problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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4. Given in rectangle $A B C D$, $A B=72, A D=$ 56. If $A B$ is divided into 72 equal parts, and parallel lines to $A D$ are drawn through each division point; and if $A D$ is divided into 56 equal parts, and parallel lines to $A B$ are drawn through each division point, then these parallel lines divide the entire rect... | 4. D.
According to the problem, establish a Cartesian coordinate system such that $A(0,0)$, $B(72,0)$, and $D(0,56)$. Then, $C(72,56)$.
Since $AC$ intersects with every horizontal line (including $AB$ and $DC$) and every vertical line (including $AD$ and $BC$), there are $57 + 73 = 130$ intersection points (including... | 120 | Geometry | MCQ | Yes | Yes | cn_contest | false |
7. Calculate $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}}(n \geqslant 2$, $n \in \mathbf{N})$ 的值为
The value of $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}... | 7. 100 .
$$
\begin{array}{l}
\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}} \\
=\sqrt[n]{\left(10^{n}-1\right)^{2}+2 \times 10^{n}-1} \\
=\sqrt[n]{10^{2 n}}=100 .
\end{array}
$$ | 100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. If $p$ is a prime number, and $p+3$ divides $5p$, then the last digit of $p^{2009}$ is $\qquad$ . | 8. 2 .
Since the positive divisors of $5p$ are $1, 5, p, 5p$, and $(p+3)$ divides $5p$, therefore, $p=2$.
Thus, $p^{2009}=2^{2009}=\left(2^{4}\right)^{502} \times 2=2 \times 16^{502}$.
Given that the last digit of $16^{502}$ is 6, it follows that the last digit of $p^{2009}$ is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. As shown in Figure 2, in quadrilateral $A B C D$, $\angle A C B=$ $\angle B A D=105^{\circ}, \angle A B C=\angle A D C=45^{\circ}$. If $A B$ $=2$, then the length of $C D$ is $\qquad$. | 9. 2 .
As shown in Figure 6, draw $EA \perp AB$ intersecting the extension of $BC$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle AEB=45^{\circ} \\
\quad=\angle ADC, \\
AE=AB=2 . \\
\text { Also } \angle DAC=\angle DAB-\angle CAB \\
=\angle DAB-\left(180^{\circ}-\angle ABC-\angle ACB\right) \\
=75^{\circ}=180... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. As shown in Figure 3, fill in the 10 spaces in the annulus with the numbers $1,2, \cdots, 10$. Add up the absolute values of the differences between all adjacent cells (cells that share a common edge). If this sum is to be maximized, then the maximum value is $\qquad$ | 10.50.
Let the two numbers in adjacent cells be $a$ and $b$ ($a > b$). Then $|a-b| = a-b$, and there are 10 differences. To maximize the sum of these 10 differences, the 10 minuends $a$ should be as large as possible, and the 10 subtrahends $b$ should be as small as possible. Since each number is adjacent to two other... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Given $\frac{x y}{x+y}=2, \frac{x z}{x+z}=3, \frac{y z}{y+z}=4$. Find the value of $7 x+5 y-2 z$. | Three, 11. Given
$$
\begin{array}{l}
\frac{1}{2}=\frac{1}{x}+\frac{1}{y}, \frac{1}{3}=\frac{1}{x}+\frac{1}{z}, \\
\frac{1}{4}=\frac{1}{y}+\frac{1}{z} .
\end{array}
$$
Solving simultaneously, we get $x=\frac{24}{7}, y=\frac{24}{5}, z=24$.
Therefore, $7 x+5 y-2 z=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The smallest positive integer $a$ that makes the inequality
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}<a-2007 \frac{1}{3}
$$
hold for all positive integers $n$ is $\qquad$ | 4. 2009 .
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$.
Obviously, $f(n)$ is monotonically decreasing. From the maximum value of $f(n)$, $f(1)<a-2007 \frac{1}{3}$, we get $a=2009$. | 2009 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. A station has exactly one bus arriving between $8:00 \sim 9:00$ and $9:00 \sim 10:00$ every morning, but the arrival times are random, and the two arrival times are independent of each other, as shown in Table 1. A passenger arrives at the station at $8:20$. What is the expected waiting time for the passenger (round... | 8. 27 .
The distribution of passenger waiting times is shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
Waiting Time \\
(min)
\end{tabular} & 10 & 30 & 50 & 70 & 90 \\
\hline Probability & $\frac{1}{2}$ & $\frac{1}{3}$ & $\frac{1}{6} \times \frac{1}{6}$ & $\frac{1}{2} \times \frac{1}... | 27 | Other | math-word-problem | Yes | Yes | cn_contest | false |
1. (14 points) Let the line $l: y=k x+m(k, m \in$ Z) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Question: Does there exist a line $l$ such that the vector $\overrightarrow{A C}... | Given the system of equations:
$$
\left\{\begin{array}{l}
y=k x+m, \\
\frac{x^{2}}{16}+\frac{y^{2}}{12}=1,
\end{array}\right.
$$
eliminating \( y \) and simplifying, we get:
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0.
$$
Let \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). Then:
$$
\beg... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. (15 points) Find the maximum and minimum values of the function
$$
y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}
$$ | 3. The domain of the function is $[0,13]$.
$$
\begin{array}{l}
\text { Given } y=\sqrt{x}+\sqrt{x+27}+\sqrt{13-x} \\
=\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \\
\geqslant \sqrt{27}+\sqrt{13}=3 \sqrt{3}+\sqrt{13},
\end{array}
$$
we know that the equality holds when $x=0$.
Thus, the minimum value of $y$ is $3 \sqrt{3}+\s... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10.3. The function $f(x)=\prod_{i=1}^{2000} \cos \frac{x}{i}$ changes sign how many times in the interval $\left[0, \frac{2009 \pi}{2}\right]$? | 10.3. 75 times.
Let $n=2009$. Consider the function $\cos \frac{x}{k}$. It changes sign at $x=\frac{k(2 m+1) \pi}{2}$. This indicates that the zeros of $f(x)$ are $x_{i}=\frac{i \pi}{2}(1 \leqslant i \leqslant n)$.
We only need to consider the sign change of $f(x)$ at $x_{i}(i=1,2, \cdots, n-1)$.
$\cos \frac{x}{k}$ c... | 75 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
10. 4. In a regular 2009-gon, a non-negative integer not exceeding 100 is placed at each vertex. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k$ su... | 10. 4. $k_{\min }=100400$.
Let the numbers at each vertex be $a_{1}, a_{2}, \cdots, a_{2009}$. Let $N=100400$.
$$
\begin{array}{l}
\text { (1) Let } a_{2}=a_{4}=\cdots=a_{2008}=100, \\
a_{1}=a_{3}=\cdots=a_{2008}=0, \\
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right) .
\end{arr... | 100400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.5. Let $k$ be a positive integer. An infinite strictly increasing sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integer $n$, we have $a_{n+1}-a_{n} \leqslant k$, and $a_{n}$ is a multiple of 1005 or 1006, but not a multiple of 97. Find the smallest possible value of $k$. | 10.5. $k_{\text {min }}=2010$.
Given a positive integer $N$, such that
$$
a_{1}<1005 \times 1006 \times 97 \times N=D \text {. }
$$
Since $D$ is a multiple of 97, it is not any term in the sequence $a_{n}$, hence there exists a positive integer $n$, such that $a_{n}<D<a_{n+1}$.
The largest multiple of 1005 or 1006 l... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given three non-zero real numbers $a, b, c$, the set $A=$ $\left\{\frac{a+b}{c}, \frac{b+c}{a}, \frac{c+a}{b}\right\}$. Let $x$ be the sum of all elements in set $A$, and $y$ be the product of all elements in set $A$. If $x=2 y$, then the value of $x+y$ is $\qquad$ | $-1 .-6$.
From the problem, we have
$$
x=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}
$$
$$
\begin{aligned}
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-3, \\
y & =\frac{a+b}{c} \cdot \frac{b+c}{a} \cdot \frac{c+a}{b} \\
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-1 .
\end{aligned}
$$
Then $x=y-2$.
Also, $x=2y$, solving this gives $y=-2$.
Thus... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The total score of an exam consists of six 6-point questions, six 9-point questions, and twelve 5-point questions. Therefore, the number of different scores that can be formed by this test paper is $\qquad$ | 8. 136.
Since there are scores, and the scores can only be generated from the 150 positive integers from $1 \sim 150$, and the sum of the scores and the deducted points is also 150, we only need to consider the number of positive integers from $1 \sim 75$ that can be represented by $6x + 9y + 5z$ (where $x, y, z$ are ... | 136 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. (14 points) Let $n$ be a positive integer greater than 16, and let $A(n, k)$ and $B(n, k) (k \geqslant 3)$ be the number of $k$-term monotonic arithmetic sequences and geometric sequences with integer common ratios, respectively, in the set $\{1,2, \cdots, n\}$.
(1) Find the value of $A(2009,50)$;
(2) Prove: $n-4 \s... | 1. (1) First, find the general $A(n, \dot{k})$.
Let the set $\{1,2, \cdots, n\}$ have a monotonically increasing arithmetic sequence $\left\{a_{i}\right\}$, with common difference $d$, and $a_{k}=a_{1}+(k-1) d$. Then
$$
\begin{array}{l}
1+(k-1) d \leqslant a_{1}+(k-1) d \leqslant n \\
\Rightarrow d \leqslant \frac{n-1}... | 80360 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Let $A=\{i \in \mathbf{N} \mid 1 \leqslant i \leqslant$ $2880\}, B \subseteq A,|B| \geqslant 9$. If all elements in set $A$ can be represented by the sum of no more than 9 different elements from $B$, find $\min |B|$, and construct a set corresponding to the minimum $|B|$.
| Let $|B|=k \geqslant 9$.
According to the problem, we should have $S_{k}=\sum_{i=1}^{9} \mathrm{C}_{k}^{i} \geqslant 2880$.
Notice that $S_{9}=2^{9}-1, S_{10}=2^{10}-2$,
$$
\begin{array}{l}
S_{11}=2^{11}-2-\mathrm{C}_{11}^{10}=2035, \\
S_{12}=2^{12}-2-\mathrm{C}_{12}^{10}-\mathrm{C}_{12}^{11}=4016>2880 .
\end{array}
$$... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Initially 261 the perimeter of an integer-sided triangle is 75, and squares are constructed on each side. The sum of the areas of the three squares is 2009. Find the difference between the longest and shortest sides of this triangle. | Solution: Given that the sum of three numbers is 75, their average is 25.
Let $2009=(25+a)^{2}+(25+b)^{2}+(25+c)^{2}$, where $a+b+c=0$. It is easy to see that
$a^{2}+b^{2}+c^{2}=134$.
Decomposing 134 into the sum of squares of three positive integers, we find the following four cases:
(1) $a^{2}+b^{2}+c^{2}=134=121+9+4... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let point $O$ be outside $\triangle A B C$, and
$$
\overrightarrow{O A}-2 \overrightarrow{O B}-3 \overrightarrow{O C}=0 \text {. }
$$
Then $S_{\triangle A B C}: S_{\triangle O B C}=$ | $-1.4$.
As shown in Figure 4, let $D$ and $E$ be the midpoints of sides $AB$ and $BC$, respectively, and connect $CD$. Then
$$
\begin{array}{l}
\overrightarrow{OA}+\overrightarrow{OB}=2 \overrightarrow{OD}, \\
\overrightarrow{OB}+\overrightarrow{OC}=2 \overrightarrow{OE} .
\end{array}
$$
(1) - (2) $\times 3$ gives
$$
\... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. (14 points) In the Cartesian coordinate system $x O y$, points with both integer coordinates are called integer points. Given $O(0,0), A(2,1)$, and $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$. If $S_{\triangle O M M}=3$, find the number of integer points $M$ that satisfy this con... | 9. Connect $O A$. It is easy to know that there are two integer points $M_{1}(-6,0)$ and $M_{2}(6,0)$ on the $x$-axis inside the ellipse that satisfy the problem.
Draw two lines $l_{1}$ and $l_{2}$ parallel to the line $O A$ through points $M_{1}$ and $M_{2}$, respectively.
According to the principle that triangles w... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Two, (50 points) The four-digit numbers $m$ and $n$ are reverse positive integers of each other, and $m+n=18 k+9\left(k \in \mathbf{N}_{+}\right), m$ and $n$ have 16 and 12 positive divisors (including 1 and themselves), respectively. The prime factors of $n$ are also prime factors of $m$, but $n$ has one fewer prime f... | Let $m=\overline{a b c d}, a d \neq 0$. Then $n=\overline{d c b a}$.
Given $m+n=9(2 k+1)$, then $9 \mid (m+n)$.
Thus, $9 \mid [(1000 a+100 b+10 c+d)+$
$$
\begin{array}{c}
(1000 d+100 c+10 b+a)], \\
9 \mid 2(a+b+c+d), 9 \mid (a+b+c+d).
\end{array}
$$
Therefore, $9 \mid m, 9 \mid n$.
Since $m+n$ is odd, it follows that ... | 1998 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Does there exist 4098 sets
\[
\begin{aligned}
B_{i} & =\left\{a_{i 1}, a_{i 2}, \cdots, a_{i n}\right\}(i=1,2, \cdots, 4098, \\
a_{i j} \in A_{j} & =\{3 j-2,3 j-1,3 j\}, j=1,2, \cdots, 12)
\end{aligned}
\]
satisfying the following three conditions:
(1) \( B_{i} \cap B_{i+1}=\varnothing(i=1,2, \cdots,... | Let $(s, t)$ denote a set circle with $s$ sets $A_{1}, A_{2}, \cdots, A_{s}$, and $t$ sets $B_{1}, B_{2}, \cdots, B_{t}$ that satisfy the given conditions. Use $\left(B_{i}, p\right)$ to denote a set formed by all elements of $B_{i}$ and the $p$-th element of $A_{i+1}$.
If $B_{1}, B_{2}, \cdots, B_{t}$ satisfy the give... | 4098 | Combinatorics | proof | Yes | Yes | cn_contest | false |
1. Calculate $\frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9$ The value equals $\qquad$ . | $\begin{array}{l} \text { II.1.3 481. } \\ \frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9 \\ = \frac{(45.1-13.9)\left(45.1^{2}+45.1 \times 13.9+13.9^{2}\right)}{45.1-13.9}+ \\ 45.1 \times 13.9 \\ = 45.1^{2}+2 \times 45.1 \times 13.9+13.9^{2} \\ =(45.1+13.9)^{2}=59^{2}=3481 .\end{array}$ | 3481 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]$ equals $\qquad$ (where [x] denotes the greatest integer not exceeding the real number $x$). | 2. 6 .
Given $0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$, then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]=0$ or 1. From the problem, we know that 18 of them are equal to 1. Therefore,
$$
\begin{array}{l}
{\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 1, in $\square A B C D$, $A D=a, C D=b$, draw the heights $h_{a}, h_{b}$ from point $B$ to sides $A D$ and $C D$ respectively. Given $h_{a} \geqslant a, h_{b} \geqslant b$, and the diagonal $A C=20$. Then the area of $\square A B C D$ is $\qquad$ | 3. 200 .
Given $h_{a} \geqslant a, h_{b} \geqslant b$, in addition, $a \geqslant h_{b}, b \geqslant h_{a}$, then $h_{a} \geqslant a \geqslant h_{b} \geqslant b \geqslant h_{a}$, i.e., $h_{a}=a=h_{b}=b$, which means that quadrilateral ABCD is a square.
Since the diagonal of this square $AC=20$, the area of $\square ABC... | 200 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}=
$$
$\qquad$ . | 5. 2 .
Notice that
$$
\begin{array}{l}
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \\
= \sqrt{(x-1)+2 \sqrt{x-1}+1}+ \\
\sqrt{(x-1)-2 \sqrt{x-1}+1} \\
= \sqrt{(\sqrt{x-1}+1)^{2}}+\sqrt{(\sqrt{x-1}-1)^{2}} \\
=|\sqrt{x-1}+1|+|\sqrt{x-1}-1| .
\end{array}
$$
Since $1 \leqslant x \leqslant 2$, we have $\sqrt{x-1}-1 \leq... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given an integer $n \geqslant 3$. Find the smallest positive integer $k$, such that there exists a $k$-element set $A$ and $n$ pairwise distinct real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1}+x_{2}, x_{2}+x_{3}, \cdots$, $x_{n-1}+x_{n}, x_{n}+x_{1}$ all belong to $A$. (Xiong Bin provided) | 2. Let $x_{1}+x_{2}=m_{1}, x_{2}+x_{3}=m_{2}, \cdots \cdots$
$x_{n-1}+x_{n}=m_{n-1}, x_{n}+x_{1}=m_{n}$.
First, $m_{1} \neq m_{2}$, otherwise, $x_{1}=x_{3}$, which is a contradiction.
Similarly, $m_{i} \neq m_{i+1}\left(i=1,2, \cdots, n, m_{n+1}=m_{1}\right)$.
Thus, $k \geqslant 2$.
If $k=2$, let $A=\{a, b\}(a \neq b)$... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. There are $n (n>12)$ people participating in a mathematics invitational competition. The test consists of fifteen fill-in-the-blank questions, with each correct answer worth 1 point and no answer or a wrong answer worth 0 points. Analyzing every possible score situation, it is found that as long as the sum of the sc... | 7. The minimum possible value of $n$ is 911.
(1) First, prove: 911 satisfies the condition.
If each student answers at least three questions correctly, since the number of different ways a student can answer three questions correctly is $\mathrm{C}_{15}^{3}=455$, then if there are 911 students participating, by the pig... | 911 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Place 3 identical white balls, 4 identical red balls, and 5 identical yellow balls into three different boxes, allowing some boxes to contain balls of different colors. The total number of different ways to do this is (answer in numbers).
允许有的盒子中球的颜色不全的不同放法共有种 (要求用数字做答).
Allowing some boxes to contain balls of diff... | $\begin{array}{l}\text { 9. } 3150 \text {. } \\ \mathrm{C}_{5}^{2} \cdot \mathrm{C}_{6}^{2} \cdot \mathrm{C}_{7}^{2}=3150 \text {. }\end{array}$ | 3150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$ | 11. 9 .
Since $(\sqrt{2}+\sqrt{3})^{2010}+(\sqrt{2}-\sqrt{3})^{2010}$ is an integer, therefore, the fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is
$$
\begin{array}{l}
1-(\sqrt{2}-\sqrt{3})^{2010} . \\
\text { Also } 0<(\sqrt{2}-\sqrt{3})^{2010}<0.2^{1005}<(0.008)^{300} \text {, then } \\
0.9<1-(\sqrt{2}-\sqrt{3})^... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11.6. A $10 \times 10$ chessboard has $k$ rooks. A square on the board that can be attacked by a rook is called "dangerous" (the square occupied by the rook itself is also considered dangerous). If removing any rook results in at least one dangerous square becoming safe, find the maximum possible value of $k$. | 11. 6. $k_{\max }=16$.
Consider a chessboard with $k$ rooks satisfying the problem's conditions. There are two cases to consider.
(1) Each row (column) has a rook.
In this case, all squares are dangerous. If there is a row (column) with at least two rooks, removing one of these rooks will still leave all squares dange... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. The first 24 digits of $\pi$ are
3. 14159265358979323846264 .
Let $a_{1}, a_{2}, \cdots, a_{24}$ be any permutation of these 24 digits. Then
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\left(a_{23}-a_{24}\right) \\
\equiv \quad(\bmod 2) .
\end{array}
$$
$(\bmod 2)$. | 3. 0 .
In the first 24 digits of $\pi$, there are 13 odd numbers. When these 13 odd numbers are placed into 12 parentheses, by the pigeonhole principle, there must be two in the same parenthesis, making the difference in this parenthesis even. Thus, the product $\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If a positive integer is written on each face of a cube, and then a number is written at each vertex, which is equal to the product of the two integers on the faces passing through that vertex, then, when the sum of the numbers at the vertices of the cube is 290, the sum of the numbers on the faces of the cube is | 3. 36 .
Let the numbers on each face of the cube be $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, $x_{5}$, $x_{6}$, and the numbers written at the vertices be
$$
\begin{array}{l}
x_{1} x_{2} x_{5}, x_{2} x_{3} x_{5}, x_{3} x_{4} x_{5}, x_{4} x_{1} x_{5}, x_{1} x_{2} x_{6}, \\
x_{2} x_{3} x_{6}, x_{3} x_{4} x_{6}, x_{4} x_{1} x_... | 36 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Given an isosceles triangle with a vertex angle less than $60^{\circ}$, the lengths of its three sides are all positive integers. Construct a square outward on each side, such that the sum of the areas of the three squares is 2009. Find the perimeter of this isosceles triangle. | Let the length of the legs of the isosceles triangle be $a$, and the length of the base be $b$. According to the problem, we have
$$
2 a^{2}+b^{2}=2009 \text{. }
$$
Therefore, $a^{2}=\frac{2009-b^{2}}{2}\frac{2009}{3}>669$.
Thus, $669<a^{2}<1005 \Rightarrow 25<a<32$.
Since $a$ and $b$ are positive integers, it is veri... | 77 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a, b, c \in \mathbf{R}$, and $a+b+c=3$. Then the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is $\qquad$ | 3. 9 .
Since the function $y=3^{x}$ is an increasing function on $(-\infty,+\infty)$, by the property of increasing functions, for any $x_{1}$, $x_{2}$, we have $\left(x_{1}-x_{2}\right)\left(3^{x_{1}}-3^{x_{2}}\right) \geqslant 0$, so,
$$
(a-1)\left(3^{a}-3^{1}\right) \geqslant 0,
$$
which means $3^{a} a-3^{\circ} \... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | From the second equation, we get $z(x+y)=23$, and since 23 is a prime number, we can obtain $\left\{\begin{array}{l}z=1, \\ x+y=23\end{array}\right.$ or $\left\{\begin{array}{l}z=23, \\ x+y=1 \text {. }\end{array}\right.$ According to the problem, $x+y=1$ is not valid, so we discard it. Substituting $z=1$ into the give... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Given $[x]$ represents the greatest integer not exceeding the real number $x$. If $P=\sum_{i=1}^{2010}\left[\frac{6^{i}}{7}\right]$, then the remainder when $P$ is divided by 35 is | 6. 20 .
First, consider $S=\frac{6}{7}+\frac{6^{2}}{7}+\cdots+\frac{6^{2010}}{7}$.
In equation (1), no term is an integer, but the sum of any two adjacent terms is an integer (since $\frac{6^{k}}{7}+\frac{6^{k+1}}{7}=6^{k}$ $(k \in \mathbf{Z})$ is an integer).
If the sum of two non-integer numbers is an integer, then... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $P_{n}(k)$ denote the number of permutations of $\{1,2, \cdots, n\}$ with $k$ fixed points. Let $a_{t}=\sum_{k=0}^{n} k^{t} P_{n}(k)$. Then
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=
\end{array}
$$ | 8. 0 .
Notice
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1} \\
=\sum_{k=0}^{n}\left(k^{5}-10 k^{4}+35 k^{3}-50 k^{2}+25 k\right) P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] . \\
\frac{n!}{(n-k)!k!} P_{n-k}(0) \\
=\sum_{k=0}^{n}\left[\frac{n!... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (15 points) Given seven different points on a circle, vectors are drawn from any one point to another (for points $A$ and $B$, if vector $\overrightarrow{A B}$ is drawn, then vector $\overrightarrow{B A}$ is not drawn). If the four sides of a convex quadrilateral determined by any four points are four consecutive v... | 11. Let the seven points on the circumference be $P_{1}, P_{2}, \cdots, P_{7}$. The number of vectors starting from point $P_{i} (i=1,2, \cdots, 7)$ is $x_{i} (i=1,2, \cdots, 7)$, then $0 \leqslant x_{i} \leqslant 6$, and
$$
\sum_{i=1}^{7} x_{i}=\mathrm{C}_{7}^{2}=21 \text {. }
$$
First, find the minimum number of "no... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Try to find the smallest positive integer $M$, such that the sum of all positive divisors of $M$ is 4896. | Let $M=\prod_{i=1}^{n} p_{i}^{\alpha_{i}}\left(p_{i}\right.$ be a prime, $\alpha_{i} \in \mathbf{N}_{+}, i=1,2, \cdots, n)$. Denote
$$
f(p, \alpha)=\sum_{k=0}^{\alpha} p^{k}\left(p\right.$ be a prime, $\left.\alpha \in \mathbf{N}_{+}\right)$.
From the problem, we know
$$
\prod_{i=1}^{n} f\left(p_{i}, \alpha_{i}\right)=... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $n$ be a positive integer, and let $S_{n}$ be a subset of the set $A_{n}=\left\{m \mid m \in \mathbf{N}_{+}\right.$, and $\left.m \leqslant n\right\}$. Moreover, the difference between any two numbers in $S_{n}$ is not equal to 4 or 7. If the maximum number of elements in $S_{n}$ is denoted by $M_... | It is known that the difference between any two numbers among $1,4,6,7,9$ is not 4 or 7. Adding 11 to each of these numbers gives $12, 15, 17, 18, 20$, which clearly also have the same property, and the difference between any of these numbers and any of the first five numbers is also not 4 or 7. By this reasoning, for ... | 922503 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
265 It is known that for all positive integers $n$,
$$
\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right) \geqslant \frac{k}{2} \sqrt[3]{19 n+8}
$$
always holds. Try to find the maximum value of $k$. | Let $T_{n}=\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right)$. Then
$$
\frac{T_{n+1}}{T_{n}}=1+\frac{1}{3 n+2}=\frac{3 n+3}{3 n+2} \text {. }
$$
Given $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$, we have $k \leqslant \frac{2 T_{n}}{\sqrt[3]{19 n+8}}$.
Let $f(n)=\frac{2 T_{n}}{\sqrt[3]{19 n+8}}$. Then
$$
\begin{array}{l... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Given real numbers $a, b \neq 0$, let
$$
x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{a b}{|a b|} \text {. }
$$
Then the sum of the maximum and minimum values of $x$ is $\qquad$ [1] | Question 1 Original Solution ${ }^{[1]}$ According to the number of negative numbers in $a$ and $b$, there are three cases:
(1) If $a$ and $b$ are both positive, then $x=3$;
(2) If $a$ and $b$ are one positive and one negative, then $x=-1$;
(3) If $a$ and $b$ are both negative, then $x=-1$.
In summary, the maximum valu... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given $a b c<0$, let
$$
P=\frac{a}{|a|}+\frac{|b|}{b}+\frac{c}{|c|}+\frac{|a b|}{a b}+\frac{a c}{|a c|}+\frac{|b c|}{b c} \text {. }
$$
Find the value of $a P^{3}+b P^{2}+c P+2009$. | Given $a b c<0$, we get $\frac{a b c}{|a b c|}=-1$, and at least one of $a, b, c$ is negative. According to the corollary, we can change the position of the absolute value symbol, then
$$
\begin{aligned}
P & =\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b}{|a b|}+\frac{a c}{|a c|}+\frac{b c}{|b c|} \\
& =\left(1+\... | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a, b, c$ satisfy
$$
\frac{a b}{a+b}=\frac{1}{3}, \frac{b c}{b+c}=\frac{1}{4}, \frac{c a}{c+a}=\frac{1}{5} \text {. }
$$
then $a b+b c+c a=$ $\qquad$ | $$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4, \frac{1}{c}+\frac{1}{a}=5 \\
\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6 \\
\Rightarrow \frac{1}{c}=3, \frac{1}{a}=2, \frac{1}{b}=1 \\
\Rightarrow a=\frac{1}{2}, b=1, c=\frac{1}{3} \\
\Rightarrow a b+b c+c a=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that the altitude to the hypotenuse of right $\triangle A B C$ is 4. Then the minimum value of the area of $\triangle A B C$ is $\qquad$ . | 2. 16 .
Let the lengths of the two legs of the right triangle $\triangle ABC$ be $a$ and $b$, and the length of the hypotenuse be $c$. From the area relationship, we have $ab = 4c$.
By the Pythagorean theorem, we know $c^2 = a^2 + b^2 \geq 2ab = 8c$.
Thus, $c \geq 8$. Therefore, $S_{\triangle ABC} = 2c \geq 16$.
When ... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In rectangle $A B C D$, $A B=12, A D=3, E$ and $F$ are points on $A B$ and $D C$ respectively. Then the minimum length of the broken line $A F E C$ is $\qquad$ . | 3. 15 .
As shown in Figure 5, construct the symmetric points $A_{1}$ and $C_{1}$ of $A$ and $C$ with respect to $DC$ and $AB$, respectively. Connect $A_{1}C_{1}$ to intersect $AB$ and $DC$ at points $E_{1}$ and $F_{1}$, respectively, and connect $A_{1}F$ and $C_{1}E$. Draw a perpendicular from $A_{1}$ to the extension... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ are six different positive integers, taking values from $1, 2, 3, 4, 5, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{alig... | 4. 10 .
Since equation (1) is a cyclic expression about $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{array}{l}
S \geqslant\left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
\quad\left|\left(x_{j+1}-1\right)+\left(x_{j+... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the value of $\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}$. | $\begin{array}{l}\text { Solve the original expression }=\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}\left(1+5^{1998}\right)}{7^{1998}\left(1+5^{1998}\right)}} \\ =\left(\frac{7}{3}\right)^{999} \times\left(\frac{3}{7}\right)^{999}=\left(\frac{7}{3} \times \frac{3}{7}\right)^{999}=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Person A writes down the positive integers $1, 2, \cdots$, 2009 on the blackboard, then turns away from the blackboard, and asks Person B to erase some of these numbers and then add the remainder of the sum of the erased numbers when divided by 7. After several such operations, only two numbers remain on the blackbo... | 3.5.
Since $1+2+\cdots+2009 \equiv 0(\bmod 7)$, therefore, the single digit $a$ satisfies $100+a \equiv 0(\bmod 7)$. Hence $a \equiv 5(\bmod 7)$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. The number of solutions to the equation $\pi^{x-1} x^{2}+\pi^{x^{2}} x-\pi^{x^{2}}=x^{2}+x-1$ is $\qquad$ ( $\pi$ is the ratio of a circle's circumference to its diameter). | 4. 2 .
The original equation is transformed into
$$
x^{2}\left(\pi^{x-1}-1\right)+(x-1)\left(\pi^{x^{2}}-1\right)=0 \text {. }
$$
When $x \neq 0,1$, $x^{2}\left(\pi^{x-1}-1\right)$ and $(x-1)\left(\pi^{x^{2}}-1\right)$ have the same sign.
Therefore, $x=0,1$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given a regular 200-gon $A_{1} A_{2} \cdots A_{200}$, connect the diagonals $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, where $A_{i+200}=$ $A_{i}(i=1,2, \cdots, 9)$. Then these 200 diagonals have $\qquad$ different intersection points inside the regular 200-gon. | 8. 1600.
For each diagonal $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, $A_{i+1}, A_{i+2}, \cdots, A_{i+8}$ each draw two diagonals intersecting with $A_{i} A_{i+9}$. And each such intersection point is counted twice, so the number of different intersection points of these 200 diagonals inside the regular 200-gon does not exc... | 1600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ that satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}(n=2,3, \cdots, m$ is a real parameter $)$. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$. | 10. Since $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}$,
$$
\begin{array}{l}
a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}, a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, \\
a_{6}=\frac{\frac{a_{0}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $x=2 \frac{1}{5}, y=-\frac{5}{11}, z=-2 \frac{1}{5}$. Then $x^{2}+x z+2 y z+3 x+3 z+4 x y+5=$ | Solve: From the given, we have
$$
x y=-1, x+z=0, y z=1 \text {. }
$$
Therefore, the expression to be found is
$$
\begin{array}{l}
=x(x+z)+2 y z+3(x+z)+4 x y+5 \\
=2 \frac{1}{5} \times 0+2 \times 1+3 \times 0+4 \times(-1)+5=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (15 points) Divide the sides of an equilateral $\triangle ABC$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points where the sides of $\triangle ABC$ and these parallel lines intersect are called grid points. If $n$ grid points are cho... | 11. $n_{\min }=5$.
Let the two equal division points on side $AB$ from point $A$ to $B$ be $D$ and $E$, the two equal division points on side $BC$ from point $B$ to $C$ be $F$ and $G$, and the two equal division points on side $CA$ from point $C$ to $A$ be $H$ and $I$, with the central grid point being $K$.
If the mi... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let a tangent line of the circle $x^{2}+y^{2}=1$ intersect the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then the minimum value of $|AB|$ is $\qquad$ . | 3. 2 .
By symmetry, without loss of generality, assume the point of tangency is
$$
P(\cos \theta, \sin \theta)\left(0<\theta<\frac{\pi}{2}\right) \text {. }
$$
Then $|P A|=\tan \theta,|P B|=\cot \theta$.
Thus, $|A B|=|P A|+|P B|=\tan \theta+\cot \theta \geqslant 2$.
Equality holds if and only if $\tan \theta=\cot \th... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)$ be an odd function defined on $\mathbf{R}$, $f(1)=2$, when $x>0$, $f(x)$ is an increasing function, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$. Then the maximum value of the function $f(x)$ on the interval $[-3,-2]$ is | 5. -4 .
Since $f(x)$ is an odd function and is increasing on $(0,+\infty)$, therefore, $f(x)$ is also increasing on $(-\infty, 0)$. Thus, $f(-3) \leqslant f(x) \leqslant f(-2)$.
$$
\begin{array}{l}
\text { Also, } f(2)=f(1)+f(1)=4, \text { then } \\
f(-2)=-f(2)=-4 .
\end{array}
$$
Therefore, the maximum value of the ... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. As shown in Figure $2, A B$ is the diameter of the semicircle $\odot O$, and $C, D$ are two moving points on the semicircle, with $C D / / A B$. If the radius of the semicircle $\odot O$ is 1, then the maximum value of the perimeter of trapezoid $A B C D$ is $\qquad$ . | 7.5.
In Figure 2, connect $A C$, and draw $C H \perp A B$ at point $H$. Let $\angle A B C=\theta\left(0<\theta<\frac{\pi}{2}\right)$. Then $A D=B C=A B \cos \theta=2 \cos \theta$.
Thus, $B H=B C \cos \theta=2 \cos ^{2} \theta$.
Therefore, $C D=A B-2 B H=2-4 \cos ^{2} \theta$.
Hence, the perimeter of trapezoid $A B C D... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 3, in $\triangle A B C$, $A B=3, A C=5$. If $O$ is the circumcenter of $\triangle A B C$, then the value of $\overrightarrow{A O} \cdot \overrightarrow{B C}$ is | 8. 8 .
Let $D$ be the midpoint of side $B C$, and connect $O D, A D$. Then $O D \perp B C$. Therefore,
$$
\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{D O} \cdo... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the sum:
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{60}\right)+\left(\frac{2}{3}+\frac{2}{4}+\cdots+\frac{2}{60}\right)+ \\
\left(\frac{3}{4}+\frac{3}{5}+\cdots+\frac{3}{60}\right)+\cdots+\left(\frac{58}{59}+\frac{59}{60}\right)
\end{array}
$$ | $$
\begin{aligned}
= & \frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\cdots+ \\
& \left(\frac{1}{60}+\frac{2}{60}+\cdots+\frac{59}{60}\right) \\
= & \frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\cdots+\frac{59}{2} \\
= & \frac{1}{2}(1+2+\cdots+59) \\
= & \frac{1}{2} \times \fr... | 885 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the set
$$
M=\left\{x|5-| 2 x-3 \mid \in \mathbf{N}_{+}\right\} \text {. }
$$
Then the number of all non-empty proper subsets of $M$ is ( ).
(A) 254
(B) 255
(C) 510
(D) 511 | - 1. C.
Given $5-|2 x-3| \in \mathbf{N}_{+}$, i.e., $|2 x-3|=0,1, 2,3,4$, we have
$$
M=\left\{-\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{7}{2}\right\} .
$$
Since $M$ contains 9 elements, the number of all non-empty proper subsets of $M$ is $2^{9}-2=510$. | 510 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 10 Calculate:
$$
\frac{2009^{2}-2008^{2}}{19492009^{2}-19492008 \times 19492010+2 \times 2008} .
$$ | Let $19492009=a, 2008=b$. Then the original expression $=\frac{(b+1)^{2}-b^{2}}{a^{2}-(a-1)(a+1)+2 b}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Let the number of ways to choose $k$ pairwise coprime numbers from the set $\{1,2, \cdots, 28\}$ be $T(k)$. Find $T(2)+T(3)+\cdots+T(12)$. | Obviously, the prime numbers in $\{1,2, \cdots, 28\}$ are 2, 3, $5,7,11,13,17,19,23$, a total of 9.
When $k \geqslant 11$, the $k$ numbers taken out, except for 1, must include at least 10 numbers. By the pigeonhole principle, there must be two numbers divisible by the same prime number, which means these two numbers ... | 15459 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Distribute 2010 red cards and 2010 white cards arbitrarily to 2010 players participating in the game, with each person receiving two cards. All players sit in a circle facing down. The game rule is that each operation requires each player to simultaneously follow the following principle: if a player h... | In 2009.
Let $n=2$ 010. A wheel (divided into $n$ sections) can be placed in a circle of $n$ players. Each player, according to the game rules, when the section they face contains a white card and they hold a red card, they place the red card in the section and take back the white card; in all other cases, it is assume... | 2009 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
The initial 267 four-digit number $w_{1}$ and the sum of its four digits is the four-digit number $w_{2}, w_{2}$ and the sum of its four digits is the four-digit number $w_{3}, w_{3}$ and the sum of its four digits is the four-digit number $w_{4}, w_{4}$ and the sum of its four digits is the four-digit number $w_{5}, w... | Let $w_{5-i}=a_{i} b_{i} c_{i} d_{i}(i=0,1, \cdots, 4)$. Then $1001 a_{0}+101 b_{0}+11 c_{0}+2 d_{0}=2009$.
Obviously, $a_{0}=1$ (otherwise, if $a_{0}=2$, then $101 b_{0}+11 c_{0}+2 d_{0}=7$, which only has the solution $b_{0}=c_{0}=0, d_{0}=\frac{7}{2}$, which is not an integer).
Thus, $101 b_{0}+11 c_{0}+2 d_{0}=1008... | 1921 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given real numbers $x, y, z$ satisfy $x-y=8$, $xy+z^{2}=-16$. Then $x+y+z=$ $\qquad$ . | Let $x=a+b, y=a-b$. Then $x-y=2b \Rightarrow b=4$.
Since $xy+z^{2}=-16$, we have $(a+b)(a-b)+z^{2}+16=0$. Also, $-b^{2}=-16$, then $a^{2}+z^{2}=0$. By the property of non-negative numbers, we know $a=0, z=0$. Therefore, $x=4, y=-4$. Hence, $x+y+z=4-4+0=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 1, it is known that rectangle $A B C D$ can be exactly divided into seven small rectangles of the same shape and size. If the area of the small rectangle is 3, then the perimeter of rectangle $A B C D$ is | $-1.19$.
Let the length of the small rectangle be $a$ and the width be $b$. Then $3a = 4b \Rightarrow a = \frac{4}{3}b \Rightarrow ab = \frac{4}{3}b^2 = 3$. Solving this, we get $b = \frac{3}{2}, a = 2$.
Therefore, the perimeter of rectangle $ABCD$ is $2(4a + b) = 8a + 2b = 16 + 3 = 19$. | 19 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an "integer point". The number of integer points on the graph of the function $y=\frac{x+12}{2 x-1}$ is $\qquad$ | 2. 6 .
Notice that $2 y=1+\frac{25}{2 x-1}$.
Since $y$ is an integer, $\frac{25}{2 x-1}$ must also be an integer. Therefore, $2 x-1= \pm 1, \pm 5, \pm 25$.
Solving for $x$ gives $x=-12,-2,0,1,3,13$.
Correspondingly, $y=0,-2,-12,13,3,1$.
Thus, there are 6 integer points. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $a$ and $b$ are integers. Then the number of ordered pairs $(a, b)$ that satisfy $a + b + ab = 2008$ is $\qquad$ groups. | 4. 12 .
$$
\begin{array}{r}
\text { Since }(a+1)(b+1) \\
=2009=41 \times 7^{2} \text {, and given }
\end{array}
$$
that $a, b$ take integer values, so, $a+1, b+1$ have 12 factor combinations, i.e., $1 \times 2009, 7 \times 287, 41 \times 49$ and their negatives as well as swaps. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Arrange the numbers $1,2, \cdots, 10$ in a row in some order, such that the sum of any three consecutive numbers does not exceed $n$. Answer the following:
(1) When $n=10$, can it be arranged? Please explain your reasoning;
(2) When it can be arranged, what is the minimum value of $n$? | 8. (1) Suppose $n=10$ is already arranged. Then the sum of the last nine numbers is less than or equal to 30. Thus, the first number is not less than 25, which is a contradiction. Therefore, it cannot be arranged.
(2) With the same consideration as (1).
When $n=11,12,13,14$, none can be arranged.
When $n=15$, the sum o... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $x_{1}, x_{2}, \cdots, x_{n}$ take values 7 or -7, and satisfy
(1) $x_{1}+x_{2}+\cdots+x_{n}=0$;
(2) $x_{1}+2 x_{2}+\cdots+n x_{n}=2009$.
Determine the minimum value of $n$. | 9. First, when $n=34$,
$$
\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{16}=x_{18}=-7, \\
x_{17}=x_{19}=x_{20}=\cdots=x_{34}=7
\end{array}
$$
satisfies the conditions of the problem.
Second, divide both sides of condition (2) by 7, and let $y_{i}=\frac{x_{i}}{7}(i=1,2, \cdots, n)$. Then
$$
\begin{array}{l}
y_{i}= \pm 1, \\
y... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. (15 points) Fill seven different perfect squares into seven consecutive boxes so that the sum of any three adjacent boxes is greater than 100. Find the minimum possible value of the sum of these seven perfect squares.
| 3. Since the three perfect squares at positions $1, 4$, and $7$ are all different, considering the smallest possible values, one of these numbers must be at least 9. After positioning 9, the remaining six positions can be divided into two groups, each with three consecutive numbers whose sum is at least 101. Therefore,... | 211 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let convex quadrilateral $ABCD$ satisfy $AB=AD=1$, $\angle A=160^{\circ}, \angle C=100^{\circ}$. Then the range of the length of diagonal $AC$ is $\qquad$ . | 2. $\{1\}$.
Since $\angle C=180^{\circ}-\frac{\angle A}{2}$, therefore, points $B$, $C$, and $D$ lie on a circle with $A$ as the center and a radius of 1. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. If the three-digit decimal number $n=abc$ satisfies that $a$, $b$, $c$ form an arithmetic sequence, then the maximum possible value of a prime factor of $n$ is $\qquad$ | 3. 317 .
$$
\begin{array}{l}
31 n, 999=27 \times 37, 987=3 \times 7 \times 47, \\
963=9 \times 107, 951=3 \times 317 .
\end{array}
$$ | 317 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. The sum of the radii of all circles passing through point $A(1505,1008)$ and tangent to the lines $l_{1}: y=0$ and $l_{2}: y=\frac{4}{3} x$ is $\qquad$ | 5. 2009 .
From $\tan 2 \theta=\frac{4}{3} \Rightarrow \tan \theta=\frac{1}{2}$, the center of the circle lies on the line $y=\frac{x}{2}$, and the center is $(2 r, r)$ (where $r$ is the radius of the circle).
$$
\begin{array}{l}
\text { Hence }(1505-2 r)^{2}+(1008-r)^{2}=r^{2} \\
\Rightarrow 4 r^{2}-8036 r+1505^{2}+10... | 2009 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the lines $l_{1} / / l_{2}$, and take 10 points $A_{1}, A_{2}, \cdots, A_{10}$ and $B_{1}, B_{2}, \cdots, B_{10}$ on $l_{1}$ and $l_{2}$ respectively. Then the line segments $A_{1} B_{1}, A_{2} B_{2}, \cdots, A_{10} B_{10}$ can divide the strip region between $l_{1}$ and $l_{2}$ into at most $\qquad$ non-overlap... | 8. 56 .
A line segment divides the original region into two parts. The $k$-th line segment can be divided into at most $k$ segments by the previous $k-1$ line segments, and it can add at most $k$ parts. Therefore, $k$ line segments can divide the region into at most $2+2+3+\cdots+k=\frac{k^{2}+k+2}{2}$ parts. | 56 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Simplify:
$$
\frac{(\sqrt{x}-\sqrt{y})^{3}+2 x \sqrt{x}+y \sqrt{y}}{x \sqrt{x}+y \sqrt{y}}+\frac{3 \sqrt{x y}-3 y}{x-y} .
$$ | Let $\sqrt{x}=a+b, \sqrt{y}=a-b$. Then
$$
\sqrt{x y}=a^{2}-b^{2}, \sqrt{x}+\sqrt{y}=2 a, \sqrt{x}-\sqrt{y}=2 b \text {. }
$$
Therefore, the original expression is
$$
\begin{aligned}
= & \frac{(2 b)^{3}+2(a+b)^{3}+(a-b)^{3}}{(a+b)^{3}+(a-b)^{3}}+ \\
& \frac{3\left(a^{2}-b^{2}\right)-3(a-b)^{2}}{4 a b} \\
= & \frac{3\le... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. (22 points) Suppose a set of planar points $S$ has the following properties:
(1) No three points are collinear;
(2) The distance between any two points is unique.
For two points $A$ and $B$ in $S$, if there exists a point $C \in S$ such that $|A C|<|A B|<|B C|$, then $A B$ is called a "middle edge" of $S$. For thre... | 4. Color all middle edges of $S$ red, and color other edges blue.
When $n \geqslant 6$, according to Ramsey's theorem, there must exist a monochromatic triangle, which must have a middle edge, and it must be a middle edge triangle.
The following set of five points with properties (1) and (2) does not have a middle ed... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n+2}}$. Then $\sum_{n=1}^{2010} a_{n}=$ $\qquad$ . | 4.5 829 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$;
When $n \equiv 2,4,8(\bmod 10)$,
$$
\begin{array}{l}
(n+1)^{n+2} \equiv 1(\bmod 4) \\
\Rightarrow a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10) ;
\end{array}
$$
When $n \equiv 3,7,9(\bmod 10)$,
$$
\begin{array}... | 5829 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. The line $l: x+y=t$ intersects the circle $\odot O: x^{2}+y^{2}=20$ at points $A$ and $B$, and $S_{\triangle O A B}$ is an integer. Then the number of all positive integer values of $t$ that satisfy the condition is $\qquad$ . | 7.2.
Let $\angle A O B=2 \alpha, O C \perp A B$, with the foot of the perpendicular being $C$. Then $O C=\sqrt{20} \cos \alpha=\frac{t}{\sqrt{1+1}}=\frac{t}{\sqrt{2}}$ $\Rightarrow \cos \alpha=\frac{t}{2 \sqrt{10}}$.
Also, $S_{\triangle O A B}=10 \sin 2 \alpha \leqslant 10$, so $\sin 2 \alpha=\frac{k}{10}(k \in\{1,2, ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given the function $f:\{0,1, \cdots, 2010\} \rightarrow \mathbf{N}$. If for all possible integers $x$, we have
$$
\begin{array}{l}
f(4 x+2)=f(4 x+1), \\
f(5 x+3)=f(5 x+2), \\
f(7 x+5)=f(7 x+4),
\end{array}
$$
then $f(x)$ can take at most $\qquad$ different values. | 8. 1033 .
For the function $f$, construct the function $g$, defined as:
$$
\begin{array}{l}
g(0)=1, \\
g(i+1)=\left\{\begin{array}{l}
1, f(i+1) \neq f(i) ; \\
0, f(i+1)=f(i),
\end{array}\right.
\end{array}
$$
where $i=0,1, \cdots, 2009$.
Then $g$ is a mapping from $\{0,1, \cdots, 2010\}$ to $\{0,1\}$, and for all pos... | 1033 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $\left(2+x-2 x^{2}\right)^{1005}=\sum_{k=0}^{2010} a_{k} x^{k}$. Then
$$
\begin{array}{l}
a_{1}+3 a_{3}+5 a_{5}+\cdots+2009 a_{2009} \\
=
\end{array}
$$ | 3. 1005.
Let $f(x)=\left(2+x-2 x^{2}\right)^{1005}$. Then $f^{\prime}(x)=1005\left(2+x-2 x^{2}\right)^{1004}(1-4 x)$. Therefore, $f^{\prime}(1)=1005 \times(-3)$, $f^{\prime}(-1)=1005 \times 5$.
Given $f(x)=\sum_{k=0}^{2010} a_{k} x^{k}$, we have
$$
\begin{array}{l}
f^{\prime}(x)=\sum_{k=1}^{2010} k a_{k} x^{k-1} . \\
... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If the polynomial $f(x)=x^{3}-6 x^{2}+a x+a$ has three roots $x_{1}, x_{2}, x_{3}$ that satisfy
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0 \text {, }
$$
then the value of the real number $a$ is .. $\qquad$ | 6. -9 .
From the problem, we have
$$
\begin{array}{l}
g(t)=f(t+3) \\
=(t+3)^{3}-6(t+3)^{2}+a(t+3)+a \\
=t^{3}+3 t^{2}+(a-9) t+4 a-27
\end{array}
$$
The three roots \( t_{1}, t_{2}, t_{3} \) satisfy \( t_{1}^{3}+t_{2}^{3}+t_{3}^{3}=0 \).
By the relationship between roots and coefficients, we get
$$
\begin{array}{l}
\l... | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given an ellipse centered at the origin, with foci on the $x$-axis, the length of the major axis is twice the length of the minor axis, and it passes through the point $M(2,1)$. A line $l$ parallel to $OM$ has a $y$-intercept of $m (m<0)$, and intersects the ellipse at two distinct points $A$ and $B$. F... | 10. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then $\left\{\begin{array}{l}a=2 b, \\ \frac{4}{a^{2}}+\frac{1}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=8, \\ b^{2}=2 \text {. }\end{array}\right.\right.$
Therefore, the equation of the ellipse is $\frac{x^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Try to find $\left[a_{2009}^{2}\right]$. | 11. From $a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}$, we get $a_{n+1}^{2}=\frac{a_{n}^{2}}{n^{2}}+\frac{n^{2}}{a_{n}^{2}}+2$. Given $a_{1}=1$, we have $a_{2}^{2}=4, a_{3}^{2}=4, a_{4}^{2}=4+\frac{25}{36}$.
Next, we use mathematical induction to prove:
When $n \geqslant 4$, $n+\frac{2}{n}\frac{n+1}{n^{2}}+\frac{n^{2}}{n+1... | 2009 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}, x y z>0$,
and
$$
\begin{array}{l}
\sqrt[3]{\frac{2010}{x^{2}}+\frac{2011}{y^{2}}+\frac{2012}{z^{2}}} \\
=\sqrt[3]{2010}+\sqrt[3]{2011}+\sqrt[3]{2012} .
\end{array}
$$
Find the value of $x+y+z$. | Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}=k$, obviously $k \neq 0$.
Then $2010=x^{3} k, 2011=y^{3} k, 2012=z^{3} k$.
From the given, we have
$$
\sqrt[3]{x k+y k+z k}=\sqrt[3]{x^{3} k}+\sqrt[3]{y^{3} k}+\sqrt[3]{z^{3} k},
$$
which means $\sqrt[3]{k} \sqrt[3]{x+y+z}=\sqrt[3]{k}(x+y+z)$.
Since $k \neq... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find all positive integers $n$, such that $2^{n-1} n+1$ is a perfect square.
(2004, Slovenia IMO National Team Selection Test | Let $2^{n-1} n+1=m^{2}\left(m \in \mathbf{N}_{+}\right)$. Then $2^{n-1} n=(m+1)(m-1)$.
When $n=1,2,3,4$, $2^{n-1} n+1$ is not a perfect square.
Therefore, $n \geqslant 5, 16 \mid (m+1)(m-1)$.
Since $m+1$ and $m-1$ have the same parity, both $m+1$ and $m-1$ are even, and $m$ is odd.
Let $m=2 k-1\left(k \in \mathbf{N}_{+... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Proof: There exist positive integers $a_{i}(1 \leqslant i \leqslant 8)$, such that
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2}-1}}+\cdots+ \\
\sqrt{\sqrt{a_{8}}-\sqrt{a_{8}-1}}=2 .
\end{array}
$$ | Prove the construction of the identity:
$$
\begin{array}{l}
\sqrt{(2 i+1)^{2}}-\sqrt{(2 i+1)^{2}-1} \\
=2 i+1-2 \sqrt{i(i+1)} \\
=(\sqrt{i+1}-\sqrt{i})^{2} .
\end{array}
$$
Take \( a_{i}=(2 i+1)^{2} \) for \( i=1,2, \cdots, 8 \), then
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
7. Given that $a$ is a root of the equation $x^{2}-3 x+1=0$. Then the value of the fraction $\frac{2 a^{6}-6 a^{4}+2 a^{5}-a^{2}-1}{3 a}$ is $\qquad$ - | II. 7. -1.
According to the problem, we have $a^{2}-3 a+1=0$.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 a^{3}\left(a^{2}-3 a+1\right)-\left(a^{2}+1\right)}{3 a} \\
=-\frac{a^{2}+1}{3 a}=-1 .
\end{array}
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Robots A and B simultaneously conduct a $100 \mathrm{~m}$ track test at a uniform speed, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the finish line. After calc... | 8. 1 .
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙 }}$, respectively. When 甲 is $1 \mathrm{~m}$ away from the finish line, the time spent is $t$. Then $v_{\text {乙 }} t=x-2$.
Thus, $\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}$.
... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The positive numbers $b_{1}, b_{2}, \cdots, b_{60}$ are arranged in sequence and satisfy $\frac{b_{2}}{b_{1}}=\frac{b_{3}}{b_{2}}=\cdots=\frac{b_{60}}{b_{59}}$.
Determine the value of $\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right)$. | Given that $\left\{b_{k}\right\}$ is a geometric sequence with a common ratio of $q$, then $b_{k}=b_{1} q^{k-1}$. Therefore, $b_{k} b_{61-k}=b_{1} b_{60}$. Hence,
$$
\begin{array}{l}
\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right) \\
=\log _{b_{1} b_{60}}\left(b_{1} b_{60}\right)^{30}=30 .
\end{array}
$$ | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. $[x]$ represents the greatest integer not exceeding the real number $x$. If
$$
\left[\log _{3} 6\right]+\left[\log _{3} 7\right]+\cdots+\left[\log _{3} n\right]=2009 \text {, }
$$
determine the value of the positive integer $n$. | 4. For $3^{k} \leqslant i \leqslant 3^{k+1}-1$, we have $\left[\log _{3} i\right]=k$. Therefore, $\sum_{i=3^{k}}^{3 k+1}\left[\log _{3} i\right]=2 \times 3^{k} k$.
Also, $3+\sum_{k=2}^{4} 2 \times 3^{k} k<2009<3+\sum_{k=2}^{5} 2 \times 3^{k} k$, hence $3^{5}<n<3^{6}-1$.
From $3+\sum_{k=2}^{4} 2 \times 3^{k} k+5\left(n... | 474 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. The graphs of the functions $f(x)=2 x^{2}-2 x-1$ and $g(x)=$ $-5 x^{2}+2 x+3$ intersect at two points. The equation of the line passing through these two points is $y=a x+b$. Find the value of $a-b$. | 6. The graphs of the functions $f(x)$ and $g(x)$ intersect at two points, which are the solutions to the system of equations $\left\{\begin{array}{l}y=2 x^{2}-2 x-1, \\ y=-5 x^{2}+2 x+3\end{array}\right.$. Therefore,
$$
\begin{array}{l}
7 y=5\left(2 x^{2}-2 x-1\right)+2\left(-5 x^{2}+2 x+3\right) \\
=-6 x+1 .
\end{arra... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given that the 6027-digit number $\frac{a b c a b c \cdots a b c}{2000 \uparrow a b c}$ is a multiple of 91. Find the sum of the minimum and maximum values of the three-digit number $\overline{a b c}$. | 7. From $91=7 \times 13, 1001=7 \times 11 \times 13$, we know 91 | 1001.
And $\overline{a b c a b c}=1001 \times \overline{a b c}$, thus, $91 \mid \overline{a b c a b c}$. $2009 \uparrow a b c$
$$
\overline{a_{2009 \uparrow a b c}^{a b c a b c}}=\underset{2 \times 1004 \uparrow a b c}{\overline{a b c a b c \cdots a b c... | 1092 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. A three-digit number has 3 digits, none of which are 0, and its square is a six-digit number that has exactly 3 digits as 0. Write down one such three-digit number: $\qquad$ | 8.448 or 548 or 949 | 448 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { Three. (20 points) (1) Prove: } \\
\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=4 \sin ^{2} 2 x-3(x \in \mathbf{R}) ;
\end{array}
$$
(2) Find the value: $\prod_{t=0}^{2^{8}}\left(4 \sin ^{2} \frac{t \pi}{2^{9}}-3\right)$. | $$
\begin{array}{l}
=(1)\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=16 \sin ^{2} x \cdot \cos ^{2} x-12\left(\sin ^{2} x+\cos ^{2} x\right)+9 \\
=4 \sin ^{2} 2 x-3
\end{array}
$$
(2) Let $A_{k}=\prod_{i=0}^{2 k-1}\left(4 \sin ^{2} \frac{t \pi}{2^{k}}-3\right)$.
When $k \geqslant 2$, by (1) we have
$$
... | -3 | Algebra | proof | Yes | Yes | cn_contest | false |
For a positive integer $n$, let $t_{n}=\frac{n(n+1)}{2}$. Writing down the last digits of $t_{1}=1, t_{2}=3, t_{3}=6, t_{4}=10, t_{5}=15 \cdots \cdots$ can form an infinite repeating decimal: $0.13605 \cdots$. Find the length of the repeating cycle of this decimal. | $$
\begin{array}{l}
t_{n+20}-t_{n}=\frac{(n+20)(n+20+1)}{2}-\frac{n(n+1)}{2} \\
=20 n+210=10(2 n+21)
\end{array}
$$
That is, the last digit of $t_{n+20}$ is the same as that of $t_{n}$.
Therefore, 20 is the length of the repeating cycle of this repeating decimal. This repeating decimal is
$$
0.13605186556815063100 .
$... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$ (for example, $f(3)$ is the sum of the digits of $3 \times 3^{2}+3+1=31$, i.e., $f(3)=4$).
(1) Prove that for any positive integer $n, f(n) \neq 1$, and $f(n) \neq 2$;
(2) Try to ... | (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, hence $f(n) \neq 1$.
If $f(n)=2$, then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e., $3 n^{2}+n+1=10^{k}+1$ (where $k$ is an integer greater than 1).
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Since $(n, ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a three-digit number $x y z(1 \leqslant x \leqslant 9,0 \leqslant y, z$ $\leqslant 9)$. If $x y z=x!+y!+z!$, then the value of $x+y+z$ is | 7. 10 .
Since $6!=720$, we have $0 \leqslant x, y, z \leqslant 5$.
And $1!=1, 2!=2, 3!=6, 4!=24, 5!=120$.
By observation, we get $x=1, y=4, z=5$. Therefore,
$$
x+y+z=10 \text {. }
$$ | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0
\end{array}
$$
$\left(a 、 b \in \mathbf{N}_{+}\right)$ have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^... | Given the known equations, we have $a \neq 1, b \neq 1$. Therefore, $a > 1, b > 1$, and $a \neq b$. Let $x_{0}$ be the common root of the two equations. It is easy to see that $x_{0} \neq 1$. By the definition of the roots of the equation, $a$ and $b$ are the two distinct real roots of the equation
$$
\left(1-x_{0}\rig... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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