problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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Example 6 If $x, y$ are real numbers, find the minimum value of $S=5 x^{2}-4 x y$ $+y^{2}-10 x+6 y+5$. | Solution: From the given, we have
$$
\begin{array}{l}
5 x^{2}-(10+4 y) x+y^{2}+6 y+5-S=0 . \\
\Delta_{x}=-4\left(y^{2}+10 y-5 S\right) \geqslant 0 . \\
\text { Hence, } 5 S \geqslant y^{2}+10 y=(y+5)^{2}-25 \\
\quad \geqslant-25 .
\end{array}
$$
Thus, $5 S \geqslant y^{2}+10 y=(y+5)^{2}-25$
$\therefore S \geqslant-5$,... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Let $x, y$ be positive numbers, and $x+y=1$. Find the minimum value of the function $W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$. (Adapted from the 3rd Canadian Mathematical Competition) | Solution: Without loss of generality, let $x \geqslant y$, and set $x=\frac{1}{2}+t$, then $y=$
$$
\begin{array}{l}
\frac{1}{2}-t, 0 \leqslant t<\frac{1}{2} . \\
\therefore W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \\
=\frac{3+2 t}{1+2 t} \cdot \frac{3-2 t}{1-2 t}=\frac{9-4 t^{2}}{1-4 t^{2}} \\
\geqslant ... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 13 Given that $x, y, z$ are positive numbers, and $4x + 5y + 8z = 30$. Find the minimum value of $W = 8x^2 + 15y^2 + 48z^2$. | Solution: Let $x=3+t_{1}, y=2+t_{2}, z=1+t_{3}$, substituting into the given equation yields $4 t_{1}+5 t_{2}+8 t_{3}=0$.
$$
\begin{aligned}
W= & 8 x^{2}+15 y^{2}+48 z^{2} \\
= & 180+12\left(4 t_{1}+5 t_{2}+8 t_{3}\right)+8 t_{1}^{2} \\
& +15 t_{2}^{2}+48 t_{3}^{2} \\
\geqslant & 180,
\end{aligned}
$$
(when and only wh... | 180 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. Find the smallest integer \( n (n \geq 4) \), such that from any \( n \) integers, four different numbers \( a, b, c, d \) can be selected, making \( a + b - c - d \) divisible by 20. | Solution: First, we consider the different residue classes modulo 20. For a set with $k$ elements, there are $\frac{1}{2} k(k - 1)$ integer pairs. If $\frac{1}{2} k(k - 1) > 20$, i.e., $k \geq 7$, then there exist two pairs $(a, b)$ and $(c, d)$ such that $a + b \equiv c + d \pmod{20}$, and $a, b, c, d$ are all distinc... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
17. The integer sequence $a_{1}, a_{2}, a_{3}, \cdots$ is defined as follows: $a_{1}=1$, for $n \geqslant 1, a_{n+1}$ is the smallest integer greater than $a_{n}$, and for all $i, j, k \in\{1,2, \cdots, n+1\}$, it satisfies $a_{i}+a_{j}=3 a_{k}$. Find $a_{1998}$. | Solution: We first find the initial few $a_{n}$.
. $a_{3}=1$.
Since $1+2=3 \times 1$, then $a_{2} \neq 2 . a_{2}=3, a_{3}=4$.
Since $4+5=3+6=3 \times 3$, then $a_{4} \neq 5, a_{4} \neq 6 . a_{4}=$
7.
Since $1+8=3 \times 3,3+9=3 \times 4$, then $a_{5} \neq 8, a_{5} \neq$ 9. $u:=10$
Since $1+11=3 \times 4$, then $a_{6} ... | 4494 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example: 120 farm workers cultivate 50 hectares of land, this
crop name workers required per hectare estimated output value per hectare
\begin{tabular}{lll}
vegetables & $\frac{1}{2}$ & 11000.00 yuan \\
cotton & $\frac{1}{3}$ & 7500.00 yuan \\
rice & $\frac{1}{4}$ & 6000.00 yuan
\end{tabular}
Question: How should it... | Let the land for growing vegetables, cotton, and rice be $x$ hectares, $y$ hectares, and $z$ hectares, respectively, and the expected total output value be $W$ yuan. According to the given conditions, we have
$$
\begin{array}{l}
x+y+z=50, \\
\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z=20 . \\
W=11000 x+7500 y+6000 z .
\e... | 450000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
刭 2 A real estate company owns a "defective rectangular" vacant land $A B C D E$, with side lengths and directions as shown in Figure 1. It is intended to build an apartment with a rectangular foundation running east-west on this land. Please draw the foundation of this land and calculate the maximum area of the founda... | Solution: Establish a rectangular coordinate system with line $BC$ and $AE$ as the $x$-axis and $y$-axis, respectively. The positive directions are along $BC$ and $AE$, with the unit of length being meters $(\mathrm{m})$.
The equation of line $AB$ is
$$
\frac{x}{30}+\frac{y}{20}=1 \text {. }
$$
Obviously, one vertex o... | 6017 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. In the border desert area, patrol vehicles travel 200 kilometers per day, and each vehicle can carry enough gasoline to travel for 14 days. There are 5 patrol vehicles that set out from base $A$ simultaneously, complete their tasks, and then return along the same route to the base. To allow three of the vehicles to ... | 5. Suppose the patrol car travels to point $B$ in $x$ days, and three of the cars take $y$ days to travel from $B$ to the farthest point, then we have
$$
2[3(x+y)+2 x]=14 \times 5,
$$
which simplifies to $5 x+3 y=35$.
From the problem, we know that $x>0, y>0$ and $14 \times 5-(5+2) x \leqslant 14 \times 3$, which mean... | 1800 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6. If a merchant sells goods that cost 8 yuan each at 10 yuan each, he can sell 100 pieces per day. Now he is using the method of raising the price and reducing the purchase quantity to increase profits. Given that the price of each item is set to how much, can maximize the daily profit? And find the maximum profit. | 6. When the price is set at 14 yuan, the daily profit is maximized, with the maximum profit being 360 yuan. | 360 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. A company has 16 machines in inventory at locations $A$ and $B$. Now, these machines need to be transported to locations 甲 and 乙, with 15 machines required at 甲 and 13 machines required at 乙. It is known that the transportation cost from $A$ to 甲 is 500 yuan per machine, and to 乙 is 400 yuan per machine; the transpo... | 7. Send 3 units from $A$ to location 甲, 13 units to location 乙; send 12 units from $B$ to location 甲, 0 units to location 乙, to minimize the total transportation cost, the minimum cost is 10300 (yuan). | 10300 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Based on market research analysis, a home appliance manufacturing company has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be at least... | 8. Every week, 30 air conditioners, 270 color TVs, and 60 refrigerators should be produced to maximize output value, with the maximum output value being 1050 thousand yuan. | 1050 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $x$ be a real number, and $f(x)=|x+1|+|x+2|+|x+3|+|x+4|+|x+5|$. Find the minimum value of $f(x)$. | Analysis: According to the geometric meaning of absolute value, draw the points $A, B, C, D, E$ corresponding to the real numbers $-1, -2, -3, -4, -5$ on the number line, as shown in Figure 1. Let $x$ correspond to the moving point $P$, then $f(x)=|PA|+|PB|+|PC|+|PD|+|PE| \geqslant |CB|+|CD|+|CA|+|CE|=2+4=6$, that is, ... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $f(x)=|1-2 x|, x \in[0$, 1]. How many real solutions does the equation $f(f(f(x)))=\frac{x}{2}$ have? | Solution: To find the number of real solutions using the graphical method. First, draw the graph of $f(x) = |2x - 1|$, and then double the y-coordinates of all points while keeping the x-coordinates unchanged. Next, shift the obtained graph down by 1 unit (as shown in Figure 2), and then reflect the part of the graph b... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 For all real numbers $x$, such that $|x-1|+|x-2|+|x-10|+|x-11| \geqslant m$ always holds. Then the maximum value of $m$ is $\qquad$ | Analysis: This type of problem is actually an extension and generalization of the function extremum problem in Example 1. It is easy to know that the function $f(x) = |x-1| + |x-2| + |x-10| + |x-11|$ has a minimum value of 18 when $x \in [2,10]$. Therefore, when $m \leqslant 18$, the inequality $f(x) \geqslant m$ alway... | 18 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let the function $f_{3}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed part of the figure enclosed by the graph of the function $y=f_{2}(x)$ and the $x$-axis is $\qquad$ | Analysis: First, draw the graph of $f_{0}(x)=|x|$ and translate it downward by 1 unit to get the graph of $y=f_{0}(x)-1$, from which we obtain the graph of $f_{1}(x)=\left|f_{0}(x)-1\right|$ (as shown in Figure 7).
Next, translate the graph of $f_{1}(x)$ downward by 2 units to get the graph of $y=f_{1}(x)-2$, and refl... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7: A town has five primary schools along a circular road, sequentially named First, Second, Third, Fourth, and Fifth Primary Schools. They have 15, 7, 11, 3, and 14 computers, respectively. To make the number of computers equal in each school, some computers need to be transferred to neighboring schools: First ... | Analysis: Let $A, B, C,$
$D, E$ represent the five primary schools in a clockwise order, and let them sequentially transfer $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ computers to their neighboring schools, as shown in Figure 9. Then,
\[
\begin{array}{c}
7 + x_{1} - x_{2} = 11 + x_{2} \\
- x_{3} = 3 + x_{3} - x_{4} = 14 + x_{... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 On a circular road, there are four middle schools arranged in sequence: $A_{1}, A_{2}, A_{3}, A_{4}$. They have 15, 8, 5, and 12 color TVs, respectively. To make the number of color TVs in each school the same, some schools are allowed to transfer color TVs to adjacent schools. How should the TVs be transferr... | Solution: Let $A_{1}$ high school transfer $x_{1}$ color TVs to $A_{2}$ high school (if $x_{1}$ is negative, it means $A_{2}$ high school transfers $r_{1}$ color TVs to $A_{1}$ high school. The same applies below), $A_{2}$ high school transfers $x_{2}$ color TVs to $A_{3}$ high school, $A_{3}$ high school transfers $x_... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 (Fill in the blank Question 1) Given $\frac{1}{4}(b-c)^{2}=$ $(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$ . | Solution 1: From the given, we have
$$
\begin{aligned}
0 & =\frac{1}{4}[-(a-b)-(c-a)]^{2}-(a-b)(c-a) \\
& =\frac{1}{4}\left[(a-b)^{2}+(c-a)^{2}-2(a-b)(c-a)\right] \\
& =\frac{1}{4}[(a-b)-(c-a)]^{2} \\
& =\frac{1}{4}(2 a-b-c)^{2} .
\end{aligned}
$$
Thus, $\frac{b+c}{a}=2$.
Solution 2: The given condition indicates that... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $[x]$ represents the greatest integer not exceeding $x$. Then the number of solutions to the equation
$$
3^{2 x}-\left[10 \times 3^{x+1}\right]+\sqrt{3^{2 x}-10 \times 3^{x+1}+82}=-80
$$
is | 5.2.
The original equation can be transformed into
$$
\begin{array}{l}
3^{2 x}-\left[10 \times 3^{x+1}\right]+82 \\
+\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-2=0 . \\
\therefore\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}+2\right) \\
\quad \cdot\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-1... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{r}
\text { 5. Given a sequence } z_{0}, z_{1}, \cdots, z_{n}, \cdots \text { satisfying } z_{0}=0, z_{1} \\
=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2,
\end{array}
$$
5. Given a sequence of complex numbers $z_{0}, z_{1}, \cdots, z_{n}, \cdots$ satisfying $z_... | 5.5.
Since $z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right)=\alpha^{2}\left(z_{n-1}-z_{n-2}\right)$ $=\cdots=\alpha^{n}\left(z_{1}-z_{0}\right)=\alpha^{n}$, therefore,
$$
z_{n}-z_{n-1}=\alpha^{n-1}, \cdots, z_{1}-z_{0}=\alpha^{0}=1 \text {. }
$$
Adding the above $n$ equations, we get
$$
z_{n}=\alpha^{n-1}+\cdots+\alph... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}, B_{2}, \cdots, B_{k}\right.$ be a family of non-empty subsets of $A$, and when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Then the maximum value of $k$ is $\qquad$ | 6.175 .
It is not hard to see that the family of all subsets of $A$ containing at most three elements meets the conditions of the problem, where the number of subsets is $\mathrm{C}_{10}^{1}+\mathrm{C}_{10}^{e}+\mathrm{C}_{10}^{3}=175$.
Suppose there is another family of subsets $C$ that meets the conditions, and the... | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 20 points) For every pair of real numbers $x, y$, the function $f(t)$ satisfies $f(x+y)=f(x)+f(y)+xy+1$. If $f(-2)=-2$, find the number of integer solutions $a$ that satisfy $f(a)=a$.
Translate the above text into English, please retain the original text's line breaks and format, and output the tran... | Let $x=y=0$, we get $f(0)=-1$; let $x=y=-1$, from $f(-2)=-2$ we get $f(-1)=-2$. Also, let $x=1$, $y=-1$ to get $f(1)=1$. Then let $x=1$, we have
$$
f(y+1)=f(y)+y+2 \text {. }
$$
Therefore, $f(y+1)-f(y)=y+2$, which means when $y$ is a positive integer, $f(y+1)-f(y)>0$.
From $f(1)=1$, we know that for all positive inte... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 20 points) A county is located in a desert area, and people have been engaged in a tenacious struggle against nature for a long time. By the end of 1998, the county's greening rate had reached $30 \%$. Starting from 1999, the following situation will occur every year: $16 \%$ of the original desert ar... | (1) Let the current desert area be $b_{1}$, and after $n$ years, the desert area be $b_{n+1}$. Thus, $a_{1}+b_{1}=1, a_{n}+b_{n}=1$.
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} a_{n}$, which is $\frac{96}{1... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}$ $+b x^{16}+1$, then $a=$ $\qquad$ | $=1.987$.
Let $p, q$ be the roots of the equation $x^{2}-x-1=0$, then
$$
p+q=1, pq=-1 \text{. }
$$
Given that $p, q$ are also the roots of the equation $a x^{17}+b x^{16}+1=0$, we have $\left\{\begin{array}{l}a p^{17}+b p^{16}+1=0, \\ a q^{17}+b q^{16}+1=0 .\end{array}\right.$
$$
\begin{array}{l}
\text{(1) } \times q^... | 987 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the incircle $\odot O$ of $\triangle A B C$ touch $B C$ at point $D$, and draw the diameter $D E$ through $D$. Connect $A E$ and extend it to intersect $B C$ at point $F$. If $B F+C D=1998$, then $B F+2 C D=$ | 2. 2997.
As shown in Figure 6, let
$\odot O$ be tangent to $AB$ and
$AC$ at points $M$ and $N$, respectively. Draw
a line $GH \parallel BC$ through
point $E$, intersecting
$AB$ and $AC$ at
points $G$ and $H$, respectively. Then $GH$ is tangent
to $\odot O$ at point $E$, and
$\triangle A G E \backsim \triangle A B F, \... | 2997 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given three real
numbers $x_{1}, x_{2}, x_{3}$, any one of them plus five times the product of the other two equals 6. The number of such triples $\left(x_{1}, x_{2}\right.$, $x_{3}$ ) is. | 4.5
From the problem, we have
$$
\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{1} x_{3}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}
$$
(1) - (2) gives $\left(x_{1}-x_{2}\right)\left(5 x_{3}-1\right)=0$.
Thus, $x_{1}=x_{2}$ or $x_{3}=\frac{1}{5}$.
Similarly, $x_{2}=x_{3}$ or $x_{1}=\frac{1}{5}$, $x_{3}=x_{1}$ or... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $S$ be a set of numbers composed of some numbers from $1,2,3, \cdots, 50$ (a set of numbers), such that the sum of any two numbers in $S$ cannot be divisible by 7. How many numbers from $1,2,3, \cdots, 50$ can $S$ contain at most (the maximum number of elements in $S$)? Prove your conclusion. | Three, divide the 50 numbers $1,2,3, \cdots, 50$ into seven different sets $F_{0}, F_{1}, \cdots, F_{6}$ based on their remainders when divided by 7:
$F_{0}$ consists of $7,14,21, \cdots, 49$ (divisible by 7), $F_{1}$ consists of $1,8,15, \cdots, 50$ (remainder 1 when divided by 7), $\qquad$
$F_{6}$ consists of $6,13,2... | 23 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. Given the function $f(x+1)=\frac{1999^{2 x+1}}{1999^{2 x+1}-1}$. Then the value of the sum $\sum_{i=1}^{4000} f\left(\frac{i}{4001}\right)$ is $\qquad$ | $$
\begin{array}{l}
=\frac{1999^{2 x-1}}{1999^{2 x-1}-1}+\frac{1999^{1-2 x}}{1999^{1-2 x}-1}=1 \\
\therefore \sum_{i=1}^{4000} f\left(\frac{i}{4001}\right)=\sum_{i=1}^{2000}\left[f\left(\frac{i}{4001}\right)+f\left(\frac{4001-i}{4001}\right)\right] \\
=\sum_{i=1}^{2000}\left[f\left(\frac{i}{4001}\right)+f\left(1-\frac{... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five, let $f(x)$ be a function that satisfies the following conditions:
(1) If $x>y$ and $f(x)+x \geqslant w \geqslant f(y)+y$, then there exists a real number $z \in[y, x]$, such that $f(z)=w-z$;
(2) The equation $f(x)=0$ has at least one solution, and among these solutions, there is one that is not greater than all t... | Let $F(x)=f(x)+x$, then $F(0)=1$.
Let $u$ be the smallest root of $f(x)=0$, then $F(u)=u$.
If $u0$
For any real number $x$, by (5) we have
$$
\begin{aligned}
0 & =f(x) f(u)=f(x f(u)+u f(x)+x u) \\
& =f(u f(x)+u u) .
\end{aligned}
$$
$\therefore u f(x)+x u$ is a root of $f(x)=0$.
$\because u$ is the smallest root of $f(... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One TV station has $n$ ad breaks in a day, during which a total of $m$ ads were broadcast. In the first ad break, one ad and $\frac{1}{8}$ of the remaining $(m-1)$ ads were broadcast. In the second ad break, 2 ads and $\frac{1}{8}$ of the remaining ads were broadcast. This pattern continued for each subsequent ad break... | Let's denote the number of ads left to be broadcast after $k$ broadcasts as $a_{k}$. Then, the number of ads broadcast at the $k$-th time is:
$$
k+\frac{1}{8}\left(a_{k-1}-k\right)=\frac{1}{8} a_{k-1}+\frac{7}{8} k .
$$
Thus, $a_{k}=a_{k-1}-\left(\frac{1}{8} a_{k-1}+\frac{7}{8} k\right)=\frac{7}{8} a_{k-1}-\frac{7}{8}... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, it is known that the natural number $N$ has 20 positive integer factors (including 1 and itself), which are sequentially denoted as $d_{1}, d_{2}, d_{3}, \cdots, d_{21}$, and the factor with index $d_{8}$ is $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$. Find the natural number $N$. | Three, because $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$ is a factor of $N$, so, $d_{1}+d_{4}+d_{6}$ and $d_{4}+d_{8}+d_{13}$ are factors of $N$. Therefore,
$$
\begin{array}{l}
d_{1}+d_{4}+d_{6} \geqslant d_{7}, d_{4}+d_{8}+d_{13} \geqslant d_{14} . \\
\therefore d_{d 8}=\left(d_{1}+d_{4}+d_{6}\r... | 2000 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (Full marks 50 points) Let $S$ be a subset of $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements in $S$ is not divisible by 7. What is the maximum number of elements that $S$ can have? | Divide $\{1,2, \cdots, 50\}$ into 7 classes according to modulo 7:
$$
\begin{array}{l}
k_{1}=\{1,8,15,22,29,36,43,50\}, \\
k_{2}=\{2,9,16,23,30,37,44\}, \\
k_{3}=\{3,10,17,24,31,38,45\}, \\
k_{4}=\{4,11,18,25,32,39,46\}, \\
k_{5}=\{5,12,19,26,33,40,47\}, \\
k_{6}=\{6,13,20,27,34,41,48\}, \\
k_{0}=\{7,14,21,28,35,42,49\... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 If $f(x)=\left(2 x^{5}+2 x^{4}-53 x^{3}-\right.$ $57 x+54)^{1998}$, then $f\left(\frac{\sqrt{1}-\frac{1}{2}-i}{2}\right)=$ $\qquad$ . | Let $\frac{\sqrt{111}-1}{2}=x$, then we have
$$
\begin{array}{l}
2 x^{2}+2 x-55=0 \\
\because 2 x^{5}+2 x^{4}-53 x^{3}-57 x+54 \\
=x^{3}\left(2 x^{2}+2 x-55\right)+x\left(2 x^{2}+2 x\right. \\
\quad-55)-\left(2 x^{2}+2 x-55\right)-1 \\
\therefore f\left(\frac{\sqrt{111}-1}{2}\right)=(-1)^{1998}=1 .
\end{array}
$$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $f(x)=\frac{2 x}{1+x}$. Find
$$
\begin{array}{l}
f(i)+f(2)+\cdots+f(100)+f\left(\frac{1}{2}\right) \\
+f\left(\frac{2}{2}\right)+\cdots+f\left(\frac{100}{2}\right)+\cdots+f\left(\frac{1}{100}\right) \\
+f\left(\frac{2}{100}\right)+\cdots+f\left(\frac{100}{100}\right)=
\end{array}
$$ | Given $f(x)=\frac{2 x}{1+x}$, we have
$$
f(x)+f\left(\frac{1}{x}\right)=2, f(1)=1 \text {. }
$$
In the required expression, there are 100 $f(1)$ terms, and the remaining $f(x)$ and $f\left(\frac{1}{x}\right)$ appear in pairs, totaling $\frac{1}{2}\left(100^{2}-\right.$ 100 ) pairs.
Therefore, the required sum is
$$
10... | 10000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, after expansion and combining like terms, the sum of the coefficients of the odd powers of $x$ is
保留源文本的换行和格式，直接输出翻译结果如下：
```
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, ... | Solution: Let $f(x)=\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}\right.$
$$
\begin{aligned}
& -3 x-3)^{2001} \\
= & a_{0}+a_{1} x+a_{2} x^{2}+\cdots \\
& +a_{4001} x^{4001}+a_{4002} x^{4002} .
\end{aligned}
$$
Let $x=1$ and $x=-1$, we get
$$
\begin{array}{l}
a_{0}+a_{1}+a_{2}+\cdots+a_{4001}+a_{4002} \\
=f(1)=0, \\
a_{... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given $P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+$ $a_{3} x^{2}+a_{4} x+a_{5}$, and when $k=1,2,3,4$, $P(k)$ $=k \times 1$ 997. Then $P(10)-P(-5)=$ $\qquad$ | Solution: Let $Q(x)=P(x)-1997 x$, then when $k=$ $1,2,3,4$, $Q(k)=P(k)-1997 k=0$.
Therefore, $1,2,3,4$ are roots of $Q(x)=0$.
Thus, we can set $Q(x)=(x-1)(x-2)(x-$ $3)(x-4)(x-r)$, then
$$
\begin{aligned}
& P(10)=Q(10)+1997 \times 10 \\
= & 9 \times 8 \times 7 \times 6 \times(10-r)+1997 \times 10, \\
& P(-5)=Q(-5)+1997 ... | 75315 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+1994^{2}$. | Solution: Since this problem only requires the unit digit of the sum, we only need to consider the unit digit of each number. Thus, the original problem simplifies to finding the unit digit of
$$
\underbrace{i^{2}+2^{2}+3^{2}+4^{2}+\cdots+9^{2}}_{\text {199 groups }}+1^{2}+2^{2}+3^{2}+4^{2}
$$
The unit digits follow a... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Let
$$
\begin{aligned}
S= & \sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+\cdots \\
& +\sqrt{1+\frac{1}{1997^{2}}+\frac{1}{1998^{2}}} .
\end{aligned}
$$
Then the integer closest to $S$ is ( ).
(A) 1997
(B) 1998
(C) 1999
(D) 2000 | $$
\begin{array}{l}
\text { Solution: } \because \sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(1+\frac{1}{n}\right)^{2}-\frac{2}{n}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(\frac{n+1}{n}\right)^{2}-2 \times \frac{n+1}{n} \times \frac{1}{n+1}+\frac{1}{(n+1)^{2}}} \\
=\sqrt{\left(\frac{n+1}{n}-\frac{1}{n+1}\ri... | 1998 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 10 Let $1995 x^{3}=1996 y^{3}=1997 z^{3}$,
$$
\begin{array}{l}
x y z>0, \text { and } \sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}} \\
=\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997} . \\
\quad \text { Then } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=
\end{array}
$$
Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$ | Solution: Let $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, obviously $k \neq 0$. Then we have
$$
1995=\frac{k}{x^{3}}, 1996=\frac{k}{y^{3}}, 1997=\frac{k}{z^{3}} .
$$
From the given, we get
$$
\sqrt{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}}>0 .
$$
Thus... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given real numbers $a, b$ satisfy $3 a^{4}+2 a^{2}-4=0$ and $b^{4}+b^{2}-3=0$. Then $4 a^{-4}+b^{4}=$ ( ).
(A) 7
(B) 8
(C) 9
(D) 10 | Solution: Transform the equation $3 a^{4}+2 a^{2}-4=0$ into $\frac{4}{a^{4}}-$ $\frac{2}{a^{2}}-3=0$, and it is known that $-\frac{2}{a^{2}}$ and $b^{2}$ are the two distinct real roots of the equation $x^{2}+x-3=0$.
Let $-\frac{2}{a^{2}}=x_{1}, b^{2}=x_{2}$. By Vieta's formulas, we have $x_{1}+x_{2}=-1, x_{1} x_{2}=-3... | 7 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 13 Given $a+b+c=0, a^{3}+b^{3}+c^{3}$ $=0$. Find the value of $a^{15}+b^{15}+c^{15}$. | Solution: From $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$ - $\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)$, we have $a b c=0$. Therefore, at least one of $a, b, c$ is 0.
Assume $c=0$, then $a, b$ are opposites,
$$
\therefore a^{15}+b^{15}+c^{15}=0 \text {. }
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example: 15 Given positive integers $x, y, z$ satisfy $x^{3}-y^{3}-z^{3}=3 x y z, x^{2}=2(y+z)$. Find the value of $x y+y z+z x$.
---
The translation is provided as requested, maintaining the original text's line breaks and format. | $$
\begin{array}{l}
\text { Solution: } x^{3}-y^{3}-z^{3}-3 x y z \\
=x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
=(x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z\right. \\
\left.+z^{2}\right)-3 y z(x-y-z) \\
=(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right) \\
=\frac{1}{2}(x-y-z)\left[(x+y)^{2}+(y-z)^{2}\right. \\
\left.+(... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Question: Find the number of five-digit numbers formed by the digits $1, 2, 3, 4, 5, 6$ such that at least three digits are different, and $1, 6$ are not adjacent. | Solution 1: Let $A$ be the set of five-digit numbers formed by $1, 2, 3, 4, 5, 6$ with at least three different digits, $B$ be the set of five-digit numbers formed by $1, 2, 3, 4, 5, 6$ with no $1, 6$ adjacent, and $R$ be the set of five-digit numbers formed by $1, 2, 3, 4, 5, 6$. By the set theory formula, we have
$$
... | 5880 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Six, in a certain exam, there are 5 multiple-choice questions, each with 4 different answers to choose from. Each person selects exactly 1 answer for each question. In 2000 answer sheets, it is found that there exists an $n$, such that in any $n$ answer sheets, there are 4 sheets where each pair of sheets has at most 3... | Solution: The minimum possible value of $n$ is 25.
Let the 4 possible answers to each question be denoted as $1,2,3,4$. The answers on each test paper are denoted as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. Let
$$
\begin{array}{l}
\{(1, h, i, j, k),(2, h, i, j, k),(3, h, i, j, k), \\
(4, h, i, j, k)\},... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, let $S=\{1,2, \cdots, 15\}$. From $S$, take $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$ that satisfy the following conditions:
(i) $\left|A_{i}\right|=7, i=1,2, \cdots, n$;
(ii) $\left|A_{i} \cap A_{j}\right| \leqslant 3,1 \leqslant i<j \leqslant n$;
(iii) For any three-element subset $M$ of $S$, there exists some... | Let $\mathscr{A}=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be any family of sets that satisfies the conditions of the problem. For any $a \in S$, let the number of sets in the family $\mathscr{A}$ that contain $a$ be denoted by $r(a)$. These $r(a)$ sets each contain $C_{6}^{2}:=15$ three-element subsets that include ... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, let $a, b, c, d$ be four distinct real numbers such that
$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$, and $a c=b d$.
Find the maximum value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}$. | Three, let $x=\frac{a}{b}, y=\frac{b}{c}$, then from $a c=b d$ we know $\frac{c}{d}=\frac{b}{a}=\frac{1}{x}, \frac{d}{a}=\frac{c}{b}=\frac{1}{y}$. Thus, the problem becomes finding the maximum value of $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}$ under the constraint $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$.
... | -12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Simplify $\frac{a+1}{a+1-\sqrt{1-a^{2}}}+\frac{a-1}{\sqrt{1-a^{2}}+a-1}$ $(0<|a|<1)$ The result is $\qquad$ | $\begin{array}{l}\text { Original expression }=\frac{(\sqrt{a+1})^{2}}{\sqrt{a+1}(\sqrt{a+1}-\sqrt{1-a})} \\ \quad+\frac{-(\sqrt{1-a})^{2}}{\sqrt{1-a}(\sqrt{1+a}-\sqrt{1-a})}=1 .\end{array}$ | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (Full score 20 points) A newly built oil tank was found to be leaking oil uniformly from the bottom after being filled. To ensure safety and minimize losses, the oil needs to be pumped out before repairs can be made. There are several small oil pumps of the same power available. If 5 pumps work together, it takes ... | Let the oil extraction rate of the oil pump be $x$ units/hour, the internal oil storage be $y$ units, and the oil leakage rate be $z$ units/hour. Suppose $n$ oil pumps are needed to extract all the oil within 3 hours. Then we have
$$
\left\{\begin{array}{l}
50 x = y - 10 z, \\
56 x = y - 8 z, \\
3 n x \geqslant y - 3 z... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (Full marks 25 points) The product of $n$ consecutive natural numbers from 1 to $n$ is called the factorial of $n$, denoted as $n$! (For example, 5! $=5 \times 4 \times 3 \times 2 \times 1$). How many consecutive zeros are at the end of 1999! ? Explain your reasoning.
---
Note: The original text uses "畍居" whic... | Three, consider the prime factorization of 1999!
$$
1999!=2^{\alpha_{1}} \cdot 3^{\sigma_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \ldots
$$
where the various $\alpha_{i}$ are non-negative integers. The number of trailing zeros in 1999! depends on $\alpha_{1}$ and $\alpha_{3}$, and is equal to the smaller ... | 496 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $(x-1)^{2}$ divides the polynomial $x^{4}+a x^{3}-$ $3 x^{2}+b x+3$ with a remainder of $x+1$. Then $a b=$ $\qquad$ . | II, 1.0.
From the given, we have
$$
\begin{array}{l}
x^{4}+a x^{3}-3 x^{2}+b x+3 \\
=(x-1)^{2}\left(x^{2}+\alpha x+\beta\right)+x+1 \\
=x^{4}+(\alpha-2) x^{3}+(\beta+1-2 \alpha) x^{2}+(1+\alpha- \\
2 \beta) x+1+\beta .
\end{array}
$$
Then $a=\alpha-2$,
$$
\begin{array}{l}
-3=\beta+1-2 \alpha, \\
b=1+\alpha-2 \beta, \\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $a$, $b$, and $c$ be the lengths of the sides of $\triangle ABC$, and suppose they satisfy $a^{2}+b^{2}=m c^{2}$. If $\frac{\cot C}{\cot A+\cot B}=$ 999, then $m=$ . $\qquad$ | 2.1999.
$$
\begin{array}{l}
\because \frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}} \\
=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin C(\sin B \cdot \cos A+\cos B \cdot \sin A)} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin C \cdot \sin (A+B)} \\
=\cos C \cdot \frac{... | 1999 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Let $n$ be a natural number, $\alpha_{n} \backslash \beta_{n}\left(\alpha_{n}>\beta_{n}\right)$ are the integer parts of the roots of the quadratic equation $x^{2}-2(n+2) x+3(n+1)=0$. Find the value of $\frac{\alpha_{1}}{\beta_{1}}+\frac{\alpha_{2}}{\beta_{2}}+\cdots+\frac{\alpha_{99}}{\beta_{99}}$. | Solution: Let the two roots of the original equation be $x_{n}, x_{n}^{\prime}$, then by the root formula we get
$$
\begin{array}{l}
x_{n}=n+2+\sqrt{n^{2}+n+1}, \\
x_{n}^{\prime}=n+2-\sqrt{n^{2}+n+1} . \\
\because n<\sqrt{n^{2}+n+1}<n+1, \\
2 n+2<n+2+\sqrt{n^{2}} \overline{n+1}<2 n+3,
\end{array}
$$
E. $\quad<n+2-\sqrt... | 10098 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 There is a pile of goods stacked in a regular manner, with each layer arranged in a rectangle. The bottom layer has 30 items on one side and 15 items on the other, and each subsequent layer has one less item on each side, until the top layer is a straight line. If this pile of goods is restacked into a square... | Solution: The total number of items in this pile of goods is
$$
\begin{array}{l}
30 \times 15 + 29 \times 14 + 28 \times 13 + \cdots + 16 \times 1 \\
= (15 + 15) \times 15 + (15 + 14) \times 14 + \cdots \\
+ (15 + 1) \times 1 \\
= 15 \times (15 + 14 + \cdots + 1) + 15^2 + 14^2 \\
+ \cdots + 1^2 \\
= 3040 .
\end{array}... | 21 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3: A certain area currently has 10,000 hectares of arable land. It is planned that in 10 years, the grain yield per unit area will increase by $22\%$, and the per capita grain possession will increase by $10\%$. If the annual population growth rate is $1\%$, try to find the maximum number of hectares by which t... | Solution: Let the average annual reduction of arable land be $x$ hectares, and let the current population of the region be $p$ people, with a grain yield of $M$ tons/hectare. Then
$$
\begin{array}{l}
\frac{M \times(1+22 \%) \times\left(10^{4}-10 x\right)}{p \times(1+1 \%)^{10}} \\
\geqslant \frac{M \times 10^{4}}{p} \t... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $D$ be a point on the side $AB$ of $\triangle ABC$, point $D$ moves along a direction parallel to $BC$ to point $E$ on side $AC$; then from point $E$ along a direction parallel to $AB$ to point $F$ on side $BC$; then from point $F$ along a direction parallel to $CA$ to point $G$ on side $AB$, $\cdots \cdo... | Analysis: We can estimate through graphical experiments that point $D$ can return to the original starting point after 6 times.
In fact, as shown in
Figure 1, by the intercept theorem of parallel lines,
we get
$$
\begin{array}{l}
\frac{A D}{B D}=\frac{A E}{E C} . \\
=\frac{B F}{F C}=\frac{B G}{A G} \\
=\frac{C H}{A H}... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 The function $f(n)$ is defined on the set of positive integers and takes non-negative integer values, and for all $m, n$ we have
$$
\begin{aligned}
f(m+n)-f(m)-f(n) & =0 \text{ or } 1, \\
\text{and } f(2)=0, f(3)>0, f(9999) & =3333 .
\end{aligned}
$$
Find $f(1982)$. | Solution: Let $m=n=1$, then
$f(2)=2 f(1)+(0$ or 1$)$.
Since $f(2)=0, f(1)$ is a non-negative integer, hence $f(1)$ $=0$.
Let $m=2, n=1$, then
$f(3)=f(2)+f(1)+(0$ or 1$)$.
By $f(3)>0, f(2)=0, f(1)=0$ we get $f(3)$ $=1$.
Next, we prove: For $kk$, then $f(3 k) \geqslant k+1$,
$$
\begin{aligned}
f(9999) & \geqslant f(9999-... | 660 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For positive integers $a, n$, define $F_{n}(a)=q+r$, where $q, r$ are non-negative integers, $a=q n+r$, and 0 $\leqslant r<n$. Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we have
$$
F_{n_{6}}\left(F_{... | To find $A$, we first prove a lemma.
Lemma: If there exists a positive integer $n$, such that for all positive integers $a \leqslant B$ we have $F_{n}(a) \leqslant 2 k$ (where $k$ is a given positive integer), then the largest positive integer $13=k^{2}+3 k$.
Proof: Since $a=q n+r, F_{n}(a)=q+r \leqslant 2 k$, we fir... | 53590 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $R_{x}$ denote the positive integer in decimal notation consisting of $x$ ones. Determine whether $Q=\frac{R_{24}}{R_{4}}$ is a decimal number consisting of several 1s and 0s, and find the number of 0s in $Q$.
(44th American High School AHSME) | Solution: By reversing the formula for the sum of a geometric series, we have
$$
\begin{aligned}
Q & =\frac{R_{24}}{R_{4}}=\frac{9 R_{24}}{9 R_{4}} \\
& =\frac{10^{24}-1}{10^{4}-1}=\frac{\left(10^{4}\right)^{6}-1}{10^{4}-1} \\
& =10^{20}+10^{16}+10^{12}+10^{8}+10^{4}+1 \\
& =100010001000100010001 .
\end{aligned}
$$
Th... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. If $s=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{10^{6}}}$. Try to find the integer part of $s$. | $\begin{array}{l}\text { (Hint: } s=1+\frac{2}{2 \sqrt{2}}+\frac{2}{2 \sqrt{3}}+\cdots+\frac{2}{2 \sqrt{10^{6}}} \\ \\ 2\left(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\cdots+\frac{1}{\sqrt{10^{6}+1}+\sqrt{10^{6}}}\right)= \\ \left.2\left(-1+\sqrt{10^{6}+1}\right)>2\left(-1+10^{3}\right)=1998 .\right)\end{array... | 1998 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given three real numbers $x_{1}, x_{2}, x_{3}$, any one of these numbers plus five times the product of the other two always equals 6. The number of such triples $\left(x_{1}, x_{2}, x_{3}\right)$ is $\qquad$.
$(1995$, Dongfang Airlines Cup - Shanghai Junior High School Mathematics Competition) | Solving the system of equations given by the problem, we have:
$$
\left\{\begin{array}{l}
x_{1}+5 x_{2} x_{3}=6, \\
x_{2}+5 x_{3} x_{1}=6, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.
$$
(1) - (2) gives
$$
\left(x_{1}-x_{2}\right)\left(1-5 x_{3}\right)=0 \text {. }
$$
(2) - (3) gives
$$
\left(x_{2}-x_{3}\right)\left(1... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. The positive integer $n$ is less than 100, and satisfies the equation $\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=n$, where $[x]$ denotes the greatest integer not exceeding $x$. How many such positive integers $n$ are there?
(A)2
(B) 3
(C) 12
(D) 16 | 4. (D).
From $\frac{n}{2}+\frac{n}{3}+\frac{n}{6}=n$, and if $x$ is not an integer, then $[x]<x$, we know that $2|n, 3| n, 6 \mid n$, which means $n$ is a multiple of 6. Therefore, the number of such positive integers less than 100 is $\left[\frac{100}{6}\right]=16$. | 16 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
2. A fruit store is conducting a promotional sale, with the following combinations: Combination A: 2 kg of fruit $A$, 4 kg of fruit $B$; Combination B: 3 kg of fruit $A$, 8 kg of fruit $B$, 1 kg of fruit $C$; Combination C: 2 kg of fruit $A$, 6 kg of fruit $B$, 1 kg of fruit $C$. It is known that fruit $A$ costs 2 yuan... | 2.150.
Let the number of sets of fruit A, B, and C sold on that day be $x$, $y$, and $z$ respectively. According to the problem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
2(2 x+3 y+2 z)=116, \\
8.8 x+25.6 y+21.2 z=441.2,
\end{array}\right. \\
\therefore\left\{\begin{array}{l}
2 x+3 y+2 z=58, \\
22 x+64 y+53 ... | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Real numbers $x, y$ satisfy $x \geqslant y \geqslant 1$ and $2 x^{2}-x y$ $-5 x+y+4=0$. Then $x+y=$ $\qquad$ | 3.4 .
From the given equation, we know that $2 x^{2}-5 x+4=y(x-1) \leqslant x(x-1)$, which leads to $x^{2}-4 x+4 \leqslant 0$, or $(x-2)^{2} \leqslant 0$. Therefore, $x=2$.
Substituting $x=2$ into the given equation yields $y=2$, hence $x+y=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $n$ be a natural number, and $a_{n}=\sqrt[3]{n^{2}+2 n+1}$ $+\sqrt[3]{n^{2}-1}+\sqrt[3]{n^{2}-2 n+1}$. Then the value of $\frac{1}{a_{1}}+\frac{1}{a_{3}}+\frac{1}{a_{5}}$ $+\cdots+\frac{1}{a_{997}}+\frac{1}{a_{999}}$ is ( ).
(A) 3
(B) 4
(C) 5
(D) 6 | 4. (C).
$$
\begin{array}{l}
a_{n}=(\sqrt[3]{n+1})^{2}+\sqrt[3]{n+1} \cdot \sqrt[3]{n-1}+(\sqrt[3]{n-1})^{2}, \\
\frac{1}{a_{n}}=\frac{1}{2}(\sqrt[3]{n+1}-\sqrt[3]{n-1}) . \\
\therefore \sum_{i=1}^{300} \frac{1}{a_{2 i}-1}=\frac{1}{2}[(\sqrt[3]{2}-0)+(\sqrt[3]{4}-\sqrt[3]{2}) \\
\quad+\cdots+(\sqrt[3]{1000}-\sqrt[3]{998... | 5 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Given that $a, b, c, d$ are all real numbers, and $a+b+c+d=4, a^{2}+b^{2}+c^{2}$ $+d^{2}=\frac{16}{3}$. Then the maximum value of $a$ is $\qquad$ . | 4. 2 .
Construct the function
$$
y=3 x^{2}-2(b+c+d) x+\left(b^{2}+c^{2}+d^{2}\right) \text {. }
$$
Since $y(x-b)^{2}+(x-c)^{2}+(x-d)^{2} \geqslant 0$, and the graph
is a parabola opening upwards, we have
$$
\Delta=4(b+c+d)^{2}-12\left(b^{2}+c^{2}+d^{2}\right) \leqslant 0,
$$
which simplifies to $(4-a)^{2}-3\left(\fr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In the permutation $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ of $1,2,3,4,5$, the permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that satisfy $a_{1}a_{3}, a_{3}a_{5}$ are ( ) kinds.
(A) 8
(B) 16
(C) 24
(D) 10 | 4. (B).
In the permutations $a_{1}, a_{2}, \cdots, a_{5}$ that satisfy the conditions, 5 can only be in the second or fourth position, i.e., $5=a_{2}$ or $a_{4}$.
When $a_{2}=5$, $a_{1}a_{3}$ naturally satisfies the condition, $a_{1}$ can be any number from $1,2,3,4$, and $a_{3}, a_{4}, a_{5}$ are the remaining three... | 16 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
Example 1 Given $x+y=3, x^{2}+y^{2}-x y=4$. Then the value of $x^{4}+y^{4}+x^{3} y+x y^{3}$ is $\qquad$
(13th Junior High School Mathematics Competition of Jiangsu Province) | $$
\begin{array}{l}
\text { Solution: } x^{4}+y^{4}+x^{3} y+x y^{3} \\
=(x+y)^{2}\left(x^{2}+y^{2}-x y\right)
\end{array}
$$
Substitute $x+y=3, x^{2}+y^{2}-x y=4$, we get
$$
x^{4}+y^{4}+x^{3} y+x y^{3}=3^{2} \times 4=36 .
$$ | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $a$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Then the value of $\frac{a^{3}-1}{a^{5}+a^{4}-a^{3}-a^{2}}$ is $\qquad$ .
$(1995$, National Junior High School Mathematics League) | Solution: According to the problem, we have $a^{2}+a=\frac{1}{4}$.
$$
\therefore \text { the original expression }=\frac{(a-1)\left(a^{2}+a+1\right)}{(a-1)\left(a^{2}+a\right)^{2}}=20 \text {. }
$$ | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$. Then $m^{5}+n^{5}=$ $\qquad$ .
(From the riverbed Jiangsu Province Junior High School Mathematics Competition) | Solution: Construct a quadratic equation $x^{2}-x-1=0$. From the given conditions, we know that $m$ and $n$ are the two roots of the equation $x^{2}-x-1=0$, so $m+n=1, m n=-1$. Then
$$
\begin{array}{l}
m^{2}+n^{2}=(m+n)^{2}-2 m n=3 . \\
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2}=7 \text {. } \\
\therefore m... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (Full marks 16 points) Given that $x$ and $y$ are positive integers, and they satisfy the conditions $x y + x + y = 71$, $x^{2} y + x y^{2} = 880$. Find the value of $x^{2} + y^{2}$.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation r... | $$
\begin{array}{l}
(x+1)(y+1)=72, \\
x y(x+y)=2^{4} \times 5 \times 11 .
\end{array}
$$
From equation (1), we know that at least one of $x$ and $y$ is odd. Without loss of generality, let $x$ be odd.
From equation (2), $x$ can only be 5, 11, or 55.
However, when $x=55$, $x+y \geqslant 56, x y(x+y) \geqslant 55 \times... | 146 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Six. (Full marks 16 points)
As shown in Figure 4, for the pentagon
$A B C D E$, each side is
translated 4 units outward
along the direction perpendicular to that side,
resulting in a new pentagon
$A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}$.
(1) Can the 5 shaded
parts in the figure be assembled into a penta... | Six, (1) The 5 shaded parts in Figure 4 can form a small pentagon.
$$
\begin{array}{c}
\because B F=A G=A H=E I=E K=D L=D M \\
=C N=C O=B P=4, \\
\angle B F B^{\prime}=\angle A G A^{\prime}=90^{\circ}, \\
\angle C O C^{\prime}=\angle B P B^{\prime}=90^{\circ}, \\
\angle D M D^{\prime}=\angle C N C^{\prime}=90^{\circ}, ... | 25 | Geometry | proof | Yes | Yes | cn_contest | false |
2. Given $a, b, c$ are non-zero real numbers, and $a+b+c$ $=0$. Then the value of $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ is . $\qquad$ | 2. -3 | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Five. (Full marks 30 points) There are several weights of 9 grams and 13 grams each. To weigh an object of 3 grams on a balance, what is the minimum number of such weights needed? Prove your conclusion.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the... | On the three pans of a balance, assuming that when the balance is level, 9 grams are used, and the number is an integer. Similarly, assuming that 13 grams of weights are used $|y|$ times. Therefore, when the balance is level and measures a 3-gram object, there should be
$$
9 x+13 y=3 \text {. }
$$
The problem becomes ... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. (15 points) As shown in Figure 4,
there are two mounds $A$ and
$B$, a depression $E$,
and a pond $F$. The volumes of the two mounds are 781 cubic meters and
1584 cubic meters, respectively. The depression $E$ needs
to be filled with 1025 cubic meters of soil, and the pond $F$
can be filled with 1390 cubic meters o... | $\vdots 1$. "cubic meter・meter" as the unit of labor cost for transporting soil, let the amount of soil transported from $A$ to $E$ be $x_{1}$, and to $F$ be $y_{1}$; the amount of soil transported from $B$ to $E$ be $x_{2}$, and to $F$ be $y_{2}$. The total "cubic meter・meter" of soil transported is $W$. According to ... | 207170 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. The integer part of $\frac{1}{3-\sqrt{7}}$ is $a$, and the fractional part is $b$. Then $a^{2}+(1+\sqrt{7}) a b=$ $\qquad$ . | 3.10 . From $\frac{1}{3-\sqrt{7}}=\frac{3+\sqrt{7}}{2}$, we know that $2<\frac{3+\sqrt{7}}{2}<3$, thus $a=2, b=\frac{1}{3-\sqrt{7}}-2=\frac{\sqrt{7}-1}{2}$. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. The coefficient of $x^{3}$ in the algebraic expression $1+x+x(1+x)+x(1+x)^{2}$ $+\cdots+x(1+x)^{1999}$ is $\qquad$
(Answer with a number) | Ni.1.1331334000.
Starting from the 4th term, there is $x^{3}$, the sum of its coefficients is
$$
\begin{array}{l}
C_{2}^{2}+C_{3}^{2}+\cdots+C_{1999}^{2}=C_{2000}^{3} \\
=\frac{2000 \times 1999 \times 1998}{6}=2000 \times 1999 \times 333 \\
=666000 \times(2000-1)=1331334000 .
\end{array}
$$ | 1331334000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In flood control and rescue operations, a depression near the river dike has experienced a pipe burst, with $x$ cubic meters of river water already rushing in, and water continues to flow in at a rate of $y$ cubic meters per minute. Now, a water extraction and plugging project needs to be carried out. If 1 water pum... | 5.4
Let at least $n$ water pumps be required. According to the problem, we have
$$
\left\{\begin{array}{l}
x+30 y=30 z, \\
x+10 y=20 z, \\
x+5 y=n \cdot 5 z .
\end{array}\right.
$$
From equations (1) and (2), we can solve for $x=15 z, y=0.5 z$.
Substituting into equation (3), we get
$15 z+0.5 z \leqslant 5 n z$, henc... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Choose any two non-adjacent numbers from $1,2, \cdots, 10$ and multiply them. The sum of all such products is $\qquad$ | 6.990.
The sum of the products of any two different numbers taken is
$$
\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right) \text {. }
$$
The sum of the products of any two adjacent numbers is $\sum_{k=1}^{9} k(k+1)$.
The sum that satisfies the condition is
$$
\begin{array}{l}
\frac{1}{2} \sum_{k=1}^{10} k... | 990 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what is the maximum number of pairs of lines that are perpendicular to each other? (Each pair consists of two lines)
保留源文本的换行和格式，翻译结果如下：
Four. (Total 20 points) On the 6 lines of the three edges of the tetrahedron $V-A B C$, what ... | Four, prove in three steps that there are at most 6 pairs of mutually perpendicular lines.
(1) 6 pairs can be achieved.
When $V A \perp$ plane $A B C$ and $A B \perp A C$, by the property of line perpendicular to a plane, we have $V A \perp A B, V A \perp B C, V A \perp C A$.
By the theorem of three perpendiculars, we ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. Let $m$ be a real number not less than -1, such that the equation $x^{2}+2(m-2) x+m^{2}-3 m+3=0$ has two distinct real roots $x_{1}$ and $x_{2}$.
(1) If $x_{1}^{2}+x_{2}^{2}=6$. Find the value of $m$;
(2) Find the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$. | $$
\begin{aligned}
\Delta & =4(m-2)^{2}-4\left(m^{2}-3 m+3\right) \\
& =-4 m+4>0 .
\end{aligned}
$$
Then $m<1$.
Combining the given conditions, we have $-1 \leqslant m<1$.
$$
\begin{array}{l}
\text { (1) } \because x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=4(m-2)^{2}-2\left(m^{2}-3 m+3\right) ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. A 33-story building has an elevator that starts on the first floor. It can accommodate a maximum of 32 people and can only stop once at one of the floors from the 2nd to the 33rd. For each person, walking down one floor of stairs results in 1 point of dissatisfaction, and walking up one floor of stairs results in 3... | 15. It is easy to see that these 32 people live on each of the 2nd to 33rd floors, one person per floor. For each person who takes the elevator to go up or down, the floor they live on must be no less than the floor of the person who walks up directly. In fact, if a person living on the $s$-th floor takes the elevator,... | 316 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Five numbers $a, b, c, d, e$, their pairwise sums are $183, 186, 187, 190, 191, 192, 193, 194, 196, 200$. If $a<b<c<d<e$, then the value of $a$ is $\qquad$ | Analysis: From the problem, we know that $a+b=183, a+c=$ $186, \cdots, d+e=200$. If we can find the value of $a+b+c+$ $d+e$, and note that these ten numbers are the sums of $a$, $b$, $c$, $d$, and $e$ taken two at a time, and that $a$, $b$, $c$, $d$, and $e$ each appear 4 times, then the sum of these ten numbers, 1912,... | 91 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (16 points) An engineering
team contracted two projects, Project A and Project B, with the workload of Project A being twice that of Project B. In the first half of the month, all workers worked on Project A, and in the second half of the month, the workers were divided into two equal groups, one group continued... | Three, suppose this construction team has $x$ people, and the monthly work volume per person is 1. Then the work volume for site A is $\frac{\bar{x}}{2}+\frac{1}{2}$ $\frac{x}{2}$, and the work volume for site B is $\frac{x}{2} \times \frac{1}{2}+1$. According to the problem, we get $\frac{x}{2}+\frac{x}{4}=2\left(\fra... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (20 points) A four-digit number, the sum of this four-digit number and the sum of its digits is 1999. Find this four-digit number and explain your reasoning. | Let this number be abcd. According to the problem, we have
$$
\begin{array}{l}
1000 a+100 b+10 c+d+a+b+c+d \\
=1999,
\end{array}
$$
which simplifies to $1001 a+101 b+11 c+2 d=1999$.
(1) Clearly, $a=1$. Otherwise, $1001 a>2000$. Subtracting $1001$ from both sides, we get
$$
101 b+11 c+2 d=998 \text{. }
$$
(2) The maxim... | 1976 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Given $\left(x^{2}-x+1\right)^{6}=a_{12} x^{12}+$
$$
\begin{array}{l}
a_{11} x^{11}+\cdots+a_{2} x^{2}+a_{1} x+a_{0} \text {. Then } a_{12}+a_{10} \\
+\cdots+a_{2}+a_{0}=\ldots
\end{array}
$$
(Fifth National Partial Provinces Junior High School Mathematics Correspondence Competition) | Solution: In the given equation, let $x=1$, we get
$$
a_{12}+a_{11}+a_{10}+\cdots+a_{2}+a_{1}+a_{0}=1 \text {; }
$$
Let $x=-1$, we get
$$
a_{12}-a_{11}+a_{10}-\cdots+a_{2}-a_{1}+a_{0}=3^{6} \text {. }
$$
Adding the above two equations, we get
$$
\begin{array}{l}
2\left(a_{12}+a_{10}+\cdots+a_{2}+a_{0}\right)=1+3^{6} ... | 365 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Find the sum of all solutions to the equation $[3 x+1]=2 x-\frac{1}{2}$.
(1987, All-China Junior High School Mathematics Competition) | Solution: We can consider $[3 x+1]$ as a whole, for example, an integer $t$, thus transforming the given equation into an inequality related to $t$.
Let $[3 x+1]=t$, then $t$ is an integer, and
$$
0 \leqslant(3 x+1)-t<1 \text {. }
$$
Thus, the original equation becomes
$$
t=2 x-\frac{1}{2} \text {, }
$$
which means $... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Find the smallest positive integer $n$ such that the equation $\left[\frac{10^{n}}{x}\right]=1989$ has an integer solution $x$.
(1989, Soviet Union Mathematical Olympiad) | (Hint: Using the inequality $\frac{10^{n}}{x}-1<\left[\frac{10^{n}}{x}\right] \leqslant \frac{10^{n}}{x}$, solve for $x=5026$, at this point $n=7$.) | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the largest positive integer $n$ that satisfies the following condition: $n$ is divisible by all positive integers less than $\sqrt[3]{n}$. (1998, Asia Pacific Mathematical Olympiad) | Analysis and Solution: The problem involves radicals, which are inconvenient to handle. If we set the largest integer less than $\sqrt[3]{n}$ as $k$, then the original problem can be transformed into the following equivalent form:
Let $k$ and $n$ be positive integers, satisfying
$$
k^{3} < n < (k+1)^{3},
$$
i.e., it h... | 420 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
In 1935, the famous mathematicians P. Erdos and G. Szekeres proved a well-known proposition:
For any positive integer $n \geqslant 3$, there exists a number $f(n)$, such that if and only if $m \geqslant f(n)$, any set of $m$ points in the plane, with no three points collinear, contains $n$ points that form a convex $n... | Lemma 1: Let the vertex set of an arbitrary convex hull $K$ be denoted as $S$. If there are $m$ points inside the convex hull forming a convex $m$-gon, and this convex $m$-gon has a side $A_{2} A_{3}$, with its adjacent sides $A_{1} A_{2}$ and $A_{3} A_{4}$ (where $A_{1}, A_{2}, A_{3}, A_{4}$ are all vertices), and if ... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
4. A magician has one hundred cards, each with a number from 1 to 100. He places these one hundred cards into three boxes, one red, one white, and one blue. Each box must contain at least one card.
A participant selects two of the three boxes and then picks one card from each of the selected boxes, announcing the sum ... | Solution: There are 12 different methods. Consider the integers from 1 to 100. For simplicity, define the color of the box in which integer $i$ is placed as the color of that integer. Use $\mathrm{r}$ to represent red, $\mathrm{w}$ to represent white, and $\mathrm{b}$ to represent blue.
(1) There exists some $i$ such t... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $S$ be the set of all prime numbers $p$ that satisfy the following condition: the number of digits in the smallest repeating block of the decimal part of $\frac{1}{p}$ is a multiple of 3, i.e., for each $p \in S$, there exists the smallest positive integer $r=r(p)$, such that
$$
\frac{1}{p}=0 . a_{1} a_{2} \cdot... | (1) Proof: The length of the smallest repeating cycle of $\frac{1}{p}$ is the $d$ that satisfies $10^{d}-1$.
Let $\varphi$ be a prime, $N_{q}=10^{2 q}+10^{q}+1$, then $N_{q} \equiv 3 \pmod{9}$. Let $p_{q}$ be a prime factor of $\frac{N_{q}}{3}$, then $p_{q}$ is not divisible by 3. Since $N_{q}$ is a factor of $10^{3 q... | 19 | Number Theory | proof | Yes | Yes | cn_contest | false |
1. Consider the permutations $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}$ formed by $\{1,2,3,4,5,6\}$, which can be transformed into 123456 with no more than 100 operations. Find the number of permutations that satisfy this condition.
(12th Korean Mathematical Olympiad) | Solution: For any permutation $a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}$, take $a_{1}$. If $a_{1}$ is not in position $a_{1}$, take the number $a_{1}^{\prime}$ in position $a_{1}$; if $a_{1}^{\prime}$ is not in position $a_{1}^{\prime}$, take the number $a_{1}^{\prime \prime}$ in position $a_{1}^{\prime}$, and continue unti... | 600 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. When $m$ takes all real values from 0 to 5, the number of integer $n$ that satisfies $3 n=m(3 m-8)$ is $\qquad$ . | 2. 13 .
Let $n$ be a quadratic function of $m$, i.e.,
$$
n=m^{2}-\frac{8}{3} m \text {, }
$$
then
$$
n=\left(m-\frac{4}{3}\right)^{2}-\frac{16}{9} \text {. }
$$
From the graph, when $0 \leqslant m \leqslant 5$,
$$
\begin{array}{l}
-\frac{16}{9} \leqslant n \leqslant \frac{35}{3} . \\
\therefore n=-1,0,1,2,3,4,5,6,7,... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. In the expansion of $(a+b)^{n}$, there are $n+1$ different terms. Then, the expansion of $(a+b+c+d)^{21}$ has different terms. | 5.2024 .
$$
\begin{array}{l}
\because(a+b+c)^{n} \\
=(a+b)^{n}+\mathrm{C}_{n}^{1}(a+b)^{n-1} \cdot c+\cdots+\mathrm{C}_{n}^{n} \cdot c^{n}
\end{array}
$$
There are $n+1$ terms, but $(a+b)^{i}$ has $i+1$ terms, $\therefore(a+b+c)^{n}$ has the number of different terms as
$$
\begin{array}{l}
\sum_{i=0}^{n}(i+1)=\frac{1}... | 2024 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Five. (20 points) Let the functions $f(x)$ and $g(x)$ be defined as
$$
\begin{array}{l}
f(x)=12^{x}, g(x)=2000^{x} . \\
a_{1}=3, a_{n+1}=f\left(a_{n}\right)(n \in \mathbf{N}), \\
b_{1}=2000, b_{n+1}=g\left(b_{n}\right)(n \in \mathbf{N}) .
\end{array}
$$
Find the smallest positive integer $m$ such that $b_{m}>a_{2000}$... | Obviously, $12^{3}8 b_{n} \text {. }$
When $n=1$, $a_{3}=12^{16^{3}}>8 b_{1}$.
Assume that equation (1) holds for $n=k$, i.e., $a_{k+2}>8 b_{k}$.
When $n=k+1$,
$$
\begin{array}{l}
a_{k+3}=12^{a_{k+2}}>12^{8 b_{k}}=\left(12^{8}\right)^{b_{k}} \\
>(8 \times 2000)^{b_{k}}>8 \times 2000^{b_{k}}=8 b_{k+1} .
\end{array}
$$
... | 1999 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 䒴 $\frac{4 x}{x^{2}-4}=\frac{a}{x+2}+\frac{b}{x-2}$, then the value of $a^{2}+$ $b^{2}$ is $\qquad$
(1996, Hope Cup National Mathematics Invitational Competition) | Solution: From the problem, we know that the given equation is not an equation in $x$, but an identity in $x$. From
$$
\frac{a}{x+2}+\frac{b}{x-2}=\frac{(a+b) x-2(a-b)}{x^{2}-4}
$$
we get the identity
$$
\begin{array}{l}
\frac{4 x}{x^{2}-4}=\frac{(a+b) x-2(a-b)}{x^{2}-4}, \\
4 x=(a+b) x-2(a-b) .
\end{array}
$$
Accord... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $a$ is a root of the equation $x^{2}+x-\frac{1}{4}=0$. Then the value of $\frac{a^{3}-1}{a^{5}+a^{4}-a^{3}-a^{2}}$ is $\qquad$ .
(1995, National Junior High School Mathematics League) | Solution: The method of solving for the root $a$ by setting up an equation and then substituting to find the value is too cumbersome. It is better to use an overall processing method for this problem. From the given, we have
$$
a^{2}+a=\frac{1}{4} \text { or } a^{2}=\frac{1}{4}-a \text {. }
$$
Using equation (1), we s... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Suppose $1995 x^{3}=1996 y^{3}=1997 z^{3}$, $x y z>0$, and $\sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}}=$ $\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997}$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$
(1996, National Junior High School Mathematics League) | Given that there are three independent equations (equations) and exactly three letters, it should be possible to solve for $x, y, z$, but in practice, it is quite difficult. Inspired by the "chain equality" type problem from the previous question, we can introduce the letter $k$ - try.
Assume $1995 x^{3}=1996 y^{3}=19... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Given that $x, y, z$ are 3 non-negative rational numbers, and satisfy $3x+2y+z=5, x+y-z=2$. If $s=2x+y-z$, then what is the sum of the maximum and minimum values of $s$?
(1996, Tianjin Junior High School Mathematics Competition) | The given equations and the expression for $s$ can be viewed as a system of equations in $x$, $y$, and $z$:
$$
\left\{\begin{array}{l}
3 x+2 y+z=5, \\
x+y-z=2, \\
2 x+y-z=s .
\end{array}\right.
$$
Solving this system, we get $\left\{\begin{array}{l}x=s-2, \\ y=5-\frac{4 s}{3}, \\ z=-\frac{s}{3}+1 .\end{array}\right.$
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Write an $n$-digit number using the digits 1 or 2, where any two adjacent positions are not both 1. Let the number of $n$-digit numbers be $f(n)$. Find $f(10)$.
(Jiangsu Province Second Mathematical Correspondence Competition) | Solution: The $n$-digit numbers that meet the conditions can be divided into two categories:
I. The first digit is 2, then the number of $(n-1)$-digit numbers that meet the conditions is $f(n-1)$;
II. The first digit is 1, then the second digit should be 2, and the number of $(n-2)$-digit numbers is $f(n-2)$. Therefore... | 144 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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