problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
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3. Let $M=\{1,2, \cdots, 2009\}$. If $n \in M$, such that $S_{n}=\frac{1}{n}\left(1^{3}+2^{3}+\cdots+n^{3}\right)$ is a perfect square, then the number of such $n$ is $\qquad$.
| 3. 44 .
Notice that
$$
S_{n}=\frac{1}{n}\left[\frac{n(n+1)}{2}\right]^{2}=\frac{n(n+1)^{2}}{4} \text {. }
$$
Since $n$ and $n+1$ are coprime, if $S_{n}$ is a perfect square, then $n$ must be a perfect square. When $n$ is an even perfect square, $\frac{n}{4}$ is a perfect square integer; when $n$ is an odd perfect squ... | 44 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange 4 identical red balls and 4 identical blue balls in a row, with the positions numbered $1,2, \cdots$, 8 from left to right. If balls of the same color are indistinguishable, the number of arrangements where the sum of the positions of the 4 red balls is less than the sum of the positions of the 4 blue balls ... | 8.31.
The sum of 8 numbers is 36. Dividing the numbers $1,2, \cdots, 8$ into two groups, there are $\frac{C_{8}^{4}}{2}=35$ ways to do so. When the sums of the two groups are not equal, the smaller sum of 4 numbers is considered the sequence of red balls, and the larger sum of 4 numbers is considered the sequence of bl... | 31 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $a_{1}, a_{2}, \cdots, a_{20}$ be 20 distinct positive integers, and the set $\left\{a_{i}+a_{j} \| \leqslant i, j \leqslant 20\right\}$ contains 201 distinct elements. Find the minimum possible number of distinct elements in the set
$$
\left.|| a_{i}-a_{j}|| 1 \leqslant i, j \leqslant 20\right\}.... | The minimum number of elements in the given set is
100.
Example: Let
$$
\begin{array}{l}
a_{i}=10^{11}+10^{i}, \\
a_{10+i}=10^{11}-10^{i}(i=1,2, \cdots, 10) .
\end{array}
$$
Then $\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 20\right\}$ contains
$$
(20+19+\cdots+1)-10+1=201
$$
different elements.
$$
\b... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $a_{1}=1, a_{2}=2$, for $n \geqslant 2$ we have
$$
a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} .
$$
If for all positive integers $n \geqslant m$, we have $a_{n}>2+$ $\frac{2008}{2009}$, then the smallest positive integer $m$ is $\qquad$ . | 8.4019 .
According to the problem, the recursive relationship becomes
$$
\begin{array}{l}
a_{n+1}-a_{n}=\frac{n-1}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
=\frac{n-1}{n+1}\left(a_{n}-a_{n-1}\right)(a \geqslant 2) .
\end{array}
$$
Therefore, when $n \geqslant 3$,
$$
\begin{array}{l}
a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n... | 4019 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) If all the coefficients of a polynomial are natural numbers, it is called a "natural polynomial". How many different natural polynomials $P(x)$ are there such that $P(2)=20$? | 10. For a positive integer $n$, let $A(n)$ denote the number of distinct natural polynomials $P(x)$ that satisfy $P(2) = n$.
It is easy to prove: for any positive integer $m$, we have
$$
\begin{array}{l}
A(2 m+1)=A(2 m) \\
=A(2 m-1)+A(m) .
\end{array}
$$
In fact, for any natural polynomial $P(x)$ that satisfies $P(2)=... | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If the three-digit number $\overline{a b c}$ satisfies $1 \leqslant a \leqslant b \leqslant c \leqslant 9$, then $\overline{a b c}$ is called an "uphill number". Then, the number of uphill numbers is $\qquad$ | 5. 165 .
For a fixed $b$, the number $c$ can take $10-b$ values, namely $b, b+1, \cdots, 9$; and for a fixed number $a$, the number $b$ can take $a, a+1, \cdots, 9$. Therefore, the number of three-digit increasing numbers is
$$
\begin{array}{l}
\sum_{a=1}^{9} \sum_{b=a}^{9}(10-b)=\sum_{a=1}^{9} \sum_{k=1}^{10-a} k \\
... | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Sure, here is the translation:
---
II. (50 points) Let $f(x)$ be a function defined on the set of natural numbers $\mathbf{N}$ and taking values in $\mathbf{N}$, satisfying: for any two distinct natural numbers $a$ and $b$,
$$
f(a) + f(b) - f(a + b) = 2009.
$$
(1) Find $f(0)$;
(2) Let $a_{1}, a_{2}, \cdots, a_{100}$ ... | (1) Take $a=0, b=1$, then $a \neq b$.
By the condition, we have
$$
f(0)+f(1)-f(0+1)=2.009 \text {. }
$$
So, $f(0)=2009$.
(2) Proof: For any natural number $n$ greater than 1, and $n$ distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\begin{array}{l}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\lef... | 198891 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The afternoon class schedule needs to arrange 5 classes: Physics, Chemistry, Biology, and two self-study periods. If the first class cannot be Biology, and the last class cannot be Physics, then, the number of different ways to arrange the schedule is $\qquad$ kinds. | 2.39.
According to the principle of inclusion-exclusion, the number of different ways to schedule classes is $\frac{5!}{2}-2 \times \frac{4!}{2}+\frac{3!}{2}=39$. | 39 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6.50 The sum of 50 positive numbers is 231, and the sum of their squares is 2009. Then, the maximum value of the largest number among these 50 numbers is $\qquad$ . | 6. 35 .
Let the maximum number be $x$, and the other 49 numbers be $a_{1}, a_{2}$, $\cdots, a_{49}$, and $x \geqslant a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{49}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{49} a_{i}=231-x, \\
2009=x^{2}+\sum_{i=1}^{49} a_{i}^{2} \geqslant x^{2}+\frac{1}{49}\left(\sum_{i=1}^{49}... | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If $n \in \mathbf{N}$, and $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$ is a positive integer, then $n=$ $\qquad$ | 6.5.
Notice that $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$
$$
=\frac{33}{\sqrt{n^{2}+24}+\sqrt{n^{2}-9}},
$$
thus $\sqrt{n^{2}+24}+\sqrt{n^{2}-9}$ could be $1,3,11,33$. Therefore, the solution is $n=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. For some integer $k$, there are exactly 2008 positive integers $n_{1}, n_{2}, \cdots, n_{2008}$, satisfying
$$
k=\left[\sqrt[3]{n_{1}}\right]=\left[\sqrt[3]{n_{2}}\right]=\cdots=\left[\sqrt[3]{n_{2008}}\right],
$$
and $k \m... | 7.668.
If $\sqrt[3]{n}-1n \text {. }$
Also, if $k \mid n$, then $n$ can take $k^{3}, k^{3}+k, \cdots, k^{3}+$ $3 k^{2}+3 k$, a total of $3 k+4$.
Therefore, $3 k+4=2008 \Rightarrow k=668$. | 668 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a, b \in \mathbf{N}_{+}$, when $a^{2}+b^{2}$ is divided by $a+b$, the quotient is $q$, and the remainder is $r$. Then the number of pairs $(a, b)$ that satisfy $q^{2}+r=2009$ is $\qquad$ pairs. | -1.0 .
From $2\left(a^{2}+b^{2}\right) \geqslant(a+b)^{2}$, we get $\frac{a^{2}+b^{2}}{a+b} \geqslant \frac{a+b}{2}$.
Then $q+1>q+\frac{r}{a+b}=\frac{a^{2}+b^{2}}{a+b}$ $\geqslant \frac{a+b}{2} \geqslant \frac{r+1}{2}$.
Thus, $r<2 q+1$.
From $q^{2}+r=2009<(q+1)^{2}$, we get $q \geqslant 44$. From $q^{2}+r=2009 \geqslan... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $a, b, c, d \in \mathbf{Z}$,
$$
f(x)=a x^{3}+b x^{2}+c x+d \text {. }
$$
Then among $A(1,1)$, $B(2009,1)$, $C(-6020,2)$, $D(2,-6020)$, at most $\qquad$ of them can be on the curve $y=f(x)$. | 4.3.
If $A$ and $C$ are both on $y=f(x)$, then $f(1)=1, f(-6020)=2$.
It is easy to see that $\left(x_{1}-x_{2}\right) \mid \left(f\left(x_{1}\right)-f\left(x_{2}\right)\right)$. Therefore, $[1-(-6020)] \mid (f(1)-f(-6020))=-1$, which is a contradiction.
Thus, $A$ and $C$ cannot both be on $y=f(x)$.
Similarly, $B$ and ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $M=\{1,2, \cdots, 17\}$. If there are four distinct numbers $a, b, c, d \in M$, such that $a+b \equiv c+d(\bmod 17)$, then $\{a, b\}$ and $\{c, d\}$ are called a “balanced pair” of set $M$. The number of balanced pairs in set $M$ is $\qquad$. | 8.476.
Divide the circumference into 17 equal parts, and label the points in clockwise order as $A_{1}, A_{2}, \cdots, A_{17}$. Then
$$
\begin{array}{l}
m+n \equiv k+l(\bmod 17) \\
\Leftrightarrow A_{m} A_{n} / / A_{k} A_{l} .
\end{array}
$$
Note that none of the chords connecting any pair of the 17 equally divided p... | 476 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the number of positive integers $t$ not exceeding 2009, such that for all natural numbers $n$, $\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}$ is coprime with 2009.
---
The translation maintains the original text's line breaks and formatting. | $$
\begin{array}{l}
\text { Three, let } \\
\begin{array}{l}
x= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}=\frac{1}{\sqrt{t}} \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}(\sqrt{t})^{2 k+1}, \\
\begin{array}{l}
y= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k} t^{k}=\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k}(\sqrt{t})^{2 k} .... | 980 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. The positive integer $n$ ensures that in every $n$-element subset of the set $\{1,2, \cdots, 2008\}$, there are 2 elements (which can be the same), whose sum is a positive integer power of 2. Then the minimum value of $n$ is $\qquad$ | 8.1003.
First, if $n=1002$, take
$$
\begin{array}{l}
A=\{5,6,7\}, B=\{17,18, \cdots, 24\}, \\
C=\{33,34, \cdots, 39\}, \\
D=\{1025,1026, \cdots, 2008\} .
\end{array}
$$
Then $S=A \cup B \cup C \cup D$.
Thus, $|S|=3+8+7+984=1002$.
It is easy to verify that the sum of any two (possibly the same) elements in $S$ is not ... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. (16 points) On the Cartesian plane, a point whose both coordinates are rational numbers is called a rational point. Find the smallest positive integer $k$ such that: for every circle that contains $k$ rational points on its circumference, the circle must contain infinitely many rational points on its circumference. | 12. First, prove: If a circle's circumference contains 3 rational points, then the circumference must contain infinitely many rational points.
Let $\odot C_{0}$ in the plane have 3 rational points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ on its circumference, with the center $C_{0}\left(x_{0}, y_{0}\right)$.
Since t... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $a, b, c \geqslant 0$, and $a b+b c+c a=\frac{1}{3}$. Prove:
$$
S=\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-a c+1}+\frac{1}{c^{2}-a b+1} \leqslant 3 .
$$ | Explanation: When $a=b=c=\frac{1}{3}$, $S=3$.
Also, when $a=0, b=\frac{\sqrt{3}}{3}, c=\frac{\sqrt{3}}{3}$, $S=3$ as well.
Therefore, the maximum value 3 is both conventional and unconventional, which is quite rare. Moreover, $S$ has no other extremum (the lower bound of $S$ is $\frac{5}{2}$, $S>\frac{5}{2}$ but cannot... | 3 | Inequalities | proof | Yes | Yes | cn_contest | false |
Example 4 Given that $a, b, c$ are positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
(2005, National High School Mathematics League, Fujian Province Preliminary Contest) | Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2} + b x + c = 0$. By Vieta's formulas, we have
$$
x_{1} + x_{2} = -\frac{b}{a}, \quad x_{1} x_{2} = \frac{c}{a}.
$$
Thus, $x_{1}9$.
Therefore, the roots $x_{1}$ and $x_{2}$ of $a x^{2} + b x + c = 0$ are in the interval $\left(-\frac{1}{3}, 0\right)$. Hence, w... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
(1) Find the minimum value of the smallest side length of a peculiar triangle;
(2) Prove that there are infinitely many isosceles peculiar triangles;
(3) How many non-isosceles peculiar triangles are there? | (1) Let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of a peculiar triangle. Then, by Heron's formula, we have
$$
\begin{aligned}
16 \Delta^{2}= & (a+b+c)(a+b-c) . \\
& (a-b+c)(-a+b+c) .
\end{aligned}
$$
Since $(a, b, c)=1$, at least one of $a, b, c$ must be odd. If there is an odd number of odd numbers a... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. In trapezoid $A B C D$, it is known that $A D / / B C(B C>$ $A D), \angle D=90^{\circ}, B C=C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$
(2004, "Xinli Cup" National Junior High School Mathematics Competition) | (提示: Extend $D A$ to point $F$, complete the trapezoid into a square $C B F D$, then rotate $\triangle A B F$ $90^{\circ}$ around point $B$ to the position of $\triangle B C G$. It is easy to see that $\triangle A B E \cong \triangle G B E$. Therefore, $A E$ $=E G=E C+A F=10$. Let $E C=x$, then $A F=10-$ $x, D E=12-x, ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. In trapezoid $A B C D$, it is known that $A D / / B C, A D \perp$ $C D, B C=C D=2 A D, E$ is a point on side $C D$, $\angle A B E=45^{\circ}$. Then $\tan \angle A E B=$ $\qquad$
(2007, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | (提示: Extend $D A$ to point $F$, such that $A F=A D$, to get the square $B C D F$. Then extend $A F$ to point $G$, such that $F G=C E$. Then $\triangle B F G \cong \triangle B C E$. It is also easy to prove that $\triangle A B G \cong \triangle A B E$. Let $C E=x, B C=2 a$. Then $F G=x, A F=A D=a, A E=A G=x+a, D E=2 a-x... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
(2008, National High School Mathematics Competition) | Explanation: By the symmetry of the parabola, we can assume
$$
P\left(2 t^{2}, 2 t\right)(t>0) \text {. }
$$
Since the equation of the circle is $x^{2}+y^{2}-2 x=0$, the equation of the chord of contact $M N$ is
$$
2 t^{2} x+2 t y-\left(x+2 t^{2}\right)=0,
$$
which simplifies to $\left(2 t^{2}-1\right) x+2 t y-2 t^{2... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Quadratic Function
$$
f(x)=a x^{2}+b x+c(a, b \in \mathbf{R} \text {, and } a \neq 0)
$$
satisfies the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and
$$
f(x) \geqslant x \text {; }
$$
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\math... | Explanation: First, derive the analytical expression of $f(x)$ from the given conditions, then classify and discuss $m$ and $t$ to determine the value of $m$.
Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about the line $x=-1$. Therefore,
$$
-\frac{b}{2 a}=-1 \Rightarrow b=2 a \text {. }
$$
From condi... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 A $6 \times 6$ chessboard has 36 squares. Placing chess pieces in the squares, if there are 4 pieces of the same color in a straight line (horizontal, vertical, or at a $45^{\circ}$ diagonal), it is called a "four-in-a-row". Player A places white pieces, and Player B places black pieces. If A goes first, what... | Solution: At least 10 chess pieces need to be placed.
This is because, to prevent the four-in-a-row formation by the black pieces placed subsequently by player B, player A must place at least one white piece in each $1 \times 4$ rectangle, meaning that at least 8 white pieces must be placed in the $2 \times 4$ rectangl... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 4, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5... | Analysis: The difficulty ratio of this problem has significantly increased from 1, mainly due to its "apparent" asymmetry (the number of rows and columns are inconsistent). Therefore, the key to solving the problem is to find the symmetry hidden behind this asymmetry.
Inspired by Example 1, we will also start from the ... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A bicycle tire, if installed on the front wheel, will wear out after the bicycle has traveled $5000 \mathrm{~km}$; if installed on the rear wheel, it will wear out after the bicycle has traveled $3000 \mathrm{~km}$. After traveling a certain distance, the front and rear tires can be swapped. If the front and rear ti... | II. 6.3750.
Let the total wear of each new tire when it is scrapped be $k$. Then the wear per $1 \mathrm{~km}$ for a tire installed on the front wheel is $\frac{k}{5000}$, and the wear per $1 \mathrm{~km}$ for a tire installed on the rear wheel is $\frac{k}{3000}$. Suppose a pair of new tires travels $x \mathrm{~km}$ b... | 3750 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five distinct integers satisfying the condition
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=9
$$
If $b$ is an integer root of the equation
$$
\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\left(x-a_{4}\right)\left(x-a_{5}\right)=2009
$$
then the value of $b$ is... | 8. 10 .
Notice
$$
\left(b-a_{1}\right)\left(b-a_{2}\right)\left(b-a_{3}\right)\left(b-a_{4}\right)\left(b-a_{5}\right)=2009,
$$
and $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five different integers, so all $b-a_{1}, b-a_{2}, b-a_{3}, b-a_{4}, b-a_{5}$ are also five different integers.
$$
\begin{array}{l}
\text { Also, ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. B. Let the positive integer $a$ satisfy $192 \mid\left(a^{3}+191\right)$, and $a<2$ 009. Find the sum of all possible positive integers $a$ that meet the condition. | 12. B. From the problem, we have $192 \mid\left(a^{3}-1\right)$.
$$
\begin{array}{l}
\text { Also, } 192=3 \times 2^{6}, \text { and } \\
a^{3}-1=(a-1)[a(a+1)+1] \\
=(a-1) a(a+1)+(a-1) .
\end{array}
$$
Since $a(a+1)+1$ is odd, we have
$$
2^{6}\left|\left(a^{3}-1\right) \Leftrightarrow 2^{6}\right|(a-1) \text {. }
$$
... | 10571 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. A. $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following conditions:
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. | 14. A. Let in $a_{1}, a_{2}, \cdots, a_{n}$, after removing $a_{i}(i=1$, $2, \cdots, n)$, the arithmetic mean of the remaining $n-1$ numbers is a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Therefore, for any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
b_{i... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $x \in \mathbf{R}$. Then the function
$$
f(x)=\sqrt{x^{2}+1}+\sqrt{(x-12)^{2}+16}
$$
has a minimum value of $\qquad$ . | 8. 13 .
As shown in Figure 1, take $A$ as the origin of the number line, $A B=$ 12, and then construct the perpendicular lines $A C$ and $B D$ such that
$$
\begin{array}{l}
A C=1, \\
B D=4 .
\end{array}
$$
On the number line, take point $P$ such that $A P=x$. Then
$$
f(x)=|C P|+|D P| \text {. }
$$
When points $C$, $... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for each $n \in \mathbf{N}_{+}, a_{n} 、 a_{n+1}$ are the two roots of the equation $x^{2}+3 n x+b_{n}=0$. Then $\sum_{k=1}^{20} b_{k}=$ $\qquad$ | 11. 6385.
For each $n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}=-3 n, \\
a_{n} a_{n+1}=b_{n} .
\end{array}
$$
Rewrite equation (1) as
$$
a_{n+1}+\frac{3(n+1)}{2}-\frac{3}{4}=-\left(a_{n}+\frac{3 n}{2}-\frac{3}{4}\right) \text {. }
$$
Therefore, $\left\{a_{n}+\frac{3 n}{2}-\frac{3}{4}\right\}$ is... | 6385 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. From the set $M=\{1,2, \cdots, 2008\}$ consisting of the first 2008 positive integers, take a $k$-element subset $A$, such that the sum of any two numbers in $A$ cannot be divisible by their difference. Then the maximum value of $k$ is $\qquad$ | 12.670.
First, take 670 yuan set $A=\{1,4,7, \cdots, 2008\}$, where the sum of any two numbers in $A$ cannot be divisible by 3, but their difference is a multiple of 3.
Then, divide the numbers in $M$ from smallest to largest into 670 segments, each containing three numbers:
$$
1,2,3 ; 4,5,6 ; \cdots ; 2005,2006,2007... | 670 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
15. For the $2n$-element set $M=\{1,2, \cdots, 2n\}$, if the $n$-element sets
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}
$$
satisfy $A \cup B=M, A \cap B=\varnothing$, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$, then $A \cup B$ is called an "equal-sum partition" ... | 15. Solution 1: Without loss of generality, let $12 \in A$. Since when set $A$ is determined, set $B$ is uniquely determined, we only need to consider the number of sets $A$.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{6}\right\}, a_{6}$ be the largest number.
By $1+2+\cdots+12=78$, we know
$a_{1}+a_{2}+\cdots+a_{6}=39, a_{... | 29 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x_{1}, x_{2}, \cdots, x_{6}$ be positive integers, and they satisfy the relation
$$
\begin{aligned}
x_{6} & =2288, \\
x_{n+3} & =x_{n+2}\left(x_{n+1}+2 x_{n}\right)(n=1,2,3) .
\end{aligned}
$$
Then $x_{1}+x_{2}+x_{3}=$ | 3. 8 .
Substituting $n=1,2,3$ into the relation
$$
\begin{array}{l}
x_{n+3}=x_{n+2}\left(x_{n+1}+2 x_{n}\right) . \\
\text { we get } x_{4}=x_{3}\left(x_{2}+2 x_{1}\right), \\
x_{5}=x_{4}\left(x_{3}+2 x_{2}\right)=x_{3}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right), \\
x_{6}=x_{3}^{2}\left(x_{2}+2 x_{1}\right)\l... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $f(x)$ is a monotonically increasing function on $\mathbf{R}$, and for any $x \in \mathbf{R}$, $f(f(f(x)))=x$. Then $f(2009)=$ $\qquad$ | 5.2 009.
If $f(2009)2009$, then we have
$$
f(f(f(2009)))>2009,
$$
contradiction. | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. $A, B$ two people are playing a table tennis match, with the rule that the first to win 3 games more than the opponent wins. After 13 games, $A$ finally wins with a record of 8 wins and 5 losses. Then the number of all different possible outcomes of these 13 games is $\qquad$ | 6. 243.
As shown in Figure 3, use the grid point $(m, n)$ to represent the moment when $m+n$ games have been played (with $A$ winning $m$ games and $B$ winning $n$ games). The problem then transforms into finding the number of grid paths from $(0,0)$ to $(8,5)$, where the points $(x, y)$ passed through satisfy $|x-y|<... | 243 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
(2003, National Junior High School Mathematics Competition) | Solution: (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$.
From the problem, we have $a>0$, and $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$
Therefore,
$$
\begin{array}{l}
\Delta=[-(2... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For any real numbers $x, y$, the inequality
$$
|x-1|+|x-3|+|x-5| \geqslant k(2-|y-9|)
$$
always holds. Then the maximum value of the real number $k$ is $\qquad$ | 3. 2 .
From the geometric meaning of absolute value, we know that the minimum value of $|x-1|+|x-3|+|x-5|$ is 4 (at this time $x=3$), and the maximum value of $2-|y-9|$ is 2 (at this time $y=9$). According to the condition, we get $4 \geqslant k \cdot 2$, so, $k \leqslant 2$. Therefore, the maximum value of $k$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+p_{5}^{2}=p_{6}^{2}$ has $\qquad$ groups of positive prime solutions $\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)$. | 7.5.
Obviously, $p_{6} \neq 2$.
Since $p^{2} \equiv 1$ or $4(\bmod 8)$ (where $p$ is a prime), and $p^{2} \equiv 4(\bmod 8)$ if and only if $p=2$, it follows that among $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$, there must be 4 twos (let's assume they are $p_{1}$, $p_{2}$, $p_{3}$, and $p_{4}$), and the other on... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A classroom has desks arranged in 6 rows and 7 columns, with 40 students. Two positions in the last row are left empty, and the rest of the students are seated based on their height and vision. There are 24 students who are tall, 18 students who have good vision, and 6 students who have both conditions. It is known ... | 8. 35 .
The number of students who are short and have poor eyesight is 4, the number of students who are tall and have good eyesight is 6, and the remaining students number 30 people.
The number of ways to leave two seats empty in the last row is $C_{7}^{2}$, where the power of 2 is 0;
The number of ways to arrange ... | 35 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) On a plane, there are 32 points, with no three points being collinear. Prove: Among these 32 points, there are at least 2135 sets of four points that form the vertices of a convex quadrilateral. | Three, let $n=32$.
Since no three points among the $n$ points are collinear, $\mathrm{C}_{n}^{3}$ triangles can be formed by any three points among the $n$ points, among which the one with the largest area is denoted as $\triangle A B C$.
Draw lines parallel to the opposite sides through points $A, B, C$, intersecting... | 2135 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. If
$$
\begin{array}{l}
a+b-2 \sqrt{a-1}-4 \sqrt{b-2} \\
=3 \sqrt{c-3}-\frac{1}{2} c-5,
\end{array}
$$
then, the value of $a+b+c$ is $\qquad$ | (Tip: Refer to Example 3. Answer: 20. ) | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If $a$, $b$, $c$ are all real numbers, and $a+b+c=0$, $abc=2$, then the minimum value that $|a|+|b|+|c|$ can reach is $\qquad$ . | (Tip: Refer to Example 3. Answer: 4. ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Find the maximum value of the function $y=\sqrt{2 x^{2}+3 x+1}+\sqrt{7-2 x^{2}-3 x}$. | ( Hint: Refer to Example 8. Answer: The maximum value is 4 . ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Each point in the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such th... | Explanation: If $n=4$, then a colored plane can be constructed such that: there is exactly one point on a circle with three colors, and the rest of the points are of another color, satisfying the problem's conditions and ensuring no four points of different colors are concyclic. Therefore, $n \geqslant 5$.
When $n=5$,... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Suppose the lengths of the three sides of a triangle are integers $l$, $m$, and $n$, and $l > m > n$. It is known that
$$
\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\},
$$
where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceedin... | Solution: Note that, when $a \equiv b(\bmod m)(0 \leqslant b < m)$, we get $500+2 n>1000+n$.
So, $n>500$.
Thus, $n \geqslant 501, m \geqslant 1001, l \geqslant 1501$.
Therefore, $m+n+l \geqslant 3003$, which means the minimum perimeter of the triangle is 3003. | 3003 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the minimum value of $y=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$. | Solution: As shown in Figure 2, construct Rt $\triangle P A C$ and Rt $\triangle P B D$ such that $A C=1, B D=2, C P=x, P D=4-x$. Thus, the original problem is transformed into: finding a point $P$ on line $l$ such that the value of $P A + P B$ is minimized.
Since $y = P A + P B \geqslant A B$, the minimum value of $y... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Consider rectangles in the plane with sides parallel to the coordinate axes (both length and width are greater than 0), and call such a rectangle a "box". If two boxes have a common point (including common points on the interior or boundary of the boxes), then the two boxes are said to "intersect". Find the largest ... | 1. The maximum value that satisfies the condition is 6.
An example is shown in Figure 1.
Below is the proof: 6 is the maximum value.
Assume the boxes $B_{1}$, $B_{2}, \cdots, B_{n}$ satisfy the condition. Let the closed intervals corresponding to the projections of $B_{k}$ on the $x$-axis and $y$-axis be $I_{k}$ and ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Consider the set $S$ of all integer points (points with integer coordinates) on the coordinate plane. For each positive integer $k$, if there exists a point $C \in S$ such that the area of $\triangle A B C$ is $k$, then two distinct points $A, B \in S$ are called “$k$-friends”; if any two points in $T$ are $k$-frien... | 3. First, consider the point $B \in S$ that is a $k$-friend of the point $(0,0)$. Let $B(u, v)$. Then $B$ is a $k$-friend of $(0,0)$ if and only if there exists a point $C(x, y) \in S$ such that $\frac{1}{2}|u y - v x| = k$ (here we use the area formula of $\triangle A B C$, where $A$ is the origin).
There exist integ... | 180180 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ is a root of the equation $x^{2}-5 x+1=0$. Then the unit digit of $a^{4}+a^{-4}$ is $\qquad$ . | ニ、1.7.
Obviously, $a^{-1}$ is another root of the equation. Then $a+a^{-1}=5$.
Thus, $a^{2}+a^{-2}=\left(a+a^{-1}\right)^{2}-2=23$, $a^{4}+a^{-4}=\left(a^{2}+a^{-2}\right)^{2}-2=527$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In a convex quadrilateral $A B C D$, the diagonals $A C$ and $B D$ intersect at point $O$. If $S_{\triangle O A D}=4, S_{\triangle O B C}=9$, then the minimum value of the area of the convex quadrilateral $A B C D$ is | 2. 25 .
Let $S_{\triangle A O B}=x, S_{\triangle C O D}=y$. Then
$$
\frac{4}{x}=\frac{O D}{O B}=\frac{y}{9} \Rightarrow x y=36 \text {. }
$$
Thus, $x+y \geqslant 2 \sqrt{x y}=12$.
Equality holds if and only if $x=y=6$.
Therefore, $S_{\text {quadrilateral } A B C D}=4+9+x+y \geqslant 25$.
Equality holds if and only if... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. In the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$, after combining like terms, the coefficient of $x^{150}$ is $\qquad$ (answer with a number). | 11.7651 .
By the polynomial multiplication rule, the problem can be transformed into finding the number of natural number solutions of the equation
$$
s+t+r=150
$$
that do not exceed 100. Clearly, the number of natural number solutions of equation (1) is $\mathrm{C}_{152}^{2}$.
Next, we find the number of natural nu... | 7651 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the maximum value of the function $y=\sqrt{5-2 x}+\sqrt{3+2 x}$.
untranslated part:
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。
This part is not translated as it contains instructions for the translation task itself. Here is the requested translation above. | Solution: From the given condition, we know $y>0$.
Notice that the variance of the two numbers $\sqrt{5-2 x}$ and $\sqrt{3+2 x}$ is
$$
\begin{array}{l}
S^{2}=\frac{1}{2}\left[(\sqrt{5-2 x})^{2}+(\sqrt{3+2 x})^{2}-2\left(\frac{y}{2}\right)^{2}\right] \\
\geqslant 0 .
\end{array}
$$
Solving this, we get $16-y^{2} \geqsl... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. In a $4 \times 4$ chessboard, 8 small squares are to be colored black such that each row and each column has exactly two black squares. Then there are $\qquad$ different ways to do this (answer with a number). | 13. 90 .
The first row can be colored with 2 black cells in $\mathrm{C}_{4}^{2}$ ways. After the first row is colored, there are the following three scenarios:
(1) The black cells in the second row are in the same columns as those in the first row, in which case, the remaining rows have only one way to be colored;
(2)... | 90 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Given that $a, b$ are positive numbers, and the parabolas $y=x^{2}+a x+2 b$ and $y=x^{2}+2 b x+a$ both have common points with the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
$(2000$, National Junior High School Mathematics League) | Solution: From the given conditions, we have
$$
a^{2}-8 b \geqslant 0,4 b^{2}-4 a \geqslant 0 \text {. }
$$
Thus, $a^{4} \geqslant 64 b^{2} \geqslant 64 a$, which means $a \geqslant 4$.
Furthermore, $b^{2} \geqslant a \geqslant 4$. Therefore, $a^{2}+b^{2} \geqslant 20$.
Moreover, when $a=4, b=2$, the parabolas $y=x^{2... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the polynomial be
$$
\begin{aligned}
P(x)= & x^{15}-2008 x^{14}+2008 x^{13}-2008 x^{12}+ \\
& 2008 x^{11}-\cdots+2008 x^{3}-2008 x^{2}+2008 x .
\end{aligned}
$$
Then \( P(2007) = \) | 8. 2007 .
$$
\begin{array}{c}
P(x)=(x-2007)\left(x^{14}-x^{13}+x^{12}- \\
x^{11}+\cdots+x^{2}-x\right)+x .
\end{array}
$$
Therefore, $P(2007)=2007$. | 2007 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. For a positive integer $n \leqslant 500$, it has the property: when an element $m$ is randomly selected from the set $\{1,2, \cdots, 500\}$, the probability that $m \mid n$ is $\frac{1}{100}$. Then the maximum value of $n$ is $\qquad$ | 12. 81 .
From the problem, we know that $n$ has exactly 5 positive divisors. Let the prime factorization of $n$ be $n=p_{1}^{\alpha_{1}} p_{1}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$.
Then the number of positive divisors of $n$ is
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)=5 ... | 81 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. As shown in Figure 1, given that the circumradius $R$ of the acute triangle $\triangle ABC$ is $1$, $\angle BAC = 60^{\circ}$, and the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively. The line segment $OH$ intersects the extension of $BC$ at point $P$. Find:
(1) The area of the concave ... | 14. (1) As shown in Figure 2, connect $A H$, and draw $O D \perp B C$ at point D.
Since $O$ is the
circumcenter of
$\triangle A B C$ and $\angle B A C$ $=60^{\circ}$, we have
$$
\begin{array}{l}
\angle B O C \\
=2 \angle B A C \\
=120^{\circ}, \\
O D=O C \cos 60^{\circ}=\frac{1}{2} .
\end{array}
$$
By the property of ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$. | 15. When $a=b=c=1$, we can get $k \geqslant 2$.
Below is the proof: the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Given
$$
x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}=1 \text {. }
$$
then the value of $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}$ is | $$
\text { II, 1. }-1 \text {. }
$$
Notice that
$$
\begin{array}{l}
x \sqrt{x^{2}+3 x+18}+x \sqrt{x^{2}-6 x+18} \\
=\frac{x^{2}\left(x^{2}+3 x+18\right)-x^{2}\left(x^{2}-6 x+18\right)}{x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}} \\
=\frac{x^{2} \cdot 9 x}{1}=9 x^{3} .
\end{array}
$$
Subtracting the above equation fr... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Does there exist a positive integer such that its cube plus 101 is exactly a perfect square? Prove your conclusion. | 3. If there exists a positive integer $x$ that satisfies the conditions of the problem, then $x^{3}+101=y^{2}\left(y \in \mathbf{N}_{+}\right)$.
If $x$ is even, then $y$ is odd.
Let $x=2 n, y=2 m+1(n, m \in \mathbf{N})$. Then $2 n^{3}+25=m^{2}+m$,
the left side of this equation is odd, while the right side is even, whi... | 95 | Number Theory | proof | Yes | Yes | cn_contest | false |
4. Given
$$
\begin{array}{l}
\frac{1}{1 \times \sqrt{2}+2 \sqrt{1}}+\frac{1}{2 \sqrt{3}+3 \sqrt{2}}+\cdots+ \\
\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}
\end{array}
$$
is greater than $\frac{19}{20}$ and less than $\frac{20}{21}$. Then the difference between the maximum and minimum values of the positive integer $n$ is | 4.39.
Notice that
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}}=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{(k+1)^{2} k-k^{2}(k+1)} \\
=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{k^{2}+k}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} .
\end{array}
$$
Then the original expression $=1-\frac{1}{\sqrt{n+1}}$.
Given $\frac{19}{20}<1... | 39 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The number of positive integer values of $n$ that satisfy $\left|\sqrt{\frac{n}{n+2009}}-1\right|>\frac{1}{1005}$ is $\qquad$ | 1.1008015.
By
$$
\begin{array}{l}
1-\sqrt{\frac{n}{n+2009}}>\frac{1}{1005} \Rightarrow \sqrt{\frac{n}{n+2009}}<\frac{1004}{1005} \\
\Rightarrow \frac{n}{n+2009}<\frac{1004^2}{1005^2} \Rightarrow n<\frac{1004^2}{1005^2-1004^2} \cdot 2009=1008014.25
\end{array}
$$
Thus, the number of positive integer values of $n$ for w... | 1008015 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $n$ be a positive integer. If in the $2 n+1$ consecutive natural numbers including 2009, the sum of the squares of the first $n+1$ numbers is equal to the sum of the squares of the last $n$ numbers, then the value of $n$ is $\qquad$. | 10. 31 .
Let $m, m+1, \cdots, m+2 n$ be $2 n+1$ consecutive natural numbers that satisfy the above conditions. Then
$$
\begin{array}{l}
m^{2}+(m+1)^{2}+\cdots+(m+n)^{2} \\
=(m+n+1)^{2}+(m+n+2)^{2}+\cdots+(m+2 n)^{2} . \\
\text { Let } S_{k}=1^{2}+2^{2}+\cdots+k^{2} \text {. Then } \\
S_{m+n}-S_{m-1}=S_{m+2 n}-S_{m+n} ... | 31 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given twelve red points on a circle. Find the minimum value of $n$, such that there exist $n$ triangles with red points as vertices, so that every chord with red points as endpoints is a side of one of these triangles.
（Supplied by Tao Pingsheng） | 4. Let the set of red points be $A=\left\{A_{1}, A_{2}, \cdots, A_{12}\right\}$.
There are 11 chords passing through point $A_{1}$, and any triangle containing vertex $A_{1}$ contains exactly two chords passing through point $A_{1}$. Therefore, these 11 chords passing through point $A_{1}$ must be distributed among at... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set of all permutations $X=(x_{1}, x_{2}, \cdots, x_{9})$ of $1,2, \cdots, 9$ be $A$. For any $X \in A$, let
\[
\begin{array}{l}
f(X)=x_{1}+2 x_{2}+\cdots+9 x_{9}, \\
M=\{f(X) \mid X \in A\} .
\end{array}
\]
Find $|M|$ (where $|M|$ denotes the number of elements in the set $M$).
(Xiong Bin) | 5. Generally prove: when $n \geqslant 4$, for all permutations $X_{n}=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ of the first $n$ positive integers $1,2, \cdots, n$, if
$$
\begin{array}{l}
f\left(X_{n}\right)=x_{1}+2 x_{2}+\cdots+n x_{n}, \\
M_{n}=\{f(X)|X \in A|,
\end{array}
$$
$$
\text { then }\left|M_{n}\right|=\frac... | 121 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. In an $8 \times 8$ grid, what is the minimum number of small squares that need to be removed so that it is impossible to cut out a complete "T-shaped" pentomino from the remaining grid?
（Sun Wenxian, problem contributor）
Will the above translation meet your needs? | 8. At least 14 small squares must be removed from Figure 6.
Figure 6
Figure 7
As shown in Figure 7, an $8 \times 8$ chessboard is divided into five regions. The central region must have at least two small squares removed to prevent a T-shaped pentomino from being placed. The two crossed positions are not equivalent; o... | 14 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) A captain and three sailors together received 2,009 coins of the same denomination. The four agreed to distribute the coins according to the following rules: Sailor 1, Sailor 2, and Sailor 3 each write down a positive integer, denoted as \( b_{1}, b_{2}, b_{3} \), satisfying \( b_{1} \geqslant b_{2} \... | Four, the maximum value is 673.
First, the captain can ensure getting no less than 673 gold coins.
In fact, when the captain divides the gold coins into 3 piles with numbers $671, 670, 668$,
(1) If $b_{1} \geqslant 671$, then the captain can get no less than $671+2=673$ coins;
(2) If $b_{1} < 671$, then the captain ca... | 673 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the three vertices $A$, $B$, and $C$ of the right triangle $\triangle ABC$ are all on the parabola $y=x^{2}$, and the hypotenuse $AB$ is parallel to the $x$-axis. Then the height $h$ from the hypotenuse is $\qquad$. | 3. 1 .
Let point $A\left(a, a^{2}\right)$ and $C\left(c, c^{2}\right)(|c|<a)$. Since $|AB|=|AC|$, we have $a^{2}-c^{2}=1$.
Therefore, the altitude from $C$ to the hypotenuse $AB$ is $a^{2}-c^{2}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2, in $\triangle A B C$, it is known that $A B=9, B C$ $=8, C A=7, A D$ is the angle bisector, and a circle is drawn with $A D$ as a chord,
touching $B C$ and intersecting $A B$ and $A C$ at points $M$ and $N$, respectively. Then $M N=$
$\qquad$ | 4. 6 .
As shown in Figure 4, connect $D M$.
$$
\begin{array}{l}
\text { By } \angle B D M \\
=\angle B A D \\
=\angle C A D \\
=\angle D M N,
\end{array}
$$
we get $M N \parallel B C$.
Thus, $\triangle A M N \sim \triangle A B C$.
It is easy to know $B D=\frac{9}{9+7} \times 8=\frac{9}{2}, B M \cdot B A=B D^{2}$
$\Ri... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. If the function $f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x$ has a range of $[n, M]$ on the interval $[-k, k](k>0)$, then $M+n$ $=$ $\qquad$ | 2. 2 .
Notice that
$$
f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x=1+\frac{2^{x}-1}{2^{x}+1}+\sin x \text {. }
$$
Let $g(x)=f(x)-1$. Then
$$
g(-x)=\frac{2^{-x}-1}{2^{-x}+1}+\sin (-x)=\frac{1-2^{x}}{1+2^{x}}-\sin x=-g(x)
$$
is an odd function.
Let the maximum value of $g(x)$ be $g\left(x_{0}\right)$. By the given information,... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The 29th Summer Olympic Games were held in Beijing on August 8, 2008, forming a memorable number 20080808. The number of different positive divisors of 20080808 divided by 8 is $\qquad$ | 3. 8 .
From $20080808=2^{3} \times 11 \times 17 \times 31 \times 433$, we know that the number of different positive divisors of 20080808 is
$$
N=(3+1)(1+1)^{4}=8^{2} \text { (divisors). }
$$
Dividing by 8 gives 8. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 As shown in Figure 3, let $F$ be the focus of the parabola $y^{2}=4 x$, and $A, B$ be two points on the parabola different from the origin $O$, satisfying $\overrightarrow{F A} \cdot \overrightarrow{F B}=0$. Extend $A F, B F$ to intersect the parabola at points $C, D$ respectively. Find the minimum value of t... | As shown in Figure 3, with $F$ as the pole and $F x$ as the polar axis, the equation of the parabola is
$$
\rho=\frac{2}{1-\cos \theta} \text {. }
$$
Let point $A\left(\rho_{1}, \theta\right)(0<\theta<\pi)$. Then
$$
\begin{array}{l}
B\left(\rho_{2}, \theta+\frac{\pi}{2}\right) 、 C\left(\rho_{3}, \theta+\pi\right) 、 \\... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. A line $l$ passing through the right focus $F$ of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$, and there are exactly 3 such lines, find $\lambda$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | (The polar equation of the hyperbola is
$$
\rho=\frac{2}{1-\sqrt{3} \cos \theta} \text {. }
$$
Let $A B$ be a chord passing through the right focus and intersecting only the right branch.
Then $|A B|=\rho_{1}+\rho_{2}$
$$
\begin{array}{l}
=\frac{2}{1-\sqrt{3} \cos \theta}+\frac{2}{1-\sqrt{3} \cos (\theta+\pi)} \\
=\fr... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 or 8. How many different ways are there to select such a group of eight points? Explain your reasoning.
(2001, China Mathematical Oly... | Solution: Number the points in sequence as $1, 2, \cdots, 24$, and list a $3 \times 8$ number table (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
\hline 9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
\hline 17 & 20 & 23 & 2 & 5 & 8 & 11 & 14 \\
\hline
\end{tabular}
In th... | 258 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 A deck of tri-color cards, totaling 32 cards, includes 10 cards of each color: red, yellow, and blue, numbered $1,2, \cdots, 10$; there are also a big joker and a small joker, each numbered 0. Several cards are drawn from this deck, and their scores are calculated according to the following rule: each card nu... | Solution: This problem is referred to as the "Two Kings Problem." If an additional joker (called the "Middle King") with a value of 0 is added, and each card is still assigned a value as described, it is called the "Three Kings Problem."
Since $2004 < 2^{11}$, adding some cards with values $2^{11}, 2^{12}, \cdots$ doe... | 1006009 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Inputting a positive integer $n$ into a machine will produce a positive integer $\frac{n(n+1)}{2}$. If 5 is input into the machine, and then the number produced is input into the machine again, the final number produced by the machine is $\qquad$ . | 2. 120 .
The first result is $\frac{5(5+1)}{2}=15$;
The second result is $\frac{15(15+1)}{2}=120$. | 120 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. $A, B, C$ three people pick watermelons. The sum of the number of watermelons picked by $A$ and $B$ is 6 less than that picked by $C$; the sum of the number of watermelons picked by $B$ and $C$ is 16 more than that picked by $A$; the sum of the number of watermelons picked by $C$ and $A$ is 8 more than that picked b... | 3. 60.
Let the number of watermelons collected by $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. Then
\[
\begin{array}{l}
a+b-c=-6, \\
b+c-a=16, \\
c+a-b=8 .
\end{array}
\]
By adding (1) + (2) + (3), we get $a+b+c=18$.
Thus, we have $c=12, b=5, a=1 \Rightarrow a b c=60$. | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. There are 19 children with ages ranging from $1 \sim 19$ years old standing in a circle. Record the difference in ages between each pair of adjacent children. The maximum possible sum of these 19 differences is . $\qquad$ | 5. 180 .
19 differences are derived from 19 equations, where each integer from $1 \sim 19$ appears twice. To maximize the sum of the differences, the larger numbers 11 and 19 should all be used as minuends, and the smaller numbers $1 \sim 9$ should all be used as subtrahends. The number 10 should be used once as a subt... | 180 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. In a certain exam, the passing score is 65 points. The overall average score of the class is 66 points, and the average score of all students who passed is 71 points, while the average score of all students who failed is 56 points. To reduce the number of failing students, the teacher adds 5 points to each student's... | 7. 24 .
Let the class have $n$ students. If before the bonus points, $x$ students passed the exam, then
$$
71 x+56(n-x)=66 n \Rightarrow x=\frac{2 n}{3},
$$
which means $n$ is a multiple of 3.
Let $y$ be the number of students who passed after the bonus points. Then
$$
75 y+59(n-y)=71 n \Rightarrow y=\frac{3 n}{4},
$... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 2, given that $A, B, C, D$ are four points on a plane that are not concyclic,
$$
\begin{array}{l}
\triangle A B D, \triangle A D C \text {, } \\
\triangle B C D, \triangle A B C
\end{array}
$$
have circumcenters
$$
\text { as } E, F, G, H \text {, }
$$
respectively. The line segments $E G, F H$
... | 8.4.
Since points $E$ and $G$ lie on the perpendicular bisector of $BD$, $EG$ is the perpendicular bisector of segment $BD$. Similarly, $FH$ is the perpendicular bisector of segment $AC$. Since $I$ is the intersection of $EG$ and $FH$, it follows that $CI = AI = 4$. | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. There is a six-digit number, the sum of its digits is divisible by 26. When this six-digit number is increased by 1, the sum of the digits of the resulting number is also divisible by 26. The smallest six-digit number that satisfies the above conditions is $\qquad$ | 10. 898999.
Obviously, this six-digit number will have a carry when 1 is added.
To make the sum of the digits a multiple of 26, at least three positions need to carry over. Therefore, the last three digits of this six-digit number are all 9, and the sum of the digits of the first three is 25.
Thus, the smallest number... | 898999 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. On a circle, there is 1 red point and 2009 blue points. Xiao Dan calculates the number of convex polygons where all vertices are blue points, while Xiao Dong calculates the number of convex polygons where one vertex is the red point. The difference between the two numbers they get is $\qquad$ | 11. 2017036.
For every convex $n$-sided polygon $(n \in \mathbf{N}_{+})$ calculated by Xiao Dan, there is a corresponding convex $(n+1)$-sided polygon calculated by Xiao Dong. Xiao Dong's convex $(n+1)$-sided polygon has one more red vertex than Xiao Dan's convex $n$-sided polygon. This correspondence does not hold wh... | 2017036 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. In an international chess tournament with 10 participants, each player must play exactly one game against every other player. After several games, it is found that among any three players, at least two have not yet played against each other. How many games have been played at most by this point? | 2. Suppose there are 5 men and 5 women among the 10 players, and all matches so far have been between men and women, thus satisfying the condition, i.e., 25 matches have been played.
Next, we prove: 25 is indeed the maximum value.
Let $k$ be the number of matches played by the player who has played the most matches, de... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Find the smallest positive integer that can be expressed as the sum of the squares of four positive integers and can divide some integer of the form $2^{n}+15\left(n \in \mathbf{N}_{+}\right)$. | 3. The smallest five positive integers that can be expressed as the sum of four positive integer squares are
$$
\begin{array}{l}
4=1+1+1+1, \\
7=4+1+1+1, \\
10=4+4+1+1, \\
12=9+1+1+1, \\
13=4+4+4+1 .
\end{array}
$$
Obviously, since $2^{n}+15$ is odd, the smallest positive integer cannot be 4, 10, or 12.
Also, $2^{n} ... | 13 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. (40 points) There is a stack of cards numbered $1 \sim 15$. After arranging them in a certain order, the following two operations are performed: Place the top card on the table, then place the second card at the bottom of the stack. These two operations are repeated until all 15 cards are sequentially placed on the ... | 1. Assuming the cards are arranged in descending order from top to bottom as shown from left to right in Table 1. The first card originally becomes the bottom card, so it must be 15. The third card must be 14. And so on.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & $\alpha$ & 14 & $\beta$ & 13 &... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. (40 points) In each game of bridge, four players play together. It is known that the number of games played is exactly equal to the number of players, and any two players have played at least one game together. Try to find the maximum number of players.
| 9. There are $n$ players, and each game can produce 6 pairs of players.
From $\mathrm{C}_{n}^{2} \leqslant 6 n$ we get
$$
n \leqslant 13 \text {. }
$$
Next, we prove that $n=13$ is feasible.
As shown in Figure 13, number the 13
players from $0 \sim 12$.
The players in each game are as follows:
$$
\begin{array}{l}
(0,2... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. In a $10 \times 10$ grid, there is a shape composed of $4 n$ $1 \times 1$ small squares, which can be covered by $n$ "田" shaped figures, or by $n$ "世" or "円" shaped figures (which can be rotated). Find the minimum value of the positive integer $n$.
(Zhu Huaiwei, problem contributor) | 7. Let the given shapes be denoted as type $A$ and type $B$.
First, we prove that $n$ is even.
Color the $10 \times 10$ grid as shown in Figure 6.
Regardless of which 4 squares type $A$ covers, the number of black squares must be even, while for type $B$ it is odd. If $n$ is odd, the number of black squares covered b... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9.1. A simplest fraction is equal to the sum of two simplest fractions with denominators 600 and 700, respectively. Find the smallest possible value of the denominator of such a simplest fraction. | 9. 1. $2^{3} \times 3 \times 7=168$.
Let the two simplest fractions be $\frac{a}{600}$ and $\frac{b}{700}$. Then $(a, 6)=(b, 7)=1$.
Thus, the sum $\frac{7 a+6 b}{4200}$ has a numerator that is coprime with 6 and 7.
Since $4200=2^{3} \times 3 \times 7 \times 5^{2}$, after canceling out the common factors, the denominat... | 168 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9.4. In a regular 100-gon, each vertex is covered by a cloth. It is known that exactly one of the cloths hides a coin. The following action is called an operation: choose any four cloths to check if there is a coin underneath, and after each operation, the cloths are returned to their original positions, while the coin... | 9.4.33 times.
Assume that a regular 100-gon is placed on a rotatable horizontal circular table, and the vector from the center to one vertex points due north. The initial positions of the vertices of the polygon are defined from due north in a counterclockwise direction as $0, 1, \cdots, 99$. Each operation and coin t... | 33 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
9.7. A diagonal of eight squares on a chessboard is called a "fence". A rook starts from a square outside the fence on the chessboard and moves, satisfying the following conditions:
(1) It stays on any square of the chessboard at most once;
(2) It never stays on a square of the fence.
Find the maximum number of times ... | 9.7.47 times.
Let the square at the $i$-th row and $j$-th column be denoted as $(i, j)$. Suppose the eight squares occupied by the fence are $(i, i) (i=1,2, \cdots, 8)$. The non-fence squares are divided into four categories $A, B, C, D$:
$$
\begin{aligned}
A= & \{(i, j) \mid 2 \leqslant j+1 \leqslant i \leqslant 4\} ... | 47 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. As shown in Figure 1, given $A(-3,0)$, $B(0,-4)$, and $P$ is any point on the hyperbola $y=\frac{12}{x} (x > 0)$. A perpendicular line from point $P$ to the x-axis meets at point $C$, and a perpendicular line from point $P$ to the y-axis meets at point $D$. Then the minimum value of the area $S$ of quadrilateral $A ... | 2. C.
Let $P\left(x, \frac{12}{x}\right)$. Then $C(x, 0)$ and $D\left(0, \frac{12}{x}\right)$.
It is easy to see that $C A=x+3, D B=\frac{12}{x}+4$, so
$$
S=\frac{1}{2} C A \cdot D B=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \text {. }
$$
Simplifying, we get $S=2\left(x+\frac{9}{x}\right)+12$.
Since $x>0, \frac{9}{... | 24 | Algebra | MCQ | Yes | Yes | cn_contest | false |
2. As shown in Figure 2, the first polygon is "expanded" from an equilateral triangle, with the number of sides denoted as $a_{3}$, the second polygon is "expanded" from a square, with the number of sides denoted as $a_{4}, \cdots \cdots$ and so on. The polygon "expanded" from a regular $n(n \geqslant 3)$-sided polygon... | 2. 2009 .
From the extended definition, we know
$$
\begin{array}{l}
a_{3}=12=3 \times 4, \\
a_{4}=20=4 \times 5, \\
\cdots \cdots \\
a_{n}=n(n+1) . \\
\text { Also, } \frac{1}{a_{n}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \\
\Rightarrow \frac{1}{a_{3}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{n}} \\
=\left(\frac{1}{3}-\... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $41^{x}=2009,7^{y}=2009$. Then the value of $\frac{1}{x}+\frac{2}{y}$ is . $\qquad$ | 4. 1 .
From $41^{x}=2009$, we get $41^{x y}=2009^{y}$.
$$
\begin{array}{l}
\text { Also } 7^{y}=2009 \Rightarrow 7^{2 y}=2009^{2} \\
\Rightarrow 49^{y}=2009^{2} \Rightarrow 49^{x y}=2009^{2 x} .
\end{array}
$$
$$
\text { Also } 41 \times 49=2009 \Rightarrow 41^{x y} \times 49^{x y}=2009^{x y} \text {. }
$$
Therefore ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Among the natural numbers from 1 to 144, the number of ways to pick three numbers that form an increasing geometric progression with an integer common ratio is $\qquad$ . | 3. 78.
Let the three numbers $a$, $a q$, and $a q^{2}$ form an increasing geometric sequence. Then
$$
1 \leqslant a < a q < a q^{2} \leqslant 144 \text{.}
$$
From this, we have $2 \leqslant q \leqslant 12$.
When $q$ is fixed, the number of integers $a$ such that the three numbers $a$, $a q$, and $a q^{2}$ are integer... | 78 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. (14 points) Let positive real numbers $x, y, z$ satisfy $xyz=1$. Try to find the maximum value of
$$
f(x, y, z)=(1-yz+z)(1-xz+x)(1-xy+y)
$$
and the values of $x, y, z$ at that time. | $$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 - y z + z < 0 , } \\
{ 1 - x z + x < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+x z<1, \\
y+x y<1
\end{array}\right.\right. \\
\Rightarrow(x+x z)(y+x y)<1 \\
\Leftrightarrow x+x y+x^{2} y<0,
\end{array}
$$
Contradiction. Therefore, among $1-y z+z$, $1-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The number of integer points within the square (including the four sides) formed by the four lines $y=x+10, y=-x+10, y=$ $x-10, y=-x-10$ in the Cartesian coordinate system is $\qquad$ . | (Tip: By plotting, it is known that the number of integer points in each quadrant is 45. The number of integer points on the coordinate axes is 41, so, the number of integer points is $(4 \times 45+41=) 221$. | 221 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Question Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$
$(2007$, National High School Mathematics League Jiangxi Province Preliminary Contest) | Another solution: Considering that $a_{n}$ is not continuous in decimal, and the digits of $a_{n}$ are all no greater than 3 (i.e., $0, 1, 2, 3$), then the form of $a_{n}$ is a sequence of consecutive integers in quaternary, thus simplifying the problem.
Let the set $\left\{\left(a_{n}\right)_{4}\right\}=\{$ consecuti... | 133113 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, the five-digit numbers formed by the digits $1, 2, 3, 4, 5$ with repetition, arranged in ascending order. Ask:
(1) What are the positions of 22435 and 43512?
(2) What is the 200th number? ${ }^{[1]}$ | Analysis: Let the $n$-th number after sorting be denoted as $a_{n}$. Since the numbers do not contain 0, we cannot directly use the quinary system. Therefore, subtract 11111 from all $a_{n}$, and denote the result as $b_{n}$. Thus, we can draw the following conclusions:
(i) Each number in $b_{n}$ consists of some of th... | 12355 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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