problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
5. (8 points) It is defined that $1 ※ 2=0.1+0.2=0.3, 2 ※ 3=0.2+0.3+0.4=0.9, 5 ※ 4=0.5+0.6+0.7+0.8=2.6$. If $a ※ 15=16.5$, then $a$ equals $\qquad$ . | 5. (8 points) Given $1 ※ 2=0.1+0.2=0.3, 2 ※ 3=0.2+0.3+0.4=0.9, 5 ※ 4=0.5+0.6+0.7+0.8=2.6$, if $a ※ 15=16.5$, then $a$ equals 4.
【Solution】Solution: According to the operation rule,
16.5 is the sum of an arithmetic sequence, 15 is the number of terms in the arithmetic sequence, the common difference is $0.1, a$ is 10 ti... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8 Let $a, b, c, d$ be non-negative real numbers, satisfying
$$
\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c},
$$
then $\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=$ | 8 Obviously $a+b+c+d \neq 0$, since
$$
\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c},
$$
we have
$$
\frac{1}{b+c+d}=\frac{1}{a+c+d}=\frac{1}{a+b+d}=\frac{1}{a+b+c} .
$$
Thus, $a=b=c=d$, hence
$$
\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=4
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, among the six diagonals on the lateral faces $A B_{1}, A_{1} B, B C_{1}, B_{1} C, C_{1} A, C A_{1}$, if one pair $A B_{1} \perp B C_{1}$, how many other pairs $k$ are also perpendicular, then $k$ equals
A. 2
B. 3
C. 4
D. 5 | 5. D As shown in the figure, take the midpoint $M$ of $B C$, and connect $B_{1} M$. Then, the projection of $A B_{1}$ on the plane $B B_{1} C_{1} C$ is $B_{1} M$. Since $A B_{1} \perp B C_{1}$, by the inverse of the three perpendiculars theorem, we have $B_{1} M \perp B C_{1}$. Taking the midpoint $N$ of $B_{1} C_{1}$,... | 5 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 9 For $n \in \mathbf{N}$, let $S_{n}=\min \left(\sum_{\mathrm{k}=1}^{\mathrm{n}} \sqrt{(2 \mathrm{k}-1)^{2}+\mathrm{a}_{\mathrm{k}}^{2}}\right)$, where $\mathrm{a}_{1}$, $\mathrm{a}_{2}, \cdots, \mathrm{a}_{\mathrm{n}} \in \mathbf{R}^{+}, \sum_{i=1}^{n} a_{n}=17$, if there exists a unique $n$ such that $S_{n}$ ... | $$
\begin{array}{l}
\text { Solution: By } \sqrt{(2 k-1)^{2}+a_{i}^{2}}=\left|(2 k-1)+a_{i} \mathrm{i}\right|,(i=1,2, \cdots, n) . \\
\quad \sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}} \\
=\left|1+a_{1} \mathrm{i}\right|+\left|3+a_{2} \mathrm{i}\right|+\cdots+\left|(2 n-1)+a_{n} \mathrm{i}\right| \\
\geqslant\left|[1+3+... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
31. Calculate: $(-1)^{4}+(-2)^{3}+(-3)^{2}+(-4)^{1}=$ | Answer: -2.
Solution: Original expression $=1-8+9-4=-2$. | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Find all values of $a$ such that the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy $\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0$. | 9. Let the variable substitution be $y=x-3$. By Vieta's formulas, and given that $y_{1}^{3}+y_{2}^{3}+y_{3}^{3}=0$, we solve to get $a=-9$. | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. On the Cartesian plane, the number of lattice points (i.e., points with both integer coordinates) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ . | Moving the center of the circle to the origin $O(0,0)$, the equation of the circle becomes $x^{2}+y^{2}=199^{2}$, at this point the integer points on the circumference are $(199,0),(0,199),(-199,0),(0,-199)$. If there is an integer point $(x, y)$ on the circle with $x y \neq 0$, then $|x|,|y|, 199$ form a Pythagorean t... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
The Yangtze Farm has a pasture, where grass grows evenly every day. If 24 cows are grazed on the pasture, the grass will be eaten up in 6 days; if only 21 cows are grazed, it will take 8 days to finish the grass. How many days will it take to finish the grass if 36 cows are grazed? | 【Analysis and Solution】
The problem of cows eating grass.
Assume 1 cow eats 1 portion per day;
24 cows eat $24 \times 6=144$ portions in 6 days;
21 cows eat $21 \times 8=168$ portions in 8 days;
Grass grows $(168-144) \div(8-6)=12$ portions per day;
The original amount of grass in the pasture is $(24-12) \times 6=72$ o... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The integer closest to $\frac{\sqrt{2}+1}{\sqrt{2}-1}-\frac{\sqrt{2}-1}{\sqrt{2}+1}$ is | 2. 6 .
Notice that,
$$
\begin{array}{l}
\frac{\sqrt{2}+1}{\sqrt{2}-1}-\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)^{2}-(\sqrt{2}-1)^{2}}{(\sqrt{2}-1)(\sqrt{2}+1)} \\
=\frac{(3+2 \sqrt{2})-(3-2 \sqrt{2})}{2-1}=4 \sqrt{2} .
\end{array}
$$
Given $1.4<\sqrt{2}<1.5$, we know $5.6<4 \sqrt{2}<6$.
Therefore, the integer ... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (20 points) Given the positive sequence $\left\{a_{n}\right\}$ satisfies:
$$
\begin{array}{l}
a_{1}=1, \\
(n+1) a_{n}^{2}-2 n a_{n+1}^{2}+\sqrt{n^{2}+n} a_{n} a_{n+1}=0, \\
z_{n}=\prod_{k=1}^{n}\left(1-\frac{i}{a_{k}}\right) \text { (i is the imaginary unit). }
\end{array}
$$
Find the value of $\left|z_{2019}-z_{2... | 10. Dividing both sides of the given equation by $n(n+1)$, we get
$$
\begin{array}{l}
\frac{a_{n}^{2}}{n}-\frac{2 a_{n+1}^{2}}{n+1}+\frac{a_{n+1} a_{n}}{\sqrt{n(n+1)}}=0 \\
\Rightarrow\left(\frac{a_{n+1}}{\sqrt{n+1}}-\frac{a_{n}}{\sqrt{n}}\right)\left(\frac{a_{n+1}}{\sqrt{n+1}}+\frac{2 a_{n}}{\sqrt{n}}\right)=0 .
\end{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{n}+a_{n+1}=1\left(n \in \mathbf{N}_{+}\right)$, let $S_{n}$ be the sum of the first $n$ terms of the sequence $a_{n}$, then the value of $S_{2017}-$ $2 S_{2018}+S_{2019}$ is $\qquad$ | $$
\begin{array}{l}
a_{1}=2, a_{2}=-1, a_{3}=2, a_{4}=-1, \cdots, \text { so } S_{2017}-2 S_{2018}+S_{2019} \\
=-\left(S_{2018}-S_{2017}\right)+\left(S_{2019}-S_{2018}\right)=-a_{2018}+a_{2019}=3 .
\end{array}
$$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Wolf Gray and its brothers caught many sheep. If each wolf gets 3 sheep, there will be 2 sheep left over; if each wolf gets 8 sheep, there will be 8 sheep short. So, including Wolf Gray, there are
wolves sharing
sheep. | 【Analysis】A simple profit and loss problem. According to the profit and loss formula, we get $(8+2) \div(8-3)=2$ (wolves). | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$10 \cdot 64$ in the sequence $1,3,2, \cdots$, after the first two terms, each term is equal to the difference of the two preceding terms (the preceding term minus the term before it). The sum of the first 100 terms of this sequence is
(A) 5 .
(B) 4 .
(C) 2 .
(D) 1 .
(E) -1 .
(26th American High School Mathematics Exam... | [Solution] According to the problem, the first eight terms of this sequence are $1, 3, 2, -1, -3, -2, 1$, 3. Since the seventh and eighth terms are the same as the first and second terms, the ninth term will be the same as the third term, and so on; that is, this sequence repeats every 6 terms, i.e.,
$a_{k}=a_{k+6}(k=1... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
10. Given $f(x)=x^{5}-10 x^{3}+a x^{2}+b x+c$. If the roots of the equation $f(x)=0$ are all real, and $m$ is the largest of these five real roots, then the maximum value of $m$ is $\qquad$ | 10. 4 .
Let the five real roots of $f(x)=0$ be
$$
x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant m \text {. }
$$
By Vieta's formulas, we have
$$
\begin{array}{l}
m+x_{1}+x_{2}+x_{3}+x_{4}=0, \\
m \sum_{i=1}^{4} x_{i}+\sum_{1 \leqslant i<j \leqslant 4} x_{i} x_{j}=-10 .
\end{array}
$$
Thus, $\sum_{1 ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. (3 points) It is known that when 2008 is divided by some natural numbers, the remainder is always 10. There are $\qquad$ such natural numbers.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 6. (3 points) It is known that when 2008 is divided by some natural numbers, the remainder is always 10. There are $\qquad$ 11 such natural numbers.
【Solution】Solution: $2008-10=1998$ must be divisible by these numbers, and these numbers must be greater than 10, $1998=2 \times 3 \times 3 \times 3 \times 37$.
The total... | 11 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
6. On the Cartesian plane, the number of integer points (i.e., points with both coordinates as integers) on the circumference of a circle centered at $(199,0)$ with a radius of 199 is $\qquad$ | 4
6.【Analysis and Solution】Let $A(x, y)$ be a point on the circle, then it satisfies the equation $y^{2}+(x-199)^{2}=199^{2}$. Clearly, $(0,0),(199,199),(199,199),(389,0)$ are four sets of integer solutions. Their characteristic is that one of $y$ and $x-199$ is 0.
If there are other solutions such that $y$ and $x-199$... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (3 points) There are four points on a plane, and no three points lie on a straight line. Connecting these four points with six line segments, the number of triangles that can be formed using these segments as sides is ( ).
A. 3
B. 4
C. 6
D. 8 | 【Analysis】According to the definition of a triangle (a figure formed by three line segments that are not on the same straight line, connected end to end in sequence is called a triangle) fill in the blank.
【Solution】Solution: As shown in the figure below: Assume there are four points $A, B, C, D$ on a plane, where no ... | 4 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
3. If $P$ is a point inside the cube $A B C D-E F G H$, and satisfies $P A=P B=\frac{3 \sqrt{3}}{2}$, $P F=P C=\frac{\sqrt{107}}{2}$, then the edge length of the cube is | 3.5
Analysis: Let the side length of the cube be $a(a>0)$. Establish a coordinate system with $A$ as the origin, and $A B, A D, A E$ as the positive axes of $x, y, z$ respectively. Then $A(0,0,0), B(a, 0,0), C(a, a, 0), F(a, 0, a)$. Since $P A=P B$, $P$ lies on the perpendicular bisector plane of segment $A B$, which ... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$33 \cdot 14$ In the set of real numbers, let $x$ be
$$
\left(\frac{\sqrt{(a-2)(|a-1|-1)}+\sqrt{(a-2)(1-|a|)}}{1+\frac{1}{1-a}}+\frac{5 a+1}{1-a}\right)^{1988},
$$
then the unit digit of $x$ is
(A) 1 .
(B) 2 .
(C) 4 .
(D) 6 .
(China Junior High School Mathematics League, 1988) | [Solution] For the two radicals to be meaningful, it is necessary that
$$
(a-2)(|a|-1) \geqslant 0 \text {, and } \quad(a-2)(1-|a|) \geqslant 0 \text {. }
$$
But \quad(a-2)(1-|a|)=-(a-2)(|a|-1),
so
$$
(a-2)(|a|-1)=0,
$$
Solving this gives \(a_{1}=2, a_{2}=1, a_{3}=-1\).
If \(a_{1}=2\), then \(1+\frac{1}{1-a}=0\);
If ... | 6 | Algebra | MCQ | Yes | Yes | olympiads | 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 $\overrig... | $11-5$ Hint: Let $A(-a, 0), B(a, 0), P\left(x_{1}, y_{1}\right), Q\left(x_{2}\right.$, $\left.y_{2}\right)$, from $\overrightarrow{A P}+\overrightarrow{B P}=\lambda(\overrightarrow{A M}+\overrightarrow{B M})$ we know, points $O, P, M$ are collinear, thus we can obtain
$$
k_{1}+k_{2}=\frac{y_{1}}{x_{1}+a}+\frac{y_{1}}{x... | -5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. If three angles $x, y, z$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan x \cdot \tan y + \tan y \cdot \tan z + \tan z \cdot \tan x$ $=$ . $\qquad$ | 6. -3 .
From the problem, we know that
$1+\tan x \cdot \tan y=\frac{\tan y-\tan x}{\tan (y-x)}=\frac{\tan y-\tan x}{\sqrt{3}}$.
Similarly, $1+\tan y \cdot \tan z=\frac{\tan z-\tan y}{\sqrt{3}}$,
$1+\tan z \cdot \tan x=\frac{\tan z-\tan x}{-\sqrt{3}}$.
Adding the above three equations, we get
$\tan x \cdot \tan y+\tan y... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. $2019^{\ln \ln 2019}-(\ln 2019)^{\ln 2019}$ $=$ . $\qquad$
Fill in the blank (8 points per question, total 64 points)
1. $2019^{\ln \ln 2019}-(\ln 2019)^{\ln 2019}$ $=$ . $\qquad$ | \begin{array}{l}\text {1.0. } \\ 2019^{\ln \ln 2019}=\left(e^{\ln 2019}\right)^{\ln \ln 2019} \\ =\left(e^{\ln \ln 2019}\right)^{\ln 2019}=(\ln 2019)^{\ln 2019} .\end{array} | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. As shown in the figure, $C, D$ are two points on the line segment $A B$, $M, N$ are the midpoints of $A C, D B$ respectively. If $A B=10, C D=2$, then the length of $M N$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | Answer: D.
Solution: Since $M, N$ are the midpoints of $A C, D B$ respectively, let $A M=x, D N=y$.
Then
$$
2 x+2 y=10-2=8,
$$
Therefore
$$
x+y=4 .
$$
So
$$
M N=x+2+y=4+2=6 \text {. }
$$
Hence, the answer is D. | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
33.22 $[x],[y],[z]$ represent the greatest integers not exceeding $x, y, z$ respectively. If $[x]=5,[y]=-3,[z]=-1$, then the number of possible values for $[x-y-z]$ is
(A) 3.
(B) 4.
(C) 5.
(D) 6.
(China Shanxi Province Taiyuan City Junior High School Mathematics Competition, 1993) | [Solution] $\because 5 \leqslant x<6,-3 \leqslant y<-2,-1 \leqslant z<0$, we have $\quad 2<-y \leqslant 3,0<-z \leqslant 1$.
$$
\therefore 7<x-y-z<10 \text {. }
$$
Thus, $[x-y-z]$ can take the values $7,8,9$. Therefore, the answer is $(A)$. | 3 | Number Theory | MCQ | Yes | Yes | olympiads | false |
2. Given the function $y=a \cos x+b$ has a maximum value of 1 and a minimum value of $-7$, the maximum value of $a \cos x+b \sin x$ is ( ).
A. 1
B. 3
C. 5
D. 7 | 2. C.
From the conditions, we know that $y_{\max }=|a|+b=1, y_{\min }=-|a|+b=-7$, so we solve to get $|a|=4, b=3$. Then the maximum value of $a \cos x+b \sin x$ is $\sqrt{a^{2}+b^{2}}=5$. | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
13. (5 points)
The distance from Earth to Planet Sadala is 80 light years (Note: A light year is a unit of length). Vegeta and Goku depart from Earth to Planet Sadala. Vegeta departs 1 day earlier than Goku. If Vegeta and Goku fly in a straight line, they can both fly 10 light years per day. How many days after Goku's ... | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) Calculate: $\frac{3.2-2.95}{0.25 \times 2+\frac{1}{4}}+\frac{2 \times 0.3}{2.3-1 \frac{2}{5}}=$ | 【Analysis】Simplify the expressions in the numerator and denominator of the complex fraction, then solve according to the basic properties of fractions.
【Solution】Solve: $\frac{3.2-2.95}{0.25 \times 2+\frac{1}{4}}+\frac{2 \times 0.3}{2.3-1 \frac{2}{5}}$
$$
\begin{array}{l}
=\frac{0.25}{0.75}+\frac{0.6}{0.9} \\
=\frac{1}... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. Find the smallest positive integer $n$ such that: if for any finite set of points $A$ in the plane, any $n$ points can be covered by two lines, then there exist two lines that can cover all points in $A$. | 13. Let $A$ be a set of 6 points consisting of the three vertices and the midpoints of the three sides of a triangle. Then any 5 points in $A$ can be covered by 2 lines, but the 6 points in $A$ cannot be covered by any two lines. Therefore, the smallest positive integer $n$ $\geqslant 6$. Next, assume that any 6 points... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 37$ A fraction has both its numerator and denominator as positive integers, and the numerator is 1 less than the denominator. If both the numerator and the denominator are decreased by 1, the resulting fraction is a positive number less than $\frac{6}{7}$. Then the number of fractions that satisfy the above c... | [Solution] Let $a$ be a positive integer, then the fraction can be expressed as $\frac{a}{a+1}$. According to the problem, $0<\frac{a-1}{a}<\frac{6}{7}$, which means $1<a<7$, so $a=2,3,4,5,6$. Therefore, the fractions that meet the conditions are $\frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}$. Hence,... | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
$2 \cdot 63$ Inserting "+" or "-" between $1^{2}, 2^{2}, 3^{2}, \cdots, 1989^{2}$, what is the smallest non-negative number that can be obtained from the resulting sum expression? | [Solution] Since there are 995 odd numbers in $1^{2}, 2^{2}, 3^{2}, \cdots, 1989^{2}$, the result of the expression obtained by adding “+” and “-” signs will always be odd, no matter how they are added. Therefore, the smallest non-negative value sought is no less than 1.
Since the squares of 4 consecutive natural numb... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. The number of trees with exactly 7 vertices (indistinguishable) is ( ). In graph theory, a tree is a connected graph without cycles, simply understood as a graph that connects \( n \) vertices with \( n-1 \) edges.
A. 9
B. 10
C. 11
D. 12 | 4. C As shown in the figure, use the exhaustive method.
We classify and list them in the order of the length of the longest chain in each tree:
Therefore, there are 11 trees that have exactly 7 vertices (undistinguished).
| 11 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
$4 \cdot 52$ for some natural number $n, 2^{n}$ and $5^{n}$ have the same digit in their highest place, what is this digit?
Translating the text into English while preserving the original formatting and line breaks, the result is as follows:
$4 \cdot 52$ for some natural number $n, 2^{n}$ and $5^{n}$ have the same ... | [Solution] Let this number be $a$. Since
$$
2^{n} \cdot 5^{n}=10^{n} \text {, }
$$
we have $a^{2}<10<(a+1)^{2}$.
Thus, we get $\quad a=3$. (For example, $2^{5}=32,5^{5}=3125$ ) | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. For all $m, n$ satisfying $1 \leqslant n \leqslant m \leqslant 5$, the number of different hyperbolas represented by the polar equation $\rho=\frac{1}{1-\mathrm{C}_{m}^{n} \cos \theta}$ is
(A) 15
(B) 10
(C) 7
(D) 6 | (D)
3.【Analysis and Solution】The polar equation of the conic section $\rho=\frac{p}{1-e \cos \theta}$ represents a hyperbola if and only if $e>1$. Therefore, the equation in this problem must satisfy $\mathrm{C}_{m}^{n}>1$. Given $1 \leqslant n \leqslant m \leqslant 5$, we can obtain 6 different values: $\mathrm{C}_{3}... | 6 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
6. Given that the equation $x^{3}+a x^{2}+b x+c=$ 0 has three non-zero real roots forming a geometric sequence. Then $a^{3} c-b^{3}=$ $\qquad$ | 6. 0 .
Solution: Let the three roots be $d, d q, d q^{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
d+d q+d q^{2}=-a \\
d^{2} q+d^{2} q^{2}+d^{2} q^{3}=b \\
d^{3} q^{3}=-c
\end{array}\right.
$$
(2) ÷ (1) gives $d q=-\frac{b}{a}$, substituting into (3) yields $\left(-\frac{b}{a}\right)^{3}=-c$, hence $a^... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8.95 Bus Route 95 departs from Shanghai Railway Station (the starting station) 11 times between 6:00 AM and 7:00 AM (with one bus departing at 6:00 and one at 7:00). It is known that the interval between the departure of adjacent buses is equal. Therefore, a Bus Route 95 departs from the starting station every ( ) minu... | 【Answer】6 minutes.
【Analysis】Key point: Application of tree planting problems in life, corresponding lecture in Xueersi: Tree Planting Problem in the 3rd grade summer class (Lecture 4). From 6:00 to 7:00, there are a total of 60 minutes, and there are ten intervals between the 11 buses, so one interval is: $60 \div(11-... | 6 | Other | math-word-problem | Yes | Yes | olympiads | false |
8. Given the sequence $\left\{a_{n}\right\}$ with the first term being 2, and satisfying $6 S_{n}=3 a_{n+1}+4^{n}-1$. Then the maximum value of $S_{n}$ is | $$
\begin{array}{l}
6 S_{n}=3\left(S_{n+1}-S_{n}\right)+4^{n}-1 \Rightarrow S_{n+1}=3 S_{n}-\frac{4^{n}-1}{3} \\
\Rightarrow S_{n+1}+\lambda \cdot \frac{4^{n+1}-1}{3}+b=3\left(S_{n}+\lambda \cdot \frac{4^{n}-1}{3}+b\right) \\
\Rightarrow S_{n+1}+4 \lambda \cdot \frac{4^{n}-1}{3}+\lambda+b=3 S_{n}+3 \lambda \cdot \frac{... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 96$ The number of real roots of the equation $x|x|-3|x|-4=0$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(2nd China Tianjin "Chinese Youth Cup" Mathematics Invitational, 1987) | [Solution] When $x \geqslant 0$, the equation can be transformed into $x^{2}-3 x-4=0$, solving which yields $x=4$ or -1 (which does not meet the initial condition, so it is discarded).
When $x<0$, the equation can be transformed into $-x^{2}+3 x-4=0$, i.e.,
$$
x^{2}-3 x+4=0 \text {. }
$$
Since its discriminant $\Delt... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
11. Xiao Ming brought 24 handmade souvenirs to sell at the London Olympics. In the morning, each souvenir was sold for 7 pounds, and the number of souvenirs sold was less than half of the total. In the afternoon, he discounted the price of each souvenir, and the discounted price was still an integer. In the afternoon, ... | 【Analysis】The maximum number sold in the morning is 11
$$
\begin{aligned}
& 120=11 \times 7+43=11 \times 7+13 \times \frac{43}{13} \\
= & 10 \times 7+50=10 \times 7+14 \times \frac{25}{7} \\
= & 9 \times 7+57=9 \times 7+15 \times 3.8 \\
= & 8 \times 7+64=8 \times 7+16 \times 4 \\
= & 7 \times 7+71=7 \times 7+17 \times ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. If the sum of the first $n$ terms of the arithmetic sequence $\left\{a_{n}\right\}$ is $S_{n}, S_{6}=36, S_{12}=144$, $S_{6 n}=576$, then $n$ equals ( ).
A. 3
B. 4
C. 5
D. 6 | 6. B. Since the sum of every consecutive $n$ terms in an arithmetic sequence forms an arithmetic sequence, according to the problem, the sum of the first 6 terms is 36, and the sum of the next 6 terms is 108. According to the properties of an arithmetic sequence, this forms an arithmetic sequence with the first term be... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
17. Given the set $S \subseteq\{1,2, \cdots, 12\},|S|=6$ ( $|S|$ represents the number of elements in the set $S$). If $a, b \in$ $S, a<b$, then $b$ is not a multiple of $a$. Then, the smallest element in the set $S$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
(E) 7 | 17. C.
Obviously, $1 \notin S$, and among $\{2,4,8\}$, $\{3,6,12\}$, and $\{5,10\}$, at most one from each set can be in $S$.
Thus, $\{7,9,11\} \subset S \Rightarrow 3 \notin S$
$\Rightarrow 6$ or $12$ is in $S$ $\Rightarrow 2 \notin S$.
Upon verification, $\{4,5,6,7,9,11\}$ satisfies the requirements. | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
9. (14 points) Given the function $f(x)=a x^{2}+b x+c$ $(a, b, c \in \mathbf{R})$, when $x \in[-1,1]$, $|f(x)| \leqslant 1$.
(1) Prove: $|b| \leqslant 1$;
(2) If $f(0)=-1, f(1)=1$, find the value of $a$. | 9. Solution: (1) $\because f(1)=a+b+c, f(-1)=a-b+c$, then $b=\frac{1}{2}(f(1)-f(-1))$,
According to the problem, $|f(1)| \leqslant 1,|f(-1)| \leqslant 1$,
$$
\begin{array}{l}
\quad \therefore b=\frac{1}{2}|f(1)-f(-1)| \leqslant \frac{1}{2}(|f(1)|- \\
|f(-1)|) \leqslant 1 .
\end{array}
$$
(2) From $f(0)=-1, f(1)=1$, we ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 31$ Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there always exist 3 numbers, the sum of any two of which is still irrational. | [Solution] Clearly, any 3 numbers among the 4 numbers $\{\sqrt{2}, \sqrt{2},-\sqrt{2},-\sqrt{2}\}$ include a pair of opposite numbers, whose sum is 0, which is of course not an irrational number. Therefore, the smallest positive integer $n \geqslant 5$.
Let $x, y, z, u, v$ be 5 irrational numbers. We use 5 points to r... | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) Square $ABCD$ and rectangle $BEFG$ are placed as shown in the figure, with $AG=CE=2$ cm. The area of square $ABCD$ is larger than the area of rectangle $BEFG$ by $\qquad$ square centimeters. | 【Solution】Solution: According to the analysis, the common part in the figure is rectangle $G H C B$, so:
the area of square $A B C D$ - the area of rectangle $B E F G$ = the area of rectangle $A D H G$ - the area of rectangle $E F H C$ = $A G \times A D - C E \times C H = 2 \times A D - 2 \times C H = 2 \times (A D - C... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 11 Let $S=\{1,2,3,4,5,6,7,8,9,10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$, satisfying
(1) $\left|A_{i}\right|=5(1 \leqslant i \leqslant k)$;
(2) $\left|A_{i} \cap A_{j}\right| \leqslant 2(1 \leqslant i<j \leqslant k)$.
Find the maximum value of the number of such subsets $k$.
(1994 National Training T... | Solution 1: Construct a $10 \times k$ table, where the element in the $i$-th row and $j$-th column is
$$
a_{i j}=\left\{\begin{array}{l}
1\left(\text { if } i \in A_{j}\right) \\
0\left(\text { if } i \notin A_{j}\right)
\end{array}(i=1,2, \cdots 10, j=1,2, \cdots, k)\right. \text {. }
$$
Thus, the sum of the elements... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (8 points) There are 10 cards, each with a number written on it: $1,2,3, \cdots, 9,10$. How many cards must be drawn to ensure that among the drawn cards, there are two cards whose numbers add up to 11? | 【Answer】Solution:
$$
\begin{array}{l}
10 \div 2=5 \text { (pieces) } \\
5+1=6 \text { (pieces) }
\end{array}
$$
So the answer is 6 | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
15. In the 6 circles below, fill in $1,2,3,4,5,6$, using each number only once, so that the sum of the three numbers on each side is equal. | 【Analysis】Let the 3 vertices be $a, b, c, 1+2+3+4+5+6+a+b+c=21+a+b+c$ is the sum of the 3 edges, $a+b+c$ is a multiple of 3. When $a+b+c=1+2+3=6$, the sum of each edge is $(21+6) \div 3=9$. By rotating the above two numbers, 4 more results can be obtained. | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. If the sum of the volumes of $n$ cubes with side lengths as positive integers is $2002^{2005}$. Find the minimum value of $n$.
| 8. Solution: Let the side lengths of $n$ cubes be $x_{1}, x_{2}, \cdots, x_{n}$, then $x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2005}$.
$$
\begin{array}{l}
\because 2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9), \\
\therefore 2002^{2005} \equiv 4^{2005} \equiv 4(\bmod 9), \\
\because x^{3} \equiv 0, \pm 1(\bmod 9),... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 23$ Among the first 100 positive integers, the number of integers that can be divided by $2,3,4,5$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(32nd American High School Mathematics Examination, 1981) | [Solution 1] The positive integers not exceeding 100 and divisible by 5 are $5,10,15, \cdots 100$. Among these, only $20,40,60,80$ are divisible by 4.
Among these, only 60 is divisible by 3. Since 60 is divisible by 4, it is naturally divisible by 2.
Therefore, there is only one number that meets the condition.
So the ... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
6. (10 points) An exam consists of 6 multiple-choice questions, with the following scoring rules: Each person starts with 6 points, gets 4 points for each correct answer, loses 1 point for each wrong answer, and gets 0 points for unanswered questions. There are 51 students taking the exam. Therefore, at least ( ) stude... | 【Analysis】According to this scoring method, the highest possible score is (30 points), and the lowest is 0 points after a deduction of 6 points, totaling $30+1=$ 31 (different scores). Since each wrong answer results in a deduction of: $1+4=5$ points, not answering a question results in a maximum deduction of 4 points,... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
Example 5 On the Cartesian plane, the number of integer points on the circumference of a circle centered at $(199,0)$ with a radius of 199 is | Let $A(x, y)$ be any integer point on circle $O$, then its equation is $(x-199)^{2}+y^{2}=199^{2}$.
Obviously, (0,0), $(199,199),(199,-199),(398,0)$ are 4 solutions to the equation. However, when $y \neq 0, \pm 199$, $(y, 199)=1$ (since 199 is a prime number), at this time, $199, y,|199-x|$ form a Pythagorean triplet, ... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Fingerprint recognition has very wide applications in daily life. Before using fingerprint recognition, fingerprints need to be recorded first. Yingchun Electronics Factory has developed a fingerprint recording system. The first press can record $\frac{4}{5}$ of the pattern, and each subsequent press can record half... | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(2) (2007 - Ningxia Hainan) Let the function $f(x)=\frac{(x+1)(x+a)}{x}$ be an odd function, then $a=$ | From the given, we have $f(-x)=-f(x)$, which means $\frac{(-x+1)(-x+a)}{-x}=-\frac{(x+1)(x+a)}{x}$, or equivalently, $-(a+1) x=(a+1) x$ for any real number $x \neq 0$. Therefore, $a=-1$.
| -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. A wire of length $150 \mathrm{~cm}$ is to be cut into $n(n>2)$ segments, each of which has an integer length of no less than $1(\mathrm{~cm})$. If no three segments can form a triangle, then the maximum value of $n$ is $\qquad$ . | 7. 10 First, $n$ should be as large as possible, which means the length of each segment must be as small as possible. The lengths of these small segments can only be $1,1,2,3,5,8$, $13,21,34,55,89, \cdots$, but $1+1+2+\cdots+34+55=143$ 150 , so the maximum value of $n$ is 10. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Given for any $x_{1}, x_{2}, \cdots, x_{2020} \in[0,4]$,
the equation
$$
\left|x-x_{1}\right|+\left|x-x_{2}\right|+\cdots+\left|x-x_{2020}\right|=2020 a
$$
has at least one root in the interval $[0,4]$. Then $a$ equals
$\qquad$ . | 3. 2 .
$$
\text { Let } \begin{aligned}
f(x)= & \left|x-x_{1}\right|+\left|x-x_{2}\right|+\cdots+ \\
& \left|x-x_{2020}\right|-2020 a .
\end{aligned}
$$
Then $f(0)=x_{1}+x_{2}+\cdots+x_{2020}-2020 a$,
$$
f(4)=8080-\left(x_{1}+x_{2}+\cdots+x_{2000}\right)-2020 a \text {. }
$$
Therefore, $f(0)+f(4)=8080-4040 a$
$$
=404... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (1993 National High School Mathematics Competition) What are the last two digits of the integer $\left[\frac{10^{93}}{10^{31}+3}\right]$? (Write the tens digit first, then the units digit) | 1. Let $10^{31}=t$, then
$$
\begin{aligned}
{\left[\frac{10^{93}}{10^{31}+3}\right] } & =\left[\frac{t^{3}}{t+3}\right]=\left[\frac{t^{3}+3^{3}}{t+3}-\frac{3^{3}}{t+3}\right]=t^{2}-3 t+3^{2}+\left[-\frac{3^{3}}{t+3}\right]=t^{2}-3 t+3^{2}-1 \\
& =t(t-3)+8=10^{31}\left(10^{31}-3\right)+8 .
\end{aligned}
$$
Therefore, t... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. In the Cartesian coordinate system, the area of the region formed by points $(x, y)$ satisfying $|x|+|y|+|x-2| \leqslant 4$ is
$\qquad$ (Supplied by An Zhenping) | 2. 12 .
Obviously, the region is symmetric about the $x$-axis. Therefore, we only need to consider the graph represented by the equation $|x|+|y|+|x-2|=4$ when $y \geqslant 0$.
Draw the graph of the function $y=4-|x|-|x-2|(-1 \leqslant x \leqslant 3)$, and then reflect it about the $x$-axis, to obtain the graph repre... | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$10 \cdot 67$ Given the sequence $10^{\frac{1}{11}}, 10^{\frac{2}{11}}, 10^{\frac{3}{11}}, 10^{\frac{4}{11}}, \cdots, 10^{\frac{n}{11}}$. The positive integer $n$ that makes the product of the first $n$ terms of the sequence equal to 1000000 is
(A) does not exist.
(B) 8 .
(C) 9 .
(D) 10 .
(E) 11 .
(22nd American High S... | [Solution] Since $1000000=10^{6}$, the sum of the exponents of the first $n$ terms of the sequence should be 6.
That is, $\frac{n(n+1)}{22}=6$, or $n(n+1)=132$,
$$
\therefore \quad n=11 \text {. }
$$
Therefore, the answer is $(E)$. | 11 | Algebra | MCQ | Yes | Yes | olympiads | false |
7・15 Draw n segments sequentially on a plane, where the terminal point of the nth segment coincides exactly with the starting point of the 1st segment. Each segment is called a "line segment." If the starting point of one line segment coincides exactly with the terminal point of another line segment, these two line seg... | [Solution] We prove that the maximum value of $n$ is 12.
First, we can draw a simple zigzag loop for $n=12$ (as shown in the figure). If $n>12$, since the $n$ segments are distributed on 6 lines, by the pigeonhole principle, at least one line $l$ contains at least three segments. These at least three segments are not a... | 12 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
17 If $2^{n+2} \cdot 3^{n}+5 n-a$ can be divisible by 25, then the smallest positive value of $a$ is | 17 4. $\quad 4 \times 6^{n}+5 n-a=4\left(5^{n}+C_{n}^{1} \cdot 5^{n-1}+\cdots+C_{n}^{n-1} \times 5+1\right)+5 n-a$, so $25 \mid 4-a, a=4$ | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Given that the real number $x$ satisfies $20 \sin x=22 \cos x$, then the greatest integer not exceeding the real number $\left(\frac{1}{\sin x \cos x}-1\right)^{7}$ is $\qquad$ . | $$
\tan x=\frac{11}{10} \Rightarrow \sin 2 x=\frac{2 \tan x}{1+\tan ^{2} x}=\frac{\frac{11}{5}}{1+\frac{121}{100}}=\frac{220}{221} \text {, }
$$
Then $\left(\frac{1}{\sin x \cos x}-1\right)^{7}=\left(\frac{2}{\sin 2 x}-1\right)^{7}=\left(1+\frac{1}{110}\right)^{7} \in(1,2)$, so the answer is 1. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. In $\triangle A B C$, the three sides $a, b, c$ form an arithmetic sequence. Find the value of $5 \cos A-4 \cos A \cos C+5 \cos C$. | 8. From the given condition, $2 b=a+c$, so $2 \sin B=\sin A+\sin C$
Given $A+B+C=\pi$, then $2 \sin (A+C)=\sin A+\sin C$
Therefore, $4 \sin \frac{A+C}{2} \cos \frac{A+C}{2}=2 \sin \frac{A+C}{2} \cos \frac{A-C}{2}$
Since $\sin \frac{A+C}{2} \neq 0$, we have $2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2}$
Thus, $5 \cos A-4 \c... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then the equation $x^{2}-[x]-2=0$ has $(\quad)$ distinct real roots.
(A) 1
(B) 2
(C) 3
(D) 4 | 3. C.
$$
\begin{array}{l}
\text { Given }[x] \leqslant x, x^{2}-[x]-2=0 \\
\Rightarrow x^{2}-x-2 \leqslant 0 \Rightarrow x \in[-1,2] \\
\Rightarrow[x] \text { can only be }-1, 0, 1, 2 \\
\Rightarrow x=-1, \sqrt{3}, 2 .
\end{array}
$$ | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
【Question 6】
Weiwei is 8 years old this year, and his father is 34 years old. In $\qquad$ years, his father's age will be three times Weiwei's age. | 【Analysis and Solution】
Age problem.
The age difference remains constant, at $34-8=26$ years;
When the father's age is three times Weiwei's age, Weiwei is $26 \div(3-1)=13$ years old; therefore, in $13-8=5$ years, the father's age will be three times Weiwei's age. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Fangfang's piggy bank contains many 1 yuan, 5 jiao, and 1 jiao coins. She wants to take out 2 yuan at once from the bank, there are $\qquad$ different ways to do so. | $9$ | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. If the six edges of a tetrahedron are all powers of 2, then in this tetrahedron, the lengths of the edges can have at most $\qquad$ values.
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | 5. 3 .
First, prove that if the three sides of a triangle are all powers of 2, then the triangle must be isosceles. Otherwise, let the three sides of the triangle be $a=2^{u}, b=2^{v}, c=2^{w}$, and assume $a>b>c$, then $u \geqslant v+1, u>w+1$, thus $b+c=2^{v}+2^{w}y$ and $x>z(y \neq z)$, then there are three differe... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. In the tetrahedron $ABCD$, $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$; if a sphere passing through $D$ is tangent to the plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron with a radius of 1, then the radius of the circumscribed sphere of the tetrahedr... | (12) 3 Hint: Draw a perpendicular from point $D$ to plane $A B C$, with the foot of the perpendicular being $H$. Draw $D E \perp A B$, $D F \perp A C$, with the feet of the perpendiculars being $E$ and $F$, respectively. Then $H E \perp A B, H F \perp A C$ and
$$
A E=A F=A D \cos 45^{\circ}=\sqrt{6},
$$
Since $\triang... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 42$ Suppose the $n$ digits of a certain $n$-digit positive integer are a permutation of $\{1,2, \cdots, n\}$, and if the integer formed by its first $k$ digits is divisible by $k$, where $k=1,2, \cdots, n$, then this $n$-digit number is called a “good number”. For example, 321 is a three-digit “good number”. ... | [Solution] Let the six-digit number be $\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}$, where $a_{i} \in\{1,2,3,4,5,6\}, i=$ $1,2,3,4,5,6$.
According to the problem, we have
$1\left|a_{1}, 2\right| \overline{a_{1} a_{2}}, 3\left|\overline{a_{1} a_{2} a_{3}}, 4\right| \overline{a_{1} a_{2} a_{3} a_{4}}$,
$51 \overline{... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
7. Definition: $u(n)$ represents the unit digit of the natural number $n$, for example $u(13)=3$. Let $a_{n}=u(7 n)-u(n)$, then $a_{1}+a_{2}+a_{3}+\cdots+a_{2020}=$ $\qquad$ | $0$ | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$6 \cdot 59$ If $\log _{M} N=\log _{N} M$, and $M \neq N, \quad M N>0$, also $M \neq 1$, $N \neq 1$, then $M N$ equals
(A) $\frac{1}{2}$.
(B) 1 .
(C) 2 .
(D) 10 .
(E) a number greater than 2 and less than 10.
(17th American High School Mathematics Examination, 1966) | [Solution] By $\log _{M} N=\frac{1}{\log _{N} M}$, hence from the given condition, $\left(\log _{N} M\right)^{2}=1$, that is,
$$
\log _{N} M= \pm 1 .
$$
If $\log _{N} M=1$, then $M=N$, which contradicts the given condition;
If $\log _{N} M=-1$, then $M=N^{-1}$, i.e., $M N=1$. Therefore, the answer is $(B)$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. In $\triangle A B C$, $A>90^{\circ}, B=20^{\circ}$, draw $A D \perp A B$ intersecting $B C$ at $D$. Given $A B=1, C D=4$, let $S$ be the area of $\triangle A B C$, then the sum of the numerator and denominator of $S^{2}$ in its simplest form is $\qquad$ . | \begin{array}{l}\text { Solution } \\ \text { As shown in the figure, } B D=\frac{1}{\cos 20^{\circ}}, B C=\frac{1}{\cos 20^{\circ}}+4, \\ \text { then } S=\frac{1}{2}\left(\frac{1}{\cos 20^{\circ}}+4\right) \sin 20^{\circ} \\ =\frac{\sin 20^{\circ}+4 \cos 20^{\circ} \sin 20^{\circ}}{2 \cos 20^{\circ}}=\frac{\sin 20^{\... | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
34. The construction cost of a project must not exceed 2.3 million yuan. If the project is undertaken by Company A, it will take 18 days, with a daily fee of 150,000 yuan; if undertaken by Company B, it will take 27 days, with a daily fee of 80,000 yuan. To shorten the construction period, it is decided that Company A ... | Answer: 4 | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 9 $\triangle A B C$ is inscribed in a unit circle, the angle bisectors of the three interior angles $A, B, C$ are extended to intersect this circle at points $A_{1}, B_{1}, C_{1}$, respectively. Find the value of $\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\s... | Using the Law of Sines to convert sides into angles and handle them as trigonometric expressions.
Solution: As shown in Figure 12-7, connect $B A_{1}$. Then, $A A_{1}=2 \sin \left(B+\frac{A}{2}\right)=2 \cos \frac{B-C}{2}$, so $A A_{1} \cos \frac{A}{2}=2 \cos \frac{B-C}{2} \cos \frac{A}{2}=\sin C+\sin B$. Similarly, $... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 1 Let the set $A=\left\{a^{2}, a+1,-3\right\}, B=\left\{a-3,2 a-1, a^{2}+1\right\}$, and $A \cap$ $B=\{-3\}$, find the value of the real number $a$. | Since $A \cap B=\{-3\}$, we have $a-3=-3$ or $2a-1=-3$, which means $a=0$ or $a=-1$.
When $a=0$, $A=\{0,1,-3\}, B=\{-1,1,-3\}, A \cap B=\{1,-3\}$, which contradicts the given information, so we discard it.
Therefore, $a=-1$. | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. At the entrance of Sheep Village, there are three lights - yellow, red, and blue (the positions of the lights cannot be changed), used to indicate signals. For example, "yellow light on, red light off, blue light on" is one signal, and "yellow light off, red light off, blue light off" is another signal. There are $\... | $8$ | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
6. Squirrel mom collects pine nuts, on sunny days she can collect 20 every day, and on rainy days she can only collect 12 every day. She collected 112 pine nuts in a row for several days, averaging 14 per day. How many of these days were rainy? | 6.【Solution】The squirrel collected for: $112 \div 14=8$ (days)
Assuming all 8 days were sunny, the number of pine cones collected would be: $20 \times 8=160$ (cones)
Actually, only 112 cones were collected, which means 48 fewer cones were collected: $160-112=48$ (cones)
Each rainy day results in 8 fewer cones collected... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
```
Given real numbers $a, b, c$ are all not equal to 0, and
\[
\begin{array}{l}
a+b+c=m, a^{2}+b^{2}+c^{2}=\frac{m^{2}}{2} . \\
\text { Find } \frac{a(m-2 a)^{2}+b(m-2 b)^{2}+c(m-2 c)^{2}}{a b c}
\end{array}
\]
``` | The three roots.
Given $a+b+c=m, a^{2}+b^{2}+c^{2}=\frac{m^{2}}{2}$, we know
$$
\begin{array}{l}
a b+b c+c a \\
=\frac{1}{2}\left((a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right)=\frac{m^{2}}{4} .
\end{array}
$$
Thus, from equation (1) we get
$$
\begin{array}{l}
t^{3}-m t^{2}+\frac{m^{2} t}{4}-a b c=0 \\
\Rightarrow ... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$17 \cdot 50$ In $\triangle A B C$, if $\angle A=2 \angle B$, side $b=4, c=5$, then side $a$ equals
(A)6.
(B) 7 .
(C) $3 \sqrt{5}$.
(D) 5 .
(Shaanxi Province, China Junior High School Mathematics Competition, 1997) | [Solution] As shown in the figure, draw the angle bisector $AD$ of $\angle A$. According to the Angle Bisector Theorem, we have
$$
\begin{array}{l}
C D=\frac{4}{9} a, B D=\frac{5}{9} a . \\
\text { Also } \angle C D A=2 \angle D A B=\angle A . \\
\therefore \quad \triangle A D C \backsim \triangle B A C .
\end{array}
... | 6 | Geometry | MCQ | Yes | Yes | olympiads | false |
18. There are 30 students standing in a row, and from left to right, they are wearing hats in the order of "red, yellow, blue, red, yellow, blue......". The PE teacher asks the students to count off from left to right as "1, 2, 1, 2, 1, 2.....", and those who count off 2 step forward. Among the remaining students, thos... | $5$ | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate: $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{99}+\sqrt{100}}=$ | $9$ | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{n}=\frac{2^{n}}{2^{2^{n}}+1}\left(n \in \mathbf{N}_{+}\right)$, let $A_{n}=\sum_{i=1}^{n} a_{i}, B_{n}=\prod_{i=1}^{n} a_{i}$. Prove: $3 A_{n}+B_{n} \cdot 2^{\frac{(1+n)(2-n)}{2}}$ is a constant. | Given $a_{1}=\frac{2}{5}, a_{2}=\frac{4}{17}$, then when $n=1$, $3 \cdot \frac{2}{5}+\frac{2}{5} \cdot 2^{1}=2$;
when $n=2$, $3 \cdot\left(\frac{2}{5}+\frac{4}{17}\right)+\frac{2}{5} \cdot \frac{4}{17} \cdot 2^{0}=2$. Conjecture: $3 A_{n}+B_{n} \cdot 2^{\frac{(1+n)(2-n)}{2}}=2$.
When $n=1$, the conclusion is obviou... | 2 | Algebra | proof | Yes | Yes | olympiads | false |
$31 \cdot 20$ Let the equation $x^{2}-y^{2}=1988$ have $n$ sets of integer solutions. Then $n$ equals
(A) 4.
(B) 6.
(C) 8.
(D) 12.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1988) | 【Solution】Given $(x-y)(x+y)=2^{2} \cdot 7 \cdot 71$, when $x, y$ are both odd or both even, $x+y, x-y$ are both even; when $x, y$ are one odd and one even, $x+y, \quad x-y$ are both odd.
Since 1988 is even, $x+y$ and $x-y$ can only both be even. Thus $\left\{\begin{array}{l}x+y= \pm 2 \cdot 7, \\ x-y= \pm 2 \cdot 71 ;\... | 8 | Number Theory | MCQ | Yes | Yes | olympiads | false |
9. Find the smallest positive integer $n$, such that when the positive integer $k \geqslant n$, in the set $M=$ $\{1,2, \cdots, k\}$ of the first $k$ positive integers, for any $x \in M$, there always exists another number $y \in M(y \neq x)$, such that $x+y$ is a perfect square. | II. 9. It is easy to see that when $n \leqslant 6$, in $M=\{1,2, \cdots, 6\}$, the number 2 added to any other number is never a square number.
We now prove that the minimum value of $n$ is 7.
If positive integers $x, y (x \neq y)$ satisfy $x+y$ being a square number, then $\{x, y\}$ is called a "square pair".
Obvious... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
4. Let $p(x)$ be a polynomial of degree $2n$, $p(0)=p(2)=\cdots=p(2n)=0, p(1)=p(3)=\cdots=$ $p(2n-1)=2, p(2n+1)=-30$. Find $n$. | 4. From $\Delta^{2 n+1} p(0)=\sum_{b=0}^{2 n+1}(-1)^{k} \cdot C_{2 n+1}^{k} \cdot p(2 n+1-k)=0$, we have $(-30)+\sum_{j=1}^{n} C_{2 n+1}^{2 j} \cdot 2=0$, which simplifies to $(-30)+2 \cdot\left(\frac{1}{2} \cdot 2^{2 n+1}-1\right)=0$, solving for $n$ yields $n=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. (10 points) A boat travels 16 kilometers downstream and 8 kilometers upstream in 4 hours; it also travels 12 kilometers downstream and 10 kilometers upstream in the same amount of time. Question: How many hours did it take for the boat to travel 24 kilometers downstream and then return? | 【Solution】Solution: According to the problem, the time taken to travel downstream $16-12=4$ kilometers is equal to the time taken to travel upstream $10-8=2$ kilometers, which means in the same time, 4 kilometers were traveled downstream and 2 kilometers upstream, so the downstream speed is twice the upstream speed. As... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Fill the nine blanks in Figure 2 with the numbers $1,2, \cdots, 9$, requiring that each row from left to right and each column from top to bottom increase sequentially. When 3 and 4 are fixed in the positions shown in the figure, the number of ways to fill the blanks is . $\qquad$ | 10.6.
The first two cells from left to right in the first row can only be arranged with $1, 2$, and the cell at the bottom right corner can only be 9. This leaves us to choose any two numbers from the remaining four numbers $5, 6, 7, 8$ to arrange in the top two cells of the right column (in ascending order), and the ... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$7 \cdot 103$ A function $y=f(x)$ defined on $R$ has the following properties:
(1) For any $x \in R$, there is $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1}, x_{2} \in R, \quad x_{1} \neq x_{2}$, there is $f\left(x_{1}\right) \neq f\left(x_{2}\right)$. Then the value of $f(0)+f(1)+f(-1)$ is
(A) 0.
(B) 1.
(C) -1.
(... | [Solution] By the given condition, we notice that $f(0)=f^{3}(0)$, which implies
$$
f(0)[f(0)-1][f(0)+1]=0 .
$$
Thus,
$$
f(0)=0,1 \text {, or }-1 \text {. }
$$
Also, $f(1)=f^{3}(1)$, similarly we know $f(1)=0,1$, or -1.
Furthermore, by $f(-1)=f^{3}(-1)$, we also know $f(-1)=0,1$, or -1. However, $f(-1), f(0), f(1)$ a... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
75. At the bus station, some people have already queued up before ticket sales begin. Now, assuming that after ticket sales start, the number of people added per minute is fixed. If 21 ticket windows are opened, the queue will just disappear after 12 minutes; if 23 ticket windows are opened, the queue will just disappe... | Reference answer: 6 | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Calculate: $\frac{\log _{2} 80}{\log _{40} 2}-\frac{\log _{2} 160}{\log _{20} 2}=(\quad$.
(A)0
(B) 1
(C) $\frac{5}{4}$
(D) 2
(E) $\log _{2} 5$ | 9. D.
$$
\begin{array}{l}
\text { Original expression }=\frac{\frac{\lg 80}{\lg 2}}{\frac{\lg 2}{\lg 40}}-\frac{\frac{\lg 160}{\lg 2}}{\frac{\lg 2}{\lg 20}} \\
=\frac{\lg 80 \times \lg 40-\lg 160 \times \lg 20}{\lg ^{2} 2} \\
=\frac{(3 \lg 2+1)(2 \lg 2+1)-(4 \lg 2+1)(\lg 2+1)}{\lg ^{2} 2} \\
=\frac{2 \lg ^{2} 2}{\lg ^{... | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
【Question 12】A and B take turns marking numbers from 1 to 17, with the rules: you cannot mark a number that has already been marked; you cannot mark a number that is twice a marked number; you cannot mark a number that is half of a marked number; the player who cannot mark a number loses. Now A marks 8 first, to ensure... | Analysis:
According to the condition “cannot mark the double of a marked number; cannot mark the half of a marked number,” we can categorize the numbers from 1 to 17 into the following four categories:
\begin{tabular}{|l|l|}
\hline Five elements & $(16,8,4,2,1)$ \\
\hline Three elements & $(12,6,3)$ \\
\hline Two eleme... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9、A, B, C, D and Xiaoqiang, five classmates, are competing in a chess tournament, where every two of them play one match. So far, A has played 4 matches, B has played 3 matches, C has played 2 matches, and D has played 1 match. How many matches has Xiaoqiang played? | 9.【Solution】 "Jia has played 4 games", which means Jia has played 1 game each with Yi, Bing, Ding, and Xiaoqiang (Xiaoqiang has played 1 game with Jia).
"Ding has played 1 game", which confirms that Ding has only played with Jia.
"Yi has played 3 games", which means Yi has played 1 game each with Jia, Bing, and Xiaoqia... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Select several numbers from $1, 2, 3, \cdots, 9, 10$ such that each of the 20 numbers $1, 2, 3, \cdots, 19, 20$ is equal to one of the selected numbers or the sum of two selected numbers (which can be the same). How many numbers at least need to be selected? $\qquad$ | 【Solution】The solution is as follows:
$$
\begin{array}{l}
1=1 ; 2=2 ; 3=1+2 ; 4=2+2 ; 5=5 ; 6=1+5 ; 7=2+5 ; 8=8 ; 9=9 ; 10=10 ; 11=1+10 ; \\
12=2+10 ; 13=5+8 ; 14=7+7 ; 15=5+10 ; 16=8+8 ; 17=8+9 ; 18=8+10 ; 19=9+10 ;
\end{array}
$$
By observation, it can be seen that selecting several numbers from $1, 2, 3, \cdots, 9,... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. If the positive number $n$ is a root of the equation $x^{2}+3 x-28=0$, then
$$
\frac{1}{\sqrt{n}+\sqrt{n+2}}+\frac{1}{\sqrt{n+2}+\sqrt{n+4}}+\cdots+\frac{1}{\sqrt{n+58}+\sqrt{n+60}}=
$$
$\qquad$ | $3$ | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given $a=\sqrt[3]{9}+\sqrt[3]{3}+1$, then $\frac{8}{a^{3}}+\frac{12}{a^{2}}+\frac{6}{a}+4=$ | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 57$ Distribute the four cards $2, 4, 7, K$ to four people, each person scores according to the number on the card ($K$ is counted as 13), then the cards are collected, reshuffled, and distributed and scored again, $\cdots$, after several rounds, it is found that the four people have accumulated scores of $16,... | [Solution] From the problem, we know that the total score of the four people each time is
$$
2+4+7+13=26 \text{, }
$$
After several rounds, the cumulative scores of the four people are
$$
16+17+21+24=78 \text{. }
$$
It can be seen that the cards were dealt three times $(78 \div 26=3)$. The person who scored 16 points... | 7 | Logic and Puzzles | MCQ | Yes | Yes | olympiads | false |
24. What is the unit digit of $1^{3}+2^{3}+3^{3}+\ldots+2017^{3}+2018^{3}$? (Note: $a^{3}=a \times a \times a$) | Reference answer: 1 | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10. Given the sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=\frac{3}{2}$, and $a_{n}=\frac{3 n a_{n-1}}{2 a_{n-1}+n-1}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) For all positive integers $n$, the inequality $a_{1} a_{2} a_{3} \cdo... | 10. Solution: (1) From $a_{n}=\frac{3 n a_{n-1}}{2 a_{n-1}+n-1} \Rightarrow \frac{a_{n}}{n}=\frac{3 a_{n-1}}{2 a_{n-1}+n-1} \Rightarrow \frac{n}{a_{n}}=\frac{2}{3}+\frac{n-1}{3 a_{n-1}} \Rightarrow \frac{n}{a_{n}}-1=\frac{1}{3}\left(\frac{n-1}{a_{n-1}-1}-1\right)$,
$\therefore$ the sequence $\left\{\frac{n}{a_{n}}-1\ri... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. There are 4 identical squares. By joining them edge-to-edge without overlapping, they can form $\qquad$ different plane figures. (Figures that can coincide after translation, rotation, or reflection are considered the same figure) | $5$ | 5 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. In the Cartesian coordinate system $x O y$, the curve $y=x^{3}-a x$ has two parallel tangent lines. If the slopes of these two tangent lines are both 1, and the distance between the two tangent lines is 8, then the value of the real number $a$ is $\qquad$. | Solving
$y^{\prime}=3 x^{2}-a=1 \Rightarrow 3 x^{2}=a+1 \Rightarrow x_{1,2}= \pm \sqrt{\frac{a+1}{3}}$, then the equation of the tangent line passing through $\left(x_{1}, y_{1}\right)$ is
$$
y-\left(x_{1}^{3}-a x_{1}\right)=x-x_{1} \Rightarrow y=x+x_{1}^{3}-(a+1) x_{1} \Rightarrow y=x-\frac{2(a+1)}{3} x_{1} \text {, }... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate: $30-29-28+27+26-25-24+23+22-21-20+19=$ | Answer: 0
Exam Point: Smart Calculation (Grouping Method)
Analysis: Group according to the sign pattern “+- - +” with four as a group, the result of each group is 0, so the final result is 0. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.44 Let $a, b, c, d$ be odd numbers, $0<a<b<c<d$, and $a d=b c$. Prove that if $a+d=2^{k}, b+c=2^{m}$, where $k, m$ are integers, then $a=1$.
(25th International Mathematical Olympiad, 1984) | [Proof] First, prove $k>m$.
From $ac-b$.
Then
$$
\begin{array}{l}
(a+d)^{2}=4 a d+(a-d)^{2} \\
=4 b c+(a-d)^{2} \\
>4 b c+(c-b)^{2} \\
=(b+c)^{2} \text {. } \\
2^{k}>2^{m}, k>m \text {. } \\
\end{array}
$$
Thus
Since $a d=b c$, then
$$
\begin{array}{l}
a\left(2^{k}-a\right)=b\left(2^{m}-b\right) \\
2^{m} b-2^{k} a=(b-... | 1 | Number Theory | proof | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.