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B2. What is the largest integer $n$ such that the quantity
$$
\frac{50!}{(5!)^{n}}
$$
is an integer?
Note: Here $k!=1 \times 2 \times 3 \times \cdots \times k$ is the product of all integers from 1 to $k$. For example, $4!=1 \times 2 \times 3 \times 4=24$. | Solution : We can calculate $5!=2^{3} \times 3 \times 5$, thus $(5!)^{n} \mid 50$ ! if
$$
\begin{array}{l}
2^{3 n} \mid 50! \\
3^{n} \mid 50!\text { and } \\
5^{n} \mid 50!
\end{array}
$$
(The notation $b \mid a$ means integer number $a$ is divisible by integer number $b$.)
Observe that among integers from 1 to 50 , ev... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. Given a linear equation in $x$, $7 x+(3+x)=k+b x$, the equation has a non-zero solution, and this solution is $\frac{1}{3}$ of the solution to the equation $7 x+(3+x)=k-b x$. What is the value of $b$? $\qquad$ - | $-4$ | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 18 (2008 Southeast Mathematical Olympiad Problem) Let $n$ be a positive integer, and $f(n)$ denote the number of $n$-digit numbers (called wave numbers) $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfy the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and $a_{i} \neq a_{i+1}, i=1,2, \cdots$;
(i... | Let $n \geqslant 2$, and call an $n$-digit waveform number $a_{1} a_{2} \cdots a_{n}$ that satisfies $a_{1}a_{2}$ a $B$-type number. By symmetry, when $n \geqslant 2$, the number of such numbers is also $g(n)$, so $f(n)=2 g(n)$.
Now we find $g(n)$: Let $m_{k}(i)$ represent the number of $k$-digit $A$-type waveform numb... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (1990 Hungarian Mathematical Olympiad) For any positive integer $q_{0}$, consider the sequence $q_{1}, q_{2}, \cdots, q_{n}$ defined by $q_{i}=\left(q_{i-1}-1\right)^{3}+$ $3(i=1,2, \cdots, n)$. If each $q_{i}(i=1,2, \cdots, n)$ is a prime, find the largest possible value of $n$. | 9. Since $m^{3}-m=m(m-1)(m+1) \equiv 0(\bmod 3)$, we have
$$
q_{i}=\left(q_{i-1}-1\right)^{3}+3 \equiv\left(q_{i-1}-1\right)^{3} \equiv q_{i-1}-1(\bmod 3) \text {. }
$$
Therefore, among $q_{1}, q_{2}, q_{3}$, there must be one that is divisible by 3, and this number should be a power of 3.
If $3 \mid(q-1)^{3}+3$, then... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Given positive real numbers $a, b, c$ satisfy
$$
x=\frac{a}{2 b+3 c}, y=\frac{2 b}{3 c+a}, z=\frac{3 c}{a+2 b} \text {. }
$$
Then the value of $\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}$ is ( ).
(A) 1
(B) $\frac{3}{2}$
(C) 2
(D) 3 | - 1. A.
Notice that, $\frac{x}{1+x}=\frac{a}{a+2 b+3 c}$,
$$
\begin{array}{l}
\frac{y}{1+y}=\frac{2 b}{a+2 b+3 c}, \\
\frac{z}{1+z}=\frac{3 c}{a+2 b+3 c} .
\end{array}
$$
Therefore, $\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}=1$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
Consider a pair of four-digit numbers, where swapping the first and last digits of one number results in the other (e.g., 1234 and 4231). How many such pairs of numbers have a sum that is a four-digit square number? | Let $\overline{a b c d}, \overline{d b c a}$ satisfy the conditions, and $a \leqslant d$. Then the sum
$$
\overline{a b c d}+\overline{d b c a}=1001 \cdot(a+d)+20(10 b+c)
$$
is a four-digit number, so $2 \leqslant a+d \leqslant 9$. The sum is also a perfect square $x^{2}$, and its last digit $a+d$ can only be $4,5,6,9... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
9. From the set $\{1,2,3, \cdots, 1000\}$, randomly and without replacement, take 3 numbers $a_{1}, a_{2}, a_{3}$, and from the remaining 997 numbers in the set, randomly and without replacement, take another 3 numbers $b_{1}, b_{2}, b_{3}$. Let the probability of the following event be $p$: a rectangular brick of dime... | 9. Solution: A rectangular brick of size $a_{1} \times a_{2} \times a_{3}$ can fit into a rectangular box of size $b_{1} \times b_{2} \times b_{3}$ if $a_{1}<b_{1}, a_{2}<b_{2}, a_{3}<b_{3}$. Given $c_{0}<c_{6}$, $c_{1}$ must be one dimension of the brick, and $c_{6}$ must be one dimension of the box. Here, we refer to... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 1 Let $m, n$ be positive integers, find the minimum value of $\left|12^{m}-5^{m}\right|$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Solution: First, when $m=n=1$, $\left|12^{m}-5^{n}\right|=7$. The following uses the method of exhaustion to prove that 7 is the minimum value sought.
(1) Clearly, $\left|12^{m}-5^{n}\right|$ is an odd number, so it cannot be $2,4,6$.
(2) Note that $3 \mid 12^{m}, 3 \nmid 5^{n}$ and $5 \nmid 12^{m}$, it is easy to see ... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
56. If $a$ is the integer part of $\sqrt{17}-2$, $b-1$ is the square root of 9, and $|a-b|=b-a$, then the value of $a+b$ is $\qquad$. | answer: 6 | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is
(A) 1;
(B) 2;
(C) $\pi$;
(D) 4. | 4. (B)
For expressions containing absolute values, discuss cases to remove the absolute value symbols, resulting in four line segments.
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x - 1 \geqslant 0 } \\
{ y - 1 \geqslant 0 } \\
{ x + y \geqslant 3 , }
\end{array} \quad \left\{\begin{array}{l}
x-1 \leqslant 0 \\
... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. (5 points) $2016 \times 2014-2013 \times 2015+2012 \times 2015-2013 \times 2016=$ | 【Analysis】According to the distributive law of multiplication, the common factor can be extracted for simplification.
【Solution】Solve: $2016 \times 2014-2013 \times 2015+2012 \times 2015-2013 \times 2016$
$$
\begin{array}{l}
=2016 \times 2014-2013 \times 2016-2013 \times 2015+2012 \times 2015 \\
=2016 \times(2014-2013)... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given the function $y=-\frac{x-a}{x-a-1}$, the graph of its inverse function is symmetric with respect to the point $(-1,3)$. Then the real number $a$ equals ( ).
(A) 2
(B) 3
(C) -2
(D) -3 | 6 Method one: Take a point $\left(1, \frac{1}{a}-1\right)$ on the original function graph, then the point $\left(\frac{1}{a}-1,1\right)$ is on the graph of its inverse function. The point symmetric to it about the point $(-1,3)$ is $\left(-1-\frac{1}{a}, 5\right)$, thus the point $\left(5,-1-\frac{1}{a}\right)$ is on t... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. Given the complex number $z=\frac{4+\sqrt{2} \mathrm{i}}{1-\mathrm{i}}$. Then $|z|=$ $\qquad$ . | $\begin{array}{l}\text { II.7.3. } \\ |z|=\left|\frac{4+\sqrt{2} i}{1-i}\right|=\frac{|4+\sqrt{2} i|}{|1-i|}=\frac{\sqrt{18}}{\sqrt{2}}=3\end{array}$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.65 Given the quadratic equation $x^{2}-4 a x+5 a^{2}-6 a=0$ has two real roots, and the absolute value of the difference between these two roots is 6. Then, the value of $a$ is
(A) -1.
(B) 1.
(C) 3.
(D) 5.
(China Beijing Junior High School Mathematics Competition, 1991) | ［Solution］Let the two roots of the equation be $x_{1}, x_{2}$. By Vieta's formulas, we have
$$
x_{1}+x_{2}=4 a \text {, and } x_{1} x_{2}=5 a^{2}-6 a \text {. }
$$
Then $\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}}$
$$
\begin{array}{l}
=\sqrt{16 a^{2}-4\left(5 a^{2}-6 a\right)} \\
=2 \sqr... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. Let $A=\{1,2, \cdots, 10\}$, if the equation $x^{2}-b x-c=0$ satisfies $b, c \in A$, and the equation has at least one root $a \in A$, then the equation is called a "beautiful equation". The number of "beautiful equations" is $\qquad$. | 5. 12 Detailed Explanation: From the problem, we know that the two roots of the equation are integers, one positive and one negative. When one root is -1, there are 9 "beautiful equations" that meet the requirements. When one root is -2, there are 3 "beautiful equations" that meet the requirements. In total, there are ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. In a cyclic quadrilateral $ABCD$, there is a point $X$ on side $AB$ such that diagonal $BD$ bisects $CX$ and $AC$ bisects $DX$. Find the minimum value of $\frac{AB}{CD}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7. As shown in Figure 1, let the midpoints of $C X$ and $D X$ be $M$ and $N$ respectively,
$$
\begin{array}{l}
\angle B A C=\angle B D C=\alpha, \angle C A D=\angle C B D=\beta . \\
\quad \text { By } S_{\triangle A X N}=S_{\triangle A D N} \\
\Rightarrow A X \cdot A N \sin \alpha=A D \cdot A N \sin \beta \\
\Rightarro... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
13. In tetrahedron $ABCD$, $AB=2$, $AC=3$, $AD=4$, $BC=\sqrt{13}$, $BD=2\sqrt{5}$, $CD=5$. Then the volume of tetrahedron $ABCD$ is ( ).
(A) 3
(B) $2 \sqrt{3}$
(C) 4
(D) $3 \sqrt{3}$
(E) 6 | 13. C.
As shown in Figure 4, from the problem statement we know
$$
\begin{array}{l}
A B^{2}+A C^{2}=B C^{2} \\
\Rightarrow A B \perp A C .
\end{array}
$$
Similarly, $A B \perp A D$, $A C \perp A D$.
$$
\begin{array}{l}
\text { Therefore, } V_{\text {tetrahedron } B-A C D}=\frac{1}{3} S_{\triangle A C D} A B \\
=\frac... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 6 (1990 National High School Mathematics League Second Test) A city has $n$ high schools, the $i$-th high school sends $C_{i}$ students to watch a ball game in the gym $\left(1 \leqslant C_{i} \leqslant 39, i=1,2, \cdots, n\right)$, the total number of all students is $\sum_{i=1}^{n} C_{i}=1990$. Each row in th... | Thought analysis: Utilize $1 \leqslant C_{i} \leqslant 39$, to analyze the minimum number of students that can sit in each row or the maximum number of empty seats that can be left in each row, to determine the minimum number of rows required.
Solution: Since $1 \leqslant C_{i} \leqslant 39$, each row can seat at leas... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. Given the function $f(x)=a x^{2}+b x(a b \neq 0)$, if $f\left(x_{1}\right)=f\left(x_{2}\right)\left(x_{1} \neq x_{2}\right)$, then the value of $f\left(x_{1}+x_{2}\right)$ is $\qquad$ . | 8. 0 . Axis of symmetry $x=\frac{x_{1}+x_{2}}{2}=-\frac{b}{2 a}$, then $x_{1}+x_{2}=-\frac{b}{a}$, then $f\left(x_{1}+x_{2}\right)=f\left(-\frac{b}{a}\right)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
67. Person A, B, and C are sawing wooden sticks of different lengths but the same thickness. A saws sticks that are all 8 meters long, B saws sticks that are all 10 meters long, and C saws sticks that are all 6 meters long. They are required to saw them into 2-meter segments. If in the same amount of time, A, B, and C ... | answer: 2 | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
20.47 In a triangle, the area and its perimeter are numerically equal, the radius of its inscribed circle is
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(E) 6. | [Solution] Let the inradius of $\triangle A B C$ be $r$ (as shown in the figure), then
$$
\begin{array}{c}
S_{\triangle A B C}=\frac{1}{2} \cdot A B \cdot r+\frac{1}{2} \cdot B C \cdot r+\frac{1}{2} \cdot A C \cdot r \\
=\frac{1}{2}(A B+B C+A C) r,
\end{array}
$$
If we set $p=A B+B C+A C$, from the given and the above... | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
5. (10 points) From the 16 integers from 1 to 16, at least how many numbers must be taken to ensure that there are two numbers, one of which is twice the other. | 12
【Analysis】According to the worst-case principle, take out the numbers in ascending order: $1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16$. Already 11 numbers have been taken. According to the pigeonhole principle, taking one more number will satisfy the condition; thus, the answer is given accordingly.
【Solution】According to... | 12 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. (10 points) As shown in the figure, 15 points can be arranged into a triangular array or a trapezoidal array in 3 different ways (rule: the number of points in adjacent rows differs by 1). Therefore, the number of different ways to arrange 2014 points into a triangular array or a trapezoidal array (at least two lay... | 【Solution】Solution: Since the number of layers and the number of points per layer are integers, and the numbers of each layer form an arithmetic sequence with a common difference of 1.
If the number of layers is odd, the total number $=$ number of layers $\times$ number of points in the middle layer.
Since the total n... | 3 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
1. Given the function $f(x)=a \ln x+x^{2}$ has an extremum at $x=1$. Then the value of the real number $a$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2 | - 1. A.
It is known that $f^{\prime}(x)=\frac{a}{x}+2 x$.
Given that $f^{\prime}(1)=0$, solving yields $a=-2$. | -2 | Calculus | MCQ | Yes | Yes | olympiads | false |
Five, (15 points) Draw five lines $l_{1}, l_{2}$, $\cdots, l_{5}$ on a plane, such that no two lines are parallel and no three lines intersect at the same point.
(1) How many intersection points do these five lines have? How many intersection points are there on each line? How many line segments are there on these five... | (1) As shown in Figure 5.
There are 10 intersection points formed by five lines, with exactly four intersection points on each line. Since there are 6 line segments on each line, the five lines together have
30 line segments.
(2) The maximum number of isosceles triangles that can be formed with these line segments as s... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Question 66, Given $x, y>0, x^{2}+y \geq x^{3}+y^{2}$, find the maximum value of $x^{2}+y^{2$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Note: The note about the format and the translation request is not part of ... | Question 66, Answer: Since $x(x-1)^{2} \geq 0 \Rightarrow x-x^{2} \geq x^{2}-x^{3}$, and according to the condition, $x^{2}+y \geq x^{3}+y^{2} \Rightarrow x^{2}-x^{3} \geq y^{2}-y$, so $x-x^{2} \geq y^{2}-y \Rightarrow x^{2}+y^{2} \leq x+y \Rightarrow\left(x^{2}+y^{2}\right)^{2} \leq(x+y)^{2} \leq 2\left(x^{2}+y^{2}\ri... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Real numbers $x, y$ satisfy $\tan x=x, \tan y=y$, and $|x| \neq|y|$, then the value of $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}$ is . $\qquad$ | 1. 0 Detailed Explanation: From the problem, we get $\frac{\sin (x+y)}{x+y}=\frac{\sin (x+y)}{\tan x+\tan y}=\frac{\sin (x+y)}{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}=\cos x \cdot \cos y$, similarly $\frac{\sin (x-y)}{x-y}$ $=\cos x \cdot \cos y$, therefore $\frac{\sin (x+y)}{x+y}-\frac{\sin (x-y)}{x-y}=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. (12 points) Fill in the squares on the right with the ten digits $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ without repetition. Adjacent squares in the same row or column can form a two-digit number from left to right or from top to bottom (0 cannot be the first digit), then, among these two-digit numbers, the maximum number of... | 【Answer】Solution:
The prime numbers in this table are $41$, $61$, $19$, $83$, $37$, $13$, and $97$.
$1$, $3$, $5$, $7$ can only be in three positions at the end of two numbers (see the green part in the figure), so the maximum number of prime numbers is 7. Therefore, the answer is 7. | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate the following expressions
(1) $1+2-3+4+5-6+\ldots+220+221-222=$ $\qquad$ .
(2) $(22+23+24+25+26) \times(25+26+27)-(22+23+24+25+26+27)$
$\times(25+26)=$ $\qquad$
(3) $1^{2}+3^{2}+5^{2}+\ldots .+19^{2}=$ $\qquad$
(4) $\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+\ldots+\frac{2019}{... | 【Answer
(1) 8103 ;
(2) 1863 ;
(3) 1330;
(4) 1020605 ;
(5) 2
【Explanation】(1) Original expression $=(1+2-3)+(4+5-6)+\ldots+(220+221-222)=0+3+6+\ldots+219=$
$(0+219) \times 74 \div 2=8103$
(2) Let $a=22+23+24+25+26, b=25+26+27$; Original expression
$$
=a \times(b+27)-(a+27) b=a b+27 a-a b-27 b=27(a-b)=27 \times(22+23+24... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
The 71st question: In a $2018 \times 4$ grid, each cell is to be colored either red or blue, with each cell being colored exactly one color, such that in each row the number of red and blue cells is the same, and in each column the number of red and blue cells is also the same. Let $\mathrm{M}$ be the number of colorin... | The 71st question:
Answer: Take 1009 squares in the first column and color them red, which can be done in $\mathrm{C}_{2018}^{1009}$ ways. Without loss of generality, assume the first 1009 squares in the first column are red.
In the second column, suppose exactly $\mathrm{k}(0 \leq \mathrm{k} \leq 1009)$ of the first ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. A rectangular prism with a square cross-section has a surface area of 2448 square centimeters. A largest possible cube is sawed off from one end, and the surface area of the cube is 216 square centimeters. The maximum number of such cubes that can be sawed from this piece of wood is $\qquad$ (Note: Each cut results ... | 【Answer】11
【Analysis】Surface area of cuboids and cubes
The cross-sectional area can be calculated as $216 \div 6=36$ square centimeters, so the height of the cuboid is $2448 \div 36=68$ centimeters; since each cut results in a loss of 2 millimeters, or 0.2 centimeters, $(68+0.2) \div(6+0.2)=11$ (tree planting problem),... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7.87 For the grid shown in the figure with positive and negative signs, the following operation is allowed: change all the signs in a certain row or column to the opposite of their original signs, and after each operation, record the absolute value of the difference in the number of the two types of signs. Try to find ... | [Solution] Because with each operation, the number of positive signs in the row or column changes by an even number, the parity of the number of positive signs in the grid remains unchanged throughout the operations. Since the original table has 37 positive signs, there will always be an odd number of positive signs in... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. Given that a line passing through the focus $F$ of the parabola $y^{2}=4 x$ intersects the parabola at points $M$ and $N$, and $E(m, 0)$ is a point on the $x$-axis. The extensions of $M E$ and $N E$ intersect the parabola at points $P$ and $Q$ respectively. If the slopes $k_{1}$ and $k_{2}$ of $M N$ and $P Q$ satisf... | 9.3.
When $M P$ is not perpendicular to the $x$-axis, let $l_{\text {MP }}: y=k(x-m)$.
Substitute into $y^{2}=4 x$ to get
$$
x^{2}-\left(\frac{4}{k^{2}}+2 m\right) x+m^{2}=0 \text {. }
$$
Then $x_{M} x_{P}=m^{2} \Rightarrow y_{M} y_{P}=-4 m$ $\Rightarrow y_{P}=\frac{-4 m}{y_{M}}$.
When $M P \perp x$-axis, the conclus... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
12. In the tetrahedron $ABCD$, there is a point $O$ inside such that the lines $AO, BO, CO, DO$ intersect the faces $BCD, ACD, ABD, ABC$ at points $A_1, B_1, C_1, D_1$ respectively, and $\frac{AO}{OA_1}=\frac{BO}{OB_1}=\frac{CO}{OC_1}=\frac{DO}{OD_1}=k$. Then $k=$ $\qquad$ . | 12. $3 \frac{V_{A B C D}}{V_{O B C D}}=\frac{A A_{1}}{O A_{1}}=\frac{A O}{O A_{1}}+\frac{O A_{1}}{O A_{1}}=1+k$, thus $\frac{V_{O B C D}}{V_{A B C D}}=\frac{1}{1+k}$.
Similarly, $\frac{V_{(X \cdot D A}}{V_{A B C D}}=\frac{V_{O D A B}}{V_{A B C D}}=\frac{V_{O A B C}}{V_{A B C D}}=\frac{1}{1+k}$.
Since $V_{O B C D}+V_{\... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(14) From the left focus $F$ of the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$, draw a tangent to the circle $x^{2}+y^{2}=9$, with the point of tangency being $T$. Extend $F T$ to intersect the right branch of the hyperbola at point $P$. If $M$ is the midpoint of segment $F P$, and $O$ is the origin, find the value... | (14) Without loss of generality, place point $P$ in the first quadrant. Let $F^{\prime}$ be the right focus of the hyperbola, and connect $P F^{\prime}$. Since $M$ and $O$ are the midpoints of $F P$ and $F F^{\prime}$ respectively, we have $|M O|=\frac{1}{2}\left|P F^{\prime}\right|$. By the definition of the hyperbola... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13. Given that $b$ and $c$ are positive integers, and the three quadratic equations $x^{2}+b x+c=k$ (for $k=0,1,2$) all have real roots but no integer roots, find the minimum value of $b+c$. | 13. The discriminant of $x^{2}+b x+c=k$ is $D_{k}=b^{2}-4 c+4 k(k=0,1,2)$. According to the problem, $D_{0} \geqslant 0$ and $\sqrt{D_{0}} \cdot \sqrt{D_{1}}=$ $\sqrt{D_{0}+4}, \sqrt{D_{2}}=\sqrt{D_{0}+8}$ are not integers.
$$
\therefore D_{0} \neq 1,4,5,8,9,12 \text {. }
$$
(1) If $D_{0}=b^{2}-4 c=2$ or 6 or 10, then ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$35 \cdot 48$ On a plane, there is a set of points $M$ and seven different circles $C_{1}, C_{2}, \cdots, C_{7}$, where circle $C_{7}$ passes through exactly 7 points in $M$, circle $C_{6}$ passes through exactly 6 points in $M$, $\cdots$, circle $C_{1}$ passes through exactly 1 point in $M$. The minimum number of poin... | [Solution] For circle $C_{7}$, set $M$ needs to have 7 points $A_{1}, A_{2}, \cdots, A_{7}$ (see figure);
For circle $C_{6}$, $M$ needs to add 4 points $A_{8}, A_{9}$, $A_{10}, A_{11}$ (the other two points are $A_{3}, A_{6}$ );
For circle $C_{5}$, $M$ only needs to add one point $A_{12}$; when considering circles $C... | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
25. One day, the owner put more than 60 identical chess pieces into some identical boxes, with only one box being empty. The mischievous Tom cat took one piece from each of the boxes that had pieces and put them into the empty box. The clever Jerry mouse rearranged the boxes, and as a result, the position of each box a... | $12$ | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. Find the unit digit of $47^{47^{174}}$, where there are $k(>1)$ 47s. | 1. Since $47 \equiv 7(\bmod 10), 47^{2} \equiv 7^{2} \equiv 49 \equiv-1(\bmod 10)$, $47^{4} \equiv(-1)^{2} \cong 1(\bmod 10)$.
(*)
Now consider $47^{47^{\cdots 47}}$ ( $k$ 47s) with the exponent $47^{\cdots 47}(k-1$ 47s) modulo 4.
Since $47 \equiv-1(\bmod 4)$, we have
$$
47^{47^{\cdots} 47}(k-1 \text { 47s}) \equiv-1 \... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $a_{n}=6^{n}+8^{n}$. Then the remainder when $a_{2018}$ is divided by 49 is $\qquad$ . | \begin{array}{l}\text { 4.2. } \\ a_{2018}=(7-1)^{2018}+(7+1)^{2018} \\ =2 \sum_{k=0}^{1009} \mathrm{C}_{2018}^{2 k} 7^{2 k} \equiv 2\left(\bmod 7^{2}\right)\end{array} | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Given a regular $n$-sided prism where the side edge length is equal to the base edge length, then the maximum value of $n$ is
(A) 4
(B) 5
(C) 6
(D) no maximum value | 3. (B).
Let the projection of the vertex $S$ of a regular $n$-pyramid onto the base $A_{1} A_{2} \cdots A_{n}$ be $O$. Then for $n \geqslant 6$, we have $S A_{1}>O A_{2} \geqslant A_{1} A_{2}$, so the $n$ that satisfies the condition is $n \leqslant 5$. When $n=5$, if the angle between the lateral edge and the base is... | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
1. Let $x, y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1, \\
(y-1)^{3}+1997(y-1)=1 .
\end{array}\right.
$$
Then $x+y=$ $\square$ . | 1. The original system of equations is transformed into
$$
\left\{\begin{array}{l}
(x-1)^{3}+1997(x-1)=-1 \\
(1-y)^{3}+1997(1-y)=-1 .
\end{array}\right.
$$
Since $f(t)=t^{3}+1997 t$ is monotonically increasing on $(-\infty,+\infty)$, and $f(x-1) = f(1-y)$, it follows that $x-1=1-y$, i.e., $x+y=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$19 \cdot 30$ If a convex polygon has exactly three obtuse interior angles, the maximum number of sides this polygon can have is
(A) 4 .
(B) 5 .
(C) 6 .
(D) 7 .
(E) 8 .
(36th American High School Mathematics Examination, 1985) | [Solution 1] Since the sum of the exterior angles of a convex polygon is $360^{\circ}$, the number of non-acute exterior angles is no more than 3 (and given that it has three obtuse angles, there are already three acute angles among the exterior angles), which means the number of non-obtuse interior angles is no more t... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
7. (10 points) As shown in the figure, in isosceles trapezoid $A B C D$, the upper base $A B$ is 4 cm, the lower base $C D$ is 12 cm, and the angle between the leg $A D$ and the lower base $D C$ is $45^{\circ}$. If $A F=F E=E D$ and $B C=4 B G$, then the area of $\triangle E F G$ is $\qquad$ square centimeters. | 【Solution】Solution: According to the analysis, draw the height of the trapezoid and mark the relevant data, as shown in the figure:
From the characteristics of the isosceles trapezoid, we know that $D M=A M=B N=C N, A B=M N$, so the length of $D M$ is:
$(12 - 4) \div 2=4$ (cm) Therefore, $A M=4$ (cm),
The area of trape... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
24.1.3 ** In a finite sequence of real numbers, the sum of any 3 consecutive terms is negative, and the sum of any 4 consecutive terms is positive. Find the maximum number of terms \( r \) in this sequence. | Parse, the maximum value of $r$ is 5.
On one hand, construct a 5-term sequence: $2,2,-5,2,2$, where the sum of any 3 consecutive terms is -1, and the sum of any 4 consecutive terms is 1. Therefore, when $r=5$, there exists a 5-term sequence that satisfies the conditions. On the other hand, prove that for any finite seq... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 7 (1994 National Training Team Question) Let $S=$ $\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$, satisfying the conditions:
(1) $\left|A_{i}\right|=5(i=1,2, \cdots, k)$;
(2) $\left|A_{i} \cap A_{j}\right| \leqslant 2(1 \leqslant i<j \leqslant k)$.
Find the maximum value of $K$. | Thought analysis: We should derive an upper bound for $K$ based on the conditions that the subsets must satisfy. Let's make the following exploration: Let $A_{1}=\{1,2,3,4,5\}$, and the intersection of any two subsets has at most 2 common elements. Let $A_{2}=\{1,2,6,7,8\}$. At this point, if $\{1,2\} \subseteq A_{3}$,... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. Xiaohong and Xiaoliang play the "Rock, Paper, Scissors" game, agreeing that if they win, they go up three steps, and if they lose, they go down two steps. They start playing from the 9th step, and after 12 rounds (each round has a winner and a loser), Xiaohong is on the 10th step. How many times did Xiaohong win? | 【Analysis】If Xiao Hong wins all 12 times, then she should climb 36 steps. In fact, Xiao Hong only climbed 1 step. If one win is changed to one loss, then she needs to go down $3+2=5$ steps. Therefore, Xiao Hong lost $(36-(1) \div 3+2=7$ times, so Xiao Hong won 5 times. | 5 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
7.56 On the table, there are 1989 equally sized circular paper pieces placed without overlapping. How many different colors are needed at least to ensure that each paper piece can be painted a color so that any two touching circular paper pieces are painted different colors, regardless of their positions? | ［Solution］Consider the case of 11 circular pieces of paper as shown in the figure. Suppose that the 6 circles on the left have been colored. Clearly, the 3 circles $A, B, E$ can only be colored with 1 or 3, and $A$ is one color, while $B$ and $E$ are the other color. If there are only 3 colors, then no matter what colo... | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 2 Solve the system of simultaneous equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3 .
\end{array}\right.
$$
Find all real or complex roots. | Let $T_{n}=x^{n}+y^{n}+z^{n} \ (n=1,2, \cdots)$, then $T_{1}=T_{2}=T_{5}=3$. Suppose $x, y, z$ are the three roots of the cubic polynomial
$$
f(t)=(t-x)(t-y)(t-z)=t^{3}-b_{1} t^{2}+b_{2} t-b_{3}
$$
By the given conditions, we have $b_{1}=T_{1}=3, b_{2}=\frac{1}{2}\left(T_{1}^{2}-T_{2}\right)=3$. Now, let's calculate $... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. In the interval $[0, \pi]$, the number of solutions to the trigonometric equation $\cos 7 x=\cos 5 x$ is $\qquad$ | $$
\cos 7 x-\cos 5 x=-2 \sin 6 x \sin x=0 \Rightarrow 6 x=k \pi \Rightarrow x=\frac{k \pi}{6}, k=0,1, \cdots, 6 \text {. }
$$
So the number of solutions in the interval $[0, \pi]$ is 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
66 The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, \sqrt{\frac{1}{a_{n}^{2}}+4}=\frac{1}{a_{n+1}}$, and let $S_{n}=\sum_{i=1}^{n} a_{i}^{2}$. If $S_{2 n+1}-S_{n} \leqslant \frac{t}{30}$ holds for any $n \in \mathbf{N}^{*}$, then the minimum value of the positive integer $t$ is
(A) 10
(B) 9
(C) 8
(D) 7 | 6) A Hint: Given $\frac{1}{a_{n+1}^{2}}-\frac{1}{a_{n}^{2}}=4$, we can find $a_{n}^{2}=\frac{1}{4 n-3}$. Let $g(n)=S_{2 n+1}-S_{n}$, then
$$
\begin{aligned}
g(n+1)-g(n) & =a_{2 n+2}^{2}+a_{2 n+3}^{2}-a_{n+1}^{2} \\
& =\frac{1}{8 n+5}+\frac{1}{8 n+9}-\frac{1}{4 n+1} \\
& <0,
\end{aligned}
$$
i.e., $g(n)$ is a decreasin... | 10 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. (3 points) If the natural numbers $a, b, c, d$ when divided by 6 all leave a remainder of 4, then the remainder when $a+b+c+d$ is divided by 3 is | 【Solution】Solution: Since the natural numbers $a, b, c, d$ when divided by 6 all leave a remainder of 4, $a, b, c, d$ can all be expressed as: 6 $\times$ integer +4,
The sum of four such numbers is: $6 \times$ integer +16, when divided by 3 leaves a remainder of 1, the resulting remainder is 1.
The answer is 1. | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6 A regular tetrahedron $D-ABC$ has a base edge length of 4 and a side edge length of 8. A section $\triangle AEF$ is made through point $A$ intersecting side edges $DB$ and $DC$. What is the minimum perimeter of $\triangle AEF$? $\qquad$ . | 6 11 Hint: Draw the lateral development of the triangular pyramid, and from the minimum perimeter of $\triangle A E F$, we get $A$, $E, F, A_{1}$ are collinear. It is easy to see that $E F / / B C$, thus the isosceles $\triangle D A B \sim \triangle A E B$, $A E=A B=4$,
$$
\frac{B E}{A B}=\frac{A B}{D A}=\frac{1}{2},
$... | 11 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. As shown in the figure, the perimeter of hexagon $A B C D E F$ is 16 cm, all six angles are $120^{\circ}$, if $A B=B C=C D=3$ cm, then $E F=$ $\qquad$cm. | 9. As shown in the figure, the perimeter of hexagon $A B C D E F$ is 16 cm, and each of the six angles is $120^{\circ}$. If $A B=B C=C D=3$ cm, then $E F=5$ cm.
【Solution】Solution: As shown in the figure, extend and reverse extend $A F, B C, D E$, intersecting at points $G, H, N$, respectively. Since each angle of hexa... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13. Among the thirteen natural numbers from $1 \sim 13$, select twelve natural numbers to fill in the 12 empty cells in the figure, so that the sum of the four numbers in each row is equal, and the sum of the three numbers in each column is equal. What is the number that is not filled in the cells? And please fill in t... | 【Analysis】Considering the row, we have 3 and $=1+2+\cdots+13-a=91-a$, which means $a$ leaves a remainder of 1 when divided by 3. Considering the column, we have 4 and $=1+2+\cdots+13-a=91-a$, which means $a$ leaves a remainder of 3 when divided by 4. Thus, $a=7$, and the sum of the row is 28, and the sum of the column ... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
1. (5 points) Calculate: $3 \frac{3}{4} \times 1.3 + 3 \div 2 \frac{2}{3}=$ | $$
\begin{array}{l}
\text { [Solution] Solution: } 3 \frac{3}{4} \times 1.3+3 \div 2 \frac{2}{3} \\
=3.75 \times 1.3+3 \times \frac{3}{8} \\
=0.375 \times 13+3 \times \frac{3}{8} \\
=\frac{3}{8} \times 13+3 \times \frac{3}{8} \\
=(13+3) \times \frac{3}{8} \\
=16 \times \frac{3}{8} \\
=6
\end{array}
$$
Therefore, the a... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=a_{2}=\frac{1}{2}, a_{n+1}=2 a_{n}+a_{n-1}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then $\left[\sum_{k=2}^{2016} \frac{1}{a_{k-1} a_{k+1}}\right]=$ $\qquad$ . | 5.1.
Given, the sequence $\left\{a_{n}\right\}$ is an increasing sequence of positive numbers.
From $a_{k+1}-a_{k-1}=2 a_{k}$, we know
$$
\begin{array}{l}
\frac{1}{a_{k+1} a_{k-1}}=\frac{1}{2}\left(\frac{1}{a_{k-1} a_{k}}-\frac{1}{a_{k} a_{k+1}}\right) \\
\Rightarrow \sum_{k=2}^{2016} \frac{1}{a_{k-1} a_{k+1}}=\frac{1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.167 Two players, A and B, play a game using the first 100 natural numbers: A first writes these 100 numbers in some order as a sequence without showing it to B. Then B can write down any 50 of these 100 natural numbers and ask A to write them in the order they appear in the sequence. After seeing the results of the p... | [Solution] Party B first proposes the array $M_{1}=\{1,2, \cdots, 50\}$, and the second time proposes the array $M_{2}=\{51,52, \cdots, 100\}$, asking Party A to write out the 50 numbers in $M_{1}$ and $M_{2}$ in the order of the sequence. Then, B records the first 25 numbers in the order of the sequence from $M_{1}$ a... | 5 | Combinatorics | proof | Yes | Yes | olympiads | false |
Consider the following problem: Find the smallest $k \in \mathbf{N}^{\boldsymbol{*}}$, such that for any integer-coefficient polynomial $f(x)$ (with leading coefficient 1) of degree $n > k$, if there are at least $\left[\frac{n}{2}\right]+1$ distinct integers $x$ for which $|f(x)|=1$, then $f(x)$ cannot be factored int... | Solution: First, if $f(x)=g(x) h(x), g, h \in Z[x]$, then $n \leqslant 7$.
Assume without loss of generality that $\partial(g(x)) \leqslant \partial(h(x))$, then $\partial(g(x)) \leqslant\left[\frac{n}{2}\right]$.
Let $\left|f\left(x_{i}\right)\right|=1,1 \leqslant i \leqslant m, m \geqslant\left[\frac{n}{2}\right]+1$,... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$18 \cdot 20$ Among the following four propositions:
(1) If $a, b$ are real numbers, then $(a+b)^{0}=1$;
(2) If $a a b^{4}>a b^{2}$;
(3) If a quadrilateral is divided into four triangles of equal area by its two diagonals, then this quadrilateral is a parallelogram;
(4) If $P$ is any point inside rectangle $A B C D$, a... | [Solution] (1) The proposition is not true when $a=-b$;
(2) From the given, we know $a b>0$, and $a b^{4}a b^{2}$. Therefore, we have $a b>a b^{4}>a b^{2}$;
(3) If the diagonals of quadrilateral $A B C D$ intersect at $O$, and divide the quadrilateral into four triangles of equal area.
$$
\begin{array}{ll}
\because & S... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
1. Given $F_{1} 、 F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$, and $A$ is a moving point on the ellipse. Circle $C$ is tangent to the extension of $F_{1} A$, the extension of $F_{1} F_{2}$, and the line segment $A F_{2}$. If $M(t, 0)$ is a point of tangency, then $t=$ $\qquad$ ... | 1.2.
Analysis: Circle $C$ is the excircle of $\triangle A F_{1} F_{2}$ corresponding to side $A F_{2}$. Let circle $C$ touch $F_{1} A$ at $M_{1}$, then $F_{1} M_{1}=F_{1} M$ and $F_{1} M+F_{1} M_{1}=F_{1} F_{2}+A F_{2}+A F_{1}=6$, which gives $F_{1} M=3$, thus $M(2,0)$, i.e., $t=2$. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The 59th question: Given any positive integer $a$, define an integer sequence $x_{1}, x_{2}, x_{3}, \ldots$, such that $x_{1}=a, x_{n+1}=2 x_{n}+1$, and let $y_{n}=2^{x_{n}}-1$. Determine the largest possible integer $\mathrm{k}$ such that for some positive integer $a, y_{1}, y_{2}, \ldots, y_{k}$ are all prime numbers... | The 59th question:
Answer: The $\mathrm{k}_{\max }=2$ satisfies the conditions of the problem. The proof is given below.
We first prove a lemma: If $y=2^{x}-1$ is a prime number, then $\mathrm{x}$ is also a prime number.
Indeed, if $x=1$, then $y=1$, which is not a prime number. If $x$ is a composite number, take any p... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. As shown in the figure, the area of parallelogram $A B C D$ is $60, D P$ intersects $A C$ at $M, A M=\frac{1}{4} A C$. The area of $\triangle A P D$ is _. $\qquad$ | $10$ | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$15 \cdot 8$ for $\{1,2\} \subseteq x \subseteq\{1,2,3,4,5\}$ (where $x$ is a subset of $\{1,2,3,4,5\}$) the number of solutions is
(A) 2.
(B) 4.
(C) 6.
(D) 8.
(E) None of these.
(23rd American High School Mathematics Examination, 1972) | [Solution] The 8 different subsets of $\{1,2\}$ and $\{3,4,5\}$, when combined with each other, form the different solutions for $x$, and there are no other solutions. Specifically, the subsets $x$ are
$$
\begin{array}{l}
\{1,2\}, \quad\{1,2,3\}, \quad\{1,2,4\}, \quad\{1,2,5\}, \quad\{1,2,3,4\}, \\
\{1,2,3,5\}, \quad\{... | 8 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
6. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=a_{2}=1, a_{3}=-1, a_{n}=a_{n-1} a_{n-3}$, find $a_{1964}$. | 3 items are enough to show that the sequence is a periodic sequence. Since $1964=7 \times 280+4$, we have $a_{1964}=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
One, (20 points) Given that $a$ and $b$ are positive integers. Find the smallest positive integer value that
$$
M=3 a^{2}-a b^{2}-2 b-4
$$
can take. | Given that $a$ and $b$ are positive integers, to make
$$
M=3 a^{2}-a b^{2}-2 b-4
$$
a positive integer, it is evident that $a \geqslant 2$.
When $a=2$, $b$ can only be 1, in which case, $M=4$.
When $a=3$, $b$ can only be 1 or 2.
If $b=1$, then $M=18$; if $b=2$, then $M=7$.
When $a=4$, $b$ can only be 1, 2, or 3.
If $b... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let the first term and common difference of an arithmetic sequence be non-negative integers, the number of terms be no less than 3, and the sum of all terms be $97^{2}$, then the number of such sequences is
A. 2
B. 3
C. 4
D. 5 | 5. C Let the first term of the arithmetic sequence be $a$, and the common difference be $d$. From the given, we have: $n a + \frac{1}{2} n(n-1) d = 97^{2}$, which simplifies to $[2 a + (n-1) d] n = 2 \times 97^{2}$. Since $a \geqslant 0, d \geqslant 0$, it follows that $2 \times 97^{2} = n(n-1) d + 2 a n \geqslant n(n-... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
【Question 6】Person A and Person B stand facing each other 30 meters apart, playing "Rock, Paper, Scissors". The winner moves forward 3 meters, the loser moves back 2 meters, and in the event of a tie, both move forward 1 meter. After 15 rounds, Person A is 17 meters from the starting point, and Person B is 2 meters fro... | Analysis:
(Method 1) In games with a winner and a loser, the distance between the two people shortens by 1 meter; in a tie, the distance shortens by 2 meters. After 15 games, the distance between the two people will shorten by 15 to 30 meters.
(1) If both people retreat in the end, the distance between them will not sh... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
4. Two vectors $\overrightarrow{O A}$ and $\overrightarrow{O B}$ on a plane satisfy $|\overrightarrow{O A}|=a,|\overrightarrow{O B}|=b$, and $a^{2}+b^{2}=4, \overrightarrow{O A} \cdot \overrightarrow{O B}=0$. If the vector $\overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mat... | 4. 2 .
Since $|\overrightarrow{O A}|=a,|\overrightarrow{O B}|=b$, and $a^{2}+b^{2}=4, O A \perp O B$, therefore, points $O, A, B$ lie on a circle with the midpoint $M$ of $A B$ as the center and 1 as the radius.
Also, $\overrightarrow{O M}=\frac{1}{2}(\overrightarrow{O A}+\overrightarrow{O B}), \overrightarrow{O C}=\l... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Let any real numbers $x_{0}>x_{1}>x_{2}>x_{3}>0$, to make $\log _{\frac{x_{0}}{x_{1}}} 1993+$ $\log _{\frac{x_{1}}{x_{2}}} 1993+\log _{\frac{x_{2}}{x_{3}}} 1993 \geqslant k \log _{\frac{x_{0}}{x_{3}}} 1993$ always hold, then the maximum value of $k$ is | 5. Rewrite the original inequality as
$$
\frac{1}{\log _{1993} \frac{x_{0}}{x_{1}}}+\frac{1}{\log _{1993} \frac{x_{1}}{x_{2}}}+\frac{1}{\log _{1993} \frac{x_{2}}{x_{3}}} \geqslant k \frac{1}{\log _{1993} \frac{x_{0}}{x_{1}} \cdot \frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}}},
$$
i.e.,
$$
\begin{array}{c}
\frac{1}{\lo... | 9 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
51 If $a, b, c \in \mathbf{R}^{+}, a+b+c=1$, then the integer part of $M=\sqrt{3 a+1}+\sqrt{3 b+1}+\sqrt{3 c+1}$ is
A. 3
B. 4
C. 5
D. 6 | 51 B. By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
M & =1 \cdot \sqrt{3 a+1}+1 \cdot \sqrt{3 b+1}+1 \cdot \sqrt{3 c+1} \\
& \leqslant \sqrt{(1+1+1)[(3 a+1)+(3 b+1)+(3 c+1)]} \\
& =3 \sqrt{2} \quad \text{for} \quad a > a^2, \, b > b^2, \, c > c^2, \, \text{so}
\end{aligned}
$$
$$
\begin{aligned}
M & >\sq... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
4. Let $F_{1}, F_{2}$ be the two foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and $P$ be a point on the ellipse such that $\left|P F_{1}\right|:\left|P F_{2}\right|=$ $2: 1$. Then the area of $\triangle P F_{1} F_{2}$ is $\qquad$. | 4. 4 Let the lengths of the major axis, minor axis, and focal distance of the ellipse be $2a, 2b, 2c$, respectively, then $a=3, b=2, c=\sqrt{5},\left|P F_{1}\right|+\left|P F_{2}\right|=$ $2a=6$, and $\left|P F_{1}\right|:\left|P F_{2}\right|=2: 1$, so $\left|P F_{1}\right|=4,\left|P F_{2}\right|=2$. In $\triangle P F_... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Let $n$ be a natural number, for any real numbers $x, y, z$ there is always $\left(x^{2}+y^{2}+z^{2}\right) \leqslant$ $n\left(x^{4}+y^{4}+z^{4}\right)$, then the minimum value of $n$ is $\qquad$ | 3. Let $a=x^{2}, b=y^{2}, c=z^{2}$, the given inequality becomes
$$
(a+b+c)^{2} \leqslant n\left(a^{2}+b^{2}+c^{2}\right) .
$$
On one hand,
$$
\begin{aligned}
& (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c \\
\leqslant & a^{2}+b^{2}+c^{2}+\left(a^{2}+b^{2}\right)+\left(b^{2}+c^{2}\right)+\left(a^{2}+c^{2}\right) \\... | 3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
15. As shown in the figure, points $A$ and $A^{\prime}$ are on the $x$-axis and are symmetric with respect to the $y$-axis. A line passing through point $A^{\prime}$ and perpendicular to the $x$-axis intersects the parabola $y^{2}=2 x$ at points $B$ and $C$. Point $D$ is a moving point on segment $A B$, and point $E$ i... | Proof: (1) Let $A\left(-2 a^{2}, 0\right), A^{\prime}\left(2 a^{2}, 0\right)$, so we can get $B\left(2 a^{2}, 2 a\right), C\left(2 a^{2},-2 a\right)$. Let $D\left(x_{1}, y_{1}\right), \overrightarrow{A D}=\lambda \overrightarrow{A B}$, then $\overrightarrow{C E}=\lambda \overrightarrow{C A}$.
Thus, $\left(x_{1}+2 a^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$8 \cdot 131$ Let the sequence $\left\{x_{n}\right\}$ satisfy: $x_{1}=1, x_{2}=-1$, and
$$
x_{n+2}=x_{n+1}^{2}-\frac{1}{2} x_{n}, n \geqslant 1 \text {. }
$$
Prove that $\lim _{n \rightarrow \infty} x_{n}$ converges and find this limit. | [Proof] It can be proven that if there exists $0<\delta<\frac{1}{2}$ and a natural number $k$ such that $\left|x_{k}\right| \leqslant \delta$, $\left|x_{k+1}\right| \leqslant \delta$, then $\lim _{n \rightarrow \infty} x_{n}=0$. In fact, from the recursive formula, we have
$\left|x_{k+2}\right| \leqslant\left|x_{k+1}\r... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
10. Given that $f(x)$ is a function defined on $\mathbf{R}$. $f(1)=1$ and for any $x \in \mathbf{R}$, we have $f(x+5) \geqslant f(x)+5, \quad f(x+1) \leqslant f(x)+1$. If $g(x)=f(x)+1-x$, then $g(2002)=$ $\qquad$ | 10.1
From $g(x)=f(x)+1-x$ we get: $f(x)=g(x)+x-1$, so
$$
\begin{array}{l}
g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5, \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 .
\end{array}
$$
That is, $g(x+5) \geqslant g(x), g(x+1) \leqslant g(x)$.
Therefore, $g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \leqslant g(x+2) \leqs... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.57 Let $M$ be a finite set of points in the plane. For any two points $A$ and $B$ in $M$, there exists a third point $C$ such that $\triangle A B C$ is an equilateral triangle. Find the maximum number of elements in $M$. | [Solution] Consider the distance between any two points in $M$, and let $A B$ be the line segment connecting the two points with the greatest distance.
By the problem's condition, there exists a point $C$ such that $\triangle A B C$ is an equilateral triangle.
Construct arcs with $A, B, C$ as centers and $A B$ as the r... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
69 If $11 z^{10}+10 i z^{9}+10 i z-11=0$, then $|z|=$
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
However, the translation is already provided in the response, so there's no need for further action. The text is a mathemati... | 69 1. From the problem, we have
$$
z^{9}=\frac{11-10 i z}{11 z+10 i} .
$$
Let $z=a+b i ，(a, b \in \mathbf{R})$, then
$$
\begin{aligned}
\left|z^{9}\right| & =\left|\frac{11-10 i z}{11+10 i}\right|=\frac{\sqrt{11^{2}+220 b+10^{2}\left(a^{2}+b^{2}\right)}}{\sqrt{11^{2}\left(a^{2}+b^{2}\right)+220 b+10^{2}}} \\
& =\frac{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. If $x^{3}+x^{2}+2 x+2=0$, then
$$
x^{-2008}+x^{-2006}+\cdots+x^{-4}+x^{-2}+1+x+x^{3}+\cdots+x^{2005}+x^{2007}+x^{2009}
$$ | Answer: 0 | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. (15 points) Two identical squares overlap. The upper square is moved 3 cm to the right and then 5 cm downward, resulting in the figure shown. It is known that the area of the shaded part is 57 square centimeters. Find the side length of the square.
| 【Analysis】As shown in the figure, divide the shaded area into two rectangles. One rectangle has a length equal to the side length of the square and a width of 3 cm. The other rectangle has a length equal to the side length of the square minus 3 cm and a width of 5 cm. Let the side length of the square be $a$. According... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 17 As shown in Figure 1.4.19, a semicircle is constructed inside trapezoid $ABCD$, tangent to the upper base and the two non-parallel sides, with its diameter lying on the lower base. If $AB=2, CD=3$, find the length of the lower base $BC$.
untranslated text remains the same as the original, only the content ... | Complete the circle, draw a tangent line $E F$ of circle $O$ parallel to $A D$, intersecting the extensions of $A B, D C$ at $E, F$ respectively. It is easy to prove that $B C$ is the midline of trapezoid $A E F D$, then
$$
A E=2 A B=4, D F=2 D C=6
$$
Since trapezoid $A E F D$ is circumscribed about circle $O$, then
$... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
(5) The number of rational solutions $(x, y, z)$ for the system of equations $\left\{\begin{array}{l}x+y+z=0, \\ x y z+z=0, \\ x y+y z+x z+y=0\end{array}\right.$ is
(A) 1
(B) 2
(C) 3
(D) 4 | (5) If $z=0$, then
$$
\left\{\begin{array}{l}
x+y=0 \\
x y+y=0
\end{array}\right.
$$
Solving, we get
$$
\left\{\begin{array} { l }
{ x = 0 , } \\
{ y = 0 }
\end{array} \text { or } \left\{\begin{array}{l}
x=-1, \\
y=1 .
\end{array}\right.\right.
$$
If $z \neq 0$, then from $x y z+z=0$ we get
$$
x y=-1 \text {. }
$$
... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. Put the operation symbols " +", " -", " ×", and " ÷" into the four positions in the figure, and fill the numbers " $1, 2, 3, 4, 5, 6, 7, 8$ " into the eight blank spaces in the figure, so that the four equations formed by the quadrilateral all hold true. How many ways are there to do this? | 【Answer】 4
【Analysis】 Number puzzle, breakthrough, multiplication and division. There are 3 different numbers from 1 to 8 in the multiplication: $2 \times 3=6, 2 \times 4=8$. The numbers not involved are $1, 5, 7$. 2 can only be filled in the lower left corner or the upper right corner, with adjacent sides forming a mu... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
29. There are 4 cards, each with a different number written on it. The doctor first distributes these 4 cards to four children: Damao, Ermao, Sanmao, and Simaoyi, then collects them back and redistributes them. This process is repeated 3 times. If the sum of the numbers on the cards received by Damao over the three rou... | answer: 7 | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
10. For $\triangle A B C$, the lengths of its sides are denoted as $\alpha, \beta, \gamma$, and the distances from its centroid to each vertex are denoted as $\lambda, \mu, v$. Find the value of $\frac{\alpha^{2}+\beta^{2}+\gamma^{2}}{\lambda^{2}+\mu^{2}+v^{2}}$. | 10. Original expression $=\frac{\sum\left|z_{A}-z_{_{B}}\right|^{2}}{\sum\left|\frac{1}{3}\left(z_{A}+z_{B}+z_{C}\right)-z_{1}\right|^{2}}$ Let's assume $z_{A}=x_{1}+y_{1} \mathrm{i} \quad z_{B}=x_{2}+y_{2} \mathrm{i} \quad z_{C}=x_{3}+y_{3} \mathrm{i}$, so the original expression $=\frac{2\left(\sum x_{i}^{2}+\sum y_{... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Let the function be
$$
f(x)=\sqrt{2 x^{2}+2 x+41}-\sqrt{2 x^{2}+4 x+4}(x \in \mathbf{R}) \text {. }
$$
Then the maximum value of $f(x)$ is $\qquad$ | 3. 5 .
Notice,
$$
\begin{array}{l}
f(x)=\sqrt{2 x^{2}+2 x+41}-\sqrt{2 x^{2}+4 x+4} \\
=\sqrt{(x+5)^{2}+(x-4)^{2}}-\sqrt{x^{2}+(x+2)^{2}} .
\end{array}
$$
Let point $A(-5,6), O(0,0), P(x, x+2)$. Then point $P$ lies on the line $y=x+2$.
It is easy to know that the symmetric point of $O$ with respect to the line $y=x+2... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. The sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{1}{2}, x_{k+1}=x_{k}^{2}+x_{k}$, then the integer part of $\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\cdots+\frac{1}{x_{2009}+1}$ is . $\qquad$ | 5.1 Detailed Explanation: From the given, we know that $\left\{x_{n}\right\}$ is an increasing sequence, and from $x_{k+1}=x_{k}^{2}+x_{k}$, we have $\frac{1}{x_{k+1}}=\frac{1}{x_{k}^{2}+x_{k}}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, which gives $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$. Therefore, $S=\sum_{k=1... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.2 Place several points on the unit sphere such that the distance between any two points is
(1) at least $\sqrt{2}$.
(2) greater than $\sqrt{2}$.
Determine the maximum number of points and prove your conclusion. | [Solution] (1) The maximum number of points is 6.
If $A$ is one of the points, let the position of $A$ on the sphere be the North Pole, then the remaining points must all be in the equator or the southern hemisphere.
If there is a point $B$ at the South Pole, then the rest of the points are all on the equator, at this... | 4 | Geometry | proof | Yes | Yes | olympiads | 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 $\mathrm{A}, \mathrm{B}$. $\mathrm{D}(a, \theta)$ is a point to the right of $F_{1}$, and lines $\mathrm{AD}$ and $\mathrm{BD}$ intersect the left directrix of the ellipse... | Given: $F_{1}(-3,0)$, the equation of the left directrix is $x=-\frac{25}{3}$; the equation of AB is $y=k(x+3)$ (where $k$ is the slope).
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} \Rightarrow\left(16+25 k^{2}\rig... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
9. In $\triangle A B C$, if $b=c, A D$ is the altitude, $B E$ is the angle bisector, $E F \perp B C$ at $F, G$ is a point on $B C$, and $E G \perp B E$, prove that $B G=4 D F$.
---
The translation is provided as requested, maintaining the original formatting and structure. | 9. As shown in Figure 9. $\frac{B G}{D F}=\frac{B E \sec \frac{B}{2}}{A E \sin \frac{A}{2}}=\frac{\sin A \sec \frac{B}{2}}{\sin \frac{B}{2} \sin \frac{A}{2}}=\frac{2 \cos \frac{A}{2}}{\sin \frac{B}{2} \cos \frac{B}{2}}$
$$
=\frac{4 \cos \frac{A}{2}}{\sin B}=4 \text {. }
$$ | 4 | Geometry | proof | Yes | Yes | olympiads | false |
5. (10 points) The total value of 12 coins is 90 cents, consisting only of 5-cent and 10-cent coins. How many of each type are there?
A. 4
B. 5
C. 6
D. 7 | 【Answer】Solution: The number of 5-fen coins:
$$
\begin{aligned}
& (12 \times 1-9) \div(1-0.5) \\
= & 3 \div 0.5 \\
= & 6 \text { (pieces); }
\end{aligned}
$$
The number of 1-jiao coins: $12-6=6$ (pieces).
Answer: Each type of coin is 6.
Therefore, the correct choice is: $C$. | 6 | Algebra | MCQ | Yes | Yes | olympiads | false |
5. To build a building, if construction teams A and B work together, the completion time is 40 days less than if team A builds it alone; if the first 3 floors are built by team A alone, and then team B joins, the completion time is 30 days less than if team A builds it alone. This building has $\qquad$ floors.
(A and B... | $12$ | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Given positive numbers $a, b, c$ satisfying $a+b+c=1$,
then $\sqrt{\frac{(a+b c)(b+c a)}{c+a b}}+\sqrt{\frac{(b+c a)(c+a b)}{a+b c}}+\sqrt{\frac{(c+a b)(a+b c)}{b+c a}}=$ $\qquad$ | $2$ | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Given $A=\left\{x \left\lvert\, \frac{1}{81}<3^{x-1} \leqslant 3\right., x \in \mathbf{Z}\right\}, B=\left\{x \left\lvert\, \frac{x+2}{x-3}<0\right., x \in \mathbf{N}\right\}$, then the number of elements in the set $C=\{m \mid m=x y, x \in A, y \in B\}$ is $\qquad$ | $$
3^{-4}<3^{x-1} \leqslant 3 \Rightarrow -3<x \leqslant 2 \Rightarrow A=\{-2,-1,0,1,2\} ;
$$
Also, $B=\{0,1,2\} \Rightarrow C=\{0, \pm 1, \pm 2, \pm 4\}$, so the number of elements in set $C$ is 7. | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (2005 Hunan Province Competition Question) For $x \in \mathbf{R}$, the function $f(x+2)+f(x-2)=f(x)$, then it is a periodic function, and one of its periods is
A. 4
B. 6
C. 8
D. 12 | 3. D Substitute $x-2$ for $x$ in the equation, then we have $f(x)+f(x-4)=f(x-2)$. Thus, $f(x+2)=-f(x-4)$, which leads to $f(x+6)=-f(x)$. Therefore, $f(x+12)=f(x)$, so the answer is D.
Note that the problem states 12 is the smallest positive period of $f(x)$, which is incorrect. For example, when $f(x)=0$. 12 is just on... | 12 | Algebra | MCQ | Yes | Yes | olympiads | false |
67. From the prime numbers within 15, choose any 3 different numbers as the lengths of the three sides of a triangle, the number of triangles that can be formed is
$\qquad$ . | Reference answer: 5 | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
5. Given that $A D$ is the angle bisector of $\triangle A B C$, the three sides of $\triangle A B C$ are all integers, $A B=2016, A C=B D$. Then the number of $\triangle A B C$ that meet the conditions is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 5. D.
Let $A C=B D=m, C D=n$.
By the Angle Bisector Theorem, we have
$$
\frac{2016}{m}=\frac{m}{n} \Rightarrow m=12 \sqrt{14 n} \text {. }
$$
Since $m$ and $n$ are integers, we have $n=14 k^{2}$ (where $k$ is a positive integer).
Thus, $m=168 k$.
By the triangle inequality, we have
$$
\left\{\begin{array}{l}
(m+n)+m>2... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
6. Given the complex number $z$ satisfies
$$
(a-2) z^{2018}+a z^{2017} \mathrm{i}+a z \mathrm{i}+2-a=0 \text {, }
$$
where, $a<1, \mathrm{i}=\sqrt{-1}$. Then $|z|=$ $\qquad$ . | 6. 1 .
Notice that,
$$
\begin{array}{l}
z^{2017}((a-2) z+a \mathrm{i})=a-2-a z \mathrm{i} . \\
\text { Hence }|z|^{2017}|(a-2) z+a \mathrm{i}|=|a-2-a z \mathrm{i}| .
\end{array}
$$
Let $z=x+y \mathrm{i}(x, y \in \mathbf{R})$.
$$
\begin{array}{l}
\text { Then }|(a-2) z+\left.a \mathrm{i}\right|^{2}-|a-2-a z \mathrm{i}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Three. (25 points) If the pair of positive integers $(a, x)$ satisfies
$$
\sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}} \neq x,
$$
find all positive integers $a$ that meet the requirement. | $$
\text { Three, let } \sqrt{\frac{a-x}{1+x}}=\frac{a-x^{2}}{1+x^{2}}=t \text {. }
$$
Then $t^{2} x+x+t^{2}-a=0$,
$$
(t+1) x^{2}+t-a=0 \text {. }
$$
$t \times(2)-x \times(1)$ gives
$t x^{2}-x^{2}-t^{2} x+t^{2}-a t+a x=0$
$\Rightarrow(t-x)(t+x-t x-a)=0$
$\Rightarrow t=x$ (discard) or $t=\frac{a-x}{1-x}$.
Thus $\frac{a... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 67$ The 2nd "Chinese Youth" Cup Junior High School Mathematics Competition, if there are 148 students whose scores are between 100 and 120 points (scores are recorded as positive integers), and students with the same score are listed as the same rank, assuming the maximum number of students in the same rank i... | [Solution] There are 21 positive integers between 100 and 120, noting that
$$
148=21 \cdot 7+1,
$$
thus the minimum value of $x$ is $7+1=8$. Therefore, the answer is (B). | 8 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
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