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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
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {. }
$$
(1) Find the number of $n$-ary Boolean functions;
(2) Let $g$ be an $n$-ary Boolean function, satisfying
$$
g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j}(\bmod 2),
$$
Find the number of elements in the set $D_{n}(g)$, and find the largest positive integer $n$, such that
$$
\sum_{\left(x_{1}, x_{2}, \cdots, \cdots, x_{n}\right) \in D_{n}(g)}\left(x_{1}+x_{2}+\cdots+x_{n}\right) \leqslant 2017 .
$$
|
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, $|D_{1}(g)|=1, |D_{2}(g)|=1$.
$$
\begin{array}{l}
\text { Also, } g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j} \\
=\left(1+x_{1}\left(1+\sum_{i=2}^{n} \prod_{j=2}^{i} x_{j}\right)\right)(\bmod 2),
\end{array}
$$
$$
\begin{aligned}
& \text { Then }|D_{n}(g)|=2^{n-1}-|D_{n-1}(g)|(n=3,4, \cdots) . \\
& \text { Hence }|D_{n}(g)|=2^{n-1}-2^{n-2}+|D_{n-2}(g)|=\cdots \\
& =\sum_{k=0}^{n-1}(-1)^{k} \cdot 2^{n-1-k} \\
= & \frac{2^{n}+(-1)^{n+1}}{3} .
\end{aligned}
$$
Let $c_{n}=\sum_{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in D_{n}(g)}\left(x_{1}+x_{2}+\cdots+x_{n}\right)$.
Notice that,
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 0\right) \equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2) \\
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}+x_{1} x_{2} \cdots x_{n-1}(\bmod 2)
\end{array}
$$
If $x_{1} x_{2} \cdots x_{n-1}=1$, then
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \equiv n+1(\bmod 2) \\
=\left\{\begin{array}{l}
0, n \text { is odd; } \\
1, n \text { is even. }
\end{array}\right.
\end{array}
$$
If $x_{1} x_{2} \cdots x_{n-1}=0$, then
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2)=0 \\
\Leftrightarrow g\left(x_{1}, x_{2}, \cdots, x_{n-1}\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2)=0
\end{array}
$$
Therefore, when $n \geqslant 2$,
$$
c_{n}=\left\{\begin{array}{l}
n+2 c_{n-1}+|D_{n-1}(g)|, n \text { is odd; } \\
2 c_{n-1}+|D_{n-1}(g)|-n, n \text { is even. }
\end{array}\right.
$$
Obviously, $c_{1}=1, c_{2}=1$.
When $n=2 m$,
$$
\begin{array}{l}
c_{2 m}=2 c_{2 m-1}+\frac{2^{2 m-1}+(-1)^{2 m}}{3}-2 m \\
=-\frac{1}{3}+\frac{4}{3} \times 2^{2 m-2}+(2 m-2)+4 c_{2 m-2} \\
=\frac{(3 m+1) 4^{m}-(6 m+1)}{9}
\end{array}
$$
Similarly, when $n=2 m+1$,
$$
\begin{array}{l}
c_{2 m+1}=2 m+1+2 c_{2 m}+\frac{2^{2 m}+(-1)^{2 m+1}}{3} \\
=\frac{1}{3}+\frac{4}{3} \times 2^{2 m-1}-(2 m-1)+4 c_{2 m-1} \\
=\frac{(6 m+5) 4^{m}+6 m+4}{9} .
\end{array}
$$
Therefore, from equations (2), (3), and
$$
\begin{array}{l}
c_{9}=828<3986=c_{11}, \\
c_{10}=1817<8643=c_{12},
\end{array}
$$
we know that the largest positive integer $n$ for which equation (1) holds is $n=10$.
|
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 $T$. A set $S \subset A$ of some finite sequences satisfies: for any infinite sequence $M$ in set $A$, there exists a unique sequence $T$ in $S$ such that $M$ starts with $T$. Which of the following three conclusions is true:
(1) $S$ must be a finite set;
(2) $S$ must be an infinite set;
(3) $S$ can be either a finite set or an infinite set? Prove your conclusion.
(2017, Peking University Middle School Mathematics Science Summer Camp)
|
【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 $T_{2}$ starts with $T_{1}$. Otherwise, consider an infinite sequence $M$ in set $A$ that starts with $T_{2}$.
Then $M$ can start with both $T_{2}$ in $S$ and $T_{1}$ in $S$, which contradicts the uniqueness in the problem.
Next, we prove that $S$ cannot be an infinite set, i.e., (3) does not hold.
If $S$ is an infinite set, then by the pigeonhole principle, there must exist some $b_{1} \in\{1,2, \cdots, 2017\}$ such that there are infinitely many sequences in $S$ that start with the sequence $\left\{b_{1}\right\}$, and by the previous analysis, we know that $\left\{b_{1}\right\} \notin S$; similarly, by the pigeonhole principle, there must exist some $b_{2} \in\{1,2, \cdots, 2017\}$ such that there are infinitely many sequences in $S$ that start with the sequence $\left\{b_{1}, b_{2}\right\}$, and $\left\{b_{1}, b_{2}\right\} \notin S$; and so on, for any given positive integer $n$, there exists a sequence $\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ such that there are infinitely many sequences in $S$ that start with $\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ and the sequence $\left\{b_{1}, b_{2}, \cdots, b_{n}\right\} \notin S$.
Thus, we can obtain an infinite sequence
$$
M^{\prime}=\left\{b_{1}, b_{2}, \cdots, b_{n}, \cdots\right\} \text {, }
$$
satisfying that for any positive integer $n$, the sequence
$$
\left\{b_{1}, b_{2}, \cdots, b_{n}\right\} \notin S \text {. }
$$
Therefore, this infinite sequence $M^{\prime}$ cannot start with any finite sequence in set $S$, leading to a contradiction.
In summary, only conclusion (1) holds.
|
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 Olympiad)
|
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 number of elements in the set $S$ sought is 8.
|
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 $\frac{B G}{G E}$.
(2010-2011, Hungarian Mathematical Olympiad)
|
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
$$
\begin{array}{l}
\frac{AM}{MB}=\frac{S_{\triangle ACD}}{S_{\triangle BCD}}=\frac{AC \cdot AD \sin \angle CAD}{BC \cdot BD \sin \angle CBD} \\
=\frac{AC}{BC} \cos \angle BCA=\left(\frac{AC}{BC}\right)^{2} .
\end{array}
$$
Applying Menelaus' Theorem to $\triangle BAE$ and the transversal $GMC$ yields
$$
\frac{BG}{GE}=\frac{CA}{CE} \cdot \frac{BM}{MA}=\frac{CA}{CE} \cdot \frac{BC^2}{AC^2}=2 \cdot \frac{CB}{CA} \cdot \frac{CO}{CE}=2 .
$$
|
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 segments $p_{i} p_{j} \backslash p_{h} p_{k}$ in the graph, the number of triangles in $S$ with $p_{i} p_{j}$ as a side is equal to the number of triangles in $S$ with $p_{h} p_{k}$ as a side. Find the smallest positive integer $n$, such that in the set $S$ there are always two triangles, each with three vertices of the same color.
(2007, China National Training Team Test)
|
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:
$$
\begin{array}{l}
\triangle p_{1} p_{2} p_{3} 、 \triangle p_{1} p_{3} p_{4} 、 \triangle p_{1} p_{4} p_{5} 、 \triangle p_{1} p_{5} p_{6} 、 \\
\triangle p_{1} p_{2} p_{6} 、 \triangle p_{2} p_{3} p_{5} 、 \triangle p_{2} p_{4} p_{5} 、 \triangle p_{2} p_{4} p_{6} 、 \\
\triangle p_{3} p_{4} p_{6} 、 \triangle p_{3} p_{5} p_{6} .
\end{array}
$$
It is easy to verify that for any line segment $p_{i} p_{j}$ $(1 \leqslant i<j \leqslant 6)$, there are exactly two triangles, but only $\triangle p_{1} p_{2} p_{3}$ has all three vertices of the same color.
Therefore, the set $S$ of these ten triangles does not satisfy the conditions of the problem.
When $n=7$, color $p_{1} 、 p_{2} 、 p_{4}$ red, and $p_{3}$ 、 $p_{5} 、 p_{6} 、 p_{7}$ blue, and consider the triangles formed by the cyclic permutation arrangement:
$\triangle p_{1} p_{2} p_{4} 、 \triangle p_{2} p_{3} p_{5} 、 \triangle p_{3} p_{4} p_{6} 、 \triangle p_{4} p_{5} p_{7} 、$
$\triangle p_{5} p_{6} p_{1} 、 \triangle p_{6} p_{7} p_{2} 、 \triangle p_{7} p_{1} p_{3}$.
It is easy to verify that the 21 edges of the complete graph of order 7 are exactly the 21 edges of these seven triangles. But among the seven triangles, only $\triangle p_{1} p_{2} p_{4}$ has all three vertices of the same color.
Therefore, the set $S$ of these seven triangles does not satisfy the conditions of the problem.
When $n=8$, call the line segment connecting two points of the same color a same-color edge, and the line segment connecting two points of different colors a different-color edge. Thus, there are at most 16 different-color edges in the graph. At this point, there are four red and four blue points among the eight vertices, with 12 same-color edges and 28 edges in total in the graph.
Since 3 does not divide $28 、 56 、 112 、 140$, but divides $84 、 168$, to satisfy the conditions of the problem, each edge can only be a common edge of three triangles or six triangles.
If $k=3$, i.e., each edge is a common edge of three triangles, then $|S|=28$.
Note that each different-color triangle has exactly two different-color edges, and there are at most 48 different-color edges in the graph (each edge is counted three times). Therefore, there are at most 24 different-color triangles. Thus, the set $S$ must contain at least four same-color triangles.
If $k=6$, then $|S|=56$. Similarly, there must be at least eight same-color triangles.
This indicates that when $n=8$, the conditions of the problem are satisfied.
In summary, the smallest positive integer $n=8$.
|
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 $a=3, b=2, c=1$, the equality holds.
Therefore, the minimum value of $\frac{a b c}{a+b+c}$ is 1.
|
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
$$
\begin{array}{l}
s+\frac{1}{t}=-\frac{99}{19}, \frac{s}{t}=\frac{1}{19} . \\
\text { Hence } \frac{s t+4 s+1}{t}=\left(s+\frac{1}{t}\right)+4 \cdot \frac{s}{t} \\
=-\frac{99}{19}+\frac{4}{19}=-5 .
\end{array}
$$
|
-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 $\sqrt{2} x_{1}^{2}=1, \sqrt{2} x_{2}^{2}=2, \sqrt{2} x_{3}^{2}=3$ respectively.
Therefore, $x_{1}^{2}=\frac{\sqrt{2}}{2}, x_{2}^{2}=\sqrt{2}, x_{3}^{2}=\frac{3 \sqrt{2}}{2}$.
The squares of the lengths of the three sides are
$$
\left(2 x_{1}\right)^{2}=2 \sqrt{2},\left(2 x_{2}\right)^{2}=4 \sqrt{2},\left(2 x_{3}\right)^{2}=6 \sqrt{2} \text {, }
$$
Thus, the triangle formed by the three line segments of lengths $2 x_{1}, 2 x_{2}, 2 x_{3}$ is a right triangle.
Let the area of this right triangle be $S$. Then
$$
\begin{array}{l}
S^{2}=\left(\frac{1}{2} \times 2 x_{1} \times 2 x_{2}\right)^{2}=\frac{2 \sqrt{2} \times 4 \sqrt{2}}{4}=4 \\
\Rightarrow S=2 .
\end{array}
$$
|
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.
$$
\text{Therefore, } \sum_{k=0}^{201}(10 k+7)^{k+1} \equiv 7^{1}+7^{2} \equiv 6(\bmod 10) \text{. }
$$
|
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 the smallest positive integer $n$ that satisfies $\sum_{i=1}^{n} a_{i}>1000$ is $n=$ $\qquad$.
|
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, $\left|\overrightarrow{P_{n} P_{n+1}}\right|=O P_{n}$,
$\left|\overrightarrow{P_{n+1} P_{n+2}}\right|=\left|O P_{n+1}\right|=\sqrt{2}\left|O P_{n}\right|$.
Therefore, $a_{n}=\left|O P_{n}\right|^{2}=2\left|O P_{n-1}\right|^{2}=2^{n-1}$.
Also, $\sum_{i=1}^{n} a_{i}=2^{n}-1>1000$, hence $n \geqslant 10$.
|
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
$$
\begin{array}{l}
S_{n+2}S_{n+2}+\lambda S_{n} \geqslant 2 \sqrt{\lambda S_{n+2} S_{n}} \\
\Rightarrow \frac{S_{n+2}}{S_{n+1}}<\frac{\lambda}{4} \cdot \frac{S_{n+1}}{S_{n}} \\
\Rightarrow 1<\frac{S_{n+1}}{S_{n}}<\left(\frac{\lambda}{4}\right)^{n-1} \frac{S_{2}}{S_{1}} .
\end{array}
$$
Since the above inequality holds for any $n$, hence $\lambda \geqslant 4$. When $\lambda=4$, taking $a_{n}=10^{2 n}$ satisfies the conditions.
In conclusion, the minimum value of $\lambda$ is 4.
|
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 element $x_{i}$ is $k_{i}(i=1,2, \cdots, n)$, and let $m=$ $\max \left\{k_{1}, k_{2}, \cdots, k_{n}\right\}$. Find the minimum value of $m$.
|
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 two elements, we have
$$
\begin{array}{l}
\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2} \geqslant 2 \mathrm{C}_{10}^{2}=90 . \\
\text { Hence } 180 \leqslant \sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \\
\leqslant \sum_{i=1}^{n} k_{i}(m-1)=45(m-1) \\
\Rightarrow m \geqslant 5 .
\end{array}
$$
Below is the construction for $m=5$.
$$
A_{i}=\{\bar{i}, \overline{i+1}, \overline{i+2}, \overline{i+4}, \overline{i+7}\},
$$
where $i=1,2, \cdots, 10, \bar{i} \equiv i(\bmod 10)$, and $\bar{i} \in$ $\{1,2, \cdots, 10\}$.
|
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 sequence $\left\{x_{n}\right\}$ and all positive integers $m$, we have
$$
\frac{1}{m} \sum_{i=1}^{m} x_{i}^{2} \leqslant \sum_{i=1}^{m} \lambda^{m-i} y_{i}^{2} .
$$
|
【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 $y_{1}=1, y_{n}=0(n \geqslant 2)$.
Substitute into equation (1) to get
$$
\begin{array}{l}
\frac{1}{m} \cdot 2^{m-1} \leqslant \lambda^{m-1} \\
\Rightarrow \frac{2}{\lambda} \leqslant \sqrt[m-1]{m} .
\end{array}
$$
When $m \rightarrow \infty$, $\frac{2}{\lambda} \leqslant 1 \Rightarrow \lambda \geqslant 2$.
Next, use mathematical induction to prove: for sequences $\left\{x_{n}\right\}$ and $\left\{y_{n}\right\}$ that satisfy the conditions and for all positive integers $m$, equation (1) holds. When $m=1$, $x_{1}^{2}=y_{1}^{2} \leqslant y_{1}^{2}$, the conclusion holds.
Assume the conclusion holds for $m=k$.
Then when $m=k+1$, let
$$
\begin{array}{l}
f(k+1)=\sum_{i=1}^{k+1} 2^{k+1-i} y_{i}^{2}-\frac{1}{k+1} \sum_{i=1}^{k+1} x_{i}^{2}, \\
f(k)=\sum_{i=1}^{k} 2^{k-i} y_{i}^{2}-\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2} .
\end{array}
$$
Thus, $f(k+1)-f(k)$
$$
\geqslant y_{k+1}^{2}+\frac{1}{k} \sum_{i=1}^{k} x_{i}^{2}-\frac{1}{k+1} x_{k+1}^{2}+f(k) \text {. }
$$
By the given conditions and the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
x_{k+1}^{2}=\left(y_{k+1}+\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{k}^{2}}\right)^{2} \\
\leqslant\left((k+1) y_{k+1}^{2}+\frac{k+1}{k} \sum_{i=1}^{k} x_{i}^{2}\right)^{2}\left(\frac{1}{k+1}+\frac{k}{k+1}\right) \\
\leqslant(k+1) y_{k+1}^{2}+\left(1+\frac{1}{k}\right) \sum_{i=1}^{k} x_{i}^{2} .
\end{array}
$$
By the induction hypothesis, we get
$$
f(k+1) \geqslant f(k) \geqslant 0 \text {. }
$$
Therefore, the conclusion holds when $m=k+1$.
In summary, the minimum value of the positive number $\lambda$ is 2.
|
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 monotonically decreasing when $t \in (-\infty,-1)$, and monotonically increasing when $t \in \left(-1, \frac{1}{4}\right]$.
Therefore, when $t=-1$, $f(t)$ has a maximum value of 4, i.e., $\left(x^{3}+1\right)\left(y^{3}+1\right)$ has a maximum value of 4.
|
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 set of points $\left\{A_{n}\right\}(n \in \mathbf{N})$ connected in sequence forms a polygonal line that encloses the feasible region.
It is easy to prove that the points $A_{n}$ are all above $l_{m}(m>n, m \in \mathbf{N})$.
In this problem, let $41 x+2 y=z$, then $y=-\frac{41}{2} x+\frac{1}{2} z$.
Since $-\frac{41}{2} \in\left[-4^{2},-5^{2}\right]$, when the line $y=-\frac{41}{2} x+\frac{1}{2} z$ passes through the point $A_{4}\left(\frac{1}{9}, \frac{20}{9}\right)$,
$$
z=41 x+2 y=41 \times \frac{1}{9}+2 \times \frac{20}{9}=9 .
$$
|
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 $x>3$,
$$
f(x)=\log _{2}\left(1-\frac{1}{x-2}\right)+\cos x \pi,
$$
Thus, $f(x)$ is monotonically increasing in the interval $(3,4)$, and
$$
f(4)=0 \text {. }
$$
By the monotonicity of $\log _{2}\left(1-\frac{1}{x-2}\right)$, we know that when $x>4$,
$$
f(x)>\log _{2} \frac{1}{2}-1=-2 \text {. }
$$
Therefore, $\beta \in(3,4)$ and is unique.
Similarly, $\alpha$ is also unique.
Thus, $\alpha+\beta=\frac{5}{2} \times 2=5$.
|
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 probability that she wins is $\frac{p}{q}\left((p, q)=1, p 、 q \in \mathbf{Z}_{+}\right)$. Then the value of $q-p$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
|
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 probability of Amelia winning is $\frac{1}{3}$.
In the first round, the probability of Amelia not winning is $1 - \frac{1}{3} = \frac{2}{3}$, and the probability of Brian not winning is $1 - \frac{2}{5} = \frac{3}{5}$. Therefore, the probability of both not winning is $\frac{2}{3} \times \frac{3}{5} = \frac{2}{5}$.
Thus, $P_{0} = \frac{1}{3} + \frac{2}{5} P_{0} \Rightarrow P_{0} = \frac{5}{9}$.
Therefore, the answer is $9 - 5 = 4$.
|
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 represent the cup mouth facing up, and 0 to represent the cup mouth facing down. The following flips can satisfy the condition:
$$
\begin{array}{l}
(1,1,1,1) \xrightarrow{1}(0,0,0,1) \xrightarrow{2}(0,1,1,0) \\
\xrightarrow{3}(1,0,1,1) \xrightarrow{4}(0,0,0,0) .
\end{array}
$$
Therefore, the minimum value of the positive integer $n$ is 4.
|
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
$$
\frac{4}{\log _{2} a}+\log _{2} a \geqslant 2 \sqrt{\frac{4}{\log _{2} a} \log _{2} a}=4 \text {, }
$$
with equality holding if and only if $a=4$.
|
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 $x_{0}$ and rearranging gives
$$
\begin{array}{l}
3 b^{2}-4 a b+1=0 \Rightarrow a=\frac{3 b^{2}+1}{4 b} . \\
\text { Hence }|z|^{2}=a^{2}+b^{2}=\left(\frac{3 b^{2}+1}{4 b}\right)^{2}+b^{2} \\
=\frac{25}{16} b^{2}+\frac{1}{16 b^{2}}+\frac{3}{8} \geqslant \frac{5}{8}+\frac{3}{8}=1,
\end{array}
$$
with equality holding if and only if $b= \pm \frac{\sqrt{5}}{5}$.
|
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 sequence;
(2) For any $n$, it holds that
$$
\prod_{i=1}^{n}\left(S_{i}+1\right) \geqslant k n,
$$
find the maximum value of $k$.
|
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, $\frac{1}{S_{1}-1}=\frac{1}{2-1}=1$, so $\left\{\frac{1}{S_{n}-1}\right\}$ is an arithmetic sequence with the first term and common difference both equal to 1.
(2) From (1), we know
$$
\frac{1}{S_{n}-1}=n \Rightarrow S_{n}=\frac{n+1}{n} \text {. }
$$
From the condition, we get
$$
k \leqslant\left(\frac{1}{n} \prod_{i=1}^{n}\left(S_{i}+1\right)\right)_{\min } \text {. }
$$
Let $f(n)=\frac{1}{n} \prod_{i=1}^{n}\left(S_{i}+1\right)$. Then
$$
\frac{f(n+1)}{f(n)}=\frac{n\left(S_{n+1}+1\right)}{n+1}=\frac{n(2 n+3)}{(n+1)^{2}}>1 \text {. }
$$
Therefore, $f(n)_{\min }=f(1)=\frac{S_{1}+1}{1}=3, k_{\max }=3$.
|
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 $P Q$ satisfy $k_{1}=3 k_{2}$, then the value of the real number $m$ is . $\qquad$
|
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 conclusion also holds.
Similarly, $y_{M} y_{N}=-4, y_{Q}=\frac{-4 m}{y_{N}}$.
Thus, $k_{1}=\frac{4}{y_{M}+y_{N}}, k_{2}=\frac{4}{y_{P}+y_{Q}}=\frac{1}{m} k_{1}$.
Also, $k_{2}=\frac{1}{3} k_{1}$, therefore, $m=3$.
|
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.,
$$
\begin{aligned}
& 1-a=0,2-(b+2)+(1-a)=0 \\
\Rightarrow & a=1, b=0 .
\end{aligned}
$$
Therefore, $8 a+4 b+|8 a+4 b-5|=11$.
In summary, the value of the algebraic expression is 5 or 11.
|
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
$$
\left\{\begin{array}{l}
a_{n+1}=a_{n}+b_{n}+\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}} \\
b_{n+1}=a_{n}+b_{n}-\sqrt{a_{n}^{2}-a_{n} b_{n}+b_{n}^{2}}
\end{array} .\right.
$$
(1) Find the general terms of the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$;
(2) Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $S_{n}=\sum_{i=1}^{n}\left[a_{i}\right], T_{n}=\sum_{i=1}^{n}\left[b_{i}\right]$. Find the smallest $n \in \mathbf{Z}_{+}$ such that
$$
\sum_{k=1}^{n}\left(S_{k}+T_{k}\right)>2017 .
$$
|
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 that, $a_{n} 、 b_{n}>0$ and $a_{n}>b_{n}$.
Thus, $a_{n}=2^{n}+\sqrt{4^{n}-3^{n}}, b_{n}=2^{n}-\sqrt{4^{n}-3^{n}}$.
(2) From (1), we know $a_{n}+b_{n}=2^{n+1}$.
Then $2^{n+1}-2<\left[a_{n}\right]+\left[b_{n}\right] \leqslant 2^{n+1}$.
We will prove by contradiction that when $n \geqslant 2$, there does not exist $n$ such that $\sqrt{4^{n}-3^{n}} \in \mathbf{Z}_{+}$.
If there exist $n 、 t \in \mathbf{Z}_{+}$ such that $4^{n}-3^{n}=t^{2}$, then
$$
\left(2^{n}+t\right)\left(2^{n}-t\right)=3^{n} \text {. }
$$
When 3 divides $\left(2^{n}+t\right)$ and 3 divides $\left(2^{n}-t\right)$, $3 \mid t$, which implies $3 \mid 4^{n}$, a contradiction.
Thus, only $2^{n}-t=1 \Rightarrow t=2^{n}-1$.
Substituting into equation (1) gives $2^{n+1}=3^{n}+1$.
However, $2^{n+1}-3^{n}=8 \times 2^{n-2}-9 \times 3^{n-2}<0$, a contradiction.
Therefore, only when $n=1$, $a_{n} 、 b_{n} \in \mathbf{Z}_{+}$, at this time, $a_{1}+b_{1}=4$;
When $n \geqslant 2$, $\left[a_{n}\right]+\left[b_{n}\right]=2^{n+1}-1$.
Then $S_{n}+T_{n}=4+\sum_{i=2}^{n}\left(2^{i+1}-1\right)$ $=2^{n+2}-n-3$.
$$
\text { Hence } \sum_{k=1}^{n}\left(S_{k}+T_{k}\right)=2^{n+3}-8-\frac{n(n+1)}{2}-3 n \text {. }
$$
In particular, when $n=8$,
$$
\sum_{k=1}^{8}\left(S_{k}+T_{k}\right)=2^{11}-8-\frac{8(8+1)}{2}-24=1980 \text {. }
$$
Therefore, the smallest $n=9$.
|
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 on segment $A C$, satisfying $\frac{|C E|}{|C A|}=\frac{|A D|}{|A B|}$.
(1) Prove that the line $D E$ intersects the parabola at exactly one point.
(2) Let the intersection point of line $D E$ and the parabola be $F$. Denote the areas of $\triangle B C F$ and $\triangle A D E$ as $S_{1}$ and $S_{2}$, respectively. Find the value of $\frac{S_{1}}{S_{2}}$.
|
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, $\left(x_{1}+2 a^{2}, y_{1}\right)=\lambda\left(4 a^{2}, 2 a\right)$.
Hence, $D\left((4 \lambda-2) a^{2}, 2 \lambda a\right)$.
Let $E\left(x_{2}, y_{2}\right)$. From $\overrightarrow{C E}=\lambda \overrightarrow{C A}$, we get $\left(x_{2}-2 a^{2}, y_{2}+2 a\right)=\lambda\left(-4 a^{2}, 2 a\right)$.
Hence, $E\left((2-4 \lambda) a^{2},(2 \lambda-2) a\right)$.
When $\lambda=\frac{1}{2}$, the line $D E$ is the $y$-axis, and the conclusion is obviously true.
When $\lambda \neq \frac{1}{2}$, the slope of line $D E$ exists, and
$$
k_{D E}=\frac{2 a}{(8 \lambda-4) a^{2}}=\frac{1}{(4 \lambda-2) a} \text {. }
$$
Thus, $l_{D E}: y-2 \lambda a=\frac{1}{(4 \lambda-2) a}\left(x-(4 \lambda-2) a^{2}\right)$
$$
\Rightarrow x=2 a(2 \lambda-1) y-2 a^{2}(2 \lambda-1)^{2} \text {. }
$$
Combining with the parabola equation, we get
$$
\begin{array}{l}
y^{2}-4 a(2 \lambda-1) y+4 a^{2}(2 \lambda-1)^{2}=0 \\
\Rightarrow y=2 a(2 \lambda-1) .
\end{array}
$$
Substituting into equation (1) gives $x=2 a^{2}(2 \lambda-1)^{2}$.
Therefore, the line $D E$ intersects the parabola at exactly one point $F\left(2 a^{2}(2 \lambda-1)^{2}, 2 a(2 \lambda-1)\right)$.
(2) Note that,
$$
S_{1}=\frac{1}{2} \cdot 4 a\left(2 a^{2}-x_{F}\right)=4 a^{3}\left(4 \lambda-4 \lambda^{2}\right) .
$$
Let the line $D E$ intersect the $x$-axis at point $G$. Setting $y=0$, substituting into equation (1) gives
$$
\begin{array}{l}
x=-2 a^{2}(2 \lambda-1)^{2} . \\
\text { Then }|A G|=2 a^{2}-2 a^{2}(2 \lambda-1)^{2} \\
=2 a^{2}\left(4 \lambda-4 \lambda^{2}\right) . \\
\text { Hence } S_{2}=\frac{1}{2}|A G|\left|y_{D}-y_{E}\right| \\
=a^{2}\left(4 \lambda-4 \lambda^{2}\right)|2 \lambda a-(2 \lambda-2) a| \\
=2 a^{3}\left(4 \lambda-4 \lambda^{2}\right) .
\end{array}
$$
Therefore, $\frac{S_{1}}{S_{2}}=2$.
|
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
$$
\begin{array}{l}
x>0, y>0, u=x-y>0 . \\
\text { Then }(y+u)^{2}-2 y^{2}-8=0 \\
\Rightarrow y^{2}-2 u y-u^{2}+8=0 \\
\Rightarrow \Delta=(2 u)^{2}-4\left(-u^{2}+8\right) \geqslant 0 \\
\Rightarrow u \geqslant 2 .
\end{array}
$$
When $x=4, y=2$, the equality holds.
Therefore, the minimum value of $|x-y|$ is 2.
|
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,
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2} \geqslant \frac{x_{2}^{2}}{2}, \\
x_{3}^{2}+\left(x_{2}+x_{3}\right)^{2} \geqslant \frac{x_{2}^{2}}{2} .
\end{array}
$$
Therefore, $x_{2}^{2} \leqslant 4$, which means $\left|x_{2}\right| \leqslant 2$.
When $x_{1}=x_{3}=-1, x_{2}=2$, we have $\left|x_{2}\right|_{\text {max }}=2$.
|
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
$$
4^{2}=b^{2}+(8-b)^{2}-2 b(8-b) \frac{3}{5}
$$
$\Rightarrow b=3$ or 5, corresponding to $c=5$ or 3.
Therefore, $S_{\triangle A B C}=\frac{1}{2} b c \sin A$
$$
=\frac{1}{2} \times 3 \times 5 \times \frac{4}{5}=6 \text {. }
$$
|
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,
\[
a^{2}+(b-4)^{2} \geqslant x^{2}+1 \geqslant 2 \text {, }
\]
When $x=1, a=-1, b=3$, $a^{2}+(b-4)^{2}$ achieves its minimum value of 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 $x=8$.
Therefore, the maximum value of $f(x)$ is 11.
|
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 $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ all lie on the graph of the function $y=f(x)$, and satisfy
$$
x_{1}=\frac{\pi}{6}, x_{n+1}-x_{n}=\frac{\pi}{2}\left(n \in \mathbf{Z}_{+}\right) .
$$
Find the value of $y_{1}+y_{2}+\cdots+y_{2018}$.
|
(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 $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\frac{\pi}{6}=\frac{\pi}{2}$.
Then $t_{n+1}-t_{n}=\pi \Rightarrow t_{n}=\left(n-\frac{1}{2}\right) \pi$.
Thus, $y_{n}=2 \sin \frac{2 n-1}{2} \pi$
$$
=\left\{\begin{array}{ll}
2, & n=2 k-1 ; \\
-2, & n=2 k
\end{array}(k \in \mathbf{Z}) .\right.
$$
Therefore, $y_{1}+y_{2}+\cdots+y_{2018}=0$.
|
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, $f(x)=0$ has and only has one root $x=9$.
From $f\left(n^{2}-8 n-2018\right) \geqslant 0$
$$
\Rightarrow 0<n^{2}-8 n-2018 \leqslant 9
$$
$\Rightarrow n$ has no integer solutions.
Thus, $M=\varnothing$, and the number of subsets of $M$ is 1.
|
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 $y_{1}$ always exist.
So $y_{1}-y_{2}=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}$
$$
=\sqrt{b^{2}-\left(b^{2}-25\right)}=5 \text {. }
$$
|
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
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}} \\
= & \mathrm{C}_{4}^{2}|\overrightarrow{P Q}|^{2}+\overrightarrow{P O} \cdot 3 \sum_{1 \leqslant i \leqslant 4} \overrightarrow{O A_{i}}+\sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{O A_{i}} \cdot \overrightarrow{O A_{j}} \\
= & 6|\overrightarrow{P Q}|^{2}-1 \geqslant-1 .
\end{aligned}
$$
When point $P$ is at the center of the rhombus, the equality holds.
|
-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 $P(x)=\prod_{j=1}^{5}\left(x-r_{j}\right)$.
$$
\begin{array}{l}
\text { Then } \prod_{j=1}^{5} Q\left(r_{j}\right)=\left(\prod_{j=1}^{5}\left(r_{j}+\mathrm{i}\right)\right)\left(\prod_{j=1}^{5}\left(r_{j}-\mathrm{i}\right)\right) \\
=P(\mathrm{i}) P(-\mathrm{i}) \\
=\left(\mathrm{i}^{5}-\mathrm{i}^{2}+1\right)\left((-\mathrm{i})^{5}-(-\mathrm{i})^{2}+1\right) \\
=(\mathrm{i}+2)(-\mathrm{i}+2)=5 .
\end{array}
$$
|
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 tetrahedron (which is the inradius of the larger tetrahedron) be $r$.
It is easy to see that, $r=\frac{1}{3} R$.
Therefore, the maximum edge length of the smaller tetrahedron is $\frac{1}{3} \times 6=2$.
|
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
$$
\begin{array}{l}
\sum_{i=0}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right) . \\
\text { Therefore, } \sum_{i=1}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)-4^{2 n} \\
\equiv(-1)^{2 n-1}-1^{2 n} \equiv 1(\bmod 3) .
\end{array}
$$
|
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 different if and only if the colors at corresponding positions are different in at least one place. If any two 3-order color sequences are different, then the maximum value of $n$ is . $\qquad$
|
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 a counterclockwise direction as “red, red, red, blue, blue, blue, red, blue” meets the conditions.
|
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
$$
R(x)=(x+1)^{n}-1-P(x)\left(x^{2}+1\right) \text {. }
$$
Then $R(x)=3 Q(x)$, where $P(x)$ is the quotient and $R(x)$ is the remainder with coefficients all being multiples of 3.
Now consider $R(\mathrm{i})$.
Since the coefficients of the odd and even powers of $R(x)$ are all divisible by 3, the real and imaginary parts of $R(\mathrm{i})$ are also divisible by 3.
By calculation, we get
$$
\begin{array}{l}
(1+i)^{2}=2 i,(1+i)^{4}=-4, \\
(1+i)^{6}=-8 i,(1+i)^{8}=16 .
\end{array}
$$
Thus, $(1+\mathrm{i})^{8}-1=15$ has a real part and an imaginary part both divisible by 3.
Since the even powers of $1+\mathrm{i}$ are either purely real or purely imaginary and their coefficients are not divisible by 3, multiplying by $1+\mathrm{i}$ still results in an imaginary part that is not divisible by 3.
Testing $n=8$, we have
$$
\begin{aligned}
& (x+1)^{8}-1 \\
= & x^{8}+8 x^{7}+28 x^{6}+56 x^{5}+70 x^{4}+ \\
& 56 x^{3}+28 x^{2}+8 x \\
\equiv & x^{8}-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x \\
\equiv & \left(x^{2}+1\right)\left(x^{6}-x^{5}+x^{2}-x\right)(\bmod 3) .
\end{aligned}
$$
Therefore, the smallest positive integer $n$ that satisfies the condition is $n=8$.
|
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 the two inequalities, we get
$$
\begin{array}{l}
\frac{27}{4}=x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y} \geqslant \frac{51}{4}+\left(\frac{3}{4 y}-\frac{15}{x}\right) \\
\Rightarrow P=\frac{15}{x}-\frac{3}{4 y} \geqslant 6 .
\end{array}
$$
Equality holds if and only if $x=2, y=\frac{1}{2}$.
Therefore, the minimum value of $P$ is 6.
|
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 remainder 2 corresponds to a term whose units digit is 4. Therefore,
$$
\begin{array}{l}
b_{2018}=a_{3364}=3364^{3}-3364 \\
\equiv 4^{3}-4 \equiv 4(\bmod 7) .
\end{array}
$$
|
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 sequence, then $q-p=$
|
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
$$
\begin{array}{l}
b_{n}=-2 b_{n-1}-3 \\
\Rightarrow b_{n}+1=-2\left(b_{n-1}+1\right) \\
\quad=(-2)^{n-1}\left(b_{1}+1\right)=(-2)^{n} \\
\Rightarrow b_{n}=(-2)^{n}-1 \\
\Rightarrow a_{n}=2^{n}-(-1)^{n} .
\end{array}
$$
Assume there exists a positive integer $c$, such that $a_{1} 、 a_{n} 、 a_{n+c}$ form an arithmetic sequence. Then
$$
\begin{array}{l}
2\left(2^{n}-(-1)^{n}\right)=2^{n+c}-(-1)^{n+c}+3 \\
\Rightarrow 2^{n+1}\left(2^{c-1}-1\right)+(-1)^{n}\left(2-(-1)^{c}\right)+3=0 .
\end{array}
$$
Clearly, the above equation holds if and only if $c=1$, and $n$ is a positive odd number.
Thus, $p=n, q=n+1 \Rightarrow q-p=1$.
|
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 results, we find the maximum value is 3.
|
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, $3 a+b=2\left(\frac{5}{2} a+b\right)-(2 a+b) \leqslant 3$.
When $a=4, b=-9$, the equality holds.
|
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 that a traveler can always choose one of the airlines and complete a journey starting from a city, passing through $n-1$ cities (excluding the departure and arrival cities), and finally arriving at the $n$-th city.
|
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 integer $x_{i}$ such that there is a sequence of non-negative integers $k_{0}<k_{1}<\cdots<k_{x_{i}-1}<i$, satisfying that starting from city $c_{k_{0}}$, passing through cities $c_{k_{1}}, c_{k_{2}}$, $\cdots, c_{k_{x-1}}$ in sequence to reach $c_{i}$, and all routes belong to airline $X$.
|
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
$$
\begin{array}{l}
\frac{1}{2} a c \sin 120^{\circ} \\
=\frac{1}{2} a \times 1 \times \sin 60^{\circ}+\frac{1}{2} c \times 1 \times \sin 60^{\circ} \\
\Rightarrow \frac{1}{a}+\frac{1}{c}=1 .
\end{array}
$$
Therefore, $4 a+c=(4 a+c)\left(\frac{1}{a}+\frac{1}{c}\right)$
$$
=5+\frac{c}{a}+\frac{4 a}{c} \geqslant 5+2 \sqrt{\frac{c}{a} \cdot \frac{4 a}{c}}=9 \text {. }
$$
Equality holds if and only if $c=2 a=3$. Therefore, the minimum value of $4 a+c$ is 9.
|
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
$$
\begin{array}{l}
nc > \frac{a_{1}+a_{2}}{a_{1}} + \frac{a_{1}+a_{2}+a_{3}}{a_{2}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}} \\
= n + \frac{a_{2}}{a_{1}} + \left(\frac{a_{1}}{a_{2}} + \frac{a_{3}}{a_{2}}\right) + \cdots + \\
\left(\frac{a_{1}}{a_{n}} + \frac{a_{2}}{a_{n}} + \cdots + \frac{a_{n+1}}{a_{n}}\right) \\
= n + \left(\frac{a_{2}}{a_{1}} + \frac{a_{1}}{a_{2}}\right) + \left(\frac{a_{3}}{a_{2}} + \frac{a_{2}}{a_{3}}\right) + \cdots + \\
\left(\frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}}\right) + \frac{a_{n+1}}{a_{n}} + \frac{a_{1}}{a_{3}} + \\
\frac{a_{1}+a_{2}}{a_{4}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n-2}}{a_{n}} \\
\geqslant n + 2(n-1) + \frac{a_{n+1}}{a_{n}} + \frac{a_{1}}{a_{3}} + \\
\frac{a_{1}+a_{2}}{a_{4}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n-2}}{a_{n}}.
\end{array}
$$
If there exists \( n \geqslant 1 \) such that \( \frac{a_{n+1}}{a_{n}} \geqslant 2 \), then
$$
\begin{array}{l}
nc > n + 2(n-1) + 2 = 3n \\
\Rightarrow c > 3 \Rightarrow c \geqslant 4.
\end{array}
$$
If for any \( n \geqslant 1 \), we have \( \frac{a_{n+1}}{a_{n}} < 2 \), then
$$
\begin{array}{l}
nc > n + 2(n-1) + \frac{1}{2^2} + \frac{1}{2^2} + \cdots + \frac{1}{2^2}.
\end{array}
$$
Taking \( n-2 > 2^2 \), we get
$$
nc > n + 2(n-1) + 2 = 3n \\
\Rightarrow c \geqslant 4.
$$
Thus, the smallest integer \( c \) is 4.
Below is a construction: take \( a_{n} = 2^{n-1} \),
$$
\begin{array}{l}
a_{1} + a_{2} + \cdots + a_{n+1} = 1 + 2 + \cdots + 2^n \\
= 2^{n+1} - 1 \\
4a_{n} = 4 \times 2^{n-1} = 2^{n+1},
\end{array}
$$
Clearly, \( a_{1} + a_{2} + \cdots + a_{n+1} < 4a_{n} \) holds.
|
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, $\tan \alpha \cdot \tan \beta=\frac{\tan ^{2} \beta-\sqrt{3} \tan \beta}{1+\sqrt{3} \tan \beta}$, $\tan \beta \cdot \tan \gamma=\frac{\tan ^{2} \beta+\sqrt{3} \tan \beta}{1-\sqrt{3} \tan \beta}$, $\tan \gamma \cdot \tan \alpha=\frac{\tan ^{2} \beta-3}{1-3 \tan ^{2} \beta}$.
Thus, $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $=\frac{9 \tan ^{2} \beta-3}{1-3 \tan ^{2} \beta}=-3$.
|
-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
$$
\begin{array}{l}
y=k x, y=x^{3}-x, \sqrt{x^{2}+y^{2}}=2 \\
\Rightarrow x=\sqrt{k+1}, y=k \sqrt{k+1} . \\
\text { Substituting into } \sqrt{x^{2}+y^{2}}=2 \text { gives } \\
(k+1)+k^{2}(k+1)=4 \\
\Rightarrow(k-1)\left(k^{2}+2 k+3\right)=0 \\
\Rightarrow k=1 .
\end{array}
$$
|
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 quadratic formula for $z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}=0$ yields $z=3+2 \mathrm{i}$ or $2+3 \mathrm{i}$.
Thus, the four roots of $P(z)=0$ are $1, \mathrm{i}, 3+2 \mathrm{i}, 2+3 \mathrm{i}$, which form a rectangle with a length of $2 \sqrt{2}$ and a width of $\sqrt{2}$, giving an area of $\frac{1}{2} \sqrt{2} \times 2 \sqrt{2}=2$.
|
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
$$
\begin{array}{l}
a_{n, m}=\sum_{k=0}^{m-1} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} \\
=\sum_{k \geqslant 0} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} .
\end{array}
$$
Verification: When $k=0$, $\mathrm{C}_{m}^{0}=\mathrm{C}_{m-1}^{0}$ holds, and with the identity $\mathrm{C}_{m}^{k}=\mathrm{C}_{m-1}^{k}+\mathrm{C}_{m-1}^{k-1}(k \geqslant 1)$, we can prove by induction.
Let $m=128$, by Lucas' Theorem, we have
$$
\begin{array}{l}
\mathrm{C}_{m-1}^{k}=\mathrm{C}_{127}^{k} \equiv 1(\bmod 2)(1 \leqslant k \leqslant 127), \\
\mathrm{C}_{127}^{k} \equiv 0(\bmod 5)(3 \leqslant k \leqslant 124)
\end{array}
$$
Then $a_{128,1}=\sum_{k=0}^{127} \mathrm{C}_{127}^{k}(k+1)^{k+1}$
$$
\begin{array}{l}
\equiv \sum_{k=0}^{127}(k+1)^{k+1} \equiv 0(\bmod 2), \\
a_{128,1}=\sum_{k \in\{0,1,2\}}^{127} \\
\equiv 4(\bmod 5) .
\end{array}
$$
Therefore, $a_{128,1} \equiv 4(\bmod 10)$.
|
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 the number of switches pressed (Note: The switches return to their original state after each operation). The questions are:
(1) What is the maximum number of different operations Charlie can perform so that Laura can accurately determine the correspondence between the lamps and the switches?
(2) If Laura operates the switches herself, what is the minimum number of operations she needs to find the correspondence between the lamps and the switches?
|
(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 performs $2^{2009}+1$ different operations, consider a particular switch $A$. Each operation either presses $A$ or leaves it untouched, dividing the operations into two categories, one of which must contain at least $2^{2008}+1$ operations (let's assume it's the category where $A$ is pressed). Consider the set of lights turned on by these operations, where one light is always on, indicating it is controlled by $A$.
(2) Since $2010<2^{11}$, we can use 11-bit binary numbers to number all the switches (from 00000000001 to 11111011010). Use 11 operations to number the lights (also using 11-bit binary numbers): the $k$-th operation ($k=1,2, \cdots, 11$) involves pressing the switches whose $k$-th bit is 1. Write 1 in the $k$-th bit of the number for all lights that are on, and 0 for those that are off. Thus, switches with the same number control the same light, establishing a one-to-one correspondence. Therefore, the switch-light correspondence can be determined in at most 11 operations.
|
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 $4. B(1,2,3)$ has no elements.$
$$\begin{aligned}
3 & =B^{(0)}=B(0)=A^{(0)}-A^{(1)}+A^{(2)}-A^{(3)} \\
& =10-(3+3+3)+(0+1+1)-0=3
\end{aligned}$$
This is consistent with the conclusion of Theorem 7. Similarly, the conclusion of Theorem 9 can be verified (left to the reader).
|
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 whether Theorem 7 holds.
|
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 elements. \square$
$B^{(0)}=B(0)$ is $5,5,11 ; B(1)$ has no elements (because property $P_{1}$ is the same as property $P_{4}$); $B(2)$ is $1,7 ; B(3)$ is $2 ; B(4)$ has no elements. $B(1,2)$ has no elements; $B(1,3)$ has no elements; $B(1,4)$ is $3,3 ; B(2,3)$ is $4 ; B(2,4)$ has no elements; $B(3,4)$ has no elements. $B(1,2,3)$ has no elements; $B(1,2,4)$ has no elements; $B(1,3,4)$ is $6 ; B(2,3,4)$ has no elements. $B(1,2,3,4)$ has no elements.
$$\begin{aligned}
3= & B^{(0)}=B(0)=A^{(0)}-A^{(1)}+A^{(2)}-A^{(3)}+A^{(4)} \\
= & 10-(3+3+3+3)+(0+1+3+1+0+1) \\
& -(0+0+1+0)+0=3
\end{aligned}$$
This is consistent with the conclusion of Theorem 7. Similarly, the conclusion of Lemma 8 can be verified (left to the reader).
|
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 particular solution (please verify this yourself). By formula (15), the number of solutions is
$$\begin{array}{l}
-[-11 \cdot 3189 / 37]-[30 \cdot 3189 / 101]-1 \\
\quad=949-947-1=1
\end{array}$$
That is, the equation has exactly one positive solution. Please find this solution yourself.
|
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
$$3^{404} \equiv 1(\bmod 10)$$
Therefore, $3^{406} \equiv 3^{404} \cdot 3^{2} \equiv 9(\bmod 10)$. So the unit digit is 9.
|
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
$$\begin{aligned}
\left(\frac{137}{227}\right) & =\left(\frac{-90}{227}\right)=\left(\frac{-1}{227}\right)\left(\frac{2 \cdot 3^{2} \cdot 5}{227}\right) \\
& =(-1)\left(\frac{2}{227}\right)\left(\frac{3^{2}}{227}\right)\left(\frac{5}{227}\right) \\
& =(-1)\left(\frac{2}{227}\right)\left(\frac{5}{227}\right) .
\end{aligned}$$
By Theorem 3, we get
$$\left(\frac{2}{227}\right)=-1$$
By Theorem 5, Theorem 1, and Theorem 3, we get
$$\left(\frac{5}{227}\right)=\left(\frac{227}{5}\right)=\left(\frac{2}{5}\right)=-1$$
From the above three equations, we get
$$\left(\frac{137}{227}\right)=-1$$
This indicates that the congruence equation $x^{2} \equiv 137(\bmod 227)$ has no solution.
|
-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
$$x^{7}-x=\left(x^{3}-x^{2}+3 x-3\right)\left(x^{3}+x^{2}-2 x-2\right) x+7 x\left(x^{2}-1\right) .$$
Therefore, the number of solutions to the original congruence equation is 3.
|
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:
$$\begin{array}{c}
2^{2} \equiv 4(\bmod 23) \\
2^{11} \equiv\left(2^{4}\right)^{2} \cdot 2^{3} \equiv(-7)^{2} \cdot 8 \equiv 3 \cdot 8 \equiv 1(\bmod 23)
\end{array}$$
So $\delta_{23}(2)=11,2$ is not a primitive root modulo 23. Next, find $\delta_{23}(3)$:
$$\begin{array}{c}
3^{2} \equiv 9(\bmod 23), \quad 3^{3} \equiv 4(\bmod 23) \\
3^{11} \equiv\left(3^{3}\right)^{3} \cdot 3^{2} \equiv 4^{3} \cdot 9 \equiv(-5) \cdot 9 \equiv 1(\bmod 23)
\end{array}$$
So, $\delta_{23}(3)=11,3$ is not a primitive root modulo 23. Next, find $\delta_{23}(4)$.
$$4^{2} \equiv-7(\bmod 23), \quad 4^{11} \equiv\left(4^{4}\right)^{2} \cdot 4^{3} \equiv 3^{2} \cdot(-5) \equiv 1(\bmod 23)$$
So $\delta_{23}(4)=11,4$ is not a primitive root modulo 23. Next, find $\delta_{23}(5)$.
$$\begin{aligned}
5^{2} & \equiv 2(\bmod 23) \\
5^{11} & \equiv\left(5^{4}\right)^{2} \cdot 5^{3} \equiv 4^{2} \cdot 10 \\
& \equiv 4 \cdot(-6) \equiv-1(\bmod 23) \\
5^{22} & \equiv 1(\bmod 23)
\end{aligned}$$
So, $\delta_{23}(5)=22,5$ is a primitive root modulo 23, and it is the smallest positive primitive root.
To further find the primitive roots modulo $p^{\alpha}, 2 p^{\alpha}$, we need to verify whether equation (8) holds when $g=5, p=23$. This is essentially finding the residue of $g^{p-1}$ modulo $p^{2}$. Here $23^{2}=529$.
$$\begin{aligned}
5^{2} & \equiv 25\left(\bmod 23^{2}\right) \\
5^{8} & \equiv(23+2)^{4} \equiv 4 \cdot 23 \cdot 2^{3}+2^{4} \equiv 10 \cdot 23-7\left(\bmod 23^{2}\right) \\
5^{10} & \equiv(10 \cdot 23-7)(23+2) \equiv 13 \cdot 23-14 \\
& \equiv 12 \cdot 23+9\left(\bmod 23^{2}\right) \\
5^{20} & \equiv(12 \cdot 23+9)^{2} \equiv 216 \cdot 23+81 \\
& \equiv 13 \cdot 23-11\left(\bmod 23^{2}\right) \\
5^{22} & \equiv(13 \cdot 23-11)(23+2) \equiv 15 \cdot 23-22 \\
& \equiv 1+14 \cdot 23\left(\bmod 23^{2}\right)
\end{aligned}$$
Since $23 \nmid 14$ and 5 is odd, it proves that 5 is a primitive root for all moduli $23^{a}, 2 \cdot 23^{a}$.
|
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 $d \mid 20, d<20$ when $2^{d} \not \equiv 1(\bmod 41)$, we have $\delta_{41}(2)=20$. Therefore, 2 is not a primitive root.
$$3^{2} \equiv 9(\bmod 41), \quad 3^{4} \equiv-1(\bmod 81), \quad 3^{8} \equiv 1(\bmod 41)$$
Similarly, $\delta_{41}(3)=8$. So 3 is also not a primitive root.
Notice that $\left[\delta_{41}(2), \delta_{41}(3)\right]=[20,8]=40$. Therefore, we do not need to calculate the orders of $4,5, \cdots$ one by one, but can use property XI in $\S 1$ to find the primitive root. Notice that
$$\delta_{41}(2)=4 \cdot 5, \quad \delta_{41}(3)=1 \cdot 8$$
By property XI, $c=2^{4} \cdot 3=48$ is a primitive root. Therefore, 7 is also a primitive root modulo 41. This method of finding primitive roots is essentially the proof 1 of Theorem 2, so proof 1 can be a method to find primitive roots in certain cases.
To find the primitive roots of $41^{\alpha}, 2 \cdot 41^{\alpha}$, we need to calculate $7^{40}$ modulo $41^{2}=1681$.
$$\begin{array}{l}
7^{2} \equiv 41+8\left(\bmod 41^{2}\right) \\
7^{4} \equiv(41+8)^{2} \equiv 18(41-1)\left(\bmod 41^{2}\right) \\
7^{5} \equiv 3 \cdot(41+1)(41-1) \equiv-3\left(\bmod 41^{2}\right) \\
7^{10} \equiv 9\left(\bmod 41^{2}\right) \\
7^{20} \equiv 81(\bmod 41)^{2} \\
7^{40} \equiv(2 \cdot 41-1)^{2} \equiv 1+(-4) \cdot 41\left(\bmod 41^{2}\right)
\end{array}$$
Since $41 \nmid-4$ and 7 is odd, it follows that 7 is a primitive root for all moduli $41^{\alpha}, 2 \cdot 41^{\alpha}$.
|
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 to the above equations. First, solve
$$u+4 v=1$$
Since $4=3+1$, we get $1=-3+4$. Therefore, $u=-3, v=1$ is a set of integer solutions to $u+4 v=1$. Thus,
$$y=8 \times(-3)=-24, z=8 \times 1=8$$
is a set of integer solutions to $y+4 z=8$. The complete set of integer solutions is
$$y=-24-4 t, \quad z=8+t, \quad t=0, \pm 1, \pm 2, \cdots$$
Therefore,
$$x=10-y-z=26+3 t$$
According to the problem, we need $x \geqslant 0, y \geqslant 0, z \geqslant 0$. From $x=26+3 t \geqslant 0$, we get $t \geqslant-\frac{26}{3}$; from $y=-24-4 t \geqslant 0$, we get $t \leqslant-6$; and from $z=8+t \geqslant 0$, we get $t \geqslant-8$. Therefore, $-8 \leqslant t \leqslant-6$ (taking $t=-8, -7, -6$) corresponds to the following three sets of solutions:
$$\left\{\begin{array} { l }
{ x = 2 } \\
{ y = 8 } \\
{ z = 0 , }
\end{array} \quad \left\{\begin{array}{l}
x=5 \\
y=4 \\
z=1
\end{array},\left\{\begin{array}{l}
x=8 \\
y=0 \\
z=2
\end{array}\right.\right.\right.$$
Therefore, there are three different ways to take the coins, which are the three sets of solutions above.
|
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, $(48,84,120)=2^{2} \times 3=12$.
(ii) Solution: Decomposing each number into prime factors, we get
$$\begin{array}{l}
360=2^{3} \times 3^{2} \times 5 \\
810=2 \times 3^{4} \times 5 \\
1260=2^{2} \times 3^{2} \times 5 \times 7 \\
3150=2 \times 3^{2} \times 5^{2} \times 7
\end{array}$$
Therefore, $(360,810,1260,3150)=2 \times 3^{2} \times 5=90$.
|
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 be 3 jin.
|
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{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{r}
117317 \\
90950
\end{tabular}
\end{tabular} & 5 \\
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
90950 \\
79101
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
26367 \\
23698
\end{tabular}
\end{tabular} & 3 \\
\hline 2 & \begin{tabular}{l}
\begin{tabular}{l}
11849 \\
10676
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
2669 \\
2346
\end{tabular}
\end{tabular} & 4 \\
\hline 2 & \begin{tabular}{l}
\begin{tabular}{r}
1173 \\
969
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
323 \\
204
\end{tabular}
\end{tabular} & 3 \\
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
204 \\
119
\end{tabular}
\end{tabular} & \begin{tabular}{l}
119 \\
85
\end{tabular} & 1 \\
\hline 1 & \begin{tabular}{l}
\begin{tabular}{l}
85 \\
68
\end{tabular}
\end{tabular} & \begin{tabular}{l}
\begin{tabular}{l}
34 \\
34
\end{tabular}
\end{tabular} & 2 \\
\hline 2 & 17 & 0 & \\
\hline
\end{tabular}
Therefore, $(435785667,131901878)=17$.
(ii) Solution: Since $15959989=3989 \times 4001$,
the two numbers 3989 and 4001 on the right are both prime, and they are clearly not factors of 7733, so $(15959989,7738)=1$.
|
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:
$$2 \times 2 \times 3=12$$
To find the greatest common divisor of several positive integers, first decompose these positive integers into prime factors, then take the common prime factors (the same prime factors are taken according to the number of times they are common) and multiply them.
|
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 above three numbers, we can see that the prime factors 2, 2, 3 (or $2^{2}, 3$) are common to all three numbers, and their product is the greatest common divisor of these three numbers:
$$2 \times 2 \times 3=12 \text { or } 2^{2} \times 3=12 \text {. }$$
For the sake of clarity, let's first discuss the greatest common divisor and the least common multiple, and then discuss the Fundamental Theorem of Arithmetic. To find the greatest common divisor of several numbers, first decompose these numbers into prime factors and write them in exponential form; then, among the common prime factors, take the powers with the smallest exponents and multiply them.
|
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,
$$\left(\frac{-286}{p}\right)=-\left(\frac{143}{p}\right)$$
Since $143=4 \times 35+3, p=3(\bmod 4)$, by Theorem 3 we have
$$\left(\frac{143}{p}\right)=-\left(\frac{p}{143}\right)$$
From $p=143 \times 29880+103$, we get
$$\left(\frac{p}{143}\right)=\left(\frac{103}{143}\right)$$
By Theorem 3 and $103=3(\bmod 4), 143 \equiv 3(\bmod 4)$, we have
$$\begin{array}{l}
\left(\frac{103}{143}\right)=-\left(\frac{143}{103}\right)=-\left(\frac{40}{103}\right)=-\left(\frac{2^{2} \times 2 \times 5}{103}\right) \\
\quad=-\left(\frac{2 \times 5}{103}\right)=-\left(\frac{2}{103}\right)\left(\frac{5}{103}\right)=-\left(\frac{5}{103}\right) \\
=-\left(\frac{103}{5}\right)=-\left(\frac{3}{5}\right)=1
\end{array}$$
Therefore,
$$\left(\frac{-286}{p}\right)=1$$
This means that equation (51) has a solution.
|
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 the modular inverse of $n$ modulo $p$. From the properties of the Legendre symbol, it is easy to see that
$$\begin{aligned}
\left(\frac{n(n+1)}{p}\right) & =\left(\frac{n\left(n+n r_{n}\right)}{p}\right)=\left(\frac{n^{2}\left(1+r_{n}\right)}{p}\right) \\
& =\left(\frac{n}{p}\right)^{2}\left(\frac{1+r_{n}}{p}\right)=\left(\frac{1+r_{n}}{p}\right)
\end{aligned}$$
We will prove that for $n \neq m(\bmod p), p \nmid n m$, it must also be true that
$$r_{n} \not\equiv r_{m}(\bmod p)$$
That is, different numbers in the reduced residue system modulo $p$ must correspond to different inverses. We use proof by contradiction. If $r_{n} \equiv r_{m}(\bmod p)$, multiplying both sides by $n m$ gives
$$n\left(m r_{m}\right) \equiv m\left(n r_{n}\right)(\bmod p)$$
Then, by the definition of the inverse, we get
$$n \equiv m(\bmod p)$$
This leads to a contradiction. This proves that when $n$ runs through $1,2, \ldots, p-1$, the corresponding inverse $r_{n}$ also runs through $1,2, \ldots, p-1(\bmod p)$, just in a different order. Also, note that from $p-1 \equiv-1(\bmod p)$, we immediately get
$$(p-1)^{2} \equiv(-1)^{2} \equiv 1(\bmod p)$$
So $r_{p-1}=p-1$. Therefore, when $n$ takes $1,2, \ldots, p-2$, the inverse $r_{n}$ of $n$ also takes $1,2, \ldots, p-2(\bmod p)$, just in a different order. Thus, we get
$$\begin{array}{l}
\left(\frac{1.2}{p}\right)+\left(\frac{2.3}{p}\right)+\ldots+\left(\frac{(p-2)(p-1)}{p}\right) \\
= \sum_{n=1}^{p-2}\left(\frac{1+r_{n}}{p}\right)=\sum_{r=1}^{p-2}\left(\frac{1+r}{p}\right) \\
=\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)-\left(\frac{1}{p}\right)
\end{array}$$
Since in a reduced residue system modulo $p$, there are exactly $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ quadratic non-residues, we have $\sum_{r=1}^{p-1}\left(\frac{r}{p}\right)=0$. Therefore, the required sum is $-\left(\frac{1}{p}\right)=-1$.
|
-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
$$\begin{aligned}
& \mathrm{C}_{a p}^{b p}-\mathrm{C}_{a}^{b}=\frac{(a p)(a p-1) \cdot \cdots \cdot((a-b) p+1)}{(b p)!}-\frac{a!}{b!(a-b)!} \\
= & \frac{a(a-1) \cdot \cdots \cdot(a-b+1) \prod_{k=a-b}^{a-1}(k p+1)(k p+2) \cdots \cdots \cdot(k p+(p-1))}{b!\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdots \cdots(k p+(p-1))} \\
& -\frac{a!}{b!(a-b)!} \\
= & \frac{a!}{b!(a-b)!} \cdot \frac{1}{\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))}\left\{\prod_{k=a-b}^{a-1}(k p+1)\right. \\
& \left.\cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))-\prod_{k=0}^{b-1}(k p+1)(k p+2) \cdot \cdots \cdot(k p+(p-1))\right\},
\end{aligned}$$
Thus, we only need to prove that $p^{3} \mid A$, where
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1)) \\
& -\prod_{k=0}^{b-1}(k p+1) \cdot(k p+2) \cdot \cdots \cdot(k p+(p-1))
\end{aligned}$$
For this, we set
$$\begin{aligned}
f(x) & =(x+1)(x+2) \cdot \cdots \cdot(x+(p-1)) \\
& =x^{p-1}+\alpha_{p-2} x^{p-2}+\cdots+\alpha_{1} x+(p-1)!
\end{aligned}$$
Then, by the conclusion of Example 1 in Section 2.4, we know that $p^{2} \mid \alpha_{1}$. Therefore,
$$\begin{aligned}
A= & \prod_{k=a-b}^{a-1} f(k p)-\prod_{k=0}^{b-1} f(k p) \\
\equiv & ((p-1)!)^{b-1} \sum_{k=a-b}^{a-1} \alpha_{1} k p+((p-1)!)^{b} \\
& -((p-1)!)^{b-1} \sum_{k=0}^{b-1} \alpha_{1} k p-((p-1)!)^{b} \\
\equiv & 0\left(\bmod p^{3}\right) .
\end{aligned}$$
Therefore, the maximum integer $c=3$.
|
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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