problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
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2. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {.... | 2. (1) Same as Question 1 (1) of Grade 1.
(2) Let $|D_{n}(g)|$ denote the number of elements in the set $D_{n}(g)$.
Obviously, $|D_{1}(g)|=1, |D_{2}(g)|=1$.
$$
\begin{array}{l}
\text { Also, } g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j} \\
=\left(1+x_{1}\left(1+\sum_{i=2}^{n}... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $A$ denote the set of all sequences (of arbitrary finite or infinite length) $\left\{a_{1}, a_{2}, \cdots\right\}$ formed by the elements of $\{1,2, \cdots, 2017\}$. If the first several consecutive terms of sequence $M$ are the terms of sequence $T$, then we say that sequence $M$ "starts with" sequence $... | 【Analysis】First, give an example where (2) does not hold.
Take 2017 sequences each with only one term: $1,2, \cdots$, 2017. Then each sequence in set $A$ must start with one of these 2017 sequences, so (2) does not hold.
First, in set $S$, there cannot exist two different sequences $T_{1}$ and $T_{2}$ such that $T_{2}... | 1 | Combinatorics | proof | Yes | Yes | cn_contest | false |
2. Given $X_{1}, X_{2}, \cdots, X_{100}$ as a sequence of non-empty subsets of a set $S$, and all are distinct. For any $i \in \{1,2, \cdots, 99\}$, we have $X_{i} \cap X_{i+1}=\varnothing, X_{i} \cup X_{i+1} \neq S$.
Find the minimum number of elements in the set $S$.
(45th United States of America Mathematical Olympi... | First use mathematical induction to prove: when $n \geqslant 4$, a subset sequence of $2^{n-1}+1$ subsets that meets the requirements can be constructed for $S=\{1,2, \cdots, n\}$; then prove that when $|S|=7$, the number of subsets in a subset sequence that meets the requirements does not exceed 100.
The minimum numbe... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $\{x\}$ denote the fractional part of the real number $x$. Given $a=(5 \sqrt{2}+7)^{2017}$. Then $a\{a\}=$ $\qquad$ . | ,- 1.1 .
Let $b=(5 \sqrt{2}-7)^{2017}$.
Then $0<b<1$, and $a b=1$.
Notice that,
$a-b=\sum_{k=0}^{1008} 2 \mathrm{C}_{2017}^{2 k+1}(5 \sqrt{2})^{2016-2 k} \times 7^{2 k+1} \in \mathbf{Z}$.
Since $a=(a-b)+b(a-b \in \mathbf{Z}, 0<b<1)$, it follows that $b=\{a\} \Rightarrow a\{a\}=a b=1$. | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $a+b+c=1$,
$$
b^{2}+c^{2}-4 a c+6 c+1=0 \text{. }
$$
Find the value of $a b c$. | Solve: From equation (1), we get $a=1-b-c$.
Substitute into equation (2) and rearrange to get
$$
\begin{array}{l}
b^{2}+5 c^{2}+4 b c+2 c+1=0 \\
\Rightarrow(b+2 c)^{2}+(c+1)^{2}=0 \\
\Rightarrow b+2 c=c+1=0 \Rightarrow b=2, c=-1 \\
\Rightarrow a=1-b-c=0 \Rightarrow a b c=0 .
\end{array}
$$ | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Define the operation " $*$ ":
$$
a * b=\log _{2}\left(2^{a}+2^{b}\right)(a 、 b \in \mathbf{R}) \text {. }
$$
Let $A=(1 * 3) * 5, B=(2 * 4) * 6$. Then the value of $1 *(A * B)$ is $\qquad$ | 1.7 .
From the definition of $*$, we know that $2^{a * b}=2^{a}+2^{b}$. Therefore, $2^{A}=2^{1 * 3}+2^{5}=2^{1}+2^{3}+2^{5}$, $2^{B}=2^{2}+2^{4}+2^{6}$.
Thus, $2^{1 *(A * B)}=2^{1}+2^{A}+2^{B}$ $=2+2^{1}+2^{2}+\cdots+2^{6}=2^{7}$.
Hence, $1 *(A * B)=7$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let real numbers $a$ and $b$ satisfy
$$
\begin{array}{l}
a^{2}\left(b^{2}+1\right)+b(b+2 a)=40, \\
a(b+1)+b=8 .
\end{array}
$$
Find the value of $\frac{1}{a^{2}}+\frac{1}{b^{2}}$.
(2014, National Junior High School Mathematics League) | Hint: Transform the two equations in the conditions to get $a^{2} b^{2}+(a+b)^{2}=40, a b+(a+b)=8$. Let $x=a+b, y=a b$.
Thus, $x^{2}+y^{2}=40, x+y=8$.
Solving yields $(x, y)=(2,6)$ (discard) or $(6,2)$. Therefore, $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{(a+b)^{2}-2 a b}{a^{2} b^{2}}=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $x, y$ satisfy
$$
\frac{4}{x^{4}}-\frac{2}{x^{2}}=3, y^{4}+y^{2}=3 \text {. }
$$
Then the value of $\frac{4}{x^{4}}+y^{4}$ is $\qquad$
(2008, "Mathematics Weekly Cup" National Junior High School Mathematics Competition) | By observation, we know that $-\frac{2}{x^{2}}$ and $y^{2}$ are two different solutions of the equation $m^{2}+m=3$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\frac{4}{x^{4}}+y^{4}=m_{1}^{2}+m_{2}^{2}=\left(m_{1}+m_{2}\right)^{2}-2 m_{1} m_{2} \\
=(-1)^{2}-2(-3)=7 .
\end{array}
$$ | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If in the real number range there is
$$
x^{3}+p x+q=(x-a)(x-b)(x-c),
$$
and $q \neq 0$, then $\frac{a^{3}+b^{3}+c^{3}}{a b c}=$ $\qquad$ | Obviously, $a$, $b$, $c$ are the roots of the equation $x^{3} + p x + q = 0$.
By Vieta's formulas, we have $a + b + c = 0$.
By formula (2), we get
$$
\frac{a^{3} + b^{3} + c^{3}}{a b c} = \frac{a^{3} + b^{3} + c^{3} - 3 a b c}{a b c} + 3 = 3.
$$ | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Given that $D$ is the intersection of the tangents to the circumcircle of $\triangle A B C$ at $A$ and $B$, the circumcircle of $\triangle A B D$ intersects the line $A C$ and the segment $B C$ at another point $E$ and $F$ respectively, and $C D$ intersects $B E$ at point $G$. If $\frac{B C}{B F}=2$, find $\frac{B G... | Using Property 1, let the circumcenter of $\triangle ABC$ be $O$. Then $O, A, D, B$ are concyclic. If point $O$ is not on side $BC$, then $FO \perp BC \Rightarrow \angle OAB=90^{\circ}$, which is a contradiction. Therefore, point $O$ is on $BC$.
Let $CD$ intersect $AB$ at point $M$. Then
$$
\begin{array}{l}
\frac{AM}{M... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Question 3 Given $n$ points $p_{1}, p_{2}, \cdots, p_{n}$ in the plane, with no three points collinear. Each point $p_{i}(i=1,2, \cdots, n)$ is arbitrarily colored red or blue. Let $S$ be a set of some triangles with vertex set $\left\{p_{1}, p_{2}, \cdots, p_{n}\right\}$, and has the property: for any two line segment... | When $n \in\{1,2, \cdots, 5\}$, it is clearly impossible to satisfy the conditions of the problem.
When $n=6$, color $p_{1} 、 p_{2} 、 p_{3}$ red, and $p_{4}$ 、 $p_{5} 、 p_{6}$ blue, and consider the triangles:
$$
\begin{array}{l}
\triangle p_{1} p_{2} p_{3} 、 \triangle p_{1} p_{3} p_{4} 、 \triangle p_{1} p_{4} p_{5} 、... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $a, b, c$ be distinct positive integers. Then the minimum value of $\frac{a b c}{a+b+c}$ is . $\qquad$ | 11.1.
Assume $a>b>c$. Then $a \geqslant 3, b \geqslant 2, c \geqslant 1$.
Thus, $a b \geqslant 6, b c \geqslant 2, c a \geqslant 3$.
Therefore, $\frac{a+b+c}{a b c}=\frac{1}{b c}+\frac{1}{c a}+\frac{1}{a b}$
$\leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$
$\Rightarrow \frac{a b c}{a+b+c} \geqslant 1$.
When $a=3, b=2... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the complex number $z$ satisfy
$$
\frac{1017 z-25}{z-2017}=3+4 \text { i. }
$$
Then $|z|=$ $\qquad$ | 2.5.
Let $w=3+4 \mathrm{i}$. Then
$$
\begin{aligned}
z & =\frac{2017 w-25}{w-2017}=\frac{2017 w-\bar{w} w}{w-2017} \\
& =\frac{2017-\bar{w}}{2017-w} w .
\end{aligned}
$$
Therefore, $|z|=|w|=5$. | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Let real numbers $s, t$ satisfy the equations
$$
\begin{array}{l}
19 s^{2}+99 s+1=0, t^{2}+99 t+19=0, \text { and } s t \neq 1 . \\
\text { Then } \frac{s t+4 s+1}{t}=
\end{array}
$$
(1999, "Mathematics Weekly Cup" National Junior High School Mathematics Competition) | Solution: Clearly, $t \neq 0$.
Thus, $19\left(\frac{1}{t}\right)^{2}+99 \cdot \frac{1}{t}+1=0$.
By comparison, $s$ and $\frac{1}{t}$ are the two distinct roots of the equation $19 x^{2}+99 x+1=0$ (since $s t \neq 1$, i.e., $s \neq \frac{1}{t}$). Therefore, by Vieta's formulas, we have
$$
\begin{array}{l}
s+\frac{1}{t}=... | -5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the parabola $y=\sqrt{2} x^{2}$ intersects with the lines $y=1, y=2, y=3$ to form three line segments. The area of the triangle formed by these three line segments is $\qquad$ . | 2. 2 .
The intersection points of the lines $y=1, y=2, y=3$ with the parabola $y=\sqrt{2} x^{2}$ are all symmetric about the $y$-axis, so the lengths of the three line segments are $2 x_{1}, 2 x_{2}, 2 x_{3}$, where $x_{1}, x_{2}, x_{3}$ are the positive roots of the equations $\sqrt{2} x_{1}^{2}=1, \sqrt{2} x_{2}^{2}... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. For any $x \in[0,1]$, we have $|a x+b| \leqslant 1$.
Then the maximum value of $|b x+a|$ is $\qquad$ | 3. 2 .
Let $f(x)=a x+b$. Then
$$
\begin{array}{l}
b=f(0), a=f(1)-f(0) . \\
\text { Hence }|b x+a|=|f(0) x+f(1)-f(0)| \\
=|f(0)(x-1)+f(1)| \\
\leqslant|f(0)||x-1|+|f(1)| \\
\leqslant 1+1=2 .
\end{array}
$$
When $a=2, b=-1, x=0$, the maximum value 2 is obtained. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. The last digit of $\sum_{k=0}^{201}(10 k+7)^{k+1}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7.6.
It is easy to see that $\sum_{k=0}^{201}(10 k+7)^{k+1} \equiv \sum_{k=0}^{201} 7^{k+1}(\bmod 10)$.
Notice that, for any natural number $n$, the last digits of $7^{4 n+1}$, $7^{4 n+2}$, $7^{4 n+3}$, and $7^{4 n+4}$ are sequentially $7$, $9$, $3$, and $1$, and the last digit of their sum is 0.
$$
\text{Therefore, }... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$, where $x_{1}=1, y_{1}=0, x_{n+1}=$ $x_{n}-y_{n}, y_{n+1}=x_{n}+y_{n}\left(n \in \mathbf{Z}_{+}\right)$. If $a_{n}=$ $\overrightarrow{P_{n} P_{n+1}} \cdot \overrightarrow{P_{n+1} P_{n+2}}$, then t... | 4. 10 .
It is known that $\overrightarrow{O P_{n+1}}$ is obtained by rotating $\overrightarrow{O P_{n}}$ counterclockwise by $\frac{\pi}{4}$ and stretching it to $\sqrt{2}$ times its original length.
Thus, $\left|\overrightarrow{P_{n} P_{n+1}}\right|=O P_{n}$,
$\left|\overrightarrow{P_{n+1} P_{n+2}}\right|=\left|O P_{... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Find the smallest real number $\lambda$, such that there exists a sequence $\left\{a_{n}\right\}$ with all terms greater than 1, for which $\prod_{i=1}^{n+1} a_{i}<a_{n}^{\lambda}$ holds for any positive integer $n$. | Given $a_{n}>1$, so,
$$
\begin{array}{l}
\prod_{i=1}^{n+1} a_{i}0\right), S_{n}=\sum_{i=1}^{n} b_{i}\left(S_{n}>0\right) .
\end{array}
$$
For equation (1) to hold, then $\lambda>0$.
From equation (1) we get
$$
\begin{array}{l}
S_{n+2}S_{n+2}+\lambda S_{n} \geqslant 2 \sqrt{\lambda S_{n+2} S_{n}} \\
\Rightarrow \frac{S... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) Given a five-element set $A_{1}, A_{2}, \cdots, A_{10}$, any two of these ten sets have an intersection of at least two elements. Let $A=\bigcup_{i=1}^{10} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets among $A_{1}, A_{2}, \cdots, A_{10}$ that contain the... | Four, it is easy to get $\sum_{i=1}^{n} k_{i}=50$.
The $k_{i}$ sets containing $x_{i}$ form $\mathrm{C}_{k_{i}}^{2}$ set pairs, $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all set pairs, which contain repetitions.
From the fact that the intersection of any two sets among $A_{1}, A_{2}, \cdots, A_{10}$ has at least... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 For any real number sequence $\left\{x_{n}\right\}$, define the sequence $\left\{y_{n}\right\}:$
$$
y_{1}=x_{1}, y_{n+1}=x_{n+1}-\left(\sum_{i=1}^{n} x_{i}^{2}\right)^{\frac{1}{2}}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Find the smallest positive number $\lambda$, such that for any real number sequen... | 【Analysis】First estimate the upper bound of $\lambda$ from the limit perspective, then try to construct a recurrence relation to solve it.
First, prove that $\lambda \geqslant 2$.
In fact, start from simple and special cases.
Take $x_{1}=1, x_{n}=\sqrt{2^{n-2}}(n \geqslant 2)$. Then $y_{1}=1, y_{n}=0(n \geqslant 2)$.
S... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Given real numbers $x, y$ satisfy $x+y=1$. Then, the maximum value of $\left(x^{3}+1\right)\left(y^{3}+1\right)$ is | 5.4.
$$
\begin{array}{l}
\text { Given }\left(x^{3}+1\right)\left(y^{3}+1\right) \\
=(x y)^{3}+x^{3}+y^{3}+1 \\
=(x y)^{3}-3 x y+2,
\end{array}
$$
let $t=x y \leqslant\left(\frac{x+y}{2}\right)^{2}=\frac{1}{4}$, then
$$
f(t)=t^{3}-3 t+2 \text {. }
$$
Also, by $f^{\prime}(t)=3 t^{2}-3$, we know that $y=f(t)$ is monoto... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $x, y \in \mathbf{R}$, for any $n \in \mathbf{Z}_{+}$, $n x+\frac{1}{n} y \geqslant 1$. Then the minimum value of $41 x+2 y$ is $\qquad$ | 8.9.
Let the line $l_{n}: n x+\frac{1}{n} y=1$, and call $l_{n} 、 l_{n+1}$ two adjacent lines. Then the intersection point of the two lines is
$$
A_{n}\left(\frac{1}{2 n+1}, \frac{n^{2}+n}{2 n+1}\right) \text {. }
$$
If the intersection point of the line $x+y=1$ and the line $y=0$ is denoted as $A_{0}(1,0)$, then the... | 9 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. If $a, b, c$ are distinct integers, then
$$
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a
$$
the minimum value is . $\qquad$ | 8. 6 .
Notice that,
$$
\begin{array}{l}
3 a^{2}+2 b^{2}+4 c^{2}-a b-3 b c-5 c a \\
=\frac{1}{2}(a-b)^{2}+\frac{3}{2}(b-c)^{2}+\frac{5}{2}(c-a)^{2} .
\end{array}
$$
Since \(a, b, c\) are distinct integers, when \(a-b=2, a-c=1, c-b=1\), the original expression achieves its minimum value of 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the function $f(x)=\log _{2} \frac{x-3}{x-2}+\cos \pi x$. If $f(\alpha)=10, f(\beta)=-10$, then $\alpha+\beta=$ $\qquad$ | 2. 5 .
It is easy to know that the domain of $f(x)$ is $(-\infty, 2) \cup(3,+\infty)$.
Then $f(5-x)=\log _{2} \frac{5-x-3}{5-x-2}+\cos (5-x) \pi$ $=-f(x)$.
Therefore, $f(x)$ is centrally symmetric about the point $\left(\frac{5}{2}, 0\right)$.
Also, when $x>3$,
$$
f(x)=\log _{2}\left(1-\frac{1}{x-2}\right)+\cos x \pi,
... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
18. Amelia tosses a coin, with the probability of landing heads up being $\frac{1}{3}$; Brian also tosses a coin, with the probability of landing heads up being $\frac{2}{5}$. Amelia and Brian take turns tossing the coins, and the first one to get heads wins. All coin tosses are independent. Starting with Amelia, the p... | 18. D.
Let $P_{0}$ be the probability of Amelia winning.
Notice,
$P_{0}=P($ Amelia wins in the first round $)+$ $P($ both fail to win in the first round $) \cdot P_{0}$, where, if both fail to win in the first round, it still starts with Amelia, and her probability of winning remains $P_{0}$.
In the first round, the p... | 4 | Algebra | MCQ | Yes | Yes | cn_contest | false |
9. There are four teacups with their mouths facing up. Now, each time three of them are flipped, and the flipped teacups are allowed to be flipped again. After $n$ flips, all the cup mouths are facing down. Then the minimum value of the positive integer $n$ is $\qquad$ . | 9.4 .
Let $x_{i}$ be the number of times the $i$-th cup ($i=1,2,3,4$) is flipped when all cup mouths are facing down, then $x_{i}$ is odd.
From $x_{1}+x_{2}+x_{3}+x_{4}=3 n$, we know that $n$ is even.
It is easy to see that when $n=2$, the condition is not satisfied, hence $n \geqslant 4$.
When $n=4$, use 1 to represe... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a>1$. Then the minimum value of $\log _{a} 16+2 \log _{4} a$ is $\qquad$ . | 3.4.
From the operation of logarithms, we get
$$
\begin{array}{l}
\log _{a} 16+2 \log _{4} a=4 \log _{a} 2+\log _{2} a \\
=\frac{4}{\log _{2} a}+\log _{2} a .
\end{array}
$$
Since $a>1$, we have $\log _{2} a>0$.
By the AM-GM inequality, we get
$$
\frac{4}{\log _{2} a}+\log _{2} a \geqslant 2 \sqrt{\frac{4}{\log _{2} ... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $\log _{\sqrt{7}}(5 a-3)=\log _{\sqrt{a^{2}+1}} 5$. Then the real number
$$
a=
$$
. $\qquad$ | 2. 2 .
Simplify the original equation to
$$
\log _{7}(5 a-3)=\log _{a^{2}+1} 5 \text {. }
$$
Since $f(x)=\log _{7}(5 x-3)$ is an increasing function for $x>\frac{3}{5}$, and $g(x)=\log _{5}\left(x^{2}+1\right)$ is also an increasing function, and $f(2)=$ $g(2)=1$, therefore, $a=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $S$ be the set of all rational numbers in the interval $\left(0, \frac{5}{8}\right)$, for the fraction $\frac{q}{p} \in S, (p, q)=1$, define the function $f\left(\frac{q}{p}\right)=\frac{q+1}{p}$. Then the number of roots of $f(x)=\frac{2}{3}$ in the set $S$ is $\qquad$ | 6.5.
Since $f(x)=\frac{2}{3}$, let $q=2 m-1, p=3 m\left(m \in \mathbf{Z}_{+}\right)$.
Then, $0<\frac{2 m-1}{3 m}<\frac{5}{8} \Rightarrow \frac{1}{2}<m<8$. Upon verification, the number of roots of the equation is 5. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. Given $z \in \mathbf{C}$. If the equation in terms of $x$
$$
4 x^{2}-8 z x+4 \mathrm{i}+3=0
$$
has real roots. Then the minimum value of $|z|$ is $\qquad$ | 9. 1 .
Let $z=a+b \mathrm{i}(a, b \in \mathbf{R}), x=x_{0}$ be the real root of the given equation. Then
$$
\begin{array}{l}
4 x_{0}^{2}-8(a+b \mathrm{i}) x_{0}+4 \mathrm{i}+3=0 \\
\Rightarrow\left\{\begin{array}{l}
4 x_{0}^{2}-8 a x_{0}+3=0, \\
-8 b x_{0}+4=0 .
\end{array}\right.
\end{array}
$$
Eliminating $x_{0}$ a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (16 points) Given the sequence $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=2, a_{n+1}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}}\left(n \in \mathbf{Z}_{+}\right) \text {, }
$$
where $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$.
(1) Prove: $\left\{\frac{1}{S_{n}-1}\right\}$ is an arithmetic seque... | 14. (1) When $n \geqslant 1$, from the condition we get
$$
\begin{array}{l}
S_{n+1}-S_{n}=-\frac{\left(S_{n}-1\right)^{2}}{S_{n}} \\
\Rightarrow S_{n+1}-1=\frac{S_{n}-1}{S_{n}} .
\end{array}
$$
Thus, $\frac{1}{S_{n+1}-1}-\frac{1}{S_{n}-1}=\frac{S_{n}}{S_{n}-1}-\frac{1}{S_{n}-1}=1$.
Also, $\frac{1}{S_{1}-1}=\frac{1}{2-... | 3 | Algebra | proof | Yes | Yes | cn_contest | false |
4. Let the function be
$$
f(x)=\left(\frac{1}{2}\right)^{x}+\left(\frac{2}{3}\right)^{x}+\left(\frac{5}{6}\right)^{x}(x \in[0,+\infty)) \text {. }
$$
Then the number of integer points on the graph of the function is $\qquad$ | 4.3.
It is known that the function $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and
$$
f(0)=3, f(1)=2, f(3)=1 .
$$
When $x>3$, we have
$$
0<f(x)<f(3)=1 \text {. }
$$
Therefore, the number of integer points on the graph of the function $y=f(x)(x \in[0,+\infty))$ is 3. | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. Given that a line passing through the focus $F$ of the parabola $y^{2}=4 x$ intersects the parabola at points $M$ and $N$, and $E(m, 0)$ is a point on the $x$-axis. The extensions of $M E$ and $N E$ intersect the parabola at points $P$ and $Q$ respectively. If the slopes $k_{1}$ and $k_{2}$ of $M N$ and $P Q$ satisf... | 9.3.
When $M P$ is not perpendicular to the $x$-axis, let $l_{\text {MР }}: y=k(x-m)$.
Substitute into $y^{2}=4 x$ to get
$$
x^{2}-\left(\frac{4}{k^{2}}+2 m\right) x+m^{2}=0 \text {. }
$$
Then $x_{M} x_{P}=m^{2} \Rightarrow y_{M} y_{P}=-4 m$ $\Rightarrow y_{P}=\frac{-4 m}{y_{M}}$.
When $M P \perp x$-axis, the conclus... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, and $c$ are three non-zero real numbers, and $x^{2}-1$ is a factor of the polynomial $x^{3}+a x^{2}+b x+c$. Then the value of $\frac{a b+3 a}{c}$ is ( ).
(A) -2
(B) -1
(C) 1
(D) 2 | - 1. A.
From the fact that $\pm 1$ are roots of the given polynomial, we have
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 + a + b + c = 0 , } \\
{ - 1 + a - b + c = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=-c, \\
b=-1
\end{array}\right.\right. \\
\Rightarrow \frac{a b+3 a}{c}=-2 .
\end{array}
$$ | -2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
One, (20 points) Let $a, b$ be real numbers, and the equation with respect to $x$
$$
\frac{x}{x-1}+\frac{x-1}{x}=\frac{a+b x}{x^{2}-x}
$$
has no real roots. Find the value of the expression $8 a+4 b+|8 a+4 b-5|$. | One, the original equation can be transformed into
$$
2 x^{2}-(b+2) x+(1-a)=0 \text {. }
$$
Thus, $\Delta=(b+2)^{2}+8 a-8$.
(1) When $\Delta>0$, equation (1) has two distinct real roots.
Since the original equation has no solution, it follows that the two distinct real roots of equation (1) are 0 and 1, i.e.,
$$
\beg... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. (20 points) Let the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
a_{1}=3, b_{1}=1
$$
and for any $n \in \mathbf{Z}_{+}$, we have
$$
\left\{\begin{array}{l}
a_{n+1}=a_{n}+b_{n}+\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}} \\
b_{n+1}=a_{n}+b_{n}-\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}}
\end{array}... | 11. (1) From the problem, we have
$$
\begin{array}{l}
a_{n+1}+b_{n+1}=2\left(a_{n}+b_{n}\right), \\
a_{n+1} b_{n+1}=3 a_{n} b_{n} . \\
\text { Also, } a_{1}+b_{1}=4, a_{1} b_{1}=3 \text {, then } \\
a_{n}+b_{n}=\left(a_{1}+b_{1}\right) 2^{n-1}=2^{n+1}, \\
a_{n} b_{n}=a_{1} b_{1} 3^{n-1}=3^{n} .
\end{array}
$$
Notice t... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. As shown in Figure 1, points $A$ and $A^{\prime}$ are on the $x$-axis and are symmetric with respect to the $y$-axis. A line passing through point $A^{\prime}$ and perpendicular to the $x$-axis intersects the parabola $y^{2}=2 x$ at points $B$ and $C$. Point $D$ is a moving point on segment $A B$, and point $E$ is ... | 15. (1) Let $A\left(-2 a^{2}, 0\right), A^{\prime}\left(2 a^{2}, 0\right)$. Then $B\left(2 a^{2}, 2 a\right), C\left(2 a^{2},-2 a\right)$.
Let $D\left(x_{1}, y_{1}\right), \overrightarrow{A D}=\lambda \overrightarrow{A B}$. Then $\overrightarrow{C E}=\lambda \overrightarrow{C A}$.
Thus, $\left(x_{1}+2 a^{2}, y_{1}\righ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $[x]$ denote the greatest integer not exceeding the real number $x$. If
$$
\begin{array}{l}
a=\frac{\sqrt{6+\sqrt{6+\cdots+\sqrt{6}}},}{2016 \text { nested radicals }}, \\
b=\frac{\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{6}}},}{2 \text { 2017 nested radicals }},
\end{array}
$$
then $[a+b]=$ . $\qquad$ | 3. 4 .
Notice that,
$2.4<\sqrt{6}<a<\frac{\sqrt{6+\sqrt{6+\cdots+\sqrt{9}}}}{2016 \text{ levels }}=3$,
1. $8<\sqrt[3]{6}<b<\frac{\sqrt[3]{6+\sqrt[3]{6+\cdots+\sqrt[3]{8}}}}{2017 \text{ layers }}=2$.
Therefore, $[a+b]=4$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Place a regular tetrahedron with a volume of 1 inside a cube, then the minimum volume of this cube is $\qquad$ | 3. 3 .
Considering in reverse, for a cube with edge length $a$ (volume $a^{3}$), its largest inscribed regular tetrahedron has vertices formed by the cube's vertices that do not share an edge, and its volume is $\frac{a^{3}}{3}$.
$$
\text { Let } \frac{a^{3}}{3}=1 \text {, then } a^{3}=3 \text {. }
$$ | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the set $M=\{1,99,-1,0,25,-36, -91,19,-2,11\}$, let the non-empty subsets of $M$ be $M_{i}(i=1,2, \cdots, 1023)$. If the product of all elements in each $M_{i}$ is $m_{i}$, then $\sum_{i=1}^{1023} m_{i}=$ $\qquad$ . | 2. -1 .
Let the set $M=\left\{a_{i} \mid i=1,2, \cdots, n\right\}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{2^{n}-1} m_{i}=\prod_{i=1}^{n}\left(a_{i}+1\right)-1 . \\
\text { Given }-1 \in M, \text { we know } \sum_{i=1}^{1023} m_{i}=-1 .
\end{array}
$$ | -1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. If $P(x, y)$ is a point on the hyperbola $\frac{x^{2}}{8}-\frac{y^{2}}{4}=1$, then the minimum value of $|x-y|$ is . $\qquad$ | 9. 2 .
From the condition, we know that $x^{2}-2 y^{2}-8=0$.
By symmetry, without loss of generality, assume
$$
\begin{array}{l}
x>0, y>0, u=x-y>0 . \\
\text { Then }(y+u)^{2}-2 y^{2}-8=0 \\
\Rightarrow y^{2}-2 u y-u^{2}+8=0 \\
\Rightarrow \Delta=(2 u)^{2}-4\left(-u^{2}+8\right) \geqslant 0 \\
\Rightarrow u \geqslant ... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
11. Given real numbers $x_{1}, x_{2}, x_{3}$ satisfy
$$
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{1} x_{2}+x_{2} x_{3}=2 \text {. }
$$
Then the maximum value of $\left|x_{2}\right|$ is $\qquad$ | 11. 2 .
From the condition, we have
$$
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+x_{3}^{2}=4 \text {. }
$$
Notice that,
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2} \geqslant \frac{x_{2}^{2}}{2}, \\
x_{3}^{2}+\left(x_{2}+x_{3}\right)^{2} \geqslant \frac{x_{2}^{2}}{2} .
\end{arr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. In $\triangle A B C$, the side lengths opposite to $\angle A 、 \angle B 、 \angle C$ are $a 、 b 、 c$, respectively, and
$$
\begin{array}{l}
\sin C \cdot \cos \frac{A}{2}=(2-\cos C) \sin \frac{A}{2}, \\
\cos A=\frac{3}{5}, a=4 .
\end{array}
$$
Then the area of $\triangle A B C$ is . $\qquad$ | 7.6 .
From equation (1) we know
$$
\begin{array}{l}
2 \sin \frac{A}{2}=\sin \left(C+\frac{A}{2}\right) \\
\Rightarrow 2 \sin A=2 \sin \left(C+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Thus, $c+b=2a$.
Also, $a^{2}=b^{2}+c^{2}-2 b c \cos A$, which means
$$
4^{2}=b^{2}+(8-b)^{2}-2 b(8-... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. If the equation with respect to $x$
$$
x^{2}+a x+b-3=0(a, b \in \mathbf{R})
$$
has real roots in the interval $[1,2]$, then the minimum value of $a^{2}+(b-4)^{2}$ is . $\qquad$ | 8. 2 .
From the problem, we know that $b=-x^{2}-a x+3$.
Then $a^{2}+(b-4)^{2}$
\[
\begin{array}{l}
=a^{2}+\left(-x^{2}-a x-1\right)^{2} \\
=x^{2}(x+a)^{2}+2\left(x^{2}+a x\right)+a^{2}+1 \\
=\left(x^{2}+1\right)(x+a)^{2}+x^{2}+1 .
\end{array}
\]
Also, $x \in[1,2]$, so,
\[
a^{2}+(b-4)^{2} \geqslant x^{2}+1 \geqslant 2... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. The function
$$
f(x)=\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x}
$$
has a maximum value of $\qquad$ | 9.11.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x})^{2} \\
\leqslant(3+2+6)\left(\frac{2 x-7}{3}+\frac{12-x}{2}+\frac{44-x}{6}\right) \\
=11^{2},
\end{array}
$$
The equality holds if and only if $\frac{9}{2 x-7}=\frac{4}{12-x}=\frac{36}{44-x}$, which is when $x=8$... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) Given the function
$$
f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1(x \in \mathbf{R}) \text {. }
$$
(1) Find the intervals where the function $f(x)$ is monotonically increasing;
(2) Let points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ a... | (1) From the problem, we have
$$
f(x)=\cos 2 x+\sqrt{3} \sin 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) \text {. }
$$
Therefore, the monotonic increasing interval of $f(x)$ is
$$
\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) \text {. }
$$
(2) Let $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\f... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x$. Then the set
$$
M=\left\{n \mid f\left(n^{2}-8 n-2018\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$
the number of subsets of $M$ is $\qquad$. | 2. 1.
For any $0< x_2 < x_1$, we have
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right)=\lg \frac{x_{1}}{x_{2}}-\frac{\lg \frac{x_{1}}{x_{2}}}{\lg 9}>0 .
\end{array}
$$
Thus, $f(x)$ is a decreasing function on the interval $(0,+\infty)$. Note that, $f(9)=0$.
Therefore, when $x>9$, $f(x)f(9)=0$.
Hence, $f(x)... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. In $\triangle A B C$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively, satisfying $a^{2}+b^{2}=4-\cos ^{2} C, a b=2$. Then $S_{\triangle A B C}=$ $\qquad$ | $-1.1$
From the problem, we have $(a-b)^{2}+\cos ^{2} C=0$.
Solving, we get $a=b=\sqrt{2}, \cos C=0$.
Therefore, $S_{\triangle A B C}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $x, y>0$. If
$$
f(x, y)=\left(x^{2}+y^{2}+2\right)\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \text {, }
$$
then the minimum value of $f(x, y)$ is | 3.4.
By completing the square, we get
$$
x^{2}+y^{2}+2 \geqslant(x+y)+(x y+1) \text {. }
$$
Then $f(x, y)$
$$
\begin{array}{l}
\geqslant((x+y)+(x y+1))\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \\
\geqslant(1+1)^{2}=4 .
\end{array}
$$
Equality holds if and only if $x=y=1$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let non-zero real numbers $a, b$ satisfy $a^{2}+b^{2}=25$. If the function $y=\frac{a x+b}{x^{2}+1}$ has a maximum value $y_{1}$ and a minimum value $y_{2}$, then $y_{1}-y_{2}=$ $\qquad$. | 3.5.
From
$$
\begin{aligned}
y & =\frac{a x+b}{x^{2}+1} \Rightarrow y x^{2}-a x+y-b=0 \\
& \Rightarrow \Delta=a^{2}-4 y(y-b) \geqslant 0 .
\end{aligned}
$$
Thus, $y_{2}$ and $y_{1}$ are the two roots of $a^{2}-4 y(y-b)=0$, at this point,
$$
\Delta_{1}=16 b^{2}+16 a^{2}=400>0 \text {. }
$$
Therefore, $y_{2}$ and $y_{... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let the side length of rhombus $A_{1} A_{2} A_{3} A_{4}$ be $1, \angle A_{1} A_{2} A_{3}=$ $\frac{\pi}{6}, P$ be a point in the plane of rhombus $A_{1} A_{2} A_{3} A_{4}$. Then the minimum value of $\sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}}$ is $\qquad$ | 6. -1 .
Let the center of the rhombus be $O$. Then
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}} \\
= & \mathrm{C}_{4}^{2}|\overrightarrow{P Q}|^{2}+\overrightarrow{P O} \cdot 3 \sum_{1 \leqslant i \leqslant 4} \overrightarrow{O A_{i}}+\sum_{1 \leqslant... | -1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $P(x)=x^{5}-x^{2}+1$ have five roots $r_{1}$, $r_{2}, \cdots, r_{5}$, and $Q(x)=x^{2}+1$. Then
$$
\begin{array}{l}
Q\left(r_{1}\right) Q\left(r_{2}\right) Q\left(r_{3}\right) Q\left(r_{4}\right) Q\left(r_{5}\right) \\
=
\end{array}
$$ | 7.5.
Given that $P(x)=\prod_{j=1}^{5}\left(x-r_{j}\right)$.
$$
\begin{array}{l}
\text { Then } \prod_{j=1}^{5} Q\left(r_{j}\right)=\left(\prod_{j=1}^{5}\left(r_{j}+\mathrm{i}\right)\right)\left(\prod_{j=1}^{5}\left(r_{j}-\mathrm{i}\right)\right) \\
=P(\mathrm{i}) P(-\mathrm{i}) \\
=\left(\mathrm{i}^{5}-\mathrm{i}^{2}+... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given a large regular tetrahedron with an edge length of 6, a smaller regular tetrahedron is placed inside it. If the smaller tetrahedron can rotate freely within the larger one, the maximum edge length of the smaller tetrahedron is . $\qquad$ | 2. 2 .
Given that a smaller regular tetrahedron can rotate freely inside a larger regular tetrahedron, the maximum edge length of the smaller tetrahedron occurs when it is inscribed in the insphere of the larger tetrahedron.
Let the circumradius of the larger tetrahedron be $R$, and the circumradius of the smaller te... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. If $(2 \dot{x}+4)^{2 n}=\sum_{i=0}^{2 n} a_{i} x^{i}\left(n \in \mathbf{Z}_{+}\right)$, then the remainder of $\sum_{i=1}^{n} a_{2 i}$ when divided by 3 is $\qquad$ | 6.1.
Let $x=0$, we get $a_{0}=4^{2 n}$.
By setting $x=1$ and $x=-1$ respectively, and adding the two resulting equations, we get
$$
\begin{array}{l}
\sum_{i=0}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right) . \\
\text { Therefore, } \sum_{i=1}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)-4^{2 n} \\
\equiv(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Divide a circle into a group of $n$ equal parts and color each point either red or blue. Starting from any point, record the colors of $k(k \leqslant n)$ consecutive points in a counterclockwise direction, which is called a “$k$-order color sequence” of the circle. Two $k$-order color sequences are considered differ... | 8. 8 .
In a 3rd-order color sequence, since each point has two color choices, there are $2 \times 2 \times 2=8$ kinds of 3rd-order color sequences.
Given that $n$ points can form $n$ 3rd-order color sequences, we know $n \leqslant 8$.
Thus, $n=8$ can be achieved.
For example, determining the colors of eight points in ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ modulo 3 is divisible by $x^{2}+1$.
$(2015$, Harvard-MIT Mathematics Tournament) | 【Analysis】More explicitly, the polynomial $(x+1)^{n}-1$ needs to satisfy the equivalent condition:
There exist integer-coefficient polynomials $P$ and $Q$ such that
$$
(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x) \text {. }
$$
Assume without loss of generality that
$$
R(x)=(x+1)^{n}-1-P(x)\left(x^{2}+1\right) \text {.... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Let positive real numbers $x, y$ satisfy
$$
x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y}=\frac{27}{4} \text {. }
$$
Then the minimum value of $P=\frac{15}{x}-\frac{3}{4 y}$ is | 7.6.
By the AM-GM inequality for three terms, we have
$$
\begin{array}{l}
x^{2}+\frac{1}{x}=\left(x^{2}+\frac{8}{x}+\frac{8}{x}\right)-\frac{15}{x} \geqslant 12-\frac{15}{x} \\
y^{2}+\frac{1}{y}=\left(y^{2}+\frac{1}{8 y}+\frac{1}{8 y}\right)+\frac{3}{4 y} \geqslant \frac{3}{4}+\frac{3}{4 y} .
\end{array}
$$
Adding th... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=n^{3}-n$ $\left(n \in \mathbf{Z}_{+}\right)$, and the terms in this sequence whose unit digit is 0, arranged in ascending order, form the sequence $\left\{b_{n}\right\}$. Then the remainder when $b_{2} 018$ is divided by 7 is $\qquad$ . | 8. 4 .
Since $a_{n}=n^{3}-n=n(n-1)(n+1)$, therefore, $a_{n}$ has a units digit of 0 if and only if the units digit of $n$ is $1, 4, 5, 6, 9, 0$. Hence, in any consecutive 10 terms of the sequence $\left\{a_{n}\right\}$, there are 6 terms whose units digit is 0.
Since $2018=336 \times 6+2,336 \times 10=3360$, the rema... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. If the function
$$
f(x)=x^{2}-2 a x-2 a|x-a|+1
$$
has exactly three zeros, then the value of the real number $a$ is $\qquad$. | 2. 1.
Let $t=|x-a|(t \geqslant 0)$.
Then the original problem is equivalent to the equation $t^{2}-2 a t+1-a^{2}=0$ having two roots $t_{1}=0, t_{2}>0$.
Upon verification, $a=1$. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and
$$
a_{1}=3, S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right) \text {. }
$$
If $\left\{a_{n}\right\}$ contains three terms $a_{1} 、 a_{p} 、 a_{q}(p 、 q \in$ $\left.\mathbf{Z}_{+}, 1<p<q\right)$ that form an arithmetic seque... | 4. 1 .
Given $S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right)$
$$
\Rightarrow S_{n-1}=2 a_{n-1}+\frac{3}{2}\left((-1)^{n-1}-1\right)(n \geqslant 2) \text {. }
$$
Subtracting the two equations yields $a_{n}=2 a_{n-1}-3(-1)^{n}(n \geqslant 2)$.
Let $b_{n}=\frac{a_{n}}{(-1)^{n}}$. Then
$$
\begin{array}{l}
b_{n}=-2 b_{n... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let integers $x, y$ satisfy $x^{2}+y^{2}4$. Then the maximum value of $x^{2}-2 x y-3 y$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | 3. 3 .
Using a TI calculator, we can obtain the region satisfying
$$
\left\{\begin{array}{l}
x^{2}+y^{2}4
\end{array},\right.
$$
and since we need to find the maximum value of $x^{2}-2 x y-3 y$, the integer point is in the third quadrant.
Substituting $(-2,-3)$ and $(-3,-2)$ into $x^{2}-2 x y-3 y$ and comparing the ... | 3 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Given the inequality $\left|a x^{2}+b x+a\right| \leqslant x$ holds for $x \in$ $[1,2]$. Then the maximum value of $3 a+b$ is $\qquad$ | 6. 3 .
From the problem, we know that $\left|a\left(x+\frac{1}{x}\right)+b\right| \leqslant 1$.
Given $x \in[1,2]$, we have $t=x+\frac{1}{x} \in\left[2, \frac{5}{2}\right]$.
Thus, $|2 a+b| \leqslant 1$, and $\left|\frac{5}{2} a+b\right| \leqslant 1$.
Therefore, $3 a+b=2\left(\frac{5}{2} a+b\right)-(2 a+b) \leqslant 3$... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) A planet has 1000 cities $c_{1}$, $c_{2}, \cdots, c_{1000}$, and three airlines $X$, $Y$, and $Z$ provide flights between these cities. For any $1 \leqslant i<j \leqslant 1000$, exactly one airline operates a one-way flight from city $c_{i}$ to city $c_{j}$. Find the largest positive integer $n$, such... | Four, the maximum value of $n$ is 9.
For each city $c_{i}(i=1,2, \cdots, 1000)$, define a triplet of non-negative integers $\left(x_{i}, y_{i}, z_{i}\right)$ according to the following rules:
If there are no flights from company $X$ arriving at city $c_{i}$, set $x_{i}=0$; otherwise, there exists a largest positive in... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, $\angle A B C=120^{\circ}$, the angle bisector of $\angle A B C$ intersects $A C$ at point $D$, and $B D=1$. Then the minimum value of $4 a+c$ is $\qquad$ | 4.9.
From the problem, we know that $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle B C D}$.
By the angle bisector property and the formula for the area of a triangle, we get
$$
\begin{array}{l}
\frac{1}{2} a c \sin 120^{\circ} \\
=\frac{1}{2} a \times 1 \times \sin 60^{\circ}+\frac{1}{2} c \times 1 \times \sin ... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
One, (40 points) Find the smallest integer $c$, such that there exists a sequence of positive integers $\left\{a_{n}\right\}(n \geqslant 1)$ satisfying:
$$
a_{1}+a_{2}+\cdots+a_{n+1}<c a_{n}
$$
for all $n \geqslant 1$. | Given the problem, we have
$$
c>\frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}}.
$$
For any \( n \geqslant 1 \), we have
$$
\begin{array}{l}
nc > \frac{a_{1}+a_{2}}{a_{1}} + \frac{a_{1}+a_{2}+a_{3}}{a_{2}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}} \\
= n + \frac{a_{2}}{a_{1}} + \left(\frac{a_{1}}{a_{2}} + \frac{a_{... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. If three angles $\alpha, \beta, \gamma$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $\qquad$ | 4. -3 .
According to the problem, $\alpha=\beta-\frac{\pi}{3}, \gamma=\beta+\frac{\pi}{3}$. Therefore, $\tan \alpha=\frac{\tan \beta-\sqrt{3}}{1+\sqrt{3} \tan \beta}, \tan \gamma=\frac{\tan \beta+\sqrt{3}}{1-\sqrt{3} \tan \beta}$.
Then, $\tan \alpha \cdot \tan \beta=\frac{\tan ^{2} \beta-\sqrt{3} \tan \beta}{1+\sqrt{3... | -3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the function $f(x)=x+\frac{2}{x}$ on the interval $[1,4]$, the maximum value is $M$, and the minimum value is $m$. Then the value of $M-m$ is $\qquad$ | ニ.7.4.
Since $f(x)$ is monotonically decreasing on the interval $[1,3]$ and monotonically increasing on the interval $[3,4]$, the minimum value of $f(x)$ is $f(3)=6$.
Also, $f(1)=10, f(4)=\frac{25}{4}$, so the maximum value of $f(x)$ is $f(1)=10$.
Therefore, $M-m=10-6=4$. | 4 | Calculus | math-word-problem | Yes | Yes | cn_contest | false |
11. Let the line $y=k x+b$ intersect the curve $y=x^{3}-x$ at three distinct points $A, B, C$, and $|A B|=|B C|=2$. Then the value of $k$ is | 11. 1.
Given that the curve is symmetric about the point $(0,0)$, and
$$
|A B|=|B C|=2 \text {, }
$$
we know that the line $y=k x+b$ must pass through the origin.
Thus, $b=0$.
Let $A(x, y)$. Then
$$
\begin{array}{l}
y=k x, y=x^{3}-x, \sqrt{x^{2}+y^{2}}=2 \\
\Rightarrow x=\sqrt{k+1}, y=k \sqrt{k+1} . \\
\text { Substi... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $f(x)=\frac{10}{x+1}-\frac{\sqrt{x}}{3}$. Then the number of elements in the set $M=\left\{n \in \mathbf{Z} \mid f\left(n^{2}-1\right) \geqslant 0\right\}$ is $\qquad$. | ,- 1.6 .
From the problem, we know that $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and $f(9)=0$.
Then $f\left(n^{2}-1\right) \geqslant f(9) \Rightarrow 1 \leqslant n^{2} \leqslant 10$.
Thus, the number of elements in set $M$ is 6. | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) Let
$$
\begin{array}{l}
P(z)=z^{4}-(6 \mathrm{i}+6) z^{3}+24 \mathrm{iz}^{2}- \\
(18 \mathrm{i}-18) z-13 .
\end{array}
$$
Find the area of the convex quadrilateral formed by the four points in the complex plane corresponding to the four roots of $P(z)=0$. | II. 9. Notice that, $P(1)=0$.
Then $P(z)=(z-1)\left(z^{3}-(6 \mathrm{i}+6) z^{2}+24 \mathrm{iz}^{2}+\right.$ $(18 \mathrm{i}-5) z+13)$.
Let $Q(z)=z^{3}-(6 \mathrm{i}+5) z^{2}+(18 \mathrm{i}-5) z+13$.
Then $Q(\mathrm{i})=0$.
Hence $Q(z)=(z-\mathrm{i})\left(z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}\right)$.
Using the quadr... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The integer sequence $\left\{a_{i, j}\right\}(i, j \in \mathbf{N})$, where,
$$
\begin{array}{l}
a_{1, n}=n^{n}\left(n \in \mathbf{Z}_{+}\right), \\
a_{i, j}=a_{i-1, j}+a_{i-1, j+1}(i, j \geqslant 1) .
\end{array}
$$
Then the unit digit of the value taken by $a_{128,1}$ is | 8. 4 .
By the recursive relation, we have
$$
\begin{array}{l}
a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}, \\
a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2} .
\end{array}
$$
Accordingly, by induction, we get
$$
\begin{array}{l}
a_{n, m}=\sum_{k=0}^{m-1} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} \\
=\sum_{k \geqslant 0} \mathrm{C}_{m-1}^{k}... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Laura has 2010 lamps and 2010 switches in front of her, with different switches controlling different lamps. She wants to find the correspondence between the switches and the lamps. For this, Charlie operates the switches. Each time Charlie presses some switches, and the number of lamps that light up is the same as ... | (1) Charlie selects a pair of switches $(A, B)$, each operation involves pressing them simultaneously or leaving them untouched, while other switches can be chosen arbitrarily, resulting in $2^{2009}$ different operations, but Laura cannot determine which lights are controlled by switches $A$ and $B$.
If Charlie perfo... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Find the smallest positive integer $n$ with $\tau(n)$ equal to
a) 1
d) 6
b) 2
e) 14
c) 3
f) 100 . | 6. a) 1
b) 2
c) 4
d) 12
e) 192
f) 45360 | 4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Find the first ten terms of the sequence of pseudo-random numbers generated by the linear congruential method with $x_{0}=6$ and $x_{n+1} \equiv 5 x_{n}+2(\bmod 19)$. What is the period length of this generator? | 2. $6.13,10,14,15,1,7,18,16,6,13, \ldots$ period length is 9 | 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 4 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, and property $P_{3}$ is that the element is divisible by 2. Try to verify whether Theorem 7 holds. | $A(1)$ is $3,3,6; A(2)$ is $1,4,7; A(3)$ is $2,4,6. A(1,2)$ has no elements; $A(1,3)$ is $6; A(2,3)$ is $4. A(1,2,3)$ has no elements.$
$$B^{(0)}=B(0) \text { is } 5,5,11; B(1) \text { is } 3,3; B(2) \text { is } 1,7; B(3) \text { is } 2 \text {. }$$
$B(1,2)$ has no elements; $B(1,3)$ is $6; B(2,3)$ is $4. B(1,2,3)$ ha... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 5 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, Property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, Property $P_{3}$ is that the element is divisible by 2, Property $P_{4}$ is that the element is divisible by 3. Try to verify whet... | Solve $A(1)$ is $3,3,6 ; A(2)$ is $1,4,7 ; A(3)$ is $2,4,6 ; A(4)$ is $3,3,6. A(1,2)$ has no elements; $A(1,3)$ is $6 ; A(1,4)$ is $3,3,6 ; A(2,3)$ is $4 ; A(2,4)$ has no elements; $A(3,4)$ is $6. A(1,2,3)$ has no elements; $A(1,2,4)$ has no elements; $A(1,3,4)$ is $6 ; A(2,3,4)$ has no elements; $A(1,2,3,4)$ has no el... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
4. Let $n$ be any given positive integer. Find the value of $\mu(n) \mu(n+1) \mu(n+2) \mu(n+3)$. | 4. 0 , because $n, n+1, n+2, n+3$ must include one number that is divisible by 4. | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Find the smallest positive integer $n$ that satisfies $\tau(n)=6$.
| 5. $n=p_{1}^{a_{1}} \cdots p_{s}^{a_{s}}, p_{1}<\cdots<p_{3} . \tau(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{s}+1\right)=6$. It must be that $\alpha_{1}=1$, $\alpha_{2}=2$; or $\alpha_{1}=2, \alpha_{2}=1$. Therefore, the smallest $n=2^{2} \cdot 3^{1}=12$. | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example: How many zeros are at the end of the decimal representation of 320! ?
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Solve for the positive integer $k$ such that $10^{k} \| 20$!. From equation (12), we know that we only need to find $\alpha(5,20)$, and from the previous example, we know that $k=4$, which is the power of 5. Therefore, there are four zeros at the end. | 4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 8 Proof: $101 x_{1}+37 x_{2}=3189$ has positive integer solutions. | Here, $c=3189<a_{1} a_{2}=101 \cdot 37$, so from the conclusion of Theorem 5, we cannot determine whether the equation has a positive solution (of course, it can be deduced that there is at most one). Therefore, we need to use formula (15) (or (14)). It can be found that $x_{1}=11 \cdot 3189, x_{2}=-30 \cdot 3189$ is a... | 1 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 1 Find the units digit of $3^{406}$ when it is written in decimal form. | Solve: According to the problem, we are required to find the smallest non-negative remainder $a$ when $3^{406}$ is divided by 10, i.e., $a$ satisfies
$$3^{406} \equiv a(\bmod 10), \quad 0 \leqslant a \leqslant 9$$
Obviously, we have $3^{2} \equiv 9 \equiv-1(\bmod 10), 3^{4} \equiv 1(\bmod 10)$, and thus
$$3^{404} \equ... | 9 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
18. Find the integer $n$ that satisfies $n \equiv 1(\bmod 4), n \equiv 2(\bmod 3)$. | 18. $n=4 k+1 \equiv 2(\bmod 3), k=1, n=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Calculate $\left(\frac{137}{227}\right)$. | To determine if 227 is a prime number, by Theorem 1 we get
$$\begin{aligned}
\left(\frac{137}{227}\right) & =\left(\frac{-90}{227}\right)=\left(\frac{-1}{227}\right)\left(\frac{2 \cdot 3^{2} \cdot 5}{227}\right) \\
& =(-1)\left(\frac{2}{227}\right)\left(\frac{3^{2}}{227}\right)\left(\frac{5}{227}\right) \\
& =(-1)\left... | -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Let $n \geqslant 1$. Prove: $(n!+1,(n+1)!+1)=1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. | 5. $(n!+1,(n+1)!+1)=(n!+1,-n)=(1,-n)=1$.
The translation is as follows:
5. $(n!+1,(n+1)!+1)=(n!+1,-n)=(1,-n)=1$. | 1 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 1 Determine whether the congruence equation $2 x^{3}+5 x^{2}+6 x+1 \equiv 0(\bmod 7)$ has three solutions. | Solve: Here the coefficient of the first term is 2. By making an identity transformation, we can know that the original equation has the same solutions as
$$4\left(2 x^{3}+5 x^{2}+6 x+1\right) \equiv x^{3}-x^{2}+3 x-3 \equiv 0(\bmod 7)$$
Performing polynomial division, we get
$$x^{7}-x=\left(x^{3}-x^{2}+3 x-3\right)\l... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
8. Let $p$ be a prime, $\delta_{p}(a)=3$. Prove: $\delta_{p}(1+a)=6$.
| 8. From $\delta_{p}(a)=3$, we get $a \neq \pm 1(\bmod p), a^{2}+a+1 \equiv 0(\bmod p)$. Therefore, $1+a \neq 1(\bmod p),(1+a)^{2} \equiv 1+2 a+a^{2} \equiv a \not \equiv 1(\bmod p),(1+a)^{3} \equiv-1(\bmod p)$. Hence, $\delta_{p}(1+a)=6$ | 6 | Number Theory | proof | Yes | Yes | number_theory | false |
Example 1 Find the primitive root of $p=23$.
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 for the translator and should not be included in the final translation. The correct output should be as follows:)
Example 1 Find the primitive root of $p=23$.
| Since the index of $a$ modulo $p$ must be a divisor of $p-1$, to find the index, we just need to compute the residue of $a^{d}$ modulo $p$, where $d \mid p-1$.
Here $p-1=22=2 \cdot 11$, its divisors $d=1,2,11,22$. First, find the index of $a=2$ modulo 23:
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 23) \\
2^{11} \equiv\le... | 5 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find a primitive root modulo 41. | Solve $41-1=40=2^{3} \cdot 5$, divisors $d=1,2,4,8,5,10,20,40$. Now we will find the orders of $a=2,3, \cdots$.
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 41), 2^{4} \equiv 16(\bmod 41), 2^{5} \equiv-9(\bmod 41) \\
2^{10} \equiv-1(\bmod 41), \quad 2^{20} \equiv 1(\bmod 41)
\end{array}$$
So $\delta_{41}(2) \mid 20$. Since... | 7 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
11. Under the notation of the previous question, find $j$ that satisfies
(i) $0 \bmod 3 \cap 0 \bmod 5=j \bmod 15$;
(ii) $1 \bmod 3 \cap 1 \bmod 5=j \bmod 15$;
(iii) $-1 \bmod 3 \cap -2 \bmod 5=j \bmod 15$. | 11. (i) $j=0$;
(ii) $j=1$;
(iii) $j=8$. | 0 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
5. Take a total of ten coins of one cent, two cents, and five cents, to pay eighteen cents. How many different ways are there to do this?
保留源文本的换行和格式，直接输出翻译结果如下：
5. Take a total of ten coins of one cent, two cents, and five cents, to pay eighteen cents. How many different ways are there to do this? | 5. Solution: Let $x, y, z$ represent the number of 1-cent, 2-cent, and 5-cent coins, respectively. Therefore, we have the following equations:
$$\begin{array}{l}
x+2 y+5 z=18 \\
x+y+z=10
\end{array}$$
Subtracting the second equation from the first, we get $y+4 z=8$.
We need to find the non-negative integer solutions t... | 3 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false |
8. Find the greatest common divisor using the method of prime factorization:
(i) $48,84,120$.
(ii) $360,810,1260,3150$. | 8.
(i) Solution: Decomposing each number into prime factors, we get
$$\begin{array}{l}
48=2 \times 2 \times 2 \times 2 \times 3=2^{4} \times 3 \\
84=2 \times 2 \times 3 \times 7=2^{2} \times 3 \times 7 \\
120=2 \times 2 \times 2 \times 3 \times 5=2^{3} \times 3 \times 5
\end{array}$$
Therefore, $(48,84,120)=2^{2} \tim... | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
17. A box of grenades, assuming the weight of each grenade is an integer greater than one pound, the net weight after removing the weight of the box is 201 pounds. After taking out several grenades, the net weight is 183 pounds. Prove that the weight of each grenade is 3 pounds. | 17. Proof: Since 201 jin and 183 jin are both the weights of an integer number of grenades, the weight of each grenade must be a common divisor of them. Their greatest common divisor is
$$(201,183)=3 \times(67,61)=3,$$
Since 3 is a prime number, and the divisors of 3 are only 1 and 3, the weight of each grenade must b... | 3 | Number Theory | proof | Yes | Yes | number_theory | false |
27. Find the greatest common divisor of the following:
(i) $435785667,131901878$.
(ii) $15959989,7738$. | 27.
(i) Solution: Since
$3\left|\begin{array}{r|r|r}435785667 \\ 395705634 & 131901878 \\ \hline 40080033 & 120240099 \\ 34985337 & 101861779 \\ \hdashline & 5094696 & 1472387 \\ 4417161 & 1355070\end{array}\right| 3$
\begin{tabular}{|c|c|c|c|}
\hline 2 & \begin{tabular}{l}
\begin{tabular}{l}
677535 \\
586585
\end{tabu... | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 1 Find the greatest common divisor of 36 and 24. | Solve: Decompose these two numbers into prime factors
$$36=2 \times 2 \times 3 \times 3, \quad 24=2 \times 2 \times 2 \times 3$$
By comparing the prime factors of these two numbers, we can see that the prime factors $2,2,3$ are common to both numbers. Their product is the greatest common divisor of these two numbers:
... | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 2 Find the greatest common divisor of $48$, $60$, and $72$. | Solve: Decompose the three numbers into prime factors respectively:
$$\begin{array}{l}
48=2 \times 2 \times 2 \times 2 \times 3=2^{4} \times 3 \\
60=2 \times 2 \times 3 \times 5=2^{2} \times 3 \times 5 \\
72=2 \times 2 \times 2 \times 3 \times 3=2^{3} \times 3^{2}
\end{array}$$
By comparing the prime factors of the ab... | 12 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
Example 6 Discuss the congruence equation
$$x^{2} \equiv -286 \pmod{4272943}$$
whether it has a solution, where 4272943 is a prime number. | Let $p=4272943$, by Lemma 7 we have
$$\left(\frac{-286}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)\left(\frac{143}{p}\right)$$
Since $4272943 \equiv 7(\bmod 8)$, we have
$$\left(\frac{-1}{p}\right)=-1,\left(\frac{2}{p}\right)=1$$
Thus,
$$\left(\frac{-286}{p}\right)=-\left(\frac{143}{p}\right)$$
Sinc... | 1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
15. Let $p \geqslant 3$, try to calculate the value of the following expression:
$$\left(\frac{1 \cdot 2}{p}\right)+\left(\frac{2 \cdot 3}{p}\right)+\cdots+\left(\frac{(p-2)(p-1)}{p}\right)$$ | 15. Solution: To solve this problem, we need to study the properties of the Legendre symbol with the general term $\left(\frac{n(n+1)}{p}\right)$, where $(n, p)=1$.
By $(n, p)$ being coprime, we know there must exist an integer $r_{n}$ such that $p \nmid r_{n}$ and $n r_{n} \equiv 1(\bmod p)$. This $r_{n}$ is called t... | -1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
29. Let $p$ be a prime, $a, b \in \mathbf{N}^{*}$, satisfying: $p>a>b>1$. Find the largest integer $c$, such that for all $(p, a, b)$ satisfying the conditions, we have
$$p^{c} \mid\left(\mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}\right)$$ | 29. When taking $p=5, a=3, b=2$, we should have $5^{c} \mid 3000$, so $c \leqslant 3$. Below, we prove that for any $p, a, b$ satisfying the conditions, we have $p^{3} \mid \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}$.
In fact, notice that
$$\begin{aligned}
& \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}=\frac{(a p)(a p-1) ... | 3 | Number Theory | math-word-problem | Yes | Yes | number_theory | false |
2. Let $m, n \in \mathbf{N}^{*}, m$ be an odd number. Prove: $\left(2^{m}-1,2^{n}+1\right)=1$.
| 2. Let $\left(2^{m}-1,2^{n}+1\right)=d$, then
$$1 \equiv\left(2^{m}\right)^{n}=\left(2^{n}\right)^{m} \equiv(-1)^{m}=-1(\bmod d)$$
This leads to $d \mid 2$. Combining this with $d$ being odd, we conclude $d=1$. | 1 | Number Theory | proof | Yes | Yes | number_theory | false |
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