2 .
\end{array}
$$
Since $t$ and $k$ are both positive integers, we have $(t, k)=(3,4)$ or $(4,3)$.
Therefore, there are two right triangles $\triangle O A B$ that satisfy the conditions. | 2 | Other | math-word-problem | Yes | Yes | cn_contest | false |
For integers $n \geqslant 3$, let
$$
f(n)=\log _{2} 3 \times \log _{3} 4 \times \cdots \times \log _{n-1} n \text {. }
$$
Then $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)=$ $\qquad$ | 2. 54
Notice,
$$
\begin{array}{l}
f(n)=\log _{2} 3 \times \frac{\log _{2} 4}{\log _{2} 3} \times \cdots \times \frac{\log _{2} n}{\log _{2}(n-1)} \\
=\log _{2} n .
\end{array}
$$
Thus, $f\left(2^{k}\right)=k$.
Therefore, $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)$
$$
=2+3+\cdots+10=54 \text {. }
$$ | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of the white balls is. $\qquad$ | 3. 126.
First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.
Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by swapping the positions of the red and white balls, we get an arrangement where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls, and vice versa.
Therefore, the number of arrangements where the sum of the numbers on the red balls is greater than the sum of the numbers on the white balls is equal to the number of arrangements where the sum of the numbers on the red balls is less than the sum of the numbers on the white balls. Thus, the number of arrangements we are looking for is $\frac{1}{2} \mathrm{C}_{10}^{5}=126$. | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given real numbers $a, b$ satisfy
$$
a+\lg a=10, b+10^{b}=10 \text {. }
$$
Then $\lg (a+b)=$ $\qquad$ . | 8.1.
Since $\lg a=10-a, 10^{b}=10-b$, therefore, $a$ is the x-coordinate of the intersection point of $y=\lg x$ and $y=10-x$, and $b$ is the y-coordinate of the intersection point of $y=10^{x}$ and $y=10-x$.
Also, $y=\lg x$ and $y=10^{x}$ are symmetric about the line $y=x$, and $y=10-x$ is symmetric about the line $y=x$, so the two intersection points are symmetric about the line $y=x$.
Thus $10-b=a \Rightarrow \lg (a+b)=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. $6^{11}+C_{11}^{1} 6^{10}+C_{11}^{2} 6^{9}+\cdots+C_{11}^{10} 6-1$ when divided by 8 yields a remainder of $\qquad$ . | 3.5.
Notice that,
$$
\begin{array}{l}
6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\mathrm{C}_{11}^{2} 6^{9}+\cdots+\mathrm{C}_{11}^{10} 6-1 \\
=\mathrm{C}_{10}^{0} 6^{11}+\mathrm{C}_{11}^{1} 6^{10}+\cdots+\mathrm{C}_{11}^{10} 6+\mathrm{C}_{11}^{11} 6^{0}-2 \\
=(6+1)^{11}-2=7^{11}-2 \\
\equiv(-1)^{11}-2 \equiv 5(\bmod 8) .
\end{array}
$$
Therefore, the required result is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
12. (14 points) A flower shop purchases a certain number of roses from a farm at a price of 5 yuan per stem every day, and then sells them at a price of 10 yuan per stem. If they are not sold on the same day, the remaining roses are treated as garbage.
(1) If the flower shop purchases 16 stems of roses one day, find the function expression for the profit $y$ (unit: yuan) of that day in terms of the demand $n$ (unit: stems, $n \in \mathbf{N}$).
(2) The flower shop recorded the daily demand for roses (unit: stems) over 100 days, and organized the data as shown in Table 1.
Table 1
\begin{tabular}{|c|l|l|l|l|l|l|l|}
\hline Daily demand $n$ & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline Frequency & 10 & 20 & 16 & 16 & 15 & 13 & 10 \\
\hline
\end{tabular}
The frequencies of each demand over 100 days are used as the probabilities of each demand.
(i) If the flower shop purchases 16 stems of roses one day, and $X$ represents the profit (unit: yuan) of that day, find the distribution of $X$, its expected value, and variance.
(ii) If the flower shop plans to purchase 16 or 17 stems of roses one day, do you think they should purchase 16 or 17 stems? Please explain your reasoning. | 12. (1) When the daily demand $n \geqslant 16$, the profit $y=80$. When the daily demand $n<16$, the profit $y=10 n-80$. Therefore, the function expression of $y$ with respect to the positive integer $n$ is
$$
y=\left\{\begin{array}{ll}
10 n-80, & n<16 ; \\
80, & n \geqslant 16 .
\end{array}\right.
$$
(2) (i) The possible values of $X$ are $60, 70, 80$, and
$$
\begin{array}{l}
P(X=60)=0.1, P(X=70)=0.2, \\
P(X=80)=0.7 .
\end{array}
$$
Thus, the distribution of $X$ is as shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|}
\hline$X$ & 60 & 70 & 80 \\
\hline$P$ & 0.1 & 0.2 & 0.7 \\
\hline
\end{tabular}
The expected value of $X$ is
$$
\mathrm{E}(X)=60 \times 0.1+70 \times 0.2+80 \times 0.7=76 \text {; }
$$
The variance of $X$ is
$$
\begin{array}{l}
\mathrm{D}(X)=(60-76)^{2} \times 0.1+(70-76)^{2} \times 0.2+ \\
\quad(80-76)^{2} \times 0.7 \\
=44 .
\end{array}
$$
(ii) Method 1 The flower shop should purchase 16 roses per day. If the flower shop purchases 17 roses per day, let $Y$ represent the profit for the day (unit: yuan), then the distribution of $Y$ is as shown in Table 3.
Table 3
\begin{tabular}{|c|c|c|c|c|}
\hline$Y$ & 55 & 65 & 75 & 85 \\
\hline$P$ & 0.1 & 0.2 & 0.16 & 0.54 \\
\hline
\end{tabular}
The expected value of $Y$ is
$$
\begin{array}{l}
\mathrm{E}(Y)=55 \times 0.1+65 \times 0.2+ \\
75 \times 0.16+85 \times 0.54 \\
=76.4 ;
\end{array}
$$
The variance of $Y$ is
$$
\begin{aligned}
\mathrm{D}(Y)= & (55-76.4)^{2} \times 0.1+ \\
& (65-76.4)^{2} \times 0.2+ \\
& (75-76.4)^{2} \times 0.16+ \\
& (85-76.4)^{2} \times 0.54 \\
= & 112.04 .
\end{aligned}
$$
Thus, $\mathrm{D}(X)<\mathrm{D}(Y)$, meaning the profit fluctuation is smaller when purchasing 16 roses.
Additionally, although $\mathrm{E}(X)<\mathrm{E}(Y)$, the difference is not significant.
Therefore, the flower shop should purchase 16 roses per day.
Method 2 The flower shop should purchase 17 roses per day.
From Method 1, we know that $\mathrm{E}(X)<\mathrm{E}(Y)$, meaning the average profit is higher when purchasing 17 roses compared to purchasing 16 roses.
Therefore, the flower shop should purchase 17 roses per day. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Problem 5: In an $n \times n$ grid, 101 cells are colored blue. It is known that there is a unique way to cut the grid along the grid lines into some rectangles, such that each rectangle contains exactly one blue cell. Find the minimum possible value of $n$.
(2015, Bulgarian Mathematical Olympiad) | The minimum possible value of $n$ is 101.
First, prove the more general conclusion below.
Lemma Given an $n \times n$ grid $P$ with $m$ cells colored blue. A "good" partition of the grid $P$ is defined as: if the grid is divided along the grid lines into $m$ rectangles, and each rectangle contains exactly one blue cell. Then the grid $P$ has a unique good partition if and only if the $m$ blue cells form a rectangle.
Proof On the one hand, if the $m$ blue cells form a $m_{1} \times m_{2}\left(m_{1} 、 m_{2} \geqslant 1\right)$ rectangle $A$, cut the four corner cells of $A$ to form rectangles with the four corner cells of the grid $P$; the cells on the four sides of $A$ form $1 \times s(s \geqslant 1)$ rectangles with the cells at the top and bottom (left and right) ends of their respective rows (columns); each cell inside the rectangle $A$ forms a rectangle by itself. This constitutes a good partition $S$ of the grid $P$, and it is easy to see that the grid $P$ has only this unique good partition.
On the other hand, consider the grid $P$ has a unique good partition $S$. Call a grid line a "partition line" if it can divide the grid $P$ into two rectangles, and each rectangle contains at least one blue cell.
We now prove: every rectangle $A$ containing at least one blue cell has a good partition.
We use mathematical induction on the number of blue cells $k$ in the rectangle $A$ to prove conclusion (1).
If $k=1$, then by the definition of a good partition, conclusion (1) is obviously true.
Assume conclusion (1) holds for $k \leqslant t-1$.
Consider the case $k=t$. In this case, there exists a partition line that divides the rectangle $A$ into two rectangles $A_{1} 、 A_{2}$, and each rectangle contains at least one blue cell. By the induction hypothesis, $A_{1} 、 A_{2}$ each have a good partition. Therefore, the rectangle $A$ has a good partition, and conclusion (1) holds.
By conclusion (1) and the definition of a partition line, every partition line $l$ divides the grid $P$ into two rectangles, each of which has a good partition. Since $S$ is the unique good partition of the grid $P$, the partition $S$ includes all partition lines in the grid $P$.
Let $l_{1}, l_{2}, \cdots, l_{p}$ be all the vertical partition lines from left to right, and $m_{1}, m_{2}, \cdots, m_{q}$ be all the horizontal partition lines from bottom to top. Then they divide the grid $P$ into some rectangles, and each rectangle contains at most one blue cell. Since $S$ divides the grid $P$ using these partition lines, it results in $m$ rectangles, each containing exactly one blue cell.
Let $l_{0}$ be the vertical grid line closest to $l_{1}$ on the left with no blue cells to its left, $l_{p+1}$ be the vertical grid line closest to $l_{p}$ on the right with no blue cells to its right, and $m_{0} 、 m_{q+1}$ be similarly defined horizontal grid lines.
From the above discussion, the distance between $l_{i}$ and $l_{i+1}$
$d\left(l_{i}, l_{i+1}\right)=1(i=0,1, \cdots, p)$;
otherwise, a new partition line can be added between two partition lines.
Similarly, $d\left(m_{j}, m_{j+1}\right)=1(j=0,1, \cdots, q)$.
Therefore, the blue cells form a rectangle with $l_{0} 、 l_{p+1} 、 m_{0} 、 m_{q+1}$ as boundary lines.
The lemma is proved.
By the lemma, since the grid $P$ has a unique good partition, the 101 blue cells must form a rectangle. Therefore, it can only be a $1 \times 101$ rectangle.
Hence, the minimum value of $n$ is 101. | 101 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 2, point $A$ is on the positive $y$-axis, point $B$ is on the positive $x$-axis, $S_{\triangle A O B}=9$, segment $A B$ intersects the graph of the inverse proportion function $y=\frac{k}{x}$ at points $C$ and $D$. If $C D=$ $\frac{1}{3} A B$, and $A C=B D$, then $k=$ . $\qquad$ | 8. 4 .
Let point $A\left(0, y_{A}\right), B\left(x_{B}, 0\right)$.
Given $C D=\frac{1}{3} A B, A C=B D$, we know that $C$ and $D$ are the trisection points of segment $A B$.
Thus, $x_{C}=\frac{1}{3} x_{B}, y_{C}=\frac{2}{3} y_{A}$.
Therefore, $k=x_{C} y_{C}=\frac{2}{9} x_{B} y_{A}=\frac{4}{9} S_{\triangle A O B}=4$. | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. If $4^{a}=6^{b}=9^{c}$, then
$$
\frac{1}{a}-\frac{2}{b}+\frac{1}{c}=
$$
$\qquad$ | 8. 0 .
Let $4^{a}=6^{b}=9^{c}=k$.
Then $a=\log _{k} 4, b=\log _{k} 6, c=\log _{k} 9$
$$
\begin{array}{l}
\Rightarrow \frac{1}{a}-\frac{2}{b}+\frac{1}{c}=\log _{k} 4-2 \log _{k} 6+\log _{k} 9 \\
\quad=\log _{k} 1=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. In rectangle $A B C D$, $A B=3, A D=4, P$ is a point on the plane of rectangle $A B C D$, satisfying $P A=2$, $P C=\sqrt{21}$. Then $\overrightarrow{P B} \cdot \overrightarrow{P D}=$ | 11.0.
As shown in Figure 3, let $A C$ and $B D$ intersect at point $E$, and connect $P E$. Then $E$ is the midpoint of $A C$ and $B D$.
Notice that,
$$
\begin{array}{l}
\overrightarrow{P B} \cdot \overrightarrow{P D}=\frac{1}{4}\left[(\overrightarrow{P B}+\overrightarrow{P D})^{2}-(\overrightarrow{P B}-\overrightarrow{P D})^{2}\right] \\
=\frac{1}{4}\left[(2 \overrightarrow{P E})^{2}-(\overrightarrow{D B})^{2}\right] \\
=|\overrightarrow{P E}|^{2}-\frac{1}{4}|\overrightarrow{D B}|^{2} .
\end{array}
$$
Similarly, $\overrightarrow{P A} \cdot \overrightarrow{P C}=|\overrightarrow{P E}|^{2}-\frac{1}{4}|\overrightarrow{C A}|^{2}$.
Since $|\overrightarrow{D B}|=|\overrightarrow{C A}|$, we have,
$$
\overrightarrow{P B} \cdot \overrightarrow{P D}=\overrightarrow{P A} \cdot \overrightarrow{P C}
$$
Given $A C=B D=\sqrt{3^{2}+4^{2}}=5=\sqrt{P A^{2}+P C^{2}}$
$$
\Rightarrow \angle A P C=90^{\circ} \text {. }
$$
Therefore, $\overrightarrow{P B} \cdot \overrightarrow{P D}=\overrightarrow{P A} \cdot \overrightarrow{P C}=0$. | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given
$$
5 x+16 y+33 z \geqslant 136\left(x 、 y 、 z \in \mathbf{R}_{+}\right) \text {. }
$$
then the minimum value of $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$ is | 3. 50 .
Notice that,
$$
\begin{array}{l}
x^{3}+x^{2}-5 x+3=(x+3)(x-1)^{2} \geqslant 0 \\
\Rightarrow x^{3}+x^{2} \geqslant 5 x-3, \\
y^{3}+y^{2}-16 y+20=(y+5)(y-2)^{2} \geqslant 0 \\
\Rightarrow y^{3}+y^{2} \geqslant 16 y-20, \\
z^{3}+z^{2}-33 z+63=(z+7)(z-3)^{2} \geqslant 0 \\
\Rightarrow z^{3}+z^{2} \geqslant 33 z-63 .
\end{array}
$$
Therefore, $x^{3}+y^{3}+z^{3}+x^{2}+y^{2}+z^{2}$
$$
\geqslant 5 x+16 y+33 z-86=50 \text {. }
$$
The minimum value 50 is achieved when and only when $x=1, y=2, z=3$. | 50 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that all positive integers are in $n$ sets, satisfying that when $|i-j|$ is a prime number, $i$ and $j$ belong to two different sets. Then the minimum value of $n$ is $\qquad$ | 6. 4 .
It is easy to see that $n \geqslant 4$ (2, 4, 7, 9 must be in four different sets).
Also, when $n=4$, the sets
$$
A_{i}=\left\{m \in \mathbf{Z}_{+} \mid m \equiv i(\bmod 4)\right\}(i=0,1,2,3)
$$
satisfy the condition.
Therefore, the minimum value of $n$ is 4. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the ellipse $C$ passes through the point $M(1,2)$, with two foci at $(0, \pm \sqrt{6})$, and $O$ is the origin, a line $l$ parallel to $OM$ intersects the ellipse $C$ at points $A$ and $B$. Then the maximum value of the area of $\triangle OAB$ is $\qquad$ | 7. 2 .
According to the problem, let $l_{A B}: y=2 x+m$.
The ellipse equation is $\frac{y^{2}}{8}+\frac{x^{2}}{2}=1$. By solving the system and eliminating $y$, we get $16 x^{2}+8 m x+2\left(m^{2}-8\right)=0$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{8 m}{16}, x_{1} x_{2}=\frac{2\left(m^{2}-8\right)}{16}$.
Then $|A B|=\sqrt{5\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]}$
$$
=\frac{\sqrt{5\left(16-m^{2}\right)}}{2} \text {. }
$$
The distance from point $O$ to line $A B$ is $d=\frac{|m|}{\sqrt{5}}$, hence
$$
S_{\triangle O A B}=\frac{1}{2}|A B| d=\frac{\sqrt{m^{2}\left(16-m^{2}\right)}}{4} \leqslant 2 .
$$
When and only when $m= \pm 2 \sqrt{2}$, the area of $\triangle O A B$ reaches its maximum value of 2. | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given positive integers $a, b, c, x, y, z$ satisfying $a \geqslant b \geqslant c \geqslant 1, x \geqslant y \geqslant z \geqslant 1$,
and $\left\{\begin{array}{l}2 a+b+4 c=4 x y z, \\ 2 x+y+4 z=4 a b c .\end{array}\right.$
Then the number of six-tuples $(a, b, c, x, y, z)$ that satisfy the conditions is $\qquad$ groups. | 8. 0 .
Assume $x \geqslant a$. Then
$$
\begin{array}{l}
4 x y z=2 a+b+4 c \leqslant 7 a \leqslant 7 x \\
\Rightarrow y z \leqslant \frac{7}{4} \Rightarrow y z \leqslant 1 \Rightarrow(y, z)=(1,1) \\
\Rightarrow\left\{\begin{array}{l}
a a+b+4 c=4 x, \\
2 x+5=4 a b c
\end{array}\right. \\
\Rightarrow 2 a+b+4 c+10=8 a b c .
\end{array}
$$
If $b \geqslant 2$, then
$$
2 a+b+4 c+10 \leqslant 12 a<8 a b c,
$$
a contradiction.
If $b=1$, then $\left\{\begin{array}{l}2 a+5=4 x, \\ 2 x+5=4 a\end{array}\right.$ has no positive integer solutions. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $0<a<1$, and satisfies
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]=$ $\qquad$ | Solve:
$$
\begin{array}{l}
0<a<1 \\
\Rightarrow 0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2 \\
\Rightarrow\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right] \text { is either }
\end{array}
$$
0 or 1.
By the problem, we know that 18 of them are equal to 1, and 11 are equal to 0.
$$
\begin{array}{l}
\text { Therefore, }\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\left[a+\frac{11}{30}\right]=0, \\
{\left[a+\frac{12}{30}\right]=\left[a+\frac{13}{30}\right]=\cdots=\left[a+\frac{29}{30}\right]=1 .}
\end{array}
$$
Thus, $0<a+\frac{11}{30}<1,1 \leqslant a+\frac{12}{30}<2$.
Solving this, we get $18 \leqslant 30 a<1 \Rightarrow 6 \leqslant 10 a<\frac{19}{3}$.
Therefore, $[10 a]=6$. | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $O$ is the circumcenter of acute $\triangle A B C$, $\angle B A C$ $=60^{\circ}$, extend $C O$ to intersect $A B$ at point $D$, extend $B O$ to intersect $A C$ at point $E$. Then $\frac{B D}{C E}=$ $\qquad$ | $=, 7.1$.
Connect $O A, D E$.
$$
\begin{array}{l}
\text { Since } \angle B O C=2 \angle B A C=120^{\circ} \\
\Rightarrow \angle C O E=60^{\circ}=\angle D A E \\
\Rightarrow A, D, O, E \text { are concyclic } \\
\Rightarrow \angle D E B=\angle D A O=\angle D B E \\
\Rightarrow D B=D E .
\end{array}
$$
Similarly, $C E=D E$.
Thus, $\frac{B D}{C E}=\frac{D E}{D E}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
For any real number $x$, $[x]$ represents the greatest integer not exceeding $x$, and $\{x\}$ represents the fractional part of $x$. Then
$$
\begin{array}{l}
\left\{\frac{2014}{2015}\right\}+\left\{\frac{2014^{2}}{2015}\right\}+\cdots+\left\{\frac{2014^{2014}}{2015}\right\} \\
=
\end{array}
$$ | $-, 1.1007$.
Notice that,
$$
\frac{2014^{n}}{2015}=\frac{(2015-1)^{n}}{2015}=M+\frac{(-1)^{n}}{2015} \text {, }
$$
where $M$ is an integer.
Then $\left\{\frac{2014^{n}}{2015}\right\}=\left\{\begin{array}{ll}\frac{1}{2015}, & n \text { is even; } \\ \frac{2014}{2015}, & n \text { is odd. }\end{array}\right.$
Therefore, the original expression $=1007$. | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a_{1}, a_{2}, \cdots, a_{9}$ as any permutation of $1,2, \cdots, 9$. Then the minimum value of $a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9}$ is $\qquad$ | 2. 214.
Since $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, 9)$, by the AM-GM inequality, we have
$$
\begin{array}{l}
a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9} \\
\geqslant 3 \sqrt[3]{a_{1} a_{2} \cdots a_{9}}=3 \sqrt[3]{9!} \\
=3 \sqrt[3]{(2 \times 5 \times 7) \times(1 \times 8 \times 9) \times(3 \times 4 \times 6)} \\
=3 \sqrt[3]{70 \times 72 \times 72} \\
>3 \times 71=213 . \\
\text { Also, } 214=70+72+72 \\
=(2 \times 5 \times 7)+(1 \times 8 \times 9)+(3 \times 4 \times 6),
\end{array}
$$
Therefore, its minimum value is 214. | 214 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+2015(x-1)=-1, \\
(y-1)^{3}+2015(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ | 4. 2 .
Notice that the function $f(z)=z^{3}+2015 z$ is a monotonically increasing function on $(-\infty,+\infty)$.
From the given condition, we have $f(x-1)=f(1-y)$.
Therefore, $x-1=1-y \Rightarrow x+y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let real numbers $x, y$ satisfy
$$
\left\{\begin{array}{l}
x-y+1 \geqslant 0 \\
y+1 \geqslant 0 \\
x+y+1 \leqslant 0 .
\end{array}\right.
$$
Then the maximum value of $2 x-y$ is $\qquad$ | 8. 1 .
When $x=0, y=-1$, $2 x-y$ takes the maximum value 1. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$.
Then the number of such $n$ is $\qquad$. | From $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right] \leqslant \frac{n}{3}+\frac{n}{6}=\frac{n}{2}$, knowing that equality holds, we find that $n$ is a common multiple of $3$ and $6$, i.e., a multiple of $6$. Therefore, the number of such $n$ is $\left[\frac{2006}{6}\right]=334$. | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
23. For any real numbers $a$, $b$, $c$, define an operation ※ with the following properties:
(1) $a ※(b ※ c)=(a ※ b) \cdot c$,
(2) $a ※ a=1$,
where “$\cdot$” denotes multiplication.
If the solution to the equation $2016 \times(6 ※ x)=100$ is $x=\frac{p}{q}(p, q$ are positive integers, $(p, q)=1)$, then the value of $p+q$ is $(\quad)$.
(A) 109
(B) 201
(C) 301
(D) 3049
(E) 33601 | 23. A.
$$
\begin{array}{l}
\text { Given } 2016 ※(6 ※ x)=(2016 ※ 6) \cdot x \\
\Rightarrow(2016 ※ 6) \cdot 6=\frac{600}{x} \\
\Rightarrow 2016 ※(6 ※ 6)=\frac{600}{x} .
\end{array}
$$
Since $a ※ a=1$, we have,
$$
\begin{array}{l}
6 ※ 6=2016 ※ 2016=1 . \\
\text { Therefore } 2016 ※(2016 ※ 2016)=\frac{600}{x} \\
\Rightarrow(2016 ※ 2016) \cdot 2016=\frac{600}{x} \\
\Rightarrow 1 \cdot 2016=\frac{600}{x} \Rightarrow x=\frac{25}{84} \\
\Rightarrow p+q=25+84=109 .
\end{array}
$$ | 109 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b, c (a < b < c)$ form a geometric sequence, and
$$
\log _{2016} a+\log _{2016} b+\log _{2016} c=3 .
$$
Then the maximum value of $a+b+c$ is $\qquad$ | One, 1.4066 273.
From the given, we know
$$
\begin{array}{l}
b^{2}=a c, a b c=2016^{3} \\
\Rightarrow b=2016, a c=2016^{2} .
\end{array}
$$
Since $a$, $b$, and $c$ are positive integers, when $a=1$, $c=2016^{2}$, $a+b+c$ reaches its maximum value
$$
2016^{2}+2017=4066273 .
$$ | 4066273 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
492 Select five subsets $A_{1}, A_{2}, \cdots, A_{5}$ from the set $\{1,2, \cdots, 1000\}$ such that $\left|A_{i}\right|=500(i=1,2$, $\cdots, 5)$. Find the maximum value of the number of common elements in any three of these subsets. | Let $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ be five 1000-dimensional vectors, and let $\alpha_{i}(j)$ denote the $j$-th element of $\alpha_{i}$. Then $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ satisfy
$$
\alpha_{i}(j)=\left\{\begin{array}{l}
1, j \in A_{i} ; \\
0, j \notin A_{i} .
\end{array}\right.
$$
Let the maximum number of common elements in any three subsets be $r$.
For any three vectors
$$
\begin{array}{l}
\boldsymbol{\beta}_{1}=\left(b_{1}, b_{2}, \cdots, b_{t}\right), \boldsymbol{\gamma}_{1}=\left(c_{1}, c_{2}, \cdots, c_{t}\right), \\
\boldsymbol{\delta}_{1}=\left(d_{1}, d_{2}, \cdots, d_{t}\right),
\end{array}
$$
define the ternary inner product operation
$$
\left(\boldsymbol{\beta}_{1}, \boldsymbol{\gamma}_{1}, \boldsymbol{\delta}_{1}\right)=b_{1} c_{1} d_{1}+b_{2} c_{2} d_{2}+\cdots+b_{t} c_{t} d_{t} .
$$
Clearly, this operation is linear and satisfies the commutative law.
By the problem, $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right)=500$.
For $1 \leqslant i<j<k \leqslant 5,\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right)$ equals the number of common elements in the sets $A_{i}, A_{j}, A_{k}$.
Let $\boldsymbol{\alpha}=\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{4}+\boldsymbol{\alpha}_{5}$. Then
$(\boldsymbol{\alpha}, \boldsymbol{\alpha}, \boldsymbol{\alpha})$
$$
\begin{aligned}
= & \sum_{i=1}^{5}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right)+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+ \\
& 6 \sum_{i<j<k}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right) \\
= & 2500+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+6 \sum_{i<j<k}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right) \\
\leqslant & 2500+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+60 r .
\end{aligned}
$$
Also, $\boldsymbol{\alpha}$ is a 1000-dimensional vector, and
$$
\boldsymbol{\alpha}=\left(a_{1}, a_{2}, \cdots, a_{1000}\right)\left(a_{i} \in\{0,1, \cdots, 5\}\right),
$$
satisfying $\sum_{i=1}^{1000} a_{i}=2500$.
At this point, $(\boldsymbol{\alpha}, \boldsymbol{\alpha}, \boldsymbol{\alpha})=\sum_{i=1}^{1000} a_{i}^{3}$, and $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)$ represents the number of common elements in the sets $A_{i}$ and $A_{j}$, so $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)=\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)$, where $(\boldsymbol{a}, \boldsymbol{b})$ denotes the usual inner product of vectors $\boldsymbol{a}$ and $\boldsymbol{b}$.
$$
\begin{array}{l}
\text { Hence } \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)=\sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right) \\
=(\boldsymbol{\alpha}, \boldsymbol{\alpha})-\sum_{i=1}^{5}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right) \\
=\sum_{i=1}^{1000} a_{i}^{2}-2500 .
\end{array}
$$
At this point,
$$
\begin{array}{l}
60 r \geqslant \sum_{i=1}^{1000} a_{i}^{3}-3\left(\sum_{i=1}^{1000} a_{i}^{2}-2500\right)-2500 \\
=\sum_{i=1}^{1000}\left(a_{i}^{3}-3 a_{i}^{2}\right)+5000 . \\
\text { Also, }\left(a_{i}-2\right)\left(a_{i}-3\right)\left(a_{i}+2\right) \geqslant 0 \\
\Leftrightarrow a_{i}^{3}-3 a_{i}^{2}-4 a_{i}+12 \geqslant 0 .
\end{array}
$$
Thus, for any $1 \leqslant i \leqslant 1000$,
$$
a_{i}^{3}-3 a_{i}^{2} \geqslant 4 a_{i}-12 \text {. }
$$
Therefore, $60 r \geqslant \sum_{i=1}^{1000}\left(4 a_{i}-12\right)+5000$
$$
=4 \times 2500-12000+5000=3000 \text {. }
$$
Thus, $r \geqslant 50$.
Below is a construction for $r=50$.
Divide the 1000 elements into 20 sets, each containing 50 elements. These 20 sets are divided into two groups, denoted as
$$
\begin{array}{l}
B_{123}, B_{124}, B_{125}, B_{134}, B_{135}, B_{145}, B_{234}, B_{235}, \\
B_{245}, B_{345} ; \\
C_{12}, C_{13}, C_{14}, C_{15}, C_{23}, C_{24}, C_{25}, C_{34}, C_{35}, C_{45} .
\end{array}
$$
Define $A_{i}=\{$ the union of all sets whose indices contain $i\}$.
At this point, for any $1 \leqslant i<j<k \leqslant 5$,
$$
A_{i} \cap A_{j} \cap A_{k}=B_{i j k},
$$
which has exactly 50 elements.
(Chen Kai, Xueersi Peiyou Beijing Branch, 100086) | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given that $a$, $b$, and $c$ are three distinct real numbers. If in the quadratic equations
$$
\begin{array}{l}
x^{2}+a x+b=0, \\
x^{2}+b x+c=0, \\
x^{2}+c x+a=0
\end{array}
$$
any two of these equations have exactly one common root, find the value of $a^{2}+$ $b^{2}+c^{2}$. | Let the equations (1) and (3) have only one common root \( x_{1} \), equations (1) and (2) have only one common root \( x_{2} \), and equations (2) and (3) have only one common root \( x_{3} \). Therefore, the roots of equation (1) are \( x_{1} \) and \( x_{2} \), the roots of equation (2) are \( x_{2} \) and \( x_{3} \), and the roots of equation (3) are \( x_{3} \) and \( x_{1} \).
Since \( x_{1} \) is a root of equations (1) and (3), we have
\[
\left\{\begin{array}{l}
x_{1}^{2} + a x_{1} + b = 0, \\
x_{1}^{2} + c x_{1} + a = 0
\end{array} \Rightarrow x_{1} = \frac{a - b}{a - c} .\right.
\]
Similarly, \( x_{2} = \frac{b - c}{b - a} \) and \( x_{3} = \frac{c - a}{c - b} \).
Thus, \( x_{1} x_{2} x_{3} = -1 \).
Clearly, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are distinct. Otherwise, assume \( x_{2} = x_{3} \).
\[
\begin{array}{l}
\text{Hence } \frac{b - c}{b - a} = \frac{c - a}{c - b} \\
\Rightarrow a^{2} + b^{2} + c^{2} - b c - c a - a b = 0 \\
\Rightarrow (b - c)^{2} + (c - a)^{2} + (a - b)^{2} = 0 \\
\Rightarrow a = b = c,
\end{array}
\]
which contradicts the given conditions.
Thus, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are distinct.
By Vieta's formulas, we have
\[
\begin{array}{l}
a = -x_{1} - x_{2} = x_{3} x_{1}, \\
b = -x_{2} - x_{3} = x_{1} x_{2}, \\
c = -x_{3} - x_{1} = x_{2} x_{3}. \\
\text{From the above three equations, we get} \\
-2(x_{1} + x_{2} + x_{3}) = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2}, \text{ (4) } \\
x_{1}(1 + x_{3}) x_{2}(1 + x_{1}) x_{3}(1 + x_{2}) \\
= (-x_{2})(-x_{3})(-x_{1}). \\
\text{Noting that } x_{1} x_{2} x_{3} = -1. \\
\text{Thus } (1 + x_{3})(1 + x_{1})(1 + x_{2}) = -1 \\
\Rightarrow x_{1} + x_{2} + x_{3} + x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -1. \text{ (5) } \\
\text{From equations (4) and (5), we get} \\
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2
\end{array}\right. \\
\Rightarrow \left\{\begin{array}{l}
a + b + c = -2, \\
b c + c a + a b = -1 \\
a b c = 1
\end{array}\right., \\
\Rightarrow a^{2} + b^{2} + c^{2} \\
= (a + b + c)^{2} - 2(b c + c a + a b) \\
= (-2)^{2} - 2(-1) = 6. \\
\end{array}
\]
Let's explore further.
From the above solution, we also obtained
\[
\left\{\begin{array}{l}
x_{1} + x_{2} + x_{3} = 1, \\
x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2 \\
x_{1} x_{2} x_{3} = -1.
\end{array}\right.
\]
Therefore, \( a \), \( b \), and \( c \) are the roots of the equation \( x^{3} + 2 x^{2} - x - 1 = 0 \); simultaneously, \( x_{1} \), \( x_{2} \), and \( x_{3} \) are the roots of the equation \( x^{3} - x^{2} - 2 x + 1 = 0 \). If we substitute \( x \rightarrow -\frac{1}{x} \) in the equation \( x^{3} + 2 x^{2} - x - 1 = 0 \), we get the equation \( x^{3} - x^{2} - 2 x + 1 = 0 \). This is an interesting conclusion.
If we use conclusion (6), we can derive many symmetric expressions involving \( a \), \( b \), and \( c \). However, if we need to find the values of \( a^{2} b + b^{2} c + c^{2} a \) and \( a b^{2} + b c^{2} + c a^{2} \), the above solution is not straightforward. We will use a different approach to find these values.
Clearly, the six roots of these three equations must be the roots of the equation \( \left(x^{2} + a x + b\right)\left(x^{2} + b x + c\right)\left(x^{2} + c x + a\right) = 0 \),
which is \( x^{6} + (a + b + c) x^{5} + (a + b + c + b c + c a + a b) x^{4} + (a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c) x^{3} + (a^{2} b + b^{2} c + c^{2} a + b c + c a + a b) x^{2} + (a b^{2} + b c^{2} + c a^{2}) x + a b c = 0 \).
Let \( s_{1} = x_{1} + x_{2} + x_{3} \),
\( s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} \), \( s_{3} = x_{1} x_{2} x_{3} \). By the relationship between the roots and coefficients, and according to conclusion (6), we have
\[
\begin{array}{l}
\left\{\begin{array}{l}
s_{1} = x_{1} + x_{2} + x_{3} = 1, \\
s_{2} = x_{2} x_{3} + x_{3} x_{1} + x_{1} x_{2} = -2, \\
s_{3} = x_{1} x_{2} x_{3} = -1.
\end{array}\right. \\
\left\{\begin{array}{l}
a + b + c = -2 s_{1}, \\
a + b + c + b c + c a + a b = s_{1}^{2} + 2 s_{2}, \\
a^{2} + b^{2} + c^{2} + b c + c a + a b + a b c \\
= -2 s_{1} s_{2} - 2 s_{3}, \\
a^{2} b + b^{2} c + c^{2} a + b c + c a + a b \\
= s_{2}^{2} + 2 s_{1} s_{3}, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3}, \\
a b c = s_{3}^{2}.
\end{array}\right.
\end{array}
\]
From equations (7) and (8), we get
\[
\begin{array}{l}
a^{2} b + b^{2} c + c^{2} a = s_{2}^{2} + 2 s_{1} s_{3} - (b c + c a + a b) \\
= (-2)^{2} + 2 \times 1(-1) - (-1) = 3, \\
a b^{2} + b c^{2} + c a^{2} = -2 s_{2} s_{3} \\
= -2(-2)(-1) = -4.
\end{array}
\] | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: A rope of length 2009 is operated as follows: first, it is divided into two ropes of positive integer lengths, and the lengths of the two ropes are recorded, then the above operation is repeated on one of the ropes, ... until 2009 ropes of length 1 are obtained. If the lengths of the two ropes obtained in a certain operation are not equal, then this operation is called "good".
(1) Find the maximum and minimum values of the number of good operations;
(2) Prove: In all operation processes where the number of good operations reaches the minimum value, the number of different lengths of ropes recorded is the same.
| 【Analysis】(1) It is easy to know that a rope of length 2 can only be divided into two segments of length 1, meaning the last operation on the rope must not be a good operation. A rope of length 2009 can be divided into 2009 segments of length 1 after exactly 2008 operations, so the number of good operations is no more than
$2008-1=2007$.
Therefore, the maximum number of good operations is 2007.
The following is such an operation: each time, cut a segment of length 1, then the first 2007 operations are all good operations.
Generalize the problem: For a rope of length $n$, let $f(n)$ be the minimum number of good operations.
By conducting mathematical experiments with simple positive integers $n$ and summarizing the experience, the following two lemmas can be discovered.
Lemma 1: $f(n)=0 \Leftrightarrow n=2^{k}$.
Lemma 2: Suppose $n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{1}}\left(k_{1}>k_{2}\right.$ $\left.>\cdots>k_{l} \geqslant 0\right)$. Let $l=S_{2}(n) \triangleq$ the sum of the digits of $n$ in binary. Then $f(n)=S_{2}(n)-1$.
Proof of Lemma 2: Since after $l-1$ good operations, we can obtain $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{1}}$, then
$f(n) \leqslant l-1$.
We will prove by mathematical induction:
$f(n) \geqslant S_{2}(n)-1$.
When $n=1$, it is Lemma 1, and the conclusion is obviously true.
Assume that for $n \leqslant k$, $f(n) \geqslant S_{2}(n)-1$.
When $n=k+1$, by the definition of $f$, there exists an operation process containing $f(k+1)$ good operations and several non-good operations, which cuts a rope of length $k+1$ into $k+1$ segments of length 1. If the first operation in the above process is a good operation, cutting a rope of length $k+1$ into segments of lengths $a$ and $b$, i.e., $k+1=a+b, a \neq b$, then
$$
\begin{array}{l}
f(k+1)=f(a)+f(b)+1 \\
\geqslant\left(S_{2}(a)-1\right)+\left(S_{2}(b)-1\right)+1 \\
=S_{2}(a)+S_{2}(b)-1 \\
\geqslant S_{2}(a+b)-1=S_{2}(k+1)-1,
\end{array}
$$
The equality holds if and only if there is no carry in the binary addition of $a+b$.
If the first operation in the above process is not a good operation, i.e., $k+1=2a$, then
$$
f(k+1)=2 f(a) \geqslant 2\left(S_{2}(a)-1\right) .
$$
Since $S_{2}(2a) \leqslant S_{2}(a)+S_{2}(a)-1$, the equality holds if and only if $S_{2}(a)=1$, i.e.,
$$
2 S_{2}(a)-2 \geqslant S_{2}(2a)-1 .
$$
Using this result, we get
$$
\begin{array}{l}
f(k+1) \geqslant 2\left(S_{2}(a)-1\right) \\
\geqslant S_{2}(2a)-1=S_{2}(k+1)-1,
\end{array}
$$
The conclusion holds, and the equality holds if and only if $S_{2}(a)=1$.
In summary, Lemma 2 is proved.
Therefore, using the binary representation of 2009, we get
$$
f(2009)=S_{2}(2009)-1=7 \text {. }
$$
(2) If the binary representation of $n$ is
$$
n=2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{i}} \text {, }
$$
we can set the set $M=\left\{k_{1}, k_{2}, \cdots, k_{l}\right\}$.
If $l \geqslant 2$, then in any operation process where the number of good operations reaches the minimum, the first operation is a good operation, i.e., there exists a non-empty proper subset $L$ of $M$ such that the rope of length $n$ is cut into segments of lengths $a$ and $b$, where $a=\sum_{k \in L} 2^{k}, b=\sum_{k \in M-L} 2^{k}$. Performing $l-1$ good operations in succession produces $l$ different lengths $2^{k_{1}}, 2^{k_{2}}, \cdots, 2^{k_{i}}$.
Then, the non-good operations produce lengths $2^{k_{1}-1}, 2^{k_{1}-2}, \cdots, 2,2^{0}$, and so on. Therefore, the number of different lengths recorded is $l-1+k_{1}$.
In solving this problem, it is found through mathematical experiments that the minimum number of good operations $f(n)$ is related to the binary representation of the positive integer $n$, and a basic result about the sum of digits is used:
$$
S_{p}(a+b)=S_{p}(a)+S_{p}(b)-k(p-1),
$$
where $k$ is the number of carries in the addition of $a+b$ in base $p$. | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. For a finite set $A$ consisting of positive integers, if $A$ is divided into two non-empty disjoint subsets $A_{1}$ and $A_{2}$, and the least common multiple (LCM) of the elements in $A_{1}$ equals the greatest common divisor (GCD) of the elements in $A_{2}$, then such a partition is called "good". Find the minimum value of the positive integer $n$ such that there exists a set of $n$ positive integers with exactly 2015 good partitions. | 3. 3024.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\left(a_{1}<a_{2}<\cdots<a_{n}\right)$.
For any non-empty finite set of positive integers $B$, let $\operatorname{lcm} B$ and $\operatorname{gcd} B$ denote the least common multiple and greatest common divisor of the elements in $B$, respectively.
Consider any good partition $\left(A_{1}, A_{2}\right)$ of $A$. By definition, there exists a positive integer $d$ such that
$\operatorname{lcm} A_{1}=d=\operatorname{gcd} A_{2}$.
For any $a_{i} \in A_{1}$ and $a_{j} \in A_{2}$, we have
$a_{i} \leqslant d \leqslant a_{j}$.
Thus, there exists a positive integer $k(1 \leqslant k<n)$ such that
$A_{1}=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$,
$A_{2}=\left\{a_{k+1}, a_{k+2}, \cdots, a_{n}\right\}$.
Therefore, each good partition is determined by an element $a_{k}$ $(1 \leqslant k<n)$, which we call a "separator".
Define $l_{k}=\operatorname{lcm}\left(a_{1}, a_{2}, \cdots, a_{k}\right)$,
$g_{k}=\operatorname{gcd}\left(a_{k+1}, a_{k+2}, \cdots, a_{n}\right)$,
where $1 \leqslant k \leqslant n-1$. Then $a_{k}$ is a separator if and only if $l_{k}=g_{k}$.
To prove some properties of separators, we need the following lemma.
Lemma If $a_{k-1}$ and $a_{k}$ $(2 \leqslant k \leqslant n-1)$ are both separators, then $g_{k-1}=g_{k}=a_{k}$.
Proof Assume $a_{k-1}$ and $a_{k}$ are both separators.
$$
\begin{array}{c}
\text { By } l_{k-1}=g_{k-1} \Rightarrow l_{k-1} \mid a_{k} \\
\Rightarrow g_{k}=l_{k}=\operatorname{lcm}\left(l_{k-1}, a_{k}\right)=a_{k}, \\
g_{k-1}=\operatorname{gcd}\left(a_{k}, g_{k}\right)=a_{k} .
\end{array}
$$
The lemma is proved.
Property 1 For each $k=2,3, \cdots, n-2$, at least one of $a_{k-1}$, $a_{k}$, and $a_{k+1}$ is not a separator.
Proof of Property 1 By contradiction.
Assume $a_{k-1}$, $a_{k}$, and $a_{k+1}$ are all separators.
By the lemma, we have $a_{k+1}=g_{k}=a_{k}$, which is a contradiction.
Property 2 $a_{1}$ and $a_{2}$ cannot both be separators, and $a_{n-2}$ and $a_{n-1}$ cannot both be separators.
Proof of Property 2 Assume $a_{1}$ and $a_{2}$ are both separators.
By the lemma, we have $a_{2}=g_{1}=l_{1}=\operatorname{lcm}\left(a_{1}\right)=a_{1}$, which is a contradiction.
Similarly, assume $a_{n-2}$ and $a_{n-1}$ are both separators. By the lemma, we have $a_{n-1}=g_{n-1}=\operatorname{gcd}\left(a_{n}\right)=a_{n}$, which is a contradiction.
Properties 1 and 2 are proved.
Suppose a set $A$ with $n$ elements has exactly 2015 good partitions. Clearly, $n \geqslant 5$.
By Property 2, at most one element in $\left\{a_{1}, a_{2}\right\}$ and $\left\{a_{n-2}, a_{n-1}\right\}$ can be a separator.
Let $\lfloor x\rfloor$ denote the greatest integer not exceeding the real number $x$, and $\lceil x\rceil$ denote the smallest integer not less than the real number $x$.
By Property 1, at least $\left\lfloor\frac{n-5}{3}\right\rfloor$ elements in $\left\{a_{3}, a_{4}, \cdots, a_{n-3}\right\}$ are not separators. Therefore, $A$ has at most
$$
(n-1)-2-\left\lfloor\frac{n-5}{3}\right\rfloor=\left\lceil\frac{2(n-2)}{3}\right\rceil
$$
separators.
$$
\text { Therefore, }\left\lceil\frac{2(n-2)}{3}\right\rceil \geqslant 2015 \Rightarrow n \geqslant 3024 \text {. }
$$
Finally, we prove that there exists a set $A$ with 3024 elements that has exactly 2015 separators.
In fact,
$$
A=\left\{2 \times 6^{i}, 3 \times 6^{i}, 6^{i+1} \mid 10 \leqslant i \leqslant 1007\right\}
$$
satisfies $|A|=3024$, and $3 \times 6^{i} (0 \leqslant i \leqslant 1007)$ and $6^{i} (1 \leqslant i \leqslant 1007)$ are all separators.
In summary, the minimum value of $n$ is 3024. | 3024 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $m$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=m, a_{1}=\varphi(m), \\
a_{2}=\varphi^{(2)}(m)=\varphi(\varphi(m)), \cdots, \\
a_{n}=\varphi^{(n)}(m)=\varphi\left(\varphi^{(n-1)}(m)\right),
\end{array}
$$
where $\varphi(m)$ is the Euler's totient function.
If for any non-negative integer $k$, we have $a_{k+1} \mid a_{k}$, find the largest positive integer $m$ not exceeding 2016.
(Weng Shiyou, problem contributor) | 3. It is known that if $a \mid b$, then $\varphi(a) \mid \varphi(b)$.
Moreover, when $m>2$, $\varphi(m)$ is even.
Therefore, it is only necessary to find integers $m$ such that $a_{1} \mid a_{0}$, which implies $a_{k+1} \mid a_{k}$.
If $m$ is an odd number greater than 2, it is impossible for $a_{1} \mid m$.
Let $m=2^{\alpha} \prod_{k=1}^{s} p_{k}^{\alpha_{k}}$, where $p_{1}, p_{2}, \cdots, p_{s}$ are all odd primes. Then
$$
\varphi(m)=m\left(1-\frac{1}{2}\right) \prod_{k=1}^{s}\left(1-\frac{1}{p_{k}}\right) .
$$
Thus, only when $s=1$ and $p_{1}=3$, we have $a_{1} \mid a_{0}$.
Therefore, $m$ must be of the form $2^{\alpha} \times 3^{\beta}$.
Hence, the largest positive integer value not exceeding 2016 is
$$
m=2^{3} \times 3^{5}=1944 \text {. }
$$ | 1944 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the set
$$
I=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\}\right\} \text {, }
$$
$A$ is a subset of $I$ and satisfies: for any
$$
\left(x_{1}, x_{2}, x_{3}, x_{4}\right) 、\left(y_{1}, y_{2}, y_{3}, y_{4}\right) \in A \text {, }
$$
there exist $i 、 j(1 \leqslant i<j \leqslant 4)$ such that
$$
\left(x_{i}-x_{j}\right)\left(y_{i}-y_{j}\right)<0 \text {. }
$$
Determine the maximum value of $|A|$, where $|A|$ denotes the number of elements in the set $A$. (Provided by Zhang Limin) | 8. First, consider the set satisfying $x_{1}+x_{2}+x_{3}+x_{4}=24$
$$
\begin{aligned}
A= & \left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{i} \in\{1,2, \cdots, 11\},\right. \\
& i=1,2,3,4\},
\end{aligned}
$$
The number of its elements is
$$
\begin{array}{l}
\mathrm{C}_{23}^{3}-4\left(\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}+\cdots+\mathrm{C}_{12}^{2}\right) \\
=\mathrm{C}_{23}^{3}-4 \mathrm{C}_{13}^{3}=891,
\end{array}
$$
Any two different vectors in this set satisfy the problem's conditions.
Next, consider the vector set
$$
\begin{array}{c}
B=\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) \mid x_{1}, x_{2} \in\{1,2, \cdots,\right. \\
\left.11\}, x_{3}=x_{4}=a \in\{1,2, \cdots, 11\}\right\},
\end{array}
$$
For each vector $\left(x_{1}, x_{2}, a, a\right)$, define the vector group as follows:
If $a=1$, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right),\left(x_{1}, x_{2}, 1,2\right), \cdots, \\
\left(x_{1}, x_{2}, 1,11\right),\left(x_{1}, x_{2}, 2,11\right), \cdots, \\
\left(x_{1}, x_{2}, 11,11\right) ;
\end{array}
$$
If $a=11$, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right),\left(x_{1}, x_{2}, 2,1\right), \cdots, \\
\left(x_{1}, x_{2}, 11,1\right),\left(x_{1}, x_{2}, 11,2\right), \cdots, \\
\left(x_{1}, x_{2}, 11,11\right) ;
\end{array}
$$
If $a$ is odd and greater than 1 and less than 11, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right), \cdots,\left(x_{1}, x_{2}, 1, a-1\right), \\
\left(x_{1}, x_{2}, 1, a\right),\left(x_{1}, x_{2}, 2, a\right), \cdots, \\
\left(x_{1}, x_{2}, a-1, a\right),\left(x_{1}, x_{2}, a, a\right) \\
\left(x_{1}, x_{2}, a, a+1\right), \cdots,\left(x_{1}, x_{2}, a, 11\right), \\
\left(x_{1}, x_{2}, a+1,11\right), \cdots,\left(x_{1}, x_{2}, 11,11\right) ;
\end{array}
$$
If $a$ is even, the vector group is
$$
\begin{array}{l}
\left(x_{1}, x_{2}, 1,1\right), \cdots,\left(x_{1}, x_{2}, a-1,1\right), \\
\left(x_{1}, x_{2}, a, 1\right),\left(x_{1}, x_{2}, a, 2\right), \cdots, \\
\left(x_{1}, x_{2}, a-1, a\right),\left(x_{1}, x_{2}, a, a\right), \\
\left(x_{1}, x_{2}, a+1, a\right), \cdots,\left(x_{1}, x_{2}, 11, a\right), \\
\left(x_{1}, x_{2}, 11, a+1\right),\left(x_{1}, x_{2}, 11,11\right) .
\end{array}
$$
For the vector set
$$
\begin{array}{c}
C=\left\{\left(a, a, x_{3}, x_{4}\right) \mid x_{3}, x_{4} \in\{1,2, \cdots,\right. \\
\left.11\}, x_{1}=x_{2}=a \in\{1,2, \cdots, 11\}\right\}
\end{array}
$$
Similarly, define the vector group.
For each vector in set $A$, it belongs to exactly one vector group, and different vectors belong to different vector groups, determining these 891 vector groups. Any 892 vectors must have two belonging to the same vector group, hence no longer possessing the property set by the problem.
Therefore, the maximum value sought is 891. | 891 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ and $b$ are integers, $\frac{127}{a}-\frac{16}{b}=1$. Then the maximum value of $b$ is $\qquad$ . | $$
\text { Two, 1.2 } 016 .
$$
The original equation is transformed into $b=\frac{16 a}{127-a}=\frac{127 \times 16}{127-a}-16$. When and only when $127-a=1$, $b$ reaches its maximum value, at this point, $b=127 \times 16-16=2016$. | 2016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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