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Example 10 When $a$ takes all real values from 0 to 5, the number of integer $b$ that satisfies $3 b=a(3 a-8)$ is $\qquad$
(1997, National Junior High School Competition)
|
Analysis: From $3 b=a(3 a-8)$, we have $b=a^{2}-$ $\frac{8}{3} a$. This is a quadratic function, and its graph is a parabola. When $a$ takes all real numbers from 0 to 5, finding the number of integer $b$ is equivalent to finding the number of integers between the maximum and minimum values of $b$.
Solution: First, draw the graph of $b=a^{2}-\frac{8}{3} a$ (note $0 \leqslant a \leqslant 5$). From the graph, we know that when $0 \leqslant a \leqslant 5$, the minimum value of $b$ is $\frac{-\left(-\frac{8}{3}\right)^{2}}{4}$ $=-\frac{16}{9}$, and the maximum value of $b$ is $f(5)=\frac{35}{3}$. There are 13 integers between $-\frac{16}{9}$ and $\frac{35}{3}$. Therefore, the number of integer $b$ is 13.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. For a real number $a>1$, written as a reduced fraction $a=\frac{q}{p}$, $(p, q)=1$. The number of values of $a$ that satisfy $p q=30!$ is $\qquad$.
|
5.512.
30! has 10 prime factors, $\left(30!=2^{26} \times 3^{14} \times 5^{7} \times 7^{7} \times 11^{2} \times 13^{2} \times 17 \times 19 \times 23 \times 29\right),(p, q)=$ 1. Each prime factor $p_{i}^{\circ}$ is either all in the numerator or all in the denominator, which gives $2^{10}$ cases in total, with those greater than 1 accounting for exactly half. Therefore, the number of $a$'s where $a > 1$ is $2^{9}=512$.
|
512
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In $\triangle A B C$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given $a^{2}+b^{2}-7 c^{2}=0$, then $\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=$
|
6.3 .
$$
\begin{array}{l}
\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\cos C \sin A \sin B}{\sin (A+B) \sin C} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^{2}} \\
=\frac{7 c^{2}-c^{2}}{2 c^{2}}=3 .
\end{array}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. On the front and back of four cards, 0 and 1, 0 and 2, 3 and 4, 5 and 6 are written respectively. By placing any three of them side by side to form a three-digit number, a total of $\qquad$ different three-digit numbers can be obtained.
|
6.124.
When the hundreds digit is 1 (or 2): $P_{3}^{2} \times 2 \times 2 \times 2=48$.
When the hundreds digit is 3 (or $4,5,6$), and the tens or units digit uses 0 pieces, there are $(05,06,50,60,00)$ 5 repetitions, so
$$
\left(P_{3}^{2} \times 2 \times 2-5\right) \times 4=76 .
$$
Therefore, the total number of different three-digit numbers is $48+76=124$.
|
124
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{c}
\text { II. (50 points) }(1) \text { For } 0 \leqslant x \leqslant 1, \text { find the range of the function } \\
h(x)=(\sqrt{1+x}+\sqrt{1-x}+2) . \\
\left(\sqrt{1-x^{2}}+1\right)
\end{array}
$$
(2) Prove: For $0 \leqslant x \leqslant 1$, there exists a positive number $\beta$ such that the inequality
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2-\frac{x^{a}}{\beta}
$$
holds for the smallest positive number $\alpha=2$. And find the smallest positive number $\beta$ at this time.
|
$$
\begin{array}{l}
=(1) 00$, the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{a}}{\beta}(x \in[0,1])$ does not hold. Conversely, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{0}}{\beta}$, which means $x^{2 \cdots} \geqslant \frac{h(x)}{2 \beta}$ holds.
Since $2-\alpha>0$, let $x \rightarrow 0$, we get
$$
0 \geqslant \frac{h(0)}{2 \beta} \text {. }
$$
Estimate $h(0)=8$.
This is impossible. This indicates that $\alpha=2$ is the smallest positive number that satisfies the condition.
Find the smallest $\beta>0$ such that the inequality $\sqrt{1+x}+\sqrt{1-x}-2 \leqslant-\frac{x^{2}}{\beta}(x \in[0,1])$, i.e., $-\frac{2 x^{2}}{h(x)} \leqslant-\frac{x^{2}}{\beta}$
holds, which is equivalent to finding
$$
\beta=\max _{0 \leqslant x \leqslant 1} \frac{1}{2} h(x) \text {. }
$$
$\because$ For any real numbers $u, v$, we have
$$
\sqrt{u}+\sqrt{v} \leqslant \sqrt{2} \bar{u}+v) \text {. }
$$
Let $u=1+x, v=1-x$, we get
$$
\sqrt{1+x}+\sqrt{1-x} \leqslant 2 .
$$
Thus, when $x \in[0,1]$,
$$
\begin{aligned}
\frac{1}{2} h(x) & =\frac{1}{2}(\sqrt{1+x}+\sqrt{1-x}+2)\left(\sqrt{1-x^{2}}+1\right) \\
& \leqslant 2\left(\sqrt{1-x^{2}}+1\right) \leqslant 4 .
\end{aligned}
$$
And $\frac{1}{2} h(0)=4$. Therefore, the maximum value of the function $\frac{1}{2} h(x)$ on $[0,1]$ is 4, i.e., the smallest positive number $\beta$ that satisfies condition (2) is 4.
$$
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For a constant $p \in N$, if the indeterminate equation $x^{2}+y^{2}=$ $p(x y-1)$ has positive integer solutions, prove that $p=5$ must hold.
|
Proof: Let $x_{0}, y_{0}$ be positive integer solutions of the equation. If $x_{0}=y_{0}$, substituting gives $p=(p-2) x_{0}^{2}$.
$$
\therefore x_{0}^{2}=\frac{p}{p-2}=1+\frac{2}{p-2} \in \mathbb{N} \text { : }
$$
Then $x_{0}^{2}=2$ or 3, which contradicts $x_{0} \in \mathbb{N}$.
Assume $x_{0}>y_{0} \geqslant 2$. Consider the original equation as a quadratic equation in $x$: $x^{2} - p y_{0} x + y_{0}^{2} + p = 0$. Its roots $x_{1}, x_{0}$ satisfy $x_{1} \geqslant x_{0} > y_{0} \geqslant 2$, and
$$
\begin{array}{l}
x_{1} + x_{0} = p y_{0}, \\
x_{1} x_{0} = y_{0}^{2} + p .
\end{array}
$$
It is easy to prove: $x_{1} x_{0} > 2 x_{1} \geqslant x_{1} + x_{0}$.
$$
\therefore y_{0}^{2} + p > p y_{0} \text {. }
$$
Thus, $py_{0} + 1$, which is a contradiction.
\end{array}
\end{array}
$$
In summary, it must be that $x_{0} > y_{0} = 1$. Substituting into the equation gives
$$
\begin{array}{l}
x_{0}^{2} - p x_{0} + p + 1 = 0 . \\
\therefore p = \frac{x_{0}^{2} + 1}{x_{0} - 1} = x_{0} + 1 + \frac{2}{x_{0} - 1} .
\end{array}
$$
When $x_{0} = 2$ or 3, $p = 5$.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 15 Find the minimum value of $\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Solution: Construct right triangles $\triangle P A C$ and $\triangle P B I$ as shown in Figure 1, such that
$$
\begin{array}{l}
A C=1, B D=2, P C \\
=x, C I=4, \text { and }
\end{array}
$$
$P C$ and $P D$ lie on line $l$. Then the problem of finding the minimum value is converted to "finding a point $I$ on line $l$ such that $P A+P B$ is minimized."
Take the symmetric point $A^{\prime}$ of $A$ with respect to $l$. Clearly,
$$
\text { the original expression }=P A+P B \geqslant A^{\prime} B=5 \text {. }
$$
Therefore, the required minimum value is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 16 The greatest integer not exceeding $(\sqrt{7}+\sqrt{5})^{6}$ is $\qquad$ .
|
Solution: Let $a=\sqrt{7}+\sqrt{5}, b=\sqrt{7}-\sqrt{5}$. Then
$$
\begin{array}{l}
a+b=2 \sqrt{7}, a b=2 . \\
a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b) \\
=44 \sqrt{7}, \\
a^{6}+b^{6}=\left(a^{3}+b^{3}\right)^{2}-2(a b)^{3}=13536 . \\
\because 0<b<1, \\
\therefore 0<b^{6}<1 .
\end{array}
$$
Therefore, the integer part of $a^{6}$ is 13535.
|
13535
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Define a positive integer $n$ to be a "tail" of a factorial if there exists a positive integer $m$ such that the decimal representation of $m$! ends with exactly $n$ zeros. How many positive integers less than 1992 are not tails of a factorial?
|
Solution: Clearly, the number of trailing zeros in $m!$ = the number of times 10 is a factor in $m!$ = the number of times 5 is a factor in $m!$.
Thus, we have $f(m)=\sum_{k=1}^{+\infty}\left[\frac{m}{5^{k}}\right]$.
From 1 Cΰ₯S $1=\sum_{k=1}^{\infty}\left[\frac{m}{5^{k}}\right]7964$.
It is also easy to see that when $5 \mid m$, $f(m)=f(m+1)$ $=f(m+2)=f(m+3)=f(m+4)<f(m+5)$. Therefore, from (1) we calculate
$$
\begin{array}{l}
f(7965)=1988, \\
f(7970)=1989, \\
f(7975)=1991, \\
f(7979)=1991 .
\end{array}
$$
Thus, for $m=0,1,2, \cdots, 7979$, the values of $f(m)$ are sequentially
$$
0,0,0,0,0,1,1,1,1,1, \cdots, 1989,1989 \text {, }
$$
1989, 1989, 1989, 1991, 1991, 1991, 1991, 1991
There are 7980 terms in total, with each number appearing 5 times, so there are $7980 \div 5=1596$ distinct numbers. Therefore, among the positive integers less than 1992, the number of integers that are not "factorial tails" is $1992-1596=396$.
|
396
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 If the fractional parts of $9+\sqrt{13}$ and $9-\sqrt{13}$ are $a$ and $b$ respectively, then $a b-4 a+3 b-2=$ $\qquad$
|
Solution: $\because 3<\sqrt{13}<4$,
$\therefore 9+\sqrt{13}$ has an integer part of 12, and a decimal part
$$
\begin{array}{l}
a=\sqrt{13}-3 . \\
\because-4<-\sqrt{13}<-3,
\end{array}
$$
i.e., $0<4-\sqrt{13}<1$,
$\therefore 9-\sqrt{13}$ has an integer part of 5, and a decimal part $b$ $=4-\sqrt{13}$.
Thus, $a b-4 a+3 b-2$
$$
\begin{array}{l}
=(a+3)(b-4)+10 \\
=\sqrt{13} \times(-\sqrt{13})+10=-3 .
\end{array}
$$
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 We notice that $6!=8 \cdot 9 \cdot 10$. Try to find the largest positive integer $n$ such that $n!$ can be expressed as the product of $n-3$ consecutive natural numbers.
|
Solution: According to the requirements of the problem, write $n!$ as
$$
n!=n(n-1) \cdot \cdots \cdot 5 \cdot 24 \text {. }
$$
Therefore, $n+1 \leqslant 24, n \leqslant 23$. Hence, the maximum value of $n$ is 23. At this point, we have
$$
23!=24 \times 23 \times \cdots \times 5 .
$$
The right-hand side of the above equation is the product of 20 consecutive natural numbers.
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Write a given rational number as a reduced fraction, and calculate the product of the resulting numerator and denominator. How many rational numbers between 0 and 1, when processed this way, yield a product of 20!?
|
5. Solution: There are 8 prime numbers within 20: $2,3,5,7,11$. $13,17,19$. Each prime number either appears in the numerator or the denominator, but not in both. Therefore, we can form $2^{\times}$ irreducible fractions $\frac{p}{q}$ such that $pq=20!$. Since in $\frac{p}{q}$ and $\frac{q}{p}$, only one lies in the interval $(0,1)$, the number of fractions that meet the criteria is $2^{7}=128$.
|
128
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Simplify $\frac{2 b-a-c}{(a-b)(b-c)}$
$$
+\frac{2 c-a-b}{(b-c)(c-a)}+\frac{2 a-b-c}{(c-a)(a-b)} .
$$
|
$\begin{array}{l}\text { Solution: Original expression }=\frac{(b-c)-(a-b)}{(a-b)(b-c)} \\ \quad+\frac{(c-a)-(b-c)}{(b-c)(c-a)}+\frac{(a-b)-(c-a)}{(c-a)(a-b)} \\ =\frac{1}{a-b}-\frac{1}{b-c}+\frac{1}{b-c}-\frac{1}{c-a} \\ \quad+\frac{1}{c-a}-\frac{1}{a-b}=0 .\end{array}$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given that $\alpha$ is a root of the equation $x^{2}-x-1=0$. Try to find the value of $\alpha^{18}+323 \alpha^{-6}$.
Translating the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Given: $\because \alpha$ is a root of $x^{2}-x-1=0$,
$$
\therefore \alpha^{2}-\alpha-1=0 \text {. }
$$
Thus, since $\alpha \neq 0$, we have $\alpha-\alpha^{-1}=1$.
Therefore, $\alpha^{2}+\alpha^{-2}=\left(\alpha-\alpha^{-1}\right)^{2}+2=3$,
$$
\begin{array}{l}
\alpha^{4}+\alpha^{-4}=7, \alpha^{6}+\alpha^{-6}=18, \\
\alpha^{12}+\alpha^{-12}=322 .
\end{array}
$$
$$
\begin{aligned}
\text { Hence, the original expression } & =\alpha^{18}+\left(\alpha^{12}+\alpha^{-12}+1\right) \alpha^{-6} \\
& =\alpha^{18}+\alpha^{6}+\alpha^{-18}+\alpha^{-6} \\
& =\left(\alpha^{6}+\alpha^{-6}\right)\left(\alpha^{12}+\alpha^{-12}\right) \\
& =18 \times 322=5796 .
\end{aligned}
$$
|
5796
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 If $x=\sqrt{19-8 \sqrt{3}}$. Then the fraction $\frac{x^{4}-6 x^{3}-2 x^{2}+18 x+23}{x^{2}-8 x+15}=$
Preserve the original text's line breaks and format, output the translation result directly.
|
Given: From the above, $x=\sqrt{(4-\sqrt{3})^{2}}=4-\sqrt{3}$, thus $x-4=-\sqrt{3}$.
Squaring and simplifying the above equation yields
$$
x^{2}-8 x+13=0 \text {. }
$$
Therefore, the denominator $=\left(x^{2}-8 x+13\right)+2=2$.
By long division, we get
the numerator $=\left(x^{2}-8 x+13\right)\left(x^{2}+2 x+1\right)+10$.
Substituting (1) into the above equation gives
the numerator $=10$.
Thus, the original fraction $=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { 2. Compare the sizes: } \frac{1}{2}(\sqrt{1998}-\sqrt{2000}) \\
\sqrt{1998}-\sqrt{1999} .
\end{array}
$$
3. Given that $\alpha$ is a root of the equation $x^{2}-3 x-1=0$. Then $\alpha^{6}+120 \alpha^{-2}=$ $\qquad$ .
|
Answer: >Answer:1309
|
1309
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. $n(\geqslant 5)$ football teams participate in a round-robin tournament. Each pair of teams plays one match, with the winning team getting 3 points, the losing team getting 0 points, and each team getting 1 point in the case of a draw. The team that finishes in 3rd to last place scores fewer points than all the teams above it, but more than the last two teams; it also plays more matches than the teams above it, but fewer than the last two teams. Find the minimum value of the number of teams $n$.
The text is translated while preserving the original line breaks and formatting.
|
Solution: Let the $n$ teams be $A_{1}, A_{2}, \cdots, A_{n-3}, B, C_{1}$, $C_{2}$.
Since the $A$ group teams score more but have fewer wins, and the $C$ group teams score less but have more wins, each team in the $A$ group must have at least 8 draws, team $B$ must have at least 1 win, 3 losses, and 4 draws, and each team in the $C$ group must have at least 2 wins and 6 losses. The last 3 teams must have at least 15 losses, and the matches among these 3 teams can be at most 3. Therefore, each team in the $A$ group must have at least 12 wins.
If $n \leqslant 14$, then the number of teams in the $A$ group $\leqslant 11$, which means some team in the $A$ group must have 2 wins and 8 draws, so $n \geqslant 11$.
Assume $n=11$. Then, the $A$ group has 8 teams, each with at least 8 draws, totaling 64 draws. The 8 teams in the $A$ group play $C_{8}^{2}=28$ matches. Each team in the $A$ group must draw at least 8 matches, including at least 4 draws with the $C$ group. Each team in the $C$ group must have at least 2 draws. Thus, each team in the $A$ group must have at least 2 wins and 10 draws, leading to a contradiction.
Assume $n=12$. Then, the $A$ group has 9 teams. Each team in the $C$ group must have at least 6 losses, and team $B$ must have 3 losses. The total number of losses for teams $B$ and $C$ is at least 15, so each team in the $A$ group must win at least 12 matches against teams in other groups. Thus, each team in the $A$ group must win at least 2 matches (if some teams in the $A$ group win 2 matches and others win 1 match, the teams that win 1 match must have 3 additional draws, leading to each team in the $C$ group having at least 9 losses and 4 wins, which contradicts $n=12$). Therefore, each team in the $C$ group must win at least 4 matches, and each team in the $A$ group must have some losses. This leads to each team in the $A$ group having 2 wins, 1 loss, and 8 draws or 2 wins and 9 draws, team $B$ having 3 wins, 4 losses, and 4 draws, and each team in the $C$ group having 4 wins and 7 losses. The total number of wins is $18+3+8=29$ matches, and the total number of losses is $\leqslant 9+4+14=27$, leading to a contradiction.
Assume $n=13$. The $A$ group has 10 teams, each with 2 wins, 1 loss, and 9 draws, team $B$ with 3 wins, 5 losses, and 4 draws, and each team in the $C$ group with 4 wins and 8 losses. The table is as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $A_{1}$ & $A_{2}$ & $A_{3}$ & $A_{4}$ & $A_{5}$ & $A_{6}$ & $A_{7}$ & $A_{8}$ & $A_{9}$ & $A_{10}$ & $B$ & $C_{1}$ & $C_{2}$ & $\cdots$ & Score \\
\hline$A_{1}$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $0: 3$ & $3: 0$ & $3: 0$ & 2 & 15 \\
\hline$A_{2}$ & $1: 1$ & & $1: 1$ & $0: 3$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & 2 & 15 \\
\hline$A_{3}$ & $1: 1$ & $1: 1$ & & $1: 1$ & $0: 3$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & 2 & 15 \\
\hline$A_{4}$ & $1: 1$ & $3: 0$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & 0.3 & $3: 0$ & 2 & 15 \\
\hline$A_{5}$ & $1: 1$ & $1: 1$ & $3: 0$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $0: 3$ & $3: 0$ & 2 & 15 \\
\hline$A_{6}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $0: 3$ & $3: 0$ & 2 & 15 \\
\hline$A_{7}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & $0: 3$ & 2 & 15 \\
\hline$A_{8}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $1: 1$ & $3: 0$ & $3: 0$ & $0: 3$ & 2 & 15 \\
\hline$A_{4}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $1: 1$ & $3: 0$ & $3: 0$ & $0: 3$ & 2 & 15 \\
\hline$A_{10}$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & & $3: 0$ & $3: 6$ & $0: 3$ & 2 & 15 \\
\hline$B$ & $3: 0$ & $1: 1$ & $1: 1$ & $1: 1$ & $1: 1$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & & $3: 0$ & $3: 0$ & 3 & 13 \\
\hline$C_{1}$ & $0: 3$ & $0: 3$ & $0: 3$ & $3: 0$ & $3: 0$ & $2: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & $0: 3$ & & $3: 0$ & 4 & 12 \\
\hline$C_{2}$ & $0: 3$ & $0: 3$ & $0: 2$ & $0: 3$ & $0: 3$ & $1: 3$ & $3: 0$ & $2: 0$ & $3: 0$ & $3: 0$ & $0: 3$ & $0: 3$ & & 4 & 12 \\
\hline
\end{tabular}
In summary, the minimum value of $n$ is 13.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a, b, c, d$ be integers, and $a<2b, b<3c, c<4d$. Given $d<100$, then the maximum possible value of $a$ is ( ).
(A) 2367
(B) 2375
(C) 2391
(D) 2399
|
$-.1 .(\mathrm{A})$.
Then, if $d=99$, we have $c<4 \times 99=395$. Taking $c=395$, we get $b<3 \times 395=1185$. Taking $b=1184$, we get $a<2368$. Therefore, the greatest integer for $a$ is 2367.
|
2367
|
Inequalities
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. For a $4 n+2$-sided polygon $A_{1} A_{2} \cdots A_{4 n-2}$ (where $n$ is a natural number), each interior angle is an integer multiple of $30^{\circ}$, and $\angle A_{1}=\angle A_{2}=\angle A_{3}=90^{\circ}$, then the possible values of $n$ are $\qquad$
|
4. 1 .
From the problem, we get
$$
\begin{array}{l}
90^{\circ} \times 3+(4 n+2-3) \times 30^{\circ} \mathrm{k} \\
=(4 n+2-2) \times 180^{\circ},
\end{array}
$$
where $k \geqslant 4 n-1$.
Simplifying, we get $(4 n-1) k=24 n-9$,
which means $24 n-9 \geqslant(4 n-1)^{2}$.
Thus, $8 n^{2}-16 n+5 \leqslant 0$.
Clearly, $0<n<2$, so the integer $n$ can only be:
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. $a_{1}, a_{2}, \cdots, a_{6}$ are six distinct natural numbers whose sum is 23. Then, $a_{1} a_{2}+a_{2} a_{3}+\cdots+$ $a_{5} a_{6}+a_{6} a_{1}$ has the minimum value of ( ).
(A) 62
(B) 64
(C) 65
(D) 67
|
5. (B).
For 6 numbers whose sum is 21, arranged as shown in Figure 2, i.e., $a_{1}=1, a_{2}=6$, $\cdots, a_{6}=5$, the required "sum value" $a_{1} a_{2} +\cdots+a_{6} a_{1}$ is the smallest. At this time, the largest number 6 is surrounded by 1,2; the second largest number 5 is surrounded by 1,3; $\cdots$. Any adjustment would increase the "sum value".
Now, it is required that the sum of 6 numbers is 23, and they are all different. The added value $23-21=2$ is most suitable to be added entirely to the largest number 6: it satisfies the requirement of being all different and makes the added value $2 \times(1+2)=6$ the smallest. At this time, the "sum value" is
$$
1 \times 8+8 \times 2+2 \times 4+4 \times 3+3 \times 5+5 \times 1=64 \text {. }
$$
|
64
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 117. For all such $m$ and $n$, the maximum value of $3m + 2n$ is $\qquad$ .
|
3.37.
Let the $m$ positive even numbers be $a_{i}$, then
$$
\sum_{i=1}^{m} a_{i} \geqslant 2+4+\cdots+2 m=m(m+1) \text {. }
$$
Let the $n$ positive odd numbers be $b_{j}$, then
$$
\sum_{j=1}^{n} b_{j} \geqslant 1+3+\cdots+(2 n-1)=n^{2} .
$$
Thus, the problem reduces to finding the maximum value of $y=3 m+2 n$ under the condition $m(m+1)+n^{2} \leqslant 117$.
Let $m=-\frac{1}{2}+k \cos \theta, n=k \sin \theta$, then
$$
\begin{aligned}
k^{2} & =\left(m+\frac{1}{2}\right)^{2}+n^{2} \\
& =m(m+1)+n^{2}+\frac{1}{4} \leqslant 117+\frac{1}{4} .
\end{aligned}
$$
Thus,
$$
\begin{array}{l}
3 m+2 n=k(3 \cos \theta+2 \sin \theta)-\frac{3}{2} \\
\leqslant \sqrt{13} \cdot k \cdot \sin (\varphi+\theta)-\frac{3}{2} \\
\leqslant \sqrt{13} \times \sqrt{117+\frac{1}{4}}-\frac{3}{2}<37.6 .
\end{array}
$$
Therefore, the maximum value of $3 m+2 n$ is 37.
Since $m$ is an odd number,
by trial, when $m=9, n=5$, $9 \times 10+5^{2}<117$.
|
37
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The solution set of the inequality $\frac{1}{x-1}+\frac{2}{x-2} \geqslant \frac{3}{2}$, is the union of some non-overlapping intervals with a total length of $\qquad$.
|
5.2.
For $g(x)=(x-1)(x-2)>($ or $\text { (or } \leqslant \text { ) } 0 \text {. }$
The graph of $y=f(x)$ is a parabola opening upwards, and the graph of $y=$ $g(x)$ is a parabola opening upwards.
For $f(x):$ when $x=1$, $y0$.
Therefore, the graph intersects the
$x$-axis at two points
$$
\begin{array}{l}
\left(x_{1}, 0\right),\left(x_{2}, 0\right) \text { and } \\
12,
\end{array}
$$
as shown in Figure 6.
The solution to the inequality is
the set of $x$ values for which both functions are either both positive or both negative,
thus the solution set is $\left(1, x_{1}\right] ;\left(2, x_{2}\right]$.
The total length $L$ of the solution set is
$$
L=\left(x_{1}-1\right)+\left(x_{2}-2\right)=\left(x_{1}+x_{2}\right)-(1+2) \text {. }
$$
Given $x_{1}+x_{2}=1+2+\frac{2}{3}(1+2)$,
$$
\therefore L=\frac{2}{3}(1+2)=2 \text {. }
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}\right.$, $\left.B_{2}, \cdots, B_{k}\right\}$ is a family of non-empty subsets of $A$, when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Find the maximum value of $k$.
|
Solution: There are 175 subsets of one, two, and three elements, which clearly meet the requirements, so $k \geqslant 175$.
Assume $k>175$, then among $\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$, there is at least one subset with more than 3 elements. Without loss of generality, let $B_{i}$ have more than 3 elements, and $a \in B_{i}$. Denote $B_{i} \backslash\{a\}$ as the set obtained by removing $a$ from $B_{i}$, then $B_{i} \cap B_{i} \backslash\{a\}$ has at least 3 elements, so $B_{i} \backslash\{a\} \bar{E}\left\{B_{1}, B_{2}, \cdots, B_{k}\right\}$.
Now, replace $B_{i}$ with $B_{i} \backslash\{a\}$, the resulting family of subsets still meets the requirements. Continuously performing this replacement can ensure that each subset in the family has at most 3 elements, without changing the number of subsets, hence the maximum value of $k$ is 175.
|
175
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $1998^{n} \mid 1999$!, then the maximum value of the positive integer $n$ is
|
$=1.55$.
$$
\because 1998=2 \times 3^{3} \times 37,3^{3}<37 .
$$
$\therefore 1999$ ! contains the factor of the form $37^{m}$, where the maximum value of $m$ is the maximum value of $n$.
$$
\therefore n \leqslant \sum_{i=1}^{\infty}\left[\frac{1999}{37^{i}}\right]=55 \text {. }
$$
Therefore, the maximum value of $n$ is 55.
|
55
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 The first 100 natural numbers are arranged in a certain order, then the sum of every three consecutive numbers is calculated, resulting in 98 sums, of which the maximum number of sums that can be odd is how many?
(21st Russian Mathematical Olympiad)
|
Solution: First, it is impossible for all 98 sums to be odd; otherwise, the arrangement of these 100 natural numbers could only be one of the following:
(1) odd odd odd odd....$;$
(2) odd even even odd even even
(3) even odd even even odd even
(4) even even odd even even odd
These four cases contradict the sequence of 100 consecutive natural numbers.
By arranging the natural numbers from $1 \sim 100$ in the following order:
odd even even odd even evenβ¦β¦odd even even odd oddβ¦β¦odd
25 odd numbers 50 even numbers
25 odd numbers
We can obtain 97 odd sums. Therefore, the maximum number of sums that are odd is $97$.
|
97
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the equation $x+8 y+8 z=n(n \in \mathbf{N})$ has 666 sets of positive integer solutions $(x, y, z)$. Then, the maximum value of $n$ is . $\qquad$
|
4.304.
When $m>1, m \in \mathbf{N}$,
$y+z=m$ has $m-1$ sets of positive integer solutions.
The original equation also has $m-1$ sets of positive integer solutions.
$$
\begin{array}{l}
\because 1+2+\cdots+(m-1)=666, \\
\therefore \frac{1}{2} m(m-1)=666 .
\end{array}
$$
Solving gives $m=37$ or -36 (discard).
$$
\therefore 1+8 \times 37 \leqslant n \leqslant 8+8 \times 37=304 \text {. }
$$
|
304
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that for every real number $x$ and $y$, the function $f(x)$ satisfies $f(x)+f(y)=f(x+y)+x y$. If $f(1)=m$, then the number of positive integer pairs $(m, n)$ that satisfy $f(n)=1998$ is $\qquad$.
|
6.16.
Let $y=1$, we get
$$
\begin{array}{l}
f(x)+f(1)=f(x+1)+x, \\
\therefore f(x+1)-f(x)=f(1)-x, \\
\therefore f(x)-f(x-1)=f(1)-(x-1), \\
\quad f(x-1)-f(x-2)=f(1)-(x-2), \\
\quad \cdots \cdots . \\
\quad f(2)-f(1)=f(1)-1 .
\end{array}
$$
Adding the above $x-1$ equations, we get
$$
\begin{array}{l}
f(x)-f(1)=(x-1) f(1)-\frac{1}{2} x(x-1) \text {. } \\
\therefore f(x)=x f(1)-\frac{1}{2} x(x-1) \\
=x m-\frac{1}{2} x(x-1) \text {. } \\
\because f(n)=1998 \text {, } \\
\therefore 1998=m n-\frac{1}{2} n(n-1) \text {. } \\
\therefore n(2 m-n+1)=4 \times 3^{3} \times 37 \text {. } \\
\because n+(2 m-n+1)=2 m+1 \text { is odd. } \\
\end{array}
$$
$\therefore n$ and $2 m-n+1$ are one odd and one even.
When $n$ is even, the number of odd values for $2 m-n+1$ is $(3+1)(1+1)=8$ (values), i.e., there are 8 solutions.
When $n$ is odd, $n$ can take 8 odd values, also giving 8 solutions. Therefore, the number of positive integer pairs $(m, n)$ is 16.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, for what real number $x$ does $y=x^{2}-x+1+$ $\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}}$ have a minimum value? What is the minimum value?
|
As shown in Figure 5, let \( P\left(x, x^{2}\right) \) be a point on the parabola \( y=x^{2} \). The distance from \( P \) to the line \( y=x-1 \) is
\[
\begin{aligned}
|P Q| & =\frac{\left|x^{2}-x+1\right|}{\sqrt{(-1)^{2}+1^{2}}} \\
& =\frac{\sqrt{2}}{2} \left| x^{2}-x+1 \right| \\
& =\frac{\sqrt{2}}{2}\left(x^{2}-x+1\right).
\end{aligned}
\]
The distance from \( P \) to \( M(-3,5) \) is
\[
|P M|=\sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}}.
\]
When \( |P Q|+|P M| \) is minimized, points \( M, P, Q \) are collinear,
and this line is the line through \( M \) and perpendicular to the line \( y=x-1 \), which is \( y=-x+2 \).
The minimum value of \( |P Q|+|P M| \) is the distance from point \( M \) to the line \( y=x-1 \): \(\frac{|-3-5-1|}{\sqrt{1^{2}+(-1)^{2}}}=\frac{9}{2} \sqrt{2}\).
Since
\[
\begin{aligned}
y & =x^{2}-x+1+\sqrt{2(x+3)^{2}+2\left(x^{2}-5\right)^{2}} \\
& =\sqrt{2}\left[\frac{\sqrt{2}}{2}\left(x^{2}-x+1\right)+\right. \\
& \left.\sqrt{(x+3)^{2}+\left(x^{2}-5\right)^{2}}\right] \\
& = \sqrt{2}(|P Q|+|P M|).
\end{aligned}
\]
And since the minimum value of \( |P Q|+|P M| \) is \(\frac{9}{2} \sqrt{2}\),
the minimum value of \( y \) is 9.
Since the line \( y=-x+2 \) intersects the parabola \( y=x^{2} \) at points \( P_{1}(-2,4) \) and \( P_{2}(1,1) \),
when \( x=-2 \) or 1, \( y \) has a minimum value of 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. Find the last four digits of $2^{1999}$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
$$
\varphi(625)=\varphi\left(5^{4}\right)=5^{4} \times\left(1-\frac{1}{5}\right)=500 \text {. }
$$
Since $(2,625)=1$, by Euler's theorem, we get $2^{500} \equiv 1(\bmod 625)$.
$$
\begin{array}{l}
\therefore 2^{2000} \equiv 1(\bmod 625) . \\
\therefore \text { Let } 2^{2000}=625 m+1(m \in \mathbf{N}) .
\end{array}
$$
Also, $2^{5} \mid 2^{2000}$,
so let $2^{2000}=2^{5} \cdot n(n \in \mathbf{N})$.
From (1) and (2), we get $625 m+1=2^{5} \cdot n$.
Solving this indeterminate equation, we get
$$
\left\{\begin{array}{l}
m=32 k+15, \\
n=625 k+293
\end{array}\right.
$$
$k \in \mathbf{Z}$.
Substituting $n=625 k+293$ into (2), we get
$$
\begin{array}{l}
2^{2000}=32(625 k+293)=20000 k+9376 . \\
\therefore 2^{1999}=10000 k+4688 .
\end{array}
$$
Therefore, the last four digits of $2^{1999}$ are 4688.
|
4688
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
74. Let the set $A=\{1,2,3, \cdots, 1997\}$, for any 999-element subset $X$ of $A$, if there exist $x, y \in X$, such that $x<y$ and $x \mid y$, then $X$ is called a good set. Find the largest natural number $a(a \in A)$, such that any 999-element subset containing $a$ is a good set.
|
Solution: We prove that $\max a=665$.
First, we prove that $a \leqslant 665$. Clearly, the 999-element subset $X_{0}$ $=\{999,1000,1001, \cdots, 1997\}$ does not contain any $x, y \in$ $X_{0}$ such that $x1997$, i.e., larger than the maximum element of $X_{0}$, which holds. Thus, $a$ cannot be any of the numbers $999,1000,1001, \cdots$, 1997. 998.
Next, we prove that $2 \neq 666$ construct
$0, \frac{1}{x} \frac{1}{52}$.
For $B_{i}$, $(666+i) \times 3 \geqslant 1998$, and $(666$ $+i) \times 2 \bar{\epsilon} B_{i}$, so no other multiples of $666+i$ are in $B_{i}$.
We have already proven that no non-multiples of any element in $X_{0}$ are in $X_{0}$, and $666 + i < 999 (i=0,1,2, \cdots, 332)$, so no multiples of any element in $X_{0}$ can be $666+i (i=0,1, \cdots, 332)$. This ensures that no multiples (other than itself) of any element in $B_{i}$ are in $B_{i}$. Thus, there do not exist two elements in $B_{i}$ such that $x<y$ and $x \mid y$. Since $B_{i}$ $(i=0,1, \cdots, 332)$ includes $666,667, \cdots$, 998, $a \neq 666,667, \cdots, 998$. Therefore, $a \geqslant 665$.
Next, we prove that 665 is achievable. Suppose there exists a 999-element subset $X$ containing 665, such that there do not exist $x, y \in X, x<y$ with $x \mid y$, then $665 \times 2,665 \times 3 \in X$.
Construct the following 997 drawers, which include all elements of $A$ except $665,665 \times 2,665 \times 3$, and each element appears only once:
$$
\begin{array}{l}
\left\{1,1 \times 2,1 \times 2^{2}, \cdots, 1 \times 2^{10}\right\} \\
\left\{3,3 \times 2,3 \times 2^{2}, \cdots, 3 \times 2^{9}\right\} \\
\left\{5,5 \times 2,5 \times 2^{2}, \cdots, 5 \times 2^{8}\right\} \\
\cdots \cdots \\
\{663,663 \times 2,663 \times 3\} \\
\{667,667 \times 2\} \\
\{669,669 \times 2\} \\
\cdots \cdots \cdots
\end{array}
$$
\{1991\} \{1993\} \{1997\}
The 998 elements of $X$ other than 665 are placed into these 997 drawers, so there must be two elements in the same drawer, and the elements in the same drawer are multiples of each other, which is a contradiction. Proof complete.
(Xiao Mingyu, Class 981, Department of Computational Mathematics and Applied Software, Central South University of Technology, 410012)
|
665
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, 610 people go to the bookstore to buy books, it is known that
(1) each person bought three books;
(2) any two people have at least one book in common.
How many people at most could have bought the book that was purchased by the fewest people?
(1993, China Mathematical Olympiad)
|
Solution: Let's assume that person A bought three books, and since A has at least one book in common with each of the other 9 people, among A's three books, the book purchased by the most people is bought by no less than 4 people.
If the book purchased by the most people is bought by 4 people, then all three books bought by A are each purchased by 4 people, and each book bought by the other 9 people is also purchased by 4 people. Therefore, the total number of books bought by the 10 people is a multiple of 4. That is, $4 \mid 30$, which is a contradiction. Thus, the book purchased by the most people is bought by at least 5 people.
Consider the following purchasing method:
$$
\begin{array}{l}
\left\{B_{1}, B_{2}, B_{3}\right\},\left\{B_{1}, B_{2}, B_{4}\right\},\left\{B_{2}, B_{3}, B_{5}\right\}, \\
\left\{B_{1}, B_{3}, B_{6}\right\},\left\{B_{1}, B_{4}, B_{5}\right\},\left\{B_{2}, B_{4}, B_{6}\right\}, \\
\left\{B_{3}, B_{4}, B_{5}\right\},\left\{B_{1}, B_{5}, B_{6}\right\},\left\{B_{2}, B_{5}, B_{6}\right\}, \\
\left\{B_{3}, B_{4}, B_{6}\right\},
\end{array}
$$
It can be seen that the book purchased by the most people is bought by at least 5 people.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let $S=\{1,2,3,4\}$. An $n$-term sequence: $q_{1}, q_{2}, \cdots, q_{n}$ has the following property: For any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted by $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
(1997, Shanghai High School Mathematics Competition)
|
Solution: First, each number in $S$ appears at least twice in the sequence $q_{1}, q_{2}$, $\cdots, q_{n}$, otherwise, since there are 3 binary subsets containing a certain number, but in the sequence, the number of adjacent pairs containing this number is at most 2, therefore, $n \geqslant 8$.
Moreover, the 8-term sequence: $3,1,2,3,4,1,2,4$ satisfies the condition exactly, hence the minimum value of $n$ is 8.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number in the $i$-th row and $j$-th column is denoted as $x_{ij}$, as shown in Table 1. Then, the numbers in each column of Table 1 are rearranged in non-increasing order as $x_{1j}^{\prime} \geqslant x_{2j}^{\prime} \geqslant \cdots \geqslant x_{100j}^{\prime}$, forming Table 2.
Find the smallest natural number $k$, such that if the numbers filled in Table 1 satisfy $\sum_{j=1}^{25} x_{ij} \leqslant 1$ for $i=1,2, \cdots, 100$, then when $i \geqslant k$, in Table 2, it is guaranteed that: $\sum_{j=1}^{25} x_{ij}^{\prime} \leqslant 1$ holds.
(1997, National High School Mathematics Competition)
|
Solution: First, consider a row $x_{r 1}, x_{r 2}$, $\cdots, x_{r 25}$ in Table 1, which must appear in the first few items of Table 2, because $100 \times 24 = 2400$, while $97 \times 25 = 2425$. Therefore, a row in Table 1 must appear in the first 97 items of Table 2.
Thus, when $i \geqslant 97$, $x_{i j}^{\prime} \leqslant x_{97 j}^{\prime} \leqslant x_{i j} (j=1$, $2, \cdots, 25)$,
so when $i \geqslant 97$, $\sum_{j=1}^{25} x_{i j}^{\prime} \leqslant \sum_{j=1}^{25} x_{r j} \leqslant 1$.
On the other hand, take
$$
x_{i j}=\left\{\begin{array}{ll}
0 & (4(j-1)+1 \leqslant i \leqslant 4 j), \\
\frac{1}{24} & (\text { otherwise } i) .
\end{array}\right.
$$
$(j=1,2, \cdots, 25)$, at this time $\sum_{j=1}^{25} x_{i j}=1(i=1,2$, $\cdots, 100)$,
after rearrangement
$$
\begin{array}{l}
x_{i j}^{\prime}=\left\{\begin{array}{ll}
\frac{1}{24} & (1 \leqslant i \leqslant 96, j=1,2, \cdots, 25), \\
0 & (97 \leqslant i \leqslant 100) .
\end{array}\right. \\
\text { we have } \sum_{j=1}^{25} x_{i j}^{\prime}=25 \times \frac{1}{24}>1(1 \leqslant i \leqslant 96) .
\end{array}
$$
Therefore, $k \geqslant 97$, i.e., the minimum value of $k$ is 97.
5 Comprehensive Estimation
To solve discrete extremum problems, it often requires the integrated use of various strategic ideas for extremum estimation.
|
97
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Let $S=\{1,2,3, \cdots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers.
(32nd IMO)
|
Solution: Let $A_{1}=\{S$ where the numbers are divisible by 2 $\}, A_{2}=$ $\{S$ where the numbers are divisible by 3 $\}, A_{3}=\{S$ where the numbers are divisible by 5 $\}, A_{4}=\{S$ where the numbers are divisible by 7 $\}$, and denote $A=$ $A_{1} \cup A_{2} \cup A_{3} \cup A_{4}$. It is easy to see that $|A|=216$.
Since in $A$ any 5 numbers must have two numbers in the same $A_{i}$, and not all are the same, hence $n \geqslant 217$.
On the other hand, let $B_{1}=\{1$ and all prime numbers in $S\}, B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}, B_{3}=\{2 \times$ $131,3 \times 89,5 \times 53,7 \times 37,11 \times 23,13 \times 19\}$, $B_{4}=\{2 \times 127,3 \times 83,5 \times 47,7 \times 31,11 \times 19$, $13 \times 17\}, B_{5}=\{2 \times 113,3 \times 79,5 \times 43,7 \times$ $29,11 \times 17\}, B_{6}=\{2 \times 109,3 \times 73,5 \times 41,7$ $\times 23,11 \times 13\}$.
Denote $B=B_{1} \cup B_{2} \cup \cdots \cup B_{6}$, then $|B|=88$. Thus, $S \backslash B$ contains 192 elements. In $S$ any 217 numbers are taken, since $217-192=25$, there must be 25 elements in $B$. Therefore, there must be 5 numbers belonging to the same $B_{i}$, and it is clear that they are pairwise coprime.
Hence, the minimum value of $n$ is 217.
|
217
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For $n \in \mathbf{N}$, let $S_{n}$ be
$$
\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}
$$
the minimum value, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=17$. If there exists a unique $n$ such that $S_{n}$ is also an integer, find the value of $n$.
|
Solution: Based on the structure of $\sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we construct the complex number $z_{k}=(2 k-1)+\sigma_{k} i, k=1,2, \cdots, n$.
Then $\sqrt{(2 k-1)^{2}+a_{k}^{2}}=\left|(2 k-i)+a_{k} i\right|$.
Thus $\sum_{k=1}^{n} \vee \sqrt{(2 k-1)^{2}+a_{k}^{2}}$
$=\sum_{k=1}^{n}\left|(2 k-1)+a_{k} i\right|$
$=\sum_{k=1}^{n}\left|z_{k}\right| \geqslant\left|\sum_{k=1}^{n} z_{k}\right|$
$=\mid[1+3+\cdots+(2 n-1)]$
$+\left(a_{1}+a_{2}+\cdots+a_{n}\right) i \mid$
$=\left|n^{2}+17 i\right|=\sqrt{n^{4}+17^{2}}$.
Given that $S_{n}$ is the minimum value of $\sum_{k=1}^{n} \sqrt{(2 k-1)^{2}+a_{k}^{2}}$, we have
$$
S_{n}=\sqrt{n^{4}+17^{2}} \text {. }
$$
Since $S_{n}$ is an integer, let $S_{n}=m$. Then we have
$$
\begin{array}{l}
n^{4}+17^{2}=m^{2}, \\
\left(m-n^{2}\right)\left(m+n^{2}\right)=289 . \\
\therefore m-n^{2}=1, m+n^{2}=289 .
\end{array}
$$
Solving these equations, we get $n=12$.
2 Using $|z|^{2}=z \bar{z}$, convert the modulus operation to complex number operations.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let the complex numbers $z_{1}$ and $z_{2}$ satisfy
$$
\left|z_{1}\right|=\left|z_{1}+z_{2}\right|=3,\left|z_{1}-z_{2}\right|=3 \sqrt{3} \text {. }
$$
Find the value of $\log _{3}\left|\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}\right|$.
(1991, National High School Mathematics Competition)
|
Solution: From the given, we have
$$
\begin{aligned}
9 & =\left|z_{1}+z_{2}\right|^{2}=\left|z_{1}\right|^{2} \\
& =\left(z_{1}+z_{2}\right)\left(\overline{z_{1}+z_{2}}\right) \\
& =\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right), \\
27 & =\left|z_{1}-z_{2}\right|^{2}=\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-z_{2}\right) \\
& =\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}\right) .
\end{aligned}
$$
Given $\left|z_{1}\right|=3$, from the above two equations, we get $\left|z_{2}\right|=3, z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=-9$, and $\left|z_{1} \bar{z}_{2}\right|=\left|\bar{z}_{1} z_{2}\right|=9$.
Let $z_{1} \bar{z}_{2}=9(\cos \theta+i \sin \theta)$, then $\bar{z}_{1} z_{2}=$ $9(\cos \theta-i \sin \theta)$. From $-9=z_{1} \bar{z}_{2}+\bar{z}_{1} z_{2}=$ $18 \cos \theta$, we get $\cos \theta=-\frac{1}{2}, \theta=\frac{2 \pi}{3}$. Therefore, $\left(z_{1} \bar{z}_{2}\right)^{2000}+\left(\bar{z}_{1} z_{2}\right)^{2000}$ $=9^{2000} \cos 2000 \theta=-9^{2000}$.
Thus, the original expression $=4000$.
|
4000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Six, let a large cube of $4 \times 4 \times 4$ be composed of 64 unit cubes. Select 16 of these unit cubes to be painted red, such that in the large cube, each $1 \times 1 \times 4$ small rectangular prism composed of 4 unit cubes contains exactly 1 red cube. How many different ways are there to select the 16 red cubes? Explain your reasoning.
|
Solution: Choose a $4 \times 4$ side of the $4 \times 4 \times 4$ large cube as the base plane. Divide the large cube into four layers parallel to the base plane (each layer being $4 \times 4 \times 1$ in size), and number these layers as $1, 2, 3, 4$. Project each red unit cube onto the base plane and write the layer number of the red unit cube in the corresponding grid. This way, we obtain a $4 \times 4$ grid, where each cell contains a number from $\{1, 2, 3, 4\}$, and cells with the same number are neither in the same row nor in the same column. Such a $4 \times 4$ grid is called a 4th-order Latin square. Conversely, each 4th-order Latin square uniquely determines a coloring method that meets the problem's requirements, and different Latin squares determine different coloring methods. Therefore, the problem is reduced to determining: how many different 4th-order Latin squares are there.
We agree to label the positions of the individual cells in the $4 \times 4$ grid as shown in Figure 1. For any permutation $(a, b, c, d)$ of $(1, 2, 3, 4)$, we will examine the number of Latin squares as shown in Figure 2. For this purpose, we first examine the number of Latin squares as shown in Figure 3.
The $(2,2)$ cell in Figure 3 can be filled with any number from $\{a, c, d\}$. We will examine these three cases separately.
(i) Suppose the $(2,2)$ cell in Figure 3 is filled with $a$.
In this case, the $(2,3)$ and $(3,2)$ cells can only be filled with $d$; then the $(2,4)$ and $(4,2)$ cells can only be filled with $c$; then the $(3,3)$ cell can be filled with $a$ or $b$. Once the $(3,3)$ cell is filled, the remaining three cells are uniquely determined. Therefore, there are exactly 2 Latin squares corresponding to this case.
(ii) Suppose the $(2,2)$ cell in Figure 3 is filled with $c$.
In this case, the $(2,4)$ and $(4,2)$ cells can only be filled with $a$; then the $(2,3)$ and $(3,2)$ cells can only be filled with $d$; then the $(3,4)$ and $(4,3)$ cells can be filled with $b$. Once these cells are filled, the remaining cells are uniquely determined. Therefore, there is exactly 1 Latin square corresponding to this case.
(iii) Suppose the $(2,2)$ cell in Figure 3 is filled with $d$.
Similar to case (ii), there is only one unique filling method for this case, i.e., there is exactly 1 Latin square.
Based on the above discussion, there are exactly 4 Latin squares as shown in Figure 3. Each such Latin square can generate $3!$ different Latin squares as shown in Figure 2 by permuting the 2nd, 3rd, and 4th rows. Therefore, there are exactly $4 \times 3!$ Latin squares as shown in Figure 2. Finally, since there are $4!$ permutations of $(a, b, c, d)$, the total number of 4th-order Latin squares is $(4!)^2$.
Thus, the total number of ways to select 16 red cubes is $(4!)^2 = 576$.
Note: Consider a more general problem, in an $n \times n \times n$ large cube, select $n^2$ unit cubes to be painted red, such that any $1 \times 1 \times n$ rectangular prism composed of $n$ unit cubes contains exactly one red unit cube. How many different ways are there to select $n^2$ red unit cubes that meet the above requirements?
Following the solution above, the problem can be reduced to finding the number of different $n$-order Latin squares $L_n$. The number of $n$-order Latin squares with the first row and first column being the standard permutation $(1, 2, \cdots, n)$ is denoted as $l_n$. Thus,
$$
L_n = (n!)((n-1)!) l_n.
$$
As $n$ increases, $l_n$ and $L_n$ grow rapidly. People have only calculated $l_n$ and $L_n$ for smaller $n$.
\begin{tabular}{|c||c|c|c|c|c|c|c|c|c|}
\hline $n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline $l_n$ & 1 & 1 & 1 & 4 & 56 & 9408 & 16942080 & $\cdots$ & $\cdots$ \\
\hline
\end{tabular}
For $n \geq 5$, a large number of cases must be analyzed to accurately calculate the large number $l_n$ without repetition or omission, which is difficult to achieve in a short time. Therefore, choosing the case $n=4$ can be considered appropriate.
(Provided by the Examination Committee, written by Zhang Zhusheng)
|
576
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Real numbers $a, b$ satisfy $\sqrt{a^{2}}-2 a+1+$ $\sqrt{36-12 a+a^{2}}=10-|b+3|-|b-2|$. Then the maximum value of $a^{2}+b^{2}$ is $\qquad$
|
3. 45
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 15 points) For the rectangle $\triangle B C(1)$, $A B=20$ cm, $B C=10$ cm. If points $M, N$ are taken on $A C$ and $A B$ (as shown in Figure 2), to make the value of $B M+M N$ the smallest, find this minimum value.
|
Three, as shown in Figure 1, construct the symmetric point $B'$ of $B$ with respect to $AC$, and connect $AB'$. Then the symmetric point of $N$ with respect to $AC$ is the point $N'$ on $AB$. At this time, the minimum value of $B$ and $M$ to $V$ is equal to the minimum value of $B \rightarrow M \rightarrow N'$, which is equal to the distance from $B$ to $AB'$, $B'H'$, and the minimum value of $BM + MN$ is $B'H'$. Now, we need to find $B'H'$. Connect $B$ with the intersection point $P$ of $AB'$ and $IC$. Then the area of $\triangle ABP$ is $\frac{1}{2} \times 20 \times 10 = 100$ square centimeters. Note that $PA = PC$ $(\because \angle PAC \stackrel{\text{symmetry}}{=} \angle BAC \stackrel{\text{alternate}}{=} \angle PAC)$, let $AP = x$. Then $P^2 - x, DP = 20 - x$.
So, $x^2 = (20 - x)^2 + 10^2, x = 12.5$.
$B'H' = \frac{100 \times 2}{12.5} = 16$ (centimeters).
Answer: The minimum value of $BM + MN$ is 16 centimeters.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 The elements of set $A$ are all integers, the smallest of which is 1, and the largest is 100. Except for 1, each element is equal to the sum of two numbers (which can be the same) in set $A$. Find the minimum number of elements in set $A$.
|
Solution: Construct a set with as many elements as possible to satisfy the conditions, such as $\{1,2,3,5,10,20,25,50, 9\}$.
Extend $\left\{1,2, x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, 100\right\}$ to also satisfy the conditions, then $x_{1} \leqslant 4, x_{2} \leqslant 8, x_{3} \leqslant 16, x_{4} \leqslant 32, x_{5}$ $\leqslant 64$.
$$
\text { but } \begin{array}{l}
x_{4}+x_{5} \leqslant 96<100, \therefore x_{5}=50, \\
x_{3}+x_{4} \leqslant 48<50, \therefore x_{4}=25, \\
x_{2}+x_{3} \leqslant 24<25, \therefore 25=2 x_{3} .
\end{array}
$$
Contradiction.
Therefore, the minimum number of elements in $A$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (Full marks 15 points) $f(n)$ is a strictly increasing function defined on $\mathbf{N}$ and taking integer values (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$). When $m, n$ are coprime, $f(m n)=f(m) f(n)$, and $f(19)=19$. Try to find $f(f(19) f(98))$.
|
II. H problem, $j(n)$ is defined on $\mathbf{N}$ with integer values and is a strictly increasing function.
$\because f(19)=f(1 \times 19)=f(1) \cdot f(19)$, but $f(19)=$ $19 \neq 0$,
$$
\therefore f(1)=1 \text {. }
$$
At this point, by $f(1)<f(2)<\cdots<f(19), f(1)=1$, $f(19)=19$, and $f(2), f(3), \cdots, f(18)$ are all natural numbers, so $f(n)=n(1 \leqslant n \leqslant 19, n \in \mathbf{N})$.
And $f(95)=f(5 \times 19)=f(5) f(19)=5 \times 19=$ 95,
$f(99)=f(9 \times 11)=f(9) f(11)=9 \times 11=99$.
Thus, $f(96)=96, f(97)=97, f(98)=98$.
Therefore, $f(f(19) f(98))=f(19 \times 98)$
$$
\begin{array}{l}
=f(19) f(98) \\
=19 \times 98=1862 .
\end{array}
$$
|
1862
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 15 points) There are 1998 athletes with numbers from 1 to 1998. Select some of these athletes to participate in the honor guard, but ensure that among the remaining athletes, no one's number is equal to the product of the numbers of any two other athletes. How many athletes can be selected for the honor guard at a minimum? Provide your selection plan and briefly explain your reasoning.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Five, the selectable numbers are $2,3,4, \cdots, 43,44$, these 43 athletes can be chosen as flag bearers to meet the requirements of the problem. The reason is as follows:
After selecting these 43 athletes as flag bearers, the product of any two numbers among the remaining athletes' numbers (excluding number 1) will be greater than $45^{2}=2025>1998$. And 1 multiplied by any number will not equal a third different number. Therefore, among the remaining athletes, there is no one whose number is equal to the product of the numbers of two other athletes.
Now we prove that 43 is the minimum number of athletes that can be selected as flag bearers. That is, we need to prove that when 42 athletes are selected as flag bearers, there will be at least three numbers among the remaining athletes, where the product of two of these numbers equals the third number. For this, we examine the triplets:
Since the function $x(89-x)$ is increasing in the interval $2 \leqslant x \leqslant 44$, the numbers written are distinct and do not exceed $44 \times 45=1980<1998$. There are 43 such triplets. If the number of selected athletes is less than 43, at most 42, at least one of these 43 triplets will have all three numbers unselected. Then these three numbers, as the numbers of the remaining athletes, will have one number equal to the product of the other two, which does not meet the problem's requirements. Therefore, the number of athletes selected as flag bearers must be at least 43.
(Supplied by Zhou Chunli, Department of Mathematics, Capital Normal University)
|
43
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 For a finite set $A$, there exists a function $f: N \rightarrow$ $A$, with the following property: if $i, j \in N$, and $|i-j|$ is a prime number, then $f(i) \neq f(j)$. How many elements must the set $A$ have at least?
|
Solution: Since the absolute value of the difference between any two numbers among $1,3,6,8$ is a prime number, according to the problem, $f(1)$, $f(3)$, $f(6)$, and $f(8)$ are four distinct elements in $A$, thus $|A| \geqslant 4$.
On the other hand, if we let $A=\{0,1,2,3\}$, and the mapping $f: N \rightarrow A$ is defined as: if $x \in N, x=4 k+r$, then $f(x)=r$, where $k \in N \cup\{0\}, r=0,1,2,3$. For any $x, y \in N$, if $|x-y|$ is a prime number, assume $f(x) = f(y)$, then $x \equiv y(\bmod 4)$, thus $4 \mid x-y\rfloor$, which contradicts the fact that $|x-y|$ is a prime number.
Therefore, $A$ contains at least 4 distinct elements.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 1, in trapezoid $ABCD$, $DC \parallel AB$, $DC : AB = 1 : 2$, $MN \parallel BD$ and bisects $AC$. If the area of trapezoid $ABCD$ is $\overline{ab}$, and $S_{\triangle AMN} = \overline{ba}$, then $\bar{a} + \bar{b} + \overline{ba} =$
|
γγ1.99.
$\triangle A M N \backsim \triangle A B D$, similarity ratio $=\frac{\frac{1}{2} A C}{\frac{2}{3} A C}=\frac{3}{4}$,
$$
\begin{array}{l}
S_{\triangle A M N}=\frac{9}{16} S_{\triangle A B D}=\frac{9}{16} \times \frac{2}{3} S_{A B C D}=\frac{3}{8} S_{A B C D} \Rightarrow(10 b \\
+a)=\frac{3}{8}(10 a+b) \Rightarrow 77 b=22 a, 7 b=2 a, \text { when } a=7
\end{array}
$$
when $a=7$, $b=2 . \therefore \overline{a b}+\overline{b a}=99$.
|
99
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is $\qquad$ .
|
Let $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$, then we have
$$
\left\{\begin{array}{l}
5^{y}=3^{x}+4^{x}, \\
4^{y}=5^{x}-3^{x}
\end{array}\right.
$$
(1) + (2) gives $5^{y}+4^{y}=5^{x}+4^{x}$.
However, the function $f(x)=5^{x}+4^{x}$ is an increasing function, so from (3) we have
$$
f(y)=f(x) \Leftrightarrow y=x .
$$
Substituting into (1) or (2) gives
$$
5^{x}=3^{x}+4^{x} \text {, }
$$
which simplifies to $\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1=\left(\frac{3}{5}\right)^{2}+\left(\frac{4}{5}\right)^{2}$.
Since the function $g(x)=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$ is a decreasing function, we have
$$
g(x)=g(2) \Leftrightarrow x=2 \text {. }
$$
Thus, from the above, we get $x=2$.
Substituting back into the original equation for verification, the solution set of the equation is $\{2\}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the pure imaginary numbers $x_{1}, x_{2}, \cdots, x_{1999}$ have a modulus of 1. Then the remainder when $x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1998} x_{1999}+x_{1999} x_{1}$ is divided by 4 is $\qquad$
|
4.1.
Obviously, if $x_{k}$ takes $i$ or $-i$, we have
$$
\begin{array}{l}
\left(i+x_{k}\right)\left(i-x_{k+1}\right) \\
=\left\{\begin{array}{l}
0, \text{ when } x_{k}=-i \text{ or } x_{k+1}=i; \\
-4, \text{ when } x_{k}=i \text{ and } x_{k+1}=-i
\end{array}\right.
\end{array}
$$
Both are multiples of 4, so the sum is also a multiple of 4. Therefore,
$$
\begin{aligned}
4 m= & \left(i+x_{1}\right)\left(i-x_{2}\right)+\left(i+x_{2}\right)\left(i-x_{3}\right) \\
& +\cdots+\left(i+x_{1999}\right)\left(i-x_{1}\right) \\
= & -1999+\left(x_{1}+x_{2}+\cdots+x_{1999}\right) i \\
& -\left(x_{1}+x_{2}+\cdots+x_{1999}\right) i \\
& -\left(x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1999} x_{1}\right) .
\end{aligned}
$$
Thus,
$$
\begin{array}{l}
x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{1999} x_{1} \\
=-4 m-1999=4(-m-500)+1 .
\end{array}
$$
It is evident that the above expression, when divided by 4, leaves a remainder of 1.
Note: As a fill-in-the-blank question, we can take $x_{k}=(-1)^{k-1} i$, then
$$
\text { original expression }=1+1+\cdots+1-1=1997=4 \times 499+1 \text {. }
$$
Or take $x_{1}=x_{2}=\cdots=x_{1999}=i$, then
$$
\text { original expression }=-1999=4 \times(-500)+1 \text {. }
$$
From this, we conjecture that the original expression, when divided by 4, leaves a remainder of 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. There are 10 people with distinct heights standing in a row, and each person is either taller or shorter than the person in front of them. The number of arrangements that meet this condition is $\qquad$ (answer with a number).
ε°δΈι’ηζζ¬ηΏ»θ―ζθ±ζοΌθ―·δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌη΄ζ₯θΎεΊηΏ»θ―η»ζγ
Note: The last sentence is a repetition of the instruction and should not be included in the translated text. Here is the final version:
5. There are 10 people with distinct heights standing in a row, and each person is either taller or shorter than the person in front of them. The number of arrangements that meet this condition is $\qquad$ (answer with a number).
|
5.512.
According to the problem, the person standing in the last position must be the tallest or the shortest. Otherwise, there would be people both taller and shorter than the 10th person, which contradicts the given conditions. Therefore, there are two ways to arrange the 10th position.
Let the number of arrangements for 10 people be $a_{10}$. After arranging the 10th position, there are $a_{9}$ ways to arrange the first 9 positions. Thus, we have
$$
\left\{\begin{array}{l}
a_{10}=2 a_{9} \\
a_{1}=1 .
\end{array}\right.
$$
By analogy, $a_{1}, a_{2}, \cdots, a_{9}, a_{10}$ form a geometric sequence with a common ratio $q=2$, so $a_{10}=a_{1} q^{9}=2^{9}=512$.
|
512
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Using $1,2, \cdots, n$ to form an $n$-digit number without repeating digits, where 2 cannot be adjacent to 1 or 3, a total of 2400 different $n$-digit numbers are obtained. Then $n=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
6.7 .
Obviously, $n \geqslant 4$, otherwise 2 must be adjacent to 1 or 3.
When 2 is adjacent to 1, there are $2 \cdot(n-1)$! arrangements, and when 2 is adjacent to 3, there are also $2 \cdot(n-1)$! arrangements. When 2 is adjacent to both 1 and 3, there are $2 \cdot(n-2)$! arrangements. Therefore, the number of arrangements that meet the conditions is
$$
\begin{array}{l}
n!-2 \cdot(n-1)!-2 \cdot(n-1)!+2 \cdot(n-2)! \\
=(n-2)!(n-2)(n-3) \\
=2400<7!.
\end{array}
$$
Moreover, $(n-2)$! $<7$!.
Thus, $n-2<7$. Therefore, $4 \leqslant n<9$.
Substituting $n=4,5,6,7,8$ into (1) in sequence, only when $n=7$
$$
5!\times 5 \times 4=2400
$$
satisfies the condition, hence $n=7$.
|
7
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (Full marks 20 points) As shown in Figure 8, point $O$ represents the sun, $\triangle A B C$ represents a triangular sunshade, and points $A$ and $B$ are two fixed points on the ground in the north-south direction. The sunlight $O C D$ from the due west direction casts the shadow of the sunshade onto the ground, forming the shadow $\triangle A B D$. It is known that the sunlight $O C D$ makes an acute angle $\theta$ with the ground.
(1) At what angle should the sunshade make with the ground to maximize the area of the shadow $\triangle A B D$?
(2) When $A C=3, B C=4, A B=5, \theta=30^{\circ}$, find the maximum area of the shadow $\triangle A B D$.
|
(1) As shown in Figure 17, draw the perpendicular line $O H$ from $O$ to the ground, connect $H D$ to intersect $A B$ at $E$, and connect $C E$. Then $H D$ is the projection of the oblique line $O D$ on the ground, and $\angle C D E = \theta$. Given that $A B$ is in the north-south direction and $C D$ is in the east-west direction, we have
$A B \perp C D$. (Implicit condition)
According to the inverse of the Three Perpendiculars Theorem, $D E \perp A B$. By the property theorem of line perpendicular to a plane, $A B \perp O H$. This implies that $A B$ is perpendicular to the intersecting lines $O H$ and $O D$, so $A B$ is perpendicular to the plane $C D E$, and thus,
$A B \perp C E$.
Hence, $C E$ is the altitude of $\triangle A B C$ on side $A B$, and $D E$ is the altitude of $\triangle A B D$ on side $A B$. The dihedral angle between the sunshade and the ground is $\angle C E D$.
In $\triangle A B D$, to maximize the area, with $A B$ and $\theta$ fixed, $D E$ should be maximized.
In $\triangle C E D$, we have
$$
\begin{aligned}
D E & =\frac{C E}{\sin \theta} \cdot \sin \angle C E D=\frac{2 S_{\triangle A B C}}{A B \sin \theta} \cdot \sin \angle C E D \\
& \leqslant \frac{2 S_{\triangle A B C}}{A B \sin \theta} \text { (a constant). }
\end{aligned}
$$
Therefore, when $\angle D C E = 90^{\circ}$ (the sunshade is perpendicular to the sunlight), $D E$ is maximized. Thus, when $\angle C E D = 90^{\circ} - \theta$, the area of the shadow $\triangle A B D$ is maximized.
(2) When $A C = 3$, $B C = 4$, $A B = 5$, and $\theta = 30^{\circ}$, $\triangle A B C$ is a right triangle, and $S_{\text {SMIX: }} = \frac{1}{2} A C \cdot B C = 6$.
Substituting (1) into the area formula, we get the maximum area as
$$
S_{\triangle A W B} = \frac{1}{2} A B \cdot D E = \frac{S_{\triangle A B C}}{\sin 30^{\circ}} = 12.
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 The sum of two positive integers is 1997 less than their product, and one of them is a perfect square. Then the larger number and the smaller (1997, Tianjin Junior High School Mathematics Competition)
|
Solution: Let the two positive integers be $\alpha, \beta$, and $\alpha>\beta$. Then
$$
\begin{array}{l}
\alpha+\beta+1997=\alpha \beta . \\
\therefore \alpha \beta-\alpha-\beta+1=1998,
\end{array}
$$
which means
$$
\begin{array}{l}
(\alpha-1)(\beta-1) \\
=1998=999 \times 2=333 \times 3 \times 2 .
\end{array}
$$
Since one of $\alpha, \beta$ is a perfect square, then
$$
\alpha-1=3, \beta-1=666
$$
or $\beta-1=3, \alpha-1=666$.
$$
\begin{array}{l}
\therefore \alpha=4, \beta=667, \text { or } \beta=4, \alpha=667 . \\
\because \alpha>\beta, \\
\therefore \alpha=667, \beta=4 .
\end{array}
$$
Therefore, the difference between the larger number and the smaller number is 663.
|
663
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given $u+v=96$, and the quadratic equation $x^{2} + u x + v = 0$ has integer roots, then its largest root is $\qquad$
(1996, Anhui Province Partial Areas Junior High School Mathematics League)
|
Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$, then we have
$$
\begin{array}{l}
x_{1}+x_{2}=-u, x x_{2}=v . \\
\because u+v=96, \\
\therefore x_{1} x_{2}-\left(x_{1}+x_{2}\right)=v+u=96 .
\end{array}
$$
Thus, $x_{1} x_{2}-x_{1}-x_{2}+1=97$,
which means $\left(x_{1}-1\right)\left(x_{2}-1\right)=97$.
Since 97 is a prime number, then
$x_{1}-1=97$ or $x_{2}-1=97$.
$\therefore x_{1}=98$ or $x_{2}=98$.
Therefore, the largest root of the equation is 98.
|
98
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $S$ be a set with 6 elements. In how many ways can two (not necessarily distinct) subsets of $S$ be chosen so that the union of the two subsets is $S$? The order of selection does not matter. For example, the pair of subsets $\{a, c\}, \{b, c, d, e, f\}$ and the pair of subsets $\{b, c, d, e, f\}, \{a, c\}$ represent the same selection.
(11th American Invitational Mathematics Examination)
|
Solution: Let $S=A \cup B$, and without loss of generality, assume $|A| \leqslant|B|$ (denote the number of elements in set $S$ as $|S|$).
If $|A|=0$, then $A=\varnothing, B=S$, there is only 1 way to choose;
If $|A|=1$, then $|B|=6,5$. At this time, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{6}^{6} +\mathrm{C}_{6}^{1} \mathrm{C}_{5}^{5}=12$ ways to choose;
If $|A|=2$, then $|B|=6,5,4$. At this time, there are $\mathrm{C}_{6}^{2} \mathrm{C}_{6}^{6}+\mathrm{C}_{6}^{2} \mathrm{C}_{2}^{1} \mathrm{C}_{4}^{4}+\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{1}=60$ ways to choose;
If $|A|=3$, then $|B|=6,5,4,3$. At this time, there are $\mathrm{C}_{6}^{3} \mathrm{C}_{6}^{6}+\mathrm{C}_{6}^{3} \mathrm{C}_{6}^{2}+\mathrm{C}_{6}^{3} \mathrm{C}_{3}^{1}+\frac{1}{2} \mathrm{C}_{6}^{3} \mathrm{C}_{3}^{3}=150$ ways to choose; $\square$
If $|A| \vdots=4$, then $|B|=6,5,4$. At this time, there are $\mathrm{C}_{6}^{4} \mathrm{C}_{6}^{6}+\mathrm{C}_{5}^{4} \mathrm{C}_{4}^{3}+\frac{\mathrm{i}}{2} \mathrm{C}_{6}^{4} \mathrm{C}_{4}^{2}=120$ ways to choose;
If $|A|=5$, then $|B|=6,5$. At this time, there are $\mathrm{C}_{6}^{5} \mathrm{C}_{6}^{6} +\frac{1}{2} \mathrm{C}_{6}^{5} \mathrm{C}_{5}^{4}=21$ ways to choose.
If $|A|=6$, then $|B|=6$. At this time, there is only one way to choose.
In summary, the number of ways to choose that satisfy the problem is
$$
1+12+60+150+120+21+1=365 \text { (ways). }
$$
|
365
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $M=\{1,2,3, \cdots, 1995\}, A$ be a subset of $M$ and satisfy the condition: if $x \in A, 15 x \notin A$, then the maximum number of elements in $A$ is $\qquad$
(1995, National High School Mathematics Competition)
|
Solution: Construct subset $A$ as follows:
$1995 \div 15=133$, let $A_{1}=\{134,135, \cdots$, 1995\}, then $\left|A_{1}\right|=1862$ (elements);
$133 \div 15=8$ remainder 13, let $A_{2}=\{9,10, \cdots$, 133\}, then $\left|A_{2}\right|=125$ (elements);
Let $A_{3}=\{1,2,3,4,5,6,7,8\}$, then $\left|A_{3}\right|=$ 8 (elements).
Obviously, $\bigcup_{j=1}^{3} A_{j}=\{1,2,3, \cdots, 1995\}, A_{i} \cap$ $A_{j}=\varnothing, i \neq j$, and $\left\{15 a \mid a \in A_{i+1}\right\} \subset A_{i}, i=$ $1,2,3$.
Let $A=A_{1} \cup A_{3}$, then, $A$ is the required subset and
$$
|A|=\left|A_{1}\right|+\left|A_{3}\right|=1862+8=1870 \text {. }
$$
|
1870
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let the set $M=\{1,2,3, \cdots, 1000\}$. For any non-empty subset $X$ of $M$, let $\alpha_{X}$ denote the sum of the largest and smallest numbers in $X$. Then the arithmetic mean of all such $\alpha_{X}$ is $\qquad$
(1991, National High School Mathematics Competition)
|
Solution: Construct the subset $X^{\prime}=\{1001-x \mid x \in X\}$, then all non-empty subsets can be divided into two categories: $X^{\prime}=X$ and $X^{\prime} \neq X$.
When $X^{\prime}=X$, it must be that $X^{\prime}=X=M$, thus, $\alpha_{X}=1001$.
When $X^{\prime} \neq X$, let $x$ and $y$ be the maximum and minimum numbers in $X$, respectively, then $1001-x$ and $1001-y$ are the minimum and maximum numbers in $X^{\prime}$, respectively. Therefore, $a_{X}=x+y$, $\alpha_{X^{\prime}}=2002-x-y$. Thus, $\frac{a_{X}}{2} \frac{\alpha_{X^{\prime}}}{2}=1001$.
Hence, the arithmetic mean of the required $\alpha_{x}$ is 1001.
4 Solution using the principle of inclusion-exclusion
Using the principle of inclusion-exclusion to solve certain subset competition problems can directly calculate the result of the problem.
|
1001
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 In hexagon $A B C D E F$, $\angle A=\angle B$ $=\angle C=\angle D=\angle E=\angle F$, and $A B+B C=$ $11, F A-C D=3$. Then $B C+D E=$ ?
(1994, Beijing Junior High School Mathematics Competition)
|
Solution: As shown in Figure 7, each angle of the hexagon is equal, all being $120^{\circ}$, so each exterior angle is $60^{\circ}$. Let the lines $A B, C D$, and $E F$ intersect at $P$, $Q$, and $R$, respectively, forming four equilateral triangles, i.e., $\triangle R A F$, $\triangle R P Q$, $\triangle B P C$, and $\triangle E D Q$ are all equilateral triangles.
$$
\begin{array}{l}
\therefore B C+D E=P Q-C D=P R-C D \\
=F A+A B+B C-C D \\
=(A B+B C)+(F A-C D) \\
=14 .
\end{array}
$$
Here, the key is to use the fact that each exterior angle of the hexagon is $60^{\circ}$, constructing equilateral triangles, which makes the solution very straightforward.
|
14
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Let the plane region $D$ be represented by $N(D)$, which denotes the number of all integer points (i.e., points on the xoy plane where both coordinates $x$ and $y$ are integers) belonging to $D$. If $A$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=10$, $y=1$ (including the boundaries), and $B$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=1$, $y=100$ (including the boundaries), then $N(A \cup B)+N(A \cap B)=$ $\qquad$
$(1992$, Shanghai Senior High School Mathematics Competition)
|
Solution: Draw the figure in the Cartesian coordinate system, and it is easy to calculate from the figure
$$
\begin{array}{l}
N(A)=1^{2}+2^{2}+\cdots+10^{2} \\
=\frac{1}{6} \times 10(10+1)(2 \times 10+1)=385, \\
N(B)=\left(101-1^{2}\right)+\left(101-2^{2}\right)+\cdots \\
+\left(101-10^{2}\right) \\
=101 \times 10-\left(1^{2}+2^{2}+\cdots+10^{2}\right) \\
=1010-385=625. \\
\end{array}
$$
$$
N(A \cap B)=10.
$$
By $N(A \cup B)=N(A)+N(B)-N(A \cap B)$, we get
$$
\begin{array}{l}
N(A \cup B)=385+625-10=1000. \\
\therefore N(A \cup B)+N(A \cap B)=1010.
\end{array}
$$
|
1010
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The perimeter of $\triangle A B C$ is $24, M$ is the midpoint of $A B$. $M C=M A=5$. Then the area of $\triangle A B C$ is ( ).
(A) 12
(E) 16
(C) 24
(D) 30
|
2. (C)
$$
\begin{array}{l}
\because M A=M B=M C=5, \\
\therefore \angle A C B=90^{\circ} .
\end{array}
$$
Given the perimeter is 24, then $A C+B C=14, A C^{2}+B C^{2}=$ $10^{2}$
$$
\begin{array}{l}
\begin{aligned}
\therefore 2 A C \cdot B C & =(A C+B C)^{2}-\left(A C^{2}+B C^{2}\right) \\
& =14^{2}-10^{2}=4 \times 24 . \\
\therefore S_{\triangle A B C}= & \frac{1}{2} A C \cdot B C=24 .
\end{aligned}
\end{array}
$$
|
24
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $\frac{1}{4}(b-c)^{2}=(a-b)(c-a)$ and $a \neq 0$. Then $\frac{b+c}{a}=$ $\qquad$
|
$$
\begin{array}{l}
\because(b-c)^{2}=4(a-b)(c-a), \\
b^{2}-2 b c+c^{2}=4 a c-4 b c+4 a b-4 a^{2}, \\
\therefore(b+c)^{2}-4 a(b+c)+4 a^{2}=0 .
\end{array}
$$
Therefore, $[2 a-(b+c)]^{2}=0$, which means $2 a=b+c$.
$$
\therefore \frac{b+c}{a}=2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a$ and $b$ are integers, and satisfy
$$
\left(\frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}-\frac{\frac{1}{b}}{\frac{1}{a}+\frac{1}{b}}\right)\left(\frac{1}{a}-\frac{1}{b}\right) \frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=\frac{2}{3} \text {. }
$$
Then $a+b=$ $\qquad$ .
|
3.3.
$$
\begin{array}{l}
\text { Left side }=\frac{a b}{a-b}=\frac{2}{3}, \\
\therefore(3 b-2)(3 a-2)=4 .
\end{array}
$$
Given that $a \neq b$ and they are integers, hence $3 b-2, 3 a-2$ can only take the values 1, 4 or -1, -4.
(1) Suppose $3 b-2=1, 3 a-2=4$.
Solving gives $b=1, a=2$. Therefore, $a+b=3$.
(2) Suppose $3 b-2=-1, 3 a-2=-4$.
This results in $a, b$ being fractions, which we discard.
Thus, $a+b=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) A class participated in an intelligence competition, with three questions: $a$, $b$, and $c$. Each question either scores full marks or 0 points, where question $a$ is worth 20 points, and questions $b$ and $c$ are each worth 25 points. The competition results show that each student got at least one question correct, one person got all three questions correct, and 15 people got two questions correct. The number of students who got question $a$ correct plus the number of students who got question $b$ correct is 29; the number of students who got question $a$ correct plus the number of students who got question $c$ correct is 25; the number of students who got question $b$ correct plus the number of students who got question $c$ correct is 20. What is the average score of the class?
|
Let $x_{a}, x_{b}, x_{c}$ respectively represent the number of people who answered questions $a$, $b$, and $c$ correctly, then we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{a}+x_{b}=29, \\
x_{a}+x_{c}=25, \\
x_{b}+x_{c}=20 .
\end{array}\right. \\
\therefore x_{a}+x_{b}+x_{c}=37 .
\end{array}
$$
Solving, we get $x_{a}=17, x_{b}=12, x_{c}=8$.
The number of people who answered 1 question correctly is $37-1 \times 3-2 \times 15=4$.
The total number of students in the class is $1+4+15=20$.
Answer: The average score of the class is 42 points.
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the 5th term of the expansion of $\left(x \sqrt{x}-\frac{1}{x}\right)^{6}$ is $\frac{15}{2}$, then $\lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-a}\right)=$
|
$\begin{array}{l}\text { II, 1.1. } \\ \because T_{5}=C_{6}^{4}(x \sqrt{x})^{2} \cdot\left(-\frac{1}{x}\right)^{4}=\frac{15}{x}, \\ \text { from } \frac{15}{x}=\frac{15}{2} \text {, we get } x^{-1}=\frac{1}{2} . \\ \therefore \lim _{n \rightarrow \infty}\left(x^{-1}+x^{-2}+\cdots+x^{-n}\right) \\ =\frac{\frac{1}{2}}{1-\frac{1}{2}}=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the functions $f(x)$ and $g(x)$ are defined on $\mathbf{R}$, and
$$
\begin{array}{l}
f(x-y)=f(x) g(y)-g(x) f(y), f(-2) \\
=f(1) \neq 0, \text { then } g(1)+g(-1)=\ldots
\end{array}
$$
(Answer with a number).
|
4. -1 .
$$
\begin{array}{l}
\because f(x-y)=f(x) g(y)-g(x) f(y), \\
\therefore f(y-x)=f(y) g(x)-g(y) f(x) \\
\quad=-[f(x) g(y)-g(x) f(y)]
\end{array}
$$
There is $f(x-y)=-f(y-x)$
$$
=-f[-(x-y)] \text {. }
$$
Then $f(-x)=-f(x)$, i.e., $f(x)$ is an odd function.
$$
\text { Hence } \begin{aligned}
f(1) & =f(-2)=f(-1-1) \\
& =f(-1) g(1)-g(-1) f(1) \\
& =-f(1)[g(1)+g(-1)] . \\
\therefore g(1) & +g(-1)=-1 .
\end{aligned}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Seven, (Full marks 12 points) Given a sequence where each term is 1 or 2, the first term is 1, and there are $2^{k-1}$ twos between the $k$-th 1 and the $(k+1)$-th 1, i.e., $1,2,1,2,2,1,2,2,2,2,1,2,2,2,2,2,2,2,2,1, \cdots$
(1) Find the sum of the first 1998 terms of the sequence, $S_{1998}$;
(2) Does there exist a positive integer $n$ such that the sum of the first $n$ terms of the sequence, $S_{n}=2001$? If $n$ exists, find the value of $n$; if $n$ does not exist, prove your conclusion.
|
For the $k$-th segment, we take the $k$-th 1 and then follow it with $2^{k-1}$ 2's, making the $2^{k-1}+i$-th term the $k$-th segment of the sequence.
Suppose the 1958th term is in the $k$-th segment. Then $k$ is the smallest positive integer satisfying
$$
\dot{\kappa}+\left(1+2+2^{2}+\cdots+2^{k-1}\right)=2^{k}+k-1 \geqslant 1998
$$
\[
\begin{array}{l}
\because \quad 2^{10}+10-1=10331998, \\
\therefore k=11 .
\end{array}
\]
Thus, $S_{1998}=2 \times 1998-11=3985$.
(2) If there exists $n \in \mathbb{N}$ such that $S_{n}=2001$. Suppose the $n$-th term is in the $k$-th segment. Then we have
$$
2 n-k=2001 \text {. }
$$
This implies that $k$ is an odd number.
From (1), we know $k \leqslant 11$, so $10001006>n,
\end{aligned}
$$
$\therefore$ the $n$-th term should be in the 10th segment, i.e., $k=10$, which contradicts the fact that $k$ is an odd number. Therefore, $n$ does not exist.
(Provided by Cao Youwen)
|
3985
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{4}+a_{6}+a_{8}+a_{10}+$ $a_{12}=120$, then the value of $2 a_{9}-a_{10}$ is ( ).
(A) 20
(B) 22
(C) 24
(D) 28
|
\begin{array}{l}-1 .(\mathrm{C}) \\ \because a_{4}+a_{6}+a_{8}+a_{10}+a_{12} \\ =5 a_{1}+(3+5+7+9+11) d \\ =5\left(a_{1}+7 d\right)=120 \Rightarrow a_{1}+7 d=24, \\ \therefore 2 a_{9}-a_{10}=2\left(a_{1}+8 d\right)-\left(a_{1}+9 d\right) \\ =a_{1}+7 d=24 .\end{array}
|
24
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. If $c=10, \frac{\cos A}{\cos B}=\frac{b}{a}=\frac{4}{3}, P$ is a moving point on the incircle of $\triangle A B C$, and $d$ is the sum of the squares of the distances from $P$ to the vertices $A, B, C$, then $d_{\text {min }}+d_{\text {max }}=$ $\qquad$
|
5. 160.
From the given information, we have $\angle C=90^{\circ}, a=$ $6, b=8, c=10$, and the inradius $r$ $=2$. Establish a rectangular coordinate system as shown, and let $P(x, y)$. Clearly, $(x-2)^{2}+(y-$ $2)^{2}=4$. Therefore, we have
$$
\begin{aligned}
d^{2} & =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]+76-2 y \\
& =88-4 y .
\end{aligned}
$$
$\because 0 \leqslant y \leqslant 4$, and $f(y)=88-4 y$ is monotonically decreasing on $[0,4]$,
$$
\begin{array}{l}
\therefore d_{\min }=f(4)=72, d_{\max }=f(0)=88 . \\
\therefore d_{\min }+d_{\max}=160 .
\end{array}
$$
|
160
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: In a certain year, the total coal production of a coal mine, apart from a certain amount of coal used for civilian, export, and other non-industrial purposes each year, the rest is reserved for industrial use. According to the standard of industrial coal consumption of a certain industrial city in that year, it can supply 3 such industrial cities for 6 years, or 4 such industrial cities for 5 years (of course, the fixed amount of coal for non-industrial use is deducted each year). How many years can it supply if it only supplies the industrial coal for this one city?
(1978, Chongqing Mathematical Competition)
|
Solution: Let the total coal production of the mine for the year be $x$, the annual non-industrial coal quota be $y$, and the industrial coal consumption of each industrial city for the year be $z$. Let $p$ be the number of years the coal can supply only one city. According to the problem, we have the system of equations
$$
\left\{\begin{array}{l}
x=6 \times(3 z)+6 y, \\
x=5 \times(4 z)+5 y, \\
p=\frac{x-p y}{2},
\end{array}\right.
$$
which is $\left\{\begin{array}{l}x=18 z+6 y, \\ x=20 z+5 y, \\ p=\frac{x-p y}{2} .\end{array}\right.$
From (1) and (2), we get $y=2 z$.
Then, by eliminating $x$, $y$, and $z$ using equations (1), (3), and (4), we get
$p=30-2 p$. Solving this, we get
$p=10$.
Therefore, the coal can supply only one city for 10 years.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 12 points) At the foot of the mountain is a pond, the scene: a steady flow (i.e., the same amount of water flows into the pond from the river per unit time) continuously flows into the pond. The pond contains a certain depth of water. If one Type A water pump is used, it will take exactly 1 hour to pump out all the water in the pond; if two Type A water pumps are used, it will take exactly 20 minutes to pump out all the water in the pond. If three Type A water pumps are used simultaneously, how long will it take to pump out all the water in the pond exactly?
|
Three, let the amount of water flowing into the pond from the spring every minute be $x$ m$^{3}$, each water pump extracts $y$ m$^{3}$ of water per minute, and the pond originally contains $z$ m$^{3}$ of water. It takes $t$ minutes for three water pumps to drain the pond. According to the problem, we have
$$
\left\{\begin{array}{l}
60 y-60 x=z, \\
20 y+20 y-20 x=z, \\
3 t y-t x=z .
\end{array}\right.
$$
Solving this, we get $t=12$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (Full marks 15 points) The unit digit of an $n$-digit number is 6. Moving the 6 to the front of the number while keeping the other digits in place results in a new $n$-digit number, which is 4 times the original $n$-digit number. Find the smallest positive number that satisfies the above conditions.
---
untranslated text:
(ζ»‘ε 15 ε) ζ $n$ δ½ζ°ηδΈͺδ½ζ°εζ― 6 , ε° 6 移ε°θ―₯ζ°ηι¦δ½γε
Άδ½εδ½ζ°εδΈε¨, εΎε°ζ°ηδΈδΈͺ $n$ δ½ζ°, θΏδΈͺζ°ζ°ι«Άε $n$ δ½ζ°η 4 ε, ζ±εΊζ»‘εδΈθΏ°ζ‘δ»Άηζε°ζ£ζ°γ
translated text:
(Full marks 15 points) The unit digit of an $n$-digit number is 6. Moving the 6 to the front of the number while keeping the other digits in place results in a new $n$-digit number, which is 4 times the original $n$-digit number. Find the smallest positive number that satisfies the above conditions.
|
From the problem, we know that the first digit must be 1.
Expression:
$$
\begin{array}{r}
1 \cdots E D C B A 6 \\
\hline 6 \cdots E D C B A
\end{array}
$$
We get $A=4, B=8, C=3, D=5, F=1$. $\therefore$ The smallest number is 15384.
|
15384
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 15 points) Expand new numbers according to the following rules:
Given two numbers $a$ and $b$, a new number can be expanded according to the rule $c = ab + a + b$. Any two numbers among $a$, $b$, and $c$ can then be used to expand another new number according to the rule, and so on. Each expansion of a new number is called an operation.
Starting with the numbers 1 and 4,
(1) Find the maximum new number that can be obtained by performing the above rule three times;
(2) Can the new number 1999 be expanded according to the above rule? Explain your reasoning.
|
(1) The first time, you can only get $1 \times 4+4+1=9$.
Since the goal is to get the maximum new number, the $\cdots$ time, take 4 and 9, to get $4 \times 9+4+9=49$.
Similarly, the second time, take 9 and 49, to get
$9 \times 49+9+49=441$.
(2) $\because a=a \cdot a+b=(a+1)(b+1)-1$,
$$
\therefore a+1=(a+1)(b+1).
$$
Taking numbers $a, c$ to get the new number
$$
\begin{array}{l}
d=(a+1)(a+1)-1=(a+1)(b+1)(a+1)-1 \\
=(a+1)^{2}(b+1)-1, \\
\therefore d+1=(a+1)^{2}(b+1).
\end{array}
$$
Taking numbers $b, a$ to get the new number
$$
\begin{aligned}
e & =(b+1)(c+1)-1=(b+1)(a+1)(b+1)-1 \\
& =(b+1)^{2}(a+1)-1.
\end{aligned}
$$
Let the final new number be $x$. Then it can always be expressed as
$$
x+1=(a+1)^{m} \cdot(b+1)^{n}.
$$
Where $m, n$ are integers.
$$
\begin{array}{l}
\text { When } a=1, b=4, x+1=2^{n'} \times 5^{n}. \\
x \because 1999+1=2000=2^{4} \times 5^{3}.
\end{array}
$$
Therefore, 1999 can be obtained through the aforementioned rules.
|
441
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$2.1999^{2000}$ divided by $10^{10}$, the remainder is $\qquad$
|
$$
\begin{array}{l}
2.5996000001 . \\
1999^{2000}=(1-2000)^{2000} \\
=1-2000 \times 2000+\frac{1}{2} \times 2000 \times 1999 \times 2000^{2} \\
\quad-\frac{1}{16} \times 2000 \times 1999 \times 1998 \times 2000^{3}+\cdots \\
\quad+2000^{2000} .
\end{array}
$$
In the expression on the right side of the equation above, each term from the fourth term onward can be divided by $10^{10}$, so the remainder when $1999^{2000}$ is divided by $10^{10}$ is the remainder when the first three terms of the expression on the right side are divided by $10^{10}$. It is easy to calculate that this remainder is 5996000001.
|
5996000001
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given two points $A(0,1), B(6,9)$. If there is an integer point $C$ (Note: A point with both coordinates as integers is called an integer point), such that the area of $\triangle A B C$ is minimized. Then the minimum value of the area of $\triangle A B C$ is $\qquad$
|
5.1.
Since the slope of line $AB$ is $k=\frac{3}{4}$, the line passing through point $C$ and parallel to $AB$ can be expressed as $3 y-4 x=m$. Let the coordinates of point $C$ be $\left(x_{0}, y_{0}\right)$, where $x_{0}, y_{0}$ are integers. Therefore, we have
$$
3 y_{0}-4 x_{0}=m \text {. }
$$
Thus, $m$ is an integer.
The equation of line $AB$ is
$$
3 y-4 x-3=0 \text {. }
$$
The distance from point $C$ to line $AB$ is
$$
d=\frac{\left|3 y_{0}-4 x_{0}-3\right|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{1}{5}|m-3| \text {. }
$$
When $d$ is at its minimum, the area of $\triangle ABC$ is minimized.
At this time, $m=2$ or 4.
Thus, $d_{\text {min }}=\frac{1}{5}$.
$$
S_{\triangle A B C}=\frac{1}{2} \cdot A B \cdot d_{\min }=\frac{1}{2} \times 10 \times \frac{1}{5}=1
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (Full marks 20 points) The sum of \( m \) different positive even numbers divisible by 5 and \( n \) different positive odd numbers divisible by 3 is \( M \). For all such \( m \) and \( n \), the maximum value of \( 5m + 3n \) is 123. What is the maximum value of \( M \)? Please prove your conclusion.
---
The sum of \( m \) different positive even numbers divisible by 5 and \( n \) different positive odd numbers divisible by 3 is \( M \). For all such \( m \) and \( n \), the maximum value of \( 5m + 3n \) is 123. We need to find the maximum value of \( M \).
### Step-by-Step Solution:
1. **Identify the sequences:**
- The sequence of positive even numbers divisible by 5 is: \( 10, 20, 30, 40, \ldots \)
- The sequence of positive odd numbers divisible by 3 is: \( 3, 9, 15, 21, \ldots \)
2. **Sum of the sequences:**
- The sum of the first \( m \) terms of the sequence \( 10, 20, 30, \ldots \) is:
\[
S_m = 10 + 20 + 30 + \cdots + 10m = 10(1 + 2 + 3 + \cdots + m) = 10 \cdot \frac{m(m+1)}{2} = 5m(m+1)
\]
- The sum of the first \( n \) terms of the sequence \( 3, 9, 15, \ldots \) is:
\[
S_n = 3 + 9 + 15 + \cdots + 3n = 3(1 + 3 + 5 + \cdots + (2n-1)) = 3 \cdot n^2
\]
3. **Total sum \( M \):**
\[
M = 5m(m+1) + 3n^2
\]
4. **Maximize \( 5m + 3n \):**
- Given \( 5m + 3n = 123 \), we need to find the maximum value of \( M \).
5. **Express \( n \) in terms of \( m \):**
\[
n = \frac{123 - 5m}{3}
\]
- \( n \) must be a positive integer, so \( 123 - 5m \) must be divisible by 3.
6. **Check divisibility:**
- \( 123 \equiv 0 \pmod{3} \)
- \( 5m \equiv 0 \pmod{3} \) implies \( m \equiv 0 \pmod{3} \)
- Let \( m = 3k \) for some integer \( k \).
7. **Substitute \( m = 3k \):**
\[
n = \frac{123 - 5(3k)}{3} = \frac{123 - 15k}{3} = 41 - 5k
\]
- \( n \) must be a positive integer, so \( 41 - 5k > 0 \)
- \( k < \frac{41}{5} \approx 8.2 \)
- The largest integer \( k \) is 8.
8. **Calculate \( m \) and \( n \):**
- \( m = 3 \times 8 = 24 \)
- \( n = 41 - 5 \times 8 = 1 \)
9. **Calculate \( M \):**
\[
M = 5m(m+1) + 3n^2 = 5 \cdot 24 \cdot 25 + 3 \cdot 1^2 = 3000 + 3 = 3003
\]
### Conclusion:
The maximum value of \( M \) is \(\boxed{3003}\).
|
From the problem, we know
$$
\begin{aligned}
M \geqslant & 5(2+4+\cdots+2 m) \\
& +3[1+3+5+\cdots+(2 n-1)] \\
= & 5 m(m+1)+3 n^{2} .
\end{aligned}
$$
That is, \( 5\left(m^{2}+m\right)+3 n^{2} \leqslant M \),
$$
5\left(m+\frac{1}{2}\right)^{2}+3 n^{2} \leqslant M+\frac{5}{4} \text {. }
$$
By the Cauchy-Schwarz inequality, we know
$$
\begin{array}{l}
5\left(m+\frac{1}{2}\right)+3 n \\
\leqslant \sqrt{(\sqrt{5})^{2}+(\sqrt{3})^{2}} \\
\quad \sqrt{\left[\sqrt{5}\left(m+\frac{1}{2}\right)\right]^{2}+(\sqrt{3} n)^{2}} \\
\leqslant \sqrt{8} \cdot \sqrt{M+\frac{5}{4}}=\sqrt{8 M+10}
\end{array}
$$
Thus, \( 5 m+3 n \leqslant \sqrt{8 M+10}-\frac{5}{2} \).
Since the maximum value of \( 5 m+3 n \) is 123,
$$
\begin{array}{l}
\therefore 123 \leqslant \sqrt{8 M+10}-\frac{5}{2}<124 . \\
\therefore 1967.5312 \leqslant M<1999.0312 .
\end{array}
$$
Thus, the maximum value of \( M \) does not exceed 1999.
When \( 5 m+3 n=123 \),
$$
\begin{array}{l}
m=3 k, n=41-5 k(k \in \mathbf{N}) . \\
\because 5 m^{2} \leqslant M<2000,3 n^{2} \leqslant M<2000, \\
\therefore m<20, n<26 . \\
\therefore 3 k<20,41-5 k<26 .
\end{array}
$$
Solving this, we get \( 3<k<\frac{20}{3} \).
Thus, \( k \) is 4, 5, 6.
Then the three pairs of values for \( (m, n) \) are
\( (12,21) \cdot(15,16) \cdot(18,11) \).
It is easy to verify that the pairs \( (12,21) \) and \( (18,11) \) do not satisfy the inequality
$$
5 m(m+1)+3 n^{2} \leqslant M<2000 .
$$
Only \( (15,16) \) satisfies:
$$
5 \times 15(15+1)+3 \times 16^{2}=1968<2000 \text {. }
$$
And it is possible to select 15 positive even numbers divisible by 5 and 16 positive odd numbers divisible by 3 such that their sum is 1998. For example:
$$
\begin{array}{l}
5(2+4+\cdots+28+30)+3(1+3+\cdots+29+31) \\
=1968, \\
5(2+4+\cdots+28+30)+3(1+3+\cdots+29+41) \\
=1998 .
\end{array}
$$
There do not exist integers \( m \) and \( n \) such that \( M \) is 1999. Otherwise, we would have
$$
5 \times 2 m_{0}+3 \times 2 n_{0}=1999-1968=31 \text {. }
$$
This equation has no integer solutions.
Therefore, the maximum value of \( M \) is 1998.
|
1998
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) Given $0<\alpha_{i}<\frac{\pi}{4}(i=1$, $2,3,4)$, and $\sum_{i=1}^{4} \sin ^{2} \alpha_{i}=1$. Prove: $\sum_{i=1}^{4} \frac{\sin ^{2} \alpha_{i}}{\cos 2 \alpha_{i}} \geqslant$ 2.
|
Let $a_{i}=\sin ^{2} \alpha_{i}$, then $0<a_{i}<\frac{1}{2}$,
$$
\begin{array}{l}
\cos 2 \alpha_{i}=1-2 \sin ^{2} \alpha_{i}=1-2 a_{i} . \\
\frac{\sin ^{2} \alpha_{i}}{\cos 2 a_{i}}=\frac{a_{i}}{1-2 a_{i}}=2\left(\frac{a_{i}^{2}}{1-2 a_{i}}+\frac{a_{i}}{2}\right) \\
=\frac{2}{1-2 a_{i}}\left(a_{i}-\frac{1-2 a_{i}}{2}\right)^{2}+2 a_{i}-\frac{1-2 a_{i}}{2}+a_{i} \\
=\frac{2}{1-2 a_{i}}\left(a_{i}-\frac{1-2 a_{i}}{2}\right)^{2}+4 a_{i}-\frac{1}{2}
\end{array}
$$
$\geqslant 4 a_{i}-\frac{1}{2}$. (Equality holds if and only if $a_{i}=\frac{1}{4}$).
$$
\begin{array}{l}
\text { Hence } \sum_{i=1}^{4} \frac{\sin ^{2} \alpha_{i}}{\cos 2 a_{i}} \geqslant \sum_{i=1}^{4}\left(4 a_{i}-\frac{1}{2}\right) \\
=4 \sum_{i=1}^{4} a_{i}-2=2 \text {. } \\
\end{array}
$$
|
2
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 50 points) If the hundreds digit of an $n$-digit natural number $N$ is 9, and the sum of its digits is $M$, where $n>3$, when the value of $\frac{N}{M}$ is the smallest, what is $N$?
---
Please note that the translation retains the original formatting and structure of the text.
|
$$
\begin{array}{l}
\text { Three, (1) When } n=4 \text {, let } N=\overline{a 9 b c} \text {. Then } \\
\frac{N}{M}=\frac{1000 a+900+10 b+c}{a+9+b+c} \\
=1+\frac{999 a+891+9 b}{a+9+b+c} \\
\geqslant 1+\frac{999 a+891+9 b}{a+9+b+9} \text { (equality holds when } c=9 \text {) } \\
=1+\frac{9(a+b+18)+990 a+729}{a+b+18} \\
=10+\frac{990 a+729}{a+b+18} \\
\geqslant 10+\frac{990 a+729}{a+9+18} \text { (equality holds when } b=9 \text {) } \\
=10+\frac{990(a+27)-20 \text { aip1 }}{a+27} \\
=1000-\frac{26001}{a+27} \\
\geqslant 1000-\frac{26001}{1+27} \text { (equality holds when } a=1 \text {) } \\
=\frac{1999}{28} \text {. } \\
\end{array}
$$
Therefore, when $\frac{N}{M}$ reaches its minimum value, $N$ is 1999.
(2) When $n>4$, prove by mathematical induction.
$$
10^{n-1}-648 n>0 \text {. }
$$
$1^{\circ}$ When $n=5$, it is obviously true.
$2^{\circ}$ Assume that when $n=k$, the proposition is true. That is,
$$
10^{k-1}-648 k>0 \text {. }
$$
Then, when $n=k+1$, we have
$$
\begin{array}{l}
10^{(k+1)-1}-648(k+1) \\
=10 \times 10^{k-1}-648 k-648 \\
=10^{k-1}-648 k+9 \times 10^{k-1}-648 \\
>9 \times 10^{k-1}-648>9 \times 10^{4}-648>0 .
\end{array}
$$
Therefore, when $n=k+1$, the proposition is true.
Thus, when $n>4$, $10^{k-1}-648 n>0$ always holds.
Hence, $\frac{N}{M} \geqslant \frac{10^{n-1}}{9 n}>\frac{648 n}{9 n}=72>\frac{1999}{28}$.
Therefore, when $\frac{N}{M}$ reaches its minimum value $\frac{1999}{28}$, $N$ is 1999.
(Xie Wenxiao, Huanggang High School, Hubei Province, 436100)
$$
|
1999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 A store sells goods that cost 10 yuan each at 18 yuan each, and can sell 60 of them per day. After conducting a market survey, the store manager found that if the selling price of the goods (based on 18 yuan each) is increased by 1 yuan, the daily sales volume will decrease by 5; if the selling price of the goods (based on 18 yuan each) is decreased by 1 yuan, the daily sales volume will increase by 10. To maximize daily profit, what should the selling price of the goods be set at per unit?
(Tenth ZΗ ChΕngzhΔ« Cup Junior High School Mathematics Invitational Competition)
|
Solution: Let the selling price of each item be $x$ yuan, and the daily profit be $s$ yuan.
$$
\begin{array}{l}
\text { When } x \geqslant 18 \text {, we have } \\
s=[60-5(x-18)](x-10) \\
=-5(x-20)^{2}+500,
\end{array}
$$
That is, when the selling price of the item is increased to $x=20$ yuan, the daily profit $s$ is maximized, with the maximum daily profit being 500 yuan.
$$
\begin{array}{l}
\text { When } x \leqslant 18 \text {, we have } \\
s=[60+10(18-x)](x-10) \\
=-10(x-17)^{2}+490, \\
\end{array}
$$
That is, when the selling price of the item is reduced to $x=17$ yuan, the daily profit is maximized, with the maximum daily profit being 490 yuan.
In summary, the selling price of this item should be set at 20 yuan each.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $x=\frac{1}{\sqrt{3}+\sqrt{2}}, y=\frac{1}{\sqrt{3}-\sqrt{2}}$. Then, $x^{2}+y^{2}$ is
untranslated part: 轩ιΉ
Note: The term "轩ιΉ" does not have a clear meaning in this context and has been left untranslated. If you can provide more context or clarify the term, I can attempt to translate it accurately.
|
$=, 7.10$.
Let $x=\sqrt{3}-\sqrt{2}, y=\sqrt{3}+\sqrt{2}$, so,
$$
x^{2}+y^{2}=(\sqrt{3}-\sqrt{2})^{2}+(\sqrt{3}+\sqrt{2})^{2}=10 .
$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 A public bus company, after two technological innovations, adjusted the seating from New Year's Day 1957, allowing each bus to carry 6 more people, so that the number of people each 5 trips could carry exceeded 270. From New Year's Day 1958, a trailer was added, allowing each bus to carry 98 more people than in 1957, so that the number of people each 3 trips could carry exceeded the number each 8 trips could carry in 1957. How many times more people can each trip carry after New Year's Day 1958 compared to before New Year's Day 1957?
(1964, Chengdu City Senior High School Mathematics Competition)
|
Solution: Let the number of people each car can carry before New Year's Day 1957 be $x$, then on New Year's Day 1957 and 1958, each car can carry $x+6$ and $x+98$ people, respectively. According to the problem, we have $\left\{\begin{array}{l}5(x+6)>270, \\ 3(x+98)>8(x+6) .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}x>48, \\ x<49.2 .\end{array}\right.$
Therefore, $x=49$ (people).
Thus, on New Year's Day 1958, each car can carry $49+98=147$ (people). Hence, on New Year's Day 1958, each trip can carry twice as many people as before New Year's Day 1957.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. A pipe burst occurred in a low-lying area by the riverbank, and river water is continuously gushing out, assuming the water gushing out per minute is constant. If two water pumps are used to pump out the water, it takes 40 minutes to finish; if four water pumps are used, it takes 16 minutes to finish. If the water needs to be pumped out within 10 minutes, then at least $\qquad$ water pumps are required.
|
12.6 .
Let the amount of water that has already gushed out before the pumping starts be $u$ cubic meters, the rate of water gushing out per minute be $b$ cubic meters, and the amount of water each pump can extract per minute be $c$ cubic meters $(c \neq 0)$. From the given conditions, we have
$$
\left\{\begin{array}{l}
a+40 b-2 \times 40 a, \\
a: 16 b=4 \times 16
\end{array}\right\}\left\{\begin{array}{l}
a=\frac{160}{3} c, \\
b=\frac{2}{3}
\end{array}\right.
$$
The number of pumps is
$$
\frac{a+10 b}{10 c}=\frac{\frac{100}{3} c+\frac{20}{3} c}{10 c}=6 .
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (Full marks 10 points) As shown in the figure, there is a natural number on each face of the cube. It is known that the sum of the two numbers on opposite faces is equal. Given that the numbers opposite to $12$, $9$, and $3$ are $a$, $b$, and $c$ respectively, find the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$.
|
Final 1. From the problem, we have $13+a=9+b=3+c$, which gives $a-b=-4, b-c=-6, c-a=10$. Therefore,
$$
\begin{array}{l}
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] \\
=\frac{1}{2}(16+36+100)=76 .
\end{array}
$$
|
76
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (Full marks 10 points) The numbers A, B, and C are 312, $270$, and 211, respectively. When these three numbers are divided by a natural number $A$, the remainder of dividing A is twice the remainder of dividing B, and the remainder of dividing B is twice the remainder of dividing C. Find this natural number $A$.
---
Note: In the problem, the numbers are referred to as η² (A), δΉ (B), and δΈ (C), but for clarity in English, I have used A, B, and C to represent the numbers 312, 270, and 211, respectively.
|
Three, from the problem, we have
$$
270 \times 2-312=228, \quad 211 \times 2-270=152 \text {. }
$$
Then we have
\begin{tabular}{|c|c|c|}
\hline \multicolumn{2}{|c|}{$2 \mid \quad 228$} & 152 \\
\hline $2 T$ & 114 & 76 \\
\hline 19 & 57 & 38 \\
\hline & 3 & 2 \\
\hline
\end{tabular}
Therefore, 4 could be $2, 4, 19, 38, 19 \times 2^{2}$.
Upon verification, only 19 meets the requirement.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Based on market research analysis, a home appliance manufacturing company has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be at least 60 units. It is known that the labor hours required per unit and the output value per unit of these home appliances are as follows:
How many units of air conditioners, color TVs, and refrigerators should be produced each week to maximize the output value? What is the highest output value (in thousand yuan)?
(1997, Jiangsu Province Junior High School Mathematics Competition)
|
5. Weekly
Producing 30 air conditioners, 270 color TVs, and 60 refrigerators will maximize the output value, with the highest output value being 1050 thousand yuan.
|
1050
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, let $x, y$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, and satisfy the condition: when $x>y$, $x-y \geqslant \frac{x}{15}$. Find the maximum number of these natural numbers $n$.
|
$$
\begin{array}{l}
\text { Let's assume } a_{1}=1, d_{4} \geqslant 2 ; \\
a_{5}=a_{4}+d_{4} \geqslant 6, d_{5} \geqslant \frac{6^{2}}{19-6}>2, d_{5} \geqslant 3 ; \\
a_{6} \geqslant 9, d_{6} \geqslant \frac{9^{2}}{19-9}>8, d_{6} \geqslant 9 ; \\
a_{7} \geqslant 18, d_{7} \geqslant \frac{18^{2}}{19-18}=324 ; \\
a_{8} \geqslant 342>19 .
\end{array}
$$
From this, we can see that $n-1 \leqslant 7, n \leqslant 8$.
On the other hand, the 8 natural numbers $1,2,3,4,6,9,18,342$ meet the conditions of the problem.
Therefore, the maximum value of $n$ is 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Suppose $a^{2}+2 a-1=0, b^{4}-2 b^{2}-1=$ 0 , and $1-a b^{2} \neq 0$. Then the value of $\left(\frac{a b^{2}+b^{2}+1}{a}\right)^{1990}$ is (1990, Hefei Junior High School Mathematics Competition)
|
Given $a^{2}+2 a-1=0$, we know $a \neq 0$.
$$
\therefore\left(\frac{1}{a}\right)^{2}-2\left(\frac{1}{a}\right)-1=0 \text {. }
$$
From $b^{4}-2 b^{2}-1=0$, we get
$$
\left(b^{2}\right)^{2}-2\left(b^{2}\right)-1=0 \text {. }
$$
From $1-a b^{2} \neq 0$, we know $\frac{1}{a} \neq b^{2}$.
From (1) and (2), we know $\frac{1}{a}$ and $b^{2}$ are the two roots of the equation $x^{2}-2 x-1=0$, so,
$$
\frac{1}{a}+b^{2}=2, \frac{b^{2}}{a}=-1 \text {. }
$$
Therefore, $\frac{a b^{2}+b^{2}+1}{a}=\left(\frac{1}{a}+b^{2}\right)+\frac{b^{2}}{a}$
$$
=2-1=1 \text {. }
$$
Thus, the value of the original expression is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let $A B C D E F$ be a hexagon, and a frog starts at vertex $A$. It can jump to one of the two adjacent vertices at each step. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different jumping sequences are possible from the start to the stop?
(1997, National High School Mathematics Competition)
|
Analysis: Grasping the possible scenarios of the frog's jumping is the key to solving the problem.
Solution: According to the conditions, the frog's jumping methods can only result in two situations:
(1) Jump 3 times to reach point $D$, with 2 ways of jumping.
(2) Stop after 5 jumps (the first 3 jumps do not reach point $D$). Note that among the $2^{3}$ ways of the first 3 jumps, 2 ways reach point $D$, so there are $2^{3}-2=6$ ways for the first 3 jumps, and the last 2 jumps have $2^{2}$ ways, thus there are $6 \times 2^{2}=24$ ways.
From (1) and (2), we know that there are a total of $2+24=26$ ways of jumping.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 There are 20 teams participating in the national football championship finals. To ensure that in any group of 3 teams, at least two teams have played against each other, how many matches must be played at minimum?
(1969, All-Soviet Union Mathematical Olympiad)
|
Analysis: The only possible competition schemes that meet the conditions are:
(1) No grouping, any two teams play a match.
(2) Divide into two groups, Group A and Group B, containing 1 team and 19 teams respectively, any two teams in Group B play a match.
(3) Divide into two groups, containing $k(2 \leqslant k \leqslant 18)$ teams and $20-k$ teams respectively, any two teams in each group play a match.
Obviously, the number of matches in schemes (1) and (2) is more than that in scheme (3), so we only need to consider scheme (3).
Solution: According to scheme (3), the total number of matches is
$$
C_{k}^{2}+C_{20-k}^{2}=(k-10)^{2}+90 \geqslant 90 .
$$
Therefore, at least 90 matches need to be played, and the number of matches is minimized when each group contains 10 teams.
|
90
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 In a sequence of coin tosses, the number of times a tail is followed by a head (denoted as "tail-head"), a head is followed by a tail (denoted as "head-tail"), a head is followed by a head (denoted as "head-head"), and a tail is followed by a tail (denoted as "tail-tail") can be counted. How many different sequences of 15 coin tosses contain 2 "head-head", 3 "head-tail", 5 "tail-tail", and 4 "tail-head"?
(4th AIME)
|
Analysis: From "ε" being 1 more than "ζ£", we know that the sequence starts and ends with "ε" and "ζ£" respectively. According to the conditions, it is easy to know that the count of "ζ£" equals $2 \times 2 + 3 + 4 = 11$, and the count of "ε" is $5 \times 2 + 3 + 4 = 17$. Since the "ε" and "ζ£" at the beginning and end are each counted once, while the other "ζ£" and "ε" are each counted twice, there are 6 "ζ£" and 9 "ε" in the sequence.
The above analysis is the key to solving this example.
Solution: Construct the sequence model as follows:
ε0000000000000ζ£ where the empty positions "β" are sequentially denoted as $a_{i}(i=1,2, \cdots, 13)$.
Notice that there are three possible cases for 2 "ζ£ζ£":
(1) If $a_{13}$ is "ε", then the remaining 5 "ζ£" can only be divided into $(2,2,1)$ or $(3,1,1)$. When the 5 "ζ£" are divided into $(2,2,1)$, there are still 7 "ε" in $a_{i}$ (where $1 \leqslant i \leqslant 12$), and the 5 "ζ£" can be inserted into 3 of the 8 gaps between the 7 "ε", like:
"βεβεζ£εβεβεβεβεβ" hence there are $\mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.
Similarly, when the 5 "ζ£" are divided into $(3,1,1)$, there are also $\mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.
Therefore, when $a_{13}$ is "ε", there are $2 \mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.
(2) If $a_{13}$ is "ζ£" and $a_{12}$ is "ε", then the remaining 4 "ζ£" can be divided into $(2,1,1)$, yielding $\mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1}$ different sequences.
(3) If both $a_{13}$ and $a_{12}$ are "ζ£", then the remaining 3 "ζ£" can only be divided into $(1,1,1)$, and $a_{11}$ must be "ε", resulting in $\mathrm{C}_{8}^{3}$ different sequences.
In summary, the total number of different sequences is
$$
2 \mathrm{C}_{8}^{3} \mathrm{C}_{3}^{1} + \mathrm{C}_{8}^{3} \mathrm{C}_{8}^{1} + \mathrm{C}_{8}^{3} = 560 \text{ (sequences). }
$$
|
560
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given that $n$ is a positive integer, and $n^{2}-71$ is divisible by $7 n+55$. Try to find the value of $n$.
(1994-1995, Chongqing and Four Other Cities Mathematics Competition)
|
Solution: Let $\frac{n^{2}-71}{7 n+55}=k$ (where $k$ is an integer),
then $n^{2}-7 k n-(55 k+71)=0$.
$$
\begin{aligned}
\Delta & =49 k^{2}+4(55 k+71) \\
& =49 k^{2}+220 k+284 .
\end{aligned}
$$
When $\Delta$ is a perfect square, the $n$ in equation (1) is an integer. And
$$
\begin{array}{l}
(7 k+15)^{2}<\Delta=49 k^{2}+220 k+284 \\
<(7 k+17)^{2} . \\
\therefore \Delta=(7 k+16)^{2} .
\end{array}
$$
Thus, $(7 k+16)^{2}=49 k^{2}+220 k+284$.
Solving this, we get $k=7$.
Substituting $k=7$ into (1), we get $n=57$. (discard -8)
|
57
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the real number $x$ satisfies $\sqrt{x^{2}-x^{3}}=x$. $\sqrt{1-x}$. Then the range of values for $x$ is
|
2.15.
From $2 x^{2}-6 x+y^{2}=0$ we get
$$
y^{2}=-2 x^{2}+6 x \geqslant 0 \text {. }
$$
Solving this, we get $0 \leqslant x \leqslant 3$.
Let $z=x^{2}+y^{2}+2 x$. Then
$$
\begin{array}{l}
z=x^{2}-2 x^{2}+6 x+2 x \\
=-x^{2}+8 x \\
=-(x-4)^{2}+16 . \\
\because 0 \leqslant x \leqslant 3,
\end{array}
$$
$\therefore$ when $x=3$, $z$ has a maximum value of 15.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in the figure, circle O is inscribed in $\triangle ABC$ touching sides $AC$, $BC$, and $AB$ at points $D$, $E$, and $F$ respectively, with $\angle C=90^{\circ}$. The area of $\triangle ABC$ is 6. Then $AF \cdot BF=$ $\qquad$
|
4.6 .
Let $A C=b, B C=a, A B=c, \odot O$ have a radius of $r$. Then
$$
\begin{array}{l}
r=\frac{a+b-c}{2} \cdot a^{2}+i^{2}=c^{2} . \\
\therefore A F \cdot B F=A C \cdot B E \\
=(b-r)(a-r) \\
=\frac{b+c-a}{2} \cdot \frac{a+c-b}{2} \\
=\frac{1}{4}\left[c^{2}-(a-b)^{2}\right] \\
=\frac{1}{2} a b=6 .
\end{array}
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Full marks 25 points) A natural number that can be expressed as the difference of the squares of two natural numbers is called a "smart number." For example, $16=5^{2}-3^{2}$ is a "smart number." Try to answer the following questions:
(1) Is 1998 a "smart number"? Explain your reasoning.
(2) Arranged in ascending order, what is the 1998th "smart number"?
|
Three, (1) 1998 is not a "smart number". If 1998 were a "smart number", we could set \(1998 = m^2 - n^2\) (where \(m\) and \(n\) are natural numbers), then \((m+n)(m-n) = 2 \times 999\).
Since \(m+n\) and \(m-n\) have the same parity, \((m+n)(m-n)\) is either odd or divisible by 4, hence it cannot be 1998. Therefore, 1998 is not a "smart number".
(2) Let \(k\) be a natural number. Using the method in (1), we can prove that natural numbers of the form \(4k + 2\) are not "smart numbers".
\[
\begin{array}{l}
\text{Also, } 4k = (2k+1)^2 - (2k-1)^2, \\
2k+1 = (k+1)^2 - k^2,
\end{array}
\]
Thus, natural numbers of the form \(4k\) and \(2k+1\) (excluding 1) are all "smart numbers". Arranging "smart numbers" in ascending order:
\[
3, 4, 5, 7, 8, 9, 11, 12, 13, \cdots
\]
Therefore, the \(n\)-th "smart number" is \(n + 2 + \left[\frac{n-1}{3}\right]\) (where \([ \] denotes the floor function). When \(n = 1998\), this "smart number" is
\[
1998 + 2 + \left[\frac{1998-1}{3}\right] = 2665.
\]
(Xu Jianguo, Nanchang Experimental High School, 330006)
|
2665
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $f(x)=x^{2}+(\lg a+2) x+\lg b$, and $f(-1)=-2$, also $f(x) \geqslant 2 x$ for all $x \in R$. Then $a+b=(\quad)$.
(A) 100
(B) 110
(C) 120
(D) 130
|
$-, 1 .(B)$
Since $f(-1)=-2$, then
$$
\begin{array}{l}
1-(\lg a+2)+\lg b=-2 \\
\Rightarrow \lg a=\lg b+1 \Rightarrow a=10 b .
\end{array}
$$
Since for all $x \in \mathbf{R}$, $f(x) \geqslant 2 x$,
$$
x^{2}+(\lg a) x+\lg b \geqslant 0 .
$$
Since the coefficient of $x^{2}$ is $1>0$, then
$$
\begin{array}{l}
\Delta=\lg ^{2} a-4 \lg b \leqslant 0, \\
\lg ^{2} a-4(\lg a-1) \leqslant 0 \Rightarrow(\lg a-2)^{2} \leqslant 0 .
\end{array}
$$
Since $(\lg a-2)^{2} \geqslant 0$, then $\lg a=2$.
$$
\therefore a=100, b=10, a+b=110 .
$$
|
110
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x$ and $y$ be real numbers, and satisfy
$$
\left\{\begin{array}{l}
(x-1)^{5}+1999(x-1)=2519, \\
(y-1)^{5}+1999(y-1)=-2519 .
\end{array}\right.
$$
Then the value of $x+y$ is $\qquad$
|
1.2.
Let $f(t)=t^{5}+1999 t$. Then $f(t)$ is increasing on $R$, and $f(x-1)=-f(y-1)$. Since $f(y-1)=-f(1-y)$, then
$$
\begin{array}{l}
f(x-1)=f(1-y) . \\
\therefore x-1=1-y \Rightarrow x+y=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the complex number $z=\cos \theta+i \sin \theta\left(0^{\circ} \leqslant \theta \leqslant\right.$ $\left.180^{\circ}\right)$, and the complex numbers $z, (1+i)z, 2\bar{z}$ correspond to three points $P, Q, R$ on the complex plane. When $P, Q, R$ are not collinear, the fourth vertex of the parallelogram formed by segments $PQ, PR$ is $S$. Then the maximum distance from point $S$ to the origin is
|
5.3.
Let the complex number $w$ correspond to point $S$. Since $Q P R S$ is a parallelogram, we have
$$
\begin{aligned}
w+z & =2 \bar{z}+(1+\mathrm{i}) z, \text { i.e., } w=2 \bar{z}+\mathrm{i} z . \\
\therefore|w|^{2} & =(2 \bar{z}+\mathrm{i} z)(2 z-\mathrm{i} \bar{z}) \\
& =4+1+2 \mathrm{i}\left(z^{2}-\bar{z}^{2}\right)=5-4 \sin 2 \theta \\
& \leqslant 5+4=9(\text { when } \theta=135 \text {, the equality holds }) .
\end{aligned}
$$
Thus, $|w|_{\text{max}}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (Full marks 20 points) There is a quantity $W$, after "modeling" the relationship is given by
$$
W=\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1-t^{2}}}\right),
$$
where $a, b, c, u, t$ are all positive, $u<1, t<1$, and satisfy $a t+b u=c, a^{2}+2 b c u=b^{2}+c^{2}$. Please design a method to find the minimum value of the quantity $W$.
|
Given $a^{2}=b^{2}+c^{2}-2 b c u$. All $a, b, c, u$ are positive, and $ub^{2}+c^{2}-2 b c=(b-c)^{2}$.
From this, we get $|b-c|<a<b+c$.
Therefore, the positive numbers $a, b, c$ can be the lengths of the three sides of a triangle. Let the vertices opposite these sides be $A, B, C$, respectively. By transforming the constraint, we get
$$
u=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\cos A.
$$
Substituting into the other constraint, we get
$$
a t+b \cos A=c.
$$
Using the projection theorem $c=a \cos B+b \cos A$, we get
$$
t=\cos B.
$$
Since $u$ and $t$ are positive, $\angle A$ and $\angle B$ are acute angles, so,
$$
\begin{array}{l}
\sqrt{1-u^{2}}=\sqrt{1-\cos ^{2} A}=\sin A, \\
\sqrt{1-t^{2}}=\sqrt{1-\cos ^{2} B}=\sin B.
\end{array}
$$
Let the circumradius of $\triangle ABC$ be $R$, then
$$
\begin{array}{l}
a=2 R \sin A, b=2 R \sin B, c=2 R \sin C. \\
\begin{aligned}
\therefore W & =\frac{1}{c}\left(\frac{3 a}{\sqrt{1-u^{2}}}+\frac{b}{\sqrt{1}-\overline{t^{2}}}\right) \\
& =\frac{1}{2 R \sin C}\left(\frac{6 R \sin A}{2}+\frac{2 R \sin B}{\sin B}\right) \\
& =\frac{8 R}{2 R \sin C}=\frac{4}{\sin C} \geqslant 4.
\end{aligned}
\end{array}
$$
The equality holds if and only if $\angle C$ is a right angle, i.e., $a^{2}+b^{2}=c^{2}$. Therefore, the minimum value of $W$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
79. (1) Prove that 1998 cannot be expressed as the sum of any number of consecutive odd numbers;
(2) If the numbers from 1 to 1997 are connected using β+β and β-β signs, then no matter how many β-β signs are used, the result cannot be 1998;
(3) If the numbers from 1 to 1999 are connected using β+β and β-β signs, then what is the minimum number of β-β signs needed to make the result exactly 2000.
|
Solution: (1) If it can be expressed, let $n_{0}$ be the first odd number, then
$$
\begin{array}{l}
1998= n_{0}+\left(n_{0}+2\right)+\cdots+\left(n_{0}+2 k\right) \\
=\left(n_{0}+2 k\right)+\left(n_{0}+2 k-2\right)+\cdots \\
+\left(n_{0}+2\right)+n_{0} . \\
\therefore 2 \times 1998=(k+1)\left(2 n_{0}+2 k\right) .
\end{array}
$$
Thus, $1998=(k+1)\left(n_{0}+k\right)$.
When $k$ is odd, $k+1$ is even, and $n_{0}+k$ is also even, so $4 \mid (k+1)\left(n_{0}+k\right)$. But $4 \nmid 1998$.
When $k$ is even, $k+1$ and $n_{0}+k$ are both odd, so $(k+1)\left(n_{0}+k\right)$ is odd. This implies $1998=(k+1)\left(n_{0}+k\right)$ does not hold.
Therefore, 1998 cannot be expressed as the sum of any number of consecutive odd numbers.
(2) Since $1+2+\cdots+1997=\frac{1997 \times 1998}{2}=1997 \times 999$ is odd, changing one β+β sign to a β-β sign in the above sum is equivalent to subtracting an even number from $1997 \times 999$. Since the sum of any number of even numbers is even, and the difference between an odd number and an even number is odd, the result cannot be 1998 regardless of how many β-β signs are used.
(3) Since $1+2+\cdots+1999=1999000$, to change the signs of some terms to make the result exactly 2000 with the fewest negative signs, the terms with changed signs should be as large as possible, so that the number 1999000 can quickly become 2000. And
$$
\begin{array}{c}
2(1415+1416+\cdots+1999)=1997190, \\
\text { and } 1999000-1997190=1810 \text {, and } 1810=2 \times 905, \\
\therefore 1+2+\cdots+903+904-905+906+907+\cdots \\
+1413+1414-1415-1416-\cdots-1997-1998-
\end{array}
$$
$1999=2000$. A total of 586 negative signs.
(Yu Xianfeng, Henan Normal University Affiliated High School, 453002)
|
586
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $x_{1}, x_{2}, \cdots, x_{9}$ all be positive integers, and $x_{1}<x_{2}<\cdots<x_{9}, x_{1}+x_{2}+\cdots+x_{9}=220$. Then, when the value of $x_{1}+x_{2}+\cdots+x_{5}$ is maximized, the minimum value of $x_{9}-x_{1}$ is $\qquad$
$(1992$, National Junior High School Mathematics League)
|
Solution: From $x_{1}+x_{2}+\cdots+x_{9}=220$,
we know $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}>110$,
or $x_{1}+x_{2}+\cdots+x_{5} \leqslant 110$.
From (1), then $x_{5} \geqslant 25$. Thus, $x_{6} \geqslant 26, x_{7} \geqslant$ $27, x_{8} \geqslant 28, x_{9} \geqslant 29$. We get
$$
\begin{aligned}
\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)+\left(x_{6}+x_{7}+\right. \\
\left.x_{8}+x_{9}\right)>110+110=220 .
\end{aligned}
$$
This contradicts the given condition.
From (2), we can take $x_{1}=20, x_{2}=21, x_{3}=22$, $x_{4}=23, x_{5}=24$, so that $x_{1}+x_{2}+\cdots+x_{5}=110$ reaches its maximum value, and $x_{1}$ also reaches its maximum value. At this point, taking $x_{6}$ $=26, x_{7}=27, x_{8}=28, x_{9}=29$ satisfies all conditions. Therefore, the minimum value of $x_{9}-x_{1}$ is $29-20=9$.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $p^{3}+q^{3}=2$, where $p, q$ are real numbers. Then the maximum value of $p+q$ is $\qquad$
(1987, Jiangsu Province Junior High School Mathematics Competition)
|
Solution: Let $s=p+q$.
From $p^{3}+q^{3}=2$ we get
$(p+q)\left(p^{2}+q^{2}-p q\right)=2$,
$(p+q)^{2}-3 p q=\frac{2}{s}$,
Thus, $p q=\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)$.
From (1) and (2), $p$ and $q$ are the two real roots of the equation
$$
x^{2}-s x+\frac{1}{3}\left(s^{2}-\frac{2}{s}\right)=0
$$
We know $\Delta=s^{2}+\frac{4}{s}\left(s^{2}-\frac{2}{s}\right) \geqslant 0$.
Simplifying, we get $s^{2} \leqslant 8$.
Therefore, $s \leqslant 2$.
Taking $p=q=1$, the maximum value of $p+q$ is 2.
4 Using the definition of absolute value
That is, $|x|= \pm x,|x|^{2}=x^{2}$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, $5 x, y, z$ are real numbers, and satisfy $x+y+z=0, xyz=2$. Find the minimum value of $|x|+|y|+|z|$.
(1990, Beijing Junior High School Mathematics Competition).
|
Solution: From the problem, we know that among $x, y, z$, there are 2 negative numbers and 1 positive number. Without loss of generality, let $x>0, y<0, z<0$. When $x>0, -y>0, -z>0$, we have
$(-y)(-z) \leqslant \left[\frac{(-y)+(-z)}{2}\right]^{2}=\frac{x^{2}}{4}$,
which means $2 x=\frac{4}{(-y)(-z)} \geqslant \frac{4}{\frac{(-y)+(-z)}{2}}$
$$
=\frac{16}{x^{2}}.
$$
Thus, $x^{3} \geqslant 8$, solving which gives $x \geqslant 2$.
Therefore, $|x|+|y|+|z|=2 x \geqslant 4$,
which means the minimum value of $|x|+|y|+|z|$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 In hexagon $A B C D E F$, $\angle A=\angle B$ $=\angle C=\angle D=\angle E=\angle F$, and $A B+B C=$ $11, F A-C D=3$. Find $B C+D E$.
(1994, Beijing Junior High School Mathematics Competition)
|
Solution: As shown in Figure 5, let $F A$ and $C B$ intersect at $P$, and $F E$ and $C D$ intersect at $Q$. It is easy to see that quadrilateral $P C Q F$ is a parallelogram, and $\triangle A P B$ and $\triangle Q E D$ are both equilateral triangles. Therefore,
$$
\begin{aligned}
& F A + A B \\
= & F P = Q C \\
= & D E + C D .
\end{aligned}
$$
It is known that
$$
\begin{aligned}
& D E - A B \\
= & F A - C D \\
= & 3, \\
& A B + B C \\
= & 11 .
\end{aligned}
$$
Adding the two equations, we get $D E + B C = 14$.
|
14
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Find the minimum value of the function with real variables $x$ and $y$
$$
u(x, y)=x^{2}+\frac{81}{x^{2}}-2 x y+\frac{18}{x} \sqrt{2-y^{2}}
$$
(2nd Hope Cup for High School Grade 2)
|
Solution: Completing the square, we get
$$
u=(x-y)^{2}+\left(\frac{9}{x}+\sqrt{2-y^{2}}\right)^{2}-2 \text {. }
$$
Consider points $P_{1}\left(x, \frac{9}{x}\right)$, $P_{2}\left(y,-\sqrt{2-y^{2}}\right)$ on the plane. When $x \in \mathbf{R}, x \neq 0$, the trajectory of $P_{1}$ is a hyperbola with the two coordinate axes as asymptotes. When $y \in \mathbf{R},|y| \leqslant \sqrt{2}$, the trajectory of $P_{2}$ is the lower half of a circle centered at the origin with a radius of $\sqrt{2}$. Geometrically, it is evident that the minimum value of $\left|P_{1} P_{2}\right|$ is $3 \sqrt{2}-\sqrt{2}$, from which we can derive
$$
u_{\text {min }}=(3 \sqrt{2}-\sqrt{2})^{2}-2=6 \text {. }
$$
Note: A slight variation of this example can yield the 1983 Putnam Mathematical Competition problem for college students:
Find the minimum value of $f(u, v)=(u-v)^{2}+\left(\sqrt{2-u}-\frac{9}{v}\right)^{2}$.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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