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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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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}
$$
Since $b$ is an integer, we have $a+1= \pm 1$.
Also, $a \neq 0$, so $a=-2, b=4, c=16$. Therefore, $a+b+c=18$. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Arrange all positive integers that are coprime with 70 in ascending order. The 2017th term of this sequence is $\qquad$ . | 10. 5881.
It is easy to know that the number of positive integers not exceeding 70 and coprime with 70 is
$$
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}
$$ | 5881 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. In two regular tetrahedrons $A-OBC$ and $D-OBC$ with their bases coinciding, $M$ and $N$ are the centroids of $\triangle ADC$ and $\triangle BDC$ respectively. Let $\overrightarrow{OA}=\boldsymbol{a}, \overrightarrow{OB}=\boldsymbol{b}, \overrightarrow{OC}=\boldsymbol{c}$. If point $P$ satisfies $\overrightarrow{OP}=x \boldsymbol{a}+y \boldsymbol{b}+z \boldsymbol{c}, \overrightarrow{MP}=2 \overrightarrow{PN}$, then the real number $9 x+81 y+729 z=$ $\qquad$
(Adapted from the 2016 National High School Mathematics League Zhejiang Regional Preliminary Contest) | Take $O$ as the origin and the line $O B$ as the $x$-axis, establishing a spatial rectangular coordinate system as shown in Figure 13.
Let $B(1,0,0)$. Then
$$
\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 {. }
$$ | 439 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n-1}}$. Then $\sum_{n=1}^{2018} a_{n}=$ $\qquad$ . | 5. 5857 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$; when $n \equiv 2,4,8(\bmod 10)$, $(n+1)^{n+2} \equiv 1(\bmod 4)$, then $a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10)$; when $n \equiv 3,7,9(\bmod 10)$, $(n+1)^{n+2} \equiv 0(\bmod 4)$, then $a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k} \equiv 1(\bmod 10)$. Thus, $a_{n}$ is a periodic sequence, with the smallest positive period being 10.
$$
\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 .
$$ | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Let $S$ be a set of positive integers with the property: for any $x \in S$, the arithmetic mean of the remaining numbers in $S$ after removing $x$ is a positive integer, and it satisfies $1 \in S$, 2016 is the largest element in $S$. Find the maximum value of $|S|$. | Let the elements of set $S$ be
$$
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 |
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 |
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 |
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 |
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 |
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 |
4. Choose any two numbers from $2, 4, 6, 7, 8, 11, 12, 13$ to form a fraction. Then, there are $\qquad$ irreducible fractions among these fractions. | 4. 36 .
Among 7, 11, 13, choose one number and among $2, 4, 6, 8, 12$, choose one number to form a reduced fraction, there are $2 \mathrm{C}_{3}^{1} \mathrm{C}_{5}^{1}$ $=30$ kinds; among $7, 11, 13$, choose two numbers to form a reduced fraction, there are $\mathrm{A}_{3}^{2}=6$ kinds.
There are a total of 36 different reduced fractions. | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 1, let $P\left(x_{p}, y_{p}\right)$ be a point on the graph of the inverse proportion function $y=\frac{2}{x}$ in the first quadrant of the Cartesian coordinate system $x O y$. Draw lines parallel to the $x$-axis and $y$-axis through point $P$, intersecting the graph of $y=\frac{10}{x}$ in the first quadrant at points $A$ and $B$, respectively. Then the area of $\triangle A O B$ is ( ).
(A) 26
(B) 24
(C) 22
(D) 20 | 3. B.
Connect $O P$.
Then $S_{\triangle A O B}=S_{\triangle A O P}+S_{\triangle P O B}+S_{\triangle A P B}$
$$
\begin{aligned}
= & \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right) y_{p}+\frac{1}{2}\left(\frac{10}{x_{p}}-y_{p}\right) x_{p}+ \\
& \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right)\left(\frac{10}{x_{p}}-y_{p}\right) \\
= & \frac{1}{2} \cdot \frac{100}{x_{p} y_{p}}-\frac{1}{2} x_{p} y_{p} \\
= & 24 .
\end{aligned}
$$ | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. Arrange natural numbers whose digits sum to 11 in ascending order to form a sequence. The $m$-th number is 2018. Then $m$ is ( ).
(A) 134
(B) 143
(C) 341
(D) 413 | 5. A.
Among single-digit numbers, there are no numbers whose digit sum is 11.
Among two-digit numbers, there are 8 numbers: $29, 38, 47, 56, 65, 74, 83, 92$.
For three-digit numbers $\overline{x y z}$, when $x=1$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of 9 numbers; when $x=2$, $y$ can take 10 numbers: $0, 1, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 0$, a total of 10 numbers; similarly, when $x=3, 4, \cdots, 9$, the numbers whose digit sum is 11 are $12-x$ each. Therefore, in three-digit numbers $\overline{x y z}$, the numbers whose digit sum is 11 total $9+10+9+\cdots+3=61$.
For four-digit numbers $\overline{1 x y z}$, the numbers whose digit sum is 11 are equivalent to finding three-digit numbers $\overline{x y z}$ whose digit sum is 10. When $x=0$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of 9 numbers. Similarly, when $x=1, 2, \cdots, 9$, the three-digit numbers $\overline{x y z}$ whose digit sum is 10 have $11-x$ each. Therefore, in four-digit numbers $\overline{1 x y z}$, the numbers whose digit sum is 11 total $9+10+9+\cdots+2=63$.
In $\overline{2 x y z}$, the numbers whose digit sum is 11, from smallest to largest, are $2009, 2018$, a total of 2 numbers.
Therefore, 2018 is the
$$
8+61+63+2=134
$$
number in the sequence. | 134 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
4. Given that $x_{1}, x_{2}, \cdots, x_{n}$ where $x_{i}(i=1,2, \cdots, n)$ can only take one of the values $-2, 0, 1$, and satisfy
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=-17, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=37 .
\end{array}
$$
Then $\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)^{2}$ is $\qquad$ | 4. 5041.
Let $x_{1}, x_{2}, \cdots, x_{n}$ have $p$ values of $x_{i}$ equal to 1, $q$ values of $x_{i}$ equal to -2, and the rest of the $x_{i}$ equal to 0. We can obtain
$$
\left\{\begin{array} { l }
{ p - 2 q = - 1 7 } \\
{ p + 4 q = 3 7 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=1, \\
q=9 .
\end{array}\right.\right.
$$
Thus, among $x_{i}(i=1,2, \cdots, n)$, one takes the value 1, nine take the value -2, and the rest take the value 0.
$$
\begin{array}{l}
\text { Therefore, }\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)^{2} \\
=\left(1 \times 1^{3}+9 \times(-2)^{3}\right)^{2} \\
=5041 .
\end{array}
$$ | 5041 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Among the $n$ positive integers from 1 to $n$, those with the most positive divisors are called the "prosperous numbers" among these $n$ positive integers. For example, among the positive integers from 1 to 20, the numbers with the most positive divisors are $12, 18, 20$, so $12, 18, 20$ are all prosperous numbers among the positive integers from 1 to 20. Then, the least common multiple of all the prosperous numbers among the positive integers from 1 to 100 is $\qquad$ | 5. 10080.
First, in the prime factorization of the first 100 positive integers, the maximum number of different prime factors is three. This is because the product of the smallest four primes is $2 \times 3 \times 5 \times 7=210$, which exceeds 100.
Second, to maximize the number of divisors, the prime factors should be as small as possible. The number with the maximum number of positive divisors can be determined through trial and classification enumeration, and the number with the maximum number of positive divisors is 12:
For numbers with only one prime factor, the one with the most positive divisors is $2^{6}=64$, which has seven positive divisors, less than 12;
For numbers with two different prime factors, those with 12 positive divisors are $2^{3} \times 3^{2}$ and $2^{5} \times 3$;
For numbers with three different prime factors, those with 12 positive divisors are
$$
2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5, 2^{2} \times 3 \times 7 \text{. }
$$
It can be seen that among the first 100 positive integers, the numbers with 12 positive divisors are only five, and their least common multiple is
$$
2^{5} \times 3^{2} \times 5 \times 7=10080 \text{. }
$$ | 10080 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Divide the set of positive even numbers $\{2,4, \cdots\}$ into groups in ascending order, with the $n$-th group containing $3 n-2$ numbers:
$$
\{2\},\{4,6,8,10\},\{12,14, \cdots, 24\}, \cdots \text {. }
$$
Then 2018 is in the group. | - 1. 27 .
Let 2018 be in the $n$-th group. Since 2018 is the 1009th positive even number and according to the problem, we have
$$
\begin{array}{l}
\sum_{i=1}^{n-1}(3 i-2)<1009 \leqslant \sum_{i=1}^{n}(3 i-2) \\
\Rightarrow \frac{3(n-1)^{2}-(n-1)}{2}<1009 \leqslant \frac{3 n^{2}-n}{2} \\
\Rightarrow n=27 .
\end{array}
$$
Therefore, 2018 is located in the 27th group. | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $2 S_{n}-n a_{n}=n\left(n \in \mathbf{Z}_{+}\right)$, and $a_{2}=3$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{a_{n} \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_{n}}}$, and $T_{n}$ be the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$. Find the smallest positive integer $n$ such that $T_{n}>\frac{9}{20}$. | (1) From $2 S_{n}-n a_{n}=n$, we get
$$
2 S_{n+1}-(n+1) a_{n+1}=n+1 \text {. }
$$
Subtracting the above two equations yields
$$
2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1 \text {. }
$$
Thus, $n a_{n}-(n-1) a_{n+1}=1$,
$$
(n+1) a_{n+1}-n a_{n+2}=1 \text {. }
$$
Subtracting (2) from (1) and rearranging gives
$$
a_{n}+a_{n+2}=2 a_{n+1} \text {. }
$$
Hence, $\left\{a_{n}\right\}$ is an arithmetic sequence.
From $2 S_{1}-a_{1}=1$ and $a_{2}=3$, we get $a_{1}=1$.
Thus, the common difference of the sequence $\left\{a_{n}\right\}$ is $d=2$.
Therefore, $a_{n}=1+2(n-1)=2 n-1$.
(2) From (1), we know
$$
\begin{array}{l}
b_{n}=\frac{1}{(2 n-1) \sqrt{2 n+1}+(2 n+1) \sqrt{2 n-1}} \\
=\frac{1}{2} \times \frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{\sqrt{2 n-1} \sqrt{2 n+1}} \\
=\frac{1}{2}\left(\frac{1}{\sqrt{2 n-1}}-\frac{1}{\sqrt{2 n+1}}\right) .
\end{array}
$$
Thus, $T_{n}=\frac{1}{2} \sum_{k=1}^{n}\left(\frac{1}{\sqrt{2 k-1}}-\frac{1}{\sqrt{2 k+1}}\right)$
$$
\begin{array}{l}
=\frac{1}{2}\left(1-\frac{1}{\sqrt{2 n+1}}\right)>\frac{9}{20} \\
\Rightarrow n \geqslant \frac{99}{2} .
\end{array}
$$
Therefore, the smallest positive integer $n$ that satisfies the condition is $n=50$. | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Let $M$ be a set composed of a finite number of positive integers
$$
\begin{array}{l}
\text { such that, } M=\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}, \\
A_{i} \neq \varnothing, B_{i} \neq \varnothing(i=1,2, \cdots, 20),
\end{array}
$$
and satisfies:
(1) For any $1 \leqslant i<j \leqslant 20$,
$$
A_{i} \cap A_{j}=\varnothing, B_{i} \cap B_{j}=\varnothing \text {; }
$$
(2) For any $1 \leqslant i \leqslant 20,1 \leqslant j \leqslant 20$, if $A_{i} \cap B_{j}=\varnothing$, then $\left|A_{i} \cup B_{j}\right| \geqslant 18$.
Find the minimum number of elements in the set $M$ ( $|X|$ denotes the number of elements in the set $X$). | 15. Let $\min _{1 \leqslant i \leqslant 20}\left\{\left|A_{i}\right|,\left|B_{i}\right|\right\}=t$.
Assume $\left|A_{1}\right|=t$,
$$
\begin{array}{l}
A_{1} \cap B_{i} \neq \varnothing(i=1,2, \cdots, k) ; \\
A_{1} \cap B_{j}=\varnothing(j=k+1, k+2, \cdots, 20) .
\end{array}
$$
Let $a_{i} \in A_{1} \cap B_{i}(i=1,2, \cdots, k)$.
By property (1), we know that $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, $\left|A_{1}\right| \geqslant k$, i.e., $t \geqslant k$.
By property (2), we know that when $j=k+1, k+2, \cdots, 20$, $\left|A_{1}\right|+\left|B_{j}\right|=\left|A_{1} \cup B_{j}\right| \geqslant 18$, i.e., $\left|B_{j}\right| \geqslant 18-t$.
$$
\begin{array}{l}
\text { Then }|M|=\left|\bigcup_{i=1}^{20} B_{i}\right|=\sum_{j=1}^{20}\left|B_{j}\right| \\
\geqslant k t+(20-k)(18-t) \\
=360+2 k t-18 k-20 t \\
=180+2(k-10)(t-9) .
\end{array}
$$
If $t \leqslant 9$, then
$k \leqslant t \leqslant 9$,
$$
|M|=180+2(k-10)(t-9) \geqslant 180 \text {; }
$$
If $t \geqslant 10$, then $|M| \geqslant 20 t \geqslant 200$.
Therefore, $|M| \geqslant 180$ always holds.
Take $A_{i}=B_{i}=\{9(i-1)+j \mid j=1,2, \cdots, 9\}$, where $i=1,2, \cdots, 20$.
Then $M=\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}=\{1,2, \cdots, 180\}$ satisfies the requirement.
In this case, $|M|=180$.
In summary, the minimum number of elements in the set $M$ is
180. | 180 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In an $8 \times 8$ chessboard, how many ways are there to select 56 squares such that: all the black squares are selected, and each row and each column has exactly seven squares selected? ? ${ }^{[1]}$
(2014, Irish Mathematical Olympiad) | The problem is equivalent to selecting eight white squares on the chessboard, with exactly one square selected from each row and each column.
The white squares on the chessboard are formed by the intersections of rows $1, 3, 5, 7$ and columns $1, 3, 5, 7$, resulting in a $4 \times 4$ submatrix, as well as the intersections of rows $2, 4, 6, 8$ and columns $2, 4, 6, 8$, resulting in another $4 \times 4$ submatrix. Since there are $4!$ ways to choose four white squares from each submatrix such that they are in different rows and columns, the total number of ways to make the selection is $(4!)^{2}=576$. | 576 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For the four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c$ 、 $d \leqslant 9)$ : if $a>b, bd$, then $\overline{a b c d}$ is called a $P$ class number; if $ac, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbers and $Q$ class numbers, respectively. Then the value of $N(P)-N(Q)$ is $\qquad$ [3] (2015, National High School Mathematics Joint Competition) | Let the set of all numbers of type $P$ and type $Q$ be denoted as $A$ and $B$, respectively. Further, let the set of all numbers of type $P$ that end in zero be denoted as $A_{0}$, and the set of all numbers of type $P$ that do not end in zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \in A_{1}$, map it to the four-digit number $\overline{d c b a}$.
Note that, $a > b$, and $bd \geqslant 1$.
Thus, $\overline{d c b a} \in B$.
Conversely, each $\overline{d c b a} \in B$ uniquely corresponds to an element $\overline{a b c d} \in A_{1}$.
Therefore, a one-to-one correspondence is established between $A_{1}$ and $B$.
Hence, $N(P) - N(Q) = |A| - |B|$
$$
= \left|A_{0}\right| + \left|A_{1}\right| - |B| = \left|A_{0}\right| \text{.}
$$
For any four-digit number $\overline{a b c 0} \in A_{0}$, $b$ can take values 0, 1, ..., 9. For each $b$, since $b < a \leqslant 9$ and $b < c \leqslant 9$, $a$ and $c$ each have $9 - b$ possible values. Therefore,
$$
\left|A_{0}\right| = \sum_{b=0}^{9} (9 - b)^{2} = \sum_{k=1}^{9} k^{2} = \frac{9 \times 10 \times 19}{6} = 285 \text{.}
$$
Thus, $N(P) - N(Q) = 285$. | 285 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers in the set $\{1,3, \cdots, 2013\}$. ${ }^{[4]}$
(2013, China Southeast Mathematical Olympiad) | First, point out two obvious conclusions:
(1) If $m \in \mathbf{Z}_{+}, x \in \mathbf{R}$, then $\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right]$;
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$
In (1), let $m=k!(k=1,2, \cdots, 2013)$ and
sum to get
$$
f(x)=\sum_{k=1}^{2013}\left[\frac{x}{k!}\right]=\sum_{k=1}^{2013}\left[\frac{[x]}{k!}\right]=f([x]) .
$$
This shows that the equation $f(x)=n$ has a real solution if and only if it has an integer solution. Therefore, we only need to consider the case where $x$ is an integer.
$$
\begin{array}{l}
\text { By } f(x+1)-f(x) \\
=[x+1]-[x]+\sum_{k=2}^{2013}\left(\left[\frac{x+1}{k!}\right]-\left[\frac{x}{k!}\right]\right) \\
\quad \geqslant 1,
\end{array}
$$
we know that $f(x)(x \in \mathbf{Z})$ is strictly increasing.
Next, find integers $a$ and $b$ such that
$$
\begin{array}{l}
f(a-1)2013,
\end{array}
$$
so $b=1173$.
Thus, the good numbers in $\{1,3, \cdots, 2013\}$ are the odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$.
In (1), let $x=2 l(l=0,1, \cdots, 586)$.
By conclusion (2), we have
$$
\left[\frac{2 l+1}{k!}\right]=\left[\frac{2 l}{k!}\right](2 \leqslant k \leqslant 2013) \text {. }
$$
Then $f(2 l+1)-f(2 l)$
$$
=1+\sum_{k=2}^{2013}\left(\left[\frac{2 l+1}{k!}\right]-\left[\frac{2 l}{k!}\right]\right)=1 \text {. }
$$
This shows that exactly one of $f(2 l)$ and $f(2 l+1)$ is odd. Therefore, there are $\frac{1174}{2}=587$ odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$, i.e., there are 587 good numbers in the set $\{1,3, \cdots, 2013\}$. | 587 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Equation
$$
x^{2}-31 x+220=2^{x}\left(31-2 x-2^{x}\right)
$$
The sum of the squares of all real roots is $\qquad$ . | 2. 25 .
Let $y=x+2^{x}$. Then the original equation is equivalent to
$$
\begin{array}{l}
y^{2}-31 y+220=0 \\
\Rightarrow y_{1}=11, y_{2}=20 \\
\Rightarrow x_{1}+2^{x_{1}}=11 \text { and } x_{2}+2^{x_{2}}=20 .
\end{array}
$$
Since $f(x)=x+2^{x}$ is a monotonically increasing function, each equation has at most one real root, which are $3$ and $4$, respectively. Therefore, the sum of the squares of all real roots of the original equation is 25. | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In the plane, there are 200 points, no three of which are collinear, and each point is labeled with one of the numbers $1, 2, 3$. All pairs of points labeled with different numbers are connected by line segments, and each line segment is labeled with a number 1, 2, or 3, which is different from the numbers at its endpoints. As a result, each of the numbers 1, 2, or 3 written on the plane appears exactly $n$ times. Then the value of $n$ is $\qquad$ | 5.199.
Let the points labeled with $1, 2, 3$ be $a, b, c$ respectively.
Thus, $a+b+c=200$, and the number of line segments labeled with $1, 2, 3$ are $bc, ca, ab$ respectively.
Then $n=a+bc=b+ca=c+ab$.
Therefore, $(a+bc)-(b+ca)=(a-b)(1-c)=0$.
Similarly, $(b-c)(1-a)=(c-a)(1-b)=0$.
If at least two of $a, b, c$ are not 1, then $a=b=c$, which contradicts $3+200$, and it is also impossible that $a=b=c=1$.
Thus, two of $a, b, c$ are 1, and the third one is 198. In this case, the equation holds, and $n=199$. | 199 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The sequence $\left\{a_{n}\right\}$ has nine terms, $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. The number of such sequences is $\qquad$
(2013, National High School Mathematics League Competition) | Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leqslant i \leqslant 8)$, mapping each sequence $\left\{a_{n}\right\}$ that meets the conditions to a unique eight-term sequence $\left\{b_{n}\right\}$, where $\prod_{i=1}^{8} b_{i}=\frac{a_{9}}{a_{1}}=1$, and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.
From the sign, we know that $b_{i}(1 \leqslant i \leqslant 8)$ contains an even number of $-\frac{1}{2}$. Suppose there are $2 k$ terms of $-\frac{1}{2}$, then there are $2 k$ terms of 2, and the remaining $8-4 k$ terms are 1, where $k$ can be $0, 1, 2$.
Therefore, the number of ways to choose $\left\{b_{n}\right\}$ is
$$
\sum_{k=0}^{2} \mathrm{C}_{8}^{2 k} \mathrm{C}_{8-2 k}^{2 k}=491,
$$
The number of ways to choose $\left\{a_{n}\right\}$ is also 491. | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n$ be a three-digit positive integer without the digit 0. If the digits of $n$ in the units, tens, and hundreds places are permuted arbitrarily, the resulting three-digit number is never a multiple of 4. Find the number of such $n$.
(54th Ukrainian Mathematical Olympiad) | Hint: Classify by the number of even digits (i.e., $2,4,6,8$) appearing in the three-digit code of $n$. It is known from the discussion that the number of $n$ satisfying the condition is $125+150+0+8=283$.
| 283 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $x \in\left(0, \frac{\pi}{2}\right)$. Then the minimum value of the function $y=\frac{1}{\sin ^{2} x}+\frac{12 \sqrt{3}}{\cos x}$ is $\qquad$ . | 7. 28 .
Notice that,
$$
\begin{array}{l}
y=16\left(\frac{1}{16 \sin ^{2} x}+\sin ^{2} x\right)+16\left(\frac{3 \sqrt{3}}{4 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16\left(\frac{3 \sqrt{3}}{8 \cos x}+\frac{3 \sqrt{3}}{8 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16 \times \frac{9}{4}-16=28 .
\end{array}
$$
When $x=\frac{\pi}{6}$, $y$ achieves its minimum value of 28. | 28 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a, b, c$ in $S$ such that $a+b, b+c, c+a$ are all in the subset $A$.
(2012, China Mathematical Olympiad) | 【Analysis】The condition of the problem is equivalent to being able to find three numbers $x, y, z$ in the subset $A$ such that
$$
\frac{-x+y+z}{2}, \frac{x-y+z}{2}, \frac{x+y-z}{2}
$$
are three distinct positive integers. For this, $x, y, z$ must be distinct, satisfy the triangle inequality, and their sum must be even.
A simple estimation shows that taking all the odd numbers can yield a 1006-element subset, where the sum of any three elements is odd, which is a larger counterexample to the required subset $A$. Further, it is found that adding the element 2 can result in a 1007-element counterexample. Therefore, we conjecture that $k=1008$.
Next, we prove this conjecture.
Here, we use a simple principle in combinatorial extremal problems: "The smallest $k$ for which all subsets $A$ of at least $k$ elements satisfy condition $P$" minus 1 is exactly "the largest $n$ for which there exists an $n$-element subset that does not satisfy condition $P$." Let condition $P$ be: the sum of three different numbers is even and satisfies the triangle inequality.
First, we use the pigeonhole principle to prove: any 1008-element subset $A$ of $S$ always contains three elements that satisfy condition $P$.
Since $|A|>\frac{|S|}{2}$ ensures that some pigeonhole $T$ contains more than $\frac{|T|}{2}$ elements of $A$, thus, the pigeonhole $T$ should satisfy the condition: any more than $\frac{|T|}{2}$ elements in $T$ contain three elements that satisfy condition $P$. Then $\frac{|T|}{2} \geqslant 3$, i.e., $|T| \geqslant 6$.
Next, we prove: from the set $\{3,4, \cdots, 8\}$, any four numbers contain three numbers that satisfy condition $P$.
The simplest and most direct proof is by enumeration.
List all three-element subsets of $\{3,4, \cdots, 8\}$ that satisfy condition $P$, to prove that all four-element subsets contain some three-element subset that satisfies condition $P$, it is only necessary to prove that the complements of all three-element subsets that satisfy condition $P$ contain all possible two-element subsets. Thus, find the complements of all three-element subsets that satisfy condition $P$, and list all two-element subsets of the complements, and then, check that all two-element subsets of the six-element set. Using vertices to represent the six elements and edges to represent the two-element subsets, it is easy to verify this by drawing a graph.
By translation, it can be proven that for any positive integer $n$,
$$
\{2 n+3,2 n+4, \cdots, 2 n+8\}
$$
is also a pigeonhole that satisfies the condition: the translation step is even, ensuring that the parity of the numbers at corresponding positions is the same as in the original pigeonhole, and the condition that the sum is even remains unchanged; all numbers increase, and the corresponding triangle inequality condition weakens.
Divide $S=\{1,2, \cdots, 2012\}$ into $\{1,2\}$, $\{6 n+3,6 n+4, \cdots, 6 n+8\}(n=0,1, \cdots, 334)$.
When $A \subset S$ and $A$ contains at least 1008 elements, $A$ contains at least four numbers in some pigeonhole $\{6 n+3,6 n+4, \cdots, 6 n+8\}$, thus, there are three numbers that satisfy condition $P$, and the required $a, b, c$ can be solved.
Therefore, the smallest $k=1008$. | 1008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
A=\{1,2, \cdots, 99\}, \\
B=\{2 x \mid x \in A\}, \\
C=\{x \mid 2 x \in A\} .
\end{array}
$$
Then the number of elements in $B \cap C$ is $\qquad$ | 1. 24 .
From the conditions, we have
$$
\begin{array}{l}
B \cap C \\
=\{2,4, \cdots, 198\} \cap\left\{\frac{1}{2}, 1, \frac{3}{2}, 2, \cdots, \frac{99}{2}\right\} \\
=\{2,4, \cdots, 48\} .
\end{array}
$$
Therefore, the number of elements in $B \cap C$ is 24 . | 24 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the integer sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy:
$$
a_{10}=3 a_{1}, a_{2}+a_{8}=2 a_{5} \text {, }
$$
and $a_{i+1} \in\left\{1+a_{i}, 2+a_{i}\right\}(i=1,2, \cdots, 9)$. Then the number of such sequences is $\qquad$ | 8. 80 .
$$
\begin{array}{l}
\text { Let } b_{i}=a_{i+1}-a_{i} \in\{1,2\}(i=1,2, \cdots, 9) \text {. } \\
\text { Then } 2 a_{1}=a_{10}-a_{1}=b_{1}+b_{2}+\cdots+b_{9}, \\
b_{2}+b_{3}+b_{4}=a_{5}-a_{2}=a_{8}-a_{5} \\
=b_{5}+b_{6}+b_{7} .
\end{array}
$$
Let $t$ represent the number of terms with value 2 among $b_{2}, b_{3}, b_{4}$. From equation (2), we know that $t$ is also the number of terms with value 2 among $b_{5}, b_{6}, b_{7}$, where $t \in\{0,1,2,3\}$.
Thus, the number of ways to choose $b_{2}, b_{3}, \cdots, b_{7}$ is
$$
\left(\mathrm{C}_{3}^{0}\right)^{2}+\left(\mathrm{C}_{3}^{1}\right)^{2}+\left(\mathrm{C}_{3}^{2}\right)^{2}+\left(\mathrm{C}_{3}^{3}\right)^{2}=20 \text {. }
$$
After fixing $b_{2}, b_{3}, \cdots, b_{7}$, we can arbitrarily assign values to $b_{8}, b_{9}$, which gives $2^{2}=4$ ways.
From equation (1), we know that $b_{1} \in\{1,2\}$ should be chosen such that $b_{1}+b_{2}+\cdots+b_{9}$ is even, and the choice of $b_{1}$ is unique, which also determines the value of the integer $a_{1}$.
Therefore, the sequence $b_{1}, b_{2}, \cdots, b_{9}$ uniquely corresponds to a sequence $a_{1}, a_{2}, \cdots, a_{10}$ that satisfies the conditions.
In summary, the number of sequences that satisfy the conditions is
$$
20 \times 4=80 \text {. }
$$ | 80 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let set $S \subset\{1,2, \cdots, 200\}, S$ such that the difference between any two elements is not 4, 5, or 9. Find the maximum value of $|S|$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note to the translator and should not be included in the translated text. Here is the final translation:
4. Let set $S \subset\{1,2, \cdots, 200\}, S$ such that the difference between any two elements is not 4, 5, or 9. Find the maximum value of $|S|$. | Hint Consider the maximum number of numbers from $S$ that can be contained in any sequence of 13 consecutive numbers. Answer: 64. | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In the expansion of $(\sqrt{3}+i)^{10}$, the sum of all odd terms is $\qquad$ . | 3. 512 .
It is known that in the expansion of $(\sqrt{3}+\mathrm{i})^{10}$, the sum of all odd terms is the real part of the complex number.
$$
\begin{array}{l}
\text { Therefore, }(\sqrt{3}+i)^{10}=\left((-2 i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^{10} \\
=(-2 i)^{10}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{10} \\
=(-1024)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) \\
=512-512 \sqrt{3} \mathrm{i}
\end{array}
$$ | 512 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure 2, two equal circles with a radius of 5 are externally tangent to each other, and both are internally tangent to a larger circle with a radius of 13, with the points of tangency being $A$ and $B$. Let $AB = \frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)\right. = 1)$. Then the value of $m+n$ is ( ).
(A) 21
(B) 29
(C) 58
(D) 69
(E) 93 | 15. D.
As shown in Figure 5, let the centers of the three circles be $X$, $Y$, and $Z$.
Then points $A$ and $B$ are on the extensions of $XY$ and $XZ$, respectively, satisfying
$$
\begin{array}{l}
XY = XZ \\
= 13 - 5 = 8, \\
YZ = 5 + 5 = 10, \\
XA = XB = 13.
\end{array}
$$
Since $YZ \parallel AB \Rightarrow \triangle XYZ \sim \triangle XAB$
$$
\Rightarrow \frac{XA}{AB} = \frac{XY}{YZ} \Rightarrow AB = \frac{XA \cdot YZ}{XY} = \frac{65}{4} \text{.}
$$
Thus, $m = 65$, $n = 4$, and $m + n = 69$. | 69 | Geometry | MCQ | Yes | Yes | cn_contest | false |
7. Let $x, y, z$ be complex numbers, and
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x y+y z+z x, \\
|x+y+z|=21,|x-y|=2 \sqrt{3},|x|=3 \sqrt{3} .
\end{array}
$$
Then $|y|^{2}+|z|^{2}=$ . $\qquad$ | 7. 132.
It is easy to see that the figures corresponding to $x$, $y$, and $z$ on the complex plane form an equilateral triangle, and note that
\[
\begin{array}{l}
|x-y|^{2}+|y-z|^{2}+|z-x|^{2}+|x+y+z|^{2} \\
=3\left(|x|^{2}+|y|^{2}+|z|^{2}\right),
\end{array}
\]
Combining the conditions and $|x-y|=|y-z|=|z-x|$, we can substitute and solve. | 132 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In $1,2, \cdots, 100$ these 100 positive integers, remove 50 so that in the remaining positive integers, any two different $a, b$ have $a \nmid b$. Find the maximum possible value of the sum of all removed positive integers. | Let the remaining numbers be $a_{i}=2^{3} t_{i}$, where $i \in \{1,2, \cdots, 50\}$, $s_{i}$ is a natural number, and $t_{i}$ is an odd number.
Since for any $1 \leqslant i \neq j \leqslant 50$, we have $a_{i} \nmid a_{j}$, it follows that $t_{i} \neq t_{j}$, meaning $t_{1}, t_{2}, \cdots, t_{50}$ are 50 distinct odd numbers.
The numbers $1,2, \cdots, 100$ can be expressed in the form $2^{s} t$, where $t$ has only 50 possibilities, namely $1,3, \cdots, 99$. Therefore,
$$
\left\{t_{1}, t_{2}, \cdots, t_{50}\right\}=\{1,3, \cdots, 99\}.
$$
Without loss of generality, let $t_{i}=2 i-1$.
For convenience, we can change the indices of the remaining numbers.
Let $a_{i}=2^{s} i$, where $i \in \{1,3, \cdots, 99\}$ and $s_{i}$ is a natural number.
Set $S_{0}=\left\{a_{99}, a_{97}, \cdots, a_{35}\right\}$,
$S_{1}=\left\{a_{33}, a_{31}, \cdots, a_{13}\right\}$,
$S_{2}=\left\{a_{11}, \cdots, a_{5}\right\}$, $S_{3}=\left\{a_{3}\right\}$, $S_{4}=\left\{a_{1}\right\}$.
First, we prove that for any $a_{i} \in S_{k}$, we have
$a_{i} \geqslant 2^{k} i$.
We use mathematical induction on $k$.
When $k=0$, $a_{i}=2^{s} t \geqslant 2^{0} i$ holds.
Assume that for $k=m (0 \leqslant m \leqslant 3)$, the conclusion holds.
Then for $k=m+1$, if $a_{i}=2^{s} i \in S_{m+1}$, we have $a_{3 i} \in S_{m}$.
Since $a_{i} \nmid a_{3 i}$, by the induction hypothesis,
$a_{3 i}=2^{s_{3 i}} 3 i (s_{3 i} \geqslant m)$.
Thus, $s_{i} \geqslant s_{3 i}+1 \geqslant m+1$, which means $a_{i} \geqslant 2^{m+1} i$, and the conclusion holds.
By mathematical induction, equation (1) is proven.
Next, we prove that equation (1) holds with equality when the conditions of the problem are satisfied.
If $a_{i}, a_{j} \in S_{k}$, then $a_{i}=2^{k} i, a_{j}=2^{k} j$.
Assume $a_{i} \mid a_{j}$. Then $i \mid j$.
Since $i, j$ are both odd, it follows that $j \geqslant 3 i$.
By $a_{i}, a_{j} \in S_{k}$, we know $3 i > j$, which is a contradiction. Hence, $a_{i} \nmid a_{j}$.
Similarly, $a_{j} \nmid a_{i}$.
If $a_{i} \in S_{k}, a_{j} \in S_{l}$, without loss of generality, let $l > k$, then $a_{i}=2^{k} i, a_{j}=2^{l} j$.
Since $l > k$, it follows that $a_{i} \nmid a_{j}$.
Assume $a_{j} \mid a_{i}$. Then $j \mid i$.
By $l > k$, we know $j > i$, which is a contradiction. Hence, $a_{j} \nmid a_{i}$.
In summary, the sum of the remaining numbers is minimized when $a_{i}=2^{k} i$.
Thus, the minimum value is
$2211 + 506 + 128 + 24 + 16 = 2885$.
Therefore, the maximum possible value of the sum of all removed positive integers is $5050 - 2885 = 2165$. | 2165 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Color each cell of a $5 \times 5$ grid with one of five colors, such that the number of cells of each color is the same. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
Color each cell of a $5 \times 5$ grid with one of five colors, such that the number of cells of each color is the same. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges. | As shown in Figure 4, a $5 \times 5$ grid is divided into five parts, each colored with one of 5 colors, and at this point, there are 16 dividing edges.
Below is the proof that the number of dividing edges is at least 16.
First, let the 5 colors be denoted as $1, 2, \cdots, 5$, and let the number of rows and columns occupied by the cells of the $i$-th color be $a_{i}$ and $b_{i}$, respectively, and the length of the boundary be $c_{i}$.
For $i \in \{1, 2, \cdots, 5\}$, the $i$-th color is exactly applied to 5 cells, and it occupies $a_{i}$ rows and $b_{i}$ columns, so
$$
a_{i} b_{i} \geqslant 5.
$$
Since for each row of the $a_{i}$ rows occupied by the $i$-th color, the left and right boundaries of the $i$-th color cells in this row must be the boundaries of all the $i$-th color cells, and for each column of the $b_{i}$ columns occupied by the $i$-th color, the upper and lower boundaries of the $i$-th color cells in this column must also be the boundaries of all the $i$-th color cells, we have
$$
c_{i} \geqslant 2\left(a_{i}+b_{i}\right).
$$
Let the number of dividing edges be $f$. Since each dividing edge must be the boundary between two colors, and the boundary of the $i$-th color cells, excluding the boundary of the $5 \times 5$ grid, must be a dividing edge, we have
$$
f=\frac{1}{2}\left(\sum_{i=1}^{5} c_{i}-5 \times 4\right).
$$
From $a_{i} b_{i} \geqslant 5$, we know $a_{i}+b_{i} \geqslant 2 \sqrt{a_{i} b_{i}} \geqslant 2 \sqrt{5}$.
Since $a_{i}, b_{i} \in \mathbf{Z}$, we have
$$
a_{i}+b_{i} \geqslant 5 \Rightarrow c_{i} \geqslant 10,
$$
and equality holds if and only if the $i$-th color cells are arranged as shown in Figure 5, denoted as Figure A.
$$
\begin{array}{l}
\text { Hence } f \geqslant \frac{1}{2}(5 \times 10-5 \times 4) \\
=15,
\end{array}
$$
and the equality condition is that there are 5 Figure A's.
Notice that 5 $2 \times 2$ grids cannot be arranged without overlapping in a $5 \times 5$ grid.
Thus, the equality in (1) does not hold.
Therefore, $f \geqslant 16$. | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. In the Cartesian coordinate system, color the set of points
$$
\left\{(m, n) \mid m, n \in \mathbf{Z}_{+}, 1 \leqslant m, n \leqslant 6\right\}
$$
red or blue. Then the number of different coloring schemes where each unit square has exactly two red vertices is $\qquad$ kinds. | 8. 126 .
Dye the first row (points with a y-coordinate of 6), there are $2^{6}$ ways to do this, which can be divided into two cases.
(1) No two same-colored points are adjacent (i.e., red and blue alternate), there are 2 ways, and the second row can only be dyed in 2 ways, each row has only 2 ways, totaling $2^{6}$ ways;
(2) At least two adjacent same-colored points exist, there are $2^{6}-2$ ways to dye, in this case, when dyeing the second row, it can be found that the satisfactory dyeing method is unique, totaling $2^{6}-2$ ways.
In summary, the number of dyeing methods that meet the conditions is $2^{6}+2^{6}-2=126$ | 126 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. A meeting was attended by 24 representatives, and between any two representatives, they either shook hands once or did not shake hands at all. After the meeting, it was found that there were a total of 216 handshakes, and for any two representatives $P$ and $Q$ who shook hands, among the remaining 22 representatives, no more than 10 representatives shook hands with exactly one of $P$ or $Q$. A "friend circle" refers to a set of three representatives who all shook hands with each other. Find the minimum possible number of friend circles among the 24 representatives.
(Chang An) | 7. Let the 24 representatives be $v_{1}, v_{2}, \cdots, v_{24}$, and for $i=1,2, \cdots, 24$, let $d_{i}$ denote the number of people who have shaken hands with $v_{i}$.
Define the set $E=\left\{\left\{v_{i}, v_{j}\right\} \mid v_{i}\right.$ has shaken hands with $v_{j} \}$.
For any $e=\left\{v_{i}, v_{j}\right\} \in E$, let $t(e)$ be the number of friend circles that include the set $e$.
By the given condition,
$$
\begin{array}{l}
\left(d_{i}-1\right)+\left(d_{j}-1\right) \leqslant 2 t(e)+10 \\
\Rightarrow \sum_{e=\left|v_{i}, v_{j}\right| \in E}\left(d_{i}+d_{j}\right) \leqslant \sum_{e \in E}(2 t(e)+12) .
\end{array}
$$
In $\sum_{e=\left|v_{i}, v_{j}\right| \in E}\left(d_{i}+d_{j}\right)$, the number of handshakes $d_{i}$ of representative $v_{i}$ is counted $d_{i}$ times, so the left side of equation (1) equals $\sum_{i=1}^{24} d_{i}^{2}$. Let the number of friend circles be $T$, noting that each friend circle is counted 3 times in $\sum_{e \in E} t(e)$, and the total number of edges is 216, then the right side of equation (1) equals $6 T+12 \times 216$.
$$
\begin{array}{l}
\text { Hence } \sum_{i=1}^{24} d_{i}^{2} \leqslant 6 T+12 \times 216 . \\
\text { By } \sum_{i=1}^{24} d_{i}^{2} \geqslant \frac{1}{24}\left(\sum_{i=1}^{24} d_{i}\right)^{2}=\frac{(2 \times 216)^{2}}{24} \\
\Rightarrow T \geqslant \frac{1}{6}\left(\frac{(2 \times 216)^{2}}{24}-12 \times 216\right)=864 .
\end{array}
$$
Divide the 24 representatives into four groups, each with 6 people, and let any two representatives from different groups shake hands once, while any two representatives from the same group do not shake hands.
The total number of handshakes is calculated as
$$
\frac{1}{2} \times 24 \times 18=216 \text {. }
$$
For any two representatives $P$ and $Q$ who have shaken hands, among the remaining 22 representatives, exactly 10 representatives have shaken hands with either $P$ or $Q$, which meets the condition. In this case, the number of friend circles is
$$
C_{4}^{3} \times 6^{3}=864 \text {. }
$$
In summary, the minimum possible number of friend circles among the 24 representatives is 864. | 864 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Arrange 10 flowers in a row using red, yellow, and blue flowers (assuming there are plenty of each color), and yellow flowers cannot be adjacent. How many different arrangements are there (the 10 flowers can be of one color or two colors)? | 11. Let $x_{n}$ be the number of different arrangements of $n$ flowers that meet the requirements. Then
$$
x_{1}=3, x_{2}=3^{2}-1=8 \text {. }
$$
When $n \geqslant 3$, let the arrangement of $n$ flowers be $a_{1}, a_{2}, \cdots, a_{n}$. If $a_{1}$ is a red or blue flower, then $a_{2}, a_{3}, \cdots, a_{n}$ is an arrangement of $n-1$ flowers that meets the conditions; if $a_{1}$ is a yellow flower, then $a_{2}$ must be a red or blue flower, in which case $a_{3}, a_{4}, \cdots, a_{n}$ is an arrangement of $n-2$ flowers that meets the conditions.
$$
\begin{array}{l}
\text { Thus, } x_{n}=2 x_{n-1}+2 x_{n-2} . \\
\text { Hence } x_{3}=2 x_{2}+2 x_{1}=22, \\
x_{4}=2 x_{3}+2 x_{2}=60, \\
x_{5}=2 x_{4}+2 x_{3}=164, \\
x_{6}=2 x_{5}+2 x_{4}=448, \\
x_{7}=2 x_{6}+2 x_{5}=1224, \\
x_{8}=2 x_{7}+2 x_{6}=3344, \\
x_{9}=2 x_{8}+2 x_{7}=9136, \\
x_{10}=2 x_{9}+2 x_{8}=24960 .
\end{array}
$$
【Note】The general term of $\left\{x_{n}\right\}$ in this problem is
$$
x_{n}=\frac{9+5 \sqrt{3}}{6}(1+\sqrt{3})^{n-1}+\frac{9-5 \sqrt{3}}{6}(1-\sqrt{3})^{n-1} .
$$ | 24960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y$ satisfy $x^{2}+y^{2}=20$. Then the maximum value of $x y+8 x+y$ is $\qquad$ . | 4. 42 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x y+8 x+y)^{2} \\
\leqslant\left(x^{2}+8^{2}+y^{2}\right)\left(y^{2}+x^{2}+1^{2}\right) \\
=84 \times 21=42^{2} .
\end{array}
$$
Therefore, the maximum value sought is 42. | 42 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. In the sequence
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]
$$
there are $\qquad$ distinct integers ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.1515.
Let the $k$-th term of the known sequence be $\left[\frac{k^{2}}{2019}\right]$. Then, when $(k+1)^{2}-k^{2} \leqslant 2019$, i.e., $k \leqslant 1009$,
$$
\begin{array}{l}
\frac{(k+1)^{2}}{2019}=\frac{k^{2}}{2019}+\frac{2 k+1}{2019} \leqslant \frac{k^{2}}{2019}+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right] \leqslant 1+\left[\frac{k^{2}}{2019}\right] \\
\Rightarrow 0 \leqslant\left[\frac{(k+1)^{2}}{2019}\right]-\left[\frac{k^{2}}{2019}\right] \leqslant 1 .
\end{array}
$$
Since $\left[\frac{1009^{2}}{2019}\right]=504$, therefore, in
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{1009^{2}}{2019}\right]
$$
the integers $0,1, \cdots, 504$ appear without omission, and these 505 integers are all distinct.
$$
\begin{array}{l}
\text { Also, when } k \geqslant 1010, \\
2 k+1=(k+1)^{2}-k^{2}>2019 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right]=\left[\frac{k^{2}+2 k+1}{2019}\right] \\
\quad \geqslant\left[\frac{k^{2}+2019}{2019}\right]=\left[\frac{k^{2}}{2019}\right]+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right]-\left[\frac{k^{2}}{2019}\right] \geqslant 1 .
\end{array}
$$
Thus, $\left[\frac{1010^{2}}{2019}\right],\left[\frac{1011^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]$ are
distinct integers, and all are greater than 504.
Therefore, the number of distinct integers in the original sequence is
$$
505+1010=1515 .
$$ | 1515 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. On each face of a cube, randomly fill in one of the numbers 1, 2, $\cdots$, 6 (the numbers on different faces are distinct). Then number the eight vertices such that the number assigned to each vertex is the product of the numbers on the three adjacent faces. The maximum value of the sum of the numbers assigned to the eight vertices is $\qquad$ | 8. 343 .
Let the numbers on the six faces be $a, b, c, d, e, f$, and $(a, b), (c, d), (e, f)$ be the numbers on the opposite faces. Thus, the sum of the numbers at the eight vertices is
$$
\begin{array}{l}
(a+b)(c+d)(e+f) \\
\leqslant\left(\frac{(a+b)+(c+d)+(e+f)}{3}\right)^{3} \\
=7^{3}=343 .
\end{array}
$$
When $a=1, b=6, c=2, d=5, e=3, f=4$, the equality holds.
Therefore, the maximum value sought is 343 . | 343 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given four positive integers
$a, b, c, d$ satisfy:
$$
a^{2}=c(d+20), b^{2}=c(d-18) \text {. }
$$
Then the value of $d$ is $\qquad$ | 4. 180 .
Let $(a, b)=t, a=t a_{1}, b=t b_{1}$.
Then $\frac{d+20}{d-18}=\frac{c(d+20)}{c(d-18)}=\frac{a^{2}}{b^{2}}=\frac{a_{1}^{2}}{b_{1}^{2}}$ (simplest fraction).
Let $d+20=k a_{1}^{2}, d-18=k b_{1}^{2}$.
Eliminating $d$ yields
$$
\begin{array}{l}
k\left(a_{1}+b_{1}\right)\left(a_{1}-b_{1}\right)=2 \times 19 \\
\Rightarrow k=2, a_{1}+b_{1}=19, a_{1}-b_{1}=1 \\
\Rightarrow a_{1}=10, b_{1}=9 \\
\Rightarrow d=k a_{1}^{2}-20=180 .
\end{array}
$$
At this point, there exist $a=20, b=18, c=2$ that satisfy the conditions.
Therefore, $d=180$. | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a, b, c, d$ are positive integers, and $\log _{a} b=\frac{3}{2}, \log _{c} d=\frac{5}{4}, a-c=9$.
Then $a+b+c+d=$ $\qquad$ | 3. 198 .
Given $a=x^{2}, b=x^{3}, c=y^{4}, d=y^{5}$.
$$
\begin{array}{l}
\text { Given } a-c=x^{2}-y^{4}=9 \\
\Rightarrow\left(x+y^{2}\right)\left(x-y^{2}\right)=9 \\
\Rightarrow x+y^{2}=9, x-y^{2}=1 \\
\Rightarrow x=5, y^{2}=4 \\
\Rightarrow a=25, b=125, c=16, d=32 \\
\Rightarrow a+b+c+d=198 .
\end{array}
$$ | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) In the sequence $\left\{a_{n}\right\}$, let $S_{n}=\sum_{i=1}^{n} a_{i}$ $\left(n \in \mathbf{Z}_{+}\right)$, with the convention: $S_{0}=0$. It is known that
$$
a_{k}=\left\{\begin{array}{ll}
k, & S_{k-1}<k ; \\
-k, & S_{k-1} \geqslant k
\end{array}\left(1 \leqslant k \leqslant n, k 、 n \in \mathbf{Z}_{+}\right)\right. \text {. }
$$
Find the largest positive integer $n$ not exceeding 2019 such that
$$
S_{n}=0 .
$$ | 10. Let the indices $n$ that satisfy $S_{n}=0$ be arranged in ascending order, denoted as the sequence $\left\{b_{n}\right\}$, then $b_{1}=0$.
To find the recurrence relation that $\left\{b_{n}\right\}$ should satisfy.
In fact, without loss of generality, assume $S_{b_{k}}=0$.
Thus, by Table 1, it is easy to prove by mathematical induction:
$$
\left\{\begin{array}{l}
S_{b_{k}+2 i-1}=b_{k}+2-i, \\
S_{b_{k}+2 i}=2 b_{k}+i+2
\end{array}\left(i=1,2, \cdots, b_{k}+2\right)\right. \text {. }
$$
Table 1
\begin{tabular}{|c|c|c|}
\hline$n$ & $S_{n}$ & $a_{n}$ \\
\hline$b_{k}$ & 0 & $/$ \\
\hline$b_{k}+1$ & $b_{k}+1$ & $b_{k}+1$ \\
\hline$b_{k}+2$ & $2 b_{k}+3$ & $b_{k}+2$ \\
\hline$b_{k}+3$ & $b_{k}$ & $-\left(b_{k}+3\right)$ \\
\hline$b_{k}+4$ & $2 b_{k}+4$ & $b_{k}+4$ \\
\hline$b_{k}+5$ & $b_{k}-1$ & $-\left(b_{k}+5\right)$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ \\
\hline
\end{tabular}
Let $b_{k}+2-i=0$, we have
$$
b_{k}+2 i-1=3 b_{k}+3 \text {, }
$$
which means the sequence $\left\{b_{n}\right\}$ satisfies:
$$
\begin{array}{l}
b_{k+1}=3 b_{k}+3, b_{1}=0 \\
\Rightarrow b_{n}=\frac{3^{n}-3}{2} .
\end{array}
$$
Thus, $b_{7}=1092$ is the desired result. | 1092 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $x, y, z \in \mathbf{R}_{+}$, satisfying $x+y+z=x y z$. Then the function
$$
\begin{array}{l}
f(x, y, z) \\
=x^{2}(y z-1)+y^{2}(z x-1)+z^{2}(x y-1)
\end{array}
$$
has the minimum value of $\qquad$ | 5. 18 .
According to the conditions, we have
$$
y+z=x(y z-1) \Rightarrow y z-1=\frac{y+z}{x} \text{. }
$$
Similarly, $z x-1=\frac{z+x}{y}, x y-1=\frac{x+y}{z}$.
From $x y z=x+y+z \geqslant 3 \sqrt[3]{x y z} \Rightarrow x y z \geqslant 3 \sqrt{3}$, thus
$$
\begin{array}{l}
f(x, y, z)=2(x y+y z+z x) \\
\geqslant 2 \times 3 \sqrt[3]{(x y z)^{2}} \geqslant 18,
\end{array}
$$
When $x=y=z=\sqrt{3}$, the equality holds. | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The sequence of positive integers $\left\{a_{n}\right\}: a_{n}=3 n+2$ and $\left\{b_{n}\right\}$ $b_{n}=5 n+3(n \in \mathbf{N})$ have a common number of terms in $M=\{1,2, \cdots, 2018\}$ which is $\qquad$ | 6. 135.
It is known that 2018 is the largest common term of the two sequences within $M$. Excluding this common term, subtract 2018 from the remaining terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively, to get
$$
\left\{\overline{a_{n}}\right\}=\{3,6,9, \cdots, 2016\},
$$
which are all multiples of 3 within $M$; and
$$
\left\{\overline{b_{n}}\right\}=\{5,10,15, \cdots, 2015\},
$$
which are all multiples of 5 within $M$.
Clearly, the common terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ correspond one-to-one to the common terms of $\left\{\overline{a_{n}}\right\}$ and $\left\{\overline{b_{n}}\right\}$, and these common terms are all multiples of 15 within $M$, which are $\left[\frac{2018}{15}\right]=134$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. Therefore, the number of common terms sought is $134+1=135$. | 135 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits as $p(n)$. If $s(n) +$ $p(n) = n$ holds, then $n$ is called a "coincidence number". Therefore, the sum of all coincidence numbers is | 8.531.
Let \( n = \overline{a_{1} a_{2} \cdots a_{k}} \left(a_{1} \neq 0\right) \).
From \( n - s(n) = p(n) \), we get
\[
\begin{array}{l}
a_{1}\left(10^{k-1}-1\right) + a_{2}\left(10^{k-2}-1\right) + \cdots + a_{k-1}(10-1) \\
= a_{1} a_{2} \cdots a_{k},
\end{array}
\]
which simplifies to \( a_{1}\left(10^{k-1}-1-a_{2} a_{3} \cdots a_{k}\right) + m = 0 \),
where \( m = a_{2}\left(10^{k-2}-1\right) + \cdots + a_{k-1}(10-1) \geqslant 0 \).
If \( k \geqslant 3 \), since
\[
10^{k-1}-1-a_{2} a_{3} \cdots a_{k} \geqslant 10^{k-1}-1-9^{k-1} > 0,
\]
this contradicts equation (1). Therefore, \( k \leqslant 2 \).
When \( k = 1 \), \( n = s(n) \), which does not meet the condition, so \( k = 2 \).
Thus, we can set \( n = \overline{a_{1} a_{2}} = 10 a_{1} + a_{2} \).
Combining \( a_{1} + a_{2} + a_{1} a_{2} = 10 a_{1} + a_{2} \), we get
\[
9 a_{1} = a_{1} a_{2} \Rightarrow a_{2} = 9 \left(a_{1} \in \{1, 2, \cdots, 9\}\right),
\]
so the complete set of coincidental numbers is \( 19, 29, \cdots, 99 \), and their sum is 531. | 531 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $a_{k}$ as the number of integer terms in $\log _{2} k, \log _{3} k, \cdots, \log _{2018} k$. Then $\sum_{k=1}^{2018} a_{k}=$ $\qquad$ | 8.4102 .
Let $b_{m}$ be the number of integer terms in $\log _{m} 1, \log _{m} 2, \cdots, \log _{m} 2018$.
Then, $\sum_{k=1}^{2018} a_{k}=\sum_{m=2}^{2018} b_{m}$.
Notice that, $\log _{m} t$ is an integer if and only if $t$ is a power of $m$.
\[
\begin{array}{l}
\text { Then } b_{2}=11, b_{3}=7, b_{4}=6, b_{5}=b_{6}=5, \\
b_{7}=b_{8}=\cdots=b_{12}=4, \\
b_{13}=b_{14}=\cdots=b_{44}=3, \\
b_{45}=b_{46}=\cdots=b_{2018}=2 . \\
\text { Hence } \sum_{m=2}^{2018} b_{m} \\
=11+7+6+5 \times 2+4 \times 6+3 \times 32+2 \times 1974 \\
=4102 .
\end{array}
\]
Therefore, $\sum_{k=1}^{2018} a_{k}=4102$. | 4102 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that the three sides of a triangle are consecutive natural numbers. If the largest angle is twice the smallest angle, then the perimeter of the triangle is $\qquad$ . | -,1.15.
Assuming the three sides of a triangle are $n-1$, $n$, and $n+1$, with the largest angle being $2 \theta$ and the smallest angle being $\theta$.
Then, by the Law of Sines, we have
$$
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \cos \theta=\frac{n+1}{2(n-1)} \text {. }
$$
By the Law of Cosines, we get
$$
\begin{array}{l}
\cos \theta=\frac{(n+1)^{2}+n^{2}-(n-1)^{2}}{2 n(n+1)} \\
=\frac{n+4}{2(n+1)} .
\end{array}
$$
Thus, $\cos \theta=\frac{n+1}{2(n-1)}=\frac{n+4}{2(n+1)}$
$$
\Rightarrow n=5 \text {. }
$$
Therefore, the perimeter of the triangle is 15. | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. There are five cities in a line connected by semi-circular roads, as shown in Figure 1. Each segment of the journey is from one city to another along a semi-circle. If the journey can be repeated, the total number of possible ways to start from city 5 and return to city 5 after four segments is $\qquad$.
保留源文本的换行和格式，所以翻译结果如下：
1. There are five cities in a line connected by semi-circular roads, as shown in Figure 1. Each segment of the journey is from one city to another along a semi-circle. If the journey can be repeated, the total number of possible ways to start from city 5 and return to city 5 after four segments is $\qquad$. | ,- 1.80 .
After four segments, there are five possible ways to start from city 5 and return to city 5:
$$
\begin{array}{l}
5 \rightarrow 1 \rightarrow 5 \rightarrow 1 \rightarrow 5, \\
5 \rightarrow 1 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 5 \rightarrow 1 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 4 \rightarrow 2 \rightarrow 5 .
\end{array}
$$
For each of the four segments, one can choose the upper semicircle or the lower semicircle, each with two options. Therefore, the total number of possible ways is $5 \times 16=80$. | 80 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let positive integers $m, n$ satisfy
$$
m(n-m)=-11 n+8 \text {. }
$$
Then the sum of all possible values of $m-n$ is $\qquad$ | 2. 18 .
From the problem, we have
$$
n=\frac{m^{2}+8}{m+11}=m-11+\frac{129}{m+11} \in \mathbf{Z}_{+} \text {. }
$$
Then $(m+11) \mid 129$
$$
\Rightarrow m+11=1,3,43,129 \text {. }
$$
Also, $m \in \mathbf{Z}_{+}$, checking we find that when $m=32,118$, the corresponding $n$ is a positive integer.
$$
\text { Hence }(m, n)=(32,24) \text { or }(118,108) \text {. }
$$
Therefore, the sum of the possible values is
$$
(32-24)+(118-108)=18 \text {. }
$$ | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Person A tosses a fair coin twice, and Person B tosses the same coin three times. If the probability that they end up with the same number of heads is written as a simplified fraction, the sum of the numerator and the denominator is $\qquad$ . (Romania) | 3.21.
Let the outcomes of a coin landing heads up and tails up be denoted as $\mathrm{H}$ and $\mathrm{T}$, respectively.
Jia has four equally probable outcomes: HH, HT, TH, TT;
Yi has eight equally probable outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
The outcomes that match $\mathrm{HH}$ are $\mathrm{HHT}$, $\mathrm{HTH}$, $\mathrm{THH}$; the outcomes that match $\mathrm{HT}$ or $\mathrm{TH}$ are HTT, THT, TTH,
and the outcome that matches TT is TTT.
Thus, the required probability is
$$
\frac{1}{4} \times \frac{3}{8} + \frac{1}{2} \times \frac{3}{8} + \frac{1}{4} \times \frac{1}{8} = \frac{5}{16},
$$
The sum of the numerator and the denominator is $5 + 16 = 21$. | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. The product $1!\cdot 2!\cdot 3!\cdot \cdots \cdot 99!\cdot 100!$ ends with $\qquad$ consecutive 0s.
(Hong Kong, China, provided) | 6. 1124.
Since in the product $1!\cdot 2!\cdot 3!\cdots \cdots 99!\cdot 100!$, there are a large number of factor 2s, the number of consecutive 0s at the end is determined by the number of factor 5s. The number of factor 5s in each factorial $x$! is shown in Table 1.
Thus, the total number of factor 5s is
$$
\begin{array}{l}
0+5(1+2+3+4+6+7+\cdots+10+ \\
12+13+\cdots+16+18+19+\cdots+22)+24 \\
=5\left(\frac{22 \times 23}{2}-5-11-17\right)+24 \\
=1124 .
\end{array}
$$
Therefore, the required value is 1124. | 1124 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[3]}$
(2012, Girls' Mathematical Olympiad) | Notice that, $2012=4 \times 503$, where $p=503$ is a prime number, and
$$
\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1} .
$$
If $p \nmid k$, then $p \mid \mathrm{C}_{2012}^{k}$;
If $p \mid k$, then $k \in\{0, p, 2 p, 3 p, 4 p\}$.
Notice that, $\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$,
$$
\begin{array}{l}
\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}=4 \mathrm{C}_{4 p-1}^{p-1} \\
= 4 \times \frac{(4 p-1)(4 p-2) \cdots(3 p+1)}{(p-1)(p-2) \cdots 1} \\
\equiv 4(\bmod p), \\
\mathrm{C}_{4 p}^{2 p}=2 \mathrm{C}_{4 p-1}^{2 p-1} \\
= 2 \times \frac{(4 p-1)(4 p-2) \cdots(3 p+1)}{(2 p-1)(2 p-2) \cdots(p+1)} \times 3 \times \\
\frac{(3 p-1)(3 p-2) \cdots(2 p+1)}{(p-1)(p-2) \cdots 1} \\
\equiv 6(\bmod p),
\end{array}
$$
neither of which is a multiple of $p$.
If the binary representation of $n$ is
$$
n=\left(a_{r} a_{r-1} \cdots a_{0}\right)_{2}=\sum_{t=0}^{r} a_{t} 2^{t},
$$
then the power of 2 in $n!$ is
$$
\begin{array}{l}
\sum_{s=1}^{\infty}\left[\frac{n}{2^{s}}\right]=\sum_{s=1}^{\infty} \sum_{t=s}^{r} a_{t} 2^{t-s}=\sum_{t=1}^{r} \sum_{s=1}^{t} a_{t} 2^{t-s} \\
=\sum_{t=1}^{r} a_{t}\left(2^{t}-1\right)=n-\sum_{t=0}^{r} a_{t}=n-S(n),
\end{array}
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$, and $S(n)$ denotes the sum of the binary digits of $n$ (i.e., the number of 1s).
$$
\begin{array}{l}
\text { Hence } 4 \nmid \mathrm{C}_{2012}^{k} \Leftrightarrow 4 \nmid \frac{2012!}{k!\cdot(2012-k)!} \\
\Leftrightarrow 4 \nmid \frac{(k+m)!}{k!\cdot m!}(m=2012-k) \\
\Leftrightarrow k+m-S(k+m) \\
\quad=k-S(k)+m-S(m)+\varepsilon(\varepsilon \in\{0,1\}) \\
\Leftrightarrow S(k+m)=S(k)+S(m)-\varepsilon(\varepsilon \in\{0,1\}) . \\
\Leftrightarrow k+m=2012=(11111011100)_{2} \text { at most }
\end{array}
$$
has one carry.
(1) If $k+m=2012$ has no carry, then the three 0s correspond to 0s in the digits of $k$ and $m$, and the eight 1s correspond to one 0 and one 1 in the digits of $k$ and $m$. Hence, there are $2^{8}$ cases, in which $\mathrm{C}_{2012}^{k}$ is odd.
(2) If $k+m=2012$ has exactly one carry, it must occur at one of the two 10s (i.e., from the 1st and 5th positions to the 2nd and 6th positions), i.e., $01+01=10$, and the other digits follow the conclusion of (1). Hence, there are $2 \times 2^{7}=2^{8}$ cases, in which $\mathrm{C}_{2012}^{k}$ is even but not a multiple of 4.
Thus, among $\mathrm{C}_{2012}^{0}, \mathrm{C}_{2012}^{1}, \cdots, \mathrm{C}_{2012}^{2012}$, there are
$$
2013-2^{8}-2^{8}=1501
$$
multiples of 4.
Additionally, $\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$ is not a multiple of 4,
$\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}$ has a power of 2 of
$$
\begin{array}{l}
S(p)+S(3 p)-S(4 p) \\
=S(3 p)=S(1509)>2,
\end{array}
$$
$\mathrm{C}_{4 p}^{2 p}$ has a power of 2 of
$$
\begin{array}{l}
S(2 p)+S(2 p)-S(4 p)=S(p) \\
=S(503)=S(2012)=8,
\end{array}
$$
among which, there are three multiples of 4.
Therefore, among $\mathrm{C}_{2012}^{0}, \mathrm{C}_{2012}^{1}, \cdots, \mathrm{C}_{2012}^{2012}$, there are
$$
1501-3=1498
$$
multiples of 2012. | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are real coefficients. Assume that
$$
P(1)=7, P(2)=52, P(3)=97 \text {, }
$$
then $\frac{P(9)+P(-5)}{4}=$ $\qquad$ . (Vietnam) | 7. 1202.
Notice that, $52-7=97-52=45$,
$$
\begin{array}{l}
7=45 \times 1-38,52=45 \times 2-38, \\
97=45 \times 3-38 .
\end{array}
$$
Let $Q(x)=P(x)-45 x+38$. Then $Q(x)$ is a fourth-degree polynomial with a leading coefficient of 1, and
$$
Q(1)=Q(2)=Q(3)=0 \text {. }
$$
Thus, for some $r$,
$$
\begin{array}{l}
Q(x)=(x-1)(x-2)(x-3)(x-r) . \\
\text { Therefore, } \frac{1}{4}(P(9)+P(-5)) \\
=\frac{1}{4}(Q(9)+Q(-5))+26 \\
=\frac{1}{4}(8 \times 7 \times 6(9-r)+6 \times 7 \times 8(5+r))+26 \\
=\frac{1}{4}(6 \times 7 \times 8 \times 14)+26=1202 .
\end{array}
$$ | 1202 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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