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6.3. Several boys and 5 girls are sitting around a round table, and there are 30 pieces of bread on the plate on the table. Each girl takes 1 piece of bread from the plate for each boy she knows, and then each boy takes 1 piece of bread from the plate for each girl he does not know, at which point the bread on the plate is all gone. How many boys are there?
|
6.3. There are $n$ boys, so there are $5 n$ different “boy-girl” pairs. In each such “pair”, 1 piece of bread was taken, because if the 2 people in the pair know each other, the girl took 1 piece of bread for the boy; and if they do not know each other, the boy took 1 piece of bread for the girl. Therefore, $5 n=$ $30, n=6$.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: Two vegetable bases, A and B, supply the same type of vegetables to three farmers' markets, A, B, and C, according to the signed contracts. They are to supply 45 tons to A, 75 tons to B, and 40 tons to C. Base A can arrange 60 tons, and Base B can arrange 100 tons. The distances in kilometers between A and B and markets A, B, and C are shown in Table 1. Assuming the transportation cost is 1 yuan/(km·t). How should the supply be arranged to minimize the total transportation cost? Find the minimum total transportation cost.
Table 1
\begin{tabular}{|c|c|c|c|}
\hline & A & B & C \\
\hline A & 10 & 5 & 6 \\
\hline B & 4 & 8 & 15 \\
\hline
\end{tabular}
|
Solution: Let the base B supply $x \mathrm{t}$ to $A$, $y \mathrm{t}$ to $B$, and $[100-(x+y)] \mathrm{t}$ to $C$, then base A supplies $(45-x) \mathrm{t}$ to $A$, $(75-y) \mathrm{t}$ to $B$, and $[40-(100-x-y)] \mathrm{t}=(x+y-60) \mathrm{t}$ to $C$.
Let the total transportation cost be $W$ yuan. According to the problem, we have
$$
\begin{array}{l}
W= 10(45-x)+5(75-y)+ \\
6(x+y-60)+4 x+8 y+ \\
15[100-(x+y)] \\
= 1965-15 x-6 y \\
= 1965-6(x+y)-9 x . \\
\text { Because }\left\{\begin{array}{l}
x \geqslant 0, \\
y \geqslant 0, \\
100-(x+y) \geqslant 0, \\
45-x \geqslant 0, \\
75-y \geqslant 0, \\
x+y-60 \geqslant 0,
\end{array}\right.
\end{array}
$$
Then, $60 \leqslant x+y \leqslant 100,0 \leqslant x \leqslant 45$.
Therefore, when and only when $x+y=100, x=45$, $W$ has the minimum value, and
$$
W_{\text {min }}=1965-6(x+y)-9 x=960 .
$$
When $x+y=100, x=45$, $y=55$.
Thus, the optimal distribution plan is:
Base B supplies $45 \mathrm{t}$ to $A$, $55 \mathrm{t}$ to $B$, and does not supply to $C$; Base A does not supply to $A$, supplies 20 $\mathrm{t}$ to $B$, and 40 $\mathrm{t}$ to $C$.
|
960
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 2,
there are soil piles $A$ and $B$, a depression $E$, and a pond $F$ at a construction site.
The volumes of soil piles $A$ and $B$ are $781 \mathrm{~m}^{3}$ and $1584 \mathrm{~m}^{3}$, respectively. Depression $E$ requires $1025 \mathrm{~m}^{3}$ of soil to be filled, and pond $F$ can accommodate $1390 \mathrm{~m}^{3}$ of soil. Now, it is required to remove the two soil piles and use the soil to first fill the depression $E$, and then use the remaining soil to fill the pond $F$. How should the soil transportation plan be arranged to minimize labor?
|
Solution: Let " $\mathrm{m}^{3} \cdot \mathrm{m}$ " be the unit for the labor cost of transporting soil. Suppose the volume of soil transported from $A$ to $E$ is $x$, and the volume of soil transported to $F$ is $781-x$; the volume of soil transported from $B$ to $E$ is $1025-x$, and the volume of soil transported to $F$ is 1584 $(1025-x)=559+x$. Let the total $\mathrm{m}^{3} \cdot \mathrm{m}^{\prime}$ of labor cost be $W$. According to the problem, we have
$$
\begin{array}{l}
W=150(781-x)+50 x+30(1025-x)+ \\
=2140(559+x) \\
\text { Since }\left\{\begin{array}{l}
x \geqslant 0, \\
781-x \geqslant 0, \\
1025-x \geqslant 0, \\
599+x \geqslant 0,
\end{array}\right.
\end{array}
$$
Therefore, $0 \leqslant x \leqslant 781$.
Since $W$ is a linear function of $x$, and $k=-10<0$, $W$ decreases as $x$ increases.
Thus, when $x$ takes its maximum value 781, $W_{\text {min }}=207170$.
When $x=781$,
$$
781-x=0,1025-x=244 \text {. }
$$
Therefore, the most labor-saving soil transportation plan is: all $781 \mathrm{~m}^{3}$ of soil from pile $A$ is transported to depression $E$, and 244 $\mathrm{m}^{3}$ of soil from pile $B$ is transported to depression $E$, with the rest being transported to pond $F$.
|
207170
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The terms of a sequence are all 3 or 5, the first term is 3, and there are $2^{k-1}$ fives between the $k$-th 3 and the $(k+1)$-th 3, i.e., $3,5,3,5,5,3,5,5,5,5,3, \cdots$. Then the sum of the first 2004 terms of this sequence $S_{2004}=$ $\qquad$ .
|
8. 9998 .
The $k$-th 3 and the $2^{k-1}$ 5s following it, totaling $\left(2^{k-1}+1\right)$ terms, are considered as one group. Suppose the 2004th term is in the $k$-th group, then $k$ is the smallest positive integer satisfying
$$
k+1+2+\cdots+2^{k-1} \geqslant 2004
$$
It is easy to find that $k=11$.
Thus, $S_{2004}=5 \times 2004-2 \times 11=9998$.
|
9998
|
Other
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given that $f(x)$ is defined on $(-1,1)$, $f\left(\frac{1}{2}\right)=-1$, and satisfies $x, y \in (-1,1)$, we have $f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right)$.
(1) The sequence $\{x_n\}$ satisfies
$$
x_1 = \frac{1}{2}, \quad x_{n+1} = \frac{2x_n}{1 + x_n^2}.
$$
Let $a_n = f(x_n)$. Find the general term formula for $\{a_n\}$;
(2) Let $b_n = n^2 + 3n + 1$. Find
$$
1 + f\left(\frac{1}{b_1}\right) + f\left(\frac{1}{b_2}\right) + \cdots + f\left(\frac{1}{b_{2002}}\right) + f\left(\frac{1}{2004}\right)
$$
the value of the above expression.
|
$$
\begin{array}{l}
\text { Therefore, } 1+f\left(\frac{1}{b_{1}}\right)+f\left(\frac{1}{b_{2}}\right)+\cdots+f\left(\frac{1}{b_{2000}}\right)+f\left(\frac{1}{2004}\right) \\
=1+\left[f\left(\frac{1}{2}\right)-f\left(\frac{1}{3}\right)\right]+\left[f\left(\frac{1}{3}\right)-f\left(\frac{1}{4}\right)\right]+ \\
\quad \cdots+\left[f\left(\frac{1}{2003}\right)-f\left(\frac{1}{2004}\right)\right]+f\left(\frac{1}{2004}\right) \\
=1+f\left(\frac{1}{2}\right)=0 .
\end{array}
$$
Three, 13. (1) Let $y=x$, we get $2 f(x)=f\left(\frac{2 x}{1+x^{2}}\right)$.
Since $x_{n+1}=\frac{2 x_{n}}{1+x_{n}^{2}}$, then $f\left(x_{n+1}\right)=2 f\left(x_{n}\right)$. Hence, $a_{n+1}=2 a_{n}$.
Also, $a_{1}=f\left(\frac{1}{2}\right)=-1$, so $\left\{a_{n}\right\}$ is a geometric sequence with the first term -1 and common ratio 2.
Therefore, $a_{n}=-2^{n-1}$.
(2) Let $x=y=0$, we get $f(0)+f(0)=f(0)$.
Then $f(0)=0$.
Let $y=-x$, we get $f(x)+f(-x)=f(0)=0$.
Then $f(-x)=-f(x)$.
$$
\begin{aligned}
& \text { And } f\left(\frac{1}{b_{n}}\right)=f\left(\frac{1}{n^{2}+3 n+1}\right)=f\left(\frac{1}{(n+1)(n+2)-1}\right) \\
= & f\left(\frac{\frac{1}{(n+1)(n+2)}}{1-\frac{1}{(n+1)(n+2)}}\right)=f\left(\frac{\frac{1}{n+1}-\frac{1}{n+2}}{1-\frac{1}{(n+1)(n+2)}}\right) \\
= & f\left(\frac{1}{n+1}\right)+f\left(-\frac{1}{n+2}\right)=f\left(\frac{1}{n+1}\right)-f\left(\frac{1}{n+2}\right),
\end{aligned}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a, b, c$ are real numbers, and $2a + b + c = 5, b - c = 1$. Then the maximum value of $ab + bc + ca$ is
|
From $2 a+b+c=5, b-c=1$, we can obtain
$$
b=3-a, c=2-a \text {. }
$$
Therefore, $a b+b c+c a$
$$
\begin{array}{l}
=a(3-a)+(3-a)(2-a)+a(2-a) \\
=-a^{2}+6 \leqslant 6 .
\end{array}
$$
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 2, quadrilateral $ABCD$ is inscribed in $\odot O$, with $BD$ being the diameter of $\odot O$, and $\overparen{AB}=\overparen{AD}$. If $BC + CD = 4$, then the area of quadrilateral $ABCD$ is $\qquad$ .
|
3.4 .
From $\overparen{A B}=\overparen{A D}$, we know $A B=A D$.
Also, $B D$ is the diameter of $\odot 0$, so $\angle A=\angle C=90^{\circ}$.
Let $A B=A D=x, B C=y, C D=z$.
By the Pythagorean theorem, we have
$$
x^{2}+x^{2}=y^{2}+z^{2} \text {. }
$$
Thus, $2 x^{2}=y^{2}+z^{2}=(y+z)^{2}-2 y z$.
Also, $S_{\text {quadrilateral } A B C D}=\frac{1}{2} A B \cdot A D+\frac{1}{2} B C \cdot C D$ $=\frac{1}{2} x^{2}+\frac{1}{2} y z$,
From equation (1), we know $S_{\text {quadrilateral } A B C D}=\frac{1}{4}(y+z)^{2}=4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Renovate a pedestrian path, changing the ground originally paved with $n^{2}-96$ identical square tiles to be paved with larger identical square tiles, requiring a total of $5 n+51$ tiles to complete the paving. If $n^{2}-96$ is divisible by $5 n+51$, find the value of the positive integer $n$.
|
Given the problem, let $\frac{n^{2}-96}{5 n+51}=k, k$ be a positive integer, then
$$
\begin{array}{l}
n^{2}-5 k n-(51 k+96)=0 . \\
\text { Therefore, } \Delta=25 k^{2}+4(51 k+96) \\
=25 k^{2}+204 k+384 .
\end{array}
$$
Since $n$ is an integer, $\Delta$ must be a perfect square.
$$
\text { And }(5 k+19)^{2}<25 k^{2}+204 k+384<(5 k+21)^{2} \text {, }
$$
Therefore, $\Delta=(5 k+20)^{2}=25 k^{2}+204 k+384$.
Solving for $k$ gives $k=4$.
Substituting $k=4$ into equation (1) yields
$$
n^{2}-20 n-300=0 \text {. }
$$
Solving for $n$ gives $n=30$.
|
30
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given that $a$ is a positive integer not greater than 2005, and $b, c$ are integers, the parabola $y=a x^{2}+b x+c$ is above the $x$-axis and passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$. Find the minimum value of $a-b+c$.
The parabola $y=a x^{2}+b x+c$ passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$.
Substituting these points into the equation of the parabola, we get:
1. For point $A(-1,4 a+7)$:
\[ 4a + 7 = a(-1)^2 + b(-1) + c \]
\[ 4a + 7 = a - b + c \]
\[ 3a + 7 = -b + c \]
\[ c - b = 3a + 7 \quad \text{(1)} \]
2. For point $B(3,4 a-1)$:
\[ 4a - 1 = a(3)^2 + b(3) + c \]
\[ 4a - 1 = 9a + 3b + c \]
\[ -5a - 1 = 3b + c \]
\[ c + 3b = -5a - 1 \quad \text{(2)} \]
We now have the system of linear equations:
\[ c - b = 3a + 7 \quad \text{(1)} \]
\[ c + 3b = -5a - 1 \quad \text{(2)} \]
Subtract equation (1) from equation (2):
\[ (c + 3b) - (c - b) = (-5a - 1) - (3a + 7) \]
\[ 4b = -8a - 8 \]
\[ b = -2a - 2 \]
Substitute \( b = -2a - 2 \) into equation (1):
\[ c - (-2a - 2) = 3a + 7 \]
\[ c + 2a + 2 = 3a + 7 \]
\[ c = a + 5 \]
Now, we need to find \( a - b + c \):
\[ a - b + c = a - (-2a - 2) + (a + 5) \]
\[ a + 2a + 2 + a + 5 \]
\[ 4a + 7 \]
Since \( a \) is a positive integer not greater than 2005, the minimum value of \( 4a + 7 \) occurs when \( a = 1 \):
\[ 4(1) + 7 = 11 \]
Thus, the minimum value of \( a - b + c \) is:
\[ \boxed{11} \]
|
Three, since $a$ is a positive integer, the parabola opens upwards.
The parabola $y=a x^{2}+b x+c$ is above the $x$-axis, so,
$$
b^{2}-4 a c<0.
$$
Given $b=-2-2a$ and $c=a+5$, we have
$$
b^{2}-4 a c=(-2-2a)^{2}-4a(a+5)=4+8a+4a^{2}-4a^{2}-20a=4-12a<0.
$$
This implies
$$
a>\frac{1}{3}.
$$
Since $a$ is a positive integer no greater than 2005, we have $a \geqslant 1$.
Thus, $b=-2-2a \leqslant-4, -b \geqslant 4, c=a+5 \geqslant 6$.
When $a=1, b=-4, c=6$, the parabola $y=x^{2}-4 x+6$ satisfies the given conditions.
Therefore, the minimum value of $a-b+c$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A cube with an edge length of a certain integer is cut into 99 smaller cubes, 98 of which are unit cubes with an edge length of 1, and the other cube also has an integer edge length. Then its edge length is $\qquad$
|
4.3.
Let the edge length of the original cube be $a$, and the edge length of the other cube after cutting be $b\left(a, b \in \mathbf{Z}_{+}, a>b\right)$. Then
$$
a^{3}-98=b^{3} \text {, }
$$
i.e., $(a-b)\left(a^{2}+a b+b^{2}\right)=98$.
Thus, $\left\{\begin{array}{l}a-b=1, \\ a^{2}+a b+b^{2}=98 ;\end{array}\right.$
$$
\left\{\begin{array} { l }
{ a - b = 2 , } \\
{ a ^ { 2 } + a b + b ^ { 2 } = 4 9 ; }
\end{array} \left\{\begin{array}{l}
a-b=7, \\
a^{2}+a b+b^{2}=14 .
\end{array}\right.\right.
$$
Upon inspection, we find that $a=5, b=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given a non-constant sequence $\left\{a_{i}\right\}$ satisfies
$$
a_{i}^{2}-a_{i-1} a_{i}+a_{i-1}^{2}=0 \text {, }
$$
and $a_{i+1} \neq a_{i-1}, i=1,2, \cdots, n$.
For a given positive integer $n, a_{1}=a_{n+1}$. Then
$$
\sum_{i=0}^{n-1} a_{i}=
$$
|
6.0 .
Given $a_{1}^{2}-a_{1-1} a_{i}+a_{1-1}^{2}=0$, so,
$$
a_{i+1}^{2}-a_{i} a_{i+1}+a_{i}^{2}=0 \text {. }
$$
Subtracting the two equations, we get
$$
\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}\right)-a_{i}\left(a_{i+1}-a_{i-1}\right)=0 .
$$
That is, $\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}-a_{i}\right)=0$.
Since $a_{i+1} \neq a_{i-1}$, we have,
$$
a_{i+1}+a_{i-1}-a_{i}=0 .
$$
Therefore, $a_{0}=a_{1}-a_{2}, a_{1}=a_{2}-a_{3}, \cdots$
$$
a_{n-2}=a_{n-1}-a_{n}, a_{n-1}=a_{n}-a_{n+1} .
$$
Thus, $\sum_{i=0}^{n-1} a_{i}=a_{1}-a_{n+1}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points) Given positive integers $x_{1}<x_{2}<\cdots<x_{n}$, and
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=2003, n \geqslant 2, \\
f(n)=n\left(x_{1}+x_{n}\right) .
\end{array}
$$
Find the minimum value of $f(n)$.
|
If $x_{1}>1$, let $x_{1}^{\prime}=1, x_{n}^{\prime}=x_{n}+x_{1}-1$, and the other $x_{i}$ values remain unchanged, then
$$
\begin{array}{l}
x_{1}^{\prime}+x_{2}+x_{3}+\cdots+x_{n-1}+x_{n}^{\prime}=2003, \\
f(n)=n\left(x_{1}+x_{n}\right)=n\left(1+x_{n}^{\prime}\right) .
\end{array}
$$
Assume $x_{1}=1$.
When $n=2$, $f(2)=4006$.
When $n \geqslant 3$, we have
$$
\begin{array}{l}
x_{2}<x_{3}<\cdots<x_{n} \\
\Rightarrow x_{n-1} \leqslant x_{n}-1, \cdots, x_{2} \leqslant x_{n}-(n-2) \\
\Rightarrow(n-1) x_{n}-[1+2+\cdots+(n-2)] \\
\quad \geqslant x_{2}+x_{3}+\cdots+x_{n}=2002 \\
\Rightarrow(n-1) x_{n}-\frac{1}{2}(n-1)(n-2) \geqslant 2002 \\
\Rightarrow x_{n} \geqslant \frac{1}{2}(n-2)+\frac{2002}{n-1} \\
\Rightarrow f(n)=n\left(1+x_{n}\right)=n+n x_{n} \\
\quad \geqslant n+\frac{1}{2} n(n-2)+\frac{2002 n}{n-1}=\frac{1}{2} n^{2}+\frac{2002 n}{n-1} .
\end{array}
$$
Let $g(n)=\frac{1}{2} n^{2}+\frac{2002 n}{n-1}$, and examine the monotonicity of $g(n)$:
When $n \leqslant 13$, $g(n)$ is monotonically decreasing;
When $n \geqslant 14$, $g(n)$ is monotonically increasing.
Also, $g(13)=2253+\frac{1}{3}, g(14)=2254$, therefore,
$$
f(n) \geqslant g(n) \geqslant 2253+\frac{1}{3} \Rightarrow f(n) \geqslant 2254 .
$$
When $n=14$, take
$$
x_{1}=1, x_{2}=148, x_{3}=149, \cdots, x_{14}=160 \text {, }
$$
then $f(n)_{\text {min }}=n\left(x_{1}+x_{n}\right)=2254$.
|
2254
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) There are 12 football teams participating in a match, with each pair of teams playing one game. The winner gets 3 points, the loser gets 0 points, and in the case of a draw, both teams get 1 point. How many points does one team need to score at a minimum to ensure that no more than 6 teams score at least as many points as this team?
|
Three, assuming there are 7 teams with the same score as a certain team, then the maximum score for each match between these 8 teams is 3 points. Therefore, their maximum total score is $3 C_{8}^{2}=84$ points, and the score of each team cannot exceed $\left[\frac{84}{8}\right]=10$ points, so each team can win at most 3 matches.
Now, divide 12 teams into two groups, one with 8 teams of the same score, and the other with the remaining 4 teams. The 8 teams in the same group are arranged in a circle, where each team wins against the 3 teams behind it, draws with the 4th team, and loses to the other 3 teams. Thus, the maximum score for each of these 8 teams is $(3 \times 3+1)+3 \times 4=22$ points (i.e., each of the 8 teams in the same group wins against 3 teams, draws with 1 team, and loses to 3 teams, and wins against the 4 teams in the other group).
Therefore, when 8 teams have the same score (with 7 teams having a score no less than that team), each team can score a maximum of 22 points. Hence, when the proposition in the question holds, the score of that team is no less than 23 points.
Below is the proof: If there is a team that scores at least 23 points, then at most 6 teams can have a score no less than that team. Otherwise, if there are 7 teams with a score no less than 23 points, then the total score of these 8 teams is at least $8 \times 23=184$ points. On the other hand, dividing 12 teams into two groups, one with 8 teams and the other with 4 teams, the maximum total score for the 8 teams in the same group is $3 \mathrm{C}_{8}^{2}+3 \times 4 \times 8=180$ points. But $180 < 184$, which is a contradiction, so the proposition holds.
|
23
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find the smallest positive integer $a$ that satisfies the following condition: there exists a positive odd number $n$, such that $203^{n} + a \cdot 198^{n}$ is a multiple of 2005.
|
Solution: Since $2005=401 \times 5$, and
$$
203+198=401,203-198=5 \text {, }
$$
Therefore,
$$
203^{n}+a \cdot 198^{n}=(401-198)^{n}+a \cdot 198^{n}
$$
is a multiple of $401 \times 5$.
Since $n$ is odd, then
$$
-198^{n}+a \cdot 198^{n}=(a-1) \cdot 198^{n}
$$
is a multiple of 401.
Since the greatest common divisor of 198 and 401 is 1, then $a-1$ is a multiple of 401. Also,
$$
203^{n}+a \cdot 198^{n}=(198+5)^{n}+a \cdot 198^{n}
$$
is a multiple of $401 \times 5$, then
$$
198^{n}+a \cdot 198^{n}=(a+1) \cdot 198^{n}
$$
is a multiple of 5.
Since the greatest common divisor of 198 and 5 is 1, then $a+1$ is a multiple of 5.
From the above, $a+1$ is a multiple of 5 and $a-1$ is a multiple of 401, so we can set
$a-1=401 t$ (where $t$ is an integer).
Thus, $a+1=401 t+2$.
Therefore, taking the smallest $t=3, a+1=1205$ is a multiple of 5.
Hence, the smallest $a$ is 1204.
|
1204
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If a store sells a certain product, which costs 100 yuan, at 120 yuan, it can sell 300 units. If the price of the product is increased by 1 yuan based on 120 yuan, it will sell 10 fewer units, and if the price is reduced by 1 yuan, it will sell 30 more units. Question: To maximize profit, what price should the store set for the product?
|
Solution: (1) If sold at 120 yuan each, 300 can be sold, and the profit is $300 \times 20 = 6000$ (yuan).
(2) If the price is increased by $x (x > 0)$ yuan, then (300 $- 10x$) can be sold. Let the profit be $y$ yuan, then
$$
\begin{array}{l}
y = (20 + x)(300 - 10x) \\
= -10x^2 + 100x + 6000 \\
= -10(x - 5)^2 + 6250 .
\end{array}
$$
Because $-100)$ yuan, then (300 $+ 30x$) can be sold. Let the profit be $y$ yuan, then
$$
\begin{array}{l}
y = (20 - x)(300 + 30x) \\
= -30x^2 + 300x + 6000 \\
= -30(x - 5)^2 + 6750 .
\end{array}
$$
Because $-30 < 0$. So,
When $x = 5$, $y_{\text{max}} = 6750$.
Therefore, when the price is reduced to 115 yuan, the maximum profit of 6750 yuan can be obtained.
In conclusion, when the price is set at 115 yuan, the store can achieve the maximum profit.
|
115
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4: On a circular road in a certain town, there are five primary schools in sequence: No.1 Primary School, No.2 Primary School, No.3 Primary School, No.4 Primary School, and No.5 Primary School. They have 15, 7, 11, 3, and 14 computers respectively. To make the number of computers in each school the same, how many computers should be transferred to the neighboring school: No.1 to No.2, No.2 to No.3, No.3 to No.4, No.4 to No.5, and No.5 to No.1? If School A gives -3 computers to School B, it means School B gives 3 computers to School A. To minimize the total number of computers moved, what arrangement should be made?
|
Solution: As shown in Figure 4, let $A, B, C, D, E$ represent the first to fifth schools, respectively, and let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the number of computers transferred to the neighboring schools. Let the total number of transfers be $y$. According to the problem, we have
$$
\begin{array}{l}
7+x_{1}-x_{2}=11+x_{2}-x_{3}=3+x_{3}-x_{4} \\
=14+x_{4}-x_{5}=15+x_{5}-x_{1}=10 .
\end{array}
$$
Thus, we have $x_{2}=x_{1}-3, x_{3}=x_{1}-2$,
$$
x_{4}=x_{1}-9, x_{5}=x_{1}-5 \text {. }
$$
According to the problem, we have
$$
\begin{aligned}
y= & \left|x_{1}\right|+\left|x_{2}\right|+\left|x_{3}\right|+\left|x_{4}\right|+\left|x_{5}\right| \\
= & \left|x_{1}\right|+\left|x_{1}-3\right|+\left|x_{1}-2\right|+\left|x_{1}-9\right|+ \\
& \left|x_{1}-5\right| \\
= & \left|x_{1}\right|+\left|x_{1}-2\right|+\left|x_{1}-3\right|+\left|x_{1}-5\right|+ \\
& \left|x_{1}-9\right| .
\end{aligned}
$$
By the geometric meaning of absolute value, when $x_{1}=3$, $y$ has the minimum value, which is 12. At this time, $x_{2}=0$, $x_{3}=1, x_{4}=-6, x_{5}=-2$.
Therefore, the transfer plan is: School 1 transfers 3 computers to School 2, School 3 transfers 1 computer to School 4, School 5 transfers 6 computers to School 4, and School 1 transfers 2 computers to School 5. This way, the total number of computers transferred is minimized, with the minimum number being 12.
Note: Using the geometric meaning of absolute value, the following conclusion can be explained:
Let $a_{1}, a_{2}, \cdots, a_{n}$ be the real numbers represented by points on the number line arranged in sequence.
When $n$ is odd, if $x=a_{\frac{n+1}{2}}$, then
$$
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|
$$
has the minimum value;
When $n$ is even, if $a_{\frac{n}{2}} \leqslant x \leqslant a_{\frac{n}{2}+1}$, then
$$
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|
$$
has the minimum value.
|
12
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 A unit spent 500,000 yuan to purchase a piece of high-tech equipment. According to the tracking survey of this model of equipment, after the equipment is put into use, if the maintenance and repair costs are averaged to each day, the conclusion is: the maintenance and repair cost on the $x$-th day is $\left[\frac{1}{4}(x-1)+500\right]$ yuan.
|
Solution: (1) If the equipment is put into use for $x$ days, and the average daily loss is $y$ yuan, then
$$
\begin{aligned}
y= & \left[500000+\left(\frac{1}{4} \times 0+500\right)+\left(\frac{1}{4} \times 1+500\right)+\right. \\
& \left.\cdots+\left(\frac{x-1}{4}+500\right)\right] \div x \\
= & \frac{1}{x}\left[500000+500 x+\frac{1}{4} \cdot \frac{x(x-1)}{2}\right] \\
= & \frac{500000}{x}+\frac{x}{8}+499 \frac{7}{8} .
\end{aligned}
$$
$$
\begin{array}{l}
\text { (2) } y=\frac{500000}{x}+\frac{x}{8}+499 \frac{7}{8} \\
\geqslant 2 \sqrt{\frac{500000}{x} \cdot \frac{x}{8}}+499 \frac{7}{8} \\
=500+499 \frac{7}{8}=999 \frac{7}{8} .
\end{array}
$$
Equality holds if and only if $\frac{500000}{x}=\frac{x}{8}$, i.e., $x=2000$,
at which point, $y$ reaches its minimum value.
Therefore, the equipment should be scrapped after being put into use for 2000 days.
|
2000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Several workers load and unload a batch of goods, with each worker having the same loading and unloading speed. If these workers work simultaneously, it will take $10 \mathrm{~h}$ to complete the loading and unloading. Now, the loading and unloading method is changed, starting with one person working, and then adding one more person every $t \mathrm{~h} (t$ is an integer). Each person who participates in the loading and unloading continues working until the task is completed, and the last person added works for a time that is $\frac{1}{4}$ of the time the first person works. Find:
(1) According to the changed loading and unloading method, how long does it take from start to finish?
(2) How many workers participate in the loading and unloading?
|
16. (1) Let the loading and unloading work take $x \mathrm{~h}$ to complete, then the first person worked for $x \mathrm{~h}$, and the last person worked for $\frac{x}{4} \mathrm{~h}$. The total work time of the two people is $\left(x+\frac{x}{4}\right) \mathrm{h}$, and the average work time per person is $\frac{1}{2}\left(x+\frac{x}{4}\right) \mathrm{h}$. According to the problem, the second person and the second-to-last person, the third person and the third-to-last person, etc., also have an average work time of $\frac{1}{2}\left(x+\frac{x}{4}\right) \mathrm{h}$. From the given conditions, we have
$$
\frac{1}{2}\left(x+\frac{x}{4}\right)=10 \text {. Solving for } x \text {, we get } x=16 \text {. }
$$
(2) Let there be $y$ people participating in the loading and unloading work. Since one person is added every $t \mathrm{~h}$, the last person works $(y-1) t \mathrm{~h}$ less than the first person. According to the problem, we have
$$
16-(y-1) t=16 \times \frac{1}{4},
$$
which simplifies to $(y-1) t=12$.
Solving this indeterminate equation, we get
$$
\begin{array}{c}
\left\{\begin{array} { l }
{ y = 2 , } \\
{ t = 1 2 ; }
\end{array} \quad \left\{\begin{array} { l }
{ y = 3 , } \\
{ t = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
y=4, \\
t=4 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l }
{ y = 5 , } \\
{ t = 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ y = 7 , } \\
{ t = 2 ; }
\end{array} \quad \left\{\begin{array}{l}
y=13, \\
t=1 .
\end{array}\right.\right.\right.
\end{array}
$$
Thus, the number of people participating is $y=2$ or 3 or 4 or 5 or 7 or 13.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, $\cdots \cdots$, and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest scores of each athlete's 9 scores is no more than 3. Let the total scores of the athletes be $c_{1}$, $c_{2}, \cdots, c_{12}$, and $c_{1} \leqslant c_{2} \leqslant \cdots \leqslant c_{12}$. Find the maximum value of $c_{1}$.
|
18.9 judges cannot give 1 point to 5 or more athletes, because among 5 or more athletes, at least one athlete must be rated no less than 5 by a judge, while according to the problem, each of these 5 athletes is rated no more than 4 by each judge, which is a contradiction. Therefore, 9 judges can give 1 point to at most 4 athletes.
Below, we discuss different scenarios.
(1) If all judges give 1 point to a single athlete, then, $c_{1}=9$.
(2) If the 9 points of 1 given by 9 judges are concentrated on two athletes, then, one of these athletes must be rated 1 by at least 5 judges. Thus, by the problem's condition, the remaining judges give this athlete a score no greater than 4, so, $c_{1} \leqslant 5 \times 1+4 \times 4$ $=21$.
(3) If the 9 points of 1 given by 9 judges are concentrated on three athletes, then, the total score of these three athletes is no more than
$9 \times 1+9 \times 3+9 \times 4=72$.
Thus, $3 c_{1} \leqslant c_{1}+c_{2}+c_{3} \leqslant 72$.
Hence, $c_{1} \leqslant 24$.
(4) If the 9 points of 1 are distributed among 4 athletes, then, the total score of these 4 athletes is
$9 \times 1+9 \times 2+9 \times 3+9 \times 4=90$.
Thus, $4 c_{1} \leqslant 90$. Hence, $c_{1}<23$.
In summary, $c_{1} \leqslant 24$.
The scenario where $c_{1}=24$ is achievable, as shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $A_{1}$ & $A_{2}$ & $A_{3}$ & $A_{4}$ & $A_{5}$ & $A_{6}$ & $A_{7}$ & $A_{8}$ & $A_{9}$ & $A_{10}$ & $A_{11}$ & $A_{12}$ \\
\hline$B_{1}$ & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline$B_{2}$ & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline$B_{3}$ & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline$B_{4}$ & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline$B_{5}$ & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline$B_{6}$ & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline$B_{7}$ & 3 & 1 & 4 & 2 & 5 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline$B_{8}$ & 3 & 1 & 4 & 5 & 2 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline$B_{9}$ & 3 & 1 & 4 & 2 & 5 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline Total & 24 & 24 & 24 & 30 & 33 & 66 & 66 & 66 & 87 & 87 & 87 & 108 \\
\hline
\end{tabular}
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. A and B are two car retailers (hereinafter referred to as A and B) who ordered a batch of cars from a car manufacturer. Initially, the number of cars A ordered was 3 times the number of cars B ordered. Later, due to some reason, A transferred 6 cars from its order to B. When picking up the cars, the manufacturer provided 6 fewer cars than the total number ordered by A and B. In the end, the number of cars A purchased was twice the number of cars B purchased. Try to find: what is the maximum number of cars that A and B finally purchased in total? And what is the minimum number of cars they purchased in total?
|
Three, 13. Let the total number of cars finally purchased by A and B be $x$ units. In the 6 fewer cars supplied by the factory, A requested $y$ fewer cars $(0 \leqslant y \leqslant 6)$, and B requested $(6-y)$ fewer cars. Then we have
$$
\begin{array}{l}
\frac{3}{4}(x+6)-6-y \\
=2\left[\frac{1}{4}(x+6)+6-(6-y)\right] .
\end{array}
$$
After rearranging, we get $x=18+12 y$.
When $y=6$, $x$ is at most 90;
When $y=0$, $x$ is at least 18.
Therefore, the total number of cars purchased by A and B is at most 90 units and at least 18 units.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. From $1,2, \cdots, 2004$, choose $k$ numbers such that among the chosen $k$ numbers, there are definitely 3 numbers that can form the side lengths of a triangle (here it is required that the three side lengths of the triangle are all different). What is the minimum value of $k$ that satisfies the condition?
|
15. This problem is equivalent to:
Selecting $k-1$ numbers from $1,2, \cdots, 2004$, such that no three of these numbers can form the lengths of the sides of a triangle (all sides being unequal). What is the maximum value of $k$ that satisfies this condition?
For any array that meets the above conditions,
when $k=4$, the smallest 3 numbers are $1,2,3$. From this, the array can be continuously expanded: as long as the number added is greater than or equal to the sum of the two largest numbers already in the array. Therefore, to maximize $k$, the number added should be equal to the sum of the two largest numbers in the array. Thus, we get:
$$
\begin{array}{l}
1,2,3,5,8,13,21,34,55,89,144, \\
233,377,610,987,1597
\end{array}
$$
There are 16 numbers in total. For any array $a_{1}, a_{2}, \cdots, a_{n}$ that meets the above conditions, it is clear that $a_{i}$ is always greater than or equal to the $i$-th number in (1). Therefore, $n \leqslant 16 \leqslant k-1$. Hence, the minimum value of $k$ is 17.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The cost price of each product is 120 yuan. During the trial sales phase, the relationship between the selling price $x$ (yuan) of each product and the daily sales volume $y$ (units) is shown in Table 2. If the daily sales volume $y$ is a linear function of the selling price $x$, to maximize profit, what should the selling price of each product be set to? What is the daily sales profit at this price? (Sales profit = Selling price - Cost price)
Table 2
\begin{tabular}{|r|r|r|r|}
\hline$x$ (yuan) & 130 & 150 & 165 \\
\hline$y($ units $)$ & 70 & 50 & 35 \\
\hline
\end{tabular}
|
(Answer: When the selling price is set to 160 yuan, the maximum daily profit is 1600 yuan.)
|
1600
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If $a+b+c=0, \frac{1}{a+1}+\frac{1}{b+2}+$ $\frac{1}{c+3}=0$, then, $(a+1)^{2}+(b+2)^{2}+(c+3)^{2}$ is equal to (.
(A) 36
(B) 16
(C) 14
(D) 3
|
8. A.
From the problem, we have
$$
\begin{array}{l}
(a+1)(b+2)+(b+2)(c+3)+(c+3)(a+1)=0 \text {. } \\
\text { Then }(a+1)^{2}+(b+2)^{2}+(c+3)^{2} \\
=(a+b+c+6)^{2}-2[(a+1)(b+2)+ \\
(b+2)(c+3)+(c+3)(a+1)]=36 .
\end{array}
$$
|
36
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$13.2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5$ can achieve the minimum value of $\qquad$ .
|
13. -10 .
Original expression $=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10$. When $x=2, y=-1$, it has the minimum value -10.
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15.1 A can of coffee is shared by person A and person B, and they finish it together in 10 days. If person A drinks it alone, it takes 12 days. 1 pound of tea is shared by person A and person B, and they finish it together in 12 days. If person B drinks it alone, it takes 20 days. Assuming that person A will never drink coffee when there is tea, and person B will never drink tea when there is coffee. How many days will it take for the two to finish 1 pound of tea and 1 can of coffee together?
|
Three, 15. It is known that A drinks $\frac{1}{30}$ of 1 catty of tea every day, and B drinks $\frac{1}{60}$ of 1 can of coffee every day.
After 30 days, A finishes the tea while B only drinks half a can of coffee, and the remaining half can of coffee is drunk by A and B together in 5 days. Therefore, it takes a total of 35 days to finish both the coffee and the tea.
|
35
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $a=1+2+\cdots+2004$. Then the remainder when $a$ is divided by 17 is
|
8.1.
$$
\begin{array}{l}
a=1+2+\cdots+2004 \\
=\frac{2004 \times 2005}{2}=2009010 .
\end{array}
$$
Then 2009010 divided by 17 gives a quotient of 118177, with a remainder of 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given $f(x)=x^{2}+x-1$. If $a b^{2} \neq 1$, and $f\left(a^{-1}\right)=f\left(b^{2}\right)=0$, then $\frac{a}{1+a b^{2}}=$ $\qquad$ .
|
9. -1 .
Given $f(x)=x^{2}+x-1, f\left(a^{-1}\right)=f\left(b^{2}\right)=0, a b^{2} \neq 1$, we know that $a^{-1}$ and $b^{2}$ are the two real roots of $f(x)=x^{2}+x-1$.
By Vieta's formulas, we get $\frac{1}{a}+b^{2}=-1, \frac{b^{2}}{a}=-1$. Thus, $\frac{1}{a}+b^{2}=\frac{b^{2}}{a}=-1$.
Therefore, we have $1+a b^{2}=-a$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Simplify $\left(\log _{3} 4+\log _{2} 9\right)^{2}-\left(\log _{3} 4-\log _{2} 9\right)^{2}$ $=$ $\qquad$
|
2.16.
$$
\begin{array}{l}
(\log 4+\log 9)^{2}-(\log 4-\log 9)^{2} \\
=4 \log 4 \cdot \log 9=4 \times \frac{\log 4}{\log 3} \times \frac{\log 9}{\log 2} \\
=4 \times \frac{2 \log 2}{\log 3} \times \frac{2 \log 3}{\log 2}=16 .
\end{array}
$$
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=2$, $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}$, and $a_{n+1} a_{n+2} \neq$ 1. Then $a_{1}+a_{2}+\cdots+a_{2004}=$ $\qquad$
|
3.4008.
Substituting $a_{1}=1, a_{2}=2$ into $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+$ $a_{n+2}$, we get $a_{3}=3$.
$$
\begin{array}{l}
\text { From } a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}, \\
a_{n+1} a_{n+2} a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3},
\end{array}
$$
Subtracting the two equations, we get $\left(a_{n+3}-a_{n}\right)\left(a_{n+1} a_{n+2}-1\right)=0$.
Since $a_{n+1} a_{n+2} \neq 1$, it follows that $a_{n+3}=a_{n}$.
Therefore, the sequence $\left\{a_{n}\right\}$ is a periodic sequence.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{2000} \\
=\left(a_{1}+a_{2}+a_{3}\right)+\left(a_{4}+a_{5}+a_{6}\right)+\cdots+ \\
\left(a_{2002}+a_{2003}+a_{2004}\right) \\
=(1+2+3) \times 668=4008 \text {. } \\
\end{array}
$$
|
4008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 In an exam, there are 5 multiple-choice questions, each with 4 different options, and each person selects exactly 1 option for each question. In 2000 answer sheets, it is found that there exists an $n$, such that in any $n$ answer sheets, there are 4 sheets where any 2 sheets have at most 3 answers the same. Find the minimum possible value of $n$.
(2000, China Mathematical Olympiad)
|
Solution: Let the 4 possible answers for each question be denoted as $1, 2, 3, 4$, and the answers on each test paper be denoted as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. Let $\{(1, h, i, j, k), (2, h, i, j, k), (3, h, i, j, k), (4, h, i, j, k)\}, h, i, j, k = 1, 2, 3, 4$, yielding a total of 256 quadruples.
Since $2000 = 256 \times 7 + 208$, by the pigeonhole principle, there are 8 test papers with answers belonging to the same quadruple. After removing these 8 test papers, there are still 8 test papers with answers belonging to the same quadruple among the remaining 1992 test papers; after removing these 8 test papers, there are still 8 test papers with answers belonging to the same quadruple among the remaining 1984 test papers. After removing these 8 test papers, a total of 24 test papers have been removed. Among these 24 test papers, any 4 of them have at least 2 with answers belonging to the same quadruple, which of course does not
meet the requirements of the problem. Therefore, $n \geqslant 25$.
Next, we prove that $n$ can be 25.
Let $S = \{(g, h, i, j, k) \mid g + h + i + j + k \equiv 0 \pmod{4}, g, h, i, j, k \in \{1,2,3,4\}\}$. Then $|S| = 256$, and any 2 answers in $S$ have at most 3 questions the same. Removing 6 elements from $S$, when each of the remaining 250 answers is chosen by exactly 8 people, a total of 2000 test papers are obtained, and among these 25 answers, there are always 4 that are different. Since they are all in $S$, they of course meet the requirements of the problem. This shows that $n = 25$ meets the requirements of the problem.
In summary, the minimum possible value of $n$ is 25.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given a set of points $P=\left\{P_{1}, P_{2}, \cdots, P_{1924}\right\}$ on a plane, and no three points in $P$ are collinear. Divide all the points in $P$ into 83 groups arbitrarily, such that each group has at least three points, and each point belongs to exactly one group. Then, connect any two points in the same group with a line segment, and do not connect points in different groups, thus obtaining a pattern $G$. Different grouping methods result in different patterns. Let the number of triangles in pattern $G$ with vertices from points in $P$ be denoted as $m(G)$.
(1) Find the minimum value $m_{0}$ of $m(G)$;
(2) Let $G^{*}$ be a pattern such that $m\left(G^{*}\right)=m_{0}$. If the line segments (referring to segments with endpoints from points in $P$) in $G^{*}$ are colored with 4 colors, and each segment is colored exactly one color. Prove: There exists a coloring scheme such that after coloring $G^{*}$, there is no triangle with vertices from points in $P$ whose three sides are of the same color.
(1994, National High School Mathematics Competition)
|
Solution: (1) Let $m(G)=m_{0}, G$ is obtained from the groups $X_{1}, X_{2}, \cdots, X_{88}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots, 83$.
Let $\left|X_{i}\right|=x_{i}, i=1,2, \cdots, 83$, then we have
$$
x_{1}+x_{2}+\cdots+x_{83}=1994 \text {, }
$$
and $m_{0}=C_{x_{1}}^{3}+C_{x_{2}}^{3}+\cdots+C_{x_{83}}^{3}$.
We now prove that, for $1 \leqslant i \neq j \leqslant 83$, we have
$$
\left|x_{i}-x_{j}\right| \leqslant 1 \text {. }
$$
In fact, if there exist $i, j (1 \leqslant i, j \leqslant 83)$ such that $\left|x_{i}-x_{j}\right| \geqslant 2$, without loss of generality, assume $x_{i}>x_{j}$. Consider a partition of the points of $P$ into groups $Y_{1}, Y_{2}, \cdots, Y_{88}$ (where $Y_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots, 83$), such that
$$
y_{k}=\left|Y_{k}\right|=\left\{\begin{array}{l}
x_{k}, \quad \text { when } k \neq i, \text { and } k \neq j \text {; } \\
x_{i}-1, \text { when } k=i \text {; } \\
x_{j}+1, \text { when } k=j \text {. }
\end{array}\right.
$$
Such a partition exists (just move one point from the original $X_{i}$ to $X_{j}$, and leave the other groups unchanged). For the pattern $G^{\prime}$ obtained from the partition $Y_{1}, Y_{2}, \cdots, Y_{88}$, we have
$$
\begin{array}{l}
m\left(G^{\prime}\right)=C_{y_{1}}^{3}+C_{y_{2}}^{3}+\cdots+C_{y_{y_{3}}}^{3} . \\
\text { Hence } m\left(G^{\prime}\right)-m_{0}=C_{y_{y_{1}}^{3}}^{3}+C_{y_{j}}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3} \\
=C_{x_{i}-1}^{3}+C_{x_{j}+1}^{3}-C_{x_{i}}^{3}-C_{x_{j}}^{3} \\
=C_{x_{x_{i}-1}-1}^{3}+C_{x_{x_{j}}^{2}}^{2}+C_{x_{j}}^{3}-C_{x_{i}-1}^{2}-C_{x_{i}-1}^{3}-C_{x_{j}}^{3} \\
=C_{x_{j}}^{2}-C_{x_{i}-1}^{2} .
\end{array}
$$
Since $x_{j}<x_{i}-1$, we have $\mathrm{C}_{x_{1}}^{2}<\mathrm{C}_{x_{1}-1}^{2}$. Therefore,
$$
m\left(G^{\prime}\right)-m_{0}<0,
$$
which contradicts the minimality of $m_{0}$.
Also, $1994=83 \times 24+2=81 \times 24+2 \times 25$,
so $m_{0}=81 C_{24}^{3}+2 C_{25}^{3}=168544$.
(2) Let the pattern $G^{*}$ be obtained from the groups $X_{1}, X_{2}, \cdots, X_{83}$, where $X_{i}$ is the set of points in the $i$-th group $(i=1,2, \cdots, 83)$.
By (1), we can assume
$\left|X_{1}\right|=\left|X_{2}\right|=\cdots=\left|X_{81}\right|=24$,
$\left|X_{82}\right|=\left|X_{83}\right|=25$.
We now provide a coloring method for $G^{*}$ such that $G^{*}$, when colored with 4 different colors, does not contain a triangle with all three sides of the same color.
Let the graph formed by the set $X_{i}$ and its connecting segments be called the $i$-th block of $G^{*}$, denoted as $G_{i}^{*}, i=1,2, \cdots, 83$.
For $G_{83}^{*}$, let
$X_{83}=Y_{1} \cup Y_{2} \cup Y_{3} \cup Y_{4} \cup Y_{5}$,
such that $Y_{\imath} \cap Y_{j}=\varnothing(1 \leqslant i \neq j \leqslant 5)$,
$\left|Y_{i}\right|=5(1 \leqslant i \leqslant 5)$.
Color the segments connecting any two points in each subset $Y_{i}(1 \leqslant i \leqslant 5)$ as shown in Figure 1, and color the segments connecting different subsets $Y_{i}$ and $Y_{j}(i \neq j)$ as shown in Figure 2, where $a, b, c, d$ represent 4 different colors. This way, the colored $G_{83}^{*}$ clearly does not contain a triangle with all three sides of the same color.
For $G_{82}^{*}$, we can use the same coloring method as for $G_{83}^{*}$. As for $G_{i}^{*}(1 \leqslant i \leqslant 81)$, we can first add 1 point and connect it to the original 24 points with a segment each, then color it according to the method for $G_{83}^{*}$, and finally remove the added point and the segments connected to it. This way, the colored $G_{i}^{*}(1 \leqslant i \leqslant 81)$ also does not contain a triangle with all three sides of the same color.
In summary, the conclusion holds.
|
168544
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. After 20 figure skaters have performed, 9 judges respectively assign them rankings from 1 to 20. It is known that the difference between any rankings given to each athlete does not exceed 3. If the sum of the rankings received by each athlete is arranged in an increasing sequence, $c_{1} \leqslant c_{2} \leqslant$ $\cdots \leqslant c_{20}$, find the maximum value of $c_{1}$.
|
(Tip: Let the set of athletes who have won first place be $A$. If $|A|=1$, then $c_{1}=9$; If $|A|=2$, then one of them wins no less than 5 first places, by the problem's condition we know $c_{1} \leqslant 5 \times 1+4 \times 4=21$; If $|A|=3$, similarly, we have $c_{1}+c_{2}+c_{3} \leqslant 1 \times 9+3 \times 9+4 \times 9=$ 72, so $c_{1} \leqslant 24$; If $|A|=4$, we have $c_{1}+c_{2}+c_{3}+c_{4} \leqslant 1 \times 9$ $+2 \times 9+3 \times 9+4 \times 9=90$, so $c_{1} \leqslant 22 ;|A| \geqslant 5$ is impossible. Therefore, $c_{1} \leqslant 24$.)
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In space, there are 10 points, all connected pairwise, and these line segments are colored with red and blue. Among them, the line segments connected from point $A$ are all red. Among the triangles formed with these 10 points as vertices, how many triangles with three sides of the same color are there at least?
In space, there are 10 points, all connected pairwise, and these line segments are colored with red and blue. Among them, the line segments connected from point $A$ are all red. Among the triangles formed with these 10 points as vertices, how many triangles with three sides of the same color are there at least?
|
(Tip: Let the number of same-colored triangles be $x$, then the number of differently colored triangles is $\mathrm{C}_{10}^{3}-x$. Let the total number of same-colored angles be $y$, then $y=3 x+\left(\mathrm{C}_{10}^{3}-\right.$ $x)=120+2 x$. When $y$ reaches its minimum value, $x$ also reaches its minimum value. The total number of same-colored angles is at least $C_{9}^{2}+9\left(C_{4}^{2}+C_{5}^{2}\right)=180$. On the other hand, let the 10 points be $A, B, C_{1}, C_{2}, C_{3}, C_{4}, D_{1}, D_{2}, D_{3}, D_{4}$. Connect $C_{i} C_{j}, D_{i} D_{j}, B D_{j}$ with red lines, and $C_{i} D_{j}, B C_{i}$ with blue lines, thus, except for point $A$, the lines connected from the other points are all 5 red 4 blue or 4 red 5 blue. In summary, the number of same-colored angles can reach 180. Therefore, the number of same-colored triangles is at least 30.)
|
30
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In a $13 \times 13$ square grid, select the centers of $k$ small squares such that no four of these points form the vertices of a rectangle (with sides parallel to those of the original square). Find the maximum value of $k$ that satisfies this condition.
|
(Let the $i$-th column have $x_{i}$ points $(i=1,2, \cdots, 13)$, then $\sum_{i=1}^{13} x_{i}=k$, the $x_{i}$ points in the $i$-th column form $\mathrm{C}_{x_{i}}^{2}$ different point pairs (if $x_{1}<2$, then $\mathrm{C}_{x_{1}}^{2}=0$). Add a column to the side of the $13 \times 13$ square, and each point pair projects onto this column. Since any 4 different points are not the vertices of a rectangle, the projections of different point pairs on the newly drawn column are different. There are $\mathrm{C}_{13}^{2}$ different point pairs on the newly drawn column, thus, $\sum_{i=1}^{13} C_{x_{i}}^{2} \leqslant C_{13}^{2}$, hence $\sum_{i=1}^{13} x_{i}^{2} \leqslant 156+k$. Since $\sum_{i=1}^{13} x_{i}^{2} \geqslant \frac{1}{13}\left(\sum_{i=1}^{13} x_{i}\right)^{2}=\frac{k^{2}}{13}$, then $k^{2} \leqslant 13 \times 156+13 k$, which gives $k \leqslant 52$. When $k=52$, a graph that meets the conditions can be constructed.)
|
52
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A test paper has 4 multiple-choice questions, each with three options (A), (B), (C). Several students take the exam, and after grading, it is found that: any 3 students have 1 question where their answers are all different. How many students can take the exam at most?
|
(提示: If 10 people take the exam, then for the 1st question, at least 7 people choose two options: for the 2nd question, at least 5 people among these 7 choose two options; for the 3rd question, at least 4 people among these 5 choose two options; for the 4th question, at least 3 people among these 4 choose two options. Therefore, these 3 people choose only two options for each question, which contradicts the problem statement. Thus, the maximum number of students taking the exam is 9. Additionally, it is easy to construct a scenario where 9 people take the exam.)
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $a_{1}, a_{2}, \cdots, a_{6} ; b_{1}, b_{2}, \cdots, b_{6} ; c_{1}$, $c_{2}, \cdots, c_{6}$ all be permutations of $1,2, \cdots, 6$. Find the minimum value of $\sum_{i=1}^{6} a_{i} b_{i} c_{i}$.
(Xiong Bin)
|
Let $S=\sum_{i=1}^{6} a_{i} b_{i} c_{i}$. By the AM-GM inequality, we have
$$
\begin{array}{l}
S \geqslant 6 \sqrt[6]{\prod_{i=1}^{6} a_{i} b_{i} c_{i}}=6 \sqrt[6]{(6!)^{3}} \\
=6 \sqrt{6!}=72 \sqrt{5}>160 .
\end{array}
$$
Next, we prove that $S>161$.
Since the geometric mean of the 6 numbers $a_{1} b_{1} c_{1}, a_{2} b_{2} c_{2}, \cdots, a_{6} b_{6} c_{6}$ is $12 \sqrt{5}$, and $26161,
\end{aligned}
$$
we have $S \geqslant 162$.
Furthermore, when $a_{1}, a_{2}, \cdots, a_{6} ; b_{1}, b_{2}, \cdots, b_{6} ; c_{1}, c_{2}, \cdots, c_{6}$ are
$$
1,2,3,4,5,6 ; \quad 5,4,3,6,1,2 ; \quad 5,4,3,1,6,2
$$
respectively, we have
$$
\begin{aligned}
S= & 1 \times 5 \times 5+2 \times 4 \times 4+3 \times 3 \times 3+4 \times 6 \times 1+ \\
& 5 \times 1 \times 6+6 \times 2 \times 2 . \\
= & 162 .
\end{aligned}
$$
Therefore, the minimum value of $S$ is 162.
|
162
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 1, there is a rectangular piece of paper $A B C D, A B=8$, $A D=6$. The paper is folded so that the edge $A D$ lies on the edge $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$ to the right, and the intersection of $A E$ and $B C$ is point $F$. The area of $\triangle C E F$ is ( ).
(A) 2
(B) 4
(C) 6
(D) 8
|
- .1.A.
From the folding process, we know $D E=A D=6, \angle D A E=\angle C E F=$ $45^{\circ}$. Therefore, $\triangle C E F$ is an isosceles right triangle, and $E C=8-6$ $=2$. Hence, $S_{\triangle C E F}=2$.
|
2
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given
$$
A=48 \times\left(\frac{1}{3^{2}-4}+\frac{1}{4^{2}-4}+\cdots+\frac{1}{100^{2}-4}\right) \text {. }
$$
The positive integer closest to $A$ is ( ).
(A) 18
(B) 20
(C) 24
(D) 25
|
4.D.
For positive integers $n(n \geqslant 3)$, we have $\frac{1}{n^{2}-4}=\frac{1}{4}\left(\frac{1}{n-2}-\frac{1}{n+2}\right)$.
Then $A=48 \times \frac{1}{4}\left[\left(1+\frac{1}{2}+\cdots+\frac{1}{8}\right)-\left(\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{R C}\right)\right]$
$$
\begin{array}{l}
=12\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{99}-\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\right) \\
=25-12\left(\frac{1}{99}+\frac{1}{100}+\frac{1}{101}+\frac{1}{102}\right) .
\end{array}
$$
Since $12\left(\frac{1}{99}+\frac{1}{100}+\frac{1}{101}+\frac{1}{102}\right)<12 \times \frac{4}{99}<\frac{1}{2}$, the positive integer closest to $A$ is 25.
|
25
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $a, b$ be positive integers, and satisfy $56 \leqslant a+b \leqslant 59, 0.9<\frac{a}{b}<0.91$. Then $b^{2}-a^{2}$ equals ( ).
(A) 171
(B) 177
(C) 180
(D) 182
|
5.B.
From the given, we have
$$
0.9 b+b56 \text {, }
$$
then $29<b<32$. Therefore, $b=30,31$.
When $b=30$, from $0.9 b<a<0.91 b$, we get $27<a<28$, there is no such positive integer $a$;
When $b=31$, from $0.9 b<a<0.91 b$, we get $27<a<29$, so $a=28$.
Thus, $b^{2}-a^{2}=177$.
|
177
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 As shown in Figure 10, in $\triangle ABC$, $AB=AC$,
$\angle BAC=120^{\circ}$,
$\triangle ADE$ is an equilateral
triangle, point $D$ is on
side $BC$. It is known that
$BD: DC=2: 3$. When the area of $\triangle ABC$ is $50 \mathrm{~cm}^{2}$, find the area of $\triangle ADE$.
(7th Japan Arithmetic Olympiad (Final))
|
Analysis: Directly solving the problem is difficult. By rotating $\triangle A B C$ counterclockwise around point $A$ by $120^{\circ}$ and $240^{\circ}$ to form a regular $\triangle M B C$ (as shown in Figure 11), the regular $\triangle A D E$ becomes regular $\triangle A D_{1} E_{1}$ and regular $\triangle A D_{2} E_{2}$. It is easy to see that the hexagon $D E D_{1} E_{1} D_{2} E_{2}$ is a regular hexagon, and $\triangle D D_{1} D_{2}$ is a regular triangle, whose area is 3 times the area of $\triangle A D E$. Therefore, by finding the area of $\triangle D D_{1} D_{2}$ from the area of the regular $\triangle M B C$ which is 150, the problem is solved.
Solution: Note that $B D: D C=C D_{1}: D_{1} M=M D_{2}: D_{2} B=2: 3$. Connecting $D M$, we have
$S_{\triangle M B D}=\frac{2}{5} S_{\triangle M B C}=60\left(\mathrm{~cm}^{2}\right)$.
And $S_{\triangle D_{2} B D}=\frac{3}{5} S_{\triangle M B D}=36\left(\mathrm{~cm}^{2}\right)$.
Similarly, $S_{\triangle W D_{1} D_{2}}=S_{\triangle \Delta D_{1}}=36\left(\mathrm{~cm}^{2}\right)$.
Therefore, $S_{\triangle D D_{1} D_{2}}=150-3 \times 36=42\left(\mathrm{~cm}^{2}\right)$.
Thus, $S_{\triangle A D E}=\frac{1}{3} S_{\triangle D D_{1} D_{2}}=14\left(\mathrm{~cm}^{2}\right)$.
|
14
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In the Cartesian coordinate system, the parabola
$$
y=x^{2}+m x-\frac{3}{4} m^{2}(m>0)
$$
intersects the $x$-axis at points $A$ and $B$. If the distances from points $A$ and $B$ to the origin are $O A$ and $O B$, respectively, and satisfy $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, then the value of $m$ is $\qquad$.
|
7. 2 .
Let the roots of the equation $x^{2}+m x-\frac{3}{4} m^{2}=0$ be $x_{1}$ and $x_{2}$, and $x_{1}<0<x_{2}$.
From $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, we know $O A>O B$.
Also, since $m>0$, the axis of symmetry of the parabola is to the left of the $y$-axis, thus,
$$
O A=\left|x_{1}\right|=-x_{1}, O B=x_{2} .
$$
Therefore, $\frac{1}{x_{2}}+\frac{1}{x_{1}}=\frac{2}{3}$,
$$
\text { then } \frac{x_{1}+x_{2}}{x_{1} x_{2}}=\frac{2}{3}=\frac{-m}{-\frac{3}{4} m^{2}} \text {. }
$$
Solving for $m$ yields $m=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given $x_{1}, x_{2}, \cdots, x_{40}$ are all positive integers, and $x_{1}+$ $x_{2}+\cdots+x_{40}=58$. If the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ is $A$, and the minimum value is $B$, then the value of $A+B$ is $\qquad$
|
10.494.
Since the number of ways to write 58 as the sum of 40 positive integers is finite, the minimum and maximum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ exist.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{40}$.
If $x_{1}>1$, then $x_{1}+x_{2}=\left(x_{1}-1\right)+\left(x_{2}+1\right)$, and
$$
\begin{array}{l}
\left(x_{1}-1\right)^{2}+\left(x_{2}+1\right)^{2} \\
=x_{1}^{2}+x_{2}^{2}+2\left(x_{2}-x_{1}\right)+2>x_{1}^{2}+x_{2}^{2} .
\end{array}
$$
Therefore, when $x_{1}>1$, $x_{1}$ can be gradually adjusted to 1, at which point $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ will increase.
Similarly, $x_{2}, x_{3}, \cdots, x_{39}$ can be gradually adjusted to 1, at which point $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ will increase.
Thus, when $x_{1}, x_{2}, \cdots, x_{39}$ are all 1 and $x_{40}=19$, $x_{1}^{2}+$ $x_{2}^{2}+\cdots+x_{40}^{2}$ reaches its maximum value, i.e.,
$$
A=\underbrace{1^{2}+1^{2}+\cdots+1^{2}}_{39 \text { terms }}+19^{2}=400 \text {. }
$$
If there exist two numbers $x_{i}$ and $x_{j}$ such that
$$
\begin{array}{l}
x_{j}-x_{i} \geqslant 2 \quad(1 \leqslant i<j \leqslant 40), \text { then } \\
\left(x_{i}+1\right)^{2}+\left(x_{j}-1\right)^{2} \\
=x_{i}^{2}+x_{j}^{2}-2\left(x_{j}-x_{i}-1\right)<x_{i}^{2}+x_{j}^{2} .
\end{array}
$$
This indicates that in $x_{1}, x_{2}, \cdots, x_{39}, x_{40}$, if the difference between any two numbers is greater than 1, then by increasing the smaller number by 1 and decreasing the larger number by 1, $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ will decrease.
Therefore, when $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ reaches its minimum value, the difference between any two numbers in $x_{1}, x_{2}, \cdots, x_{40}$ does not exceed 1.
Thus, when $x_{1}=x_{2}=\cdots=x_{22}=1$ and $x_{23}=x_{24}=\cdots=x_{40}=2$, $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ reaches its minimum value, i.e.,
$$
B=\underbrace{1^{2}+1^{2}+\cdots+1^{2}}_{22 \text { terms }}+\underbrace{2^{2}+2^{2}+\cdots+2^{2}}_{18 \text { terms }}=94 \text {. }
$$
Hence, $A+B=494$.
|
494
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. From the 200 positive integers $1, 2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a < b < c)$, we have $a b \neq c$?
|
14. First, the 193 numbers $1, 14, 15, \cdots, 205$ satisfy the conditions of the problem.
In fact, if $a, b, c (a < b < c)$, then $a b \geqslant 14 \times 15 = 210 > c$.
On the other hand, consider the following 12 sets of numbers:
$(2, 25, 2 \times 25), (3, 24, 3 \times 24), \cdots$,
$(13, 14, 13 \times 14)$.
These 36 numbers are all distinct, and the smallest number is 2, while the largest number is $13 \times 14 = 182 < 205$. Therefore, the 3 numbers in each set cannot all be selected. Thus, if the selected numbers satisfy the conditions of the problem, the number of selected numbers does not exceed 205 $-12=193$ (numbers).
In summary, from $1, 2, \cdots, 205$, the maximum number of numbers that can be selected to satisfy the conditions of the problem is 193.
|
193
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A city's No. 10 Middle School has built a new six-story enclosed student dormitory, with 15 rooms on each floor, and this dormitory has only one entrance and exit safety door. The safety inspection department has tested the normal passage of this safety door: in an emergency, the safety door can function normally for the first $1 \mathrm{~min}$, but due to the continuous crowding of students, the exit efficiency will decrease by $10 \%$ in the second $1 \mathrm{~min}$, and the exit efficiency will be $20 \%$ lower than normal afterwards. To ensure that all students in the dormitory can evacuate safely within $5 \mathrm{~min}$ in an emergency, the school stipulates that each room can accommodate 8 people. Therefore, the safety inspection department measured that the minimum number of people that this safety door can normally pass through per minute should be $\qquad$ people.
|
3.168.
Assuming that under normal circumstances, this safety door can allow an average of $x$ people to pass through per minute. According to the problem, we have
$$
8 \times 15 \times 6 \leqslant x(1+90\%+80\% \times 3) .
$$
Solving this, we get $167.4 \approx \frac{720}{4.3} \leqslant x$.
Therefore, the minimum integer value of $x$ is 168.
|
168
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1(0<b<4)$, the left focus is $F$, and line $l$ intersects the ellipse at points $A$ and $B$, with $|A B|$ $=6$. Then the maximum value of $|F A| \cdot|F B|$ is $\qquad$
|
5. 25 .
Let the right focus of the ellipse be $F^{\prime}$, and connect $A F^{\prime}$ and $B F^{\prime}$. Then,
$$
|A B| \leqslant\left|A F^{\prime}\right|+\left|B F^{\prime}\right| \text {. }
$$
When the chord $A B$ passes through the right focus $F^{\prime}$, the equality holds.
Thus, $|F A|+|F B|+|A B|$
$$
\begin{array}{l}
\leqslant|F A|+\left|F^{\prime} A\right|+|F B|+\left|F^{\prime} B\right| \\
=2 a+2 a=4 a=16 .
\end{array}
$$
Therefore, $|F A|+|F B| \leqslant 10$.
Hence, $|F A| \cdot|F B| \leqslant\left(\frac{|F A|+|F B|}{2}\right)^{2} \leqslant 25$.
For the equalities in (1) and (2) to hold simultaneously, $A B$ should pass through the right focus and be perpendicular to the $x$-axis.
Thus, $(|F A| \cdot|F B|)_{\max }=25$.
|
25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (20 points) Given $a, b, c \in \mathbf{R}_{+}$, and
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1 \text{. }
$$
Prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant 12$.
|
Given, we have
$$
\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{1}{b}}+\frac{1}{1+\frac{1}{c}}=1 \text {. }
$$
Let $x=\frac{1}{1+\frac{1}{a}}, y=\frac{1}{1+\frac{1}{b}}, z=\frac{1}{1+\frac{1}{c}}$. Then
$$
x+y+z=1 \text {. }
$$
From $\frac{1}{a}=\frac{1}{x}-1, \frac{1}{b}=\frac{1}{y}-1, \frac{1}{c}=\frac{1}{z}-1$, we get
$$
\begin{array}{l}
\frac{1}{a b c}=\frac{1-x}{x} \cdot \frac{1-y}{y} \cdot \frac{1-z}{z} \\
=\frac{y+z}{x} \cdot \frac{x+z}{y} \cdot \frac{x+y}{z} \\
\geqslant \frac{2 \sqrt{y z}}{x} \cdot \frac{2 \sqrt{x z}}{y} \cdot \frac{2 \sqrt{x y}}{z}=2^{3} .
\end{array}
$$
Thus, $a b c \leqslant 2^{-3}$.
Therefore, $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant \frac{3}{\sqrt[3]{a^{2} b^{2} c^{2}}} \geqslant \frac{3}{\sqrt[3]{2^{-6}}}=12$.
|
12
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let the positive integer $n$ be a multiple of 75, and have exactly 75 positive divisors (including 1 and itself). Find the minimum value of $n$.
|
Solution: Let the prime factorization of $n$ be
$$
n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}},
$$
where $p_{1}, p_{2}, \cdots, p_{k}$ are the distinct prime factors of $n$, and $r_{1}, r_{2}, \cdots, r_{k}$ are all positive integers.
Thus, the number of distinct positive divisors of $n$ is
$$
\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right).
$$
From the problem, we have
$$
\left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{k}+1\right)=75=3 \times 5^{2}.
$$
Therefore, $n$ can have at most 3 distinct prime factors.
To make $n$ the smallest and a multiple of 75, the prime factors of $n$ should be $2, 3, 5$, and 3 should appear at least once, and 5 should appear at least twice, i.e.,
$$
\begin{array}{l}
n=2^{r_{1}} \times 3^{r_{2}} \times 5^{r_{3}}, \\
\left(r_{1}+1\right)\left(r_{2}+1\right)\left(r_{3}+1\right)=75, \\
r_{1} \geqslant 0, r_{2} \geqslant 1, r_{3} \geqslant 2 .
\end{array}
$$
The tuples $\left(r_{1}, r_{2}, r_{3}\right)$ that satisfy the above conditions are
$$
\begin{array}{l}
(4,4,2),(4,2,4),(2,4,4),(0,4,14), \\
(0,14,4),(0,2,24),(0,24,2) .
\end{array}
$$
After calculation, it is found that when $r_{1}=r_{2}=4, r_{3}=2$, $n$ takes the minimum value, which is $n=2^{4} \times 3^{4} \times 5^{2}=32400$.
|
32400
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given that $x_{1}, x_{2}, \cdots, x_{10}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{10}=99$. Find the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{10}^{2}$.
|
Solution: Since there are only a finite number of ways to write 99 as the sum of 10 positive integers, there must exist a way that maximizes the sum of the squares of these 10 positive integers: and there must also exist a way that minimizes the sum of the squares of these 10 positive integers.
Assume $x_{1}, x_{2}, \cdots, x_{10}$ satisfy $x_{1}+x_{2}+\cdots+$ $x_{10}=99$, and make $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reach its maximum value. Without loss of generality, assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$.
If $x_{1}>1$, because
$$
x_{1}+x_{2}=\left(x_{1}-1\right)+\left(x_{2}+1\right) \text {, }
$$
and $\left(x_{1}-1\right)^{2}+\left(x_{2}+1\right)^{2}$
$=x_{1}^{2}+x_{2}^{2}+2\left(x_{2}-x_{1}\right)+2$
$>x_{1}^{2}+x_{2}^{2}$,
so, when using $x_{1}-1, x_{2}+1, x_{3}, \cdots, x_{10}$ to replace $x_{1}, x_{2}, \cdots, x_{10}$, their sum is still 99. But their sum of squares has increased. This contradicts the fact that $x_{1}, x_{2}, \cdots, x_{10}$ has already made $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reach its maximum value, hence $x_{1}=1$.
Similarly, $x_{2}=x_{3}=\cdots=x_{9}=1$.
Thus, $x_{10}=90$.
Therefore, the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is
$1^{2}+1^{2}+\cdots+1^{2}+90^{2}=8109$.
Next, find the minimum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$.
Assume $x_{1}, x_{2}, \cdots, x_{10}$ satisfy $x_{1}+x_{2}+\cdots+$ $x_{10}=99$, and make $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reach its minimum value. Without loss of generality, assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{10}$.
If there exist $x_{i} 、 x_{j}(1 \leqslant i<j \leqslant 10)$, such that $x_{j}-x_{i} \geqslant 2$, because
$$
\begin{array}{l}
\left(x_{j}-1\right)^{2}+\left(x_{i}+1\right)^{2}-\left(x_{j}^{2}+x_{i}^{2}\right) \\
=2\left(x_{i}-x_{j}\right)+2 \\
\leqslant-2<0,
\end{array}
$$
so, when using $x_{i}+1$ and $x_{j}-1$ to replace $x_{i}$ and $x_{j}$, with other numbers unchanged, their sum is still 99, but their sum of squares has decreased. This contradicts the fact that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ has already reached its minimum value.
Thus, when $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ reaches its minimum value, the absolute difference between any two of $x_{1}, x_{2}, \cdots, x_{10}$ does not exceed 1, hence, these 10 numbers can only be 1 9 and 9 10s. Therefore, the minimum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{10}^{2}$ is
$$
9^{2}+10^{2}+\cdots+10^{2}=981 .
$$
|
8109
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Among the numbers $1,2, \cdots, 2005$, what is the minimum number of numbers that need to be removed so that in the remaining numbers, no number is equal to the product of any two other numbers?
|
(Cue: Strike out the 43 numbers $2,3, \cdots, 44$. Among the remaining numbers (excluding 1), the product of any two numbers is greater than $45^{2}=2025>2005$. Also, for the following 43 triples:
$$
(2,87,2 \times 87),(3,86,3 \times 86), \cdots,(44,45,44 \times 45),
$$
these 129 numbers are all distinct, greater than 1, and less than 2005, so at least 43 numbers must be struck out.)
|
43
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Among the five-digit numbers formed by the 3 digits $1,2,3$, $1,2,3$ each appear at least once. There are $\qquad$ such five-digit numbers.
|
5.150.
Among the five-digit numbers,
if 1 appears only once, there are $\mathrm{C}_{5}^{1}\left(\mathrm{C}_{4}^{1}+\mathrm{C}_{4}^{2}+\mathrm{C}_{4}^{3}\right)=70$;
if 1 appears twice, there are $C_{5}^{2}\left(C_{3}^{1}+C_{3}^{2}\right)=60$;
if 1 appears three times, there are $\mathrm{C}_{5}^{3} \mathrm{C}_{2}^{1}=20$.
Therefore, there are a total of 150 such five-digit numbers.
|
150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (24 points)(1) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals 2005, find the minimum value of $n$ and explain the reason;
(2) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals $2002^{2000}$, find the minimum value of $n$ and explain the reason.
|
4. (1) Since $2005=1728+125+125+27=12^{3}+$ $5^{3}+5^{3}+3^{3}$, therefore, there exists $n=4$, such that $n_{\min } \leqslant 4$.
Also, $10^{3}=1000,11^{3}=1331,12^{3}=1728,13^{3}=$ 2197, and $12^{3}3 \times 8^{3}$, so the edge length of the largest cube can only be $9,10, 11$ or 12. And
$$
\begin{array}{l}
20050,
\end{array}
$$
thus $x \neq 9$;
$$
\begin{array}{l}
2005-10^{3}-10^{3}=5,2005-10^{3}-9^{3}=276, \\
2005-10^{3}-8^{3}=493,2005-10^{3}-7^{3}-7^{3}>0,
\end{array}
$$
thus $x \neq 10$;
$2005-11^{3}-9^{3}0$,
thus $x \neq 11$;
$2005-12^{3}-7^{3}0$,
thus $x \neq 12$.
Therefore, $n=3$ is not possible.
In conclusion, $n_{\min }=4$.
(2) Let the edge lengths of $n$ cubes be $x_{1}, x_{2}, \cdots, x_{n}$, then
$$
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=2002^{2005} \text {. }
$$
By $2002 \equiv 4(\bmod 9), 4^{3} \equiv 1(\bmod 9)$, we get
$$
\begin{array}{l}
2002^{2005} \equiv 4^{2005} \equiv 4^{668 \times 3+1} \\
\equiv\left(4^{3}\right)^{668} \times 4 \equiv 4(\bmod 9)
\end{array}
$$
Also, when $x \in \mathbf{N}_{+}$, $x^{3} \equiv 0, \pm 1(\bmod 9)$, so,
$$
\begin{array}{l}
x_{1}^{3} \not \equiv 4(\bmod 9), x_{1}^{3}+x_{2}^{3} \not \equiv 4(\bmod 9), \\
x_{1}^{3}+x_{2}^{3}+x_{3}^{3} \not \equiv 4(\bmod 9) .
\end{array}
$$
Taking equation (1) modulo 9, and from equations (2) and (3), we know $n \geqslant 4$.
Since $2002=10^{3}+10^{3}+1^{3}+1^{3}$, then
$$
\begin{array}{l}
2002^{2005}=2002^{2004} \times\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
=\left(2002^{668}\right)^{3} \times\left(10^{3}+10^{3}+1^{3}+1^{3}\right) \\
=\left(2002^{668} \times 10\right)^{3}+\left(2002^{668} \times 10\right)^{3}+ \\
\left(2002^{668}\right)^{3}+\left(2002^{668}\right)^{3} .
\end{array}
$$
Therefore, $n=4$ is the required minimum value.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then, $(2+a)(2+b)+b^{2}$
is equal to ( ).
(A) 4
(B) -4
(C) 2
(D) -2
|
1. A.
From the given, we have $(1+a)(1+b)=(1-a)(1-b)$, simplifying to $a+b=0$, which means $a=-b$.
Therefore, $(2+a)(2+b)+b^{2}=(2-b)(2+b)+b^{2}=4$.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Place the natural numbers $1,2, \cdots, 2 n$ randomly on a circle. It is found that among all sets of three consecutive numbers, there are $a$ sets where all three numbers are odd, $b$ sets where exactly two numbers are odd, $c$ sets where only one number is odd, and $d$ sets where all three numbers are even. If $a \neq d$, then the value of $\frac{b-c}{a-d}$ is $\qquad$ .
|
4. -3 .
If each number is counted 3 times, a total of $6 n$ numbers are counted ($3 n$ odd numbers, $3 n$ even numbers), then the system of equations can be set up as
$$
\left\{\begin{array}{l}
3 a+2 b+c=3 n, \\
2 c+b+3 d=3 n .
\end{array}\right.
$$
By eliminating $n$ from equations (1) and (2), we get
$$
\frac{b-c}{a-d}=-3 .
$$
|
-3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given that $a, b, c, d$ are all even numbers, and $0<a<b<c<d, d-a=90$. If $a, b, c$ form an arithmetic sequence, and $b, c, d$ form a geometric sequence, then the value of $a+b+c+d$ is ( ).
(A) 194
(B) 284
(C) 324
(D) 384
|
6. A.
According to the problem, we can set $a, b, c, d$ as $b-m, b, b+m, \frac{(b+m)^{2}}{b}$ (where $m$ is a positive even number, and $m<b$).
From $d-a=90$, we get $\frac{(b+m)^{2}}{b}-(b-m)=90$, which simplifies to
$$
m^{2}+3 b m-90 b=0 \text {. }
$$
Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know that $m$ is a multiple of 6, and $m<30$.
$$
\begin{array}{l}
\text { Let } m=6 k, \text { substituting into equation (1) gives } \\
36 k^{2}+18 b k-90 b=0 .
\end{array}
$$
Solving for $b$ gives $b=\frac{2 k^{2}}{5-k}$.
Substituting $k=1,2,3,4$ one by one, and combining with the given conditions, we find that only $k=4, b=32$ is valid. Thus, $m=24$.
Therefore, $a, b, c, d$ are $8, 32, 56, 98$, respectively, so
$$
a+b+c+d=194 .
$$
|
194
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given a triangle $\triangle A B C$ with side lengths $4,5,6$ respectively, the circumcircle of $\triangle A B C$ is a great circle of sphere $O$, and $P$ is a point on the sphere. If the distances from point $P$ to the three vertices of $\triangle A B C$ are all equal, then the volume of the tetrahedron $P-A B C$ is $\qquad$ .
|
1.10 .
Since $P A=P B=P C$, the projection of point $P$ on the plane $A B C$ is the circumcenter $O$ of $\triangle A B C$, i.e., $P O \perp$ plane $A B C$, and $P O$ equals the radius $R$ of sphere $O$. Therefore,
$$
V_{P-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot P O=\frac{1}{3} \cdot \frac{a b c}{4 R} \cdot R=10 \text {. }
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Three positive integers $a$, $b$, and $c$ satisfy the conditions:
(1) $a<b<c<30$;
(2) For some positive integer base, the logarithms of $a(2 b-a)$ and $c^{2}+$ $60 b-11 a$ are 9 and 11, respectively.
Then the value of $a-2 b+c$ is $\qquad$.
|
3. -4 .
Let the base be a positive integer $n(n \geqslant 2)$, then from condition (2) we get
$$
\left\{\begin{array}{l}
a(2 b-a)=n^{9} . \\
c^{2}+60 b-11 a=n^{11} .
\end{array}\right.
$$
When $n \geqslant 3$, from equation (1) we get
$$
3^{9} \leqslant n^{9}=a(2 b-a)2^{9}$, i.e., $b \geqslant 23$.
$$
Thus, we have
$$
\left\{\begin{array}{l}
a(2 b-a)=2^{9}, \\
c^{2}+60 b-11 a=2^{11} .
\end{array}\right.
$$
Combining that $a, b, c$ are positive integers, and $a<b<c<30, b \geqslant 23$, we get $a=16, b=24, c=28$.
Therefore, $a-2 b+c=-4$.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Now define an operation * :
When $m$ and $n$ are both positive odd numbers or both positive even numbers,
$$
m * n=m+n \text {; }
$$
When one of $m$ and $n$ is a positive odd number and the other is a positive even number,
$$
m * n=m \cdot n \text {. }
$$
Then, the number of elements in the set $M=\{(a, b) \mid a * b=36, m$ 、 $\left.n \in \mathbf{N}_{+}\right\}$ is $\qquad$ .
|
5.41.
Since $36=1+35=3+33=5+31=7+29=\cdots=$ $33+3=35+1$ (a total of 18) $; 36=2+34=4+32=\cdots=32$ $+4=34+2$ (a total of 17) ; $36=1 \times 36=3 \times 12=4 \times 9=9$ $\times 4=12 \times 3=36 \times 1$ (a total of 6), therefore, the number of elements in set $M$ is $18+17+6=41$.
|
41
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $\alpha, \beta, \gamma \in \mathbf{R}$,
$$
\begin{aligned}
u= & \sin (\alpha-\beta)+\sin (\beta-\gamma)+ \\
& \sin (\gamma-\alpha) .
\end{aligned}
$$
Then $u_{\text {max }}+u_{\text {min }}=$
|
6.0.
$$
\begin{aligned}
u & =\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \gamma+\sin \gamma \cdot \cos \alpha- \\
& =\left|\begin{array}{lll}
\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \gamma-\cos \gamma \cdot \sin \alpha \\
\sin \alpha & \cos \alpha & 1 \\
\sin \beta & \cos \beta & 1 \\
\sin \gamma & \cos \gamma & 1
\end{array}\right| .
\end{aligned}
$$
Let \( A(\sin \alpha, \cos \alpha) \), \( B(\sin \beta, \cos \beta) \), and \( C(\sin \gamma, \cos \gamma) \). Then \( |u| = 2 S_{\triangle ABC} \).
Clearly, \( A \), \( B \), and \( C \) are any three points on the unit circle \( x^2 + y^2 = 1 \). Since the maximum area of \( \triangle ABC \) is \( \frac{3 \sqrt{3}}{4} \) when and only when \( \triangle ABC \) is an equilateral triangle, we have \( |u| \leq \frac{3 \sqrt{3}}{2} \).
Thus, \( u_{\text{max}} = \frac{3 \sqrt{3}}{2} \) and \( u_{\text{min}} = -\frac{3 \sqrt{3}}{2} \). Therefore, \( u_{\text{max}} + u_{\text{min}} = 0 \).
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\} \text {, }
$$
where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedra. How many triangles can graph $G$ have at most?
|
Three, first prove the following conclusion:
In a space graph with $n$ points, if there are no triangles, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$.
Proof: Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the most, say there are $k$ edges: $A_{1} A_{n}, A_{1} A_{n-1}$, $\cdots, A_{1} A_{n-k+1}$. According to the condition, there are no triangles, so there are no edges between the points $A_{n}$, $A_{n-1}, \cdots, A_{n-k+1}$. Therefore, each edge in the space graph has at least one endpoint among the points $A_{1}, A_{2}, \cdots, A_{n-k}$, and each $A_{i}(1 \leqslant i \leqslant n-k)$ can have at most $k$ edges. Thus, the total number of edges is less than or equal to $k(n-k) \leqslant\left[\frac{n^{2}}{4}\right]$.
Next, prove that in a 9-point set $M$, if no 4 points are coplanar, and several line segments are connected between these 9 points, if the graph $G$ already has (at least) 28 triangles, then there is at least one tetrahedron.
Use proof by contradiction.
Assume there is no tetrahedron in the 9-point set $M=\left\{A_{1}\right.$, $\left.A_{2}, \cdots, A_{9}\right\}$. By the pigeonhole principle, there must be a point that is a vertex of at least $\left[\frac{28 \times 3}{9}\right]+1=10$ triangles. Thus, at least 5 edges are drawn from this point, let this point be $A_{1}$.
(1) If 5 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots$, $A_{1} A_{6}$ are drawn from point $A_{1}$, according to the condition, since there is no tetrahedron, the subgraph formed by the points $A_{2}, A_{3}, \cdots$, $A_{6}$ has no triangles. By the previous conclusion, this subgraph has at most $\left[\frac{5^{2}}{4}\right]=6$ edges. Therefore, the number of triangles with $A_{1}$ as a vertex is at most 6, which is a contradiction.
(2) If 6 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots$, $A_{1} A_{7}$ are drawn from point $A_{1}$, similarly to (1), the number of triangles with $A_{1}$ as a vertex is at most $\left[\frac{6^{2}}{4}\right]=9$, which is a contradiction.
(3) If 7 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots$, $A_{1} A_{8}$ are drawn from point $A_{1}$, since there is no tetrahedron, the subgraph formed by the points $A_{2}, A_{3}, \cdots, A_{8}$ has no triangles. This subgraph has at most $\left[\frac{7^{2}}{4}\right]=12$ edges. Therefore, the number of triangles with $A_{1}$ as a vertex is at most 12, and the triangles not having $A_{1}$ as a vertex must have point $A_{9}$ as a vertex. Similarly, there are at most 12 such triangles, so the total number of triangles is less than or equal to $12 \times 2=24<28$, which is a contradiction.
(4) If 8 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots$, $A_{1} A_{9}$ are drawn from point $A_{1}$, the subgraph formed by the points $A_{2}, A_{3}, \cdots, A_{9}$ has no triangles. By the previous conclusion, this subgraph has at most $\left[\frac{8^{2}}{4}\right]=16$ edges. Therefore, the original graph $G$ has at most 16 triangles, which is a contradiction.
Thus, the number of triangles that meet the requirements is at most 27.
Divide the 9-point set $M$ into three groups $\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}\right.$, $\left.A_{6}\right\},\left\{A_{7}, A_{8}, A_{9}\right\}$, such that no two points in the same group are connected, while any two points in different groups are connected. This way, there are $\mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1}=27$ triangles, of course, with no tetrahedron.
|
27
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 (1) In a $4 \times 4$ grid paper, some small squares are colored red, and then two rows and two columns are crossed out. If no matter how they are crossed out, there is at least one red small square that is not crossed out, how many small squares must be colored at least?
(2) If the “$4 \times 4$” grid paper in (1) is changed to “$n \times n$” $(n \geqslant 5)$ grid paper, with other conditions unchanged, how many small squares must be colored at least?
|
Solution: (1) If the number of colored small squares is less than or equal to 4, then we can appropriately strike out two rows and two columns to remove all the colored small squares.
If the number of colored small squares is 5, then by the pigeonhole principle, there must be at least one row with 2 or more colored small squares. Striking out this row, the remaining number of colored small squares does not exceed 3. Then, striking out one more row and two columns can remove all the colored small squares.
If the number of colored small squares is 6, then there must be at least one row with 3 colored small squares or two rows each with 2 colored small squares. Therefore, striking out two rows can remove at least 4 colored small squares, leaving no more than 2 colored small squares. Then, striking out two columns can remove them all.
Thus, the number of colored small squares must be greater than or equal to 7.
Furthermore, if the squares are colored as shown in Figure 1, then striking out any two rows and two columns cannot remove all the colored small squares.
Therefore, at least 7 small squares must be colored.
(2) If the number of colored small squares is less than or equal to 4, then striking out two rows and two columns can definitely remove all of them.
Furthermore, if the squares are colored as shown in Figure 2, then striking out any two rows and two columns cannot remove all the colored small squares.
Therefore, at least 5 small squares must be colored.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $P(x)$ be the product of the digits in the decimal representation of $x$. Try to find all positive integers $x$ such that
$$
P(x)=x^{2}-10 x-22
$$
holds.
|
Solution: Let $x=\overline{a_{n} a_{n-1} \cdots a_{1}}$.
When $n \geqslant 2$, according to (1) and (2), we have the estimate
$$
x \geqslant 10^{n-1} \cdot a_{n}>9^{n-1} \cdot a_{n} \geqslant P(x) \text {, }
$$
thus $x^{2}-10 x-22<x$.
Solving this, we get $0<x<13$.
Upon verification, $x=12$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Some middle school students in a city participated in a mathematics invitational competition, and this competition had a total of 6 problems. It is known that each problem was solved correctly by exactly 500 students, but for any two students, there is at least one problem that neither of them solved correctly. Question: What is the minimum number of middle school students from the city who participated in this mathematics invitational competition?
|
Solution: First, each student can answer at most 4 questions correctly.
In fact, according to the problem, it is impossible for any student to answer 6 questions correctly. If a student answers 5 questions correctly, by the problem's condition, all other students must have answered the same question incorrectly, which contradicts the condition that exactly 500 students answered each question correctly.
If a student answers 4 questions correctly, let's assume they answered questions 1, 2, 3, and 4 correctly. Then no student can answer both questions 5 and 6 correctly, otherwise it would contradict the problem's condition. Therefore, the number of students who answered question 5 and question 6 correctly are each 500, so the total number of students is at least $500 + 500 + 1 > 1000$.
If each student answers at most 3 questions correctly, since the total number of questions answered correctly by all students is $500 \times 6 = 3000$, the number of students is at least $3000 \div 3 = 1000$.
Below is a construction to show that 1000 people is possible. The number of people who answered the following questions correctly is 100 each:
$$
\begin{array}{l}
(1,2,3),(1,3,4),(1,4,5),(1,5,6), \\
(1,2,6),(2,4,6),(2,3,5),(2,4,5), \\
(3,4,6),(3,5,6) .
\end{array}
$$
In summary, at least 1000 people participated in this mathematics invitation competition.
|
1000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 On a plane, there are 7 points, and some line segments can be connected between them, so that any three points among the 7 points must have two points connected by a line segment. How many line segments are needed at least? Prove your conclusion.
|
Proof: First, to meet the requirements of the problem, the number of line segments to be connected must be greater than or equal to 9.
Below, we discuss in 4 cases.
(1) If among the 7 points, there exists 1 point not connected to any other points, by the problem's condition, the remaining 6 points must each be connected to every other point, in this case, at least $\frac{6 \times 5}{2}=15$ line segments need to be connected.
(2) If among the 7 points, there is 1 point that only connects to 1 line segment, then the remaining 5 points must each be connected to every other point, in this case, there are at least $\frac{5 \times 4}{2}+1=11$ line segments.
(3) If each point connects to at least 2 line segments, and there is 1 point that exactly connects to 2 line segments, let's assume this point is $A$, and the 2 line segments it connects to are $A B$ and $A C$, then the 4 points not connected to point $A$ must each be connected to every other point, requiring $\frac{4 \times 3}{2}=6$ line segments. And point $B$ must connect to at least 2 line segments, so apart from $B A$, there must be at least 1 more, thus, in this case, at least $6+2+1=9$ line segments need to be connected.
(4) If each point connects to at least 3 line segments, then at least $7 \times 3 \div 2>10$ line segments need to be connected.
In summary, the number of line segments in the graph is greater than or equal to 9.
As shown in Figure 3, an example of construction is given, which indicates that 9 line segments are sufficient.
In conclusion, the minimum number of line segments to be connected is 9.
|
9
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (16 points) Let $n$ be a positive integer, and $d_{1}<d_{2}<$ $d_{3}<d_{4}$ be the 4 smallest consecutive positive integer divisors of $n$. If $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$, find the value of $n$.
|
If $n$ is odd, then $d_{1} 、 d_{2} 、 d_{3} 、 d_{4}$ are all odd. Hence $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2} \equiv 1+1+1+1 \equiv 0(\bmod 4)$. Contradiction.
If $4 \mid n$, then $d_{1}=1, d_{2}=2$. By $d_{i}^{2} \equiv 0$ or $1(\bmod 4)$, we have $n \equiv 1+0+d_{3}^{2}+d_{4}^{2} \neq 0(\bmod 4)$. Also a contradiction.
Therefore, $n=2\left(2 n_{1}-1\right), n_{1}$ is some positive integer, and the tuple $\left(d_{1}, d_{2}, d_{3}, d_{4}\right)=(1,2, p, q)$ or $(1,2, p, 2 p)$, where $p, q$ are odd primes.
In the first case, $n=1^{2}+2^{2}+p^{2}+q^{2} \equiv 3(\bmod 4)$. Contradiction.
Thus, it can only be $n=1^{2}+2^{2}+p^{2}+(2 p)^{2}=5\left(1+p^{2}\right)$. Hence $51 n$.
If $d_{3}=3$, then $d_{4}=5$, which will return to the first case, so it can only be $d_{3}=p=5$, then $n=1^{2}+2^{2}+5^{2}+10^{2}=130$.
It is easy to verify that the 4 smallest consecutive positive divisors of 130 are $1,2,5,10$, satisfying the condition.
Therefore, $n=130$.
|
130
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Simplify $\frac{1}{4+\sqrt{59+30 \sqrt{2}}}+\frac{1}{3-\sqrt{66-40 \sqrt{2}}}$, the result is ( ).
(A) irrational number
(B) proper fraction
(C) odd number
(D) even number
|
\begin{array}{l}\text { - 1.D. } \\ \text { Original expression }=\frac{1}{4+\sqrt{(5 \sqrt{2}+3)^{2}}}+\frac{1}{3-\sqrt{(5 \sqrt{2}-4)^{2}}} \\ =\frac{1}{7+5 \sqrt{2}}+\frac{1}{7-5 \sqrt{2}}=-14 .\end{array}
|
-14
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Among the natural numbers not exceeding 100, add up all the numbers that are multiples of 3 or 5, the sum is $\qquad$ .
|
1.2418.
The sum of multiples of 3 from $1 \sim 100$ is $S_{3}=3(1+2+\cdots+33)=1683$,
The sum of multiples of 5 from $1 \sim 100$ is $S_{5}=5(1+2+\cdots+20)=1050$, the sum of multiples of 15 from $1 \sim 100$ is $S_{15}=15(1+2+\cdots+6)=315$. Therefore, $S=S_{3}+S_{5}-S_{15}=2418$.
|
2418
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If real numbers $x, y$ satisfy
$$
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \frac{x}{5^{3}+4^{2}}+\frac{y}{5^{3}+6^{3}}=1,
$$
then $x+y=$ $\qquad$ .
|
3.432 .
It is known that $3^{3}$ and $5^{3}$ are the two roots of the equation $\frac{x}{t+4^{3}}+\frac{y}{t+6^{3}}=1$, which means they are the roots of the equation
$$
t^{2}+\left(4^{3}+6^{3}-x-y\right) t+\left(4^{3} \times 6^{3}-4^{3} y-6^{3} x\right)=0
$$
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
3^{3}+5^{3}=-\left(4^{3}+6^{3}-x-y\right) . \\
\text { Therefore, } x+y=3^{3}+4^{3}+5^{3}+6^{3}=432 \text {. }
\end{array}
$$
|
432
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (25 points) $a$, $b$, $c$ are positive integers, and $a^{2}+b^{3}=$ $c^{4}$. Find the minimum value of $c$.
保留了源文本的换行和格式。
|
3. Clearly, $c>1$. From $b^{3}=\left(c^{2}-a\right)\left(c^{2}+a\right)$, if we take $c^{2}-a=b, c^{2}+a=b^{2}$, then at this time
$$
c^{2}=\frac{b(b+1)}{2} \text {. }
$$
Examining $b$ from small to large to make the right side a perfect square, we find that when $b=8$, $c=6$, and thus, $a=28$.
Next, we show that there is no smaller positive integer solution for $c$, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|}
\hline$c$ & $c^{4}$ & Cubes less than $c^{4}$ \\
\hline 2 & 16 & 1,8 \\
\hline 3 & 81 & $1,8,27,64$ \\
\hline 4 & 256 & $1,8,27,64,125,216$ \\
\hline 5 & 625 & $1,8,27,64,125,216,343,512$ \\
\hline
\end{tabular}
In each row of Table 1, $c^{4}-x^{3}$ is never a perfect square, therefore, the smallest value of $c$ is 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and its smallest positive period is 2. Then the value of $f(-1)$ is $\qquad$ .
|
1.0.
Since $f(x+2)=f(x)$, let $x=-1$, then we have $f(-1+2)=f(-1)$,
which means $f(1)=f(-1)=-f(1)$.
Therefore, $f(-1)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the set $M=\left\{x \mid x=2^{n}-2^{k}\right.$, where $n, k \in$ $\mathbf{N}$, and $n>k\}, P=\{x \mid 1912 \leqslant x \leqslant 2004$, and $x$ $\in \mathbf{N}\}$. Then, the sum of all elements in the set $M \cap P$ is $\qquad$.
|
2.3904.
Since $2^{10}=1024,2^{11}=2048$, all elements in the set $M \cap P$ must be of the form $2^{11}-2^{k}$ (where $k \in \mathbf{N}$).
Also, $1912 \leqslant 2^{11}-2^{k} \leqslant 2004$, so $44 \leqslant 2^{k} \leqslant 136$.
The only values of $k$ that satisfy this condition are 6 and 7, so the set $M \cap$ $P$ contains only two elements, which are
$$
2^{11}-2^{6}=1984 \text { and } 2^{11}-2^{7}=1920 \text {. }
$$
|
3904
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=99^{\frac{1}{99}}, a_{n}=$ $\left(a_{n-1}\right)^{a_{1}}$. When $a_{n}$ is an integer, the smallest positive integer $n$ is $\qquad$
|
4. 100 .
Obviously, for any positive integer $n$, we have $a_{n}>0$.
From $a_{n}=\left(a_{n-1}\right)^{a_{1}}$, taking the common logarithm on both sides, we get
$\lg a_{n}=a_{1} \lg a_{n-1}$.
Therefore, $\left\{\lg a_{n}\right\}$ is a geometric sequence with the first term $\lg a_{1}$ and the common ratio $a_{1}$. So, $\lg a_{n}=a_{1}^{n-1} \lg a_{1}$, which gives
$$
a_{n}=a_{1}^{a_{1}^{n-1}}=\left(99^{\frac{1}{9}}\right)^{\frac{1}{99_{9}^{9}}(n-1)}
$$
To make $a_{n}$ an integer, it is necessary that $n-1=99 k$ (where $k$ is a positive integer).
Therefore, the smallest positive integer $n$ is 100.
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For $i=1,2, \cdots, n$, we have $\left|x_{i}\right|<1$, and
$$
\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|=19+\left|x_{1}+x_{2}+\cdots+x_{n}\right| \text {. }
$$
Find the minimum value of the positive integer $n$.
|
Given $19=\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|-\mid x_{1}+x_{2}+\cdots+x_{n}|\leqslant| x_{1}|+| x_{2}|+\cdots+| x_{n} \mid<n$, it follows that $n \geqslant 20$. When $n=20$, take
$$
x_{i}=\left\{\begin{array}{cc}
0.95, & \text { when } i \text { is odd: } \\
-0.95, & \text { when } i \text { is even. }
\end{array}\right.
$$
They satisfy the given conditions, hence the minimum value of $n$ is 20.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The largest natural number that cannot be expressed as the sum of four distinct composite numbers is $\qquad$ .
翻译完成，保留了原文的格式和换行。
|
2. 26 .
The sum of the four smallest composite numbers is $4+6+8+9=27$. It is easy to prove that any natural number greater than 27 can be expressed as the sum of four different composite numbers.
|
26
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find $\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}$.
Analysis: Considering the use of identity (III), we can obtain
$$
\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \text {. }
$$
Let $l=2 n-k-1$, and combining with equation (III) again yields $\frac{l+1}{\mathrm{C}_{2 n}^{l}}$, creating a cyclic phenomenon, which leads to the solution of the problem.
|
Let the original expression be $y_{n}$, from the analysis we have
$$
y_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{C_{2 n}^{2 n-k-1}} \text {. }
$$
By making a change of the summation index, set $l=2 n-$ $k-1$, when $k$ varies from 0 to $2 n-1$, $l$ varies from $2 n-1$ to 0. Thus,
$$
\begin{array}{l}
y_{n}=\sum_{l=0}^{2 n-1}(-1)^{2 n-l} \frac{l+1}{\mathbf{C}_{2 n}^{l}} \\
=\sum_{l=0}^{2 n-1}(-1)^{l} \frac{l+1}{\mathbf{C}_{2 n}^{l}}=-y_{n} .
\end{array}
$$
Therefore, $y_{n}=0$, that is,
$$
\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}=0 .
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given that $a, b, x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$
Find the value of $a x^{5}+b y^{5}$.
|
Given the conditions, construct the recurrence relation:
$$
\begin{array}{l}
s_{n}=a x^{n}+b y^{n}, n=1,2, \cdots, \\
\text { then } s_{n+2}=a x^{n+2}+b y^{n+2} \\
=\left(a x^{n+1}+b y^{n+1}\right)(x+y)- \\
\quad x y\left(a x^{n}+b y^{n}\right) .
\end{array}
$$
Therefore, \( s_{n+2}=(x+y) s_{n+1}-x y s_{n} \). Substituting the known equations, we get the system of equations:
$$
\left\{\begin{array}{l}
16=7(x+y)-3 x y, \\
42=16(x+y)-7 x y .
\end{array}\right.
$$
Solving these, we get \( x+y=-14 \) and \( x y=-38 \).
Thus, \( s_{5}=(x+y) s_{4}-x y s_{3} \)
$$
=-14 \times 42+38 \times 16=20 \text {, }
$$
which means \( a x^{5}+b y^{5}=20 \).
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 1, in rectangle $A B C D$, $A B=2, B C=$ $\sqrt{3}, E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Line segments $D E$, $B F$, and $E F$ intersect diagonal $A C$ at points $M$, $N$, and $P$ respectively. The number of right triangles formed by the line segments in Figure 1 is ( ).
(A) 8
(B) 12
(C) 16
(D) 20
|
4.B.
Let the notation $[A]$ represent the number of right-angled triangles with $A$ as the right-angle vertex, then $[A]=2$,
Similarly, $[B]=1,[C]=2,[D]=1,[E]=3$,
$$
[F]=3,[M]=[N]=[P]=0 \text {. }
$$
Therefore, the number of right-angled triangles is
$$
2+1+2+1+3+3=12 \text {. }
$$
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Fill in $n$ distinct numbers on a circle so that for every three consecutive numbers, the middle number is equal to the product of the two numbers on its sides. Then $n=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2.6.
According to the problem, $n \geqslant 3$. Take any two adjacent numbers $a, b(a \neq b)$, and by the problem's condition, the numbers on the circle can be written in sequence as
$$
a, b, \frac{b}{a}, \frac{1}{a}, \frac{1}{b}, \frac{a}{b}, a, b, \cdots
$$
Starting from the 7th number, the sequence of numbers has already entered a cycle, so $n \leqslant 6$.
Next, we show that $n$ cannot be $3, 4, 5$.
If $n=3$, then there are only 3 numbers on the circle, so in the above sequence, we should have $a=\frac{1}{a}, b=\frac{1}{b}$, thus,
$$
a= \pm 1, b= \pm 1, \frac{b}{a}= \pm 1 \text {. }
$$
Therefore, $a, b, \frac{b}{a}$ must have two that are equal, which is a contradiction.
If $n=4$, then there are only 4 numbers on the circle (although the sequence of numbers can be formally written, the subsequent numbers actually repeat the first four numbers), so $a=\frac{1}{b}$, i.e., $b=\frac{1}{a}$. Therefore, the second and fourth numbers among the first four numbers are already equal, which is a contradiction.
If $n=5$, then there are 5 numbers on the circle, so $a=\frac{a}{b}$, i.e., $b=1$. Therefore, the $b$ and $\frac{1}{b}$ among the first five numbers are both equal to 1, which is a contradiction.
Therefore, only $n=6$.
On the other hand, when $n=6$, we can specifically give 6 numbers:
$$
2,3, \frac{3}{2}, \frac{1}{2}, \frac{1}{3}, \frac{2}{3} \text {. }
$$
|
6
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. A certain temple has three bronze bells, A, B, and C. Bell A rings once every $4 \mathrm{~s}$, Bell B rings once every $5 \mathrm{~s}$, and Bell C rings once every $6 \mathrm{~s}$. At the arrival of the New Year, the three bells ring simultaneously and stop ringing at the same time. A person hears a total of 365 bell rings. If during this period, Bell A, Bell B, and Bell C ring $x$ times, $y$ times, and $z$ times respectively, then $x+y+z=$ $\qquad$
|
3.484 .
Let the duration of the bell ringing be $60 k \mathrm{~s}$, then during this period, the number of times the first clock rings is $x=\frac{60 k}{4}+1=15 k+1$; the number of times the second clock rings is $y=\frac{60 k}{5}+1=12 k+1$; the number of times the third clock rings is $z=\frac{60 k}{6}+1=10 k+1$; the number of times the first and second clocks ring simultaneously is $\frac{60 k}{20}+1=3 k+1$; the number of times the first and third clocks ring simultaneously is $\frac{60 k}{12}+1=5 k+1$; the number of times the second and third clocks ring simultaneously is $\frac{60 k}{30}+1=2 k+1$; the number of times the first, second, and third clocks ring simultaneously is $\frac{60 k}{60}+1=k+1$. According to the problem, we have
$$
\begin{aligned}
365= & (15 k+1)+(12 k+1)+(10 k+1)- \\
& (3 k+1)-(5 k+1)-(2 k+1)+(k+1) \\
= & 28 k+1 .
\end{aligned}
$$
Therefore, $k=13$.
Substituting $k$ into the expressions for $x, y, z$ we get
$$
\begin{array}{l}
x=15 k+1=196, y=12 k+1=157, \\
z=10 k+1=131 .
\end{array}
$$
Thus, $x+y+z=484$.
|
484
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. From $1,2, \cdots, 10$ choose 3 different numbers $a, b, c$ as the coefficients of the quadratic equation $a x^{2}+b x=c$. Then the number of equations with different solutions is $\qquad$
|
4.654.
When considering $a1$, for each factor $k$ of $d$ greater than 1, if $\frac{c}{k} \geqslant 3$, then the array $\left(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}\right)$ has already been considered before and should be removed. The number of remaining arrays is denoted as $f(c)$. Thus, we have
$$
\begin{array}{l}
f(3)=1, f(4)=3, f(5)=6, f(6)=10-f(3)=9, \\
f(7)=15, f(8)=21-f(4)=18, \\
f(9)=28-f(3)=27, f(10)=36-f(5)=30 .
\end{array}
$$
Therefore, $f(3)+f(4)+\cdots+f(10)=109$.
For each group $(a, b, c)$, by swapping the positions of $a$, $b$, and $c$, 6 different solutions of the equation can be formed.
Thus, the total number of equations is $109 \times 6=654$.
|
654
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (25 points) Given that $a$ and $b$ are coprime positive integers, satisfying $a+b=2005$. Let $[x]$ denote the integer part of the number $x$, and let
$$
\begin{array}{l}
A=\left[\frac{2005 \times 1}{a}\right]+\left[\frac{2005 \times 2}{a}\right]+\cdots+\left[\frac{2005 \times a}{a}\right], \\
B=\left[\frac{2005 \times 1}{b}\right]+\left[\frac{2005 \times 2}{b}\right]+\cdots+\left[\frac{2005 \times b}{b}\right] .
\end{array}
$$
Try to find the value of $A+B$.
|
3. It is evident that the last terms of the two sums $A$ and $B$ are both 2005. Consider the remaining terms.
Since $(a, b)=1, a+b=2005$, it follows that
$(a, 2005)=1, (b, 2005)=1$.
Therefore, for $1 \leqslant m \leqslant a-1, 1 \leqslant n \leqslant b-1, \frac{2005 m}{a}$ and $\frac{2005 n}{b}$ are not integers, and they are not equal.
In fact, if $\frac{2005 m}{a}=\frac{2005 n}{b}$, then $\frac{a}{b}=\frac{m}{n}$, and $1 \leqslant m \leqslant a-1, 1 \leqslant n \leqslant b-1$, which contradicts the fact that $a$ and $b$ are coprime.
Now, consider a set of intervals of length 1 on the number line:
$(1,2),(2,3), \cdots,(2003,2004)$, each of which is called a "unit interval".
Since $\frac{2005}{a}>1$, then $\frac{2005(a-1)}{a}k a, 2005 n>k b$
and $2005 mk$.
By adding the two equations in (2), we get $m+n<k+1$.
Therefore, $k<m+n<k+1$, which contradicts the fact that the interval $(k, k+1)$ does not contain any integer. Thus,
$$
\begin{array}{r}
\frac{2005 \times 1}{a}, \frac{2005 \times 2}{a}, \cdots, \frac{2005 \times(a-1)}{a} \\
\text { and } \quad \frac{2005 \times 1}{b}, \frac{2005 \times 2}{b}, \cdots, \frac{2005 \times(b-1)}{b}
\end{array}
$$
contain a total of $a+b-2(=2003)$ numbers, each falling into different intervals $(1,2)$, $(2,3), \cdots,(2003,2004)$. Each interval contains exactly one of these numbers. When a number $x$ belongs to the interval $(k, k+1)$, we have $[x]=k$. Therefore,
$$
\begin{array}{l}
A+B=(1+2+3+\cdots+2003)+2005+2005 \\
=\frac{2003 \times 2004}{2}+4010=2011016 .
\end{array}
$$
(Provided by Ping Sheng)
|
2011016
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. (16 points) A warehouse has 50 containers of the same specification, which are to be delivered to the dock by a transportation company. The transportation company has three types of trucks that can carry 1, 2, and 3 containers per trip, respectively. The charges for each trip are 120 yuan, 160 yuan, and 180 yuan, respectively. Now, it is required to arrange 20 trucks to transport all the containers in one trip. How many trucks of each type are needed, and how many ways can this be arranged? Which arrangement requires the least amount of freight cost? What is the minimum freight cost?
|
2. Let the number of trucks needed to transport 1, 2, and 3 containers be $x$, $y$, and $z$ respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=20, \\
x+2 y+3 z=50 .
\end{array}\right.
$$
(1) $\times 3$ - (2) gives $2 x+y=10$.
Then $\left\{\begin{array}{l}y=10-2 x, \\ z=10+x .\end{array}\right.$
Since $y \geqslant 0$, we have $0 \leqslant x \leqslant 5$.
Thus, $x$ can only take the values $0, 1, 2, 3, 4, 5$. Therefore, there are
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ x = 0 , } \\
{ y = 1 0 , } \\
{ z = 1 0 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x = 1 , } \\
{ y = 8 , } \\
{ z = 1 1 ; }
\end{array} \quad \left\{\begin{array}{l}
x=2, \\
y=6, \\
z=12 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l }
{ x = 3 , } \\
{ y = 4 , } \\
{ z = 1 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x = 4 , } \\
{ y = 2 , } \\
{ z = 1 4 ; }
\end{array} \quad \left\{\begin{array}{l}
x=5, \\
y=0, \\
z=15
\end{array}\right.\right.\right. \\
\end{array}
$$
these six arrangements.
Let the total transportation cost be $w$ yuan. Then
$$
\begin{array}{l}
w=120 x+160 y+180 z \\
=120 x+160(10-2 x)+180(10+x) \\
=3400-20 x .
\end{array}
$$
When $x=5$, the total transportation cost is the lowest, and the lowest cost is
$$
w=3400-20 \times 5=3300 \text { yuan. }
$$
|
3300
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.6. How many ways are there to divide the set
$$
\left\{2^{0}, 2^{1}, 2^{2}, \cdots, 2^{2000}\right\}
$$
into two non-empty disjoint subsets $A$ and $B$, such that the equation $x^{2}-S(A) x+S(B)=0$ has integer roots? Here, $S(M)$ denotes the sum of all elements in the set $M$.
|
9.6.1 003 types.
Let $x_{1} \leqslant x_{2}$ be the roots of the equation, then
$$
x_{1}+x_{2}=S(A), x_{1} x_{2}=S(B),
$$
and $x_{1}, x_{2} \in \mathbf{Z}_{+}$. Therefore,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right)=S(B)+S(A)+1 \\
=1+2+4+\cdots+2^{200}+1=2^{2006},
\end{array}
$$
which means $x_{1}+1=2^{k}, x_{2}+1=2^{2000-k}, k=1,2, \cdots, 1003$.
Conversely, when $x_{1}+1=2^{k}, x_{2}+1=2^{2000-k}$, they are the roots of the equation $x^{2}-p x+q=0$, where $p=2^{k}+2^{2000-k}-2$, $q=2^{2006}-1-p$. The number $p$ has a unique binary representation, in which the highest power of 2 does not exceed 2005. Since $p+q=2^{2006}-1$, in the binary representation of $p$, where there is a 1, there is a 0 in the binary representation of $q$, and vice versa. Therefore, for each $k(1 \leqslant k \leqslant 1003)$, there exists a unique partition $(A, B)$ such that the roots of the equation are exactly $x_{1}, x_{2}$.
|
1003
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.1. Try to find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers.
|
10.1.11.
From the problem, we have
$$
\begin{array}{l}
1=\frac{4-2}{4-2}, 3=\frac{8-2}{4-2}, 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2}, \\
7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2}, \\
2=2 \times 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \times 5=\frac{2^{6}-2^{2}}{2^{3}-2} .
\end{array}
$$
Assume $11=\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$.
Without loss of generality, we can assume $a>b, c>d$. Let $m=a-b$, $n=c-d$, $k=b-d$. Then, we have
$$
11\left(2^{n}-1\right)=2^{k}\left(2^{m}-1\right) \text {. }
$$
The left side of the above equation is odd, so $k=0$.
It is easy to see that $n=1$ cannot satisfy the equation. If $m>n>1$, then $2^{n}-1$ and $2^{m}-1$ both have a remainder of 3 when divided by 4, thus, the left side of the equation has a remainder of 1 when divided by 4, while the right side has a remainder of 3. This is a contradiction.
|
11
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.3. On the back of 2005 cards, there are 2005 different real numbers written. Each time you ask a question, you can point to any three cards and ask about the set of 3 numbers written on them. How many times do you need to ask at minimum to definitely find out what numbers are written on the back of each card?
|
10.3.1003 times.
Assume that $N$ questions have been proposed.
Obviously, each card should participate in at least one question; otherwise, it would be impossible to determine the number on that card.
Suppose there are $k$ cards that each participate in exactly one question. Then, in the same question, it is impossible to encounter two such cards. In fact, if two such cards participate in the same question, swapping the numbers written on these two cards would not change the answer received, making it impossible to determine which number is written on each card. Therefore, $k \leqslant N$. The remaining cards each participate in at least two questions. If we sum the number of questions each card participates in, we get:
$$
3 N \geqslant k + 2(2005 - k) = 4010 - k \geqslant 4010 - N.
$$
Thus, $2 N \geqslant 2005$, which means $N \geqslant 1003$.
Next, we will show that the goal can be achieved with 1003 questions.
Remove 1 card and divide the remaining cards into 334 groups, each containing 6 cards. Number the cards in each group from 1 to 6, and ask three questions for each group: $(1,2,3)$, $(3,4,5)$, $(5,6,1)$. This way, in each group, the cards numbered 1, 3, and 5 each appear in two questions, allowing their numbers to be uniquely determined; the numbers on the cards numbered 2, 4, and 6 can also be determined. Thus, through $\frac{2004}{6} \times 3 = 1002$ questions, the numbers on 2004 cards can be determined. The remaining one question is used to determine the number on the card that was set aside (by asking it along with any two other cards).
|
1003
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.5. Place 16 rooks on an $8 \times 8$ chessboard. How many pairs of rooks can attack each other (a pair of rooks can attack each other if they are in the same row or column and there are no other rooks between them)?
|
10.5.16 Correct.
It should be noted that if there are $a$ rooks in a row, then there are $a-1$ pairs of rooks that can attack each other in that row. Therefore, the number of rooks that can attack each other horizontally will not be less than 8 pairs.
Similarly, the number of rooks that can attack each other vertically will not be less than 8 pairs.
There exists an arrangement where exactly 16 pairs of rooks can attack each other, for example, by placing all the rooks on the two main diagonals (top-left to bottom-right, top-right to bottom-left).
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11.1. Let $a_{1}, a_{2}, \cdots, a_{50}, b_{1}, b_{2}, \cdots, b_{50}$ be distinct numbers such that the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{50}\right|
\end{array}
$$
has a finite number of roots. How many roots can there be at most?
|
11.1.49.
$$
\begin{array}{l}
\text { Let } f(x)=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right|- \\
\left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{50}\right| \text {. } \\
\end{array}
$$
Thus, the original equation is $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{100}$ be the elements of the set
$$
\left\{a_{1}, a_{2}, \cdots, a_{50}, b_{1}, b_{2}, \cdots, b_{50}\right\}
$$
arranged in increasing order. In the intervals
$$
\left(-\infty, c_{1}\right],\left[c_{1}, c_{2}\right], \cdots,\left[c_{99}, c_{100}\right],\left[c_{100},+\infty\right)
$$
the function $f(x)$ is linear in each of the 101 intervals, and in the interval $\left(-\infty, c_{1}\right]$, $f(x)=a_{1}+a_{2}+\cdots+a_{50}-b_{1}-b_{2}-\cdots-b_{50}=m$, and in the interval $\left[c_{100},+\infty\right)$, $f(x)=-m$.
Since the number of roots of the equation is finite, $m \neq 0$.
Next, move along the number line from left to right. Initially, the coefficient of $x$ in $f(x)$ is 0. Each time a $c_{i}$ is crossed, the opening method of one of the absolute values in $f(x)$ changes, causing the coefficient of $x$ to change by $\pm 2$ (increase by 2 or decrease by 2). This indicates that the coefficient of $x$ is always even and does not change sign before it becomes 0. Therefore, the coefficient of $x$ in any two adjacent intervals is either both non-negative or both non-positive. Thus, $f(x)$ is either non-increasing or non-decreasing in such a union of intervals. As a result, if $f(x)=0$ has only a finite number of roots, then it has at most one root in each of the intervals $\left[c_{1}, c_{3}\right], \cdots,\left[c_{97}, c_{99}\right]$, $\left[c_{99}, c_{100}\right]$. Additionally, since $f\left(c_{1}\right)$ and $f\left(c_{100}\right)$ have different signs, and $f(x)$ changes sign at each root, $f(x)=0$ has an odd number of roots, and thus, no more than 49 roots.
On the other hand, if
$$
\begin{array}{l}
\left\{a_{1}, a_{2}, \cdots, a_{50}\right\} \\
=\{1,4,5,8,9,12, \cdots, 89,92,93,96,98,99\} \text {, } \\
\left\{b_{1}, b_{2}, \cdots, b_{50}\right\} \\
=\{2,3,6,7,10,11, \cdots, 90,91,94,95,97,101\} \text {, } \\
\text { then } f(1)=-1, f(2)=f(3)=1 \text {, } \\
f(4)=f(5)=-1, f(6)=f(7)=1, \cdots \text {, } \\
f(92)=f(93)=-1, f(94)=f(95)=1 \text {, } \\
f(96)=f(97)=-1, f(98)=f(99)=-3 \text {, } \\
f(101)=1 \text {. } \\
\end{array}
$$
Therefore, the equation $f(x)=0$ has exactly 49 roots.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4.6 Four table tennis players are dressed in 4 different colors for a performance match, with 2 in red, 2 in yellow, 1 in blue, and 1 in black. Each performance selects 3 players to appear, and matches are only played between players with different colored outfits, with the specific rules being:
(1) If the "3-person group" on stage has all different colored outfits, then each pair of players will play 1 match, and players who have already played against each other in different "3-person groups" will still play again if they meet;
(2) If the "3-person group" on stage has 2 players with the same colored outfit, then these 2 players will not play against each other, and only 1 of them will play 1 match against the other player.
According to these rules, when all different "3-person groups" have appeared, a total of $\qquad$ matches have been played.
|
4.44.
Represent the 2 athletes in red as two parallel lines $l_{1}$, $l_{2}$; the 2 athletes in yellow as two other parallel lines $l_{3}$, $l_{4}$; and the athletes in blue and black as intersecting lines $l_{5}$, $l_{6}$, which intersect with $l_{1}$, $l_{2}$, $l_{3}$, $l_{4}$. This results in Figure 6, where there are no "three lines intersecting at a single point".
(1) When the "group of 3" have different colored outfits, according to the rules, this corresponds to three lines intersecting pairwise, and the number of matches is exactly the number of line segments in the figure (Figure 7).
(2) When the "group of 3" has 2 athletes with the same colored outfits, according to the rules, this corresponds to "two parallel lines being intersected by a third line", and the number of matches is exactly the number of line segments in the figure (Figure 8).
Thus, we only need to count the number of line segments in the figure.
First, $l_{1}$, $l_{2}$, $l_{3}$, $l_{4}$ each have 4 intersection points, and each line has 6 line segments, totaling $6 \times 4 = 24$ line segments, corresponding to 24 matches.
Second, $l_{5}$, $l_{6}$ each have 5 intersection points, and each line has 10 line segments, totaling $10 \times 2 = 20$ line segments, corresponding to 20 matches.
Adding the two cases together, the total number of matches is $24 + 20 = 44$.
|
44
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) A tetrahedron has 6 edges and 4 triangular faces. Given that the 6 edge lengths are exactly the positive integers $n, n+$ $1, n+2, n+3, n+4, n+5$. If the perimeter of a certain triangular face is a multiple of 3, then this triangle is painted red; otherwise, it is painted yellow. Question: What is the maximum number of yellow triangles? Prove your conclusion.
|
Three, at most 3 yellow triangles.
Since 6 consecutive positive integers divided by 3 must have remainders of 2 zeros, 2 ones, and 2 twos, denoted as $a, a, b, b, c, c$. If there are 4 yellow triangles, take any yellow $\triangle ABC$, then it must have two sides with the same remainder when divided by 3, denoted as $a, a$ for sides $AB \equiv AC \equiv a(\bmod 3)$. Since the two $a$s are already used, $AD$ must not have a remainder of $a$ when divided by 3, denoted as $AD \equiv b(\bmod 3)$, then among $BD, CD$,
at most one can have a remainder of $b$ when divided by 3 (as shown in Figure 10).
When $BD \equiv c(\bmod 3)$
$\triangle ABD$ is a red triangle;
When $CD \equiv c(\bmod 3)$
$\triangle ACD$ is a red triangle.
Therefore, there cannot be 4 yellow triangles.
Furthermore, when $AB=3, BC=4, CA=5, AD=6, CD=8, BD=7$, except for $\triangle ABC$, there are 3 yellow triangles.
Thus, the maximum number of yellow triangles is 3.
(Luo Zengru, Department of Mathematics, Shaanxi Normal University, 710062)
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The last four digits of the number $7^{355}$ are $\qquad$
|
II. 1.1943.
This problem is to find the remainder of $7^{355}$ divided by $10^{4}$.
$$
\begin{array}{l}
7^{355}=7 \times 49^{1 \pi}=7 \times 49 \times(50-1)^{2 \times 88} \\
=7 \times 49 \times(2400+1)^{28} \\
\equiv 7 \times 49 \times\left(1+2400 \mathrm{C}_{88}^{1}\right) \\
\equiv 343 \times[1+2400 \times(100-12)] \\
\equiv 343 \times(1-2400 \times 12) \equiv 343-343 \times 12 \times 2400 \\
\equiv 343-43 \times 12 \times 2400 \equiv 343-103200 \times 12 \\
\equiv 343-3200 \times 12 \equiv 343-38400 \\
\equiv 343-8400 \equiv-8057 \equiv 1943\left(\bmod 10^{4}\right) .
\end{array}
$$
Therefore, the last four digits of $7^{355}$ are 1943.
|
1943
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$.
|
Explanation: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, with $x_{1}<x_{2}$, and let $C(0, c)$, where $c \neq 0$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1}<0<x_{2}$, and it must be that $\angle A C B=90^{\circ}$. The graph can only be as shown in Figure 2 or Figure 3. In either case, we have $O C^{2}=O A \cdot O B$, which means
$$
c^{2}=-x_{1} x_{2}=-\frac{c}{a} .
$$
Therefore, $a c=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 7 Let $a, b, c$ be positive real numbers. Prove:
$$
\begin{array}{l}
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+a+c)^{2}}{2 b^{2}+(c+a)^{2}}+ \\
\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8^{[3]} .
\end{array}
$$
|
Prove: Let $s=a+b+c$,
$$
f(t)=\frac{(t+s)^{2}}{2 t^{2}+(s-t)^{2}}, t \in[0, s) \text {. }
$$
Since $f(t)=\frac{1}{3}+\frac{2}{3} \times \frac{4 s t+s^{2}}{3\left(t-\frac{s}{3}\right)^{2}+\frac{2}{3} s^{2}}$
$$
\leqslant \frac{1}{3}+\frac{4 s t+s^{2}}{s^{2}}=4\left(\frac{1}{3}+\frac{t}{s}\right),
$$
thus $f(a)+f(b)+f(c)$
$$
\leqslant 4\left(\frac{1}{3} \times 3+\frac{a+b+c}{s}\right)=8 \text {. }
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
a_{n}=2^{n}+3^{n}+6^{n}-1(n=1,2,3, \cdots) \text {. }
$$
Find all positive integers that are coprime with every term of this sequence.
(Poland provided)
|
4. The only positive integer that satisfies the condition is 1.
The following is the proof: For any prime number $p$, it must be a divisor of some term in the sequence $\left\{a_{n}\right\}$.
For $p=2$ and $p=3$, they are divisors of $a_{2}=48$.
For every prime number $p$ greater than 3, since
$$
(2, p)=1,(3, p)=1,(6, p)=1 \text {, }
$$
by Fermat's Little Theorem, we have
$$
\begin{array}{l}
2^{p-1} \equiv 1(\bmod p), \\
3^{p-1} \equiv 1(\bmod p), \\
6^{p-1} \equiv 1(\bmod p) .
\end{array}
$$
Then $3 \times 2^{p-1}+2 \times 3^{p-1}+6^{p-1}$
$$
\equiv 3+2+1=6(\bmod p) \text {, }
$$
i.e., $6 \times 2^{p-2}+6 \times 3^{p-2}+6 \times 6^{p-2} \equiv 6(\bmod p)$.
Thus, $a_{p-2}=2^{p-2}+3^{p-2}+6^{p-2}-1 \equiv 0(\bmod p)$.
That is, $p \mid a_{p-2}$.
For any positive integer $n$ greater than 1, it must have a prime factor $p$.
If $p \in\{2,3\}$, then $\left(n, a_{2}\right)>1$;
If $p \geqslant 5$, then $\left(n, a_{p-2}\right)>1$.
Therefore, positive integers greater than 1 do not meet the requirement.
Since 1 is coprime with all positive integers, the only positive integer that satisfies the condition is 1.
Note: The average score for this problem is 3.78.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. There are two people, A and B. A, on a car, notices that B is walking in the opposite direction. $1 \mathrm{~min}$ later, A gets off the car to chase B. If A's speed is twice that of B, but slower than the car's speed by $\frac{4}{5}$, then the time it takes for A to catch up with B after getting off the car is $\mathrm{min}$.
|
8.11.
Let the speed of A be $x \mathrm{~m} / \mathrm{min}$, then according to the problem, the speed of B is $\frac{x}{2} \mathrm{~m} / \mathrm{min}$, and the speed of the car is $5 x \mathrm{~m} / \mathrm{min}$.
Let the time it takes for A to catch up with B be $t \mathrm{~min}$. According to the problem, we have $\frac{x}{2}+5 x=x t-\frac{x}{2} t$.
Solving for $t$ yields $t=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. If a convex $n$-sided polygon has exactly 4 obtuse interior angles, then the maximum number of sides $n$ of this polygon is $\qquad$ .
|
10.7.
Since a convex $n$-sided polygon has exactly 4 obtuse interior angles, the sum of these 4 angles is greater than $360^{\circ}$ and less than $720^{\circ}$. The other $n-4$ angles are right angles or acute angles, so the sum of these $n-4$ angles is no more than $(n-4) \times 90^{\circ}$ and greater than $0^{\circ}$. Therefore, we have the inequality
$$
360^{\circ}+0^{\circ}<(n-2) \times 180^{\circ}<720^{\circ}+(n-4) \times 90^{\circ} \text {. }
$$
Solving this, we get $4<n<8$. Therefore, the number of sides $n$ is at most 7.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. If in a $4 \times 4$ grid of 16 cells, each cell is filled with a number such that the sum of the numbers in all adjacent cells of each cell is 1, then the sum of the 16 numbers in the grid is $\qquad$
(Note: Adjacent cells refer to cells that share exactly one edge)
|
11.6.
First, fill the 16 squares with the English letters $A, B, C$, $\cdots, O, P$ (as shown in Figure 8). Given that for each square, the sum of the numbers in all adjacent squares is 1, so,
$$
\begin{aligned}
16= & 2(A+D+M+P)+3(B+C+E+H+ \\
& I+L+N+O)+4(F+G+J+K) \\
= & 2(A+B+C+\cdots+O+P)+(B+C+E+ \\
& H+I+L+N+O)+2(F+G+J+K) \\
= & 2(A+B+C+\cdots+O+P)+(B+E+J+
\end{aligned}
$$
$$
\begin{array}{l}
G)+(C+F+K+H)+(F+I+N+K)+ \\
(G+J+O+L) .
\end{array}
$$
$$
\begin{array}{l}
\text { and } B+E+J+G=C+F+K+H \\
=F+I+N+K=G+J+O+L=1,
\end{array}
$$
Therefore, 16 $=2(A+B+C+\cdots+O+P)+4$. Hence, the sum of the 16 numbers in the grid is
$$
A+B+C+\cdots+O+P=6
$$
|
6
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given that $p$, $q$, $\frac{2 q-1}{p}$, and $\frac{2 p-1}{q}$ are all integers, and $p>1$, $q>1$. Find the value of $p+q$.
|
13. If $\frac{2 q-1}{p} \geqslant 2, \frac{2 p-1}{q} \geqslant 2$, then
$$
2 q-1 \geqslant 2 p, 2 p-1 \geqslant 2 q \text {. }
$$
Adding the two inequalities gives $2 p+2 q-2 \geqslant 2 p+2 q$. This is clearly a contradiction.
Therefore, at least one of $\frac{2 q-1}{p}$ and $\frac{2 p-1}{q}$ must be less than 2. Suppose $\frac{2 q-1}{p}<2, q>1$, then $\frac{2 q-1}{p}=1$, i.e., $2 q-1=p$.
Also, $\frac{2 p-1}{q}=\frac{4 q-3}{q}$ is an integer, i.e., $4-\frac{3}{q}$ is an integer, so $q=1$ or $q=3$.
Since $q>1$, then $q=3$, and simultaneously $p=2 q-1=5$, hence $p+q=8$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. A box contains balls of red, yellow, and white. If the number of white balls is at most $\frac{1}{2}$ of the number of yellow balls, and at least $\frac{1}{3}$ of the number of red balls, and the total number of yellow and white balls is no more than 55, then the box can contain at most $\qquad$ red balls.
|
11.54.
Suppose there are $x$ white balls, $y$ yellow balls, and $z$ red balls. According to the problem, we can set up the following system of equations:
$$
\left\{\begin{array}{l}
x \leqslant \frac{1}{2} y, \\
x \geqslant \frac{1}{3} z, \\
y + x \leqslant 55 .
\end{array}\right.
$$
From equation (1), we get $y \geqslant 2 x$.
From equations (3) and (4), we get $3 x \leqslant 55$, which means $x \leqslant \frac{55}{3}$. Therefore, the integer $x \leqslant 18$. From equation (2), we get $z \leqslant 3 x$, which means $z \leqslant 54$.
Hence, there are at most 54 red balls in the box.
|
54
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In a football tournament with 10 teams, a double round-robin system is used, meaning any two teams play each other twice, once at home and once away. A win earns 3 points, a draw earns 1 point, and a loss earns 0 points.
(1) How many matches were played in total?
(2) If every match results in the maximum points, what is the total points accumulated by all teams? If every match results in the minimum points, what is the total points accumulated by all teams?
(3) If the teams are ranked by their points at the end of the tournament, what is the maximum point difference between two teams with adjacent rankings?
|
(1) Using a list or formula calculation, it can be known that a total of 90 matches were played in this competition.
(2) The highest score for a single match is 3 points. If the maximum score is achieved in every match, the total points of all teams in this competition would be $3 \times 90 = 270$ (points): The lowest score for a single match is 2 points. If the minimum score is achieved in every match, the total points of all teams in this competition would be $2 \times 90 = 180$ (points).
(3) Let the points difference between the $k$-th and $(k+1)$-th ranked teams be the maximum. Since the total score of each match is at most 3 points and at least 2 points, the maximum score for the top $k$ teams playing against each other is
$$
3 \times 2 \times \frac{k(k-1)}{2} = 3 k(k-1) \text{ (points); }
$$
The maximum score for the top $k$ teams playing against the remaining $10-k$ teams is
$$
3 \times 2 \times k(10-k) = 6 k(10-k) \text{ (points). }
$$
Therefore, the total score of the top $k$ teams is at most $3 k(19-k)$ points. Thus, the maximum score for the $k$-th ranked team is $3(19-k)$ points.
For the remaining $10-k$ teams, the minimum total score for their matches against each other is
$$
2 \times 2 \times \frac{(10-k)(9-k)}{2} = 2(10-k)(9-k) \text{ (points), }
$$
and their score from matches against the top $k$ teams is 0.
Therefore, the minimum score for the $(k+1)$-th ranked team is
$$
\frac{2(10-k)(9-k)}{10-k} = 2(9-k) \text{ (points). }
$$
Thus, when $k=1$, the point difference reaches its maximum value, with the two teams differing by $3(19-k) - 2(9-k) = 38$ points.
|
38
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) As shown in Figure 5, a particle moves within the region $\{(x, y) \mid x \geqslant 0, y \geqslant 0\}$. In the first second, it moves from the origin to point $B_{1}(0,1)$, then from point $B_{1} \rightarrow C_{1}$ $\rightarrow A_{1}$, and subsequently moves along the $x$-axis, $y$-axis, and their parallel lines in the direction indicated by the arrows in the figure, moving 1 unit length per second. Find the time required for the particle to move from the origin to point $P(16,44)$.
|
Three, suppose the time taken for a particle to reach points $A_{n}, B_{n}, C_{n}$ from the origin are $a_{n}, b_{n}, c_{n}$, respectively. Then it is clear that
$$
\begin{array}{l}
a_{1}=3, a_{2}=a_{1}+1, \\
a_{3}=a_{1}+12=a_{1}+3 \times 4, a_{4}=a_{3}+1, \\
a_{5}=a_{3}+20=a_{3}+5 \times 4, a_{6}=a_{5}+1, \\
\cdots \cdots \\
a_{2 n-1}=a_{2 n-3}+(2 n-1) \times 4, a_{2 n}=a_{2 n-1}+1 . \\
\text { Hence } a_{2 n-1}=a_{1}+4[3+5+\cdots+(2 n-1)]=4 n^{2}-1, \\
a_{2 n}=a_{2 n-1}+1=4 n^{2}, \\
b_{2 n-1}=a_{2 n-1}-2(2 n-1)=4 n^{2}-4 n+1, \\
b_{2 n}=a_{2 n}+2 \times 2 n=4 n^{2}+4 n, \\
c_{2 n-1}=b_{2 n-1}+(2 n-1)=4 n^{2}-2 n \\
=(2 n-1)^{2}+(2 n-1), \\
c_{2 n}=a_{2 n}+2 n=4 n^{2}+2 n=(2 n)^{2}+2 n .
\end{array}
$$
Therefore, $c_{n}=n^{2}+n$.
From Figure 5, we know that the time taken for the particle to move from the origin to point $P(16,44)$ is the time taken to reach point $C_{4}$ plus $44-16$ $=28$ (s), so,
$$
t=44^{2}+44+28=2008(\mathrm{~s}) \text {. }
$$
|
2008
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Define the sequence $\left\{a_{n}\right\}: a_{n}=4+n^{3}, n \in \mathbf{N}_{+}$. Let $d_{n}=\left(a_{n}, a_{n+1}\right)$, i.e., $d_{n}$ is the greatest common divisor of $a_{n}$ and $a_{n+1}$. Then the maximum value of $d_{n}$ is $\qquad$
|
5.433.
Given) $d_{n} 1\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(4+n^{3}, 3 n^{2}+3 n+1\right)$.
Thus, we have
$$
d_{n} \mid\left[-3\left(4+n^{3}\right)+n\left(3 n^{2}+3 n+1\right), 3 n^{2}+3 n+1\right] \text {, }
$$
which means
$$
d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}+3 n+1\right) \text {. }
$$
Therefore, $d_{n} \mid \left(2 n+13,3 n^{2}+3 n+1\right)$.
Thus, $d_{n} \mid (33 n-2,2 n+13)$.
Hence, $d_{n} \mid [-2(33 n-2)+33(2 n+13)]$, which means $d_{n} \mid 433$.
Therefore, $\left(d_{n}\right)_{\text {max }} \leqslant 433$.
Let $n=\frac{27 \times 4^{2}-3 \times 4}{2}=210$.
From $4+\left(\frac{27 \times 4^{2}-3 \times 4}{2}\right)^{3}$
$$
=\left[\left(27 \times 4^{2}+1\right)^{2}-3\left(27 \times 4^{2}+1\right) \times 13+3 \times 13^{2}-5\right] \times \frac{433}{8}
$$
we know $433 \mid \left(4+210^{3}\right)$.
Thus, $433 \mid \left[4+211^{3}-\left(4+210^{3}\right)\right]$.
Therefore, $\left(d_{n}\right)_{\max }=433$.
|
433
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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