problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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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{... | 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... | 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_{\... | 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
$$
\begin{array}{l}
\left(1+z^{2}\right... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b, c, d$ be real numbers, satisfying
$$
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] \geqs... | 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, $-... | -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 $a^{3}+b^{3}=8 b^{3} \sin ^{3} 10^{\circ}+b^{3}$
$$
\begin{array}{l}
=8 b^{3} \cdot \frac{1}{4}\left(3 \sin 10^{\circ}-\sin 30^{\circ}\right)+b^{3}=6 b^{3} \sin 10^{\circ} \\
\Rightarrow \frac{a^{3}+b^{3}}{a b^{2}}=\frac{6 b^{3} \sin 10^{\circ}}{2 b \sin 10^{\circ} \cdot b^{2}}... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a_{n}=2^{n}, b_{n}=5 n-1\left(n \in \mathbf{Z}_{+}\right)$,
$$
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\}$ ... | 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 inscrib... | 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, $... | 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 t... | 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... | 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 {, }
$$
i.e., when $a=\frac{7}{... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the line $l$ passing through the origin intersect the graph of the function $y=|\sin x|$ $(x \geqslant 0)$ at exactly three points, with $\alpha$ being the largest of the x-coordinates of these intersection points. Then
$$
\frac{\left(1+\alpha^{2}\right) \sin 2 \alpha}{2 \alpha}=
$$
$\qquad$ . | 3. 1 .
As shown in Figure 2, let the line $l$ be tangent to the function $y=|\sin x|(x \geqslant 0)$
at point $P(\alpha,-\sin \alpha)$, and $k_{l}=-\cos \alpha$.
Then the line $l: y+\sin \alpha=-(x-\alpha) \cos \alpha$.
Substituting $(0,0)$, we get $\alpha=\tan \alpha$.
Therefore, $\frac{\left(1+\alpha^{2}\right) \sin... | 1 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P(1,4,5)$ is a fixed point in the rectangular coordinate system $O-x y z$, a plane is drawn through $P$ intersecting the positive half-axes of the three coordinate axes at points $A$, $B$, and $C$ respectively. Then the minimum value of the volume $V$ of all such tetrahedrons $O-A B C$ is $\qquad$ | 4. 90 .
Let the plane equation be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where the positive numbers $a$, $b$, and $c$ are the intercepts of the plane on the $x$-axis, $y$-axis, and $z$-axis, respectively.
Given that point $P$ lies within plane $ABC$, we have $\frac{1}{a}+\frac{4}{b}+\frac{5}{c}=1$.
From $1=\frac{1}{... | 90 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the number of solutions to the equation
$$
|| \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|... | 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 $x^{2}+y^{2}=5$.
$$
\begin{array}{l}
\text { Also, } 2 x y=(x+y)^{2... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups.
The integer solutions $(x, y)$ of the indeterminate equation $x^{2}+y^{2}=x y+2 x+2 y$ are in total groups. | 7.6 .
Rewrite the given equation as
$$
\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}
$$
Since $y$ is an integer, thus, $y=0,1,2,3,4$.
Upon calculatio... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) From the 2015 positive integers 1, 2, $\cdots$, 2015, select $k$ numbers such that the sum of any two different numbers is not a multiple of 50. Find the maximum value of $k$.
untranslated part:
在 1,2 , $\cdots, 2015$ 这 2015 个正整数中选出 $k$ 个数,使得其中任意两个不同的数之和均不为 50 的倍数. 求 $k$ 的最大值.
translated part:
From ... | 10. Classify $1 \sim 2015$ by their remainders when divided by 50.
Let $A_{i}$ represent the set of numbers from $1 \sim 2015$ that have a remainder of $i$ when divided by 50, where $i=0,1, \cdots, 49$. Then $A_{1}, A_{2}, \cdots, A_{15}$ each contain 41 numbers, and $A_{0}, A_{16}, A_{17}, \cdots, A_{49}$ each contai... | 977 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) As shown in Figure 4, the three sides of $\triangle ABC$ are all positive integers, and the perimeter is 35. $G$ and $I$ are the centroid and incenter of $\triangle ABC$, respectively, and $\angle GIC = 90^{\circ}$. Find the length of side $AB$. | 11. Extend $G I$, intersecting $C B$ and $C A$ (or their extensions) at points $P$ and $Q$ respectively.
Since $C I$ is the angle bisector of $\angle C$ and $\angle G I C=90^{\circ}$, we know that $\triangle C P Q$ is an isosceles triangle.
Draw $G E \perp P C$ and $G F \perp C Q$ from point $G$, and draw $I R \perp ... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the function $f: \mathbf{N} \rightarrow \mathbf{N}$ defined as follows:
$$
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{. }
$$
Then $f(f(f(x)))<8$, which does not... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Real numbers $x, y, a$ satisfy $x+y=a+1$ and $xy=a^{2}-7a+16$. Then the maximum value of $x^{2}+y^{2}$ is $\qquad$ | 5. 32 .
Notice that,
$$
\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}
$$... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a triangle with sides as three consecutive natural numbers, the largest angle is twice the smallest angle. Then the perimeter of the triangle is $\qquad$ | II, 7.15.
Let the three sides of $\triangle ABC$ be $AB=n+1, BC=n, AC=n-1$, and the angle opposite to side $AC$ be $\theta$, and the angle opposite to side $AB$ be $2\theta$.
According to the Law of Sines and the Law of Cosines, we have
$$
\begin{array}{l}
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \... | 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 ... | 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 sati... | 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(\overrigh... | 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 $d=4$.
The... | -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}=\l... | 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... | 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 $A$, $P$, a... | 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 c... | (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 ... | 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
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right|=\left|\frac{1}{k} \sum_{i=1}^{k} x_{i}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}\right| \\
=\left|\frac{1}{k(k+1)}\left(\sum_{i=1}^{k} x_{i}-k x_{k+1}\right)\right| \\
\leqslant \frac{1}{k(k+1)} \sum_{i=1}^{k} i\left|x_{i}-x_{i+1}\ri... | 1990 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let positive integers $a_{1}, a_{2}, \cdots, a_{31}, b_{1}, b_{2}, \cdots, b_{31}$ satisfy
$$
\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=\lef... | 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 $S=S_{1}+S_{2}$.... | 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$P$ & $\frac{1}{7}$... | 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 $[10,12)$ ... | 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, $x+2 y+3 z \geqslant-12$... | -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,
$$
\begin{array}{l}
2 \sin \frac{\pi}{18}\left(\sum_{k=1}^{n} \sin \frac{k \pi}{9}\right) \\
=\sum_{k=1}^{n}\left(\cos \frac{(2 k-1) \pi}{18}-\cos \frac{(2 k+1) \pi}{18}\right) \\
=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \tan \frac{4 \pi}{9} \cdot \sin \frac{\pi}{18}=\cos \frac{\pi... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. In the coordinate plane, points with both coordinates as integers are called integer points, and triangles with all three vertices as integer points are called integer point triangles. Find the number of integer point right triangles $OAB$ with a right angle at the origin $O$ and with $I(2015,7 \times 2015)$ as the... | 14. Let point $A$ be in the first quadrant.
Let $\angle x O I=\alpha$.
Then $\tan \alpha=7$,
$k_{O A}=\tan \left(\alpha-\frac{\pi}{4}\right)=\frac{\tan \alpha-1}{1+\tan \alpha}=\frac{3}{4}$
$\Rightarrow k_{O B}=-\frac{4}{3}$.
Since $A$ and $B$ are integer points, set
$$
A\left(4 t_{1}, 3 t_{1}\right), B\left(-3 t_{2},... | 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 requir... | 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
$$
\begin{array}{l}
M_{0}+N... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. After removing all perfect squares from the sequence of positive integers $\{1,2, \cdots\}$, the remaining numbers form a sequence $\left\{a_{n}\right\}$ in their original order. Then $a_{2015}=$ $\qquad$ . | 6.2060.
Let $k^{2}2014
\end{array}\right. \\
\Rightarrow k=45, a_{2015}=2060 .
\end{array}
$ | 2060 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and
$$
|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}... | 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 $a_{2}>a_{1}$ a... | 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) $2-\sqrt{3}$ | $-1 . \mathrm{C}$.
$$
\begin{array}{l}
\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}} \times \\
\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}} \\
=\sqrt{2-\sqrt{3}} \times \sqrt{2-\sqrt{2-\sqrt{3}}} \times \\
\quad \sqrt{2^{2}-\left(\sqrt{2-\sqrt{2-\sqrt{3}})^{2}}\right.} \\
=\sqrt{... | 1 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given positive integers $a$, $b$, $c$, $d$ satisfy $a^{2}=c(d+29)$, $b^{2}=c(d-29)$. Then the value of $d$ is $\qquad$. | $=, 1.421$
Let $(a, b)=d^{\prime}, a=d^{\prime} a_{1}, b=d^{\prime} b_{1},\left(a_{1}, b_{1}\right)=1$.
From the given conditions, dividing the two equations yields
$$
\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... | 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_... | 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,
$\frac{a}{b}-\frac{1}{5}=0.2015 \cdots-0.2=0.0015$
$$
125... | 129 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the sequence $\left\{a_{n}\right\}$ :
$$
\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) \\... | 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... | 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,
$$
\begin{array}{l}
1980=2^{2} \times 3^{2} \times 5 \times 11, \\
v_{11}(1980!)=\sum_{i \geqslant 1}\left[\frac{1980}{11^{i}}\right]=197 . \\
\text { Since }\left[\frac{m}{5^{i}}\right] \geqslant\left[2 \times \frac{m}{11^{i}}\right] \geqslant 2\left[\frac{m}{11^{i}}\right], \text { hence } \\
v_{5}(m!) \geqsl... | 197 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. As shown in Figure 3, in $\triangle A B C$, $O$ is the midpoint of side $B C$, and a line through point $O$ intersects lines $A B$ and $A C$ at two distinct points $M$ and $N$. If
$$
\begin{array}{l}
\overrightarrow{A B}=m \overrightarrow{A M}, \\
\overrightarrow{A C}=n \overrightarrow{A N},
\end{array}
$$
then th... | 14. 2 .
Since $O$ is the midpoint of side $B C$, we have
$$
\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 ... | 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 ... | 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
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{1}-2\right)+\left(x_{2}+1\right)\left(x_{2}... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let $a_{1} \in \mathbf{Z}_{+}$, and $a_{1} \leqslant 18$, define the sequence $\left\{a_{n}\right\}:$
$$
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.
$$
Find the maximum number of elements in the set $M=\left\{a_{n} \mid ... | 11. Given the positive integer $a_{1} \leqslant 18$ and the recursive formula for the sequence $\left\{a_{n}\right\}$, the following properties can be derived:
(1) $a_{n+1} \equiv 2 a_{n}(\bmod 36)$, i.e.,
$a_{n+1} \equiv 2 a_{n}(\bmod 4)$, and $a_{n+1} \equiv 2 a_{n}(\bmod 9)$;
(2) All terms are positive integers, and... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $S=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{980100}}$. Find the greatest positive integer $[S]$ that does not exceed the real number $S$. | Notice that, $\frac{1}{\sqrt{n}}=\frac{2}{\sqrt{n}+\sqrt{n}}$.
Therefore, $\frac{1}{\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n-1}}=2(\sqrt{n}-\sqrt{n-1})$,
$2(\sqrt{n+1}-\sqrt{n})=\frac{2}{\sqrt{n+1+\sqrt{n}}}<\frac{1}{\sqrt{n}}$.
Thus, $1977<2(\sqrt{980101}-\sqrt{2})$
$$
<S<2(\sqrt{980100}-1)=1978 \text {. }
$$
Hence, $[S]=... | 1977 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 For integer pairs $(a, b)(0<a<b<1000)$, a set $S \subseteq\{1,2, \cdots, 2003\}$ is called a "jump set" of the pair $(a, b)$: if for any element pair $\left(s_{1}, s_{2}\right)$, $s_{1} 、 s_{2} \in$ $S,\left|s_{1}-s_{2}\right| \notin\{a, b\}$.
Let $f(a, b)$ be the maximum number of elements in a jump set of ... | 【Analysis】This method is quite typical, the minimum value is obtained using the greedy idea, while the maximum value is derived using the pigeonhole principle. Unfortunately, during the exam, most candidates only answered one part correctly.
For the minimum value, take $a=1, b=2$, then
$$
\begin{array}{l}
\{1,2,3\},\{4... | 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 t... | 【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 $1 \leqslant i \leqslant 31$, ... | 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 $1<S=1+\frac{1}{2^{3}}+\frac{1}{3^{3}}+\cdots+\frac{1}{2011^{3}}$
$$
<1+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2011 \times 2012}\right)<\frac{5}{4} \text {. }
$$
... | 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, ... | 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
$$
\left(6-\pi^{2} x^{2}\right) \sin \pi x-2 \pi x \cos \pi x>0.
$... | 8 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the set $M=\{1,2, \cdots, 12\}$, and the three-element set $A=$ $\{a, b, c\}$ satisfies $A \subset M$, and $a+b+c$ is a perfect square. Then the number of sets $A$ is $\qquad$. | ,- 1.26 .
From $6 \leqslant a+b+c \leqslant 33$, we know that the square numbers within this range are $9, 16, 25$. Let's assume $a<b<c$.
If $a+b+c=9$, then the possible values for $c$ are 6, 5, 4, in which case,
$$
A=\{1,2,6\},\{1,3,5\},\{2,3,4\} \text {; }
$$
If $a+b+c=16$, then $7 \leqslant c \leqslant 12$, we can... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The volume of a rectangular prism is 8 cubic centimeters, and the total surface area is 32 square centimeters. If the length, width, and height form a geometric sequence, then the sum of all the edges of this rectangular prism is $\qquad$ | 4.32 cm.
Let the common ratio be $q$, and the length, width, and height be $q a, a, \frac{a}{q}$, respectively. Then $a^{3}=8 \Rightarrow a=2$.
$$
\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}
$$
Therefore, the sum... | 32 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) If the subset $A$ of the set $M=\{1,2, \cdots, 200\}$ consists of elements each of which can be expressed as the sum of the squares of two natural numbers (allowing the same number), find the maximum number of elements in the set $A$.
| 11. Notice that, the squares not exceeding 200 are $0^{2}, 1^{2}, 2^{2}, \cdots, 14^{2}$.
First, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form $k^{2}+0^{2}$, and there are 14 such numbers; while the sum of each pair of numbers in $1^{2}$, $2^{2}, \cdots, 10^{2}$ (allowing the same n... | 79 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If $2016+3^{n}$ is a perfect square, then the positive integer $n=$ . $\qquad$ | 4. 2 .
Obviously, $2016+3^{n}$ is an odd perfect square.
So $2016+3^{n} \equiv 1(\bmod 8)$
$$
\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 {. }
$$
So $224+3^{2 k}=224+\left(3^{k}\right)^{2}$ is a perfect square... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\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 {. }
$$
$t \times$ (2) $-x \times$ (1) gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\f... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In the plane, $m$ points have no three points collinear, and their convex hull is an $n$-sided polygon. By appropriately connecting lines, a grid region composed of triangles can be obtained. Let the number of non-overlapping triangles be $f(m, n)$. Then $f(2016,30)=$ $\qquad$ | One, 1.4000.
Since no three points are collinear, we have
$$
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.
$$
and a fixed point $B(a, b)$ satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$ | 2. 2 .
According to the problem, for any $(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking $(x, y)=(1,0),(0,1)$, we get the fixed point $B(a, b)$ satisfying the necessary conditions $\left\{\begin{array}{l}a \leqslant 1, \\ b \leqslant 1,\end{array}\right.$, which implies $a+b \leqslant 2$.
Thus, $(a+b)_{\text ... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the set $T=\{1,2, \cdots, 2010\}$, for each non-empty subset of $T$, calculate the reciprocal of the product of all its elements. Then the sum of all such reciprocals is $\qquad$ | 6. 2010 .
Notice that, in the expansion of the product
$$
(1+1)\left(1+\frac{1}{2}\right) \cdots\left(1+\frac{1}{2010}\right)
$$
the $2^{2010}$ terms are all the reciprocals of positive integers, and each is precisely the reciprocal of the product of the numbers in one of the $2^{2010}$ subsets of set $T$. Removing t... | 2010 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the incenter $I(-1,7)$ of a right-angled triangle $\triangle O A B$ with all three vertices as integer points, and the origin $O$ as the right-angle vertex. The number of such right-angled triangles $\triangle O A B$ is $\qquad$
Translate the above text into English, please retain the original text's line bre... | 8.2.
As shown in Figure 4.
Let $\angle x O A=\alpha, \angle x O I=\beta$.
Then $\alpha=\beta-\frac{\pi}{4}$.
Also, $\tan \beta=-7$, so
$$
\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) \\
... | 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, $f\left(2^{2}\right)+f\left(2^{3}\right)+\cdots+f\left(2^{10}\right)$
$$
=2+3+\cdots+10=54 \text ... | 54 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. There are ten small balls of the same size, five of which are red and five are white. Now, these ten balls are arranged in a row arbitrarily, and numbered from left to right as $1,2, \cdots, 10$. Then the number of arrangements where the sum of the numbers of the red balls is greater than the sum of the numbers of t... | 3. 126.
First, the sum of the numbers is $1+2+\cdots+10=$ 55. Therefore, the sum of the numbers on the red balls cannot be equal to the sum of the numbers on the white balls.
Second, if a certain arrangement makes the sum of the numbers on the red balls greater than the sum of the numbers on the white balls, then by ... | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given real numbers $a, b$ satisfy
$$
a+\lg a=10, b+10^{b}=10 \text {. }
$$
Then $\lg (a+b)=$ $\qquad$ . | 8.1.
Since $\lg a=10-a, 10^{b}=10-b$, therefore, $a$ is the x-coordinate of the intersection point of $y=\lg x$ and $y=10-x$, and $b$ is the y-coordinate of the intersection point of $y=10^{x}$ and $y=10-x$.
Also, $y=\lg x$ and $y=10^{x}$ are symmetric about the line $y=x$, and $y=10-x$ is symmetric about the line $y... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. $6^{11}+C_{11}^{1} 6^{10}+C_{11}^{2} 6^{9}+\cdots+C_{11}^{10} 6-1$ when divided by 8 yields a remainder of $\qquad$ . | 3.5.
Notice that,
$$
\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{... | 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 th... | 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 possi... | 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... | 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,
$$
\begin{array}{l}
\overrightarrow{P B} \cdot \overrightarrow{P D}=\frac{1}{4}\left[(\overrightarrow{P B}+\overrightarrow{P D})^{2}-(\overrightarrow{P B}-\overrightarrow... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given
$$
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-6... | 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}=-\fra... | 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 $\qquad$ gro... | 8. 0 .
Assume $x \geqslant a$. Then
$$
\begin{array}{l}
4 x y z=2 a+b+4 c \leqslant 7 a \leqslant 7 x \\
\Rightarrow y z \leqslant \frac{7}{4} \Rightarrow y z \leqslant 1 \Rightarrow(y, z)=(1,1) \\
\Rightarrow\left\{\begin{array}{l}
a a+b+4 c=4 x, \\
2 x+5=4 a b c
\end{array}\right. \\
\Rightarrow 2 a+b+4 c+10=8 a b c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $0<a<1$, and satisfies
$$
\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}
$$
0 or 1.
By the problem, we know that 18 of them are equal to 1, and 11 ... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $O$ is the circumcenter of acute $\triangle A B C$, $\angle B A C$ $=60^{\circ}$, extend $C O$ to intersect $A B$ at point $D$, extend $B O$ to intersect $A C$ at point $E$. Then $\frac{B D}{C E}=$ $\qquad$ | $=, 7.1$.
Connect $O A, D E$.
$$
\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}
$$
Similarly, $C E=D... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
For any real number $x$, $[x]$ represents the greatest integer not exceeding $x$, and $\{x\}$ represents the fractional part of $x$. Then
$$
\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 {, }
$$
where $M$ is an integer.
Then $\left\{\frac{2014^{n}}{2015}\right\}=\left\{\begin{array}{ll}\frac{1}{2015}, & n \text { is even; } \\ \frac{2014}{2015}, & n \text { is odd. }\end{array}\right.$
Therefore,... | 1007 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a_{1}, a_{2}, \cdots, a_{9}$ as any permutation of $1,2, \cdots, 9$. Then the minimum value of $a_{1} a_{2} a_{3}+a_{4} a_{5} a_{6}+a_{7} a_{8} a_{9}$ is $\qquad$ | 2. 214.
Since $a_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, 9)$, by the AM-GM inequality, we have
$$
\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 \tim... | 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... | 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,
$$
\begin{array}{l}
6 ※ 6=2016 ※ 2016=1 . \\
\text { Therefore } 2016 ※(2016 ※ 2016)=\frac{600}{x} \\
\Rightar... | 109 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given positive integers $a, b, c (a < b < c)$ form a geometric sequence, and
$$
\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.
$$
Let the ... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given that $a$, $b$, and $c$ are three distinct real numbers. If in the quadratic equations
$$
\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} ... | 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 ... | 【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 ... | 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... | 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... | 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'... | 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^... | 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... | 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
$$
\begin{array}{l}
\mathrm{C}_{23}^{3}-4\left(\mathrm{C}_{2}^{2}+\mathrm{... | 891 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ and $b$ are integers, $\frac{127}{a}-\frac{16}{b}=1$. Then the maximum value of $b$ is $\qquad$ . | $$
\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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