problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 2 43 | problem_type stringclasses 8
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8. From $1,2, \cdots, 2013$, select $3 k$ different numbers to form $k$ triples $\left(a_{i}, b_{i}, c_{i}\right)(i=1,2$, $\cdots, k)$. If $a_{i}+b_{i}+c_{i}(i=1,2, \cdots, k)$ these $k$ numbers are all distinct and less than 2013, then the maximum value of $k$ is | 8. 402.
On the one hand,
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \leqslant \sum_{i=1}^{k}(2013-i) \\
=2013 k-\frac{k(k+1)}{2},
\end{array}
$$
and
$$
\begin{array}{l}
\sum_{i=1}^{k}\left(a_{i}+b_{i}+c_{i}\right) \geqslant 1+2+\cdots+3 k \\
=\frac{3 k(3 k+1)}{2} .
\end{array}
$$
Thus, $2013 k-... | 402 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given the sequence $\left\{a_{n}\right\}(n \in \mathbf{N})$ satisfies: $a_{1}=1$, and for any non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$
Find the value of $\left[\frac{a_{2013}}{2012}\right]$ (where $[x]$ denotes ... | 10. Let $m=n$, we get $a_{0}=1$.
Let $n=0$, we get $a_{2 m}=4 a_{m}+2 m-3$.
In particular, $a_{2}=3$.
Let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right)=2 a_{m}+m \\
\Rightarrow a_{m+1}-a_{m}=a_{m}-a_{m-1}+2 .
\end{array}
$$
Thus, $a_{m+1}-a_{m}=2 m(m \geqslant 1)$,
$$
\be... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the smallest positive integer $n$, such that the sum of the squares of all positive divisors of $n$ is $(n+3)^{2}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note and not part of the problem statement, so it is provided in its original form as it is ... | Three, let $16 n+8, \\
\end{array}
$$
Contradiction.
Thus, $k \leqslant 5$.
And $k=0$ clearly does not meet the problem's conditions, so $1 \leqslant k \leqslant 5$.
Let $n=\prod_{i=1}^{i} p_{i}^{\alpha_{i}}$. Then
$$
\prod_{i=1}^{t}\left(\alpha_{i}+1\right)=k+2 \in[3,7] .
$$
Thus, $1 \leqslant t \leqslant 2$.
We dis... | 287 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers in the ... | 5. First, give two obvious conclusions:
(1) If $m$ is a positive integer and $x$ is a real number, then
$$
\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right] ;
$$
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$
Returning to the ori... | 587 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $A$ be a set of ten real-coefficient polynomials of degree five. It is known that there exist $k$ consecutive positive integers $n+1$, $n+2, \cdots, n+k$, and $f_{i}(x) \in A(1 \leqslant i \leqslant k)$, such that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence. Find the maximum possible... | 6. Given that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence, we know there exist real numbers $a$ and $b$ such that
$$
f_{i}(n+i)=a i+b .
$$
Notice that, for any fifth-degree polynomial $f$, the equation
$$
f(n+x)=a x+b
$$
has at most five real roots. Therefore, each polynomial in $A$ appea... | 50 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that $m$, $n$, and $p$ are real numbers. If $x-1$ and $x+4$ are both factors of the polynomial $x^{3}+m x^{2}+n x+p$, then
$$
2 m-2 n-p+86=
$$
$\qquad$. | 2. 100 .
From the divisibility property of polynomials, we know
$$
\left\{\begin{array} { l }
{ 1 + m + n + p = 0 , } \\
{ - 6 4 + 1 6 m - 4 n + p = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=12-4 m, \\
n=3 m-13 .
\end{array}\right.\right.
$$
Therefore, $2 m-2 n-p+86$
$$
\begin{array}{l}
=2 m-2(3 m-13)-(1... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
During the Teachers' Day, 200 teachers at a school sent text messages to greet each other, with each teacher sending exactly 1 text message to another teacher. Now, from them, the maximum number of $k$ teachers can be selected to attend an award ceremony, such that none of them has sent a text message to any of the oth... | Let the 200 teachers be denoted as $A_{1}, A_{2}, \cdots, A_{200}$, and let $G=\left\{A_{1}, A_{2}, \cdots, A_{200}\right\}$. If $A_{i}$ sends a message to $A_{j}$, then we mark an arrow from $A_{i}$ to $A_{j}$ as
$A_{i} \rightarrow A_{j}(1 \leqslant i \neq j \leqslant 200)$.
First, construct an instance for $k=67$.
In... | 67 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (50 points) There are 2014 points distributed on a circle, which are arbitrarily colored red and yellow. If starting from a certain point and moving around the circle in any direction to any position, the number of red points (including the starting point) is always greater than the number of yellow points, then... | Three, the advantages must be red points.
First, consider the simple case.
When there are $1, 2, 3, 4, 5, 6, 7$ points on the circumference, if there is at least one advantage point on the circumference, then the maximum number of yellow points on the circumference are $0, 0, 0, 1, 1, 1, 2$. From this, we can derive a ... | 671 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Given 361, find the smallest positive integer $n$, such that 16 numbers can be selected from $1,2, \cdots, n$ and filled into a $4 \times 4$ grid so that the product of the numbers in each row and each column is equal. | The smallest value of the positive integer $n$ is 27.
First, when $n=27$, a valid arrangement is shown in Figure 3.
Next, we prove that the smallest value of $n$ is 27.
Consider the prime factorization of the product $P$ of each row or column.
(1) If $P$ has only one prime factor, then $n \geqslant 2^{15} > 27$.
(2) If... | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find the last three digits of $2013^{2013^{2013}}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Notice,
$$
\begin{array}{l}
2013 \equiv 3(\bmod 5), 2013^{2} \equiv-1(\bmod 5), \\
2013^{4} \equiv 1(\bmod 5), 2013 \equiv 13(\bmod 25), \\
2013^{2} \equiv-6(\bmod 25), 2013^{4} \equiv 11(\bmod 25), \\
2013^{8} \equiv-4(\bmod 25), 2013^{12} \equiv 6(\bmod 25), \\
2013^{20} \equiv 1(\bmod 25) .
\end{array}
$$
Since $20... | 053 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Find the number of polynomials $f(x)=a x^{3}+b x$ that satisfy the following two conditions:
(1) $a, b \in\{1,2, \cdots, 2013\}$;
(2) The difference between any two numbers in $f(1), f(2), \cdots, f(2013)$ is not a multiple of 2013. (Wang Bin) | 4. It is known that the prime factorization of $2013=3 \times 11 \times 61$.
Let $p_{1}=3, p_{2}=11, p_{3}=61$.
For $a, b \in\{1,2, \cdots, 2013\}$, let
$a \equiv a_{i}\left(\bmod p_{i}\right), b \equiv b_{i}\left(\bmod p_{i}\right)$,
where $i=1,2,3$.
By the Chinese Remainder Theorem, we know that $(a, b)$ and $\left(... | 7200 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Let $x_{k} \in[-2,2](k=1,2, \cdots, 2013)$,
and
$x_{1}+x_{2}+\cdots+x_{2013}=0$. Try to find
$$
M=x_{1}^{3}+x_{2}^{3}+\cdots+x_{2013}^{3}
$$
the maximum value.
(Liu Kangning) | Given $x_{i} \in[-2,2](i=1,2, \cdots, 2013)$, we know that $x_{i}^{3}-3 x_{i}=\left(x_{i}-2\right)\left(x_{i}+1\right)^{2}+2 \leqslant 2$.
The equality holds if and only if $x_{i}=2$ or -1.
Noting that, $\sum_{i=1}^{2013} x_{i}=0$.
Thus, $M=\sum_{i=1}^{2013} x_{i}^{3}=\sum_{i=1}^{2013}\left(x_{i}^{3}-3 x_{i}\right)$
$$... | 4026 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Write 2013 different real numbers on 2013 cards. Place the cards face down on the table. Two players, A and B, play the following game: in each round, A can arbitrarily select ten cards, and B will tell A one of the ten numbers written on these cards (B does not tell A which card the number is written on). Find the ... | 4. The maximum value of $t$ is $1986=2013-27$.
Let $A_{1}, A_{2}, \cdots, A_{2013}$ be these 2013 cards.
First, note that player B has a strategy to prevent player A from determining the number written on any of the cards $A_{1}, A_{2}, \cdots, A_{27}$.
B divides $T=\{1,2, \cdots, 27\}$ into nine groups
$T_{i}=\{3 i-2... | 1986 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
8. $\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}$ modulo 31 is | 8. 16 .
Notice,
$$
\begin{array}{l}
\sum_{0 \leqslant i<j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}=\frac{1}{2} \sum_{0 \leqslant i \neq j \leqslant 50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j} \\
=\frac{1}{2}\left[\sum_{i=0}^{50} \sum_{j=0}^{50} \mathrm{C}_{50}^{i} \mathrm{C}_{50}^{j}-\sum_{j=0}^{50}\left(... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Given real numbers
$$
x_{i} \in[-6,10](i=1,2, \cdots, 10), \sum_{i=1}^{10} x_{i}=50 \text {. }
$$
Find the maximum value of $\sum_{i=1}^{10} x_{i}^{2}$, and the conditions that should be satisfied when the maximum value is achieved. | 10. Let $a_{i}=x_{i}+6$. Then $a_{i} \in[0,16]$, and
$$
\begin{array}{l}
\sum_{i=1}^{10} a_{i}=110, \\
\sum_{i=1}^{10} a_{i}^{2}=\sum_{i=1}^{10} x_{i}^{2}+12 \sum_{i=1}^{10} x_{i}+360 \\
=\sum_{i=1}^{10} x_{i}^{2}+960 .
\end{array}
$$
When there are at least two numbers in $a_{i}$ that are not both 0 and not both 16, ... | 772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (allowing repetition). What is the minimum number of distinct numbers in $a_{2}, \cdots, a_{2006}$? ${ }^{[1]}$
(2006, China Mathematical Olympiad) | This problem is solved using combinatorial geometry methods.
Let the positive integers $a_{1}, a_{2}, \cdots, a_{n}$ be such that $\frac{a_{i}}{a_{i+1}} (1 \leqslant i \leqslant n-1)$ are all distinct. Then the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{n}$ is $k=\left\lceil\frac{1+\sqrt{4 n-7}}{2}... | 46 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given a sequence $\left\{a_{n}\right\}$ with nine terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in \left\{2,1,-\frac{1}{2}\right\}$. Then the number of such sequences is | 8.491.
Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leqslant i \leqslant 8)$. Then for each sequence $\left\{a_{n}\right\}$ that meets the conditions, it satisfies
$$
\prod_{i=1}^{8} b_{i}=\prod_{i=1}^{8} \frac{a_{i+1}}{a_{i}}=\frac{a_{9}}{a_{1}}=1,
$$
and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.... | 491 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the maximum number of elements in a set $S$ that satisfies the following conditions:
(1) Each element in set $S$ is a positive integer not exceeding 100;
(2) For any two distinct elements $a, b$ in set $S$, there exists an element $c$ in $S$ such that
$$
(a, c)=(b, c)=1 \text {; }
$$
(3) For any two dist... | Each positive integer can be expressed as
$$
n=2^{k_{1}} \times 3^{k_{2}} \times 5^{k_{3}} \times 7^{k_{4}} \times 11^{k_{5}} q,
$$
where $q$ is coprime with 2, 3, 5, 7, and 11, and $k_{1}, k_{2}, \cdots, k_{5}$ are non-negative integers.
Let $A=\left\{n \leqslant 100 \mid k_{1}, k_{2}, \cdots, k_{5}\right.$ have exac... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $S=\{1,2, \cdots, 98\}$. Find the smallest positive integer $n$, such that in any $n$-element subset of $S$, one can always select 10 numbers, and no matter how these 10 numbers are divided into two groups, there is always one group in which there is a number that is coprime with the other four numbers, a... | The subset of 49 even numbers in set $S$ obviously does not satisfy the coprime condition, hence $n \geqslant 50$.
To prove that any 50-element subset $T$ of set $S$ contains 10 numbers that satisfy the condition, we use the following facts.
Lemma 1 If nine elements in $T$ have a common factor greater than 1, and the... | 50 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that $P(-2,3)$ is a point on the graph of the inverse proportion function $y=\frac{k}{x}$, $Q$ is a moving point on the branch of the hyperbola in the fourth quadrant. A line is drawn through point $Q$ such that it intersects the hyperbola $y=\frac{k}{x}$ at only one point, and intersects the $x$-axis and $y$-... | 4. 48.
It is known that $k=-6, y=-\frac{6}{x}, A(-4,0), B(0,6)$.
Let the line passing through point $Q$ be $y=a x+b$.
Then $\left\{\begin{array}{l}y=a x+b, \\ y=-\frac{6}{x}\end{array}\right.$ has only one solution.
Eliminating and rearranging gives $a x^{2}+b x+6=0$.
By $\Delta=b^{2}-24 a=0 \Rightarrow b^{2}=24 a$.
T... | 48 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$ satisfying $(a+b) \mid a b$.
$(1996$, China Mathematical Olympiad) | Let the greatest common divisor of $a$ and $b$ be $d$, and $a = a_{1}d$, $b = b_{1}d$. Then $\left(a_{1}, b_{1}\right) = 1$.
Substituting into $(a+b) \mid ab$, we get
$$
\left(a_{1} + b_{1}\right) \mid a_{1} b_{1} d \Rightarrow \left(a_{1} + b_{1}\right) \mid d.
$$
Let $d = k\left(a_{1} + b_{1}\right)$, we obtain all ... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If a number, from the highest digit to the lowest digit, does not decrease at each digit, it is called a "positive number" (such as $12$, $22$, $566$, $1448$, $123456789$, etc.); if a number, from the highest digit to the lowest digit, does not increase at each digit, it is called a "negative number" (such as $21$, ... | 4.525.
According to the definition of a wavy number, we discuss in two cases:
(1) The hundreds, tens, and units digits are in the pattern "small-large-small": there are
$$
2 \times 1+3 \times 2+\cdots+9 \times 8=240 \text { (numbers). }
$$
(2) The hundreds, tens, and units digits are in the pattern "large-small-large"... | 525 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Let the 10 complex roots of the equation $x^{10}+(13 x-1)^{10}=0$ be $x_{1}, x_{2}, \cdots, x_{10}$. Then
$$
\frac{1}{x_{1} \overline{x_{1}}}+\frac{1}{x_{2} \overline{x_{2}}}+\cdots+\frac{1}{x_{5} \overline{x_{5}}}=
$$
$\qquad$ | 7. 850 .
Let $\varepsilon=\cos \frac{\pi}{10}+\mathrm{i} \sin \frac{\pi}{10}$. Then $\varepsilon^{10}=-1$.
Given that the 10 complex roots of the equation $x^{10}+(13 x-1)^{10}=0$ are $x_{1}, x_{2}, \cdots, x_{10}$, we can assume them to be $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$, $\overline{x_{1}}, \overline{x_{2}}, \ove... | 850 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) 11 interest classes, several students participate (can participate repeatedly), and each interest class has the same number of students (full, unknown number). It is known that any nine interest classes include all students, while any eight interest classes do not include all students. Find the minimu... | Let the set of students in 11 interest classes be $A_{1}$, $A_{2}, \cdots, A_{11}$.
By the problem, we know $\left|A_{1}\right|=\left|A_{2}\right|=\cdots=\left|A_{11}\right|=x$.
Let $T=A_{1} \cup A_{2} \cup \cdots \cup A_{11}$.
By the problem, we know the union of any nine sets is $T$, and the union of any eight sets i... | 165 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let non-negative real numbers $x_{1}, x_{2}, \cdots, x_{6}$ satisfy
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{6}=1, x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6} \geqslant \frac{1}{540} . \\
\text { If } \max \left\{x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{6}+\right. \\
\left.x_{5} x_{6} x_{1}+x_{6... | Let $r=x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6}$,
$$
\begin{aligned}
s= & x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+ \\
& x_{4} x_{5} x_{6}+x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2} .
\end{aligned}
$$
By the AM-GM inequality,
$$
\begin{array}{l}
r+s=\left(x_{1}+x_{4}\right)\left(x_{2}+x_{5}\right)\left(x_{3}+x_{6}\ri... | 559 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. In $\triangle A B C$, it is known that $\angle A=2 \angle B, C D$ is the angle bisector of $\angle C$, $A C=16, A D=8$. Then $B C=$ | 3. 24 .
From $\angle A=2 \angle B$, we know $B C>A C$.
As shown in Figure 5, take a point $E$ on side $B C$ such that $E C=A C$,
and connect $D E$.
Then $\triangle C E D \cong \triangle C A D$
$$
\begin{array}{l}
\Rightarrow E D=A D, \\
\angle C E D=\angle C A D . \\
\text { Hence } \angle B D E=\angle C E D-\angle D... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given prime numbers $p$ and $q$, such that $p^{3}-q^{5}=(p+q)^{2}$. Then $\frac{8\left(p^{2013}-p^{2010} q^{5}\right)}{p^{2011}-p^{2009} q^{2}}=$ $\qquad$. | 4. 140.
If $p$ and $q$ have the same remainder when divided by 3, and since $p$ and $q$ are both prime numbers:
If the remainder is 0 when divided by 3, then $p=q=3$, at this time,
$$
p^{3}-q^{5}=0 \Rightarrow 3^{3}>q^{5} \Rightarrow q^{5}<27,
$$
such a prime number $q$ does not exist.
Therefore, it can only be $q=3$... | 140 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. From the set $A=\{1,2, \cdots, 30\}$, select five different numbers such that these five numbers form an arithmetic sequence. The number of different arithmetic sequences obtained is $\qquad$ . | 7. 196.
Let the five numbers be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, with a common difference of $d$.
From $a_{5}-a_{1}=4d$, we know $a_{1} \equiv a_{5}(\bmod 4)$.
Divide the set $A=\{1,2, \cdots, 30\}$ into four categories based on the remainder modulo 4:
$$
\begin{array}{l}
B=\{1,5,9, \cdots, 29\}, \\
C=\{2,6,10, \c... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x_{n}=\sum_{k=1}^{2013}\left(\cos \frac{k!\pi}{2013}\right)^{n}$. Then $\lim _{n \rightarrow \infty} x_{n}=$ | 4. 1953.
Notice that, $2013=3 \times 11 \times 61$.
Therefore, when $1 \leqslant k \leqslant 60$,
$$
\frac{k!}{2013} \notin \mathbf{Z}, \cos \frac{k!\pi}{2013} \in(-1,1).
$$
Hence $\lim _{n \rightarrow \infty} \cos ^{n} \frac{k!\pi}{2013}=0$.
And when $k \geqslant 61$, $\frac{k!}{2013}$ is an integer, and always an e... | 1953 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If the sum of all elements in a non-empty subset of $\{1,2, \cdots, 9\}$ is a multiple of 3, the subset is called a "Jin state subset". Then the number of such Jin state subsets is $\qquad$ . | 5. 175.
If a proper subset is a peculiar subset, then its complement is also a peculiar subset. Therefore, we only need to consider peculiar subsets with fewer than or equal to 4 elements.
Thus, there are 3 peculiar subsets with only 1 element; there are $\mathrm{C}_{3}^{2}+3 \times 3=12$ peculiar subsets with exactl... | 175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. The function $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for all $n \in \mathbf{N}$, we have
$$
\begin{array}{c}
f(f(n))+f(n)=2 n+3 \text {, and } f(0)=1 . \\
\text { Then } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=
\end{array}
$$ | 3. 2772 .
Substituting $n=0$ into equation (1) yields
$$
f(1)+1=3 \Rightarrow f(1)=2 \text {. }
$$
Furthermore, let $n=1$.
From equation (1), we get $f(2)=3$.
Similarly, $f(n)=n+1(n=3,4, \cdots, 10)$.
$$
\text { Hence } \frac{f(6) f(7) f(8) f(9) f(10)}{f(1)+f(2)+f(3)+f(4)+f(5)}=2772 \text {. }
$$ | 2772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. If $x, y$ are two different real numbers, and
$$
x^{2}=2 x+1, y^{2}=2 y+1 \text {, }
$$
then $x^{6}+y^{6}=$ $\qquad$ . | 4. 198 .
$$
\text { Let } S_{n}=x^{n}+y^{n} \text {. }
$$
From $x^{2}=2 x+1, y^{2}=2 y+1$, we get
$$
x^{n+2}=2 x^{n+1}+x^{n}, y^{n+2}=2 y^{n+1}+y^{n} \text {. }
$$
Thus, $S_{n+2}=2 S_{n+1}+S_{n}$.
Also, $S_{1}=2, S_{2}=6$, so
$$
S_{3}=14, S_{4}=34, S_{5}=82, S_{6}=198
$$ | 198 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If the remainder of $\underbrace{11 \cdots 1}_{n+1 \uparrow} 1$ divided by 3102 is 1, then the smallest positive integer $n$ is $\qquad$ . | 8. 138 .
Notice that, $3102=2 \times 3 \times 11 \times 47$.
From $\underbrace{11 \cdots 1}_{n+1 \uparrow}=3102 k+1(k \in \mathbf{Z})$, we know $\underbrace{11 \cdots 10}_{n \uparrow}=3102 k$.
Thus, $\underbrace{11 \cdots 10}_{n \uparrow}$ is divisible by $2, 3, 11, 47$.
(1) For any positive integer $n$, obviously, $\... | 138 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. Place 11 identical balls into six distinct boxes so that at most three boxes are empty. The number of ways to do this is $\qquad$. | 10. 4212 .
If there are no empty boxes, there are $\mathrm{C}_{10}^{5}$ ways to place them; if there is one empty box, there are $\mathrm{C}_{6}^{1} \mathrm{C}_{10}^{4}$ ways to place them; if there are two empty boxes, there are $\mathrm{C}_{6}^{2} \mathrm{C}_{10}^{3}$ ways to place them; if there are three empty box... | 4212 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4: There are 2013 balls placed around a circle, numbered $1, 2, \cdots, 2013$ in a clockwise direction. Starting from a certain ball $a$, and then taking every other ball in a clockwise direction, this process continues until only one ball remains on the circle. For what value of the number of ball $a$ will the... | First, consider a special case.
When the ball-picking procedure starts with the ball numbered 1, then when the person picking the balls reaches the next ball to be taken (numbered 3), it is essentially as if the ball numbered 2 has been placed at the farthest point in the direction of his movement. And when he takes th... | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. For a positive integer $n$, let $1 \times 2 \times \cdots \times n=n$!. If $M=1!\times 2!\times \cdots \times 10!$, then the number of perfect cubes among the positive divisors of $M$ is ( ).
(A) 468
(B) 684
(C) 846
(D) 648 | 2. A.
Notice that,
$$
\begin{array}{l}
M=1! \times 2! \times \cdots \times 10! \\
=2^{9} \times 3^{8} \times 4^{7} \times 5^{6} \times 6^{5} \times 7^{4} \times 8^{3} \times 9^{2} \times 10 \\
=2^{9+2 \times 7+5+3 \times 3+1} \times 3^{8+5+2 \times 2} \times 5^{6+1} \times 7^{4} \\
=2^{38} \times 3^{17} \times 5^{7} \... | 468 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
2. The number of all positive integer solutions $(x, y, z)$ for the equation $x y+1000 z=2012$ is. $\qquad$ | 2. 18 .
First consider the positive integer solutions for $x \leqslant y$.
When $z=1$, $x y=1012=4 \times 253=2^{2} \times 11 \times 23$. Hence, $(x, y)=(1,1012),(2,506),(4,253)$,
$$
(11,92),(22,46),(23,44) \text {. }
$$
When $z=2$, $x y=12=2^{2} \times 3$. Hence, $(x, y)=(1,12),(2,6),(3,4)$.
Next, consider the posit... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Xiao Li and Xiao Zhang are running at a constant speed on a circular track. They start from the same place at the same time. Xiao Li runs clockwise and completes a lap every 72 seconds, while Xiao Zhang runs counterclockwise and completes a lap every 80 seconds. At the start, Xiao Li has a relay baton, and each time... | 3. 720 .
The time taken from the start to the first meeting is
$$
\frac{1}{\frac{1}{72}+\frac{1}{80}}=\frac{720}{19}
$$
seconds, during which Xiao Li runs $\frac{10}{19}$ laps, and Xiao Zhang runs $\frac{9}{19}$ laps.
Divide the circular track into 19 equal parts, and number the points clockwise from the starting poin... | 720 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: At each vertex of a regular 2009-gon, a non-negative integer not exceeding 100 is placed. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k... | Solve $k_{\min }=100400$.
Let the numbers at vertices $A_{1}, A_{2}, \cdots, A_{2009}$ be $a_{1}, a_{2}, \cdots, a_{2009}$ respectively.
First, take $a_{2}=a_{4}=a_{2008}=100, a_{1}=a_{3}=a_{2009}=0$.
For each operation, assign
$$
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right)... | 100400 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given that $n$ is a positive integer. If in any permutation of $1,2, \cdots, n$, there always exists a sum of six consecutive numbers greater than 2013, then $n$ is called a "sufficient number". Find the smallest sufficient number.
---
Please note that the term "温饱数" is translated as "sufficient nu... | Three, the smallest well-fed number is 671.
First, prove: 671 is a well-fed number.
For any permutation $b_{1}, b_{2}, \cdots, b_{671}$ of $1,2, \cdots, 671$, divide it into 112 groups, where each group contains six numbers, and the last two groups share one number, i.e.,
$$
\begin{array}{l}
A_{1}=\left(b_{1}, b_{2}, \... | 671 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. From $1,2, \cdots, 100$ choose three different numbers such that they cannot form the three sides of a triangle. The number of different ways to do this is. | 2. 82075.
Let these three numbers be $i, j, k(1 \leqslant i<j<k \leqslant 100)$. Then $k \geqslant i+j$.
Therefore, the number of different ways to choose these three numbers is
$$
\begin{array}{l}
\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100}\left(\sum_{k=i+j}^{100} 1\right)\right]=\sum_{i=1}^{100}\left[\sum_{j=i+1}^{100-... | 82075 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{1,2, \cdots, 6\}(i=0,1,2,3)$. If positive integers $m, n \in A$, and $m+n=2014(m>n)$, then the number of pairs of positive integers $(m, n)$ that satisfy the condition is $\qquad$.
| 3. 551.
Notice that, $2014=5 \times 7^{3}+6 \times 7^{2}+5$.
Given $m, n \in A$, let
$$
\begin{array}{l}
m=a \times 7^{3}+b \times 7^{2}+c \times 7+d, \\
n=a^{\prime} \times 7^{3}+b^{\prime} \times 7^{2}+d^{\prime},
\end{array}
$$
where $a, b, c, d, a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime} \in\{1,2, \cdots, 6\}... | 551 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { II. (40 points) Given } x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\} \\
(i=1,2, \cdots, 2013) \text {, let } \\
S=x_{1} x_{2}+x_{2} x_{3}+\cdots+x_{2012} x_{2013} .
\end{array}
$$
Find the number of different integer values that $S$ can take. | Because $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, (\sqrt{2}-1)(\sqrt{2}+1)=1,
$$
Therefore, $x_{i} x_{i+1} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ instances of $3+2 \sqrt{2}$, $b$ instances of $3-2 \sqrt{2}$, and $c$ instances of 1. Then $a, b, c \in \mathbf{N}$, and... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the set $X=\{1,2, \cdots, 100\}$, and the function $f: X \rightarrow X$ satisfies the following conditions:
(1) For any $x \in X$, $f(x) \neq x$;
(2) For any 40-element subset $A$ of $X$, $A \cap f(A) \neq \varnothing$.
Find the smallest positive integer $k$ such that for any function $f$ satisfying the above c... | 5. First consider the function $f: X \rightarrow X$ defined as follows:
For $i=1,2, \cdots, 30, j=91,92, \cdots, 99$, define
$f(3 i-2)=3 i-1, f(3 i-1)=3 i$,
$f(3 i)=3 i-2, f(j)=100, f(100)=99$.
Clearly, the function $f$ satisfies condition (1).
For any 40-element subset $A$ of the set $X$, either there exists an integ... | 69 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. A three-digit number that is divisible by 35 and whose digits sum to 15 is | 5. 735.
Since the sum of the digits is 15, the three-digit number is a multiple of 3. Also, since the three-digit number is a multiple of 35, this three-digit number is a multiple of $35 \times 3=105$.
Starting from 105, list the three-digit multiples of 105:
$$
105,210,315,420,525,630,735,840, 945 \text {, }
$$
Amon... | 735 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $a, b, c$ are distinct positive integers. If the set
$$
\{a+b, b+c, c+a\}=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\} \text {, }
$$
where $n \in \mathbf{Z}_{+}$. Then the minimum value of $a^{2}+b^{2}+c^{2}$ is | 6. 1297 .
From $n^{2}+(n+1)^{2}+(n+2)^{2}=2(a+b+c)$ being even, we know that among $n, n+1, n+2$, there are two odd numbers and one even number, which means $n$ is odd.
Obviously, $n>1$.
Without loss of generality, let $a<b<c$.
If $n=3$, then
$$
\begin{array}{l}
a+b=9, a+c=16, b+c=25 \\
\Rightarrow a+b+c=25 .
\end{arr... | 1297 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. The minimum value of the function $f(x)=\sum_{k=1}^{2013}|x-k|$ is | 7.1013042.
Notice that, the median of $1,2, \cdots, 2013$ is 1007.
Therefore, when $x=1007$, the function reaches its minimum value
$$
\begin{array}{l}
f(1007)=2(1+2+\cdots+1006) \\
=1006 \times 1007=1013042
\end{array}
$$ | 1013042 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If three non-zero and distinct real numbers $a, b, c$ satisfy $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$, then $a, b, c$ are called "harmonic"; if they satisfy $a+c=2b$, then $a, b, c$ are called "arithmetic".
Given the set $M=\{x|| x | \leqslant 2013, x \in \mathbf{Z}\}$, the set $P$ is a three-element subset of set $M... | 3.1006.
If $a, b, c$ are both harmonic and arithmetic, then
$$
\left\{\begin{array}{l}
\frac{1}{a} + \frac{1}{b} = \frac{2}{c}, \\
a + c = 2b
\end{array} \Rightarrow \left\{\begin{array}{l}
a = -2b \\
c = 4b
\end{array}\right.\right.
$$
Thus, the good set is of the form $\{-2b, b, 4b\}(b \neq 0)$.
Since the good set ... | 1006 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y$ satisfy $x y+1=4 x+y$, and $x>1$. Then the minimum value of $(x+1)(y+2)$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 4.27.
From the problem, we know $y=\frac{4 x-1}{x-1}$.
$$
\begin{array}{l}
\text { Therefore, }(x+1)(y+2)=(x+1)\left(\frac{4 x-1}{x-1}+2\right) \\
\quad= \frac{3(x+1)(2 x-1)}{x-1} .
\end{array}
$$
Let $x-1=t>0$. Then
$$
\begin{array}{l}
(x+1)(y+2)=\frac{3(t+2)(2 t+1)}{t} \\
=6\left(t+\frac{1}{t}\right)+15 \geqslant ... | 27 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
13. Given a convex quadrilateral $ABCD$ whose four interior angles form an arithmetic sequence. In $\triangle ABD$ and $\triangle BCD$, if $\angle ABD = \angle CDB$, $\angle DAB = \angle CBD$, and the three interior angles of these two triangles also form an arithmetic sequence, then the maximum possible value of the s... | 13. D.
Assume the four interior angles of the quadrilateral are
$$
\alpha, \alpha+d, \alpha+2d, \alpha+3d \text{.}
$$
Since the sum of the interior angles of a quadrilateral is $360^{\circ}$, we have
$$
\alpha+\alpha+3d=\alpha+d+\alpha+2d=180^{\circ} \text{.}
$$
Given that $\angle ABD = \angle CDB$ and $\angle DAB =... | 240 | Geometry | MCQ | Yes | Yes | cn_contest | false |
19. In $\triangle A B C$, it is known that $A B=13, B C=14, C A=$ 15, points $D, E, F$ are on sides $B C, C A, D E$ respectively, and $A D \perp$ $B C, D E \perp A C, A F \perp B F$. If the length of segment $D F$ is $\frac{m}{n}$ $\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$, then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24... | 19. B.
As shown in Figure 3, with $D$ as the origin, the line $BC$ as the $x$-axis, and the line $AD$ as the $y$-axis, we establish a Cartesian coordinate system $x D y$.
It is easy to know that $p_{\triangle ABC}=13+14+15=42$,
$S_{\triangle ABC}=\sqrt{21(21-13) \times(21-14) \times(21-15)}=84$.
Thus, $AD=\frac{2 S_{\... | 21 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=$ $\qquad$ . | 10.2013.
Obviously, when $k \geqslant 11$, $\sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=0$.
$$
\begin{array}{l}
\text { Hence } \sum_{k=0}^{2013}\left[\frac{2013+2^{k}}{2^{k+1}}\right]=\sum_{k=0}^{10}\left[\frac{2013+2^{k}}{2^{k+1}}\right] \\
= 1007+503+252+126+63+ \\
31+16+8+4+2+1 \\
= 2013 .
\end{array... | 2013 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If every prime factor of 2013 is a term in a certain arithmetic sequence $\left\{a_{n}\right\}$ of positive integers, then the maximum value of $a_{2013}$ is $\qquad$ | $-, 1.4027$.
Notice that, $2013=3 \times 11 \times 61$.
If $3, 11, 61$ are all terms in a certain arithmetic sequence of positive integers, then the common difference $d$ should be a common divisor of $11-3=8$ and $61-3=58$. To maximize $a_{2013}$, the first term $a_{1}$ and the common difference $d$ should both be as ... | 4027 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If $a, b, c > 0, \frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$, then the minimum value of $a+2b+3c$ is . $\qquad$ | 2. 36 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a+2 b+3 c=(a+2 b+3 c)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$ | 36 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. If the distances from the center of the ellipse to the focus, the endpoint of the major axis, the endpoint of the minor axis, and the directrix are all positive integers, then the minimum value of the sum of these four distances is $\qquad$ .
| 5.61.
Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the distances from the center $O$ of the ellipse to the endpoints of the major axis, the endpoints of the minor axis, the foci, and the directrices are $a$, $b$, $c$, $d$ respectively, and satisfy
$$
c^{2}=a^{2}-b^{2}, d=\frac... | 61 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Given a composite number $k(1<k<100)$. If the sum of the digits of $k$ is a prime number, then the composite number $k$ is called a "pseudo-prime". The number of such pseudo-primes is . $\qquad$ | 7.23.
Let $S(k)$ denote the sum of the digits of $k$, and $M(p)$ denote the set of composite numbers with a pseudo-prime $p$.
When $k \leqslant 99$, $S(k) \leqslant 18$, so there are 7 prime numbers not exceeding 18, which are $2, 3, 5, 7, 11, 13, 17$.
The composite numbers with a pseudo-prime of 2 are $M(2)=\{20\}$.... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Make a full permutation of the elements in the set $\{1,2, \cdots, 8\}$, such that except for the number at the far left, for each number $n$ on the right, there is always a number to the left of $n$ whose absolute difference with $n$ is 1. The number of permutations that satisfy this condition is $\qquad$ | 8. 128 .
Suppose for a certain permutation that satisfies the conditions, the first element on the left is $k(1 \leqslant k \leqslant 8)$. Then, among the remaining seven numbers, the $8-k$ numbers greater than $k$, $k+1, k+2, \cdots, 8$, must be arranged in ascending order; and the $k-1$ numbers less than $k$, $1,2, ... | 128 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $g(x)=\sum_{i=1}^{n} \mathrm{C}_{n}^{i} \frac{i x^{i}(1-x)^{n-i}}{n}$. Then $g(2014)=$ $\qquad$ | $$
-, 1.2014
$$
From $\frac{r}{n} \mathrm{C}_{n}^{r}=\mathrm{C}_{n-1}^{r-1}$, we get
$$
g(x)=x \sum_{k=0}^{n-1} \mathrm{C}_{n-1}^{k} x^{k}(1-x)^{n-k}=x \text {. }
$$
Thus, $g(2014)=2014$. | 2014 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ distinct positive integers, none of whose decimal representations contain the digit 9. Prove:
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<30 .
$$ | Prove that if all positive integers in decimal representation without the digit 9 are arranged as $b_{1}, b_{2}, \cdots$, then
$$
\begin{array}{l}
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<\frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots \\
\leqslant c+\frac{9}{10} c+\left(\frac{9}{10}\right)^{2} c+\left(\frac{9}{10... | 30 | Number Theory | proof | Yes | Yes | cn_contest | false |
Let the function $f(x)$ satisfy $f(1)=1, f(4)=7$, and for any $a, b \in \mathbf{R}$, we have
$$
f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3} .
$$
Then $f(2014)=$ $\qquad$
(A) 4027
(B) 4028
(C) 4029
(D) 4030 | Solution 1 From $f\left(\frac{a+2 b}{3}\right)=\frac{f(a)+2 f(b)}{3}$, we get
$f(2)=f\left(\frac{4+2 \times 1}{3}\right)=\frac{f(4)+2 f(1)}{3}=3$,
$f(3)=f\left(\frac{1+2 \times 4}{3}\right)=\frac{f(1)+2 f(4)}{3}=5$.
Thus, we conjecture that $f(n)=2 n-1\left(n \in \mathbf{Z}_{+}\right)$.
We prove this by mathematical in... | 4027 | Algebra | MCQ | Yes | Yes | cn_contest | false |
15. Given
$$
3^{p}+3^{4}=90,2^{r}+44=76,5^{3}+6^{3}=1421 \text {. }
$$
Then $p r s=(\quad)$.
(A) 27
(B) 40
(C) 50
(D) 70
(E) 90 | 15. B. From the known information, we get $p=2, r=5, s=4$. Therefore, prs $=40$. | 40 | Algebra | MCQ | Yes | Yes | cn_contest | false |
3. A science and technology innovation competition sets first, second, and third prizes (all participants will receive an award), and the probabilities of winning the corresponding prizes form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prizes form an arithmetic sequence with... | 3. 500 .
Let the prize money obtained be $\xi$ yuan. Then $\xi=700,560,420$. From the problem, we know
$$
\begin{array}{l}
P(\xi=700)=a, P(\xi=560)=2 a, \\
P(\xi=420)=4 a .
\end{array}
$$
From $7 a=1$, we get $a=\frac{1}{7}$.
Therefore, $E \xi=700 \times \frac{1}{7}+560 \times \frac{2}{7}+420 \times \frac{4}{7}$ $=50... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $l_{1}, l_{2}, \cdots, l_{100}$ are 100 distinct and coplanar lines. If the lines numbered $4 k\left(k \in \mathbf{Z}_{+}\right)$ are parallel to each other, and the lines numbered $4 k-1$ all pass through point $A$, then the maximum number of intersection points of these 100 lines is $\qquad$ . | 7.4351 .
According to the problem, the number of combinations of any two lines out of 100 lines is
$$
\mathrm{C}_{100}^{2}-\mathrm{C}_{25}^{2}-\mathrm{C}_{25}^{2}+1=4351 .
$$ | 4351 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Seven balls of different colors are placed into three boxes numbered 1, 2, and 3. It is known that the number of balls in each box is not less than its number. The number of different ways to place the balls is $\qquad$ | 5.455.
(1) If the number of balls placed in boxes 1, 2, and 3 are 2, 2, and 3 respectively, then the number of different ways to place them is $\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2}=210$;
(2) If the number of balls placed in boxes 1, 2, and 3 are 1, 3, and 3 respectively, then the number of different ways to place them... | 455 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A_{1} A_{2} \cdots A_{101}$ be a regular 101-gon. Each vertex is colored either red or blue. Let $N$ be the number of obtuse triangles that satisfy the following conditions: the three vertices of the triangle are vertices of the 101-gon, the two acute vertices have the same color, and the color of the obtuse ve... | 2. Let $x_{i}=0$ or 1 represent $A_{i}$ being red or blue, respectively. For an obtuse triangle $\triangle A_{i-a} A_{i} A_{i+b}$, where vertex $A_{i}$ is the obtuse angle vertex, i.e., $a+b \leqslant 50$.
The coloring of the three vertices satisfies the condition if and only if
$$
\left(x_{i}-x_{i-a}\right)\left(x_{i}... | 32175 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
16. Given that one side of the square $A B C D$ lies on the line $y=2 x-17$, and the other two vertices are on the parabola $y=x^{2}$. Then the minimum value of the area of the square is $\qquad$ . | 16. 80 .
From the problem, we know that the distance from a point $\left(x, x^{2}\right)$ on the parabola to the line $l$ is
$$
d=\frac{\left|2 x-x^{2}-17\right|}{\sqrt{5}}>0 .
$$
When $x=1$, $d$ reaches its minimum value $\frac{16}{\sqrt{5}}$.
To minimize the area of the square, and by the symmetry of the square, th... | 80 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the sequence $\left\{a_{n}\right\}_{n \geqslant 1}$ satisfies
$$
a_{n+2}=a_{n+1}-a_{n} \text {. }
$$
If the sum of the first 1000 terms of the sequence is 1000, then the sum of the first 2014 terms is $\qquad$ . | 3. 1000 .
From $a_{n+2}=a_{n+1}-a_{n}$, we get
$$
a_{n+3}=a_{n+2}-a_{n+1}=\left(a_{n+1}-a_{n}\right)-a_{n+1}=-a_{n} \text {. }
$$
Therefore, for any positive integer $n$ we have
$$
a_{n}+a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}+a_{n+5}=0 \text {. }
$$
Also, $2014=1000(\bmod 6)$, so
$$
S_{2014}=S_{1000}=1000 \text {. }
$$ | 1000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. A middle school has 35 lights on each floor. To save electricity while ensuring the lighting needs of the corridors, the following requirements must be met:
(1) Two adjacent lights cannot be on at the same time;
(2) Any three consecutive lights cannot be off at the same time.
If you were to design different lighting... | 8. 31572 .
Notice that, $a_{n+3}=a_{n}+a_{n+1}, a_{3}=4, a_{4}=5$. Therefore, $a_{35}=31572$. | 31572 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $[x]$ denote the greatest integer not exceeding the real number $x$,
$$
a_{k}=\left[\frac{2014}{k}\right](k=1,2, \cdots, 100) \text {. }
$$
Then, among these 100 integers, the number of distinct integers is | 6.69.
When $01$, $[a+b]>[a]$.
$$
\begin{array}{l}
\text { Also } \frac{2014}{k}-\frac{2014}{k+1}=\frac{2014}{k(k+1)} \\
\Rightarrow \frac{2014}{k}=\frac{2014}{k+1}+\frac{2014}{k(k+1)} .
\end{array}
$$
When $k \leqslant 44$, $\frac{2014}{k(k+1)}>1$;
When $k \geqslant 45$, $\frac{2014}{k(k+1)}<1$.
Therefore, when $k \l... | 69 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $A$ and $B$ be two different subsets of the set $\{a, b, c, d, e\}$, such that set $A$ is not a subset of set $B$, and $B$ is not a subset of set $A$. Then the number of different ordered pairs $(A, B)$ is | 7.570.
Notice that, the set $\{a, b, c, d, e\}$ has $2^{5}$ subsets, and the number of different ordered pairs $(A, B)$ is $2^{5}\left(2^{5}-1\right)$.
If $A \subset B$, and suppose the set $B$ contains $k(1 \leqslant k \leqslant 5)$ elements, then the number of ordered pairs $(A, B)$ satisfying $A \subset B$ is
$$
\... | 570 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
385 A certain school's 2014 graduates numbered 2014 students. The school's six leaders must sign each student's graduation album. It is known that each leader must and can only use one of the three designated colored pens, and the color choices for the six leaders signing 2014 albums can be represented as a $6 \times 2... | Let 1, 2, 3 represent three colors, and the colors used by six school leaders for signing in each memorial album be represented by a six-element ordered array
$$
\left(x_{1}, x_{2}, \cdots, x_{6}\right)\left(x_{i} \in\{1,2,3\}\right)
$$
This array is referred to as the sequence group $\left(x_{1}, x_{2}, \cdots, x_{6}... | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Real-coefficient polynomials $f_{i}(x)=a_{i} x^{2}+b_{i} x+c_{i}$ $\left(a_{i}>0, i=1,2, \cdots, 2011\right)$, and $\left\{a_{i}\right\} 、\left\{b_{i}\right\} 、\left\{c_{i}\right\}$ are all arithmetic sequences. If $F(x)=\sum_{i=1}^{204} f_{i}(x)$ has real roots, then at most how many polynomials in $\left\{f... | 【Analysis】From $\left\{a_{i}\right\}$ being an arithmetic sequence, we know
$$
\sum_{i=1}^{2011} a_{i}=(2 \times 1005+1) a_{1006}=2011 a_{1006} \text {. }
$$
Similarly, $\sum_{i=1}^{2011} b_{i}=2011 b_{1006}$,
$$
\sum_{i=1}^{2011} c_{i}=2011 c_{1006} \text {. }
$$
Therefore, $F(x)=\sum_{i=1}^{200} f_{i}(x)=2011 f_{10... | 1005 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given
$$
\begin{array}{l}
A \cup B \cup C=\{a, b, c, d, e\}, A \cap B=\{a, b, c\}, \\
c \in A \cap B \cap C .
\end{array}
$$
Then the number of sets $\{A, B, C\}$ that satisfy the above conditions is. $\qquad$ | 7. 100 .
As shown in Figure 3, the set $A \cup B \cup C$ can be divided into seven mutually exclusive regions, denoted as $1, 2, \cdots, 7$.
It is known that element $c$ is in region 7, elements $a, b$ can appear in each of the regions 4, 7, with 4 possibilities; elements $d, e$ can appear in each of the regions $1, ... | 100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Use five different colors to color the six vertices of the triangular prism $A B C-D E F$, requiring each point to be colored with one color, and the two endpoints of each edge to be colored with different colors. Then the number of different coloring methods is $\qquad$.
| 9. 1920.
Transform the adjacency relationship of the vertices of a triangular prism into the regional relationship as shown in Figure 4, then the coloring methods for vertices $A$, $B$, and $C$ are $\mathrm{A}_{5}^{3}$ kinds.
For each determined coloring scheme of points $A$, $B$, and $C$, if we temporarily do not co... | 1920 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Determine the least possible value of the largest term in an arithmetic sequence composed of seven distinct primes. ${ }^{[4]}$
(2005, British Mathematical Olympiad) | 【Analysis】Let these seven different prime numbers be $p_{i}(i=1,2$, $\cdots, 7)$, and
$$
\begin{array}{l}
p_{2}=p_{1}+d, p_{3}=p_{1}+2 d, p_{4}=p_{1}+3 d, \\
p_{5}=p_{1}+4 d, p_{6}=p_{1}+5 d, p_{7}=p_{1}+6 d,
\end{array}
$$
where the common difference $d \in \mathbf{Z}_{+}$.
Since $p_{2}=p_{1}+d$ and $p_{3}=p_{1}+2 d$... | 907 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{aligned}
f(x+2) & =f(x)+2, \\
\text { then } \sum_{k=1}^{2014} f(k) & =
\end{aligned}
$$ | 14.2029105.
Notice that, $f(0)=0$.
By $f(x+2)=f(x)+2$, let $x=-1$. Then $f(1)=f(-1)+2 \Rightarrow f(1)=1$.
$$
\begin{array}{l}
\text { Also, } f(2 n)=\sum_{k=1}^{n}(f(2 k)-f(2 k-2))+f(0) \\
=2 n, \\
f(2 n-1)=\sum_{k=2}^{n}(f(2 k-1)-f(2 k-3))+f(1) \\
=2 n-1, \\
\text { Therefore, } \sum_{k=1}^{2014} f(k)=\sum_{k=1}^{20... | 2029105 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
17. A courier company undertakes courier services between 13 cities in a certain area. If each courier can take on the courier services for at most four cities, to ensure that there is at least one courier between every two cities, the courier company needs at least $\qquad$ couriers. | 17. 13.
From the problem, we know that there are $\mathrm{C}_{13}^{2}$ types of express delivery services between 13 cities. Each courier can handle at most $\mathrm{C}_{4}^{2}$ types of express delivery services between four cities. Therefore, at least $\frac{\mathrm{C}_{13}^{2}}{\mathrm{C}_{4}^{2}}=13$ couriers are ... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
21. Among the 100 integers from $1 \sim 100$, arbitrarily select three different numbers to form an ordered triplet $(x, y, z)$. Find the number of triplets that satisfy the equation $x+y=3z+10$. | (1) When $3 z+10 \leqslant 101$, i.e., $z \leqslant 30$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
S=\sum_{k=1}^{30}(3 k+9)=1665 \text{. }
$$
(2) When $3 z+10 \geqslant 102$, i.e., $31 \leqslant z \leqslant 63$, the number of ternary tuples satisfying $x+y=3 z+10$ is
$$
\begin{aligned}
T & =\sum_{k=31}... | 3194 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Fill the four characters “马” “上” “成” “功” in a $5 \times 5$ grid, with at most one character in each small square. “马” “上” must be filled in from left to right, and “成” “功” must also be filled in from left to right. “马” “上” must be in the same row or in the same column from top to bottom, or “成” “功” must be in the sa... | 8.42100.
The problem is equivalent to:
Filling 2 $a$s and 2 $b$s in a $5 \times 5$ grid, with at most one letter in each small square, such that at least one pair of the same letters is in the same row or column.
First, consider the case where the same letters are neither in the same row nor in the same column.
The nu... | 42100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 The sequence $\left\{a_{n}\right\}$:
$1,1,2,1,1,2,3,1,1,2,1,1,2,3,4, \cdots$.
Its construction method is:
First, give $a_{1}=1$, then copy this item 1 and add its successor number 2, to get $a_{2}=1, a_{3}=2$;
Next, copy all the previous items $1,1,2$, and add the successor number 3 of 2, to get
$$
a_{4}=1, ... | From the construction method of the sequence $\left\{a_{n}\right\}$, it is easy to know that
$$
a_{1}=1, a_{3}=2, a_{7}=3, a_{15}=4, \cdots \cdots
$$
In general, we have $a_{2^{n}-1}=n$, meaning the number $n$ first appears at the $2^{n}-1$ term, and if $m=2^{n}-1+k\left(1 \leqslant k \leqslant 2^{n}-1\right)$, then
$... | 3952 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 There are 800 points on a circle, numbered $1,2, \cdots, 800$ in a clockwise direction. They divide the circumference into 800 gaps. Choose one point and color it red, then proceed to color other points red according to the following rule: if the $k$-th point has been colored red, then move $k$ gaps in a cloc... | Consider a circle with $2n$ points in general.
(1) On a circle with $2n$ points, if the first red point is an even-numbered point, such as the $2k$-th point, then according to the coloring rule, every red point colored afterward will also be an even-numbered point. In this case, if we rename the 2nd, 4th, ..., $2k$-th,... | 25 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 2 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.
Try to find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value, and explain the reason. | Given that $f(x)=x^{12}$ is a convex function, then $\sum_{i=1}^{197} x_{i}^{12}$ achieves its maximum value when at least 1996 of $x_{1}, x_{2}, \cdots, x_{1997}$ are equal to $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$.
Assume that there are $t$ instances of $-\frac{1}{\sqrt{3}}$ and $1996-t$ instances of $\sqrt{3}$. Then t... | 189548 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If $a, b$ are positive integers, and satisfy $5a+7b=50$, then $ab=$ . $\qquad$ | 3. 15 .
From the problem, we know
$$
a=\frac{50-7 b}{5}=10-b-\frac{2 b}{5} \text { . }
$$
Therefore, $a=3, b=5$.
Thus, $a b=15$. | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Ed and Ann have lunch together, Ed and Ann order medium and large lemonade drinks, respectively. It is known that the large size has a capacity 50% more than the medium size. After both have drunk $\frac{3}{4}$ of their respective drinks, Ann gives the remaining $\frac{1}{3}$ plus 2 ounces to Ed. After lunch, they f... | 6. D.
Let the size of the medium lemon drink be $x$ ounces. According to the problem, we have
$$
x+\frac{3}{2} \times \frac{1}{4} \times \frac{1}{3} x+2=\frac{3}{2} x-\frac{3}{2} \times \frac{1}{4} \times \frac{1}{3} x-2 \text {. }
$$
Solving for $x$, we get $x=16$.
Therefore, the total amount they drank is $16+\frac... | 40 | Algebra | MCQ | Yes | Yes | cn_contest | false |
24. Let pentagon $A B C D E$ be a cyclic pentagon, $A B=$ $C D=3, B C=D E=10, A E=14$, the sum of the lengths of all diagonals is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$. Then $m+n=$ ( ).
(A) 129
(B) 247
(C) 353
(D) 391
(E) 421 | 24. D.
As shown in Figure 3, from the given conditions, we know that quadrilaterals $ABCD$ and $BCDE$ are both isosceles trapezoids.
Thus,
$$
\begin{array}{c}
AC=BD=CE. \\
\text{Let } AC=BD=CE \\
=x, AD=y, BE=z.
\end{array}
$$
By Ptolemy's theorem, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x^{2}=10 y+9, \\
... | 391 | Geometry | MCQ | Yes | Yes | cn_contest | false |
11. (20 points) Natural numbers $a, b$ make $4a + 7b$ and $5a + 6b$ both multiples of 11, and $a + 21b \geq 792$. Find the minimum value of $T = 21a + b$. | 11. Let $4a + 7b = 11x$, $5a + 6b = 11y$. Then
$$
\begin{array}{l}
a = 7y - 6x \geqslant 0, \\
b = 5x - 4y \geqslant 0.
\end{array}
$$
Thus, $T = 11(13y - 11x)$.
Let $13y - 11x = r$.
From $a + 21b \geqslant 792$, we get
$$
9x - 7y \geqslant 72.
$$
Substituting equation (3) into equation (1) to eliminate $y$ gives $x ... | 44 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1.2014 Arrange chairs in a circle, with $n$ people sitting on the chairs, such that when one more person sits down, they will always sit next to one of the original $n$ people. Then the minimum value of $n$ is $\qquad$ | -、1. 672 .
From the problem, we know that after $n$ people sit down, there are at most two empty chairs between any two people.
If we can arrange for there to be exactly two empty chairs between any two people, then $n$ is minimized.
Thus, if we number the chairs where people are sitting, we easily get the arithmetic s... | 672 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. It is known that 99 wise men are seated around a large round table, each wearing a hat of one of two different colors. Among them, 50 people's hats are of the same color, and the remaining 49 people's hats are of the other color. However, they do not know in advance which 50 people have the same color and which 49 p... | 8. Sure.
Suppose there are 50 white hats and 49 black hats. Those who see 50 white hats and 48 black hats are all wearing black hats. Thus, they can all write down the correct color, and there are 49 such people.
Next, let those who see 49 white hats and 49 black hats follow this strategy:
Observe which color of hat... | 74 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
6. Each point on a circle is colored one of three colors: red, yellow, or blue, and all three colors appear. Now, $n$ points are chosen from the circle. If among these points, there always exist three points that form a triangle with vertices of the same color and an obtuse angle, then the minimum possible value of $n$... | 6.13.
First, we state that \( n \geqslant 13 \).
If \( n \geqslant 13 \), then by the pigeonhole principle, among these \( n \) points, there must be \(\left[\frac{13-1}{3}\right]+1=5\) points of the same color (let's assume they are red).
Take a diameter, whose endpoints do not belong to these five points.
By the pig... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the largest positive integer $x$, such that for every positive integer $y$, we have $x \mid\left(7^{y}+12 y-1\right)$.
| When $y=1$, $7^{y}+12 y-1=18$.
Let 18 I $\left(7^{y}+12 y-1\right)$. Notice that,
$$
\begin{array}{l}
7^{y+1}+12(y+1)-1 \\
=6 \times\left(7^{y}+2\right)+\left(7^{y}+12 y-1\right) .
\end{array}
$$
Since $7^{y}+2 \equiv 1+2 \equiv 0(\bmod 3)$, therefore,
$$
18 \mid\left[7^{y+1}+12(y+1)-1\right] \text {. }
$$
Thus, for ... | 18 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find the last two digits of $\left[(\sqrt{29}+\sqrt{21})^{2012}\right]$.
(Adapted from the 25th IMO Preliminary Question) | Let $a_{n}=(\sqrt{29}+\sqrt{21})^{2 n}+(\sqrt{29}-\sqrt{21})^{2 n}$
$$
=(50+2 \sqrt{609})^{n}+(50-2 \sqrt{609})^{n} \text {. }
$$
Then the characteristic equation of the sequence $\left\{a_{n}\right\}$ is
$$
x^{2}-100 x+64=0 \text {. }
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has the recurrence relation
$$
a... | 31 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Given $x, y \in(0,1)$, and $3x+7y$, $5x+y$ are both integers. Then there are $\qquad$ pairs of $(x, y)$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | 【Analysis】Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v .\end{array}\right.$
We can first determine the integer values of $u$ and $v$ based on $x, y \in(0,1)$.
Solution Let $\left\{\begin{array}{l}3 x+7 y=u, \\ 5 x+y=v,\end{array}\right.$ where the possible values of $u$ and $v$ are $u \in\{1,2, \cdots, 9\}, v \in\... | 31 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Below is the inscription by the famous Chinese mathematician, Academician Wang Yuan:
数棈妻好
If different Chinese characters represent different digits from $0 \sim 9$, and assuming “数学竞赛好” represents the largest five-digit number that is a perfect square, formed by different digits. Then this five-digit number is $\q... | Notice that, $300^{2}=90000,310^{2}=96100$,
$$
320^{2}=102400>100000 \text {. }
$$
Thus, this five-digit number with distinct digits might be among the nine square numbers between $310^{2} \sim$ $320^{2}$.
$$
\begin{array}{l}
\text { Also, } 311^{2}=96721,312^{2}=97344,313^{2}=97969, \\
314^{2}=98596,315^{2}=99225,316... | 96721 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. On each face of a cube, write a positive integer, and at each vertex, write the product of the positive integers on the three faces meeting at that vertex. If the sum of the numbers written at the eight vertices is 2014, then the sum of the numbers written on the six faces is $\qquad$ .
| 6. 74 .
As shown in Figure 4, let the pairs of positive integers written on the opposite faces of the cube be $(a, b),(c, d),(e, f)$.
Then the sum of the numbers written at each vertex is
$$
\begin{array}{l}
a c f+a d f+a c e+a d e+b c f+b c e+b d f+b d e \\
=(a+b)(c+d)(e+f) \\
=2014=2 \times 19 \times 53 .
\end{arra... | 74 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given $A, B$ are digits in the set $\{0,1, \cdots, 9\}$, $r$ is a two-digit integer $\overline{A B}$, $s$ is a two-digit integer $\overline{B A}$, $r, s \in\{00,01$, $\cdots, 99\}$. When $|r-s|=k^{2}$ ( $k$ is an integer), the number of ordered pairs $(A, B)$ is $\qquad$. | 7. 42 .
Notice,
$$
|(10 A+B)-(10 B+A)|=9|A-B|=k^{2} \text {. }
$$
Then $|A-B|$ is a perfect square.
When $|A-B|=0$, there are 10 integer pairs:
$$
(A, B)=(0,0),(1,1), \cdots,(9,9) \text {; }
$$
When $|A-B|=1$, there are 18 integer pairs:
$$
(A, B)=(0,1),(1,2), \cdots,(8,9) \text {, }
$$
and their reverse numbers;
W... | 42 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Let $a, b$ be real numbers, for any real number $x$ satisfying $0 \leqslant x \leqslant 1$ we have $|a x+b| \leqslant 1$. Then the maximum value of $|20 a+14 b|+|20 a-14 b|$ is . $\qquad$ | 9.80 .
Let $x=0$, we know $|b| \leqslant 1$;
Let $x=1$, we know $|a+b| \leqslant 1$.
Therefore, $|a|=|a+b-b| \leqslant|a+b|+|b| \leqslant 2$, when $a=2, b=-1$, the equality can be achieved.
If $|20 a| \geqslant|14 b|$, then
$$
\begin{array}{l}
|20 a+14 b|+|20 a-14 b| \\
=2|20 a|=40|a| \leqslant 80 ;
\end{array}
$$
If... | 80 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1.200 people stand in a circle, some of whom are honest people, and some are liars. Liars always tell lies, while honest people tell the truth depending on the situation. If both of his neighbors are honest people, he will definitely tell the truth; if at least one of his neighbors is a liar, he may sometimes tell the ... | 1. There can be at most 150 honest people.
Since liars do not say they are liars, the 100 people who say they are liars are all honest people, and what they say is false. This indicates that they are all adjacent to liars.
Since each liar is adjacent to at most two honest people, there must be at least 50 liars. This ... | 150 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
8. A $3 \times 3$ grid with the following properties is called a "T-grid":
(1) Five cells are filled with 1, and four cells are filled with 0;
(2) Among the three rows, three columns, and two diagonals, at most one of these eight lines has three numbers that are pairwise equal.
Then the number of different T-grids is $... | 8. 68.
First, the number of all ways to fill a $3 \times 3$ grid with five 1s and four 0s is $\mathrm{C}_{9}^{4}=126$.
Next, consider the number of ways that do not satisfy property (2), i.e., methods that result in at least two lines of three equal numbers (hereafter referred to as good lines).
We will count these ... | 68 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 3, two equilateral triangles overlap to form a six-pointed star. Now, the positive integers 1, 2, $\cdots, 12$ are to be filled in the 12 nodes of the figure, such that the sum of the four numbers on each straight line is equal.
(1) Try to find the minimum value of the sum of the numbers at... | (1) Solution: For any arrangement that satisfies the conditions, the number filled at point $a_i$ $(i=1,2, \cdots, 12)$ is still denoted as $a_i$. If the sum of the four numbers on each line is $s$, then
$6 s=2(1+2+\cdots+12) \Rightarrow s=26$.
Therefore, in $\triangle a_{1} a_{3} a_{5}$ and $\triangle a_{2} a_{4} a_{6... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Solve the equation
$$
x^{2}-x-1000 \sqrt{1+8000 x}=1000 .
$$ | Solution: Clearly, $x \neq 0$. The original equation can be transformed into
$$
\begin{array}{l}
x^{2}-x=1000(1+\sqrt{1+8000 x}) \\
=\frac{8 \times 10^{6} x}{\sqrt{1+8000 x}-1} .
\end{array}
$$
Then $x-1=\frac{8 \times 10^{6}}{\sqrt{1+8000 x}-1}$.
Let $a=10^{3}, t=\sqrt{1+8 a x}-1$.
Thus, $x=\frac{t^{2}+2 t}{8 a}$, an... | 2001 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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