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3. A math competition has a total of 15 questions. The table below shows the number of people who got $n$ $(n=0,1,2, \cdots, 15)$ questions correct:
\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
$n$ & 0 & 1 & 2 & 3 & $\cdots$ & 12 & 13 & 14 & 15 \\
\hline Number of people who got $n$ questions correct & 7 & 8 & 10 & 21 & $\cdots$ & 15 & 6 & 3 & 1
\end{tabular}
If it is also known that the average number of questions answered correctly by students who answered 4 or more questions correctly is 6, and the average number of questions answered correctly by students who answered 10 or fewer questions correctly is 4. How many people does this table at least include?
|
(Tip: The number of people who got at most 3 questions correct is $7+8+10+21$ $=46$ people.
The number of people who got at least 12 questions correct is $15+6+3+1=25$ people.
The total number of questions correctly answered by those who got at most 3 questions correct is 91 questions,
The total number of questions correctly answered by those who got at least 12 questions correct is 315 questions.
Let the total number of people be $y$, and the number of people who got 11 questions correct be $x_{11}$, then the number of people who got at least 4 questions correct is $y-46$, and the number of people who got at most 10 questions correct is $y-25-x_{11}$. Therefore, the total number of questions correctly answered by those who got 5 to 11 questions correct is
$$
\begin{array}{l}
6(y-46)-315=4\left(y-x_{11}-25\right)+11 x_{11}-91 \\
y=200+3.5 x_{11}\left(x_{11} \geqslant 0\right) . \\
\text { When } \left.x_{11}=0 \text {, } y=200 \text { is the minimum. }\right)
\end{array}
$$
|
200
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Find all positive integers $n$ such that the sum $1+2+3+\cdots+n$ is a three-digit number composed of the same digit.
|
5. $n=36$.
$1+2+\cdots+n=\frac{n(n+1)}{2}$, which means solving the equation $\frac{n(n+1)}{2} = 111,222, \cdots, 999$, or equivalently $n(n+1)=2 \times 3 \times 37 \times k, k = 1,2, \cdots, 9$. It can be seen that $n=36, k=6$ is the only integer solution.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Question: Among $1,2,3, \cdots, 1999,2000,2001$, what is the maximum number of numbers that can be chosen such that the sum of any three chosen numbers is divisible by 21?
|
10. Take at most 95.
Let $a, b, c, d$ be 4 of the numbers taken. According to the problem, $a+b+c$ and $b+c+d$ are both multiples of 21. Therefore, the difference $a - d$ is also a multiple of 21. By the arbitrariness of $a$ and $d$, it can be concluded that the difference between any two of the numbers taken is a multiple of 21.
Assume the form of the numbers taken is
$$
\begin{array}{l}
d, d+21, d+42, \cdots, d+21(n-1), d+21 n, \\
d, n \in \mathbf{Z} .
\end{array}
$$
First, the first three terms of this sequence satisfy the condition: $211(3 d+63)$. That is, 7 divides $d$. To make the sequence as long as possible, $d$ should take the smallest value, so $d=7$.
Also, $7+21 n \leqslant 2001$, so the maximum value of $n$ is 94.
Therefore, the sequence of numbers taken is $7, 28, 49, \cdots, 1960, 1981$, a total of 95 numbers.
|
95
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $\triangle A B C$ be an acute triangle, and construct isosceles triangles $\triangle D A C$, $\triangle E A B$, and $\triangle F B C$ outside $\triangle A B C$ such that $D A = D C$, $E A = E B$, and $F B = F C$, with $\angle A D C = 2 \angle B A C$, $\angle B E A = 2 \angle A B C$, and $\angle C F B = 2 \angle A C B$. Let $D^{\prime}$ be the intersection of line $D B$ and $E F$, $E^{\prime}$ be the intersection of $E C$ and $D F$, and $F^{\prime}$ be the intersection of $F A$ and $D E$. Find the value of $\frac{D B}{D D^{\prime}} + \frac{E C}{E E^{\prime}} + \frac{F A}{F F^{\prime}}$.
|
Solution: Since $\angle ABC$ is an acute triangle, then $\angle ADC$, $\angle BEA$, $\angle CFB < \pi$. Therefore,
$$
\begin{array}{l}
\angle DAC = \frac{\pi}{2} - \frac{1}{2} \angle ADC = \frac{\pi}{2} - \angle BAC. \\
\angle BAE = \frac{\pi}{2} - \frac{1}{2} \angle BEA = \frac{\pi}{2} - \angle ABC. \\
\text{Thus, } \angle DAE = \angle DAC + \angle BAC + \angle BAE \\
= \pi - \angle ABC < \pi. \\
\end{array}
$$
Similarly, $\angle EBF < \pi$, $\angle FCD < \pi$.
Therefore, the hexagon $DAEBCF$ is a convex hexagon, and
$$
\begin{array}{l}
\angle ADC + \angle BEA + \angle CFB \\
= 2(\angle BAC + \angle ABC + \angle ACB) = 2\pi.
\end{array}
$$
Let $w_1$, $w_2$, $w_3$ be the circles with centers at $D$, $E$, $F$ and radii $DA$, $EB$, $FC$ respectively. Since $\angle ADC + \angle BEA + \angle CFB = 2\pi$, these three circles intersect at a common point. Let this point be $O$, as shown in Figure 7. Then $O$ is the reflection of $C$ over $DF$.
Similarly, $O$ is also the reflection of $A$ over $DE$ and $B$ over $EF$. Therefore,
$$
\begin{array}{l}
\frac{DB}{DD'} = \frac{DD' + D'B}{DD'} = 1 + \frac{D'B}{DD'} \\
= 1 + \frac{S_{\triangle AKB}}{S_{\triangle FDF}} = 1 + \frac{S_{\triangle EOF}}{S_{\triangle DHF}}. \\
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\frac{EC}{EE'} = 1 + \frac{S_{\triangle DFO}}{S_{\triangle DEF}} \cdot \frac{FA}{FF'} = 1 + \frac{S_{\triangle LDE}}{S_{\triangle UEF}}. \\
\text{Thus, } \frac{DB}{DD'} + \frac{EC}{EE'} + \frac{FA}{FF'} \\
= 3 + \frac{S_{\triangle OEF} + S_{\triangle ODF} + S_{\triangle UOF}}{S_{\triangle OHF}} = 4. \\
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and
$$
a_{n}=\left|a_{n-1}-a_{n-2}\right|+\left|a_{n-2}-a_{n-3}\right|, n \geqslant 4 \text {. }
$$
Find $a_{4^{4}}$.
|
For $n \geqslant 2$, define $s_{n}=\left|a_{n}-a_{n-1}\right|$. Then for $n \geqslant 5, a_{n}=s_{n-1}+s_{n-2}, a_{n-1}=s_{n-2}+s_{n-3}$. Thus, $s_{n}=$ $\left|s_{n-1}-s_{n-3}\right|$. Since $s_{n} \geqslant 0$, if $\max \left\{s_{n}, s_{n+1}, s_{n+2}\right\} \leqslant T$, then for all $m \geqslant n$, we have $s_{m} \leqslant T$. In particular, the sequence $\left\{s_{n}\right\}$ is bounded. We will prove the following proposition:
If for some $i, \max \left|s_{i}, s_{i+1}, s_{i+2}\right|=T \geqslant 2$, then $\max \left\{s_{1+6}, s_{1+7}, s_{1+8}\right\} \leqslant T-1$.
Proof by contradiction. If the above conclusion does not hold, then for $j=i, i+1, \cdots, i+6$, we have $\max \left\{s_{j}, s_{j+1}, s_{j+2}\right\}=T \geqslant 2$. For $s_{t}, s_{t+1}$, or $s_{t+2}=T$, take $j=i, i+1$, or $i+2$ accordingly, so the sequence $s_{j}, s_{j+1}, s_{j+2}, \cdots$ has the form $T, x, y, T-y, \cdots$, where $0 \leqslant x, y \leqslant T, \max \{x, y, T-y\}=T$. Therefore, either $x=T$, or $y=T$, or $y=0$.
(1) If $x=T$, then the sequence has the form $T, T, y, T-y, y, \cdots$. Thus $\max \mid y, T-y, y\}=T$. So $y=T$ or $y=0$.
(2) If $y=T$, then the sequence has the form $T, x, T, 0, x, T-x, \cdots$. Since $\max \mid 0, x, T-x=T$, so $x=0$ or $x=T$.
(3) If $y=0$, then the sequence has the form $T, x, 0, T, T-x, T-x, x, \cdots$. Since $\max \{T-x, T-x, x\}=T$, so $x=0$ or $x=T$.
In each of the above cases, $x$ and $y$ are either 0 or $T$. In particular, $T$ must divide every term in $s_{j}, s_{j+1}$, and $s_{j+2}$. Since $\max \left\{s_{2}, s_{3}, s_{4}\right\}=s_{3}=T$, for $n \geqslant 4$, $T$ divides $s_{n}$. However, when $s_{4}=11^{11}6(N-1)+4$, $a_{n}=s_{n-1}+s_{n-2}=0,1$ or 2. In particular, $a_{M}=0,1$ or 2. Since $M$ is a multiple of 7, $a_{M} \equiv a_{7} \equiv 1(\bmod 2)$, hence $a_{M}=1$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 The three edge lengths of an isosceles tetrahedron are 3, $\sqrt{10}$, and $\sqrt{13}$. Then the radius of the circumscribed sphere of this tetrahedron is
|
Analysis: According to Basic Conclusion 10, the triangular pyramid can be expanded into a rectangular parallelepiped, making the known three edges the diagonals of the faces of the rectangular parallelepiped. At this point, the original triangular pyramid and the rectangular parallelepiped have the same circumscribed sphere. Let the three edges at the same vertex of the rectangular parallelepiped be $x$, $y$, and $z$, respectively. Then, from the conditions, we have $y^{2}+z^{2}=9$, $z^{2}+x^{2}=10$, and $x^{2}+y^{2}=13$. According to this, the radius of the required circumscribed sphere is
$$
\frac{1}{2} \sqrt{x^{2}+y^{2}+z^{2}}=2 \text {. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Third Question: Before the World Cup, the coach of country $F$ plans to evaluate seven players, $A_{1}, A_{2}, \cdots, A_{7}$, by having them play in three training matches (each 90 minutes long). Assume that at any moment during the matches, exactly one of these players is on the field, and the total playing time (in minutes) for $A_{1}, A_{2}, A_{3}, A_{4}$ can all be divided by 7, while the total playing time (in minutes) for $A_{5}, A_{6}, A_{7}$ can all be divided by 13. If there are no limits on the number of substitutions per match, how many different scenarios are there for the total playing time of each player?
|
Solution: Let the times for $A_{1}, A_{2}, A_{3}, A_{4}$ be $7 k_{1}, 7 k_{2}$, $7 k_{3}, 7 k_{4}$, and their sum be $x: A_{5}, A_{6}, A_{7}$ have times $13 k_{5}$, $13 k_{6}, 13 k_{7}$, and their sum be $y$.
Then the original problem is to find the number of positive integer solutions $\left(k_{1}, k_{2}, \cdots, k_{7}\right)$ that satisfy $7\left(k_{1}+k_{2}+k_{3}+k_{4}\right)+13\left(k_{5}+\right.$ $\left.k_{6}+k_{7}\right)=3 \times 90=270$. For $7 x+13 y=270, x=\frac{270-13 y}{7} \geqslant 4$ we get
$$
\left\{\begin{array} { l }
{ x = 3 3 , } \\
{ y = 3 ; }
\end{array} \quad \left\{\begin{array} { l }
{ x = 2 0 , } \\
{ y = 1 0 ; }
\end{array} \left\{\begin{array}{l}
x=7, \\
y=17 .
\end{array}\right.\right.\right.
$$
Thus, there are only three pairs $(x, y)$ that satisfy the conditions.
Solving the system of equations $\left\{\begin{array}{l}k_{1}+k_{2}+k_{3}+k_{4}=33, \\ k_{5}+k_{6}+k_{7}=3 ;\end{array}\right.$,
$$
\left\{\begin{array} { l }
{ k _ { 1 } + k _ { 2 } + k _ { 3 } + k _ { 4 } = 2 0 , } \\
{ k _ { 5 } + k _ { 6 } + k _ { 7 } = 1 0 ; }
\end{array} \quad \left\{\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4}=7, \\
k_{5}+k_{6}+k_{7}=17 .
\end{array}\right.\right.
$$
There are a total of $\mathrm{C}_{32}^{3} \mathrm{C}_{2}^{2}+\mathrm{C}_{19}^{3} \mathrm{C}_{9}^{2}+\mathrm{C}_{6}^{3} \mathrm{C}_{16}^{2}=42244$ positive integer solutions, and each solution corresponds to a time distribution.
Therefore, there are 42244 cases in total.
|
42244
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, try to find all positive integers $k$, such that for any positive numbers $a, b, c$ satisfying the inequality
$$
k(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right)
$$
there must exist a triangle with side lengths $a, b, c$.
|
Three, due to $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$,
thus $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$.
It can be known that $k>5$. Noting that $k$ is a positive integer, therefore, $k \geqslant 6$.
Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have
$$
k(1 \times 1+1 \times 2+1 \times 2) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right),
$$
which means $k \leqslant 6$.
The following proves that $k=6$ satisfies the problem's requirements.
Assume without loss of generality that $a \leqslant b \leqslant c$, then
$$
6(a b+b c+c a)>5\left(a^{2}+b^{2}+c^{2}\right),
$$
that is
$$
\begin{array}{l}
5 c^{2}-6(a+b) c+5 a^{2}+5 b^{2}-6 a b5\left(a^{2}+b^{2}+c^{2}\right)
\end{array}
$$
a contradiction. Hence $c<a+b$.
Method two: Construct a function
$$
f(x)=5 x^{2}-6(a+b) x+5 a^{2}+5 b^{2}-6 a b \text {. }
$$
Then $f(c)<0$.
Since $f(x)$ is increasing in the interval $\left[\frac{3}{5}(a+b),+\infty\right)$, and
$$
\begin{aligned}
f(a+b)= & 5(a+b)^{2}-6(a+b)(a+b)+5 a^{2}+ \\
& 5 b^{2}-6 a b=4(a-b)^{2} \geqslant 0,
\end{aligned}
$$
thus $c<a+b$.
|
6
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Lift Your Veil
A 101-digit natural number $A=\underbrace{88 \cdots 8}_{\text {S0 digits }} \square \underbrace{99 \cdots 9}_{\text {S0 digits }}$ is divisible by 7. What is the digit covered by $\square$?
|
Explanation: β $\square$ β is a veil, covering the number to be found. To lift the veil, whether approaching from the front or the back, there are 50 digits in between, which is too many. The key is to find a way to reduce the number of digits.
The difference between two multiples of 7 is still a multiple of 7.
We already know that $A$ is a multiple of 7.
Find another multiple of 7, $B$, such that all its digits are 1. Through trial division, we find
$$
B=111111=7 \times 15873 .
$$
From this, we see that $8 B=888888$ and $9 B=999999$ are also multiples of 7.
Removing 888888 from the front of the number $A$ is equivalent to subtracting $88888800 \cdots 0$ (95 zeros) from $A$. Therefore, the resulting number is still a multiple of 7.
Continuing this process, we can remove 6 eights from the front each time, until a total of 48 eights have been removed.
Similarly, we can remove 6 nines from the back of $A$ each time, until a total of 48 nines have been removed.
Finally, a 5-digit number 88 $\square$ 99 remains, which is a multiple of 7.
The problem is simplified to finding a 5-digit number $C$, which starts with 88 and ends with 99, and is a multiple of 7. The result is
$$
\begin{aligned}
C & =(49+350)+(84000+4200) \\
& =88599 .
\end{aligned}
$$
Therefore, the digit covered by $\square$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Eight, let $A_{1}, A_{2}, \cdots, A_{8}$ be any 8 points taken on a plane. For any directed line $l$ taken on the plane, let the projections of $A_{1}, A_{2}, \cdots, A_{8}$ on this line be $P_{1}, P_{2}, \cdots, P_{8}$, respectively. If these 8 projections are pairwise distinct and are arranged in the direction of line $l$ as $P_{1_{1}}, P_{2_{2}}, \cdots, P_{i_{8}}$, then we obtain a permutation $i_{1}, i_{2}, \cdots, i_{8}$ of $1,2, \cdots, 8$ (in Figure 1, this permutation is $2,1,8,3,7,4$, $6,5$). Let the number of different permutations obtained by projecting these 8 points onto all directed lines on the plane be $N_{8}=N\left(A_{1}, A_{2}, \cdots, A_{8}\right)$. Try to find the maximum value of $N_{8}$.
|
Proof 3: First, prove that $\triangle D E F$ is the triangle with the shortest perimeter among all inscribed triangles in the acute $\triangle A B C$.
Let point $D^{\prime}$ be any fixed point on side $B C$, as shown in Figure 4. Construct the reflection of point $D^{\prime}$ about $A B$ and $A C$.
Assume the number of all lines passing through point $O$ and perpendicular to the line segments connecting any two points is $k$. It is evident that $k \leqslant \mathrm{C}_{8}^{2} = 28$. This generates $2 k$ directed lines, arranged in a counterclockwise direction, denoted as $l_{1}, l_{2}, \cdots, l_{2 h}$, as shown in Figure 6.
For any directed line $l$ (different from $l_{1}, \cdots, l_{24}$), there must exist two adjacent directed lines $l_{j}, l_{j+1}$ such that $l_{j}, l, l_{j+1}$ are arranged in a counterclockwise direction. It is evident that for a fixed $j$, the corresponding arrangement generated by such an $l$ must be the same.
If for two directed lines $l, l^{\prime}$ (different from $l_{1}, \cdots, l_{2 k}$), there does not exist a $j$ such that both $l_{j}, l, l_{j+1}$ and $l_{j}, l^{\prime}, l_{j+1}$ satisfy the requirement mentioned in the previous paragraph, then there must be a $j$ such that $l^{\prime}, l_{j}, l$ are arranged in a counterclockwise direction. Suppose $l$ is perpendicular to the line segment connecting $A_{j_{1}}$ and $A_{j_{2}}$. It is evident that the projections of points $A_{J_{1}}$ and $A$ on the directed lines $l, l^{\prime}$ must have different orders, resulting in different corresponding arrangements.
As described above, the number of different arrangements is $2 k$. Noting that $k=\mathrm{C}_{8}^{2}$ can be achieved, we have $N_{8}=56$.
Eight, for two parallel and co-directional directed lines, the projection order of $A_{1}, A_{2}, \cdots, A_{8}$ must be the same. Therefore, we only need to consider all directed lines passing through a fixed point $O$.
If the chosen directed line is perpendicular to the line segment connecting any two points, the projections of these two points must coincide, and thus no corresponding arrangement is generated. Otherwise, the projections of $A_{1}, A_{2}, \cdots, A_{8}$ must be distinct, and therefore, there is a corresponding arrangement.
|
56
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, 17 football fans plan to go to Cao Guo to watch the World Cup football matches, and they have selected a total of 17 matches. The booking of tickets meets the following conditions:
(i) Each person can book at most one ticket per match;
(ii) The tickets booked by any two people have at most one match in common;
(iii) Only one person has booked 6 tickets.
How many tickets can these fans book at most? Explain your reasoning.
|
Solution: Draw a $17 \times 17$ grid, with 17 columns representing 17 matches and 17 rows representing 17 people. If the $i$-th person has booked a ticket for the $j$-th match, then the center of the cell at the intersection of the $i$-th row and the $j$-th column is marked with a red dot. Thus, the problem is transformed into finding the maximum number of red dots in the grid such that no four red dots form the vertices of a rectangle with sides parallel to the grid lines, given that one row has 6 red dots.
Assume the first row has red dots in the first 6 cells.
Divide the $17 \times 17$ grid into two parts: a $17 \times 6$ and a $17 \times 11$. In the first part, the first row has 6 red dots, so the other 16 rows can have at most 1 red dot each. Therefore, this part can have at most 22 red dots. In the second part, the first row has no red dots, so we need to determine the maximum number of red dots in a $16 \times 11$ grid.
Let $x_i$ be the number of red dots in the $i$-th row, and consider a pair of red dots in the same row as a "red dot pair." Thus, the $i$-th row generates $\mathrm{C}_{x_i}^{2}$ red dot pairs (where $\mathrm{C}_{1}^{2}=\mathrm{C}_{0}^{2}=0$). Since no four red dots can form a rectangle with sides parallel to the grid lines, we have:
$$
\mathrm{C}_{x_1}^{2} + \mathrm{C}_{x_2}^{2} + \cdots + \mathrm{C}_{x_{16}}^{2} \leq \mathrm{C}_{11}^{2} = 55.
$$
It is easy to see that the sum on the left is minimized when $x_1, x_2, \cdots, x_{16}$ are as evenly distributed as possible (differing by at most 1). Therefore, when $x_1, x_2, \cdots, x_{16}$ include two 4s and fourteen 3s, $\mathrm{C}_{x_1}^{2} + \mathrm{C}_{x_2}^{2} + \cdots + \mathrm{C}_{x_{16}}^{2} = 54$, and $x_1 + x_2 + \cdots + x_{16} = 50$. It is clear that if the grid has 51 red dots, it cannot meet the requirements. Thus, the $17 \times 17$ grid can have at most 72 red dots.
In the following grid, there are 71 red dots, with the first 6 cells in the first row being red dots, and no four red dots form a rectangle with sides parallel to the grid lines. This indicates that the maximum number of red dots is at least 71.
The key is whether 72 red dots can meet the requirements. Suppose there are 72 red dots that meet the requirements. Then, the first 6 columns have 22 red dots, and the last 11 columns have 50 red dots. From the inequality and subsequent derivation, the distribution of 50 red dots in the $16 \times 11$ grid can only be one of two scenarios:
(1) Two rows each have 4 red dots, and the other 14 rows each have 3 red dots;
(2) Three rows each have 4 red dots, one row has 2 red dots, and the other 12 rows each have 3 red dots.
First, consider (1). In the $17 \times 17$ grid, let the red dots in the first row be numbered 1, 2, 3, 4, 5, 6, and the red dots in the second row be numbered 1, 7, 8, 9, 10. The third row also has 5 red dots, and the last 14 rows each have 4 red dots.
Examine the distribution of red dots in columns 7, 8, 9, 10. In this case, each of the last 15 rows can have at most 1 red dot in these 4 columns. If a row has 1 red dot, then the number of red dots in the last 7 columns of that row is 2 or 3 (only one row has 3). Since each of these 4 columns can form 7 different "red dot pairs" with the last 7 columns, each column can have at most 3 red dots in the last 15 rows, so the $17 \times 4$ grid can have at most 16 red dots.
Removing the first 2 rows and the first 10 columns, at most 38 red dots are removed, leaving at least 34 red dots in the $15 \times 7$ grid. Since $34 = 3 \times 4 + 2 \times 11$, these red dots form at least
$$
3 \times 4 + 11 = 23
$$
different "red dot pairs," and $23 > 21 = \mathrm{C}_{7}^{2}$, which leads to a contradiction.
Next, consider (2). Let the red dots in the first row be numbered 1, 2, 3, 4, 5, 6, the red dots in the second row be numbered 1, 7, 8, 9, 10, the third and fourth rows each have 5 red dots, and the last row has 3 red dots, with the other 12 rows each having 4 red dots.
Again, examine the distribution of red dots in columns 7, 8, 9, 10. If there are at most 16 red dots, it leads to a contradiction as in (1). However, since the last row has only 3 red dots, one of which is in the first 6 cells. If none of the cells 7, 8, 9, 10 have a red dot, it is the same as the previous case; if one of the cells 7, 8, 9, 10 has a red dot, then the last 7 cells have only one red dot, leading to 17 red dots in the $17 \times 4$ grid. Removing the last row, the remaining $14 \times 7$ grid has 32 red dots. Since $32 = 3 \times 4 + 2 \times 10$, the number of "red dot pairs" is at least
$$
3 \times 4 + 10 = 22 > 21 = \mathrm{C}_{7}^{2}.
$$
This leads to a contradiction.
In conclusion, the maximum number of tickets that 17 people can book is 71.
|
71
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A bus, a truck, and a car are traveling in the same direction on the same straight line. At a certain moment, the truck is in the middle, the bus is in front, and the car is behind, and the distances between them are equal. After 10 minutes, the car catches up with the truck; after another 5 minutes, the car catches up with the bus. How many more minutes will it take for the truck to catch up with the bus?
|
Ni.1.15.
As shown in Figure 4, let the distances between the truck and the bus, and the truck and the car be $S$. The speeds of the car, truck, and bus are approximately $a, b, c$. It is given that the truck catches up with the bus after $x$ minutes. Therefore, we have
$$
\begin{array}{l}
10(a-b)=S, \\
15(a-c)=2S, \\
x(b-c)=S .
\end{array}
$$
From equation (1), we get
$$
a-b=\frac{1}{10} S, \quad a-c=\frac{2}{15} S \text{. }
$$
Subtracting the two equations, we get $b-c=\frac{1}{30} S$, which means $S=30(b-c)$.
Substituting into equation (3), we get $x(b-c)=30(b-c)$.
Since $b-c \neq 0$, we get $x=30.30-10-5=15$ (minutes).
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the polynomial $P=2 a^{2}-8 a b+17 b^{2}-16 a-4 b$ +2070, then the minimum value of $P$ is
|
$$
\begin{array}{l}
2.2002 \\
P=2 a^{2}-8(b+2) a+17 b^{2}-4 b+2070 \\
=2\left[a^{2}-4(b+2) a+4(b+2)^{2}\right]-8(b+2)^{2} \\
\quad+17 b^{2}-4 b+2070 \\
=2(a-2 b-4)^{2}+9 b^{2}-36 b+36+2002 \\
=2(a-2 b-4)^{2}+9(b-2)^{2}+2002
\end{array}
$$
When $a-2 b-4=0$ and $b-2=0$, $P$ reaches its minimum value of 2002.
|
2002
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure $1, \angle A O B=30^{\circ}$, within $\angle A O B$ there is a fixed point $P$, and $O P$ $=10$, on $O A$ there is a point $Q$, and on $O B$ there is a fixed point $R$. To make the perimeter of $\_P Q R$ the smallest, the minimum perimeter is
|
3.10 .
As shown in Figure 5, construct the symmetric points $P_{1}, P_{2}$ of $P$ with respect to the sides $O A, O B$ of $\angle A O B$. Connect $P_{1} P_{2}$, intersecting $O A, O B$ at $Q, R$. Connect $P Q, P R$. It is easy to see that
$$
P Q=P_{1} Q, P R=P_{2} R,
$$
which implies
$$
\begin{aligned}
l_{\triangle P Q N} & =P_{1} Q+Q R+P_{2} R \\
& =P_{1} P_{2} .
\end{aligned}
$$
Thus, $\triangle P Q R$ is the triangle with the minimum perimeter.
Given that $O P=O P_{1}=O P_{2}=10$,
$$
\angle P O Q=\angle P_{1} O Q, \angle P O R=\angle P_{2} O R,
$$
we have
$$
\begin{aligned}
\angle P_{1} O P_{2} & =2 \angle P O Q+2 \angle P O R \\
& =2(\angle P O Q+\angle P O R)=2 \angle A O B=60^{\circ} .
\end{aligned}
$$
Therefore, $\triangle P_{1} O P_{2}$ is an equilateral triangle,
Thus, $P_{1} P_{2}=P_{1} O=10$.
Hence, the minimum perimeter of $\triangle P Q R$ is 10.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given theorem: βFor any prime number $n$ greater than 3, $b$ and $c$ satisfy the equation $2a + 5b = c$. Then $a + b + c$ is how many times the integer $n$? Prove your conclusion.β
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Three, the maximum possible value of $n$ is 9.
First, we prove that $a+b+c$ is divisible by 3.
In fact, $a+b+c=a+b+2a+5b=3(a+2b)$.
Thus, $a+b+c$ is a multiple of 3.
Let the remainders when $a$ and $b$ are divided by 3 be $r_{a}$ and $r_{b}$, respectively, with $r_{a} \neq 0$ and $r_{b} \neq 0$.
If $r_{a} \neq r_{b}$, then $r_{a}=1, r_{b}=2$ or $r_{a}=2, r_{b}=1$. In this case, $2a+5b$ must be a multiple of 3, meaning $c$ is a composite number. This is a contradiction.
Therefore, $r_{a}=r_{b}$, so $r_{a}=r_{b}=1$ or $r_{a}=r_{b}=2$. In this case, $a+2b$ must be a multiple of 3, and thus $a+b+c$ is a multiple of 9.
Next, we prove that 9 is the largest.
Since $2 \times 11 + 5 \times 5 = 47$ and $11 + 5 + 47 = 63$,
and $2 \times 13 + 5 \times 7 = 61$ and $13 + 7 + 61 = 81$,
and $(63, 81) = 9$,
therefore, 9 is the maximum possible value.
|
9
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Multiplicative Magic Square
Figure 1 shows a partially filled magic square. Fill in the following nine numbers: $\frac{1}{4}, \frac{1}{2}, 1,2,4,8,16,32,64$ in the grid so that the product of the numbers in each row, column, and diagonal is the same. The number that should be filled in the β $x$ β cell is $\qquad$ .
|
Explanation: Label the unfilled cells with letters, as shown in Figure 2.
The product of the nine known numbers is
$$
\frac{1}{4} \times \frac{1}{2} \times 8 \times 1 \times 2 \times 32 \times 4 \times 16 \times 64=64^{3} \text {. }
$$
Therefore, the fixed product of the three numbers in each row, each column, and each diagonal is 64.
Since the fixed product is 64, we get from the second column and the third row respectively
$$
a c=1, e f=1 .
$$
Thus, $a, c, e, f$ are each one of the numbers $\frac{1}{4}, \frac{1}{2}, 2, 4$.
Considering the product of the first row, we get $a x=2$.
This means $x$ can only be 1 or 8.
Considering the product of the diagonal, we get $64=c e x$.
If $x=1$, then $c e=64$, which is impossible.
The only possibility is $x=8$. The filling method is shown in Figure 3.
The answer to this problem is: The number to be filled in the β$x$β cell should be 8.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The value of the 1000th term in the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6, \cdots$ is ( ).
(A) 42
(B) 45
(C) 48
(D) 51
|
5. (B).
The key is to determine which number segment $a_{1000}$ falls into. Suppose $a_{1000}$ is in the $k$-th number segment, then $a_{100}=k$. Since the $k$-th number segment contains $k$ numbers, $k$ should satisfy the inequality $\frac{(k-1) k}{2}=1+2+3+\cdots+(k-1)<1000 \leqslant 1+2+3+\cdots+k=\frac{k(k+1)}{2}$, i.e., $k^{2}-k<2000 \leqslant k^{2}+k$. Thus, $k=45$. Therefore, $a_{1000}=45$.
|
45
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. The remainder when the number $7^{99}$ is divided by 2550 is
|
3.343 .
Notice that $2550=50 \times 51$. First, we find the remainders when $7^{*})$ is divided by 5 and 51, respectively.
First, $7^{199}=7 \times 7^{48}=7 \times 49^{19} \equiv 7 \times(-2)^{459}$
$\equiv-14 \times 16^{12} \equiv-14 \times 256^{6} \equiv-14 \times 1^{6}$
$\equiv-14(\bmod 51)$;
Additionally, $7^{(9)}=7 \times 49^{19} \equiv 7 \times(-1)^{49} \equiv-7(\bmod 50)$.
Thus, $\left\{\begin{array}{l}50 x \equiv-14 \times 50, \\ 51 x \equiv-7 \times 51 .\end{array}(\bmod 2550)\right.$
Subtracting the two equations, we get
$$
x \equiv-7 \times 51-(-14) \times 50 \equiv 343(\bmod 2550) \text {. }
$$
Therefore, the remainder when $7^{* x}$ is divided by 2550 is 343.
|
343
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
120. In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $1^{\circ}$. Find the maximum value of the difference between the largest and smallest interior angles.
|
Solution: Let the largest interior angle be $\alpha$, and the smallest interior angle be $\alpha-k$ (omitting the degree symbol for simplicity, the same applies below). Therefore, what we are looking for is the maximum value of $k$. Hence, the interior angles should also include $\alpha-1, \alpha-2, \alpha-3, \cdots$, $\alpha-(k-1)$, etc. Since the process of changing from $\alpha$ to $\alpha-k$ can be done in two directions, clockwise and counterclockwise, the above $k-1$ data points should each appear at least twice.
In addition, the other $n-2k$ interior angles $\beta_{1}, \beta_{2}, \cdots, \beta_{n-2k}$ should also take values from the above $k+1$ data points. Suppose that among the aforementioned $2k$ angles, there are $p$ angles $\left.\mid \beta_{1}\right\}(i=1,2, 3, \cdots, n-2k)$ inserted between two adjacent interior angles $\alpha_{1}$ and $\alpha_{2}$. Assume these are $\beta_{1}, \beta_{2}, \beta_{3}, \cdots, \beta_{p}$. Since $[x \pm 1]=[x] \pm 1$ (where [] is the notation for the floor function), the parity of the integer parts of any two adjacent interior angles is exactly different.
In the above changes, $\left[\alpha_{1}\right],\left[\beta_{1}\right],\left[\beta_{2}\right],\left[\beta_{3}\right], \cdots,\left[\beta_{p}\right]$, $\left[\alpha_{2}\right]$ alternate in parity, and since $\left[\alpha_{2}\right]=\left[\alpha_{1} \pm 1\right]$ has a different parity from $\left[\alpha_{1}\right]$, $p$ should be even. At the same time, since no two adjacent angles are the same, the sum of the two angles should not exceed $\alpha+(\alpha-1)$, so the sum of these $p$ angles should not exceed $\frac{p}{2}[\alpha+(\alpha-1)]$. Clearly, the same property applies to the other angles, meaning the sum of these $n-2k$ angles should not exceed $\frac{n-2k}{2}[\alpha+(\alpha-1)]$.
By the interior angle sum theorem, we have
$$
\begin{array}{l}
(n-2) \cdot 180 \\
= \alpha+2[(\alpha-1)+(\alpha-2)+(\alpha-3)+\cdots \\
\quad+(\alpha-(k-1))]+(\alpha-k)+\left(\beta_{1}+\beta_{2}+\beta_{3}+\right. \\
\left.\cdots+\beta_{n-2 k}\right) \\
= \alpha+2 \times \frac{(k-1)[\alpha-1+\alpha-(k-1)]}{2} \\
\quad+\alpha-k+\left(\beta_{1}+\beta_{2}+\cdots+\beta_{n-2 k}\right) \\
\leqslant 2 k \alpha-k^{2}+\frac{n-2 k}{2}[\alpha+(\alpha-1)] .
\end{array}
$$
Rearranging and combining with $\alpha > 0$,
we know that a convex polygon satisfying $k_{\max }=18$ exists.
In summary, the maximum value sought is $18^{\circ}$.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. $x, y$ are real numbers, and $\left(x+\sqrt{x^{2}+1}\right)(y+$ $\left.\sqrt{y^{2}+1}\right)=1$. Then $x+y=$ $\qquad$ .
|
δΊγ1.0.
Multiply both sides of the original equation by $\sqrt{x^{2}+1}-x$, we get $y+\sqrt{y^{2}+1}=\sqrt{x^{2}+1}-x$; multiply both sides of the original equation by $\sqrt{y^{2}+1}-y$, we get $x+\sqrt{x^{2}+1}=\sqrt{y^{2}+1}-y$. Adding the two equations immediately yields $x+y=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A triangle with all sides as integers, and the longest side being 11, has
$\qquad$ possibilities.
|
Ni, 1.36.
Let the three sides be $x, y, z$, and $x \leqslant y \leqslant z, z=11, x+$ $y>z$, then
$$
6 \leqslant y \leqslant 11 .
$$
When $y=6, x=6$, it forms 1 triangle;
When $y=7, x$ is $5, 6, 7$, it forms 3 triangles; .....
When $y=11, x$ is $1, 2, \cdots, 11$, it forms 11 triangles. Therefore, there are a total of $1+3+5+\cdots+11=36$ triangles.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $\left|\log \frac{a}{\pi}\right|<2$, then the number of $a$ for which the function $y=\sin (x+a)+\cos (x-a)(x \in \mathbf{R})$ is an even function is $\qquad$.
|
2.10.
$$
\begin{aligned}
\because-2 & <\log \frac{\alpha}{\pi}<2, \therefore \frac{1}{\pi}<\alpha<\pi^{3} . \\
f(x)= & \sin (x+\alpha)+\cos (x+\alpha) \\
& =\sqrt{2} \sin \left(x+\alpha+\frac{\pi}{4}\right) .
\end{aligned}
$$
Since $f(x)$ is an even function,
$$
\therefore \sin \left(\frac{\pi}{4}+\alpha+x\right)=\sin \left(-x+\alpha+\frac{\pi}{4}\right) \text {, }
$$
i.e., $\alpha=k \pi+\frac{\pi}{4}, k \in \mathbf{Z}$. Therefore, $\frac{1}{\pi}<k \pi+\frac{\pi}{4}<\pi^{3}$. Thus, $\frac{1}{\pi^{2}}-\frac{1}{4}<k<\pi^{2}-\frac{1}{4}$,
Hence $0 \leqslant k \leqslant 9, k \in \mathbf{Z}$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. From $\{1,2,3, \cdots, 20\}$, select 3 numbers such that no two numbers are adjacent, there are $\qquad$ different ways.
|
4.816.
17 numbers produce 16 spaces plus 2 ends making a total of 18 spaces. According to the problem, there are $\mathrm{C}_{18}^{3}$ different ways to choose.
|
816
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The largest positive integer $n$ for which the inequality $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$ holds for a unique integer $k$ is $\qquad$ .
|
6.112 .
Obviously, $\frac{13}{7}<\frac{n+k}{n}<\frac{15}{8} \Leftrightarrow \frac{6}{7}<\frac{k}{n}<\frac{7}{8}$
$$
\Leftrightarrow 48 n<56 k<49 n \text {. }
$$
There are at least $n-1$ integers between $48 n+1$ and $49 n-1$.
If $n-1 \geqslant 2 \times 56$, then there must be at least 2 multiples of 56 in this range, so, $n-1<2 \times 56$, which means $n \leqslant 112$.
Also, $48 \times 112=56 \times 96<56 \times 98=112 \times 49$, hence $k=97, n=112$.
|
112
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Total 20 points) The sequence $\left\{x_{n}\right\}$ satisfies
$$
\begin{array}{l}
x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, n \in \mathbf{N}, y_{n}=\frac{1}{1+x_{n}}, \\
S_{n}=y_{1}+y_{2}+\cdots+y_{n}, P_{n}=y_{1} y_{2} \cdots y_{n} .
\end{array}
$$
Find $P_{n}+\frac{1}{2} S_{n}$.
|
$$
\begin{array}{l}
\text { Three, } \because x_{1}=\frac{1}{2}, x_{n+1}=x_{n}^{2}+x_{n}, \\
\therefore x_{n+1}>x_{n}>0, x_{n+1}=x_{n}\left(1+x_{n}\right) . \\
\therefore y_{n}=\frac{1}{1+x_{n}}=\frac{x_{n}^{2}}{x_{n} x_{n+1}}=\frac{x_{n+1}-x_{n}}{x_{n} x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{x_{n+1}} . \\
\therefore P_{n}=y_{1} y_{2} \cdots y_{n}=\frac{x_{1}}{x_{2}} \cdot \frac{x_{2}}{x_{3}} \cdots \cdots \frac{x_{n}}{x_{n+1}}=\frac{1}{2 x_{n+1}}, \\
S_{n}=\frac{1}{x_{1}}-\frac{1}{x_{2}}+\frac{1}{x_{2}}-\frac{1}{x_{3}}+\cdots+\frac{1}{x_{n}}-\frac{1}{x_{n+1}} \\
\quad=2-\frac{1}{x_{n+1}} .
\end{array}
$$
Therefore, $P_{n}+\frac{1}{2} S_{n}=\frac{1}{2 x_{n+1}}+1-\frac{1}{2 x_{n+1}}=1$ :
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (Total 50 points) 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 the above requirement.
|
Three, 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 $x_{i}<2$, then $\mathrm{C}_{x_{i}}^{2}=0$).
If an additional column is added to the side of a $13 \times 13$ square, and each point pair is projected 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 $C_{13}^{2}$ different point pairs on the newly drawn column, thus we get $\sum_{i=1}^{13} \mathrm{C}_{x_{i}}^{2} \leqslant \mathrm{C}_{13}^{2}$, i.e., $\sum_{i=1}^{13} x_{i}\left(x_{i}-1\right) \leqslant 13 \times 12$.
Hence $\sum_{i=1}^{13} x_{i}^{2} \leqslant 156+k$.
$\because \sum_{i=1}^{13} x_{i}^{2} \geqslant \frac{1}{13}\left(\sum_{i=1}^{13} x_{i}\right)^{2}=\frac{k^{2}}{13}$,
$\therefore k^{2} \leqslant 13 \times 156+13 k$, which gives $-39 \leqslant k \leqslant 52$.
When $k=52$, a graph that meets the conditions can be constructed (as shown in Figure 3).
In summary, the maximum value of $k$ is 52.
|
52
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in $S$ is a positive integer not exceeding 100;
(2) For any two different elements $a, b$ in $S$, there exists an element $c$ in $S$ such that the greatest common divisor (gcd) of $a$ and $c$ is 1, and the gcd of $b$ and $c$ is also 1;
(3) For any two different elements $a, b$ in $S$, there exists an element $d$ in $S$ different from $a$ and $b$ such that the gcd of $a$ and $d$ is greater than 1, and the gcd of $b$ and $d$ is also greater than 1.
|
II. 72.
Express each positive integer $n$ not exceeding 100 as
$$
n=2^{\alpha_{1}} \cdot 3^{a_{2}} \cdot 5^{a_{3}} \cdot 7^{a_{4}} \cdot 11^{a_{5}} \cdot q \text {. }
$$
where $q$ is a positive integer not divisible by $2, 3, 5, 7, 11$, and $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-negative integers.
We select the positive integers that satisfy the condition that exactly 1 or 2 of $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}$ are non-zero to form the set $S$. Thus, $S$ includes 50 even numbers $2, 4, \cdots, 98, 100$, but excludes $2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5, 2 \times 3 \times 7, 2^{2} \times 3 \times 7, 2 \times 5 \times 7, 2 \times 3 \times 11$; 17 odd multiples of 3: $3 \times 1, 3 \times 3, \cdots, 3 \times 33$; 7 numbers with the smallest prime factor 5: $5 \times 1, 5 \times 5, 5 \times 7, 5 \times 11, 5 \times 13, 5 \times 17, 5 \times 19$; 4 numbers with the smallest prime factor 7: $7 \times 1, 7 \times 7, 7 \times 11, 7 \times 13$; and the prime number 11. Therefore, $S$ contains a total of $(50-7)+17+7+4+1=72$ numbers.
We now prove that the constructed $S$ satisfies the given conditions.
Condition (1) is obviously satisfied.
For condition (2), note that among the prime factors of $[a, b]$, at most 4 of $2, 3, 5, 7, 11$ appear. Let the missing prime factor be $p$. Clearly, $p \in S$, and
$$
\begin{array}{l}
(p, a) \leqslant(p,[a, b])=1, \\
(p, b) \leqslant(p,[a, b])=1 .
\end{array}
$$
Thus, taking $c=p$ suffices.
For condition (3), when $(a, b)=1$, take the smallest prime factor $p$ of $a$ and the smallest prime factor $q$ of $b$. Clearly, $p \neq q$, and $p, q \in \{2, 3, 5, 7, 11\}$. Thus, $pq \in S$, and
$$
(pq, a) \geqslant p > 1, (pq, b) \geqslant q > 1.
$$
The fact that $a$ and $b$ are coprime ensures that $pq$ is different from $a$ and $b$. Thus, taking $c=pq$ suffices.
When $(a, b)=e>1$, take $p$ as the smallest prime factor of $e$ and $q$ as the smallest prime such that $q \times [a, b]$. Clearly, $p \neq q$, and $p, q \in \{2, 3, 5, 7, 11\}$. Thus, $pq \in S$, and
$$
\begin{array}{l}
(pq, a) \geqslant (p, a) = p > 1, \\
(pq, b) \geqslant (p, b) = p > 1.
\end{array}
$$
The condition $q \times [a, b]$ ensures that $pq$ is different from $a$ and $b$. Thus, taking $d=pq$ suffices.
We now prove that the number of elements in any set $S$ satisfying the given conditions does not exceed 72.
Clearly, $1 \notin S$. For any two primes $p, q$ greater than 10, since the smallest number not coprime with both $p$ and $q$ is $pq$, which is greater than 100, by condition (3), at most one of the 21 primes between 10 and 100, $11, 13, \cdots, 89, 97$, can be in $S$. Let the set of the remaining 78 natural numbers not exceeding 100, excluding 1 and these 21 primes, be $T$. We claim that at least 7 numbers in $T$ are not in $S$, thus $S$ can have at most $78-7+1=72$ elements.
(i) When there is a prime $p$ greater than 10 in $S$, the smallest prime factor of all numbers in $S$ can only be $2, 3, 5, 7$, and $p$. Using condition (2), we can derive the following conclusions:
(1) If $7p \in S$, since $2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5$ include all the smallest prime factors, by condition (2), $2 \times 3 \times 5, 2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5 \notin S$; if $7p \notin S$, note that $2 \times 7p > 100$, and $p \in S$, so by condition (3), $7 \times 1, 7 \times 7, 7 \times 11, 7 \times 13 \notin S$.
(2) If $5p \in S$, then $2 \times 3 \times 7, 2^{2} \times 3 \times 7 \notin S$; if $5p \notin S$, then $5 \times 1, 5 \times 5 \notin S$.
(3) $2 \times 5 \times 7$ and $3p$ cannot both be in $S$.
(4) $2 \times 3p$ and $5 \times 7$ cannot both be in $S$.
(5) If $5p, 7p \notin S$, then $5 \times 7 \notin S$.
When $p=11$ or 13, by (1), (2), (3), (4), we can respectively conclude that at least $3, 2, 1, 1$ numbers in $T$ are not in $S$, totaling 7 numbers; when $p=17$ or 19, by (1), (2), (3), we can respectively conclude that at least $4, 2, 1$ numbers in $T$ are not in $S$, totaling 7 numbers; when $p>20$, by (1), (2), (3), we can respectively conclude that at least $4, 2, 1$ numbers in $T$ are not in $S$, totaling 7 numbers.
(ii) If no prime greater than 10 is in $S$, then the smallest prime factor of numbers in $S$ can only be $2, 3, 5, 7$. Thus, the following 7 pairs of numbers cannot both be in $S$:
$$
\begin{array}{r}
(3, 2 \times 5 \times 7), (5, 2 \times 3 \times 7), (7, 2 \times 3 \times 5), (2 \times 3, \\
5 \times 7), (2 \times 5, 3 \times 7), (2 \times 7, 3 \times 5), (2^{2} \times 7, 3^{2} \times 5)
\end{array}
$$
Thus, at least 7 numbers in $T$ are not in $S$.
In summary, the answer to this problem is 72.
|
72
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let real numbers $a, b, c, d$ satisfy $a^{2}+b^{2}+c^{2}+d^{2}=5$. Then the maximum value of $(a-b)^{2}+(a-c)^{2}+(a-d)^{2}+(b-c)^{2}+(b-d)^{2}$ $+(c-d)^{2}$ is $\qquad$ .
|
6. 20
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (This question is worth 18 points)
On a piece of paper, there are $1,2, \cdots, n$ these $n$ positive integers. In the first step, the first 4 numbers $1,2,3,4$ are crossed out, and the sum of the 4 crossed-out numbers, 10, is written at the end of $n$; in the second step, the first 4 numbers $5,6,7,8$ are crossed out, and the sum of the 4 crossed-out numbers, 26, is written at the end; and so on (i.e., in each step, the first 4 numbers are crossed out, and the sum of the 4 crossed-out numbers is written at the end).
(1) If only one number is left at the end, what is the necessary and sufficient condition for $n$?
(2) Taking $n=2002$, what is the total sum of all the numbers written (including the original $1,2, \cdots, 2002$) until only one number is left?
|
(1) With each step, the number of numbers on the paper decreases by 3. If $n$ numbers are reduced to 1 number after $p$ steps, then $n-3p=1$, i.e., $n=3p+1$.
$\therefore$ The necessary and sufficient condition for $n$ is that $n$ leaves a remainder of 1 when divided by 3.
(2) Suppose there are $4^{k}$ numbers initially, with their sum being $S$. When these $4^{k}$ numbers are completed, it takes $4^{k-1}$ steps, and $4^{k-1}$ numbers remain on the paper, and the sum of these $4^{k-1}$ numbers equals the original sum $S$ of the $4^{k}$ numbers. Therefore, when only 1 number remains, the total sum of the numbers written is $(k+1)S$.
$2002=1024+978=4^{5}+978$. Originally, 2002 numbers, after $\frac{978}{3}=326$ steps, leave $4^{5}$ numbers, and the numbers removed are $1,2,3$ $\cdots, 1304$ (note that $1304=4 \times 326$), and the sum of the $4^{5}$ numbers remaining on the paper is $1+2+\cdots+2002$. Therefore, when only 1 number remains, the total sum of the numbers written is $(1+2+\cdots+1304)+6(1$ $+2+3+\cdots+2002)=12880878$.
|
12880878
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 A home appliance manufacturing company, based on market research analysis, has decided to adjust its production plan and is preparing to produce a total of 360 units of air conditioners, color TVs, and refrigerators per week (calculated at 120 labor hours). Moreover, the production of refrigerators should be at least 60 units. It is known that the labor hours required per unit and the output value per unit of these home appliances are as shown in Table 1.
Table 1
\begin{tabular}{|l|c|c|c|}
\hline Home Appliance Name & Air Conditioner & Color TV & Refrigerator \\
\hline Labor Hours & $\frac{1}{2}$ & $\frac{1}{3}$ & $\frac{1}{4}$ \\
\hline Output Value (thousand yuan) & 4 & 3 & 2 \\
\hline
\end{tabular}
How many units of air conditioners, color TVs, and refrigerators should be produced each week to maximize the output value? What is the maximum output value (in thousand yuan)?
(12th Jiangsu Province Junior High School Mathematics Competition)
|
Solution: Let the number of air conditioners produced each week be $x$, the number of color TVs be $y$, and the number of refrigerators be $z$, with the total output value being $w$ thousand yuan. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=360, \\
\frac{1}{2} x+\frac{1}{3} y+\frac{1}{4} z=120, \\
w=4 x+3 y+2 z, \\
z \geqslant 60 .
\end{array}\right.
$$
From (1) and (2), we get
$$
x=\frac{z}{2}, y=360-\frac{3 z}{2} \text {, }
$$
Substituting into (3), we get
$$
w=-\frac{1}{2} z+1080 .
$$
Since $k=-\frac{1}{2}<0$,
$w$ increases as $z$ decreases.
Given $z \geqslant 60$, when $z=60$, $w$ is maximized, with the maximum value being 1050.
At this point, $x=30, y=270$.
Therefore, to maximize the output value, 30 air conditioners, 270 color TVs, and 60 refrigerators should be produced each week, with the highest output value being 1050 thousand yuan.
|
1050
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The number of sets $X$ that satisfy the condition $\{1,2,3\} \subseteq X \subseteq\{1,2,3,4,5,6\}$ is $\qquad$ .
|
ii. 7.8 items.
$X$ must include the 3 elements $1,2,3$, while the numbers 4,5,6 may or may not belong to $X$. Each number has 2 possibilities, so the total number of different $X$ is $2^{3}=8$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
οΌ1οΌFor any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1} γ x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right)$ $\neq f\left(x_{2}\right)$.
Then the value of $f(0)+f(1)+f(-1)$ is $\qquad$
|
10.0 .
From $f(0)=f^{3}(0)$, we know $f(0)[1-f(0)][1+f(0)]=0$,
thus, $f(0)=0$ or $f(0)=1$, or $f(0)=-1$;
from $f(1)=f^{3}(1)$, similarly $f(1)=0$ or 1 or 1;
from $f(-1)=f^{3}(-1)$, similarly $f(-1)=0$ or 1 or -1.
However, $f(0)$, $f(1)$, and $f(-1)$ are pairwise distinct,
so $\{f(0), f(1), f(-1)\}=\{0,1,-1\}$.
Therefore, $f(0)+f(1)+f(-1)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. (13 points) Let the quadratic equation in $x$, $2 x^{2}-t x-2=0$, have two roots $\alpha, \beta(\alpha<\beta)$.
(1) If $x_{1} γ x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove that $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1}$, and the maximum and minimum values of $f(x)$ on the interval $[\alpha, \beta]$ be $f_{\max }$ and $f_{\min }$, respectively. Let $g(t)=f_{\max }-f_{\min }$. Find the minimum value of $g(t)$.
|
$$
\begin{aligned}
14. (1) & \text{ From the conditions, we have } \alpha+\beta=\frac{t}{2}, \alpha \beta=-1. \\
& \text{ Without loss of generality, assume } \alpha \leqslant x_{1}4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+x_{2}\right)+4 \alpha \beta=4 x_{1} x_{2} \\
& -t\left(x_{1}+x_{2}\right)-4 .
\end{aligned}
$$
Therefore,
$$
4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-40, f_{\min }=f(\alpha)<0 . \\
g(t) & =f(\beta)-f(\alpha) \\
& =|f(\beta)|+|f(\alpha)| \\
& \geqslant 2 \sqrt{|f(\beta)| \cdot|f(\alpha)|}=4 .
$$
Equality holds if and only if $f(\beta)=-f(\alpha)=2$,
i.e., $\frac{8}{\sqrt{t^{2}+16}+t}=2$, which implies $t=0$.
Thus, the minimum value of $g(t)$ is 4.
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (13 points) Given $a_{1}=1, a_{2}=3, a_{n+2}=(n+3) a_{n+1}$ $-(n+2) a_{n}$. If for $m \geqslant n$, the value of $a_{m}$ can always be divided by 9, find the minimum value of $n$.
|
15. From $a_{n+2}-a_{n+1}=(n+3) a_{n+1}-(n+2) a_{n}$
$$
\begin{array}{l}
-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)=(n+2)(n+1) \cdot \\
\left(a_{n}-a_{n-1}\right)=\cdots=(n+2) \cdot(n+1) \cdot n \cdots+4 \cdot 3 \cdot\left(a_{2}-\right. \\
\left.a_{1}\right)=(n+2)!, \\
\text { Therefore, } a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right) \\
\quad=1+2!+3!+\cdots+n!(n \geqslant 1) .
\end{array}
$$
Therefore, $a_{n}=a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\cdots+\left(a_{n}-a_{n-1}\right)$
Given $a_{1}=1, a_{2}=3, a_{3}=9, a_{4}=33, a_{5}=153$, at this point 153 is divisible by 9.
When $m \geqslant 5$, $a_{n}=a_{5}+\sum_{k=6}^{m} k!$, and when $k \geqslant 6$, $k!$ is divisible by 9.
Thus, when $m \geqslant 5$, $a_{n}$ is divisible by 9. Therefore, the smallest value of $n$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given the sequence $\left\{a_{n}\right\}$, where $a_{n}$ is a real number, and for $n \geqslant 3, n \in$ $\mathbf{N}$, we have $a_{n}=a_{n-1}-a_{n-2}$. If the sum of the first 1985 terms is 1000, and the sum of the first 1995 terms is 4000, then the sum of the first 2002 terms is $\qquad$.
|
8.3000
Let $a_{1}=a, a_{2}=b$, then the first 6 terms of the sequence are $a, b, b-a$, $-a, -b, a-b$. Also, $a_{7}=a, a_{8}=b$, it is easy to verify that the sum of any consecutive 6 terms of this sequence is 0.
$$
\begin{aligned}
\therefore S_{1995} & =S_{332 \times 6+3}=0+a+b+(b-a) \\
& =2 b=4000 . \\
S_{1 \text { ges }} & =S_{380 \times 6+5} \\
& =0+a+b+(b-a)+(-a)+(-b) \\
& =b-a=1000 .
\end{aligned}
$$
From equations (1) and (2), we solve to get $b=2000, a=1000$.
$$
\text { Then } \begin{aligned}
S_{2002} & =S_{333 \times 6+4}=0+a+b+(b-a)+(-a) \\
& =2 b-a=3000 .
\end{aligned}
$$
|
3000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. The expansion of $(a+b+c)^{10}$, after combining like terms, has
$\qquad$ terms.
|
12.66.
The form of each term after merging is $a^{k_{1}} b^{k_{2}} c^{k_{3}}$, where $k_{1}+k_{2}+k_{3}=10\left(k_{1} γ k_{2} γ k_{3} \in \mathrm{N}\right)$, and when $k_{1} γ k_{2}$ are determined, $k_{3}$ is uniquely determined.
If $k_{1}=10, k_{2}=0$, there is 1 way;
If $k_{1}=9, k_{2}=0$ or 1, there are 2 ways;
If $k_{1}=8, k_{2}=0 γ 1 γ 2$, there are 3 ways;
......
If $k_{1}=1, k_{2}=0 γ 1 γ 2 γ \cdots γ 9$, there are 10 ways;
If $k_{1}=0, k_{2}=0 γ 1 γ 2 γ \cdots γ 10$, there are 11 ways.
In total, there are $1+2+3+\cdots+10+11=66$ ways, each
way corresponds to one like term, hence there are 66 terms.
|
66
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Given $p$ is a prime number, $r$ is the remainder when $p$ is divided by 210. If $r$ is a composite number and can be expressed as the sum of two perfect squares, find
(52nd Belarusian Mathematical Olympiad (Final C Category))
|
Let $p=210n+r$, then $07$.
Let $q$ be the smallest prime dividing $r$, i.e., $r=qm, q \leqslant m$, we get $210>r=qm \geqslant q^{2}$, so, $q \leqslant 13$.
On the other hand, $r$ cannot be divisible by 2, 3, 5, 7, otherwise $p$ would also be divisible by the same number, but $p$ is a prime, so $q>7$, hence $q=11$ or $q=13$. At this point, $r=a^{2}+b^{2}$, where $a$ and $b$ are positive integers.
If $q=11$, then $a^{2}+b^{2}$ is divisible by 11. Write down all the remainders of perfect squares when divided by 11: $0,1,4,9,5,3$, to get the sum of the remainders of two perfect squares to be divisible by 11, it must be that both $a$ and $b$ are divisible by 11, i.e., $a^{2}$ and $b^{2}$ are divisible by 121. Thus, $r=a^{2}+b^{2} \geqslant 121+121 \geqslant 242>210$. Contradiction.
Therefore, $q=13, m<\frac{121}{q}<17$, i.e., $m \leqslant 16$.
Since the smallest prime factor of $m$ is at least 13, hence $m=13$.
Thus, $r=mq=13^{2}=169$, and it satisfies $169=12^{2}+5^{2}$.
|
169
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 A certain project, if contracted by Team A and Team B, can be completed in $2 \frac{2}{5}$ days, costing 180000 yuan; if contracted by Team B and Team C, it can be completed in $3 \frac{3}{4}$ days, costing 150000 yuan; if contracted by Team A and Team C, it can be completed in $2 \frac{6}{7}$ days, costing 160000 yuan. Now the project is to be contracted to a single team, under the condition that it must be completed within a week, which team would have the lowest cost?
(2002, National Junior High School Mathematics Competition)
|
Solution: Let the time required for A, B, and C to complete the work alone be $x$, $y$, and $z$ days, respectively, then
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=\frac{5}{12} \\
\frac{1}{y}+\frac{1}{z}=\frac{4}{15} \\
\frac{1}{z}+\frac{1}{x}=\frac{7}{20}
\end{array} .\right.
$$
Solving, we get $x=4, y=6, z=10$.
Let the cost for A, B, and C to work one day alone be $u$, $v$, and $w$ yuan, respectively, then
$$
\left\{\begin{array}{l}
\frac{12}{5}(u+v)=180000, \\
\frac{15}{4}(v+w)=150000, \\
\frac{20}{7}(w+u)=160000 .
\end{array}\right.
$$
Solving, we get $u=45500, v=29500, w=10500$.
Thus, the cost for A to complete the work alone is
$$
45500 \times 4=182000 \text { (yuan); }
$$
The cost for B to complete the work alone is
$$
29500 \times 6=177000 \text { (yuan); }
$$
While C cannot complete the work within a week.
By comparison, it is easy to see that the cost for B to complete the work is the lowest.
|
177000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In a mathematics competition,
(i) the number of problems is $n(n \geqslant 4)$;
(ii) each problem is solved by exactly 4 people;
(iii) for any two problems, exactly 1 person solves both problems.
If the number of participants is greater than or equal to $4 n$, find the minimum value of $n$ such that there always exists a person who solves all the competition problems.
(15th Korean Mathematical Olympiad)
|
Solution: Let $n \geqslant 14$, and represent each problem as a rectangle, with its 4 vertices representing the 4 contestants who solved the problem. Let $P$ be any problem.
Each problem different from $P$ shares a common vertex with $P$, since $n \geqslant 14$. By the pigeonhole principle, there is a vertex of $P$, denoted as $S$, that is a vertex of at least 4 (different from $P$) rectangles.
Assume there exists a rectangle $P$ that does not share vertex $S$ with $P$, then there are at least 5 rectangles with vertex $S$, and each of these rectangles shares only one common vertex with $P$. By the pigeonhole principle, a vertex $S^{\prime}$ of $P$ is a vertex of at least 2 rectangles that have $S$ as a vertex.
If $S \neq S^{\prime}$, then there are at least 2 rectangles that share both vertex $S$ of $P$ and vertex $S^{\prime}$ of $P$, which contradicts condition (iii).
Hence, $S=S^{\prime}$, meaning all rectangles share vertex $S$.
Table 1 is an example when $n=13$ where no one solved all the problems. Number the problems as $1,2, \cdots, 13$, and number the contestants as 1, $2, \cdots$. In Table 1, all contestants except for contestants 1 to 13 did not solve any problems. The first row represents the problem numbers, and the $i$-th column $(1 \leqslant i \leqslant 13)$ represents the numbers of the 4 contestants who solved the $i$-th problem.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Competition & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\
\hline \multirow{4}{*}{\begin{tabular}{l}
Contestants who \\
solved the problem
\end{tabular}} & 1 & 1 & 1 & 1 & 2 & 2 & 2 & 3 & 3 & 3 & 4 & 4 & 4 \\
\hline & $?$ & 5 & 8 & 11 & 5 & 6 & 7 & 5 & 6 & 7 & 5 & 6 & 7 \\
\hline & 5 & 6 & 9 & 12 & 8 & 9 & 10 & 9 & 10 & 8 & 10 & 8 & 9 \\
\hline & 4 & 7 & 10 & 13 & 11 & 12 & 13 & 13 & 11 & 12 & 12 & 13 & 11 \\
\hline
\end{tabular}
For $4 \leqslant n \leqslant 12$, similar examples can be obtained by removing some columns (problems). Therefore, the minimum value is 14.
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. 14 people participate in a Japanese chess round-robin tournament, where each person plays against the other 13 people, and there are no draws in the matches. Find the maximum number of "triangles" (here, a "triangle" refers to a set of three people where each person has one win and one loss).
(2002, Japan Mathematical Olympiad (First Round))
|
Solution: Let the 14 people be represented as $A_{1}, A_{2}, \cdots, A_{14}$, and let $A_{4}$ win $a_{4}$ games. In any group of three people (i.e., selecting 3 out of 14 people), if there is no "double corner," then one person must have won the other two. Therefore, the number of groups of three that do not form a "triple corner" is $\sum_{k=1}^{14} \mathrm{C}_{a_{k}}^{2}$ (here, define $C_{1}^{2}=C_{0}^{2}=0$). The number of "triple corners" is $C_{14}^{3}-$ $\sum_{k=1}^{14} C_{a_{k}}^{2}$, which means we need to find the minimum value of $\sum_{k=1}^{14} C_{a_{k}}^{2}$.
It is known that $\sum_{k=1}^{14} a_{k}=C_{14}^{2}=91$. Below, we prove:
(1) For integers $x_{k} \geqslant 0$, and satisfying $\sum_{i=1}^{14} x_{k}=91$. The minimum value of $\sum_{k=1}^{14} \mathrm{C}_{x_{k}}^{2}$ is 252;
(2) There exists a win-loss scenario that satisfies $\sum_{i=1}^{14} \mathrm{C}_{a_{k}}^{2}=252$.
First, we prove (1). Since each $x_{i}$ has only a finite number of possible values, the minimum value of $\sum_{k=1}^{14} \mathrm{C}_{x_{k}}^{2}$ exists. Suppose there exist $1 \leqslant i, j \leqslant 14$ such that $x_{1}-x_{j} \geqslant 2$. Define
$$
y_{k}=\left\{\begin{array}{ll}
x_{i}-1 & (k=i), \\
x_{j}+1 & (k=j), \\
x_{k} & (k \neq i, j) .
\end{array}\right.
$$
Then $y_{k} \geqslant 0, \sum_{k=1}^{14} y_{k}=91$, and
$$
\begin{array}{l}
\sum_{k=1}^{14} \mathrm{C}_{x_{k}}^{2}-\sum_{k=1}^{14} \mathrm{C}_{y_{k}}^{2}=\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{j}}^{2}-\mathrm{C}_{x_{i}-1}^{2}-\mathrm{C}_{x_{j}+1}^{2} \\
= \frac{x_{1}\left(x_{i}-1\right)}{2}+\frac{x_{j}\left(x_{j}-1\right)}{2}-\frac{\left(x_{i}-1\right)\left(x_{i}-2\right)}{2}- \\
\frac{\left(x_{j}+1\right) x_{j}}{2} \\
= x_{i}-x_{j}-1>0 .
\end{array}
$$
Therefore, $\sum_{k=1}^{14} C_{x_{k}}^{2}$ is not the minimum value. Thus, the $x_{k}$ that makes $\sum_{k=1}^{14} C_{x_{k}}^{2}$ reach its minimum value should satisfy: for all $1 \leqslant i, j \leqslant 14,\left|x_{1}-x_{j}\right| \leqslant 1$. Clearly,
$$
\begin{array}{l}
\left\{x_{1}, x_{2}, \cdots, x_{14}\right\} \\
=\{6,6,6,6,6,6,6,7,7,7,7,7,7,7\} .
\end{array}
$$
Thus, the minimum value of $\sum_{k=1}^{14} \mathrm{C}_{\mathrm{r}_{k}}^{2}$ is $\mathrm{C}_{6}^{2} \times 7+\mathrm{C}_{7}^{2} \times 7=252$.
Next, we prove (2).
The win-loss situation in each cycle is as shown in Table 2.
Here, in the $(i, j)$ table, βOβ indicates $A_{1}$ wins $A_{j}$, β$\bullet$β indicates $A_{i}$ loses to $A_{j}$, and β$\times$β indicates that no match was played between $A_{i}$ and $A_{j}$.
Therefore, the maximum number of "double corners" is
$$
C_{14}^{3}-252=112 .
$$
|
112
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. Given a set of 2002 points on the $x O y$ plane, denoted as $S$, and no two points in $S$ are connected by a line parallel to the coordinate axes. For any two distinct points $P, Q$ in $S$, consider the rectangle with $P Q$ as its diagonal, and whose sides are parallel to the coordinate axes. Let $W_{P Q}$ denote the number of points in $S$ within this rectangle (excluding $P, Q$).
When the statement "There exists at least one pair of points $P, Q$ in $S$ such that $W_{P Q} \geqslant N$, regardless of how the points in $S$ are distributed on the coordinate plane" is true, find the maximum value of $N$.
(2002, Japan Mathematical Olympiad (Second Round))
|
Solution: Here we only discuss rectangles with sides parallel to the coordinate axes. For any two points $P, Q$ in $S$, $R_{P Q}$ denotes the rectangle with $P Q$ as its diagonal, and $W_{P Q}$ denotes the number of points in $S$ within the rectangle $R_{P Q}$. Here, $P$ and $Q$ do not have to be distinct; if $P$ and $Q$ are the same, then there are no points in the rectangle $R_{P Q}$, and we denote $W_{P Q}=0$.
We will prove that the maximum value of $N$ is 400.
(1) Regardless of how the 2002 points in $S$ are placed, there always exist $P, Q \in S$ such that $W_{P Q} \geqslant 400$;
(2) If the 2002 points in $S$ are placed in a certain way, then for $\forall P, Q \in S$, we have $W_{P Q} \leqslant 400$.
First, we prove (1). Given a point set $S$ with 2002 points, let the 4 points in $S$ with the smallest x-coordinate, smallest y-coordinate, largest x-coordinate, and largest y-coordinate be denoted as $A, B, C, D$, respectively. These 4 points may coincide with each other.
Thus, we can assert that any point in $S$ other than $A, B, C, D$ (at least 1998 points) is contained in at least one of the rectangles $R_{A B}, R_{B C}, R_{C D}, R_{D A}, R_{A C}$. This conclusion is derived from the following three cases.
(i) For a point $M$ with a y-coordinate less than those of $A$ and $C$. If the x-coordinate of $M$ is less than the x-coordinate of $B$, then $M$ is in $R_{A B}$; otherwise, $M$ is in $R_{B C}$.
(ii) For a point $M$ with a y-coordinate greater than those of $A$ and $C$. If its x-coordinate is less than the x-coordinate of $D$, then $M$ is in $R_{D A}$; otherwise, $M$ is in $R_{C D}$.
(iii) For a point with a y-coordinate between those of $A$ and $C$, this point is in $R_{A C}$.
This indicates that at least one of the 5 rectangles $R_{A B}, R_{B C}, R_{C D}, R_{D A}, R_{A C}$ contains more than 399 points in $S$. Otherwise, $S$ would have at most $399 \times 5 = 1995$ points other than $A, B, C, D$, which contradicts the fact that $S$ has 2002 points. Therefore, (1) is established.
Next, we prove (2).
Place the 2002 points in $S$ as shown in Figure 7, divided into 5 groups of points: $E, F, G, H, I$. Then we have
(i) For any rectangle $R_{P Q} (P, Q \in S)$, it contains at most one point group;
(ii) This point group is either $I$ or a group that includes at least one of $P, Q$.
This means that a rectangle with a diagonal formed by two points in $S$ contains at most 400 points in $S$.
(1) and (2) are thus proven. Therefore, the maximum value of $N$ is 400.
|
400
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 1, the area of square $A B C D$ is 256, point $F$ is on $A D$, and point $E$ is on the extension of $A B$. The area of right triangle $\triangle C E F$ is 200. Then the length of $B E$ is $(\quad)$.
(A) 10
(B) 11
(C) 12
(D) 15
|
2. (C).
It is easy to prove that $\mathrm{Rt} \triangle C D F \cong \mathrm{Rt} \triangle C B E$. Therefore, $C F=C E$.
Since the area of Rt $\triangle C E F$ is 200, i.e., $\frac{1}{2} \cdot C F \cdot C E=200$, hence $C E=20$.
And $S_{\text {square } B C D}=B C^{2}=256$, so $B C=16$.
By the Pythagorean theorem, $B E=\sqrt{C E^{2}-B C^{2}}=12$.
|
12
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) In a square with an area of 1, construct a smaller square as follows: divide each side of the unit square into $n$ equal parts, then connect each vertex to the nearest division point of its opposite vertex, as shown in Figure 5. If the area of the smaller square is exactly $\frac{1}{3281}$, find the value of $n$.
|
I. Draw $C_{1} P \perp A_{1} C$, with the foot of the perpendicular being $P$. Since the area of the square $A B C D$ is 1, we have
$$
A B=B C=C D=1, A_{1} B=\frac{n-1}{n}, C C_{1}=\frac{1}{n} .
$$
By the Pythagorean theorem,
$$
A_{1} C=\sqrt{B C^{2}+A_{1} B^{2}}=\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}} .
$$
It is easy to see that $\mathrm{Rt} \triangle A_{1} B C \backsim \mathrm{Rt} \triangle C P C_{1}$, so
$$
\frac{C_{1} P}{B C}=\frac{C C_{1}}{A_{1} C} \text {. }
$$
Thus, $C_{1} P=\frac{B C \cdot C C_{1}}{A_{1} C}=\frac{1 \times \frac{1}{n}}{\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}}}$
$$
=\frac{1}{\sqrt{2 n^{2}-2 n+1}} \text {. }
$$
Since the area of the small shaded square is $\frac{1}{3281}$, we have
$$
C_{1} P^{2}=\frac{1}{2 n^{2}-2 n+1}=\frac{1}{3281} \text {. }
$$
Simplifying and factoring, we get
$$
(n-41)(n+40)=0 \text {, }
$$
Given $n>0$, we find $n=41$.
|
41
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the sequence of positive integers, starting from 1, certain numbers are painted red according to the following rules. First, paint 1, then paint 2 even numbers $2, 4$; then paint the 3 nearest consecutive odd numbers after 4, which are $5, 7, 9$; then paint the 4 nearest consecutive even numbers after 9, which are $10, 12, 14, 16$; then paint the 5 nearest consecutive odd numbers after 16, which are $17, 19, 21, 23, 25$. Continue this process to get a red subsequence $1, 2, 4, 5, 7, 9, 10, 12, 14, 16, 17, \cdots$. In this red subsequence, the 2003rd number starting from 1 is ( ).
(A) 3844 (B) 3943 (C) 3945 (D) 4006
|
6. (B).
The first time, 1 number is painted red: $1,1=1^{2}$;
The second time, 2 numbers are painted red: $2,4,4=2^{2}$;
The third time, 3 numbers are painted red: $5,7,9,9=3^{2}$.
It is easy to see that the last number painted red in the $k$-th time is $k^{2}$, then the $k+1$ numbers painted red in the $k+1$-th time are:
$$
k^{2}+1, k^{2}+3, k^{2}+5, \cdots, k^{2}+2 k-1, k^{2}+2 k+1,
$$
where the last number is $(k+1)^{2}$. According to this rule, the first $k$ segments have a total of
$$
1+2+3+\cdots+(k-1)+k=\frac{k(k+1)}{2}
$$
numbers painted red.
Solving the inequality $\frac{k(k+1)}{2} \leqslant 2003$, we get $k \leqslant 62$.
At this point, the first 62 segments have a total of $\frac{62 \times 63}{2}=1953$ numbers, where the last number in the 62nd segment is $62^{2}=3844$, which is the 1953rd red number. The 2003rd red number should be in the 63rd segment, and all numbers in the 63rd segment are odd.
Since $2003-1953=50$, the 2003rd red number is
$$
3844+1+(50-1) \times 2=3943 .
$$
|
3943
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure $11, \angle A O B=$ $30^{\circ}, \angle A O B$ contains a fixed point $P$, and $O P=10, O A$ has a point $Q, O B$ has a fixed point $R$. If the perimeter of $\triangle P Q R$ is minimized, find its minimum value.
|
(Tip: Draw auxiliary lines with $O A$ and $O B$ as axes of symmetry. Answer:
10.)
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let the function $f_{0}(x)=|x|, f_{1}(x)=$ $\left|f_{0}(x)-1\right|, f_{2}(x)=\left|f_{1}(x)-2\right|$. Then the area of the closed figure formed by the graph of $y$ $=f_{2}(x)$ and the $x$-axis is $\qquad$
(1989, National High School Mathematics Competition)
|
Analysis: If it is not easy to directly draw the graph of $y=f_{2}(x)$, but we know the relationship between the graphs of $y=f(x)$ and $y=|f(x)|$. If we follow the sequence
$$
\begin{array}{l}
f_{0}(x)=|x| \rightarrow y=f_{0}(x)-1 \\
\rightarrow f_{1}(x)=\left|f_{0}(x)-1\right| \rightarrow y=f_{1}(x)-2 \\
\rightarrow f_{2}(x)=\left|f_{1}(x)-2\right|
\end{array}
$$
to perform graphical transformations, it is easier to draw the graph of $y=f_{2}(x)$.
It is easy to get the answer as 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 The sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{2 n}, a_{2 n+1}$ forms an arithmetic sequence, and the sum of the terms with odd indices is 60, while the sum of the terms with even indices is 45. Then the number of terms $n=$ $\qquad$
|
Analysis: From $\left\{\begin{array}{l}a_{1}+a_{3}+\cdots+a_{2 n+1}=60, \\ a_{2}+a_{4}+\cdots+a_{2 n}=45\end{array}\right.$ $\Rightarrow\left\{\begin{array}{l}(n+1) a_{n+1}=60, \\ n a_{n+1}=45\end{array} \Rightarrow \frac{n+1}{n}=\frac{4}{3}\right.$, solving gives $n=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Given the three side lengths $a$, $b$, and $c$ of $\triangle ABC$ satisfy: (1) $a>b>c$, (2) $2b=a+c$, (3) $b$ is an integer, (4) $a^{2}+b^{2}+c^{2}=84$. Then the value of $b$ is $\qquad$
|
Analysis: Starting from $2 b=a+c$, let $a=b+d$, $c=b-d(d>0)$, then
$$
(b+d)^{2}+b^{2}+(b-d)^{2}=84 .
$$
That is, $3 b^{2}+2 d^{2}=84$.
Obviously, $2 d^{2}$ is a multiple of 3. Also, by $b+c>a \Rightarrow$ $b>2 d$, substituting into (4) gives $2 d^{2}<12$. Therefore, $2 d^{2}=3,6,9$. Verification shows $b=5$.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 13 On the coordinate plane, points with both integer horizontal and vertical coordinates are called integer points. For any natural number $n$, connect the origin $O$ with $A_{n}(n, n+3)$, and let $f(n)$ denote the number of integer points on the line segment $O A_{n}$, excluding the endpoints. Then
$$
f(1)+f(2)+\cdots+f(2001)=
$$
$\qquad$
(Adapted from the 1990 National High School Mathematics Competition)
|
Analysis: It is easy to obtain the equation of $O A_{n}$ as
$$
y=\frac{n+3}{n} x=x+\frac{3}{n} x \quad(0 \leqslant x \leqslant n) .
$$
Obviously, for integer points, $n$ must be a multiple of 3. Let $n=3 k$, then
$$
y=\frac{k+1}{k} x \text {. }
$$
When $x=k$ and $2 k$, we get the integer points $(k, k+1)$, $(2 k, 2(k+1))$.
Therefore, $f(1)+f(2)+\cdots+f(2001)=1334$.
|
1334
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 14 Let $f(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are constants. If $f(1)=10$, $f(2)=20$, $f(3)=30$, then $f(10)+f(-6)=$ $\qquad$
(1998, Zhongshan City, Guangdong Province Mathematics Competition)
|
Analysis: Based on $f(1)=10, f(2)=20, f(3)=$ 30, construct
$$
\begin{array}{c}
f(x)=(x-1)(x-2)(x-3)(x-t)+10 x . \\
\text { Hence } f(10)+f(-6)=9 \times 8 \times 7(10-t)+ \\
10 \times 10+[(-7) \times(-8) \times(-9)(-6-t)- \\
60]=8104 .
\end{array}
$$
|
8104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 19 Let $f(x)=|1-2 x|, x \in[0,1]$. Then, the number of solutions to the equation $f\{f[f(x)]\}=\frac{1}{2} x$ is $\qquad$ .
|
Analysis: Let $y=f(x), z=f(y), w=f(z)$, use β $\rightarrow$ β to indicate the change, we have
$$
\begin{array}{l}
x: 0 \rightarrow 1, y: 1 \rightarrow 0 \rightarrow 1, \\
z: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1, \\
w: 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 \rightarrow 0 \rightarrow 1 .
\end{array}
$$
Since the β $\rightarrow$ β change is linear and occurs within the non-negative real number range, the number of intersections between the graph of $w=f\{f[f(x)]\}$ and $w=\frac{x}{2}$ is 8, which means the number of solutions is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 As shown in Figure 4, in rectangle $A B C D$, $A B=$ $20 \text{ cm}, B C=10 \text{ cm}$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition)
|
Solution: As shown in Figure 4, construct $\angle B_{1} A C = \angle B A C$, draw $B Q \perp A C$ at $Q$, and extend $B Q$ to intersect $A B_{1}$ at $B_{1}$; draw $N P \perp A C$ at $P$, and extend $N P$ to intersect $A B_{1}$ at $N_{1}$, then points $B_{1}$ and $B$, and points $N_{1}$ and $N$ are symmetric with respect to $A C$. Draw $B H \perp A B_{1}$ at $H$.
It is easy to see that $\triangle M P N \cong \triangle M P N_{1}$.
Thus, $M N_{1} = M N$.
Therefore, $B M + M N = B M + M N_{1}$
$$
\geqslant B N_{1} \geqslant B H \text {. }
$$
Hence, the minimum value of $B M + M N$ is the distance from point $B$ to $A B_{1}$.
In the right triangle $\triangle A B C$,
$$
B Q = \frac{A B \cdot B C}{A C} = 4 \sqrt{5}, B B_{1} = 2 B Q = 8 \sqrt{5} \text {. }
$$
In the right triangle $\triangle A B Q$,
$$
A Q = \sqrt{A B^{2} - B Q^{2}} = 8 \sqrt{5} \text {. }
$$
In $\triangle A B B_{1}$,
$$
B H = \frac{A Q \cdot B B_{1}}{A B_{1}} = \frac{8 \sqrt{5} \times 8 \sqrt{5}}{20} = 16 \text {. }
$$
Therefore, the minimum value of $B M + M N$ is $16 \text{ cm}$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 If $x \in \mathbf{R}$, find the maximum value of $F(x)=\min \{2 x+1$, $x+2,-x+6\}$.
(38th AHSME)
|
Solution: As shown in Figure 1, draw the graph of $F(x)$ (the solid part). From the graph, we can see that the maximum value of $F(x)$ is equal to the y-coordinate of the intersection point of $y=x+2$ and $y=-x+6$.
Therefore, the maximum value of $F(x)$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given positive numbers $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}$, $\cdots, b_{n}$ satisfying
$$
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1 .
$$
Find the maximum value of $F=\min \left\{\frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \cdots, \frac{a_{n}}{b_{n}}\right\}$.
(1979, Guangdong Province High School Mathematics Competition)
|
Solution: It is easy to see that when all the letters are equal, the value of $F$ is 1.
Below we prove that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$; $b_{1}, b_{2}, \cdots, b_{n}$, we have $F \leqslant 1$.
If not, then $F>1$.
$$
\text { Hence } \frac{a_{1}}{b_{1}}>1, \frac{a_{2}}{b_{2}}>1, \cdots, \frac{a_{n}}{b_{n}}>1 \text {, }
$$
which implies $a_{1}^{2}>b_{1}^{2}, a_{2}^{2}>b_{2}^{2}, \cdots, a_{n}^{2}>b_{n}^{2}$.
Thus, $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}>b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}$, contradicting the given condition.
Therefore, the maximum value of $F$ is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If real numbers $x, y, z$ satisfy $x+\frac{1}{y}=4, y+\frac{1}{z}=1, z+$ $\frac{1}{x}=\frac{7}{3}$, then the value of $x y z$ is $\qquad$.
|
7.1 .
Since $4=x+\frac{1}{y}=x+\frac{1}{1-\frac{1}{z}}=x+\frac{z}{z-1}$
$=x+\frac{\frac{7}{3}-\frac{1}{x}}{\frac{7}{3}-\frac{1}{x}-1}=x+\frac{7 x-3}{4 x-3}$, then
$4(4 x-3)=x(4 x-3)+7 x-3$,
which simplifies to $(2 x-3)^{2}=0$.
Thus, $x=\frac{3}{2}$.
Therefore, $z=\frac{7}{3}-\frac{1}{x}=\frac{5}{3}, y=1-\frac{1}{z}=\frac{2}{5}$.
Hence, $x y z=\frac{3}{2} \times \frac{2}{5} \times \frac{5}{3}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given the quadratic function $y=a x^{2}+b x+c$ (where $a$ is a positive integer) whose graph passes through the points $A(-1,4)$ and $B(2,1)$, and intersects the $x$-axis at two distinct points. Then the maximum value of $b+c$ is $\qquad$ .
|
10. -4 .
From the given, we have $\left\{\begin{array}{l}a-b+c=4, \\ 4 a+2 b+c=1,\end{array}\right.$ solving this yields $\left\{\begin{array}{l}b=-a-1, \\ c=3-2 a .\end{array}\right.$
Since the graph of the quadratic function intersects the $x$-axis at two distinct points, we have,
$$
\begin{array}{l}
\Delta=b^{2}-4 a c>0, \\
(-a-1)^{2}-4 a(3-2 a)>0,
\end{array}
$$
which simplifies to $(9 a-1)(a-1)>0$.
Since $a$ is a positive integer, we have $a>1$, so $a \geqslant 2$.
Since $b+c=-3 a+2 \leqslant-4$, and when $a=2, b=$ $-3, c=-1$, the conditions are satisfied, hence the maximum value of $b+c$ is -4.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
|
14. (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$. From the problem, we know $a>0$, and $b+c=2-a, bc=\frac{4}{a}$. Therefore, $b, c$ are the two real roots of the quadratic equation $x^{2}-(2-a)x+\frac{4}{a}=0$, then
$$
\begin{array}{l}
\Delta=(2-a)^{2}-4 \times \frac{4}{a} \geqslant 0, \\
a^{3}-4 a^{2}+4 a-16 \geqslant 0, \\
\left(a^{2}+4\right)(a-4) \geqslant 0 .
\end{array}
$$
Thus, $a \geqslant 4$.
When $a=4, b=c=-1$, the conditions are satisfied.
Therefore, the minimum value of the maximum among $a, b, c$ is 4.
(2) Since $abc>0$, $a, b, c$ are either all positive or one positive and two negative.
(i) If $a, b, c$ are all positive, then from (1), the maximum of $a, b, c$ is not less than 4, which contradicts $a+b+c=2$.
(ii) If $a, b, c$ are one positive and two negative, let $a>0, b<0, c<0$, then
$$
\begin{array}{l}
|a|+|b|+|c|=a-b-c \\
=a-(2-a)=2a-2 .
\end{array}
$$
From (1), we know $a \geqslant 4$, hence $2a-2 \geqslant 6$.
When $a=4, b=c=-1$, the conditions are satisfied and the equality holds. Therefore, the minimum value of $|a|+|b|+|c|$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a_{1}, a_{2}, \cdots, a_{2002}$ have values of 1 or -1, let $S$ be the sum of the pairwise products of these 2002 numbers.
(1) Find the maximum and minimum values of $S$, and specify the conditions under which these maximum and minimum values can be achieved;
(2) Find the smallest positive value of $S$, and specify the conditions under which this smallest positive value can be achieved.
|
$$
\begin{array}{l}
\text { 3. (1) }\left(a_{1}+a_{2}+\cdots+a_{200}\right)^{2} \\
=a_{1}^{2}+a_{2}^{2}+\cdots+a_{2002}^{2}+2 m=2002+2 m, \\
m=\frac{\left(a_{1}+a_{2}+\cdots+a_{200}\right)^{2}-2002}{2} .
\end{array}
$$
When $a_{1}=a_{2}=\cdots=a_{200}=1$ or -1, $m$ reaches its maximum value of 2003001.
When $a_{1}, a_{2}, \cdots, a_{2002}$ contain exactly 1001 1s and 1001 -1s, $m$ reaches its minimum value of -1001.
(2) Since the smallest perfect square greater than 2002 is $45^{2}=$ 2025, and $a_{1}+a_{2}+\cdots+a_{20 \mathrm{~m}}$ must be even, therefore, when
$$
a_{1}+a_{2}+\cdots+a_{2002}=46 \text { or }-46 \text {; }
$$
i.e., when $a_{1}, a_{2}, \cdots, a_{2002}$ contain exactly 1024 1s and 978 -1s or exactly 1024 -1s and 978 1s, $m$ reaches its minimum value
$$
\frac{1}{2}\left(46^{2}-2002\right)=57 .
$$
|
57
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $0 \leqslant a-b \leqslant 1,1 \leqslant a+b \leqslant 4$. Then, when $a-2 b$ reaches its maximum value, the value of $8 a+2002 b$ is $\qquad$ .
|
3.8.
$$
\begin{array}{l}
\text { Let } 0 \leqslant a-b \leqslant 1, \\
1 \leqslant a+b \leqslant 4, \\
\text { and } m(a-b)+n(a+b)=a-2 b .
\end{array}
$$
By comparing the coefficients of $a$ and $b$ on both sides, we get the system of equations and solve to find
$$
m=\frac{3}{2}, n=-\frac{1}{2} \text {. }
$$
Thus, $a-2 b=\frac{3}{2}(a-b)-\frac{1}{2}(a+b)$.
From (1) and (2), we have $-2 \leqslant a-2 b \leqslant 1$.
Therefore, the maximum value of $a-2 b$ is 1, at which point $b=\frac{a-1}{2}$.
Substituting into (1) and (2) gives $0 \leqslant a \leqslant 1, 1 \leqslant a \leqslant 3$.
This implies $a=1, b=0$. Hence, $8 a+2002 b=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If a positive integer is equal to 4 times the sum of its digits, then we call this positive integer a quadnumber. The sum of all quadnumbers is $\qquad$ .
|
4.120.
Case analysis.
There are no four-composite numbers that are single-digit numbers.
For two-digit four-composite numbers, they satisfy
$$
\overline{a b}=4(a+b) \text {. }
$$
Thus, the four-composite numbers can be found to be $12, 24, 36, 48$, and their total sum is 120. For three-digit four-composite numbers, they satisfy
$$
100 a+10 b+c=4 a+4 b+4 c,
$$
which has no solution. This indicates that there are no three-digit four-composite numbers. Similarly, it can be analyzed that there are no four-composite numbers with more than three digits.
|
120
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $A$ be a $k$-element subset of the set $\{1,2,3, \cdots, 16\}$, and any two subsets of $A$ have different sums of elements. For any $(k+1)$-element subset $B$ of the set $\{1,2,3, \cdots, 16\}$ that contains $A$, there exist two subsets of $B$ whose sums of elements are equal.
(1) Prove: $k \leqslant 5$;
(2) Find the maximum and minimum values of the sum of elements of the set $A$.
(2002, Bulgarian Winter Mathematical Competition)
|
Proof: (1) Since $A$ has $2^{k}$ subsets (including the empty set), and the sum of elements of any two subsets is not equal, the sum of elements of $A$ must have at least $2^{k}-1$ different values.
If $k \geqslant 7$, then $2^{k}-1>16 k$, which is impossible.
If $k=6$, consider the one-, two-, three-, and four-element subsets of $A$, there are $C_{6}^{1}+C_{6}^{2}+C_{6}^{3}+C_{6}^{4}=56$ different sums, and the smallest sum is 1, the largest sum is $16+15+14+12=57$ (where $16+13=15+14$, so these 4 numbers cannot all be in $A$).
If $1 \in A$, then the largest sum is $16+14+12+9=51$ (where $16=15+1$, so these 3 numbers cannot all be in $A$; $14=13+1$, so these 3 numbers cannot all be in $A$; $12=11+1$, so these 3 numbers cannot all be in $A$; $16+10=14+12$, so these 4 numbers cannot all be in $A$), there are at most 51 different sums, which is a contradiction.
If $2 \in A$, then the largest sum is $16+15+12+9=52$, there are at most 51 different sums, which is a contradiction.
If $1 \notin A, 2 \notin A$, then there are at most 55 sums in $[3,57]$, which is a contradiction.
Therefore, $k \leqslant 5$.
Solution: (2) If the sum of elements of $A=\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ is less than 16, then the set $\left\{a_{1}, a_{2}, \cdots, a_{k}, 16\right\}$ has no subsets with equal sums, which contradicts the definition of $A$. $A=\{1,2,4,9\}$ satisfies the condition, and the sum of its elements 16 is the smallest.
If $16 \notin A$, then $15+14+13+11+8=61$.
If $16 \in A, 15 \notin A$, then $16+14+13+12+8=63$;
If $15 \in A, 14 \notin A$, then $16+15+13+11+7=62$.
Assume $\{14,15,16\} \subset A$, then $13 \notin A$.
If $12 \notin A$, then $16+15+14+11+8=64$.
If $12 \in A$. Since $11+16=12+15,10+16=12+14$, assume $9 \in A$, therefore, the set $A=\{9,12,14,15,16\}$ satisfies the condition, and the sum of its elements 66 is the largest.
|
66
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. A $9 \times 9$ grid of squares is colored in two colors, black and white, such that the number of black squares adjacent to each white square is greater than the number of white squares, and the number of white squares adjacent to each black square is greater than the number of black squares (squares sharing a common edge are considered adjacent). Find the maximum difference in the number of black and white squares.
(53rd Belarusian Mathematical Olympiad (Final D Class))
|
Solution: A coloring scheme of a square matrix that satisfies the problem's conditions is called a good scheme. For a good coloring scheme, it is easy to prove the following conclusions:
(1) If two adjacent cells have the same color and are in different rows (or columns), then any two adjacent cells in these two rows (columns) have the same color, and these colors alternate between black and white (as shown in Figure 2).
(2) Three consecutive cells cannot have the same color.
(3) If a corner cell is white (black), then the adjacent cell is black (white).
(4) If there are two adjacent cells of the same color in a certain column (row), then it is impossible to have two adjacent cells of the same color in any other row (column).
From these conclusions, we can deduce that if there are no two adjacent cells of the same color in a certain column (row), then the colors of the other cells in the matrix depend solely on the coloring of this column (row). Therefore, we only need to check the color distribution of the first column (row). It is easy to see that if the matrix is colored like a chessboard, the maximum difference between the number of black and white cells is 1.
Consider the case where there are two adjacent cells of the same color in the first row. From (1) to (4), we can deduce that the number of black and white cells in any two adjacent rows is the same, so the difference between the number of black and white cells is equal to the difference in the first row. It is easy to see that the difference in the number of black and white cells in the first row is 3 (as shown in Figure 3).
When there are two adjacent cells of the same color in the first column, we can draw the same conclusion. From (4), we know that any good coloring scheme cannot have two adjacent cells of the same color in both the first row and the first column simultaneously.
In summary, the maximum difference between the number of black and white cells is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12.12 People are sitting around a table in a circle. How many ways are there for 6 pairs of people to shake hands so that no arms cross?
(2001, British Mathematical Olympiad (Round 1))
|
Solution: According to the problem, handshakes only occur on the table, not behind other people.
Let $N_{p}$ denote the total number of ways for $2p$ people sitting around a circular table to shake hands without crossing arms, and each person shakes hands only once. Consider each person as a point. Without loss of generality, let the point where Arthur is located be denoted as A. Since handshakes occur in pairs, the number of people on either side of Arthur and the person he shakes hands with must be 0 or even.
If there are 2 people, there is only one way to shake hands, so $N_{1}=1$.
If there are 4 people, Arthur can shake hands with his adjacent person (such as B or D in Figure 4), but not with the opposite person C, otherwise, it would cause a crossing. Therefore, $N_{2}=2$.
If there are 6 people, Arthur can shake hands with his adjacent person (such as B or F in Figure 5), and the remaining 4 people have $N_{2}$ ways to shake hands. Arthur can also shake hands with the opposite person D, and on both sides of AD, there are 2 people, BC and FE each have $N_{1}$ ways to shake hands. Therefore, $N_{3}=2 N_{2}+N_{1}^{2}=5$.
Using this method, we can gradually find $N_{6}$.
When there are 8 people, Arthur can shake hands with his adjacent person (such as B or H in Figure 6). The remaining 6 people on one side have $N_{3}$ ways to shake hands. Arthur can also shake hands with D or F, then the 2 people on one side have $N_{1}$ ways to shake hands, and the 4 people on the other side have $N_{2}$ ways to shake hands. Therefore, $N_{4}=2 N_{3}+2 N_{1} \times N_{2}=14$.
Similarly, we get
$$
\begin{array}{l}
N_{5}=2 N_{4}+2 N_{3} \times N_{1}+N_{2}^{2}=42, \\
N_{6}=2 N_{5}+2 N_{4} \times N_{1}+2 N_{3} \times N_{2}=132 .
\end{array}
$$
Therefore, the solution to this problem is 132.
Note: The sequence $1,2,5,14,42,132, \cdots$ is the famous Catalan number. If we define $N_{0}=1$, then the general term of this sequence is
$$
N_{k}=\sum_{r=0}^{k-1}\left(N, \times N_{k-r-1}\right) .
$$
|
132
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. Given a wire of length $150 \mathrm{~cm}$, it is to be cut into $n(n>2)$ smaller segments, each of which has an integer length of no less than $1(\mathrm{~cm})$. If no three segments can form a triangle, find the maximum value of $n$, and how many ways there are to cut the wire into $n$ segments that satisfy the condition.
|
17. Since the sum of $n$ segments is a fixed value of $150(\mathrm{~cm})$, to make $n$ as large as possible, the length of each segment must be as small as possible. Given that the length of each segment is no less than $1(\mathrm{~cm})$, and no three segments can form a triangle, the lengths of these segments can only be $1,1,2,3,5,8$, $13,21,34,55,89, \cdots$.
But $1+1+2+\cdots+34+55=143<150$,
hence the maximum value of $n$ is 10.
There are 7 ways to divide a wire of length $150(\mathrm{~cm})$ into 10 segments that meet the conditions:
$$
\begin{array}{l}
1,1,2,3,5,8,13,21,34,62 ; \\
1,1,2,3,5,8,13,21,35,61 ; \\
1,1,2,3,5,8,13,21,36,60 ; \\
1,1,2,3,5,8,13,21,37,59 ; \\
1,1,2,3,5,8,13,22,35,60 ; \\
1,1,2,3,5,8,13,22,36,59 ; \\
1,1,2,5,8,14,22,36,58
\end{array}
$$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For all positive integers $n$ greater than 2, the greatest common divisor of the number $n^{5}-5 n^{3}+4 n$ is
untranslated part:
untranslated part remains the same as it is a mathematical expression.
|
4.120 .
$$
n^{5}-5 n^{3}+4 n=(n-2)(n-1) n(n+1)(n+2) \text {. }
$$
For any positive integer $n$ greater than 2, the number $n^{5}-5 n^{3}+4 n$ contains the common divisor $1 \times 2 \times 3 \times 4 \times 5=120$. Therefore, the greatest common divisor of these numbers is 120.
|
120
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (20 points) A scientific expedition team went to the upper reaches of a certain river to investigate an ecological area. After setting out, they advanced at a speed of $17 \mathrm{~km}$ per day, traveling upstream along the riverbank for several days before reaching their destination. They then spent several days investigating the ecological area. After completing their mission, they returned at a speed of 25 $\mathrm{km}$ per day. On the 60th day after departure, the expedition team traveled $24 \mathrm{~km}$ and returned to the starting point. How many days did the scientific expedition team spend investigating the ecological area?
|
Let the expedition team took $x$ days to reach the ecological area, $y$ days to return, and spent $z$ days on the investigation. Then,
$$
x+y+z=60,
$$
and $17 x-25 y=-1$ or $25 y-17 x=1$.
(Here $x, y$ are positive integers)
First, find a special solution $\left(x_{0}, y_{0}\right)$ of equation (1) (here $x_{0}, y_{0}$ can be negative integers). Using the Euclidean algorithm, we have
$$
25=1 \times 17+8,17=2 \times 8+1 \text {. }
$$
Thus, $1=17-2 \times 8=17-2 \times(25-17)$
$$
=3 \times 17-2 \times 25 \text {. }
$$
Comparing with the left side of equation (1), we get
$$
x_{0}=-3, y_{0}=-2 \text {. }
$$
Next, find the solution of equation (1) that meets the problem's requirements.
It is easy to see that all integer solutions of equation (1) can be expressed as
$$
x=-3+25 t, y=-2+17 t \text {. }
$$
Then $x+y=42 t-5, t$ is an integer.
According to the problem, $0<x+y<60$, so only when $t=1$ does it meet the requirements, at this time $x+y=42-5=37$.
Thus, $z=60-(x+y)=23$.
|
23
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Mr. Huang's home phone number is an eight-digit number. The sum of the number formed by the first four digits and the number formed by the last four digits is 14405. The sum of the number formed by the first three digits and the number formed by the last five digits is 16970. Mr. Huang's home phone number is
|
14. 82616144 .
Let the phone number be $100000 x+10000 y+z$, where $x$, $y$, and $z$ are all natural numbers, and $100 \leqslant x \leqslant 999, 0 \leqslant y \leqslant 9, 1000 \leqslant z \leqslant 9999$. Then
$$
\left\{\begin{array}{l}
10 x+y+z=14405, \\
x+10000 y+z=16970
\end{array} \Rightarrow 1111 y-x=285 .\right.
$$
From $100 \leqslant x \leqslant 999, y \geqslant 0$,
we get $y=1, x=826, z=6144$.
|
82616144
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. A bouncy ball falls from point $A$ to the ground, bounces up to point $B$, then falls to a platform $20 \mathrm{~cm}$ high, bounces up to point $C$, and finally falls to the ground. Each time it bounces, the height it reaches is $80 \%$ of the height from which it fell. It is known that point $A$ is $68 \mathrm{~cm}$ higher above the ground than point $C$. Find the height of point $C$ above the ground.
|
Three, 15. Let the height of point $C$ above the ground be $x \mathrm{~cm}$, then $\frac{\frac{x-20}{80 \%}+20}{80 \%}-68=x$. Solving this, we get $x=132$. Answer:η₯.
Note: The word "η₯" at the end is not translated as it seems to be a placeholder or abbreviation in the original text, possibly indicating that the detailed steps or further explanation are omitted. If you need a specific translation for "η₯", please provide more context.
|
132
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. The teachers and students of a township primary school went to the county town for a visit. It was stipulated that the bus would depart from the county town and arrive at the school at 7:00 AM to pick up the visiting teachers and students and immediately head to the county town. However, the bus broke down on its way to the school and had to stop for repairs. The teachers and students at the school waited until 7:10 AM but still did not see the bus, so they started walking towards the county town. On their way, they met the repaired bus, immediately got on, and headed to the county town, arriving 30 minutes later than the originally scheduled time. If the speed of the bus is 6 times the walking speed, how long did the bus spend on the road for repairs?
|
15. Suppose it took \( x \) minutes to fix the car. As shown in Figure 8, let point \( A \) be the location of the county town, point \( C \) be the location of the school, and point \( B \) be the place where the teachers and students met the car on the way.
In the 30 minutes that the teachers and students were late to the county town, 10 minutes were due to the late departure, and another 20 minutes were due to walking from \( C \) to \( B \) instead of taking the car.
The 30 minutes that the car was late were partly due to the \( x \) minutes spent fixing the car, and partly due to the time saved by not making a round trip between \( B \) and \( C \).
Given that the car's speed is 6 times the walking speed, and that walking from \( C \) to \( B \) takes 20 minutes longer than taking the car, it follows that the car should take \(\frac{20}{6-1}=4\) minutes to travel from \( C \) to \( B \). A round trip saves 8 minutes.
Therefore, \( x - 8 = 30 \), so \( x = 38 \).
That is, it took 38 minutes to fix the car on the way.
|
38
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $a$ and $b$ are both prime numbers, and $a^{2}+b=2003$, then the value of $a+b$ is ( ).
(A) 1999
(B) 2000
(C) 2001
(D) 2002
|
$-γ 1 .(\mathrm{C})$.
Since 2003 is an odd number, one of $a$ or $b$ must be the even prime number 2. If $b=2$, then $a^{2}=2001$, but 2001 is not a perfect square. Therefore, $a=2, b=1999$, and we have $a+b=2001$.
|
2001
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. The three sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $b+c=$ $8, bc=a^{2}-12a+52$. Then the perimeter of $\triangle A B C$ is ( ).
(A) 10
(B) 14
(C) 16
(D) cannot be determined
|
3. (B).
Since $(b-c)^{2}=64-4\left(a^{2}-12 a+52\right)$
$$
=-4(a-6)^{2} \geqslant 0 \text {, }
$$
Therefore, $a=6$, and $b=c$.
Also, $b+c=8$, then $b=c=4$. Thus, $a+b+c=14$.
|
14
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$, $b$, and $c$ are real numbers, and the polynomial $x^{3}+$ $a x^{2}+b x+c$ can be divided by $x^{2}+3 x-4$. Then $2 a-2 b$ $-c=$ $\qquad$ .
|
$=γ 1.14$.
From the problem, -4 and 1 are roots of the equation $x^{3}+a x^{2}+b x+c=0$, which means
$$
\begin{array}{l}
16 a-4 b+c=64, \\
a+b+c=-1 .
\end{array}
$$
(1) $\times 6$ - (2) gives $2 a-2 b-c=14$.
|
14
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If positive integers $a$, $b$, and $c$ satisfy $a b + b c = 518$, $a b - a c = 360$, then the maximum value of $a b c$ is $\qquad$ .
|
2. 1008 .
From $ab+bc=518, ab-ac=360$, subtracting the two equations, we get $c(a+b)=2 \times 79$.
Upon verification, taking $c=2, a+b=79$, then $ab=518-bc=518-2b$.
Upon inspection, when $b=7$, $ab$ reaches its maximum value of 504. Therefore, the maximum value of $abc$ is 1008.
|
1008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $a b c=1, \frac{x}{1+a+a b}+\frac{x}{1+b+b c}+$ $\frac{x}{1+c+a c}=2003$, then $x=$ $\qquad$
|
3. 2003 .
Given $a b c=1$, we have
$$
\frac{1}{1+a+a b}+\frac{1}{1+b+b c}+\frac{1}{1+c+c a}=1,
$$
thus $x=2003$.
|
2003
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
δΏηζΊζζ¬ηζ’θ‘εζ ΌεΌοΌηΏ»θ―η»ζε¦δΈοΌ
```
Three, (25 points) Let $a$, $b$, $c$ be distinct integers from 1 to 9. Find the maximum possible value of $\frac{a+b+c}{a b c}$.
```
|
Three, let $P=\frac{a+b+c}{a b c}$.
In equation (1), let $a$ and $b$ remain unchanged, and only let $c$ vary, where $c$ can take any integer from 1 to 9.
Then, from $P=\frac{a+b+c}{a b c}=\frac{1}{a b}+\frac{a+b}{a b c}$, we know that
when $c=1$, $P$ reaches its maximum value, so $c=1$.
Thus, $P=\frac{a+b+1}{a b}=\frac{1}{a}+\frac{a+1}{a b}$.
In equation (2), let $a$ remain unchanged, and only let $b$ vary, where $b$ can take any integer from 2 to 9.
Then, from equation (2), we know that when $b=2$, $P$ reaches its maximum value, so $b=2$.
At this point, $P=\frac{1}{2}+\frac{3}{2 a}$.
In equation (3), $P$ reaches its maximum value when $a$ takes its minimum value, and $a$ can take any integer from 3 to 9, so $a=3$.
Therefore, when $a=3, b=2, c=1$ (the letters can be interchanged), $P$ reaches its maximum value of 1.
(Shen Xueming, Changqiao Middle School, Wuzhong District, Suzhou City, Jiangsu Province, 215128)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successful" fillings.
In a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)(n$ $\geqslant 2)$ grid, each cell is filled with 1 or -1. If the number in any cell is equal to the product of the numbers in the cells that share an edge with it, then this filling method is called "successful". Find the total number of "successful" fillings.
|
Three, assuming there exists some successful filling method that contains -1. First, prove: if this successful filling method is symmetric about the middle column (row), then the middle column (row) is entirely 1.
Let $a_{0}=1, a_{2}^{n}=1$.
If $a_{1}=1$, by $a_{1}=a_{0} \times a_{2} \times 1$, we get $a_{2}=1$.
Similarly, $a_{3}=1, \cdots$, $a_{2^{n}-1}=1$.
If $a_{2}^{n}-1=1$, similarly, we get $a_{1}=a_{2}=\cdots=a_{2}-2=1$.
Therefore, if $a_{1}=-1$, then $a_{2}^{n}-1=-1$. This way,
$$
\begin{array}{l}
a_{1}=-1, a_{2}=-1, a_{3}=1, a_{4}=-1, a_{5}=-1, a_{6} \\
=1, \cdots, a_{2}^{n}-2=-1, a_{2}^{n}-1=-1, a_{2}^{n}=1 \text {. Thus } 312^{2},
\end{array}
$$
a contradiction.
Therefore, the middle column (row) is entirely 1.
Secondly, if this successful filling method is not symmetric about the middle column, first rotate this filling method 180Β° along the middle column, to get another filling method, then multiply the numbers in the same positions of the two filling methods, to get a successful filling method symmetric about the middle column, which contains -1. This successful filling method can be further transformed into a successful filling method that is symmetric about both the middle row and the middle column, which contains -1.
For such a $\left(2^{n}-1\right) \times\left(2^{n}-1\right)$ successful filling method, removing all 1s in the middle row and the middle column, we can get 4 $\left(2^{n-1}-1\right) \times\left(2^{n-1}-1\right)$ successful filling methods, at least one of which contains -1. For this successful filling method, repeat the above operation, and eventually, we can get a $3 \times 3$ successful filling method, which contains -1, but the middle row and the middle column are all 1, which contradicts the definition of a successful filling method.
Therefore, for any successful filling method, it must not contain -1. Hence, there is only 1 successful filling method, which is to fill each cell with 1.
(Anqing No.1 High School, Anhui Province, 246004)
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. 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 $a c$ $=$ $\qquad$ .
|
-1 .
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right)$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1} γ x_{2}$ must have opposite signs. Thus, $x_{1} x_{2}=\frac{c}{a}<0$.
By the projection theorem, we know $|O C|^{2}=|A O| \cdot|B O|$, i.e., $c^{2}=\left|x_{1}\right| \cdot\left|x_{2}\right|=\left|\frac{c}{a}\right|$. Therefore, $|a c|=1, a c=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $m$ be an integer, and the two roots of the equation $3 x^{2}+m x-2=0$ are both greater than $-\frac{9}{5}$ and less than $\frac{3}{7}$. Then $m=$ $\qquad$ .
|
2.4 .
From the problem, we have
$$
\left\{\begin{array}{l}
3 \times\left(-\frac{9}{5}\right)^{2}+m \times\left(-\frac{9}{5}\right)-2>0, \\
3 \times\left(\frac{3}{7}\right)^{2}+m \times\left(\frac{3}{7}\right)-2>0 .
\end{array}\right.
$$
Solving this, we get \(3 \frac{8}{21}<m<4 \frac{13}{45}\). Therefore, \(m=4\).
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given positive integers $a$ and $b$ differ by 120, their least common multiple is 105 times their greatest common divisor. The larger of $a$ and $b$ is $\qquad$
|
4.225.
Let $(a, b)=d$, and $a=m d, b=n d$, where $m>n, m$ and $n$ are coprime. Thus, the least common multiple of $a$ and $b$ is $m n d$. According to the problem, we have
$$
\left\{\begin{array}{l}
m d-n d=120, \\
\frac{m n d}{d}=105,
\end{array}\right.
$$
which simplifies to
$$
\left\{\begin{array}{l}(m-n) d=2^{3} \times 3 \times 5, \\ m n=3 \times 5 \times 7 .\end{array}\right.
$$
Since $m>n$, from equation (2) we get
$$
\left\{\begin{array}{l}
m=105, \\
n=1;
\end{array} \quad \left\{\begin{array}{l}
m=35, \\
n=3;
\end{array} \quad \left\{\begin{array}{l}
m=21, \\
n=5;
\end{array} \quad \left\{\begin{array}{l}
m=15, \\
n=7 .
\end{array}\right.\right.\right.\right.
$$
According to equation (1), only $\left\{\begin{array}{l}m=15, \\ n=7 .\end{array}\right.$ is valid, and we can find $d=15$.
Therefore, the larger number is $m d=225$.
|
225
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given that the area of quadrilateral $ABCD$ is $32$, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is $16$.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
|
(1) As shown in Figure 7, let $AB = a$, $CD = b$, $AC = l$, and let the height from $A$ to $BC$ in $\triangle ABC$ be $h_1$, and the height from $D$ to $AC$ in $\triangle ADC$ be $h_2$. Then,
$$
\begin{array}{l}
S_{\text{quadrilateral } BCD} = S_{\triangle ABC} + S_{\triangle ADC} \\
= \frac{1}{2} (h_1 a + h_2 b) \leq \frac{1}{2} l(a + b),
\end{array}
$$
with equality holding if and only if $h_1 = h_2 = l$. That is, in quadrilateral $ABCD$, equality holds when $AC \perp AB$ and $AC \perp CD$.
From the given information, we have $64 \leq l(a + b)$.
Given that $a + b = 16 - l$, we get
$$
64 \leq l(16 - l) = 64 - (l - 8)^2 \leq 64.
$$
Thus, $l = 8$, $a + b = 8$, and at this point, $AC \perp AB$ and $AC \perp CD$. Therefore, there are the following 4 quadrilaterals:
$$
\begin{array}{l}
a = 1, b = 7, l = 8; a = 2, b = 6, l = 8; \\
a = 3, b = 5, l = 8; a = b = 4, l = 8.
\end{array}
$$
These are all trapezoids or parallelograms with $AC$ as the height.
(2) Given $AB = a$, $CD = 8 - a$, then
$$
BC^2 = 8^2 + a^2, \quad AD^2 = 8^2 + (8 - a)^2.
$$
Therefore, the sum of the squares of the sides of such a quadrilateral is
$$
2a^2 + 2(8 - a)^2 + 128 = 4(a - 4)^2 + 192.
$$
Thus, when $a = b = 4$, the sum of the squares is minimized, and it is 192.
|
192
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. An editor uses the digits $0 \sim 9$ to number the pages of a book. If a total of 636 digits were written, then the book has $\qquad$ pages.
|
$$
\text { Ni, 7. } 248 \text {. }
$$
Assuming the book has $x$ pages, then we have
$$
(x-99) \times 3+90 \times 2+9=636 \text {. }
$$
Solving for $x$ gives $x=248$.
|
248
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (18 points) On a plane, there are 7 points, and some line segments can be connected between them, so that any 3 points among the 7 points must have 2 points connected by a line segment. How many line segments are needed at least? Prove your conclusion.
|
(1) If one of the 7 points is isolated (i.e., it is not connected to any other points), then the remaining 6 points must be connected in pairs, which requires at least $\frac{6 \times 5}{2}=15$ lines.
(2) If one of the 7 points is connected to only one other point, then the remaining 5 points must be connected in pairs, which requires at least $1+\frac{5 \times 4}{2}=11$ lines.
(3) If each point is connected to at least 3 other points, then at least $\frac{7 \times 3}{2}$ lines are required. Since the number of lines must be an integer, at least 11 lines are required in this case.
(4) If each point is connected to at least 2 other points, and there is one point (denoted as $A$) that is connected to only 2 other points $A B$ and $A C$, then the 4 points not connected to $A$ must be connected in pairs, requiring $\frac{4 \times 3}{2}=6$ lines. The lines extending from $B$ must be at least 2, i.e., in addition to $B A$, there must be at least one more. Therefore, at least $6+2+1=9$ lines are required in this case.
Figure 6 shows a situation with 9 lines connected.
Combining (1) and (4), at least 9 lines are required to meet the requirements.
|
9
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. (16 points) Two vegetable bases, A and B, supply the same type of vegetables to three farmers' markets, $A$, $B$, and $C$. According to the signed contract, 45 t should be supplied to $A$, 75 t to $B$, and 40 t to $C$. Base A can arrange for 60 t, and Base B can arrange for 100 t. The distance (in km) between A and B and markets $A$, $B$, and $C$ is shown in Table 1. The transportation cost is 1 yuan/(kmΒ·t). How should the supply be arranged to minimize the total transportation cost? What is 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}
|
2. Let base B supply $x \text{t}$ to $A$, $y \text{t}$ to $B$, and $[100-(x+y)] \text{t}$ to $C$. Then base A supplies $(45-x) \text{t}$ to $A$, $(75-y) \text{t}$ to $B$, and $[40-(100-x-y)] \text{t} = [(x+y)-60] \text{t}$ to $C$. According to the problem, the total transportation cost is
$$
\begin{aligned}
W= & 10(45-x)+5(75-y)+6[(x+y)-60]+ \\
& 4 x+8 y+15[100-(x+y)] \\
= & 1965-3[2(x+y)+3 x] .
\end{aligned}
$$
Since $0 \leqslant x+y \leqslant 100, 0 \leqslant x \leqslant 45$, the minimum value of $W$ is achieved when and only when $x+y = 100, x=45$, then
$$
W_{\min }=1965-3(200+135)=960 \text{ (yuan). }
$$
|
960
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $\frac{3 x+4}{x^{2}-x-2}=\frac{A}{x-2}-\frac{B}{x+1}$, where $A$ and $B$ are constants. Then the value of $4 A-B$ is ( ).
(A) 7
(B) 9
(C) 13
(D) 5
|
2. (C).
From $\frac{A}{x-2}-\frac{B}{x+1}=\frac{(A-B) x+A+2 B}{x^{2}-x-2}$, we get $A-B=3, A+2 B=4$.
Thus, $4 A-B=3(A-B)+(A+2 B)=13$.
|
13
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
8. If 4 lines in a plane intersect each other pairwise and no three lines are concurrent, then there are $\qquad$ pairs of consecutive interior angles.
|
8.24.
Each line intersects with 3 other lines, forming 3 intersection points. Each 2 intersection points determine a line segment, making a total of 3 line segments. Since each line has a pair of consecutive interior angles on both sides, there are $3 \times 4=12$ line segments, and a total of 24 pairs of consecutive interior angles.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given
$$
\left(x+\sqrt{x^{2}+2002}\right)\left(y+\sqrt{y^{2}+2002}\right)=2002 .
$$
Then $x^{2}-3 x y-4 y^{2}-6 x-6 y+58=$ $\qquad$ .
|
11.58
$$
\begin{array}{l}
\text { Given }\left(x+\sqrt{x^{2}+2002}\right)\left(\sqrt{x^{2}+2002}-x\right) \\
=2002,
\end{array}
$$
we get $\sqrt{x^{2}+2002}-x=\sqrt{y^{2}+2002}+y$.
Similarly, $\sqrt{y^{2}+2002}-y=\sqrt{x^{2}+2002}+x$.
Adding these two equations gives $x+y=0$.
Therefore, $x^{2}-3 x y-4 y^{2}-6 x-6 y+58$
$$
=(x+y)(x-4 y)-6(x+y)+58=58 \text {. }
$$
|
58
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. As shown in Figure 4, line $AB$ intersects $\odot O$ at points $A$ and $B$, point $O$ is on $AB$, and point $C$ is on $\odot O$, with $\angle AOC=40^{\circ}$. Point $E$ is a moving point on line $AB$ (not coinciding with point $O$), and line $EC$ intersects $\odot O$ at another point $D$. The number of points $E$ that satisfy $DE = DO$ is $\qquad$ .
|
12.3.
Consider the position of point $E$: Point $E$ can be on the extension of line segment $O A$; Point $E$ can be on line segment $O A$ (excluding point $O$); Point $E$ can be on the extension of line segment $O B$; but point $E$ cannot be on line segment $O B$ (excluding point $O$). Therefore, point $E$ has a total of 3 positions.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. There are two equations:
Good + Good $=$ Wonderful, Wonderful $\times$ GoodGood $\times$ ReallyGood $=$ WonderfulProblemProblemWonderful, where each Chinese character represents a digit from $0 \sim 9$, the same character represents the same digit, and different characters represent different digits. Therefore, the number of all factors of the four-digit number "WonderfulProblemProblemWonderful" is $\qquad$
|
13.16.
From the addition formula, we get βε₯½β $<5$, βε¦β $\neq 0$, so βε₯½β $=1$, βε¦β $=2$ or βε₯½β $=2$, βε¦β $=4$ or βε₯½β $=3$, βε¦β $=6$ or βε₯½β $=4$, βε¦β $=8$. Clearly, the middle two scenarios do not satisfy the multiplication formula, so it can only be:
(1) βε₯½β $=1$, βε¦β $=2$, thus the multiplication formula becomes
$2 \times 11 \times($ η $\times 10+1)=2002+$ ι’ $\times 110$, i.e., η $\times 10+1=91+$ ι’ $\times 5$.
The left side $\leqslant 91$, the right side $\geqslant 91$, so both sides equal 91. Therefore, βηβ $=9$, βι’β $=0$, βε¦ι’ι’ε¦β $=2002$.
(2) βε₯½β $=4$, βε¦β $=8$, the multiplication formula is
$8 \times 44 \times($ η $\times 10+4)=8008+$ ι’ $\times 110$,
i.e., $704+1760 \times$ η $=4004+$ ι’ $\times 55$.
In the range $0 \sim 9$, only βηβ $=2$, βι’β $=4$ satisfy the above equation, but in this case, βε₯½β and βι’β represent the same digit, which contradicts the problem statement.
Therefore, the four-digit number βε¦ι’ι’ε¦β has a unique solution 2002.
Since $2002=2 \times 7 \times 11 \times 13$, the number of all divisors of 2002 is $2^{4}=16$.
|
16
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given real numbers $a, b, c$, satisfying $a+b+c=0, a^{2}+b^{2}+c^{2}=6$. Then the maximum value of $a$ is
|
14.2.
From the problem, we get $c=-(a+b)$, thus,
$$
a^{2}+b^{2}+[-(a+b)]^{2}=6,
$$
which simplifies to $b^{2}+a b+a^{2}-3=0$.
Since $b$ is a real number, the above equation, which is a quadratic equation in $b$, must have real roots, hence,
$$
\Delta=a^{2}-4\left(a^{2}-3\right) \geqslant 0, \quad -2 \leqslant a \leqslant 2.
$$
When $a=2$, $b=c=-1$, which satisfies the problem's conditions.
Therefore, the maximum value of $a$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. A group of 17 middle school students went to several places for a summer social survey, with a budget for accommodation not exceeding $x$ yuan per person per day. One day, they arrived at a place with two hostels, $A$ and $B$. $A$ has 8 first-class beds and 11 second-class beds; $B$ has 10 first-class beds, 4 second-class beds, and 6 third-class beds. It is known that the daily rates for first-class, second-class, and third-class beds are 14 yuan, 8 yuan, and 5 yuan, respectively. If the entire group stays in one hostel, they can only stay at $B$ according to the budget. Then the integer $x=$ $\qquad$.
|
4. $x=10$.
If staying at location $A$, even choosing the most economical beds, the average accommodation cost per person will exceed 10 yuan (since $8 \times 11 + 14 \times 6 = 172$ (yuan), $172 \div 17 \approx 10.12$ (yuan)).
If staying at location $B$, with a reasonable choice of beds, the budget can be met, and the most economical value is
$$
\begin{array}{l}
5 \times 6 + 8 \times 4 + 14 \times 7 = 160 \text{ (yuan) }, \\
160 \div 17 \approx 9.41 \text{ (yuan). }
\end{array}
$$
Since $9.41 \leqslant x < 10.12$, and $x$ is an integer, therefore, $x = 10$.
|
10
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that the lengths of the four sides of a rectangle are all integers less than 10, these four length numbers can form a four-digit number, where the thousand's digit and the hundred's digit of this four-digit number are the same, and this four-digit number is a perfect square. Find the area of this rectangle.
|
(Tip: Let the two adjacent sides of the rectangle be $a$ and $b$, then $\overline{a a b b}=$ $11(100 a+b)$ is a perfect square, which implies 11 divides $(a+b)$. Thus, $a + b = 11$. Upon verification, only $a=7, b=4$. Answer: 28.)
|
28
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the polynomial $x^{2}-x+1$ can divide another polynomial $x^{3}+x^{2}+a x+b(a, b$ are constants). Then $a+b$ equals ( ).
(A) 0
(B) -1
(C) 1
(D) 2
|
2. (C).
By synthetic division, it is easy to obtain the remainder $r(x)=(a+1) x+b-2$. Since it can be divided exactly, we have $a+1=0$ and $b-2=0$, i.e., $a=-1, b=2$. Therefore, $a+b=1$.
|
1
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. In decimal, if a positive integer with at least two digits has all its digits, except the leftmost one, smaller than the digit to their left, then it is called a descending positive integer. The total number of such descending positive integers is ( ).
(A) 1001
(B) 1010
(C) 1011
(D) 1013
|
5.(D).
Let the leftmost digit be $a$, then the digits $a-1$, $a-2, \cdots, 2,1,0$ that are smaller than $a$ can each appear at most once to the right of $a$. Therefore, the number of decreasing positive integers starting with $a$ is equal to the number of non-empty subsets of $\{a-1, a-2, \cdots, 2,1,0\}$, which is $2^{a}-1$. Since $a=1,2,3, \cdots, 9$, the total number of decreasing positive integers is
$$
\sum_{a=1}^{9}\left(2^{a}-1\right)=\left(\sum_{a=1}^{9} 2^{a}\right)-9=2^{10}-2-9=1013 \text { (numbers). }
$$
|
1013
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. $f(x)=\frac{x^{2}}{8}+x \cos x+\cos (2 x)(x \in \mathbf{R})$'s minimum value is $\qquad$ .
|
4. -1 .
$$
\begin{array}{l}
f(x)=\frac{x^{2}}{8}+x \cos x+2 \cos ^{2} x-1 \\
=\frac{1}{8}(x+4 \cos x)^{2}-1 \geqslant-1 .
\end{array}
$$
Since the equation $\cos x=-\frac{x}{4}$ (as can be seen from the graph) has a solution, we have
$$
f(x)_{\min }=-1
$$
|
-1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Try to find the smallest possible value of the positive integer $k$, such that the following proposition holds: For any $k$ integers $a_{1}, a_{2}, \cdots, a_{k}$ (equality is allowed), there must exist corresponding $k$ integers $x_{1}, x_{2}, \cdots, x_{k}$ (equality is also allowed), and $\left|x_{i}\right| \leqslant 2(i=1,2, \cdots, k),\left|x_{1}\right|+\left|x_{2}\right|+\cdots+$ $\left|x_{k}\right| \neq 0$, such that 2003 divides $x_{1} a_{1}+x_{2} a_{2}+\cdots$ $+x_{k} a_{k}$.
|
Three, first prove that the proposition holds when $k=7$. For this, consider the sum
$$
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)=y_{1} a_{1}+y_{2} a_{2}+\cdots+y_{7} a_{7} \text {, }
$$
where $y_{i} \in\{-1,0,1\}$.
There are $3^{7}=2187$ such sums, and since $2187>2003$, by the pigeonhole principle, there must be two different sums $S\left(y_{1}, y_{2}, \cdots, y_{7}\right)$ and $S\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots, y_{7}^{\prime}\right)$ that have the same remainder when divided by 2003.
Therefore, 2003 divides
$$
\begin{array}{l}
S\left(y_{1}, y_{2}, \cdots, y_{7}\right)-S\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots, y_{7}^{\prime}\right) \\
=\left(y_{1}-y_{1}^{\prime}\right) a_{1}+\left(y_{2}-y_{2}^{\prime}\right) a_{2}+\cdots+\left(y_{7}-y_{7}^{\prime}\right) a_{7} .
\end{array}
$$
where $y_{i}-y_{i}^{\prime} \in\{-2,-1,0,1,2\}, i=1,2, \cdots, 7$ (since $\left.y_{i} γ y_{i} \in\{-1,0,1\}, i=1,2, \cdots, 7\right)$, and at least one $y_{i}-y_{i}^{\prime} \neq 0$ (because the arrays $\left(y_{1}, y_{2}, \cdots, y_{7}\right) \neq\left(y_{1}^{\prime}, y_{2}^{\prime}, \cdots\right.$, $\left.y_{7}^{\prime}\right)$ ). At this point, taking $x_{i}=y_{i}-y_{i}^{\prime}, i=1,2, \cdots, 7$, satisfies the requirement.
Thus, the proposition holds when $k=7$.
Next, prove that the proposition does not hold when $k=6$. For this, we provide a counterexample.
Take $a_{1}=3^{1}, a_{2}=3^{2}, a_{3}=3^{3}, a_{4}=3^{4}, a_{5}=3^{5}, a_{6}=$ $3^{6}$, then for any 6 integers $x_{1}, x_{2}, \cdots, x_{6},\left|x_{i}\right| \leqslant$ $2,\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{6}\right| \neq 0$, the sum
$$
S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{6} x_{6}
$$
is always a multiple of 3.
$$
\begin{array}{l}
\text { Then } S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \text { | } \\
\leqslant 2\left(3^{1}+3^{2}+\cdots+3^{6}\right) \\
=3\left(3^{6}-1\right)=3\left(3^{3}-1\right)\left(3^{3}+1\right) \\
=3 \times 26 \times 28=3 \times 728 .
\end{array}
$$
Assume without loss of generality that the largest index $k_{1}$ of the non-zero elements $x_{1}, x_{2}, \cdots, x_{6}$ is $x_{k_{1}}$, i.e., $x_{k_{1}} \neq 0$, and $x_{k_{1}+1}=x_{k_{1}+2}=\cdots=x_{6}=$ 0, then
$$
\begin{array}{l}
S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{k_{1}} x_{k_{1}} \\
=3^{1} x_{1}+3^{2} x_{2}+\cdots+3^{k_{1}} x_{k_{1}} .
\end{array}
$$
Additionally, assume without loss of generality that $x_{k_{1}}>0$ (if $x_{k_{1}}<0$, the argument is similar).
If $k_{1} \geqslant 2$, then
If $k_{1}=1$, then
$$
x_{1} \neq 0, S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}>0 \text {. }
$$
In summary,
$$
\begin{array}{l}
3 \mid S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \text {, and } S\left(x_{1}, x_{2}, \cdots, x_{6}\right) \neq 0 \text {, } \\
\left|S\left(x_{1}, x_{2}, \cdots, x_{6}\right)\right| \leqslant 3 \times 728 .
\end{array}
$$
Clearly, 2003 and 3 are coprime.
If 2003 divides $S\left(x_{1}, x_{2}, \cdots, x_{6}\right)$, then $S\left(x_{1}\right.$, $\left.x_{2}, \cdots, x_{6}\right)=3 t$, where $t$ is an integer, and $1 \leqslant|t| \leqslant 728$. Thus, $2003 \mid t$, which contradicts $1 \leqslant|t| \leqslant 728$. Therefore, when taking $a_{i}=3^{i}$, $i=1,2, \cdots, 6$, it is impossible to have $x_{1}, x_{2}, \cdots, x_{6} \in \mathbf{Z},\left|x_{i}\right|$ $\leqslant 2, i=1,2, \cdots, 6,\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{6}\right| \neq 0$, such that
$2003$ divides $S\left(x_{1}, x_{2}, \cdots, x_{6}\right)=a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{6} x_{6}$. This counterexample shows that the proposition does not hold when $k=6$.
From the above two steps, the smallest positive integer $k$ is 7.
(Wu Weizhao, Department of Mathematics, College of Science, Guangzhou University, 510405)
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 A factory plans to arrange 214 workers to produce 6000 $A$ components and 2000 $B$ components. It is known that the time it takes for each worker to produce 5 $A$ components can produce 3 $B$ components. Now the workers are divided into two groups to produce these two types of components, and they start at the same time. How should the groups be divided to complete the task as quickly as possible?
|
Solution: Let $x$ be the number of people producing component $A$, then the number of people producing component $B$ is $(214-x)$. Taking the time for one person to produce 5 $A$ components as the unit, the time required to produce 6000 $A$ components and 2000 $B$ components is respectively
$$
y_{1}=\frac{6000}{5 x} \text { and } y_{2}=\frac{2000}{3(214-x)} \text {. }
$$
The total time to complete the task is
$$
T(x)=\max \left\{y_{1}, y_{2}\right\} \text {. }
$$
Note that when $x>0$, the inverse proportion function $y_{1}$ is a decreasing function, and when $0<x<214$, the inverse proportion function $y_{2}$ is an increasing function.
First, consider the equation $y_{1}=y_{2}$, solving it yields $x=137 \frac{4}{7}$.
Then, by the monotonicity of the above $y_{1}$ and $y_{2}$, we get
$$
T_{\min }=\min \{T(137), T(138)\} .
$$
Calculating, we find $T(137)=\frac{6000}{685}<T(138)=\frac{2000}{228}$.
Therefore, arranging 137 people to produce $A$ components and 77 people to produce $B$ components will complete the task the fastest.
|
137
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $n$ be a given positive integer. Try to find non-negative integers $k, l$, satisfying $k+l \neq 0$, and $k+l \neq n$, such that
$$
s=\frac{k}{k+l}+\frac{n-k}{n-(k+l)}
$$
takes the maximum value.
|
If $l=0$, then $s=2$.
If $l>0$, let $x=k+l$, then $00 .
\end{array}
$
Therefore, $f(1)1$ when, $s$ reaches the maximum value 2;
(2) If $n=2$, then when $l=1, k=0$ or $l=0, 0 < k \neq 2$, $s$ reaches the maximum value 2;
(3) If $n \geqslant 3$, then when $l=n-1, k=0$, $s$ reaches the maximum value $n$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 There are 1988 unit cubes, and they (all or part of them) are arranged into 3 "squares" (i.e., 3 one-layer rectangular prisms with dimensions $a \times a \times 1, b \times b \times 1, c \times c \times 1$ where $a \leqslant b \leqslant c$) $A, B$, and $C$. Now, place square $C$ in the first quadrant of the $x O y$ plane, with one of its vertices at the origin, then place $B$ on $C$ so that each small block of $B$ is exactly on a small block of $C$, with the sides of $B$ parallel to the corresponding sides of $C$, but the perimeter of $B$ does not align with the perimeter of $C$; then place $A$ in the same manner on $B$, thus forming a "three-story building". For what values of $a, b$, and $c$ will the number of different forms of the three-story building relative to the origin be maximized?
|
Solution: According to the problem, the condition for $B$ to be placed on $C$ is $b \leqslant c - 2$, and there are $(c-b-1)^{2}$ different ways to place it; the condition for $A$ to be placed on $B$ is $a \leqslant b-2$, and there are $(b-a-1)^{2}$ different ways to place it. Thus, the problem is transformed into:
Finding $a, b, c$ that satisfy the conditions
$$
1 \leqslant a \leqslant b-2 \leqslant c-4 \left(a, b, c \in \mathbf{N}_{+}\right)
$$
and $a^{2}+b^{2}+c^{2} \leqslant 1988$, such that the function
$$
f=(b-a-1)^{2}(c-b-1)^{2}
$$
reaches its maximum value.
Obviously, when $a=1$, $f$ reaches its maximum value.
Thus, the problem is further transformed into:
Under the conditions $3 \leqslant b \leqslant c-2$ and $b^{2}+c^{2} \leqslant 1987$, find the maximum value of $g=(b-2)^{2}(c-b-1)^{2}$, or the maximum value of $p=(b-2)(c-b-1)$.
When $c$ is fixed, since
$$
(b-2)+(c-b-1)=c-3,
$$
$p$ is maximized when $|(b-2)-(c-b-1)|=|c-2b+1|$ is minimized (refer to Example 2, Solution 2).
Since $c^{2} \leqslant 1987-b^{2} \leqslant 1987-9=1978$, it follows that $c \leqslant 44$.
We will discuss different values of $c$:
(1) If $c=44$, then
$$
b^{2} \leqslant 1987-44^{2}=51, 3 \leqslant b \leqslant 7.
$$
At this time, $p=(b-2)(c-b-1)=(43-b)(b-2)$ reaches its maximum value 180 when $b=7$ (at this time $|c-2b+1|$ is minimized).
(2) If $c=43$, then $3 \leqslant b \leqslant 11$. At this time, $p$ reaches its maximum value 279 when $b=11$.
(3) If $c=42$, then $3 \leqslant b \leqslant 14$. At this time, $p$ reaches its maximum value 324 when $b=14$.
(4) If $c=41$, then $3 \leqslant b \leqslant 17$. At this time, $p$ reaches its maximum value 345 when $b=17$.
(5) If $c=40$, then $3 \leqslant b \leqslant 19$. At this time, $p$ reaches its maximum value 340 when $b=19$.
(6) If $c \leqslant 39$, then
$$
\begin{aligned}
p & =(b-2)(c-b-1) \\
& \leqslant\left(\frac{c-3}{2}\right)^{2} \leqslant 18^{2}=324.
\end{aligned}
$$
In summary, when $a=1, b=17, c=41$, the number of different forms of the "three-story building" is the maximum, which is 345.
|
345
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are 1989 points in space, where no three points are collinear. Divide them into 30 groups with different numbers of points. From any 3 different groups, take one point each to form a triangle. To maximize the total number of such triangles, how many points should each group have?
|
(Tip: Let the number of points in the $i$-th group be $x_{i}$, then the problem is transformed into: when $x_{1}$ $+x_{2}+\cdots+x_{30}=1989$, and $x_{1}, x_{2}, \cdots, x_{30}$ are all different, find the maximum value of $M=\sum_{1<i<j<k \leqslant 30} x_{i} x_{j} x_{k}$. Without loss of generality, assume $x_{1}<x_{2}$ $<\cdots<x_{30}$, as in Example 3, it can be proved that when $M$ is maximized, (1) $x_{i+1}-x_{i}$ $\leqslant 2(i=1,2, \cdots, 29)$; (2) the number of $i$ such that $x_{i+1}-x_{i}=2$ is no more than 1. From (1) and (2), we know that the values of $x_{1}, x_{2}, \cdots, x_{30}$ can only be two types: $\alpha, \alpha+1, \alpha+2, \cdots, \alpha+29$ or $\alpha, \alpha+1, \cdots, \alpha+i-1, \alpha$ $+i+1, \cdots, \alpha+30$. The first type has no solution, and for the second type, $\alpha$ $=51, i=6$.
|
51
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 In $\triangle A B C$, $A B=37, A C=58$, a circle with center $A$ and radius $A B$ intersects $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$.
In $\triangle A B C$, $A B=37, A C=58$, a circle with center $A$ and radius $A B$ intersects $B C$ at point $D$, and $D$ is between $B$ and $C$. If the lengths of $B D$ and $D C$ are both integers, find the length of $B C$.
|
Explanation: A triangle with two known sides is indeterminate, but note that the lengths of $B D$ and $D C$ are both integers, so the length of $B C$ is also an integer. Therefore, we can start by determining the range of possible values for $B C$, and then find the value of $B C$ that meets the requirements.
As shown in Figure 2, let the line $A C$ intersect $\odot A$ at points $E$ and $F$. By the secant theorem, we have
$$
\begin{array}{l}
D C \cdot B C=C E \cdot C F \\
=(A C-A E)(A C+A F) \\
=(A C-A B)(A C+A B) \\
=A C^{2}-A B^{2}=58^{2}-37^{2} \\
=21 \times 95=3 \times 5 \times 7 \times 19 .
\end{array}
$$
Figure 2
Since $A C - A B < B C$, point $D$ is not between $B$ and $C$, and should be discarded. Therefore, $B C=57$, at which time $D C=35, B D=22$.
Note: This problem is from "Mathematical Olympiad Problems" in the March 1996 issue of "Intermediate Mathematics". The solution above is more in line with students' thinking patterns.
|
57
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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