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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, $5 S \geqslant y^{2}+10 y=(y+5)^{2}-25$
$\therefore S \geqslant-5$, so the minimum value of $S$ is -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 $y=$
$$
\begin{array}{l}
\frac{1}{2}-t, 0 \leqslant t<\frac{1}{2} . \\
\therefore W=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \\
=\frac{3+2 t}{1+2 t} \cdot \frac{3-2 t}{1-2 t}=\frac{9-4 t^{2}}{1-4 t^{2}} \\
\geqslant \frac{9-36 t^{2}}{1-4 t^{2}}=9 .
\end{array}
$$
When $t=0$, i.e., $x=y$, the minimum value sought is 9.
|
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 when $t_{1}=t_{2}=t_{3}=0$)
That is, when $x=3, y=2, z=1$, the minimum value of $W$ is 180.
|
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 distinct.
In general, consider a set of 9 different elements. If this set has 7 or more elements belonging to different residue classes modulo 20, then from the previous derivation, we can find four distinct numbers $a, b, c, d$ such that $a + b \equiv c + d \pmod{20}$. Assuming that in this set, at most six elements belong to the same residue class modulo 20, we can still find $a, b, c, d$ such that $a + b - c - d$ is divisible by 20.
For a set of 8 elements, such as
$$
\{0, 20, 40, 1, 2, 4, 7, 12\},
$$
we prove that it does not satisfy the required property. The residues of these numbers modulo 20 are $0, 0, 0, 1, 2, 4, 7, 12$. These residues have the property that each number is greater than the sum of any two smaller numbers, and the sum of any two numbers is less than 20. Let $a, b, c, d$ be four distinct numbers from this set, treating 20 and 40 as 0. Assume $a$ is the largest of the four chosen numbers. Then $a + b - c - d < a + b - c < a + b < 20$. Therefore, $a + b - c - d$ is not divisible by 20.
The minimum value of $n$ is 9.
|
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 $a_{6} \neq 11 \cdot a_{6}=12, a_{7}=13$.
Since $7+14=3 \times 7,15+15=3 \times 10$, then $a_{8} \neq 14$,
$a_{8} \neq 15 \cdot u_{8}: 16$.
By: $1.3 \div 17=3 \times 10.18+3=3 \times 7$, then $a_{9} \neq 17$, $a_{9} \neq 18 \cdot a_{9}=19$
Since $1+20=3 \times 7$, then $a_{10} \neq 20 \cdot a_{10}=21, a_{11}=$ 22 .
Since $7+23=3 \times 10,24+24=3 \times 16$, then $a_{1} \neq$ $23, a_{1} \neq 24, a_{1=}=25$. Next:
$u_{n}: 13$ 个 10 i2 131619212225
If we consider every four terms as a group, then each term in a group can be obtained by adding 9 to each term in the previous group. Below, we will verify that $a_{n}$ satisfies the following formula:
$$
\begin{array}{l}
a_{4 r+1}=9 r+1, a_{4 r+2}=9 r+3, \\
a_{4 r+3}=9 r+4, a_{4 r+4}=9 r+7,
\end{array}
$$
where $r=0,1,2, \cdots$.
When $n=0,1,2$, the conclusion holds.
Assume that for $r=0,1, \cdots, m(m \geqslant 2)$ the conclusion holds, then $a_{1}$, $a_{2}, \cdots, a_{4 m+4}$ do not have any term that is congruent to 2 modulo 3, and in the interval $[1,9 m$ $+7]$, all numbers that are congruent to 1 modulo 3 appear in the above sequence.
Since $(9 m+8)+4=3(3 m+4)$, and $3 m+4=$ $1(\bmod 3)$ is in the above sequence, so $a_{4 m+5} \neq 9 m+8$. Similarly, since $(9 m+9)+3=3(3 m+4)$, so $a_{4 m+5}$ $\neq 9 m+9$. In fact, $a_{4 m+5}=9 m+10$. Because if $(9 m+$ $10)+y=3 z$, then $y \equiv 2(\bmod 3)$, and $y$ is not in the above sequence, which is a contradiction.
Since $(9 m+11)+1=3(3 m+4)$, so $a_{4 m+6} \neq$ $9 m+11$. In fact, $a_{4 m+6}=9 m+12$. Because if $(9 m+$ $12)+y=3 z$, then $y \equiv 0(\bmod 3)$, hence $y=9 r+3$. And $(9 m$ $+12)+(9 r+3)=3(3 m+3 r+5)$, but $3 m+3 r+5$ $\equiv 2(\bmod 3)$ is not in the above sequence, which is a contradiction.
Since there is no $y \equiv 2(\bmod 3)$ in the above sequence, so $a_{4 m+7}=9 m+13$.
Since $(9 m+14)+7=3(3 m+7)$, and $3 m+7 \equiv$ $1(\bmod 3)$ is in the above sequence, so $a_{4 m+8} \neq 9 m+14$. Similarly, since $(9 m+15)+(9 m+15)=3(6 m+10)$,
and $6 m+10 \equiv 1(\bmod 3)$ is in the above sequence, so $a_{\mathrm{s} m+8}$ $\Rightarrow 9 m+15$.
Finally, since there is no $y \equiv 2(\bmod 3)$ in the above sequence, so $a_{4 m+8}=9 m+16$.
By mathematical induction, the formula for $a_{n}$ holds.
Since $1998=4 \times 499+2$, then
$a_{1 \mathrm{VMSM}}=9 \times 499+3=4494$.
|
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 be arranged so that every hectare of land is planted with crops, all workers are employed, and the total estimated output value of agricultural products is maximized?
|
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
$$
\begin{array}{l}
x+y+z=50, \\
\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z=20 . \\
W=11000 x+7500 y+6000 z .
\end{array}
$$
From (1) and (2), we get $y=90-3 x, z=2 x-40$.
Substituting into (3), we get $W=50 x+435000$.
By $x \geqslant 0, y=90-3 x \geqslant 0, z=2 x-40 \geqslant$ 0, we get
$$
20 \leqslant x \leqslant 30 \text {. }
$$
$\therefore$ When $x=30$, $W$ reaches its maximum value of 450000 yuan, at which point $y=0, z=20$.
Thus, planting 30 hectares of vegetables and 20 hectares of rice will maximize the expected total output value, which can reach 450000 yuan.
|
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 foundation (accurate to $1 \mathrm{~m}^{2}$ ).
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
---
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 foundation (accurate to $1 \mathrm{~m}^{2}$ ).
|
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 of the largest rectangle should be $D$. First, consider the case where the non-adjacent vertex $F$ is on $AB$. Then $F\left(x, 20-\frac{2}{3} x\right), 0 \leqslant x \leqslant 30$.
Thus, the area of the rectangle is
$$
\begin{aligned}
S & =(100-x)\left[80-\left(20-\frac{2}{3} x\right)\right] \\
& =-\frac{2}{3} x^{2}+\frac{20}{3} x+6000 \\
& =-\frac{2}{3}(x-5)^{2}+6016 \frac{2}{3} .
\end{aligned}
$$
$\therefore$ When $x=5, y=20-\frac{2}{3} x \approx 17$, the maximum
rectangular area is $6000 \mathrm{~m}^{2}$ and $5600 \mathrm{~m}^{2}$.
Answer: Select point $F(5,17)$, the maximum area is 6017 $\mathrm{m}^{2}$.
|
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 patrol as far as possible (and then return together), vehicles 甲 and 乙 travel to a point $B$ along the way, leaving only enough gasoline for their return to the base, and giving the remaining gasoline to the other three vehicles. How far can the other three vehicles travel in kilometers?
|
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 means $x \geqslant 4$. From (1), we get
$$
y=\frac{35-5 x}{3} \text {. }
$$
To maximize the distance traveled, we need to find the maximum value of $y$, which occurs when $x$ is at its minimum value, i.e., $x=4$. Thus, $y=5$.
Therefore, $200 \times(4+5)=1800$ (km), meaning the farthest distance the other three cars can travel is 1800 km.
|
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 transportation cost from $B$ to 甲 is 300 yuan per machine, and to 乙 is 600 yuan per machine. What transportation plan should be designed to minimize the total transportation cost?
|
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 60 units. It is known that the labor hours required per unit and the output value per unit of home appliances are as follows:
How many units of air conditioners, color TVs, and refrigerators should be produced each week to maximize output value? What is the maximum output value? (in thousand yuan units)
| Product | Labor Hours per Unit | Output Value per Unit (thousand yuan) |
|---------|----------------------|--------------------------------------|
| Air Conditioner | 2 | 4 |
| Color TV | 1 | 3 |
| Refrigerator | 3 | 5 |
|
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, when point $P$ coincides with $C$, the sum is minimized to 6.
Therefore, when $x=-3$, the minimum value of the function $f(x)$ is
Figure 1 6.
|
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 below the x-axis to above the x-axis, to get the graph of $f(f(x)) = |2f(x) - 1|$ (as shown in Figure 3).
Using the same method, we can draw the graph of the function $f(f(f(x))) = |2f(f(x)) - 1|$ (as shown in Figure 4), which intersects the line $y = \frac{x}{2}$ at 8 points in the interval $[0,1]$. Therefore, the original equation has 8 real solutions.
|
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 $f(x) \geqslant m$ always holds, i.e., the maximum value of $m$ is 18.
|
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 reflect the part of the graph below the $x$-axis to above the $x$-axis, to obtain the graph of $f_{2}(x)=\mid f_{1}(x)-2 |$ (as shown in Figure 8). The area of the closed figure formed with the $x$-axis is
$$
\begin{array}{l}
\quad S=S_{\text {and }}-S \\
=\frac{6+2}{2} \times 2-\frac{1}{2} \times 2 \times 1 \\
=7 .
\end{array}
$$
|
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 to Second, Second to Third, Third to Fourth, Fourth to Fifth, and Fifth to First. If School A gives -3 computers to School B, it means School B gives 3 computers to School A. To minimize the total number of computers moved, how should the transfers be arranged?
|
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,
\[
\begin{array}{c}
7 + x_{1} - x_{2} = 11 + x_{2} \\
- x_{3} = 3 + x_{3} - x_{4} = 14 + x_{4} - x_{5} = 15 \div x_{5} \\
- x_{1} = \frac{1}{5}(15 + 7 + 11 + 3 + 14). \\
\therefore x_{2} = x_{1} - 3, x_{3} = x_{1} - 2, \\
x_{4} = x_{1} - 9, x_{5} = x_{1} - 5.
\end{array}
\]
Since the total number of computers transferred should be minimized, we need to find the minimum value of the function
\[
\begin{aligned}
y = & \left|x_{1}\right| + \left|x_{2}\right| + \left|x_{3}\right| + \left|x_{4}\right| + \left|x_{5}\right| \\
= & \left|x_{1}\right| + \left|x_{1} - 3\right| + \left|x_{1} - 2\right| + \left|x_{1} - 9\right| \\
& + \left|x_{1} - 5\right|
\end{aligned}
\]
From the conclusion in Example 1, we know that when $x_{1} = 3$, the function $y_{\text {min }} = 12$. At this point, we have $x_{2} = 0, x_{3} = 1, x_{4} = -6, x_{5} = -2$.
Therefore, Primary School 1 transfers 3 computers to Primary School 2, Primary School 2 transfers 1 computer to Primary School 4, Primary School 5 transfers 6 computers to Primary School 4, and Primary School 1 transfers 2 computers to Primary School 5. This way, the total number of computers transferred is minimized and is 12.
|
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 transferred to minimize the total number of TVs transferred? Find the minimum total number of TVs transferred.
|
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 $x_{3}$ color TVs to $A_{4}$ high school, and $A_{4}$ high school transfers $x_{4}$ color TVs to $A_{1}$ high school. Since there are 40 color TVs in total, with an average of 10 per school, therefore,
$$
\begin{array}{c}
15-x_{1}+x_{4}=10,8-x_{2}+x_{1}=10, \\
5-x_{3}+x_{2}=10,12-x_{4}+x_{3}=10 . \\
\therefore x_{4}=x_{1}-5, x_{1}=x_{2}+2, x_{2}=x_{3}+5, \\
x_{3}=x_{4}-2, x_{3}=\left(x_{1}-5\right)-2=x_{1}-7, x_{2}= \\
\left(x_{1}-7\right)+5=x_{1}-2 .
\end{array}
$$
This problem requires finding the minimum value of $y=\left|x_{1}\right|+\left|x_{2}\right|+\left|x_{3}\right|+$ $\left|x_{4}\right|=\left|x_{1}\right|+|x_{1}-2|+| x_{1}-7 |+$ $\left|x_{1}-5\right|$. Here, $x_{1}$ is an integer satisfying $-8 \leqslant x_{1} \leqslant 15$.
Let $x_{i}=x$.
Consider the function defined on $-8 \quad 15$
$$
y=|x|+|x-2|+|x-7|+|x-5| \text {. }
$$
$|x|+|x-7|$ represents the sum of the distances from $x$ to 0 and 7. When $0 \leq x \leq 7$, $|x|+|x-7|$ takes the minimum value of 7; similarly, when $2 \leqslant x \leqslant 5$, $|x-2|+|x-5|$ takes the minimum value of 3.
Therefore, when $2 \leqslant x \leqslant 5$, $y$ takes the minimum value of 10. In other words, when $x_{1}=2,3,4,5$, $\left|x_{1}\right|+\left|x_{1}-2\right|+$ $\left|x_{1}-7\right|+\left|x_{1}-5\right|$ takes the minimum value of 10.
Thus, the minimum total number of color TVs to be transferred is 10, and the transfer schemes are as follows:
|
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 the quadratic equation with roots $\frac{a-b}{2}, \frac{c-a}{2}$ is
$$
\begin{array}{l}
\left(x-\frac{a-b}{2}\right)\left(x-\frac{c-a}{2}\right)=0 \\
\Leftrightarrow x^{2}+\frac{1}{2}(b-c) x+\frac{1}{4}(a-b)(c-a)=0 .
\end{array}
$$
We have $\Delta=\frac{1}{4}(b-c)^{2}-(a-b)(c-a)=0$.
Therefore, the two roots are equal, i.e., $\frac{a-b}{2}=\frac{c-a}{2}$.
Thus, $\frac{b+c}{a}=2$.
Analysis: Let $x=a-b, y=c-a$. The background knowledge of this problem is: in the identity
$$
\frac{1}{4}(x+y)^{2}-x y=\frac{1}{4}(x-y)^{2}
$$
if the left side is 0, then the right side is also 0.
|
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
$$
\begin{array}{l}
3^{2 x}-\left[10 \times 3^{x+1}\right]+82 \\
+\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-2=0 . \\
\therefore\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}+2\right) \\
\quad \cdot\left(\sqrt{3^{2 x}-\left[10 \times 3^{x+1}\right]+82}-1\right)=0 . \\
\therefore 3^{2 x}-\left[10 \times 3^{x+1}\right]+81=0,
\end{array}
$$
It is only necessary that $3^{2 x}-10 \times 3^{x+1}+81 \leqslant 0$,
which means $\left(3^{x}-3\right)\left(3^{x}-27\right) \leqslant 0 \Rightarrow 3 \leqslant 3^{x} \leqslant 27$.
$$
\therefore 1 \leqslant x \leqslant 3 \text {. }
$$
When $x=1,3$, the original equation holds. When $x=2$, the original equation has no solution. Therefore, the original equation has two solutions.
|
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 $z_{0}=0, z_{1}$ $=1, z_{n+1}-z_{n}=\alpha\left(z_{n}-z_{n-1}\right), \alpha=1+\sqrt{3} \mathrm{i}, n=1,2$, $\cdots$. The number of $z_{n}$ contained within the circle $\mid z \mid=10$ is
|
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
$$
z_{n}=\alpha^{n-1}+\cdots+\alpha+1=\frac{\alpha^{n}-1}{\alpha-1} \text {. }
$$
Since $\alpha=1+\sqrt{3} i=2\left(\cos \frac{\pi}{3}+\sin \frac{\pi}{3}\right)$,
we have, from $\left|z_{n}\right|<10$,
$$
\left|\frac{2^{n}\left(\cos \frac{n \pi}{3}+i \sin \frac{n \pi}{3}\right)-1}{\sqrt{3} \mathrm{i}}\right|<10,
$$
which means $\left|2^{n} \cos \frac{n \pi}{3}-1+\mathrm{i} 2^{n} \sin \frac{n \pi}{3}\right|<10 \sqrt{3}$.
Thus, $2^{2 n}-2^{n+1} \cos \frac{n \pi}{3}+1<300$.
Noting that $\left(2^{n}-1\right)^{2}=2^{2 n}-2^{n+1}+1 \leqslant 2^{2 n}-2^{n+1}$ $\cdot \cos \frac{n \pi}{3}+1 \leqslant 300$, we can conclude that $n \leqslant 4$, meaning there are 5 points $z_{n}$ inside the circle $|z|<10$, namely $z_{0}, z_{1}, z_{2}, z_{3}$, $z_{4}$.
|
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 there is a subset $B \in C$, where subset $B$ has more than 3 elements. Let $a \in B$, since $B \cap (B \setminus \{a\})$ has more than 2 elements, hence $B \setminus \{a\} \notin C$. Then, when replacing $B$ in $C$ with $B \setminus \{a\}$, the new family of subsets still meets the conditions of the problem (since $B \setminus \{a\}$ has one fewer element than $B$). Repeating this process, we can eventually ensure that each subset in the resulting family has no more than 3 elements. Note that the number of subsets in the family does not increase after each replacement, so the number of subsets in $C$ is $\leqslant 175$.
|
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 translation result directly.
|
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 integers $y, f(y)>0$. Thus, when $y \in \mathbf{N}$, $f(y+1)=f(y)+y+2>y+1$, meaning for all positive numbers $t$ greater than 1, $f(t)>t$.
Also from equation $(1), f(-3)=-1, f(-4)=1$.
Next, we prove that when the integer $t \leqslant-4$, $f(t)>0$. Since $t \leqslant$
-4, hence $-(t+2)0 \text {, }
$$
which means $\square$
$$
\begin{array}{l}
f(-5)-f(-4)>0, \\
f(-6)-f(-5)>0, \cdots, \\
f(t+1)-f(t+2)>0, \\
f(t)-f(t+1)>0 .
\end{array}
$$
Adding them up, we get $f(t)-f(-4)>0$,
which means $f(t)>f(-4)=1>0$.
Since $t \leqslant-4$, hence $f(t)>t$.
In summary, the integers that satisfy $f(t)=t$ are only $t=1,-2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) A county is located in a desert area, and people have been engaged in a tenacious struggle against nature for a long time. By the end of 1998, the county's greening rate had reached $30 \%$. Starting from 1999, the following situation will occur every year: $16 \%$ of the original desert area will be greened, while at the same time, due to various reasons, $4 \%$ of the original greened area will be re-sandified.
(1) Let the total area of the county be 1, and the total greened area at the end of 1998 be $a_{1}=\frac{3}{10}$. After $n$ years, the total greened area is $a_{n+1}$. Prove: $a_{n+1} = \frac{4}{5} a_{n} + \frac{4}{25}$.
(2) How many years of effort are needed at least to achieve a county-wide greening rate?
|
(1) Let the current desert area be $b_{1}$, and after $n$ years, the desert area be $b_{n+1}$. Thus, $a_{1}+b_{1}=1, a_{n}+b_{n}=1$.
According to the problem, $a_{n+1}$ consists of two parts: one part is the remaining area of the original oasis $a_{n}$ after being eroded by $\frac{4}{100} a_{n}$, which is $\frac{96}{100} a_{n}$; the other part is the newly greened area $\frac{16}{100} b_{n}$. Therefore,
$$
\begin{array}{l}
a_{n+1}=\frac{96}{100} a_{n}+\frac{16}{100} b_{n} \\
=\frac{96}{100} a_{n}+\frac{16}{100}\left(1-a_{n}\right)=\frac{4}{5} a_{n}+\frac{4}{25} .
\end{array}
$$
(2) From (1), we get
$$
\begin{array}{l}
a_{n+1}-\frac{4}{5}=\frac{4}{5}\left(a_{n}-\frac{4}{5}\right)=\left(\frac{4}{5}\right)^{2}\left(a_{n-1}-\frac{4}{5}\right) \\
=\cdots=\left(\frac{4}{5}\right)^{n}\left(a_{1}-\frac{4}{5}\right)=-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n} .
\end{array}
$$
Therefore, $a_{n+1}=\frac{4}{5}-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n}$.
Note that $a_{n+1} \geqslant 60 \%$, i.e., $\frac{4}{5}-\frac{1}{2} \times\left(\frac{4}{5}\right)^{n}>\frac{3}{5}$. Simplifying, we get $\left(\frac{4}{5}\right)^{n}\frac{1-2 \lg 2}{1-3 \lg 2}>4$. Therefore, it will take at least 5 years to make the county's greening rate exceed $60 \%$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}$ $+b x^{16}+1$, then $a=$ $\qquad$
|
$=1.987$.
Let $p, q$ be the roots of the equation $x^{2}-x-1=0$, then
$$
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^{16} \text{ gives} \\
a p^{17} q^{16}+b p^{16} q^{16}+q^{16}=0 . \\
\therefore a p+b=-q^{16} .
\end{array}
$$
(2) $\times p^{16}$ gives $a q+b=-p^{16}$.
Solving (3) and (4) yields
$$
\begin{aligned}
a & =(p+q)\left(p^{2}+q^{2}\right)\left(p^{4}+q^{4}\right)\left(p^{8}+q^{8}\right) \\
& =987 .
\end{aligned}
$$
|
987
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the incircle $\odot O$ of $\triangle A B C$ touch $B C$ at point $D$, and draw the diameter $D E$ through $D$. Connect $A E$ and extend it to intersect $B C$ at point $F$. If $B F+C D=1998$, then $B F+2 C D=$
|
2. 2997.
As shown in Figure 6, let
$\odot O$ be tangent to $AB$ and
$AC$ at points $M$ and $N$, respectively. Draw
a line $GH \parallel BC$ through
point $E$, intersecting
$AB$ and $AC$ at
points $G$ and $H$, respectively. Then $GH$ is tangent
to $\odot O$ at point $E$, and
$\triangle A G E \backsim \triangle A B F, \triangle A G H \backsim \triangle A B C$.
Let the perimeters of $\triangle A G H$ and $\triangle A B C$ be $2 p^{\prime}$ and $2 p$, respectively. Then
$$
\begin{aligned}
& A G+G E=A G+G M=A M=A N=A H+H N \\
= & A H+H E=p^{\prime} .
\end{aligned}
$$
Thus, $\frac{p^{\prime}}{p}=\frac{2 p^{\prime}}{2 p}=\frac{A G}{A B}=\frac{G E}{B F}=\frac{A G+G E}{A B+B F}$
$$
=\frac{p^{\prime}}{A B+B F} \text {. }
$$
$\therefore p=A B+B F$. Hence $B F=p-A B=C D$.
$$
\therefore B F+2 C D=1998+\frac{1}{2} \times 1998=2997 \text {. }
$$
|
2997
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given three real
numbers $x_{1}, x_{2}, x_{3}$, any one of them plus five times the product of the other two equals 6. The number of such triples $\left(x_{1}, x_{2}\right.$, $x_{3}$ ) is.
|
4.5
From the problem, we have
$$
\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}
$$
(1) - (2) gives $\left(x_{1}-x_{2}\right)\left(5 x_{3}-1\right)=0$.
Thus, $x_{1}=x_{2}$ or $x_{3}=\frac{1}{5}$.
Similarly, $x_{2}=x_{3}$ or $x_{1}=\frac{1}{5}$, $x_{3}=x_{1}$ or $x_{2}=\frac{1}{5}$.
Therefore, $\left(x_{1}, x_{2}, x_{3}\right)$ has 5 sets.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $S$ be a set of numbers composed of some numbers from $1,2,3, \cdots, 50$ (a set of numbers), such that the sum of any two numbers in $S$ cannot be divisible by 7. How many numbers from $1,2,3, \cdots, 50$ can $S$ contain at most (the maximum number of elements in $S$)? Prove your conclusion.
|
Three, divide the 50 numbers $1,2,3, \cdots, 50$ into seven different sets $F_{0}, F_{1}, \cdots, F_{6}$ based on their remainders when divided by 7:
$F_{0}$ consists of $7,14,21, \cdots, 49$ (divisible by 7), $F_{1}$ consists of $1,8,15, \cdots, 50$ (remainder 1 when divided by 7), $\qquad$
$F_{6}$ consists of $6,13,20, \cdots, 48$ (remainder 6 when divided by 7), $S$ can contain at most one number from $F_{0}$, because the sum of any two numbers in $F_{0}$ is divisible by 7.
Furthermore, $S$ can include all numbers from $F_{1}, F_{2}, \cdots, F_{6}$. However, since the sum of any two numbers in $S$ cannot be divisible by 7, $S$ cannot simultaneously include numbers from $F_{1}$ and $F_{6}$, or $F_{2}$ and $F_{5}$, or $F_{3}$ and $F_{4}$.
Therefore, $S$ can contain at most 1 number from $F_{0}$, 8 numbers from $F_{1}$, 7 numbers from $F_{2}$ (or $F_{5}$), and 7 numbers from $F_{3}$ (or $F_{4}$).
Thus, $S$ can contain at most $1+8+7+7=23$ numbers.
|
23
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the function $f(x+1)=\frac{1999^{2 x+1}}{1999^{2 x+1}-1}$. Then the value of the sum $\sum_{i=1}^{4000} f\left(\frac{i}{4001}\right)$ is $\qquad$
|
$$
\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{i}{4001}\right)\right] \\
=\sum_{i=1}^{2000} 1=2000 . \\
2.2000 .
\end{array}
$$
It is easy to see that $f(x)=\frac{1999^{2 x-1}}{1999^{2 x-1}-1}$.
Also, $f(x)+f(1-x)$
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, let $f(x)$ be a function that satisfies the following conditions:
(1) If $x>y$ and $f(x)+x \geqslant w \geqslant f(y)+y$, then there exists a real number $z \in[y, x]$, such that $f(z)=w-z$;
(2) The equation $f(x)=0$ has at least one solution, and among these solutions, there is one that is not greater than all the others;
(3) $f(0)=1$;
(4) $f(-1999) \leqslant 2000$;
(5)
$$
\begin{array}{l}
f(x) f(y) \\
=f(x f(y)+y f(x)+x y) .
\end{array}
$$
Find the value of $f(-1999)$.
|
Let $F(x)=f(x)+x$, then $F(0)=1$.
Let $u$ be the smallest root of $f(x)=0$, then $F(u)=u$.
If $u0$
For any real number $x$, by (5) we have
$$
\begin{aligned}
0 & =f(x) f(u)=f(x f(u)+u f(x)+x u) \\
& =f(u f(x)+u u) .
\end{aligned}
$$
$\therefore u f(x)+x u$ is a root of $f(x)=0$.
$\because u$ is the smallest root of $f(x)=0$,
$$
\begin{array}{l}
\therefore u f(x)+x u \geqslant u . \\
\text { Thus, } f(x)+x \geqslant 1,
\end{array}
$$
i.e., $f(x) \geqslant 1-x$.
$$
\begin{array}{l}
\therefore f(-1999) \geqslant 2000 . \\
\text { Also, } \because f(-1999) \leqslant 2000, \\
\therefore f(-1999)=2000 .
\end{array}
$$
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One TV station has $n$ ad breaks in a day, during which a total of $m$ ads were broadcast. In the first ad break, one ad and $\frac{1}{8}$ of the remaining $(m-1)$ ads were broadcast. In the second ad break, 2 ads and $\frac{1}{8}$ of the remaining ads were broadcast. This pattern continued for each subsequent ad break, with the last ad break, the $n$-th one, broadcasting the remaining ads $(n>1)$. How many ad breaks were there that day? And what is the number of ads $m$?
|
Let's denote the number of ads left to be broadcast after $k$ broadcasts as $a_{k}$. Then, the number of ads broadcast at the $k$-th time is:
$$
k+\frac{1}{8}\left(a_{k-1}-k\right)=\frac{1}{8} a_{k-1}+\frac{7}{8} k .
$$
Thus, $a_{k}=a_{k-1}-\left(\frac{1}{8} a_{k-1}+\frac{7}{8} k\right)=\frac{7}{8} a_{k-1}-\frac{7}{8} k$.
$$
\therefore a_{k-1}=k+\frac{8}{7} a_{k} \text {. }
$$
From this recursive formula, we get
$$
\begin{aligned}
m= & 1+\frac{8}{7} a_{1}=1+\frac{8}{7}\left(2+\frac{8}{7} a_{2}\right) \\
= & 1+2 \times \frac{8}{7}+\left(\frac{8}{7}\right)^{2} \cdot a_{2} \\
= & 1+2 \times \frac{8}{7}+\left(\frac{8}{7}\right)^{2} \cdot\left(3+\frac{8}{7} a_{3}\right) \\
= & 1+2 \times \frac{8}{7}+3 \times\left(\frac{8}{7}\right)^{2}+\left(\frac{8}{7}\right)^{3} \times a_{3} \\
= & \cdots \cdots \\
= & 1+2 \times \frac{8}{7}+3 \times\left(\frac{8}{7}\right)^{2}+\cdots+n \times\left(\frac{8}{7}\right)^{n-1} \\
& +\left(\frac{8}{7}\right)^{n} \cdot a_{n} .
\end{aligned}
$$
By the definition of $a_{k}$, we know that $a_{n}=0$. Therefore,
$$
m=\sum_{i=1}^{n} i\left(\frac{8}{7}\right)^{i-1} \text {. }
$$
$$
\therefore \frac{8}{7} m=\sum_{i=1}^{n} i\left(\frac{8}{7}\right)^{i} \text {. }
$$
Subtracting (1) from (2) gives
$$
\begin{array}{l}
-\frac{1}{7} m= 1+\frac{8}{7}+\left(\frac{8}{7}\right)^{2}+\cdots+\left(\frac{8}{7}\right)^{n-1} \\
-n \cdot\left(\frac{8}{7}\right)^{n} \\
= \frac{\left(\frac{8}{7}\right)^{n}-1}{\frac{8}{7}-1}-n \cdot\left(\frac{8}{7}\right)^{n} . \\
\therefore m=49+(n-7) \cdot \frac{8^{n}}{7^{n-1}} .
\end{array}
$$
For $n>1$, it is clear that
$$
|n-7|<7^{n-1},\left(8^{n}, 7^{n-1}\right)=1 \text {. }
$$
Since $m$ is an integer, it must be that $n-7=0$, i.e., $n=7$. Thus, $m=49$.
Therefore, there are 7 broadcasts, and a total of 1,9 ads are broadcast.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, it is known that the natural number $N$ has 20 positive integer factors (including 1 and itself), which are sequentially denoted as $d_{1}, d_{2}, d_{3}, \cdots, d_{21}$, and the factor with index $d_{8}$ is $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$. Find the natural number $N$.
|
Three, because $\left(d_{1}+d_{4}+d_{6}\right)\left(d_{4}+d_{8}+d_{13}\right)$ is a factor of $N$, so, $d_{1}+d_{4}+d_{6}$ and $d_{4}+d_{8}+d_{13}$ are factors of $N$. Therefore,
$$
\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}\right)\left(d_{4}+d_{8}+d_{13}\right) \\
\geqslant d_{7} \cdot d_{14}=N . \\
\because d_{d 8} \leqslant N, \\
\therefore d_{d 8}=N . \\
\therefore d_{8}=20 . \text{ At this time } d_{1}+d_{4}+d_{6}=d_{7}, d_{4}+d_{8}+ \\
d_{i 3}= d_{14} .
\end{array}
$$
From $d_{8}=20$ we know that $N$ contains the six positive integer factors $1,2,4,5,10,20$, so $N$ must contain the two prime factors 2 and 5.
Since $N$ has 20 positive factors, $20=2 \times 2 \times 5=2 \times 10=4 \times 5$, hence $N$ can be set as $2^{4} \times 5 \times p$ (where $p$ is a prime number not equal to 2 and 5), $2^{9} \times 5$, $2^{3} \times 5^{4}$, or $2^{4} \times 5^{3}$.
(1) When $N=2^{4} \times 5 \times p$,
(1) When $p=3$, $d_{1}, d_{2}, \cdots, d_{8}$ are sequentially $1,2,3,4$, $5,6,8,10$. At this time $d_{8}=10$, which contradicts $d_{8}=20$.
(2) When $p=7$, $d_{1}, d_{2}, \cdots, d_{8}$ are sequentially $1,2,4,5$, $7,8,10,14$. At this time $d_{8}=14$, which contradicts $d_{8}=20$.
(3) When $7<p<23$, $p \geqslant 11$, $d_{1}, d_{2}, \cdots, d_{8}$ are sequentially $1,2,4,5,8,10, p, 16$ or $1,2,4,5,8,10,16, p$, which contradicts $d_{8}=20$.
(4) When $p \geqslant 23$, the positive factors of $N$ are $1,2,4,5,8,10$,
$$
\begin{array}{l}
16,20,40,80, p, 2 p, 4 p, 5 p, 8 p, \cdots . \\
\quad \therefore d_{4}=5, d_{8}=20, d_{13}=4 p, d_{14}=5 p . \\
\therefore 5+20+4 p=5 p .
\end{array}
$$
Thus, $p=25$, which is not a prime number.
Therefore, $N \neq 2^{4} \times 5 p$.
(2) When $N=2^{9} \times 5$, $d_{4}=5, d_{8}=20, d_{13}=128$, $d_{14}=160$. It does not satisfy $d_{4}+d_{8}+d_{13}=d_{14}$.
(3) When $N=2^{3} \times 5^{4}$, $d_{8}=25$, which contradicts $d_{8}=20$.
(4) When $N=2^{4} \times 5^{3}$, $d_{1}=1, d_{4}=5, d_{6}=10$, $d_{7}=16, d_{8}=20, d_{13}=100, d_{14}=125$. Clearly, it satisfies $d_{1}+d_{4}+d_{6}=d_{7}, d_{4}+d_{8}+d_{13}=d_{14}$.
$$
\therefore N=2000 \text{. }
$$
Therefore, the natural number $N$ is 2000.
|
2000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (Full marks 50 points) Let $S$ be a subset of $\{1,2, \cdots, 50\}$ with the following property: the sum of any two distinct elements in $S$ is not divisible by 7. What is the maximum number of elements that $S$ can have?
|
Divide $\{1,2, \cdots, 50\}$ into 7 classes according to modulo 7:
$$
\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\} .
\end{array}
$$
We will prove that $S=k_{1} \cup k_{2} \cup k_{3} \cup\{7\}$ is the largest set that meets the requirements.
First, for $a, b \in S, a \neq b$, there are 3 possibilities:
(1) $a, b \in k_{i}(1 \leqslant i \leqslant 3)$, then
$$
a+b \equiv 2 i(\bmod 7),
$$
which means $a+b$ cannot be divisible by 7.
(2) $a \in k_{i}, b \in k_{j}(1 \leqslant i \neq j \leqslant 3)$, then
$$
a+b \equiv i+j(\bmod 7),
$$
which means $a+b$ cannot be divisible by 7.
(3) $a \in k_{i}, b=7(1 \leqslant i \leqslant 3)$, then
$$
a+b=i(\bmod 7),
$$
which means $a+b$ cannot be divisible by 7.
In summary, the sum of any two elements in $S$ cannot be divisible by 7.
Next, we prove that if an element $c$ is added to $S$, there must exist an element in $S$ whose sum with $c$ is divisible by 7.
The added $c$ has 4 possibilities:
(1) $c \in k_{4}$, then the sum of $c$ and an element in $k_{3}$ can be divisible by 7.
(2) $c \in k_{5}$, then the sum of $c$ and an element in $k_{2}$ can be divisible by 7.
(3) $c \in k_{6}$, then the sum of $c$ and an element in $k_{1}$ can be divisible by 7.
(4) $c \in k_{0}$, then the sum of $c$ and 7 can be divisible by 7.
In summary, no more elements can be added to $S$. Therefore, the maximum number of elements in $S$ is
$$
|S|=\left|k_{1}\right|+\left|k_{2}\right|+\left|k_{3}\right|+1=23 .
$$
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 If $f(x)=\left(2 x^{5}+2 x^{4}-53 x^{3}-\right.$ $57 x+54)^{1998}$, then $f\left(\frac{\sqrt{1}-\frac{1}{2}-i}{2}\right)=$ $\qquad$ .
|
Let $\frac{\sqrt{111}-1}{2}=x$, then we have
$$
\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 {. }
$$
In the required expression, there are 100 $f(1)$ terms, and the remaining $f(x)$ and $f\left(\frac{1}{x}\right)$ appear in pairs, totaling $\frac{1}{2}\left(100^{2}-\right.$ 100 ) pairs.
Therefore, the required sum is
$$
100 f(1)+\frac{1}{2}\left(100^{2}-100\right) \times 2=10000 .
$$
|
10000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, after expansion and combining like terms, the sum of the coefficients of the odd powers of $x$ is
保留源文本的换行和格式，直接输出翻译结果如下：
```
Example 3 The polynomial $\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}-\right.$ $3 x-3)^{2001}$, after expansion and combining like terms, the sum of the coefficients of the odd powers of $x$ is
```
|
Solution: Let $f(x)=\left(x^{2}+2 x+2\right)^{2001}+\left(x^{2}\right.$
$$
\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_{0}-a_{1}+a_{2}-\cdots-a_{4001}+a_{4002} \\
=f(-1)=2 .
\end{array}
$$
Subtracting the two equations, we get
$$
\begin{array}{l}
2\left(a_{1}+a_{3}+\cdots+a_{4001}\right)=-2 . \\
\therefore a_{1}+a_{3}+\cdots+a_{4001}=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given $P(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+$ $a_{3} x^{2}+a_{4} x+a_{5}$, and when $k=1,2,3,4$, $P(k)$ $=k \times 1$ 997. Then $P(10)-P(-5)=$ $\qquad$
|
Solution: Let $Q(x)=P(x)-1997 x$, then when $k=$ $1,2,3,4$, $Q(k)=P(k)-1997 k=0$.
Therefore, $1,2,3,4$ are roots of $Q(x)=0$.
Thus, we can set $Q(x)=(x-1)(x-2)(x-$ $3)(x-4)(x-r)$, then
$$
\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 \times(-5) \\
= & (-6) \times(-7) \times(-8) \times(-9) \\
\times & (-5-r)+1997 \times(-5) . \\
\therefore & P(10)-P(-5) \\
= & 9 \times 8 \times 7 \times 6 \times 5+1997 \times 15 \\
= & 75315 .
\end{aligned}
$$
|
75315
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the unit digit of the sum $1^{2}+2^{2}+3^{2}+4^{2}+\cdots+1994^{2}$.
|
Solution: Since this problem only requires the unit digit of the sum, we only need to consider the unit digit of each number. Thus, the original problem simplifies to finding the unit digit of
$$
\underbrace{i^{2}+2^{2}+3^{2}+4^{2}+\cdots+9^{2}}_{\text {199 groups }}+1^{2}+2^{2}+3^{2}+4^{2}
$$
The unit digits follow a periodic pattern:
$$
1,4,9,6,5,6,9,4,1 \text {, }
$$
The unit digit of the sum of each group is 5, so the unit digit of the sum of 199 groups is also 5. Considering the unit digit sum of the remaining four numbers is 0, the unit digit of the required sum is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Let
$$
\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}\right)^{2}}=\frac{n+1}{n}-\frac{1}{n+1} \\
=1+\frac{1}{n}-\frac{1}{n+1}, \\
\therefore S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\cdots \\
\quad \quad+\left(1+\frac{1}{1997}-\frac{1}{1998}\right) \\
=1998-\frac{1}{1998} .
\end{array}
$$
Therefore, the integer closest to $S$ is 1998, choose (B).
$$
|
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, $\sqrt[3]{k} \times \sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$
$$
\begin{array}{l}
=\sqrt[3]{k}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) . \\
\because k \neq 0, \\
\therefore \sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} .
\end{array}
$$
From the given, we have $x>0, y>0, z>0$, hence $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
|
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 $x_{1}+x_{2}=-1, x_{1} x_{2}=-3$.
$\therefore 4 a^{-4}+b^{4}=\left(-\frac{2}{a^{2}}\right)+\left(b^{2}\right)^{2}$ $=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=7$.
|
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.
|
$$
\begin{array}{l}
\text { Solution: } x^{3}-y^{3}-z^{3}-3 x y z \\
=x^{3}-(y+z)^{3}-3 x y z+3 y^{2} z+3 y z^{2} \\
=(x-y-z)\left(x^{2}+x y+x z+y^{2}+2 y z\right. \\
\left.+z^{2}\right)-3 y z(x-y-z) \\
=(x-y-z)\left(x^{2}+y^{2}+z^{2}+x y-y z+x z\right) \\
=\frac{1}{2}(x-y-z)\left[(x+y)^{2}+(y-z)^{2}\right. \\
\left.+(z-x)^{2}\right] \text {. } \\
\end{array}
$$
From the given, we have $x-y-z=0$,
$$
\therefore x=y+z \text {. }
$$
Substituting $x^{2}=2(y+z)$, we get $x=2, y+z=2$. Given $y \geqslant 1, z \geqslant 1$, we have $y=z=1$.
$$
\therefore x y+y z+z x=5 \text {. }
$$
|
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
$$
\begin{array}{l}
|A \cap B|=|R|-|\overline{A \cap B}| \\
=|R|-|\bar{A} \cup \bar{B}| \\
=|R|-|\bar{A}|-|\bar{B}|+|\bar{A} \cap \bar{B}| .
\end{array}
$$
$(|E|$ represents the number of elements in set $E$)
And $|R|=6^{5}=7776$.
$|\bar{A}|$ is the number of five-digit numbers formed by one or two digits from $1, 2, 3, 4, 5, 6$. Therefore,
$$
|\bar{A}|=6+\mathrm{C}_{6}^{2}\left(\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{2}+\mathrm{C}_{5}^{3}+\mathrm{C}_{5}^{4}\right)
$$
$$
=6+15 \times 30=456 \text {. }
$$
$|\bar{A} \cap \bar{B}|$ is the number of five-digit numbers formed by one or two digits from $1, 2, 3, 4, 5, 6$ and with $1, 6$ adjacent. Therefore, it is the number of five-digit numbers formed by $1, 6$, so,
$$
|\bar{A} \cap \bar{B}|=C_{5}^{1}+C_{5}^{2}+C_{5}^{3}+C_{5}^{4}=30 .
$$
$|\bar{B}|$ is the number of five-digit numbers with $1, 6$ adjacent. To find this number, we use a recursive method. Let $b_{n}$ be the number of $n$-digit numbers with $1, 6$ adjacent, and $c_{n}$ be the number of $n$-digit numbers with the first digit being 1 or 6 and $1, 6$ adjacent. Since all are $n$-digit numbers formed by $1, 2, 3, 4, 5, 6$, if the first digit is not $1, 6$, it must be one of $2, 3, 4, 5$. If the first digit is one of $2, 3, 4, 5$, then the remaining $n-1$ digits must contain $1, 6$ adjacent, so,
$$
b_{n}=2 c_{n}+4 b_{n-1} \text {. }
$$
This alone is not enough for recursion, so we need to find a recursive formula for $c_{n}$. $c_{n}$ includes $n$-digit numbers with the first digit being 1 or 6 and $1, 6$ adjacent, which can be in three cases:
(i) The second digit is 6 or 1, in which case the first two digits are already 1, 6 adjacent, and the remaining $n-2$ digits can be any;
(ii) The first digit is still 1 or 6, and from the second digit, the $n-1$ digits must contain 1, 6 adjacent;
(iii) The second digit is one of $2, 3, 4, 5$, then from the third digit, the $n-2$ digits must contain 1, 6 adjacent. So,
$$
\begin{array}{l}
c_{n}=6^{n-2}+c_{n-1}+4 b_{n-2} . \\
\text { Substituting equation (2) into equation (1) gives } \\
b_{n}=2 \times 5^{n-2}+2 c_{n-1}+8 b_{n-2}+4 b_{n-1} \\
=2 \times 6^{n-2}+4 b_{n-2}+5 b_{n-1} .
\end{array}
$$
This is the recursive formula for $b_{n}$. Clearly, $b_{1}=0, b_{2}=2$. Substituting into equation (3) gives
$$
\begin{array}{l}
b_{3}=2 \times 6+5 \times 2=22, \\
b_{4}=2 \times 6^{2}+4 \times 2+5 \times 22=190, \\
b_{5}=2 \times 6^{3}+4 \times 22+5 \times 190 \\
=1470=|\bar{B}| . \\
\therefore|A \cap B|=7776-456-1470+30 \\
=5880 .
\end{array}
$$
Thus, the number of five-digit numbers is 5880.
Solution 2: Let $x_{n}$ be the number of $n$-digit numbers formed by $1, 2, 3, 4, 5, 6$ with no 1, 6 adjacent and the first digit being 1 or 6, and let $y_{n}$ be the number of $n$-digit numbers formed by $1, 2, 3, 4, 5, 6$ with no 1, 6 adjacent and the first digit being 2, 3, 4, or 5.
Adding a digit in front of an $n$-digit number makes it an $(n+1)$-digit number. To ensure no 1, 6 adjacent, we just need to make sure that if the first digit is 1 or 6, the new digit added cannot be 6 or 1, so,
$$
x_{n+1}=2 y_{n}+x_{n} \text {. }
$$
This indicates that the number of $(n+1)$-digit numbers with no 1, 6 adjacent and the first digit being 1 or 6 can be obtained by adding 1 or 6 in front of the $n$-digit numbers with the first digit being 2, 3, 4, or 5, or by adding 1 or 6 in front of the $n$-digit numbers with the first digit being 1 (or 6).
$$
y_{n+1}=4\left(y_{n}+x_{n}\right) \text {. }
$$
This indicates that the number of $(n+1)$-digit numbers with no 1, 6 adjacent and the first digit not being 1 or 6 can be obtained by adding 2, 3, 4, or 5 in front of the $n$-digit numbers with the first digit being 1, 2, 3, 4, 5, or 6 but not containing 1, 6 adjacent.
$$
\begin{array}{l}
\text { Clearly } x_{1}=2, y_{1}=4 . \text { Substituting into equations (4) and (5) gives } \\
x_{2}=2 \times 4+2=10, \\
y_{2}=4 \times(2+4)=24, \\
x_{3}=2 \times 24+10=58, \\
y_{3}=4 \times(10+24)=136, \\
x_{4}=2 \times 136+58=330,
\end{array}
$$
$$
\begin{array}{l}
y_{4}=4 \times(136+58)=776, \\
x_{5}=2 \times 776+330=1882, \\
y_{5}=4 \times(776+330)=4424 . \\
\therefore x_{5}+y_{5}=6306 .
\end{array}
$$
However, $x_{5}+y_{5}$ includes the five-digit numbers formed by one or two digits, the number of which is
$$
6+\left(\mathrm{C}_{6}^{2}-1\right)\left(\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{2}+\mathrm{C}_{5}^{3}+\mathrm{C}_{5}^{4}\right)=426 .
$$
Here, $\left(\mathrm{C}_{6}^{2}-1\right)$ subtracts 1 to exclude the five-digit numbers formed by 1, 6, as they are already 1, 6 adjacent and thus not included in $x_{5}+y_{5}$. Therefore, the total number of five-digit numbers is
6306 - 426 = 5880.
|
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 questions with the same answers. Find the minimum possible value of $n$.
(Li Chengzhang, provided)
|
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
$$
\begin{array}{l}
\{(1, h, i, j, k),(2, h, i, j, k),(3, h, i, j, k), \\
(4, h, i, j, k)\}, h, i, j, k=1,2,3,4,
\end{array}
$$
This yields 256 quadruples. Since $2000=256 \times 7+208$, by the pigeonhole principle, there are 8 test papers with answers belonging to the same quadruple. After removing these 8 test papers, there are still 8 test papers with answers belonging to the same quadruple among the remaining 1992 test papers; after removing these 8 test papers, there are still 8 test papers with answers belonging to the same quadruple among the remaining 1984 test papers; after removing these 8 test papers, together with the previously removed test papers, there are 24 test papers. In these 24 test papers, any 4 of them have at least two with answers belonging to the same quadruple, which of course does not meet the requirements of the problem. Therefore, the minimum value of $n \geqslant 25$.
$$
\begin{array}{l}
\text { On the other hand, let } \\
S=\{(g, h, i, j, k) \mid g+h+i+j+k \equiv 0(\bmod 4), \\
g, h, i, j, k \in\{1,2,3,4\}\} .
\end{array}
$$
- Then $|S|=256$, and any two answers in $S$ have at most 3 questions the same. Removing 6 elements from $S$, when each of the remaining 250 answers is chosen by exactly 8 people, a total of 2000 test papers are obtained, and in any 25 of these answers, there are always 4 that are different, since they are all in $S$, of course, they meet the requirements of the problem. This shows that when $n=25$, the requirements of the problem can be met.
In summary, the minimum possible value of $n$ is 25.
|
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 $A_{k}$ such that $M \subset A_{k}$.
Find the minimum value of the number of such subsets $n$.
|
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 $a$. On the other hand, the three-element subsets of $S$ are contained in the sets of the family, so,
$$
15 r(a) \geqslant 91, \quad r(a) \geqslant 7.
$$
By calculating the total number of elements in the sets of the family $\mathscr{A}$ in two ways, we see that
$$
n \times 7=\sum_{a \in S} r(a) \geqslant 15 \times 7, \quad n \geqslant 15.
$$
Next, we construct a specific family of sets $S=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ that satisfies the conditions of the problem. First, we label the 15 elements of $S=\{1,2, \cdots, 15\}$ in a clockwise order on the 15 equally spaced points on a circle (as shown in Figure 1).
We define $A_{1}=\{1,2,4,5,6,11,13\}$, and let $A_{j}$ (for $j=2,3, \cdots, 15$) be the 7-element subset of $S$ obtained by rotating $A_{1}$ clockwise by $j-1$ arc segments. Here, an arc segment is the arc between two adjacent points on the circle divided into 15 equal parts. Figure 2 shows the diagram of $A_{1}$. The numbers marked on the inner side of the arc segments in the figure represent the number of arc segments contained in each arc segment.
We now verify that such a $\mathscr{A}=\left\{A_{1}, A_{2}, \cdots, A_{15}\right\}$ satisfies the conditions of the problem.
(i) Clearly, each set in $\mathscr{A}$ consists of exactly 7 elements of $S$.
(ii) If $1 \leqslant j-i \leqslant 7$, then rotating $A_{i}$ clockwise by $j-i$ arc segments results in $A_{j}$. Analyzing the diagram of $A_{1}$ (Figure 2), we see that there are exactly three pairs of elements for each of the distances
$$
1,2,3,4,5,6,7
$$
in the clockwise direction. Each $A_{i}$ has the same property. Rotating $A_{i}$ clockwise by $j-i$ arc segments $(1 \leqslant j-i \leqslant 7)$, exactly three elements of $A_{i}$ will move to positions where the elements are still in $A_{i}$. Therefore,
$$
\left|A_{i} \cap A_{j}\right|=3.
$$
If $8 \leqslant j-i \leqslant 14$, then rotating $A_{j}$ clockwise by $15-(j-i)$ arc segments results in $A_{i}$. Similarly, we can conclude that
$$
\left|A_{i} \cap A_{j}\right|=3.
$$
(iii) Let $M=\{u, v, w\}$ be any three-element subset of $S$. Without loss of generality, assume $u, v, w$ are arranged in a clockwise order on the circle. Consider the number of arc segments between $u$ and $v$, $v$ and $w$, and $w$ and $u$ in the clockwise direction, and denote the pair of the smaller two numbers of these arc segments as $(a, b)$ (in clockwise order). We will classify the three-element subsets of $S$ based on these pairs. Clearly,
$$
1 \leqslant a, b \leqslant 7,
$$
and except for the pair $(5,5)$, all other pairs $(a, b)$ satisfy
$$
a+b \leqslant 9.
$$
Each pair $(a, b)$ type appears exactly once in $A_{1}$:
$$
\begin{array}{l}
(1,1),(1,2),(2,1),(1,3),(3,1),(1,4),(4,1), \\
(1,5),(5,1),(1,6),(6,1),(1,7),(7,1) ; \\
(2,2),(2,3),(3,2),(2,4),(4,2),(2,5),(5,2), \\
(2,6),(6,2),(2,7),(7,2) ; \\
(3,3),(3,4),(4,3),(3,5),(5,3),(3,6,),(6,3) ; \\
(4,4),(4,5),(5,4) ; \\
(5,5) ;
\end{array}
$$
Accordingly, any three-element subset $M$ of $S$ is contained in at least one set of the family $\mathscr{A}$ (in fact, exactly one set of the family $\mathscr{A}$).
|
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 $x \neq 1, y \neq 1, x+y+\frac{1}{x}+\frac{1}{y}=4$.
Let $x+\frac{1}{x}=e, y+\frac{1}{y}=f$, then $x y+\frac{y}{x}+\frac{1}{x y}+\frac{x}{y}=e f$.
$\because$ When $t>0$, $t+\frac{1}{t} \geqslant 2$;
When $t<0, y<0$, then
$$
f \leqslant-2, e=4-f \geqslant 6, e f \leqslant-12 \text{. }
$$
When and only when $y=-1, x=3 \pm 2 \sqrt{2}$, the equality holds. Specifically, when $a=3+2 \sqrt{2}, b=1, c=-1, d=-(3+2 \sqrt{2})$, the equality holds.
$$
\therefore(e f)_{\max }=-12 \text{. }
$$
|
-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 10 hours to pump out all the oil; if 7 pumps work together, it takes 8 hours to pump out all the oil. To pump out the oil within 3 hours, what is the minimum number of pumps needed to work together?
|
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
$$
\left\{\begin{array}{l}
50 x = y - 10 z, \\
56 x = y - 8 z, \\
3 n x \geqslant y - 3 z .
\end{array}\right.
$$
From (1) and (2), we solve for $y = 80 x$ and $z = 3 x$. Substituting into (3), we get $3 n x \geqslant 71 x$, hence $n \geqslant \frac{71}{3}$. Since $n$ is an integer, the minimum value of $n$ is 24.
|
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" which seems to be a typo or an incorrect character. The correct term should be "reasoning" or "justification."
|
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 of these two numbers. The exponent of 5 in the prime factorization is (where $[x]$ denotes the greatest integer less than or equal to $x$)
$$
\begin{aligned}
\alpha_{3} & =\left[\frac{1999}{5}\right]+\left[\frac{1999}{5^{2}}\right]+\left[\frac{1999}{5^{3}}\right]+\left[\frac{1999}{5^{4}}\right]+\cdots \\
& =399+79+15+3=496 .
\end{aligned}
$$
Similarly, $\alpha_{1}=1990>\alpha_{3}$ can be calculated, so 1999! has 496 trailing zeros.
|
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 $a=\alpha-2$,
$$
\begin{array}{l}
-3=\beta+1-2 \alpha, \\
b=1+\alpha-2 \beta, \\
3=1+\beta .
\end{array}
$$
Solving, we get $\alpha=3, \beta=2, a=1, b=0$.
$$
\therefore a b=0 \text {. }
$$
|
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.
$$
\begin{array}{l}
\because \frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}} \\
=\frac{\sin A \cdot \sin B \cdot \cos C}{\sin C(\sin B \cdot \cos A+\cos B \cdot \sin A)} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin C \cdot \sin (A+B)} \\
=\cos C \cdot \frac{\sin A \cdot \sin B}{\sin ^{2} C} \\
=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^{2}} \text { (by the Law of Sines and Cosines) } \\
\quad=\frac{m-1}{2}=999 . \\
\therefore m=1999 .
\end{array}
$$
|
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. $\quad<n+2-\sqrt{n^{2}+n+1}<2$.
Therefore, the integer parts of $x_{n}$ and $x_{n}^{\prime}$ are $2 n+2$ and $1$ respectively.
Hence $\alpha_{n}=2 n+2, \beta_{n}=1$.
$$
\begin{array}{l}
\therefore \frac{\alpha_{1}}{\beta_{1}}+\frac{\alpha_{2}}{\beta_{2}}+\cdots+\frac{\alpha_{99}}{\beta_{99}} \\
=2(2+3+\cdots+100) \\
=10098 .
\end{array}
$$
|
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 pyramid (not necessarily reaching the apex), how many items should be placed on each side of the bottom layer at least?
(1997, Fujian Province Mathematics Competition)
|
Solution: The total number of items in this pile of goods is
$$
\begin{array}{l}
30 \times 15 + 29 \times 14 + 28 \times 13 + \cdots + 16 \times 1 \\
= (15 + 15) \times 15 + (15 + 14) \times 14 + \cdots \\
+ (15 + 1) \times 1 \\
= 15 \times (15 + 14 + \cdots + 1) + 15^2 + 14^2 \\
+ \cdots + 1^2 \\
= 3040 .
\end{array}
$$
Suppose after changing the stacking method, at least $x$ items are stacked on each side of the bottom layer, then we have
$$
\left\{\begin{array}{l}
x^2 + (x-1)^2 + \cdots + 1^2 \geqslant 3040, \\
(x-1)^2 + (x-2)^2 + \cdots + 1^2 \leqslant 3040 .
\end{array}\right.
$$
That is, to solve the system of inequalities:
$$
\left\{\begin{array}{l}
x(x+1)(2x+1) \geqslant 18240, \\
(x-1)x(2x-1) \leqslant 18240 .
\end{array}\right.
$$
The left side of the above two inequalities is approximately $2x^3$. From $2x^3 \approx 18240$, we have $x \approx 21$. Substituting $x=21$, both inequalities are satisfied.
When $x=20$, the left side of (1) is 17220, so it does not satisfy;
When $x=22$, the left side of (2) is 19866, which also does not satisfy.
In summary, $x=21$ is the solution, meaning at least 21 items should be placed on each side of the bottom layer.
|
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 the arable land can decrease on average each year.
|
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
$$
\begin{array}{l}
\frac{M \times(1+22 \%) \times\left(10^{4}-10 x\right)}{p \times(1+1 \%)^{10}} \\
\geqslant \frac{M \times 10^{4}}{p} \times(1+10 \%) .
\end{array}
$$
Simplifying, we get
$$
\begin{aligned}
x \leqslant & 10^{3} \times\left[1-\frac{1.1 \times(1+0.01)^{10}}{1.22}\right] \\
= & 10^{3} \times\left[1-\frac{1.1}{1.22}\left(1+\mathrm{C}_{10}^{1} \times 0.01\right.\right. \\
& \left.\left.+\mathrm{C}_{10}^{2} \times 0.01^{2}+\cdots\right)\right] \\
\approx & 10^{3} \times\left[1-\frac{1.1}{1.22} \times 1.1045\right] \approx 4.1 .
\end{aligned}
$$
Therefore, the maximum average annual reduction of arable land in hectares is 4.
|
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 $AB$, $\cdots \cdots$ each move along a side parallel to another side counts as one move. Then, at most $n$ moves, point $D$ can return to its original starting point for the first time. What is the value of $n$?
|
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
$$
\begin{array}{l}
\frac{A D}{B D}=\frac{A E}{E C} . \\
=\frac{B F}{F C}=\frac{B G}{A G} \\
=\frac{C H}{A H}=\frac{C K}{B K}=\frac{A M}{B M} .
\end{array}
$$
By the theorem of the sum of ratios, we get
$$
\frac{A D+B D}{B D}=\frac{A M+B M}{B M},
$$
which means
$$
\frac{A B}{B D}=\frac{A B}{B M} \text { . }
$$
Therefore, point $D$ coincides with point $M$, so $n=6$. Of course, if we take the midpoint of any side as the starting point, it is obvious that it only takes 3 times to return to the original starting point for the first time.
|
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 $f(3 k) \geqslant k+1$,
$$
\begin{aligned}
f(9999) & \geqslant f(9999-3 k)+f(3 k) \\
& \geqslant 3333-k+k+1>3333 .
\end{aligned}
$$
- This contradicts the problem statement, so $f(3 k)=k$.
Thus, $1982=f(3 \times 1982)$
$$
\geqslant 3 f(1982) \text {. }
$$
That is, $f(1982) \leqslant \frac{1982}{3}<661$.
$$
\begin{array}{l}
\text { Also, } f(1982) \geqslant f(3 \times 660)+f(2)=660, \\
\therefore f(1982)=660 .
\end{array}
$$
|
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
$$
F_{n_{6}}\left(F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)\right)=1 \text {, }
$$
Prove your conclusion.
|
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 first do not restrict the range of $r$, and only consider the non-negative integer decompositions of $q+r=i(i=1,2, \cdots, 2 k)$ and their corresponding $a$ values, obtaining Table 1.
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$q+r$ & \multicolumn{5}{|c|}{ Values of natural number $a$ } \\
\hline $2 k$ & $2 k$ & $n+(2 k-1)$ & $2 n+(2 k-2)$ & $\cdots$ & $(2 k-1) n+1$ & $2 k n$ \\
\hline $2 k-1$ & $2 k-1$ & $n+(2 k-2)$ & $2 n+(2 k-3)$ & $\cdots$ & $(2 k-1) n$ & \\
\hline $2 k-2$ & $2 k-2$ & $n+(2 k-3)$ & & $\cdots$ & & \\
\hline $2 k-3$ & $2 k-3$ & & & & & \\
\hline & & & & & & \\
\hline & & & $2 n$ & & & \\
\hline 1 & 1 & $n$ & & & & \\
\hline
\end{tabular}
It is easy to see that the values of $a$ in Table 1 have three important properties:
(1) The numbers in each column increase continuously from bottom to top;
(2) The $i$-th column consists of numbers of the form $(i+1) n+r$;
(3) From the second column onwards, the lowest number in each column is a multiple of $n$.
We call the numbers $q n+r$ in each column of Table 1 that make $r \geqslant n$ "pseudo" numbers. It is easy to see that after removing all the "pseudo" numbers from Table 1, the corresponding $a$ values in the new table are all different. Therefore, for a fixed $n$, the necessary condition for the $a$ values in the new table to cover all natural numbers from 1 to $B(n)$ is that the top number in the $(i+1)$-th column must be $n$ less than 2 (this is because the last number in the $i \div 2$-th column is $(i+1) n$, so there is no number of the form $i n+(n-i)$ in the table), at this time we should have
$$
(i+1)+(n-2)=2 k,
$$
i.e., the maximum value of $a$ is
$$
\begin{array}{l}
B(n)=(2 k+2-n) n+(n-2) \\
=-\left(n-\frac{2 k+3}{2}\right)^{2}+k^{2}+3 k+\frac{1}{4} \\
\leqslant k^{2}+3 k .
\end{array}
$$
Equality holds if and only if $n=k+1$ or $n=k+2$. At this time, Table 2 is given, where $n=k+1$ or $n=k+2$.
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$q+r$ & \multicolumn{2}{|c|}{\begin{tabular}{l}
$(2 k+1-n) n+$ \\
$(n-1)$
\end{tabular}} & \begin{tabular}{c}
$(2 k+2-n) n+$ \\
$(n-2)$
\end{tabular} \\
\hline $2 k$ & & & & \begin{tabular}{c}
$(2 k+1-n) n+$ \\
$(n-2)$
\end{tabular} & $\vdots$ \\
\hline $2 k-1$ & & & & & \\
\hline$\vdots$ & & $n+(n-1)$ & $2 n+(n-1)$ & & \\
\hline$n-1$ & $n-1$ & $\vdots$ & & & \\
\hline$\vdots$ & $\vdots$ & $\vdots$ & $2 n+1$ & & & \\
\hline 2 & 2 & $n+1$ & $2 n$ & & & \\
\hline 1 & 1 & $n$ & & & & \\
\hline
\end{tabular}
From Table 2, we can see that $a$ exactly takes all natural numbers from 1 to $k^{2}+3 k$. Thus, the lemma is proved.
Next, we prove that if for all $a \leqslant A$, there exists $n$ such that $F_{n}(a)=1$, then $A=2$.
In fact, $a=q n+r, q+r=1$, so $q=0, r=1$ or $q=1, r=0$, corresponding to $a=1$ or $n$. Clearly, if $n \geqslant 3$, then $A=1$;
if $n=2$, then $A=2$;
if $n=1$, then $A=1$.
Therefore, $A=2$.
Using the above conclusion and repeatedly applying the lemma, we have
$F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)$
$\leqslant 2=2 \times 1$,
$\therefore F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)$
$\leqslant 1^{2}+3 \times 1=4=2 \times 2$,
$F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)$
$\leqslant 2^{2}+3 \times 2=10=2 \times 5$,
$F_{n_{2}}\left(F_{n_{1}}(a)\right) \leqslant 5^{2}+3 \times 5=40=2 \times 20$,
$F_{n_{1}}(a) \leqslant 20^{2}+3 \times 20=460=2 \times 230$.
Therefore, $a \leqslant 230^{2}+3 \times 230=53590$.
Hence, the maximum value of $A$ is 53590, and there exist
$n_{1}=231, n_{2}=21, n_{3}=6, n_{4}=3$,
$n_{5}=2, n_{6}=2$
or $n_{1}=232, n_{2}=22, n_{3}=7, n_{4}=4$,
$n_{5}=3, n_{6}=2$,
such that for any $a \leqslant A$ we have
$F_{n_{6}}\left(F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)\right)=1$.
|
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
$$
\begin{aligned}
Q & =\frac{R_{24}}{R_{4}}=\frac{9 R_{24}}{9 R_{4}} \\
& =\frac{10^{24}-1}{10^{4}-1}=\frac{\left(10^{4}\right)^{6}-1}{10^{4}-1} \\
& =10^{20}+10^{16}+10^{12}+10^{8}+10^{4}+1 \\
& =100010001000100010001 .
\end{aligned}
$$
Therefore, $Q$ contains 15 zeros.
|
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
$$
\left(x_{2}-x_{3}\right)\left(1-5 x_{1}\right)=0 \text {. }
$$
From (1), (4), and (5), the original system of equations can be transformed into:
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x _ { 1 } = x _ { 2 } , } \\
{ x _ { 2 } = x _ { 3 } , } \\
{ x _ { 3 } + 5 x _ { 1 } x _ { 2 } = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
x_{1}=x_{2}, \\
5 x_{1}=1, \\
x_{3}+5 x_{1} x_{2}=6 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l }
{ 5 x _ { 3 } = 1 , } \\
{ x _ { 2 } = x _ { 3 } , } \\
{ x _ { 3 } + 5 x _ { 1 } x _ { 2 } = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
5 x_{3}=1, \\
5 x_{1}=1, \\
x_{3}+5 x_{1} x_{2}=6 .
\end{array}\right.\right.
\end{array}
$$
Solving these four systems of equations, we get 5 sets of solutions for $\left(x_{1}, x_{2}, x_{3}\right)$: $\square$
$$
\begin{array}{l}
(1,1,1),\left(-\frac{6}{5},-\frac{6}{5},-\frac{6}{5}\right), \\
\left(\frac{1}{5}, \frac{1}{5}, \frac{29}{5}\right),\left(\frac{1}{5}, \frac{29}{5}, \frac{1}{5}\right),\left(\frac{29}{5}, \frac{1}{5}, \frac{1}{5}\right) .
\end{array}
$$
|
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 per kg, fruit $B$ costs 1.2 yuan per kg, and fruit $C$ costs 10 yuan per kg. One day, the store earned a total of 441.2 yuan from selling these three combinations, with the sales of fruit $A$ amounting to 116 yuan. What is the sales amount for fruit $C$? $\qquad$ 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
$$
\begin{array}{l}
\left\{\begin{array}{l}
2(2 x+3 y+2 z)=116, \\
8.8 x+25.6 y+21.2 z=441.2,
\end{array}\right. \\
\therefore\left\{\begin{array}{l}
2 x+3 y+2 z=58, \\
22 x+64 y+53 z=1103 .
\end{array}\right.
\end{array}
$$
(2) - (1) $\times 11$, eliminating $x$, we get
$$
31(y+z)=465 \text {, }
$$
Thus, $y+z=15$.
Therefore, a total of 15 kg of fruit C was sold, and the sales revenue from fruit C is $15 \times 10=150$ (yuan).
|
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).
$$
\begin{array}{l}
a_{n}=(\sqrt[3]{n+1})^{2}+\sqrt[3]{n+1} \cdot \sqrt[3]{n-1}+(\sqrt[3]{n-1})^{2}, \\
\frac{1}{a_{n}}=\frac{1}{2}(\sqrt[3]{n+1}-\sqrt[3]{n-1}) . \\
\therefore \sum_{i=1}^{300} \frac{1}{a_{2 i}-1}=\frac{1}{2}[(\sqrt[3]{2}-0)+(\sqrt[3]{4}-\sqrt[3]{2}) \\
\quad+\cdots+(\sqrt[3]{1000}-\sqrt[3]{998})]=5 .
\end{array}
$$
|
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 $(4-a)^{2}-3\left(\frac{16}{3}-a^{2}\right) \leqslant 0$.
Solving this, we get $0 \leqslant a \leqslant 2$. Therefore, the maximum value of $a$ is 2.
|
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 numbers, with $a_{4}$ being the largest of these three. Therefore, when $a_{2}=5$, there are $C_{4}^{1} \times 2=8$ valid permutations.
Similarly, when $a_{4}=5$, there are also 8 valid permutations.
Therefore, the total number of valid permutations is 16.
|
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
$$
\begin{array}{l}
m^{2}+n^{2}=(m+n)^{2}-2 m n=3 . \\
m^{4}+n^{4}=\left(m^{2}+n^{2}\right)^{2}-2 m^{2} n^{2}=7 \text {. } \\
\therefore m^{5}+n^{5} \\
=(m+n)\left(m^{4}+n^{4}\right)-m n(m+n) \\
\text { - }\left[(m+n)^{2}-3 m n\right] \\
=7-(-1) \times\left[1^{2}-3 \times(-1)\right]=11 \text {. } \\
\end{array}
$$
|
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 result directly.
|
$$
\begin{array}{l}
(x+1)(y+1)=72, \\
x y(x+y)=2^{4} \times 5 \times 11 .
\end{array}
$$
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 $x=55$, $x+y \geqslant 56, x y(x+y) \geqslant 55 \times 1 \times$ $56>880$, so $x=5$ or 11.
If $x=5$, from equation (1) we know $y=11$;
If $x=11$, we know $y=5$, in this case the values of $x$ and $y$ satisfy equation (2).
Therefore, $x^{2}+y^{2}=5^{2}+11^{2}=146$.
|
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 pentagon? Explain your reasoning:
(2) Prove that the perimeter of pentagon $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}$ is at least 25 units longer than the perimeter of pentagon $A B C D E$.
|
Six, (1) The 5 shaded parts in Figure 4 can form a small pentagon.
$$
\begin{array}{c}
\because B F=A G=A H=E I=E K=D L=D M \\
=C N=C O=B P=4, \\
\angle B F B^{\prime}=\angle A G A^{\prime}=90^{\circ}, \\
\angle C O C^{\prime}=\angle B P B^{\prime}=90^{\circ}, \\
\angle D M D^{\prime}=\angle C N C^{\prime}=90^{\circ}, \\
\angle E K E^{\prime}=\angle D L D^{\prime}=90^{\circ}, \\
\angle A H A^{\prime}=\angle E I E^{\prime}=90^{\circ}, \\
\text { and ( } \left.\angle A^{\prime}+\angle B^{\prime}+\angle C^{\prime}+\angle D^{\prime}+\angle E^{\prime}\right)+(\angle 1 \\
+\angle 2+\angle 3+\angle 4+\angle 5)=5 \times 180^{\circ}, 91 j \\
\angle A^{\prime}+\angle B^{\prime}+\angle C^{\prime}+\angle D^{\prime}+\angle E^{\prime} \\
=(\angle-2) \times 180^{\circ}=3 \times 180^{\circ} . \\
\therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5=360^{\circ} .
\end{array}
$$
The 5 shaded parts in Figure 4 can form a small pentagon as shown in Figure 6 (enlarged).
(2) The additional perimeter of the polygon is equal to
Figure 6
$A^{\prime} H+A^{\prime} G+B^{\prime} F+$
$B^{\prime} P+C^{\prime} O+C^{\prime} N+D^{\prime} M+D^{\prime} L+E^{\prime} K+E^{\prime} I$.
This value is exactly the perimeter of the pentagon formed by the shaded areas in Figure 4, and it is evident that this pentagon has an inscribed circle with a radius of 4.
Since the perimeter $s$ of the pentagon is greater than the circumference of its inscribed circle, the circumference of the inscribed circle is $8 \pi, \pi>3.14$,
then $s>8 \pi>8 \times 3.14=25.12>25$.
Therefore, the perimeter of the new pentagon increases by at least 25 units.
(Provided by Tang Wenqing, Hainan Middle School, Haimen City, Jiangsu Province)
|
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 translation result directly.
|
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 finding the minimum value of $|x|+|y|$.
First, we can take any integer solution of the equation, for example, because $y=$ $\frac{3-9 x}{13}$, when $x=9$, we get $y=-6$. Using this solution, we have
$$
\left\{\begin{array}{l}
9 \times 9-13 \times 6=3, \\
9 x+13 y=3 .
\end{array}\right.
$$
Subtracting the two equations, we get $9(x-9)+13(y+6)=0$,
$$
9(x-9)=-13(y+6) \text {. }
$$
Since 9 and 13 are coprime, $x-9$ must be divisible by 13. Therefore, let $x-9=13 k$, where $k$ is an integer. At this time, we have
$9 \times 13 k=-13(y+6),-(y+6)=9 k$.
In summary, we have $\left\{\begin{array}{l}x=9+13 k, \\ y=-6-9 k .\end{array} \quad k=0, \pm 1, \pm 2, \cdots\right.$
(i) When $k=0$, $x=9, y=-6$.
$$
|x|+|y|=15 \text {; }
$$
(ii) When $k \geqslant 1$, $|x| \geqslant 22,|y|>0$, thus
$$
|x|+|y|>22 \text {; }
$$
(iii) When $k \leqslant-1$, if $k=-1$, then $x=-4, y=3$, $|x|+|y|=7$; if $k12,|x|>0$, thus $|x|+|y|>12$.
From the above, we know that at least 7 such weights are needed, with 4 of them being 9 grams and 3 being 13 grams.
(Provided by Li Shuwen, No. 40 Middle School of Benzhuang City, Shandong Province)
|
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 of soil. Now it is required to dig out the two mounds and use the soil to first fill the depression $E$, and then use the remaining soil to fill the pond $F$. How should the soil transportation plan be arranged to minimize labor?
|
$\vdots 1$. "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 the problem, we have
$$
\begin{array}{l}
x_{1}+y_{1}=781, \\
x_{2}+y_{2}=1584, \\
x_{1}+x_{2}=1025, \\
y_{1}+y_2=1340, \\
W=50 . x_{1}+150 y_{1}+30 x_{2}+120 y_{2} .
\end{array}
$$
where $0 \leqslant x_{1} \leqslant 781,0 \leqslant x_{2} \leqslant 1584$.
From equation (1), we get $y_{1}=781-x_{1}$.
From equation (3), we get $x_{2}=1025-x_{1}$.
Thus, from equation (4), we get
$$
\begin{aligned}
y_{2} & =1340-y_{1}=1340-\left(781-x_{1}\right) \\
& =559+x_{1} .
\end{aligned}
$$
Then $W=50 x_{1}+150\left(781-x_{1}\right)+30(1025-$
$$
\begin{aligned}
& \left.x_{1}\right)+120\left(559+x_{1}\right) \\
= & 214980-10 x_{1} .
\end{aligned}
$$
Therefore, when $x_{1}$ takes the maximum value of 781, $W_{\text {group }}=207170$.
Thus, the soil transportation plan that minimizes labor is:
All 781 cubic meters of soil from mound $A$ are transported to depression $E$, 244 cubic meters of soil from mound $B$ are transported to depression $E$, and the remaining soil from mound $B$ is all transported to pond $F$.
(Provided by Nan Xiuquan, Huanggang City Teaching and Research Office, Hubei Province)
|
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 pump is used, it will take 30 minutes to pump out all the water and start construction; if 2 water pumps work simultaneously, it will take 10 minutes to pump out all the water and start construction. Due to the urgent situation, the command post requires that the water be pumped out within 5 minutes to immediately start construction. At least how many water pumps need to work simultaneously? Assume that each water pump has a pumping rate of $z$ cubic meters per minute ($z>0$).
|
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
$15 z+0.5 z \leqslant 5 n z$, hence $n \geqslant 3.1$.
Since $n$ is a positive integer, the minimum value is $n=4$.
That is, at least 4 water pumps need to be organized to work simultaneously.
|
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
$$
\begin{array}{l}
\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right)-\sum_{k=1}^{9} k(k+1) \\
= \frac{1}{2} \sum_{k=1}^{9}[k(55-k)-2 k(k+1)] \\
+\frac{1}{2} \times 10 \times(55-10) \\
= \frac{1}{2} \sum_{k=1}^{9} k(53-3 k)+225 \\
= \frac{53}{2} \sum_{k=1}^{9} k-\frac{3}{2} \sum_{k=1}^{9} k^{2}+225 \\
= \frac{53 \times 45}{2}-\frac{45 \times 19}{2}+225=990
\end{array}
$$
|
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 is the maximum number of pairs of lines that are perpendicular to each other? (Each pair consists of two lines)
|
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 also have $V C \perp A B, V B \perp A C$.
This results in 6 pairs of mutually perpendicular lines (refer to Figure 11).
(2) 8 pairs are impossible.
Since a triangle can have at most one right angle and at least two acute angles, 4 triangular faces must have at least 8 acute angles. Given that 6 lines can form $C_{0}^{2}=15$ pairs of lines, and $15-3=7$, we know that the number of mutually perpendicular lines does not exceed 7.
(3) 7 pairs are impossible.
If otherwise, there are 7 pairs of perpendicular lines, and since there are only 3 pairs of skew lines, at least 4 pairs of perpendicular lines must be coplanar, meaning all 4 faces of the tetrahedron must be right-angled triangles (yielding 4 pairs). Furthermore, the 3 pairs of skew lines must also be perpendicular. In this case, the projection of one vertex of the tetrahedron onto the opposite face must be the orthocenter of that triangle, and the orthocenter of a right-angled triangle is the right-angle vertex. Therefore, one lateral edge of the tetrahedron is perpendicular to the base, and the foot of the perpendicular is the vertex of the right-angled triangle (in Figure 11, $V A \perp$ plane $A B C$). At this point, the opposite face of the right-angle vertex of the base triangle is an acute triangle ($\triangle V B C$ is an acute triangle), which contradicts the fact that all 4 faces are right-angled triangles.
In summary, the maximum number of mutually perpendicular lines is 6.
|
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 $-1 \leqslant m<1$.
$$
\begin{array}{l}
\text { (1) } \because x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2} \\
=4(m-2)^{2}-2\left(m^{2}-3 m+3\right) \\
=2 m^{2}-10 m+10, \\
\therefore 2 m^{2}-10 m+10=6, \\
\end{array}
$$
Thus, $m^{2}-5 m+2=0$. Solving this, we get $m=\frac{5 \pm \sqrt{17}}{2}$.
Since $-1 \leqslant m<1$, we have $m=\frac{5-\sqrt{17}}{2}$.
$$
\begin{array}{l}
\text { (2) } \frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}} \\
=\frac{m\left[x_{1}^{2}\left(1-x_{2}\right)+x_{2}^{2}\left(1-x_{1}\right)\right]}{\left(1-x_{1}\right)\left(1-x_{2}\right)} \\
=\frac{m\left[x_{1}^{2}+x_{2}^{2}-x_{1} x_{2}\left(x_{1}+x_{2}\right)\right]}{x_{1} x_{2}-x_{1}-x_{2}+1} \\
=\frac{m\left[\left(2 m^{2}-10 m+10\right)+\left(m^{2}-3 m+3\right)(2 m-4)\right]}{\left(m^{2}-3 m+3\right)+(2 m-4)+1} \\
=\frac{m\left(2 m^{3}-8 m^{2}+8 m-2\right)}{m^{2}-m} \\
=\frac{2 m(m-1)\left(m^{2}-3 m+1\right)}{m(m-1)} \\
=2\left(m^{2}-3 m+1\right). \\
\end{array}
$$
Let $y=2\left(m^{2}-3 m+1\right)=2\left(m-\frac{3}{2}\right)^{2}-\frac{5}{2}$, $-1 \leqslant m<1$. Since $y$ is decreasing on $-1 \leqslant m<1$, the maximum value of $y$ is 10 when $m=-1$.
Therefore, the maximum value of $\frac{m x_{1}^{2}}{1-x_{1}}+\frac{m x_{2}^{2}}{1-x_{2}}$ is 10.
|
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 points of dissatisfaction. Now, there are 32 people on the first floor, and they each live on one of the floors from the 2nd to the 33rd. Question: On which floor should the elevator stop to minimize the total dissatisfaction score of these 32 people? What is the minimum value? (Some people can choose to walk up the stairs instead of taking the elevator).
|
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, and a person living on the $t$-th floor walks up directly, with $s < t$, swapping their methods of going up, while keeping the others unchanged, will reduce the total dissatisfaction score.
Let the elevator stop at the $x$-th floor, and $y$ people on the first floor do not take the elevator but walk up directly. Then, the total dissatisfaction score is
$$
\begin{aligned}
S= & 3[1+2+\cdots+(33-x)]+3(1+2+\cdots \\
& +y)+[1+2+\cdots+(x-y-2)] \\
= & \frac{3 \times(33-x)(34-x)}{2}+\frac{3 y(y+1)}{2} \\
& +\frac{(x-y-2)(x-y-1)}{2} \\
= & 2 x^{2}-x y-102 x+2 y^{2}+3 y+1684 \\
= & 2 x^{2}-(y+102) x+2 y^{2}+3 y+1684 \\
= & 2\left(x-\frac{y+102}{4}\right)^{2}+\frac{1}{8}\left(15 y^{2}-180 y+3068\right) \\
= & 2\left(x-\frac{y+102}{4}\right)^{2}+\frac{15}{8}(y-6)^{2}+316 \\
\geqslant & 316 .
\end{aligned}
$$
Furthermore, when $x=27, y=6$, $S=316$.
Therefore, when the elevator stops at the 27th floor, the total dissatisfaction score is minimized, with the minimum value being 316 points.
(Note: Directly providing the correct answer earns 5 points.).
(Provided by Qiao Yu)
|
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, should equal $4(a+b+c+d+e)$, i.e., $a+b+c+d+e$ $=478$. From this, we get $a=(183+186+200)-478$ $=91$.
Explanation: We can also find that $b=92, c=95, d=$ $99, e=101$.
|
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 working on Project A, and the other group moved to Project B. After one month, Project A was completed, and the remaining workload of Project B was just enough for one worker to complete in one month. If each worker has the same work efficiency, how many workers are there in the engineering team?
|
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 $\frac{x}{2}+\frac{x}{4}=2\left(\frac{x}{4}+1\right)$. Solving for $x$ yields $x=8$.
|
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 maximum value of $11 c + 2 d$ is $99 + 18 = 117$, so $101 b \geq 998 - 117 = 881$, which means $b=9$. Then we have
$$
11 c+2 d=998-909=89.
$$
(3) Since $0 \leq 2 d \leq 18$, it follows that $89 - 18 = 71 \leq 11 c \leq 89$. Therefore, $c=7$ or $c=8$.
When $c=7$, $11 c+2 d=77+2 d=89$, so $d=6$;
When $c=8$, $11 c+2 d=88+2 d=89$, so $d=\frac{1}{2}$. (discard).
Thus, this four-digit number is 1976.
|
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
$$
\begin{array}{l}
2\left(a_{12}+a_{10}+\cdots+a_{2}+a_{0}\right)=1+3^{6} . \\
\text { Therefore, the original expression }=\frac{1}{2}\left(1+3^{6}\right)=365 .
\end{array}
$$
|
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 $x=\frac{1}{2} t+\frac{1}{4}$.
Substituting into inequality (1), we get
$$
0 \leqslant \frac{3}{2} t+\frac{3}{4}+1-t<1 \text {, }
$$
which simplifies to $0 \leqslant \frac{1}{2} t+\frac{7}{4}<1$.
$$
\therefore-\frac{7}{2} \leqslant t<-\frac{3}{2} \text {. }
$$
Since $t$ is an integer, the only integers satisfying inequality (3) are $t=-2, t=-3$.
Substituting the values of $t$ into equation (2), we get
$$
x_{1}=-\frac{3}{4}, x_{2}=-\frac{5}{4} \text {. }
$$
Therefore, the sum of the roots is
$$
x_{1}+x_{2}=\left(-\frac{3}{4}\right)+\left(-\frac{5}{4}\right)=-2 \text {. }
$$
|
-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 holds for integers $k \geqslant 9$.
Thus, when $k \geqslant 9$, $n$ lies between 1 and 2 times $k(k-1)(k-2)$, which cannot satisfy the given conditions.
Upon calculation, when $k=7$, the least common multiple (LCM) of 1 to $k$ is 420, which lies between $7^{3}=343$ and $8^{3}=512$. Therefore, $n=420$ is the largest solution when $k$ is odd.
When $k$ is even, $k-1, k-2, k-3$ are pairwise coprime, so $(k-1)(k-2)(k-3)$ divides $n$. In this case, we still have
$$
(k-1)(k-2)(k-3) < k^{3} < n.
$$
The inequality
$$
(k+1)^{3} < 2(k-1)(k-2)(k-3)
$$
holds for positive integers $k \geqslant 14$. Therefore, the allowed values of $k$ must be even numbers not exceeding 12.
Upon inspection, when $k=8, 10, 12$, the LCM of the positive integers from 1 to $k$ is greater than $(k+1)^{3}$, which does not meet the given conditions; and for $k \leqslant 6$, the corresponding $n$ must not exceed $7^{3} = 343$.
In summary, the answer to this problem is 420.
Note: In this problem, we used the conclusion: "For positive integers $a$ and $b$, if there exists an integer $n$ such that $na < b < (n+1)a$, then $a$ does not divide $b$."
|
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 $n$-gon. They conjectured that
$$
f(n)=2^{n-2}+1 \text {. }
$$
Below is a proof for the case $n=5$, which includes an important lemma:
|
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 the region $T$ (the shaded area in Figure 1) is formed by the extensions of $A_{1} A_{2}$ and $A_{4} A_{3}$ and the side $A_{2} A_{3}$, and there are $n$ points ($\epsilon S$) in $T$, then the convex hull formed by these $n$ points and the convex $m$-gon is a convex $(m+n)$-gon, as shown in Figure 1.
Proof: Let the region formed by the convex $m$-gon be denoted as $A$. Clearly, $A \cup T$ is a convex set. If the number of vertices of the convex hull $B$ formed by the $n$ points and the convex $m$-gon is less than $m+n$, then there must be a point inside $B$. Since the vertices of the $m$-gon are all boundary points of the convex set $A \cup T$ (i.e., $A_{2}, A_{3}$), this point must be among the $n$ points. Thus, this point is in $S$. It is also easy to see that $B$ is in $K$, so this point is also inside $K$. Since this point belongs to $S$, it must be a vertex of $K$, which is a contradiction. Therefore, the lemma is proved.
Next, we will prove the proposition.
First, it is easy to see that there are counterexamples with 8 points. For example, in Figure 2, place 4 points on each branch of a hyperbola, such that no three points are collinear, and in this case, there is no convex pentagon.
Now we prove that there must be a convex pentagon among 9 points. Let the 9 points on the plane be $A_{1}, A_{2}, \ldots, A_{9}$. Clearly, we only need to consider the following two cases:
(1) The convex hull is a quadrilateral, say $A_{1} A_{2} A_{3} A_{4}$, and the points $A_{5}, A_{6}, \ldots, A_{9}$ are inside it:
If $A_{5}, A_{6}, \ldots, A_{9}$ form a convex pentagon, the problem is solved.
Otherwise, among these five points, there must be four points such that one of them is inside the triangle formed by the other three. Without loss of generality, assume $A_{5}$ is inside $\triangle A_{6} A_{7} A_{8}$, as shown in Figure 3.
Connect $A_{5} A_{6}, A_{5} A_{7}, A_{5} A_{8}$ and extend them. Let the region enclosed by the rays $A_{5} A_{6}$ and $A_{5} A_{7}$, excluding $\triangle A_{5} A_{6} A_{7}$, be denoted as $\mathrm{I}$, and similarly define regions II and III, as shown in Figure 3. Clearly, $A_{1}, A_{2}, A_{3}, A_{4}$ are all in these three regions.
Since there are only three regions and four points ($A_{1}, A_{2}, A_{3}, A_{4}$), by the pigeonhole principle, at least one region must contain at least two of these points. Without loss of generality, assume $A_{1}, A_{2}$ are in region I. By the lemma, the points $A_{5}, A_{6}, A_{2}, A_{1}, A_{2}$ form a convex pentagon.
(2) The convex hull is a triangle, say $\triangle A_{1} A_{2} A_{3}$, and the other 6 points $A_{4}, A_{5}, \ldots, A_{9}$ are inside it. For $A_{4}, A_{5}, \ldots, A_{9}$, consider the following two cases:
1) If the convex hull is a quadrilateral, say $A_{4} A_{5} A_{6} A_{7}$, consider extending the line segment $A_{8} A_{9}$ at both ends. If it intersects the adjacent sides of the quadrilateral $A_{4} A_{5} A_{6} A_{7}$, say $A_{4} A_{5}$ and $A_{4} A_{7}$, as shown in Figure 4, then $A_{8}, A_{9}, A_{5}, A_{6}, A_{7}$ form a convex pentagon.
If extending the line segment $A_{8} A_{9}$ at both ends intersects the opposite sides of the quadrilateral $A_{4} A_{5} A_{6} A_{7}$, say $A_{5} A_{6}$ and $A_{4} A_{7}$, as shown in Figure 5, the four regions I, II, III, and IV are shaded. It is easy to see that these four regions cover the part outside the quadrilateral $A_{4} A_{5} A_{6} A_{7}$ (of course, if the rays $A_{8} A_{4}$ and $A_{9} A_{5}$ intersect or the rays $A_{8} A_{7}$ and $A_{9} A_{6}$ intersect, the four regions will overlap).
There are three points $A_{1}, A_{2}, A_{3}$ in these four regions. By the lemma, if there is no convex pentagon, then regions II and IV cannot contain any points. Regions I and III can each contain at most one point, so the four regions can contain at most two points, which contradicts the existence of three points $A_{1}, A_{2}, A_{3}$.
2) If the convex hull is a triangle, say $\triangle A_{4} A_{5} A_{6}$, as shown in Figure 6, and without loss of generality, assume extending $A_{7} A_{8}$ intersects $A_{4} A_{5}$ and $A_{4} A_{6}$. Define regions I, II, and III (it is easy to see that regions I and II overlap). Similarly, $A_{1}, A_{2}, A_{3}$ are in these three regions.
If there is no convex pentagon, then by the lemma, region III cannot contain any points, and regions I and II can each contain at most one point, which contradicts the existence of three points $A_{1}, A_{2}, A_{3}$. Therefore, there must be a convex pentagon. In conclusion, $f(5)=9$.
|
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 of the numbers on the two cards. Knowing this sum, the magician can identify which box was not selected.
How many ways are there to place the cards so that the trick always works? (Two methods are considered different if at least one card is placed in a different colored box.)
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
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 that $i, i+1, i-2$ have mutually different colors, for example, $\mathrm{wb}$. Since $i+(i+3)=(i+1)+(i+2)$, the color of $i \div 3$ cannot be the color of $i+1$, which is $\mathrm{w}$, nor can it be the color of $i+2$, which is $\mathrm{w}$. It can only be $\mathrm{r}$. Therefore, as long as three adjacent numbers have mutually different colors, the color of the next number can be determined. Furthermore, this pattern of three numbers' colors must repeat: after rwb, it must be $r$, then $w, b, \cdots$ and so on. Similarly, this process also holds for the opposite direction: before rwb, it must be $b, \cdots$ and so on.
Therefore, it is only necessary to determine the colors of 1, 2, and 3. There are 6 different ways to do this, and all 6 methods can make the magic trick successful because their sums $r+w, w+b, b+r$ give mutually different remainders modulo 3.
(2) There do not exist three consecutive numbers with mutually different colors. Assume 1 is red. Let $i$ be the smallest number that is not red, and assume $i$ is white. Let $k$ be the smallest blue number. By assumption, $i+11$ must be red, and from $t+99=(t-1)+100$, it follows that $t-1$ is blue, which contradicts the assumption that only 100 is blue.
Therefore, the colors of these numbers must be rww…wwb. This method indeed works:
If the sum of the numbers on the two selected cards $\leqslant 100$, the box from which no card was selected must be blue;
If the sum of the numbers is $101$, the box from which no card was selected must be white;
If the sum of the numbers $>101$, the box from which no card was selected must be red.
Finally, there are 6 different ways to arrange the colors in the above manner. Therefore, the answer is 12.
|
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
$$
\frac{1}{p}=0 . a_{1} a_{2} \cdots a_{3 r} a_{1} a_{2} \cdots a_{3 r} \cdots \text {. }
$$
For each $p \in S$ and any integer $k \geqslant 1$, define
$$
f(k, p)=a_{k}+a_{k+r(p)}+a_{k+2 r(p)} \text {. }
$$
(1) Prove that the set $S$ contains infinitely many elements.
|
(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 $10^{3 q}-1$, the repeating cycle length of $\frac{1}{p_{q}}$ is $3 q$, so the length of the smallest repeating cycle of $\frac{1}{p_{q}}$ is a divisor of $3 q$.
If the length of the smallest repeating cycle is $q$, then by $10^{q} \equiv 1 \pmod{p_{q}}$,
we get $N_{q}=10^{2 q}+10^{q}+1 \equiv 3 \neq 0 \pmod{p_{q}}$. Contradiction.
If the length of the smallest repeating cycle is 3, there can only be one case, i.e., $p_{8}$ is a factor of $10^{3}-1=3^{3} \times 37$, i.e., $p_{q}=37$. In this case, $N_{q}=3 \times 37 \equiv 3 \pmod{4}$.
However, $N_{q}=10^{2 q}+10^{q}+1 \equiv 1 \pmod{4}$, contradiction.
Thus, for each prime $q$, we can find a prime $p_{q}$ such that the length of the smallest repeating cycle of the decimal part of $\frac{1}{p_{q}}$ is $3 q$.
(2) Solution: Let the prime $p \in S, 3 r(p)$ be the length of the smallest repeating cycle of $\frac{1}{p}$. $p$ is a factor of $10^{3 r(p)}-1$, but not a factor of $10^{r(p)}-1$, so $p$ is a factor of $N_{r(p)}=10^{2 r(p)}+10^{r(p)}+1$.
$$
\begin{array}{l}
\text { Let } \frac{1}{p}=0 . a_{1} a_{2} a_{3} \cdots, \\
x_{j}=\frac{10^{j-1}}{p}, y_{j}=\left\{x_{j}\right\}=0 . a_{j} a_{j+1} a_{j+2} \cdots,
\end{array}
$$
where $\{x\}$ denotes the fractional part of $x$. Then $a_{j}<10 y_{j}$. Therefore,
$$
\begin{aligned}
f(k, p)= & a_{k}+a_{k+r(p)}+a_{k+2 r(p)} \\
& <10\left(y_{k}+y_{k+r(p)}+y_{k+2 r(p)}\right) .
\end{aligned}
$$
Since $x_{k}+x_{k+r(p)}+x_{k+2 r(p)}=\frac{10^{k-1} N_{r(p)}}{p}$ is an integer, $y_{k}+y_{k+r(p)}+y_{k+2 r(p)}$ is also an integer and is less than 3, i.e.,
$$
y_{k}+y_{k+r(p)}+y_{k+2 r(p)} \leqslant 2 \text {. }
$$
Thus, $f(k, p)<20$.
Therefore, the maximum value of $f(k, p)$ does not exceed 19.
Since $f(2,7)=4+8+7=19$,
the required maximum value is 19.
|
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 until the number $a_{1}$ in position $a_{1}$ is taken. Then $\left(a_{1}, a_{1}^{\prime}, a_{1}^{\prime}, \cdots, a_{1}^{\prime}\right)$ is called a chain. The length of the chain is the number of elements in the chain. Clearly, the length of the chain can be $1,2,3,4,5,6$, and a chain of length $i(i=1,2,3,4,5,6)$ requires $i-1$ swaps to achieve the goal, until all chains are divided into chains of length 1.
Now we solve this problem using the exclusion method.
Since the problem requires no more than 4 swaps, the longest chain in a permutation that meets the condition can be of length $5,4,3,2,1$, and any permutation with the longest chain of length $5,4,3,2,1$ meets the condition.
Now we examine the number of permutations with the longest chain of length 6, so the required number is $6!-5!=720-120=600$.
(Zhao Jiwei, Class 002, Zhengzhou No.1 High School, Zhengzhou, Henan Province, China)
|
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 $0 \leqslant m \leqslant 5$,
$$
\begin{array}{l}
-\frac{16}{9} \leqslant n \leqslant \frac{35}{3} . \\
\therefore n=-1,0,1,2,3,4,5,6,7,8,9,10,11 .
\end{array}
$$
|
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
$$
\begin{array}{l}
\sum_{i=0}^{n}(i+1)=\frac{1}{2}(n+1)(n+2) \text { (terms). } \\
\because(a+b+c+d)^{n}=((a+b+c)+d)^{n} \\
=(a+b+c)^{n}+C_{n}^{1}(a+b+c)^{n-1} \cdot d+\cdots+d^{n},
\end{array}
$$
There are $n+1$ terms, and $(a+b+c)^{i}$ has $\frac{1}{2}(i+1)(i+2)$ different terms.
$\therefore(a+b+c+d)^{n}$ has the number of different terms as
$$
\begin{array}{l}
\sum_{i=0}^{n} \frac{1}{2}(i+1)(i+2) \\
=\sum_{i=0}^{n} \frac{1}{2}\left(i^{2}+3 i+2\right) \\
=\frac{1}{2}\left(\frac{1}{6} n(n+1)(2 n+1)+3 \times \frac{1}{2} n(n+1)+2(n+1)\right) \\
=\frac{1}{6}(n+1)(n+2)(n+3) .
\end{array}
$$
When $n=21$, the number of terms is 2024.
|
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 $n=k+1$,
$$
\begin{array}{l}
a_{k+3}=12^{a_{k+2}}>12^{8 b_{k}}=\left(12^{8}\right)^{b_{k}} \\
>(8 \times 2000)^{b_{k}}>8 \times 2000^{b_{k}}=8 b_{k+1} .
\end{array}
$$
Therefore, equation (1) holds for all $n$.
From equation (1), we get $b_{1998} \leqslant 8 b_{1998}1998 \text {. }$
In conclusion, $m=1999$.
|
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
$$
\begin{array}{l}
\frac{4 x}{x^{2}-4}=\frac{(a+b) x-2(a-b)}{x^{2}-4}, \\
4 x=(a+b) x-2(a-b) .
\end{array}
$$
According to the conditions for the identity of polynomials, we get the system of equations about $a$ and $b$
$$
\left\{\begin{array}{l}
a+b=4, \\
a-b=0 .
\end{array}\right.
$$
Solving, we get $a=b=2$.
Therefore, $a^{2}+b^{2}=8$.
|
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 simplify the numerator and denominator of the expression to be evaluated respectively:
$$
\begin{array}{l}
a^{3}-1=(a-1)\left(a^{2}+a+1\right) \\
\quad=(a-1)\left(\frac{1}{4}+1\right)=\frac{5}{4}(a-1) ; \\
a^{5}+a^{4}-a^{3}-a^{2} \\
=a^{3}\left(a^{2}+a\right)-a\left(a^{2}+a\right) \\
=\frac{1}{4} a^{3}-\frac{1}{4} a \\
=\frac{1}{4}\left[\frac{5}{4}(a-1)+1\right)-\frac{1}{4} a \\
=\frac{1}{16}(a-1) .
\end{array}
$$
Therefore, the original expression $=20$.
|
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 $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, then
$$
x=\sqrt[3]{\frac{k}{1995}}, y=\sqrt[3]{\frac{k}{1996}}, z=\sqrt[3]{\frac{k}{1997}} .
$$
From the given conditions, $x, y, z, k$ have the same sign, and since $x y z>0$, we know $x, y, z, k>0$. Substituting into the last equation given, we get
$$
\sqrt[3]{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}} .
$$
Rearranging, we get
$$
\sqrt[3]{k}\left(\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)=0 \text{. }
$$
Since $k>0$, we know $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is a root of $\sqrt[3]{u}-u=0$.
Therefore, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. (The other two roots $0, -1$ do not meet the conditions)
|
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 $\left\{\begin{array}{l}x=s-2, \\ y=5-\frac{4 s}{3}, \\ z=-\frac{s}{3}+1 .\end{array}\right.$
Since $x$, $y$, and $z$ are non-negative rational numbers, we have $\left\{\begin{array}{l}s-2 \geqslant 0, \\ 5-\frac{4 s}{3} \geqslant 0, \\ -\frac{s}{3}+1 \geqslant 0 .\end{array} \quad\right.$ Solving these inequalities, we get $\left\{\begin{array}{l}s \geqslant 2, \\ s \leqslant \frac{15}{4}, \\ s \leqslant 3 .\end{array}\right.$
Therefore, the sum of the maximum and minimum values of $s$ is $3+2=$
5.
|
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,
$$
f(n)=f(n-1)+f(n-2) \text {. }
$$
Thus, $\{f(n)\}$ is the Fibonacci sequence.
$$
\begin{array}{l}
\because f(1)=2, \\
\therefore f(10)=144 .
\end{array}
$$
The key to solving this problem is to classify based on the first digit. If we generalize to 3 digits, the situation becomes more complex.
|
144
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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