problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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14. A unit spent 500,000 yuan to purchase a piece of high-tech equipment. According to the tracking survey of this model of equipment: after the equipment is put into use, if the maintenance and repair costs are averaged to the first day, the conclusion is: the maintenance and repair cost on the $x$-th day is
$$
\left[... | 14. (1) The equipment is put into use for $x$ days. The average daily loss is:
$$
\begin{aligned}
y= & \frac{1}{x}\left[500000+\left(\frac{1}{4} \times 0+500\right)+\left(\frac{1}{4} \times 1+500\right)\right. \\
& \left.+\left(\frac{1}{4} \times 2+500\right)+\cdots+\left(\frac{x-1}{4}+500\right)\right] \\
= & \frac{1}... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. A piece of lead wire of length $2 n$ (where $n$ is a natural number and $n \geqslant 4$) is folded into a triangle with integer side lengths. Let $(a, b, c)$ represent a triangle with side lengths $a, b, c$ such that $a \leqslant b \leqslant c$.
(1) For the cases $n=4, 5, 6$, write down all the $(a, b, c)$ that sat... | 15. (1) When $n=4$, the length of the lead wire is 8. Then the only group of $(a, b, c)$ that satisfies the condition is $(2,3,3)$;
When $n=5$, the length of the lead wire is 10. Then the groups of $(a, b, c)$ that satisfy the condition are $(2,4,4),(3,3,4)$;
When $n=6$, the length of the lead wire is 12. Then the gr... | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $b$ take odd numbers between 2 and 16, and $c$ take any natural number. Then the number of quadratic equations $3 x^{2}+$ $(b+1) x+c=0$ that can be formed with two distinct real roots is ( ).
(A) 64
(B) 66
(C) 107
(D) infinitely many | 4. (A).
$\because$ The equation $3 x^{2}+(b+1) x+c=0$ has two distinct real roots, $\therefore \Delta=(b+1)^{2}-12 c>0 \Rightarrow c<\frac{1}{12}(b+1)^{2}$.
Given $b=3,5,7,9,11,13,15$.
When $b=3$, $c<\frac{4}{3}$, there is 1 value of $c$ that meets the condition; when $b=5$, $c<3$, there are 2 values of $c$ that meet t... | 64 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. The square number $y^{2}$ is
the sum of the squares of 11 consecutive integers. Then the smallest value of the natural number $y$ is
$\qquad$ | 3.11.
Let the middle number of these 11 consecutive integers be \( a \). Then
\[
\begin{aligned}
y^{2}= & (a-5)^{2}+(a-4)^{2}+(a-3)^{2}+(a-2)^{2} \\
& +(a-1)^{2}+a^{2}+(a+1)^{2}+(a+2)^{2} \\
& +(a+3)^{2}+(a+4)^{2}+(a+5)^{2} \\
= & 11 a^{2}+2\left(5^{2}+4^{2}+3^{2}+2^{2}+1^{2}\right) \\
= & 11\left(a^{2}+10\right) .
\e... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given $a_{n}=6^{n}+8^{n}$. Then $a_{84} \equiv$ $\qquad$ $(\bmod 49)$ | 5.2.
$$
\begin{aligned}
a_{84}= & (7-1)^{84}+(7 \\
& +1)^{84} \\
= & 2\left(\mathrm{C}_{84}^{0} \cdot 7^{84}+\right. \\
& \mathrm{C}_{84}^{2} \cdot 7^{82}+\cdots \\
& \left.+\mathrm{C}_{84}^{82} \cdot 7^{2}+\mathrm{C}_{84}^{84}\right) \\
= & 45 \times M+2 . \\
a_{84}= & 2(\text { modulo } 49) .
\end{aligned}
$$ | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. The number of planes for which the ratio of distances to the 4 vertices of a regular tetrahedron is $1: 1: 1: 2$ is $\qquad$. | 6.32
Let the vertices of a regular tetrahedron be $A, B, C, D$. The plane $\alpha$ that divides the distances to these four points in the ratio $1: 1: 1: 2$ can be of two types:
(1) $A, B, C$ are on the same side of $\alpha$, there are 2 such planes (as shown in Figure 8).
(1) $\frac{A A_{1}}{A_{1} D}=\frac{B B_{1}}{B... | 32 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) The player numbers of a sports team are uniquely selected from the positive integers 1 to 100. If the number of any player is neither the sum of the numbers of any other two players nor twice the number of another player, what is the maximum number of players this sports team can have? | Second, all odd numbers $1, 3, \cdots, 99$ totaling 50 can be the numbers of the team members. Below is the proof that it is impossible to increase further. If the sports team
has 51 members, from smallest to largest, denoted as
$$
a_{1}, a_{2}, \cdots, a_{50}, a_{51} \text {. }
$$
Take the differences
$$
a_{51}-a_{1... | 50 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) In a convex $n$-sided rose garden $(n \geqslant 4)$, there is 1 red rose planted at each of the $n$ vertices. There is a straight path between every two red roses, and these paths do not have the situation of "three lines intersecting at one point."
--- They divide the garden into many non-overlappi... | Three, (1) Solution 1: The straight paths in the rose garden form a convex $n$-sided polygon and its diagonals $\left(k_{n}\right)$. It is easy to know that there are $C_{n}^{2}-n$ diagonals in the graph. Furthermore, since the diagonals do not have "three lines intersecting at one point", the number of intersection po... | 99 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$=6$ integer solutions.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-(10 a+b), \\
x_{1} x_{2}=10 b+a .
\end{array}\right.
$$
We have $10 x_{1}+10 x_{2}+x_{1} x_{2}=-99 a(1 \leqslant a \leqslant 9)$.
Thus $\left(x_{1}+10\right)\left(x_{2}+10\right)=100-99 a$.
By the p... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Through the vertex of isosceles $\triangle A B C$, draw a line intersecting the extension of the opposite side at $D$. If the resulting triangles are all isosceles, then $\triangle A B C$ has $\qquad$ such configurations. | 4.5.
Draw a straight line from the base angle to intersect with the extension of the opposite side, there are 5 $\triangle A B C$ that meet the conditions, with base angles of $45^{\circ} γ 72^{\circ} γ 36^{\circ} γ \frac{180^{\circ}}{7}$ γ $\frac{2 \times 180^{\circ}}{7}$.
Draw a straight line from the vertex angle t... | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
(20 points) On a certain day in a certain month in 2001 at midnight, teachers and students of a certain school were watching the live broadcast from Moscow of the announcement of the host city for the 2008 Summer Olympic Games. Beijing won the bid. In their excitement, they noticed that the number of people present was... | Let the number of teachers be $a$, the number of students be $b$, the number of months be $c$, the number of days be $d$, and the number of girls be $e$. From the problem, we have
$$
\left\{\begin{array}{l}
a+b=d, \\
c+p=b . \\
b+c+d-a \leqslant 30, \\
au$, so $e$ is even.
Substituting (1) into (1), we get $2 b+c \leqs... | 29 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The number of triangles with unequal integer sides and a perimeter less than 13 is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | (3)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | 3 | Number Theory | proof | Yes | Yes | cn_contest | false |
2. The smallest positive integer $x$ that satisfies the equation $\tan 19 x^{\circ}=\frac{\cos 99^{\circ}+\sin 99^{\circ}}{\cos 99^{\circ}-\sin 99^{\circ}}$ is $x=$ $\qquad$ . | 2.36.
The right side of the original equation $=\frac{1+\tan 99^{\circ}}{1-\tan 99^{\circ}}=\frac{\tan 45^{\circ}+\tan 99^{\circ}}{1-\tan 45^{\circ} \cdot \tan 99^{\circ}}$ $=\tan 144^{\circ}$.
Then $19 x=144+180 k(k \in \mathbf{N})$.
So $x=\frac{180 k+144}{19}=9 k+7+\frac{9 k+11}{19}$.
We know $19 \mid 9 k+11$. Thus... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In Pascal's Triangle, each number is the sum of the two numbers directly above it. The first few rows of this triangle are as follows:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Row 0 & \multicolumn{6}{|c|}{1} \\
\hline Row 1 & & & 1 & & 1 & \\
\hline Row 2 & & 1 & & 2 & & 1 \\
\hline Row 3 & & & 3 & & 3 & 1 \\
\hline ... | 5. Line 62.
In Pascal's Triangle, its $n$-th row consists of binomial coefficients $C_{n}^{k}$ $(k=0,1, \cdots, n)$. If the ratio of three consecutive terms in the $n$-th row of Pascal's Triangle is $3: 4: 5$, then there exists a positive integer $k$ such that
$$
\begin{array}{l}
\frac{3}{4}=\frac{C_{n}^{k-1}}{C_{n}^{... | 62 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) Given point $A(\sqrt{5}, 0)$ and the curve $y=$ $\sqrt{\frac{x^{2}}{4}-1}(2 \leqslant x \leqslant 2 \sqrt{5})$ with points $P_{1} γ P_{2}, \cdots$ γ $P_{n}$. If $\left|P_{1} A\right| γ\left|P_{2} A\right| γ \cdots γ\left|P_{n} A\right|$ form an arithmetic sequence with common difference $d \in\left(\... | The given curve is a segment of the following hyperbola, i.e.,
$$
\frac{1}{4} x^{2}-y^{2}=1(2 \leqslant x \leqslant 2 \sqrt{5}, y \geqslant 0) \text {. }
$$
$A(\sqrt{5}, 0)$ is its right focus, as shown in Figure 5 (where the line $l$ is the right directrix $x=\frac{4}{\sqrt{5}}$, and the point $P(2 \sqrt{5}, 2)$, with... | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let $P_{n}=(1+1)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{7}\right) \cdots$ $\left(1+\frac{1}{3 n-2}\right)$. Find the greatest integer part of $P_{2000}$. | $$
\begin{array}{l}
\text{First, prove the following inequality:} \\
\sqrt[3]{\frac{7 n+1}{7 n-6}}1) .
\end{array}
$$
The inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{7 n+1}{7 n-6}0 \\
\Leftrightarrow 27(n-1)^{2}+13(n-1)>0 .
\end{array}
$$
The last inequality holds for $n>1$, thus inequality (1) holds.
S... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: In a 10000-meter race on a 400-meter circular track at a school sports meet, two athletes, A and B, start running at the same time. B runs faster than A. At the 15th minute, A speeds up. At the 18th minute, A catches up with B and begins to overtake B. At the 23rd minute, A catches up with B again, and at 23... | Solution: According to the problem, draw the figure, as shown in Figure 2.
The broken line $O A B$ represents the movement of A, and the line segment $O C$ represents the movement of B. Since A catches up with B again at the 23rd minute, $E F=400$. And $B G=C K=10000$.
Since $\triangle D E F \sim \triangle D H B$, acc... | 25 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 On a street $AB$, person A walks from $A$ to $B$, and person B rides a bike from $B$ to $A$. Person B's speed is 3 times that of person A. At this moment, a public bus departs from the starting station $A$ and heads towards $B$, and a bus is dispatched every $x$ minutes. After some time, person A notices that... | Solution: According to the problem, draw the graph as shown in Figure 4.
$A C, A_{1} C_{1}, A_{2} C_{2}, A_{3} C_{3}$ represent the motion graphs of buses departing every $x$ minutes, $A D$ and $B . V$ are the motion graphs of person A and person B, respectively. $E_{1}, E_{2}$ are the points where a bus catches up wit... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 A mall has an escalator moving uniformly from the bottom to the top. Two people, A and B, are in a hurry to go upstairs. While riding the escalator, they both climb the stairs at a uniform speed. A reaches the top after climbing 55 steps, and B's climbing speed is twice that of A (the number of steps B climbs... | Solution: According to the problem, draw the graph, as shown in Figure 5.
$O C, O B, O A$ are the motion graphs of the escalator, person A riding the escalator, and person B riding the escalator, respectively.
Draw $A M \perp t$-axis, intersecting $O B$ and $O C$ at $E$ and $F$, respectively. Draw $B N \perp t$-axis, i... | 66 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) Given the algebraic expression $-x^{3}+100 x^{2}+x$, the letter $x$ is only allowed to take values within the set of positive integers. When the value of this algebraic expression reaches its maximum, what is the value of $x$? Prove your conclusion. | Let $x=k$, the value of the algebraic expression is $a_{k}$.
$$
a_{k+1}-a_{k}=-3 k^{2}+197 k+100 \text {. }
$$
When $00$;
When $k \geqslant 67$, $a_{k+1}-a_{k}<0$.
$\therefore$ When $x=67$, the algebraic expression reaches its maximum value. | 67 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Nine, (15 points) satisfying the following two conditions:
(1) For all natural numbers $x, x^{2}-2001 x+n$ $\geqslant 0$
(2) There exists a natural number $x_{0}$, such that $x_{0}^{2}-2002 x_{0}+n$ $<0$ the number of positive integers $n$ equals $\qquad$ | $Nine, 1001$ | 1001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example $8 A$ and $B$ are two fixed points on a plane. Find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles right triangle. There are $\qquad$ such points $C$. | Solution: As shown in Figure 5, the vertices of the two isosceles right triangles with $AB$ as the hypotenuse are $C_{1}$ and $C_{2}$; the vertices of the four isosceles right triangles with $AB$ as the legs are $C_{3}, C_{4}, C_{5}, C_{6}$.
Therefore, there are a total of 6 points $C$ that meet the conditions. | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
14. As shown in Figure 5, in $\square A B C D$, $P_{1}$, $P_{2}, \cdots$, $P_{n-1}$ are the $n$ equal division points on $B D$. Connect $A P_{2}$ and extend it to intersect $B C$ at point $E$, and connect $A P_{n-2}$ and extend it to intersect $C D$ at point $F$.
(1) Prove that $E F \parallel B D$;
(2) Let the area of ... | 14. (1) Since $A D / / B C, A B / / D C$. Therefore, $\left.\triangle P_{n-2} F I\right) \triangle \triangle P_{n-2} A B, \triangle P_{2} B E \subset \triangle P_{2} D . A$.
Thus, we know
$$
\frac{A P_{n-2}}{P_{n+2} F}=\frac{B P_{n}}{P_{n-2} D}=\frac{n-2}{2}, \frac{A P_{2}}{P_{2} E}=\frac{D P_{2}}{P_{2} B}=\frac{n-2}{2... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four, (18 points) The real numbers $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001 . \\
\text { Let } y_{k}=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2 .
\end{array}
$$
$\cdots, 2$ 001. Find the maximum possible value of $\sum_{k=1}^{2000}\left|y... | For $k=1,2, \cdots, 2000$, we have
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\right| \\
=\frac{\left|\left(x_{1}-x_{2}\right)+2\left(x_{2}-x_{3}\right)+\cdots+... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. The graph of the quadratic function $y=-\frac{1}{2} x^{2}+\frac{1999}{2} x+1000$ passes through ( ) integer lattice points in the first quadrant (i.e., points with positive integer coordinates).
(A) 1000
(B) 1001
(C) 1999
(D) 2001 | 6. (C).
$y=-\frac{1}{2}(x-2000)(x+1)$ The range of the independent variable in the first quadrant is $0<x<2000$. When $x$ is odd, $x+1$ is even; when $x$ is even, $x-2000$ is even. Therefore, when $x=1,2, \cdots, 1999$, $y$ takes integer values.
Thus, the number of lattice points in the first quadrant is 1999. | 1999 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. From a square iron sheet with a side length of 10 cm, circular pieces with a diameter of 1 cm can be cut out, at most $\qquad$ pieces.
| 4.106.
As shown in Figure 6, it is easy to see that $\angle \mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ is an equilateral triangle with a side length of $1 \mathrm{~cm}$, so the height $A O_{2}$ is $\frac{\sqrt{3}}{2} \mathrm{~cm}$. Assuming we can fit $k$ rows, then
$$
\frac{\sqrt{3}}{2}(k-1)+1 \leqslant 10 \text {... | 106 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Several containers are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each container does not exceed 1 ton. To ensure that these containers can be transported in one go, how many trucks with a carrying capacity of 3 tons are needed at least? | First, note that the weight of each container does not exceed 1 ton, so the weight of containers that each vehicle can carry at one time will not be less than 2 tons; otherwise, another container can be added.
Let $n$ be the number of vehicles, and the weights of the containers they carry be $a_{1}, a_{2}, \cdots, a_{... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. For any positive integer $n$, connect the origin $O$ with the point $A_{n}(n, n+3)$, and let $f(n)$ denote the number of all integer points on the line segment $O A_{n}$ except for the endpoints. Then the value of $f(1)+f(2)+f(3)+\cdots+f(2002)$ is ( ).
(A) 2002
(B) 2001
(C) 1334
(D) 667 | 3. (C).
The integer points $(x, y)$ on the line segment $O A_{n}$ satisfy $1=\frac{n+3}{n} \cdot x(0<x<n$, and $x \in \mathbf{N})$. When $n=3 k(k \in \mathbf{N})$,
$$
y=\frac{k+1}{k} \cdot x(0<x<3 k, x \in \mathbf{N}) .
$$
$\because k$ and $k+1$ are coprime,
$\therefore$ only when $x=k$ or $2 k$, $y \in \mathbf{N}$. T... | 1334 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
2. From the 99 natural numbers $1,2,3, \cdots, 99$, the number of ways to choose two different numbers such that their sum is less than 99 is $\qquad$ ways. | 2.2352 .
Let the two selected numbers be $x$ and $y$, then $x+y$ has three cases:
$$
x+y=100, x+y<100, x+y>100 \text {. }
$$
Below, we only consider the first two scenarios:
(1) When $x+y=100$, the number of ways to select is 49;
(2) When $x+y<100$. This indicates that the number of ways to select when $x+y<100$ is e... | 2352 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A rectangular piece of land enclosed by fences has a length and width of $52 \mathrm{~m}$ and $24 \mathrm{~m}$, respectively. An agricultural science technician wants to divide this land into several congruent square test plots. The land must be fully divided, and the sides of the squares must be parallel to the bou... | 6.702 pieces.
Assume the land is divided into several squares with side length $x$, then there exist positive integers $m, n$, such that
$$
\frac{24}{x}=m \text{, and } \frac{52}{x}=n \text{. }
$$
$\therefore \frac{m}{n}=\frac{6}{13}$, i.e., $m=6 k, n=13 k(k \in \mathbf{N})$.
Note that when the value of $k$ is as larg... | 702 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 In Figure 1, there are 8 vertices, each with a real number. The real number at each vertex is exactly the average of the numbers at the 3 adjacent vertices (two vertices connected by a line segment are called adjacent vertices). Find
$$
a+b+c+d-(e+f+g+h)
$$ | The following solution has appeared in a journal:
Given
$$
\begin{array}{l}
a=\frac{b+e+d}{3}, \\
b=\frac{a+f+c}{3}, \\
c=\frac{b+g+d}{3}, \\
d=\frac{c+h+a}{3}.
\end{array}
$$
Adding the four equations yields
$$
\begin{array}{l}
a+b+c+d \\
=\frac{1}{3}(2 a+2 b+2 c+2 d+e+f+g+h),
\end{array}
$$
which simplifies to $a+b... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the sum of all numbers in the following square matrix
\begin{tabular}{ccccc}
1901 & 1902 & $\cdots$ & 1949 & 1950 \\
1902 & 1903 & $\cdots$ & 1950 & 1951 \\
$\cdots$ & & & & \\
1950 & 1951 & $\cdots$ & 1998 & 1999
\end{tabular} | Many people first use formula (5) to find the sum of the first row:
$$
S_{1}=\frac{1901+1950}{2} \times 50=96275 \text {; }
$$
Then they find the sums of the 2nd, 3rd, ..., 50th rows; finally, they add these sums together. If one notices that each number in the second row is 1 greater than the corresponding number in ... | 4875000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find all real numbers $p$ such that the cubic equation $5 x^{3}$ $-5(p+1) x^{2}+(71 p-1) x+1=66 p$ has three roots that are all natural numbers. | Analysis: It can be observed that 1 is a solution to the equation. The equation can be transformed into
$$
(x-1)\left(5 x^{2}-5 p x+66 p-1\right)=0 \text {. }
$$
The problem is then reduced to: Find all real numbers $p$ such that the equation
$$
5 x^{2}-5 p x+66 p-1=0
$$
has natural number solutions.
By Vieta's formu... | 76 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2: When A was B's current age, B was 10 years old; when B was A's current age, A was 25 years old. Who is older, A or B? How many years older? | Solution: Let the age difference between A and B be $k$ years, which is an undetermined constant. When $k>0$, A is older than B; when $k<0$, A is younger than B. Then, A's age $y$ and B's age $x$ have a linear relationship:
$$
y=x+k \text {. }
$$
After designing this dynamic process, the given conditions become 3 "mom... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If $\left(1+x+x^{2}+x^{3}\right)^{5}\left(1-x+x^{2}-\right.$ $\left.x^{3}\right)^{5}=a_{30}+a_{29} x+\cdots+a_{1} x^{29}+a_{0} x^{30}$, find $a_{15}$. | Let $f(x)=\left(1+x+x^{2}+x^{3}\right)^{5}$, then the original expression is
$$
F(x)=f(x) f(-x)
$$
which is an even function. Therefore, we have
$$
\begin{array}{l}
\boldsymbol{F}(x)=\frac{1}{2}[\boldsymbol{F}(x)+\boldsymbol{F}(-x)] \\
=a_{30}+a_{28} x^{2}+\cdots+a_{2} x^{28}+a_{0} x^{30},
\end{array}
$$
where all th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 There are 5 medicine boxes, every 2 boxes contain one same medicine, each medicine appears in exactly 2 boxes, how many kinds of medicines are there? | Solution: Represent the medicine boxes as 5 points. When a medicine box contains the same medicine, draw a line segment between the corresponding points.
Since every 2 medicine boxes have one kind of the same medicine, a line should be drawn between every two points. Also, because each kind of medicine appears in exac... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find the smallest positive integer $n$ such that for this $n$, there is a unique positive integer $k$ satisfying
$$
\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13} \text {. }
$$
(Fifth American Mathematical Invitational) | Solution: Since the fractions $\frac{8}{15}$ and $\frac{7}{13}$ satisfy $15 \times 7 - 8 \times 13 = 1$, it must be that $\frac{n}{n+k} = \frac{8+7}{15+13} = \frac{15}{28}$, thus we get $n=15, k=13$. | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $a, b$ be positive integers and satisfy
$$
\frac{2}{3}<\frac{a}{b}<\frac{5}{7} \text {. }
$$
When $b$ is the minimum value, $a+b=$ $\qquad$
(Fifth "Hope Cup" National Mathematics Invitational Competition Training Question) | Solution: According to the theorem, we have
$$
\frac{a}{b}=\frac{2+5}{3+7}=\frac{7}{10} \text {. }
$$
Thus, $a=7, b=10$. Therefore, $a+b=17$.
From the above theorem, it can be seen that although addition and subtraction is a rather clever and simple method, it is not suitable for the case where $b c-a d>1$. For exampl... | 17 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
0, \underbrace{1,1, \cdots, 1}_{k i}, \underbrace{2,2, \cdots,}_{k \uparrow}, 2 \underbrace{3,3, \cdots, 3}_{k \uparrow}, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. When $k=3$, calculate the values of $a_{2002}$ and $S_{2002}$. | Solution: This problem can be regarded as an equal-segment increment sequence, the sum of the first $m$ segments is
$$
1+\underbrace{k+k+\cdots+k}_{(m-1)+}=k(m-1)+1 .
$$
It is evident that, if and only if
$$
k(m-1)+2 \leqslant n \leqslant k m+1
$$
then, $a_{n}=m$. Solving the inequality (6) for $m$, we get
$\frac{n-1... | 668334 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $a=1999 x+2000, b=1999 x+2001, c$ $=1999 x+2002$. Then the value of the polynomial $a^{2}+b^{2}+c^{2}-a b-b c-$ $c a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3 | 2. (D).
$$
\begin{array}{l}
\because a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] . \\
\text { Also } a-b=-1, b-c=-1, c-a=2, \\
\therefore \text { the original expression }=\frac{1}{2}\left[(-1)^{2}+(-1)^{2}+2^{2}\right]=3 .
\end{array}
$$ | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
11. The number of integers $n$ that satisfy $\left(n^{2}-n-1\right)^{n+2}=1$ is.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task ... | 11.4 .
From $n+2=0, n^{2}-n-1 \neq 0$, we get $n=-2$;
From $n^{2}-n-1=1$, we get $n=-1, n=2$;
From $n^{2}-n-1=-1$ and $n+2$ is even, we get $n=0$.
Therefore, $n=-1,-2,0,2$ for a total of 4. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. For a certain project, if contracted to Team A and Team B, it will be completed in $2 \frac{2}{5}$ days, costing 180,000 yuan; if contracted to Team B and Team C, it will be completed in $3 \frac{3}{4}$ days, costing 150,000 yuan; if contracted to Team A and Team C, it will be completed in $2 \frac{6}{7}$ days, cos... | Three, 13. Let the number of days required for A, B, and C to complete the task individually be \( x, y, z \) respectively. Then,
$$
\left\{\begin{array} { l }
{ \frac { 1 } { x } + \frac { 1 } { y } = \frac { 5 } { 1 2 } . } \\
{ \frac { 1 } { y } + \frac { 1 } { z } = \frac { 4 } { 1 5 } , } \\
{ \frac { 1 } { z } +... | 177000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, 18 football teams are participating in a single round-robin tournament, meaning each round the 18 teams are divided into 9 groups, with each group's two teams playing one match. In the next round, the teams are regrouped to play, for a total of 17 rounds, ensuring that each team plays one match against each of t... | ```
3. Consider the following competition program:
1.(1.2)(3.4)(5.6)(7.8)(9,18)
(10,11)(12,13)(14,15)(16,17)
2.(1,3)(2.4)(5,7)(6.9)(8,17)
(10,12)(11,13)(14,16)(15.18)
3.(1.4)(2.5)(3.6)(8.9)(7.16)
(10,13)(11,14)(12,15)(17,18)
4.(1,5)(2.7)(3.8)(4,9)(6.15)
(10.14)(11,16)(12.17)(13.18)
5.(1,6)(2,8)(3,9)(4,7... | 7 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
16. Given positive integers $x, y$, then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has ( ) solutions of the form $(x, 1)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite
(E) Infinite
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result direc... | 16. (C).
$$
\text { Given } \frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5) \Rightarrow x^{2}=\frac{50 y}{y+5} \text {. }
$$
If $(y, 5)=1$, then $(y, y+5)=1$, so $(y+5) \mid 50=5^{2} \times 2$. Since $y+5>2$, then $(y+5) \mid 5^{2}$, which contradicts $(y+5,5)=1$. Therefore, $5 \mid y$.
Let $y=5 y... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1 Elimination Method
Example 1 Given that $x, y, z$ are real numbers, and $x+2y-z=6, x-y+2z=3$. Then, the minimum value of $x^{2}+y^{2}+z^{2}$ is $\qquad$
(2001, Hope Cup Junior High School Mathematics Competition) | Solution: From the given, we can solve for $y=5-x, z=4-x$. Then
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x^{2}+(5-x)^{2}+(4-x)^{2} \\
=3(x-3)^{2}+14 . \\
\because 3(x-3)^{2} \geqslant 0, \quad \therefore x^{2}+y^{2}+z^{2} \geqslant 14 .
\end{array}
$$
Therefore, the minimum value of $x^{2}+y^{2}+z^{2}$ is 14. | 14 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2 Factorization Method
Example 2 Let $a, b, c$ be distinct natural numbers, and $a b^{2} c^{3}=1350$. Then the maximum value of $a+b+c$ is $\qquad$
(1990, Wu Yang Cup Junior High School Mathematics Competition) | Solution: $\because 1350=2 \times 5^{2} \times 3^{3}=150 \times 3^{2} \times 1^{3}$, $\therefore$ when $a=150, b=3, c=1$, the maximum value of $a+b+c$ is 154. | 154 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Let $x_{1}, x_{2}, \cdots, x_{7}$ be natural numbers. And $x_{1}$
$$
<x_{2}<\cdots<x_{6}<x_{7} \text {, and } x_{1}+x_{2}+\cdots+x_{7}=
$$
159. Find the maximum value of $x_{1}+x_{2}+x_{3}$.
(1997, Anhui Province Junior High School Mathematics Competition) | Solution: $\because 159=x_{1}+x_{2}+\cdots+x_{7}$
$$
\begin{array}{l}
\geqslant x_{1}+\left(x_{1}+1\right)+\left(x_{1}+2\right)+\cdots+\left(x_{1}+6\right) \\
=7 x_{1}+21, \\
\therefore x_{1} \leqslant 19 \frac{5}{7} .
\end{array}
$$
Therefore, the maximum value of $x_{1}$ is 19.
$$
\begin{array}{l}
\text { Also, } \b... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 3, each face of the cube is written with a natural number, and the sum of the two numbers on opposite faces is equal. If the number opposite to 10 is a prime number $a$, the number opposite to 12 is a prime number $b$, and the number opposite to 15 is a prime number $c$, then $a^{2}+b^{2}+c^{2}-a ... | 5.19.
Given that $10+a=12+b=15+c$. Therefore, $c=2$. Then $a=7, b=5$. | 19 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given in $\triangle A B C$, $\angle A, \angle B$ are acute angles, and $\sin A$ $=\frac{5}{13}, \tan B=2, A B=29 \mathrm{~cm}$. Then the area of $\triangle A B C$ is $\qquad$ $\mathrm{cm}^{2}$ | 6.145 .
Draw a perpendicular from point $C$ to $AB$. Let the foot of the perpendicular be $D$.
$\because \sin A=\frac{5}{13}=\frac{CD}{AC}$, let $m>0$,
$\therefore CD=5m, AC=13m$.
$\because \tan B=\frac{CD}{BD}=2$, we can set $n>0, CD=2n, BD=n$,
$\therefore BD=n=\frac{CD}{2}=\frac{5}{2}m$.
$\therefore AD=\sqrt{(13m)^2... | 145 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
(1996, National Junior High School Mathematics League... | Solution: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1} < 0 < x_{2}$,
then $x_{1} < 0, \\
\therefore b > 2 \sqrt{a c} . \\
\text{Also, } \because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 14 Let $a, b, c, a+b-c, a+c-b, b+c-a, a+b+c$ be seven distinct prime numbers, and the sum of two of $a, b, c$ is 800. Let $d$ be the difference between the largest and smallest of these seven prime numbers. Find the maximum possible value of $d$.
(2001, China Mathematical Olympiad) | $$
\begin{array}{l}
\text { Let } a<b<c<d \text { be prime numbers, and } a+b, a+c, b+c \text { are also prime numbers. } \\
\text { Without loss of generality, let } a< b, \\
\therefore c<a+b<a+c<b+c .
\end{array}
$$
Also, since one of $a+b$, $a+c$, $b+c$ is
800,
$$
\therefore c<800 \text {. }
$$
Since $799=17 \time... | 1594 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $[A]$ denote the greatest integer less than or equal to $A$, and set $A=38+$ $17 \sqrt{5}$. Then $A^{2}-A[A]=$ $\qquad$ . | $\begin{array}{l}\text { II.1.1. } \\ \because A=38+17 \sqrt{5}=(\sqrt{5}+2)^{3} \text {, let } B=(\sqrt{5}-2)^{3} \text {, } \\ \therefore A-B=76 . \\ \text { Also } \because 0<(\sqrt{5}-2)^{3}<1 \text {, } \\ \therefore[A]=76 \text {, then } A-[A]=B . \\ \text { Therefore } A^{2}-A[A]=A(A-[A])=A B=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$ be a real root of the equation $x^{2}-2002 x+1=0$. Then $a^{2}-2001 a+\frac{2002}{a^{2}+1}=$ $\qquad$ . | 2.2001 .
$\because \alpha$ is a real root of the equation $x^{2}-2002 x+1=0$, then
$$
\begin{array}{l}
\alpha^{2}-2002 \alpha+1=0 . \\
\therefore \alpha+\frac{1}{\alpha}=2002 .
\end{array}
$$
Therefore, $\alpha^{2}-2001 \alpha+\frac{2002}{\alpha^{2}+1}=\alpha-1+\frac{1}{\alpha}=2001$. | 2001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) There are 20 weights, all of which are integers, such that any integer weight $m(1 \leqslant m \leqslant 2002)$ can be balanced by placing it on one pan of a scale and some of the weights on the other pan. What is the smallest possible value of the heaviest weight among these 20 weights? | Let's assume the weights of these 20 weights are \(a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{20}\), where \(a_{i} (1 \leqslant i \leqslant 20)\) are positive integers. It is easy to see that
\[
a_{1}=1, a_{k+1} \leqslant a_{1}+a_{2}+\cdots+a_{k}+1 (1 \leqslant k \leqslant 19).
\]
\[
\text{Then } a_{2} \leqsla... | 146 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function $y=\frac{a-x}{x-a-1}$, the graph of its inverse function is symmetric about the point $(-1,4)$. Then the value of the real number $a$ is $\qquad$ . | 2.3.-
From the problem, we know that the graph of the function $y=\frac{a-x}{x-a-1}$ is centrally symmetric about the point $(4,-1)$.
$\because y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, we have $(y+1)[x-(a+1)]=-1$,
$\therefore$ the graph of the function is a hyperbola with its center at $(a+1,-1)$.
Also, $\because$ t... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that line segment $A D / /$ plane $\alpha$, and the distance to plane $\alpha$ is 8, point $B$ is a moving point on plane $\alpha$, and satisfies $A B=10$. If $A D=21$, then the minimum distance between point $D$ and point $B$ is $\qquad$ . | 4.17.
As shown in Figure 4, let the projections of points $A$ and $D$ on plane $\alpha$ be $O$ and $C$, respectively, then we have $A O = C D = 8$.
$$
\begin{array}{l}
\because A B = 10, \\
\therefore O B = 6 \text{ (constant). }
\end{array}
$$
Therefore, the trajectory of point $B$ in $\alpha$ is a circle with $O$ a... | 17 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $S=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots+[\sqrt{2002}]$, where $[\sqrt{n}]$ denotes the greatest integer not exceeding $\sqrt{n}$. Then the value of $[\sqrt{S}]$ is | 6.242.
Let $k^{2} \leqslant n<(k+1)^{2}$, then $k \leqslant \sqrt{n}<k+1$, so
$$
\begin{array}{l}
{[\sqrt{n}]=k .} \\
\because(k+1)^{2}-k^{2}=2 k+1,
\end{array}
$$
$\therefore$ From $k^{2}$ to $(k+1)^{2}$ there are $2 k+1$ numbers, the integer parts of whose square roots are all $k$,
$$
\begin{array}{l}
\therefore\lef... | 242 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) There are 2002 points distributed on a circle. Now, they are arbitrarily colored white or black. If starting from a certain point and moving in any direction around the circle to any point, the total number of white points (including the point itself) is always greater than the number of black points... | From the problem, we know that a good point must be white. The following discussion is for the general case: there are $3n+1$ points on the circumference, which are colored black and white. Only when the number of black points $\leqslant n$, can we ensure that there must be a good point.
We prove this by mathematical i... | 667 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant k\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
holds. | Solution: First, estimate the upper bound of $k$.
When $a=b=c=d=1$, we have $4 k \leqslant 4+4, k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, it always holds that
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
First, we prove a... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$ | Solution: First, we aim to eliminate the parameter $a$, and then it will be easier to find the minimum value of $k$. Using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n} .
\... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If $xy=1$, then the minimum value of the algebraic expression $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is $\qquad$ .
(1996, Huanggang City, Hubei Province, Junior High School Mathematics Competition) | $$
\text { Sol: } \begin{aligned}
\because & \frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\left(\frac{1}{x^{2}}\right)^{2}+\left(\frac{1}{2 y^{2}}\right)^{2} \\
& \geqslant 2 \cdot \frac{1}{x^{2}} \cdot \frac{1}{2 y^{2}}=1,
\end{aligned}
$$
$\therefore \frac{1}{x^{4}}+\frac{1}{4 y^{4}}$'s minimum value is 1.
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In right $\triangle ABC$, the area is 120, and $\angle BAC=90^{\circ}$. $AD$ is the median of the hypotenuse. A line $DE \perp AB$ is drawn from point $D$ to point $E$. Connect $CE$ to intersect $AD$ at point $F$. Then the area of $\triangle AFE$ is ( ).
(A) 18
(B) 20
(C) 22
(D) 24 | 4. (B).
As shown in Figure 6, draw $F G \perp A B$ at $G$, then $F G / / D E / / A C$. Therefore, we have
$$
\begin{array}{l}
\frac{F G}{\frac{1}{2} A C}=\frac{F G}{D E}=\frac{A G}{A E}, \\
\frac{F G}{A C}=\frac{E G}{A E} .
\end{array}
$$
(1) + (2) gives $\frac{3 F G}{A C}=\frac{A E}{A E}=1$.
Thus, $F G=\frac{1}{3} A... | 20 | Geometry | MCQ | Yes | Yes | cn_contest | false |
3. Person A and Person B go to a discount store to buy goods. It is known that both bought the same number of items, and the unit price of each item is only 8 yuan and 9 yuan. If the total amount spent by both on the goods is 172 yuan, then the number of items with a unit price of 9 yuan is $\qquad$ pieces.
Person A a... | 3.12.
Suppose each person bought $n$ items, among which $x$ items cost 8 yuan each, and $y$ items cost 9 yuan each. Then we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x + y = 2 n , } \\
{ 8 x + 9 y = 172 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=18 n-172, \\
y=172-16 n .
\end{array}\right.\right.... | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $N=23 x+92 y$ be a perfect square, and $N$ does not exceed 2,392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs.
| 4.27.
$\because N=23 x+92 y=23(x+4 y)$, and 23 is a prime number, $N$ is a perfect square not exceeding 2392,
$\therefore x+4 y=23 m^{2}$ ( $m$ is a positive integer) and
$N=23^{2} \cdot m^{2} \leqslant 2392$,
thus $m^{2} \leqslant \frac{2392}{23^{2}}=\frac{104}{23}<5$.
Solving, we get $m^{2}=1$ or 4.
When $m^{2}=1$, f... | 27 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Five, (15 points) 1 and 0 alternate to form the following sequence of numbers:
$$
101,10101,1010101,101010101, \cdots
$$
Please answer, how many prime numbers are there in this sequence? And please prove your conclusion. | Obviously, 101 is a prime number.
Below is the proof that $N=\underbrace{101010 \cdots 01}_{k \uparrow 1}(k \geqslant 3)$ are all composite numbers (with $k-1$ zeros in between).
$$
\begin{aligned}
11 N= & 11 \times \underbrace{10101 \cdots 01}_{k \uparrow 1} \\
& =\underbrace{1111 \cdots 11}_{2 k \uparrow 1}=\underbra... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Under the conditions $x+2 y \leqslant 3, x \geqslant 0, y \geqslant 0$, the maximum value that $2 x+y$ can reach is $\qquad$
(2000, Hope Cup Junior High School Mathematics Competition Second Trial) | Solution: As shown in Figure 1, draw the line $x + 2y = 3$. The set of points satisfying the inequalities $x \geqslant 0, y \geqslant 0, x + 2y \leqslant 3$ is the region $\triangle ABO$ (including the boundaries) enclosed by the line and the $x$ and $y$ axes. To find the maximum value of $s = 2x + y$, we transform $s ... | 6 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
For example, if $a$ and $b$ are positive numbers, and the parabolas $y=x^{2}+ax+2b$ and $y=x^{2}+2bx+a$ both intersect the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
(2000, National Junior High School Mathematics League) | Solution: From the problem, we have
$$
\Delta_{1}=a^{2}-8 b \geqslant 0, \Delta_{2}=4 b^{2}-4 a \geqslant 0 \text {. }
$$
Thus, $a^{2} \geqslant 8 b$ and $b^{2} \geqslant a$.
Since $a$ and $b$ are both positive numbers,
$$
\therefore a^{4} \geqslant 64 b^{2} \geqslant 64 a \text {, i.e., } a \geqslant 4 \text {. }
$$
... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. For the arithmetic sequence $\left.\mid a_{n}\right\}$, the first term $a_{1}=8$, and there exists a unique $k$ such that the point $\left(k, a_{k}\right)$ lies on the circle $x^{2}+y^{2}=10^{2}$. Then the number of such arithmetic sequences is $\qquad$. | 2.17.
Obviously, the point $(1,8)$ is inside the circle, and the points on the circle $(1, \pm \sqrt{9 a})$
do not satisfy the general term formula of the arithmetic sequence
$$
a_{k}=a_{1}+(k-1) d .
$$
However, the points $\left(k, \pm \sqrt{100-k^{2}}\right)(k=2,3, \cdots, 10)$ on the circle
$$
k^{2}+a_{k}^{2}=100... | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. There are 100 equally divided points on a circle. The number of obtuse triangles formed by these points as vertices is $\qquad$ . | $6.50 \times 49 \times 48$ (or 117600).
Let $A_{t}$ be the vertex of the obtuse angle. $\angle A_{\text{s}}$
intercepts an arc of $x$ equal parts, and the other two angles intercept arcs of $y, z$ equal parts, respectively, such that
$$
\left\{\begin{array}{l}
x+y+z=100, \\
x \geqslant 51, y \geqslant 1, z \geqslant 1 ... | 117600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) We call non-empty sets $A_{1}, A_{2}, \cdots, A_{n}$ a $n$-partition of set $A$ if:
(1) $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A$;
(2) $A_{i} \cap A_{j}=\varnothing, 1 \leqslant i<j \leqslant n$.
Find the smallest positive integer $m$, such that for any 13-partition $A_{1}, A_{2}, \cdots, A_{13}$ o... | (1) First, prove that $m \geqslant 117$.
If not, $m1+\frac{13}{104} \geqslant \frac{9}{8}$.
This contradicts $b<a \leqslant \frac{9}{8} b$. Therefore, $m \geqslant 117$.
(2) Next, prove that $m=117$ satisfies the condition.
Because, in this case, the largest 14 numbers $104,105, \cdots, 117$ are distributed among 13 s... | 117 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For any four-digit number (digits can be the same), by changing the original order of its digits, one can always obtain the smallest four-digit number. If the difference between these two four-digit numbers is 999, find the number of such four-digit numbers.
Translate the above text into English, please keep the origi... | Solution: Let the original four-digit number be $\overline{a b c d}$ (where $a, b, c, d$ represent the digits in each place), then from the given information we have
$$
\begin{array}{l}
1000 a+100 b+10 c+d-999 \\
=1000(a-1)+100 b+10 c+(d+1) .
\end{array}
$$
Obviously, when $d \leqslant 8, a>1$, $(a-1) b c(d+1)$ is the... | 48 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $c=10, \frac{\cos A}{\cos B}=\frac{b}{a}=$ $\frac{4}{3}, P$ is a moving point on the incircle of $\triangle A B C$, and $d$ is the sum of the squares of the distances from $P$ to the vertices $A, B, C$. T... | Solution: In $\triangle ABC$, from $\frac{\cos A}{\cos B}=\frac{\sin B}{\sin A}$ we can get
$$
\sin 2A=\sin 2B.
$$
Therefore, $\angle A+\angle B=\frac{\pi}{2}$, which means $\angle C=\frac{\pi}{2}$.
$$
\text{Given } \frac{b}{a}=\frac{4}{3}, c=10,
$$
it is easy to find $b=8, a=6$, and the radius of the inscribed circl... | 160 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given that a square has three vertices $A, B, C$ on the parabola $y=x^{2}$. Find the minimum value of the area of such a square.
(1998, Shanghai High School Mathematics Competition) | Solution: As shown in Figure 3, without loss of generality, assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis). Let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, an... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. When $s$ and $t$ take all real values, the minimum value that can be reached by $(s+5-3|\cos t|)^{2}$ $+(s-2|\sin t|)^{2}$ is $\qquad$
(1989, National High School Mathematics Competition) | (The original expression can be regarded as the square of the distance between any point on the line $\left\{\begin{array}{l}x=s+5, \\ y=s\end{array}\right.$ and any point on the ellipse arc $\left\{\begin{array}{l}x=3|\cos t| \\ y=2|\sin t|\end{array}\right.$. It is known that the square of the distance from the point... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given that $p$, $q$, $\frac{2p-1}{q}$, $\frac{2q-1}{p}$ are all integers, and $p>1$, $q>1$. Try to find the value of $p+q$. | Analysis: The conditions of this problem do not provide specific numbers. Yet, it requires finding the value of $p+q$, which poses a certain difficulty. If we analyze the given conditions one by one, we almost get no useful information. However, if we look at the conditions as a whole, i.e., consider $\frac{2 p-1}{q}$ ... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9.4 In Greek mythology, the "many-headed serpent" god is composed of some heads and necks, with each neck connecting two heads. With each strike of a sword, one can sever all the necks connected to a certain head $A$. However, head $A$ immediately grows new necks connecting to all the heads it was not previously connec... | 9.410 .
We will reformulate the problem using graph theory terminology, with heads as vertices, necks as edges, and a strike that cuts the necks connected to head $A$ as a "reversal" of vertex $A$. It is easy to see that if a vertex $X$ has a degree no greater than 10, then it is sufficient to perform a "reversal" on ... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}... | Analysis: If we consider the four known equations as a system of four equations in $x, y, z, w$, solving this system to find the values of $x, y, z, w$ would be quite difficult. However, if we view these four equations as a whole and temporarily treat $x, y, z, w$ as known numbers, then these four equations can be tran... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11.5 Find the smallest positive integer that can be expressed as the sum of 2002 positive integers with equal sums of their digits, and also as the sum of 2003 positive integers with equal sums of their digits. | 11.510010 .
Assume for a positive integer $n$ the following expression holds:
$$
n=a_{1}+a_{2}+\cdots+a_{2002}=b_{1}+b_{2}+\cdots+b_{2000} \text {. }
$$
Notice that the sum of the digits of $a_{1}, a_{2}, \cdots, a_{2002}$ is the same, so they have the same remainder when divided by 9. Let this remainder be $r(0 \leq... | 10010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Player B has to guess a two-digit number from Player A. If both digits are correct, or one is correct and the other is off by 1, Player A says βnear,β otherwise, Player A says βfar.β For example, if Player Aβs number is 65, and Player B guesses 65, 64, 66, 55, or 75, Player A says βnear,β otherwise, Player A says βf... | 7. (a) The first 17 guesses can exclude at most $17 \times 5=85$ numbers. There are 5 numbers left; the 18th guess cannot determine the number from the remaining 5 numbers.
(c) Construct a $10 \times 9$ table, where the cell in the $i$-th column and $j$-th row contains $(10 i+j-1)$, where $1 \leqslant i \leqslant 9,1 \... | 22 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of intersections of the function $y=x \cdot|x|-\left(4 \cos 30^{\circ}\right) x+2$ with the $x$-axis is $\qquad$ | 3.3 .
$$
y=\left\{\begin{array}{ll}
x^{2}-2 \sqrt{3} x+2, & x>0, \\
2, & x=0 . \\
-x^{2}-2 \sqrt{3} x+2, & x<0
\end{array}\right.
$$
When $x>0$, $y=x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 2 points;
When $x=0$, $y=2$ does not intersect the x-axis;
When $x<0$, $y=-x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 1 p... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always 3 numbers, where the sum of any two of them is still irrational.
| Three, take 4 irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, clearly they do not satisfy the condition, hence $n \geqslant 5$.
Consider 5 irrational numbers $a, b, c, d, e$. View them as 5 points. If the sum of two numbers is a rational number, then connect the corresponding two points with a red lin... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $x_{1}, x_{2}, \cdots, x_{67}$ are positive integers, and their sum equals 110. Find the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{67}^{2}$. | Explanation: Start with any set of 67 positive integers $x_{1}, x_{2}, \cdots, x_{67}$ whose sum is 110. Without loss of generality, assume
$$
x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{66} \leqslant x_{67}.
$$
First, freeze $x_{2}, x_{3}, \cdots, x_{66}$ and only study $x_{1}$ and $x_{67}$. Since
$$
\begin{a... | 2002 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Determine the smallest natural number $k$, such that for any $a \in [0,1]$ and any $n \in \mathbf{N}$ we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$ | Solution: By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k} .
\end{array}
$$
Thus, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only i... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that the three altitudes of $\triangle A B C$ are $A D=3, B E=4, C F=5$, and the lengths of the three sides of this triangle are all integers. Then the minimum value of the length of the shortest side is ( ).
(A) 10
(B) 12
(C) 14
(D) 16 | 5. (B).
From $S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=\frac{1}{2} C A \cdot B E=\frac{1}{2} A B \cdot C F$,
we get $3 B C=4 C A=5 A B$.
It is clear that $A B$ is the shortest side.
From $B C=\frac{5}{3} A B, C A=\frac{5}{4} A B$ and the lengths of $B C, C A, A B$ are all
integers, we know that $3 \mid A B$ and ... | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
II. (25 points) Let $a$, $b$, and $c$ be three distinct real numbers, and $c \neq 1$. It is known that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common root. Find the value of $a + b + c$. | Let the common root of the first two equations be $x_{1}$, then
$$
\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}
$$
(2)
(1) - (2) gives $(a-b) x_{1}+(1-c)=0$.
$$
\because a \neq b, \quad \therefore x_{1}=\frac{c-1}{a-b} \text {. }
$$
Similarly, the common root of the last two equations... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the system of equations $\left\{\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1, \\ x^{2}+y^{2}=50\end{array}\right.$ has only integer solutions. Then the number of real pairs $(a, b)$ that satisfy the condition is $\qquad$ . | 6.60.
It is easy to know that the equation $x^{2}+y^{2}=50$ has 12 sets of integer solutions: $(\pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$, which correspond to 12 integer points on the circle. For each pair of real numbers that satisfy the condition, there corresponds a line in the coordinate system. Therefore, th... | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For any natural numbers $m, n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Find the maximum value of $\lambda$.
| (Let $G=|(m, n)| m<\sqrt{7} n, m, n \in \mathbf{N}$. $\lambda_{\text {max }}=\min _{(m, n \in 6} 17 n^{2}-m^{2}$, then perform $\bmod 7$ analysis on $7 n^{2}-m^{2}$, obtaining $\lambda_{\text {max }}=3$. ) | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example $\mathbf{3}$ A factory's production on the first day does not exceed 20 units, and thereafter, the daily production increases every day, but the amount of increase each time does not exceed 20 units. Prove: When the daily production reaches 1995 units, the total number of products produced by the factory will n... | Explanation: Suppose the daily output on the $n$-th day after the factory starts operation reaches 1995 pieces. If the daily output on the first day is $a_{1}$ pieces, and the increase in output on the $i$-th day is $a_{i}$ pieces, then the daily output on the $n$-th day should be
$$
a_{1}+a_{2}+\cdots+a_{n}=1995 .\lef... | 100500 | Algebra | proof | Yes | Yes | cn_contest | false |
8. For the polynomial $\left(\sqrt{x}+\frac{1}{2 \sqrt[4]{x}}\right)^{n}$ expanded in descending powers of $x$, if the coefficients of the first three terms form an arithmetic sequence, then the number of terms in the expansion where the exponent of $x$ is an integer is $\qquad$ . | 8.3.
It is easy to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these three numbers form an arithmetic sequence, we have $2 \times \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get $n=8$ and $n=1$ (the latter is discarded).
When $n=8$, $T_{r+1}=\mathrm{... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given that $f(x)$ is a function defined on
$\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$ we have
$$
f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1 \text {. }
$$
If $g(x)=f(x)+1-x$, then $g(2002)=$ | 10.1.
$$
\begin{array}{l}
\text { From } g(x)=f(x)+1-x \text { we get } f(x)=g(x)+x-1 . \\
\text { Then } g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \text {, } \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 . \\
\text { Therefore, } g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \\
\therefore g(x) \leqslant g(x+5) \leqslant g(x+4... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 In the school football championship, it is required that each team must play a match against all the other teams. Each winning team gets 2 points, a draw gives each team 1 point, and a losing team gets 0 points. It is known that one team scored the most points, but it played fewer matches than any other team.... | Explanation: We call the team $A$ with the highest score the winning team.
Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must play at least $n+1$ matches, meaning they score no less than $2(n+1)$ points. Therefo... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}$, $a \neq 0$ ) satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find... | 15. Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about $x=-1$. Therefore, $-\frac{b}{2a}=-1, b=2a$.
From (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From (1), $f(1) \geqslant 1$, and from (2), $f(1) \leqslant 1$,
thus $f(1)=1$, i.e., $a+b+c=1$.
Also, $a-b+c=0$, so $b=\frac{1}{2}, a=\frac{1}{4}, c=\frac{1... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (... | Three, let the playing time of the $i$-th player be $x_{i}$ minutes $(i=1,2$, $\cdots, 7)$, the problem is to find the number of positive integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+\cdots+x_{7}=270
$$
under the conditions $71 x_{i}(1 \leqslant i \leqslant 4)$ and $131 x_{j}(5 \leqslant j \leqslant ... | 42244 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A B C D$ be a rectangle with an area of 2, and let $P$ be a point on side $C D$. Let $Q$ be the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant... | Thus, $\frac{1}{2} P A \cdot P B \sin \angle A P B=1$,
which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$.
Equality holds only when $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, and this circle should intersect with $C D$,
Therefore, when $P A \... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given an integer array consisting of 121 integers, each integer in this array takes a value between 1 and 1000 (inclusive of 1 and 1000), and repeated values are allowed. The arithmetic mean of these numbers is $m$, and there is a unique "mode" (the number that appears most frequently) $M$ in this set of numb... | Obviously, to make $D$ as large as possible, the mode $M$ should be as large as possible. For this purpose, let $M=1000$, and the mean $m$ should be as small as possible, so the other numbers should be as small as possible, i.e., $1,2,3, \cdots$.
Thus, the key lies in the frequency of the mode $M$, i.e., how many time... | 947 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. We call $A_{1}, A_{2}, \cdots, A_{n}$ a $n$-partition of set $A$ if
(1) $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A$;
(2) $A_{i} \cap A_{j} \neq \varnothing, 1 \leqslant i<j \leqslant n$.
Find the smallest positive integer $m$, such that for any 14-partition $A_{1}, A_{2}, \cdots, A_{14}$ of $A=\{1,2, \cdots, m\}$, ... | 4. (i) If $ma > b$, and $a - b \geqslant 14$. Therefore, $b \leqslant a - 141 + \frac{14}{42} = \frac{4}{3}$.
Hence, the positive integer $m \geqslant 56$.
(ii) If $m = 56$, then for any partition of $A$ into $A_{1}, A_{2}, \cdots, A_{14}$, among the numbers $42, 43, \cdots, 56$, there must be two numbers belonging to ... | 56 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
16. Given $a+b+c=0$, and $a, b, c$ are all non-zero. Then simplify $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ to | 16. -3 .
$$
\begin{aligned}
\text { Original expression }= & \left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right) \\
& +\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-\frac{a}{a}-\frac{b}{b}-\frac{c}{c} \\
= & \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-3=-3 .
\end{aligned}
$$ | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Question: How many real roots does the equation $x^{2}|x|-5 x|x|+2 x=0$ have (where $|x|$ represents the absolute value of $x$)? | 1.4 .
The original equation simplifies to $x(x|x|-5|x|+2)=0$. After removing the absolute value signs and discussing, we get
$$
x_{1}=0, x_{2}=\frac{5+\sqrt{17}}{2}, x_{3}=\frac{5-\sqrt{17}}{2}, x_{4}=\frac{5-\sqrt{33}}{2}
$$
as the four real roots of the original equation. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Team A and Team B each send out 5 players to participate in a chess broadcast tournament according to a pre-arranged order. The two teams first have their No. 1 players compete; the loser is eliminated, and the winner then competes with the No. 2 player of the losing team, β¦, until all players of one side are elimin... | 3.252 .
Due to the tournament rules, the first team to achieve 5 victories is declared the winner, even if some players have not participated. However, we can consider the players who did not participate as having lost. Thus, we get a complete ten-game result for one side: five wins and five losses. A match process, i... | 252 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x_{1}=\sqrt[3]{3}, x_{2}=\left(x_{1}\right)^{\sqrt[3]{3}}$, for $n>1$ define $x_{n+1}$ $=\left(x_{n}\right)^{\sqrt[3]{3}}$. Find the smallest positive integer $n$ such that $x_{n}=27$. | 4.7.
It is known that $x_{n}=x_{1}^{x_{1}^{n-1}}(n>1)$, and $27=(\sqrt[3]{3})^{(\sqrt[3]{3})^{6}}$. Therefore, $n-1=6$. Hence, $n=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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