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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) Find the number of $n$-ary Boolean functions;
(2) Let $g$ be a 10-ary Boolean function, satisfying
$$
g\left(x_{1}, x_{2}, \cdots, x_{10}\right) \equiv 1+\sum_{i=1}^{10} \prod_{j=1}^{i} x_{j}(\bmod 2),
$$
Find the number of elements in the set $D_{10}(g)$, and find
$$
\sum_{\left(x_{1}, x_{2}, \cdots, x_{10}\right) \in D_{10}(g)}\left(x_{1}+x_{2}+\cdots+x_{10}\right) \text {. }
$$ | 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, “ * ” represents a value that can be 0 or 1. Then
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,0, *, *, *, *, *, *, *, *)
\end{array}
$$
$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{8}=256$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,0, *, *, *, *, *, *)
\end{array}
$$
$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{6}=64$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,1,1,0, *, *, *, *)
\end{array}
$$
$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{4}=16$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,1,1,1,1,0, *, *)
\end{array}
$$
$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there are $2^{2}=4$ such cases;
$$
\begin{array}{l}
\text { when }\left(x_{1}, x_{2}, \cdots, x_{10}\right) \\
=(1,1,1,1,1,1,1,1,1,0)
\end{array}
$$
$g\left(x_{1}, x_{2}, \cdots, x_{10}\right)=0$, there is 1 such case.
Thus, the number of elements in the set $D_{10}(\mathrm{~g})$ is
$$
\begin{array}{l}
256+64+16+4+1=341 . \\
\text { Therefore, } \sum_{\left(x_{1}, x_{2}, \cdots, x_{10}\right) \in D_{10}(g)}\left(x_{1}+x_{2}+\cdots+x_{10}\right) \\
=1 \times 256+128 \times 8+3 \times 64+32 \times 6+ \\
\quad 5 \times 16+8 \times 4+7 \times 4+2 \times 2+9 \\
=1817 .
\end{array}
$$ | 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
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {. }
$$
(1) Find the number of $n$-ary Boolean functions;
(2) Let $g$ be an $n$-ary Boolean function, satisfying
$$
g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j}(\bmod 2),
$$
Find the number of elements in the set $D_{n}(g)$, and find the largest positive integer $n$, such that
$$
\sum_{\left(x_{1}, x_{2}, \cdots, \cdots, x_{n}\right) \in D_{n}(g)}\left(x_{1}+x_{2}+\cdots+x_{n}\right) \leqslant 2017 .
$$ | 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, $|D_{1}(g)|=1, |D_{2}(g)|=1$.
$$
\begin{array}{l}
\text { Also, } g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j} \\
=\left(1+x_{1}\left(1+\sum_{i=2}^{n} \prod_{j=2}^{i} x_{j}\right)\right)(\bmod 2),
\end{array}
$$
$$
\begin{aligned}
& \text { Then }|D_{n}(g)|=2^{n-1}-|D_{n-1}(g)|(n=3,4, \cdots) . \\
& \text { Hence }|D_{n}(g)|=2^{n-1}-2^{n-2}+|D_{n-2}(g)|=\cdots \\
& =\sum_{k=0}^{n-1}(-1)^{k} \cdot 2^{n-1-k} \\
= & \frac{2^{n}+(-1)^{n+1}}{3} .
\end{aligned}
$$
Let $c_{n}=\sum_{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in D_{n}(g)}\left(x_{1}+x_{2}+\cdots+x_{n}\right)$.
Notice that,
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 0\right) \equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2) \\
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}+x_{1} x_{2} \cdots x_{n-1}(\bmod 2)
\end{array}
$$
If $x_{1} x_{2} \cdots x_{n-1}=1$, then
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \equiv n+1(\bmod 2) \\
=\left\{\begin{array}{l}
0, n \text { is odd; } \\
1, n \text { is even. }
\end{array}\right.
\end{array}
$$
If $x_{1} x_{2} \cdots x_{n-1}=0$, then
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2)=0 \\
\Leftrightarrow g\left(x_{1}, x_{2}, \cdots, x_{n-1}\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2)=0
\end{array}
$$
Therefore, when $n \geqslant 2$,
$$
c_{n}=\left\{\begin{array}{l}
n+2 c_{n-1}+|D_{n-1}(g)|, n \text { is odd; } \\
2 c_{n-1}+|D_{n-1}(g)|-n, n \text { is even. }
\end{array}\right.
$$
Obviously, $c_{1}=1, c_{2}=1$.
When $n=2 m$,
$$
\begin{array}{l}
c_{2 m}=2 c_{2 m-1}+\frac{2^{2 m-1}+(-1)^{2 m}}{3}-2 m \\
=-\frac{1}{3}+\frac{4}{3} \times 2^{2 m-2}+(2 m-2)+4 c_{2 m-2} \\
=\frac{(3 m+1) 4^{m}-(6 m+1)}{9}
\end{array}
$$
Similarly, when $n=2 m+1$,
$$
\begin{array}{l}
c_{2 m+1}=2 m+1+2 c_{2 m}+\frac{2^{2 m}+(-1)^{2 m+1}}{3} \\
=\frac{1}{3}+\frac{4}{3} \times 2^{2 m-1}-(2 m-1)+4 c_{2 m-1} \\
=\frac{(6 m+5) 4^{m}+6 m+4}{9} .
\end{array}
$$
Therefore, from equations (2), (3), and
$$
\begin{array}{l}
c_{9}=828<3986=c_{11}, \\
c_{10}=1817<8643=c_{12},
\end{array}
$$
we know that the largest positive integer $n$ for which equation (1) holds is $n=10$. | 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 $P$ containing three elements are $P_{1}, P_{2}, \cdots, P_{k}$. Then $\sum_{i=1}^{k}\left|P_{i}\right|=$ $\qquad$ | 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 Mathematics League Competition) | 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 edges can be connected between any other two vertices. Hence, the subgraph $G^{\prime}$ has exactly four edges;
If there exists a vertex $A$ in $G^{\prime}$ with a degree of 3, then the three vertices connected to $A$ cannot be connected to each other, and the remaining vertex can have at most one edge. Hence, the subgraph $G^{\prime}$ has at most four edges;
If the degree of each vertex $A$ in $G^{\prime}$ is no more than 2, then the subgraph $G^{\prime}$ can have at most $\frac{5 \times 2}{2}=5$ edges.
The lemma is proved.
For the graph $G(V, E)$, let $A$ be one of the vertices in $G$ with the maximum number of edges, and let $d(A)=s$. Let the set of vertices connected to $A$ be $S$, and $|S|=s$. Then no edges can be connected between the vertices in set $S$. Let the set of vertices other than $A$ and the vertices in set $S$ be $T$.
If $s \geqslant 4$, then $|T|=9-s \leqslant 5$.
By the lemma, the vertices in set $T$ can have at most five edges.
Since each vertex in set $T$ can be connected to at most one vertex in set $S$, we have
$|E| \leqslant s+5+(9-s)=14$.
If $s \leqslant 3$, then $|E| \leqslant \frac{3 \times 10}{2}=15$.
Therefore, in either case, the graph $G(V, E)$ can have at most 15 edges.
The graph shown in Figure 1 has 15 edges and meets the requirements.
In conclusion, the maximum number of edges is 15. | 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 $C_{1}, C_{2}, \cdots, C_{2015}$, with the $x$-coordinates of their centers being in increasing order.
First, we prove a lemma.
Lemma For any integer $i>1$, among the circles $C_{1}, C_{2}, \cdots, C_{i-1}$, at most 75 circles have a common point with circle $C_{i}$.
Proof According to the positive direction of the $x$-axis, the circles $C_{1}, C_{2}, \cdots, C_{i-1}$ are to the left of circle $C_{i}$. Divide the left half of circle $\widetilde{C_{i}}$ into three equal sectors.
By assumption, each sector contains at most 25 of the centers of the circles $C_{1}, C_{2}, \cdots, C_{i-1}$. Therefore, the left half of circle $\widetilde{C_{i}}$ contains at most 75 of the centers of the circles $C_{1}, C_{2}, \cdots, C_{i-1}$. Thus, among the circles $C_{1}, C_{2}, \cdots, C_{i-1}$, at most 75 circles have a common point with circle $C_{i}$.
The lemma is proved.
Next, we place these 2015 circles into 76 sets $A_{1}, A_{2}, \cdots, A_{76}$.
Place the circles $C_{1}, C_{2}, \cdots, C_{76}$ into sets $A_{1}, A_{2}, \cdots, A_{76}$, respectively. For any circle $C_{i} (i=77,78, \cdots, 2015)$, among the circles $C_{1}, C_{2}, \cdots, C_{i-1}$, at most 75 circles have a common point with circle $C_{i}$. Therefore, when placing circle $C_{i}$, there must exist a set $A_{j} (j \in \{1,2, \cdots, 76\})$ such that $A_{j}$ does not contain any circle that has a common point with circle $C_{i}$. Place circle $C_{i}$ into set $A_{j}$.
Place these 2015 circles into the corresponding sets, ensuring that any two circles in the same set do not have a common point.
Let $[x]$ denote the greatest integer not exceeding the real number $x$.
By the pigeonhole principle, there exists a set containing at least
$\left[\frac{2015}{76}\right]+1=27$ circles. Select 27 circles from this set to form set $S$ that satisfies the condition. | 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 $T$. A set $S \subset A$ of some finite sequences satisfies: for any infinite sequence $M$ in set $A$, there exists a unique sequence $T$ in $S$ such that $M$ starts with $T$. Which of the following three conclusions is true:
(1) $S$ must be a finite set;
(2) $S$ must be an infinite set;
(3) $S$ can be either a finite set or an infinite set? Prove your conclusion.
(2017, Peking University Middle School Mathematics Science Summer Camp) | 【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 $T_{2}$ starts with $T_{1}$. Otherwise, consider an infinite sequence $M$ in set $A$ that starts with $T_{2}$.
Then $M$ can start with both $T_{2}$ in $S$ and $T_{1}$ in $S$, which contradicts the uniqueness in the problem.
Next, we prove that $S$ cannot be an infinite set, i.e., (3) does not hold.
If $S$ is an infinite set, then by the pigeonhole principle, there must exist some $b_{1} \in\{1,2, \cdots, 2017\}$ such that there are infinitely many sequences in $S$ that start with the sequence $\left\{b_{1}\right\}$, and by the previous analysis, we know that $\left\{b_{1}\right\} \notin S$; similarly, by the pigeonhole principle, there must exist some $b_{2} \in\{1,2, \cdots, 2017\}$ such that there are infinitely many sequences in $S$ that start with the sequence $\left\{b_{1}, b_{2}\right\}$, and $\left\{b_{1}, b_{2}\right\} \notin S$; and so on, for any given positive integer $n$, there exists a sequence $\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ such that there are infinitely many sequences in $S$ that start with $\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ and the sequence $\left\{b_{1}, b_{2}, \cdots, b_{n}\right\} \notin S$.
Thus, we can obtain an infinite sequence
$$
M^{\prime}=\left\{b_{1}, b_{2}, \cdots, b_{n}, \cdots\right\} \text {, }
$$
satisfying that for any positive integer $n$, the sequence
$$
\left\{b_{1}, b_{2}, \cdots, b_{n}\right\} \notin S \text {. }
$$
Therefore, this infinite sequence $M^{\prime}$ cannot start with any finite sequence in set $S$, leading to a contradiction.
In summary, only conclusion (1) holds. | 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 Olympiad) | 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 number of elements in the set $S$ sought is 8. | 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 English, please keep the original text's line breaks and format, and output the translation result directly. | 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 $55 < 66$, there must be a column without horizontal separating edges or a row without vertical separating edges.
Assume there is a column without horizontal separating edges, and all cells in this column are $A$-colored.
(1) If the number of $A$-colored separating edges is no more than 32, then there must be a row without $A$-colored separating edges (otherwise, the number of $A$-colored separating edges would be at least 33, leading to a contradiction).
Thus, this row is a monochromatic row and is $A$-colored.
Therefore, the number of horizontal or vertical $A$-colored separating edges is no more than 16.
Assume the number of $A$-colored horizontal separating edges is no more than 16, then there must be 17 columns all $A$-colored, which contradicts the fact that the number of cells of each color is equal.
(2) If the number of $A$-colored separating edges is at least 33, then
(i) There exists a column entirely $B$-colored and a column entirely $C$-colored. Between these three columns, there are at least 33 vertical separating edges in each row (at least one in each row), hence there are at least 66 separating edges, leading to a contradiction.
(ii) There exists a column entirely $B$-colored, but no column entirely $C$-colored.
Since there is no row or column entirely $C$-colored, the number of $C$-colored separating edges
$\geqslant$ the length of the projection of all $C$-colored cells in the horizontal and vertical directions
$\geqslant 39$.
Otherwise,
the number of $C$-colored cells
$\leqslant$ the product of the lengths of the projections of all $C$-colored cells in the horizontal and vertical directions
$$
\leqslant 19 \times 19 < 363,
$$
leading to a contradiction.
If a $C$-colored cell is between a column entirely $A$-colored and a column entirely $B$-colored, then the horizontal projection of this $C$-colored cell is only one line segment. In this row, at least 2 $C$-colored separating edges are generated, so the contribution of this row to the difference between the left and right sides in conclusion (1) is 1. And between the $A$-colored column and the $B$-colored column, there is one vertical separating edge in each row. Suppose there are $k$ rows where there is a $C$-colored cell between the $A$-colored column and the $B$-colored column, then,
$$
\text{Number of } C \text{-colored cells} \geqslant 39 - k + 33 + k = 72,
$$
leading to a contradiction.
(iii) There is no column entirely $B$-colored, and no column entirely $C$-colored.
Similarly to (ii), the number of $B$-colored separating edges and $C$-colored separating edges is at least 39 each, and each separating edge is at least of two colors, hence
the total number of separating edges $\geqslant \frac{33 + 39 + 39}{2} = 55.5$, leading to a contradiction.
Construct an example with 56 separating edges, as shown in Figure 5.
Therefore, the minimum value is 56. | 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 $\prod_{i=1}^{n} x_{i} \leqslant 99$.
$$
\begin{array}{l}
\text { and } \prod_{i=1}^{n} x_{i}=\prod_{i=1}^{n}\left(\left(x_{i}-1\right)+1\right) \\
\geqslant \sum_{i=1}^{n}\left(x_{i}-1\right)+1=\sum_{i=1}^{n} x_{i}-n+1, \\
\text { then } \sum_{i=1}^{n} x_{i} \leqslant \prod_{i=1}^{n} x_{i}+n-1 \leqslant n+98 \\
\Rightarrow 99(n+98) \geqslant 100 n \\
\Rightarrow n \leqslant 99 \times 98=9702 .
\end{array}
$$
Taking $x_{1}=99, x_{2}=x_{3}=\cdots=x_{9702}=1$, can make the equality in (1) hold. | 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 these cards over in their original positions, leaving only one card face-up, and then the magician enters the room, observes the cards on the table, and selects a face-down card, guessing its color. Find the minimum value of $n$, such that the magician and his assistant can complete this trick according to some predetermined strategy. | 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 trick can be completed.
For convenience, use $(a, i)$ to denote that the $a$-th card is of the $i$-th color, where $1 \leqslant a \leqslant n, 1 \leqslant i \leqslant 2017$.
For $n$ cards, note that the number of cards of each color is 1000000 (greater than $n$), so there are $2017^{n}$ different states (each position has 2017 possible colors). Each state is equivalent to a set of the form
$$
\left\{\left(1, i_{1}\right),\left(2, i_{2}\right), \cdots,\left(n, i_{n}\right)\right\}
$$
where $i_{1}, i_{2}, \cdots, i_{n} \in\{1,2, \cdots, 2017\}$.
For any $a, i$, since when the magician sees $(a, i)$, they can always guess at least one $(b, j)(b \neq a)$, and the number of states containing $(a, i)$ and $(b, j)$ is only $2017^{n-2}$, the total number of different states the magician can guess (i.e., $2017^{n}$) does not exceed $2017 \times 2017^{n-2} n$.
Thus, $n \geqslant 2017$.
This indicates that $n$ can only be 2017. Therefore, from any $(a, i)$, the $(b, j)(b \neq a)$ that can be guessed is unique.
Therefore, draw an edge from $(a, i)$ to $(b, j)$, denoted as $(a, i) \rightarrow(b, j)$, and call $a, b$ "associated" (association is mutual, i.e., $b, a$ are also associated), and any two different guesses $(a, i) \rightarrow(b, j)$ and $(c, k) \rightarrow(d, l)$
cannot correspond to the same state.
This indicates that at least one of $c, d$ belongs to $\{a, b\}$.
The following lemma is proved.
Lemma If $(a, i) \rightarrow(b, j)$, then there must exist $k \neq i$ such that $(b, j) \rightarrow(a, k)$.
Proof Suppose $(b, j) \rightarrow(c, k)$. If $c \neq a$ or $(c, k)=(a, i)$, then a state containing $(a, i), (b, j), (c, k)$ has two guesses $(a, i) \rightarrow(b, j)$ and $(b, j) \rightarrow(c, k)$, which is a contradiction.
Thus, there must exist a $k \neq i$ such that
$(b, j) \rightarrow(a, k)$.
The lemma is proved.
Assume 1 and 2 are associated.
Note that 3 must be associated with another number $a$. By the previous discussion, $a \in\{1,2\}$, assume $a=1$.
At this point, for any $b(4 \leqslant b \leqslant 2017)$, since $b$ must be associated with another number $c$, and 1 and 2 are associated, 1 and 3 are associated, thus $c \in\{1,2\}, c \in\{1,3\}$. Therefore, $c=1$, i.e., 1 and $b$ are associated.
This indicates that 2, 3, ..., 2017 are all associated with 1.
Thus, for any $t \in\{2,3, \cdots, 2017\}$, combining the lemma, there exist $i_{t}, j_{t} \in\{1,2, \cdots, 2017\}$ such that $\left(1, i_{t}\right) \rightarrow\left(t, j_{t}\right)$.
Using the lemma repeatedly, there exist $k_{t}, l_{t} \in\{1,2, \cdots, 2017\}\left(k_{t} \neq i_{t}, l_{t} \neq j_{t}\right)$ such that
$\left(t, j_{t}\right) \rightarrow\left(1, k_{t}\right) \rightarrow\left(t, l_{t}\right)$.
Thus, $i_{2}, k_{2}, i_{3}, k_{3}, \cdots, i_{2017}, k_{2017}$ are all different, but $2 \times 2016 > 2017$, which is a contradiction. This indicates that the case $n=2017$ cannot complete the magic trick.
In summary, the minimum value of $n$ is 2018. | 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 number of pieces on the chessboard. | 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 200, are all placed with blue pieces. It is easy to verify that each piece can see exactly five pieces of the opposite color. At this point, the total number of pieces is
$$
5 \times 190 \times 4 = 3800 \text{.}
$$
Now assume there is a valid placement of pieces such that the total number of pieces on the board exceeds 3800. In this case, connect each piece to the pieces it can see with edges, so that the number of edges connected to each piece is exactly 5. Thus, the total number of edges exceeds
$$
\frac{1}{2} \times 5 \times 3800 = 9500.
$$
Consider any row: if there are no pieces of the opposite color in that row, then the number of pieces in that row does not exceed 200, and the number of edges is 0; if there are pieces of the opposite color in that row, suppose there is a red piece \( R \) and a blue piece \( B \). Since the red piece \( R \) can see exactly five pieces of the opposite color, there can be at most five blue pieces in that row. Similarly, there can be at most five red pieces in that row. Therefore, the number of pieces in that row does not exceed 10, and the number of edges does not exceed 25.
If there are 191 rows containing pieces of the opposite color, then the total number of pieces on the board does not exceed
$$
191 \times 10 + 9 \times 200 < 3800,
$$
which contradicts the assumption. This indicates that there are at most 190 rows containing pieces of the opposite color. Thus, the number of edges connecting pieces of the opposite color in the same row does not exceed \( 190 \times 25 = 4750 \).
Similarly, the number of edges connecting pieces of the opposite color in the same column does not exceed 4750.
Therefore, the total number of edges does not exceed
$$
2 \times 4750 = 9500 \text{,}
$$
which also leads to a contradiction.
This shows that the total number of pieces on the board must not exceed 3800. In summary, the maximum value sought is 3800. | 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-digit mountain numbers; when the middle number is 7, there are $\mathrm{C}_{6}^{4} \mathrm{C}_{7}^{4}$ nine-digit mountain numbers; when the middle number is 6, there are $\mathrm{C}_{5}^{4} \mathrm{C}_{6}^{4}$ nine-digit mountain numbers; when the middle number is 5, there are $\mathrm{C}_{4}^{4} \mathrm{C}_{5}^{4}$ nine-digit mountain numbers. Therefore, the total number of nine-digit mountain numbers is 11875. | 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
$$
\begin{array}{l}
\overline{x_{k}}=\mathrm{e}^{\frac{-2 k \pi}{2017} \mathrm{i}}=\mathrm{e}^{\frac{2(2017-k) \pi \mathrm{i}}{2017} \mathrm{i}}=x_{2017-k} . \\
\text { Then } \frac{1}{1+x_{k}}+\frac{1}{1+x_{2017-k}} \\
=\frac{1}{1+x_{k}}+\frac{1}{1+\overline{x_{k}}} \\
=\frac{1}{1+x_{k}}+\frac{x_{k} \overline{x_{k}}}{x_{k} \overline{x_{k}}+\overline{x_{k}}}=1 . \\
\text { Therefore, } \sum_{k=1}^{2016} \frac{1}{1+x_{k}}=1008 .
\end{array}
$$ | 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 for any two cities, one can always fly from one city to the other, with the number of flights not exceeding $d$ and being even?
Note: All flight routes allow passing through a city multiple times. | 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) $\{\alpha, \beta\} \cap \{E, F, G, H\} \neq \varnothing$;
(2) $\{\alpha, \beta\} \cap \{E, F, G, H\} = \varnothing$ and they are not collinear.
For case (1), by symmetry, assume $\alpha = E$. If $\beta \neq A$ and $B$, then $A$ and $B$ are not adjacent to $I$. Thus, $A$ is adjacent to $E$ and $H$, and $B$ is adjacent to $E$ and $F$. This means $E$ has three neighbors, which is a contradiction.
Therefore, $\beta \in \{A, B\}$. Without loss of generality, assume $\beta = A$. In this case, $B$, $C$, and $D$ are not adjacent to $I$, so their neighbors on $\Gamma$ are already determined. Thus, we must have
$\Gamma = A I E B F C G D H A$.
Note that $\alpha$ has 4 choices, and $\beta$ has 2 choices, so there are $4 \times 2 = 8$ circles that satisfy case (1).
For case (2), assume $\alpha = A$ and $\beta = B$. In this case, $C$ and $D$ are not adjacent to $I$. Therefore, their neighbors on $\Gamma$ are already determined, and we have the segments $A I B$, $F C G D H$, and $E$ on $\Gamma$.
If $E$ is adjacent to $A$, then $B$ must be adjacent to $F$, giving us
$\Gamma = A I B F C G D H E A$.
Similarly, if $E$ is adjacent to $B$, then
$\Gamma = A I B E F C G D H A$.
If $E$ is not adjacent to $A$ or $B$, then $E$ must be adjacent to $F$ and $H$. In this case, the segments $A I B$ and $E F C G D H$ cannot be connected to form a circle.
Note that $\{\alpha, \beta\}$ has 4 choices, so there are $4 \times 2 = 8$ circles that satisfy case (2).
In summary, the number of circles corresponding to the bus route is 16, and each circle has two directions.
Therefore, the number of different orders in which the bus passes through the stations is $16 \times 2 = 32$. | 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 $\frac{B G}{G E}$.
(2010-2011, Hungarian Mathematical Olympiad) | 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
$$
\begin{array}{l}
\frac{AM}{MB}=\frac{S_{\triangle ACD}}{S_{\triangle BCD}}=\frac{AC \cdot AD \sin \angle CAD}{BC \cdot BD \sin \angle CBD} \\
=\frac{AC}{BC} \cos \angle BCA=\left(\frac{AC}{BC}\right)^{2} .
\end{array}
$$
Applying Menelaus' Theorem to $\triangle BAE$ and the transversal $GMC$ yields
$$
\frac{BG}{GE}=\frac{CA}{CE} \cdot \frac{BM}{MA}=\frac{CA}{CE} \cdot \frac{BC^2}{AC^2}=2 \cdot \frac{CB}{CA} \cdot \frac{CO}{CE}=2 .
$$ | 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 segments $p_{i} p_{j} \backslash p_{h} p_{k}$ in the graph, the number of triangles in $S$ with $p_{i} p_{j}$ as a side is equal to the number of triangles in $S$ with $p_{h} p_{k}$ as a side. Find the smallest positive integer $n$, such that in the set $S$ there are always two triangles, each with three vertices of the same color.
(2007, China National Training Team Test) | 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:
$$
\begin{array}{l}
\triangle p_{1} p_{2} p_{3} 、 \triangle p_{1} p_{3} p_{4} 、 \triangle p_{1} p_{4} p_{5} 、 \triangle p_{1} p_{5} p_{6} 、 \\
\triangle p_{1} p_{2} p_{6} 、 \triangle p_{2} p_{3} p_{5} 、 \triangle p_{2} p_{4} p_{5} 、 \triangle p_{2} p_{4} p_{6} 、 \\
\triangle p_{3} p_{4} p_{6} 、 \triangle p_{3} p_{5} p_{6} .
\end{array}
$$
It is easy to verify that for any line segment $p_{i} p_{j}$ $(1 \leqslant i<j \leqslant 6)$, there are exactly two triangles, but only $\triangle p_{1} p_{2} p_{3}$ has all three vertices of the same color.
Therefore, the set $S$ of these ten triangles does not satisfy the conditions of the problem.
When $n=7$, color $p_{1} 、 p_{2} 、 p_{4}$ red, and $p_{3}$ 、 $p_{5} 、 p_{6} 、 p_{7}$ blue, and consider the triangles formed by the cyclic permutation arrangement:
$\triangle p_{1} p_{2} p_{4} 、 \triangle p_{2} p_{3} p_{5} 、 \triangle p_{3} p_{4} p_{6} 、 \triangle p_{4} p_{5} p_{7} 、$
$\triangle p_{5} p_{6} p_{1} 、 \triangle p_{6} p_{7} p_{2} 、 \triangle p_{7} p_{1} p_{3}$.
It is easy to verify that the 21 edges of the complete graph of order 7 are exactly the 21 edges of these seven triangles. But among the seven triangles, only $\triangle p_{1} p_{2} p_{4}$ has all three vertices of the same color.
Therefore, the set $S$ of these seven triangles does not satisfy the conditions of the problem.
When $n=8$, call the line segment connecting two points of the same color a same-color edge, and the line segment connecting two points of different colors a different-color edge. Thus, there are at most 16 different-color edges in the graph. At this point, there are four red and four blue points among the eight vertices, with 12 same-color edges and 28 edges in total in the graph.
Since 3 does not divide $28 、 56 、 112 、 140$, but divides $84 、 168$, to satisfy the conditions of the problem, each edge can only be a common edge of three triangles or six triangles.
If $k=3$, i.e., each edge is a common edge of three triangles, then $|S|=28$.
Note that each different-color triangle has exactly two different-color edges, and there are at most 48 different-color edges in the graph (each edge is counted three times). Therefore, there are at most 24 different-color triangles. Thus, the set $S$ must contain at least four same-color triangles.
If $k=6$, then $|S|=56$. Similarly, there must be at least eight same-color triangles.
This indicates that when $n=8$, the conditions of the problem are satisfied.
In summary, the smallest positive integer $n=8$. | 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
$$
\begin{array}{l}
(x y)^{2}-4 x y+3=0 \\
\Rightarrow x y=1 \text { or } x y=3 \text { (discard). } \\
\text { Also, } x^{5}+y^{5}=\left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)-x^{2} y^{2}(x+y), \\
x^{3}+y^{3}=(x+y)\left((x+y)^{2}-3 x y\right)=18, \\
x^{2}+y^{2}=(x+y)^{2}-2 x y=7,
\end{array}
$$
Therefore, $x^{5}+y^{5}=123$. | 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 $a^{2}+a b+b^{2}$
$$
\begin{array}{l}
=(a+b)(a+b+c)-(a b+b c+c a) \\
=5(5-c)-5=5(4-c),
\end{array}
$$
Similarly, $b^{2}+b c+c^{2}=5(4-a)$,
$$
c^{2}+c a+a^{2}=5(4-b) \text {. }
$$
Therefore, the value of the required expression is
$$
\begin{array}{l}
125(4-a)(4-b)(4-c) \\
= 125(64-16(a+b+c)+ \\
4(a b+b c+c a)-a b c) \\
= 125(64-16 \times 5+4 \times 5-(-1)) \\
= 125 \times 5=625 .
\end{array}
$$ | 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 $a=3, b=2, c=1$, the equality holds.
Therefore, the minimum value of $\frac{a b c}{a+b+c}$ is 1. | 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
$$
\begin{array}{l}
(a+1)^{2}+(b+3)^{2}+(c+5)^{2} \\
=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+y z+z x) \\
=100 .
\end{array}
$$ | 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 $\boldsymbol{u} \cdot \boldsymbol{v}=-12$,
$\boldsymbol{\alpha} \cdot \boldsymbol{\beta}=-54-32-84=-170$ is the minimum value. | -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
$$
\begin{array}{l}
s+\frac{1}{t}=-\frac{99}{19}, \frac{s}{t}=\frac{1}{19} . \\
\text { Hence } \frac{s t+4 s+1}{t}=\left(s+\frac{1}{t}\right)+4 \cdot \frac{s}{t} \\
=-\frac{99}{19}+\frac{4}{19}=-5 .
\end{array}
$$ | -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 possible to have exactly
$$
2017 \text { clockwise triangles? }
$$ | 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 counterclockwise) of $\triangle P_{i} P_{j} P_{k}$ will change. It is stipulated that during the movement, only one point is allowed to cross the line connecting two other points, and this operation does not change the orientation of any other triangle. Continuously perform such operations until $P_{1}, P_{2}, \cdots, P_{100}$ are arranged clockwise on the circle. At this point, the number of clockwise triangles is $\mathrm{C}_{100}^{3}>2017$.
Notice that, after each operation, the number of clockwise triangles increases by 1. Therefore, there must be a moment when the number of clockwise triangles is exactly 2017. | 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 $\angle APB = 90^{\circ}$, the equality holds.
In fact, as shown in the figure
2, with $AB = 8$ as the
diameter, draw $\odot O$, and on one side of $AB$,
draw a parallel line at a distance
of 3 from $AB$ intersecting
$\odot O$ at points $P$ and $P_1$,
connect $AP$ and $BP$. Then in
$\triangle APB$,
$$
\angle APB = 90^{\circ}, AP \cdot PB = AB \cdot PH = 24 \text{. }
$$ | 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 $\sqrt{2} x_{1}^{2}=1, \sqrt{2} x_{2}^{2}=2, \sqrt{2} x_{3}^{2}=3$ respectively.
Therefore, $x_{1}^{2}=\frac{\sqrt{2}}{2}, x_{2}^{2}=\sqrt{2}, x_{3}^{2}=\frac{3 \sqrt{2}}{2}$.
The squares of the lengths of the three sides are
$$
\left(2 x_{1}\right)^{2}=2 \sqrt{2},\left(2 x_{2}\right)^{2}=4 \sqrt{2},\left(2 x_{3}\right)^{2}=6 \sqrt{2} \text {, }
$$
Thus, the triangle formed by the three line segments of lengths $2 x_{1}, 2 x_{2}, 2 x_{3}$ is a right triangle.
Let the area of this right triangle be $S$. Then
$$
\begin{array}{l}
S^{2}=\left(\frac{1}{2} \times 2 x_{1} \times 2 x_{2}\right)^{2}=\frac{2 \sqrt{2} \times 4 \sqrt{2}}{4}=4 \\
\Rightarrow S=2 .
\end{array}
$$ | 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
$$
\begin{array}{l}
n+1=\overline{a_{1} a_{2} \cdots a_{t-1} a_{t}^{\prime}}\left(a_{t}^{\prime}=a_{t}+1\right) . \\
\text { Hence } S(n+1)=S(n)+1 .
\end{array}
$$
From the problem, we know
$$
S(n)=11 p_{1}, S(n+1)=11 p_{2} \text {. }
$$
Thus, $11\left(p_{2}-p_{1}\right)=1$, which is a contradiction.
Therefore, $a_{t}=9$.
Assume the last $k$ digits of $n$ are all 9. Then
$$
\begin{array}{l}
S(n)=a_{1}+a_{2}+\cdots+a_{t-k}+9 k=11 p_{1}, \\
S(n+1)=a_{1}+a_{2}+\cdots+\left(a_{t-k}+1\right)+0 \times k \\
=11 p_{2} .
\end{array}
$$
From the above two equations, we get $11\left(p_{1}-p_{2}\right)=9 k-1$.
For the smallest $n$, there is the smallest natural number $k$ such that 11 divides $(9 k-1)$.
Upon inspection, $k=1,2,3,4$ do not meet the requirements.
When $k=5$, $9 \times 5-1=44$ is a multiple of 11, at this time, $p_{1}-p_{2}=4$.
Therefore, the last five digits of $n$ are all 9.
Take the smallest $n$ such that $p_{1}-p_{2}=4$, then
$$
\begin{array}{l}
p_{1}=5, p_{2}=1 \\
\Rightarrow S(n)=55, S(n+1)=11 .
\end{array}
$$
From $a_{1}+a_{2}+\cdots+a_{t-k}+9 \times 5=55$, we get
$$
a_{1}+a_{2}+\cdots+a_{t-k}=10>9 \Rightarrow t-k \geqslant 2 \text {. }
$$
Since $k=5$, the smallest $n$ is a seven-digit number, with the sum of the first two digits being 10.
Thus, $n=2899999, n+1=2900000$. | 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 $a_{1}+a_{2}+\cdots+a_{20}=$ . $\qquad$ | 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.
$$
\text{Therefore, } \sum_{k=0}^{201}(10 k+7)^{k+1} \equiv 7^{1}+7^{2} \equiv 6(\bmod 10) \text{. }
$$ | 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 the smallest positive integer $n$ that satisfies $\sum_{i=1}^{n} a_{i}>1000$ is $n=$ $\qquad$. | 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, $\left|\overrightarrow{P_{n} P_{n+1}}\right|=O P_{n}$,
$\left|\overrightarrow{P_{n+1} P_{n+2}}\right|=\left|O P_{n+1}\right|=\sqrt{2}\left|O P_{n}\right|$.
Therefore, $a_{n}=\left|O P_{n}\right|^{2}=2\left|O P_{n-1}\right|^{2}=2^{n-1}$.
Also, $\sum_{i=1}^{n} a_{i}=2^{n}-1>1000$, hence $n \geqslant 10$. | 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
$$
\begin{array}{l}
S_{n+2}S_{n+2}+\lambda S_{n} \geqslant 2 \sqrt{\lambda S_{n+2} S_{n}} \\
\Rightarrow \frac{S_{n+2}}{S_{n+1}}<\frac{\lambda}{4} \cdot \frac{S_{n+1}}{S_{n}} \\
\Rightarrow 1<\frac{S_{n+1}}{S_{n}}<\left(\frac{\lambda}{4}\right)^{n-1} \frac{S_{2}}{S_{1}} .
\end{array}
$$
Since the above inequality holds for any $n$, hence $\lambda \geqslant 4$. When $\lambda=4$, taking $a_{n}=10^{2 n}$ satisfies the conditions.
In conclusion, the minimum value of $\lambda$ is 4. | 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 element $x_{i}$ is $k_{i}(i=1,2, \cdots, n)$, and let $m=$ $\max \left\{k_{1}, k_{2}, \cdots, k_{n}\right\}$. Find the minimum value of $m$. | 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 two elements, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2} \geqslant 2 \mathrm{C}_{10}^{2}=90 . \\
\text { Hence } 180 \leqslant \sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \\
\leqslant \sum_{i=1}^{n} k_{i}(m-1)=45(m-1) \\
\Rightarrow m \geqslant 5 .
\end{array}
$$
Below is the construction for $m=5$.
$$
A_{i}=\{\bar{i}, \overline{i+1}, \overline{i+2}, \overline{i+4}, \overline{i+7}\},
$$
where $i=1,2, \cdots, 10, \bar{i} \equiv i(\bmod 10)$, and $\bar{i} \in$ $\{1,2, \cdots, 10\}$. | 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 remainders of $S_{1}, S_{2}, \cdots, S_{100}$ modulo 11, there are 19 ones, 18 threes, 18 sixes, 18 tens, 9 fours, and 18 zeros.
Therefore, the number of qualifying groups is
$$
3 \mathrm{C}_{18}^{2}+\mathrm{C}_{19}^{2}+\mathrm{C}_{18}^{2}+18=801 .
$$ | 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 sequence $\left\{x_{n}\right\}$ and all positive integers $m$, we have
$$
\frac{1}{m} \sum_{i=1}^{m} x_{i}^{2} \leqslant \sum_{i=1}^{m} \lambda^{m-i} y_{i}^{2} .
$$ | 【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 $y_{1}=1, y_{n}=0(n \geqslant 2)$.
Substitute into equation (1) to get
$$
\begin{array}{l}
\frac{1}{m} \cdot 2^{m-1} \leqslant \lambda^{m-1} \\
\Rightarrow \frac{2}{\lambda} \leqslant \sqrt[m-1]{m} .
\end{array}
$$
When $m \rightarrow \infty$, $\frac{2}{\lambda} \leqslant 1 \Rightarrow \lambda \geqslant 2$.
Next, use mathematical induction to prove: for sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ that satisfy the conditions and for all positive integers $m$, equation (1) holds. When $m=1$, $x_{1}^{2}=y_{1}^{2} \leqslant y_{1}^{2}$, the conclusion holds.
Assume the conclusion holds for $m=k$.
Then when $m=k+1$, let
$$
\begin{array}{l}
f(k+1)=\sum_{i=1}^{k+1} 2^{k+1-i} y_{i}^{2}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}^{2}, \\
f(k)=\sum_{i=1}^{k} 2^{k-i} y_{i}^{2}-\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2} .
\end{array}
$$
Thus, $f(k+1)-f(k)$
$$
\geqslant y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} x_{k+1}^{2}+f(k) \text {. }
$$
By the given conditions and the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
x_{k+1}^{2}=\left(y_{k+1}+\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}}\right)^{2} \\
\leqslant\left((k+1) y_{k+1}^{2}+\frac{k+1}{k} \sum_{i=1}^{k} x_{i}^{2}\right)^{2}\left(\frac{1}{k+1}+\frac{k}{k+1}\right) \\
\leqslant(k+1) y_{k+1}^{2}+\left(1+\frac{1}{k}\right) \sum_{i=1}^{k} x_{i}^{2} .
\end{array}
$$
By the induction hypothesis, we get
$$
f(k+1) \geqslant f(k) \geqslant 0 \text {. }
$$
Therefore, the conclusion holds when $m=k+1$.
In summary, the minimum value of the positive number $\lambda$ is 2. | 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 monotonically decreasing when $t \in (-\infty,-1)$, and monotonically increasing when $t \in \left(-1, \frac{1}{4}\right]$.
Therefore, when $t=-1$, $f(t)$ has a maximum value of 4, i.e., $\left(x^{3}+1\right)\left(y^{3}+1\right)$ has a maximum value of 4. | 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
$$
\begin{array}{l}
\left|\overrightarrow{A P_{3}}\right|+\left|\overrightarrow{P_{3} P_{2}}\right|+\left|\overrightarrow{P_{2} P_{1}}\right|+\left|\overrightarrow{P_{1} O}\right| \\
\geqslant|\overrightarrow{A O}|=2018 .
\end{array}
$$
When $x_{k}=y_{k}=0(k=1,2,3)$, the equality holds, the required minimum value is 2018. | 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 set of points $\left\{A_{n}\right\}(n \in \mathbf{N})$ connected in sequence forms a polygonal line that encloses the feasible region.
It is easy to prove that the points $A_{n}$ are all above $l_{m}(m>n, m \in \mathbf{N})$.
In this problem, let $41 x+2 y=z$, then $y=-\frac{41}{2} x+\frac{1}{2} z$.
Since $-\frac{41}{2} \in\left[-4^{2},-5^{2}\right]$, when the line $y=-\frac{41}{2} x+\frac{1}{2} z$ passes through the point $A_{4}\left(\frac{1}{9}, \frac{20}{9}\right)$,
$$
z=41 x+2 y=41 \times \frac{1}{9}+2 \times \frac{20}{9}=9 .
$$ | 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, $a_{n}=2^{n}+2 n-80$.
Hence, $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{12}\right|$
$$
\begin{array}{l}
=-\left(a_{1}+a_{2}+\cdots+a_{6}\right)+\left(a_{7}+a_{8}+\cdots+a_{12}\right) \\
=S_{12}-2 S_{6}=8010 .
\end{array}
$$ | 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 $x>3$,
$$
f(x)=\log _{2}\left(1-\frac{1}{x-2}\right)+\cos x \pi,
$$
Thus, $f(x)$ is monotonically increasing in the interval $(3,4)$, and
$$
f(4)=0 \text {. }
$$
By the monotonicity of $\log _{2}\left(1-\frac{1}{x-2}\right)$, we know that when $x>4$,
$$
f(x)>\log _{2} \frac{1}{2}-1=-2 \text {. }
$$
Therefore, $\beta \in(3,4)$ and is unique.
Similarly, $\alpha$ is also unique.
Thus, $\alpha+\beta=\frac{5}{2} \times 2=5$. | 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
$$
\begin{array}{l}
\omega^{4}-\omega^{3}+\omega^{2}-\omega+1=0 \\
\Rightarrow \omega^{5}=-1 \Rightarrow \omega^{10}=\left(\omega^{2}\right)^{5}=1 .
\end{array}
$$
Thus, $\omega^{2}$ is a fifth root of unity.
Let $z=x^{18} y^{-18}=\left(\left(\frac{x}{y}\right)^{2}\right)^{9}=\left(\omega^{2}\right)^{9}$, which is still a fifth root of unity.
$$
\text { Also, } \sum_{m=0}^{29} \sum_{n=0}^{29} x^{18 m n} y^{-18 m n}=\sum_{m=0}^{29} \sum_{n=0}^{29} z^{m n} \text {, }
$$
When $5 \mid m$, $z^{m n}=1$;
When $5 \nmid m$, $z^{m}$ is still a fifth root of unity, and
$$
\begin{array}{l}
\sum_{k=0}^{4} z^{k m}=0, \\
\text { Hence } \sum_{m=0}^{29} \sum_{n=0}^{29} z^{m n}=\sum_{5 \mid m}^{29} \sum_{n=0}^{29} z^{m n}+\sum_{5 \nmid m}^{29} \sum_{n=0}^{29} z^{m n} \\
=6 \times 30+0=180 .
\end{array}
$$ | 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 probability that she wins is $\frac{p}{q}\left((p, q)=1, p 、 q \in \mathbf{Z}_{+}\right)$. Then the value of $q-p$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5 | 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 probability of Amelia winning is $\frac{1}{3}$.
In the first round, the probability of Amelia not winning is $1 - \frac{1}{3} = \frac{2}{3}$, and the probability of Brian not winning is $1 - \frac{2}{5} = \frac{3}{5}$. Therefore, the probability of both not winning is $\frac{2}{3} \times \frac{3}{5} = \frac{2}{5}$.
Thus, $P_{0} = \frac{1}{3} + \frac{2}{5} P_{0} \Rightarrow P_{0} = \frac{5}{9}$.
Therefore, the answer is $9 - 5 = 4$. | 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) $A+B=C$.
If $B=0, A=C$, we get the numbers satisfying the property 110, $220, \cdots, 990$. Each number has two permutations. Therefore, there are $2 \times 9=18$ such numbers.
If $B=1, A+1=C$, we get the numbers satisfying the property $121,231, \cdots, 891$, where the first number has three permutations, and the other seven numbers each have six permutations. Therefore, there are $3+7 \times 6=45$ such numbers.
If $B=2, A+2=C$, we get the numbers satisfying the property $242,352, \cdots, 792$, where the first number has three permutations, and the other five numbers each have six permutations. Therefore, there are $3+5 \times 6=33$ such numbers.
If $B=3, A+3=C$, we get the numbers satisfying the property $363,473, \cdots, 693$, where the first number has three permutations, and the other three numbers each have six permutations. Therefore, there are $3+3 \times 6=21$ such numbers.
If $B=4, A+4=C$, we get the numbers satisfying the property
484, 594. The first number has three permutations, and the second number has six permutations. Therefore, there are 9 such numbers.
In total, there are $18+45+33+21+9=126$ numbers.
(2) $A+B=C+11$.
If $C=0, A+B=11$, we get the numbers satisfying the property $902, 803, 704, 605$. Each number has four permutations. Therefore, there are $4 \times 4=16$ such numbers.
If $C=1, A+B=12$, we get the numbers satisfying the property $913, 814, 715, 616$, where the first three numbers each have six permutations, and the last number has three permutations. Therefore, there are $3 \times 6+3=21$ such numbers.
If $C=2, A+B=13$, we get the numbers satisfying the property $924, 825, 726$. Each number has six permutations. Therefore, there are $3 \times 6=18$ such numbers.
If $C=3, A+B=14$, we get the numbers satisfying the property $935, 836, 737$. The first two numbers each have six permutations, and the last number has three permutations. Therefore, there are $2 \times 6+3=15$ such numbers.
If $C=4, A+B=15$, we get the numbers satisfying the property 946, 847. Each number has six permutations. Therefore, there are $2 \times 6=12$ such numbers.
If $C=5, A+B=16$, we get the numbers satisfying the property $957, 858$. The first number has six permutations, and the second number has three permutations. Therefore, there are 9 such numbers.
If $C=6, A+B=17$, we get the number 968. It has six permutations. Therefore, there are 6 such numbers.
If $C=7, A+B=18$, we get the number 979. It has three permutations. Therefore, there are 3 such numbers.
In total, there are
$$
16+21+18+15+12+9+6+3=100
$$
numbers.
Combining (1) and (2), the total number is
$126+100=226$. | 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 such that the sum of their labels is exactly $n$, this is called a "scheme for $n$". The number of different schemes for $n$ is denoted as $f(n)$. Find $P=$ $\sum_{n=1}^{1000} f(n)$.
| 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 ways to pick cards labeled with $3^{k}$ is $\mathrm{C}_{2}^{0}+\mathrm{C}_{2}^{1}+\mathrm{C}_{2}^{2}=4$, and each picking method corresponds to a unique scheme. Thus, the total number of ways to pick cards labeled with $1,3,3^{2}, \cdots, 3^{k-1}$ is $4^{k}-1$ (excluding the case where no card is picked).
Notice that, $1000=3^{6}+3^{5}+3^{3}+1$.
By repeatedly applying the above conclusion, we get
$$
\begin{aligned}
P= & \sum_{n=1}^{1000} f(n) \\
= & \sum_{n=1}^{3^{6-1}} f(n)+f\left(3^{6}\right)+\sum_{n=3^{6}+1}^{3^{6}+3^{5}-1} f(n)+ \\
& f\left(3^{6}+3^{5}\right)+\sum_{n=3^{6}+3^{5}+1}^{3^{6}+3^{5}+3^{3}-1} f(n)+ \\
& f\left(3^{6}+3^{5}+3^{3}\right)+f\left(3^{6}+3^{5}+3^{3}+1\right) \\
= & 4^{6}-1+f\left(3^{6}\right)+2\left(4^{5}-1\right)+f\left(3^{6}+3^{5}\right)+ \\
& 4\left(4^{3}-1\right)+f\left(3^{6}+3^{5}+3^{3}\right)+ \\
& f\left(3^{6}+3^{5}+3^{3}+1\right) \\
= & 4^{6}-1+2+2\left(4^{5}-1\right)+2^{2}+ \\
& 4\left(4^{3}-1\right)+2^{3}+2^{4} \\
= & 6423 .
\end{aligned}
$$
(Zheng Kai, Changshu High School, Jiangsu Province, 215500) | 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 that $S\left(a_{1}\right)=S\left(a_{2}\right)=\cdots=S\left(a_{8}\right)$. What is the minimum value of the positive integer $n$? ( ).
(A) 185
(B) 187
(C) 189
(D) 191 | 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, $1899, 1989, 1998$; and the number with a digit sum of 28 is only 1999.
Furthermore, taking seven numbers from those with a digit sum of $2,3, \cdots, 26$, and taking all the rest, does not satisfy the condition.
Therefore, $n \geqslant 4+7 \times 25+4+1+1=185$.
By the pigeonhole principle, when $n=185$, the condition is satisfied. In summary, the minimum value of the positive integer $n$ is 185. | 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 represent the cup mouth facing up, and 0 to represent the cup mouth facing down. The following flips can satisfy the condition:
$$
\begin{array}{l}
(1,1,1,1) \xrightarrow{1}(0,0,0,1) \xrightarrow{2}(0,1,1,0) \\
\xrightarrow{3}(1,0,1,1) \xrightarrow{4}(0,0,0,0) .
\end{array}
$$
Therefore, the minimum value of the positive integer $n$ is 4. | 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
$$
\frac{4}{\log _{2} a}+\log _{2} a \geqslant 2 \sqrt{\frac{4}{\log _{2} a} \log _{2} a}=4 \text {, }
$$
with equality holding if and only if $a=4$. | 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
$$
\begin{array}{l}
a_{n+1}=3 a_{n}-4^{n-1} \\
\Rightarrow a_{n+1}+4^{n}=3 a_{n}-4^{n-1}+4^{n} \\
\quad=3\left(a_{n}+4^{n-1}\right) \\
\Rightarrow a_{n}+4^{n-1}=3^{n-1}\left(a_{1}+1\right)=3^{n} \\
\Rightarrow a_{n}=3^{n}-4^{n-1} .
\end{array}
$$
Calculations show that
when $n \leqslant 4$, $a_{n}>0$;
when $n \geqslant 5$, $a_{n}<0$.
Thus, when $n=4$, $S_{n}$ reaches its maximum value of
$$
S_{4}=\sum_{i=1}^{4}\left(3^{i}-4^{i-1}\right)=35 .
$$ | 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
$$
\begin{array}{l}
2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1} . \\
\text { Then } n^{2}-(4 k+1) n+4 k^{2}-2=0 \\
\Rightarrow n_{1,2}=\frac{4 k+1 \pm \sqrt{8 k+9}}{2} .
\end{array}
$$
Since $n$ is a positive integer, let
$$
8 k+9=(2 m+1)^{2} \Rightarrow 2 k=m^{2}+m-2 \text {. }
$$
Substituting into equation (1) gives
$$
n_{1}=(m+1)^{2}-2, n_{2}=m^{2}-2 \text {. }
$$
Therefore, the maximum value of the three-digit positive integer $n$ is 959 . | 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 $0 \leqslant x \leqslant 6$, and $x$ is an integer, we get $x=3, n=11$.
Therefore, the initial number of people participating in the activity is $n+4=15$. | 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 $x_{0}$ and rearranging gives
$$
\begin{array}{l}
3 b^{2}-4 a b+1=0 \Rightarrow a=\frac{3 b^{2}+1}{4 b} . \\
\text { Hence }|z|^{2}=a^{2}+b^{2}=\left(\frac{3 b^{2}+1}{4 b}\right)^{2}+b^{2} \\
=\frac{25}{16} b^{2}+\frac{1}{16 b^{2}}+\frac{3}{8} \geqslant \frac{5}{8}+\frac{3}{8}=1,
\end{array}
$$
with equality holding if and only if $b= \pm \frac{\sqrt{5}}{5}$. | 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 sequence;
(2) For any $n$, it holds that
$$
\prod_{i=1}^{n}\left(S_{i}+1\right) \geqslant k n,
$$
find the maximum value of $k$. | 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, $\frac{1}{S_{1}-1}=\frac{1}{2-1}=1$, so $\left\{\frac{1}{S_{n}-1}\right\}$ is an arithmetic sequence with the first term and common difference both equal to 1.
(2) From (1), we know
$$
\frac{1}{S_{n}-1}=n \Rightarrow S_{n}=\frac{n+1}{n} \text {. }
$$
From the condition, we get
$$
k \leqslant\left(\frac{1}{n} \prod_{i=1}^{n}\left(S_{i}+1\right)\right)_{\min } \text {. }
$$
Let $f(n)=\frac{1}{n} \prod_{i=1}^{n}\left(S_{i}+1\right)$. Then
$$
\frac{f(n+1)}{f(n)}=\frac{n\left(S_{n+1}+1\right)}{n+1}=\frac{n(2 n+3)}{(n+1)^{2}}>1 \text {. }
$$
Therefore, $f(n)_{\min }=f(1)=\frac{S_{1}+1}{1}=3, k_{\max }=3$. | 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}
$$
Therefore, among $S_{1}, S_{2}, \cdots, S_{2017}$, the numbers that are divisible by 2 but not by 4 are $\left[\frac{2017}{8}\right]=252$ in total. | 252 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a positive geometric sequence $\left\{a_{n}\right\}$ satisfies
$$
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^{3}-1}+\frac{1}{\sqrt{q^{3}-1}}\right)^{2} \\
\geqslant 49 \times 4=196,
\end{array}
$$
When $\sqrt{q^{3}-1}=1$, i.e., $q=\sqrt[3]{2}$, the equality holds. Therefore, the minimum value of $a_{9}+a_{8}+a_{7}$ is 196. | 196 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the function be
$$
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}
$$
Since $b$ is an integer, we have $a+1= \pm 1$.
Also, $a \neq 0$, so $a=-2, b=4, c=16$. Therefore, $a+b+c=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the function be
$$
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 {. }
$$
Therefore, the number of integer points on the graph of the function $y=f(x)(x \in[0,+\infty))$ is 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that a line passing through the focus $F$ of the parabola $y^{2}=4 x$ intersects the parabola at points $M$ and $N$, and $E(m, 0)$ is a point on the $x$-axis. The extensions of $M E$ and $N E$ intersect the parabola at points $P$ and $Q$ respectively. If the slopes $k_{1}$ and $k_{2}$ of $M N$ and $P Q$ satisfy $k_{1}=3 k_{2}$, then the value of the real number $m$ is . $\qquad$ | 9.3.
When $M P$ is not perpendicular to the $x$-axis, let $l_{\text {MР }}: y=k(x-m)$.
Substitute into $y^{2}=4 x$ to get
$$
x^{2}-\left(\frac{4}{k^{2}}+2 m\right) x+m^{2}=0 \text {. }
$$
Then $x_{M} x_{P}=m^{2} \Rightarrow y_{M} y_{P}=-4 m$ $\Rightarrow y_{P}=\frac{-4 m}{y_{M}}$.
When $M P \perp x$-axis, the conclusion also holds.
Similarly, $y_{M} y_{N}=-4, y_{Q}=\frac{-4 m}{y_{N}}$.
Thus, $k_{1}=\frac{4}{y_{M}+y_{N}}, k_{2}=\frac{4}{y_{P}+y_{Q}}=\frac{1}{m} k_{1}$.
Also, $k_{2}=\frac{1}{3} k_{1}$, therefore, $m=3$. | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Arrange all positive integers that are coprime with 70 in ascending order. The 2017th term of this sequence is $\qquad$ . | 10. 5881.
It is easy to know that the number of positive integers not exceeding 70 and coprime with 70 is
$$
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}=69 . \\
\text{Let } A=\left\{a_{i} \mid 1 \leqslant i \leqslant 24, i \in \mathbf{N}\right\}.
\end{array}
$$
Notice that, for any $k, r \in \mathbf{N}$, we have
$$
(70 k+r, 70)=(r, 70) \text{.}
$$
Therefore, for each fixed non-negative integer $k$, when $r$ takes all values in the set $A$, there are 24 positive integers of the form $70 k+r$ that are coprime with 70. Thus, we obtain the 24 terms of the sequence $\left\{a_{n}\right\}$.
$$
\begin{array}{l}
\text{Also, } 2017=24 \times 84+1, \text{ therefore,} \\
a_{2017}=70 \times 84+a_{1}=5881 .
\end{array}
$$ | 5881 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, and $c$ are three non-zero real numbers, and $x^{2}-1$ is a factor of the polynomial $x^{3}+a x^{2}+b x+c$. Then the value of $\frac{a b+3 a}{c}$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2 | - 1. A.
From the fact that $\pm 1$ are roots of the given polynomial, we have
$$
\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}
$$ | -2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. In two regular tetrahedrons $A-OBC$ and $D-OBC$ with their bases coinciding, $M$ and $N$ are the centroids of $\triangle ADC$ and $\triangle BDC$ respectively. Let $\overrightarrow{OA}=\boldsymbol{a}, \overrightarrow{OB}=\boldsymbol{b}, \overrightarrow{OC}=\boldsymbol{c}$. If point $P$ satisfies $\overrightarrow{OP}=x \boldsymbol{a}+y \boldsymbol{b}+z \boldsymbol{c}, \overrightarrow{MP}=2 \overrightarrow{PN}$, then the real number $9 x+81 y+729 z=$ $\qquad$
(Adapted from the 2016 National High School Mathematics League Zhejiang Regional Preliminary Contest) | Take $O$ as the origin and the line $O B$ as the $x$-axis, establishing a spatial rectangular coordinate system as shown in Figure 13.
Let $B(1,0,0)$. Then
$$
\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}, \frac{\sqrt{3}}{6},-\frac{\sqrt{6}}{3}\right) .
\end{array}
$$
From the problem and $\overrightarrow{M P}=2 \overrightarrow{P N}$, we know $P\left(\frac{11}{18}, \frac{13 \sqrt{3}}{54},-\frac{2 \sqrt{6}}{27}\right)$.
Thus $\overrightarrow{O P}=x a+y b+z c$
$\Rightarrow x=-\frac{2}{9}, y=\frac{4}{9}, z=\frac{5}{9}$
$$
\Rightarrow 9 x+81 y+729 z=439 \text {. }
$$ | 439 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Let $a, b$ be real numbers, and the equation with respect to $x$
$$
\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 {. }
$$
Thus, $\Delta=(b+2)^{2}+8 a-8$.
(1) When $\Delta>0$, equation (1) has two distinct real roots.
Since the original equation has no solution, it follows that the two distinct real roots of equation (1) are 0 and 1, i.e.,
$$
\begin{aligned}
& 1-a=0,2-(b+2)+(1-a)=0 \\
\Rightarrow & a=1, b=0 .
\end{aligned}
$$
Therefore, $8 a+4 b+|8 a+4 b-5|=11$.
In summary, the value of the algebraic expression is 5 or 11. | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n-1}}$. Then $\sum_{n=1}^{2018} a_{n}=$ $\qquad$ . | 5. 5857 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$; when $n \equiv 2,4,8(\bmod 10)$, $(n+1)^{n+2} \equiv 1(\bmod 4)$, then $a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10)$; when $n \equiv 3,7,9(\bmod 10)$, $(n+1)^{n+2} \equiv 0(\bmod 4)$, then $a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k} \equiv 1(\bmod 10)$. Thus, $a_{n}$ is a periodic sequence, with the smallest positive period being 10.
$$
\text { Hence } \sum_{n=1}^{2018} a_{n}=201 \times 29+28=5857 \text {. }
$$ | 5857 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A five-digit number $\overline{a b c d e}$ satisfies:
$$
ac>d, dd, b>e \text {. }
$$
For example, 34 201, 49 412. If the digits of the number change in a pattern similar to the monotonicity of a sine function over one period, then the five-digit number is said to follow the "sine rule." The number of five-digit numbers that follow the sine rule is $\qquad$.
| 6. 2892.
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers, respectively, and $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$, at this point, there are $10-k$ ways to choose $(b, d)$, and $a$, $c$, and $e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ combinations. Therefore, the number of numbers that conform to the sine rule is $(k-1)^{3}(10-k)$.
Thus, the number of numbers that conform to the sine rule is
$$
\sum_{k=2}^{9}(k-1)^{3}(10-k)=2892 .
$$ | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
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} .\right.
$$
(1) Find the general terms of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$;
(2) Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $S_{n}=\sum_{i=1}^{n}\left[a_{i}\right], T_{n}=\sum_{i=1}^{n}\left[b_{i}\right]$. Find the smallest $n \in \mathbf{Z}_{+}$ such that
$$
\sum_{k=1}^{n}\left(S_{k}+T_{k}\right)>2017 .
$$ | 11. (1) From the problem, we have
$$
\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}
$$
Notice that, $a_{n} 、 b_{n}>0$ and $a_{n}>b_{n}$.
Thus, $a_{n}=2^{n}+\sqrt{4^{n}-3^{n}}, b_{n}=2^{n}-\sqrt{4^{n}-3^{n}}$.
(2) From (1), we know $a_{n}+b_{n}=2^{n+1}$.
Then $2^{n+1}-2<\left[a_{n}\right]+\left[b_{n}\right] \leqslant 2^{n+1}$.
We will prove by contradiction that when $n \geqslant 2$, there does not exist $n$ such that $\sqrt{4^{n}-3^{n}} \in \mathbf{Z}_{+}$.
If there exist $n 、 t \in \mathbf{Z}_{+}$ such that $4^{n}-3^{n}=t^{2}$, then
$$
\left(2^{n}+t\right)\left(2^{n}-t\right)=3^{n} \text {. }
$$
When 3 divides $\left(2^{n}+t\right)$ and 3 divides $\left(2^{n}-t\right)$, $3 \mid t$, which implies $3 \mid 4^{n}$, a contradiction.
Thus, only $2^{n}-t=1 \Rightarrow t=2^{n}-1$.
Substituting into equation (1) gives $2^{n+1}=3^{n}+1$.
However, $2^{n+1}-3^{n}=8 \times 2^{n-2}-9 \times 3^{n-2}<0$, a contradiction.
Therefore, only when $n=1$, $a_{n} 、 b_{n} \in \mathbf{Z}_{+}$, at this time, $a_{1}+b_{1}=4$;
When $n \geqslant 2$, $\left[a_{n}\right]+\left[b_{n}\right]=2^{n+1}-1$.
Then $S_{n}+T_{n}=4+\sum_{i=2}^{n}\left(2^{i+1}-1\right)$ $=2^{n+2}-n-3$.
$$
\text { Hence } \sum_{k=1}^{n}\left(S_{k}+T_{k}\right)=2^{n+3}-8-\frac{n(n+1)}{2}-3 n \text {. }
$$
In particular, when $n=8$,
$$
\sum_{k=1}^{8}\left(S_{k}+T_{k}\right)=2^{11}-8-\frac{8(8+1)}{2}-24=1980 \text {. }
$$
Therefore, the smallest $n=9$. | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Let $S$ be a set of positive integers with the property: for any $x \in S$, the arithmetic mean of the remaining numbers in $S$ after removing $x$ is a positive integer, and it satisfies $1 \in S$, 2016 is the largest element in $S$. Find the maximum value of $|S|$. | Let the elements of set $S$ be
$$
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 \mathbf{Z}_{+} \\
\Rightarrow(n-1) \mid\left(x_{j}-x_{i}\right) .
\end{array}
$$
In particular, we have
$$
(n-1)|(2016-1) \Rightarrow(n-1)| 5 \times 13 \times 31 \text {. }
$$
Notice that, for any $2 \leqslant j \leqslant n$, we have
$$
(n-1) \mid\left(x_{j}-1\right) \text {. }
$$
Thus, for $x_{n}-1$ we have
$$
\begin{array}{l}
x_{n}-1=\sum_{k=1}^{n-1}\left(x_{k+1}-x_{k}\right) \geqslant(n-1)^{2} \\
\Rightarrow(n-1)^{2} \leqslant 2015 \\
\Rightarrow n \leqslant 45 . \\
\text { Also, }(n-1) \mid 2015, \text { so } n \leqslant 32 . \\
\text { Let } x_{i}=31 i-30(1 \leqslant i \leqslant 31), x_{32}=2016,
\end{array}
$$
It is easy to verify that $S$ satisfies the conditions in this case.
Therefore, $|S|_{\max }=32$. | 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
\[
S(n)=\min \left\{m \left| m \in \mathbf{Z}_{+}, n \right| m!\right\},
\]
For example, \( S(2)=2, S(3)=3, S(6)=3 \).
(1) Find the values of \( S(16) \) and \( S(2016) \);
(2) If \( S(n)=7 \), find the maximum value of the positive integer \( n \);
(3) Prove that there are infinitely many composite numbers \( n \) such that \( S(n)=p \), where \( p \) is the largest prime factor of \( n \). | 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, $n \leqslant 7$.
Notice that, $S(7!)=7$.
Therefore, the maximum value of the positive integer $n$ is $7!=5040$.
(3) It is easy to know that, for any odd prime $p$, taking $n=2 p$, we always have $S(n)=p$.
Since there are infinitely many odd primes, there exist infinitely many composite numbers $n$ such that $S(n)=p$, where $p$ is the largest prime factor of $n$. | 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 on segment $A C$, satisfying $\frac{|C E|}{|C A|}=\frac{|A D|}{|A B|}$.
(1) Prove that the line $D E$ intersects the parabola at exactly one point.
(2) Let the intersection point of line $D E$ and the parabola be $F$. Denote the areas of $\triangle B C F$ and $\triangle A D E$ as $S_{1}$ and $S_{2}$, respectively. Find the value of $\frac{S_{1}}{S_{2}}$. | 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, $\left(x_{1}+2 a^{2}, y_{1}\right)=\lambda\left(4 a^{2}, 2 a\right)$.
Hence, $D\left((4 \lambda-2) a^{2}, 2 \lambda a\right)$.
Let $E\left(x_{2}, y_{2}\right)$. From $\overrightarrow{C E}=\lambda \overrightarrow{C A}$, we get $\left(x_{2}-2 a^{2}, y_{2}+2 a\right)=\lambda\left(-4 a^{2}, 2 a\right)$.
Hence, $E\left((2-4 \lambda) a^{2},(2 \lambda-2) a\right)$.
When $\lambda=\frac{1}{2}$, the line $D E$ is the $y$-axis, and the conclusion is obviously true.
When $\lambda \neq \frac{1}{2}$, the slope of line $D E$ exists, and
$$
k_{D E}=\frac{2 a}{(8 \lambda-4) a^{2}}=\frac{1}{(4 \lambda-2) a} \text {. }
$$
Thus, $l_{D E}: y-2 \lambda a=\frac{1}{(4 \lambda-2) a}\left(x-(4 \lambda-2) a^{2}\right)$
$$
\Rightarrow x=2 a(2 \lambda-1) y-2 a^{2}(2 \lambda-1)^{2} \text {. }
$$
Combining with the parabola equation, we get
$$
\begin{array}{l}
y^{2}-4 a(2 \lambda-1) y+4 a^{2}(2 \lambda-1)^{2}=0 \\
\Rightarrow y=2 a(2 \lambda-1) .
\end{array}
$$
Substituting into equation (1) gives $x=2 a^{2}(2 \lambda-1)^{2}$.
Therefore, the line $D E$ intersects the parabola at exactly one point $F\left(2 a^{2}(2 \lambda-1)^{2}, 2 a(2 \lambda-1)\right)$.
(2) Note that,
$$
S_{1}=\frac{1}{2} \cdot 4 a\left(2 a^{2}-x_{F}\right)=4 a^{3}\left(4 \lambda-4 \lambda^{2}\right) .
$$
Let the line $D E$ intersect the $x$-axis at point $G$. Setting $y=0$, substituting into equation (1) gives
$$
\begin{array}{l}
x=-2 a^{2}(2 \lambda-1)^{2} . \\
\text { Then }|A G|=2 a^{2}-2 a^{2}(2 \lambda-1)^{2} \\
=2 a^{2}\left(4 \lambda-4 \lambda^{2}\right) . \\
\text { Hence } S_{2}=\frac{1}{2}|A G|\left|y_{D}-y_{E}\right| \\
=a^{2}\left(4 \lambda-4 \lambda^{2}\right)|2 \lambda a-(2 \lambda-2) a| \\
=2 a^{3}\left(4 \lambda-4 \lambda^{2}\right) .
\end{array}
$$
Therefore, $\frac{S_{1}}{S_{2}}=2$. | 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 no isolated vertices.
Perform the following operation on graph $G_{0}$: For an edge $u v \in E_{0}$ in graph $G_{0}$, if the degrees of both $u$ and $v$ in $G_{0}$ are at least 2, delete this edge, and the resulting graph still has no isolated vertices. Continue this operation until a graph $G_{1}=\left(V, E_{1}\right)$ is obtained, which has no isolated vertices, and for any edge $u v \in E_{1}$, at least one of the vertices $u$ or $v$ has a degree of 1 in $G_{1}$.
First, consider a connected component of graph $G_{1}$, with $n$ vertices.
When $n=2$, it is $K_{1,1}$.
Assume $n \geqslant 3$. Let $u v$ be one of the edges, and the degree of $u$ is at least 2. Suppose $u$ is adjacent to $v_{1}, v_{2}, \cdots, v_{k}$. Then, $v_{1}, v_{2}, \cdots, v_{k}$ have a degree of 1 in $G_{1}$, and this connected component is a star graph $K_{1, k}(k=n-1)$. Combining the definition of good edges, we know that $v_{1}, v_{2}, \cdots, v_{n-1}$ are not adjacent in $G$. Thus, $G_{1}$ is the union of several star graphs.
Let $G_{1}$ have $m$ connected components, with $n_{1}, n_{2}, \cdots, n_{m}$ vertices respectively. Then, $n_{i} \geqslant 2, \sum_{i=1}^{m} n_{i}=100$.
Next, consider the number of edges between two connected components $K_{1, n_{i}-1}$ and $K_{1, n_{j}-1}$ in $G$.
Let the center of the first star graph be $u$, and the pendant vertices be $u_{1}, u_{2}, \cdots, u_{n_{i}-1}$; the center of the second star graph be $v$, and the pendant vertices be $v_{1}, v_{2}, \cdots, v_{n_{i}-1}$. The following is considered in $G$.
If $u v \in E$, then $u_{1}, u_{2}, \cdots, u_{n_{i}-1}$ are not adjacent to $v$, and $v_{1}, v_{2}, \cdots, v_{n_{j}-1}$ are not adjacent to $u$. Therefore, the number of edges between these two connected components is at most $\left(n_{i}-1\right)\left(n_{j}-1\right)+1$. If $u$ and $v$ are not adjacent, suppose $u$ is adjacent to $s$ vertices among $v_{1}, v_{2}, \cdots, v_{n_{j}-1}$, denoted as $v_{1}, v_{2}, \cdots, v_{s}$. Suppose $v$ is adjacent to $t$ vertices among $u_{1}, u_{2}, \cdots, u_{n_{i}-1}$, denoted as $u_{1}, u_{2}, \cdots, u_{t}$. Then, $v_{1}, v_{2}, \cdots, v_{s}$ are not adjacent to any $u_{k}\left(1 \leqslant k \leqslant n_{i}-1\right)$, and $u_{1}, u_{2}, \cdots, u_{t}$ are not adjacent to any $v_{k}(1 \leqslant k \leqslant n_{j}-1)$. Therefore, the number of edges between these two connected components is at most
$$
\begin{array}{l}
\left(n_{i}-t-1\right)\left(n_{j}-s-1\right)+t+s \\
\leqslant\left(n_{i}-1\right)\left(n_{j}-1\right)+1 .
\end{array}
$$
Then $|E|$
$$
\begin{array}{l}
\leqslant \sum_{i=1}^{m}\left(n_{i}-1\right)+\sum_{1 \leqslant i1200 .
\end{array}
$$
When $17 \leqslant m \leqslant 50$,
Thus $f(m)<3800$
For $1 \leqslant m \leqslant 16$, calculate sequentially, we know
$m \neq 8$ when, $f(m) \leqslant 3822 ; f(8)=3823$.
If $|E|=f(8)=3823$, the equality holds when $n_{1}=n_{2}=\cdots=n_{8}$, but 100 is not divisible by 8, so the equality does not hold.
Therefore, $|E| \leqslant 3822$.
Finally, construct a 100-vertex graph $G$: First, take eight disjoint star graphs, four of which are $K_{1,11}$, and the other four are $K_{1,12}$. Connect the center points of any two star graphs, and connect each non-center point of the first star graph to each non-center point of the second star graph. Then
$$
\begin{array}{l}
n_{1}=n_{2}=n_{3}=n_{4}=12, \\
n_{5}=n_{6}=n_{7}=n_{8}=13, \\
|E|=\sum_{i=1}^{8}\left(n_{i}-1\right)+\sum_{1 \leqslant i<j \in 8}\left(\left(n_{i}-1\right)\left(n_{j}-1\right)+1\right) \\
=3822 .
\end{array}
$$
Note that the edges in the initial star graphs are all good edges. Therefore, each vertex has a good edge, and graph $G$ meets the requirements.
In summary, the maximum number of edges in graph $G$ is 3822. | 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 $S_{i}$ and $S_{j}$ intersect, then there exist $x, y \in M_{i}, u, v \in M_{j}, x>y, u>v$, satisfying
$$
x-y=u-v \Rightarrow x+v=y+u.
$$
Thus, in $A_{1}, A_{2}, \cdots, A_{n}$, there are four sets
$$
\left\{x, x+d_{i}\right\}, \left\{y, y+d_{i}\right\}, \left\{u, u+d_{j}\right\}, \left\{v, v+d_{j}\right\},
$$
satisfying
$$
\left\{x, x+d_{i}\right\}+\left\{v, v+d_{j}\right\}=\left\{y, y+d_{i}\right\}+\left\{u, u+d_{j}\right\},
$$
which contradicts the condition. Hence, $S_{1}, S_{2}, \cdots, S_{m}$ are mutually disjoint.
Since the elements in set $M_{i}$ are all no more than 2017, and the elements in set $S_{i}$ are all positive integers no more than 2016, then
$$
\sum_{i=1}^{m}\left|S_{i}\right| \leqslant 2016.
$$
Let $M_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n_{i}}\right\}\left(x_{1}<x_{2}<\cdots<x_{n_{i}}\right)$. Then set $S_{i}$ contains at least $n_{i}-1$ distinct numbers
$$
x_{2}-x_{1}<x_{3}-x_{1}<\cdots<x_{n_{i}}-x_{1}.
$$
Thus, $\left|S_{i}\right| \geqslant n_{i}-1(1 \leqslant i \leqslant m)$.
Combining (1) and (2), and noting that $d\left(A_{i}\right) \leqslant 2017$, we have $m \leqslant 2017$.
Therefore, $2016 \geqslant \sum_{i=1}^{m}\left|S_{i}\right| \geqslant \sum_{i=1}^{m}\left(n_{i}-1\right)=n-m$ $\Rightarrow n \leqslant m+2016 \leqslant 2017+2016=4033$.
For $1 \leqslant i \leqslant 2017$, let $A_{i}=\{1, i+1\}$; for $2018 \leqslant j \leqslant 4033$, let $A_{j}=\{j-2016,2018\}$.
We now verify that the 4033 sets $A_{1}, A_{2}, \cdots, A_{4033}$ satisfy the requirements.
Note that, the sets with distances $1,2, \cdots, 2016$ each have 2 sets, and the set with distance 2017 has 1 set,
$$
\begin{array}{l}
M_{1}=\{1,2017\}, M_{2}=\{1,2016\}, \cdots, \\
M_{2016}=\{1,2\}, M_{2017}=\{1\}.
\end{array}
$$
For a finite set $A \subset \mathbf{Z}$, translating $A$ so that the minimum element is 0, i.e., $A-\min A$, is called the "standardization" of $A$.
For two binary sets $A, B$ with distance $d$, the standardization of $A+B$ is
$$
\{0, d\}+\{0, d\}=\{0, d, 2 d\},
$$
i.e., $A+B$ is a ternary set and is an arithmetic sequence with common difference $d$;
For two binary sets $A, B$ with distances $d_{1}, d_{2}$, $d_{1} \neq d_{2}$, the standardization of $A+B$ is
$$
\left\{0, d_{1}\right\}+\left\{0, d_{2}\right\}=\left\{0, d_{1}, d_{2}, d_{1}+d_{2}\right\},
$$
$A+B$ is a quaternary set, and the standardization of $A+B$ is uniquely determined by $d_{1}, d_{2}$.
Therefore, to verify that $A_{i}+A_{j}(1 \leqslant i \leqslant j \leqslant 4033)$ are all distinct, it suffices to verify for each $1 \leqslant i \leqslant 2017$ that $x+y$ $\left(x, y \in M_{i}, x \leqslant y\right)$ are all distinct, and for each $1 \leqslant i < j \leqslant 2017$, that $x+y\left(x \in M_{i}, y \in M_{j}\right)$ are all distinct. Direct verification shows that this is true. | 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 gives $11-a$ possible values. When both $a$ and $b$ are determined, the value of $c$ is uniquely determined. Therefore, the number of three-digit numbers is $\sum_{a=1}^{9}(11-a)=54$.
For four-digit numbers $\overline{1 a b c}$ where $a+b+c=9$, the number of non-negative integer solutions $(a, b, c)$ is $\mathrm{C}_{11}^{2}=55$; there are two four-digit numbers starting with 2, which are 2008 and 2017.
Thus, in $\{1,2, \cdots, 2017\}$, the number of numbers that satisfy the condition is $9+54+55+2=120$. | 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 from the set $\{1,2,3,5,6,7\}$, making a total of 6 cases;
(3) When $a=b=3$, $c$ has 4 cases;
(4) When $a=b=2$, $c$ has 2 cases.
In summary, there are $3 \times(40+6+4+2)+9=165$ cases in total. | 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
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=-\frac{b}{a}, x_{1} x_{2} x_{3}=-\frac{d}{a} . \\
\text { Therefore } \frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}=\frac{x_{1}+x_{2}+x_{3}}{x_{1} x_{2} x_{3}} \\
=\frac{b}{d}=1996
\end{array}
$$ | 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
$$
\begin{array}{l}
x>0, y>0, u=x-y>0 . \\
\text { Then }(y+u)^{2}-2 y^{2}-8=0 \\
\Rightarrow y^{2}-2 u y-u^{2}+8=0 \\
\Rightarrow \Delta=(2 u)^{2}-4\left(-u^{2}+8\right) \geqslant 0 \\
\Rightarrow u \geqslant 2 .
\end{array}
$$
When $x=4, y=2$, the equality holds.
Therefore, the minimum value of $|x-y|$ is 2. | 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,
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2} \geqslant \frac{x_{2}^{2}}{2}, \\
x_{3}^{2}+\left(x_{2}+x_{3}\right)^{2} \geqslant \frac{x_{2}^{2}}{2} .
\end{array}
$$
Therefore, $x_{2}^{2} \leqslant 4$, which means $\left|x_{2}\right| \leqslant 2$.
When $x_{1}=x_{3}=-1, x_{2}=2$, we have $\left|x_{2}\right|_{\text {max }}=2$. | 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 $9 \mid\left(x^{3}+y^{3}+z^{3}\right)$, it is necessary that $x, y, z$ are all congruent to 0 modulo 3, or $x, y, z$ are pairwise incongruent modulo 3.
Therefore, the number of elements in set $A$ is
$$
3^{3}+A_{3}^{3} \times 3 \times 3 \times 4=243 \text {. }
$$ | 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, $a_{n}-a_{n-1}$
$$
\begin{array}{l}
=\sum_{k=1}^{n}\left[\frac{n}{k}\right]-\sum_{k=1}^{n-1}\left[\frac{n-1}{k}\right] \\
=\sum_{k=1}^{n}\left[\frac{n}{k}\right]-\sum_{k=1}^{n}\left[\frac{n-1}{k}\right] \\
=\sum_{k=1}^{n}\left(\left[\frac{n}{k}\right]-\left[\frac{n-1}{k}\right]\right)
\end{array}
$$
is exactly the number of positive divisors of $n$.
Note that the number of positive divisors of a positive integer $n$ is odd if and only if $n$ is a perfect square. Therefore, $a_{n}$ and $a_{n-1}$ have different parities if and only if $n$ is a perfect square. Since $a_{1}=1$ is odd, $a_{2}$ and $a_{3}$ are both odd, $a_{4}$ is even, $a_{5}$, $a_{6}$, $a_{7}$, and $a_{8}$ are all even, and so on.
Since $2018 \in (44^2, 45^2)$, the number of even numbers among $a_{1}, a_{2}, \cdots, a_{2018}$ is
$$
\sum_{k=2}^{43}(-1)^{k-1} k^{2}+2018-44^{2}+1=1028
$$ | 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 $3^{5}=243$, the 96th term is $729+243=972$, and the 100th term is $972+9=981$. | 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
$$
4^{2}=b^{2}+(8-b)^{2}-2 b(8-b) \frac{3}{5}
$$
$\Rightarrow b=3$ or 5, corresponding to $c=5$ or 3.
Therefore, $S_{\triangle A B C}=\frac{1}{2} b c \sin A$
$$
=\frac{1}{2} \times 3 \times 5 \times \frac{4}{5}=6 \text {. }
$$ | 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,
\[
a^{2}+(b-4)^{2} \geqslant x^{2}+1 \geqslant 2 \text {, }
\]
When $x=1, a=-1, b=3$, $a^{2}+(b-4)^{2}$ achieves its minimum value of 2. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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