problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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7. (40 points) A cat caught 81 mice and arranged them in a circle, numbering them from $1 \sim 81$ in a clockwise direction. The cat starts counting from a certain mouse in a clockwise direction, continuously counting “$1, 2, 3$” and eating all the mice that are counted as 3. As the cat continues to count, the circle g... | 7. First, arrange the numbers of all the mice in 9 columns as follows.
\begin{tabular}{ccccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\
19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 \\
28 & 29 & 30 & 31 & 32 & 33 & 34 & 35 & 36 \\
37 & 38 & 39 & 40 & 41 & 42 & 43 & 44 & 45 \\
4... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
10. (40 points) Given positive integers $a$, $b$, and $c$ to 甲, 乙, and 丙 respectively, each person only knows their own number. They are told that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, and each is asked the following two questions:
(1) Do you know the value of $a+b+c$?
(2) Do you know the values of $a$, $b$, and $c$... | 10. First solve the equation $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.
Assume $a \geqslant b \geqslant c$. If $c=3$, then $1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{3}{c}$.
Thus, $a=b=c=3$. Otherwise, $c=2$.
Then, $\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$.
Solving this, we get $a=b=4$ or $b=3, a=6$.
The sums co... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a, b$ be positive real numbers, and
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}, \\
(a-b)^{2}=4(a b)^{3} .
\end{array}
$$
Then $\log _{a} b=$ $\qquad$ | 3. -1 .
From $\frac{1}{a}+\frac{1}{b} \leqslant 2 \sqrt{2}$, we get $a+b \leqslant 2 \sqrt{2} a b$.
$$
\begin{array}{l}
\text { Also, }(a+b)^{2}=4 a b+(a-b)^{2} \\
=4 a b+4(a b)^{3} \\
\geqslant 4 \times 2 \sqrt{a b(a b)^{3}}=8(a b)^{2},
\end{array}
$$
which means $a+b \geqslant 2 \sqrt{2} a b$.
Thus, $a+b=2 \sqrt{2}... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Now arrange for 7 students to participate in 5 sports events, requiring that students A and B cannot participate in the same event, each event must have participants, and each person can only participate in one event. The number of different arrangements that meet the above requirements is $\qquad$ (answer in number... | 5. 15000 .
According to the problem, there are two scenarios that meet the conditions:
(1) One project has 3 participants, with a total of
$$
C_{7}^{3} \times 5! - C_{5}^{1} \times 5! = 3600
$$
schemes;
(2) Two projects each have 2 participants, with a total of
$$
\frac{1}{2}\left(\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2... | 15000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given
$$
a_{n}=\mathrm{C}_{200}^{n}(\sqrt[3]{6})^{200-n}\left(\frac{1}{\sqrt{2}}\right)^{n}(n=1,2, \cdots, 95) \text {. }
$$
The number of integer terms in the sequence $\left\{a_{n}\right\}$ is $\qquad$ | 8. 15 .
Notice that $a_{n}=\mathrm{C}_{200}^{n} \times 3^{\frac{200-n}{3}} \times 2^{\frac{200-5 n}{6}}$.
To make $a_{n}(1 \leqslant n \leqslant 95)$ an integer, it must be that $\frac{200-n}{3}$ and $\frac{400-5 n}{6}$ are both integers, i.e., $61(n+4)$.
When $n=6 k+2(k=0,1, \cdots, 13)$, $\frac{200-n}{3}, \frac{400... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $A$ be a $3 \times 9$ grid, with each small cell filled with a positive integer. If the sum of all numbers in an $m \times n (1 \leqslant m \leqslant 3, 1 \leqslant n \leqslant 9)$ subgrid of $A$ is a multiple of 10, then it is called a "good rectangle"; if a $1 \times 1$ cell in $A$ is not contai... | First, we prove by contradiction that there are no more than 25 bad cells in $A$.
Assume the conclusion is not true. Then, in the grid $A$, there is at most 1 cell that is not a bad cell. By the symmetry of the grid, we can assume that all cells in the first row are bad cells.
Let the numbers filled in the $i$-th col... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given a sequence of numbers
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \cdots, \frac{k}{1}, \frac{k-1}{2}, \cdots, \frac{1}{k} \text {. }
$$
In this sequence, the index of the 40th term that equals 1 is ( ).
(A) 3120
(B) 3121
(C) 3200
(D) 3201 | - 1. B.
For the terms where the sum of the numerator and denominator is $k+1$, we denote them as the $k$-th group. According to the arrangement rule, the 40th term with a value of 1 should be the 40th number in the $2 \times 40-1=79$ group, with the sequence number being
$$
\begin{array}{l}
(1+2+\cdots+78)+40 \\
=\fra... | 3121 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{199}-\frac{1}{200}}{\frac{1}{201^{2}-1^{2}}+\frac{1}{202^{2}-2^{2}}+\cdots+\frac{1}{300^{2}-100^{2}}} \\
= \\
\end{array}
$$ | 3. 400 .
Original expression
$$
\begin{array}{l}
=\frac{\left(1+\frac{1}{2}+\cdots+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{200}\right)}{\frac{1}{202 \times 200}+\frac{1}{204 \times 200}+\cdots+\frac{1}{400 \times 200}} \\
=\frac{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}{\frac{1}{400... | 400 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. A real-coefficient polynomial $P(x)$ of degree not exceeding 2011 takes integer values for any integer $x$, and the remainders when $P(x)$ is divided by $x-1, x-2, \cdots, x-2011$ are $1, 2, \cdots, 2011$ respectively. Then $\max _{x \in \{-1, -2, \cdots, -2011\}}|P(x)|$ has the minimum value of $\qquad$ | $-1.2011$
First, $P(x)-x$. can be divided by $x-1, x-2, \cdots$, $x-2011$ respectively, so we can assume
$$
P(x)-x=q(x-1)(x-2) \cdots(x-2011) \text {, }
$$
where, $q \in \mathbf{Q}$.
If $q \neq 0$, then
$$
\begin{array}{l}
|P(-2011)|=\left|-2011-q \cdot \frac{4022!}{2011!}\right| \\
\geqslant\left|q \cdot \frac{4022!}... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $x$ is a positive integer, and $2011-x$ is a perfect cube. Then the minimum value of $x$ is $\qquad$ . | 1. 283.
From $12^{3}=1728<2011<13^{3}=2197$, we know the minimum value of $x$ is
$$
2011-1728=283 \text {. }
$$ | 283 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Divide a cube with an edge length of a positive integer into 99 smaller cubes, among which, 98 smaller cubes are unit cubes. Find the surface area of the original cube. | Let the side length of the original cube be $x$, and the side lengths of the 99 smaller cubes be 1 and $y(x>y>1)$. Then
$$
\begin{array}{l}
x^{3}-y^{3}=98 \\
\Rightarrow(x-y)\left[(x-y)^{2}+3 x y\right]=98 \\
\Rightarrow(x-y) \mid 98=7^{2} \times 2 \\
\Rightarrow x-y=1,2,7,14,49,98 . \\
\text { By }(x-y)^{3}<x^{3}-y^{3... | 150 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) As shown in Figure 4, given that $AB$ is the diameter of $\odot O$, chord $CD$ intersects $AB$ at point $E$. A tangent to $\odot O$ is drawn from point $A$ and intersects the extension of $CD$ at point $F$. Given $AC=8$, $CE:ED=6:5$, and $AE:EB=2:3$. Find the length of $AB$ and the value of $\tan \angle... | Let $C E=6 x, E D=5 x, A E=2 y$, $E B=3 y, D F=z$.
By the intersecting chords theorem, we have
$$
A E \cdot B E=C E \cdot E D \Rightarrow y=\sqrt{5} x \text {. }
$$
By the secant-tangent theorem, we have
$$
A F^{2}=D F \cdot C F=z(z+11 x) \text {. }
$$
By the Pythagorean theorem, we have
$$
\begin{array}{l}
A F^{2}=E... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the function
$$
f(x)=\frac{1+\ln (x+1)}{x} \text {, }
$$
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
untranslated part:
$k$ is a positive integer. When $x>0$, $f(x)>\frac{k}{x+1}$ always holds. Find the maximum value of $k$.
(Note: The... | 11. Since $x>0$, we have
$$
k0)
$$
always holds.
$$
\text{Let } g(x)=\frac{(x+1)[1+\ln (x+1)]}{x}(x>0) \text{.}
$$
Then $g^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
Let $h(x)=x-1-\ln (x+1)$.
Obviously, $h(2)0$.
If $h\left(x_{0}\right)=0$, then $20$ (since $x>0$), then $h(x)$ is monotonically increasing in $(0,+\infty... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the number of ordered integer pairs $(a, b)$ such that
$$
x^{2}+a x+b=167 y
$$
has integer solutions $(x, y)$, where $1 \leqslant a, b \leqslant 2004 .^{[4]}$ (2004, Singapore Mathematical Olympiad) | If $x^{2}+a x+b \equiv 0(\bmod 167)$, then completing the square gives
$$
a^{2}-4 b \equiv(2 x+a)^{2}(\bmod 167) .
$$
Therefore, for a fixed $a$, $a^{2}-4 b$ is a quadratic residue modulo 167.
Hence, by the lemma, $b$ can take $\frac{167-1}{2}+1=84$ values modulo 167.
And $\frac{2004}{167}=12$, so each $a$ correspon... | 2020032 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a=\sqrt{3}-1$. Then $a^{2012}+2 a^{2011}-2 a^{2010}=$ | 2. 0 .
Notice that $a^{2}=(\sqrt{3}-1)^{2}=4-2 \sqrt{3}$. Then $a^{2}+2 a-2=0$.
Therefore, $a^{2012}+2 a^{2011}-2 a^{2010}$ $=a^{2010}\left(a^{2}+2 a-2\right)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Figure 2 is a rectangle composed of 6 squares. If the area of the smallest square is 1, then the area of this rectangle is $\qquad$ .
| 3. 143 .
Let the side lengths of the six squares, from smallest to largest, be
$$
1, x, x, x+1, x+2, x+3 \text{.}
$$
Then, by the equality of the top and bottom sides of the rectangle, we have
$$
\begin{array}{l}
x+x+(x+1)=(x+2)+(x+3) \\
\Rightarrow x=4 .
\end{array}
$$
Thus, the length and width of the rectangle ar... | 143 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $\triangle A B C$
the lengths of the three medians are 3,
4, 5. Then $S_{\triangle M B C}$ is $\qquad$ | 4. 8 .
As shown in Figure 4, extend GD to point $D^{\prime}$, making it twice as long.
Then the side lengths of $\triangle G D^{\prime} C$ are $\frac{2}{3}$ times the lengths of the three medians of $\triangle A B C$.
Therefore, it is a right triangle, and its area is $\frac{8}{3}$.
Additionally, the area of $\triangl... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) There are $m$ regular $n$-sided polygons, and the sum of the interior angles of these $m$ regular polygons can be divided by 8. Find the minimum value of $m+n$.
| Three, from the problem, we know that the total sum of the interior angles of these $m$ regular polygons is $m(n-2) \times 180$.
From $81[180 m(n-2)]$
$\Rightarrow 21 \mathrm{~m}(n-2) \Rightarrow 21 \mathrm{mn}$.
Thus, at least one of $m$ and $n$ is even.
Also, $m \geqslant 1, n \geqslant 3$, and both are integers.
To ... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) There are several (more than enough) socks in red, yellow, blue, and white. If any two socks of the same color can make 1 pair, the question is: What is the minimum number of socks needed to ensure that 10 pairs of socks can be formed? | Solution 1 Since there are 4 colors, among 5 socks, there must be 1 pair.
After taking out 1 pair, 3 socks remain. By adding 2 more socks, another pair can be formed.
Following this logic, the number of pairs of socks $(x)$ and the number of socks needed $(y)$ have the following relationship:
$$
y=2 x+3 \text {. }
$$... | 23 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 As shown in Figure 5, in $\triangle A B C$, $\angle C=90^{\circ}$, $I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$
(2004, Beijing Middle School Mathematics Competit... | Solve As shown in Figure 5, construct the symmetric points $F, G$ of points $E, D$ with respect to $AD, BE$ respectively. Then $F, G$ lie on $AB$. Connect $IF, IG$. It is easy to know that
$$
\angle AIB=90^{\circ}+\frac{1}{2} \angle C=135^{\circ}.
$$
By the properties of axial symmetry, we have
$$
\begin{array}{l}
IF=... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the set $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can be found two elements $a, b$, such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$ (where $|M|$ denotes the number of elements in the set $M$). (Supplied by Feng Zhigang) | 2. When
$$
M=\left\{1,2,2^{2}, \cdots, 2^{10}, 3,3 \times 2,3 \times 2^{2}, \cdots, 3 \times 2^{9}\right\}
$$
it satisfies the condition, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011,
\end{array}
$$
contradiction.
In summary, the maximum value of $|M|$ is 21. | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given an integer $n \geqslant 2$.
(1) Prove: The set $\{1,2, \cdots, n\}$ can be properly arranged into subsets $A_{1}, A_{2}, \cdots, A_{2^{n}}$, such that the number of elements in $A_{i}$ and $A_{i+1}\left(i=1,2, \cdots, 2^{n}\right.$, and $\left.A_{2^{n+1}}=A_{1}\right)$ differ by exactly 1.
(2) For subsets $A_{... | 3. (1) Prove by mathematical induction: there exists a subsequence $A_{1}, A_{2}, \cdots, A_{2^{n}}$ that meets the requirements, and $A_{1}=\{1\}$, $A_{2^{n}}=\varnothing$.
When $n=2$, the sequence $\{1\}, \{1,2\}, \{2\}, \varnothing$ satisfies the requirements.
Assume that when $n=k$, there exists a subsequence $B_... | 0 | Combinatorics | proof | Yes | Yes | cn_contest | false |
1. Nine positive real numbers $a_{1}, a_{2}, \cdots, a_{9}$ form a geometric sequence, and
$$
a_{1}+a_{2}=\frac{3}{4}, a_{3}+a_{4}+a_{5}+a_{6}=15 .
$$
Then $a_{7}+a_{8}+a_{9}=$ . $\qquad$ | 1. 112.
Let the common ratio be $q$. Then, from the given conditions, we have
$$
\begin{array}{l}
a_{1}(1+q)=\frac{3}{4} \\
a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15
\end{array}
$$
Dividing the above two equations yields $q^{2}\left(1+q^{2}\right)=20$.
Thus, $q=2, a_{1}=\frac{1}{4}$.
Therefore, $a_{7}+a_{8}+a_{9}=a_{... | 112 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. The function $f(x)$ is defined on $(0, \infty)$, and satisfies
$$
f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0 \text {. }
$$
Then the minimum value of $f(x)$ is $\qquad$ | 5.3.
From $f(x)-2 x f\left(\frac{1}{x}\right)+3 x^{2}=0$, we get $f\left(\frac{1}{x}\right)-\frac{2}{x} f(x)+\frac{3}{x^{2}}=0$.
By solving the above two equations simultaneously, we obtain $f(x)=x^{2}+\frac{2}{x}$.
By the AM-GM inequality,
$$
x^{2}+\frac{2}{x}=x^{2}+\frac{1}{x}+\frac{1}{x} \geqslant 3\left(x^{2} \cdo... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The complex number $z$ satisfies
$$
|z|(3 z+2 \mathrm{i})=2(\mathrm{i} z-6) \text {. }
$$
Then $|z|$ equals $\qquad$ . | 6. 2 .
Solution 1 Direct calculation shows that
$$
|3 z+2 \mathrm{i}|^{2}-|\mathrm{i} z-6|^{2}=8\left(|z|^{2}-4\right) \text {. }
$$
From this, if $|z|>2$, then
$$
\begin{array}{l}
|3 z+2 \mathrm{i}|>|\mathrm{i} z-6| \\
\Rightarrow|| z|(3 z+2 \mathrm{i})|>|2(\mathrm{i} z-6)|,
\end{array}
$$
which contradicts the giv... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $\left(1+x-x^{2}\right)^{10}=a_{0}+a_{1} x+\cdots+a_{20} x^{20}$. Then $a_{0}+a_{1}+2 a_{2}+\cdots+20 a_{20}=$ $\qquad$ | 7. -9 .
Let $x=0$, we get $a_{0}=1$.
Differentiating both sides of the given equation, we get
$$
\begin{array}{l}
10\left(1+x-x^{2}\right)^{9}(1-2 x) \\
=a_{1}+2 a_{2} x+\cdots+20 a_{20} x^{19} . \\
\text { Let } x=1 \text {, we get. } \\
a_{1}+2 a_{2}+\cdots+20 a_{20}=-10 .
\end{array}
$$
Then $a_{0}+a_{1}+2 a_{2}+\... | -9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 10, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition) | As shown in Figure 10, construct the symmetric point $B^{\prime}$ of point $B$ with respect to line $A C$, and let $B B^{\prime}$ intersect $A C$ at point $E$. Draw $B^{\prime} N \perp A B$ at point $N$, and let $B^{\prime} N$ intersect $A C$ at point $M$. Then, $M$ and $N$ are the required points. The minimum value so... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the set of positive real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$, and the set $S=\{(a, b) \mid a \in A, b \in A, a-b \in A\}$. Then the set $S$ can have at most $\qquad$ elements. | 9.4950 .
The number of ordered pairs of real numbers $(a, b)$ formed by the elements of set $A$ is $100^{2}=10000$.
Since $a_{i}-a_{i}=0 \notin A$, we have $\left(a_{i}, a_{i}\right) \notin S(i=1,2, \cdots, 100)$.
When $\left(a_{i}, a_{j}\right) \in S$, then $\left(a_{j}, a_{i}\right) \notin S$.
Therefore, the maximum... | 4950 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. Each vertex of the convex pentagon $A B C D E$ is colored with one of five colors, such that the two endpoints of each diagonal have different colors. The number of such coloring methods is $\qquad$ (answer with a number), | 12. 1020 .
The number of coloring ways where all vertices have different colors is $A_{5}^{5}=$ 120. The number of coloring ways where two adjacent vertices have the same color and the rest of the vertices have different colors is
$$
A_{5}^{1} A_{5}^{4}=5(5 \times 4 \times 3 \times 2)=600 \text { ways. }
$$
The numbe... | 1020 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 1, in the tetrahedron $D-ABC$, it is known that $DA \perp$ plane $ABC$, and $\triangle ABC$ is an equilateral triangle with a side length of 2. Then, when the tangent value of the dihedral angle $A-BD-C$ is 2, $V_{D-ABC}=$ $\qquad$ | 4.2.
Given $D A \perp$ plane $A B C$, we know
$D A \perp A B$, plane $D A B \cdot \perp$ plane $A B C$.
As shown in Figure 3, take the midpoint $O$ of $A B$.
Then, by the problem statement, $C O \perp A B, C O$
is on plane $D A B$, and
$$
C O=\sqrt{3} \text {. }
$$
Draw $O E \perp D B$ at point $E$, and connect $C E$... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given positive integers $x, y, z$ satisfying $x y z=(14-x)(14-y)(14-z)$, and $x+y+z<28$. Then the maximum value of $x^{2}+y^{2}+z^{2}$ is . $\qquad$ | 7. 219 .
From the problem, we know that $x$, $y$, and $z$ are all positive integers less than 14. On the other hand, expanding the given equation, we get
$$
2 x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x) \text {. }
$$
Thus, $71 x y z$.
Since $x$, $y$, and $z$ are all less than 14, at least one of $x$, $y$, or $z$ must ... | 219 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. For a positive integer $n$, let $x_{n}$ be the real root of the equation
$$
n x^{3}+2 x-n=0
$$
with respect to $x$, and let
$$
a_{n}=\left[(n+1) x_{n}\right](n=2,3, \cdots),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
\frac{1}{1005}\left(a_{2}+a_{3}+\cdots+a_{2011}\right)... | 9.2013.
Let $f(x)=n x^{3}+2 x-n$.
It is easy to see that when $n$ is a positive integer, $f(x)$ is an increasing function.
When $n \geqslant 2$,
$$
\begin{array}{l}
f\left(\frac{n}{n+1}\right)=n\left(\frac{n}{n+1}\right)^{3}+2 \times \frac{n}{n+1}-n \\
=\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$.
Therefore, when $n ... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. In the Cartesian coordinate system, given the point set $I=\{(x, y) \mid x, y$ are integers, and $0 \leqslant x, y \leqslant 5\}$. Then the number of different squares with vertices in the set $I$ is . $\qquad$ | 10. 105.
It is easy to know that there are only two types of squares that meet the conditions: squares whose sides lie on lines perpendicular to the coordinate axes, called "standard squares," and squares whose sides lie on lines not perpendicular to the coordinate axes, called "oblique squares."
(1) In standard squar... | 105 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ . | $$
\begin{array}{l}
\left(x^{4}+2 x^{3}+x^{2}\right)+\left[(2+k) x^{2}+(2+k) x\right]+2 k=0 \\
\Rightarrow\left(x^{2}+x\right)^{2}+(2+k)\left(x^{2}+x\right)+2 k=0 \\
\Rightarrow\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 .
\end{array}
$$
Since the original equation has real roots, and $x^{2}+x+2=0$ has no real root... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. There are 4 colors of light bulbs (with enough of each color), and we need to install a light bulb at each vertex of the triangular prism $A B C-A_{1} B_{1} C_{1}$. The requirement is that the light bulbs at the two endpoints of the same edge must be of different colors, and each color of light bulb must be used at ... | 9.216 .
We can first install $A$, $B$, and $C$, which has $\mathrm{A}_{4}^{3}$ ways; then select one vertex from $A_{1}$, $B_{1}$, and $C_{1}$ to install the fourth color of the light bulb, which has $\mathrm{C}_{3}^{1}$ ways; finally, there are 3 ways to install the remaining two vertices.
Therefore, there are 216 di... | 216 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. The smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers is $\qquad$ . | 10. 495 .
$$
\begin{array}{l}
\text { Let } t=l+(l+1)+\cdots+(l+8) \\
=m+(m+1)+\cdots+(m+9) \\
=n+(n+1)+\cdots+(n+10)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $l=n+2+\frac{2 n+1}{9}$,
$$
m=\frac{n}{10}+n+1 .
$$
Therefore, $2 n+1 \equiv 0(\bmod 9)$,
$$
n \equiv 0(\bmod 10) \text {. }
$$
Thus, the smalle... | 495 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $A$ and $B$ be the common vertices of the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$
and the hyperbola
$$
\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)
$$
Let $P$ and $M$ be two moving points on the hyperbola and the ellipse, respectively, different from $A$ and $B$, and satisfy
$$
\... | 11. -5 .
Let $A(-a, 0), B(a, 0), P\left(x_{1}, y_{1}\right), M\left(x_{2}, y_{2}\right)$.
From $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A M}+\overrightarrow{B M})$, we know that points $O, P, M$ are collinear, and we can find that
$$
\begin{array}{l}
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_... | -5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 5, given $\triangle A B C$ is inscribed in $\odot O$, chord $A F \perp B C$ at point $H, G$ is the midpoint of $B F$. Then $\frac{A C}{O G}=$ | 4. 2 .
Connect $F O$. Then
$$
\begin{array}{l}
\text { Rt } \triangle A B H \backsim \text { Rt } \triangle O \\
\Rightarrow \frac{A H}{B H}=\frac{O G}{F G} . \\
\text { Also } \triangle A C H \backsim \triangle B F H \\
\Rightarrow \frac{A H}{B H}=\frac{A C}{B F} .
\end{array}
$$
Therefore $A C=2 O G \Rightarrow \fr... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of real numbers $a$ that make the equation
$$
x^{2}+a x+8 a=0
$$
have only integer solutions is $\qquad$ . | 2. 8 .
Let the equation (1) have integer solutions $m, n (m \leqslant n)$. Then $m+n=-a, mn=8a$.
Thus, $(m+8)(n+8)=64$.
Solving for $(m, n)$
$$
\begin{array}{c}
=(-72,-9),(-40,-10),(-24,-12), \\
(-16,-16),(-7,56),(-6,24), \\
(-4,8),(0,0) .
\end{array}
$$
Correspondingly,
$$
\begin{array}{l}
a=-(m+n) \\
=81,50,36,32,-... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. All prime numbers $p$ that make $2 p^{4}-p^{2}+36$ a perfect square are $\qquad$ . | 5.2.
When $p=2$,
$$
2 p^{4}-p^{2}+36=64
$$
is a perfect square, thus, $p=2$ is the solution.
When $p=3$,
$$
2 p^{4}-p^{2}+36=189
$$
is not a perfect square.
When $p$ is an odd prime greater than 3, let
$2 p^{4}-p^{2}+36=k^{2}$ ( $k$ is an odd positive integer).
Then $p^{2}\left(2 p^{2}-1\right)=(k-6)(k+6)$
$\Rightar... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and it satisfies
$$
S_{n+1}-5 S_{n}-4 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the last four digits of $a_{2 \text { on }}$ are $\qquad$ | 7.8124.
Given $S_{n+1}-5 S_{n}-4 n-4=0$, and
$$
S_{n}-5 S_{n+1}-4 n=0 \text {, }
$$
subtracting the two equations yields
$$
\begin{array}{l}
a_{n+1}-5 a_{n}-4=0 \\
\Rightarrow a_{n+1}+1=5\left(a_{n}+1\right) \\
\Rightarrow a_{n}=5^{n}-1 .
\end{array}
$$
Notice that, when $k \in \mathbf{N}_{+}$,
$$
\begin{aligned}
5^... | 8124 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 㷵 On a board, there is a convex 2011-gon, and Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? [3] | To prove by induction: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. The following $2n-6$ diagonals can be drawn sequentially:
$A_{2} A_{4}, A_{3} A_{5}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.
When $n=3$, the... | 4016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given a moving point $O$ inside $\triangle A B C$, rays $A O, B O, C O$ intersect the opposite sides at points $A^{\prime}, B^{\prime}, C^{\prime}$ respectively. Let $B^{\prime} C^{\prime}$ intersect $A O$ at $D$, $C^{\prime} A^{\prime}$ intersect $B O$ at $E$, and $A^{\prime} B^{\prime}$ intersect $C O$ at $... | Prove that, as shown in Figure 3, there exist three complete quadrilaterals.
From the complete quadrilateral \( A C^{\prime} O B^{\prime} B C \), the complete quadrilateral \( B A^{\prime} O C^{\prime} C A \), and the complete quadrilateral \( C B^{\prime} O A^{\prime} A B \), we get
\[
\begin{array}{l}
\frac{O D}{A D}... | 1 | Geometry | proof | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 11. If } a+b-2 \sqrt{a-1}-4 \sqrt{b-2} \\
=3 \sqrt{c-3}-\frac{1}{2} c-5 \text {, }
\end{array}
$$
then $a+b+c=$ . $\qquad$ | Ni, 11.20.
Notice,
$$
(\sqrt{a-1}-1)^{2}+(\sqrt{b-2}-2)^{2}+\frac{1}{2}(\sqrt{c-3}-3)^{2}=0 \text {. }
$$
Therefore, $a=2, b=6, c=12$.
So $a+b+c=20$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Given real numbers $a, b$ satisfy
$$
6^{a}=2010,335^{b}=2010 \text {. }
$$
Then the value of $\frac{1}{a}+\frac{1}{b}$ is $\qquad$ | 12. 1 .
Notice, $6^{a b}=2010^{b}, 335^{a b}=2010^{a}$.
Then $(6 \times 335)^{a b}=2010^{a+b}$.
Thus $a b=a+b \Rightarrow \frac{1}{a}+\frac{1}{b}=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the function $y=f(x)$ have the domain $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
2 f\left(x^{2}+x\right)+f\left(x^{2}-3 x+2\right) \\
=9 x^{2}-3 x-6 .
\end{array}
$$
Then the value of $f(60)$ is . $\qquad$ | 2. 176.
Substituting $1-x$ for $x$ in equation (1) yields
$$
\begin{array}{l}
2 f\left(x^{2}-3 x+2\right)+f\left(x^{2}+x\right) \\
=9 x^{2}-15 x .
\end{array}
$$
Combining equations (1) and (2) gives
$$
\begin{array}{l}
f\left(x^{2}+x\right)=3 x^{2}+3 x-4 \\
=3\left(x^{2}+x\right)-4 .
\end{array}
$$
Therefore, $f(60... | 176 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. $[x]$ represents the greatest integer not exceeding the real number $x$. Then, in the Cartesian coordinate system $x O y$, the area of the figure formed by all points $(x, y)$ that satisfy $[x][y]=$ 2011 is $\qquad$ | 3.4.
Let $[x]=a,[y]=b$, meaning all such points $(x, y)$ form the region
$$
a \leqslant x<a+1, b \leqslant y<b+1
$$
bounded by these inequalities, with an area of 1.
Since 2011 is a prime number, the points $(x, y)$ that satisfy
$$
[x][y]=2011
$$
form 4 regions, each with an area of 1, making the total area 4. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If the two quadratic equations
$$
x^{2}+x+m=0 \text { and } m x^{2}+x+1=0
$$
each have two distinct real roots, but one of them is a common real root $\alpha$, then the range of the real root $\alpha$ is $\qquad$. | Given that the common real root of the two equations is $\alpha$, and $m \neq 1$. Then
$$
\begin{array}{l}
\alpha^{2}+\alpha+m=0, \\
m \alpha^{2}+\alpha+1=0 .
\end{array}
$$
From equation (1), we get $m=-\alpha^{2}-\alpha$.
Substituting into equation (2), we get
$$
\begin{array}{l}
\alpha^{4}+\alpha^{3}-\alpha-1=0 \\
... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
20. Given the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$, a line is drawn through its left focus $F_{1}$ intersecting the ellipse at points $A$ and $B$. Point $D(a, 0)$ is a point to the right of $F_{1}$. Connecting $A D$ and $B D$ intersects the left directrix of the ellipse at points $M$ and $N$. If the circ... | 20. It is known that $F_{1}(-3,0)$, the equation of the left directrix is $x=-\frac{25}{3}$, and $l_{A B}: y=k(x+3)$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\left\{\begin{array}{l}y=k(x+3), \\ \frac{x^{2}}{25}+\frac{y^{2}}{16}=1\end{array}\right.$
$$
\Rightarrow\left(16+25 k^{2}\right) x^{2... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Let positive real numbers $a, b, c, d, e$ satisfy $a<b<c<d$ $<e$, and the smallest three of the 10 products of any two numbers are $28, 32, 56$, and the largest two are 128, 240. Then $e=$ $\qquad$ | 2.16.
From the problem, we know
$$
a b=28, a c=32, c e=128, d e=240 \text {. }
$$
Then $c=\frac{8}{7} b, d=\frac{15}{8} c=\frac{15}{7} b$.
Thus, $a d=\frac{15}{7} a b=60>56$.
Therefore, $b c=56 \Rightarrow \frac{8}{7} b^{2}=56 \Rightarrow b=7$.
Hence, $a=4, c=8, d=15, e=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If placing the positive integer $N$ to the left of the positive integer $n$ results in a new number that is divisible by 7, then $N$ is called a "magic number" of $n$. $M$ is a set of positive integers such that for any positive integer $n$, there exists a positive integer in set $M$ that is a magic ... | For $n=1,2, \cdots, 7$.
If $|M| \leqslant 6$, then by the pigeonhole principle, there must be a positive integer $N$ in set $M$ that is a common magic number of $i, j(1 \leqslant i<j \leqslant 7)$, i.e.,
$71(10 N+i), 7 I(10 N+j)$.
Then $7!(j-i)$, but $0<j-i \leqslant 6$, a contradiction.
Therefore, $|M| \geqslant 7$.
T... | 28 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. In the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=4, A A_{1}=A D=2$, points $E, F, G$ are the midpoints of edges $A A_{1}, C_{1} D_{1}, B C$ respectively. Then the volume of the tetrahedron $B_{1}-E F G$ is $\qquad$ | 2.3.
Take point $H$ on the extension of $D_{1} A_{1}$ such that $A_{1} H=\frac{1}{2}$. Then $H E / / B_{1} G$. Therefore, $H E / /$ plane $B_{1} F G$.
Thus, $V_{B_{1}-E F G}=V_{E-B_{1} F C}=V_{H-B_{1} F G}=V_{G-B_{1} F H}$.
And $S_{\triangle B_{1} F H}=\frac{9}{2}$, the distance from point $G$ to plane $B_{1} F H$ is ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the three sides of $\triangle A B C$ are $a, b, c$. If $a+b+c=16$, then
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
$=$ | 6. 64 .
Let the circumradius of $\triangle ABC$ be $R$. Then
$$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. An $8 \times 8$ chessboard is colored in the usual way, with 32 black squares and 32 white squares. A "path" consists of 8 white squares, one in each row, and adjacent white squares share a common vertex. The number of such paths is $\qquad$. | 8. 296
We can use the number labeling method:
\begin{tabular}{cccccccc}
1 & Black & 1 & Black & 1 & Black & 1 & Black \\
Black & 2 & Black & 2 & Black & 2 & Black & 1 \\
2 & Black & 4 & Black & 4 & Black & 3 & Black \\
Black & 6 & Black & 8 & Black & 7 & Black & 3 \\
6 & Black & 14 & Black & 15 & Black & 10 & Black \\... | 296 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) In the geometric sequence $\left\{a_{n}\right\}$ where all terms are positive, what is the maximum number of terms that can be integers between $100 \sim 1000$? | 9. Let the geometric sequence $\left\{a q^{n-1}\right\}$ satisfy
$$
100 \leqslant a q^{n-1} \leqslant 1000 \quad (a, q > 1) \text{ are integers.}
$$
Clearly, $q$ must be a rational number.
Let $q=\frac{t}{s}(t>s \geqslant 1, (t, s)=1)$.
Since $a q^{n-1}=a\left(\frac{t}{s}\right)^{n-1}$ is an integer, $a$ must be a mult... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) There are 12 points on a plane, and no three points are collinear. Using any one of these points as the starting point and another point as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of zero t... | Let the 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $C_{12}^{3}$. Let the number of vectors starting from $P_{i}(i=1,2, \cdots, 12)$ be $x_{i}$ $\left(0 \leqslant x_{i} \leqslant 11\right)$.
If a triangle with three points as vertices is a non-zero triangle, th... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given 315, find
$$
S=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\cdots+\sqrt[2011]{2011}}}}
$$
the integer part of \( S \). | Let $n=2011$, and set
$$
\begin{array}{l}
a_{1}=\sqrt[n]{n}, \\
a_{k}=\sqrt[n-k+1]{n-k+1+a_{k-1}}(2 \leqslant k \leqslant n-1) .
\end{array}
$$
Then $S=a_{n-1}$.
We will prove by induction that:
$$
1<a_{m}<2(m=1,2, \cdots, n-1) \text {. }
$$
(1) When $m=1$, since $1<n<2^{n}$, we have
$$
1<\sqrt[n]{n}<2 \text {. }
$$
(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
316 Given that the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ is a monotonically increasing function. If $f(f(n))=3 n$, find $f(2011)$. | Let $f(1)=m$. Then $f(m)=f(f(1))=3$.
Thus, $m>1 \Rightarrow f(m)>f(1) \Rightarrow 3>m$.
Therefore, $f(1)=m=2 \Rightarrow f(2)=f(f(1))=3$.
Similarly, $f(3)=6, f(6)=9$.
From $f(3)<f(4)<f(5)<f(6)$, we get
$6<f(4)<f(5)<9$.
Hence, $f(4)=7, f(5)=8$.
Furthermore, $f(7)=f(f(4))=12$,
$f(8)=f(f(5))=15$,
$f(9)=f(f(6))=18$,
$f(12)... | 3846 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given $a, b \in \mathbf{R}$, the equation about $x$
$$
x^{4}+a x^{3}+2 x^{2}+b x+1=0
$$
has one real root. Find the minimum value of $a^{2}+b^{2}$. | 【Analysis】This is a quartic equation, and parameters $a, b$ are not easy to handle. We might as well regard $a, b$ as the main variables and $x$ as a parameter. Since $a^{2}+b^{2}$ represents the square of the distance from the moving point $P(a, b)$ to the origin, and $P(a, b)$ lies on the line
$$
x^{3} a + x b + x^{4... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If the two roots of the equation
$$
x^{2}-3 x-1=0
$$
are also roots of the equation
$$
x^{4}+a x^{2}+b x+c=0
$$
then the value of $a+b-2 c$ is ( ).
(A) -13
(B) -9
(C) 6
(D) 0 | $$
\begin{array}{l}
\text { From the problem, we know that } x^{4}+a x^{2}+b x+c \text { can definitely be divided by } x^{2}-3 x-1. \\
=\left(x^{2}-3 x-1\right)\left(x^{2}+3 x+a+10\right)+ \\
{[(3 a+b+33) x+(a+c+10)], } \\
\text { then }\left\{\begin{array}{l}
3 a+b+33=0, \\
a+c+10=0
\end{array}\right. \\
\Rightarrow... | -13 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a$, $b$, and $c$ in $S$ such that $a+b$, $b+c$, and $c+a$ are all in the set $A$. | 6. Let $a<b<c$. Let
$x=a+b, y=a+c, z=b+c$.
Then $x\langle y\langle z, x+y\rangle z$, and $x+y+z$ is even. (1)
Conversely, if there exist $x, y, z \in A$ satisfying property (1), then take
$$
a=\frac{x+y-z}{2}, b=\frac{x+z-y}{2}, c=\frac{y+z-x}{2},
$$
we have $a, b, c \in \mathbf{Z}, 1 \leqslant a<b<c \leqslant 2012$, ... | 1008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. The graph of the quadratic function $y=x^{2}-a x+2$ is symmetric about the line $x=1$. Then the minimum value of $y$ is $\qquad$ . | $$
=, 1.1 .
$$
From the condition, we know that $a=2$. Then $y=(x-1)^{2}+1$ has a minimum value of 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a=\sqrt{3}-1$. Then the value of $a^{2012}+2 a^{2011}-$ $2 a^{2010}$ is $\qquad$. | 2. 0 .
From the condition we know
$$
\begin{array}{l}
(a+1)^{2}=3 \Rightarrow a^{2}+2 a-2=0 \text {. } \\
\text { Then } a^{2012}+2 a^{2011}-2 a^{2010} \\
=a^{2010}\left(a^{2}+2 a-2\right)=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In a chess tournament, there are $n$ female players and $9 n$ male players. Each player plays one game against each of the other $10 n-1$ players. The scoring system is as follows: the winner gets 2 points, the loser gets 0 points, and in the case of a draw, each player gets 1 point. After the tournament, it was fou... | 4.1.
Let the total score of the girls be $m$. Then the total score of the boys is $4 m$. According to the problem,
$$
\begin{array}{l}
\frac{10 n(10 n-1)}{2} \times 2=5 m \\
\Rightarrow m=2 n(10 n-1) .
\end{array}
$$
Since each player competes in $10 n-1$ matches, the maximum score is $2(10 n-1)$, so the total score ... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Two quadratic equations with unequal leading coefficients $(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0$, (1) and $(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0$ ( $a, b$ are positive integers) have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$. | Given the conditions $a>1, b>1, a \neq b$.
Assume the common root of equations (1) and (2) is $x_{0}$. Then
$$
\begin{array}{l}
(a-1) x_{0}^{2}-\left(a^{2}+2\right) x_{0}+\left(a^{2}+2 a\right)=0, \\
(b-1) x_{0}^{2}-\left(b^{2}+2\right) x_{0}+\left(b^{2}+2 b\right)=0 .
\end{array}
$$
(3) $\times(b-1)$ - (4) $\times(a-1... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. When $s$ and $t$ take all real values,
$$
(s+7-|\cos t|)^{2}+(s-2|\sin t|)^{2}
$$
the minimum value is $\qquad$ | 6. 18
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "6. 18" is already in a numerical form that is universal, it does not require translation. Here is the retained format:
6. 18 | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) It is known that $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 9 and common difference 7.
(1) Prove: The sequence $\left\{a_{n}\right\}$ contains infinitely many perfect squares;
(2) The 100th perfect square in the sequence $\left\{a_{n}\right\}$ is the nth term? | (1) The general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=9+7(n-1) \text {. }
$$
When $n=7 k^{2} \pm 6 k+1(k=0,1, \cdots)$, we have
$$
\begin{array}{l}
a_{n}=9+7(n-1) \\
=9+7\left(7 k^{2} \pm 6 k\right)=(7 k \pm 3)^{2} .
\end{array}
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has infinitel... | 17201 | Number Theory | proof | Yes | Yes | cn_contest | false |
1. If the equations
$$
x^{2}+b x+1=0 \text { and } x^{2}-x-b=0
$$
have a common root, find the value of $b$. | Prompt: Example 1. Use the substitution method to find the common root \( x_{0}=-1 \).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The foci of the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ are $F_{1}$ and $F_{2}$. If a point $P$ on the ellipse makes $P F_{1} \perp P F_{2}$, then the area of $\triangle P F_{1} F_{2}$ is $\qquad$ | 2. 9 .
It is known that $F_{1} F_{2}=8, P F_{1}+P F_{2}=10$.
Then $\left(P F_{1}+P F_{2}\right)^{2}=10^{2}$.
In the right triangle $\triangle P F_{1} F_{2}$, we have
$$
P F_{1}^{2}+P F_{2}^{2}=8^{2} \text{. }
$$
From equations (1) and (2), we get
$$
S_{\triangle P F_{1} F_{2}}=\frac{1}{2} P F_{1} \cdot P F_{2}=9 .
$$ | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. If $4n+1$ and $6n+1$ are both perfect squares, then the smallest positive integer $n$ is $\qquad$ | 4. 20 .
Obviously, $4 n+1, 6 n+1$ are both odd square numbers.
Let $6 n+1=(2 m+1)^{2}=4 m(m+1)+1$.
Then $3 n=2 m(m+1)$.
Since $m(m+1)$ is even, $4 \mid n$.
Let $n=4 k$. Then
$4 n+1=16 k+1, 6 n+1=24 k+1$.
When $k=1,2,3,4$, $4 n+1, 6 n+1$ are not both square numbers, but when $k=5$, i.e., $n=20$, $4 n+1=81$, $6 n+1=121$... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. If the four digits of the four-digit number $\overline{a b c d}$ satisfy $a+b=c+d$, then it is called a "good number" (for example, 2011 is a good number). Then, the number of good numbers is $\qquad$ | 8. 615 .
Let $k=a+b=c+d$.
Since $1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant 9$, then
$1 \leqslant k \leqslant 18$.
When $1 \leqslant k \leqslant 9$, in the above equation, $a$ can take any value in $\{1,2$, $\cdots, k\}$, $c$ can take any value in $\{0,1, \cdots, k\}$, and once $a 、 c$ are determined, $... | 615 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. Given an arithmetic sequence $\left\{a_{n}\right\}$, the sum of the first 15 terms $S_{15}=30$. Then $a_{1}+a_{8}+a_{15}=$ $\qquad$ | 12.6.
From $S_{15}=30 \Rightarrow a_{1}+7 d=2$.
Therefore, $a_{1}+a_{8}+a_{15}=3\left(a_{1}+7 d\right)=6$. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. Let $x, y$ be real numbers. Then
$$
\max _{S x^{2}+4 y^{2}=10 x}\left(x^{2}+y^{2}\right)=
$$
$\qquad$ | 15. 4 .
$$
\begin{array}{l}
\text { Given } 5 x^{2}+4 y^{2}=10 x \\
\Rightarrow 4 y^{2}=10 x-5 x^{2} \geqslant 0 \\
\Rightarrow 0 \leqslant x \leqslant 2 .
\end{array}
$$
Then $4\left(x^{2}+y^{2}\right)=10 x-x^{2}$
$$
\begin{array}{l}
=25-(5-x)^{2} \leqslant 25-3^{2} \\
\Rightarrow x^{2}+y^{2} \leqslant 4 .
\end{array... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: There are $n$ people registered to participate in four sports events: A, B, C, and D. It is stipulated that each person must participate in at least one event and at most two events, but events B and C cannot be registered for simultaneously. If in all different registration methods, there must be at least o... | Let the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{T}}\right)$ represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is 1 (for example, if participating in event 甲, then $a_{\text {甲 }}=1... | 172 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
7. Let $M=\{1,2,3,4,5\}$. Then the number of mappings $f: M \rightarrow M$ such that
$$
f(f(x))=f(x)
$$
is $\qquad$ | 7. 196.
For the case where $M$ contains $n$ elements, from
$$
f(f(x))=f(x) \text {, }
$$
we know that for any $a \in M$ (let $f(c)=a$), we have
$$
f(f(c))=f(a)=f(c)=a .
$$
If the range of $f$ contains $k$ elements, then the $n-k$ elements of set $M$ that are not in the range have $k^{n-k}$ possible mappings. Therefo... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Find
$$
f(x)=|x-1|+2|x-2|+\cdots+2011|x-2011|
$$
the minimum value. | 10. Obviously, when $x=2011$, $f(x)$ has no minimum value.
The following assumes $x \in[1,2011]$.
When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2010)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2011} i(i-x) \\
=\left(k^{2}+k-2011 \times 1006\right) x+ \\
\quad \frac{2011 \times 2012 \time... | 794598996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\},
$$
where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedron. Question: What is the maximum number of triangles in graph $G$? | Four, first prove a lemma.
Lemma If in a space graph with $n$ points there is no triangle, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$ ( $[x]$ represents the largest integer not exceeding the real number $x$).
Proof Let the $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, among which, the numbe... | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given a set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$
Then the number of complex numbers in $D$ whose real and imaginary parts are both intege... | Solve: Establish a complex plane, let
$$
z=x+y \mathrm{i}(x, y \in \mathbf{R}) \text {. }
$$
From $\left|z_{1}\right|=1$, we know $z_{1}^{2}$ is also on the unit circle $\odot 0$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \\
=\left|z_{1}^{2}-1\right|^{2}
\end{array}
$$
This represents the... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given that $x, y, z$ are real numbers, satisfying
$$
\begin{array}{c}
x=\sqrt{y^{2}-\frac{1}{16}}+\sqrt{z^{2}-\frac{1}{16}}, \\
y=\sqrt{z^{2}-\frac{1}{25}}+\sqrt{x^{2}-\frac{1}{25}}, \\
z=\sqrt{x^{2}-\frac{1}{36}}+\sqrt{y^{2}-\frac{1}{36}},
\end{array}
$$
and $x+y+z=\frac{m}{\sqrt{n}}\left(m, n \in \mathbf{... | Solve as shown in Figure 11, construct an acute triangle $\triangle ABC$, such that
$$
\begin{array}{l}
BC=x, \\
CA=y, \\
AB=z .
\end{array}
$$
Draw perpendiculars from $A$, $B$, and $C$ to $BC$, $CA$, and $AB$ respectively, with the feet of the perpendiculars being $D$, $E$, and $F$. Let $AD=u$, $BE=v$, and $CF=w$. T... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, 32006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find
$$
a_{1}+a_{2}+\cdots+a_{2000}
$$
the minimum value. ${ }^{[1]}$ | First, we prove: For any 119 positive integers \( b_{1}, b_{2}, \cdots, b_{119} \), there must exist a subset (at least one, or all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
\[ b_{1}, b_{1}+b_{2}, b_{1}+b_{2}+b_{3}, \cdots, b_{1}+b_{2}+\cdots+b_{119}. \]
If one of them is a... | 3910 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 In the positive integers not exceeding 2012, take $n$ numbers arbitrarily, so that there are definitely two numbers whose ratio is within the interval $\left[\frac{2}{3} ; \frac{3}{2}\right]$. Try to find the minimum value of $n$.
| Solve for
$$
\begin{array}{l}
\{1,2,4,7,11,17,26,40,61,92,139, \\
209,314,472,709,1064,1597\}
\end{array}
$$
When any two numbers are taken, their ratio is either greater than $\frac{3}{2}$ or less than $\frac{2}{3}$.
Therefore, $n \geqslant 18$.
Next, we prove: When 18 numbers are taken, there exist two numbers whose... | 18 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given two equations about $x$
$$
x^{2}-x+3 m=0, x^{2}+x+m=0(m \neq 0) \text {. }
$$
If one root of the first equation is three times a root of the second equation, then the value of the real number $m$ is $\qquad$ | - 1. -2 .
Let one root of the latter equation be $\alpha$. And the former equation has a root $3 \alpha$, then
$$
\alpha^{2}+\alpha+m=0,
$$
and
$$
\begin{array}{l}
9 \alpha^{2}-3 \alpha+3 m=0 \\
\Rightarrow 3 \alpha^{2}-\alpha+m=0 .
\end{array}
$$
(2) - (1) gives
$$
2 \alpha^{2}-2 \alpha=0 \Rightarrow \alpha=0 \text ... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the eight numbers $-7, -5, -3, -2, 2, 4, 6, 13$ as $a, b, c, d, e, f, g, h$, such that
$$
(a+b+c+d)^{2}+(e+f+g+h)^{2}
$$
is minimized. Then this minimum value is $\qquad$ | 4.34.
Let $x=a+b+c+d$. Then
$$
\begin{array}{l}
e+f+g+h=8-x, \\
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
=x^{2}+(8-x)^{2}=2(x-4)^{2}+32 .
\end{array}
$$
From the known eight numbers, the sum of any four numbers $x$ is an integer and cannot be 4, so we get
$$
\begin{array}{l}
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
\geqslant 2+32=34,
\en... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. There are 10 chess players participating in a round-robin tournament (i.e., each pair of players competes in one match). The rules state that a win earns 2 points, a draw earns 1 point for each player, and a loss earns 0 points. After the tournament, it is found that each player's score is unique, and the second-pla... | 7. 16 .
The last five contestants have to play 10 matches, so the total score of the last five contestants is greater than or equal to $2 \times 10=20$.
Thus, the score of the second-place contestant is greater than or equal to
$$
\frac{4}{5} \times 20=16 \text {. }
$$
The score of the first-place contestant is less ... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $a b c d$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ . | 8. 22 .
$$
\begin{array}{l}
\text { Let } a b c d=k+(k+1)+\cdots+(k+34) \\
=35(k+17),
\end{array}
$$
where $k$ is a positive integer. Then the prime numbers $a, b, c, d$ must include one 5 and one 7.
Assume $a=5, b=7$. Then $c d=k+17$.
When $k \geqslant 8$,
$$
c+d \geqslant 2 \sqrt{c d}=2 \sqrt{k+17} \geqslant 2 \sqrt... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given a positive integer $n$ that satisfies the following condition: In any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$. ${ }^{[2]}$ | Since $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor not exceeding 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17$,
$19,23,29,31,37,41,43$.
On one hand,
$2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}$,
$19^{2}, 23^{2}, 29^{2}... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The quadratic trinomial
$$
x^{2}+a x+b(a 、 b \in \mathbf{N})
$$
has real roots, and $a b=2^{2011}$. Then the number of such quadratic trinomials is $\qquad$. | $-1.1341$.
It is known that the numbers $a$ and $b$ are powers of 2 with non-negative integer exponents, i.e., $a=2^{k}, b=2^{2011-k}$. Therefore, we have
$$
\begin{array}{l}
\Delta=a^{2}-4 b \geqslant 0 \\
\Rightarrow 2 k \geqslant 2013-k \Rightarrow k \geqslant \frac{2013}{3}=671 .
\end{array}
$$
But $k \leqslant 20... | 1341 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (15 points) On a plane, $n$ points are called a "standard $n$-point set" if among any three of these points, there are always two points whose distance is no more than 1. To ensure that a circular paper with a radius of 1 can cover at least 25 points of any standard $n$-point set, find the minimum value of $n$.
| First, prove: $n_{\text {min }}>48$.
Draw a line segment $AB$ of length 5 on the plane, and construct two circles with radii of 0.5 centered at $A$ and $B$, respectively. Take 24 points in each circle. Then there are 48 points on the plane that satisfy the problem's condition (any three points must have at least two po... | 49 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (15 points) Given a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$ we have
$$
\begin{array}{l}
\frac{1}{2} f(x y)+\frac{1}{2} f(x z)-f(x) f(y z) \geqslant \frac{1}{4} . \\
\text { Find }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)]
\end{array}
$$
where $[a]$ denotes th... | $$
f(1)=\frac{1}{2} \text {. }
$$
Let $y=z=0$. Then
$$
-\frac{1}{2} f(0)+\frac{1}{2} f(0)-f(x) f(0) \geqslant \frac{1}{4} \text {. }
$$
Substituting $f(0)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \leqslant \frac{1}{2} \text {. }
$$
Now let $y=z=1$. Then
$$
\frac{1}{2} f(x)+\frac{1}{2} f(x)-f(x) f(1... | 1011030 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A$ and $B$ be two sets, and call $(A, B)$ a "pair". When $A \neq B$, consider $(A, B)$ and $(B, A)$ as different pairs. Then the number of different pairs $(A, B)$ that satisfy the condition
$$
A \cup B=\{1,2,3,4\}
$$
is $\qquad$ | 2. 81.
When set $A$ has no elements, i.e., $A=\varnothing$, set $B$ has 4 elements, there is 1 case; when set $A$ contains $k(k=1,2,3,4)$ elements, set $B$ contains the other $4-k$ elements besides these $k$ elements, the elements in set $A$ may or may not be in set $B$, there are $\mathrm{C}_{4}^{k} \times 2^{k}$ cas... | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies the recurrence relation
$$
a_{n+1}=2 a_{n}+2^{n}-1\left(n \in \mathbf{N}_{+}\right) \text {, }
$$
and $\left\{\frac{a_{n}+\lambda}{2^{n}}\right\}$ is an arithmetic sequence. Then the value of $\lambda$ is $\qquad$ | 6. -1 .
Notice,
$$
\begin{array}{l}
\frac{a_{n+1}+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2 a_{n}+2^{n}-1+\lambda}{2^{n+1}}-\frac{a_{n}+\lambda}{2^{n}} \\
=\frac{2^{n}-1-\lambda}{2^{n+1}} . \\
\text { From } \frac{2^{n}-1-\lambda}{2^{n+1}} \text { being a constant, we know } \lambda=-1 .
\end{array}
$$ | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given three points $A, B, C$ in a plane satisfying $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=4,|\overrightarrow{C A}|=5$.
Then $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}=$ $\qquad$ | 8. -25 .
Given that $\overrightarrow{A B} \perp \overrightarrow{B C}$. Then
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B} \\
=\overrightarrow{C A} \cdot(\overrightarrow{A B}+\overrightarrow{B C})=-\ove... | -25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Fill in each small square of a $9 \times 9$ grid with a number, with no more than four different numbers in each row and each column. What is the maximum number of different numbers that can be in this grid? ${ }^{[4]}$ | If there are 29 different numbers in this grid, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is the first row).
The remaining 25 numbers are in rows $2 \sim 9$. Similarly, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is t... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. There are 8 red, 8 white, and 8 yellow chopsticks. Without looking, how many chopsticks must be taken out to ensure that at least two pairs of chopsticks are of different colors? $\qquad$ | 10. 11.
Since among 11 chopsticks there must be a pair of the same color (let's say yellow), the number of black or white chopsticks must be at least 3, among which there must be a pair of the same color, i.e., both black or both white. Therefore, 11 chopsticks ensure success. However, if only 10 chopsticks are taken,... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $x, y$ satisfy
$$
3|x+1|+2|y-1| \leqslant 6 \text {. }
$$
Then the maximum value of $2 x-3 y$ is $\qquad$ (1) | Solve As shown in Figure 1, the figure determined by inequality (1) is the quadrilateral $\square A B C D$ and its interior, enclosed by four straight lines, where,
$$
\begin{array}{l}
A(-1,4), B(1,1), \\
C(-1,-2), D(-3,1) .
\end{array}
$$
Consider the family of lines $2 x-3 y=k$, i.e.,
$$
y=\frac{2}{3} x-\frac{k}{3} ... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the sum of the first $n$ terms of the real geometric sequence $\left\{a_{n}\right\}$ be
$S_{n}$. If $S_{10}=10, S_{30}=70$, then $S_{40}=$ $\qquad$ . | 9. 150 .
$$
\begin{array}{l}
\text { Let } b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, \\
b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30} .
\end{array}
$$
Let $q$ be the common ratio of $\left\{a_{n}\right\}$. Then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with common ratio $r=q^{10}$. Therefore,
$$
\begin{array}{l}
70=S... | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $x$ be a real number, and define $\lceil x\rceil$ as the smallest integer not less than the real number $x$ (for example, $\lceil 3.2 \rceil = 4, \lceil -\pi \rceil = -3$). Then, the sum of all real roots of the equation
$$
\lceil 3 x+1\rceil=2 x-\frac{1}{2}
$$
is equal to | 10. -4 .
Let $2 x-\frac{1}{2}=k \in \mathbf{Z}$. Then
$$
x=\frac{2 k+1}{4}, 3 x+1=k+1+\frac{2 k+3}{4} \text {. }
$$
Thus, the original equation is equivalent to
$$
\begin{array}{l}
{\left[\frac{2 k+3}{4}\right]=-1 \Rightarrow-2<\frac{2 k+3}{4} \leqslant-1} \\
\Rightarrow-\frac{11}{2}<k \leqslant-\frac{7}{2} \\
\Right... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$. | Three, 13. Let $t=\sqrt{100-m x}$. Then $x=\frac{100-t^{2}}{m}$.
Hence $y=\frac{100-t^{2}}{m}+t$
$$
=-\frac{1}{m}\left(t-\frac{m}{2}\right)^{2}+\frac{100}{m}+\frac{m}{4} \text {. }
$$
When $t=\frac{m}{2}$, $y$ has a maximum value $\frac{100}{m}+\frac{m}{4}$, that is,
$$
g(m)=\frac{100}{m}+\frac{m}{4} \geqslant 2 \sqrt... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Given the function
$$
f(x)=2\left(\sin ^{4} x+\cos ^{4} x\right)+m(\sin x+\cos x)
$$
has a maximum value of 5 for $x \in\left[0, \frac{\pi}{2}\right]$. Find the value of the real number $m$. | 14. Notice,
$$
f(x)=2-(2 \sin x \cdot \cos x)^{2}+m(\sin x+\cos x)^{4} \text {. }
$$
Let $t=\sin x+\cos x$
$$
=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \in[1, \sqrt{2}] \text {. }
$$
Thus, $2 \sin x \cdot \cos x=t^{2}-1$.
Therefore, $f(x)=2-\left(t^{2}-1\right)^{2}+m t^{4}$
$=(m-1) t^{4}+2 t^{2}+1$.
Let $u=t^{2} \i... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{aligned}
M= & |2012 x-1|+|2012 x-2|+\cdots+ \\
& |2012 x-2012|
\end{aligned}
$$
The minimum value of the algebraic expression is . $\qquad$ | 2. 1012036 .
By the geometric meaning of absolute value, we know that when
$$
1006 \leqslant 2012 x \leqslant 1007
$$
$M$ has the minimum value.
$$
\begin{array}{l}
\text { Then } M_{\text {min }}=(-1-2-\cdots-1006)+ \\
(1007+1008+\cdots+2012) \\
=1006 \times 1006=1012036 .
\end{array}
$$ | 1012036 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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