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1. If real numbers $x, y$ satisfy $(x+5)^{2}+(y-12)^{2}=14^{2}$, then the minimum value of $x^{2}+y^{2}$ is
A. 2
B. 1
C. $\sqrt{3}$
D. $\sqrt{2}$ | 1. B Hint: Combine geometric and algebraic thinking, notice that $(x+5)^{2}+(y-12)^{2}=14^{2}$ is a circle with center $C(-5,12)$ and radius 14. Let $P$ be any point on the circle, then $|O P| \geqslant|C P|-|O C|=14-13=1$. | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 1 Let $x_{1}, x_{2}, x_{3}, \cdots$ be a decreasing sequence of positive numbers such that for any natural number $n$,
$$
x_{1}+\frac{x_{4}}{2}+\frac{x_{9}}{3}+\cdots+\frac{x_{n}^{2}}{n} \leqslant 1 \text {. }
$$
Prove that for any natural number $n$, $x_{1}+\frac{x_{2}}{2}+\frac{x_{3}}{3}+\cdots+\frac{x_{n}}{... | Prove that since $x_{1}, x_{2}, x_{3}, \cdots$ is a decreasing sequence of positive numbers, for any natural number $k$, we have $\frac{x_{k}^{2}}{k^{2}}+\frac{x_{k^{2}+1}}{k^{2}+1}+\cdots+\frac{x_{(k+1)^{2}-1}}{(k+1)^{2}-1}<(2 k+1) \frac{x_{k^{2}}}{k^{2}} \leqslant 3 \frac{x_{k}{ }^{2}}{k}$. For any natural number $n$... | 3 | Inequalities | proof | Yes | Yes | olympiads | false |
20 Find the real solutions of the equation $\left(x^{2008}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2006}\right)=2008 x^{2007}$. | (20 Divide both sides of the equation by $x^{2007}$, we get
$$
\begin{array}{c}
\left(x+\frac{1}{x^{2007}}\right)\left(1+x^{2}+x^{4}+\cdots+x^{2006}\right)=2008 \\
x+x^{3}+x^{5}+\cdots+x^{2007}+\frac{1}{x^{2007}}+\frac{1}{x^{2005}}+\cdots+\frac{1}{x}=2008, \\
2008=x+x^{3}+x^{5}+\cdots+x^{2007}+\frac{1}{x^{2007}}+\frac{... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
One. (20 points) Given the function
$$
f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1(x \in \mathbf{R}) \text {. }
$$
(1) Find the intervals where the function $f(x)$ is monotonically increasing;
(2) Let points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ a... | (1) From the problem, we have
$$
f(x)=\cos 2 x+\sqrt{3} \sin 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) \text {. }
$$
Therefore, the monotonic increasing interval of $f(x)$ is
$$
\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) \text {. }
$$
(2) Let $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\f... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 5.7 How many 35-complete partitions with the smallest number of parts are there? What are their part numbers? | Solution: $35+1=36=2^{2} \cdot 3^{2}$. By Theorem 5.15, the number of 35-complete partitions with the smallest number of parts is $\frac{(2+2)!}{2!2!}=6$, and their part numbers are
$$
2(2-1)+2(3-1)=6 .
$$ | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$28.816 \div(0.40+0.41+0.42+\cdots+0.59)$ The integer part of the value is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990) | [Solution] The original expression $>16 \div(0.6 \cdot 20)=16 \div 12>1$, and $\quad$ the original expression $<16 \div(0.4 \cdot 20)=16 \div 8=2$. Therefore, the integer part of the original expression is 1. Hence, the answer is $(A)$. | 1 | Algebra | MCQ | Yes | Yes | olympiads | false |
16. Anna and Bonnie are racing on a 400-meter oval track. They start at the same time, but Anna is ahead because she is 25% faster than Bonnie. Then Anna will have run ( ) laps when she first overtakes Bonnie.
(A) $\frac{5}{4}$
(B) $\frac{10}{3}$
(C) 4
(D) 5
(E) 25 | 16. D.
Since Anna is 25% faster than Bonnie, when Bonnie finishes the first lap, Anna has run an additional $\frac{1}{4}$ lap. Thus, when Bonnie completes 4 laps, Anna has run an extra full lap. This means that when Anna completes 5 laps, she overtakes Bonnie for the first time. | 5 | Algebra | MCQ | Yes | Yes | olympiads | false |
13. (15 points) There are 60 students in the class, and their birthdays are recorded as a certain month and day. Ask each student the same two questions: How many students in the class have the same birth month as you? How many students in the class have the same birth day as you (for example, the birth day of January ... | 【Analysis】The answers for the same month and the same day number cover from 0 to 14, meaning the number of people with the same month and the same day number ranges from 1 to 15, thus leading to the analysis and solution.
【Solution】The answers include all integers from 0 to 14, which means each answer contains a numbe... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) Among all the factors of the four-digit number Shuangcheng Shuangcheng, 3 are prime numbers, and the other 39 are not prime. Then,
the four-digit number Shuangcheng Shuangcheng has $\qquad$ factors. | 【Solution】Solution: First, according to the sum of the digits in odd and even positions being equal, it must be a multiple of 11. The total number of factors is $3+39=42$ (individuals), decomposing 42 into the product of 3 numbers $42=2 \times 3 \times 7$.
$\overline{\text { double become become double }}=a \times b^{2... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Question 140, Find the smallest real number $\mathrm{r}$, such that there exists a sequence of positive real numbers $\left\{\mathrm{x}_{\mathrm{n}}\right\}$, satisfying for any $\mathrm{n} \in \mathrm{N}^{+}$, we have $\sum_{\mathrm{i}=1}^{\mathrm{n}+1} \mathrm{x}_{\mathrm{i}} \leq \mathrm{r} \cdot \mathrm{x}_{\mathrm... | Question 140, Solution: Clearly, $r \in R^{+}$.
Let $S_{n}=\sum_{i=1}^{n} x_{1}$, then $\left\{S_{n}\right\}$ is a strictly increasing sequence, and $S_{n+1} \leq r \cdot\left(S_{n}-S_{n-1}\right)$. Therefore, we know that $r \cdot S_{n} \geq S_{n+1}+r \cdot S_{n-1} \geq 2 \sqrt{S_{n+1} \cdot r \cdot S_{n-1}}$, rearran... | 4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
3. (6 points) $A, B, C, D$ four people stay in four rooms numbered $1, 2, 3, 4$, with one person per room; so that $B$ does not stay in room 2, and $B, C$ two people require to stay in adjacent numbered rooms. The number of ways to arrange this is. $\qquad$ | 【Solution】Solution: According to the problem, $B$ does not stay in room 2, so $B$ can stay in rooms $1, 3, 4$. If $B$ stays in room 1, then $C$ can stay in room 2, and the remaining 2 people can be arranged in the other 2 rooms. In this case, there are $A_{2}^{2}=2$ situations.
If $B$ stays in room 3, then $C$ can stay... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (19th "Hope Cup" Senior High School Competition Question) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=-\frac{1}{a_{n}+1}$, then $a_{2008}$ equals
A. $-\frac{4}{3}$
B. $-\frac{1}{4}$
C. 3
D. -3 | 4. $\mathrm{C} \quad a_{1}=3, a_{2}=-\frac{1}{a_{1}+1}=-\frac{1}{4}, a_{3}=-\frac{1}{a_{2}+1}=-\frac{4}{3}, a_{4}=-\frac{1}{a_{3}+1}=3, \cdots$
It can be seen that the sequence $\left\{a_{n}\right\}$ has a period of 3. Therefore, from
$$
2008=669 \times 3+1
$$
we get $a_{2008}=a_{1}=3$, hence the answer is $\mathrm{C... | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
11. 6 teams participate in a round-robin tournament (each team plays one match against each of the other teams), with 3 points awarded for a win, 1 point for a draw, and 0 points for a loss. In the end, the 6 teams have different points, the first and second place teams differ by 4 points, the fourth and fifth place te... | $8$ | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
14. (12 points) 2014 students line up in a row, and report numbers according to the following rules: If a student reports a single-digit number, the next student should report the sum of this number and 8; If a student reports a two-digit number, the next student should report the sum of the unit digit of this number a... | 14. (12 points) 2014 students line up from front to back, and report numbers according to the following rules: If a student reports a single-digit number, the next student should report the sum of this number and 8; if a student reports a two-digit number, the next student should report the sum of the unit digit of thi... | 12 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. (26th Russian Mathematical Olympiad) Let $a$, $b$, $c$ be three distinct real numbers such that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common real root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common real root. Find $a + b + c$. | 5. If $x_{1}^{2}+a x_{1}+1=0$ and $x_{1}^{2}+b x_{1}+c=0$, then $(a-b) x_{1}+(1-c)=0$. Therefore, $x_{1}=\frac{c-1}{a-b}$. Similarly, from the equations $x_{2}^{2}+x_{2}+a=0$ and $x_{2}^{2}+c x_{2}+b=0$, we derive $x_{2}=\frac{a-b}{c-1}$ (obviously $c \neq 1$). Hence, $x_{2}=\frac{1}{x_{1}}$.
On the other hand, by Vie... | -3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Given $(\sqrt{3}+1) x=\sqrt{3}-1$. Then $x^{4}-5 x^{3}+6 x^{2}-5 x+4=$ $\qquad$
(2013, Huanggang City, Hubei Province Mathematics Competition) | Solve: From $x=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}$, we get $x-2=\sqrt{3}$. Squaring it, we obtain
$$
x^{2}-4 x+1=0 \text {. }
$$
Use equation (1) to reduce the power of the target algebraic expression. Performing polynomial division, we get
$$
\begin{array}{l}
x^{4}-5 x^{3}+6 x^{2}-5 x+4 \\
=\left(x^{2}-4 x+1\ri... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) $\left[x-\frac{1}{2}\right]=3 x-5$, here $[x]$ represents the greatest integer not exceeding $x$, then $x=$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Analysis】According to the problem, for the original equation to hold, then $\left[x-\frac{1}{2}\right] \leqslant x-\frac{1}{2}, \Rightarrow 3 x-5 \leqslant x-\frac{1}{2}$, and $3 x-5$ is an integer, it is not difficult to find that $x=2$.
【Solution】Solution: According to the analysis, for the original equation to hol... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
16. (3 points) $A, B, C, D$ four boxes contain $8, 6, 3, 1$ balls respectively. The first child finds the box with the fewest balls, then takes one ball from each of the other boxes and puts it into this box; the second child also finds the box with the fewest balls, and then also takes one ball from each of the other ... | A $\cdots)$, the 6th child repeats with the 2nd, meaning a cycle of 4 groups; thus, by analogy: $(50-1) \div 4=12 \cdots 1$ (times);
That is: Excluding the first irregular group, there should be 49 repeated groups, with one remaining, so after the 50th child takes, the four boxes $A B C D$ should contain: $6,4,5,3$ ba... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a>2, a_{2017}=$ 2017, and for any positive integer $n, a_{n+1}=a_{n}^{2}-2$. Then $\left[\frac{\sqrt{a-2}}{10^{6}} a_{1} a_{2} \cdots a_{2017}\right]=$ $\qquad$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. | 5.2.
Let $a_{1}=a=2+\varepsilon(\varepsilon>0)$.
Then $a_{2}=a_{1}^{2}-2=(2+\varepsilon)^{2}-2$
$$
\begin{array}{l}
=2+4 \varepsilon+\varepsilon^{2}>2+4 \varepsilon, \\
a_{3}=a_{2}^{2}-2>(2+4 \varepsilon)^{2}-2 \\
=2+16 \varepsilon+16 \varepsilon^{2}>2+16 \varepsilon .
\end{array}
$$
Using mathematical induction, we ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Given that $A B C D$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ are two rhombuses with side lengths of $\sqrt{3}+1$. If $A C \perp A^{\prime} C^{\prime}$, $\angle A B C=\angle A^{\prime} B^{\prime} C^{\prime}=120^{\circ}$, then the perimeter of the shaded part is $\qquad$ | $8$ | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$35 \cdot 22$ A supermarket has 128 boxes of apples. Each box contains between 120 and 144 apples. Boxes with the same number of apples are called a group. What is the minimum number $n$ of boxes in the largest group?
(A) 4 .
(B) 5 .
(C) 6 .
(D) 24 .
(E) 25 .
(27th American High School Mathematics Examination, 1976) | [Solution] The number of apples that can be packed in 1 box is $120, 121, \cdots, 144$, which gives 25 different ways. If each way is used to pack 5 boxes, then at most $25 \cdot 5 = 125$ boxes.
Since there are 128 boxes now, we have $n \geqslant 6$.
Because there is the following possibility: boxes containing $k$ appl... | 6 | Combinatorics | MCQ | Yes | Yes | olympiads | false |
5. Let $S$ be a set composed of positive integers. If for any three distinct $x, y, z \in S$, at least one of $x, y, z$ is a divisor of $x+y+z$, then $S$ is called "beautiful". Prove that there exists a positive integer $N$ such that the following relationship holds: for any beautiful set $S$, there exists a positive i... | 5. The minimum value of $N$ is 6.
If a beautiful set $S$ satisfies that any two different elements are coprime, then this set is called "extremely beautiful."
First, prove that $N \leqslant 5$ does not meet the requirements.
Let $a_{1}$ and $a_{2}$ be two odd numbers not less than 3, and $a_{1}$ and $a_{2}$ are coprim... | 6 | Number Theory | proof | Yes | Yes | olympiads | false |
A heptagon with all interior angles less than 180 degrees has at least ( ) obtuse angles.
(A) 6
(B) 5
(C) 4
(D) 3 | Solution: The following solution tries to use elementary school level thinking as much as possible:
The sum of the interior angles of a heptagon is $180 \times(7-2)=900^{\circ}$
Try to have as many angles as possible be $90^{\circ}$,
1) If 6 angles are $90^{\circ}$, then the last 1 angle is greater than $180^{\circ}$
... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
4. 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 $2 a+b=-2$, and $f_{k}(x)=-243 x+244$, then $k=$ | 4. 5 Detailed Explanation: We can first find $f_{1}(x), f_{2}(x), f_{3}(x)$, and by induction we get $f_{k}(x)=a^{k} x+\frac{1-a^{k}}{1-a} \cdot b$. Substituting the known values, we can solve for the result. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is
(A) 1
(B) 2
(C) $\pi$
( D ) 4 | (B)
4.【Analysis and Solution】For expressions containing absolute values, discuss in cases to remove the absolute value symbol, obtaining four line segments.
They determine a curve that is a square with vertices at $(1,0),(2,1),(1,2),(0,1)$, with a side length of $\sqrt{2}$, thus the area is 2. | 2 | Geometry | MCQ | Yes | Yes | olympiads | false |
7. The right figure is composed of 4 regular hexagons, each with an area of 6. Using the vertices of these 4 hexagons as vertices, the number of equilateral triangles that can be formed with an area of 4 is $\qquad$ . | 8, | 8 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Find the smallest natural number $n$, such that in any two-coloring of $K_{n}$ there exist 5 edge-disjoint monochromatic triangles. | Solution: $n_{\min }=11$.
(1) When $n \leqslant 10$. We can take a part or all of the following graph, where there do not exist 5 monochromatic triangles without common edges (solid lines represent red, no lines represent blue).
(2) When $n=11$, establish the following three lemmas:
Lemma 1: In a two-coloring of $K_{6}... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
19. Given that the three sides of $\triangle A B C$ are all integers, $\cos A=\frac{11}{16}, \cos B=\frac{7}{8}, \cos C=-\frac{1}{4}$. Then the minimum possible value of the perimeter of $\triangle A B C$ is ( ).
(A) 9
(B) 12
(C) 23
(D) 27
(E) 44 | 19. A.
From the problem, we know
$$
\begin{array}{l}
\cos A=\frac{11}{16}, \sin A=\frac{3 \sqrt{15}}{16}, \\
\cos B=\frac{7}{8}, \sin B=\frac{\sqrt{15}}{8}, \\
\cos C=-\frac{1}{4}, \sin C=\frac{\sqrt{15}}{4} .
\end{array}
$$
By the Law of Sines, we get
$$
a: b: c=\sin A: \sin B: \sin C=3: 2: 4 \text {. }
$$
Therefor... | 9 | Geometry | MCQ | Yes | Yes | olympiads | false |
Example 3 Given $f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x$.
(1) Solve the equation $f(x)=0$;
(2) Find the number of subsets of the set $M=\left\{n \mid f\left(n^{2}-214 n-1998\right)\right.$ $\geqslant 0, n \in \mathbf{Z}\}$. | $$
\begin{array}{l}
\text { (1) For any } 0<x_{1}<x_{2}, \text { we have } \frac{x_{1}+1}{x_{2}+1}>\frac{x_{1}}{x_{2}}, \therefore \lg \frac{x_{1}+1}{x_{2}+1}>\lg \frac{x_{1}}{x_{2}} . \\
\therefore f\left(x_{1}\right)-f\left(x_{2}\right)>\lg \frac{x_{1}+1}{x_{2}+1}-\log _{9} \frac{x_{1}}{x_{2}} \\
\quad=\lg \frac{x_{1... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
35. Given $a=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $a^{4}-5 a^{3}+10 a^{2}-11 a+4$ is | Answer: 1
Solution: Given $a=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{3-\sqrt{5}}{2}$, we get $\quad 3-2 a=\sqrt{5}$,
Therefore,
$$
(2 a-3)^{2}=5 \text {, }
$$
which means
$$
4 a^{2}-12 a+4=0 \text {, }
$$
or equivalently
$$
a^{2}-3 a+1=0,
$$
Thus, $a^{4}-5 a^{3}+10 a^{2}-11 a+4=a^{2}\left(a^{2}-3 a+1\right)-2 a\left(a^... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Teacher Wang has 45 candies, and he decides to eat some every day. Since the candies are very delicious, starting from the second day, the number of candies he eats each day is 3 more than the previous day, and he finishes all the candies in 5 days. So, how many candies did Teacher Wang eat on the second day? ( ) | 【Analysis】Calculation, Arithmetic Sequence
Since the number of candies eaten each day is 3 more than the previous day, it forms an arithmetic sequence with a common difference of 3, and there are 5 terms. Directly find the middle term, the third day $45 \div 5=9$, so the second day is $9-3=6$ (candies). | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-b\right|<\frac{1}{125}$ is what? | 2. $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$, it can be known that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250>3^{5}$, from which we can deduce that $n-1>5... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. Place the numbers $1,2,3,4,5$ in a circle. We call a placement method a "central ring" placement if for some number $n$ in the range 1 to 15, it is impossible to select several consecutive numbers on the circle such that their sum is $n$. If two placement methods are the same after rotation or reflection, they are ... | 【Analysis】 $1,2,3,4,5$ and $15,14,13,12,11,10$ can definitely be taken; if 6 can be taken, then $15-6=9$, which cannot be taken; if 7 can be taken, then $15-7=8$, which cannot be taken; any of the above conditions being met is a valid placement method.
$$
\begin{array}{l}
6=1+5=2+4=1+2+3 ; \\
7=2+5=3+4=1+2+4 ;
\end{arr... | 2 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$29 \cdot 33$ How many integers $n$ between 1 and 100 allow $x^{2}+x-n$ to be factored into the product of two linear factors with integer coefficients
(A) 0 .
(B) 1 .
(C) 2 .
(D) 9 .
(E) 10 .
(40th American High School Mathematics Examination, 1989) | [Solution] Let $x^{2}+x+n=(x-a)(x-b)$,
then $a+b=-1$, and $ab=-n$.
That is, the absolute values of $a$ and $b$ differ by 1, and the absolute value of the negative integer is larger. Also, $ab=-n$, with $n$ ranging from 1 to 100. Thus, we have the table:
\begin{tabular}{r|rrrrrrrrr}
\hline$a$ & 1 & 2 & 3 & 4 & 5 & 6 & 7... | 9 | Algebra | MCQ | Yes | Yes | olympiads | false |
(1) Given the function $f(x)=x^{3}+a x^{2}+x+1(a \in \mathbf{R})$ is decreasing in the interval $\left(-\frac{2}{3},-\frac{1}{3}\right)$ and increasing in the interval $\left(-\frac{1}{3},+\infty\right)$, then $a=$ | (2) Hint: From the given, we know $f^{\prime}(x)=3 x^{2}+2 a x+1$, and $x=-\frac{1}{3}$ is an extremum point of the function $f(x)$, i.e.,
$$
f^{\prime}\left(-\frac{1}{3}\right)=-\frac{2}{3} a+\frac{4}{3}=0,
$$
Solving for $a$ yields $a=2$. | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
1. Given 3 feet $=1$ yard. If there is a rectangular field that is 12 feet long and 9 feet wide, then it requires ( ) square yards of grass to cover it completely.
(A) 12
(B) 36
(C) 108
(D) 324
(E) 972 | 1. A.
$$
\frac{12}{3} \times \frac{9}{3}=12 \text{.}
$$ | 12 | Geometry | MCQ | Yes | Yes | olympiads | false |
7. In the Cartesian coordinate system, if the circle with center $(r+1,0)$ and radius $r$ has a point $(a, b)$ satisfying $b^{2} \geq 4 a$, then the minimum value of $r$ is $\qquad$ . | Answer: 4.
Solution: From the condition, we have $(a-r-1)^{2}+b^{2}=r^{2}$, hence
$$
4 a \leq b^{2}=r^{2}-(a-r-1)^{2}=2 r(a-1)-(a-1)^{2}.
$$
That is, $a^{2}-2(r-1) a+2 r+1 \leq 0$.
The above quadratic inequality in $a$ has a solution, so the discriminant
$$
(2(r-1))^{2}-4(2 r+1)=4 r(r-4) \geq 0 \text{, }
$$
Solving t... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (2002 Hunan Province Competition Question) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.
Translate the above text into English, please retain the original text's line breaks and format, and output ... | 8. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1)\left(a_{n}-\right.$ $\left.a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots 4 \cdot 3 \cdot\left(a_{2}-a_{1}\right)=(n+2)!$,
it follows that $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\l... | 5 | Other | math-word-problem | Yes | Yes | olympiads | false |
1. Let $[x]$ denote the greatest integer not greater than the real number $x$. The number of real roots of the equation $\lg ^{2} x-[\lg x]-2=0$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 1. Solution: 3 From $[\lg x] \leqslant \lg x$ we get $\lg ^{2} x-\lg x-2 \leqslant 0$ which means $-1 \leqslant \lg x \leqslant 2$.
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$, substituting into the original equation gives $\lg x= \pm 1$, but $\lg x=1$ is not valid, so $\lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \l... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given point $G$ is the centroid of $\triangle A B C$, and $\overrightarrow{A B}=\boldsymbol{a}, \overrightarrow{A C}=\boldsymbol{b}$. If $P Q$ passes through point $G$ and intersects $A B, A C$ at points $P, Q$, and $\overrightarrow{A P}=m \boldsymbol{a}, \overrightarrow{A Q}=n \boldsymbol{b}$, then $\frac{1}{m}+\fr... | 1. 3 Detailed Explanation: Special Value Method. When $P Q // B C$, $m=n=\frac{2}{3}$, thus $\frac{1}{m}+\frac{1}{n}=3$. | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
13. Given the sequence $a_{1}=1, a_{n}=1-\frac{1}{4 a_{n-1}}$, and $b_{n}=\frac{2}{2 a_{n}-1}$.
(1) Prove: For $n \geqslant 1$, it always holds that $\frac{1}{2}<a_{n} \leqslant 1$;
(2) Find the general term formula for the sequence $\left\{b_{n}\right\}$;
(3) Find the value of $\lim _{n \rightarrow+\infty} \frac{a_{n}... | 13. (1) Prove the following by mathematical induction:
(1) When $n=1$, $a_{1}=1$, we have $\frac{1}{2}<a_{1} \leqslant 1$;
(2) Assume that when $n=k$, the proposition holds, i.e., $\frac{1}{2}<a_{k} \leqslant 1$, then when $n=k+1$, since $a_{k+1}=1-\frac{1}{4 a_{k}}$, and because $\frac{1}{2}<a_{k} \leqslant 1$, we hav... | 0 | Algebra | proof | Yes | Yes | olympiads | false |
6. In a regular quadrilateral pyramid $P-A B C D$, $G$ is the centroid of $\triangle P B C$. Then $\frac{V_{G-P A D}}{V_{G-P A B}}=$ $\qquad$ | As shown in the figure, $M, N, F$ are the midpoints of the corresponding sides, $M E \perp P N$, $O H \perp P F$. Since $P-A B C D$ is a regular quadrilateral pyramid, we have $\angle M N E=\angle O F H=\alpha$, thus
$$
M E=M N \sin \alpha=2 O F \sin \alpha=2 O H \text {. }
$$
This means the distance from point $M$ to... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
7. Let $a, b, c \in \mathbf{R}^{+}$. Prove that $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+a+c)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8$.
(2003 USA Olympiad Problem) | 7. Let $a+b+c=s, \frac{a}{a+b+c}=x_{1}, \frac{b}{a+b+c}=x_{2}, \frac{c}{a+b+c}=x_{3}$, then the original proposition is transformed into: Given positive real numbers $x_{1}, x_{2}, x_{3}$ satisfying $\sum_{i=1}^{3} x_{i}=1$, prove that $\sum_{i=1}^{3} \frac{\left(1+x_{i}\right)^{2}}{2 x_{i}^{2}+\left(1-x_{i}\right)^{2}... | 8 | Inequalities | proof | Yes | Yes | olympiads | false |
3. The faces of a hexahedron and the faces of a regular octahedron are all equilateral triangles with side length $a$. The ratio of the radii of the inscribed spheres of these two polyhedra is a reduced fraction $\frac{m}{n}$. Then, the product $m \cdot n$ is $\qquad$. | 6
3.【Analysis and Solution】Solution 1 Let the radius of the inscribed sphere of this hexahedron be $r_{1}$. On one hand, the volume of this hexahedron can be calculated by multiplying the radius of the inscribed sphere by the surface area and then dividing by $\frac{1}{3}$. On the other hand, it can also be calculated ... | 6 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
$\qquad$
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 【Analysis】Since $\left(101010_{2}=(42)_{10}\right.$ is divisible by 7, every 6 bits are exactly divisible by 7.
$2014 \div 6=335 \cdots 4$, so ( $\underbrace{201010 \cdots 10}_{2014 \text { years }}$ when divided by 7 has the same remainder as $\left(1010_{2}\right.$ when divided by 7.
And $(1010)_{2}=10_{10}$, the rem... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
$31 \cdot 28$ satisfies the equation $x^{4} y^{4}-10 x^{2} y^{2}+9=0$ for the number of different pairs $(x, y)$ (where $x, y$ are positive integers) is
(A) 0.
(B) 3.
(C) 4.
(D) 12.
(E) infinite.
(20th American High School Mathematics Examination, 1969) | ［Solution］The given equation is equivalent to
$$
\left(x^{2} y^{2}-1\right)\left(x^{2} y^{2}-9\right)=0 \text {, }
$$
Therefore,
$$
x y= \pm 1 \text { or } \pm 3 \text {. }
$$
Thus, the different pairs of positive integers that satisfy the equation are $(1,1),(1,3)$ or $(3,1)$. Hence, the answer is $(B)$. | 3 | Algebra | MCQ | Yes | Yes | olympiads | false |
2. There are 3 logs, each cut into 3 pieces, it took a total of 18 minutes, how many minutes does it take to saw each time? | 2. $(3-1) \times 3=6$ (times) $18 \div 6=3$ (minutes) | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
8. On the blackboard, there are $N(N \geqslant 9)$ distinct non-negative real numbers less than 1. It is known that for any eight numbers on the blackboard, there exists another number on the blackboard such that the sum of these nine numbers is an integer. Find all possible values of $N$.
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... | 8. $N=9$.
An example for $N=9$ exists. It is sufficient for the sum of these nine numbers to be an integer.
Now, let $N>9$. Let $S$ denote the sum of all numbers.
Take any seven numbers $a_{1}, a_{2}, \cdots, a_{7}$, and let $T$ denote their sum, and $A$ denote the complement of the set formed by these seven numbers.
... | 9 | Other | math-word-problem | Yes | Yes | olympiads | false |
8 If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ .
| 8 Let $z$ be a root satisfying the condition, then the original equation is equivalent to $z^{2008}(z+1)=-1$. Taking the modulus on both sides, we get $|z+1|=1$. Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle with center $(-1,0)$ and radius... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
10. For any $x \in \mathbf{R}$, the function $f(x)$ satisfies the relation $f(x+2008)=f(x+2007)+f(x+2009)$, and $f(1)=\lg \frac{3}{2}, f(2)=\lg 15$, then the value of $f(2007)$ is $\qquad$ . | 10. 1. By substituting $x-2007$ for $x$ in the given relation, we get $f(x)=f(x-1)+f(x+1)$.
Replacing $x$ with $x+1$, we get $f(x+1)=f(x)+f(x+2)$.
Adding (1) and (2) yields $f(x-1)+f(x+2)=0$.
Replacing $x$ with $x+1$, we get $f(x)+f(x+3)=0$, so $f(x+3)=-f(x), f(x+6)=f[(x+3)+3]=-f(x+3)=f(x)$. Therefore, $f(x)$ is a per... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. In the Land of Dwarfs, there are 1 yuan, 5 yuan, 10 yuan, and 50 yuan gold coins. To buy a sheep priced at 168 yuan without receiving change, the minimum number of gold coins needed is $\qquad$. | $8$ | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
5 Let $a, b$ be two positive numbers, and $a>b$. Points $P, Q$ are on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. If the line connecting point $A(-a, 0)$ and $Q$ is parallel to the line $O P$, and intersects the $y$-axis at point $R, O$ being the origin, then $\frac{|A Q| \cdot|A R|}{|O P|^{2}}=$ $\qquad$. | 5 2. Let $A Q:\left\{\begin{array}{l}x=-a+t \cos \theta \\ y=t \sin \theta\end{array}\right.$ ( $t$ is a parameter)
Substitute (1) into $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we get
$$
t=\frac{2 a b^{2} \cos \theta}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} .
$$
Therefore,
$$
|A Q|=\frac{2 a b^{2}|\cos \the... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. If $\frac{1-\cos \theta}{4+\sin ^{2} \theta}=\frac{1}{2}$, then $\left(4+\cos ^{3} \theta\right) \cdot\left(3+\sin ^{3} \theta\right)=$ | 2. 9 Detailed Explanation: From the condition, we get $2-2 \cos \theta=4+\sin ^{2} \theta, \Rightarrow(\cos \theta-1)^{2}=4, \Rightarrow \cos \theta=-1$, then $\left(4+\cos ^{3} \theta\right) \cdot\left(3+\sin ^{3} \theta\right)=9$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Given the vector $\boldsymbol{a}=(\cos \theta, \sin \theta)$, vector $\boldsymbol{b}=(\sqrt{3},-1)$, then the maximum value of $|2 \boldsymbol{a}-\boldsymbol{b}|$ is | 5. 4 Hint: $2 \boldsymbol{a}-\boldsymbol{b}=(2 \cos \theta-\sqrt{3} \cdot 1+2 \sin \theta) \quad$ so $\quad|2 \boldsymbol{a}-\boldsymbol{b}|=\sqrt{8-8 \cos \left(\theta+\frac{\pi}{6}\right)}$ Therefore, when $\cos \left(\theta+\frac{\pi}{6}\right)=-1$, $|2 \boldsymbol{a}-\boldsymbol{b}|_{\text {max }}=4$ | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 10 Let real numbers $s, t$ satisfy $19 s^{2}+99 s+1=0, t^{2}+99 t+19=0$, and $s t \neq 1$. Find the value of $\frac{s t+4 s+1}{t}$. | Solve: Given $19 s^{2}+99 s+1=0, t^{2}+99 t+19=0$, we have
$$
19 s^{2}+99 s+1=0, 19\left(\frac{1}{t}\right)^{2}+99 \cdot\left(\frac{1}{t}\right)+1=0 \text {. }
$$
Since $s t \neq 1$, we know $s \neq \frac{1}{t}$. Therefore, $s, \frac{1}{t}$ are the two distinct real roots of the equation $19 x^{2}+99 x+1=0$.
By the re... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$18.4 .7 * \star$ Among numbers of the form $36^{k}-5^{l}$, where $k$ and $l$ are natural numbers. Taking the minimum value in absolute value, prove: the number obtained in this way is certainly the smallest positive number.
Translating the text into English while preserving the original formatting and line breaks, ... | The units digit of the numbers $36^{k}$ and $5^{l}$ are 6 and 5, respectively. Therefore, the units digit of $\left|36^{k}-5^{l}\right|$ is either 1 or 9.
However, $36^{k}-5^{l}=1$ and $5^{l}-36^{k}=9$ cannot both hold. Otherwise, $36^{k}-5^{l}=1$ can be written as $5^{l} = \left(6^{k}-1\right)\left(6^{k}+1\right)$. C... | 11 | Number Theory | proof | Yes | Yes | olympiads | false |
5. In $\triangle A B C$, the lengths of the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $c-a$ equals the height $h$ from vertex $A$ to side $AC$, then the value of $\sin \frac{C-A}{2}+\cos \frac{C+A}{2}$ is
(A) 1 ;
(B) $\frac{1}{2}$;
(C) $\frac{1}{3}$;
(D) -1 . | 5. (A)
As shown in the right figure,
$$
1=c-a=h=a .
$$
At this moment,
$$
\begin{aligned}
& \sin \frac{C-A}{2}+\cos \frac{C+A}{2} \\
= & \sin 30^{\circ}+\cos 60^{\circ}=1
\end{aligned}
$$ | 1 | Geometry | MCQ | Yes | Yes | olympiads | false |
4. (42nd IMO Shortlist) Let $\triangle ABC$ be an acute-angled triangle. Construct isosceles triangles $\triangle DAC$, $\triangle EAB$, and $\triangle FBC$ outside $\triangle ABC$ such that $DA = DC$, $EA = EB$, $FB = FC$, $\angle ADC = 2 \angle BAC$, $\angle BEA = 2 \angle ABC$, and $\angle CFB = 2 \angle ACB$. Let $... | 4. Since $\triangle A B C$ is an acute triangle
Therefore, $\angle A D C, \angle B E A, \angle C F B < \pi$
So $\angle D A C = \frac{\pi}{2} - \frac{1}{2} \angle A D C = \frac{\pi}{2} - \angle B A C$
$$
\angle B A E = \frac{\pi}{2} - \frac{1}{2} \angle B E A = \frac{\pi}{2} - \angle A B C
$$
So $\angle D A E = \angle... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$. Then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is $\qquad$. | 3. $n=7$ Detailed Explanation: From the recurrence relation, we get: $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right)$, thus $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore, $S_{n}-n=\left(a_{1}-1\right)+\left(a_{2}-1\right)+\cdots+\left(a_{n}-1\right)=\fr... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. If real numbers $\alpha, \beta$ satisfy:
$$
\alpha^{3}-3 \alpha^{2}+5 \alpha=1, \beta^{3}-3 \beta^{2}+5 \beta=5 \text {, }
$$
then $\alpha+\beta=$ $\qquad$ . | 4. 2 .
Construct the function $f(x)=x^{3}+2 x$.
According to the problem, we know that $f(\alpha-1)=-2, f(\beta-1)=2$.
By the odd-even property and monotonicity of $f(x)$, we get
$$
(\alpha-1)+(\beta-1)=0 \text {. }
$$
Therefore, $\alpha+\beta=2$. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
$4 \cdot 36$ Find the smallest positive integer $n$, such that in any two-coloring of $K_{n}$, there exist 3 monochromatic triangles, each pair of which has no common edge.
untranslated text preserved the line breaks and format. | [Solution] In the given two-colored $K_{8}$, there are two blue $K_{4}$, and the rest of the unshown edges are all red. It is easy to see that there are no red triangles and exactly 8 blue triangles in the graph. However, among any 3 blue triangles, there are always 2 triangles in the same blue $K_{4}$, sharing 1 commo... | 9 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
$32 \cdot 25$ Among the 101 integers $1,2,3, \cdots, 101$, the numbers that can be divided by both 3 and 5 and leave a remainder of 1 are
(A) 6.
(B) 7.
(C) 8.
(D) 9.
(2nd "Jinyun Cup" Junior High School Mathematics Invitational, 1985) | 〔Solution〕 Obviously, numbers that leave a remainder of 1 when divided by both 3 and 5 (note that 3 and 5 are both prime numbers) can be expressed as
$$
3 \cdot 5 k+1=15 k+1 \quad(k=0,1,2 \cdots),
$$
Thus, among the first 101 natural numbers, there are $1,16,31,46,61,76,91$, a total of 7 numbers that meet the conditio... | 7 | Number Theory | MCQ | Yes | Yes | olympiads | false |
$17 \cdot 102$ In the figure, $D$ is the midpoint of side $BC$ of $\triangle ABC$, $E$ is an internal point on side $AC$, $AC=3CE$, $BE$ and $AD$ intersect at $G$. Then $AG: GD$ is
(A) 2 .
(B) 3 .
(C) 3 or 4 .
(D) 4 . | $\begin{array}{l}\text { [Solution] Draw } D F / / B E \text { intersecting } A C \text { at } F \text {. } \\ \text { Since } D \text { is the midpoint of } B C \text {, then } F \text { is the midpoint of } E C \text {. } \\ \text { Given } A C=3 C E \text {, we have } A E=2 C E=4 E F \text {, } \\ \text { and } D F ... | 4 | Geometry | MCQ | Yes | Yes | olympiads | false |
52 Given $\left(a x^{4}+b x^{3}+c x^{2}+d x+e\right)^{5} \cdot\left(a x^{4}-b x^{3}+c x^{2}-d x+e\right)^{5}=a_{0}+a_{1} x+$ $a_{2} x^{2}+\cdots+a_{41} x^{10}$, then $a_{1}+a_{3}+a_{5}+\cdots+a_{39}=$ $\qquad$. | 520 . Let $F(x)=\left(a x^{4}+b x^{3}+c x^{2}+d x+e\right)^{5} \cdot\left(a x^{4}-b x^{3}+c x^{2}-d x+e\right)^{5}$, from $F(-x)=F(x)$ we know that $F(x)$ is an even function, so the expansion of $F(x)$ does not contain odd powers of $x$, i.e.,
$$
a_{1}=a_{3}=\cdots=a_{39}=0,
$$
thus
$$
a_{1}+a_{3}+a_{5}+\cdots+a_{39}... | 0 | Algebra | math-word-problem | 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 4 mooncakes per pack. It is required that no mooncakes are 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 |
9. As shown in the figure, in $\triangle A B C$, $\cos \frac{C}{2}=\frac{2 \sqrt{5}}{5}, \overrightarrow{A H} \cdot \overrightarrow{B C}=0, \overrightarrow{A B} \cdot(\overrightarrow{C A}+\overrightarrow{C B})=0$, then the eccentricity of the hyperbola passing through point $C$, with $A, H$ as the two foci is $\qquad$ ... | Given, $\cos C=2 \cos ^{2} \frac{C}{2}-1=\frac{3}{5}, \sin C=\frac{4}{5}$, let's assume $A C=B C=5$, then $A H=4, C H=3$. Therefore, for the hyperbola, $2 a=2,2 c=4 \Rightarrow e=\frac{c}{a}=2$. | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
10.3.11 ** Given $\triangle A B C$ with corresponding side lengths $a, b, c$, and $P$ is a point in the plane of $\triangle A B C$. If $P A=p, P B=q, P C=r$, then $\frac{p q}{a b}+\frac{q r}{b c}+\frac{r p}{a c} \geqslant 1$. | Let $A$, $B$, $C$ correspond to the complex numbers $t_{1}$, $t_{2}$, $t_{3}$, respectively, and let $P$ correspond to the origin $O$. Then $\frac{p q}{a b}=\frac{\left|t_{1} t_{2}\right|}{\left|\left(t_{3}-t_{1}\right)\left(t_{3}-t_{2}\right)\right|}$, and similarly for the other two expressions. Therefore,
$$
\frac{p... | 1 | Inequalities | proof | Yes | Yes | olympiads | false |
3. Given a positive real number $x$ satisfies
$$
x^{3}+x^{-3}+x^{6}+x^{-6}=2754 \text {. }
$$
Then $x+\frac{1}{x}=$ $\qquad$ . | 3. 4 .
Transform the given equation into
$$
\begin{array}{l}
\left(x^{3}+x^{-3}\right)^{2}+\left(x^{3}+x^{-3}\right)-2756=0 \\
\Rightarrow\left(x^{3}+x^{-3}+53\right)\left(x^{3}+x^{-3}-52\right)=0 .
\end{array}
$$
Notice that, $x^{3}+x^{-3}+53>0$.
Thus, $x^{3}+x^{-3}=52$.
Let $b=x+x^{-1}$.
$$
\begin{array}{l}
\text {... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. In space, there are $n(n \geqslant 3)$ planes, among which any three planes do not have a common perpendicular plane. There are the following four conclusions:
(1) No two planes are parallel to each other;
(2) No three planes intersect in a single line;
(3) Any two lines of intersection between planes are not parall... | 3. 4 .
If two planes $\alpha_{1}, \alpha_{2}$ are parallel, then the common perpendicular plane of $\alpha_{1}, \alpha_{3}$ is also perpendicular to $\alpha_{2}$, which contradicts the condition that any three planes have no common perpendicular plane. Conclusion (1) is true.
If three planes intersect at a line, then... | 4 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
21. Let $a$, $b$, $c$ be non-zero real numbers, and $a+b+c=0$. Then the value of $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b c}{|a b c|}$ could be ( ).
(A) 0
(B) $\pm 1$
(C) $\pm 2$
(D) 0 or $\pm 2$
(E) 0 or $\pm 1$ | 21. A.
Assume $a \leqslant b \leqslant c$.
Since $a, b, c$ are non-zero real numbers, and $a+b+c=0$, thus, $a<0$, at this time,
$$
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b c}{|a b c|}=0 \text { ; }
$$
When $b>0$, $a b c<0$, at this time,
$$
\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b c}{|a b c|}=0 ... | 0 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. In the plane region
$$
M=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}
0 \leqslant y \leqslant 2-x, \\
0 \leqslant x \leqslant 2
\end{array}\right\}\right.\right.
$$
take any $k$ points, it is always possible to divide these $k$ points into two groups $A$ and $B$, such that the sum of the x-coordinates of the ... | 8. 11.
Since there exist 12 points in the plane region $M$:
$$
\begin{array}{l}
P_{t}\left(1-10^{-4}, 1+10^{-t-3} t\right)(t=1,2, \cdots, 5), \\
P_{t}\left(1+6 \times 10^{-4}, 1-10^{-4} t\right)(t=6,7, \cdots, 12),
\end{array}
$$
these points cannot be divided into two groups as required by the problem, so $k \leqsla... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
8.78 On the blackboard, all natural numbers from 1 to 1988 are written. Operations $A$ and $B$ are alternately performed on these numbers, i.e., first $A$, then $B$, then $A$ again, then $B$, and so on. Operation $A$ involves subtracting the same natural number from each number on the blackboard (the number subtracted ... | ［Solution］Since operation $A$ does not decrease the number of numbers on the blackboard, while operation $B$ reduces the number of numbers by 1 each time, after operation $A$ and operation $B$ are each performed 1987 times, only 1 number remains on the blackboard.
Let $d_{k}$ be the natural number subtracted during th... | 1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
5. Given positive numbers $a, b, x, y, z$ satisfying $a \leqslant x, y, z \leqslant b$, and the difference between the maximum and minimum values of $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$ is $\frac{9}{5}$, then $\frac{2 b}{a}=$ $\qquad$ . | $5$ | 5 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
2. Let $\triangle A B C$ have internal angles $A, B, C$ with opposite sides $a, b, c$ respectively, and satisfy $a \cos B - b \cos A = \frac{3}{5} c$. Then the value of $\frac{\tan A}{\tan B}$ is | Solution: 4.
[Method 1] From the given and the cosine rule, we have
$$
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \text {, i.e., } a^{2}-b^{2}=\frac{3}{5} c^{2} \text { . }
$$
Thus, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cdot \frac{c^{2}+a... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Example 3 Let $S=\{1,2, \cdots, 10\}, A_{1}, A_{2}, \cdots, A_{k}$ be subsets of $S$ and satisfy (1) $\left|A_{i}\right|=5, i=1,2, \cdots, k$; (2) $\left|A_{i} \cap A_{j}\right| \leqslant 2,1 \leqslant i<j \leqslant k$. Find the maximum value of $k$. | Solution: Let $k$ be the number of subsets satisfying conditions (1) and (2), and let $i$ belong to $x_{i}$ of these $k$ subsets, where $i=1,2, \cdots, 10$. If $i \in A_{j}, i \in A_{k}, j \neq k$, then $i$ is called a repeated pair. Thus, the number of repeated pairs caused by the number $i$ is $\mathrm{C}_{x_{i}}^{2}... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
(9) (14 points) Let the line $l: y=k x+m$ (where $k, m$ are integers) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Does there exist a line $l$ such that the vector $\overrightarr... | 9 From $\left\{\begin{array}{l}y=k x+m, \\ \frac{x^{2}}{16}+\frac{y^{2}}{12}=1,\end{array}\right.$ eliminating $y$ and simplifying, we get
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0 .
$$
Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$, then $x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}$.
$$
\Delta_... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. Given that point $P(x, y)$ lies on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$, find the maximum value of $2 x-y$.
untranslated text remains the same as the source text. | 4. Let $\vec{a}=\left(\frac{x}{2}, \frac{y}{3}\right), \vec{b}=(4,-3)$, then
$$
1=\frac{x^{2}}{4}+\frac{y^{2}}{9}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{(2 x-y)^{2}}{4^{2}+(-3)^{2}}=\frac{(2 x-y)^{2}}{25} \text {. }
$$
$(2 x-y)^{2} \leqslant 25, 2x-y$'s maximum value is 5. | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. Given: $\left\{\begin{array}{l}5 x+10 y \leqslant 30 \\ 2 x-y \leqslant 3 \\ x \cdot y \in N\end{array}\right.$ Then the maximum value of $x+y$ is ().
A. 6
B. 5
C. 4
D. 3 | 6. C
From $\left\{\begin{array}{l}5 x+10 y=30 \\ 2 x-y=3\end{array}\right.$ we get the intersection point $D\left(\frac{12}{5}, \frac{9}{5}\right)$. Substituting into $x+y$ gives: $x^{1}+y^{1}=\frac{21}{5}=4 \frac{1}{5}$, so $x+y \leqslant 4 \frac{1}{5}$. Since $x, y \in N$, we have $x+y \leqslant 4$. When $x=y=2$, $(... | 4 | Inequalities | MCQ | Yes | Yes | olympiads | false |
Example 7: In $1^{2}, 2^{2}, 3^{2}, \cdots, 2005^{2}$, add a “+” or “-” sign before each number to make their algebraic sum the smallest non-negative number, and write out the equation. | Analysis: First, we explore the general pattern.
Solution: Since there are 1003 odd numbers among the 2005 numbers from $1^{2}$ to $2005^{2}$, the sum of these 2005 numbers is odd. Changing some of the “+” signs to “-” signs does not change the parity of the result, i.e., the algebraic sum of these 2005 numbers is odd.... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Test D Let $M$ be the set of all polynomials of the form
$$
P(x)=a x^{3}+b x^{2}+c x+d \quad(a, b, c, d \in \mathbf{R}),
$$
and which satisfy $|p(x)| \leqslant 1$ for $x \in[-1,1]$. Prove: there must be a number $k$, such that for all $p(x) \in M$, we have $a \leqslant k$, and find the smallest $k$. | Prove that for any four values $x_{0}, x_{1}, x_{2}, x_{3}$ chosen in the interval $[-1,1]$ satisfying the condition
$$
0<\left|x_{i}-x_{j}\right| \leqslant 1 \quad(0 \leqslant i \leqslant j \leqslant 3)
$$
then by Lemma 1, we have
$$
|a| \leqslant \max _{0 \leqslant 3 \leqslant 3}\left\{\left|p\left(x_{i}\right)\righ... | 4 | Algebra | proof | Yes | Yes | olympiads | false |
$32 \cdot 32$ A two-digit natural number has a one-digit number (including 0) inserted in the middle, turning it into a three-digit number. Some two-digit numbers, when a one-digit number is inserted in the middle, become a three-digit number that is exactly 9 times the original two-digit number. The number of such two... | [Solution] If a "three-digit number" is 9 times a "two-digit number", then the sum of such a "three-digit number" and a "two-digit number" should be 10 times the "two-digit number".
Therefore, the unit digit of such a "two-digit number" can only be 0 and 5.
Also, because the inserted digit should be less than the unit ... | 4 | Number Theory | MCQ | Yes | Yes | olympiads | false |
3. Let the complex numbers $z_{1}=(2-a)+(1-b) \mathrm{i}, z_{2}=(3+2 a)+(2+3 b) \mathrm{i}, z_{3}=(3-a)+(3-2 b) \mathrm{i}$, where $a, b \in \mathbf{R}$. When $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ is minimized, $3 a+4 b=$ $\qquad$ . | 3. 12 .
It is easy to find that $z_{1}+z_{2}+z_{3}=8+6 \mathrm{i}$, thus $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|=10$. $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right|$ achieves its minimum value if and only if
$$
\frac{2-a}{1-b}=\frac{3+2 a}{2+3 b}=\fr... | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. The equation
$$
\lg ^{2} x-[\lg x]-2=0
$$ | 3. From $[\lg x] \leqslant \lg x$ we get $\lg ^{2} x-\lg x-2 \leqslant 0$ which means $-1 \leqslant \lg x \leqslant 2$
When $-1 \leqslant \lg x<0$, we have $[\lg x]=-1$, substituting into the original equation gives $\lg x= \pm 1$, but $\lg x=1$ is not valid, $\therefore \lg x=-1, x_{1}=\frac{1}{10}$.
When $0 \leqsla... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. Given a function $f(x)$ defined on $\mathbf{R}$ whose graph is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, and satisfies $f(x)=-f\left(x+\frac{3}{2}\right), f(-1)=1, f(0)=-2$, then the value of $f(1)+f(2)+\cdots+f(2008)$ is $\qquad$ | 12. From the fact that the graph of the function $f(x)$ is centrally symmetric about the point $\left(-\frac{3}{4}, 0\right)$, we know that $f(x)=-f\left(-x-\frac{3}{2}\right)$.
Given that $f(x)=-f\left(x+\frac{3}{2}\right)$, it follows that $f\left(-x-\frac{3}{2}\right)=f\left(x+\frac{3}{2}\right)$.
Therefore, $f(-x)... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
14. Let the quadratic equation in $x$ be $2 x^{2}-t x-2=0$ with two roots $\alpha$ and
(Problem 13 interval) $\beta(\alpha<\beta)$.
(1) If $x_{1} 、 x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove: $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1}$, and the maxim... | 14. (I) From the conditions, we have: $\alpha+\beta=\frac{t}{2}, \alpha \beta=-1$.
Assume $x_{1}4\left(x_{1}-\alpha\right)\left(x_{2}-\beta\right)=4 x_{1} x_{2}-4\left(\alpha x_{2}+\beta x_{1}\right)+4 \alpha \beta=4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+\right.$ $\left.x_{2}\right)+4 \alpha \beta+2(\alpha-\beta)\lef... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
12. When $s$ and $t$ take all real values, the minimum value that $(s+5-3|\cos t|)^{2}+(s-2|\sin t|)^{2}$ can reach is $\qquad$ | Solve the system $\left\{\begin{array}{l}x=s+5, \\ y=s\end{array} \Rightarrow x-y-5=0,\left\{\begin{array}{l}x=3|\cos t|, \\ y=2|\sin t|\end{array} \Rightarrow \frac{x^{2}}{9}+\frac{y^{2}}{4}=1(x, y \geqslant 0)\right.\right.$, thus the original expression represents the square of the minimum distance from points on th... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
16. The function $f(x)=a x^{2}+b x+c, a, b, c \in \mathbf{Z}$, and $f(x)$ has two distinct roots in $(0,1)$. Find the smallest positive integer $a$ that satisfies the above conditions. | 16. Let the roots of the equation $f(x)=0$ be $r$ and $s(0<s<1<r)$. Then, $f(x)=a(x-r)(x-s)=a[x^2-(r+s)x+rs]$. Since $0<s<1<r$, we have $f(0)=c=ars>0, f(1)=a+b+c=a(1-r)(1-s)>$ 0. Given $0<s<1<r, c \in \mathbf{Z}$, then $c \geqslant 1$, i.e., $f(0) \geqslant 1$. Also, $a+b+c \geqslant 1$, i.e., $f(1) \geqslant 1$, so $1... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. As shown in the figure, the graph of the linear function $y=k x+b$ passes through the point $P(1,3)$, and intersects the positive halves of the $x$-axis and $y$-axis at points $A, B$, respectively. Point $O$ is the origin. The minimum value of the area of $\triangle A O B$ is $\qquad$ | $6$ | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. (10 points) Four people went to a bookstore to buy books. Each person bought 4 different books, and every two people have exactly 2 books in common. Therefore, these 4 people bought at least $\qquad$ kinds of books. | 【Analysis】From simple properties: If there are only two people, to meet the requirements of the problem, there must be 6 different books. Number these 6 books: $1, 2, 3, 4, 5, 6$. Suppose A buys $1, 2, 3, 4$, and B buys $1, 2, 5, 6$. At this point, C arrives. To meet the requirements, C can choose not to buy any other ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. The number of solutions for $x$ in the interval $(-\pi, \pi)$ of the equation $(a-1)(\sin 2 x+\cos x)+(a-1)(\sin x-\cos 2 x)=0(a<0)$ is ( ).
A. 2
B. 3
C. 4
D. 5 or more than 5 | 3. C. Reason: The original equation can be transformed into
$$
\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)\left[\cos \left(\frac{3 x}{2}-\frac{\pi}{4}\right)+\frac{a+1}{a-1} \sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)\right]=0 .
$$
(1) The equation $\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)=0$ has a solution $x=-\f... | 4 | Algebra | MCQ | Yes | Yes | olympiads | false |
8. There is a pair of tightly adjacent rubber drive wheels, each with a mark line passing through the center (as shown in the figure below).
The radius of the driving wheel is 105 cm, and the radius of the driven wheel is 90 cm. When they start to rotate, the mark lines on both wheels are in a straight line. Question:... | 8. The driving gear has rotated 3 turns
8.【Solution】The least common multiple of 105 and 90 is $630. 630 \div 105=6$,
So the driving gear has rotated 6 half-turns, which is 3 turns, and the marks on both gears are aligned again | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2.1.15 ** In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$. | From the given, we have
$$
\begin{array}{c}
\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B \\
\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C=3 \sin B \\
\sin A+\sin C+\sin (A+C)=3 \sin B \\
\sin A+\sin C=2 \sin B \\
2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2}=4 \sin \frac{B}... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Let the left focus of the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ be $F$, and let the line $l$ passing through $(1,1)$ intersect the ellipse at points $A$ and $B$. When the perimeter of $\triangle FAB$ is maximized, the area of $\triangle FAB$ is $\qquad$ | 11.3.
Let $F_{1}$ be the right focus.
Notice,
$$
\begin{array}{l}
F A+F B+A B \\
=2 a-F_{1} A+2 a-F_{1} B+A B \\
=4 a-\left(F_{1} A+F_{1} B-A B\right) \leqslant 4 a,
\end{array}
$$
The equality holds when $l$ passes through point $F_{1}$.
Thus, $l$ is perpendicular to the $x$-axis, at this moment, $S_{\triangle F A B... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. For positive integer $a$ and integers $b, c$, in the rectangular coordinate system $O-xyz$, points $O(0,0,0)$, $A(a, b, c)$, and $B\left(x^{2}, x, 1\right)$ satisfy $=\frac{\pi}{2}$. The real number $x$ has exactly two distinct real solutions $x_{1}, x_{2} \in (0,1)$. The minimum value of $a$ is $\qquad$.
| 4. 5 Detailed Explanation: For the equation $a x^{2}+b x+c=0$ to have two distinct real roots $x_{1}, x_{2}$ in the open interval $(0,1)$, assume $0<c>0, a \geqslant 2$; from $x_{1}+x_{2}=-\frac{b}{a}>0$ we know $b<0$, and $x_{1} x_{2}=\frac{c}{a}>0$, thus $c>0$. Therefore, $4 a c+1 \leqslant(a+c-1)^{2}$, which simplif... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (5 points) A team of 23 people participated in the 16th Guangzhou Asian Games. They were arranged in descending order of height. The average height of the top 5 players is 3 cm more than the average height of the top 8 players; the average height of the last 15 players is 0.5 cm less than the average height of the ... | 【Answer】Solution: The average height of positions $6 \sim 8$ is less than the first 5 positions by: $3 \div 3 \times 8=8$ (cm); therefore, the average height of positions $6 \sim 8$ is more than the average height of positions $9 \sim 23$ by $0.5 \div 3 \times 18=3$ cm. The average height of the first 8 positions is mo... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (20 points) Given a function $f(x)$ defined on $[0,1]$, $f(0)=0, f(1)=1$, and satisfying the following conditions:
(a) For any $x \in[0,1]$, $f(x) \geqslant 0$;
(b) For any two real numbers $x_{1}, x_{2}$ satisfying $x_{1} \geqslant 0, x_{2} \geqslant 0, x_{1}+x_{2} \leqslant 1$, $f\left(x_{1}+x_{2}\right) \geqslan... | 11. For any $0 \leqslant x \leqslant y \leqslant 1$, we have $0 \leqslant y-x \leqslant 1$. By condition (a), we have $f(y-x) \geqslant 0$. Therefore, by condition (b), we have
$$
f(y)=f(y-x+x) \geqslant f(y-x)+f(x) \geqslant f(x) .
$$
Thus, for any $0 \leqslant x \leqslant y \leqslant 1$, we have
$$
f(x) \leqslant f(... | 2 | Inequalities | math-word-problem | Yes | Yes | olympiads | false |
Example 14 Given $n$ points $p_{1}, p_{2}, \cdots, p_{n}$ in a plane, no three of which are collinear, each point $p_{i}(1 \leqslant i \leqslant n)$ is arbitrarily colored red or blue. Let $S$ be the set of triangles whose vertex set is $\left\{p_{1}, p_{2}, \cdots, p_{n}\right\}$, and has the property: for any two lin... | Let the number of triangles in $S$ with $p_{i} p_{j}$ as an edge be $k$, then $k$ is a positive integer and is independent of $i, j$. Since there are $C_{n}^{2}$ edges $p_{i} p_{j}$, there are $k \cdot C_{n}^{2}$ triangles (including repeated counts). Each triangle has three edges, and each triangle is counted 3 times ... | 8 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
11. Let the plane point sets be
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(y-x)\left(y-\frac{18}{25 x}\right) \geqslant 0\right.\right\}, \\
B=\left\{(x, y) \mid(x-1)^{2}+(y-1)^{2} \leqslant 1\right\} .
\end{array}
$$
If $(x, y) \in A \cap B$, find the minimum value of $2 x-y$. | Let $z=2 x-y$. Then $y=2 x-z,-z$ represents the y-intercept of the line $y=2 x-z$.
It is easy to know that when the line $y=2 x-z$ passes through point $P$ in the shaded area, $z=2 x-y$ reaches its minimum value.
Point $P$ is on the circle $(x-1)^{2}+(y-1)^{2}=1$, with coordinates $(1+\cos \theta, 1+\sin \theta)$. Fr... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
62. Arrange 7 small cubes to form a "T" shape, then paint the surface red, and finally separate the small cubes. The number of small cubes with 3 faces painted red is $\qquad$.
Arrange 7 small cubes to form a "T" shape, then paint the surface red, and finally separate the small cubes. The number of small cubes with 3 ... | Answer: 1 | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{n}=(\sqrt{2}+1)^{n}-(\sqrt{2}-1)^{n}(n \in \mathbf{N})$, and $[x]$ represents the greatest integer not exceeding the real number $x$, then the unit digit of $\left[a_{2017}\right]$ is
A. 2
B. 4
C. 6
D. 8 | Let $b_{n}=(1+\sqrt{2})^{n}+(1-\sqrt{2})^{n} \Rightarrow b_{1}=2, b_{2}=6$, and $x=1+\sqrt{2} \Rightarrow(x-1)^{2}=2$ $\Rightarrow x^{2}-2 x-1=0$, thus $b_{n+2}=2 b_{n+1}+b_{n}$. The pattern of the last digits is $2,6,4,4,2,8,8,4,6,6,8,2$, $2,6, \cdots$, with a period of 12, hence the last digit of $b_{2017}$ is 2, so ... | 2 | Number Theory | MCQ | Yes | Yes | olympiads | false |
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