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14. A unit spent 500,000 yuan to purchase a piece of high-tech equipment. According to the tracking survey of this model of equipment: after the equipment is put into use, if the maintenance and repair costs are averaged to the first day, the conclusion is: the maintenance and repair cost on the $x$-th day is
$$
\left[\frac{1}{4}(x-i)+5 \sim 0\right] \text { yuan. }
$$
(1) If the total maintenance and repair costs from the start of use to the scrapping of the equipment, as well as the cost of purchasing the equipment, are averaged to each day, this is called the daily average loss. Please express the daily average loss $y$ (yuan) as a function of the usage days $x$ (days);
(2) According to the technical and safety management requirements of this industry, when the average loss of this equipment reaches its minimum value, it should be scrapped. How many days should the equipment be used before it should be scrapped?
[Note]: In solving this problem, you may need to use the following two mathematical points (if needed, you can directly cite the following conclusions):
(A) For any integer $n$, the following equation must hold:
$$
1+2+3+4+\cdots+n=\frac{n(n+1)}{2} \text {. }
$$
(B) For any positive constants $a$ and $b$, and for any positive real number $x$, the following inequality must hold:
$\frac{a}{x}+\frac{x}{b} \geqslant 2 \sqrt{\frac{a x}{x b}}=2 \sqrt{\frac{a}{b}}$. It can be seen that $2 \sqrt{\frac{a}{b}}$ is a constant. That is, the function $y=\frac{a}{x}+\frac{x}{b}$ has a minimum value of $2 \sqrt{\frac{a}{b}}$, and this minimum value is achieved when $\frac{a}{x}=\frac{x}{b}$.
|
14. (1) The equipment is put into use for $x$ days. The average daily loss is:
$$
\begin{aligned}
y= & \frac{1}{x}\left[500000+\left(\frac{1}{4} \times 0+500\right)+\left(\frac{1}{4} \times 1+500\right)\right. \\
& \left.+\left(\frac{1}{4} \times 2+500\right)+\cdots+\left(\frac{x-1}{4}+500\right)\right] \\
= & \frac{1}{x}\left[500000+500 x+\frac{1}{4} \cdot \frac{x(x-1)}{2}\right] \\
= & \frac{500000}{x}+\frac{x}{8}+499 \frac{7}{8} .
\end{aligned}
$$
(2)
$$
\begin{aligned}
y & =\frac{5000000}{x}+\frac{x}{8}+499 \frac{7}{8} \\
& \geqslant 2 \sqrt{\frac{500000}{x} \times \frac{x}{8}}+499 \frac{7}{8} \\
& =500+499 \frac{7}{8}=999 \frac{7}{8} .
\end{aligned}
$$
When and only when $\frac{500000}{r}=\frac{x}{8}$, i.e., $x=2000$, the equality holds.
Therefore, the equipment should be scrapped after being put into use for 2000 days.
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. A piece of lead wire of length $2 n$ (where $n$ is a natural number and $n \geqslant 4$) is folded into a triangle with integer side lengths. Let $(a, b, c)$ represent a triangle with side lengths $a, b, c$ such that $a \leqslant b \leqslant c$.
(1) For the cases $n=4, 5, 6$, write down all the $(a, b, c)$ that satisfy the conditions;
(2) Someone, based on the conclusions in (1), conjectured that when the length of the lead wire is $2 n$ ($n$ is a natural number and $n \geqslant 4$), the number of corresponding $(a, b, c)$ is always $n-3$. In fact, this is an incorrect conjecture; write down all the $(a, b, c)$ for $n=12$, and state the number of $(a, b, c)$;
(3) Try to classify all the $(a, b, c)$ that satisfy the conditions for $n=12$ according to at least two different criteria.
|
15. (1) When $n=4$, the length of the lead wire is 8. Then the only group of $(a, b, c)$ that satisfies the condition is $(2,3,3)$;
When $n=5$, the length of the lead wire is 10. Then the groups of $(a, b, c)$ that satisfy the condition are $(2,4,4),(3,3,4)$;
When $n=6$, the length of the lead wire is 12. Then the groups of $(a, b, c)$ that satisfy the condition are $(2,5,5),(3,4,5),(4,4,4)$.
(2) When $n=12$, the length of the lead wire is 24. According to the problem,
$$
a+b+c=24 \text {, and }\left\{\begin{array}{l}
a+b>c \\
a \leqslant b \leqslant c
\end{array}\right. \text {. }
$$
From this, we get $8 \leqslant c \leqslant 11$, i.e., $c=8,9,10,11$.
Therefore, the groups of $(a, b, c)$ that satisfy the condition are 12:
$$
\begin{array}{l}
A(2,11,11) ; B(3,10,11) ; C(4,9,11) ; \\
D(5,8,11) ; E(6,7,11) ; F(4,10,10) ; \\
G(5,9,10) ; H(6,8,10) ; I(7,7,10) ; \\
J(6,9,9) ; K(7,8,9) ; L(8,8,8) .
\end{array}
$$
(3) Different classification criteria determine different classifications. Here are some examples:
(1) Classify by the value of the largest side $c$, there are four categories:
i) $c=11$, there are five: $A, B, C, D, E$;
ii) $c=10$, there are four: $F, G, H, I$;
iii) $c=9$, there are two: $J, K$;
iv) $c=8$, there is only one: $L$.
(2) Classify by whether they are equilateral or isosceles triangles, there are three categories:
i) Equilateral triangles: only $L$;
ii) Isosceles but not equilateral triangles: $A, F, I, J$ four;
iii) Neither equilateral nor isosceles triangles: $B, C, D, E, G, H, K$ seven.
(3) Classify by the relationship between the largest angle and the right angle, there are three categories:
i) Acute triangles: $A, F, G, I, J, K, L$ seven;
ii) Right triangles: only $H$;
iii) Obtuse triangles: $B, C, D, E$ four.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $b$ take odd numbers between 2 and 16, and $c$ take any natural number. Then the number of quadratic equations $3 x^{2}+$ $(b+1) x+c=0$ that can be formed with two distinct real roots is ( ).
(A) 64
(B) 66
(C) 107
(D) infinitely many
|
4. (A).
$\because$ The equation $3 x^{2}+(b+1) x+c=0$ has two distinct real roots, $\therefore \Delta=(b+1)^{2}-12 c>0 \Rightarrow c<\frac{1}{12}(b+1)^{2}$.
Given $b=3,5,7,9,11,13,15$.
When $b=3$, $c<\frac{4}{3}$, there is 1 value of $c$ that meets the condition; when $b=5$, $c<3$, there are 2 values of $c$ that meet the condition; when $b=7$, $c<\frac{16}{3}$, there are 5 values of $c$ that meet the condition; when $b=9$, $c<\frac{25}{3}$, there are 8 values of $c$ that meet the condition; when $b=11$, $c<12$, there are 11 values of $c$ that meet the condition; when $b=13$, $c<\frac{49}{3}$, there are 16 values of $c$ that meet the condition; when $b=15$, $c<\frac{64}{3}$, there are 21 values of $c$ that meet the condition. Therefore, a total of $1+2+5+8+11+16+21=64$ (quadratic equations) are formed.
|
64
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. The square number $y^{2}$ is
the sum of the squares of 11 consecutive integers. Then the smallest value of the natural number $y$ is
$\qquad$
|
3.11.
Let the middle number of these 11 consecutive integers be \( a \). Then
\[
\begin{aligned}
y^{2}= & (a-5)^{2}+(a-4)^{2}+(a-3)^{2}+(a-2)^{2} \\
& +(a-1)^{2}+a^{2}+(a+1)^{2}+(a+2)^{2} \\
& +(a+3)^{2}+(a+4)^{2}+(a+5)^{2} \\
= & 11 a^{2}+2\left(5^{2}+4^{2}+3^{2}+2^{2}+1^{2}\right) \\
= & 11\left(a^{2}+10\right) .
\end{aligned}
\]
Thus, when \( a^{2}+10=11 \), i.e., \( a= \pm 1 \), \( y^{2} \) is minimized. At this point, \( v_{\text {min }}=11 \).
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $a_{n}=6^{n}+8^{n}$. Then $a_{84} \equiv$ $\qquad$ $(\bmod 49)$
|
5.2.
$$
\begin{aligned}
a_{84}= & (7-1)^{84}+(7 \\
& +1)^{84} \\
= & 2\left(\mathrm{C}_{84}^{0} \cdot 7^{84}+\right. \\
& \mathrm{C}_{84}^{2} \cdot 7^{82}+\cdots \\
& \left.+\mathrm{C}_{84}^{82} \cdot 7^{2}+\mathrm{C}_{84}^{84}\right) \\
= & 45 \times M+2 . \\
a_{84}= & 2(\text { modulo } 49) .
\end{aligned}
$$
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The number of planes for which the ratio of distances to the 4 vertices of a regular tetrahedron is $1: 1: 1: 2$ is $\qquad$.
|
6.32
Let the vertices of a regular tetrahedron be $A, B, C, D$. The plane $\alpha$ that divides the distances to these four points in the ratio $1: 1: 1: 2$ can be of two types:
(1) $A, B, C$ are on the same side of $\alpha$, there are 2 such planes (as shown in Figure 8).
(1) $\frac{A A_{1}}{A_{1} D}=\frac{B B_{1}}{B_{1} D}=\frac{C C_{1}}{C_{1} D}=\frac{1}{2}$
(2) $\frac{A_{1} A}{A D}=\frac{B_{1} B}{B D}=\frac{C_{1} C}{C D}=1$
Figure 8
(2) $A, B, C$ are on opposite sides of $\alpha$, there are 6 such planes (as shown in Figure 9).
(1) $\frac{A A_{1}}{A_{1} D}=\frac{B B_{1}}{B_{1} D}=\frac{1}{2}$
(2) $\frac{C_{1}}{C_{1} D}=\frac{1}{2}$
(3) $\frac{A C_{1}}{C_{1} D}=\frac{1}{2}$
$\frac{A D_{1}}{D_{1} C}=\frac{B C_{1}}{C_{1} C}=1$
$\frac{A A_{1}}{A_{1} C}=\frac{B B_{1}}{B_{1} C}=1$
$\frac{A A_{1}}{A_{1} B}=\frac{A B_{1}}{B_{1} C}=1$
(4) $\frac{A A_{1}}{A_{1} B}=\frac{A B_{1}}{B_{1} C}=1$
(5) $\frac{A A_{1}}{A_{1} B}=\frac{B B_{1}}{B_{1} C}=1$
(6) $\frac{A A_{1}}{A_{1} B}=\frac{B B_{1}}{B_{1} C}=1$
$\frac{B D_{1}}{D_{1} D}=\frac{O C_{1}}{C_{1} D}=\frac{1}{2}$
$\frac{B C_{1}}{C_{1} D}=\frac{1}{2}$
$\frac{A D_{1}}{D_{1} D}=\frac{C C_{1}}{C_{1} D}=\frac{1}{2}$
By permuting $A, B, C, D$, a total of $8 \times 4=32$ planes can be obtained.
|
32
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) The player numbers of a sports team are uniquely selected from the positive integers 1 to 100. If the number of any player is neither the sum of the numbers of any other two players nor twice the number of another player, what is the maximum number of players this sports team can have?
|
Second, all odd numbers $1, 3, \cdots, 99$ totaling 50 can be the numbers of the team members. Below is the proof that it is impossible to increase further. If the sports team
has 51 members, from smallest to largest, denoted as
$$
a_{1}, a_{2}, \cdots, a_{50}, a_{51} \text {. }
$$
Take the differences
$$
a_{51}-a_{1}, a_{51}-a_{2}, \cdots, a_{51}-a_{50},
$$
this results in $51+50=101$ numbers. Since the numbers in (1) and (2) do not exceed the range of 1 to 100, by the pigeonhole principle, there must be two numbers that are equal. However, the numbers in (1) are all distinct, and the numbers in (2) are also all distinct, so it can only be that a number $a_{i}$ in (1) equals a number $a_{51}-a_{j}$ in (2). This gives
$$
a_{i}=a_{51}-a_{j},
$$
thus $a_{i}+a_{j}=a_{51}$.
This means that the number $a_{51}$ is equal to the sum of two other numbers (when $i \neq j$) or equal to twice a certain member's number (when $i=j$), which contradicts the condition. Therefore, the maximum number of members is 50.
|
50
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) In a convex $n$-sided rose garden $(n \geqslant 4)$, there is 1 red rose planted at each of the $n$ vertices. There is a straight path between every two red roses, and these paths do not have the situation of "three lines intersecting at one point."
--- They divide the garden into many non-overlapping regions (triangles, quadrilaterals, ...), and each region has one white rose (or black rose).
(1) Find the expression for the total number of roses $f(n)$ in the rose garden.
(2) Can there be exactly 99 roses in the garden? Explain your reasoning.
|
Three, (1) Solution 1: The straight paths in the rose garden form a convex $n$-sided polygon and its diagonals $\left(k_{n}\right)$. It is easy to know that there are $C_{n}^{2}-n$ diagonals in the graph. Furthermore, since the diagonals do not have "three lines intersecting at one point", the number of intersection points of the diagonals inside the shape is $\mathrm{C}_{n}^{4}$. Now, let's first find the number of regions $g(n)$ inside the $n$-sided polygon, which can be expressed as
$$
f(n)=g(n)+n \text {. }
$$
Now, remove diagonal $l_{1}$, and let $l_{1}$ intersect with other diagonals at $d_{1}$ points, then the graph $K_{n}$ is reduced by $d_{1}+1$ regions. Let the remaining graph be $G_{1}$.
In $G_{1}$, remove diagonal $l_{2}$, and let $l_{2}$ intersect with other diagonals (excluding $l_{1}$) at $d_{2}$ points in $G_{1}$, then the graph $G_{1}$ is reduced by $d_{2}+1$ regions. Let the remaining graph be $G_{2}$.
Proceeding in this manner, each time a diagonal is removed from the remaining graph. When the last diagonal is removed, it no longer intersects with any remaining diagonals, and the graph is reduced by 1 region, leaving 1 region. Thus, $d_{C_{n}^{2}-n}^{2}+1=1$. Summing up,
$$
\begin{aligned}
g(n) & =\sum_{m=1}^{c_{n}^{2}-n}\left(d_{m}+1\right)+1 \\
& =\sum_{m=1}^{c_{n}^{2}-n} d_{m}+\left(C_{n}^{2}-n\right)+1 \\
& =C_{n}^{4}+C_{n}^{2}-n+1=C_{n}^{4}+C_{n-1}^{2} .
\end{aligned}
$$
Therefore, $f(n)=g(n)+n=\dot{C}_{n}^{4}+\mathrm{C}_{n-1}^{2}+n$.
Solution 2: Connect an arc between every two vertices, which adds $n$ new bow-shaped regions. Imagine transplanting a red rose to each bow-shaped region at the vertices, then the number of roses is the number of regions. This problem is the same as a problem in the 1999 No. 3 issue of the High School Mathematical Olympiad training questions, and a new proof is provided here.
Consider the recursive relationship between $f(k)$ and $f(k+1)$ (as shown in Figure 14).
For the case of $k$ points, when point $A_{k+1}$ is added, $A_{k+1}$ connects to the previous $k$ points with $k$ lines (dashed lines in Figure 14). At this point, the graph has $\mathrm{C}_{k+1}^{4}$ diagonal intersection points, and the number of new intersection points added is $C_{k+1}^{4}-C_{k}^{4}$. Since each line $A_{k+1} A_{i}(1 \leqslant i \leqslant k)$ intersects with the sides or diagonals of the original $k$-sided polygon at $d_{i}$ points $\left(d_{i} \geqslant 0\right)$, these intersection points divide $A_{k+1} A_{i}$ into $d_{i}+1$ non-overlapping small segments. Each such small segment divides the region it is in into two, so the line $A_{k+1} A_{i}$ adds $d_{i}+1$ regions. Summing up,
$$
\begin{array}{c}
f(k+1)-f(k)=\sum_{i=1}^{k}\left(d_{i}+1\right) \\
=\sum_{i=1}^{k} d+k=\left(C_{k+1}^{4}-C_{k}^{4}\right)+k=C_{k}^{3}+C_{k}^{1} \\
\text { Therefore, } f(n)=\sum_{k=5}^{n}(f(k)-f(k-1))+f(4) \\
=\sum_{k=5}^{n}\left(C_{k-1}^{3}+C_{k}^{1} 1\right)+8 \quad(f(4)=8) \\
=\left(C_{n}^{3}+\cdots+C_{3}^{3}\right)-C_{3}^{3}+\left(C_{n}^{1} \quad+\cdots+C_{2}^{1}+\right.
\end{array}
$$
Therefore, $f(n)=\sum_{k=5}^{n}(f(k)-f(k-1))+f(4)$
$$
\begin{aligned}
& \left.\mathrm{C}_{1}^{1}\right)-\mathrm{C}_{3}^{1}-\mathrm{C}_{2}^{1}-\mathrm{C}_{1}^{1}+8 \\
= & \mathrm{C}_{n}^{4}+\mathrm{C}_{n}^{2}+1
\end{aligned}
$$
Note: This problem has the background of Euler's theorem, and the formula $F=E-V+2$ can be directly substituted to solve it.
(2) The general formula can be rewritten as
$$
f(n)=\frac{1}{24}\left(n^{4}-6 n^{3}+23 n^{2}-18 n+24\right)
$$
The problem is transformed into solving the equation
$$
\frac{1}{24}\left(n^{4}-6 n^{3}+23 n^{2}-18 n+24\right)=99
$$
which simplifies to
$$
n^{4}-6 n^{3}+23 n^{2}-18 n-2352=0,
$$
or $(n-8)\left(n^{3}+2 n^{2}+39 n+294\right)=0$
Since $n$ is a positive integer, we have
$$
n^{3}+2 n^{2}+39 n+294>0 \text {. }
$$
Solving this, we get $n=8$
Therefore, when $n=8$, there are exactly 99 roses in the rose garden.
(Luo Zengru, Shaanxi Normal University, 710062)
|
99
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$=6$ integer solutions.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-(10 a+b), \\
x_{1} x_{2}=10 b+a .
\end{array}\right.
$$
We have $10 x_{1}+10 x_{2}+x_{1} x_{2}=-99 a(1 \leqslant a \leqslant 9)$.
Thus $\left(x_{1}+10\right)\left(x_{2}+10\right)=100-99 a$.
By the problem, $x_{1}<0, x_{2}<0$, and
$$
\overline{a b}=-\left(x_{1}+x_{2}\right), \overline{b a}=x_{1} x_{2}=99 \text {. }
$$
Let $\left|x_{1}\right| \leqslant\left|x_{2}\right|$, by $\left|x_{1}\right| \cdot\left|x_{2}\right| \leqslant 99$ we know $\left|x_{1}\right| \leqslant 9$.
Obviously, $x_{1} \neq 0$, then
$$
9 \geqslant x_{1}+10 \geqslant 1 \text {. }
$$
Discussion: (1) When $a=1$.
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=1 \text {. }
$$
By (2), only
$$
\left\{\begin{array}{l}
x_{1}+10=1, \\
x_{2}+10=1 .
\end{array}\right.
$$
That is, $x_{1}=x_{2}=-9$.
Substituting into equation (1) gives $\overline{a b}=18, \overline{b a}=81$.
(2) When $a=2$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-98 \text {. }
$$
By (2) we have,
$$
x_{2}+10<0 \text {. }
$$
Thus $-98=k \cdot\left(-\frac{98}{k}\right) \cdot\left(k, \frac{98}{k}\right.$ are integers, $\left.1 \leqslant k \leqslant 9\right)$
$$
\begin{array}{l}
\therefore\left(x_{1}+10\right)\left(x_{2}+10\right) \\
=1 \times(-98)=2 \times(-49)=7 \times(-14) .
\end{array}
$$
Solving gives $\left\{\begin{array}{l}x_{1}=-9, \\ x_{2}=-108 ;\end{array}\left\{\begin{array}{l}x_{1}=-8, \\ x_{2}=-59 ;\end{array}\left\{\left\{\begin{array}{l}x_{1}=-3, \\ x_{2}=-24\end{array}\right.\right.\right.\right.$.
Substituting into equation (1) gives $\overline{a b}=27, \overline{b a}=72$.
At this time, $x_{1}=-3, x_{2}=-24$.
(3) When $a=3$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-197=1 \times(-197) \text {. }
$$
By (2) we know there is no solution.
(4) When $a=4$,
$$
\begin{array}{l}
\left(x_{1}+10\right)\left(x_{2}+10\right) \\
=-296=1 \times(-296)=2 \times(-148) \\
=4 \times(-74)=8 \times(-37) .
\end{array}
$$
By (2) we know $k=1,2,4$, there is no solution.
Solving gives $x_{1}=-2, x_{2}=-47$.
Substituting into equation (1) gives $\overline{a b}=49, \overline{b a}=94$.
(5) When $a=5$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-395=5 \times(-75)$, there is no solution.
(6) When $a=6$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-494=2 \times(-247) \text {, no solution. }
$$
(7) When $a=7$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-593=1 \times(-593)$, there is no solution.
(8) When $a=8$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-692=4 \times(-173)$, there is no solution.
(9) When $a=9$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-791=7 \times(-113)$, there is no solution.
In summary, when $\overline{a b}=18,27,49$, the integer solutions of the equation are: $\left\{\begin{array}{l}x_{1}=-9, \\ x_{2}=-9 ;\end{array}\left\{\begin{array}{l}x_{1}=-3, \\ x_{2}=-24 ;\end{array}\left\{\begin{array}{l}x_{1}=-2, \\ x_{2}=-47 .\end{array}\right.\right.\right.$
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Through the vertex of isosceles $\triangle A B C$, draw a line intersecting the extension of the opposite side at $D$. If the resulting triangles are all isosceles, then $\triangle A B C$ has $\qquad$ such configurations.
|
4.5.
Draw a straight line from the base angle to intersect with the extension of the opposite side, there are 5 $\triangle A B C$ that meet the conditions, with base angles of $45^{\circ} 、 72^{\circ} 、 36^{\circ} 、 \frac{180^{\circ}}{7}$ 、 $\frac{2 \times 180^{\circ}}{7}$.
Draw a straight line from the vertex angle to intersect with the extension of the opposite side, there is 1 $\triangle A B C^{\prime}$ that meets the conditions, with a base angle of $72^{\circ}$, which is repeated above.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(20 points) On a certain day in a certain month in 2001 at midnight, teachers and students of a certain school were watching the live broadcast from Moscow of the announcement of the host city for the 2008 Summer Olympic Games. Beijing won the bid. In their excitement, they noticed that the number of people present was the day number, the number of boys was the month number, and the numbers of teachers, students, month, and day were all prime numbers. The number of boys was the highest, and the number of teachers was the lowest. After calculation, the sum of the number of students, the month number, and the day number, minus the number of teachers, was equal to the number of the 2008 Olympic Games, which was no more than 30. Try to answer:
(1) On which day and month did Beijing win the bid for the 2008 Olympic Games?
(2) Which edition of the Olympics was the 2008 Games?
|
Let the number of teachers be $a$, the number of students be $b$, the number of months be $c$, the number of days be $d$, and the number of girls be $e$. From the problem, we have
$$
\left\{\begin{array}{l}
a+b=d, \\
c+p=b . \\
b+c+d-a \leqslant 30, \\
au$, so $e$ is even.
Substituting (1) into (1), we get $2 b+c \leqslant 30$.
Since $c<b$, we have $3 c<2 b+c \leqslant 30$.
Given $c<10$, we know $c=3,5,7$.
When $c=3$, from $2<e<3$, there is no solution.
When $c=5$, from $2<p<5, e=4$, and $b=9$, which does not meet the problem's conditions.
When $c=7$, from $2<e<7, e=4$ or $e=6$. If $e=6$, then $b=13, d=15$. This does not meet the problem's conditions. If $e=4$, then $b=11, d=13$.
In summary, $a=2, b=11, c=7, d=13$.
Thus, $b+c+d-a=29$.
Answer: Beijing won the right to host the 2008 Olympic Games on July 13, 2001, and the 2008 Olympics was the 29th.
|
29
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of triangles with unequal integer sides and a perimeter less than 13 is $\qquad$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
(3)
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. The smallest positive integer $x$ that satisfies the equation $\tan 19 x^{\circ}=\frac{\cos 99^{\circ}+\sin 99^{\circ}}{\cos 99^{\circ}-\sin 99^{\circ}}$ is $x=$ $\qquad$ .
|
2.36.
The right side of the original equation $=\frac{1+\tan 99^{\circ}}{1-\tan 99^{\circ}}=\frac{\tan 45^{\circ}+\tan 99^{\circ}}{1-\tan 45^{\circ} \cdot \tan 99^{\circ}}$ $=\tan 144^{\circ}$.
Then $19 x=144+180 k(k \in \mathbf{N})$.
So $x=\frac{180 k+144}{19}=9 k+7+\frac{9 k+11}{19}$.
We know $19 \mid 9 k+11$. Thus, $k_{\operatorname{man}}=3$, hence,
$$
\begin{array}{l}
x=9 \times 3+7+ \\
\frac{9 \times 3+11}{19}=36 .
\end{array}
$$
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In Pascal's Triangle, each number is the sum of the two numbers directly above it. The first few rows of this triangle are as follows:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Row 0 & \multicolumn{6}{|c|}{1} \\
\hline Row 1 & & & 1 & & 1 & \\
\hline Row 2 & & 1 & & 2 & & 1 \\
\hline Row 3 & & & 3 & & 3 & 1 \\
\hline Row 4 & 1 & 4 & & 6 & & 4 \\
\hline
\end{tabular}
Then, in Pascal's Triangle, three consecutive numbers with a ratio of $3: 4: 5$ will appear in the $\qquad$ row.
|
5. Line 62.
In Pascal's Triangle, its $n$-th row consists of binomial coefficients $C_{n}^{k}$ $(k=0,1, \cdots, n)$. If the ratio of three consecutive terms in the $n$-th row of Pascal's Triangle is $3: 4: 5$, then there exists a positive integer $k$ such that
$$
\begin{array}{l}
\frac{3}{4}=\frac{C_{n}^{k-1}}{C_{n}^{k}}=\frac{k!(n-k)!}{(k-1)!(n-k+1)!}=\frac{k}{n-k+1}, \\
\frac{4}{5}=\frac{C_{n}^{k}}{C_{n}^{k+1}}=\frac{(k+1)!(n-k-1)!}{k!(n-k)!}=\frac{k+1}{n-k} .
\end{array}
$$
This leads to $\left\{\begin{array}{l}3 n-7 k=-3, \\ 4 n-9 k=5 .\end{array}\right.$
Solving the system of equations yields $k=27, n=62$.
|
62
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) Given point $A(\sqrt{5}, 0)$ and the curve $y=$ $\sqrt{\frac{x^{2}}{4}-1}(2 \leqslant x \leqslant 2 \sqrt{5})$ with points $P_{1} 、 P_{2}, \cdots$ 、 $P_{n}$. If $\left|P_{1} A\right| 、\left|P_{2} A\right| 、 \cdots 、\left|P_{n} A\right|$ form an arithmetic sequence with common difference $d \in\left(\frac{1}{5}, \frac{1}{\sqrt{5}}\right)$, find the maximum value of $n$.
|
The given curve is a segment of the following hyperbola, i.e.,
$$
\frac{1}{4} x^{2}-y^{2}=1(2 \leqslant x \leqslant 2 \sqrt{5}, y \geqslant 0) \text {. }
$$
$A(\sqrt{5}, 0)$ is its right focus, as shown in Figure 5 (where the line $l$ is the right directrix $x=\frac{4}{\sqrt{5}}$, and the point $P(2 \sqrt{5}, 2)$, with the eccentricity $\left.e=\frac{\sqrt{5}}{2}\right)$.
It is easy to see that $\left|P_{n} A\right|_{\text {min }}=\sqrt{5}-2$,
$$
\left|P_{n} A\right|_{\text {max }}=e|P H|=\frac{\sqrt{5}}{2}\left(2 \sqrt{5}-\frac{4}{\sqrt{5}}\right)=3 \text {. }
$$
According to the problem, we can set the first term of the arithmetic sequence $a_{1}=\sqrt{5}-2$, and the $n$-th term $a_{n}=3$. Then $3=(\sqrt{5}-2)+(n-1) d$.
This gives $d=\frac{5-\sqrt{5}}{n-1}(n>1)$.
By the problem, $\frac{1}{5}<d<\frac{1}{\sqrt{5}}$, i.e., $\frac{1}{5}<\frac{5-\sqrt{5}}{n-1}<\frac{1}{\sqrt{5}}$.
This yields $5 \sqrt{5}-4<n<26-5 \sqrt{5}$.
And $7=5 \times 2.2-4<5 \sqrt{5}-4$,
and $26-5 \sqrt{5}<26-5 \times 2.2=15$,
thus $7<n<15$.
Therefore, $n$ can be at most 14.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $P_{n}=(1+1)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{7}\right) \cdots$ $\left(1+\frac{1}{3 n-2}\right)$. Find the greatest integer part of $P_{2000}$.
|
$$
\begin{array}{l}
\text{First, prove the following inequality:} \\
\sqrt[3]{\frac{7 n+1}{7 n-6}}1) .
\end{array}
$$
The inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{7 n+1}{7 n-6}0 \\
\Leftrightarrow 27(n-1)^{2}+13(n-1)>0 .
\end{array}
$$
The last inequality holds for $n>1$, thus inequality (1) holds.
Similarly, inequality (2) ( (2) $\Leftrightarrow n>1$ ) can also be proven to hold.
Next, find $\left[P_{2000}\right]$.
From inequalities (1) and (2), we get
$$
\begin{array}{l}
P_{2000}>(1+1) \times \sqrt[3]{\frac{15}{8}} \times \sqrt[3]{\frac{22}{15}} \times \ldots \times \\
\sqrt[3]{\frac{72000+1}{7 \times 2000-6}} \\
=\sqrt[3]{14001} \doteq 24.1, \\
P_{2000}<(1+1) \times \sqrt[3]{\frac{2}{1}} \times \sqrt[3]{\frac{3}{2}} \times \ldots \times \sqrt[3]{2000} \\
=\sqrt[3]{16000} \doteq 25.2 .
\end{array}
$$
Therefore, $\left[P_{2000}\right]=25$.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: In a 10000-meter race on a 400-meter circular track at a school sports meet, two athletes, A and B, start running at the same time. B runs faster than A. At the 15th minute, A speeds up. At the 18th minute, A catches up with B and begins to overtake B. At the 23rd minute, A catches up with B again, and at 23 minutes and 50 seconds, A reaches the finish line. Therefore, the time B takes to complete the entire course is $\qquad$ minutes.
(Ninth Hope Cup National Mathematics Invitational Competition)
|
Solution: According to the problem, draw the figure, as shown in Figure 2.
The broken line $O A B$ represents the movement of A, and the line segment $O C$ represents the movement of B. Since A catches up with B again at the 23rd minute, $E F=400$. And $B G=C K=10000$.
Since $\triangle D E F \sim \triangle D H B$, according to the ratio of corresponding heights of similar triangles equals the similarity ratio, we get
$\frac{E F}{B H}=\frac{M V}{M G}$, which is $\frac{400}{B H}=\frac{5}{5 \frac{5}{6}}$.
Solving this, we get $B H=\frac{1400}{3}$.
Thus, $G H=B G-B H=10000-\frac{1400}{3}$
$$
=\frac{28600}{3} \text {. }
$$
From $\triangle O G H \sim \triangle O K C$, we get
$\frac{G H}{C K}=\frac{O G}{O K}$, which is $\frac{28600}{10000}=\frac{23 \frac{5}{6}}{O K}$.
Therefore, $O K=25$.
Hence, the time B takes to complete the entire course is 25 minutes.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 On a street $AB$, person A walks from $A$ to $B$, and person B rides a bike from $B$ to $A$. Person B's speed is 3 times that of person A. At this moment, a public bus departs from the starting station $A$ and heads towards $B$, and a bus is dispatched every $x$ minutes. After some time, person A notices that a bus catches up to him every 10 minutes, while person B feels that he encounters a bus every 5 minutes. Therefore, the interval time $x=$ $\qquad$ for the dispatch of buses at the starting station.
(9th Hope Cup National Mathematics Invitational for Junior High School)
|
Solution: According to the problem, draw the graph as shown in Figure 4.
$A C, A_{1} C_{1}, A_{2} C_{2}, A_{3} C_{3}$ represent the motion graphs of buses departing every $x$ minutes, $A D$ and $B . V$ are the motion graphs of person A and person B, respectively. $E_{1}, E_{2}$ are the points where a bus catches up with person A every 10 minutes. $F_{1}, F_{2}, F_{3}, F_{4}$ are the points where person B encounters a bus every 5 minutes.
Draw $E_{1} G \perp t$-axis through $E_{1}$, draw $F_{2} H / / t$-axis intersecting $O C$ at $H$, and draw $F_{1} M \perp H F_{2}$, then $A G=$ 10. $M F_{2}=5, H F_{2}=A A_{1}=x$.
Let $\angle E_{1} A G=\alpha, \angle F_{1} H M=\angle E_{1} A_{1} G=$ $\beta, \angle F_{1} F_{2} M=\gamma$. Also,
$\tan \alpha=\frac{E_{1} G}{A G}=\frac{E_{1} G}{10}$.
In $\triangle E_{1} A_{1} G$,
$\tan \beta=\frac{E_{1} G}{A_{1} G}=\frac{E_{1} G}{A G-A A_{1}}=\frac{E_{1} G}{10-x}$.
In $\triangle F_{1} H M$,
$\tan \beta=\frac{F_{1} M}{H M}=\frac{F_{1} M}{H F_{2}-M F_{2}}=\frac{F_{1} M}{x-5}$.
$\tan \gamma=\frac{F_{1} M}{M F_{2}}=\frac{F_{1} M}{5}$.
(1) $\div$ (2) gives $\frac{\tan \alpha}{\tan \beta}=\frac{10-x}{10}$.
(3) $\div$ (4) gives $\frac{\tan \beta}{\tan \gamma}=\frac{5}{x-5}$.
Since person B's speed is 3 times that of person A, then $\tan \gamma=3 \tan \alpha$.
(3) $\times$ (6) gives $\frac{1}{3}=\frac{10-x}{2(x-5)}$. Solving for $x$ yields $x=8$.
Therefore, the interval between departures at the starting station is 8 minutes.
The key here is equations (3) and (6). Their practical significance is the same as in Example 2.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 A mall has an escalator moving uniformly from the bottom to the top. Two people, A and B, are in a hurry to go upstairs. While riding the escalator, they both climb the stairs at a uniform speed. A reaches the top after climbing 55 steps, and B's climbing speed is twice that of A (the number of steps B climbs per unit time is twice that of A), and he reaches the top after climbing 60 steps. Therefore, the number of steps from the bottom to the top of the escalator is $\qquad$
(1998, Beijing Middle School Mathematics Competition for Grade 8 Preliminary)
|
Solution: According to the problem, draw the graph, as shown in Figure 5.
$O C, O B, O A$ are the motion graphs of the escalator, person A riding the escalator, and person B riding the escalator, respectively.
Draw $A M \perp t$-axis, intersecting $O B$ and $O C$ at $E$ and $F$, respectively. Draw $B N \perp t$-axis, intersecting $O C$ at $G$.
Assume that the escalator from the ground floor to the upper floor has $x$ steps. Then $A M = B N = x$, $F M, E F, A F$ are the number of steps the escalator rises, the number of steps person A climbs, and the number of steps person B climbs after time $t_1$, respectively. $B G$ is the number of steps person A climbs after time $t_2$. Therefore,
\[
\begin{array}{l}
B C = 55, A F = 60, \\
F M = x - 60, G N = x - 55.
\end{array}
\]
Since person B climbs the stairs at twice the speed of person A, $A F = 2 E F$. Thus, $E F = 30$.
\[
\begin{aligned}
& \because \triangle O E F \sim \triangle O B G, \\
\therefore & \frac{E F}{B G} = \frac{O M}{O N}. \\
\because & \triangle O M F \sim \triangle O N G, \\
\therefore & \frac{F M}{G N} = \frac{O M}{O N}.
\end{aligned}
\]
From (1) and (2), we get $\frac{E F}{B G} = \frac{F M}{G N}$,
which is $\frac{30}{55} = \frac{x - 60}{x - 55}$. Solving for $x$ gives $x = 66$.
Therefore, the escalator from the ground floor to the upper floor has 66 steps.
The key here is equation (3), which can be transformed into $\frac{E F}{F M} = \frac{B G}{G N}$, meaning that when the speed is constant, the ratio of the distances traveled by the two people in the same time is equal.
In summary, using a "distance-time" graph to solve travel application problems can transform implicit conditions into explicit conditions, making abstract concepts more intuitive, thus converting abstract thinking into visual thinking.
|
66
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) Given the algebraic expression $-x^{3}+100 x^{2}+x$, the letter $x$ is only allowed to take values within the set of positive integers. When the value of this algebraic expression reaches its maximum, what is the value of $x$? Prove your conclusion.
|
Let $x=k$, the value of the algebraic expression is $a_{k}$.
$$
a_{k+1}-a_{k}=-3 k^{2}+197 k+100 \text {. }
$$
When $00$;
When $k \geqslant 67$, $a_{k+1}-a_{k}<0$.
$\therefore$ When $x=67$, the algebraic expression reaches its maximum value.
|
67
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Nine, (15 points) satisfying the following two conditions:
(1) For all natural numbers $x, x^{2}-2001 x+n$ $\geqslant 0$
(2) There exists a natural number $x_{0}$, such that $x_{0}^{2}-2002 x_{0}+n$ $<0$ the number of positive integers $n$ equals $\qquad$
|
$Nine, 1001$
|
1001
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example $8 A$ and $B$ are two fixed points on a plane. Find a point $C$ on the plane such that $\triangle A B C$ forms an isosceles right triangle. There are $\qquad$ such points $C$.
|
Solution: As shown in Figure 5, the vertices of the two isosceles right triangles with $AB$ as the hypotenuse are $C_{1}$ and $C_{2}$; the vertices of the four isosceles right triangles with $AB$ as the legs are $C_{3}, C_{4}, C_{5}, C_{6}$.
Therefore, there are a total of 6 points $C$ that meet the conditions.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. As shown in Figure 5, in $\square A B C D$, $P_{1}$, $P_{2}, \cdots$, $P_{n-1}$ are the $n$ equal division points on $B D$. Connect $A P_{2}$ and extend it to intersect $B C$ at point $E$, and connect $A P_{n-2}$ and extend it to intersect $C D$ at point $F$.
(1) Prove that $E F \parallel B D$;
(2) Let the area of $\square A B C D$ be $S$. If $S_{\triangle A E F} = \frac{3}{8} S$, find the value of $n$.
|
14. (1) Since $A D / / B C, A B / / D C$. Therefore, $\left.\triangle P_{n-2} F I\right) \triangle \triangle P_{n-2} A B, \triangle P_{2} B E \subset \triangle P_{2} D . A$.
Thus, we know
$$
\frac{A P_{n-2}}{P_{n+2} F}=\frac{B P_{n}}{P_{n-2} D}=\frac{n-2}{2}, \frac{A P_{2}}{P_{2} E}=\frac{D P_{2}}{P_{2} B}=\frac{n-2}{2},
$$
which means $\frac{A P_{n-2}}{P_{n} \frac{{ }_{2}}{} F}=\frac{A P_{2}}{P_{2} E}$; hence $E F / / B D$.
(2) From (1), we know $\frac{A P_{2}}{A E}=\frac{n-2}{n}, \frac{S_{\triangle A P_{2} P_{n-2}}}{S_{\text {ZAFF }}}=$ $\left(\frac{A P_{2}}{A E}\right)^{2}$. Therefore,
$$
\begin{aligned}
& S_{\triangle_{2} P_{n-2}}=\left(\frac{n-2}{n}\right)^{2} \cdot S_{\text {IAE }} \\
= & \left(\frac{n-2}{n}\right)^{2} \cdot \frac{3}{8} S .
\end{aligned}
$$
Also, since $S_{\triangle A A S}=\frac{1}{2} S_{\triangle A / \times D}=\frac{1}{2} S$, and $P_{1}, P_{2}, \cdots$ 。 $P_{n}$ 。are the $n$ equal division points of $B D$. Therefore,
$$
\begin{array}{l}
S_{-N_{2} p_{n}}=\left(\frac{n-4}{n}\right) \cdot S_{\text {U(S) }} \\
=\left(\frac{n-4}{n}\right) \cdot \frac{1}{2} S .
\end{array}
$$
From (1) and (2), we get $\left(\frac{n-2}{n}\right)^{2} \cdot \frac{3}{8} S=\left(\frac{n-4}{n}\right) \cdot \frac{1}{2} S$,
which simplifies to $n^{2}-4 n-12=0$.
Solving, we get $n=6, n=-2$ (discard). Therefore, $n=6$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (18 points) The real numbers $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy
$$
\begin{array}{l}
\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001 . \\
\text { Let } y_{k}=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2 .
\end{array}
$$
$\cdots, 2$ 001. Find the maximum possible value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$.
|
For $k=1,2, \cdots, 2000$, we have
$$
\begin{array}{l}
\left|y_{k}-y_{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}}{k}-\frac{x_{1}+x_{2}+\cdots+x_{k}+x_{k+1}}{k+1}\right| \\
=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\right| \\
=\frac{\left|\left(x_{1}-x_{2}\right)+2\left(x_{2}-x_{3}\right)+\cdots+k\left(x_{k}-x_{k+1}\right)\right|}{k(k+1)} \\
\leqslant \frac{\left|x_{1}-x_{2}\right|+2\left|x_{2}-x_{3}\right|+\cdots+k\left|x_{k}-x_{k+1}\right|}{k(k+1)} .
\end{array}
$$
By the identity
$$
\begin{array}{l}
\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{(n-1) n} \\
=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right) \\
=1-\frac{1}{n}
\end{array}
$$
and its result
$$
\begin{array}{l}
\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}+\cdots+\frac{1}{(n-1) n} \\
=\frac{1}{k}\left(1-\frac{k}{n}\right),
\end{array}
$$
we can obtain $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$
$$
\begin{aligned}
\leqslant & \left|x_{1}-x_{2}\right|\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{2000 \cdot 2001}\right) \\
& +2\left|x_{2}-x_{3}\right|\left(\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{2000 \cdot 2001}\right) \\
& \quad+\cdots+2000 \cdot\left|x_{2000}-x_{2001}\right| \cdot \frac{1}{2000 \cdot 2001} \\
= & \left|x_{1}-x_{2}\right|\left(1-\frac{1}{2001}\right)+\left|x_{2}-x_{3}\right|\left(1-\frac{2}{2001}\right) \\
& \quad+\cdots+\left|x_{20000}-x_{2001}\right|\left(1-\frac{2000}{2001}\right) \\
\leqslant & \sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|\left(1-\frac{1}{2001}\right)=2000 .
\end{aligned}
$$
Equality holds if and only if $\left|x_{1}-x_{2}\right|=2001, x_{2}=x_{3}=\cdots=x_{2001}$, especially when $x_{1}=2001, x_{2}=x_{3}=\cdots=x_{2001}=0$, the equality can be achieved.
Therefore, the maximum value of $\sum_{k=1}^{2000}\left|y_{k}-y_{k+1}\right|$ is 2000.
(Proposed by: Li Dayuan, Liu Hongkun, Xiong Bin, Ye Shengyang, Yu Yinglong)
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The graph of the quadratic function $y=-\frac{1}{2} x^{2}+\frac{1999}{2} x+1000$ passes through ( ) integer lattice points in the first quadrant (i.e., points with positive integer coordinates).
(A) 1000
(B) 1001
(C) 1999
(D) 2001
|
6. (C).
$y=-\frac{1}{2}(x-2000)(x+1)$ The range of the independent variable in the first quadrant is $0<x<2000$. When $x$ is odd, $x+1$ is even; when $x$ is even, $x-2000$ is even. Therefore, when $x=1,2, \cdots, 1999$, $y$ takes integer values.
Thus, the number of lattice points in the first quadrant is 1999.
|
1999
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. From a square iron sheet with a side length of 10 cm, circular pieces with a diameter of 1 cm can be cut out, at most $\qquad$ pieces.
|
4.106.
As shown in Figure 6, it is easy to see that $\angle \mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ is an equilateral triangle with a side length of $1 \mathrm{~cm}$, so the height $A O_{2}$ is $\frac{\sqrt{3}}{2} \mathrm{~cm}$. Assuming we can fit $k$ rows, then
$$
\frac{\sqrt{3}}{2}(k-1)+1 \leqslant 10 \text {. }
$$
Solving this, we get $k \leqslant 1+6 \sqrt{3}$, with the integer solution being $k \leqslant 11$.
Thus, 6 rows have 10 circles, and 5 rows have 9 circles, totaling 105 circles. In fact, if we arrange the last three rows to each have 10 circles, and the 9 circles are only in 4 rows, then the total is $\frac{1}{2} \sqrt{3} \times 8+1 \times 3=4 \sqrt{3}+3<10$.
Therefore, the maximum number of circles that can be placed is $9 \times 4+10 \times 7=106$.
|
106
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Several containers are unloaded from a cargo ship, with a total weight of 10 tons, and the weight of each container does not exceed 1 ton. To ensure that these containers can be transported in one go, how many trucks with a carrying capacity of 3 tons are needed at least?
|
First, note that the weight of each container does not exceed 1 ton, so the weight of containers that each vehicle can carry at one time will not be less than 2 tons; otherwise, another container can be added.
Let $n$ be the number of vehicles, and the weights of the containers they carry be $a_{1}, a_{2}, \cdots, a_{n}$, then $2 \leqslant a_{1} \leqslant 3(i=1,2, \cdots, n)$.
Let the total weight of all the containers carried be $s$, then
$$
2 n \leqslant s=a_{1}+a_{2}+\cdots+a_{n} \leqslant 3 n \text {. }
$$
That is, $2 n \leqslant 10 \leqslant 3 n$.
Thus, $\frac{10}{3} \leqslant n \leqslant 5$, which means $n=4$ or 5.
In fact, 4 vehicles are not enough.
Suppose there are 13 containers, each weighing $\frac{10}{13}$ tons. Since $\frac{10}{13} \times 3 < 3$, each vehicle can only carry 3 containers, so 4 vehicles cannot carry all of them.
Therefore, at least 5 trucks are needed.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For any positive integer $n$, connect the origin $O$ with the point $A_{n}(n, n+3)$, and let $f(n)$ denote the number of all integer points on the line segment $O A_{n}$ except for the endpoints. Then the value of $f(1)+f(2)+f(3)+\cdots+f(2002)$ is ( ).
(A) 2002
(B) 2001
(C) 1334
(D) 667
|
3. (C).
The integer points $(x, y)$ on the line segment $O A_{n}$ satisfy $1=\frac{n+3}{n} \cdot x(0<x<n$, and $x \in \mathbf{N})$. When $n=3 k(k \in \mathbf{N})$,
$$
y=\frac{k+1}{k} \cdot x(0<x<3 k, x \in \mathbf{N}) .
$$
$\because k$ and $k+1$ are coprime,
$\therefore$ only when $x=k$ or $2 k$, $y \in \mathbf{N}$. Thus, $f(3 k)=2$.
When $n=3 k+1(k \in \mathbf{N})$,
$$
y=\frac{3 k+4}{3 k+1} \cdot x(0<x<3 k+1, x \in \mathbf{N}) \text {. }
$$
$\because 3 k+4$ and $3 k+1$ are coprime,
$$
\therefore f(3 k+1)=0 \text {. }
$$
When $n=3 k-1(k \in \mathbf{N})$, similarly, $f(3 k-1)=0$.
Therefore, $f(1)+f(2)+f(3)+\cdots+f(2002)$
$$
=2\left[\frac{2002}{3}\right]=1334 .
$$
|
1334
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. From the 99 natural numbers $1,2,3, \cdots, 99$, the number of ways to choose two different numbers such that their sum is less than 99 is $\qquad$ ways.
|
2.2352 .
Let the two selected numbers be $x$ and $y$, then $x+y$ has three cases:
$$
x+y=100, x+y<100, x+y>100 \text {. }
$$
Below, we only consider the first two scenarios:
(1) When $x+y=100$, the number of ways to select is 49;
(2) When $x+y<100$. This indicates that the number of ways to select when $x+y<100$ is equal to the number of ways when $x+y>100$, and their values are both $\frac{\mathrm{C}_{49}^{2}-49}{2}$.
In the case of $x+y<100$, there are $x+y=99$ and $x+y<99$, and the selection method for $x+y=99$ should be excluded, which is 49.
Therefore, the number of ways that satisfy the condition is $\frac{\mathrm{C}_{99}^{2}-49}{2}-49=2352$.
|
2352
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A rectangular piece of land enclosed by fences has a length and width of $52 \mathrm{~m}$ and $24 \mathrm{~m}$, respectively. An agricultural science technician wants to divide this land into several congruent square test plots. The land must be fully divided, and the sides of the squares must be parallel to the boundaries of the land. There are $2002 \mathrm{~m}$ of fencing available. The land can be divided into a maximum of $\qquad$ square test plots.
|
6.702 pieces.
Assume the land is divided into several squares with side length $x$, then there exist positive integers $m, n$, such that
$$
\frac{24}{x}=m \text{, and } \frac{52}{x}=n \text{. }
$$
$\therefore \frac{m}{n}=\frac{6}{13}$, i.e., $m=6 k, n=13 k(k \in \mathbf{N})$.
Note that when the value of $k$ is as large as possible, the number of experimental plots reaches its maximum. The total length of the fence used to divide the land into several squares with side length $x \mathrm{~m}$ is
$$
L=(m-1) \times 52+(n-1) \times 24=624 k-76 .
$$
Since the maximum length of the fence available is $2002 \mathrm{~m}$. Therefore, we have
$624 k-76 \leqslant 2002$, which gives $k \leqslant \frac{2078}{624} \approx 3.33$.
Thus, $k_{\max }=3$. At this point, the total number of squares is $m n=702$ (pieces).
|
702
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 In Figure 1, there are 8 vertices, each with a real number. The real number at each vertex is exactly the average of the numbers at the 3 adjacent vertices (two vertices connected by a line segment are called adjacent vertices). Find
$$
a+b+c+d-(e+f+g+h)
$$
|
The following solution has appeared in a journal:
Given
$$
\begin{array}{l}
a=\frac{b+e+d}{3}, \\
b=\frac{a+f+c}{3}, \\
c=\frac{b+g+d}{3}, \\
d=\frac{c+h+a}{3}.
\end{array}
$$
Adding the four equations yields
$$
\begin{array}{l}
a+b+c+d \\
=\frac{1}{3}(2 a+2 b+2 c+2 d+e+f+g+h),
\end{array}
$$
which simplifies to $a+b+c+d-(e+f+g+h)=0$.
The above solution, though clever, has two drawbacks.
First, it is too coincidental. Adding the equations to get $a+b+c+d$ is expected, but the fact that it simplifies to $a+b+c+d-(e+f+g+h)=0$ is purely coincidental. If the problem were to find $2a+b+c+d-(2e+f+g+h)$, the above method would no longer work.
Second, the solution fails to reveal the essence of the problem, which is the relationship among the 8 numbers $a, b, c, d, e, f, g, h$.
To solve this problem, one should first consider a specific example. To make each vertex number equal to the average of the adjacent vertex numbers, the simplest example is to take
$$
a=b=c=d=e=f=g=h.
$$
In fact, this is the only possibility. Without loss of generality, assume $a$ is the largest of the 8 numbers. Since $a$ is the average of $b, e, d$, $b, e, d$ must also be the largest, i.e., equal to $a$. By the same reasoning, all 8 numbers must be equal. Thus, whether it is $a+b+c+d-(e+f+g+h)$ or $2a+b+c+d-(2e+f+g+h)$, the result is clearly 0.
Therefore, equations (1), (2), etc., are just superficial, while the equality of the 8 numbers is the true essence.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the sum of all numbers in the following square matrix
\begin{tabular}{ccccc}
1901 & 1902 & $\cdots$ & 1949 & 1950 \\
1902 & 1903 & $\cdots$ & 1950 & 1951 \\
$\cdots$ & & & & \\
1950 & 1951 & $\cdots$ & 1998 & 1999
\end{tabular}
|
Many people first use formula (5) to find the sum of the first row:
$$
S_{1}=\frac{1901+1950}{2} \times 50=96275 \text {; }
$$
Then they find the sums of the 2nd, 3rd, ..., 50th rows; finally, they add these sums together. If one notices that each number in the second row is 1 greater than the corresponding number in the previous row, then the sums of the rows are:
$$
\begin{array}{l}
S_{2}=96275+50, \\
S_{3}=96275+50 \times 2, \\
\cdots \cdots \\
S_{50}=96275+50 \times 49 .
\end{array}
$$
Thus, one can use (5) again to find $S_{1}+S_{2}+\cdots+S_{50}$. However, a simpler approach is to apply the idea of averaging. By averaging each number with the number symmetrically opposite to it about the center, each average is $\frac{1901+1999}{2}=1950$.
Thus, the total sum is
$$
1950 \times 50 \times 50=4875000 \text {. }
$$
It is evident that even a simple and common example contains a lot of content, and it is not the case that a single abstraction can strip it of all its essence and spirit.
|
4875000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find all real numbers $p$ such that the cubic equation $5 x^{3}$ $-5(p+1) x^{2}+(71 p-1) x+1=66 p$ has three roots that are all natural numbers.
|
Analysis: It can be observed that 1 is a solution to the equation. The equation can be transformed into
$$
(x-1)\left(5 x^{2}-5 p x+66 p-1\right)=0 \text {. }
$$
The problem is then reduced to: Find all real numbers $p$ such that the equation
$$
5 x^{2}-5 p x+66 p-1=0
$$
has natural number solutions.
By Vieta's formulas, $p$ is the sum of the roots of the equation, i.e., $p$ is a natural number. Following Example 2, we get
$$
\begin{array}{l}
\Delta=(5 p-132)^{2}-17404 . \\
\text { Let }(5 p-132)^{2}-17404=n^{2}(n>0, n \text { is a natural number}). \\
\end{array}
$$
Rearranging and factoring, we get
$$
\begin{array}{l}
(5 p-132+n)(5 p-132-n) \\
=2^{2} \times 19 \times 229 . \\
\text { Also, }(5 p-132+n),(5 p-132-n) \text { have the same parity, }
\end{array}
$$
Therefore,
$$
\left\{\begin{array}{l}
5 p-132+n=2 \times 229 \\
5 p-132-n=2 \times 19
\end{array}\right.
$$
Solving, we get $p=76$.
Note: On the surface, $p$ in this problem is any real number, but by Vieta's formulas, it is actually a natural number, so the method used above can be applied to find it.
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: When A was B's current age, B was 10 years old; when B was A's current age, A was 25 years old. Who is older, A or B? How many years older?
|
Solution: Let the age difference between A and B be $k$ years, which is an undetermined constant. When $k>0$, A is older than B; when $k<0$, A is younger than B. Then, A's age $y$ and B's age $x$ have a linear relationship:
$$
y=x+k \text {. }
$$
After designing this dynamic process, the given conditions become 3 "moments" of this process:
(1) The current ages of B and A are $(x, y)$;
(2) "When A was B's current age, B was 10 years old", which is equivalent to taking $(10, x)$, i.e.,
$$
x=10+k \text {; }
$$
(3) "When B was A's current age, A was 25 years old", which is equivalent to taking $(y, 25)$, i.e.,
$$
25=y+k .
$$
By combining (1)+(2)+(3) and eliminating $x, y$, we get
$$
k=5 \text {. }
$$
Answer: A is 5 years older than B.
2. Solve local problems in a global context
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If $\left(1+x+x^{2}+x^{3}\right)^{5}\left(1-x+x^{2}-\right.$ $\left.x^{3}\right)^{5}=a_{30}+a_{29} x+\cdots+a_{1} x^{29}+a_{0} x^{30}$, find $a_{15}$.
|
Let $f(x)=\left(1+x+x^{2}+x^{3}\right)^{5}$, then the original expression is
$$
F(x)=f(x) f(-x)
$$
which is an even function. Therefore, we have
$$
\begin{array}{l}
\boldsymbol{F}(x)=\frac{1}{2}[\boldsymbol{F}(x)+\boldsymbol{F}(-x)] \\
=a_{30}+a_{28} x^{2}+\cdots+a_{2} x^{28}+a_{0} x^{30},
\end{array}
$$
where all the coefficients of the odd powers of $x$ are zero, hence $a_{15}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 There are 5 medicine boxes, every 2 boxes contain one same medicine, each medicine appears in exactly 2 boxes, how many kinds of medicines are there?
|
Solution: Represent the medicine boxes as 5 points. When a medicine box contains the same medicine, draw a line segment between the corresponding points.
Since every 2 medicine boxes have one kind of the same medicine, a line should be drawn between every two points. Also, because each kind of medicine appears in exactly 2 medicine boxes, there is exactly one line between every two points. This results in Figure 2. The number of lines in Figure 2 is the number of different medicines, so there are 10 kinds of medicines.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Find the smallest positive integer $n$ such that for this $n$, there is a unique positive integer $k$ satisfying
$$
\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13} \text {. }
$$
(Fifth American Mathematical Invitational)
|
Solution: Since the fractions $\frac{8}{15}$ and $\frac{7}{13}$ satisfy $15 \times 7 - 8 \times 13 = 1$, it must be that $\frac{n}{n+k} = \frac{8+7}{15+13} = \frac{15}{28}$, thus we get $n=15, k=13$.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $a, b$ be positive integers and satisfy
$$
\frac{2}{3}<\frac{a}{b}<\frac{5}{7} \text {. }
$$
When $b$ is the minimum value, $a+b=$ $\qquad$
(Fifth "Hope Cup" National Mathematics Invitational Competition Training Question)
|
Solution: According to the theorem, we have
$$
\frac{a}{b}=\frac{2+5}{3+7}=\frac{7}{10} \text {. }
$$
Thus, $a=7, b=10$. Therefore, $a+b=17$.
From the above theorem, it can be seen that although addition and subtraction is a rather clever and simple method, it is not suitable for the case where $b c-a d>1$. For example, the rational number with the smallest denominator between $\frac{2}{5}$ and $\frac{7}{9}$ is not $\frac{2+7}{5+9}=\frac{9}{14}$, but $\frac{1}{2}$. Therefore, the method of addition and subtraction does not have universal applicability.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
0, \underbrace{1,1, \cdots, 1}_{k i}, \underbrace{2,2, \cdots,}_{k \uparrow}, 2 \underbrace{3,3, \cdots, 3}_{k \uparrow}, \cdots \text {. }
$$
Find $a_{n}$ and $S_{n}$. When $k=3$, calculate the values of $a_{2002}$ and $S_{2002}$.
|
Solution: This problem can be regarded as an equal-segment increment sequence, the sum of the first $m$ segments is
$$
1+\underbrace{k+k+\cdots+k}_{(m-1)+}=k(m-1)+1 .
$$
It is evident that, if and only if
$$
k(m-1)+2 \leqslant n \leqslant k m+1
$$
then, $a_{n}=m$. Solving the inequality (6) for $m$, we get
$\frac{n-1}{k} \leqslant m \leqslant \frac{n+k-2}{k}$, the same term has $m(m=1,2, \cdots)$ terms.
$$
\because 0<\frac{n+k-2}{k}-\frac{n-1}{h}<1 \text {, }
$$
$\therefore m$ is the integer part of ${ }^{n}+h-2$, denoted as
$$
m=\left[\frac{n+k-2}{k}\right] \text {. }
$$
Thus, $a_{n}=\left[\frac{n+k-2}{k}\right](k \geqslant 2)$.
When $k=3$,
$$
a_{n}=\left[\frac{n+1}{3}\right], a_{2003}=\left[\frac{2003}{3}\right]=667 \text {. }
$$
Therefore, the sum of the first $n$ segments of the sequence $a_{n}$ is
$$
\begin{array}{l}
S_{n \in \mathrm{E}}=0+1 \cdot k+2 \cdot k+\cdots+(n-1) k \\
=k[1+2+\cdots+(n-1)]=\frac{k}{2} n(n-1) .
\end{array}
$$
Replacing $n$ with $m$, the sum of the first $m$ segments is
$$
\begin{array}{l}
S_{k(m-1)+1}=\frac{1}{2} k m(m-1) . \\
\because a_{n}=\left[\frac{n+k-2}{k}\right],
\end{array}
$$
$\therefore$ the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is
$$
\begin{array}{l}
S_{n}=S_{k(m-1)+1}+[(n-1)-k(m-1)] m \\
=\frac{1}{2} k(m-1) m+m n-m-k m(m-1) \\
=m n-m-\frac{1}{2} k m(m-1),
\end{array}
$$
where $m=\left[\frac{n+k-2}{k}\right]$.
When $k=3$, $S_{n}=m n-m-\frac{3}{2} m(m-1)$.
$$
\begin{array}{l}
\text { Also, } \because m=\left[\frac{2003}{3}\right]=667, \\
\begin{array}{l}
\therefore S_{2002}= 667 \times 2002-667 \\
-\frac{3}{2} \times 667 \times(667-1) \\
=1335334-667-666333=668334 .
\end{array}
\end{array}
$$
|
668334
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $a=1999 x+2000, b=1999 x+2001, c$ $=1999 x+2002$. Then the value of the polynomial $a^{2}+b^{2}+c^{2}-a b-b c-$ $c a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
|
2. (D).
$$
\begin{array}{l}
\because a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] . \\
\text { Also } a-b=-1, b-c=-1, c-a=2, \\
\therefore \text { the original expression }=\frac{1}{2}\left[(-1)^{2}+(-1)^{2}+2^{2}\right]=3 .
\end{array}
$$
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. The number of integers $n$ that satisfy $\left(n^{2}-n-1\right)^{n+2}=1$ is.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. However, if the task is to translate the problem statement itself, the translation would be:
11. The number of integers $n$ that satisfy $\left(n^{2}-n-1\right)^{n+2}=1$ is.
|
11.4 .
From $n+2=0, n^{2}-n-1 \neq 0$, we get $n=-2$;
From $n^{2}-n-1=1$, we get $n=-1, n=2$;
From $n^{2}-n-1=-1$ and $n+2$ is even, we get $n=0$.
Therefore, $n=-1,-2,0,2$ for a total of 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. For a certain project, if contracted to Team A and Team B, it will be completed in $2 \frac{2}{5}$ days, costing 180,000 yuan; if contracted to Team B and Team C, it will be completed in $3 \frac{3}{4}$ days, costing 150,000 yuan; if contracted to Team A and Team C, it will be completed in $2 \frac{6}{7}$ days, costing 160,000 yuan. Now the project is to be contracted to a single team, under the condition of ensuring completion within a week, which team will have the lowest cost?
|
Three, 13. Let the number of days required for A, B, and C to complete the task individually be \( x, y, z \) respectively. Then,
$$
\left\{\begin{array} { l }
{ \frac { 1 } { x } + \frac { 1 } { y } = \frac { 5 } { 1 2 } . } \\
{ \frac { 1 } { y } + \frac { 1 } { z } = \frac { 4 } { 1 5 } , } \\
{ \frac { 1 } { z } + \frac { 1 } { x } = \frac { 7 } { 2 0 } . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
x=4, \\
y=6, \\
z=10 .
\end{array}\right.\right.
$$
Let the daily cost for A, B, and C working individually be \( u, r, \mathbb{u} \) yuan respectively. Then,
$$
\left\{\begin{array} { l }
{ \frac { 1 2 } { 5 } ( u + v ) = 1 8 0 0 0 0 , } \\
{ \frac { 1 5 } { 4 } ( r + u ) = 1 5 0 0 0 0 , } \\
{ \frac { 2 0 } { 7 } ( u + u ) = 1 6 0 0 0 0 . }
\end{array} \text { Solving, we get } \left\{\begin{array}{l}
u=45500, \\
r=29500 \\
u=10500 .
\end{array}\right.\right.
$$
Thus, the cost for A to complete the task alone is
$$
45500 \times 4=182000 \text { (yuan); }
$$
The cost for B to complete the task alone is
$$
29500 \times 6=177000 \text { (yuan); }
$$
And C cannot complete the task within a week.
Therefore, the cost for B to complete the task is the lowest:
|
177000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, 18 football teams are participating in a single round-robin tournament, meaning each round the 18 teams are divided into 9 groups, with each group's two teams playing one match. In the next round, the teams are regrouped to play, for a total of 17 rounds, ensuring that each team plays one match against each of the other 17 teams. After $n$ rounds of the tournament according to any feasible schedule, there always exist 4 teams that have played a total of only 1 match among themselves. Find the maximum possible value of $n$.
(Li Chengzhang, contributor)
|
```
3. Consider the following competition program:
1.(1.2)(3.4)(5.6)(7.8)(9,18)
(10,11)(12,13)(14,15)(16,17)
2.(1,3)(2.4)(5,7)(6.9)(8,17)
(10,12)(11,13)(14,16)(15.18)
3.(1.4)(2.5)(3.6)(8.9)(7.16)
(10,13)(11,14)(12,15)(17,18)
4.(1,5)(2.7)(3.8)(4,9)(6.15)
(10.14)(11,16)(12.17)(13.18)
5.(1,6)(2,8)(3,9)(4,7)(5,14)
(10,15)(11,17)(12,18)(13.16)
6.(1,7)(2.9)(3.5)(6,8)(4,13)
(10,16)(11,18)(12,14)(15,17)
7.(1.8)(2.6)(4.5)(7.9)(3.12)
(10.17)(11,15)(13,14)(16,18)
8. (1,9)(3,7)(4,6)(5,8)(2,11)
(10,18)(12,16)(13.15)(14.17)
9.(2.3)(4.8)(5.9)(6.7)(1.10)
(11,12)(13,17)(14,18)(15,16)
10.(1,11)(2.12)(3,13)(4,14)(5,15)
(6,16)(7,17)(8,18)(9,10)
11.(1.12)(2,13)(3,14)(4,15)(5,16)
(6,17)(7,18)(8,10)(9,11)
12.(1,13)(2.14)(3,15)(4,16)(5,17)
(6,18)(7.10)(8.11)(9.12)
.....
17.(1,18)(2,10)(3,11)(4,12)(5,13)
(6,14)(7,15)(8,16)(9,17)
The first 9 teams are called Group A, and the last 9 teams are called Group B. It is easy to see that after 9 rounds, any two teams in the same group have already played against each other. Therefore, any 4 teams have already played at least two matches, which of course does not meet the requirements of the problem.
If the above program is reversed and the matches are played in the new order, then after 8 rounds, any two teams in the same group have not played against each other. Each team has played 1 match against 8 of the 9 teams in the other group. At this point, any 4 teams in the same group have not played against each other, and any 4 teams not all in the same group have played at least two matches. Of course, none of these meet the requirements of the problem.
The above discussion shows that the maximum possible value of \( n \) is no more than 7.
Assume that 7 rounds of matches have been played and no 4 teams meet the requirements of the problem.
Select two teams \( A_1 \) and \( A_2 \) that have played one match. Thus, each team has played against 6 other teams. The two teams have played against at most 12 other teams. Therefore, there are at least 4 teams \( B_1, B_2, B_3, B_4 \) that have not played against \( A_1 \) and \( A_2 \). By the assumption of contradiction, \( B_1, B_2, B_3, \) and \( B_4 \) have all played against each other.
Teams \( B_1 \) and \( B_2 \) have each played 3 matches among the 4 teams \( \{B_1, B_2, B_3, B_4\} \). Thus, each team has played 1 match against 4 of the other 14 teams. This leads to at least 6 teams \( C_1, C_2, \cdots, C_6 \) that have not played against \( B_1 \) and \( B_2 \). By the assumption of contradiction, \( C_1, C_2, \cdots, C_6 \) have all played against each other.
Teams \( C_1 \) and \( C_2 \) have each played 5 matches among the 6 teams \( \{C_1, C_2, \cdots, C_6\} \). Thus, each team has played 1 match against 2 of the other 12 teams. Therefore, there are at least 8 teams \( D_1, D_2, \cdots, D_8 \) that have not played against \( C_1 \) and \( C_2 \). By the assumption of contradiction, these 8 teams have all played against each other. This way, \( D_1 \) and \( D_2 \) have not played against 10 other teams. Since only 7 rounds have been played, there must be two teams \( E_1 \) and \( E_2 \) among the 10 other teams that have not played against each other. Thus, \( D_1, D_2, E_1, \) and \( E_2 \) have played a total of 1 match, which contradicts the assumption of contradiction.
In summary, the maximum number of rounds that can be played is 7.
```
|
7
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Given positive integers $x, y$, then $\frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5}$ has ( ) solutions of the form $(x, 1)$.
(A) 0
(B) 1
(C) 2
(D) More than 2, but finite
(E) Infinite
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
16. (C).
$$
\text { Given } \frac{10}{x^{2}}-\frac{1}{y}=\frac{1}{5} \Rightarrow 50 y=x^{2}(y+5) \Rightarrow x^{2}=\frac{50 y}{y+5} \text {. }
$$
If $(y, 5)=1$, then $(y, y+5)=1$, so $(y+5) \mid 50=5^{2} \times 2$. Since $y+5>2$, then $(y+5) \mid 5^{2}$, which contradicts $(y+5,5)=1$. Therefore, $5 \mid y$.
Let $y=5 y_{0}$, then $x^{2}=\frac{50 y_{0}}{y_{0}+1}=\frac{5^{2} \cdot 2 y_{0}}{y_{0}+1} \geqslant 5^{2}$.
In (I), from $y+5>y$ we have $x^{2}<50$.
Therefore, $x$ can only be 5, 6, or 7. Upon verification, $(x, y)$ has only two positive integer solutions: $(5,5),(7,245)$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1 Elimination Method
Example 1 Given that $x, y, z$ are real numbers, and $x+2y-z=6, x-y+2z=3$. Then, the minimum value of $x^{2}+y^{2}+z^{2}$ is $\qquad$
(2001, Hope Cup Junior High School Mathematics Competition)
|
Solution: From the given, we can solve for $y=5-x, z=4-x$. Then
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x^{2}+(5-x)^{2}+(4-x)^{2} \\
=3(x-3)^{2}+14 . \\
\because 3(x-3)^{2} \geqslant 0, \quad \therefore x^{2}+y^{2}+z^{2} \geqslant 14 .
\end{array}
$$
Therefore, the minimum value of $x^{2}+y^{2}+z^{2}$ is 14.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2 Factorization Method
Example 2 Let $a, b, c$ be distinct natural numbers, and $a b^{2} c^{3}=1350$. Then the maximum value of $a+b+c$ is $\qquad$
(1990, Wu Yang Cup Junior High School Mathematics Competition)
|
Solution: $\because 1350=2 \times 5^{2} \times 3^{3}=150 \times 3^{2} \times 1^{3}$, $\therefore$ when $a=150, b=3, c=1$, the maximum value of $a+b+c$ is 154.
|
154
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Let $x_{1}, x_{2}, \cdots, x_{7}$ be natural numbers. And $x_{1}$
$$
<x_{2}<\cdots<x_{6}<x_{7} \text {, and } x_{1}+x_{2}+\cdots+x_{7}=
$$
159. Find the maximum value of $x_{1}+x_{2}+x_{3}$.
(1997, Anhui Province Junior High School Mathematics Competition)
|
Solution: $\because 159=x_{1}+x_{2}+\cdots+x_{7}$
$$
\begin{array}{l}
\geqslant x_{1}+\left(x_{1}+1\right)+\left(x_{1}+2\right)+\cdots+\left(x_{1}+6\right) \\
=7 x_{1}+21, \\
\therefore x_{1} \leqslant 19 \frac{5}{7} .
\end{array}
$$
Therefore, the maximum value of $x_{1}$ is 19.
$$
\begin{array}{l}
\text { Also, } \because 19+x_{2}+x_{3}+\cdots+x_{7}=159, \\
\therefore 140 \geqslant x_{2}+\left(x_{2}+1\right)+\cdots+\left(x_{2}+5\right) \\
=6 x_{2}+15. \\
\therefore x_{2} \leqslant 20 \frac{5}{6}. \\
\end{array}
$$
Therefore, the maximum value of $x_{2}$ is 20.
$$
\begin{array}{l}
\text { From } 20+x_{3}+\cdots+x_{7}=140, \text { we have } \\
\begin{aligned}
120 & \geqslant x_{3}+\left(x_{3}+1\right)+\cdots+\left(x_{3}+4\right) \\
& =5 x_{3}+10 .
\end{aligned}
\end{array}
$$
Then $x_{3} \leqslant 22$,
Therefore, the maximum value of $x_{3}$ is 22.
Thus, the maximum value of $x_{1}+x_{2}+x_{3}$ is
$$
19+20+22=61.
$$
|
61
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. As shown in Figure 3, each face of the cube is written with a natural number, and the sum of the two numbers on opposite faces is equal. If the number opposite to 10 is a prime number $a$, the number opposite to 12 is a prime number $b$, and the number opposite to 15 is a prime number $c$, then $a^{2}+b^{2}+c^{2}-a b-a c-b c=$ $\qquad$ .
|
5.19.
Given that $10+a=12+b=15+c$. Therefore, $c=2$. Then $a=7, b=5$.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given in $\triangle A B C$, $\angle A, \angle B$ are acute angles, and $\sin A$ $=\frac{5}{13}, \tan B=2, A B=29 \mathrm{~cm}$. Then the area of $\triangle A B C$ is $\qquad$ $\mathrm{cm}^{2}$
|
6.145 .
Draw a perpendicular from point $C$ to $AB$. Let the foot of the perpendicular be $D$.
$\because \sin A=\frac{5}{13}=\frac{CD}{AC}$, let $m>0$,
$\therefore CD=5m, AC=13m$.
$\because \tan B=\frac{CD}{BD}=2$, we can set $n>0, CD=2n, BD=n$,
$\therefore BD=n=\frac{CD}{2}=\frac{5}{2}m$.
$\therefore AD=\sqrt{(13m)^2-(5m)^2}=12m$.
Thus, $AB=AD+BD=12m+\frac{5}{2}m=\frac{29}{2}m$.
From $29=\frac{29}{2}m$, we get $m=2$. Then $CD=5m=10$.
Therefore, $S_{\triangle ABC}=\frac{1}{2} AB \cdot CD=\frac{1}{2} \times 29 \times 10=145(\text{cm}^2)$.
|
145
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 Given that $a$, $b$, and $c$ are all positive integers, and the parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at two distinct points $A$ and $B$. If the distances from $A$ and $B$ to the origin are both less than 1, find the minimum value of $a+b+c$.
(1996, National Junior High School Mathematics League)
|
Solution: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, and $x_{1} < 0 < x_{2}$,
then $x_{1} < 0, \\
\therefore b > 2 \sqrt{a c} . \\
\text{Also, } \because |O A| = |x_{1}| > 1$. Therefore, the parabola opens upwards, and when $x = -1$, $y > 0$, so $a(-1)^{2} + b(-1) + c > 0$, which means $b < a + c + 1$. Since $b > 2 \sqrt{a c}$, we have $2 \sqrt{a c} + 1 < a + c$,
i.e., $(\sqrt{a} - \sqrt{c})^{2} > 1$.
From equation (2), we get $\sqrt{a} - \sqrt{c} > 1$,
i.e., $\sqrt{a} > \sqrt{c} + 1$.
Thus, $a > (\sqrt{c} + 1)^{2} \geq (\sqrt{1} + 1)^{2} = 4$.
Solving, we get $a \geq 5$.
$$
\begin{array}{l}
\text{Also, } \because b > 2 \sqrt{a c} \geq 2 \sqrt{5 \times 1} > 4, \\
\therefore b \geq 5 .
\end{array}
$$
When $a = 5, b = 5, c = 1$, the parabola $y = 5 x^{2} + 5 x + 1$ satisfies the given conditions.
Therefore, the minimum value of $a + b + c$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14 Let $a, b, c, a+b-c, a+c-b, b+c-a, a+b+c$ be seven distinct prime numbers, and the sum of two of $a, b, c$ is 800. Let $d$ be the difference between the largest and smallest of these seven prime numbers. Find the maximum possible value of $d$.
(2001, China Mathematical Olympiad)
|
$$
\begin{array}{l}
\text { Let } a<b<c<d \text { be prime numbers, and } a+b, a+c, b+c \text { are also prime numbers. } \\
\text { Without loss of generality, let } a< b, \\
\therefore c<a+b<a+c<b+c .
\end{array}
$$
Also, since one of $a+b$, $a+c$, $b+c$ is
800,
$$
\therefore c<800 \text {. }
$$
Since $799=17 \times 47$ and 798 are not prime numbers, but 797 is a prime number, we have
$$
c \leqslant 797, d \leqslant 1594 .
$$
On the other hand, when $a+b=800$, note that
$$
\begin{array}{l}
a=5, b=795, \\
a=7, b=793=13 \times 61, \\
a=11, b=789=3 \times 263
\end{array}
$$
are not all prime numbers, thus cannot meet the requirements of the problem.
However, when $a=13, b=787$, both are prime numbers. At this time, $a+b-c=3, a+c-b=23$ are also prime numbers. It is easy to verify that $b+c-a=1571$ and $a+b+c=1597$ are also prime numbers.
In summary, the maximum possible value of $d$ is 1594.
|
1594
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $[A]$ denote the greatest integer less than or equal to $A$, and set $A=38+$ $17 \sqrt{5}$. Then $A^{2}-A[A]=$ $\qquad$ .
|
$\begin{array}{l}\text { II.1.1. } \\ \because A=38+17 \sqrt{5}=(\sqrt{5}+2)^{3} \text {, let } B=(\sqrt{5}-2)^{3} \text {, } \\ \therefore A-B=76 . \\ \text { Also } \because 0<(\sqrt{5}-2)^{3}<1 \text {, } \\ \therefore[A]=76 \text {, then } A-[A]=B . \\ \text { Therefore } A^{2}-A[A]=A(A-[A])=A B=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $a$ be a real root of the equation $x^{2}-2002 x+1=0$. Then $a^{2}-2001 a+\frac{2002}{a^{2}+1}=$ $\qquad$ .
|
2.2001 .
$\because \alpha$ is a real root of the equation $x^{2}-2002 x+1=0$, then
$$
\begin{array}{l}
\alpha^{2}-2002 \alpha+1=0 . \\
\therefore \alpha+\frac{1}{\alpha}=2002 .
\end{array}
$$
Therefore, $\alpha^{2}-2001 \alpha+\frac{2002}{\alpha^{2}+1}=\alpha-1+\frac{1}{\alpha}=2001$.
|
2001
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) There are 20 weights, all of which are integers, such that any integer weight $m(1 \leqslant m \leqslant 2002)$ can be balanced by placing it on one pan of a scale and some of the weights on the other pan. What is the smallest possible value of the heaviest weight among these 20 weights?
|
Let's assume the weights of these 20 weights are \(a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{20}\), where \(a_{i} (1 \leqslant i \leqslant 20)\) are positive integers. It is easy to see that
\[
a_{1}=1, a_{k+1} \leqslant a_{1}+a_{2}+\cdots+a_{k}+1 (1 \leqslant k \leqslant 19).
\]
\[
\text{Then } a_{2} \leqslant 2, a_{3} \leqslant 4, a_{4} \leqslant 8, a_{5} \leqslant 16, a_{6} \leqslant 32,
\]
\[
\begin{array}{l}
a_{7} \leqslant 64, a_{8} \leqslant 128. \\
\text{Also, since } a_{1}+a_{2}+\cdots+a_{20} \geqslant 2002, \\
\therefore 12 a_{20} \geqslant a_{9}+a_{10}+\cdots+a_{20},
\end{array}
\]
or \(12 a_{20} \geqslant (a_{1}+a_{2}+\cdots+a_{20}) - (a_{1}+a_{2}+\cdots+a_{8})\)
\[
\geqslant 2002 - (1+2+4+\cdots+128) \geqslant 1747.
\]
Thus, \(a_{20} \geqslant 145 \frac{7}{12}\).
Since \(a_{20}\) is an integer, \(\therefore a_{20} \geqslant 146\).
Let \(a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=8, a_{5}=16, a_{6}=32\). \(a_{7}=64, a_{8}=128, a_{9}=a_{10}=a_{11}=a_{12}=a_{13}=145, a_{14}\)
\[
=a_{15}=\cdots=a_{20}=146.
\]
Using \(a_{1}, a_{2}, \cdots, a_{8}\) these 8 weights, we can measure integer weights from 1 to 255. Adding \(a_{9}\), we can measure integer weights from 1 to 400. Adding \(a_{10}\), we can measure integer weights from 1 to 545. And so on, 20 weights can measure integer weights from 1 to 2002.
Therefore, the minimum value of the heaviest weight is 146.
(Yang Jin, No. 13 Middle School, Wuhu City, Anhui Province, 241002)
|
146
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the function $y=\frac{a-x}{x-a-1}$, the graph of its inverse function is symmetric about the point $(-1,4)$. Then the value of the real number $a$ is $\qquad$ .
|
2.3.-
From the problem, we know that the graph of the function $y=\frac{a-x}{x-a-1}$ is centrally symmetric about the point $(4,-1)$.
$\because y=\frac{a-x}{x-a-1}=-1-\frac{1}{x-(a+1)}$, we have $(y+1)[x-(a+1)]=-1$,
$\therefore$ the graph of the function is a hyperbola with its center at $(a+1,-1)$.
Also, $\because$ the graph of the hyperbola is centrally symmetric about its center,
$\therefore a+1=4$, which means $a=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that line segment $A D / /$ plane $\alpha$, and the distance to plane $\alpha$ is 8, point $B$ is a moving point on plane $\alpha$, and satisfies $A B=10$. If $A D=21$, then the minimum distance between point $D$ and point $B$ is $\qquad$ .
|
4.17.
As shown in Figure 4, let the projections of points $A$ and $D$ on plane $\alpha$ be $O$ and $C$, respectively, then we have $A O = C D = 8$.
$$
\begin{array}{l}
\because A B = 10, \\
\therefore O B = 6 \text{ (constant). }
\end{array}
$$
Therefore, the trajectory of point $B$ in $\alpha$ is a circle with $O$ as the center and 6 as the radius.
$$
\begin{array}{l}
\because C D \perp \alpha, \text{ then } \\
B D^{2} = B C^{2} + C D^{2} \\
= B C^{2} + 8^{2},
\end{array}
$$
$\therefore$ The minimum value of $B D$ is to find the minimum value of $B C$.
Since point $C$ is outside the circle $\odot O$, connect $O C$ intersecting $\odot O$ at $B_{0}$. By plane geometry knowledge, we easily get
$$
\begin{array}{l}
(B C)_{\min} = B_{0} C = C O - O B_{0} = 21 - 6 = 15. \\
\text{ Hence } (B D)_{\min} = \sqrt{(B C)_{\min}^{2} + 8^{2}} = 17.
\end{array}
$$
|
17
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $S=[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\cdots+[\sqrt{2002}]$, where $[\sqrt{n}]$ denotes the greatest integer not exceeding $\sqrt{n}$. Then the value of $[\sqrt{S}]$ is
|
6.242.
Let $k^{2} \leqslant n<(k+1)^{2}$, then $k \leqslant \sqrt{n}<k+1$, so
$$
\begin{array}{l}
{[\sqrt{n}]=k .} \\
\because(k+1)^{2}-k^{2}=2 k+1,
\end{array}
$$
$\therefore$ From $k^{2}$ to $(k+1)^{2}$ there are $2 k+1$ numbers, the integer parts of whose square roots are all $k$,
$$
\begin{array}{l}
\therefore\left[\sqrt{k^{2}}\right]+\left[\sqrt{k^{2}+1}\right]+\cdots+\left[\sqrt{\left(k^{2}+1\right)^{2}-1}\right] \\
=k(2 k+1)=2 k^{2}+k . \\
\text { Also, } \because 44^{2}<2002=44^{2}+66<45^{2},
\end{array}
$$
$\therefore$ Between 1 and 2002, there are 44 perfect squares
$$
\begin{array}{l}
1^{2}, 2^{2}, \cdots, 44^{2} . \\
\therefore S=[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{2002}] \\
\quad=2\left(1^{2}+\cdots+43^{2}\right)+(1+\cdots+43)+44 \times 67 \\
\quad=58762 .
\end{array}
$$
Thus $[\sqrt{S}]=[\sqrt{58762}]=242$.
|
242
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) There are 2002 points distributed on a circle. Now, they are arbitrarily colored white or black. If starting from a certain point and moving in any direction around the circle to any point, the total number of white points (including the point itself) is always greater than the number of black points, then the point is called a good point. To ensure that there is at least one good point on the circle, find the maximum number of black points that can be colored.
|
From the problem, we know that a good point must be white. The following discussion is for the general case: there are $3n+1$ points on the circumference, which are colored black and white. Only when the number of black points $\leqslant n$, can we ensure that there must be a good point.
We prove this by mathematical induction on $n$.
(1) When $n=1$, there are 4 points on the circumference. The black and white coloring is one black point and three white points. Among the three consecutive white points, take the middle one, and it is easy to see that it is a good point. Therefore, the conclusion holds when $n=1$.
(2) Suppose the proposition holds when $n=k$. Then, when $n=k+1$, take any one of the $k+1$ black points and denote it as $A$. Take the nearest white points to $A$ on both sides and denote them as $B$ and $C$. Remove these three points $A$, $B$, and $C$ from the circumference, leaving $3k+1$ points, of which $k$ are black. By the induction hypothesis, there must be a good point among these $3k+1$ points, denoted as $D$ (a white point). Then, put $A$, $B$, and $C$ back onto the circumference, resulting in $3(k+1)+1$ points.
Now we prove that $D$ is still a good point.
Indeed, since $D$ is a white point, point $D$ must be outside the arc $\overparen{BAC}$. Therefore, when moving from point $D$ along the points on the circumference to reach the points within $\overparen{BA}$ (or $\overparen{CA}$) (excluding point $A$), the difference between the number of white points and black points is increased by 1 compared to the original difference. Thus, when reaching point $A$, the difference between the number of white points and black points must be greater than 0, which means $D$ is still a good point. Therefore, the conclusion holds when $n=k+1$.
On the other hand, when the total number of black points is $n+1$, there is a black and white coloring such that no good point exists: $n+1$ black points divide the circumference into $n+1$ small arcs. Then, place the remaining $2n$ white points into these $n+1$ small arcs so that no arc has more than 2 white points, which is always possible. This coloring ensures that no good point exists.
$$
\because 2002=3 \times 667+1 \text {, }
$$
$\therefore$ To ensure the existence of a good point, the total number of black points $\leqslant 667$. In summary, the maximum number of black points is 667.
(Xu Jingling, Anhui Province, Susong Middle School, 246501)
|
667
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the largest constant $k$, such that for all real numbers $a, b, c, d$ in $[0,1]$, the inequality
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant k\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
holds.
|
Solution: First, estimate the upper bound of $k$.
When $a=b=c=d=1$, we have $4 k \leqslant 4+4, k \leqslant 2$.
Next, we prove that for $a, b, c, d \in[0,1]$, it always holds that
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
First, we prove a lemma.
Lemma If $x, y \in[0,1]$, then
$$
x^{2} y+1 \geqslant x^{2}+y^{2} \text {. }
$$
This inequality is equivalent to $(y-1)\left(x^{2}-y-1\right) \geqslant 0$.
Thus, inequality (2) holds. By substituting $a, b$ and $b, c$ and $c, d$ and $d, a$ for $x, y$ in inequality (1), we get
$$
\begin{array}{l}
a^{2} b+1 \geqslant a^{2}+b^{2}, b^{2} c+1 \geqslant b^{2}+c^{2}, \\
c^{2} d+1 \geqslant c^{2}+d^{2}, d^{2} a+1 \geqslant d^{2}+a^{2} .
\end{array}
$$
Adding the above four inequalities, we obtain
$$
\begin{array}{l}
a^{2} b+b^{2} c+c^{2} d+d^{2} a+4 \\
\geqslant 2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) .
\end{array}
$$
In conclusion, the maximum value of $k$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the smallest positive integer $k$, such that for all $a$ satisfying $0 \leqslant a \leqslant 1$ and all positive integers $n$, we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$
|
Solution: First, we aim to eliminate the parameter $a$, and then it will be easier to find the minimum value of $k$. Using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{k+n}=\frac{k}{k+n} .
\end{array}
$$
Therefore, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only if $a=\frac{k(1-a)}{n}$, i.e., $a=\frac{k}{n+k}$.
Thus, we need to find the smallest positive integer $k$ such that for any positive integer $n$, we have
$$
\frac{k^{k} n^{n}}{(n+k)^{n+k}}<\frac{1}{(n+1)^{3}}.
$$
When $k=1$, taking $n=1$ leads to a contradiction in equation (4);
When $k=2$, taking $n=1$ leads to a contradiction in equation (4);
When $k=3$, taking $n=3$ leads to a contradiction in equation (4).
Therefore, $k \geqslant 4$.
Next, we prove that when $k=4$, equation (4) holds, i.e.,
$$
4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}.
$$
When $n=1,2,3$, it is easy to verify that equation (5) holds.
When $n \geqslant 4$, using the arithmetic-geometric mean inequality, we get
$$
\begin{array}{l}
\sqrt[n+4]{4^{4} n^{n}(n+1)^{3}} \\
=\sqrt[n+4]{16(2 n)(2 n)(2 n)(2 n) n^{n-4}(n+1)^{3}} \\
\leqslant \frac{16+4 \times 2 n+n(n-4)+3(n+1)}{n+4} \\
=\frac{n^{2}+7 n+19}{n+4}<n+4 .
\end{array}
$$
In conclusion, the minimum value of $k$ is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If $xy=1$, then the minimum value of the algebraic expression $\frac{1}{x^{4}}+\frac{1}{4 y^{4}}$ is $\qquad$ .
(1996, Huanggang City, Hubei Province, Junior High School Mathematics Competition)
|
$$
\text { Sol: } \begin{aligned}
\because & \frac{1}{x^{4}}+\frac{1}{4 y^{4}}=\left(\frac{1}{x^{2}}\right)^{2}+\left(\frac{1}{2 y^{2}}\right)^{2} \\
& \geqslant 2 \cdot \frac{1}{x^{2}} \cdot \frac{1}{2 y^{2}}=1,
\end{aligned}
$$
$\therefore \frac{1}{x^{4}}+\frac{1}{4 y^{4}}$'s minimum value is 1.
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In right $\triangle ABC$, the area is 120, and $\angle BAC=90^{\circ}$. $AD$ is the median of the hypotenuse. A line $DE \perp AB$ is drawn from point $D$ to point $E$. Connect $CE$ to intersect $AD$ at point $F$. Then the area of $\triangle AFE$ is ( ).
(A) 18
(B) 20
(C) 22
(D) 24
|
4. (B).
As shown in Figure 6, draw $F G \perp A B$ at $G$, then $F G / / D E / / A C$. Therefore, we have
$$
\begin{array}{l}
\frac{F G}{\frac{1}{2} A C}=\frac{F G}{D E}=\frac{A G}{A E}, \\
\frac{F G}{A C}=\frac{E G}{A E} .
\end{array}
$$
(1) + (2) gives $\frac{3 F G}{A C}=\frac{A E}{A E}=1$.
Thus, $F G=\frac{1}{3} A C$.
Hence, $S_{\text {LAFE }}=\frac{1}{2} A E \times F G$
$$
=\frac{1}{2}\left(\frac{1}{2} A B \times \frac{1}{3} A C\right)=\frac{1}{6} S_{\triangle A B C}=20 \text {. }
$$
|
20
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A and Person B go to a discount store to buy goods. It is known that both bought the same number of items, and the unit price of each item is only 8 yuan and 9 yuan. If the total amount spent by both on the goods is 172 yuan, then the number of items with a unit price of 9 yuan is $\qquad$ pieces.
Person A and Person B go to a discount store to buy goods,
it is known that both bought
the same number of items, and the unit price of each
item is only 8 yuan and 9 yuan. If the total amount spent by both on the goods is
172 yuan, then the number of items with a unit price of 9 yuan is $\qquad$ pieces.
|
3.12.
Suppose each person bought $n$ items, among which $x$ items cost 8 yuan each, and $y$ items cost 9 yuan each. Then we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x + y = 2 n , } \\
{ 8 x + 9 y = 172 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=18 n-172, \\
y=172-16 n .
\end{array}\right.\right. \\
\because x \geqslant 0, y \geqslant 0, \therefore\left\{\begin{array}{l}
18 n-172 \geqslant 0, \\
172-16 n \geqslant 0 .
\end{array}\right.
\end{array}
$$
Solving, we get $9 \frac{5}{9} \leqslant n \leqslant 10 \frac{3}{4}$.
Thus, $n=10$. Therefore, $y=172-160=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $N=23 x+92 y$ be a perfect square, and $N$ does not exceed 2,392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs.
|
4.27.
$\because N=23 x+92 y=23(x+4 y)$, and 23 is a prime number, $N$ is a perfect square not exceeding 2392,
$\therefore x+4 y=23 m^{2}$ ( $m$ is a positive integer) and
$N=23^{2} \cdot m^{2} \leqslant 2392$,
thus $m^{2} \leqslant \frac{2392}{23^{2}}=\frac{104}{23}<5$.
Solving, we get $m^{2}=1$ or 4.
When $m^{2}=1$, from $x+4 y=23$, we get $y=1,2,3,4$, $5, x=19,15,11,7,3$;
When $m^{2}=4$, from $x+4 y=92$, we get $y=1,2,3,4$, $5, \cdots, 22, x=88,84,80, \cdots, 4$.
Therefore, there are $(1,19),(2,15),(3,11),(4,7),(5,3)$ and $(1,88),(2,84), \cdots,(22,4)$.
Hence, the number of pairs $(x, y)$ that satisfy the conditions is $5+22=27$.
|
27
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five, (15 points) 1 and 0 alternate to form the following sequence of numbers:
$$
101,10101,1010101,101010101, \cdots
$$
Please answer, how many prime numbers are there in this sequence? And please prove your conclusion.
|
Obviously, 101 is a prime number.
Below is the proof that $N=\underbrace{101010 \cdots 01}_{k \uparrow 1}(k \geqslant 3)$ are all composite numbers (with $k-1$ zeros in between).
$$
\begin{aligned}
11 N= & 11 \times \underbrace{10101 \cdots 01}_{k \uparrow 1} \\
& =\underbrace{1111 \cdots 11}_{2 k \uparrow 1}=\underbrace{11 \cdots 1}_{k \uparrow 1} \times\left(10^{k}+1\right) .
\end{aligned}
$$
(1) When $k$ is an odd number not less than 3, according to the divisibility rule for 11, 11 does not divide $\underbrace{11 \cdots 11}_{k \uparrow 1}$, so, $111\left(10^{k}+1\right)$, i.e.,
$$
\frac{10^{k}+1}{11}=M_{1}>1 \text {. }
$$
Thus, $N=\underbrace{11 \cdots 11}_{k \uparrow 1} \times\left(\frac{10^{k}+1}{11}\right)=\underbrace{11 \cdots 11}_{k \uparrow 1} \times M_{1}$. Therefore, $N$ is a composite number.
(2) When $k$ is an even number not less than 3, it is easy to see that $11 \mid \underbrace{11 \cdots 11}_{k \uparrow 1}$.
i.e., $\frac{11 \cdots 11}{11}=M_{2}>1$.
Thus, $N=\frac{\overbrace{11 \cdots 11}^{k \uparrow 1}}{11} \times\left(10^{k}+1\right)=M_{2}\left(10^{k}+1\right)$.
Therefore, $N$ is a composite number.
In summary, when $k \geqslant 3$, $N=\underbrace{10101 \cdots 01}_{k \uparrow 1}$ must be a composite number.
Therefore, in the sequence $101,10101,1010101,101010101, \cdots \cdots$, only 101 is a prime number.
(Provided by Zhou Chunluo, Capital Normal University)
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Under the conditions $x+2 y \leqslant 3, x \geqslant 0, y \geqslant 0$, the maximum value that $2 x+y$ can reach is $\qquad$
(2000, Hope Cup Junior High School Mathematics Competition Second Trial)
|
Solution: As shown in Figure 1, draw the line $x + 2y = 3$. The set of points satisfying the inequalities $x \geqslant 0, y \geqslant 0, x + 2y \leqslant 3$ is the region $\triangle ABO$ (including the boundaries) enclosed by the line and the $x$ and $y$ axes. To find the maximum value of $s = 2x + y$, we transform $s = 2x + y$ into $y = -2x + s$, whose corresponding image is a bundle of parallel lines with a slope of -2. To find the maximum value of $s$, it is converted into finding the maximum intercept of the parallel lines when they pass through $\triangle ABO$. Clearly, when the line $y = -2x + s$ passes through $A(3,0)$, the intercept $s$ is maximized, at which point $s = 6$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, if $a$ and $b$ are positive numbers, and the parabolas $y=x^{2}+ax+2b$ and $y=x^{2}+2bx+a$ both intersect the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
(2000, National Junior High School Mathematics League)
|
Solution: From the problem, we have
$$
\Delta_{1}=a^{2}-8 b \geqslant 0, \Delta_{2}=4 b^{2}-4 a \geqslant 0 \text {. }
$$
Thus, $a^{2} \geqslant 8 b$ and $b^{2} \geqslant a$.
Since $a$ and $b$ are both positive numbers,
$$
\therefore a^{4} \geqslant 64 b^{2} \geqslant 64 a \text {, i.e., } a \geqslant 4 \text {. }
$$
Similarly, we have $b^{2} \geqslant a \geqslant 4, b \geqslant 2$.
Therefore, $a_{\min }=4, b_{\text {min }}=2$.
$$
\text { Hence }\left(a^{2}+b^{2}\right)_{\min }=4^{2}+2^{2}=20 \text {. }
$$
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For the arithmetic sequence $\left.\mid a_{n}\right\}$, the first term $a_{1}=8$, and there exists a unique $k$ such that the point $\left(k, a_{k}\right)$ lies on the circle $x^{2}+y^{2}=10^{2}$. Then the number of such arithmetic sequences is $\qquad$.
|
2.17.
Obviously, the point $(1,8)$ is inside the circle, and the points on the circle $(1, \pm \sqrt{9 a})$
do not satisfy the general term formula of the arithmetic sequence
$$
a_{k}=a_{1}+(k-1) d .
$$
However, the points $\left(k, \pm \sqrt{100-k^{2}}\right)(k=2,3, \cdots, 10)$ on the circle
$$
k^{2}+a_{k}^{2}=100
$$
can satisfy the arithmetic sequence (1), provided that the common difference is
$$
d=\frac{ \pm \sqrt{100-k^{2}}-8}{k-1} .
$$
When $k=2,3, \cdots, 10$, 17 common differences can be derived, corresponding to 17 arithmetic sequences with the same first term.
Note: Here, the line (1) intersects the circle (2) at one point. If we solve the system (1) and (2) and calculate the discriminant to be 0, it will yield no solution, because the intersection point is not a tangent point.
|
17
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are 100 equally divided points on a circle. The number of obtuse triangles formed by these points as vertices is $\qquad$ .
|
$6.50 \times 49 \times 48$ (or 117600).
Let $A_{t}$ be the vertex of the obtuse angle. $\angle A_{\text{s}}$
intercepts an arc of $x$ equal parts, and the other two angles intercept arcs of $y, z$ equal parts, respectively, such that
$$
\left\{\begin{array}{l}
x+y+z=100, \\
x \geqslant 51, y \geqslant 1, z \geqslant 1 .
\end{array}\right.
$$
This can be rewritten as
$$
\left\{\begin{array}{l}
(x-50)+y+z=50 . \\
x-50 \geqslant 1, y \geqslant 1, z \geqslant 1 .
\end{array}\right.
$$
We can express (1) as the sum of 50 ones. There are 49 plus signs in between. By choosing any two of these 49 plus signs and merging the numbers they separate, we obtain a positive integer solution to equation (1); conversely, for any positive integer solution to equation (1), we can split it, corresponding to "choosing 2 out of 49 plus signs." This means the indeterminate equation (1) has $\mathrm{C}_{49}^{2}$ positive integer solutions.
Letting $i$ range from $1,2, \cdots, 100$ gives the number of obtuse triangles as $100 \mathrm{C}_{49}^{2}=50 \times 49 \times 48=117600$.
|
117600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) We call non-empty sets $A_{1}, A_{2}, \cdots, A_{n}$ a $n$-partition of set $A$ if:
(1) $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A$;
(2) $A_{i} \cap A_{j}=\varnothing, 1 \leqslant i<j \leqslant n$.
Find the smallest positive integer $m$, such that for any 13-partition $A_{1}, A_{2}, \cdots, A_{13}$ of $A=\{1,2, \cdots, m \}$, there must exist a set $A_{i}(1 \leqslant i \leqslant 13)$, in which there are two elements $a, b$ satisfying
$$
b<a \leqslant \frac{9}{8} b \text {. }
$$
|
(1) First, prove that $m \geqslant 117$.
If not, $m1+\frac{13}{104} \geqslant \frac{9}{8}$.
This contradicts $b<a \leqslant \frac{9}{8} b$. Therefore, $m \geqslant 117$.
(2) Next, prove that $m=117$ satisfies the condition.
Because, in this case, the largest 14 numbers $104,105, \cdots, 117$ are distributed among 13 sets. By the pigeonhole principle, there must be two numbers $a$ and $b (b < a)$ in the same set $A_{i}(1 \leqslant i \leqslant 13)$, then
$$
104 \leqslant b<a \leqslant 117=\frac{9}{8} \times 104 \leqslant \frac{9}{8} b .
$$
In summary, the smallest $m=117$.
|
117
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For any four-digit number (digits can be the same), by changing the original order of its digits, one can always obtain the smallest four-digit number. If the difference between these two four-digit numbers is 999, find the number of such four-digit numbers.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Solution: Let the original four-digit number be $\overline{a b c d}$ (where $a, b, c, d$ represent the digits in each place), then from the given information we have
$$
\begin{array}{l}
1000 a+100 b+10 c+d-999 \\
=1000(a-1)+100 b+10 c+(d+1) .
\end{array}
$$
Obviously, when $d \leqslant 8, a>1$, $(a-1) b c(d+1)$ is the smallest four-digit number after reordering.
When $d=9$, the unit digit of the smallest four-digit number after reordering is 0, and obviously, the tens and hundreds digits should also be 0, otherwise it would not be the smallest four-digit number. This means that three of the four digits are 0, which clearly does not meet the problem's requirements.
Therefore, $d \leqslant 8$.
Since the digits in $\overline{a b c d}$ and $\overline{(a-1) b c(d+1)}$ remain unchanged, it must be true that
$$
\left\{\begin{array}{l}
a=d+1, \\
d=a-1,
\end{array} \text { i.e., } a-d=1\right. \text {. }
$$
Thus, the reordered four-digit number is $\overline{d b c}$. Therefore, we also have $d \geqslant 1$. For any $d$ in $\{1,2,3,4,5,6,7,8\}$, the reordered four-digit numbers can be:
$$
\overline{d d d e}, \overline{d d a a}, \overline{d u a u}, \overline{d 00 a}, \overline{d 0 c a}, \overline{d 0 d u}
$$
(To ensure the minimality of the new number, since the positions of $b, c$ remain unchanged, we must have (1) $d \leqslant b \leqslant c \leqslant a$ or (2) $b=0, d \leqslant c \leqslant a$ or (3) $b=c=0$. $d<a$. — Editor's note)
Clearly, these 6 numbers are all different, so they correspond to 6 different four-digit numbers that meet the problem's conditions. Therefore, there are $6 \times 8=48$ four-digit numbers that satisfy the conditions of the problem.
(Note, the problem easily includes cases with the digit 0.)
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $c=10, \frac{\cos A}{\cos B}=\frac{b}{a}=$ $\frac{4}{3}, P$ is a moving point on the incircle of $\triangle A B C$, and $d$ is the sum of the squares of the distances from $P$ to the vertices $A, B, C$. Then $d_{\text {min }}+d_{\text {max }}$ $=$ $\qquad$
(Sixth Henan Province High School Mathematics League)
|
Solution: In $\triangle ABC$, from $\frac{\cos A}{\cos B}=\frac{\sin B}{\sin A}$ we can get
$$
\sin 2A=\sin 2B.
$$
Therefore, $\angle A+\angle B=\frac{\pi}{2}$, which means $\angle C=\frac{\pi}{2}$.
$$
\text{Given } \frac{b}{a}=\frac{4}{3}, c=10,
$$
it is easy to find $b=8, a=6$, and the radius of the inscribed circle $r=2$.
As shown in Figure 1, establish a Cartesian coordinate system, let the moving point $P(x, y)$, clearly $(x-2)^{2}+(y-2)^{2}=4$. Then
$$
\begin{array}{l}
d=x^{2}+(y-8)^{2}+(x-6)^{2}+y^{2}+x^{2}+y^{2} \\
=3 x^{2}+3 y^{2}-12 x-16 y+100 \\
=3\left[(x-2)^{2}+(y-2)^{2}\right]+76-4 y \\
=88-4 y.
\end{array}
$$
Since $0 \leqslant y \leqslant 4$, by the properties of linear functions, $f(y)=88-4y$ is monotonically decreasing on $[0,4]$. Therefore,
$$
d_{\text{min}}=f(4), d_{\text{max}}=f(0).
$$
Thus, $d_{\text{min}}+d_{\text{max}}=f(4)+f(0)$
$$
=(88-16)+(88-0)=160.
$$
|
160
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given that a square has three vertices $A, B, C$ on the parabola $y=x^{2}$. Find the minimum value of the area of such a square.
(1998, Shanghai High School Mathematics Competition)
|
Solution: As shown in Figure 3, without loss of generality, assume that two of the three vertices are on the right side of the $y$-axis (including the $y$-axis). Let the coordinates of points $A$, $B$, and $C$ be $\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$, respectively, and the slope of $BC$ be $k$ $(k>0)$. Then we have
$$
\begin{aligned}
y_{3}-y_{2} & =k\left(x_{3}-x_{2}\right), \\
y_{1}-y_{2} & =-\frac{1}{k}\left(x_{1}-x_{2}\right).
\end{aligned}
$$
Since points $A$, $B$, and $C$ lie on the parabola, we have $y_{1}=x_{1}^{2}$, $y_{2}=x_{2}^{2}$, and $y_{3}=x_{3}^{2}$. Substituting these into the above equations, we get
$$
x_{3}=k-x_{2}, \quad x_{1}=-\frac{1}{k}-x_{2}.
$$
Since $|AB|=|BC|$,
$$
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} = \sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}},
$$
we have $\sqrt{1+\frac{1}{k^{2}}}\left(x_{2}-x_{1}\right)=\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)$. Therefore, $x_{2}-x_{1}=k\left(x_{3}-x_{2}\right)$. From equation (1), we get
$$
\frac{1}{k}+2 x_{2}=k\left(k-2 x_{2}\right).
$$
Thus, $k^{2}-\frac{1}{k}=(2 k+2) x_{2} \geqslant 0$. Solving this, we get $k \geqslant 1$, and $x_{2}=\frac{k^{3}-1}{2 k(k+1)}$. Therefore, the side length of the square is
$$
\begin{array}{l}
\sqrt{1+k^{2}}\left(x_{3}-x_{2}\right)=\sqrt{1+k^{2}}\left(k-2 x_{2}\right) \\
=\sqrt{1+k^{2}}\left[k-\frac{k^{3}-1}{k(k+1)}\right] \\
=\frac{k^{2}+1}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \geqslant \frac{2 k}{k} \cdot \frac{\sqrt{k^{2}+1}}{k+1} \\
\geqslant \frac{2 k}{k} \cdot \frac{\sqrt{\frac{1}{2}(k+1)^{2}}}{k+1}=\sqrt{2}.
\end{array}
$$
The equality holds if and only if $k=1$, i.e., when point $B$ is at the origin. Therefore, the minimum area of the square is 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. When $s$ and $t$ take all real values, the minimum value that can be reached by $(s+5-3|\cos t|)^{2}$ $+(s-2|\sin t|)^{2}$ is $\qquad$
(1989, National High School Mathematics Competition)
|
(The original expression can be regarded as the square of the distance between any point on the line $\left\{\begin{array}{l}x=s+5, \\ y=s\end{array}\right.$ and any point on the ellipse arc $\left\{\begin{array}{l}x=3|\cos t| \\ y=2|\sin t|\end{array}\right.$. It is known that the square of the distance from the point $(3,0)$ to the line is the smallest, with a value of 2.)
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given that $p$, $q$, $\frac{2p-1}{q}$, $\frac{2q-1}{p}$ are all integers, and $p>1$, $q>1$. Try to find the value of $p+q$.
|
Analysis: The conditions of this problem do not provide specific numbers. Yet, it requires finding the value of $p+q$, which poses a certain difficulty. If we analyze the given conditions one by one, we almost get no useful information. However, if we look at the conditions as a whole, i.e., consider $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ together, we will find that at least one of $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ is less than 2. Otherwise, if $\frac{2 p-1}{q} \geqslant 2$ and $\frac{2 q-1}{p} \geqslant 2$ were both true, then we would have
$2 p-1 \geqslant 2 q, 2 q-1 \geqslant 2 p$.
Adding these two inequalities gives
$$
2 p+2 q-2 \geqslant 2 p+2 q \text {. }
$$
which simplifies to
$$
-2 \geqslant 0 \text{. }
$$
This is clearly impossible.
By observing the given conditions as a whole, we have found a way to solve the problem.
Since at least one of $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ is less than 2, let's assume $\frac{2 p-1}{q} < 2$. From $\frac{2 p-1}{q} < 2$, we get $2 p - 1 < 2 q$, or $2 p - 2 q < 1$. Since $p$ and $q$ are integers, the only possible solution is $2 p - 2 q = 0$, which implies $p = q$. However, this contradicts the condition that $\frac{2 p-1}{q} < 2$. Therefore, we need to consider the next possible integer values. From $\frac{2 p-1}{q} < 2$, we can try $p = 3$ and $q = 5$. Substituting these values, we get $\frac{2 \cdot 3 - 1}{5} = \frac{5}{5} = 1 < 2$, which satisfies the condition. Thus, $p = 3$ and $q = 5$. Therefore, $p+q=8$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.4 In Greek mythology, the "many-headed serpent" god is composed of some heads and necks, with each neck connecting two heads. With each strike of a sword, one can sever all the necks connected to a certain head $A$. However, head $A$ immediately grows new necks connecting to all the heads it was not previously connected to (each head is connected by only one neck). Only by severing the "many-headed serpent" into two mutually disconnected parts can it be defeated. Try to find the smallest natural number $N$ such that for any "many-headed serpent" god with 100 necks, it can be defeated with no more than $N$ strikes.
|
9.410 .
We will reformulate the problem using graph theory terminology, with heads as vertices, necks as edges, and a strike that cuts the necks connected to head $A$ as a "reversal" of vertex $A$. It is easy to see that if a vertex $X$ has a degree no greater than 10, then it is sufficient to perform a "reversal" on all vertices adjacent to $X$ to make vertex $X$ "isolated". If a vertex $X$ is not adjacent to at most $n (n \leqslant 9)$ vertices, then it is sufficient to first perform a "reversal" on $X$, and then perform a "reversal" on each of these $n$ vertices to make vertex $X$ "isolated".
If each vertex has at least 11 adjacent vertices and at least 10 non-adjacent vertices, then there must be at least 22 vertices. Thus, the number of edges (necks) is no less than $22 \times 11 > 100$, which is not within the scope of consideration.
We provide an example to show that 9 strikes may not be sufficient.
Assume there are two groups of 10 heads each, with each head in one group connected to each head in the other group, resulting in exactly 100 necks.
If 9 strikes are made, then there is at least one head in each group that is not struck, denoted as $A$ and $B$. For the remaining 18 heads, by the problem's conditions, any head $C$ is connected to exactly one of $A$ or $B$ at any time (during the 9 strikes), and not to the other. Since $A$ and $B$ are connected, the "many-headed snake" remains connected.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$
Find the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
|
Analysis: If we consider the four known equations as a system of four equations in $x, y, z, w$, solving this system to find the values of $x, y, z, w$ would be quite difficult. However, if we view these four equations as a whole and temporarily treat $x, y, z, w$ as known numbers, then these four equations can be transformed into a single equation in $t$ with roots $2^{2}, 4^{2}, 6^{2}, 8^{2}$:
$$
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.
$$
At the same time, according to the problem's requirements, we only need to find the value of $x^{2}+y^{2}+z^{2}+w^{2}$, so there is no need to find the values of $x, y, z, w$ individually, but only to consider their overall sum $x^{2}+y^{2}+z^{2}+w^{2}$.
Rearranging equation (1) yields:
$$
\begin{array}{l}
t^{4}-\left(x^{2}+y^{2}+z^{2}+w^{2}+84\right) t^{3}+a_{2} t^{2} \\
+a_{3} t+a_{4}=0.
\end{array}
$$
Since $2^{2}, 4^{2}, 6^{2}, 8^{2}$ are the roots of equation (1), we have:
$$
\left(t-2^{2}\right)\left(t-4^{2}\right)\left(t-6^{2}\right)\left(t-8^{2}\right)=0.
$$
Rearranging gives:
$$
\begin{array}{l}
t^{4}-\left(2^{2}+4^{2}+6^{2}+8^{2}\right) t^{3}+a_{2} t^{2}+a_{3} t \\
+a_{4}=0.
\end{array}
$$
Comparing the coefficients of $t^{3}$ in equations (2) and (3) yields:
$$
x^{2}+y^{2}+z^{2}+w^{2}+84=2^{2}+4^{2}+6^{2}+8^{2}.
$$
Solving for $x^{2}+y^{2}+z^{2}+w^{2}$ gives $x^{2}+y^{2}+z^{2}+w^{2}=36$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11.5 Find the smallest positive integer that can be expressed as the sum of 2002 positive integers with equal sums of their digits, and also as the sum of 2003 positive integers with equal sums of their digits.
|
11.510010 .
Assume for a positive integer $n$ the following expression holds:
$$
n=a_{1}+a_{2}+\cdots+a_{2002}=b_{1}+b_{2}+\cdots+b_{2000} \text {. }
$$
Notice that the sum of the digits of $a_{1}, a_{2}, \cdots, a_{2002}$ is the same, so they have the same remainder when divided by 9. Let this remainder be $r(0 \leqslant r \leqslant 8)$. Similarly, the remainder when $b_{1}, b_{2}, \cdots, b_{2003}$ are divided by 9 is the same, denoted as $s(0 \leqslant s \leqslant 8)$.
Thus, $n-2002 r$ and $n-2003 s$ are both multiples of 9. Therefore,
$$
\begin{array}{l}
(n-2002 r)-(n-2003 s) \\
=2003 s-2002 r=2003(r+s)-4005 r
\end{array}
$$
is a multiple of 9.
Since 4005 is a multiple of 9, and 2003 is coprime with 9, $r+s$ must be a multiple of 9.
If $r=s=0$, then $n \geqslant 9 \times 2003$ (because in this case $b_{1}, b_{2}, \cdots, b_{2003}$ are all divisible by 9). If $r \neq 0$, then $r+s=9$. Thus, at least one of $r$ and $s$ is no less than 5. In this case, we get $n \geqslant 5 \times 2002$ or $n \geqslant 5 \times 2003$.
Since $10010=5 \times 2002=4 \times 2002+2002 \times 1$, and 4 and 2002 have the same sum of digits, 10010 is the desired number.
|
10010
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Player B has to guess a two-digit number from Player A. If both digits are correct, or one is correct and the other is off by 1, Player A says “near,” otherwise, Player A says “far.” For example, if Player A’s number is 65, and Player B guesses 65, 64, 66, 55, or 75, Player A says “near,” otherwise, Player A says “far.”
(a) Prove that Player B cannot guarantee knowing Player A’s number in just 18 guesses:
(b) Find a method for Player B to guarantee knowing Player A’s number in 24 guesses:
(c) Is there a method for Player B to guarantee knowing Player A’s number in 22 guesses?
|
7. (a) The first 17 guesses can exclude at most $17 \times 5=85$ numbers. There are 5 numbers left; the 18th guess cannot determine the number from the remaining 5 numbers.
(c) Construct a $10 \times 9$ table, where the cell in the $i$-th column and $j$-th row contains $(10 i+j-1)$, where $1 \leqslant i \leqslant 9,1 \leqslant j \leqslant 10$.
Divide this table by bold lines (as shown in Figure 4).
There are 11 cells of the form. It contains 55 numbers, so B first guesses 21, 26,
$34,42,47,55,63,68,71,76,84$
a total of 11 times. If A says “close,” for example, 21 is “close,” B only needs to guess $11,20,31$.
to determine A's number; if “far,” then B guesses 11 times to exclude 55 numbers.
There are 7 cells of the form ㄱ․, containing 28 numbers. So B guesses $13,18,39,50,89,92,97$ a total of 7 times. If A says “close,” for example, 13 is “close,” B only needs to guess 12,14 to determine A's number; if “far,” then B guesses 7 times to exclude 28 numbers.
At this point, there are $10,30,80,90,15,59,95$ a total of 7 numbers left.
B then guesses 20,80 a total of 2 times. If 20 is “close,” then A's number must be among 10,30, guessing 10 can determine A's number; if 80 is “close,” then A's number must be among 80,90, guessing 70 can determine A's number; if “far,” then B guesses 2 times to exclude 4 numbers.
Finally, only $15,59,95$ a total of 3 numbers are left. Guessing 2 more times can determine A's number.
Therefore, it only takes $11+7+2+2=22$ guesses to determine A's number.
At this point, (b) is clearly satisfied.
|
22
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of intersections of the function $y=x \cdot|x|-\left(4 \cos 30^{\circ}\right) x+2$ with the $x$-axis is $\qquad$
|
3.3 .
$$
y=\left\{\begin{array}{ll}
x^{2}-2 \sqrt{3} x+2, & x>0, \\
2, & x=0 . \\
-x^{2}-2 \sqrt{3} x+2, & x<0
\end{array}\right.
$$
When $x>0$, $y=x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 2 points;
When $x=0$, $y=2$ does not intersect the x-axis;
When $x<0$, $y=-x^{2}-2 \sqrt{3} x+2$ intersects the x-axis at 1 point.
Therefore, the graph of this function intersects the x-axis at 3 points.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always 3 numbers, where the sum of any two of them is still irrational.
|
Three, take 4 irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, clearly they do not satisfy the condition, hence $n \geqslant 5$.
Consider 5 irrational numbers $a, b, c, d, e$. View them as 5 points. If the sum of two numbers is a rational number, then connect the corresponding two points with a red line; otherwise, connect them with a blue line.
(1) There is no red triangle. Otherwise, assume without loss of generality that $a+b, b+c, c+a$ are all rational numbers. Since $(a+b)+(c+a)-(b+c)=2a$, this contradicts the fact that $a$ is irrational.
(2) There must be a monochromatic triangle. Otherwise, the graph must contain a red cycle, with the 5 numbers at the vertices, where the sum of any two is a rational number. Suppose $a+b, b+c, c+d, d+e, e+a$ are rational numbers, then by $(a+b)-(b+c)+(c+d)-(d+e)+(e+a)=2a$ we derive a contradiction. Thus, the monochromatic triangle must be a blue triangle. Therefore, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given that $x_{1}, x_{2}, \cdots, x_{67}$ are positive integers, and their sum equals 110. Find the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{67}^{2}$.
|
Explanation: Start with any set of 67 positive integers $x_{1}, x_{2}, \cdots, x_{67}$ whose sum is 110. Without loss of generality, assume
$$
x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{66} \leqslant x_{67}.
$$
First, freeze $x_{2}, x_{3}, \cdots, x_{66}$ and only study $x_{1}$ and $x_{67}$. Since
$$
\begin{array}{l}
\left(x_{1}-1\right)^{2}+\left(x_{67}+1\right)^{2} \\
=x_{1}^{2}+x_{67}^{2}+2+2\left(x_{67}-x_{1}\right)>x_{1}^{2}+x_{67}^{2},
\end{array}
$$
this indicates that if the smallest number $x_{1}$ is decreased by 1 and the largest number $x_{67}$ is increased by 1 (the sum of the 67 positive integers remains unchanged), their sum of squares increases.
For this reason, we perform the following adjustment:
Each time, decrease $x_{1}$ by 1 and add the 1 to $x_{67}$, until $x_{1}=1$, thus completing the adjustment for $x_{1}$.
The result of this adjustment is that the sum of the 67 positive integers remains 110, while the sum of squares increases after the adjustment compared to before.
Next, unfreeze $x_{2}$ and adjust $x_{2}$, still by decreasing $x_{2}$ by 1 and adding 1 to $x_{67}$ each time, until $x_{2}=1$, thus completing the adjustment for $x_{2}$.
Continue this process, adjusting $x_{3}, x_{4}, \cdots, x_{66}$ step by step, until the set $\left(x_{1}, x_{2}, \cdots, x_{66}, x_{67}\right)$ is adjusted to $(1,1, \cdots, 1,44)$.
At this point, since
$$
1+1+\cdots+1+44=66 \times 1+44=110,
$$
and each adjustment increases the sum of squares, the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{67}^{2}$ is
$$
\underbrace{1^{2}+\cdots+1^{2}}_{66 \text{ times}}+44^{2}=2002.
$$
|
2002
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Determine the smallest natural number $k$, such that for any $a \in [0,1]$ and any $n \in \mathbf{N}$ we have
$$
a^{k}(1-a)^{n}<\frac{1}{(n+1)^{3}} .
$$
|
Solution: By the arithmetic-geometric mean inequality, we have
$$
\begin{array}{l}
\sqrt[n+k]{a^{k}\left[\frac{k}{n}(1-a)\right]^{n}} \\
\leqslant \frac{k a+n\left[\frac{k}{n}(1-a)\right]}{n+k}=\frac{k}{n+k} .
\end{array}
$$
Thus, $a^{k}(1-a)^{n} \leqslant \frac{k^{k} n^{n}}{(n+k)^{n+k}}$.
Equality holds if and only if $a=\frac{k(1-a)}{n}\left(\right.$ i.e., $\left.a=\frac{k}{n+k}\right)$.
This means that when $a=\frac{k}{n+k}(\in[0,1])$, $a^{k}(1-a)^{n}$ attains its maximum value. Therefore, the problem can be reformulated as: determine the smallest natural number $k$ such that for any $n \in \mathbf{N}$,
$$
\frac{k^{k} n^{n}}{(n+k)^{n+k}}<\frac{1}{(n+1)^{3}} .
$$
It is easy to verify that for the pairs $(k, n)=(1,1),(2,1)$, $(3,3)$, inequality (1) does not hold. Therefore, for it to hold for all $n \in \mathbf{N}$, we must have $k \geqslant 4$.
Next, we prove that when $k=4$, inequality (1) holds, i.e., for all $n \in \mathbf{N}$, we have $4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}$.
Indeed, for $n=1,2,3$, this can be directly verified.
For $n \geqslant 4$, we have
$$
\begin{array}{l}
\sqrt[n+1]{4^{4} n^{n}(n+1)^{3}} \\
=\sqrt[n+4]{16 \times(2 n)(2 n)(2 n)(2 n) n^{n-4}(n+1)^{3}} \\
\leqslant \frac{16+8 n+(n-4) n+3(n+1)}{n+4} \\
=\frac{n^{2}+7 n+19}{n+4}<\frac{n^{2}+8 n+16}{n+4}=n+4,
\end{array}
$$
i.e., $4^{4} n^{n}(n+1)^{3}<(n+4)^{n+4}$.
In summary, the smallest value of $k$ is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that the three altitudes of $\triangle A B C$ are $A D=3, B E=4, C F=5$, and the lengths of the three sides of this triangle are all integers. Then the minimum value of the length of the shortest side is ( ).
(A) 10
(B) 12
(C) 14
(D) 16
|
5. (B).
From $S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=\frac{1}{2} C A \cdot B E=\frac{1}{2} A B \cdot C F$,
we get $3 B C=4 C A=5 A B$.
It is clear that $A B$ is the shortest side.
From $B C=\frac{5}{3} A B, C A=\frac{5}{4} A B$ and the lengths of $B C, C A, A B$ are all
integers, we know that $3 \mid A B$ and $4 \mid A B$.
$$
\because(3,4)=1, \therefore 12 \mid A B \text {. }
$$
Therefore, the minimum possible value of the shortest side $A B$ is 12.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Let $a$, $b$, and $c$ be three distinct real numbers, and $c \neq 1$. It is known that the equations $x^{2} + a x + 1 = 0$ and $x^{2} + b x + c = 0$ have a common root, and the equations $x^{2} + x + a = 0$ and $x^{2} + c x + b = 0$ also have a common root. Find the value of $a + b + c$.
|
Let the common root of the first two equations be $x_{1}$, then
$$
\begin{array}{l}
x_{1}^{2}+a x_{1}+1=0, \\
x_{1}^{2}+b x_{1}+c=0 .
\end{array}
$$
(2)
(1) - (2) gives $(a-b) x_{1}+(1-c)=0$.
$$
\because a \neq b, \quad \therefore x_{1}=\frac{c-1}{a-b} \text {. }
$$
Similarly, the common root of the last two equations is $x_{2}=\frac{a-b}{c-1}$.
$$
\therefore x_{2}=\frac{1}{x_{1}} \text {. }
$$
By Vieta's formulas, the other root of the equation $x^{2}+a x+1=0$ is $\frac{1}{x_{1}}$. Therefore, $x_{2}$ is the common root of the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$. Thus, we have
$$
x_{2}^{2}+a x_{2}+1=0, \quad x_{2}^{2}+x_{2}+a=0 .
$$
Subtracting these gives $(a-1)\left(x_{2}-1\right)=0$.
When $a=1$, these two equations have no real roots, so $a \neq 1$. Hence, we have $x_{2}=1$, and thus $x_{1}=1$. Therefore, $a=-2, b+c=-1$.
$$
\therefore a+b+c=-3 .
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the system of equations $\left\{\begin{array}{l}\frac{x}{a}+\frac{y}{b}=1, \\ x^{2}+y^{2}=50\end{array}\right.$ has only integer solutions. Then the number of real pairs $(a, b)$ that satisfy the condition is $\qquad$ .
|
6.60.
It is easy to know that the equation $x^{2}+y^{2}=50$ has 12 sets of integer solutions: $(\pm 1, \pm 7), (\pm 7, \pm 1), (\pm 5, \pm 5)$, which correspond to 12 integer points on the circle. For each pair of real numbers that satisfy the condition, there corresponds a line in the coordinate system. Therefore, the problem can be transformed into finding the number of lines passing through these 12 integer points. It is easy to know that the lines satisfying the condition fall into two categories:
(1) Lines passing through one of the integer points and tangent to the circle, totaling 12 lines;
(2) Lines passing through two of the integer points, totaling $\mathrm{C}_{12}^{2}=66$ lines.
Since the given line equation is in intercept form, lines passing through the origin or parallel to the coordinate axes should be excluded. Therefore, the number of lines satisfying the condition is $12+66-3 \times 6=60$. That is, there are 60 pairs of real numbers $(a, b)$ that satisfy the condition.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For any natural numbers $m, n$ satisfying $\frac{m}{n}<\sqrt{7}$, the inequality $7-\frac{m^{2}}{n^{2}} \geqslant \frac{\lambda}{n^{2}}$ always holds. Find the maximum value of $\lambda$.
|
(Let $G=|(m, n)| m<\sqrt{7} n, m, n \in \mathbf{N}$. $\lambda_{\text {max }}=\min _{(m, n \in 6} 17 n^{2}-m^{2}$, then perform $\bmod 7$ analysis on $7 n^{2}-m^{2}$, obtaining $\lambda_{\text {max }}=3$. )
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example $\mathbf{3}$ A factory's production on the first day does not exceed 20 units, and thereafter, the daily production increases every day, but the amount of increase each time does not exceed 20 units. Prove: When the daily production reaches 1995 units, the total number of products produced by the factory will not be less than 100500 units.
|
Explanation: Suppose the daily output on the $n$-th day after the factory starts operation reaches 1995 pieces. If the daily output on the first day is $a_{1}$ pieces, and the increase in output on the $i$-th day is $a_{i}$ pieces, then the daily output on the $n$-th day should be
$$
a_{1}+a_{2}+\cdots+a_{n}=1995 .\left(01)\right.$$, at this point, we increase $a_{k}$ by 1 and correspondingly decrease $a_{1}$ by 1, adjusting to
$$
\begin{array}{l}
a_{1}^{\prime}=a_{1}-1, a_{2}^{\prime}=a_{2}, \cdots, a_{k-1}^{\prime}=a_{k-1}, \\
a_{k}^{\prime}=a_{k}+1, a_{k+1}^{\prime}=a_{k+1}, \cdots, a_{n}^{\prime}=a_{n} .
\end{array}
$$
At this point, we still have $a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{n}^{\prime}=1995$.
However, the corresponding $S^{\prime}$ value is smaller than $S$, i.e.,
$$
S-S^{\prime}=k-1 \text {. }
$$
It can be seen that the minimum value of $S$ is obtained when $0<a_{1} \leqslant 20$ and $a_{2}=a_{3}=\cdots=a_{n}=20$. Since
$$
1995=99 \times 20+15 \text {, }
$$
then $a_{1}=15, n=100$, at this point, the minimum value can be calculated as
$$
\begin{aligned}
S & =100 \times 15+(1+2+\cdots+99) \times 20 \\
& =100500 .
\end{aligned}
$$
|
100500
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. For the polynomial $\left(\sqrt{x}+\frac{1}{2 \sqrt[4]{x}}\right)^{n}$ expanded in descending powers of $x$, if the coefficients of the first three terms form an arithmetic sequence, then the number of terms in the expansion where the exponent of $x$ is an integer is $\qquad$ .
|
8.3.
It is easy to find that the coefficients of the first three terms are $1, \frac{1}{2} n, \frac{1}{8} n(n-1)$. Since these three numbers form an arithmetic sequence, we have $2 \times \frac{1}{2} n=1+\frac{1}{8} n(n-1)$. Solving this, we get $n=8$ and $n=1$ (the latter is discarded).
When $n=8$, $T_{r+1}=\mathrm{C}_{8}^{r}\left(\frac{1}{2}\right)^{r} x^{\frac{(16-3 r)}{4}}$, where $r=0$, $1, \cdots, 8$. $r$ must satisfy $4 \mid(16-3 r)$, so $r$ can only be $0,4,8$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given that $f(x)$ is a function defined on
$\mathbf{R}$, $f(1)=1$ and for any $x \in \mathbf{R}$ we have
$$
f(x+5) \geqslant f(x)+5, f(x+1) \leqslant f(x)+1 \text {. }
$$
If $g(x)=f(x)+1-x$, then $g(2002)=$
|
10.1.
$$
\begin{array}{l}
\text { From } g(x)=f(x)+1-x \text { we get } f(x)=g(x)+x-1 . \\
\text { Then } g(x+5)+(x+5)-1 \geqslant g(x)+(x-1)+5 \text {, } \\
g(x+1)+(x+1)-1 \leqslant g(x)+(x-1)+1 . \\
\text { Therefore, } g(x+5) \geqslant g(x), g(x+1) \leqslant g(x) . \\
\therefore g(x) \leqslant g(x+5) \leqslant g(x+4) \leqslant g(x+3) \\
\quad \leqslant g(x+2) \leqslant g(x+1) \leqslant g(x) . \\
\therefore g(x+1)=g(x) .
\end{array}
$$
Thus, $g(x)$ is a periodic function with a period of 1.
Given $g(1)=1$, hence $g(2002)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 In the school football championship, it is required that each team must play a match against all the other teams. Each winning team gets 2 points, a draw gives each team 1 point, and a losing team gets 0 points. It is known that one team scored the most points, but it played fewer matches than any other team. How many teams participated at least?
|
Explanation: We call the team $A$ with the highest score the winning team.
Suppose team $A$ wins $n$ matches and draws $m$ matches, then the total score of team $A$ is $2n + m$ points.
From the given conditions, every other team must play at least $n+1$ matches, meaning they score no less than $2(n+1)$ points. Therefore,
$2n + m > 2(n+1)$. This implies $m \geqslant 3$.
Thus, there must be a team that draws with the winning team $A$, and this team's score should be no less than $2(n+1) + 1$ points, i.e.,
$$
2n + m > 2(n+1) + 1, \quad m \geqslant 4.
$$
Let there be $x$ teams in total, then the winner must win at least one match. Otherwise, its score would not exceed $x-1$ points. In this case, any other team's score would be strictly less than $x-1$ points, and the total score of all participating teams would be less than $x(x-1)$ points, but the total score of $x$ participating teams is $x(x-1)$ points, leading to a contradiction.
Thus, $m \geqslant 4, n \geqslant 1$, meaning the winning team $A$ must play at least 5 matches, and therefore, there must be at least 6 teams participating.
From this, we can estimate: at least 6 teams are participating in the competition.
We can construct a match points table: Table 1 shows 6 teams (let's call them $A, B, C, D, E, F$) participating, with the winning team $A$ having the highest score and the fewest draws.
Table 1 Match Points Table
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline & $A$ & $B$ & $C$ & $D$ & $E$ & $F$ & Score \\
\hline 4 & lle & 1 & 1 & 1 & 1 & 2 & 6 \\
\hline$B$ & 1 & lle & 2 & 0 & 0 & 2 & 5 \\
\hline i & 1 & 0 & Illo & 0 & 2 & 2 & 5 \\
\hline$D$ & 1 & 2 & 2 & & 0 & 0 & 5 \\
\hline$E$ & 1 & 2 & 0 & 2 & & 0 & 5 \\
\hline $\boldsymbol{r}$ & 0 & 0 & 0 & 2 & 2 & 16 & 4 \\
\hline
\end{tabular}
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let the quadratic function $f(x)=a x^{2}+b x+c(a, b, c \in \mathbf{R}$, $a \neq 0$ ) satisfy the following conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and $f(x) \geqslant x$;
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find the largest $m(m>1)$. Such that there exists $t \in \mathbf{R}$, for any $x \in[1, m]$, we have $f(x+t) \leqslant x$.
|
15. Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about $x=-1$. Therefore, $-\frac{b}{2a}=-1, b=2a$.
From (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From (1), $f(1) \geqslant 1$, and from (2), $f(1) \leqslant 1$,
thus $f(1)=1$, i.e., $a+b+c=1$.
Also, $a-b+c=0$, so $b=\frac{1}{2}, a=\frac{1}{4}, c=\frac{1}{4}$.
Hence, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.
Assume there exists $t \in \mathbf{R}$, such that for any $x \in[1, m]$, $f(x+t) \leqslant x$.
Taking $x=1$, we have $f(t+1) \leqslant 1$, i.e.,
$\frac{1}{4}(t+1)^{2}+\frac{1}{2}(t+1)+\frac{1}{4} \leqslant 1$. Solving this, we get $-4 \leqslant t \leqslant 0$.
For a fixed $t \in[-4,0]$, taking $x=m$, we have $f(t+m) \leqslant m$, i.e., $\frac{1}{4}(t+m)^{2}+\frac{1}{2}(t+m)+\frac{1}{4} \leqslant m$. Simplifying, we get $m^{2}-2(1-t) m+\left(t^{2}+2 t+1\right) \leqslant 0$.
Solving this, we get $1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t}$.
Thus, $m \leqslant 1-t+\sqrt{-4 t} \leqslant 1-(-4)+\sqrt{-4(-4)}=9$.
When $t=-4$, for any $x \in[1,9]$, we always have
$$
\begin{array}{l}
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right) \\
=\frac{1}{4}(x-1)(x-9) \leqslant 0 .
\end{array}
$$
Therefore, the maximum value of $m$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in minutes) for $A_{1}, A_{2}, A_{3}, A_{4}$ can all be divided by 7, while the total playing time (in minutes) for $A_{5}, A_{6}, A_{7}$ can all be divided by 13. If there are no limits on the number of substitutions per match, how many different scenarios are there for the total playing time of each player?
|
Three, let the playing time of the $i$-th player be $x_{i}$ minutes $(i=1,2$, $\cdots, 7)$, the problem is to find the number of positive integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+\cdots+x_{7}=270
$$
under the conditions $71 x_{i}(1 \leqslant i \leqslant 4)$ and $131 x_{j}(5 \leqslant j \leqslant 7)$.
If $\left(x_{1}, x_{2}, \cdots, x_{7}\right)$ is a set of positive integer solutions satisfying condition (1), then it should have
$$
\sum_{i=1}^{4} x_{i}=7 m, \sum_{i=5}^{7} x_{i}=13 n, m, n \in \mathbf{N} .
$$
Thus, $m$ and $n$ are a set of positive integer solutions to the indeterminate equation
$$
7 m+13 n=270
$$
under the conditions $m \geqslant 4$ and $n \geqslant 3$.
Since $7(m-4)+13(n-3)=203$, let $m^{\prime}=m-4$, $n^{\prime}=n-3$, we have
$$
7 m^{\prime}+13 n^{\prime}=203 \text {. }
$$
Therefore, finding (2) satisfying the conditions $m \geqslant 4, n \geqslant 3$ is equivalent to finding the non-negative integer solutions to (3).
It is easy to observe that $7 \times 2+13 \times(-1)=1$. Thus, we have
$$
7 \times 406+13 \times(-203)=203 \text {, }
$$
which means $m_{0}=406, n_{0}=-203$ is a particular integer solution to (3).
Thus, the general integer solution to (3) is
$$
\begin{array}{l}
m^{\prime}=406-13 k, n^{\prime}=-203+7 k, k \in \mathbf{Z} . \\
\text { Let } m^{\prime} \geqslant 0, n^{\prime} \geqslant 0, \text { solving gives } 29 \leqslant k \leqslant 31 .
\end{array}
$$
Taking $k=29,30,31$, we get three sets of non-negative integer solutions to (3) that satisfy the conditions
$$
\left\{\begin{array} { l }
{ m ^ { \prime } = 2 9 , } \\
{ n ^ { \prime } = 0 ; }
\end{array} \quad \left\{\begin{array} { l }
{ m ^ { \prime } = 1 6 , } \\
{ n ^ { \prime } = 7 ; }
\end{array} \quad \left\{\begin{array}{l}
m^{\prime}=3, \\
n^{\prime}=14 .
\end{array}\right.\right.\right.
$$
Thus, we get three sets of positive integer solutions to (2) that satisfy the conditions
$$
\left\{\begin{array} { l }
{ m = 3 3 , } \\
{ n = 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ m = 2 0 , } \\
{ n = 1 0 ; }
\end{array} \quad \left\{\begin{array}{l}
m=7, \\
n=17 .
\end{array}\right.\right.\right.
$$
1) When $m=33, n=3$, it is clear that $x_{5}=x_{6}=x_{7}=13$ has only one possibility;
Let $x_{i}=7 y_{i}(i=1,2,3,4)$, then, by the indeterminate equation $y_{i}+y_{2}+y_{3}+y_{4}=33$, there are $\mathrm{C}_{33-1}^{4-1}=\mathrm{C}_{32}^{3}=4960$ sets of positive integer solutions. Thus, in this case, (1) has $\mathrm{C}_{32}^{3}=4960$ sets of positive integer solutions,
2) When $m=20, n=10$, let $x_{1}=7 y_{1}(i=1,2,3,4)$, $x_{j}=13 y_{j}(j=5,6,7)$. By $y_{1}+y_{2}+y_{3}+y_{4}=20$, there are $\mathrm{C}_{19}^{3}$ sets of positive integer solutions, and by $y_{5}+y_{6}+y_{7}=10$, there are $\mathrm{C}_{9}^{2}$ sets of positive integer solutions. Thus, in this case, (1) has $\mathrm{C}_{19}^{3} \cdot \mathrm{C}_{9}^{2}=34884$ sets of positive integer solutions.
3) When $m=7, n=17$, still let $x_{i}=7 y_{i}(i=1,2,3,4)$, $x_{j}=13 y_{1}(j=5,6,7)$. By $y_{1}+y_{2}+y_{3}+y_{4}=7$ and $y_{5}+y_{6}+y_{7}=17$, there are $\mathrm{C}_{6}^{3}$ and $\mathrm{C}_{16}^{2}$ sets of positive integer solutions, respectively. Thus, in this case, (1) has $\mathrm{C}_{6}^{3} \cdot \mathrm{C}_{10}^{2}=2400$ sets of positive integer solutions.
In summary, the number of sets of positive integer solutions to (1) that satisfy the conditions is
$$
C_{32}^{3}+C_{19}^{3} \cdot C_{9}^{2}+C_{6}^{3} \cdot C_{16}^{2}=42244 .
$$
|
42244
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $A B C D$ be a rectangle with an area of 2, and let $P$ be a point on side $C D$. Let $Q$ be the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant 2 B C$;
(2) Find the value of $A Q \cdot B Q$.
(Ruo Zengru)
|
Thus, $\frac{1}{2} P A \cdot P B \sin \angle A P B=1$,
which means $P A \cdot P B=\frac{2}{\sin \angle A P B} \geqslant 2$.
Equality holds only when $\angle A P B=90^{\circ}$. This indicates that point $P$ lies on the circle with $A B$ as its diameter, and this circle should intersect with $C D$,
Therefore, when $P A \cdot P B$ takes its minimum value, we should have $B C \leqslant \frac{A B}{2}$, i.e., $A B \geqslant 2 B C$.
(2) Let the inradius of $\triangle A P B$ be $r$, then
$$
\begin{array}{l}
P A \cdot P B=(r+A Q)(r+B Q) \\
=r(r+A Q+B Q)+A Q \cdot B Q .
\end{array}
$$
And $P A \cdot P B=2 S_{\triangle A P B}, r(r+A Q+B Q)=S_{\triangle A P B}$,
Thus, $A Q \cdot B Q=S_{\triangle A P B}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given an integer array consisting of 121 integers, each integer in this array takes a value between 1 and 1000 (inclusive of 1 and 1000), and repeated values are allowed. The arithmetic mean of these numbers is $m$, and there is a unique "mode" (the number that appears most frequently) $M$ in this set of numbers. Let $D=M-m$. If $D$ is as large as possible, find $[D]$ (where $[D]$ represents the greatest integer not exceeding $D$).
|
Obviously, to make $D$ as large as possible, the mode $M$ should be as large as possible. For this purpose, let $M=1000$, and the mean $m$ should be as small as possible, so the other numbers should be as small as possible, i.e., $1,2,3, \cdots$.
Thus, the key lies in the frequency of the mode $M$, i.e., how many times $M$ appears for $D$ to be maximized.
Let $M=1000$ appear $x$ times, then $1,2, \cdots$, $\frac{121-x}{x-1}$ each appear $x-1$ times (when $\frac{121-x}{x-1}$ is an integer). At this point, the arithmetic mean is
$$
\begin{aligned}
m & =\frac{(x-1)\left[1+2+\cdots+\frac{121-x}{x-1}\right]+1000 x}{121} \\
& =\frac{20}{121} \cdot \frac{50 x^{2}-53 x+363}{x-1} .
\end{aligned}
$$
Let $m_{1}=\frac{50 x^{2}-53 x+363}{x-1}$, we now find the minimum value of $m_{1}$. The equation
$$
50 x^{2}-\left(53+m_{1}\right) x+121 \times 3+m_{1}=0
$$
has real roots, then
$$
\begin{array}{l}
\Delta=\left(53+m_{1}\right)^{2}-4 \times 50\left(121 \times 3+m_{1}\right) \\
\quad \geqslant 0, \\
m_{1}^{2}-94 m_{1}-69791 \geqslant 0 .
\end{array}
$$
Thus, $m_{1} \geqslant 315$, $m \geqslant \frac{20}{121} \times 315=52 \frac{8}{121}$, i.e., $m>52$.
At this point, $D=1000-m<948$.
Therefore, $[D]=947$. At this point, $x=4$ can be obtained.
Thus, a set of integers with $[D]=947$ can be constructed:
$$
1,1,1,2,2,2, \cdots, 39,39,39,1000,1000 \text {, }
$$
$1000,1000$.
|
947
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. We call $A_{1}, A_{2}, \cdots, A_{n}$ a $n$-partition of set $A$ if
(1) $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A$;
(2) $A_{i} \cap A_{j} \neq \varnothing, 1 \leqslant i<j \leqslant n$.
Find the smallest positive integer $m$, such that for any 14-partition $A_{1}, A_{2}, \cdots, A_{14}$ of $A=\{1,2, \cdots, m\}$, there must exist some set $A_{i}(1 \leqslant i \leqslant 14)$, in which there are two elements $a$ and $b$ satisfying $b<a \leqslant \frac{4}{3} b$.
(Leng Gangsong)
|
4. (i) If $ma > b$, and $a - b \geqslant 14$. Therefore, $b \leqslant a - 141 + \frac{14}{42} = \frac{4}{3}$.
Hence, the positive integer $m \geqslant 56$.
(ii) If $m = 56$, then for any partition of $A$ into $A_{1}, A_{2}, \cdots, A_{14}$, among the numbers $42, 43, \cdots, 56$, there must be two numbers belonging to the same $A_{i}$, such that they satisfy $b \leqslant a \leqslant \frac{4}{3} b$.
In conclusion, the smallest positive integer value of $m$ is 56.
（Liu Kangning provided）
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Given $a+b+c=0$, and $a, b, c$ are all non-zero. Then simplify $a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{a}+\frac{1}{c}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)$ to
|
16. -3 .
$$
\begin{aligned}
\text { Original expression }= & \left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right) \\
& +\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-\frac{a}{a}-\frac{b}{b}-\frac{c}{c} \\
= & \frac{0}{a}+\frac{0}{b}+\frac{0}{c}-3=-3 .
\end{aligned}
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Question: How many real roots does the equation $x^{2}|x|-5 x|x|+2 x=0$ have (where $|x|$ represents the absolute value of $x$)?
|
1.4 .
The original equation simplifies to $x(x|x|-5|x|+2)=0$. After removing the absolute value signs and discussing, we get
$$
x_{1}=0, x_{2}=\frac{5+\sqrt{17}}{2}, x_{3}=\frac{5-\sqrt{17}}{2}, x_{4}=\frac{5-\sqrt{33}}{2}
$$
as the four real roots of the original equation.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Team A and Team B each send out 5 players to participate in a chess broadcast tournament according to a pre-arranged order. The two teams first have their No. 1 players compete; the loser is eliminated, and the winner then competes with the No. 2 player of the losing team, …, until all players of one side are eliminated, and the other side wins. The sequence of wins and losses of the players thus forms a match process. Given that there are no draws in each match, how many possible match processes are there?
|
3.252 .
Due to the tournament rules, the first team to achieve 5 victories is declared the winner, even if some players have not participated. However, we can consider the players who did not participate as having lost. Thus, we get a complete ten-game result for one side: five wins and five losses. A match process, in this context, is how to arrange a team's five wins and five losses.
In ten slots, filling in 5 wins can be done in $C_{10}^{5}=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=252$ ways.
|
252
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $x_{1}=\sqrt[3]{3}, x_{2}=\left(x_{1}\right)^{\sqrt[3]{3}}$, for $n>1$ define $x_{n+1}$ $=\left(x_{n}\right)^{\sqrt[3]{3}}$. Find the smallest positive integer $n$ such that $x_{n}=27$.
|
4.7.
It is known that $x_{n}=x_{1}^{x_{1}^{n-1}}(n>1)$, and $27=(\sqrt[3]{3})^{(\sqrt[3]{3})^{6}}$. Therefore, $n-1=6$. Hence, $n=7$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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