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 |
|---|---|---|---|---|---|---|---|---|
24. The blackboard has 11 39 written on it. Each time, Teacher Li erases 2 numbers and writes down their difference (the larger number minus the smaller number) on the blackboard. When only one number remains on the blackboard, the smallest possible number is $\qquad$. | $1$ | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
17. Among all the divisors of 2020, there are ( ) numbers that have more than 3 divisors (for example, 12 has six divisors, which are $1,2,3,4,6$ and 12).
(A) 6
(B) 7
(C) 8
(D) 9
(E) 10 | 17. B.
Given $2020=2^{2} \times 5 \times 101$, we know that 2020 has 12 divisors, which are $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020$. Among these, the divisors $1, 2, 4, 5, 101$ have no more than 3 divisors.
Therefore, the number of divisors that meet the requirement is $12-5=7$. | 7 | Number Theory | MCQ | Yes | Yes | olympiads | false |
Example 3-18 For the problem
$$
\begin{array}{c}
x_{1}+x_{2}+x_{3}=15 \\
0 \leqslant x_{1} \leqslant 5, \quad 0 \leqslant x_{2} \leqslant 6, \quad 0 \leqslant x_{3} \leqslant 7
\end{array}
$$
Find the number of integer solutions. | If the solution without the upper bound condition according to formula ( *) should be
$$
\binom{15+3-1}{15}=\binom{17}{15}=\binom{17}{2}=272 / 2=136
$$
For problems with upper bounds, just make a transformation
$$
\begin{array}{c}
\xi_{1}=5-x_{1}, \quad \xi_{2}=6-x_{2}, \quad \xi_{3}=7-x_{3} \\
x_{1}+x_{2}+x_{3}=5-x_{... | 10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) There are 10 cards each of the numbers “3”, “4”, and “5”. Select any 8 cards so that their sum is 33, then the maximum number of cards that can be “3” is $\qquad$.
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。 | 7. (10 points) There are 10 cards each of the numbers “3”, “4”, and “5”. Select 8 cards such that their sum is 33. The maximum number of cards that can be “3” is $\qquad$.
【Analysis】This problem requires determining the maximum number of cards that can be “3”. We can use the assumption method and consider different sc... | 3 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
Example 2 Let the odd function $f(x)$ have the domain $\mathbf{R}$. It satisfies $f(x)+f(x+2)=a, f(1)=0$, where $a$ is a constant. Try to determine how many roots the equation $f(x)=0$ has at least in the interval $(-3,7)$.
Translate the above text into English, please keep the original text's line breaks and format, ... | Analysis: Deriving new equations from abstract equations to investigate the periodicity of $f(x)$.
Solution: In $f(x)+f(x+2)=a$, replace $x$ with $x+2$, yielding $f(x+2)+f(x+4)=a$.
Therefore, $f(x)=f(x+4)$, so 4 is a period of $f(x)$. Thus, $f(5)=f(1)=0, f(3)=$ $f(-1)=-f(1)=0, f(4)=f(0)=0$. Therefore, $-1,0,1,3,4,5$ ar... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Let planar vectors $\boldsymbol{a}, \boldsymbol{b}$ satisfy: $|\boldsymbol{a}|=1,|\boldsymbol{b}|=2, \boldsymbol{a} \perp \boldsymbol{b}$. Points $O, A, B$ are three points on the plane, satisfying $\overrightarrow{O A}=$ $2 \boldsymbol{a}+\boldsymbol{b}, \overrightarrow{O B}=-3 \boldsymbol{a}+2 \boldsymbol{b}$, the... | Let $\boldsymbol{a}=(1,0), \boldsymbol{b}=(0,2)$, then $\overrightarrow{O A}=(2,2), \overrightarrow{O B}=(-3,4)$. Therefore, $S_{\triangle A O B}=\frac{1}{2}|2 \times 4+2 \times 3|=7$. | 7 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
16. In the figure below, the numbers labeled at 8 vertices: $a, b, c, d, e, f, g, h$, each of these numbers is equal to $1/3$ of the sum of the numbers at the three adjacent vertices. Find: $(a+b+c+d)-(e+f+g+h)$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note and not part ... | 16.【Solution】From the given conditions, we know
$$
\begin{array}{l}
a=\frac{b+e+d}{3}, b+e+d=3 a \\
b=\frac{c+f+a}{3}, c+f+a=3 b \\
c=\frac{d+g+b}{3}, d+g+b=3 c \text { (3) } \\
d=\frac{a+h+e}{3}, a+h+e=3 d \text { (4) } \\
\text { (1) }+ \text { (2) }+ \text { (3) }+(4), \text { then } 2(a+b+c+d)+(e+f+g+h)=3 \\
(a+b+c... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7.43 A square blackboard is divided into $n^2$ small squares of side length 1 by a network of horizontal and vertical straight lines. For what largest natural number $n$ is it always possible to select $n$ small squares such that any rectangle of area at least $n$ (with sides along the grid lines) contains at least one... | [Solution] $n=7$.
Obviously, if $n$ small squares are selected to meet the problem's conditions, then in each row and each column there is exactly one selected small square (we stipulate that $n \geqslant 3$, obviously $n=2$ cannot be the largest).
We fix a row $A$, where the first square is selected. We take another ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
21. (12 points) As shown in Figure 4, curve $C_{1}$ is part of an ellipse centered at the origin $O$ with foci $F_{1}$ and $F_{2}$. Curve $C_{2}$ is part of a parabola with vertex at $O$ and focus at $F_{2}(1,0)$. Point $A\left(\frac{3}{2}, \sqrt{6}\right)$ is the intersection of curves $C_{1}$ and $C_{2}$.
(1) Find th... | 21. (1) Let the equation of the parabola be $y^{2}=2 p x$. Given that $F_{2}(1,0)$ is the focus, we have
$$
p=2 \Rightarrow y^{2}=4 x \text {. }
$$
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-1}=1$.
Substituting the point $\left(\frac{3}{2}, \sqrt{6}\right)$ into the above equation gives... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
54. As shown in the figure, in rectangle $A B C D$, $A B=10$. The paper is folded so that side $A D$ falls on side $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$, and the intersection of $A E$ and $B C$ is point $F$. Given that the area of $\triangle A B F$ is 2, then $B C=$ $\qqua... | Answer: 6
Solution: It is easy to know that $\triangle A B F$ and $\triangle E C F$ are both isosceles right triangles, so $B F=2$. Let $B C=x$, then
$$
F C=x-2, \quad E C=10-x,
$$
By $F C=E C$, we get
$$
x=6 \text { . }
$$ | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Given the sequence $\left\{a_{n}\right\}$ with the general term formula
$$
a_{n}=2^{n}+3^{n}+6^{n}-1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the positive integers that are coprime to each term of this sequence. ${ }^{[3]}$
(46th IMO Shortlist Problem) | First, consider a prime number $p$ that is coprime with any $a_{n}$. Since $a_{2}=48$, any such $p \neq 2, 3$. Next, we prove that for any prime $p \geq 5$, $p \mid a_{p-2}$.
In fact, extend the definition to the set of negative integers.
When $n=-1$, $a_{-1}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}-1=0$.
When the prime $p... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 9 (1) When $0 \leqslant x \leqslant 1$, find the range of the function
$f(x)=(\sqrt{1+x}+\sqrt{1+x}+2) \cdot\left(\sqrt{1-x^{2}}+1\right)$.
(2) Prove that when $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequality $\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{a}}{\beta}$ hol... | Solution (1) From $1+x \geqslant 0$ and $1-x \geqslant 0$, we have $-1 \leqslant x \leqslant 1$, which also satisfies $1-x^{2} \geqslant 0$. Using the method of magnification and reduction, we have $00$, the inequality
$\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{\alpha}}{\beta}(x \in[0,1])$ does not hold.
Conversely, t... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
4. (10 points) A, B, C, D, and E are sitting around a circular table playing poker, with A having a fixed seat. If B and D cannot sit next to each other, then there are ( ) different ways to arrange the seating.
A. 10
B. 8
C. 12
D. 16 | 【Analysis】This problem is essentially arranging 4 people, Yi, Bing, Ding, and Wu, in a certain order. Therefore:
(1) The number of ways to arrange all 4 people is: $\mathbf{A}_{4}^{4}=24$.
(2) When Yi and Ding are adjacent, the arrangement can be divided into two steps: The first step is to treat the 2 people as 1 unit... | 12 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
22 、A triangle with a perimeter of 36 has three sides that are all composite numbers, then the number of such triangles is. $\qquad$ | 【Analysis】The longest side should be between 12 and 17
Since it is required that all three sides are composite numbers
I. The longest side is 16
In this case, the sum of the other two sides is 20, which can be (16, 4) (14, 6) (12, 8) (10, 10), a total of 4 kinds
II. The longest side is 15
In this case, the sum of the o... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2.20 ** Let $n$ students be such that among any 3 of them, 2 know each other, and among any 4 of them, 2 do not know each other. Find the maximum value of $n$.
| The maximum value of $n$ sought is 8.
When $n=8$, the example shown in the figure satisfies the requirements, where $A_{1}$, $A_{2}, \cdots, A_{8}$ represent 8 students, and a line connecting $A_{i}$ and $A_{j}$ indicates that $A_{i}$ and $A_{j}$ know each other; otherwise, they do not know each other.
Assume $n$ stud... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
12. (5 points)
Fairy has a magic wand that can turn “death” into “life” or “life” into “death” with one wave. One day, Fairy saw 4 trees, 2 of which were already withered, as shown in the figure below. She waved her magic wand, hoping all the trees would be in a “living” state, but unfortunately, the magic wand malfunc... | $4$ | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
67. $x$ is a positive rational number, $(x)$ represents the number of prime numbers not exceeding $x$, for example, $(5)=3$, meaning there are 3 prime numbers not exceeding 5, which are 2, 3, 5. Therefore, $(x)$ defines an operation on $x$. Then $((20) \times(1)+(7))$ is $\qquad$ | Answer: 2.
Solution: $(x)$ defines an operation for positive rational numbers $x$. The steps of the operation are:
(1) Write down the sequence of natural numbers not exceeding $x$;
(2) Mark the prime numbers in this sequence;
(3) Count the number of these prime numbers.
According to the above operation method, we calc... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(15) Let $x, y$ be real numbers, then $\max _{5 x^{2}+4 y^{2}=10 x}\left(x^{2}+y^{2}\right)=$ $\qquad$ | (15) 4 Hint: $5 x^{2}+4 y^{2}=10 x \Rightarrow 4 y^{2}=10 x-5 x^{2} \geqslant 0 \Rightarrow 0 \leqslant x \leqslant 2$,
$$
4\left(x^{2}+y^{2}\right)=10 x-x^{2}=25-(5-x)^{2} \leqslant 25-3^{2} \Rightarrow x^{2}+y^{2} \leqslant 4 .
$$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
13. (15 points) A motorcycle traveling 120 kilometers takes the same time as a car traveling 180 kilometers. In 7 hours, the distance traveled by the motorcycle is 80 kilometers less than the distance traveled by the car in 6 hours. If the motorcycle starts 2 hours earlier, and then the car starts from the same startin... | 【Analysis】First, analyze the ratio of the distances traveled by the two vehicles, which is the ratio of their speeds. The time required can be found by dividing the distance difference by the speed difference.
【Solution】Solution: According to the problem:
The speed of the motorcycle: The speed of the car $=120: 180=2: ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 6 For the function $y=(n-1) t^{2}-10 t+10, t \in(0,4]$, does there exist a positive integer $n$ such that $0<y \leqslant 30$ always holds? | Solve $0<(n-1) t^{2}-10 t+10 \leqslant 30 \Leftrightarrow 0<n t^{2}-t^{2}-10 t+10 \leqslant 30 \Leftrightarrow-\frac{10}{t^{2}}+$ $\frac{10}{t}+1<n \leqslant \frac{20}{t^{2}}+\frac{10}{t}+1, \frac{1}{t} \in\left[\frac{1}{4},+\infty\right) \Leftrightarrow-10\left(\frac{1}{t}-\frac{1}{2}\right)^{2}+\frac{7}{2}<n \leqslan... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$$
6 \cdot 40 \quad \lg \left(\tan 1^{\circ}\right)+\lg \left(\tan 2^{\circ}\right)+\lg \left(\tan 3^{\circ}\right)+\cdots+\lg \left(\tan 88^{\circ}\right)+\lg \left(\tan 89^{\circ}\right)
$$
The value is
(A) 0.
(B) $\frac{1}{2} \lg \frac{\sqrt{3}}{2}$.
(C) $\frac{1}{2} \lg 2$.
(D) 1.
(E) None of the above.
(38th Amer... | ［Solution］From the given problem and properties of logarithms, we have
$$
\begin{aligned}
\text { Original expression } & =\lg \left(\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \cdots \cdot \tan 88^{\circ} \cdot \tan 89^{\circ}\right) \\
& =\lg \left[\left(\tan 1^{\circ} \cdot \tan 89^{\circ}\right)\left(\... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
7. (10 points) If a four-digit number $5 a b 4$ is a square of a number, then $a+b=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Analysis】 $70^{2}=4900,80^{2}=6400,5000$ is a four-digit number more, so it should be the square of a two-digit number between 70 and 80. Also, its last digit is 4, so the unit digit of this two-digit number can only be 2 or 8. $72^{2}=5184$, which meets the requirement. We need to check if $78^{2}$ meets the requirem... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
11. Given $z_{1}, z_{2}, z_{3} \in \mathbb{C}$ and $\left.\mid z_{1}\right\}=\left\{z_{2} \mid=\sqrt{5}\right\} z_{3} \mid$ and $z_{1}+z_{3}=z_{2}$, then $\frac{z_{1} z_{2}}{z_{3}^{2}}=$ | 11. -5 .
$$
\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2} \Rightarrow z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2} \Rightarrow\left(\frac{\bar{z}_{1}}{z_{2}}\right)=\frac{z_{2}}{z_{1}} \text { and } z_{3}=z_{2}-z_{1}
$$
So $\frac{z_{3}^{2}}{z_{1} z_{2}}=\frac{z_{1}}{z_{2}}-2+\frac{z_{2}}{z_{1}}=\frac{z_{1}}{z_{2}}+\left(\frac{... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. In $\triangle A B C$, $B D=D C$. There is a piece of cheese on side $A C$, located at the point closest to point $C$ among the four equal segments. On $A D$, there are three mirrors $W_{1}, W_{2}, W_{3}$, which divide $A D$ into four equal segments. A very suspicious mouse is crawling on $A B$ (from $A$ to $B$), wi... | 【Answer】 5
【Solution】The key to this problem is to determine which points can see the cheese. Let the position of the cheese be $E$ (as shown in the figure below). Since $\frac{A E}{A C}=\frac{A W_{3}}{A D}=\frac{3}{4}$, it follows that $G_{3} E \parallel B C$, thus $\frac{A G_{3}}{A B}=\frac{3}{4}$. Given that $A B=40... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
11. As shown in the figure on the right, there is a circle filled with the number 1. Please fill in the blank circles with one of the numbers $2,3,4,5,6$, ensuring no repetition, and the difference between numbers in adjacent circles is at least 2. How many different ways are there to fill the circles? | 【Key Points】Counting, Enumeration
【Difficulty】Fufu
【Answer】 3
【Analysis】The difference between the numbers in adjacent circles is at least 2, so 2 can only be filled in $d$ and $e$.
(1) Fill 2 in $d$, 2's surroundings cannot have 3. So 3 can only be filled in $a$. 3's surroundings cannot have 4, so 4 can only be filled... | 3 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2.79 Two people start from points $R$ and $S$, which are 76 miles apart, and walk towards each other at the same time. The person at $R$ walks at a constant speed of $4 \frac{1}{2}$ miles per hour, while the person at $S$ walks at a speed of $3 \frac{1}{4}$ miles per hour for the first hour, $3 \frac{3}{4}$ miles per h... | ［Solution］Let the time from departure to meeting be $k$ hours. According to the problem, we have
$$
\frac{9}{2} k+\frac{1}{2} k\left[2 \cdot \frac{13}{4}+\frac{1}{2}(k-1)\right]=76,
$$
which simplifies to $k^{2}+30 k-304=0$, solving this gives $k=8$ (considering only the integer solution). Therefore, the meeting point... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. Given two triangles with side lengths of $3,4,5$, four triangles with side lengths of $4,5, \sqrt{41}$, and six triangles with side lengths of $\frac{5}{6} \sqrt{2}, 4,5$. Using these triangles as faces, we can form $\qquad$ tetrahedra. | When four identical triangles form a tetrahedron, it is called an isosceles tetrahedron. Since the isosceles tetrahedron is inscribed in a rectangular parallelepiped, its faces must be acute triangles. Clearly, the triangle with side lengths $4,5, \sqrt{41}$ is a right triangle, and the triangle with side lengths $\fra... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (6 points) Convert $\frac{15}{37}$ to a decimal, the 2016th digit from left to right in the decimal part is | 【Solution】Solution: According to the problem, we have:
$$
\begin{array}{l}
\frac{15}{37}=0.405 \\
2016 \div 3=672
\end{array}
$$
Therefore, the 2016th digit is 5. Hence, the answer is: 5 | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Question 1: Let set $A=\{1,2, m\}$, where $m$ is a real number. Let $B=\left\{a^{2} \mid a \in A\right\}, C=A \cup B$. If the sum of all elements in $C$ is 6, then the product of all elements in $\mathrm{C}$ is $\qquad$ | Question 1, Answer: According to the conditions, we know: $C=\{1,2, m\} \cup\left\{4, \mathrm{~m}^{2}\right\}$. Since $1+2+4+\mathrm{m}+\mathrm{m}^{2}>6$, $1+2+4+m^{2}>6$, it can only be that $C=\{1,2,4, m\}$, thus we know $m=-1$. Therefore, the product of all elements in $C$ is -8. | -8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. When $s$ and $t$ take all real values, the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can achieve is | 9.2 Hint: $(s+5, s)$ is a point on the line, and $\left\{\begin{array}{l}x=3|\cos t| \\ y=2|\sin t|\end{array}\right.$ is a point on the elliptical arc. Clearly, the point $(3,0)$ on it has the minimum square of the distance to the line, which is 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
74. There are $n$ pairwise distinct natural numbers, the sum of the products of these numbers taken two at a time is 239, then the maximum value of $n$ is | Answer: 7
Solution: If $n \geq 8$,
then the sum of pairwise products $\geq 0 \times(1+2+3+4+5+6+7)+1 \times(2+3+4+5+6+7)$
$$
+2 \times(3+4+5+6+7)+3 \times(4+5+6+7)+4 \times(5+6+7)+5 \times(6+7)+6 \times 7
$$
$=27+50+66+72+65+42=322$ contradiction
Therefore,
$$
n \leq 7,
$$
Then construct 7 numbers $0,1,2,3,4,6,9$ that... | 7 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
(2) Vectors $\overrightarrow{O A}=(1,0), \overrightarrow{O B}=(1,1), O$ is the origin, and a moving point $P(x, y)$ satisfies $\left\{\begin{array}{l}0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O A} \leqslant 1, \\ 0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O B} \leqslant 2,\end{array}\right.$ th... | (2) 2 Hint: From the problem, the point $P(x, y)$ satisfies $\left\{\begin{array}{l}0 \leqslant x \leqslant 1, \\ 0 \leqslant x+y \leqslant 2,\end{array}\right.$ Let $\left\{\begin{array}{l}x+y=u, \\ y=v,\end{array}\right.$ then the point $Q(u, v)$ satisfies $\left\{\begin{array}{l}0 \leqslant u-v \leqslant 1, \\ 0 \le... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Example 5 Let $A A^{\prime}$ be the major axis of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, and $P\left(x_{1}, y_{1}\right)$ be a point on the ellipse. Draw a line $l$ through $P$ with a slope of $-\frac{4 x_{1}}{9 y_{1}}$. Draw lines perpendicular to the major axis through $A$ and $A^{\prime}$, intersecting $l$... | Proof: Given that the equation of line $l$ is $4 x_{1} x+9 y_{1} y=36$, $A^{\prime}(-3,0), A(3,0),\left|A A^{\prime}\right|=6$, it is easy to find that $M\left(3, \frac{12-4 x_{1}}{3 y_{1}}\right)$, $M^{\prime}\left(-3, \frac{12+4 x_{1}}{3 y_{1}}\right)$. Therefore, $|A M| \cdot\left|A^{\prime} M^{\prime}\right|=4 \cdo... | 12 | Geometry | proof | Yes | Yes | olympiads | false |
2. (5 points) The number of different divisors of 60 (excluding 1) is | 【Analysis】First, decompose 60 into prime factors, $60=2 \times 2 \times 3 \times 5$, then write it in standard form as $2^{2} \times 3 \times 5$, and use the formula for the number of divisors, which is to add 1 to each of the exponents of the different prime factors and then multiply them, and finally subtract 1 to ge... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Example 2 In octal, a perfect square is $(a b c)_{8},(a \neq 0)$, find $c$.
Translate the above text into English, keep the original text's line breaks and format, and output the translation result directly. | Solution: Let $n^{2}=(a b c)_{2}, n \in Z^{+}$, and classify $n$ by modulo 4.
When $n=4 k\left(k \in Z^{+}\right)$, $n^{2}=16 k^{2}, 8 \mid n^{2}$, then $c=0$, and $\frac{n^{2}}{8}=2 k^{2}$ is even; but $\frac{n^{2}}{8}=(a b c)_{8}$ is odd, which is a contradiction.
When $n=4 k+2(k \geqslant 0, k \in Z)$, $n^{2}=8\lef... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (10 points) Xiao Ming has several 1 yuan, 2 yuan, and 5 yuan RMB notes. He wants to use no more than 10 RMB notes to buy a kite priced at 18 yuan, and he must use at least two different denominations. How many different payment methods are there? ( )
A. 3
B. 9
C. 11
D. 8 | 【Analysis】Discuss in three cases: no more than 10 notes
(1) 1 yuan and 2 yuan; (2) 1 yuan and 5 yuan; (3) 2 yuan and 5 yuan, discuss based on the number of notes and the final value being 18 yuan.
【Solution】Solution: (1) 1 yuan and 2 yuan: 8 notes of 2 yuan, 2 notes of 1 yuan
(2) 1 yuan and 5 yuan: 2 notes of 5 yuan, ... | 11 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
13. In the expression $9+8-7 \times 6 \div 5+4-3 \times 2 \div 1$, add parentheses arbitrarily so that the calculation result $\mathrm{N}$ is a natural number. The minimum value of $\mathrm{N}$ is | 【Analysis】 $[9+(8-7) \times 6] \div 5+4-3 \times 2 \div 1=1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Calculate: $\frac{\left(2024^{2}-2030\right)\left(2024^{2}+4045\right)}{2021 \times 2023 \times 2026 \times 2027}=$ | $1$ | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Let $S$ be the set of all rational numbers in $\left(0, \frac{5}{8}\right)$, and for a simplified fraction $\frac{q}{p} \in S, (p, q)=1$, define the function $f(q p)=\frac{q+1}{p}$. Then the number of roots of $f(x)=\frac{2}{3}$ in $S$ is $\qquad$.
| Since $\frac{2}{3}=\frac{2 k}{3 k}=\frac{q+1}{p} \Rightarrow \frac{q}{p}=\frac{2 k-1}{3 k} \in\left(0, \frac{5}{8}\right) \Rightarrow 1 \leqslant k \leqslant 7$, the corresponding $\frac{q}{p}=\frac{1}{3}, \frac{3}{6}, \frac{5}{9}, \frac{7}{12}, \frac{9}{15}, \frac{11}{18}, \frac{13}{21}$.
Removing the fractions that ... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
55 In a geometric sequence $\left\{a_{n}\right\}$, the first term $a_{1}=2^{-5}$, and the geometric mean of its first 11 terms is $2^{5}$. If, among the first 11 terms, one term is removed and the geometric mean of the remaining terms is $2^{4}$, then the removed term is
A. The 8th term
B. The 9th term
C. The 10th term... | 55 D. Let the common ratio of the geometric sequence be $q$, and the term removed is the $k$-th term, then $a_{k}=\left(2^{5}\right)^{11} \div\left(2^{4}\right)^{10}=2^{15}$. Also, $2^{-5} \cdot q^{(1+2+\cdots+10) / 11}=2^{5}$, so $q=2^{2}$. From $2^{15}=\left(2^{-5}\right) \cdot\left(2^{2}\right)^{10}=a_{1} \cdot q^{1... | 11 | Algebra | MCQ | Yes | Yes | olympiads | false |
3. From the five numbers $-3, -2, \frac{1}{2}, 1, 3$, a number is randomly selected and denoted as $a$. If the number $a$ makes the system of inequalities
$$
\left\{\begin{array}{l}
\frac{1}{3}(2 x+7) \geqslant 3, \\
x-a<0
\end{array}\right.
$$
have no solution, and makes the fractional equation
$$
\frac{x}{x-3}-\frac... | 3. B.
From the given system of inequalities, we get $x \geqslant 1$ and $x < a$. Since the system of inequalities has no solution, we know $a \leqslant 1$.
From $\frac{x}{x-3}-\frac{a-2}{3-x}=-1$, we get $x=\frac{5-a}{2}$.
Since $x=\frac{5-a}{2}$ is a positive integer, and $a \leqslant 1$, thus, $a=-3$ or 1.
Therefore... | -2 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. Pack 35 egg yolk mooncakes, there are two packaging specifications: large bags with 9 mooncakes per pack, and small bags with $\mathbf{4}$ mooncakes per pack. The requirement is that no mooncakes should be left over, so a total of $\qquad$ packs were made. | 【Answer】 5
Analysis: Let the number of large bags be $\mathrm{x}$, and the number of small bags be $\mathrm{y}$, (both $\mathrm{x}$ and $\mathrm{y}$ are integers) so $9 \mathrm{x}+4 \mathrm{y}=35$, it is easy to get $\left\{\begin{array}{l}x=3 \\ y=2\end{array}\right.$, so a total of $2+3=5$ bags were packed | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
16.5 In the plane of square $ABCD$, there is a point $P$ such that $\triangle PAB, \triangle PBC$, $\triangle PCD, \triangle PDA$ are all isosceles triangles. Then, the number of points $P$ with this property is
(A) 9.
(B) 17.
(C) 1.
(D) 5.
(China High School Mathematics Competition, 1983) | [Solution 1] In a square, if an isosceles triangle is constructed with one side as the base, then because the vertex $P$ of this isosceles triangle must lie on the perpendicular bisector of this side, the opposite side must also be the base of another isosceles triangle with $P$ as the vertex. Therefore, we can discuss... | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
10. (3 points) A and B are having a shooting competition, agreeing that each hit earns 20 points, and each miss deducts 12 points. Each of them shoots 10 times, and together they score 208 points. In the end, A scores 64 points more than B. B hits $\qquad$ shots. | 10. (3 points) A and B are having a shooting competition, agreeing that each hit earns 20 points, and each miss deducts 12 points. Each of them shoots 10 times, and they score a total of 208 points. In the end, A scores 64 points more than B, and B hits _6 times.
【Solution】Assume all shots are hits,
B misses: $(20 \tim... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let $0 \leqslant x \leqslant \pi$, and
$$
3 \sin \frac{x}{2}=\sqrt{1+\sin x}-\sqrt{1-\sin x} \text {. }
$$
Then $\tan x=$ . $\qquad$ | 11. 0 .
Notice that the given equation can be transformed into
$9 \sin ^{2} \frac{x}{2}=2-2 \sqrt{1-\sin ^{2} x}=2-2|\cos x|$.
If $0 \leqslant x \leqslant \frac{\pi}{2}$, then
$9 \sin ^{2} \frac{x}{2}=2-2 \cos x=4 \sin ^{2} \frac{x}{2}$
$\Rightarrow \sin \frac{x}{2}=0 \Rightarrow \tan x=0$;
If $\frac{\pi}{2}<x \leqslan... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 1 Calculate $\left|\frac{(\sqrt{3}+\sqrt{2} i)(\sqrt{5}+\sqrt{2} i)(\sqrt{5}+\sqrt{3} i)}{(\sqrt{2}-\sqrt{3} i)(\sqrt{2}-\sqrt{5} i)}\right|$
(1983 Shanghai High School Competition) | Solve the original expression $=|\sqrt{5}+\sqrt{3} i|^{2}=8$.
| 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
15. (2004 German Mathematics Competition Question) There is a game where at the beginning, the numbers $1,2, \cdots, 2004$ are written on the blackboard. Each step of the game includes the following steps:
(1) Choose a set of some numbers on the blackboard;
(2) Write the remainder of the sum of these numbers modulo 11 ... | 15. The other number is 4.
Since the sum of all numbers on the blackboard remains unchanged modulo 11 at each step, and \(1+2+\cdots+2004 \equiv 3(\bmod 11)\), \(1000 \equiv-1(\bmod 11)\), the other number must be congruent to 4 modulo 11. However, during the game, all numbers written on the blackboard are less than 1... | 4 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$34 \cdot 7$ For each natural number $n$, let $a_{n}=\frac{(n+9)!}{(n-1)!}$, and let $k$ be the smallest natural number such that the rightmost nonzero digit of $a_{k}$ is odd. Then the rightmost nonzero digit of $a_{k}$ is
(A) 1.
(B) 3.
(C) 5.
(D) 7.
(E) 9.
(49th American High School Mathematics Examination, 1998) | [Solution] Clearly, $a_{n}=(n+9)(n+8) \cdots(n+1) n$.
Also, the number of factor 2 in $a_{n}$ is at least $\left[\frac{10}{2}\right]+\left[\frac{10}{4}\right]+\left[\frac{10}{8}\right]=8$ times.
And $a_{n}$ has exactly 2 multiples of 5, at this time to make the rightmost non-zero digit of $a_{n}$ an odd number, $a_{n}$... | 9 | Number Theory | MCQ | Yes | Yes | olympiads | false |
66. Given $O$ is the origin, $P(m, n)(m>0)$ is a point on the function $y=\frac{k}{x}(k>0)$, a line $P A \perp O P$ is drawn through point $P$, and the line $P A$ intersects the positive x-axis at point $A(a, 0)(a>m)$. Let the area of $\triangle O P A$ be $S$, and $S=1+\frac{n^{4}}{4}$. Let $n$ be an integer less than ... | answer: 5 | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. Let $a$, $b$, $c$ be the roots of the cubic equation
$$
x^{3}+3 x^{2}+5 x+7=0
$$
The cubic polynomial $P$ satisfies
$$
\begin{array}{l}
P(a)=b+c, P(b)=a+c, \\
P(c)=a+b, P(a+b+c)=-16 .
\end{array}
$$
Then $P(0)=$ $\qquad$ | 9.11.
By Vieta's formulas, we have
$$
a+b+c=-3 \text {. }
$$
Let $Q(x)=P(x)+x+3$, then
$$
\begin{array}{l}
Q(a)=P(a)+a+3 \\
=P(a)-b-c=0, \\
Q(b)=P(b)+b+3 \\
=P(b)-a-c=0, \\
Q(c)=P(c)+c+3 \\
=P(c)-a-b=0 .
\end{array}
$$
Thus, $Q(x)=t(x-a)(x-b)(x-c)$
$$
=t\left(x^{3}+3 x^{2}+5 x+7\right) \text {. }
$$
Notice that, $P... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
【Question 8】
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 Dad's age is three times Weiwei's age, Weiwei is $26 \div(3-1)=13$ years old; therefore, in $13-8=5$ years, Dad's age will be three times Weiwei's age. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5.75 Find all values of $a$ such that the polynomial
$$
x^{3}-6 x^{2}+a x+a
$$
has roots $x_{1}, x_{2}, x_{3}$ satisfying
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0 .
$$ | [Solution] Make the variable substitution $y=x-3$, then $y_{1}=x_{1}-3, y_{2}=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial
$$
\begin{aligned}
& (y+3)^{3}-6(y+3)^{2}+a(y+3)+a \\
= & y^{3}+3 y^{2}+(a-9) y+4 a-27
\end{aligned}
$$
By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
y_{1}+y_{2}+y_{3}=-3... | -9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) If $\left[\frac{1}{1+\frac{24}{4}}-\frac{5}{9}\right] \times \frac{3}{2 \frac{5}{7}} \div \frac{2}{3 \frac{3}{4}}+2.25=4$, then the value of $A$ is | 【Analysis】First, simplify the complex fraction, derive the equation about the unknown $A$, and then solve the equation according to the properties of equations.
【Solution】Solve: $\left[\frac{1}{1+\frac{5}{\mathrm{~A}}}-\frac{5}{9}\right] \times \frac{3}{2 \frac{5}{7}} \div \frac{2}{3 \frac{3}{4}}+2.25=4$
$$
\begin{arra... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$6 \cdot 12$ If $10^{2 x}=25$, then $10^{1-x}$ equals
(A) 5 .
(B) 2 .
(C) $\frac{1}{5}$.
(D) $\frac{1}{50}$.
(China Partial Provinces and Cities Junior High School Mathematics Correspondence Competition, 1988) | [Solution] From $10^{2 x}=25$, we get $10^{x}=5$.
Therefore, $10^{(1-x)}=10 \cdot 10^{-x}=10 \cdot\left(10^{x}\right)^{-1}=10 \cdot 5^{-1}=2$. Hence, the answer is $(B)$. | 2 | Algebra | MCQ | Yes | Yes | olympiads | false |
II. (25 points) Several boxes are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each box does not exceed 1 ton. To ensure that these boxes can be transported away in one go, how many trucks with a carrying capacity of 3 tons are needed at least? | First, notice that the weight of each box does not exceed 1 ton. Therefore, the weight of boxes that each vehicle can carry at once will not be less than 2 tons. Otherwise, another box can be added.
Let $n$ be the number of vehicles needed, and the weights of the boxes carried by each vehicle be $a_{1}, a_{2}, \cdots,... | 5 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
12. $f(x)=a x^{2}+b x+c, a, b, c \in \mathbf{R}$, and when $|x| \leqslant 1$, $|f(x)| \leqslant 1$, then when $|x| \leqslant 1$, the maximum value of $|a x+b|$ is $\qquad$ | $\begin{array}{l}\text { 12. } 2 f(0)=c, f(1)=a-b+c, f(-1)=a-b+c . a=\frac{1}{2}(f(1)+f(-1)-2 f(0)), b=\frac{1}{2} \\ (f(1)-f(-1)),|a x+b|=\left|\left(\frac{1}{2} x+\frac{1}{2}\right) f(1)+\left(\frac{1}{2} x-\frac{1}{2}\right) f(-1)-x f(0)\right| \leqslant \frac{1}{2}|x+1|+ \\ \frac{1}{2}|x-1|+|x| \leqslant 2 \text {.... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
54. An exam consists of six true/false questions, with correct answers marked by $\sqrt{ }$ and incorrect answers marked by $\times$. The scoring method is: 2 points for each correct answer, 1 point for each unanswered question, and 0 points for each incorrect answer. The answers and scores of $A, B, C, D, E, F, G$ are... | Reference answer: 8
Key point: Logical reasoning -- contradiction and elimination | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Example 35 (2004 Western China Mathematical Olympiad) Find all integers $n$, such that $n^{4}+6 n^{3}+11 n^{2}+$ $3 n+31$ is a perfect square. | 奥
㷊
远
克
数
学
中
的
数
论
问
题
解 Let $A=n^{4}+6 n^{3}+11 n^{2}+3 n+31$ be a perfect square, i.e., $A=\left(n^{2}+3 n+1\right)^{2}-3(n-$ 10) is a perfect square.
When $n>10$, $A<\left(n^{2}+3 n+1\right)^{2}$.
If $n \leqslant-3$, or $10>n \geqslant 0$, then $n^{2}+3 n \geqslant 0$. Therefore,
$$
\begin{array}{l}
A \geqslant\lef... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Let $a=\sqrt{3 x+1}+\sqrt{3 y+1}+\sqrt{3 z+1}$, where $x+y+z=1, x, y, z \geqslant 0$. Then $[a]=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.4.
Notice,
$$
\begin{aligned}
a^{2}= & (3 x+1)+(3 y+1)+(3 z+1)+ \\
& 2 \sqrt{(3 x+1)(3 y+1)}+ \\
& 2 \sqrt{(3 y+1)(3 z+1)}+ \\
& 2 \sqrt{(3 z+1)(3 x+1)} \\
\leqslant & 3((3 x+1)+(3 y+1)+(3 z+1))=18 .
\end{aligned}
$$
Then $a \leqslant \sqrt{18}<5$.
Also, $0 \leqslant x, y, z \leqslant 1$, so,
$$
x \geqslant x^{2}, ... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$2 \cdot 32$ Let $A$ be a set of integers, where the smallest element is 1 and the largest element is 100. Except for 1, each element is equal to the sum of two elements in $A$ (which can be twice a single element). Find the minimum number of elements in set $A$. | [Solution] The set $\{1,2,4,8,16,32,36,64,100\}$ satisfies the requirement and has 9 elements, so the minimum value does not exceed 9.
If the set $A$ satisfies the requirement in the problem and has only 8 elements: $1=a_{1}<a_{2}<\cdots<a_{7}<a_{8}=100$, then because $a_{i+1} \leqslant 2 a_{i}$, we have $a_{i} \leqsl... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
23. In a basketball game at a certain middle school, A scored $\frac{1}{4}$ of the total points, B scored $\frac{2}{7}$ of the total points, C scored 15 points, and the remaining seven players each scored no more than 2 points. Then the total score of the remaining seven players is ( ).
(A) 10
(B) 11
(C) 12
(D) 13
(E) ... | 23. B.
Let the total score of all players be $S$. Then
$$
\frac{S}{4}, \frac{2 S}{7} \in \mathbf{Z}_{+} \Rightarrow 28 \text{ divides } S.
$$
Given that the total score of the remaining seven players is $x$, we know
$$
\begin{array}{l}
x \leqslant 14 . \\
\text { Hence } \frac{S}{4}+\frac{2 S}{7}+15+x=S \\
\Rightarro... | 11 | Algebra | MCQ | Yes | Yes | olympiads | false |
7.14 There is a $5 \times 5$ square grid chessboard, consisting of 25 $1 \times 1$ unit square cells. At the center of each unit square cell, a red dot is painted. Please find several straight lines on the chessboard that do not pass through any red dots, dividing the chessboard into several pieces (which may be of dif... | [Solution] As shown in the figure, after dividing the chessboard into several sections with 8 straight lines, there is at least one red point in each small section.
Below, we use proof by contradiction to show that it is impossible to have fewer lines that satisfy the requirements of the problem.
Assume that the numb... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Let the prime $p$ satisfy that there exist positive integers $x, y$ such that
$$
p-1=2 x^{2}, p^{2}-1=2 y^{2} \text {. }
$$
Then the number of primes $p$ that satisfy the condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 4 | 6. A.
Obviously, $p \neq 2$.
When the prime $p \geqslant 3$, obviously,
$$
p>y>x,
$$
and $p(p-1)=2(y+x)(y-x)$.
From $p \mid 2(y+x)(y-x)$, we know $p \mid(y+x)$.
Then $p \leqslant x+y<2 p \Rightarrow p=x+y$
$$
\begin{array}{l}
\Rightarrow p-1=2(y-x) \Rightarrow p=4 x-1 \\
\Rightarrow 4 x-1-1=2 x^{2} \Rightarrow x=1 \R... | 1 | Number Theory | MCQ | Yes | Yes | olympiads | false |
3 Let $n$ be a natural number, $f(n)$ be the sum of the digits of $n^{2}+1$ (in decimal), $f_{1}(n)=f(n), f_{k+1}(n)$ $=f\left(f_{k}(n)\right)$, then the value of $f_{100}(1990)$ is $\qquad$. | 311.
$$
\begin{array}{c}
f(1990)=20, f 2(1990)=f(20)=5, \\
f 3(1990)=f(5)=8, f 4(1990)=f(8)=11 \\
f_{5}(1990)=f(11)=5=f_{2}(1990)
\end{array}
$$
Therefore
$$
f_{100}(1990)=f_{4}(1990)=11 .
$$ | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. (10 points) Calculate: $\frac{1-0.75}{1.25+2 \frac{1}{4}}+\frac{4 \times 0.3+0.1}{1.8-\frac{2}{5}}=$ | 【Analysis】Simplify the expressions in the numerator and denominator of the complex fraction, then solve according to the basic properties of fractions.
【Solution】Solve: $\frac{1-0.75}{1.25+2 \frac{1}{4}}+\frac{4 \times 0.3+0.1}{1.8-\frac{2}{5}}$
$$
\begin{array}{l}
=\frac{0.25}{3.5}+\frac{1.3}{1.4} \\
=\frac{1}{14}+\fr... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (10 points) Among the three given phrases “尽心尽力”, “可拔山”, and “山穷水尽”, each Chinese character represents a number between 1 and 8, with the same character representing the same number and different characters representing different numbers. If the sum of the numbers represented by the characters in each phrase is 19, ... | 【Analysis】Through analysis, we can know:
From “尽心尽力” (exerting oneself to the utmost), “可拔山” (capable of moving mountains), and “山穷水尽” (reaching the end of one's resources), the sum of the numbers represented by the Chinese characters in each word is 19, which gives us the equations:
$$
\left\{\begin{array}{l}
\text {... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
9. Given $a \in \mathbf{R}, b \in \mathbf{R}_{+}$, $\mathrm{e}$ is the base of the natural logarithm, then the minimum value of $\left(\mathrm{e}^{a}-\ln b\right)^{2}+(a-b)^{2}$ is | The problem is to find the minimum value of the square of the distance between points $A\left(a, \mathrm{e}^{a}\right)$ and $B(b, \ln b)$. Since $y=\mathrm{e}^{x}$ and $f(x)=\ln x$ are inverse functions, the minimum value of $|A B|$ is twice the minimum distance from point $B$ to the line $y=x$.
Considering the tangent... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. (10 points) Football teams $A, B, C, D, E$ are in a round-robin tournament (each pair of teams plays one match). In each match, the winning team gets 3 points, the losing team gets 0 points, and in case of a draw, both teams get 1 point. If the total points for teams $A, B, C, D$ are $1, 4, 7, 8$, respectively, ple... | 【Analysis】In a single round-robin football match, there are a total of $4+3+2+1=10$ games. From the scoring criteria, a win or loss in a match earns 3 points, while a draw earns a total of 2 points. The question asks for the maximum and minimum scores of Team $E$, which clearly depends on the number of draws in the ent... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
17. A museum has a special clock with 20 divisions in a circle. The needle jumps once every 7 minutes, and each jump covers 9 divisions. This morning at 8 o'clock, the needle jumped from 0 to 9. What division was the needle pointing to at 8 o'clock last night? | 【Answer】Solution: Since the pointer jumps once every 7 minutes, and each jump covers 9 spaces, that is, it jumps 9 spaces every 7 minutes,
and there are 20 spaces in a full circle,
so after $7 \times 20=140$ minutes, it will have jumped exactly 9 circles and return to its original position.
From 8 PM last night to 8 A... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
7. (1984 National High School League Question) For all $m, n$ satisfying $1 \leqslant n \leqslant m \leqslant 5$, the number of different hyperbolas represented by the polar equation $\rho=$ $\frac{1}{1-C_{m}^{n} \cos \theta}$ is ( ).
A. 5
B. 10
C. 7
D. 6 | 7. D. The necessary and sufficient condition for the equation to represent a hyperbola is $e=C_{m}^{n}>1$, and $1 \leqslant m \leqslant 5$. There are 6 different values of $C_{m}^{n}>1$, namely $C_{2}^{1}, C_{3}^{1}, C_{4}^{1}, C_{1}^{2}, C_{5}^{1}, C_{5}^{2}$, which correspond to 6 different hyperbolas. Therefore, the... | 6 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
II. (20 points) Given the parabola $E: y=x^{2}$ with focus $F$, a line $l$ passing through a point $M$ on the positive $y$-axis intersects the parabola $E$ at points $A$ and $B$. $O$ is the origin, and $\overrightarrow{O A} \cdot \overrightarrow{O B}=2$.
(1) Prove that the line $l$ passes through a fixed point;
(2) Let... | (1) Let $l: y=kx+m (m>0)$. Substituting into $y=x^2$, we get $x^2 - kx - m = 0$. Let $A(x_1, y_1)$ and $B(x_2, y_2)$.
Then $x_1 x_2 = -m$.
Thus, $y_1 y_2 = x_1^2 x_2^2 = m^2$.
Since $\overrightarrow{OA} \cdot \overrightarrow{OB} = x_1 x_2 + y_1 y_2 = 2$, we have $m^2 - m - 2 = 0$.
Solving, we get $m = -1$ (discard) or... | 3 | Algebra | proof | Yes | Yes | olympiads | false |
7.98 Two people are playing a math game. Player A selects a set of 1-digit integers $x_{1}, x_{2}, \cdots, x_{n}$ as the answer, which can be positive or negative. Player B can ask: what is the sum $a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}$? Here, $a_{1}, a_{2}, \cdots, a_{n}$ can be any set of numbers. Determine the... | [Solution] When taking
$$
a_{j}=100^{j-1}, j=1,2, \cdots, n
$$
then, by asking just one question, we can determine $x_{1}, x_{2}, \cdots, x_{n}$. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
23. Define the operations: $a \bigcirc b=a^{\log _{7} b}, a \otimes b=a^{\frac{1}{\log ^{6} b}}$, where $a, b \in \mathbf{R}$, and the sequence $\left\{a_{n}\right\}(n \geqslant 4)$ satisfies:
$$
a_{3}=3 \otimes 2, a_{n}=(n \otimes(n-1)) \bigcirc a_{n-1} \text {. }
$$
Then the integer closest to $\log _{7} a_{2019}$ i... | 23. D.
From the problem, we know
$$
\begin{array}{l}
a_{3}=3^{\frac{1}{\log _{7} 2}}, a_{n}=\left(n^{\frac{1}{\log _{7}(n-1)}}\right)^{\log _{7} a_{n-1}}=n^{\log _{n-1} a_{n-1}} . \\
\text { Then } \log _{n} a_{n}=\log _{n-1} a_{n-1}=\cdots \\
=\log _{3} a_{3}=\frac{1}{\log _{7} 2}=\log _{2} 7 \\
\Rightarrow a_{n}=n^{... | 11 | Algebra | MCQ | Yes | Yes | olympiads | false |
12. For the parabola $y=a x^{2}+b x+1$, the parameters $a$ and $b$ satisfy $8 a^{2}+4 a b=b^{3}$. Then, as $a$ and $b$ vary, the trajectory equation of the vertex $(s, t)$ of the parabola is $\qquad$ | 12. $s t=1$.
Notice that, the coordinates of the vertex of the parabola $y=a x^{2}+b x+1$ are $\left(-\frac{b}{2 a}, \frac{4 a-b^{2}}{4 a}\right)$.
Let $s=-\frac{b}{2 a}, t=\frac{4 a-b^{2}}{4 a}$. Then
$$
\frac{b}{a}=-2 s, t=1-\frac{b^{2}}{4 a}=1+\frac{b s}{2} \text {. }
$$
Since $a \neq 0$, the conditions that $a, b... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. Prove that $W=(u-v)^{2}+\left(\sqrt{2-u^{2}}-\frac{v}{9}\right)^{2}$ has a minimum value of 8 on $00$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 10. Let $P\left(u, \sqrt{2-u^{2}}\right), Q=\left(v, \frac{9}{v}\right)$, then $P$ is on the circle $x^{2}+y^{2}=2$, $Q$ is on the hyperbola $y=\frac{1}{x}$, and $W=(u-$ $v)^{2}+\left(\sqrt{2-u^{2}}-\frac{9}{v}\right)^{2}=|P Q|^{2}$, and $|P Q| \geqslant|O Q|-|O P|=\sqrt{v^{2}+\left(\frac{9}{v}\right)^{2}}-\sqrt{2} \ge... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$36 \cdot 5$ Write down the natural numbers starting from 1, all the way to the 198th digit:
$$
\underbrace{123456789101112 \cdots}_{198\text{ digits}}
$$
Then the remainder of this number when divided by 9 is
(A) 4.
(B) 6.
(C) 7.
(D) None of the above.
(China Junior High School Mathematics League, 1987) | [Solution] It is easy to deduce that (the number of digits $1 \sim 9$ counts 9, two-digit numbers $10 \sim 99$ count 90, up to 198 digits) this number is
$$
1234 \cdots 9101112 \cdots 99100101102 \text{, }
$$
Since $1+2+\cdots+9$ is divisible by 9, that is,
$$
1+0+1+1+1+2+\cdots+9+9
$$
is also divisible by 9,
so the ... | 6 | Number Theory | MCQ | Yes | Yes | olympiads | false |
124.2 ** Gluing the bases of two congruent regular tetrahedra together, we get a hexahedron with all dihedral angles equal, and the shortest edge of this hexahedron is 2. Then the distance between the farthest two points is $\qquad$ . | Determine the edge length of the hexahedron. As shown in the figure, draw $C E \perp A D$, connect $E F$, it is easy to prove that $E F \perp A D$, thus $\angle C E F$ is the plane angle of the dihedral angle $C-A D-F$. Let $G$ be the midpoint of $C D$, similarly, $\angle A G B$ is the plane angle of the dihedral angle... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Let the real-coefficient polynomial function be $P(x)=a x^{3}+b x^{2}+c x+d$. Prove: If for any $|x|<1$, we have $|P(x)| \leqslant 1$, then $|a|+|b|+|c|+|d| \leqslant 7$. | 3. Since $P(x)$ is a continuous function, when $|x| \leqslant 1$, $|P(x)| \leqslant 1$.
Let $x=\lambda$ and $\frac{\lambda}{2}$ (where $\lambda= \pm 1$), we get $|\lambda a+b+\lambda c+d| \leqslant 1,\left|\frac{\lambda}{8} a+\frac{1}{4} b+\frac{\lambda}{2} c+d\right| \leqslant 1$.
And $|\lambda a+b|=\left|\frac{4}{3}... | 7 | Algebra | proof | Yes | Yes | olympiads | false |
1. The smallest positive odd number that cannot be expressed as $7^{x}-3 \times 2^{y}\left(x 、 y \in Z_{+}\right)$ is $\qquad$ . | $-1.3$
Since $x, y \in \mathbf{Z}_{+}$, therefore, $7^{x}-3 \times 2^{y}$ is always an odd number, and $7^{1}-3 \times 2^{1}=1$.
If $7^{x}-3 \times 2^{y}=3$, then $317^{x}$.
And $7^{x}=(1+6)^{x} \equiv 1(\bmod 3)$, thus, there do not exist positive integers $x, y$ such that
$$
7^{x}-3 \times 2^{y}=3 \text {. }
$$
Ther... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
B2 For each positive integer $n$, define $\varphi(n)$ to be the number of positive divisors of $n$. For example, $\varphi(10)=4$, since 10 has 4 positive divisors, namely $\{1,2,5,10\}$.
Suppose $n$ is a positive integer such that $\varphi(2 n)=6$. Determine the minimum possible value of $\varphi(6 n)$. | Solution: The answer is 8 .
Solution 1: Recall that if a positive integer $m$ has prime factorization $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{t}^{e_{t}}$, where $p_{1}, \ldots, p_{t}$ are distinct primes, then the number of positive divisors of $m$ is $\varphi(m)=\left(e_{1}+\right.$ 1) $\left(e_{2}+1\right) \ldots\lef... | 8 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
14. Let the four idioms "一家之言", "言扬行举", "举世皆知", "知行合一" each character represent one of 11 consecutive non-zero natural numbers, with the same character representing the same number and different characters representing different numbers. If the sum of the numbers represented by the four characters in each idiom is 21, ... | 【Answer】 8
【Solution】 It is not difficult to observe that the characters “一, “言”, “举”, “知”, “行” each appear twice, while the other Chinese characters appear only once. Let the numbers represented by these five characters be $a, b, c, d, e$ (all positive integers),
Assume the 11 consecutive natural numbers are $(x+1),(... | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
33 - 50 Let $a$ denote the fractional part of $\frac{1}{3-\sqrt{5}}$, then $\log _{2} a(2 a+1)$ equals
(A) -1.
(B) -2.
(C) 0.
(D) $\frac{1}{2}$.
(China Junior High School Mathematics Correspondence Competition, 1987) | [Solution] From $\frac{1}{3-\sqrt{5}}=\frac{3+\sqrt{5}}{4}$, and since $1<\frac{3+\sqrt{5}}{4}<2$, $\therefore \quad a=\frac{3+\sqrt{5}}{4}-1=\frac{\sqrt{5}-1}{4}$. Then $a(2 a+1)=\frac{\sqrt{5}-1}{4}\left(\frac{\sqrt{5}-1}{2}+1\right)$ $=\frac{(\sqrt{5}-1)(\sqrt{5}+1)}{8}=\frac{1}{2}$. $\therefore \quad \log _{2} a(2 ... | -1 | Algebra | MCQ | Yes | Yes | olympiads | false |
$8 \cdot 22$ The product of all real roots of the equation $x^{\lg x}=10$ is
(A) 1 .
(B) -1 .
(C) 10 .
(D) $10^{-1}$
(E) None of the above.
(35th American High School Mathematics Examination, 1984) | ［Sol］Take the logarithm with base 10 on both sides of the given equation
i.e. $\square$
$$
(\lg x)(\lg x)=1,
$$
then
we get
$$
\begin{array}{c}
(\lg x)^{2}=1, \\
\lg x= \pm 1, \\
x=10 \text { or } 10^{-1} .
\end{array}
$$
It can be verified that they are all roots of the original equation, so their product is 1. The... | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
$29-35$ Let $a$ be a positive integer, $a<100$, and $a^{3}+23$ is divisible by 24. Then, the number of such $a$ is
(A) 4.
(B) 5.
(C) 9.
(D) 10.
(China High School Mathematics League, 1991) | [Solution] We only need to consider $24 \mid \left(a^{3}-1\right)$, we have
$$
a^{3}-1=(a-1)\left(a^{2}+a+1\right)=(a-1)[a(a+1)+1] \text {. }
$$
Since $a(a+1)+1$ is odd, for $24 \mid \left(a^{3}-1\right)$, it must be that $2^{3} \mid (a-1)$.
If $a-1$ is not divisible by 3, then $3 \mid a(a+1)$, which means $a(a+1)+1$ ... | 5 | Number Theory | MCQ | Yes | Yes | olympiads | false |
7. Let the edge length of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ be 1, and the endpoints of the line segment $M N$ be $M$ on the ray $A A_{1}$, and point $N$ on the ray $B C$, and $M N$ intersects the edge $C_{1} D_{1}$ at point $L$. Then the minimum value of $M N$ is $\qquad$ | 7.3.
Let $A M=x, B N=y$. Then
$$
M N=\sqrt{x^{2}+y^{2}+1} \text {. }
$$
Draw $L L_{1} \perp D C$ at point $L_{1}$. Thus,
$$
\begin{array}{l}
\frac{1}{x-1}=\frac{A A_{1}}{M A_{1}}=\frac{L M}{L N}=\frac{N C}{C B}=\frac{y-1}{1} \\
\Rightarrow x y=x+y \\
\Rightarrow x^{2} y^{2} \geqslant 4 x y \\
\Rightarrow x y \geqslan... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Three. (50 points) Given an integer $n \geqslant 2$ and $n$ positive numbers $p_{1}, p_{2}, \cdots, p_{n}$ whose sum is 1. Let $x_{1}, x_{2}, \cdots, x_{n}$ be any $n$ positive numbers that are not all equal. For $k=2,3$, define
$$
\begin{aligned}
d_{k}= & p_{1} x_{1}^{k}+p_{2} x_{2}^{k}+\cdots+p_{n} x_{n}^{k}- \\
& \l... | Three, let the constant $C$ satisfy the condition.
First, take $x_{1}=x_{2}=\cdots=x_{n-1}=1, x_{n}=1+\varepsilon$ ( $\varepsilon$ is a sufficiently small positive number).
For $k=2,3$, we have
$$
\begin{array}{l}
d_{k}=\left(p_{1}+p_{2}+\cdots+p_{n-1}+p_{n}(1+\varepsilon)^{k}\right)- \\
\left(p_{1}+p_{2}+\cdots+p_{n-... | 3 | Inequalities | proof | Yes | Yes | olympiads | false |
2. (3 points) There are five gifts priced at 2 yuan, 5 yuan, 8 yuan, 11 yuan, and 14 yuan, and five gift boxes priced at 3 yuan, 6 yuan, 9 yuan, 12 yuan, and 15 yuan. One gift is paired with one gift box, resulting in $\qquad$ different prices. | 【Analysis】According to the known prices, the "list method" can be used to solve it.
【Solution】Solution:
\begin{tabular}{|l|c|c|c|c|c|c|c|}
\hline Packaging \\
\hline & 3 \\
\hline
\end{tabular}
There are 25 possible combinations, among which there are cases of repeated prices. A total of 5 yuan, 8 yuan, 11 yuan, 14 yu... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
34. For a large cube with integer edge length, several faces are fully painted, and then it is divided into small cubes with edge length 1. At this point, the ratio of the number of small cubes with at least one face painted to the number of small cubes with no faces painted is $1: 3$. What is the minimum edge length o... | $4$ | 4 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
9. (16 points) Find the sum of all elements in the set
$$
\begin{aligned}
M= & \left\{a \left\lvert\, a=\frac{x+y+z}{t}\right., 3^{x}+3^{y}+3^{z}=3^{t},\right. \\
& x 、 y 、 z 、 t 、 a \in \mathbf{Z}\}
\end{aligned}
$$ | Let $x \leqslant y \leqslant z$.
Then $3^{t}=3^{x}+3^{y}+3^{z} \leqslant 3 \times 3^{z}=3^{z+1}$
$$
\Rightarrow t \leqslant z+1 \text {. }
$$
From $3^{t}>3^{z}$, we know $t>z$.
$$
\begin{array}{l}
\text { Hence } z<t \leqslant z+1 \Rightarrow t=z+1 \\
\Rightarrow 3 \times 3^{z}=3^{z+1}=3^{x}+3^{y}+3^{z} \\
\Rightarrow... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$13 \cdot 62$ Consider the sequence $x_{1}, x_{2}, x_{3}, \cdots$, where $x_{1}=\sqrt[3]{3}, x_{2}=$ $(\sqrt[3]{3})^{\sqrt[3]{3}}$, and generally, $x_{n}=\left(x_{n-1}\right)^{\sqrt[3]{3}}$. For $n>1$, what is the smallest $n$ such that $x_{n}$ is an integer?
(A) 2 .
(B) 3 .
(C) 4 .
(D) 9 .
(E) 27 .
(36th American High... | [Solution] Let $x_{1}=\sqrt[3]{3}=3^{\frac{1}{3}}, x_{2}=(\sqrt[3]{3})^{\sqrt[3]{3}}=\left(3^{\frac{1}{3}}\right)^{3^{\frac{1}{3}}}=3^{\frac{1}{3} \cdot 3^{\frac{1}{3}}}$, $x_{3}=x_{2}^{\sqrt[3]{3}}=3^{\frac{1}{3} \cdot 3^{\frac{1}{3}} \cdot 3^{\frac{1}{3}}}, \cdots$
In general, $x_{n}=3^{\frac{1}{3} \cdot \underbrace{... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
6. (5 points) As shown in the figure, quadrilateral $ABCD$ is a square, $ABGF$ and $FGCD$ are both rectangles, point $E$ is on $AB$, and $EC$ intersects $FG$ at point $M$. If $AB=6$, and the area of $\triangle ECF$ is 12, then the area of $\triangle BCM$ is $\qquad$ | 【Solution】Solution: According to the analysis, given that the area of $\triangle E C F$ is 12, we know that $\frac{1}{2} \times F M \times B G + \frac{1}{2} \times F M \times C G = 12$,
$$
\begin{array}{l}
\Rightarrow \frac{1}{2} \times F M \times(B G+G C)=\frac{1}{2} \times F M \times B C=12 \\
\Rightarrow F M=\frac{2... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10. The figure below is a face of a Slake, $A B C D E 、 J K L I F 、 G H M N O$ are regular pentagons, $F G H I$ is a square, $A B / / F G, \triangle E K L 、 \triangle C M N 、 \triangle B O N$ have areas of 2, 3, 4, respectively, then the area of $\triangle A K J$ is $\qquad$ . | $3$ | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. The school organizes 1511 people to go on an outing, renting 42-seat coaches and 25-seat minibuses. If it is required that there is exactly one seat per person and one person per seat, then there are $\qquad$ rental options. | 【Key Point】Indeterminate Equation
【Answer】2
【Analysis】Let $a$ be the number of large buses, and $b$ be the number of medium buses.
According to the problem, set up the indeterminate equation $42 a+25 b=1511$
$1511 \div 25 \cdots 11$, then $42 a \div 25 \cdots 11, 42 a$ ends in 6, by trial when $a=8$, $b=47$; when $a=8+... | 2 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Let the polynomial be
$$
f(x)=x^{2020}+\sum_{i=0}^{2019} c_{i} x^{i}\left(c_{i} \in\{-1,0,1\}\right) \text {. }
$$
Let $N$ be the number of positive integer roots of $f(x)$ (including multiplicities). If the equation $f(x)$ has no negative integer roots, find the maximum value of $N$.
（Luo Hua | 3. First, since $c_{i} \in\{-1,0,1\}$, the absolute value of the roots of $f(x)$ cannot be greater than or equal to 2. Therefore, only the integer roots $-1$, $0$, and $1$ need to be considered.
Notice,
$$
\begin{aligned}
f(x)= & (x-1)\left(x^{3}-1\right)\left(x^{5}-1\right)\left(x^{11}-1\right) . \\
& \left(x^{21}-1\r... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. Given prime numbers $p$ and $q$ such that $p^{2}+3 p q+q^{2}$ is a perfect square of some integer. Then the maximum possible value of $p+q$ is $\qquad$ (Bulgaria) | 4. 10 .
Let $p^{2}+3 p q+q^{2}=r^{2}\left(r \in \mathbf{Z}_{+}\right)$.
If $3 \nmid r$, then $r^{2} \equiv 1(\bmod 3)$.
If $p \neq 3$ and $q \neq 3$, then
$$
p^{2}+3 p q+q^{2} \equiv 2(\bmod 3),
$$
but $r^{2} \equiv 1(\bmod 3)$, a contradiction.
Thus, without loss of generality, assume $p=3$.
Hence $q^{2}+9 q+9=r^{2}... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8 、Using 2 colors to color 4 small squares on a $2 \times 2$ chessboard, there are. $\qquad$ different coloring schemes. | 【Analysis】If the rotation of the chessboard is not considered, there are $2^{4}=16$ ways.
If the rotation of the chessboard is considered, then using one color has 2 ways; using one color for 1 and the other for 3 has 2 ways; using two colors with 2 each has 2 ways; in total, there are 6 ways. | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (12 points) Choose several different numbers from $1, 2, 3, 4, 5, 6, 7$ (the order of the selected numbers does not matter), such that the sum of the even numbers equals the sum of the odd numbers. The number of ways to select the numbers that meet the condition is $\qquad$. | 4. (12 points) Choose several different numbers from $1, 2, 3, 4, 5, 6, 7$ (the order of the selected numbers does not matter), such that the sum of the even numbers equals the sum of the odd numbers. The number of ways to do this is $\qquad$.
【Solution】Solution: Among $1,2,3,4,5,6,7$, $1,3,5,7$ are odd numbers, and $... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
17. Let $P, Q, R, S$ be four integer points on the circle $\odot O: x^{2}+y^{2}=25$, and $|P Q|, |R S|$ are both irrational numbers. Then the maximum possible value of $\frac{|P Q|}{|R S|}$ is ( ).
(A) 3
(B) 5
(C) $3 \sqrt{5}$
(D) 7
(E) $5 \sqrt{2}$ | 17. D.
From the problem, we know
$$
\begin{array}{c}
P, Q, R, S \in \{( \pm 3, \pm 4), (\pm 4, \pm 3), \\
(0, \pm 5), (\pm 5,0)\} .
\end{array}
$$
Since $|P Q|$ and $|R S|$ are both irrational numbers, points $P$ and $Q$ are selected from different quadrants, and points $R$ and $S$ are selected from the same quadrant... | 7 | Geometry | MCQ | Yes | Yes | olympiads | false |
(7) Let $f(x)=a x+b$, where $a, b$ are real numbers, $f_{1}(x)=f(x), f_{n+1}(x)=$ $f\left(f_{n}(x)\right), n=1,2,3, \cdots$, if $f_{7}(x)=128 x+381$, then $a+b=$ $\qquad$ . | (7) From the problem, we know
$$
\begin{aligned}
f_{n}(x) & =a^{n} x+\left(a^{n-1}+a^{n-2}+\cdots+a+1\right) b \\
& =a^{n} x+\frac{a^{n}-1}{a-1} \cdot b,
\end{aligned}
$$
From
$$
f_{7}(x)=128 x+381
$$
we get $a^{7}=128, \frac{a^{7}-1}{a-1} \cdot b=381$, so $a=2, b=3$, thus $a+b=5$, | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
(4) Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=2, a_{n+1}=1-\frac{1}{a_{n}}$, and let the product of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ be $P_{n}$, then the value of $P_{2009}$ is ( ).
(A) $-\frac{1}{2}$
(B) -1
(C) $\frac{1}{2}$
(D) 1 | (4) B Hint: Since
$$
a_{n+2}=1-\frac{1}{a_{n+1}}=1-\frac{1}{1-\frac{1}{a_{n}}}=\frac{-1}{a_{n}-1},
$$
then
$$
a_{n+3}=1-\frac{1}{a_{n+2}}=1-\frac{1}{\frac{-1}{a_{n}-1}}=a_{n},
$$
thus $\left\{a_{n}\right\}$ is a periodic sequence with a period of 3.
Given $a_{1}=2, a_{2}=\frac{1}{2}, a_{3}=-1$, hence $P_{3}=-1$, so,
... | -1 | Algebra | MCQ | Yes | Yes | olympiads | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.