problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
6.3. Several boys and 5 girls are sitting around a round table, and there are 30 pieces of bread on the plate on the table. Each girl takes 1 piece of bread from the plate for each boy she knows, and then each boy takes 1 piece of bread from the plate for each girl he does not know, at which point the bread on the plat... | 6.3. There are $n$ boys, so there are $5 n$ different “boy-girl” pairs. In each such “pair”, 1 piece of bread was taken, because if the 2 people in the pair know each other, the girl took 1 piece of bread for the boy; and if they do not know each other, the boy took 1 piece of bread for the girl. Therefore, $5 n=$ $30,... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: Two vegetable bases, A and B, supply the same type of vegetables to three farmers' markets, A, B, and C, according to the signed contracts. They are to supply 45 tons to A, 75 tons to B, and 40 tons to C. Base A can arrange 60 tons, and Base B can arrange 100 tons. The distances in kilometers between A and B... | Solution: Let the base B supply $x \mathrm{t}$ to $A$, $y \mathrm{t}$ to $B$, and $[100-(x+y)] \mathrm{t}$ to $C$, then base A supplies $(45-x) \mathrm{t}$ to $A$, $(75-y) \mathrm{t}$ to $B$, and $[40-(100-x-y)] \mathrm{t}=(x+y-60) \mathrm{t}$ to $C$.
Let the total transportation cost be $W$ yuan. According to the prob... | 960 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 2,
there are soil piles $A$ and $B$, a depression $E$, and a pond $F$ at a construction site.
The volumes of soil piles $A$ and $B$ are $781 \mathrm{~m}^{3}$ and $1584 \mathrm{~m}^{3}$, respectively. Depression $E$ requires $1025 \mathrm{~m}^{3}$ of soil to be filled, and pond $F$ can a... | Solution: Let " $\mathrm{m}^{3} \cdot \mathrm{m}$ " be the unit for the labor cost of transporting soil. Suppose the volume of soil transported from $A$ to $E$ is $x$, and the volume of soil transported to $F$ is $781-x$; the volume of soil transported from $B$ to $E$ is $1025-x$, and the volume of soil transported to ... | 207170 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
8. The terms of a sequence are all 3 or 5, the first term is 3, and there are $2^{k-1}$ fives between the $k$-th 3 and the $(k+1)$-th 3, i.e., $3,5,3,5,5,3,5,5,5,5,3, \cdots$. Then the sum of the first 2004 terms of this sequence $S_{2004}=$ $\qquad$ . | 8. 9998 .
The $k$-th 3 and the $2^{k-1}$ 5s following it, totaling $\left(2^{k-1}+1\right)$ terms, are considered as one group. Suppose the 2004th term is in the $k$-th group, then $k$ is the smallest positive integer satisfying
$$
k+1+2+\cdots+2^{k-1} \geqslant 2004
$$
It is easy to find that $k=11$.
Thus, $S_{2004}... | 9998 | Other | math-word-problem | Yes | Yes | cn_contest | false |
13. Given that $f(x)$ is defined on $(-1,1)$, $f\left(\frac{1}{2}\right)=-1$, and satisfies $x, y \in (-1,1)$, we have $f(x) + f(y) = f\left(\frac{x+y}{1+xy}\right)$.
(1) The sequence $\{x_n\}$ satisfies
$$
x_1 = \frac{1}{2}, \quad x_{n+1} = \frac{2x_n}{1 + x_n^2}.
$$
Let $a_n = f(x_n)$. Find the general term formula ... | $$
\begin{array}{l}
\text { Therefore, } 1+f\left(\frac{1}{b_{1}}\right)+f\left(\frac{1}{b_{2}}\right)+\cdots+f\left(\frac{1}{b_{2000}}\right)+f\left(\frac{1}{2004}\right) \\
=1+\left[f\left(\frac{1}{2}\right)-f\left(\frac{1}{3}\right)\right]+\left[f\left(\frac{1}{3}\right)-f\left(\frac{1}{4}\right)\right]+ \\
\quad \c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $a, b, c$ are real numbers, and $2a + b + c = 5, b - c = 1$. Then the maximum value of $ab + bc + ca$ is | From $2 a+b+c=5, b-c=1$, we can obtain
$$
b=3-a, c=2-a \text {. }
$$
Therefore, $a b+b c+c a$
$$
\begin{array}{l}
=a(3-a)+(3-a)(2-a)+a(2-a) \\
=-a^{2}+6 \leqslant 6 .
\end{array}
$$ | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 2, quadrilateral $ABCD$ is inscribed in $\odot O$, with $BD$ being the diameter of $\odot O$, and $\overparen{AB}=\overparen{AD}$. If $BC + CD = 4$, then the area of quadrilateral $ABCD$ is $\qquad$ . | 3.4 .
From $\overparen{A B}=\overparen{A D}$, we know $A B=A D$.
Also, $B D$ is the diameter of $\odot 0$, so $\angle A=\angle C=90^{\circ}$.
Let $A B=A D=x, B C=y, C D=z$.
By the Pythagorean theorem, we have
$$
x^{2}+x^{2}=y^{2}+z^{2} \text {. }
$$
Thus, $2 x^{2}=y^{2}+z^{2}=(y+z)^{2}-2 y z$.
Also, $S_{\text {quadri... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Renovate a pedestrian path, changing the ground originally paved with $n^{2}-96$ identical square tiles to be paved with larger identical square tiles, requiring a total of $5 n+51$ tiles to complete the paving. If $n^{2}-96$ is divisible by $5 n+51$, find the value of the positive integer $n$. | Given the problem, let $\frac{n^{2}-96}{5 n+51}=k, k$ be a positive integer, then
$$
\begin{array}{l}
n^{2}-5 k n-(51 k+96)=0 . \\
\text { Therefore, } \Delta=25 k^{2}+4(51 k+96) \\
=25 k^{2}+204 k+384 .
\end{array}
$$
Since $n$ is an integer, $\Delta$ must be a perfect square.
$$
\text { And }(5 k+19)^{2}<25 k^{2}+20... | 30 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $a$ is a positive integer not greater than 2005, and $b, c$ are integers, the parabola $y=a x^{2}+b x+c$ is above the $x$-axis and passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$. Find the minimum value of $a-b+c$.
The parabola $y=a x^{2}+b x+c$ passes through points $A(-1,4 a+7)$ an... | Three, since $a$ is a positive integer, the parabola opens upwards.
The parabola $y=a x^{2}+b x+c$ is above the $x$-axis, so,
$$
b^{2}-4 a c<0.
$$
Given $b=-2-2a$ and $c=a+5$, we have
$$
b^{2}-4 a c=(-2-2a)^{2}-4a(a+5)=4+8a+4a^{2}-4a^{2}-20a=4-12a<0.
$$
This implies
$$
a>\frac{1}{3}.
$$
Since $a$ is a positive integer ... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. A cube with an edge length of a certain integer is cut into 99 smaller cubes, 98 of which are unit cubes with an edge length of 1, and the other cube also has an integer edge length. Then its edge length is $\qquad$ | 4.3.
Let the edge length of the original cube be $a$, and the edge length of the other cube after cutting be $b\left(a, b \in \mathbf{Z}_{+}, a>b\right)$. Then
$$
a^{3}-98=b^{3} \text {, }
$$
i.e., $(a-b)\left(a^{2}+a b+b^{2}\right)=98$.
Thus, $\left\{\begin{array}{l}a-b=1, \\ a^{2}+a b+b^{2}=98 ;\end{array}\right.$
... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. Given a non-constant sequence $\left\{a_{i}\right\}$ satisfies
$$
a_{i}^{2}-a_{i-1} a_{i}+a_{i-1}^{2}=0 \text {, }
$$
and $a_{i+1} \neq a_{i-1}, i=1,2, \cdots, n$.
For a given positive integer $n, a_{1}=a_{n+1}$. Then
$$
\sum_{i=0}^{n-1} a_{i}=
$$ | 6.0 .
Given $a_{1}^{2}-a_{1-1} a_{i}+a_{1-1}^{2}=0$, so,
$$
a_{i+1}^{2}-a_{i} a_{i+1}+a_{i}^{2}=0 \text {. }
$$
Subtracting the two equations, we get
$$
\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}\right)-a_{i}\left(a_{i+1}-a_{i-1}\right)=0 .
$$
That is, $\left(a_{i+1}-a_{i-1}\right)\left(a_{i+1}+a_{i-1}-a_{i}\... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (50 points) Given positive integers $x_{1}<x_{2}<\cdots<x_{n}$, and
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=2003, n \geqslant 2, \\
f(n)=n\left(x_{1}+x_{n}\right) .
\end{array}
$$
Find the minimum value of $f(n)$. | If $x_{1}>1$, let $x_{1}^{\prime}=1, x_{n}^{\prime}=x_{n}+x_{1}-1$, and the other $x_{i}$ values remain unchanged, then
$$
\begin{array}{l}
x_{1}^{\prime}+x_{2}+x_{3}+\cdots+x_{n-1}+x_{n}^{\prime}=2003, \\
f(n)=n\left(x_{1}+x_{n}\right)=n\left(1+x_{n}^{\prime}\right) .
\end{array}
$$
Assume $x_{1}=1$.
When $n=2$, $f(2... | 2254 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) There are 12 football teams participating in a match, with each pair of teams playing one game. The winner gets 3 points, the loser gets 0 points, and in the case of a draw, both teams get 1 point. How many points does one team need to score at a minimum to ensure that no more than 6 teams score at l... | Three, assuming there are 7 teams with the same score as a certain team, then the maximum score for each match between these 8 teams is 3 points. Therefore, their maximum total score is $3 C_{8}^{2}=84$ points, and the score of each team cannot exceed $\left[\frac{84}{8}\right]=10$ points, so each team can win at most ... | 23 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Find the smallest positive integer $a$ that satisfies the following condition: there exists a positive odd number $n$, such that $203^{n} + a \cdot 198^{n}$ is a multiple of 2005. | Solution: Since $2005=401 \times 5$, and
$$
203+198=401,203-198=5 \text {, }
$$
Therefore,
$$
203^{n}+a \cdot 198^{n}=(401-198)^{n}+a \cdot 198^{n}
$$
is a multiple of $401 \times 5$.
Since $n$ is odd, then
$$
-198^{n}+a \cdot 198^{n}=(a-1) \cdot 198^{n}
$$
is a multiple of 401.
Since the greatest common divisor of ... | 1204 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 If a store sells a certain product, which costs 100 yuan, at 120 yuan, it can sell 300 units. If the price of the product is increased by 1 yuan based on 120 yuan, it will sell 10 fewer units, and if the price is reduced by 1 yuan, it will sell 30 more units. Question: To maximize profit, what price should th... | Solution: (1) If sold at 120 yuan each, 300 can be sold, and the profit is $300 \times 20 = 6000$ (yuan).
(2) If the price is increased by $x (x > 0)$ yuan, then (300 $- 10x$) can be sold. Let the profit be $y$ yuan, then
$$
\begin{array}{l}
y = (20 + x)(300 - 10x) \\
= -10x^2 + 100x + 6000 \\
= -10(x - 5)^2 + 6250 .
\... | 115 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4: On a circular road in a certain town, there are five primary schools in sequence: No.1 Primary School, No.2 Primary School, No.3 Primary School, No.4 Primary School, and No.5 Primary School. They have 15, 7, 11, 3, and 14 computers respectively. To make the number of computers in each school the same, how ma... | Solution: As shown in Figure 4, let $A, B, C, D, E$ represent the first to fifth schools, respectively, and let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be the number of computers transferred to the neighboring schools. Let the total number of transfers be $y$. According to the problem, we have
$$
\begin{array}{l}
7+x_{1}-x... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 A unit spent 500,000 yuan to purchase a piece of high-tech equipment. According to the tracking survey of this model of equipment, after the equipment is put into use, if the maintenance and repair costs are averaged to each day, the conclusion is: the maintenance and repair cost on the $x$-th day is $\left[\... | Solution: (1) If the equipment is put into use for $x$ days, and the average daily loss is $y$ yuan, then
$$
\begin{aligned}
y= & \left[500000+\left(\frac{1}{4} \times 0+500\right)+\left(\frac{1}{4} \times 1+500\right)+\right. \\
& \left.\cdots+\left(\frac{x-1}{4}+500\right)\right] \div x \\
= & \frac{1}{x}\left[500000... | 2000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. Several workers load and unload a batch of goods, with each worker having the same loading and unloading speed. If these workers work simultaneously, it will take $10 \mathrm{~h}$ to complete the loading and unloading. Now, the loading and unloading method is changed, starting with one person working, and then addi... | 16. (1) Let the loading and unloading work take $x \mathrm{~h}$ to complete, then the first person worked for $x \mathrm{~h}$, and the last person worked for $\frac{x}{4} \mathrm{~h}$. The total work time of the two people is $\left(x+\frac{x}{4}\right) \mathrm{h}$, and the average work time per person is $\frac{1}{2}\... | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
18. 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, $\cdots \cdots$, and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest ... | 18.9 judges cannot give 1 point to 5 or more athletes, because among 5 or more athletes, at least one athlete must be rated no less than 5 by a judge, while according to the problem, each of these 5 athletes is rated no more than 4 by each judge, which is a contradiction. Therefore, 9 judges can give 1 point to at most... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
13. A and B are two car retailers (hereinafter referred to as A and B) who ordered a batch of cars from a car manufacturer. Initially, the number of cars A ordered was 3 times the number of cars B ordered. Later, due to some reason, A transferred 6 cars from its order to B. When picking up the cars, the manufacturer pr... | Three, 13. Let the total number of cars finally purchased by A and B be $x$ units. In the 6 fewer cars supplied by the factory, A requested $y$ fewer cars $(0 \leqslant y \leqslant 6)$, and B requested $(6-y)$ fewer cars. Then we have
$$
\begin{array}{l}
\frac{3}{4}(x+6)-6-y \\
=2\left[\frac{1}{4}(x+6)+6-(6-y)\right] .... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. From $1,2, \cdots, 2004$, choose $k$ numbers such that among the chosen $k$ numbers, there are definitely 3 numbers that can form the side lengths of a triangle (here it is required that the three side lengths of the triangle are all different). What is the minimum value of $k$ that satisfies the condition? | 15. This problem is equivalent to:
Selecting $k-1$ numbers from $1,2, \cdots, 2004$, such that no three of these numbers can form the lengths of the sides of a triangle (all sides being unequal). What is the maximum value of $k$ that satisfies this condition?
For any array that meets the above conditions,
when $k=4$, ... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. The cost price of each product is 120 yuan. During the trial sales phase, the relationship between the selling price $x$ (yuan) of each product and the daily sales volume $y$ (units) is shown in Table 2. If the daily sales volume $y$ is a linear function of the selling price $x$, to maximize profit, what should the ... | (Answer: When the selling price is set to 160 yuan, the maximum daily profit is 1600 yuan.) | 1600 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If $a+b+c=0, \frac{1}{a+1}+\frac{1}{b+2}+$ $\frac{1}{c+3}=0$, then, $(a+1)^{2}+(b+2)^{2}+(c+3)^{2}$ is equal to (.
(A) 36
(B) 16
(C) 14
(D) 3 | 8. A.
From the problem, we have
$$
\begin{array}{l}
(a+1)(b+2)+(b+2)(c+3)+(c+3)(a+1)=0 \text {. } \\
\text { Then }(a+1)^{2}+(b+2)^{2}+(c+3)^{2} \\
=(a+b+c+6)^{2}-2[(a+1)(b+2)+ \\
(b+2)(c+3)+(c+3)(a+1)]=36 .
\end{array}
$$ | 36 | Algebra | MCQ | Yes | Yes | cn_contest | false |
$13.2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5$ can achieve the minimum value of $\qquad$ . | 13. -10 .
Original expression $=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10$. When $x=2, y=-1$, it has the minimum value -10. | -10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15.1 A can of coffee is shared by person A and person B, and they finish it together in 10 days. If person A drinks it alone, it takes 12 days. 1 pound of tea is shared by person A and person B, and they finish it together in 12 days. If person B drinks it alone, it takes 20 days. Assuming that person A will never drin... | Three, 15. It is known that A drinks $\frac{1}{30}$ of 1 catty of tea every day, and B drinks $\frac{1}{60}$ of 1 can of coffee every day.
After 30 days, A finishes the tea while B only drinks half a can of coffee, and the remaining half can of coffee is drunk by A and B together in 5 days. Therefore, it takes a total ... | 35 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $a=1+2+\cdots+2004$. Then the remainder when $a$ is divided by 17 is | 8.1.
$$
\begin{array}{l}
a=1+2+\cdots+2004 \\
=\frac{2004 \times 2005}{2}=2009010 .
\end{array}
$$
Then 2009010 divided by 17 gives a quotient of 118177, with a remainder of 1. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given $f(x)=x^{2}+x-1$. If $a b^{2} \neq 1$, and $f\left(a^{-1}\right)=f\left(b^{2}\right)=0$, then $\frac{a}{1+a b^{2}}=$ $\qquad$ . | 9. -1 .
Given $f(x)=x^{2}+x-1, f\left(a^{-1}\right)=f\left(b^{2}\right)=0, a b^{2} \neq 1$, we know that $a^{-1}$ and $b^{2}$ are the two real roots of $f(x)=x^{2}+x-1$.
By Vieta's formulas, we get $\frac{1}{a}+b^{2}=-1, \frac{b^{2}}{a}=-1$. Thus, $\frac{1}{a}+b^{2}=\frac{b^{2}}{a}=-1$.
Therefore, we have $1+a b^{2}=-... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Simplify $\left(\log _{3} 4+\log _{2} 9\right)^{2}-\left(\log _{3} 4-\log _{2} 9\right)^{2}$ $=$ $\qquad$ | 2.16.
$$
\begin{array}{l}
(\log 4+\log 9)^{2}-(\log 4-\log 9)^{2} \\
=4 \log 4 \cdot \log 9=4 \times \frac{\log 4}{\log 3} \times \frac{\log 9}{\log 2} \\
=4 \times \frac{2 \log 2}{\log 3} \times \frac{2 \log 3}{\log 2}=16 .
\end{array}
$$ | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=2$, $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}$, and $a_{n+1} a_{n+2} \neq$ 1. Then $a_{1}+a_{2}+\cdots+a_{2004}=$ $\qquad$ | 3.4008.
Substituting $a_{1}=1, a_{2}=2$ into $a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+$ $a_{n+2}$, we get $a_{3}=3$.
$$
\begin{array}{l}
\text { From } a_{n} a_{n+1} a_{n+2}=a_{n}+a_{n+1}+a_{n+2}, \\
a_{n+1} a_{n+2} a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3},
\end{array}
$$
Subtracting the two equations, we get $\left(a_{n+3}-a_{n}... | 4008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In an exam, there are 5 multiple-choice questions, each with 4 different options, and each person selects exactly 1 option for each question. In 2000 answer sheets, it is found that there exists an $n$, such that in any $n$ answer sheets, there are 4 sheets where any 2 sheets have at most 3 answers the same. ... | Solution: Let the 4 possible answers for each question be denoted as $1, 2, 3, 4$, and the answers on each test paper be denoted as $(g, h, i, j, k)$, where $g, h, i, j, k \in \{1,2,3,4\}$. Let $\{(1, h, i, j, k), (2, h, i, j, k), (3, h, i, j, k), (4, h, i, j, k)\}, h, i, j, k = 1, 2, 3, 4$, yielding a total of 256 qua... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given a set of points $P=\left\{P_{1}, P_{2}, \cdots, P_{1924}\right\}$ on a plane, and no three points in $P$ are collinear. Divide all the points in $P$ into 83 groups arbitrarily, such that each group has at least three points, and each point belongs to exactly one group. Then, connect any two points in th... | Solution: (1) Let $m(G)=m_{0}, G$ is obtained from the groups $X_{1}, X_{2}, \cdots, X_{88}$, where $X_{i}$ is the set of points in the $i$-th group, $i=1,2, \cdots, 83$.
Let $\left|X_{i}\right|=x_{i}, i=1,2, \cdots, 83$, then we have
$$
x_{1}+x_{2}+\cdots+x_{83}=1994 \text {, }
$$
and $m_{0}=C_{x_{1}}^{3}+C_{x_{2}}^{... | 168544 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. After 20 figure skaters have performed, 9 judges respectively assign them rankings from 1 to 20. It is known that the difference between any rankings given to each athlete does not exceed 3. If the sum of the rankings received by each athlete is arranged in an increasing sequence, $c_{1} \leqslant c_{2} \leqslant$ $... | (Tip: Let the set of athletes who have won first place be $A$. If $|A|=1$, then $c_{1}=9$; If $|A|=2$, then one of them wins no less than 5 first places, by the problem's condition we know $c_{1} \leqslant 5 \times 1+4 \times 4=21$; If $|A|=3$, similarly, we have $c_{1}+c_{2}+c_{3} \leqslant 1 \times 9+3 \times 9+4 \ti... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. In space, there are 10 points, all connected pairwise, and these line segments are colored with red and blue. Among them, the line segments connected from point $A$ are all red. Among the triangles formed with these 10 points as vertices, how many triangles with three sides of the same color are there at least?
In ... | (Tip: Let the number of same-colored triangles be $x$, then the number of differently colored triangles is $\mathrm{C}_{10}^{3}-x$. Let the total number of same-colored angles be $y$, then $y=3 x+\left(\mathrm{C}_{10}^{3}-\right.$ $x)=120+2 x$. When $y$ reaches its minimum value, $x$ also reaches its minimum value. The... | 30 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. In a $13 \times 13$ square grid, select the centers of $k$ small squares such that no four of these points form the vertices of a rectangle (with sides parallel to those of the original square). Find the maximum value of $k$ that satisfies this condition. | (Let the $i$-th column have $x_{i}$ points $(i=1,2, \cdots, 13)$, then $\sum_{i=1}^{13} x_{i}=k$, the $x_{i}$ points in the $i$-th column form $\mathrm{C}_{x_{i}}^{2}$ different point pairs (if $x_{1}<2$, then $\mathrm{C}_{x_{1}}^{2}=0$). Add a column to the side of the $13 \times 13$ square, and each point pair projec... | 52 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. A test paper has 4 multiple-choice questions, each with three options (A), (B), (C). Several students take the exam, and after grading, it is found that: any 3 students have 1 question where their answers are all different. How many students can take the exam at most? | (提示: If 10 people take the exam, then for the 1st question, at least 7 people choose two options: for the 2nd question, at least 5 people among these 7 choose two options; for the 3rd question, at least 4 people among these 5 choose two options; for the 4th question, at least 3 people among these 4 choose two options. ... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Let $a_{1}, a_{2}, \cdots, a_{6} ; b_{1}, b_{2}, \cdots, b_{6} ; c_{1}$, $c_{2}, \cdots, c_{6}$ all be permutations of $1,2, \cdots, 6$. Find the minimum value of $\sum_{i=1}^{6} a_{i} b_{i} c_{i}$.
(Xiong Bin) | Let $S=\sum_{i=1}^{6} a_{i} b_{i} c_{i}$. By the AM-GM inequality, we have
$$
\begin{array}{l}
S \geqslant 6 \sqrt[6]{\prod_{i=1}^{6} a_{i} b_{i} c_{i}}=6 \sqrt[6]{(6!)^{3}} \\
=6 \sqrt{6!}=72 \sqrt{5}>160 .
\end{array}
$$
Next, we prove that $S>161$.
Since the geometric mean of the 6 numbers $a_{1} b_{1} c_{1}, a_{2}... | 162 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. As shown in Figure 1, there is a rectangular piece of paper $A B C D, A B=8$, $A D=6$. The paper is folded so that the edge $A D$ lies on the edge $A B$, with the fold line being $A E$. Then, $\triangle A E D$ is folded along $D E$ to the right, and the intersection of $A E$ and $B C$ is point $F$. The area of $\tri... | - .1.A.
From the folding process, we know $D E=A D=6, \angle D A E=\angle C E F=$ $45^{\circ}$. Therefore, $\triangle C E F$ is an isosceles right triangle, and $E C=8-6$ $=2$. Hence, $S_{\triangle C E F}=2$. | 2 | Geometry | MCQ | Yes | Yes | cn_contest | false |
4. Given
$$
A=48 \times\left(\frac{1}{3^{2}-4}+\frac{1}{4^{2}-4}+\cdots+\frac{1}{100^{2}-4}\right) \text {. }
$$
The positive integer closest to $A$ is ( ).
(A) 18
(B) 20
(C) 24
(D) 25 | 4.D.
For positive integers $n(n \geqslant 3)$, we have $\frac{1}{n^{2}-4}=\frac{1}{4}\left(\frac{1}{n-2}-\frac{1}{n+2}\right)$.
Then $A=48 \times \frac{1}{4}\left[\left(1+\frac{1}{2}+\cdots+\frac{1}{8}\right)-\left(\frac{1}{5}+\frac{1}{6}+\cdots+\frac{1}{R C}\right)\right]$
$$
\begin{array}{l}
=12\left(1+\frac{1}{2}+\... | 25 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. Let $a, b$ be positive integers, and satisfy $56 \leqslant a+b \leqslant 59, 0.9<\frac{a}{b}<0.91$. Then $b^{2}-a^{2}$ equals ( ).
(A) 171
(B) 177
(C) 180
(D) 182 | 5.B.
From the given, we have
$$
0.9 b+b56 \text {, }
$$
then $29<b<32$. Therefore, $b=30,31$.
When $b=30$, from $0.9 b<a<0.91 b$, we get $27<a<28$, there is no such positive integer $a$;
When $b=31$, from $0.9 b<a<0.91 b$, we get $27<a<29$, so $a=28$.
Thus, $b^{2}-a^{2}=177$. | 177 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 6 As shown in Figure 10, in $\triangle ABC$, $AB=AC$,
$\angle BAC=120^{\circ}$,
$\triangle ADE$ is an equilateral
triangle, point $D$ is on
side $BC$. It is known that
$BD: DC=2: 3$. When the area of $\triangle ABC$ is $50 \mathrm{~cm}^{2}$, find the area of $\triangle ADE$.
(7th Japan Arithmetic Olympiad (Fina... | Analysis: Directly solving the problem is difficult. By rotating $\triangle A B C$ counterclockwise around point $A$ by $120^{\circ}$ and $240^{\circ}$ to form a regular $\triangle M B C$ (as shown in Figure 11), the regular $\triangle A D E$ becomes regular $\triangle A D_{1} E_{1}$ and regular $\triangle A D_{2} E_{2... | 14 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. In the Cartesian coordinate system, the parabola
$$
y=x^{2}+m x-\frac{3}{4} m^{2}(m>0)
$$
intersects the $x$-axis at points $A$ and $B$. If the distances from points $A$ and $B$ to the origin are $O A$ and $O B$, respectively, and satisfy $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, then the value of $m$ is $\qquad$. | 7. 2 .
Let the roots of the equation $x^{2}+m x-\frac{3}{4} m^{2}=0$ be $x_{1}$ and $x_{2}$, and $x_{1}<0<x_{2}$.
From $\frac{1}{O B}-\frac{1}{O A}=\frac{2}{3}$, we know $O A>O B$.
Also, since $m>0$, the axis of symmetry of the parabola is to the left of the $y$-axis, thus,
$$
O A=\left|x_{1}\right|=-x_{1}, O B=x_{2} ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Given $x_{1}, x_{2}, \cdots, x_{40}$ are all positive integers, and $x_{1}+$ $x_{2}+\cdots+x_{40}=58$. If the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ is $A$, and the minimum value is $B$, then the value of $A+B$ is $\qquad$ | 10.494.
Since the number of ways to write 58 as the sum of 40 positive integers is finite, the minimum and maximum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ exist.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{40}$.
If $x_{1}>1$, then $x_{1}+x_{2}=\left(x_{1}-1\rig... | 494 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. From the 200 positive integers $1, 2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a < b < c)$, we have $a b \neq c$? | 14. First, the 193 numbers $1, 14, 15, \cdots, 205$ satisfy the conditions of the problem.
In fact, if $a, b, c (a < b < c)$, then $a b \geqslant 14 \times 15 = 210 > c$.
On the other hand, consider the following 12 sets of numbers:
$(2, 25, 2 \times 25), (3, 24, 3 \times 24), \cdots$,
$(13, 14, 13 \times 14)$.
These ... | 193 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. A city's No. 10 Middle School has built a new six-story enclosed student dormitory, with 15 rooms on each floor, and this dormitory has only one entrance and exit safety door. The safety inspection department has tested the normal passage of this safety door: in an emergency, the safety door can function normally fo... | 3.168.
Assuming that under normal circumstances, this safety door can allow an average of $x$ people to pass through per minute. According to the problem, we have
$$
8 \times 15 \times 6 \leqslant x(1+90\%+80\% \times 3) .
$$
Solving this, we get $167.4 \approx \frac{720}{4.3} \leqslant x$.
Therefore, the minimum int... | 168 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5. For the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1(0<b<4)$, the left focus is $F$, and line $l$ intersects the ellipse at points $A$ and $B$, with $|A B|$ $=6$. Then the maximum value of $|F A| \cdot|F B|$ is $\qquad$ | 5. 25 .
Let the right focus of the ellipse be $F^{\prime}$, and connect $A F^{\prime}$ and $B F^{\prime}$. Then,
$$
|A B| \leqslant\left|A F^{\prime}\right|+\left|B F^{\prime}\right| \text {. }
$$
When the chord $A B$ passes through the right focus $F^{\prime}$, the equality holds.
Thus, $|F A|+|F B|+|A B|$
$$
\begin... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Four. (20 points) Given $a, b, c \in \mathbf{R}_{+}$, and
$$
\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=1 \text{. }
$$
Prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant 12$. | Given, we have
$$
\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{1}{b}}+\frac{1}{1+\frac{1}{c}}=1 \text {. }
$$
Let $x=\frac{1}{1+\frac{1}{a}}, y=\frac{1}{1+\frac{1}{b}}, z=\frac{1}{1+\frac{1}{c}}$. Then
$$
x+y+z=1 \text {. }
$$
From $\frac{1}{a}=\frac{1}{x}-1, \frac{1}{b}=\frac{1}{y}-1, \frac{1}{c}=\frac{1}{z}-1$, we get
... | 12 | Inequalities | proof | Yes | Yes | cn_contest | false |
Example 1 Let the positive integer $n$ be a multiple of 75, and have exactly 75 positive divisors (including 1 and itself). Find the minimum value of $n$. | Solution: Let the prime factorization of $n$ be
$$
n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{k}^{r_{k}},
$$
where $p_{1}, p_{2}, \cdots, p_{k}$ are the distinct prime factors of $n$, and $r_{1}, r_{2}, \cdots, r_{k}$ are all positive integers.
Thus, the number of distinct positive divisors of $n$ is
$$
\left(r_{1}+1\rig... | 32400 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $x_{1}, x_{2}, \cdots, x_{10}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{10}=99$. Find the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{10}^{2}$. | Solution: Since there are only a finite number of ways to write 99 as the sum of 10 positive integers, there must exist a way that maximizes the sum of the squares of these 10 positive integers: and there must also exist a way that minimizes the sum of the squares of these 10 positive integers.
Assume $x_{1}, x_{2}, \... | 8109 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Among the numbers $1,2, \cdots, 2005$, what is the minimum number of numbers that need to be removed so that in the remaining numbers, no number is equal to the product of any two other numbers? | (Cue: Strike out the 43 numbers $2,3, \cdots, 44$. Among the remaining numbers (excluding 1), the product of any two numbers is greater than $45^{2}=2025>2005$. Also, for the following 43 triples:
$$
(2,87,2 \times 87),(3,86,3 \times 86), \cdots,(44,45,44 \times 45),
$$
these 129 numbers are all distinct, greater than... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Among the five-digit numbers formed by the 3 digits $1,2,3$, $1,2,3$ each appear at least once. There are $\qquad$ such five-digit numbers. | 5.150.
Among the five-digit numbers,
if 1 appears only once, there are $\mathrm{C}_{5}^{1}\left(\mathrm{C}_{4}^{1}+\mathrm{C}_{4}^{2}+\mathrm{C}_{4}^{3}\right)=70$;
if 1 appears twice, there are $C_{5}^{2}\left(C_{3}^{1}+C_{3}^{2}\right)=60$;
if 1 appears three times, there are $\mathrm{C}_{5}^{3} \mathrm{C}_{2}^{1}=2... | 150 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. (24 points)(1) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals 2005, find the minimum value of $n$ and explain the reason;
(2) If the sum of the volumes of $n\left(n \in \mathbf{N}_{+}\right)$ cubes with edge lengths as positive integers equals $2... | 4. (1) Since $2005=1728+125+125+27=12^{3}+$ $5^{3}+5^{3}+3^{3}$, therefore, there exists $n=4$, such that $n_{\min } \leqslant 4$.
Also, $10^{3}=1000,11^{3}=1331,12^{3}=1728,13^{3}=$ 2197, and $12^{3}3 \times 8^{3}$, so the edge length of the largest cube can only be $9,10, 11$ or 12. And
$$
\begin{array}{l}
20050,
\e... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then, $(2+a)(2+b)+b^{2}$
is equal to ( ).
(A) 4
(B) -4
(C) 2
(D) -2 | 1. A.
From the given, we have $(1+a)(1+b)=(1-a)(1-b)$, simplifying to $a+b=0$, which means $a=-b$.
Therefore, $(2+a)(2+b)+b^{2}=(2-b)(2+b)+b^{2}=4$. | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Place the natural numbers $1,2, \cdots, 2 n$ randomly on a circle. It is found that among all sets of three consecutive numbers, there are $a$ sets where all three numbers are odd, $b$ sets where exactly two numbers are odd, $c$ sets where only one number is odd, and $d$ sets where all three numbers are even. If $a ... | 4. -3 .
If each number is counted 3 times, a total of $6 n$ numbers are counted ($3 n$ odd numbers, $3 n$ even numbers), then the system of equations can be set up as
$$
\left\{\begin{array}{l}
3 a+2 b+c=3 n, \\
2 c+b+3 d=3 n .
\end{array}\right.
$$
By eliminating $n$ from equations (1) and (2), we get
$$
\frac{b-c}{... | -3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $a, b, c, d$ are all even numbers, and $0<a<b<c<d, d-a=90$. If $a, b, c$ form an arithmetic sequence, and $b, c, d$ form a geometric sequence, then the value of $a+b+c+d$ is ( ).
(A) 194
(B) 284
(C) 324
(D) 384 | 6. A.
According to the problem, we can set $a, b, c, d$ as $b-m, b, b+m, \frac{(b+m)^{2}}{b}$ (where $m$ is a positive even number, and $m<b$).
From $d-a=90$, we get $\frac{(b+m)^{2}}{b}-(b-m)=90$, which simplifies to
$$
m^{2}+3 b m-90 b=0 \text {. }
$$
Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know th... | 194 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given a triangle $\triangle A B C$ with side lengths $4,5,6$ respectively, the circumcircle of $\triangle A B C$ is a great circle of sphere $O$, and $P$ is a point on the sphere. If the distances from point $P$ to the three vertices of $\triangle A B C$ are all equal, then the volume of the tetrahedron $P-A B C$ is... | 1.10 .
Since $P A=P B=P C$, the projection of point $P$ on the plane $A B C$ is the circumcenter $O$ of $\triangle A B C$, i.e., $P O \perp$ plane $A B C$, and $P O$ equals the radius $R$ of sphere $O$. Therefore,
$$
V_{P-A B C}=\frac{1}{3} S_{\triangle A B C} \cdot P O=\frac{1}{3} \cdot \frac{a b c}{4 R} \cdot R=10 \... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Three positive integers $a$, $b$, and $c$ satisfy the conditions:
(1) $a<b<c<30$;
(2) For some positive integer base, the logarithms of $a(2 b-a)$ and $c^{2}+$ $60 b-11 a$ are 9 and 11, respectively.
Then the value of $a-2 b+c$ is $\qquad$. | 3. -4 .
Let the base be a positive integer $n(n \geqslant 2)$, then from condition (2) we get
$$
\left\{\begin{array}{l}
a(2 b-a)=n^{9} . \\
c^{2}+60 b-11 a=n^{11} .
\end{array}\right.
$$
When $n \geqslant 3$, from equation (1) we get
$$
3^{9} \leqslant n^{9}=a(2 b-a)2^{9}$, i.e., $b \geqslant 23$.
$$
Thus, we have
$$... | -4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Now define an operation * :
When $m$ and $n$ are both positive odd numbers or both positive even numbers,
$$
m * n=m+n \text {; }
$$
When one of $m$ and $n$ is a positive odd number and the other is a positive even number,
$$
m * n=m \cdot n \text {. }
$$
Then, the number of elements in the set $M=\{(a, b) \mid a... | 5.41.
Since $36=1+35=3+33=5+31=7+29=\cdots=$ $33+3=35+1$ (a total of 18) $; 36=2+34=4+32=\cdots=32$ $+4=34+2$ (a total of 17) ; $36=1 \times 36=3 \times 12=4 \times 9=9$ $\times 4=12 \times 3=36 \times 1$ (a total of 6), therefore, the number of elements in set $M$ is $18+17+6=41$. | 41 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given $\alpha, \beta, \gamma \in \mathbf{R}$,
$$
\begin{aligned}
u= & \sin (\alpha-\beta)+\sin (\beta-\gamma)+ \\
& \sin (\gamma-\alpha) .
\end{aligned}
$$
Then $u_{\text {max }}+u_{\text {min }}=$ | 6.0.
$$
\begin{aligned}
u & =\sin \alpha \cdot \cos \beta+\sin \beta \cdot \cos \gamma+\sin \gamma \cdot \cos \alpha- \\
& =\left|\begin{array}{lll}
\cos \alpha \cdot \sin \beta-\cos \beta \cdot \sin \gamma-\cos \gamma \cdot \sin \alpha \\
\sin \alpha & \cos \alpha & 1 \\
\sin \beta & \cos \beta & 1 \\
\sin \gamma & \c... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\} \text {, }
$$
where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedra. How many triangles can graph $G$ have at most? | Three, first prove the following conclusion:
In a space graph with $n$ points, if there are no triangles, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$.
Proof: Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the most, say there are $k$ edges: $... | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 (1) In a $4 \times 4$ grid paper, some small squares are colored red, and then two rows and two columns are crossed out. If no matter how they are crossed out, there is at least one red small square that is not crossed out, how many small squares must be colored at least?
(2) If the “$4 \times 4$” grid paper ... | Solution: (1) If the number of colored small squares is less than or equal to 4, then we can appropriately strike out two rows and two columns to remove all the colored small squares.
If the number of colored small squares is 5, then by the pigeonhole principle, there must be at least one row with 2 or more colored sm... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $P(x)$ be the product of the digits in the decimal representation of $x$. Try to find all positive integers $x$ such that
$$
P(x)=x^{2}-10 x-22
$$
holds. | Solution: Let $x=\overline{a_{n} a_{n-1} \cdots a_{1}}$.
When $n \geqslant 2$, according to (1) and (2), we have the estimate
$$
x \geqslant 10^{n-1} \cdot a_{n}>9^{n-1} \cdot a_{n} \geqslant P(x) \text {, }
$$
thus $x^{2}-10 x-22<x$.
Solving this, we get $0<x<13$.
Upon verification, $x=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Some middle school students in a city participated in a mathematics invitational competition, and this competition had a total of 6 problems. It is known that each problem was solved correctly by exactly 500 students, but for any two students, there is at least one problem that neither of them solved correctl... | Solution: First, each student can answer at most 4 questions correctly.
In fact, according to the problem, it is impossible for any student to answer 6 questions correctly. If a student answers 5 questions correctly, by the problem's condition, all other students must have answered the same question incorrectly, which ... | 1000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 On a plane, there are 7 points, and some line segments can be connected between them, so that any three points among the 7 points must have two points connected by a line segment. How many line segments are needed at least? Prove your conclusion. | Proof: First, to meet the requirements of the problem, the number of line segments to be connected must be greater than or equal to 9.
Below, we discuss in 4 cases.
(1) If among the 7 points, there exists 1 point not connected to any other points, by the problem's condition, the remaining 6 points must each be connecte... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Three, (16 points) Let $n$ be a positive integer, and $d_{1}<d_{2}<$ $d_{3}<d_{4}$ be the 4 smallest consecutive positive integer divisors of $n$. If $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2}$, find the value of $n$. | If $n$ is odd, then $d_{1} 、 d_{2} 、 d_{3} 、 d_{4}$ are all odd. Hence $n=d_{1}^{2}+d_{2}^{2}+d_{3}^{2}+d_{4}^{2} \equiv 1+1+1+1 \equiv 0(\bmod 4)$. Contradiction.
If $4 \mid n$, then $d_{1}=1, d_{2}=2$. By $d_{i}^{2} \equiv 0$ or $1(\bmod 4)$, we have $n \equiv 1+0+d_{3}^{2}+d_{4}^{2} \neq 0(\bmod 4)$. Also a contradi... | 130 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Simplify $\frac{1}{4+\sqrt{59+30 \sqrt{2}}}+\frac{1}{3-\sqrt{66-40 \sqrt{2}}}$, the result is ( ).
(A) irrational number
(B) proper fraction
(C) odd number
(D) even number | \begin{array}{l}\text { - 1.D. } \\ \text { Original expression }=\frac{1}{4+\sqrt{(5 \sqrt{2}+3)^{2}}}+\frac{1}{3-\sqrt{(5 \sqrt{2}-4)^{2}}} \\ =\frac{1}{7+5 \sqrt{2}}+\frac{1}{7-5 \sqrt{2}}=-14 .\end{array} | -14 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Among the natural numbers not exceeding 100, add up all the numbers that are multiples of 3 or 5, the sum is $\qquad$ . | 1.2418.
The sum of multiples of 3 from $1 \sim 100$ is $S_{3}=3(1+2+\cdots+33)=1683$,
The sum of multiples of 5 from $1 \sim 100$ is $S_{5}=5(1+2+\cdots+20)=1050$, the sum of multiples of 15 from $1 \sim 100$ is $S_{15}=15(1+2+\cdots+6)=315$. Therefore, $S=S_{3}+S_{5}-S_{15}=2418$. | 2418 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If real numbers $x, y$ satisfy
$$
\frac{x}{3^{3}+4^{3}}+\frac{y}{3^{3}+6^{3}}=1, \frac{x}{5^{3}+4^{2}}+\frac{y}{5^{3}+6^{3}}=1,
$$
then $x+y=$ $\qquad$ . | 3.432 .
It is known that $3^{3}$ and $5^{3}$ are the two roots of the equation $\frac{x}{t+4^{3}}+\frac{y}{t+6^{3}}=1$, which means they are the roots of the equation
$$
t^{2}+\left(4^{3}+6^{3}-x-y\right) t+\left(4^{3} \times 6^{3}-4^{3} y-6^{3} x\right)=0
$$
By the relationship between roots and coefficients, we hav... | 432 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. (25 points) $a$, $b$, $c$ are positive integers, and $a^{2}+b^{3}=$ $c^{4}$. Find the minimum value of $c$.
保留了源文本的换行和格式。 | 3. Clearly, $c>1$. From $b^{3}=\left(c^{2}-a\right)\left(c^{2}+a\right)$, if we take $c^{2}-a=b, c^{2}+a=b^{2}$, then at this time
$$
c^{2}=\frac{b(b+1)}{2} \text {. }
$$
Examining $b$ from small to large to make the right side a perfect square, we find that when $b=8$, $c=6$, and thus, $a=28$.
Next, we show that ther... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and its smallest positive period is 2. Then the value of $f(-1)$ is $\qquad$ . | 1.0.
Since $f(x+2)=f(x)$, let $x=-1$, then we have $f(-1+2)=f(-1)$,
which means $f(1)=f(-1)=-f(1)$.
Therefore, $f(-1)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the set $M=\left\{x \mid x=2^{n}-2^{k}\right.$, where $n, k \in$ $\mathbf{N}$, and $n>k\}, P=\{x \mid 1912 \leqslant x \leqslant 2004$, and $x$ $\in \mathbf{N}\}$. Then, the sum of all elements in the set $M \cap P$ is $\qquad$. | 2.3904.
Since $2^{10}=1024,2^{11}=2048$, all elements in the set $M \cap P$ must be of the form $2^{11}-2^{k}$ (where $k \in \mathbf{N}$).
Also, $1912 \leqslant 2^{11}-2^{k} \leqslant 2004$, so $44 \leqslant 2^{k} \leqslant 136$.
The only values of $k$ that satisfy this condition are 6 and 7, so the set $M \cap$ $P$ c... | 3904 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=99^{\frac{1}{99}}, a_{n}=$ $\left(a_{n-1}\right)^{a_{1}}$. When $a_{n}$ is an integer, the smallest positive integer $n$ is $\qquad$ | 4. 100 .
Obviously, for any positive integer $n$, we have $a_{n}>0$.
From $a_{n}=\left(a_{n-1}\right)^{a_{1}}$, taking the common logarithm on both sides, we get
$\lg a_{n}=a_{1} \lg a_{n-1}$.
Therefore, $\left\{\lg a_{n}\right\}$ is a geometric sequence with the first term $\lg a_{1}$ and the common ratio $a_{1}$. So... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For $i=1,2, \cdots, n$, we have $\left|x_{i}\right|<1$, and
$$
\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|=19+\left|x_{1}+x_{2}+\cdots+x_{n}\right| \text {. }
$$
Find the minimum value of the positive integer $n$. | Given $19=\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|-\mid x_{1}+x_{2}+\cdots+x_{n}|\leqslant| x_{1}|+| x_{2}|+\cdots+| x_{n} \mid<n$, it follows that $n \geqslant 20$. When $n=20$, take
$$
x_{i}=\left\{\begin{array}{cc}
0.95, & \text { when } i \text { is odd: } \\
-0.95, & \text { when } i \text {... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The largest natural number that cannot be expressed as the sum of four distinct composite numbers is $\qquad$ .
翻译完成，保留了原文的格式和换行。 | 2. 26 .
The sum of the four smallest composite numbers is $4+6+8+9=27$. It is easy to prove that any natural number greater than 27 can be expressed as the sum of four different composite numbers. | 26 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find $\sum_{k=0}^{2 n-1}(-1)^{k+1}(k+1)\left(\mathrm{C}_{2 n}^{k}\right)^{-1}$.
Analysis: Considering the use of identity (III), we can obtain
$$
\frac{k+1}{\mathrm{C}_{2 n}^{k}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{k+1}}=\frac{2 n+1}{\mathrm{C}_{2 n+1}^{2 n-k}}=\frac{2 n-k}{\mathrm{C}_{2 n}^{2 n-k-1}} \text {. }... | Let the original expression be $y_{n}$, from the analysis we have
$$
y_{n}=\sum_{k=0}^{2 n-1}(-1)^{k+1} \frac{2 n-k}{C_{2 n}^{2 n-k-1}} \text {. }
$$
By making a change of the summation index, set $l=2 n-$ $k-1$, when $k$ varies from 0 to $2 n-1$, $l$ varies from $2 n-1$ to 0. Thus,
$$
\begin{array}{l}
y_{n}=\sum_{l=0... | 0 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given that $a, b, x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$
Find the value of $a x^{5}+b y^{5}$. | Given the conditions, construct the recurrence relation:
$$
\begin{array}{l}
s_{n}=a x^{n}+b y^{n}, n=1,2, \cdots, \\
\text { then } s_{n+2}=a x^{n+2}+b y^{n+2} \\
=\left(a x^{n+1}+b y^{n+1}\right)(x+y)- \\
\quad x y\left(a x^{n}+b y^{n}\right) .
\end{array}
$$
Therefore, \( s_{n+2}=(x+y) s_{n+1}-x y s_{n} \). Substit... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 1, in rectangle $A B C D$, $A B=2, B C=$ $\sqrt{3}, E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Line segments $D E$, $B F$, and $E F$ intersect diagonal $A C$ at points $M$, $N$, and $P$ respectively. The number of right triangles formed by the line segments in Figure 1 is ( ).
(... | 4.B.
Let the notation $[A]$ represent the number of right-angled triangles with $A$ as the right-angle vertex, then $[A]=2$,
Similarly, $[B]=1,[C]=2,[D]=1,[E]=3$,
$$
[F]=3,[M]=[N]=[P]=0 \text {. }
$$
Therefore, the number of right-angled triangles is
$$
2+1+2+1+3+3=12 \text {. }
$$ | 12 | Geometry | MCQ | Yes | Yes | cn_contest | false |
2. Fill in $n$ distinct numbers on a circle so that for every three consecutive numbers, the middle number is equal to the product of the two numbers on its sides. Then $n=$ $\qquad$
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directl... | 2.6.
According to the problem, $n \geqslant 3$. Take any two adjacent numbers $a, b(a \neq b)$, and by the problem's condition, the numbers on the circle can be written in sequence as
$$
a, b, \frac{b}{a}, \frac{1}{a}, \frac{1}{b}, \frac{a}{b}, a, b, \cdots
$$
Starting from the 7th number, the sequence of numbers has... | 6 | Number Theory | proof | Yes | Yes | cn_contest | false |
3. A certain temple has three bronze bells, A, B, and C. Bell A rings once every $4 \mathrm{~s}$, Bell B rings once every $5 \mathrm{~s}$, and Bell C rings once every $6 \mathrm{~s}$. At the arrival of the New Year, the three bells ring simultaneously and stop ringing at the same time. A person hears a total of 365 bel... | 3.484 .
Let the duration of the bell ringing be $60 k \mathrm{~s}$, then during this period, the number of times the first clock rings is $x=\frac{60 k}{4}+1=15 k+1$; the number of times the second clock rings is $y=\frac{60 k}{5}+1=12 k+1$; the number of times the third clock rings is $z=\frac{60 k}{6}+1=10 k+1$; the... | 484 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. From $1,2, \cdots, 10$ choose 3 different numbers $a, b, c$ as the coefficients of the quadratic equation $a x^{2}+b x=c$. Then the number of equations with different solutions is $\qquad$ | 4.654.
When considering $a1$, for each factor $k$ of $d$ greater than 1, if $\frac{c}{k} \geqslant 3$, then the array $\left(\frac{a}{k}, \frac{b}{k}, \frac{c}{k}\right)$ has already been considered before and should be removed. The number of remaining arrays is denoted as $f(c)$. Thus, we have
$$
\begin{array}{l}
f(3... | 654 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. (25 points) Given that $a$ and $b$ are coprime positive integers, satisfying $a+b=2005$. Let $[x]$ denote the integer part of the number $x$, and let
$$
\begin{array}{l}
A=\left[\frac{2005 \times 1}{a}\right]+\left[\frac{2005 \times 2}{a}\right]+\cdots+\left[\frac{2005 \times a}{a}\right], \\
B=\left[\frac{2005 \tim... | 3. It is evident that the last terms of the two sums $A$ and $B$ are both 2005. Consider the remaining terms.
Since $(a, b)=1, a+b=2005$, it follows that
$(a, 2005)=1, (b, 2005)=1$.
Therefore, for $1 \leqslant m \leqslant a-1, 1 \leqslant n \leqslant b-1, \frac{2005 m}{a}$ and $\frac{2005 n}{b}$ are not integers, and t... | 2011016 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. (16 points) A warehouse has 50 containers of the same specification, which are to be delivered to the dock by a transportation company. The transportation company has three types of trucks that can carry 1, 2, and 3 containers per trip, respectively. The charges for each trip are 120 yuan, 160 yuan, and 180 yuan, re... | 2. Let the number of trucks needed to transport 1, 2, and 3 containers be $x$, $y$, and $z$ respectively. According to the problem, we have
$$
\left\{\begin{array}{l}
x+y+z=20, \\
x+2 y+3 z=50 .
\end{array}\right.
$$
(1) $\times 3$ - (2) gives $2 x+y=10$.
Then $\left\{\begin{array}{l}y=10-2 x, \\ z=10+x .\end{array}\r... | 3300 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9.6. How many ways are there to divide the set
$$
\left\{2^{0}, 2^{1}, 2^{2}, \cdots, 2^{2000}\right\}
$$
into two non-empty disjoint subsets $A$ and $B$, such that the equation $x^{2}-S(A) x+S(B)=0$ has integer roots? Here, $S(M)$ denotes the sum of all elements in the set $M$. | 9.6.1 003 types.
Let $x_{1} \leqslant x_{2}$ be the roots of the equation, then
$$
x_{1}+x_{2}=S(A), x_{1} x_{2}=S(B),
$$
and $x_{1}, x_{2} \in \mathbf{Z}_{+}$. Therefore,
$$
\begin{array}{l}
\left(x_{1}+1\right)\left(x_{2}+1\right)=S(B)+S(A)+1 \\
=1+2+4+\cdots+2^{200}+1=2^{2006},
\end{array}
$$
which means $x_{1}+1... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.1. Try to find the smallest positive integer that cannot be expressed in the form $\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, where $a, b, c, d$ are all positive integers. | 10.1.11.
From the problem, we have
$$
\begin{array}{l}
1=\frac{4-2}{4-2}, 3=\frac{8-2}{4-2}, 5=\frac{16-1}{4-1}=\frac{2^{5}-2}{2^{3}-2}, \\
7=\frac{16-2}{4-2}, 9=2^{3}+1=\frac{2^{6}-1}{2^{3}-1}=\frac{2^{7}-2}{2^{4}-2}, \\
2=2 \times 1=\frac{2^{3}-2^{2}}{2^{2}-2}, \cdots, 10=2 \times 5=\frac{2^{6}-2^{2}}{2^{3}-2} .
\en... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10.3. On the back of 2005 cards, there are 2005 different real numbers written. Each time you ask a question, you can point to any three cards and ask about the set of 3 numbers written on them. How many times do you need to ask at minimum to definitely find out what numbers are written on the back of each card? | 10.3.1003 times.
Assume that $N$ questions have been proposed.
Obviously, each card should participate in at least one question; otherwise, it would be impossible to determine the number on that card.
Suppose there are $k$ cards that each participate in exactly one question. Then, in the same question, it is impossib... | 1003 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.5. Place 16 rooks on an $8 \times 8$ chessboard. How many pairs of rooks can attack each other (a pair of rooks can attack each other if they are in the same row or column and there are no other rooks between them)? | 10.5.16 Correct.
It should be noted that if there are $a$ rooks in a row, then there are $a-1$ pairs of rooks that can attack each other in that row. Therefore, the number of rooks that can attack each other horizontally will not be less than 8 pairs.
Similarly, the number of rooks that can attack each other vertical... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11.1. Let $a_{1}, a_{2}, \cdots, a_{50}, b_{1}, b_{2}, \cdots, b_{50}$ be distinct numbers such that the equation
$$
\begin{array}{l}
\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right| \\
=\left|x-b_{1}\right|+\left|x-b_{2}\right|+\cdots+\left|x-b_{50}\right|
\end{array}
$$
has a finite number of r... | 11.1.49.
$$
\begin{array}{l}
\text { Let } f(x)=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{50}\right|- \\
\left|x-b_{1}\right|-\left|x-b_{2}\right|-\cdots-\left|x-b_{50}\right| \text {. } \\
\end{array}
$$
Thus, the original equation is $f(x)=0$.
Let $c_{1}<c_{2}<\cdots<c_{100}$ be the elements of the... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4.6 Four table tennis players are dressed in 4 different colors for a performance match, with 2 in red, 2 in yellow, 1 in blue, and 1 in black. Each performance selects 3 players to appear, and matches are only played between players with different colored outfits, with the specific rules being:
(1) If the "3-person gr... | 4.44.
Represent the 2 athletes in red as two parallel lines $l_{1}$, $l_{2}$; the 2 athletes in yellow as two other parallel lines $l_{3}$, $l_{4}$; and the athletes in blue and black as intersecting lines $l_{5}$, $l_{6}$, which intersect with $l_{1}$, $l_{2}$, $l_{3}$, $l_{4}$. This results in Figure 6, where there ... | 44 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) A tetrahedron has 6 edges and 4 triangular faces. Given that the 6 edge lengths are exactly the positive integers $n, n+$ $1, n+2, n+3, n+4, n+5$. If the perimeter of a certain triangular face is a multiple of 3, then this triangle is painted red; otherwise, it is painted yellow. Question: What is th... | Three, at most 3 yellow triangles.
Since 6 consecutive positive integers divided by 3 must have remainders of 2 zeros, 2 ones, and 2 twos, denoted as $a, a, b, b, c, c$. If there are 4 yellow triangles, take any yellow $\triangle ABC$, then it must have two sides with the same remainder when divided by 3, denoted as $a... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The last four digits of the number $7^{355}$ are $\qquad$ | II. 1.1943.
This problem is to find the remainder of $7^{355}$ divided by $10^{4}$.
$$
\begin{array}{l}
7^{355}=7 \times 49^{1 \pi}=7 \times 49 \times(50-1)^{2 \times 88} \\
=7 \times 49 \times(2400+1)^{28} \\
\equiv 7 \times 49 \times\left(1+2400 \mathrm{C}_{88}^{1}\right) \\
\equiv 343 \times[1+2400 \times(100-12)] \... | 1943 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$. | Explanation: Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$, with $x_{1}<x_{2}$, and let $C(0, c)$, where $c \neq 0$. Since $\triangle A B C$ is a right triangle, it follows that $x_{1}<0<x_{2}$, and it must be that $\angle A C B=90^{\circ}$. The graph can only be as shown in Figure 2 or Figure 3. In either ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Question 7 Let $a, b, c$ be positive real numbers. Prove:
$$
\begin{array}{l}
\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+a+c)^{2}}{2 b^{2}+(c+a)^{2}}+ \\
\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8^{[3]} .
\end{array}
$$ | Prove: Let $s=a+b+c$,
$$
f(t)=\frac{(t+s)^{2}}{2 t^{2}+(s-t)^{2}}, t \in[0, s) \text {. }
$$
Since $f(t)=\frac{1}{3}+\frac{2}{3} \times \frac{4 s t+s^{2}}{3\left(t-\frac{s}{3}\right)^{2}+\frac{2}{3} s^{2}}$
$$
\leqslant \frac{1}{3}+\frac{4 s t+s^{2}}{s^{2}}=4\left(\frac{1}{3}+\frac{t}{s}\right),
$$
thus $f(a)+f(b)+f(... | 8 | Inequalities | proof | Yes | Yes | cn_contest | false |
4. The sequence $a_{1}, a_{2}, \cdots$ is defined as follows:
$$
a_{n}=2^{n}+3^{n}+6^{n}-1(n=1,2,3, \cdots) \text {. }
$$
Find all positive integers that are coprime with every term of this sequence.
(Poland provided) | 4. The only positive integer that satisfies the condition is 1.
The following is the proof: For any prime number $p$, it must be a divisor of some term in the sequence $\left\{a_{n}\right\}$.
For $p=2$ and $p=3$, they are divisors of $a_{2}=48$.
For every prime number $p$ greater than 3, since
$$
(2, p)=1,(3, p)=1,(6,... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. There are two people, A and B. A, on a car, notices that B is walking in the opposite direction. $1 \mathrm{~min}$ later, A gets off the car to chase B. If A's speed is twice that of B, but slower than the car's speed by $\frac{4}{5}$, then the time it takes for A to catch up with B after getting off the car is $\ma... | 8.11.
Let the speed of A be $x \mathrm{~m} / \mathrm{min}$, then according to the problem, the speed of B is $\frac{x}{2} \mathrm{~m} / \mathrm{min}$, and the speed of the car is $5 x \mathrm{~m} / \mathrm{min}$.
Let the time it takes for A to catch up with B be $t \mathrm{~min}$. According to the problem, we have $\f... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. If a convex $n$-sided polygon has exactly 4 obtuse interior angles, then the maximum number of sides $n$ of this polygon is $\qquad$ . | 10.7.
Since a convex $n$-sided polygon has exactly 4 obtuse interior angles, the sum of these 4 angles is greater than $360^{\circ}$ and less than $720^{\circ}$. The other $n-4$ angles are right angles or acute angles, so the sum of these $n-4$ angles is no more than $(n-4) \times 90^{\circ}$ and greater than $0^{\cir... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. If in a $4 \times 4$ grid of 16 cells, each cell is filled with a number such that the sum of the numbers in all adjacent cells of each cell is 1, then the sum of the 16 numbers in the grid is $\qquad$
(Note: Adjacent cells refer to cells that share exactly one edge) | 11.6.
First, fill the 16 squares with the English letters $A, B, C$, $\cdots, O, P$ (as shown in Figure 8). Given that for each square, the sum of the numbers in all adjacent squares is 1, so,
$$
\begin{aligned}
16= & 2(A+D+M+P)+3(B+C+E+H+ \\
& I+L+N+O)+4(F+G+J+K) \\
= & 2(A+B+C+\cdots+O+P)+(B+C+E+ \\
& H+I+L+N+O)+2(F... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
13. Given that $p$, $q$, $\frac{2 q-1}{p}$, and $\frac{2 p-1}{q}$ are all integers, and $p>1$, $q>1$. Find the value of $p+q$. | 13. If $\frac{2 q-1}{p} \geqslant 2, \frac{2 p-1}{q} \geqslant 2$, then
$$
2 q-1 \geqslant 2 p, 2 p-1 \geqslant 2 q \text {. }
$$
Adding the two inequalities gives $2 p+2 q-2 \geqslant 2 p+2 q$. This is clearly a contradiction.
Therefore, at least one of $\frac{2 q-1}{p}$ and $\frac{2 p-1}{q}$ must be less than 2. Sup... | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. A box contains balls of red, yellow, and white. If the number of white balls is at most $\frac{1}{2}$ of the number of yellow balls, and at least $\frac{1}{3}$ of the number of red balls, and the total number of yellow and white balls is no more than 55, then the box can contain at most $\qquad$ red balls. | 11.54.
Suppose there are $x$ white balls, $y$ yellow balls, and $z$ red balls. According to the problem, we can set up the following system of equations:
$$
\left\{\begin{array}{l}
x \leqslant \frac{1}{2} y, \\
x \geqslant \frac{1}{3} z, \\
y + x \leqslant 55 .
\end{array}\right.
$$
From equation (1), we get $y \geqs... | 54 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
13. In a football tournament with 10 teams, a double round-robin system is used, meaning any two teams play each other twice, once at home and once away. A win earns 3 points, a draw earns 1 point, and a loss earns 0 points.
(1) How many matches were played in total?
(2) If every match results in the maximum points, wh... | (1) Using a list or formula calculation, it can be known that a total of 90 matches were played in this competition.
(2) The highest score for a single match is 3 points. If the maximum score is achieved in every match, the total points of all teams in this competition would be $3 \times 90 = 270$ (points): The lowest ... | 38 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) As shown in Figure 5, a particle moves within the region $\{(x, y) \mid x \geqslant 0, y \geqslant 0\}$. In the first second, it moves from the origin to point $B_{1}(0,1)$, then from point $B_{1} \rightarrow C_{1}$ $\rightarrow A_{1}$, and subsequently moves along the $x$-axis, $y$-axis, and their p... | Three, suppose the time taken for a particle to reach points $A_{n}, B_{n}, C_{n}$ from the origin are $a_{n}, b_{n}, c_{n}$, respectively. Then it is clear that
$$
\begin{array}{l}
a_{1}=3, a_{2}=a_{1}+1, \\
a_{3}=a_{1}+12=a_{1}+3 \times 4, a_{4}=a_{3}+1, \\
a_{5}=a_{3}+20=a_{3}+5 \times 4, a_{6}=a_{5}+1, \\
\cdots \c... | 2008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Define the sequence $\left\{a_{n}\right\}: a_{n}=4+n^{3}, n \in \mathbf{N}_{+}$. Let $d_{n}=\left(a_{n}, a_{n+1}\right)$, i.e., $d_{n}$ is the greatest common divisor of $a_{n}$ and $a_{n+1}$. Then the maximum value of $d_{n}$ is $\qquad$ | 5.433.
Given) $d_{n} 1\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(4+n^{3}, 3 n^{2}+3 n+1\right)$.
Thus, we have
$$
d_{n} \mid\left[-3\left(4+n^{3}\right)+n\left(3 n^{2}+3 n+1\right), 3 n^{2}+3 n+1\right] \text {, }
$$
which means
$$
d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}+3 n+1\right) \text {. }
$$
Ther... | 433 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.