problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
1. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {.... | 1. (1) The total number of all possible values of $x_{1}, x_{2}, \cdots, x_{n}$ is $2^{n}$, and each corresponding function value can be either 0 or 1. Therefore, the number of all different $n$-ary Boolean functions is $2^{2^{n}}$.
(2) Let $\mid D_{10}(g)$ | denote the number of elements in the set $D_{10}(g)$. Below,... | 1817 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {.... | 2. (1) Same as Question 1 (1) of Grade 1.
(2) Let $|D_{n}(g)|$ denote the number of elements in the set $D_{n}(g)$.
Obviously, $|D_{1}(g)|=1, |D_{2}(g)|=1$.
$$
\begin{array}{l}
\text { Also, } g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j} \\
=\left(1+x_{1}\left(1+\sum_{i=2}^{n}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. For a finite set
$$
A=\left\{a_{i} \mid 1 \leqslant i \leqslant n, i \in \mathbf{Z}_{+}\right\}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
let $S=\sum_{i=1}^{n} a_{i}$, then $S$ is called the "sum" of set $A$, denoted as
$|A|$. Given the set $P=\{2 n-1 \mid n=1,2, \cdots, 10\}$,
all the subsets of $P$ containin... | 8. 3600.
Since $1+3+\cdots+19=100$, and each element in $1,3, \cdots, 19$ appears in the three-element subsets of set $P$ a number of times equal to $\mathrm{C}_{9}^{2}=36$, therefore, $\sum_{i=1}^{k}\left|P_{i}\right|=3600$. | 3600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given ten points in space, where no four points lie on the same plane. Connect some of the points with line segments. If the resulting figure contains no triangles and no spatial quadrilaterals, determine the maximum number of line segments that can be drawn. ${ }^{[1]}$ (2016, National High School Mathematic... | Let the graph that satisfies the conditions be $G(V, E)$.
First, we prove a lemma.
Lemma In any $n(n \leqslant 5)$-order subgraph $G^{\prime}$ of graph $G(V, E)$, there can be at most five edges.
Proof It suffices to prove the case when $n=5$.
If there exists a vertex $A$ in $G^{\prime}$ with a degree of 4, then no edg... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given 2015 circles of radius 1 in the plane. Prove: Among these 2015 circles, there exists a subset $S$ of 27 circles such that any two circles in $S$ either both have a common point or both do not have a common point. ${ }^{[1]}$
(The 28th Korean Mathematical Olympiad) | Proof that there do not exist 27 circles such that any two circles have a common point.
Select a line $l$ such that $l$ is neither parallel to the line connecting the centers of any two of the 2015 circles nor perpendicular to these lines.
Let this line be the $x$-axis, and the 2015 circles be denoted as $C_{1}, C_{2... | 27 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 5 Let $A$ denote the set of all sequences (of arbitrary finite or infinite length) $\left\{a_{1}, a_{2}, \cdots\right\}$ formed by the elements of $\{1,2, \cdots, 2017\}$. If the first several consecutive terms of sequence $M$ are the terms of sequence $T$, then we say that sequence $M$ "starts with" sequence $... | 【Analysis】First, give an example where (2) does not hold.
Take 2017 sequences each with only one term: $1,2, \cdots$, 2017. Then each sequence in set $A$ must start with one of these 2017 sequences, so (2) does not hold.
First, in set $S$, there cannot exist two different sequences $T_{1}$ and $T_{2}$ such that $T_{2}... | 1 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. Given $X_{1}, X_{2}, \cdots, X_{100}$ as a sequence of non-empty subsets of a set $S$, and all are distinct. For any $i \in \{1,2, \cdots, 99\}$, we have $X_{i} \cap X_{i+1}=\varnothing, X_{i} \cup X_{i+1} \neq S$.
Find the minimum number of elements in the set $S$.
(45th United States of America Mathematical Olympi... | First use mathematical induction to prove: when $n \geqslant 4$, a subset sequence of $2^{n-1}+1$ subsets that meets the requirements can be constructed for $S=\{1,2, \cdots, n\}$; then prove that when $|S|=7$, the number of subsets in a subset sequence that meets the requirements does not exceed 100.
The minimum numbe... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
The third question: In a $33 \times 33$ grid, each cell is colored with one of three colors, such that the number of cells of each color is equal. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
Translate the above text into Eng... | Assume the number of separating edges is no more than 55, and denote the three colors as $A$, $B$, and $C$.
If a separating edge corresponds to two cells of colors $A$ and $B$ or $A$ and $C$, then it is called an $A$-colored separating edge. Similarly, define $B$-colored and $C$-colored separating edges.
Since $55 < ... | 56 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $n$ is a positive integer, such that there exist positive integers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfying
$$
x_{1} x_{2} \cdots x_{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right)=100 n .
$$
Find the maximum possible value of $n$.
(Lin Jin, problem contributor) | 2. The maximum possible value of $n$ is 9702.
Obviously, from the given equation, we have $\sum_{i=1}^{n} x_{i} \geqslant n$. Therefore, $\prod_{i=1}^{n} x_{i} \leqslant 100$.
Since equality cannot hold, then $\prod_{i=1}^{n} x_{i} \leqslant 99$.
$$
\begin{array}{l}
\text { and } \prod_{i=1}^{n} x_{i}=\prod_{i=1}^{n}\... | 9702 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. A magician and his assistant have a deck of cards, all of which have the same back, and the front is one of 2017 colors (each color has 1000000 cards). The magic trick is: the magician first leaves the room, the audience arranges $n$ face-up cards in a row on the table, the magician's assistant then flips $n-1$ of t... | 4. When $n=2018$, the magician and the assistant can agree that for $i=1,2, \cdots, 2017$, if the assistant retains the $i$-th card face up, the magician will guess that the 2018-th card is of the $i$-th color. This way, the magic trick can be completed.
Assume for some positive integer $n \leqslant 2017$, the magic t... | 2018 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6. On a $200 \times 200$ chessboard, some cells contain a red or blue piece, while others are empty. If two pieces are in the same row or column, we say one piece can "see" the other. Assume each piece can see exactly five pieces of the opposite color (it may also see some pieces of the same color). Find the maximum nu... | 6. First, give an example with 3800 pieces.
The intersections of rows 1 to 5 and columns 11 to 200, as well as the intersections of columns 1 to 5 and rows 11 to 200, are all placed with red pieces; the intersections of rows 6 to 10 and columns 11 to 200, as well as the intersections of columns 6 to 10 and rows 11 to ... | 3800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $\{x\}$ denote the fractional part of the real number $x$. Given $a=(5 \sqrt{2}+7)^{2017}$. Then $a\{a\}=$ $\qquad$ . | ,- 1.1 .
Let $b=(5 \sqrt{2}-7)^{2017}$.
Then $0<b<1$, and $a b=1$.
Notice that,
$a-b=\sum_{k=0}^{1008} 2 \mathrm{C}_{2017}^{2 k+1}(5 \sqrt{2})^{2016-2 k} \times 7^{2 k+1} \in \mathbf{Z}$.
Since $a=(a-b)+b(a-b \in \mathbf{Z}, 0<b<1)$, it follows that $b=\{a\} \Rightarrow a\{a\}=a b=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If the digits $a_{i}(i=1,2, \cdots, 9)$ satisfy
$$
a_{9}a_{4}>\cdots>a_{1} \text {, }
$$
then the nine-digit positive integer $\overline{a_{9} a_{8} \cdots a_{1}}$ is called a "nine-digit peak number", for example 134698752. Then, the number of all nine-digit peak numbers is . $\qquad$ | 4. 11875 .
From the conditions, we know that the middle number of the nine-digit mountain number can only be 9, 8, 7, 6, 5.
When the middle number is 9, there are $\mathrm{C}_{8}^{4} \mathrm{C}_{9}^{4}$ nine-digit mountain numbers; when the middle number is 8, there are $\mathrm{C}_{7}^{4} \mathrm{C}_{8}^{4}$ nine-dig... | 11875 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that the 2017 roots of the equation $x^{2017}=1$ are 1, $x_{1}, x_{2}, \cdots, x_{2016}$. Then $\sum_{k=1}^{2016} \frac{1}{1+x_{k}}=$ $\qquad$ . | 5.1008 .
Given $x_{k}=\mathrm{e}^{\frac{2 \pi m}{2017} \mathrm{i}}(k=1,2, \cdots, 2016)$, we know
$$
\begin{array}{l}
\overline{x_{k}}=\mathrm{e}^{\frac{-2 k \pi}{2017} \mathrm{i}}=\mathrm{e}^{\frac{2(2017-k) \pi \mathrm{i}}{2017} \mathrm{i}}=x_{2017-k} . \\
\text { Then } \frac{1}{1+x_{k}}+\frac{1}{1+x_{2017-k}} \\
=... | 1008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In a country, some cities have direct two-way flights between them. It is known that one can fly from any city to any other city with no more than 100 flights, and also one can fly from any city to any other city with an even number of flights. Find the smallest positive integer $d$ such that it is guaranteed that f... | Prompt Answer: $d=200$.
First, construct an example, then prove that $d=200$ is feasible. For any two cities $A$ and $B$, consider the shortest path connecting them that has an even length, let its length be $2k$. Use proof by contradiction combined with the extremal principle to show that $k \leqslant 100$. | 200 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $a+b+c=1$,
$$
b^{2}+c^{2}-4 a c+6 c+1=0 \text{. }
$$
Find the value of $a b c$. | Solve: From equation (1), we get $a=1-b-c$.
Substitute into equation (2) and rearrange to get
$$
\begin{array}{l}
b^{2}+5 c^{2}+4 b c+2 c+1=0 \\
\Rightarrow(b+2 c)^{2}+(c+1)^{2}=0 \\
\Rightarrow b+2 c=c+1=0 \Rightarrow b=2, c=-1 \\
\Rightarrow a=1-b-c=0 \Rightarrow a b c=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the operation " $*$ ":
$$
a * b=\log _{2}\left(2^{a}+2^{b}\right)(a 、 b \in \mathbf{R}) \text {. }
$$
Let $A=(1 * 3) * 5, B=(2 * 4) * 6$. Then the value of $1 *(A * B)$ is $\qquad$ | 1.7 .
From the definition of $*$, we know that $2^{a * b}=2^{a}+2^{b}$. Therefore, $2^{A}=2^{1 * 3}+2^{5}=2^{1}+2^{3}+2^{5}$, $2^{B}=2^{2}+2^{4}+2^{6}$.
Thus, $2^{1 *(A * B)}=2^{1}+2^{A}+2^{B}$ $=2+2^{1}+2^{2}+\cdots+2^{6}=2^{7}$.
Hence, $1 *(A * B)=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let real numbers $a$ and $b$ satisfy
$$
\begin{array}{l}
a^{2}\left(b^{2}+1\right)+b(b+2 a)=40, \\
a(b+1)+b=8 .
\end{array}
$$
Find the value of $\frac{1}{a^{2}}+\frac{1}{b^{2}}$.
(2014, National Junior High School Mathematics League) | Hint: Transform the two equations in the conditions to get $a^{2} b^{2}+(a+b)^{2}=40, a b+(a+b)=8$. Let $x=a+b, y=a b$.
Thus, $x^{2}+y^{2}=40, x+y=8$.
Solving yields $(x, y)=(2,6)$ (discard) or $(6,2)$. Therefore, $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{(a+b)^{2}-2 a b}{a^{2} b^{2}}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Figure 1 is a road map of a city, where $A$, $B, \cdots, I$ represent nine bus stops. A bus departs from station $A$, travels along the roads, reaches each bus stop exactly once, and finally returns to station $A$. The number of different sequences of stops it can pass through is
| 8. 32.
The bus route can be considered as a circle $\Gamma$ (direction is not considered for now), and each station on $\Gamma$ has exactly two adjacent stations.
Let the two stations adjacent to $I$ on $\Gamma$ be $\alpha$ and $\beta$. There are two cases:
(1) $\{\alpha, \beta\} \cap \{E, F, G, H\} \neq \varnothing$... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $x, y$ satisfy
$$
\frac{4}{x^{4}}-\frac{2}{x^{2}}=3, y^{4}+y^{2}=3 \text {. }
$$
Then the value of $\frac{4}{x^{4}}+y^{4}$ is $\qquad$
(2008, "Mathematics Weekly Cup" National Junior High School Mathematics Competition) | By observation, we know that $-\frac{2}{x^{2}}$ and $y^{2}$ are two different solutions of the equation $m^{2}+m=3$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\frac{4}{x^{4}}+y^{4}=m_{1}^{2}+m_{2}^{2}=\left(m_{1}+m_{2}\right)^{2}-2 m_{1} m_{2} \\
=(-1)^{2}-2(-3)=7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If in the real number range there is
$$
x^{3}+p x+q=(x-a)(x-b)(x-c),
$$
and $q \neq 0$, then $\frac{a^{3}+b^{3}+c^{3}}{a b c}=$ $\qquad$ | Obviously, $a$, $b$, $c$ are the roots of the equation $x^{3} + p x + q = 0$.
By Vieta's formulas, we have $a + b + c = 0$.
By formula (2), we get
$$
\frac{a^{3} + b^{3} + c^{3}}{a b c} = \frac{a^{3} + b^{3} + c^{3} - 3 a b c}{a b c} + 3 = 3.
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $D$ is the intersection of the tangents to the circumcircle of $\triangle A B C$ at $A$ and $B$, the circumcircle of $\triangle A B D$ intersects the line $A C$ and the segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. If $\frac{B C}{B F}=2$, find $\frac{B G... | Using Property 1, let the circumcenter of $\triangle ABC$ be $O$. Then $O, A, D, B$ are concyclic. If point $O$ is not on side $BC$, then $FO \perp BC \Rightarrow \angle OAB=90^{\circ}$, which is a contradiction. Therefore, point $O$ is on $BC$.
Let $CD$ intersect $AB$ at point $M$. Then
$$
\begin{array}{l}
\frac{AM}{M... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Given $n$ points $p_{1}, p_{2}, \cdots, p_{n}$ in the plane, with no three points collinear. Each point $p_{i}(i=1,2, \cdots, n)$ is arbitrarily colored red or blue. Let $S$ be a set of some triangles with vertex set $\left\{p_{1}, p_{2}, \cdots, p_{n}\right\}$, and has the property: for any two line segment... | When $n \in\{1,2, \cdots, 5\}$, it is clearly impossible to satisfy the conditions of the problem.
When $n=6$, color $p_{1} 、 p_{2} 、 p_{3}$ red, and $p_{4}$ 、 $p_{5} 、 p_{6}$ blue, and consider the triangles:
$$
\begin{array}{l}
\triangle p_{1} p_{2} p_{3} 、 \triangle p_{1} p_{3} p_{4} 、 \triangle p_{1} p_{4} p_{5} 、... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given real numbers $x, y$ satisfy $x+y=3, \frac{1}{x+y^{2}}+\frac{1}{x^{2}+y}=\frac{1}{2}$.
Find the value of $x^{5}+y^{5}$. [3]
(2017, National Junior High School Mathematics League) | From the given equation, we have
$$
\begin{array}{l}
2 x^{2}+2 y+2 x+2 y^{2}=x^{3}+x y+x^{2} y^{2}+y^{3} \\
\Rightarrow 2(x+y)^{2}-4 x y+2(x+y) \\
\quad=(x+y)\left((x+y)^{2}-3 x y\right)+x y+x^{2} y^{2} .
\end{array}
$$
Substituting $x+y=3$ into the above equation, we get
$$
\begin{array}{l}
(x y)^{2}-4 x y+3=0 \\
\Ri... | 123 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given $a+b+c=5$,
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}=15, a^{3}+b^{3}+c^{3}=47 . \\
\text { Find }\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)
\end{array}
$$
the value. ${ }^{[4]}$
$(2016$, National Junior High School Mathematics League (B Volume)) | From formula (1), we know
$$
\begin{array}{l}
2(a b+b c+c a) \\
=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)=10 \\
\Rightarrow a b+b c+c a=5 .
\end{array}
$$
From formula (2), we know
$$
47-3 a b c=5(15-5) \Rightarrow a b c=-1 \text {. }
$$
And $a^{2}+a b+b^{2}$
$$
\begin{array}{l}
=(a+b)(a+b+c)-(a b+b c+c a) \\
=5(5-... | 625 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $a, b, c$ be distinct positive integers. Then the minimum value of $\frac{a b c}{a+b+c}$ is . $\qquad$ | 11.1.
Assume $a>b>c$. Then $a \geqslant 3, b \geqslant 2, c \geqslant 1$.
Thus, $a b \geqslant 6, b c \geqslant 2, c a \geqslant 3$.
Therefore, $\frac{a+b+c}{a b c}=\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b}$
$\leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$
$\Rightarrow \frac{a b c}{a+b+c} \geqslant 1$.
When $a=3, b=2... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given $a+b+c=1$,
$$
\frac{1}{a+1}+\frac{1}{b+3}+\frac{1}{c+5}=0 \text {. }
$$
Find the value of $(a+1)^{2}+(b+3)^{2}+(c+5)^{2}$. (2017, National Junior High School Mathematics League (Grade 8)) | Let $x=a+1, y=b+3, z=c+5$.
Then the given equations can be transformed into
$$
\begin{array}{l}
x+y+z=10, \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 \Rightarrow x y+y z+z x=0 .
\end{array}
$$
From formula (1), we get
$$
\begin{array}{l}
(a+1)^{2}+(b+3)^{2}+(c+5)^{2} \\
=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x) \\
=1... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the complex number $z$ satisfy
$$
\frac{1017 z-25}{z-2017}=3+4 \text { i. }
$$
Then $|z|=$ $\qquad$ | 2.5.
Let $w=3+4 \mathrm{i}$. Then
$$
\begin{aligned}
z & =\frac{2017 w-25}{w-2017}=\frac{2017 w-\bar{w} w}{w-2017} \\
& =\frac{2017-\bar{w}}{2017-w} w .
\end{aligned}
$$
Therefore, $|z|=|w|=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let planar vectors $\boldsymbol{\alpha}, \boldsymbol{\beta}$ satisfy $|\boldsymbol{\alpha}+2 \boldsymbol{\beta}|=3,|2 \boldsymbol{\alpha}+3 \boldsymbol{\beta}|=4$. Then the minimum value of $\boldsymbol{\alpha} \cdot \boldsymbol{\beta}$ is $\qquad$ . | 5. -170 .
Let $\alpha+2 \beta=u, 2 \alpha+3 \beta=v,|u|=3,|v|=4$.
From $\alpha=2 v-3 u, \beta=2 u-v$, we get
$$
\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=-6|u|^{2}-2|\boldsymbol{v}|^{2}+7 u \cdot v \text {. }
$$
When $\boldsymbol{u} \cdot \boldsymbol{v}=-12$,
$\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=-54-32-84... | -170 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let real numbers $s, t$ satisfy the equations
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, t^{2}+99 t+19=0, \text { and } s t \neq 1 . \\
\text { Then } \frac{s t+4 s+1}{t}=
\end{array}
$$
(1999, "Mathematics Weekly Cup" National Junior High School Mathematics Competition) | Solution: Clearly, $t \neq 0$.
Thus, $19\left(\frac{1}{t}\right)^{2}+99 \cdot \frac{1}{t}+1=0$.
By comparison, $s$ and $\frac{1}{t}$ are the two distinct roots of the equation $19 x^{2}+99 x+1=0$ (since $s t \neq 1$, i.e., $s \neq \frac{1}{t}$). Therefore, by Vieta's formulas, we have
$$
\begin{array}{l}
s+\frac{1}{t}=... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P_{1}, P_{2}, \cdots, P_{100}$ as 100 points on a plane, satisfying that no three points are collinear. For any three of these points, if their indices are in increasing order and they form a clockwise orientation, then the triangle with these three points as vertices is called "clockwise". Question: Is it po... | 4. Suppose $P_{1}, P_{2}, \cdots, P_{100}$ are arranged counterclockwise on a circle. At this point, the number of clockwise triangles is 0.
Now, move these points (not necessarily along the circumference). When a point $P_{i}$ crosses the line $P_{j} P_{k}$ during the movement, the orientation (clockwise or countercl... | 2017 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $AB$ is a line segment of length 8, and point $P$ is at a distance of 3 from the line containing $AB$. Then the minimum value of $AP \cdot PB$ is $\qquad$. | $-, 1.24$
Draw a perpendicular from point $P$ to the line $AB$, with the foot of the perpendicular being $H$. Then
$$
\begin{array}{l}
\frac{1}{2} AB \cdot PH = S_{\triangle PAB} = \frac{1}{2} AP \cdot PB \sin \angle APB \\
\Rightarrow AP \cdot PB = \frac{24}{\sin \angle APB} \geqslant 24 .
\end{array}
$$
When $\angle... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the parabola $y=\sqrt{2} x^{2}$ intersects with the lines $y=1, y=2, y=3$ to form three line segments. The area of the triangle formed by these three line segments is $\qquad$ . | 2. 2 .
The intersection points of the lines $y=1, y=2, y=3$ with the parabola $y=\sqrt{2} x^{2}$ are all symmetric about the $y$-axis, so the lengths of the three line segments are $2 x_{1}, 2 x_{2}, 2 x_{3}$, where $x_{1}, x_{2}, x_{3}$ are the positive roots of the equations $\sqrt{2} x_{1}^{2}=1, \sqrt{2} x_{2}^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. For any $x \in[0,1]$, we have $|a x+b| \leqslant 1$.
Then the maximum value of $|b x+a|$ is $\qquad$ | 3. 2 .
Let $f(x)=a x+b$. Then
$$
\begin{array}{l}
b=f(0), a=f(1)-f(0) . \\
\text { Hence }|b x+a|=|f(0) x+f(1)-f(0)| \\
=|f(0)(x-1)+f(1)| \\
\leqslant|f(0)||x-1|+|f(1)| \\
\leqslant 1+1=2 .
\end{array}
$$
When $a=2, b=-1, x=0$, the maximum value 2 is obtained. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the set
$$
A=\{n|n \in \mathbf{N}, 11| S(n), 11 \mid S(n+1)\},
$$
where $S(m)$ denotes the sum of the digits of the natural number $m$. Then the smallest number in set $A$ is $\qquad$ . | 5.2899999.
Let the smallest number in $A$ be $n=\overline{a_{1} a_{2} \cdots a_{t}}$,
$$
S(n)=a_{1}+a_{2}+\cdots+a_{t} \text {. }
$$
If the unit digit of $n$, $a_{t} \neq 9$, then
$$
\begin{array}{l}
n+1=\overline{a_{1} a_{2} \cdots a_{t-1} a_{t}^{\prime}}\left(a_{t}^{\prime}=a_{t}+1\right) . \\
\text { Hence } S(n+1... | 2899999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange all positive odd numbers in ascending order, take the first number as $a_{1}$, take the sum of the next two numbers as $a_{2}$, then take the sum of the next three numbers as $a_{3}$, and so on, to get the sequence $\left\{a_{n}\right\}$, that is,
$$
a_{1}=1, a_{2}=3+5, a_{3}=7+9+11, \cdots \cdots
$$
Then $... | 4. 44100 .
Notice that, $\sum_{i=1}^{20} a_{i}$ is the sum of the first $\sum_{i=1}^{20} i=210$ odd numbers. Therefore, $210^{2}=44100$. | 44100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. The last digit of $\sum_{k=0}^{201}(10 k+7)^{k+1}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7.6.
It is easy to see that $\sum_{k=0}^{201}(10 k+7)^{k+1} \equiv \sum_{k=0}^{201} 7^{k+1}(\bmod 10)$.
Notice that, for any natural number $n$, the last digits of $7^{4 n+1}$, $7^{4 n+2}$, $7^{4 n+3}$, and $7^{4 n+4}$ are sequentially $7$, $9$, $3$, and $1$, and the last digit of their sum is 0.
$$
\text{Therefore, }... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $M=\{1,2, \cdots, 2017\}$ be the set of the first 2017 positive integers. If one element is removed from the set $M$, and the sum of the remaining elements is exactly a perfect square, then the removed element is $\qquad$ . | 8. 1677 .
Notice,
$$
\begin{array}{l}
1+2+\cdots+2017=\frac{2017 \times 2018}{2} \\
>[\sqrt{2017 \times 1009}]^{2}=1426^{2} \\
=2033476 . \\
\text { Then } \frac{2017 \times 2018}{2}-2033476=1677 .
\end{array}
$$ | 1677 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$, where $x_{1}=1, y_{1}=0, x_{n+1}=$ $x_{n}-y_{n}, y_{n+1}=x_{n}+y_{n}\left(n \in \mathbf{Z}_{+}\right)$. If $a_{n}=$ $\overrightarrow{P_{n} P_{n+1}} \cdot \overrightarrow{P_{n+1} P_{n+2}}$, then t... | 4. 10 .
It is known that $\overrightarrow{O P_{n+1}}$ is obtained by rotating $\overrightarrow{O P_{n}}$ counterclockwise by $\frac{\pi}{4}$ and stretching it to $\sqrt{2}$ times its original length.
Thus, $\left|\overrightarrow{P_{n} P_{n+1}}\right|=O P_{n}$,
$\left|\overrightarrow{P_{n+1} P_{n+2}}\right|=\left|O P_{... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Find the smallest real number $\lambda$, such that there exists a sequence $\left\{a_{n}\right\}$ with all terms greater than 1, for which $\prod_{i=1}^{n+1} a_{i}<a_{n}^{\lambda}$ holds for any positive integer $n$. | Given $a_{n}>1$, so,
$$
\begin{array}{l}
\prod_{i=1}^{n+1} a_{i}0\right), S_{n}=\sum_{i=1}^{n} b_{i}\left(S_{n}>0\right) .
\end{array}
$$
For equation (1) to hold, then $\lambda>0$.
From equation (1) we get
$$
\begin{array}{l}
S_{n+2}S_{n+2}+\lambda S_{n} \geqslant 2 \sqrt{\lambda S_{n+2} S_{n}} \\
\Rightarrow \frac{S... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Given a five-element set $A_{1}, A_{2}, \cdots, A_{10}$, any two of these ten sets have an intersection of at least two elements. Let $A=\bigcup_{i=1}^{10} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets among $A_{1}, A_{2}, \cdots, A_{10}$ that contain the... | Four, it is easy to get $\sum_{i=1}^{n} k_{i}=50$.
The $k_{i}$ sets containing $x_{i}$ form $\mathrm{C}_{k_{i}}^{2}$ set pairs, $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all set pairs, which contain repetitions.
From the fact that the intersection of any two sets among $A_{1}, A_{2}, \cdots, A_{10}$ has at least... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. The sum $\sum_{i=1}^{k} a_{m+i}$ is called the sum of $k$ consecutive terms of the sequence $a_{1}, a_{2}, \cdots, a_{n}$, where $m, k \in \mathbf{N}, k \geqslant 1, m+k \leqslant n$. The number of groups of consecutive terms in the sequence $1,2, \cdots, 100$ whose sum is a multiple of 11 is $\qquad$. | 7.801.
Let $S_{k}=\sum_{i=1}^{k} i=\frac{k(k+1)}{2}$.
Notice,
$$
\begin{array}{l}
S_{k+11}=\frac{(k+11)(k+12)}{2} \\
\equiv \frac{k(k+1)}{2}=S_{k}(\bmod 11),
\end{array}
$$
and the remainders of $S_{1}, S_{2}, \cdots, S_{11}$ modulo 11 are $1,3,6$, $10,4,10,6,3,1,0,0$.
Since $100=9 \times 11+1$, thus, among the rema... | 801 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 For any real number sequence $\left\{x_{n}\right\}$, define the sequence $\left\{y_{n}\right\}:$
$$
y_{1}=x_{1}, y_{n+1}=x_{n+1}-\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the smallest positive number $\lambda$, such that for any real number sequen... | 【Analysis】First estimate the upper bound of $\lambda$ from the limit perspective, then try to construct a recurrence relation to solve it.
First, prove that $\lambda \geqslant 2$.
In fact, start from simple and special cases.
Take $x_{1}=1, x_{n}=\sqrt{2^{n-2}}(n \geqslant 2)$. Then $y_{1}=1, y_{n}=0(n \geqslant 2)$.
S... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $x, y$ satisfy $x+y=1$. Then, the maximum value of $\left(x^{3}+1\right)\left(y^{3}+1\right)$ is | 5.4.
$$
\begin{array}{l}
\text { Given }\left(x^{3}+1\right)\left(y^{3}+1\right) \\
=(x y)^{3}+x^{3}+y^{3}+1 \\
=(x y)^{3}-3 x y+2,
\end{array}
$$
let $t=x y \leqslant\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}$, then
$$
f(t)=t^{3}-3 t+2 \text {. }
$$
Also, by $f^{\prime}(t)=3 t^{2}-3$, we know that $y=f(t)$ is monoto... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $x_{k} 、 y_{k} \geqslant 0(k=1,2,3)$. Calculate:
$$
\begin{array}{l}
\sqrt{\left(2018-y_{1}-y_{2}-y_{3}\right)^{2}+x_{3}^{2}}+\sqrt{y_{3}^{2}+x_{2}^{2}}+ \\
\sqrt{y_{2}^{2}+x_{1}^{2}}+\sqrt{y_{1}^{2}+\left(x_{1}+x_{2}+x_{3}\right)^{2}}
\end{array}
$$
the minimum value is | 6. 2018.
Let $O(0,0), A(0,2018)$,
$$
\begin{array}{l}
P_{1}\left(x_{1}+x_{2}+x_{3}, y_{1}\right), P_{2}\left(x_{2}+x_{3}, y_{1}+y_{2}\right), \\
P_{3}\left(x_{3}, y_{1}+y_{2}+y_{3}\right) .
\end{array}
$$
The required is
$$
\begin{array}{l}
\left|\overrightarrow{A P_{3}}\right|+\left|\overrightarrow{P_{3} P_{2}}\righ... | 2018 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $x, y \in \mathbf{R}$, for any $n \in \mathbf{Z}_{+}$, $n x+\frac{1}{n} y \geqslant 1$. Then the minimum value of $41 x+2 y$ is $\qquad$ | 8.9.
Let the line $l_{n}: n x+\frac{1}{n} y=1$, and call $l_{n} 、 l_{n+1}$ two adjacent lines. Then the intersection point of the two lines is
$$
A_{n}\left(\frac{1}{2 n+1}, \frac{n^{2}+n}{2 n+1}\right) \text {. }
$$
If the intersection point of the line $x+y=1$ and the line $y=0$ is denoted as $A_{0}(1,0)$, then the... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. If $a, b, c$ are distinct integers, then
$$
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a
$$
the minimum value is . $\qquad$ | 8. 6 .
Notice that,
$$
\begin{array}{l}
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a \\
=\frac{1}{2}(a-b)^{2}+\frac{3}{2}(b-c)^{2}+\frac{5}{2}(c-a)^{2} .
\end{array}
$$
Since \(a, b, c\) are distinct integers, when \(a-b=2, a-c=1, c-b=1\), the original expression achieves its minimum value of 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. (15 points) In the sequence $\left\{a_{n}\right\}$,
$$
a_{n}=2^{n} a+b n-80\left(a 、 b \in \mathbf{Z}_{+}\right) \text {. }
$$
It is known that the minimum value of the sum of the first $n$ terms $S_{n}$ is obtained only when $n=6$, and $7 \mid a_{36}$. Find the value of $\sum_{i=1}^{12}\left|a_{i}\right|$. | Three, 13. Notice that, $\left\{a_{n}\right\}$ is an increasing sequence.
From the given, $a_{6}0$, that is,
$$
64 a+6 b-800 \text {. }
$$
Combining $a, b \in \mathbf{Z}_{+}$, we get
$$
a=1, b=1 \text { or } 2 \text {. }
$$
Also, $a_{36}=2^{36}+36 b-80$
$$
\equiv 1+b-3 \equiv 0(\bmod 7) .
$$
Thus, $b=2$.
Therefore, ... | 8010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function $f(x)=\log _{2} \frac{x-3}{x-2}+\cos \pi x$. If $f(\alpha)=10, f(\beta)=-10$, then $\alpha+\beta=$ $\qquad$ | 2. 5 .
It is easy to know that the domain of $f(x)$ is $(-\infty, 2) \cup(3,+\infty)$.
Then $f(5-x)=\log _{2} \frac{5-x-3}{5-x-2}+\cos (5-x) \pi$ $=-f(x)$.
Therefore, $f(x)$ is centrally symmetric about the point $\left(\frac{5}{2}, 0\right)$.
Also, when $x>3$,
$$
f(x)=\log _{2}\left(1-\frac{1}{x-2}\right)+\cos x \pi,
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Given non-zero complex numbers $x, y$ satisfy $y^{2}\left(x^{2}-x y+y^{2}\right)+x^{3}(x-y)=0$.
Find the value of $\sum_{m=0}^{29} \sum_{n=0}^{29} x^{18 m n} y^{-18 m n}$. | 11. Divide both sides of the known equation by $y^{4}$,
$$
\left(\frac{x}{y}\right)^{4}-\left(\frac{x}{y}\right)^{3}+\left(\frac{x}{y}\right)^{2}-\left(\frac{x}{y}\right)+1=0 \text {. }
$$
Let $\frac{x}{y}=\omega$, then
$$
\begin{array}{l}
\omega^{4}-\omega^{3}+\omega^{2}-\omega+1=0 \\
\Rightarrow \omega^{5}=-1 \Right... | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
18. Amelia tosses a coin, with the probability of landing heads up being $\frac{1}{3}$; Brian also tosses a coin, with the probability of landing heads up being $\frac{2}{5}$. Amelia and Brian take turns tossing the coins, and the first one to get heads wins. All coin tosses are independent. Starting with Amelia, the p... | 18. D.
Let $P_{0}$ be the probability of Amelia winning.
Notice,
$P_{0}=P($ Amelia wins in the first round $)+$ $P($ both fail to win in the first round $) \cdot P_{0}$, where, if both fail to win in the first round, it still starts with Amelia, and her probability of winning remains $P_{0}$.
In the first round, the p... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
25. Among the integers between 100 and 999, there are ( ) numbers that have the property: the digits of the number can be rearranged to form a number that is a multiple of 11 and is between 100 and 999 (for example, 121 and 211 both have this property).
(A) 226
(B) 243
( C) 270
(D) 469
(E) 486 | 25. A.
Let a three-digit number be $\overline{A C B}$. Then
11. $\overline{A C B} \Leftrightarrow 11 \mathrm{I}(A+B-C)$
$\Leftrightarrow A+B=C$ or $A+B=C+11$.
We will discuss the following scenarios.
Note that, $A$ and $B$ are of equal status, so we can assume $A \geqslant B$ (the case for $A < B$ is similar).
(1) $A+... | 226 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
In a box, there are 10 red cards and 10 blue cards, each set of cards containing one card labeled with each of the numbers $1, 3, 3^{2}, \cdots, 3^{9}$. The total sum of the numbers on the cards of both colors is denoted as $S$. For a given positive integer $n$, if it is possible to select several cards from the box su... | Let the maximum sum of the labels of two-color cards marked as $1,3,3^{2}, \cdots, 3^{k}$ be denoted as $S_{k}$. Then,
$$
S_{k}=2 \sum_{n=0}^{k} 3^{n}=3^{k+1}-1<3^{k+1} \text {. }
$$
In the sequence $1,3,3^{2}, \cdots, 3^{k}$, the sum of any subset of these numbers is not equal to $3^{m}$. Therefore, the number of way... | 6423 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the sum of the digits of a positive integer $m$ be denoted as $S(m)$, for example, $S(2017)=2+0+1+7=10$. Now, from the 2017 positive integers $1,2, \cdots$, 2017, any $n$ different numbers are taken. It is always possible to find eight different numbers $a_{1}, a_{2}, \cdots, a_{8}$ among these $n$ numbers such ... | 6. A.
Notice that, among $1,2, \cdots, 2017$, the minimum sum of digits is 1, and the maximum sum is 28.
It is easy to see that the numbers with a digit sum of 1 are $1, 10, 100, 1000$; the numbers with a digit sum of $2,3, \cdots, 26$ are no less than eight; the numbers with a digit sum of 27 are only 999, $1899, 19... | 185 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
9. There are four teacups with their mouths facing up. Now, each time three of them are flipped, and the flipped teacups are allowed to be flipped again. After $n$ flips, all the cup mouths are facing down. Then the minimum value of the positive integer $n$ is $\qquad$ . | 9.4 .
Let $x_{i}$ be the number of times the $i$-th cup ($i=1,2,3,4$) is flipped when all cup mouths are facing down, then $x_{i}$ is odd.
From $x_{1}+x_{2}+x_{3}+x_{4}=3 n$, we know that $n$ is even.
It is easy to see that when $n=2$, the condition is not satisfied, hence $n \geqslant 4$.
When $n=4$, use 1 to represe... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $x_{1}=1, x_{2}=2, x_{3}=3$ are zeros of the function
$$
f(x)=x^{4}+a x^{3}+b x^{2}+c x+d
$$
then $f(0)+f(4)=$ $\qquad$ | 2. 24 .
Let $f(x)=(x-1)(x-2)(x-3)(x-k)$. Then $f(0)+f(4)=6k+6(4-k)=24$. | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a>1$. Then the minimum value of $\log _{a} 16+2 \log _{4} a$ is $\qquad$ . | 3.4.
From the operation of logarithms, we get
$$
\begin{array}{l}
\log _{a} 16+2 \log _{4} a=4 \log _{a} 2+\log _{2} a \\
=\frac{4}{\log _{2} a}+\log _{2} a .
\end{array}
$$
Since $a>1$, we have $\log _{2} a>0$.
By the AM-GM inequality, we get
$$
\frac{4}{\log _{2} a}+\log _{2} a \geqslant 2 \sqrt{\frac{4}{\log _{2} ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the sequence $\left\{a_{n}\right\}$ with the first term being 2, and satisfying
$$
6 S_{n}=3 a_{n+1}+4^{n}-1 \text {. }
$$
Then the maximum value of $S_{n}$ is $\qquad$. | 8. 35 .
According to the problem, we have
$$
\left\{\begin{array}{l}
6 S_{n}=3 a_{n+1}+4^{n}-1 \\
6 S_{n-1}=3 a_{n}+4^{n-1}-1
\end{array}\right.
$$
Subtracting the two equations and simplifying, we get
$$
\begin{array}{l}
a_{n+1}=3 a_{n}-4^{n-1} \\
\Rightarrow a_{n+1}+4^{n}=3 a_{n}-4^{n-1}+4^{n} \\
\quad=3\left(a_{n}... | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given that the angle between vector $\boldsymbol{a}$ and $\boldsymbol{b}$ is $120^{\circ}$, and $|a|=2,|b|=5$. Then $(2 a-b) \cdot a=$ $\qquad$ | \begin{array}{l}\text { II.13. 13. } \\ (2 a-b) \cdot a=2|a|^{2}-a \cdot b=13 \text {. }\end{array} | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Given the parabola $y^{2}=a x(a>0)$ and the line $x=1$ enclose a closed figure with an area of $\frac{4}{3}$. Then, the coefficient of the $x^{-18}$ term in the expansion of $\left(x+\frac{a}{x}\right)^{20}$ is | 14. 20 .
According to the problem, we know $2 \int_{0}^{1} \sqrt{a x} \mathrm{~d} x=\frac{4}{3} \Rightarrow a=1$. Therefore, the term containing $x^{-18}$ is $\mathrm{C}_{20}^{19} x\left(\frac{1}{x}\right)^{19}$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $\log _{\sqrt{7}}(5 a-3)=\log _{\sqrt{a^{2}+1}} 5$. Then the real number
$$
a=
$$
. $\qquad$ | 2. 2 .
Simplify the original equation to
$$
\log _{7}(5 a-3)=\log _{a^{2}+1} 5 \text {. }
$$
Since $f(x)=\log _{7}(5 x-3)$ is an increasing function for $x>\frac{3}{5}$, and $g(x)=\log _{5}\left(x^{2}+1\right)$ is also an increasing function, and $f(2)=$ $g(2)=1$, therefore, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $S$ be the set of all rational numbers in the interval $\left(0, \frac{5}{8}\right)$, for the fraction $\frac{q}{p} \in S, (p, q)=1$, define the function $f\left(\frac{q}{p}\right)=\frac{q+1}{p}$. Then the number of roots of $f(x)=\frac{2}{3}$ in the set $S$ is $\qquad$ | 6.5.
Since $f(x)=\frac{2}{3}$, let $q=2 m-1, p=3 m\left(m \in \mathbf{Z}_{+}\right)$.
Then, $0<\frac{2 m-1}{3 m}<\frac{5}{8} \Rightarrow \frac{1}{2}<m<8$. Upon verification, the number of roots of the equation is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If the expansion of $(a+2 b)^{n}$ has three consecutive terms whose binomial coefficients form an arithmetic sequence, then the largest three-digit positive integer $n$ is $\qquad$ | 3. 959 .
Let the binomial coefficients $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1}(1 \leqslant k \leqslant n-1)$ of three consecutive terms in the expansion of $(a+2 b)^{n}$ satisfy
$$
\begin{array}{l}
2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1} . \\
\text { Then } n^{2}-(4 k+1) ... | 959 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In the sequence $\left\{a_{n}\right\}$, for $1 \leqslant n \leqslant 5$, we have $a_{n}=n^{2}$, and for all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n} \text {. }
$$
Then $a_{2023}=$ . $\qquad$ | 3. 17 .
For all positive integers $n$, we have
$$
a_{n+5}+a_{n+1}=a_{n+4}+a_{n}=\cdots=a_{5}+a_{1}=26 \text {. }
$$
Then $a_{n}=26-a_{n+4}=26-\left(26-a_{n+8}\right)=a_{n+8}$, which means $\left\{a_{n}\right\}$ is a sequence with a period of 8.
Therefore, $a_{2023}=a_{7}=26-a_{3}=26-9=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In a certain social event, it was originally planned that each pair of people would shake hands exactly once, but four people each shook hands twice and then left. As a result, there were a total of 60 handshakes during the entire event. Then the number of people who initially participated in the event is $\qquad$ | 5. 15 .
Let the number of people participating in the activity be $n+4$, among which, the number of handshakes among the four people who quit is $x\left(0 \leqslant x \leqslant \mathrm{C}_{4}^{2}=6\right)$.
From the problem, we have $\mathrm{C}_{n}^{2}+4 \times 2=60+x$, which simplifies to $n(n-1)=104+2 x$.
Given $0 \... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Given $z \in \mathbf{C}$. If the equation in terms of $x$
$$
4 x^{2}-8 z x+4 \mathrm{i}+3=0
$$
has real roots. Then the minimum value of $|z|$ is $\qquad$ | 9. 1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be the real root of the given equation. Then
$$
\begin{array}{l}
4 x_{0}^{2}-8(a+b \mathrm{i}) x_{0}+4 \mathrm{i}+3=0 \\
\Rightarrow\left\{\begin{array}{l}
4 x_{0}^{2}-8 a x_{0}+3=0, \\
-8 b x_{0}+4=0 .
\end{array}\right.
\end{array}
$$
Eliminating $x_{0}$ a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=2, a_{n+1}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$.
(1) Prove: $\left\{\frac{1}{S_{n}-1}\right\}$ is an arithmetic seque... | 14. (1) When $n \geqslant 1$, from the condition we get
$$
\begin{array}{l}
S_{n+1}-S_{n}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}} \\
\Rightarrow S_{n+1}-1=\frac{S_{n}-1}{S_{n}} .
\end{array}
$$
Thus, $\frac{1}{S_{n+1}-1}-\frac{1}{S_{n}-1}=\frac{S_{n}}{S_{n}-1}-\frac{1}{S_{n}-1}=1$.
Also, $\frac{1}{S_{1}-1}=\frac{1}{2-... | 3 | Algebra | proof | Yes | Yes | cn_contest | false |
2. If the function
$$
f(x)=\left(x^{2}-1\right)\left(x^{2}+a x+b\right)
$$
satisfies $f(x)=f(4-x)$ for any $x \in \mathbf{R}$, then the minimum value of $f(x)$ is $\qquad$ . | 2. -16 .
Notice that, $f(1)=f(-1)=0$.
Also, $f(x)=f(4-x)$, so $f(3)=f(5)=0$.
Therefore, $f(x)=(x^{2}-1)(x-3)(x-5)$ $=(x^{2}-4x+3)(x^{2}-4x-5)$.
Let $t=x^{2}-4x+4 \geqslant 0$, then $f(x)=(t-1)(t-9)=(t-5)^{2}-16$.
Thus, the minimum value of $f(x)$ is -16. | -16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 6. Let } a_{n}=1+2+\cdots+n\left(n \in \mathbf{Z}_{+}\right) , \\
S_{m}=a_{1}+a_{2}+\cdots+a_{m}(m=1,2, \cdots) \text {. }
\end{array}
$$
Then among $S_{1}, S_{2}, \cdots, S_{2017}$, the numbers that are divisible by 2 but not by 4 are $\qquad$ in number.
$$ | 6. 252 .
Notice that, $S_{m}=\frac{m(m+1)(m+2)}{6}$.
Thus $S_{m} \equiv 2(\bmod 4)$
$$
\begin{array}{l}
\Leftrightarrow m(m+1)(m+2) \equiv 4(\bmod 8) \\
\Leftrightarrow m \equiv 3(\bmod 8) .
\end{array}
$$
Therefore, among $S_{1}, S_{2}, \cdots, S_{2017}$, the numbers that are divisible by 2 but not by 4 are $\left[\... | 252 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{6}+a_{5}+a_{4}-a_{3}-a_{2}-a_{1}=49 \text {. }
$$
Then the minimum value of $a_{9}+a_{8}+a_{7}$ is $\qquad$ | 2. 196.
Let the common ratio be $q$. From the condition, we have
$$
\left(q^{3}-1\right)\left(a_{3}+a_{2}+a_{1}\right)=49.
$$
Clearly, $q^{3}-1>0$.
Then $a_{3}+a_{2}+a_{1}=\frac{49}{q^{3}-1}$.
Thus, $a_{9}+a_{8}+a_{7}=q^{6}\left(a_{3}+a_{2}+a_{1}\right)$
$$
\begin{array}{l}
=\frac{49 q^{6}}{q^{3}-1}=49\left(\sqrt{q^{... | 196 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the function be
$$
f(x)=x^{3}+a x^{2}+b x+c \quad (x \in \mathbf{R}),
$$
where $a, b, c$ are distinct non-zero integers, and
$$
f(a)=a^{3}, f(b)=b^{3} \text {. }
$$
Then $a+b+c=$ $\qquad$ | 3. 18 .
Let $g(x)=f(x)-x^{3}=a x^{2}+b x+c$.
From the problem, we have $g(a)=g(b)=0$.
Thus, $g(x)=a(x-a)(x-b)$
$$
\begin{array}{l}
\Rightarrow b=-a(a+b), c=a^{2} b \\
\Rightarrow b=-\frac{a^{2}}{a+1}=1-a-\frac{1}{a+1} .
\end{array}
$$
Since $b$ is an integer, we have $a+1= \pm 1$.
Also, $a \neq 0$, so $a=-2, b=4, c=1... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the function be
$$
f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}(x \in[0,+\infty)) \text {. }
$$
Then the number of integer points on the graph of the function is $\qquad$ | 4.3.
It is known that the function $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and
$$
f(0)=3, f(1)=2, f(3)=1 .
$$
When $x>3$, we have
$$
0<f(x)<f(3)=1 \text {. }
$$
Therefore, the number of integer points on the graph of the function $y=f(x)(x \in[0,+\infty))$ is 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that a line passing through the focus $F$ of the parabola $y^{2}=4 x$ intersects the parabola at points $M$ and $N$, and $E(m, 0)$ is a point on the $x$-axis. The extensions of $M E$ and $N E$ intersect the parabola at points $P$ and $Q$ respectively. If the slopes $k_{1}$ and $k_{2}$ of $M N$ and $P Q$ satisf... | 9.3.
When $M P$ is not perpendicular to the $x$-axis, let $l_{\text {MР }}: y=k(x-m)$.
Substitute into $y^{2}=4 x$ to get
$$
x^{2}-\left(\frac{4}{k^{2}}+2 m\right) x+m^{2}=0 \text {. }
$$
Then $x_{M} x_{P}=m^{2} \Rightarrow y_{M} y_{P}=-4 m$ $\Rightarrow y_{P}=\frac{-4 m}{y_{M}}$.
When $M P \perp x$-axis, the conclus... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Arrange all positive integers that are coprime with 70 in ascending order. The 2017th term of this sequence is $\qquad$ . | 10. 5881.
It is easy to know that the number of positive integers not exceeding 70 and coprime with 70 is
$$
35-(7+5)+1=24 \text{.}
$$
Let the sequence of all positive integers coprime with 70, arranged in ascending order, be $\left\{a_{n}\right\}$. Then
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, a_{3}=9, \cdots, a_{24}=6... | 5881 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, and $c$ are three non-zero real numbers, and $x^{2}-1$ is a factor of the polynomial $x^{3}+a x^{2}+b x+c$. Then the value of $\frac{a b+3 a}{c}$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2 | - 1. A.
From the fact that $\pm 1$ are roots of the given polynomial, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 + a + b + c = 0 , } \\
{ - 1 + a - b + c = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=-c, \\
b=-1
\end{array}\right.\right. \\
\Rightarrow \frac{a b+3 a}{c}=-2 .
\end{array}
$$ | -2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. In two regular tetrahedrons $A-OBC$ and $D-OBC$ with their bases coinciding, $M$ and $N$ are the centroids of $\triangle ADC$ and $\triangle BDC$ respectively. Let $\overrightarrow{OA}=\boldsymbol{a}, \overrightarrow{OB}=\boldsymbol{b}, \overrightarrow{OC}=\boldsymbol{c}$. If point $P$ satisfies $\overrightarrow{OP}... | Take $O$ as the origin and the line $O B$ as the $x$-axis, establishing a spatial rectangular coordinate system as shown in Figure 13.
Let $B(1,0,0)$. Then
$$
\begin{array}{l}
C\left(\frac{1}{2}, \frac{\sqrt{3}}{2}, 0\right), \\
A\left(\frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3}\right), \\
D\left(\frac{1}{2}, \... | 439 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Let $a, b$ be real numbers, and the equation with respect to $x$
$$
\frac{x}{x-1}+\frac{x-1}{x}=\frac{a+b x}{x^{2}-x}
$$
has no real roots. Find the value of the expression $8 a+4 b+|8 a+4 b-5|$. | One, the original equation can be transformed into
$$
2 x^{2}-(b+2) x+(1-a)=0 \text {. }
$$
Thus, $\Delta=(b+2)^{2}+8 a-8$.
(1) When $\Delta>0$, equation (1) has two distinct real roots.
Since the original equation has no solution, it follows that the two distinct real roots of equation (1) are 0 and 1, i.e.,
$$
\beg... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n-1}}$. Then $\sum_{n=1}^{2018} a_{n}=$ $\qquad$ . | 5. 5857 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$; when $n \equiv 2,4,8(\bmod 10)$, $(n+1)^{n+2} \equiv 1(\bmod 4)$, then $a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10)$; when $n \equiv 3,7,9(\bmod 10)$, $(n+1)^{n+2} \equiv 0(\bmod 4)$, then $a_{n} \equiv n^{(n+... | 5857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A five-digit number $\overline{a b c d e}$ satisfies:
$$
ac>d, dd, b>e \text {. }
$$
For example, 34 201, 49 412. If the digits of the number change in a pattern similar to the monotonicity of a sine function over one period, then the five-digit number is said to follow the "sine rule." The number of five-digit num... | 6. 2892.
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers, respectively, and $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$, at this point, there are $10-k$ ways to choose $(b, d)$, and $a$, $c$, and $e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ combinations. There... | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
a_{1}=3, b_{1}=1
$$
and for any $n \in \mathbf{Z}_{+}$, we have
$$
\left\{\begin{array}{l}
a_{n+1}=a_{n}+b_{n}+\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}} \\
b_{n+1}=a_{n}+b_{n}-\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}}
\end{array}... | 11. (1) From the problem, we have
$$
\begin{array}{l}
a_{n+1}+b_{n+1}=2\left(a_{n}+b_{n}\right), \\
a_{n+1} b_{n+1}=3 a_{n} b_{n} . \\
\text { Also, } a_{1}+b_{1}=4, a_{1} b_{1}=3 \text {, then } \\
a_{n}+b_{n}=\left(a_{1}+b_{1}\right) 2^{n-1}=2^{n+1}, \\
a_{n} b_{n}=a_{1} b_{1} 3^{n-1}=3^{n} .
\end{array}
$$
Notice t... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Let $S$ be a set of positive integers with the property: for any $x \in S$, the arithmetic mean of the remaining numbers in $S$ after removing $x$ is a positive integer, and it satisfies $1 \in S$, 2016 is the largest element in $S$. Find the maximum value of $|S|$. | Let the elements of set $S$ be
$$
1=x_{1}<x_{2}<\cdots<x_{n}=2016 \text {. }
$$
Then for $1 \leqslant j \leqslant n$, we have
$$
y_{j}=\frac{\sum_{i=1}^{n} x_{i}-x_{j}}{n-1} \in \mathbf{Z}_{+} \text {. }
$$
Thus, for $1 \leqslant i<j \leqslant n$, we have
$$
\begin{array}{l}
y_{i}-y_{j}=\frac{x_{j}-x_{i}}{n-1} \in \m... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. In 1993, American mathematician F. Smarandache proposed many number theory problems, attracting the attention of scholars both at home and abroad. One of these is the famous Smarandache function. The Smarandache function of a positive integer \( n \) is defined as
\[
S(n)=\min \left\{m \left| m \in \mathbf{Z}_{+}, ... | 14. (1) It is easy to know that $16=2^{4}$. Then $S(16)=6$.
From $2016=2^{5} \times 3^{2} \times 7$, we know
$S(2016)=\max \left\{S\left(2^{5}\right), S\left(3^{2}\right), S(7)\right\}$.
Also, $S(7)=7, S\left(3^{2}\right)=6, S\left(2^{5}\right)=8$, so $S(2016)=8$.
(2) From $S(n)=7$, we know $n \mid 7!$.
Thus, $n \leq... | 5040 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure 1, points $A$ and $A^{\prime}$ are on the $x$-axis and are symmetric with respect to the $y$-axis. A line passing through point $A^{\prime}$ and perpendicular to the $x$-axis intersects the parabola $y^{2}=2 x$ at points $B$ and $C$. Point $D$ is a moving point on segment $A B$, and point $E$ is ... | 15. (1) Let $A\left(-2 a^{2}, 0\right), A^{\prime}\left(2 a^{2}, 0\right)$. Then $B\left(2 a^{2}, 2 a\right), C\left(2 a^{2},-2 a\right)$.
Let $D\left(x_{1}, y_{1}\right), \overrightarrow{A D}=\lambda \overrightarrow{A B}$. Then $\overrightarrow{C E}=\lambda \overrightarrow{C A}$.
Thus, $\left(x_{1}+2 a^{2}, y_{1}\righ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In a chess tournament, $n$ players participate in a round-robin competition. After players A and B each played two games, they withdrew from the competition due to certain reasons. It is known that a total of 81 games were ultimately played. Then $n=$ $\qquad$ | 4. 15 .
If there is no match between A and B, then
$$
\mathrm{C}_{n-2}^{2}+2 \times 2=81 \text{, }
$$
this equation has no positive integer solution.
If A and B have a match, then
$$
\mathrm{C}_{n-2}^{2}+3=81 \Rightarrow n=15 .
$$ | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $G$ be a simple graph of order 100. It is known that for any vertex $u$, there exists another vertex $v$ such that $u$ and $v$ are adjacent, and there is no vertex adjacent to both $u$ and $v$. Find the maximum possible number of edges in graph $G$. (Provided by Yunhao Fu) | 2. Let $G=(V, E)$.
For $u v \in E$, if there is no other vertex adjacent to both $u$ and $v$, then $u v$ is called a "good edge".
Let $E_{0}$ be the set of all good edges, and $G_{0}=\left(V, E_{0}\right)$.
From the problem statement, we know that each vertex in graph $G$ has at least one good edge, i.e., $G_{0}$ has ... | 3822 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $A_{1}, A_{2}, \cdots, A_{n}$ be binary subsets of the set $\{1,2, \cdots, 2018\}$, such that the sets $A_{i}+A_{j}(1 \leqslant i \leqslant j \leqslant n)$ are all distinct, where,
$$
A+B=\{a+b \mid a \in A, b \in B\} \text {. }
$$
Find the maximum possible value of $n$.
(Qiu Zhenhua) | 4. The maximum value of $n$ is 4033.
If there exists $A_{i}=A_{j}(i \neq j)$, then $A_{i}+A_{i}=A_{j}+A_{j}$, which contradicts the condition.
Thus, $A_{1}, A_{2}, \cdots, A_{n}$ are all distinct.
For a binary set $A=\{a, b\}(a<b)$, let $S_{i}=\{x-y \mid x, y \in M_{i}, x>y\}$.
If there exists $i \neq j$ such that $S_... | 4033 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If
$$
\begin{array}{l}
a=\frac{\sqrt{6+\sqrt{6+\cdots+\sqrt{6}}},}{2016 \text { nested radicals }}, \\
b=\frac{\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{6}}},}{2 \text { 2017 nested radicals }},
\end{array}
$$
then $[a+b]=$ . $\qquad$ | 3. 4 .
Notice that,
$2.4<\sqrt{6}<a<\frac{\sqrt{6+\sqrt{6+\cdots+\sqrt{9}}}}{2016 \text{ levels }}=3$,
1. $8<\sqrt[3]{6}<b<\frac{\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{8}}}}{2017 \text{ layers }}=2$.
Therefore, $[a+b]=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If the cube of a three-digit positive integer is an eight-digit number of the form $\overline{A B C D C D A B}$, then such a three-digit number is | 5. 303 | 303 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Place a regular tetrahedron with a volume of 1 inside a cube, then the minimum volume of this cube is $\qquad$ | 3. 3 .
Considering in reverse, for a cube with edge length $a$ (volume $a^{3}$), its largest inscribed regular tetrahedron has vertices formed by the cube's vertices that do not share an edge, and its volume is $\frac{a^{3}}{3}$.
$$
\text { Let } \frac{a^{3}}{3}=1 \text {, then } a^{3}=3 \text {. }
$$ | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange all positive integers whose sum of digits is 10 in ascending order to form the sequence $\left\{a_{n}\right\}$. If $a_{n}=2017$, then $n=$ . $\qquad$ | 8. 120.
It is easy to know that there are 9 two-digit numbers $\overline{a b}$ whose digits sum to 10.
For three-digit numbers $\overline{a b c}$ whose digits sum to 10, the first digit $a$ can take any value in $\{1,2, \cdots, 9\}$. Once $a$ is determined, $b$ can take any value in $\{0,1, \cdots, 10-a\}$, which give... | 120 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the set $M=\{1,99,-1,0,25,-36, -91,19,-2,11\}$, let the non-empty subsets of $M$ be $M_{i}(i=1,2, \cdots, 1023)$. If the product of all elements in each $M_{i}$ is $m_{i}$, then $\sum_{i=1}^{1023} m_{i}=$ $\qquad$ . | 2. -1 .
Let the set $M=\left\{a_{i} \mid i=1,2, \cdots, n\right\}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{2^{n}-1} m_{i}=\prod_{i=1}^{n}\left(a_{i}+1\right)-1 . \\
\text { Given }-1 \in M, \text { we know } \sum_{i=1}^{1023} m_{i}=-1 .
\end{array}
$$ | -1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the three-digit number $n=\overline{a b c}$, where the lengths $a, b, c$ can form an isosceles (including equilateral) triangle. Then the number of such three-digit numbers $n$ is. $\qquad$ | 4. 165.
When $a=b=c$, there are 9 cases in total.
When $a=b \neq c, a=c \neq b, b=c \neq a$, the number of cases for each situation is the same. Taking $a=b \neq c$ as an example:
(1) When $a=b \geqslant 5$, $a$ and $b$ have 5 cases, and $c$ has 8 cases, making a total of 40 cases;
(2) When $a=b=4$, $c$ can be any one... | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $p(x)=a x^{3}+b x^{2}+c x+d$ is a cubic polynomial, satisfying
$$
p\left(\frac{1}{2}\right)+p\left(-\frac{1}{2}\right)=1000 p(0) \text {. }
$$
Let $x_{1} 、 x_{2} 、 x_{3}$ be the three roots of $p(x)=0$. Then the value of $\frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}$ is $\qquad$ . | 6. 1996.
From equation (1) we get
$$
\frac{1}{2} b+2 d=1000 d \Rightarrow \frac{b}{d}=1996 \text {. }
$$
By Vieta's formulas we get
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} x_{2} x_{3}=-\frac{d}{a} . \\
\text { Therefore } \frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}=\frac{x_{1}+... | 1996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. If $P(x, y)$ is a point on the hyperbola $\frac{x^{2}}{8}-\frac{y^{2}}{4}=1$, then the minimum value of $|x-y|$ is . $\qquad$ | 9. 2 .
From the condition, we know that $x^{2}-2 y^{2}-8=0$.
By symmetry, without loss of generality, assume
$$
\begin{array}{l}
x>0, y>0, u=x-y>0 . \\
\text { Then }(y+u)^{2}-2 y^{2}-8=0 \\
\Rightarrow y^{2}-2 u y-u^{2}+8=0 \\
\Rightarrow \Delta=(2 u)^{2}-4\left(-u^{2}+8\right) \geqslant 0 \\
\Rightarrow u \geqslant ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given real numbers $x_{1}, x_{2}, x_{3}$ satisfy
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{1} x_{2}+x_{2} x_{3}=2 \text {. }
$$
Then the maximum value of $\left|x_{2}\right|$ is $\qquad$ | 11. 2 .
From the condition, we have
$$
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+x_{3}^{2}=4 \text {. }
$$
Notice that,
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2} \geqslant \frac{x_{2}^{2}}{2}, \\
x_{3}^{2}+\left(x_{2}+x_{3}\right)^{2} \geqslant \frac{x_{2}^{2}}{2} .
\end{arr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Let the set $M=\{1,2, \cdots, 10\}$,
$$
\begin{aligned}
A= & \{(x, y, z) \mid x, y, z \in M, \text { and } \\
& \left.9 \mid\left(x^{3}+y^{3}+z^{3}\right)\right\} .
\end{aligned}
$$
Then the number of elements in set $A$ is $\qquad$ . | 12.243.
Notice that,
when $x \equiv 1(\bmod 3)$, $x^{3} \equiv 1(\bmod 9)$;
when $x \equiv 2(\bmod 3)$, $x^{3} \equiv-1(\bmod 9)$;
when $x \equiv 0(\bmod 3)$, $x^{3} \equiv 0(\bmod 9)$.
Thus, for $x \in \mathbf{Z}$, we have
$$
x^{3} \equiv 0,1,-1(\bmod 9) \text {. }
$$
To make $9 \mid\left(x^{3}+y^{3}+z^{3}\right)$, ... | 243 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Set $a_{n}=\sum_{k=1}^{n}\left[\frac{n}{k}\right]$. Then the number of even numbers in $a_{1}, a_{2}, \cdots, a_{2018}$ is . $\qquad$ | 8. 1028.
It is easy to see that if and only if $k \mid n$,
$$
\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=1 \text{. }
$$
Otherwise, $\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]=0$.
Thus, $a_{n}-a_{n-1}$
$$
\begin{array}{l}
=\sum_{k=1}^{n}\left[\frac{n}{k}\right]-\sum_{k=1}^{n-1}\left[\frac{n-1}{k}\rig... | 1028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given an increasing sequence composed of powers of 3 or the sum of several different powers of 3: $1,3,4,9,10,12,13, \cdots$. Then the 100th term of this sequence is $\qquad$ | 6. 981.
The terms of the sequence are given by $\sum_{i=0}^{n} 3^{i} a_{i}$, where,
$$
a_{i} \in\{0,1\}(i=1,2, \cdots, n) \text {. }
$$
When $n=5$, there are $2^{6}-1=63$ numbers that can be formed, and the 64th term is $3^{6}=729$.
Starting from the 65th term, there are $2^{5}-1=31$ terms that do not contain $3^{5}... | 981 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. In $\triangle A B C$, the side lengths opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$, respectively, and
$$
\begin{array}{l}
\sin C \cdot \cos \frac{A}{2}=(2-\cos C) \sin \frac{A}{2}, \\
\cos A=\frac{3}{5}, a=4 .
\end{array}
$$
Then the area of $\triangle A B C$ is . $\qquad$ | 7.6 .
From equation (1) we know
$$
\begin{array}{l}
2 \sin \frac{A}{2}=\sin \left(C+\frac{A}{2}\right) \\
\Rightarrow 2 \sin A=2 \sin \left(C+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Thus, $c+b=2a$.
Also, $a^{2}=b^{2}+c^{2}-2 b c \cos A$, which means
$$
4^{2}=b^{2}+(8-b)^{2}-2 b(8-... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. If the equation with respect to $x$
$$
x^{2}+a x+b-3=0(a, b \in \mathbf{R})
$$
has real roots in the interval $[1,2]$, then the minimum value of $a^{2}+(b-4)^{2}$ is . $\qquad$ | 8. 2 .
From the problem, we know that $b=-x^{2}-a x+3$.
Then $a^{2}+(b-4)^{2}$
\[
\begin{array}{l}
=a^{2}+\left(-x^{2}-a x-1\right)^{2} \\
=x^{2}(x+a)^{2}+2\left(x^{2}+a x\right)+a^{2}+1 \\
=\left(x^{2}+1\right)(x+a)^{2}+x^{2}+1 .
\end{array}
\]
Also, $x \in[1,2]$, so,
\[
a^{2}+(b-4)^{2} \geqslant x^{2}+1 \geqslant 2... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.