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5.64k
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7. Given integers $p$ and $q$ satisfy $p+q=2010$, and the quadratic equation $67 x^{2}+p x+q=0$ has two positive integer roots. Then $p=$ $\qquad$ .
|
7. -2278 .
Let the two positive integer roots of the equation be $x_{1}, x_{2}\left(x_{1} \leqslant x_{2}\right)$.
Then $x_{1}+x_{2}=-\frac{p}{67}, x_{1} x_{2}=\frac{q}{67}$.
Thus, $x_{1} x_{2}-x_{1}-x_{2}=\frac{p+q}{67}=\frac{2010}{67}=30$
$$
\begin{array}{l}
\Rightarrow\left(x_{1}-1\right)\left(x_{2}-1\right)=31 \\
\Rightarrow\left\{\begin{array}{l}
x_{1}-1=1, \\
x_{2}-1=31 .
\end{array}\right.
\end{array}
$$
Therefore, $x_{1}+x_{2}=34, p=-34 \times 67=-2278$.
|
-2278
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Place the nine digits $1,2, \cdots, 9$ into the nine small squares in Figure 4, so that the seven three-digit numbers $\overline{a b c} 、 \overline{d e f} 、 \overline{g h i} 、 \overline{a d g} 、 \overline{b e h} 、 \overline{c f i}$ and $\overline{a e i}$ are all divisible by 11. Find the maximum value of the three-digit number $\overline{c e g}$.
|
12. According to the problem, for modulo 11 we have
$$
\begin{array}{l}
a+c \equiv b, d+f \equiv e, g+i \equiv h, a+g \equiv d, \\
b+h \equiv e, c+i \equiv f, a+i \equiv e . \\
\text { Then }(a+c)+(d+f)+(g+i)+(b+h)+e \\
\equiv b+h+3 e \equiv 4 e(\bmod 11) .
\end{array}
$$
The left side of the above equation is
$$
1+2+\cdots+9=45 \equiv 1(\bmod 11) \text {. }
$$
So $4 e \equiv 1(\bmod 11), e=3$.
Thus, $d+f \equiv b+h \equiv a+i \equiv 3(\bmod 11)$.
Therefore, $\{d, f\} 、\{b, h\} 、\{a, i\}$ are permutations of $\{1,2\} 、\{6,8\}$ 、 $\{5,9\}$.
Hence, $\{c, g\}$ is $\{4,7\}$.
To maximize $\overline{c e g}$, we can take
$$
c=7, g=4 \text {. }
$$
Figure 8 is one valid arrangement that satisfies the problem.
Therefore, the maximum value of $\overline{c e g}$ is 734.
|
734
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A natural number is called a "good number" if it is exactly 2007 more than the sum of its digits. Then the sum of all good numbers is $\qquad$ .
|
5.20145.
Let $f(n)=n-S(n)$ ($S(n)$ is the sum of the digits of the natural number $n$). Then the function $f(n)$ is a non-strictly increasing function, and
$$
\begin{array}{l}
f(2009)<f(2010) \\
=f(2011)=\cdots=f(2019)=2007 \\
<f(2020) .
\end{array}
$$
Therefore, there are only 10 natural numbers that satisfy the condition, and their sum is
$$
2010+2011+\cdots+2019=20145 .
$$
|
20145
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the sum of 12 distinct positive integers is 2010. Then the maximum value of the greatest common divisor of these positive integers is . $\qquad$
|
$-1.15$.
Let the greatest common divisor be $d$, and the 12 numbers be $a_{1} d$, $a_{2} d, \cdots, a_{12} d$, where $\left(a_{1}, a_{2}, \cdots, a_{12}\right)=1$.
Let $S=\sum_{i=1}^{12} a_{i}$. Then $2010=S d$.
To maximize $d$, $S$ should be minimized.
Since $a_{1}, a_{2}, \cdots, a_{12}$ are distinct, then
$S \geqslant 1+2+\cdots+12=78$.
Also, $S \mid 2010$, and $2010=2 \times 3 \times 5 \times 67$, so the smallest positive divisor of 2010 greater than 77 is $2 \times 67=134$.
Therefore, $d \leqslant 15$, and $d=15$ can be achieved, by setting
$$
\left(a_{1}, a_{2}, \cdots, a_{11}, a_{12}\right)=(1,2, \cdots, 11,68)
$$
This completes the translation.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=1, a_{n}+a_{n+1}=-n^{2} \text {. }
$$
then $a_{15}=$ $\qquad$
|
7. -104 .
Rewrite $a_{n}+a_{n+1}=-n^{2}$ as
$$
\left(a_{n}+\frac{n^{2}}{2}-\frac{n}{2}\right)+\left[a_{n+1}+\frac{(n+1)^{2}}{2}-\frac{n+1}{2}\right]=0 \text {. }
$$
Let $b_{n}=a_{n}+\frac{n^{2}}{2}-\frac{n}{2}$. Then
$$
b_{1}=1 \text {, and } b_{n+1}=-b_{n} \text {. }
$$
Therefore, $b_{2 k-1}=1, b_{2 k}=-1(k=1,2, \cdots)$.
Thus, $b_{15}=1$, which means
$$
a_{15}=1+\frac{15}{2}-\frac{15^{2}}{2}=-104 .
$$
|
-104
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If a four-digit number $n$ contains at most two different digits among its four digits, then $n$ is called a "simple four-digit number" (such as 5555 and 3313). Then, the number of simple four-digit numbers is
|
8. 576.
If the four digits of a four-digit number are all the same, then there are nine such four-digit numbers.
If the four digits of a four-digit number have two different values, the first digit \( a \in \{1,2, \cdots, 9\} \) has 9 possible choices. After choosing \( a \), select \( b \in \{0,1, \cdots, 9\} \) and \( b \neq a \), which gives 9 ways to choose \( b \); for the remaining three positions, each can be filled with \( a \) or \( b \), but they cannot all be filled with \( a \), giving \( 2^3 - 1 = 7 \) ways to fill them. Therefore, the number of four-digit numbers with exactly two different digits is
$$
9 \times 9 \times 7 = 567 \text{ (numbers). }
$$
Thus, the total number of simple four-digit numbers is \( 9 + 567 = 576 \).
|
576
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. The capacity of a set refers to the sum of its elements. Then the total capacity of all non-empty sets $A$ that satisfy the condition “ $A \subseteq\{1,2, \cdots, 7\}$, and if $a \in A$ then $8-a \in A$ ” is
(Answer with a specific number).
|
12.224.
First, find the single-element and two-element sets that satisfy the conditions:
$$
A_{1}=\{4\}, A_{2}=\{1,7\}, A_{3}=\{2,6\}, A_{4}=\{3,5\} .
$$
Then, any combination of elements from these four sets also meets the requirements.
Therefore, the total sum of elements in all sets $A$ that satisfy the conditions is
$$
(4+8+8+8) \times 2^{3}=224 \text {. }
$$
|
224
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3: There is an $8 \times 8$ chessboard, and at the start, each of the 64 small squares contains a "castle" chess piece. If a castle chess piece can attack an odd number of other castle chess pieces still on the board, it is removed. Question: What is the maximum number of castle chess pieces that can be removed (a castle chess piece can attack other pieces in the same row or column without any obstruction)?
|
First, prove that the pieces on the four corners of the chessboard will not be taken away. Since a castle on a corner can attack at most two other castles, if this castle is taken away, it means that the castle can only attack one. Without loss of generality, we can assume that the castle in the upper left corner is the first to be taken away among these four, which means that either the first column or the first row has been completely cleared. In other words, the castle in the upper right corner or the lower left corner has been taken away, which is impossible.
When there are five castles left, the castles outside these four corners can only attack two or zero castles, so at least five castles will remain. Taking the castles in the order of the numbers in Figure 5 is one way to end up with five castles left. Therefore, the maximum number of castles that can be taken away is $64-5=59$.
|
59
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Take the subset $A_{i}=\left\{a_{i}, a_{i+1}, \cdots, a_{i+59}\right\}(i=1,2, \cdots$, 70 ) of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}$, where $a_{70+i}=a_{i}$. If there exist $k$ sets among $A_{1}, A_{2}, \cdots, A_{70}$ such that the intersection of any seven of them is non-empty, find the maximum value of $k$.
---
The problem is asking to find the maximum number of sets $k$ from the collection $\{A_1, A_2, \ldots, A_{70}\}$ such that the intersection of any seven of these sets is non-empty. Each set $A_i$ is defined as a subset of $S$ containing 60 consecutive elements, with the indices wrapping around modulo 70.
|
Given:
$$
A_{1} \cap A_{2} \cap \cdots \cap A_{60}=\left\{a_{60}\right\} \text {, }
$$
Furthermore, the intersection of any seven sets from $A_{1}, A_{2}, \cdots, A_{\infty}$ is non-empty, hence $k \geqslant 60$.
We will now prove that if $k>60$, it cannot be guaranteed that the intersection of any seven sets from the $k$ selected sets is non-empty. Therefore, the maximum value of $k$ is 60.
Divide the known sets $A_{1}, A_{2}, \cdots, A_{70}$ into 10 groups (constructing pigeonholes):
$$
B_{i}=\left\{A_{i}, A_{10+i}, \cdots, A_{60+i}\right\}(i=1,2, \cdots, 10) .
$$
Since the union of the complements $\overline{A_{j}}$ of any seven sets $A_{j}$ within each $B_{i}$ satisfies
$$
\begin{array}{l}
\overline{A_{i}} \cup \overline{A_{10+i}} \cup \cdots \cup \overline{A_{60+i}} \\
=\left\{a_{i+60}, a_{i+61}, \cdots, a_{i+69}\right\} \cup \\
\left\{a_{i+70}, a_{i+11}, \cdots, a_{i+79}\right\} \cup \cdots \cup \\
\left\{a_{i+120}, a_{i+121}, \cdots, a_{i+129}\right\} \\
=\left\{a_{1}, a_{2}, \cdots, a_{70}\right\}=S\left(a_{70+i}=a_{i}\right), \\
\end{array}
$$
it follows that the intersection of any seven sets within each $B_{i}$ is the empty set:
$$
A_{i} \cap A_{10+i} \cap \cdots \cap A_{60+i}=\varnothing \text {. }
$$
For $k>60$, we have $k=6 \times 10+r(1 \leqslant r \leqslant 9)$.
By the pigeonhole principle, any selection of $k$ sets from $A_{1}, A_{2}, \cdots, A_{70}$ must include seven sets from the same $B_{i}(1 \leqslant i \leqslant 10)$.
This contradicts the condition that the intersection of any seven sets is non-empty.
Therefore, the maximum value of $k$ is 60.
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
410 translators are invited to an international mathematics conference. Each translator is proficient in exactly two of the five languages: Greek, Slovenian, Vietnamese, Spanish, and German, and no two translators are proficient in the same pair of languages. The translators are to be assigned to five rooms, with two translators in each room, and these two translators must be proficient in the same language. How many different allocation schemes are there (considering all possible allocation schemes of 5 pairs of translators arranged in five rooms as the same scheme)?
|
Construct a graph $G$, where points $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ represent five languages, and any two points (such as $x_{1}, x_{2}$) are connected by an edge $(x_{1} x_{2})$ representing a translator proficient in these two languages.
By the problem statement, graph $G$ is a simple complete graph.
Next, orient all edges of graph $G$ according to the following rule: if the common language of the room where the translator $x_{1} x_{2}$ is located is $x_{1}$, then the edge $x_{1} x_{2}$ is oriented as $x_{2} \rightarrow x_{1}$, which can also be denoted as $\overrightarrow{x_{2} x_{1}}$.
By the problem statement, graph $G$ is a simple directed graph, and the in-degrees of all points in graph $G$ are even (0, 2, or 4).
(1) If there are two points $x_{1}, x_{2}$ with in-degree 4, then the edge $x_{1} x_{2}$ is both $x_{1} \rightarrow x_{2}$ and $x_{2} \rightarrow x_{1}$, which is a contradiction.
(2) If the in-degree of each point is 2, then the out-degree is also 2.
Take any point $x_{1}$, and draw two edges $\overrightarrow{x_{1} x_{2}}$ and $\overrightarrow{x_{4} x_{5}}$.
As shown in Figure 2, place $x_{i} (i=1,2, \cdots, 5)$ in a circle.
$x_{3} \rightarrow x_{4}, x_{3} \rightarrow x_{5};$
$x_{2} \rightarrow x_{4}, x_{3} \rightarrow x_{4};$
From $\overrightarrow{x_{3} x_{5}}$ and $\overrightarrow{x_{4} x_{5}}$, we have
$x_{5} \rightarrow x_{1}, x_{5} \rightarrow x_{2}$.
Thus, we obtain graph $G^{\prime}$, which corresponds one-to-one with a 5-cycle arrangement.
Therefore, there are $4! = 24$ allocation schemes.
(3) If there is a point $x_{1}$ with in-degree 4, then there must be another point $x_{2}$ with out-degree 4, and the remaining three points have in-degree 2.
Clearly, there are edges
$\overrightarrow{x_{i} x_{1}} (i=2,3,4,5), \overrightarrow{x_{2} x_{j}} (j=3,4,5)$.
Thus, $x_{3}, x_{4}, x_{5}$ form a unidirectional cycle, which corresponds one-to-one with a 3-cycle arrangement, and $\overrightarrow{x_{i} x_{1}}$ can be divided into two groups in three ways.
Therefore, there are $5 \times 4 \times 2! \times 3 = 120$ allocation schemes.
In summary, there are 144 allocation schemes.
|
144
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A. Let $a=\sqrt{7}-1$. Then the value of the algebraic expression $3 a^{3}+12 a^{2}-$ $6 a-12$ is ( ).
(A) 24
(B) 25
(C) $4 \sqrt{7}+10$
(D) $4 \sqrt{7}+12$
|
,- 1. A. A.
Notice
$$
a=\sqrt{7}-1 \Rightarrow a^{2}+2 a-6=0 \text {. }
$$
Then $3 a^{3}+12 a^{2}-6 a-12$
$$
=\left(a^{2}+2 a-6\right)(3 a+6)+24=24 .
$$
|
24
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. B. Given that the lengths of the two legs are integers $a$ and $b$ $(b<2011)$. Then the number of right triangles with the hypotenuse length $b+1$ is
|
6. B. 31.
By the Pythagorean theorem, we have
$$
a^{2}=(b+1)^{2}-b^{2}=2 b+1 \text{. }
$$
Given $b<2011$, we know that $a$ is an odd number in the interval $(1, \sqrt{4023})$, so $a$ must be $3, 5, \cdots, 63$.
Therefore, there are 31 right-angled triangles that satisfy the conditions.
|
31
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. A. As shown in Figure 3, in the right triangle $\triangle ABC$, the hypotenuse $AB$ is 35 units long, and the square $CDEF$ is inscribed in $\triangle ABC$ with a side length of 12. Then the perimeter of $\triangle ABC$ is $\qquad$
|
10. A. 84.
Let $BC = a, AC = b$. Then,
$$
a^{2} + b^{2} = 35^{2} = 1225.
$$
Since Rt $\triangle AFE \sim \text{Rt} \triangle ACB$, we have,
$$
\frac{FE}{CB} = \frac{AF}{AC} \Rightarrow \frac{12}{a} = \frac{b-12}{b}.
$$
Thus, $12(a + b) = ab$.
From equations (1) and (2), we get
$$
\begin{array}{l}
(a + b)^{2} = a^{2} + b^{2} + 2ab \\
= 1225 + 24(a + b).
\end{array}
$$
Solving this, we get $a + b = -25$ (discard), 49.
Therefore, $a + b + c = 49 + 35 = 84$.
|
84
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. B. If five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are randomly selected from $1,2, \cdots, n$, and one of these integers is always a prime number, find the maximum value of $n$.
|
13. B. When $n \geqslant 49$, take the integers $1, 2^{2}, 3^{2}, 5^{2}, 7^{2}$. These five integers are five pairwise coprime distinct integers, but none of them are prime.
When $n=48$, in the integers $1, 2, \cdots, 48$, take any five pairwise coprime distinct integers $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are all not prime, then at least four of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are composite (let these be $a_{1}, a_{2}, a_{3}, a_{4}$).
Let the smallest prime factors of $a_{1}, a_{2}, a_{3}, a_{4}$ be $p_{1}, p_{2}, p_{3}, p_{4}$, respectively.
Since $a_{1}, a_{2}, a_{3}, a_{4}$ are pairwise coprime, $p_{1}, p_{2}, p_{3}, p_{4}$ are pairwise distinct.
Let $p$ be the largest of $p_{1}, p_{2}, p_{3}, p_{4}$. Then $p \geqslant 7$.
Since $a_{1}, a_{2}, a_{3}, a_{4}$ are composite, there must exist an $a_{i} \geqslant p^{2} \geqslant 7^{2}=49$, which is a contradiction.
Therefore, at least one of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ must be prime.
In conclusion, the maximum value of the positive integer $n$ is 48.
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (16 points) Given a positive integer $n$ that satisfies the following condition: for each positive integer $m$ in the open interval $(0,2009)$, there always exists a positive integer $k$, such that
$$
\frac{m}{2009}<\frac{k}{n}<\frac{m+1}{2010} \text {. }
$$
Find the minimum value of such $n$.
|
12. Notice
$$
\begin{array}{l}
\frac{m}{2009}2010 k
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
m n+1 \leqslant 2009 k, \\
m n+n-1 \geqslant 2010 k
\end{array}\right. \\
\Rightarrow 2009(m n+n-1) \geqslant 2009 \times 2010 k \\
\geqslant 2010(m n+1) \\
\Rightarrow 2009 m n+2009 n-2009 \\
\geqslant 2010 m n+2010 \\
\Rightarrow \geqslant \frac{4019}{2009-m} .
\end{array}
$$
Since the above inequality holds for every positive integer $m$ in the interval $(0,2009)$, we have
$$
n \geqslant \frac{4019}{2009-2008}=4019 \text {. }
$$
On the other hand, according to the inequality “$a, b, c, d \in \mathbf{R}_{+}$, $\frac{a}{b}<\frac{c}{d} \Rightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$”, we know
$$
\frac{m}{2009}<\frac{m+(m+1)}{2009+2010}<\frac{m+1}{2010},
$$
which means $\frac{m}{2009}<\frac{2 m+1}{4019}<\frac{m+1}{2010}$
holds for every positive integer $m$ in the interval $(0,2009)$.
Therefore, the minimum value of $n$ is 4019.
|
4019
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $k_{1}<k_{2}<\cdots<k_{n}$ be non-negative integers, satisfying $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=227$.
Then $k_{1}+k_{2}+\cdots+k_{n}=$ $\qquad$
|
- 1. 19.
Notice that
$$
\begin{array}{l}
227=1+2+32+64+128 \\
=2^{0}+2^{1}+2^{5}+2^{6}+2^{7} .
\end{array}
$$
Therefore, $k_{1}+k_{2}+\cdots+k_{n}=0+1+5+6+7=19$.
|
19
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let the sequence of rational numbers $\left\{a_{n}\right\}$ be defined as follows:
$a_{k}=\frac{x_{k}}{y_{k}}$, where $x_{1}=y_{1}=1$, and
if $y_{k}=1$, then $x_{k+1}=1, y_{k+1}=x_{k}+1$;
if $y_{k} \neq 1$, then $x_{k+1}=x_{k}+1, y_{k+1}=y_{k}-1$.
How many terms in the first 2011 terms of this sequence are positive integers?
|
11. The sequence is
$$
\frac{1}{1}, \frac{1}{2}, \frac{2}{1}, \frac{1}{3}, \frac{2}{2}, \frac{3}{1}, \cdots, \frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}, \cdots \text {. }
$$
Group it as follows:
$$
\begin{array}{l}
\left(\frac{1}{1}\right),\left(\frac{1}{2}, \frac{2}{1}\right),\left(\frac{1}{3}, \frac{2}{2}, \frac{3}{1}\right), \cdots, \\
\left(\frac{1}{k}, \frac{2}{k-1}, \cdots, \frac{k}{1}\right), \cdots
\end{array}
$$
Let any number in the $k$-th group be $\frac{b}{a}(a, b \in \mathbf{N}_{+})$. Then $a+b=k+1$.
$$
\begin{array}{l}
\text { Hence } \frac{b}{a} \in \mathbf{N}_{+} \Leftrightarrow b=a t\left(t \in \mathbf{N}_{+}\right) \\
\Leftrightarrow k+1=(t+1) a .
\end{array}
$$
Thus, each positive integer term $\frac{b}{a}$ in the $k$-th group corresponds one-to-one with a divisor of $k+1$ greater than 1.
Therefore, the number of positive integer terms in the $k$-th group is the number of distinct divisors of $k+1$ greater than 1.
By calculation, the first 63 groups contain
$$
\begin{array}{l}
1 \times 18+2 \times 4+3 \times 20+4 \times 1+5 \times 10+ \\
6 \times 1+7 \times 6+8 \times 1+9 \times 1+11 \times 1 \\
=216
\end{array}
$$
positive integer terms.
Notice that
$$
1+2+\cdots+63=2016 \text {. }
$$
Thus, the 2011th term of the sequence belongs to the 63rd group, and is the 6th number from the end.
Therefore, the terms in the 63rd group that do not belong to the first 2011 terms are
$$
\left(\frac{59}{5}, \frac{60}{4}, \frac{61}{3}, \frac{62}{2}, \frac{63}{1}\right) \text {, }
$$
among which, there are three integers.
In summary, the first 2011 terms of the sequence contain 213 terms that are positive integers.
|
213
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) A dance troupe has $n(n \geqslant 5)$ actors, and they have arranged some performances, each of which is performed by four actors on stage. In one performance, they found that: it is possible to appropriately arrange several performances so that every two actors in the troupe perform on stage together exactly once during this performance. Find the minimum value of $n$.
---
The translation maintains the original formatting and structure of the source text.
|
Three, use $n$ points to represent $n$ actors.
If two actors have performed on the same stage once, then connect the corresponding points with an edge. Thus, the condition of this problem is equivalent to:
Being able to partition the complete graph $K_{n}$ of order $n$ into several complete graphs $K_{4}$ of order 4, such that each edge belongs to exactly one $K_{4}$.
First, from the problem statement, we know $\mathrm{C}_{4}^{2} I \mathrm{C}_{n}^{2}$, i.e.,
$12 \ln (n-1)$.
Therefore, $n \neq 5, 6, 7, 8, 10, 11$.
Second, consider the edges containing point $A$ (as an endpoint), there are $n-1$ such edges, and each edge belongs to exactly one $K_{4}$, thus, there are $n-1$ $K_{4}$s containing point $A$ (as an endpoint). However, each $K_{4}$ containing point $A$ has three edges containing point $A$, thus, each $K_{4}$ is counted 3 times.
Thus, $31(n-1)$.
Therefore, $n \neq 9, 12$.
Finally, represent 13 points using $0,1, \cdots, 12$. For $m=0,1, \cdots, 12$, let $m, m+1, m+5, m+11$ form a $K_{4}$ (points are considered modulo 13). Then the 13 $K_{4}$s form a valid partition.
In summary, the minimum value of $n$ is 13.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. From the numbers $1,2, \cdots, 2014$, what is the maximum number of numbers that can be selected such that none of the selected numbers is 19 times another?
|
According to the problem, if $k$ and $19 k$ cannot both appear in $1,2, \cdots, 2014$, and since $2014=19 \times 106$, and $106=5 \times 19+11$, then choose
$$
1,2,3,4,5,106,107, \cdots, 2013 \text {, }
$$
These 1913 numbers satisfy the requirement.
The numbers not chosen are $6,7, \cdots, 105,2014$, a total of 101 numbers.
Since $2014=1913+101$, if 1914 numbers are selected, then in the following 101 pairs of numbers
$$
(6,6 \times 19),(7,7 \times 19), \cdots,(106,106 \times 19)
$$
there must be a pair where both numbers are selected.
Thus, there must be one number that is 19 times the other.
Therefore, from the 2014 numbers $1,2, \cdots, 2014$, the maximum number of numbers that can be selected is 1913.
|
1913
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $T \subseteq\{1,2, \cdots, 25\}$. If for any two distinct elements $a, b (a \neq b)$ in $T$, their product $ab$ is not a perfect square, find the maximum number of elements in $T$, and the number of subsets $T$ that satisfy this condition.
|
Prompt: Divide the set $\{1,2, \cdots, 25\}$ into several subsets such that the product of any two elements in the same subset is a perfect square, and the product of any two elements in different subsets is not a perfect square.
To maximize the number of elements in $T$, let
$$
\begin{array}{l}
A_{1}=\{1,4,9,16,25\}, A_{2}=\{2,8,18\}, \\
A_{3}=\{3,12\}, A_{4}=\{5,20\}, A_{5}=\{6,24\}, \\
A_{6}=\{7\}, A_{7}=\{10\}, A_{8}=\{11\}, A_{9}=\{13\}, \\
A_{10}=\{14\}, A_{11}=\{15\}, A_{12}=\{17\}, A_{13}=\{19\}, \\
A_{14}=\{21\}, A_{15}=\{22\}, A_{16}=\{23\} .
\end{array}
$$
Taking one number from each of these 16 sets to form the set $T$ meets the requirement.
Therefore, the set $T$ can have at most 16 elements.
If the set $T$ has more than 16 elements, then there must be two elements from the same set $A_{k}(k=1,2, \cdots, 16)$, thus, the product of these two elements will be a perfect square.
Hence, the number of sets $T$ that satisfy the requirement is
$$
5 \times 3 \times 2 \times 2 \times 2=120 \text { (sets). }
$$
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Find the smallest positive integer $n$ such that for any $n$ integers, there exist at least two numbers whose sum or difference is divisible by 1991.
(1991, Australian Mathematical Olympiad)
|
Let $M=\left\{a_{i} \mid a_{i}=0,1, \cdots, 995\right\}$.
Since $a_{i}+a_{j} \leqslant 995+994=1989<1991$, $0<\left|a_{i}-a_{j}\right| \leqslant 995$,
then the sum and difference of any two numbers in $M$ are not multiples of 1991.
Therefore, $n \geqslant 997$.
Let $a_{1}, a_{2}, \cdots, a_{997}$ be any 997 integers.
If there exist $a_{i} 、 a_{j}$ such that $a_{i} \equiv a_{j}(\bmod 1991)$, then the difference $a_{i}-a_{j}$ is divisible by 1991.
If any two numbers $a_{i} 、 a_{j}$ are not congruent modulo 1991, consider the smallest absolute residues of 1991
$$
-995,-994, \cdots,-1,0,1, \cdots, 994,995 \text {. }
$$
Divide them into 996 groups:
$$
\{-995,995\},\{-994,994\}, \cdots,\{-1,1\},\{0\} \text {. }
$$
Thus, there are at least two different numbers $a_{i} 、 a_{j}$ such that
$$
a_{i}=-a_{j}(\bmod 1991) \text {. }
$$
In this case, 1991 ! $\left(a_{i}+a_{j}\right)$.
|
997
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the side lengths of a trapezoid are $3,4,5,6$. Then the area of this trapezoid is $\qquad$ .
|
3. 18.
First, determine the lengths of the two bases. As shown in Figure 7, let trapezoid $ABCD$ have $AD$ and $BC$ as the upper and lower bases, respectively. Draw $AE \parallel CD$. Then $BE$ is the difference between the upper and lower bases.
In $\triangle ABE$, $AB - AE = AB - CD < BE = BC - AD$. Therefore, $AB = 5, CD = 4, AD = 3, BC = 6$. In $\triangle ABE$, $AB = 5, BE = 3, AE = CD = 4$. Thus, $AE$ is the height of the trapezoid.
Therefore, $S_{\text{trapezoid}} = \frac{1}{2} \times (3 + 6) \times 4 = 18$.
|
18
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (25 points) For a set $M=\left\{p_{1}, p_{2}, \cdots, p_{2_{n}}\right\}$ consisting of $2 n$ prime numbers, its elements can be paired to form $n$ products, resulting in an $n$-element set. If
$$
\begin{aligned}
A & =\left\{a_{1} a_{2}, a_{3} a_{4}, \cdots, a_{2 n-1} a_{2 n}\right\} \\
\text { and } \quad B & =\left\{b_{1} b_{2}, b_{3} b_{4}, \cdots, b_{2 n-1} b_{2 n}\right\}
\end{aligned}
$$
are two $n$-element sets obtained in this way, where
$$
\left\{a_{1}, a_{2}, \cdots, a_{2 n}\right\}=\left\{b_{1}, b_{2}, \cdots, b_{2 n}\right\}=M,
$$
and $A \cap B=\varnothing$, then the set pair $\{A, B\}$ is called a "couplet" formed by $M$ (for example, from the four-element set $\{a, b, c, d\}$, three couplets can be formed:
$$
\begin{array}{l}
\{a b, c d\} \sim\{a c, b d\}, \\
\{a b, c d\} \sim\{a d, b c\}, \\
\{a c, b d\} \sim\{a d, b c\} .
\end{array}
$$
Find the number of couplets that can be formed from the six-element prime set $M=\{a, b, c, d, e, f\}$.
|
11. Six elements can form fifteen different "slips of paper," listed as follows:
$$
\begin{array}{l}
\{a b, c d, e f\},\{a b, c e, d f\},\{a b, c f, d e\}, \\
\{a c, b d, e f\},\{a c, b e, d f\},\{a c, b f, d e\}, \\
\{a d, b c, e f\},\{a d, b e, c f\},\{a d, b f, c e\}, \\
\{a e, b c, d f\},\{a e, b d, c f\},\{a e, b f, c d\}, \\
\{a f, b c, d e\},\{a f, b d, c e\},\{a f, b e, c d\} .
\end{array}
$$
Consider the slip of paper at the intersection of the $i$-th row and the $j$-th column as a coordinate point, denoted as $(i, j)$.
For the point $(1,1)$ in the first row, it pairs with two points in each of the following rows:
$$
\begin{array}{l}
(1,1) \sim(2,2),(1,1) \sim(2,3) ; \\
(1,1) \sim(3,2),(1,1) \sim(3,3) ; \\
(1,1) \sim(4,1),(1,1) \sim(4,2) ; \\
(1,1) \sim(5,1),(1,1) \sim(5,2) .
\end{array}
$$
This results in $4 \times 2=8$ pairs.
Similarly, the points $(1,2)$ and $(1,3)$ each pair with 8 points in the following four rows.
Thus, the three points in the first row form $3 \times 4 \times 2=24$ pairs with the points in the following four rows;
Each point in the second row pairs with two points in each of the following rows, resulting in $3 \times 3 \times 2=18$ pairs;
Each point in the third row pairs with two points in each of the following rows, resulting in $3 \times 2 \times 2=12$ pairs;
Each point in the fourth row pairs with two points in the following row, resulting in $3 \times 1 \times 2=6$ pairs.
Therefore, the total number of pairs is $6 \times(1+2+3+4)=60$, meaning that 60 couplets can be created from the set $M$.
|
60
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let two fixed points in the plane be $A(-3,0)$ and $B(0,-4)$, and let $P$ be any point on the curve $y=\frac{12}{x}(x>0)$. Draw $PC \perp x$-axis and $PD \perp y$-axis, with the feet of the perpendiculars being $C$ and $D$, respectively. Then the minimum value of $S_{\text{quadrilateral } ACD}$ is
|
4. 24.
Notice that
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \\
=2\left(x+\frac{9}{x}\right)+12 \geqslant 24 .
\end{array}
$$
The equality holds if and only if $x=3$. Therefore, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 24.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 There are three types of goods, A, B, and C. If you buy 3 pieces of A, 7 pieces of B, and 1 piece of C, it costs a total of 315 yuan; if you buy 4 pieces of A, 10 pieces of B, and 1 piece of C, it costs a total of 420 yuan. Question: How much would it cost to buy one piece each of A, B, and C?
|
Let the unit prices of A, B, and C be $x$, $y$, and $z$ yuan, respectively. Then, according to the problem, we have
$$
\left\{\begin{array}{l}
3 x+7 y+z=315, \\
4 x+10 y+z=420 .
\end{array}\right.
$$
The problem actually only requires finding the value of $x+y+z$, without necessarily solving for $x$, $y$, and $z$ individually. Therefore, we should try to isolate the value of $x+y+z$.
The system of equations (1) can be equivalently transformed into
$$
\left\{\begin{array}{l}
2(x+3 y)+(x+y+z)=315, \\
3(x+3 y)+(x+y+z)=420 .
\end{array}\right.
$$
It is easy to see that, $x+y+z=105$.
Therefore, purchasing one each of A, B, and C would cost 105 yuan.
|
105
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For $\triangle A B C$, squares are constructed outward on its three sides $a, b, c$, with their areas denoted as $S_{a}, S_{b}, S_{c}$ respectively. If $a+b+c=18$, then the minimum value of $S_{a}+S_{b}+S_{c}$ is $\qquad$
|
2. 108.
$$
\begin{array}{l}
\text { Given }(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0 \\
\Rightarrow 2\left(a^{2}+b^{2}+c^{2}\right) \geqslant 2(a b+b c+c a) \\
\Rightarrow 3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2} \\
\Rightarrow a^{2}+b^{2}+c^{2} \geqslant \frac{18^{2}}{3}=108 . \\
\text { Therefore, }\left(S_{a}+S_{b}+S_{c}\right)_{\text {min }}=108 .
\end{array}
$$
|
108
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If the real numbers $x, y, z, w$ satisfy
$$
\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}=1, \\
\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}
$$
then $x^{2}+y^{2}+z^{2}+w^{2}=$
|
3. 36 .
It is known that $2^{2}$ and $4^{2}$ are the two roots of the equation with respect to $t$:
$$
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}=1
$$
which means they are the two roots of the equation:
$$
t^{2}-\left(1^{2}+3^{2}+x^{2}+y^{2}\right) t+1^{2} \times 3^{2}+3^{2} x^{2}+1^{2} \times y^{2}=0
$$
Therefore,
$$
\begin{array}{l}
1^{2}+3^{2}+x^{2}+y^{2}=2^{2}+4^{2} \\
\Rightarrow x^{2}+y^{2}=2^{2}-1^{2}+4^{2}-3^{2}=10 .
\end{array}
$$
Similarly, $z^{2}+w^{2}=6^{2}-5^{2}+8^{2}-7^{2}=26$.
Thus, $x^{2}+y^{2}+z^{2}+w^{2}=36$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) In the set of numbers $1,2, \cdots, 2009$, what is the maximum number of numbers that can be selected such that the sum of any two selected numbers is divisible by 100?
|
Three, let the $n$ numbers that meet the conditions be,
$$
a_{1}, a_{2}, \cdots, a_{n} \text {, }
$$
Take any three of these numbers (let them be $a_{k} \backslash a_{m} \backslash a_{f}$). Then
$$
\begin{array}{l}
a_{k}+a_{m}=100 k_{1}, \\
a_{k}+a_{4}=100 k_{2}, \\
a_{m}+a_{4}=100 k_{3},
\end{array}
$$
where $k_{1} 、 k_{2} 、 k_{3}$ are positive integers.
$$
\begin{array}{l}
\text { (1) }+ \text { (2)-(3) gives } \\
a_{k}=50\left(k_{1}+k_{2}-k_{3}\right) .
\end{array}
$$
Similarly, $a_{m}=\mathbf{5 0}\left(k_{1}+k_{3}-k_{2}\right)$,
$$
a_{1}=50\left(k_{2}+k_{3}-k_{1}\right) \text {. }
$$
Therefore, $a_{k} 、 \grave{a}_{m} 、 a$ are all multiples of 50.
However, the multiples can only be odd or even, and they must all be odd or all even to be divisible by 100, so the numbers that meet the conditions are $50,150, \cdots, 1950$ or $100,200, \cdots, 2000$, with 20 numbers in each set.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (50 points) A company printed a batch of T-shirts, each T-shirt can have three different colors: red, yellow, and blue, and four different patterns. Now, this batch of T-shirts is to be distributed to $n$ new employees, with each employee receiving exactly 4 T-shirts with different patterns. Try to find the minimum value of $n$, such that there will always be two people who have received 4 T-shirts of the same color for two of the patterns.
---
Note: The translation maintains the original text's line breaks and formatting.
|
Four, the minimum value of $n$ is 19.
When $n=18$, the answer scenario shown in Table 1 does not meet the requirements.
[Note] In Table 1, (1), (2), (3), (4) are patterns, $A_{1}, A_{2}, \cdots, A_{18}$ are members, and $A$, $B$, $C$ represent red, yellow, and blue colors, respectively.
The proof below shows that when $n \geqslant 19$, there must exist two people who meet the requirements.
If the 4 cultural shirts of a case have the same color, then there must exist a sub-table in the form of a rectangle, where the letters (colors) in the four corner cells are the same.
If for a certain color (for example, red), let $A_{i}$ be assigned $x_{i}$ red cultural shirts. Then when $\sum_{i=1}^{n} \mathrm{C}_{x_{i}}^{2}>6$ (it is agreed that when $x_{i}<2$, $\mathrm{C}_{x_{i}}^{2}=0$), there must be two columns with the same pair; thus, there must be a rectangle with all four corners being $A$.
When $n \geqslant 19$, taking any 19 people, among all the colors of their cultural shirts, at least one color appears no less than
$$
\left[\frac{4 \times 19-1}{3}\right]+1=26 \text { times, }
$$
let's assume it is red.
Suppose $A_{i}(i=1,2, \cdots, 19)$ is assigned $x_{i}\left(x_{i}\right.$ is a non-negative integer) red cultural shirts. Then
$$
\sum_{i=1}^{19} x_{i} \geqslant 26 \text {. }
$$
By the adjustment method, it is easy to know that when $\sum_{i=1}^{19} \mathrm{C}_{x_{i}}^{2}$ takes the minimum value, for any $1 \leqslant j6 \text {. }
$$
This indicates that when $n \geqslant 19$, there must exist a sub-table in the form of a rectangle with all four corners being the same letter.
In summary, the minimum value of $n$ is 19.
In fact, if all the colors and patterns of the cultural shirts of all people are made into a table as above; if there exist two people with the same two patterns...
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 In $\triangle A B C$, it is known that $A B=A C=2$, and there are 100 different points $P_{1}, P_{2}, \cdots, P_{100}$ on side $B C$. Let $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C(i=1,2, \cdots, 100)$.
Find the value of $m_{1}+m_{2}+\cdots+m_{100}$.
|
Solve As shown in Figure 2, since $\triangle A B C$ is an isosceles triangle, applying the property we get
$$
\begin{aligned}
& A P_{i}^{2} \\
= & A B^{2}-B P_{i} \cdot P_{i} C .
\end{aligned}
$$
Therefore, $m_{i}=A P_{i}^{2}+B P_{i} \cdot P_{i} C=A B^{2}=4$.
Thus, $m_{1}+m_{2}+\cdots+m_{100}=400$.
|
400
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=3, \text { and } \\
a_{n+2}=\left|a_{n+1}\right|-a_{n}
\end{array}\left(n \in \mathbf{N}_{+}\right) .
$$
Let $\left\{a_{n}\right\}$'s sum of the first $n$ terms be $S_{n}$. Then $S_{100}=$
|
- 1. 89.
From the given, $a_{k+9}=a_{k}$.
Then $S_{100}=a_{1}+11\left(a_{1}+a_{2}+\cdots+a_{9}\right)=89$.
|
89
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $n<100$. Then the largest integer $n$ such that the expansion of $(a+b)^{n}$ has three consecutive terms with coefficients in arithmetic progression is $\qquad$ . . .
|
3. 98.
Let the coefficients of three consecutive terms in the expansion of $(a+b)^{n}$ be $\mathrm{C}_{n}^{k-1}, \mathrm{C}_{n}^{k}, \mathrm{C}_{n}^{k+1} (1 \leqslant k \leqslant n-1)$.
By the problem, we have $2 \mathrm{C}_{n}^{k}=\mathrm{C}_{n}^{k-1}+\mathrm{C}_{n}^{k+1}$.
Expanding and rearranging according to the definition of combination numbers, we get
$$
n^{2}-(4 k+1) n+4 k^{2}-2=0 \text{. }
$$
Thus, $n_{1,2}=\frac{4 k+1 \pm \sqrt{8 k^{2}+9}}{2}\left(n_{1,2} \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
8 k^{2}+9=(2 m+1)^{2} \\
\Rightarrow 2 k=m^{2}+m-2 .
\end{array}
$$
Substituting into equation (1), we get
$$
n_{1}=(m+1)^{2}-2, n_{2}=m^{2}-2 \text{. }
$$
From $(m+1)^{2}-2<100$, we know $n=98$.
|
98
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Among the positive integers less than 20, each time three numbers are taken without repetition, so that their sum is divisible by 3. Then the number of different ways to do this is $\qquad$ .
|
4.327.
Divide these 19 numbers into three categories based on the remainder when divided by 3:
$$
\begin{array}{l}
A_{1}: 3,6,9,12,15,18 ; \\
A_{2}: 2,5,8,11,14,17 ; \\
A_{3}: 1,4,7,10,13,16,19 .
\end{array}
$$
Thus, the number of ways to satisfy the conditions of the problem are only four scenarios.
(1) Choose any three numbers from $A_{1}$, there are $\mathrm{C}_{6}^{3}=20$ ways;
(2) Choose any three numbers from $A_{2}$, there are $\mathrm{C}_{6}^{3}=20$ ways;
(3) Choose any two numbers from $A_{3}$, there are $\mathrm{C}_{7}^{3}=35$ ways;
(4) Choose one number from each of $A_{1}$, $A_{2}$, and $A_{3}$, there are
$$
6 \times 6 \times 7=252
$$
ways.
Therefore, the total number of ways is
$$
20 \times 2+35+252=327 \text { (ways). }
$$
|
327
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Place 10 numbers on a given circle such that their total sum is 200, and the sum of any three consecutive numbers is not less than 58. Then the maximum value of the largest number among all sets of 10 numbers that satisfy the above requirements is $\qquad$
|
8. 26 .
Let the maximum number in all placements be $A$. Then
$$
A+3 \times 58 \leqslant 200 \Rightarrow A \leqslant 26 \text {. }
$$
In fact, $26,6,26,26,6,26,26,6,26,26$ satisfies.
|
26
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the set
$$
A=\left\{x \mid x=a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}\right\} \text {, }
$$
where, $a_{i} \in\{0,1, \cdots, 6\}(i=0,1,2,3)$, and $a_{3} \neq 0$.
If positive integers $m 、 n \in A$, and $m+n=2010(m>n)$, then the number of positive integers $m$ that satisfy the condition is $\qquad$.
|
6. 662 .
According to the problem, we know that $m$ and $n$ are four-digit numbers in base 7, and the largest four-digit number in base 7 is
$$
6 \times 7^{3}+6 \times 7^{2}+6 \times 7+6=2400,
$$
the smallest one is $1 \times 7^{3}=343$.
Since $m+n=2010(m>n)$, therefore,
$$
1006 \leqslant m \leqslant 1667 \text {. }
$$
Thus, the number of positive integers $m$ that meet the condition is
$$
1667-1006+1=662 \text { (numbers). }
$$
|
662
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Arrange the real solutions of the equation $x^{3}-3[x]=4$ in ascending order to get $x_{1}, x_{2}, \cdots, x_{k}$. Then the value of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}$ is $\qquad$ ( $[x]$ denotes the greatest integer less than or equal to the real number $x$).
|
8. 15 .
Notice that $x-1<[x] \leqslant x$.
Therefore, when $x \geqslant 3$,
$$
\begin{array}{l}
x^{3}-3[x] \geqslant x^{3}-3 x=x\left(x^{2}-3\right) \\
\geqslant 3 \times 6=18 ;
\end{array}
$$
When $x \leqslant-3$,
$$
\begin{array}{l}
x^{3}-3[x]<x^{3}-3(x-1)=x\left(x^{2}-3\right)+3 \\
\leqslant-3 \times 6+3=-15 .
\end{array}
$$
Therefore, the solution to the equation can only be within the interval $(-3,3)$, and $[x]$ can only take the values $-3, -2, -1, 0, 1, 2$.
If $[x]=-3$, then $x^{3}=-5$, which does not meet the requirement;
If $[x]=-2$, then $x^{3}=-2 \Rightarrow x=-\sqrt[3]{2}$, which meets the requirement;
If $[x]=-1$, then $x^{3}=1$, which does not meet the requirement;
If $[x]=0$, then $x^{3}=4$, which does not meet the requirement;
If $[x]=1$, then $x^{3}=7 \Rightarrow x=\sqrt[3]{7}$, which meets the requirement;
If $[x]=2$, then $x^{3}=10 \Rightarrow x=\sqrt[3]{10}$, which meets the requirement.
Thus, $x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3}=-2+7+10=15$.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. There are three numbers arranged in sequence: $3, 9, 8$. For any two adjacent numbers, the difference between the right number and the left number is written between these two numbers, resulting in a new sequence $3, 6, 9, -1, 8$, which is called the first operation; after the second similar operation, a new sequence $3, 3, 6, 3, 9, -10, -1, 9, 8$ is produced; continue this operation. Ask: Starting from the sequence 3, 9, 8, what is the sum of all numbers in the new sequence after the 100th operation?
|
For convenience, let the sequence of $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. According to the problem, the newly added numbers are $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{n}-a_{n-1}$. Therefore, the sum of the newly added numbers is
$$
\begin{array}{l}
\left(a_{2}-a_{1}\right) + \left(a_{3}-a_{2}\right) + \cdots + \left(a_{n}-a_{n-1}\right) \\
=a_{n}-a_{1} .
\end{array}
$$
The original sequence is three numbers $3, 9, 8$. After the first operation, the resulting sequence is $3, 6, 9, -1, 8$. By equation (1), the sum of the two newly added numbers is
$$
6 + (-1) = 5 = 8 - 3.
$$
After the second operation, by equation (1), the sum of the four newly added numbers is
$$
3 + 3 + (-10) + 9 = 5 = 8 - 3.
$$
Thus, after the 100th operation, the sum of all numbers in the new sequence is
$$
(3 + 9 + 8) + 100 \times (8 - 3) = 520.
$$
|
520
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
19. There are 10 red, 10 black, and 10 white balls. Now, all of them are to be placed into two bags, A and B, with the requirement that each bag must contain balls of all three colors, and the product of the number of balls of each color in bags A and B must be equal. How many ways are there to do this?
|
19. Let the number of red, black, and white balls in bag A be $x$, $y$, and $z$ respectively. Then $1 \leqslant x, y, z \leqslant 9$, and
$$
x y z=(10-x)(10-y)(10-z) \text {, }
$$
i.e., $x y z=500-50(x+y+z)+5(x y+y z+z x)$.
Thus, $5 \mid x y z$.
Therefore, one of $x, y, z$ must be 5.
Assume $x=5$, substituting into equation (1) gives $y+z=10$.
At this point, $y$ can take $1,2, \cdots, 9$ (correspondingly $z$ takes 9, 8, ..., 1), for a total of 9 ways.
Similarly, when $y=5$ or $z=5$, there are also 9 ways each.
However, when $x=y=z$, two ways are repeated, so there are a total of $9 \times 3-2=25$ ways.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. It is known that fresh shiitake mushrooms contain $90 \% \sim 99 \%$ water, while dried shiitake mushrooms contain $30 \% \sim 45 \%$ water. Then, under the influence of drying, by what maximum factor can the weight of fresh shiitake mushrooms be reduced?
|
2. Let the weights of fresh mushrooms, baked mushrooms be $m_{1} \mathrm{~g}, m_{2} \mathrm{~g}$, and the weight of dried mushrooms be $x \mathrm{~g}$ (unknown). Then the range of the proportion of dried mushrooms in fresh mushrooms and baked mushrooms is
$$
\begin{array}{l}
0.01=1-0.99 \\
\leqslant \frac{x}{m_{1}} \leqslant 1-0.90=0.10, \\
0.55=1-0.45 \\
\leqslant \frac{x}{m_{2}} \leqslant 1-0.3=0.70 .
\end{array}
$$
Therefore, the maximum multiple of the weight of fresh mushrooms to the weight of baked mushrooms is
$$
\frac{m_{1}}{m_{2}}=\frac{\frac{x}{m_{2}}}{\frac{x}{m_{1}}}=\frac{0.70}{0.01}=70 .
$$
|
70
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given two circles $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent at point $A$, circle $\Gamma$ is externally tangent to $\Gamma_{1}$ and $\Gamma_{2}$ at points $B$ and $C$ respectively. Extend the chord $B A$ of circle $\Gamma_{1}$ to intersect circle $\Gamma_{2}$ at point $D$, extend the chord $C A$ of circle $\Gamma_{2}$ to intersect circle $\Gamma_{1}$ at point $E$, and extend the chords $E B$ and $D C$ to intersect circle $\Gamma$ at points $F$ and $G$ respectively: If $B C=5, B F=12$, find the length of $B G$.
|
5. First, prove that quadrilateral $B C G F$ is a rectangle:
In fact, let $K L, B M, C N$ be the common tangents of circles $\Gamma_{1}$ and $\Gamma_{2}$, $\Gamma_{1}$ and $\Gamma$, $\Gamma_{2}$ and $\Gamma$ (Figure 2). According to the inscribed angle and the angle between a tangent and a chord, we have
$$
\angle A B E=\angle E A K=\angle C A L=\angle A D C \text {. }
$$
Then $B E / / C D$
$$
\begin{array}{l}
\Rightarrow \angle C B E+\angle B C D=180^{\circ} . \\
\text { Also } \angle B C N=\angle C B M, \\
\angle A C N=\angle C A L=\angle A B E, \\
\angle A B M=\angle B A L=\angle A C D,
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore } \angle E B C=\angle D C B=90^{\circ} \\
\Rightarrow \angle B F G=\angle B C D=90^{\circ} \text {. }
\end{array}
$$
Thus, $B G=\sqrt{5^{2}+12^{2}}=13$.
|
13
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ be a function, and for any positive integers $m, n$, we have
$$
f(f(m)+f(n))=m+n .
$$
Find the value of $f(2011)$.
|
Since $f(x)$ is a function from the set of positive integers to the set of positive integers, let
$$
f(1)=p\left(p \in \mathbf{N}_{+}\right) \text {. }
$$
If $p>1$, then $p \geqslant 2$.
$$
\text { Let } p=1+b\left(b \in \mathbf{N}_{+}\right), f(b)=c\left(c \in \mathbf{N}_{+}\right) \text {. }
$$
On one hand,
$$
\begin{array}{l}
f(f(1)+f(1))=f(p+p) \\
=f(1+b+1+b)=f(2+2 b) \\
=f(f(f(1)+f(1))+f(f(b)+f(b))) \\
=f(f(2 f(1))+f(2 c))=2 f(1)+2 c \\
=2 p+2 c=2+2 b+2 c>2 .
\end{array}
$$
On the other hand, $f(f(1)+f(1))=2$. This is a contradiction.
Therefore, $p=1$.
Thus, $f(1)=1$,
$$
f(2)=f(1+1)=f(f(1)+f(1))=2 \text {. }
$$
Conjecture: $f(n)=n$.
We will prove this by mathematical induction.
When $n=1$, $f(1)=1$, so the conclusion holds.
Assume that when $n=k$, the conclusion holds.
When $n=k+1$,
$$
f(k+1)=f(f(k)+f(1))=k+1 \text {. }
$$
Thus, when $n=k+1$, the conclusion holds.
Therefore, for all $n \in \mathbf{N}_{+}, f(n)=n$.
Hence, $f(2011)=2011$.
(Pan Tie, Tianjin Experimental High School, 300074)
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2, given that $A B$ is the diameter of $\odot O$, chord $C D$ intersects $A B$ at point $E$, a tangent line through $A$ intersects the extension of $C D$ at point $F$, and $D$ is the midpoint of $E F$. If $D E=\frac{3}{4} C E, A C=$ $8 \sqrt{5}$, then $A B=$ . $\qquad$
|
4.24.
Let $C E=4 x, A E=y$. Then
$$
D F=D E=3 x, E F=6 x \text {. }
$$
Connect $A D, B C$.
Since $A B$ is the diameter of $\odot O$ and $A F$ is the tangent of $\odot O$, we have
$$
\angle E A F=90^{\circ}, \angle A C D=\angle D A F \text {. }
$$
Since $D$ is the midpoint of the hypotenuse $E F$ of the right triangle $\triangle A E F$,
$$
\begin{array}{l}
D A=D E=D F \\
\Rightarrow \angle A F D=\angle D A F=\angle A C F \\
\Rightarrow A F=A C=8 \sqrt{5} .
\end{array}
$$
In the right triangle $\triangle A E F$, by the Pythagorean theorem,
$$
E F^{2}=A E^{2}+A F^{2} \Rightarrow 36 x^{2}=y^{2}+320 \text {. }
$$
Let $B E=z$.
Since $\triangle A D E \backsim \triangle C B E$, we have
$$
\frac{A D}{B C}=\frac{D E}{B E}=\frac{A E}{C E} \text {. }
$$
Thus, $B C=C E$, and
$$
\begin{array}{l}
C E \cdot D E=A E \cdot B E \\
\Rightarrow y z=4 x \cdot 3 x=12 x^{2} .
\end{array}
$$
Therefore, $y^{2}+320=3 y z$.
In the right triangle $\triangle A C B$, by the Pythagorean theorem,
$$
A B^{2}=A C^{2}+B C^{2} \Rightarrow(y+z)^{2}=320+z^{2} \text {. }
$$
Thus, $y^{2}+2 y z=320$.
Solving equations (1) and (2) simultaneously, we get $y=8, z=16$.
Therefore, $A B=A E+B E=24$.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. If the constant term in the expansion of $\left(a \sqrt{x}-\frac{1}{\sqrt{x}}\right)^{6}$ is -160, then $\int\left(3 x^{2}-1\right) \mathrm{d} x=$. $\qquad$
|
$$
\begin{array}{l}
T_{r+1}=\mathrm{C}_{6}^{r}(a \sqrt{x})^{6-r}\left(-\frac{1}{\sqrt{x}}\right)^{r} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{\frac{6-r}{2}-\frac{r}{2}} \\
=\mathrm{C}_{6}^{r} a^{6-r}(-1)^{r} x^{3-r} .
\end{array}
$$
Let $3-r=0$. Then $r=3$.
The constant term is
$$
\begin{array}{l}
-\mathrm{C}_{6}^{3} a^{3}=-20 a^{3}=-160 \Rightarrow a=2 . \\
\text { Hence } \int_{0}^{3}\left(3 x^{2}-1\right) \mathrm{d} x=\left.\left(x^{3}-x\right)\right|_{0} ^{3}=18 .
\end{array}
$$
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $x_{1}$ and $x_{2}$ are the two real roots of the equation
$$
x^{2}-2009 x+2011=0
$$
real numbers $m$ and $n$ satisfy
$$
\begin{array}{l}
2009 m x_{1}+2009 n x_{2}=2009, \\
2010 m x_{1}+2010 n x_{2}=2010 .
\end{array}
$$
Then $2011 m x_{1}+2011 n x_{2}=$ $\qquad$
|
2. -2009 .
From the given, we know
$$
\begin{array}{l}
=2010 \times 2009-2011 \times 2009=-2009 . \\
\end{array}
$$
|
-2009
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the set $M$ is a subset of $\{1,2, \cdots, 2011\}$, and the sum of any four elements in $M$ cannot be divisible by 3. Then $|M|_{\text {max }}=$ $\qquad$
|
- 1. 672.
Consider the set $A=\{3,6,9, \cdots, 2010\}$,
$$
\begin{array}{l}
B=\{1,4,7, \cdots, 2011\}, \\
C=\{2,5,8, \cdots, 2009\} .
\end{array}
$$
If $M \cap A \neq \varnothing$, then $|M \cap B|<3,|M \cap C|<3,|M \cap A|<4$. Therefore, $|M|<10$.
Now assume $M \cap A=\varnothing$.
If $|M \cap B| \geqslant 2$, then $|M \cap C|<2$.
$$
\begin{array}{l}
\text { Hence }|M|=|M \cap B|+|M \cap C| \\
\leqslant|B|+1=672 .
\end{array}
$$
If $|M \cap B| \leqslant 1$, then $|M| \leqslant|C|+1=671$. Therefore, $|M|_{\max }=672$.
For example, take $M=B \cup\{2\}$.
|
672
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For any positive integer $n$, let $a_{n}$ be the smallest positive integer such that $n \mid a_{n}$!. If $\frac{a_{n}}{n}=\frac{2}{5}$, then $n=$ $\qquad$ .
|
4.25.
From $\frac{a_{n}}{n}=\frac{2}{5} \Rightarrow a_{n}=\frac{2 n}{5} \Rightarrow 51 n$.
Let $n=5 k\left(k \in \mathbf{N}_{+}\right)$. If $k>5$, then $5 k \mid k!$.
Thus, $a_{\mathrm{n}} \leqslant k<\frac{2 n}{5}$, a contradiction.
Clearly, when $k=2,3,4$, $a_{n} \neq \frac{2 n}{5}$.
Also, $a_{25}=10=\frac{2}{5} \times 25$, so $n=25$.
|
25
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that on the side $C A$ of $\angle A C B$ there are 2011 points $A_{1}, A_{2}, \cdots, A_{2011}$, and on the side $C B$ there are 2011 points $B_{1}, B_{2}, \cdots, B_{2011}$, satisfying
$$
\begin{array}{l}
A_{1} A_{2}=A_{2} A_{3}=\cdots=A_{2010} A_{2011}, \\
B_{1} B_{2}=B_{2} B_{3}=\cdots=B_{2010} B_{2011} .
\end{array}
$$
If $S_{\text {quadrilateral } A_{1} A_{3} B_{3} B_{1}}=2, S_{\text {quadrilateral } A_{2} A_{4} B_{4} B_{2}}=3$, then
$$
S_{\text {quadrilateral } A_{2} A_{2011} B_{2011} B_{2009}}=
$$
. $\qquad$
|
7.2010 .
$$
\begin{array}{l}
\text { Let } \angle A C B=\alpha, C A_{1}=a, C B_{1}=b, \\
A_{i} A_{i+1}=s, B_{i} B_{i+1}=t(i=1,2, \cdots, 2010), \\
S_{\text {trapezoid } A_{i} A_{i+1} B_{i+2} B_{i}}=S_{i}(i=1,2, \cdots, 2009) . \\
\text { Then } S_{i}=\frac{1}{2}\{[a+(i+1) s][b+(i+1) t]- \\
\quad[a+(i-1) s][b+(i-1) t]\} \sin \alpha \\
=(a t+b s+2 i s t) \sin \alpha .
\end{array}
$$
Therefore, $\left\{S_{i}\right\}$ is an arithmetic sequence.
Given $S_{1}=2, S_{2}=3$, we get $S_{2009}=S_{2010}$.
|
2010
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given positive integers $a, b, c$ satisfy
$$
(a!)(b!)=a!+b!+c! \text {. }
$$
then $c\left(a^{5}+b^{5}+c^{2}\right)+3=$ $\qquad$
|
8.2011 .
$$
\begin{array}{l}
\text { Given }(a!)(b!)=a!+b!+c! \\
\Rightarrow(a!-1)(b!-1)=c!+1 \text {. }
\end{array}
$$
Assume without loss of generality that $a \geqslant b$. Clearly, $c > a$. Then $(a!-1) \mid(c!+1)$.
Also, $c!+1=\frac{c!}{a!}(a!-1)+\frac{c!}{a!}+1$, so $(a!-1) \left\lvert\,\left(\frac{c!}{a!}+1\right)\right.$
$$
\text { Hence } \frac{c!}{a!}+1 \geqslant a!-1 \Rightarrow c!\geqslant a!(a!-2) \text {. }
$$
Also, $c!=(a!-1) \backslash a!\leqslant a!(a!-2)$, combining with equation (1) we get $a=b$, and $c!=a!(a!-2)$.
Clearly, $a \geqslant 3$.
If $c \geqslant a+3$, then $\frac{c!}{a!} \equiv 0(\bmod 3)$.
But $a!-2 \equiv 1(\bmod 3)$, a contradiction.
If $c=a+2$, then
$$
\begin{array}{l}
(a+2)(a+1)=a!-2 \\
\Rightarrow a^{2}+3 a+4=a! \\
\Rightarrow a \mid 4 \Rightarrow a=4,
\end{array}
$$
which does not satisfy equation (2), a contradiction.
If $c=a+1$, then
$$
a+1=a!-2 \Rightarrow a+3=a!\Rightarrow a=3 \text {. }
$$
Thus, $a=b=3, c=4$.
In this case, $c\left(a^{5}+b^{5}+c^{2}\right)+3=2011$.
|
2011
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. At a concert, there are 20 singers who will perform. For each singer, there is a set of other singers (possibly an empty set) such that he wishes to perform later than all the singers in this set. Question: Is there a way to have exactly 2,010 ways to order the singers so that all their wishes are satisfied?
|
1. Such examples exist.
A ranking of singers that satisfies everyone's wishes is called "good".
If for a set of wishes of $k$ singers there exist $N$ good rankings, then $N$ is called "achievable by $k$ singers" (or simply "$k$-achievable").
Next, we prove that 2010 is "20-achievable".
First, we prove a lemma.
Lemma Assume $n_{1}$ and $n_{2}$ are $k_{1}$-achievable and $k_{2}$-achievable, respectively. Then $n_{1} n_{2}$ is $\left(k_{1}+k_{2}\right)$-achievable.
Proof Suppose singers $A_{1}, A_{2}, \cdots, A_{k_{1}}$ (some of whom have wishes) can achieve $n_{1}$, and singers $B_{1}, B_{2}, \cdots, B_{k_{2}}$ (some of whom have wishes) can achieve $n_{2}$, and add the wish for each singer $B_{i}$: to perform later than all singers $A_{j}$. Then the good rankings of singers have the form
$$
\left(A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k_{1}}}, B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{k_{2}}}\right),
$$
where, ( $\left.A_{i_{1}}, A_{i_{2}}, \cdots, A_{i_{k_{1}}}\right)$ is a good ranking for $A_{i}$, and $\left(B_{j_{1}}, B_{j_{2}}, \cdots, B_{j_{k_{2}}}\right)$ is a good ranking for $B_{j}$.
Conversely, every ranking of this form is clearly good, so the number of good rankings is $n_{1} n_{2}$.
Back to the original problem.
By the lemma, construct the number of achievable rankings for 4 singers, 3 singers, and 13 singers as $5, 6, 67$, respectively. Then
$$
2010=5 \times 6 \times 67
$$
is achievable by $4+3+13=20$ singers.
Examples of these three scenarios are shown in Figures 1 to 3 (the numbers in parentheses are the number of good rankings, and wishes are indicated by arrows).
In Figure 1, $c$ wishes to perform later than $a$ and $b$, and $d$ wishes to perform later than $b$. Then there are exactly 5 good rankings
$$
\begin{array}{l}
(a, b, c, d), (a, b, d, c), (b, a, c, d), \\
(b, a, d, c), (b, d, a, c).
\end{array}
$$
In Figure 2, every ranking is good, totaling 6, as they have no wishes.
In Figure 3, the order of $a_{1}, a_{2}, \cdots, a_{11}$ is fixed on this line, singer $x$ can be after each $a_{i}(i \leqslant 9)$, singer $y$ can be before each $a_{j}(j \geqslant 5)$, and when between $a_{i}$ and $a_{i+1}(5 \leqslant i \leqslant 8)$, the order of the two singers can be swapped.
Thus, the number of good rankings is
$$
9 \times 7+4=67.
$$
|
2010
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (50 points) Given that $p$ is a prime number, the fractional part of $\sqrt{p}$ is $x$, and the fractional part of $\frac{1}{x}$ is $\frac{\sqrt{p}-31}{75}$. Find all prime numbers $p$ that satisfy the condition.
|
3. Let $p=k^{2}+r$, where $k, r$ are integers, and satisfy $0 \leqslant r \leqslant 2 k$.
Since $\frac{\sqrt{p}-31}{75}$ is the fractional part of $\frac{1}{x}$, we have $0 \leqslant \frac{\sqrt{p}-31}{75} < 1$. Let
$\frac{1}{x}=\frac{1}{\sqrt{p}-k}=N+\frac{\sqrt{p}-31}{75}(N \geqslant 1)$.
Then $\frac{\sqrt{p}+k}{r}=N+\frac{\sqrt{p}-31}{75}$.
By equating the irrational and rational parts, we get
$\frac{1}{r}=\frac{1}{75}, \frac{k}{r}=\frac{75 N-31}{75}$
$\Rightarrow r=75, k=75 N-31$.
If $N \geqslant 2$, then $k \geqslant 75 \times 2-31=119$.
Thus, $[\sqrt{p}]=k \geqslant 119$, which contradicts $31 \leqslant \sqrt{p}<106$.
Therefore, $N=1, k=44$.
Hence, $p=44^{2}+75=2011$, and 2011 is a prime number.
|
2011
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. (50 points) A scientist stored the design blueprint of a time machine on a computer, setting the file opening password as a permutation of $\{1,2, \cdots, 64\}$. They also designed a program that, when eight positive integers between 1 and 64 are input each time, the computer will indicate the order (from left to right) of these eight numbers in the password. Please design an operation scheme such that the password can be determined with at most 45 inputs.
|
8. Prepare $n^{2}(n=8)$ cards, the front side of which are numbered $1,2, \cdots, n^{2}$, and the back side corresponds to the position of the number in the password (counting from the left). Of course, the operator does not know the numbers on the back side in advance.
First, divide the $n^{2}$ cards into $n$ groups (each group has $n$ cards). In the first $n$ operations, input the card numbers of each group once, so you can know the size order of the numbers on the back of each group (also recorded as the size order of the cards).
Arrange the cards in each group in ascending order and place them on the table, forming an $n \times n$ matrix $\left(a_{i j}\right)$, where $a_{i j}$ is the card number of the $i$-th card in the $j$-th group, and the corresponding number on the back is $b_{i j}$.
Second, input two more times, which are
$\left\{a_{1 j} \mid j=1,2, \cdots, n\right\},\left\{a_{n j} \mid j=1,2, \cdots, n\right\}$.
These $2 n$ cards are called “original cards” (the former is “original small card”, the latter is “original large card”). Then
$$
\begin{array}{l}
\min _{1 \leqslant j \leqslant n} b_{1 j}=\min _{1 \leqslant j \leqslant n} \min _{1 \leqslant i \leqslant n} b_{i j}=\min _{1 \leqslant i, j \leqslant n} b_{i j}, \\
\max _{1 \leqslant j \leqslant n} b_{n j}=\max _{1 \leqslant j \leqslant n} \max _{1 \leqslant i \leqslant n} b_{i j}=\max _{1 \leqslant i, j \leqslant n} b_{i j} .
\end{array}
$$
Thus, we know the card numbers of the smallest and largest numbers in $\left\{b_{i j}\right\}$ (i.e., the first and last two numbers of the password). Remove these two cards and move the card below (or above) the front (or back) one to its position (called “new small (or large) card”).
Then, input the card numbers of these two new cards and the $\frac{n}{2}-1$ smaller (or larger) cards among the original small (or large) cards, so you can know the card numbers of the smallest and largest numbers among the remaining $n^{2}-2$ $b_{i j}$ (i.e., the 2nd and $n^{2}-1$th numbers of the password).
Similarly, by following the two operation methods below, all cards can be arranged in ascending order of the numbers on their back, and the card numbers on the front form the password.
Operation A: When the number of new small and large cards is less than $\frac{n}{2}$, input $\frac{n}{2}$ smaller cards and $\frac{n}{2}$ larger cards (i.e., take $\frac{n}{2}$ cards from the 1st and $n$th rows on the table) after combining these new cards with some smaller (or larger) original cards. Then, you can know the card numbers of the smallest and largest cards on the table. Remove these two cards and move the card below (or above) the front (or back) one to its position (called new small (or large) card).
Operation B: When the number of new small (or large) cards is $\frac{n}{2}$, input the card numbers of all cards in the 1st (or $n$th) row (i.e., $\frac{n}{2}$ original small (or large) cards and $\frac{n}{2}$ new small (or large) cards, which are all renamed as original cards). Then, you can know the card number of the smallest (or largest) card on the table. Remove this card and move the card below (or above) it to its position (called new small (or large) card).
The feasibility of the operations is obvious, and except for the first $n$ operations, all subsequent operations can be considered as either operation A or B. The $n+1$th and $n+2$th operations can be considered as operation B, and the $n+3$th operation can be considered as operation A.
Assume that operations A and B are performed $x$ and $y$ times, respectively, until all cards are removed.
Since each operation A removes 2 cards, and each operation B removes 1 card, and the last operation can remove at most $n$ cards, we have
$$
2 x+y \leqslant n^{2}-n+2 \text {. }
$$
Note that each operation A can increase the number of new cards by at most 2, and each operation B can decrease the number of new cards by $\frac{n}{2}-1$.
If the last operation is operation B, then
$$
2 x+2 \geqslant\left(\frac{n}{2}-1\right)(y-2)+1 \text {; }
$$
If the last operation is operation A, then
$$
2(x-1)+2 \geqslant\left(\frac{n}{2}-1\right)(y-2) \text {. }
$$
From equations (1) and (2)
$$
\begin{array}{l}
\Rightarrow 2 x+y-1 \geqslant \frac{n}{2}(y-2) \\
\Rightarrow n^{2}-n+1 \geqslant \frac{n}{2}(y-2) \\
\Rightarrow \frac{y}{2} \leqslant n+\frac{1}{n} .
\end{array}
$$
Thus, the total number of operations is
$$
\begin{array}{l}
n+x+y=n+\frac{(2 x+y)+y}{2} \\
\leqslant n+\frac{n^{2}-n+2}{2}+n+\frac{1}{n} \\
=\frac{(n+1)(n+2)}{2}+\frac{1}{n}=45 \frac{1}{8} \\
\Rightarrow n+x+y \leqslant 45 .
\end{array}
$$
[Note] If a column is emptied ahead of time, a card can be taken from any row between the 2nd and $(n-1)$th rows to fill the gap.
(Proposed by Li Jianquan, Li Baoyi, Ding Yunlong, Pan Tie, Song Qiang)
|
45
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In the array of numbers shown in Figure 1, the three numbers in each row form an arithmetic sequence, and the three numbers in each column also form an arithmetic sequence. If $a_{22}=2$, then the sum of all nine numbers is equal to
保留源文本的换行和格式，直接输出翻译结果如下：
5. In the array of numbers shown in Figure 1, the three numbers in each row form an arithmetic sequence, and the three numbers in each column also form an arithmetic sequence. If $a_{22}=2$, then the sum of all nine numbers is equal to
|
5.18.
From the problem, we have
$$
\begin{array}{l}
a_{11}+a_{13}=2 a_{12}, a_{21}+a_{23}=2 a_{22}, \\
a_{31}+a_{33}=2 a_{32}, a_{12}+a_{32}=2 a_{22} .
\end{array}
$$
Thus, the sum of all nine numbers is $9 a_{n}=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $p$ and $q$ are both prime numbers, and $7p+q$, $2q+11$ are also prime numbers. Then $p^{q}+q^{p}=$ $\qquad$ .
|
8. 17 .
Since $7 p+q$ is a prime number, and $7 p+q>2$, $7 p+q$ must be an odd number.
Therefore, one of $p$ or $q$ must be even (which can only be 2).
Clearly, $q \neq 2$ (otherwise, $2 q+11=15$, which is not a prime number). Thus, $p=2$.
At this point, $14+q$ and $2 q+11$ are both prime numbers.
If $q=3 k+1\left(k \in \mathbf{N}_{+}\right)$, then
$$
14+q=3(k+5) \text {, }
$$
is not a prime number;
If $q=3 k+2\left(k \in \mathbf{N}_{+}\right)$, then
$$
2 q+11=3(2 k+5) \text {, }
$$
is not a prime number.
Therefore, $q=3 k\left(k \in \mathbf{N}_{+}\right)$, and the only solution is $q=3$.
Thus, $p^{q}+q^{p}=2^{3}+3^{2}=17$.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 As shown in Figure 7, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3$, the distance from point $P$ to line $A N$ is $P D=12, A D=$
30, and a line is drawn through $P$ intersecting
$A N$ and $A M$ at points
$B$ and $C$ respectively. Find the minimum
value of the area of $\triangle A B C$. ${ }^{[7]}$
|
Solve As shown in Figure 7, it can be proven: when the line moves to $B_{0} P=P C_{0}$, the area of $\triangle A B C$ is minimized.
Draw $C_{0} Q / / A N$, intersecting $C B$ at point $Q$.
Thus, $\triangle P C_{0} Q \cong \triangle P B_{0} B$.
Also, $S_{\triangle P C_{0} C} \geqslant S_{\triangle P C_{0} Q}=S_{\triangle P B_{0} B}$, then
$S_{\triangle \triangle B C} \geqslant S_{\triangle A B_{0} C_{0}}$.
Next, we find the area of $\triangle A B_{0} C_{0}$.
As shown in Figure 7, draw the altitude $C_{0} E$ from point $C_{0}$ to side $A B$, with $E$ as the foot of the perpendicular.
Since $C_{0} E / / P D$, and $C_{0} P=P B_{0}$, we get
$C_{0} E=2 P D=24$.
In the right triangle $\triangle C_{0} A E$,
$\tan \angle M A N=\frac{C_{0} E}{A E}=3$.
Therefore, $A E=8$.
Then $E D=A D-A E=B_{0} D$.
Hence $\left(S_{\triangle I B C}\right)_{\text {min }}=S_{\triangle M B_{0} C_{0}}=\frac{1}{2} A B_{0} \cdot C_{0} E$
$=\frac{1}{2} \times 52 \times 24=624$.
|
624
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. From the set $\{1,2, \cdots, 10\}$, any two non-adjacent numbers are taken and multiplied. Then the sum of all such products is equal to
|
5. 990 .
Take any two numbers, multiply them, and then find their sum:
$$
\begin{array}{l}
S_{1}=\frac{1}{2} \sum_{k=1}^{10} k\left(\sum_{i=1}^{10} i-k\right) \\
=\frac{1}{2} \sum_{k=1}^{10} k(55-k)=\frac{1}{2} \sum_{k=1}^{10}\left(55 k-k^{2}\right) \\
=\frac{1}{2}(55 \times 55-385)=1320,
\end{array}
$$
Among them, the sum that does not meet the condition is
$$
\begin{array}{l}
S_{2}=\sum_{k=1}^{9} k(k+1)=\sum_{k=1}^{2}\left(k^{2}+k\right) \\
=285+45=330 .
\end{array}
$$
Therefore, the required sum is
$$
S=S_{1}-S_{2}=1320-330=990 .
$$
|
990
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given positive integers $a_{1}, a_{2}, \cdots, a_{18}$ satisfying
$$
\begin{array}{l}
a_{1}<a_{2}<\cdots<a_{18}, \\
a_{1}+a_{2}+\cdots+a_{18}=2011 .
\end{array}
$$
Then the maximum value of $a_{9}$ is
|
7.193.
To maximize $a_{9}$, $a_{1}, a_{2}, \cdots, a_{8}$ should be as small as possible, and $a_{10}, a_{11}, \cdots, a_{18}$ should be as close to $a_{9}$ as possible. Therefore, we take $a_{1}, a_{2}, \cdots, a_{8}$ to be $1, 2, \cdots, 8$, respectively, with their sum being 36.
Let $a_{9}=n$. Then
$$
\begin{array}{l}
a_{10}=n+1, a_{11}=n+2, \cdots, a_{18}=n+9 \\
\Rightarrow n=193 .
\end{array}
$$
Thus, the maximum value of $a_{9}$ is 193.
|
193
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Define the sequence $\left\{a_{n}\right\}: a_{n}=n^{3}+4\left(n \in \mathbf{N}_{+}\right)$, let $d_{n}=\left(a_{n}, a_{n+1}\right)$. Then the maximum value of $d_{n}$ is $\qquad$
|
8.433.
Given $d_{n} \mid\left(n^{3}+4,(n+1)^{3}+4\right)$, we know $d_{n} \mid\left(n^{3}+4,3 n^{2}+3 n+1\right)$.
Then $d_{n} \mid\left[-3\left(n^{3}+4\right)+n\left(3 n^{2}+3 n+1\right)\right]$,
and $\square$
$$
\begin{array}{l}
d_{n} \mid\left(3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(3 n^{2}+n-12,3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid\left(2 n+13,3 n^{2}+3 n+1\right) \\
\Rightarrow d_{n} \mid(2 n+13,33 n-2) \\
\Rightarrow d_{n} \mid[-2(33 n-2)+33(2 n+13)] \\
\Rightarrow d_{n} \mid 433 .
\end{array}
$$
Therefore, $\left(d_{n}\right)_{\max } \leqslant 433$.
It is easy to see that, $\left(a_{210}, a_{211}\right)=433$.
Thus, $\left(d_{n}\right)_{\max }=433$.
|
433
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.3. There is a convex 2011-gon on the blackboard. Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw?
|
9.3.4016.
Use induction to prove: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. We can sequentially draw $2n-6$ diagonals as follows: $A_{2} A_{4}, A_{3} A_{5}$, $A_{4} A_{6}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.
We will use mathematical induction to prove that at most $2n-6$ diagonals can be drawn.
When $n=3$, the conclusion is obvious. Assume that the last diagonal drawn is $A_{1} A_{k}$. It can intersect at most one of the previously drawn diagonals (if it exists, let it be $d$) at an internal point. All the drawn diagonals, except for $A_{1} A_{k}$ and $d$, are entirely within the $k$-sided polygon $A_{1} A_{2} \cdots A_{k}$ and the $(n+2-k)$-sided polygon $A_{k} A_{k+1} \cdots A_{n} A_{1}$. By the induction hypothesis, we have at most
$$
\begin{array}{l}
(2 k-6)+[2(n+2-k)-6] \\
=2 n-8 \text { (diagonals). }
\end{array}
$$
|
4016
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let four distinct real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2011}-c^{2011}\right)\left(a^{2011}-d^{2011}\right)=2011, \\
\left(b^{2011}-c^{2011}\right)\left(b^{2011}-d^{2011}\right)=2011 . \\
\text { Then }(a b)^{2011}-(c d)^{2011}=(\quad) .
\end{array}
$$
Then $(a b)^{2011}-(c d)^{2011}=(\quad)$.
(A) -2012
(B) -2011
(C)2012
(D) 2011
|
4. B.
From $a \neq b$, we know $a^{2011} \neq b^{2011}$.
Thus, $a^{2011}$ and $b^{2011}$ are the two distinct real roots of the quadratic equation in $x$:
$$
\left(x-c^{2011}\right)\left(x-d^{2011}\right)=2011,
$$
which is
$$
x^{2}-\left(c^{2011} \div d^{2011}\right) x+(c d)^{2011}-2011=0.
$$
By Vieta's formulas, we get
$$
\begin{array}{l}
a^{2011} b^{2011}=(c d)^{2011}-2011 . \\
\text { Therefore, }(a b)^{2011}-(c d)^{2011}=-2011 .
\end{array}
$$
|
-2011
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Figure 2 is a part of the HZ district map. A river runs through the district, with the two banks being the broken lines $A-B-C$ and $D-O-E$, and there are two locations $M$ and $N$. Two bridges perpendicular to the riverbanks and roads are to be built to connect $M$ and $N$ to both banks of the river, making the total bridge and road length between $M$ and $N$ the shortest. If
$$
\begin{array}{l}
A(3,15), B(3,3), C(15,3), D(0,15), \\
E(15,0), M(-16,12), N(11,-1),
\end{array}
$$
then the minimum total bridge and road length between $M$ and $N$ is ( ).
(A) 33
(B) 32
(C) 31
(D) 30
|
6. B.
Translate point $M(-16,12)$ 3 units to the left to get point $M^{\prime}(-13,12)$, and translate point $N(11,-1)$ 3 units upwards to get point $N^{\prime}(11,2)$. Connect $M^{\prime} N^{\prime}$, which intersects $A B$ and $B C$ at points $P$ and $Q$ respectively.
Then the minimum length of the road is
$$
M^{\prime} N^{\prime}=\sqrt{(-13-11)^{2}+(12-2)^{2}}=26 \text {. }
$$
Therefore, the minimum length of the bridge road between $M$ and $N$ is
$$
M^{\prime} N^{\prime}+3+3=26+6=32 \text {. }
$$
The minimum value is achieved when bridges are built at points $P$ and $Q$.
|
32
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given a prime number $p$ such that $p^{3}-6 p^{2}+9 p$ has exactly 30 positive divisors. Then the smallest value of $p$ is $\qquad$ .
|
4. 23 .
Obviously, when $p=2$ or 3, it does not meet the requirements of the problem.
Therefore, $p>3$.
Also, $p^{3}-6 p^{2}+9 p=p(p-3)^{2}$, at this point,
$(p, p-3)=(p, 3)=1$.
Since $p$ has two factors, $(p-3)^{2}$ has 15 factors.
And $15=5 \times 3$, to make $p$ the smallest, $p-3$ is also even, so it can only be
$(p-3)^{2}=2^{4} \times 3^{2}$ or $2^{4} \times 5^{2}$.
Thus, $p=15$ (discard) or 23 .
Therefore, the minimum value of $p$ is 23 .
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. From the set $\{1,2, \cdots, 2011\}$, any two different numbers $a, b$ are selected such that $a+b=n$ (where $n$ is some positive integer) with a probability of $\frac{1}{2011}$. Then the minimum value of $ab$ is
|
6.2010.
Let the number of ways such that $a+b=n$ be $k$. Then
$$
\frac{k}{\mathrm{C}_{2011}^{2}}=\frac{1}{2011} \Rightarrow k=1005 \text {. }
$$
Consider $a b$ to be as small as possible, and the number of ways such that $a+b=n$ is 1005.
Take $n=2011$. Then
$$
1+2010=2+2009=\cdots=1005+1006 \text {. }
$$
At this point, the number of ways such that $a+b=2011$ is exactly 1005.
Thus, the minimum value of $a b$ is $1 \times 2010=2010$.
|
2010
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
---
The above text translated into English, preserving the original text's line breaks and format, is as follows:
Given $a$, $b$, $x$ are positive integers, and $a \neq b$, $\frac{1}{x}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$. Try to find the minimum value of $x$.
|
Solve: It is easy to know that $x=\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
Let $d=(a, b), a=d a_{0}, b=d b_{0},\left(a_{0}, b_{0}\right)=1$.
Then $x=\frac{d^{2} a_{0}^{2} b_{0}^{2}}{a_{0}^{2}+b_{0}^{2}} \in \mathbf{N}_{+}$.
By $\left(a_{0}^{2}+b_{0}^{2}, a_{0}^{2} b_{0}^{2}\right)=1$, we know $\left(a_{0}^{2}+b_{0}^{2}\right) \mid d^{2}$.
To find the minimum value of $x$, we need $a_{0}, b_{0}, d$ to be as small as possible.
So, $a_{0}=1, b_{0}=2, d=5$.
At this point, $a=5, b=10$.
Therefore, $x_{\text {min }}=20$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Find the smallest positive integer $n$, such that there exist $n$ distinct positive integers $s_{1}, s_{2}, \cdots, s_{n}$, satisfying
$$
\left(1-\frac{1}{s_{1}}\right)\left(1-\frac{1}{s_{2}}\right) \cdots\left(1-\frac{1}{s_{n}}\right)=\frac{51}{2010} .
$$
|
1. Suppose the positive integer $n$ satisfies the condition, and let $s_{1}=39$.
Thus, $n \geqslant 39$.
Below, we provide an example to show that $n=39$ satisfies the condition.
Take 39 different positive integers:
$$
2,3, \cdots, 33,35,36, \cdots, 40,67,
$$
which satisfy the given equation.
In conclusion, the minimum value of $n$ is 39.
|
39
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $a_{1}, a_{2}, \cdots$ be integers, for any positive integer $n$ we have
$$
a_{n}=(n-1)\left[\left(\frac{a_{2}}{2}-1\right) n+2\right] \text {. }
$$
If $2001 a_{199}$, find the smallest positive integer $n(n>1)$, such that $200 \mathrm{l} a_{n}$.
|
3. From the given, we have
$$
a_{\mathrm{T99}}=198\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right] \text {. }
$$
From $2001 a_{199}$, we get
$$
100 \left\lvert\, 99\left[\left(\frac{a_{2}}{2}-1\right) \times 199+2\right]\right. \text {. }
$$
Thus, $a_{2}$ is an even number.
Let $a_{2}=2 m$. Then
$$
1001[(m-1) \times 199+2] \text {, }
$$
which simplifies to $1001(m-3)$.
Hence, let $m-3=100 t$. Then
$$
\begin{array}{l}
a_{n}=(n-1)[(100 t+2) n+2] \\
=n(n-1) \times 100 t+2\left(n^{2}-1\right) .
\end{array}
$$
Since $200 \mid a_{n}$, i.e., $100 \mid\left(n^{2}-1\right)$, $n$ must be an odd number.
Let $n=2 k+1$. Then $251 k(k+1)$.
Since $25=5^{2},(k, k+1)=1$, we have $5^{2} \mid k$ or $5^{2} \mid(k+1)$.
Thus, $k_{\min }=24, n_{\text {min }}=49$.
It is easy to verify that the smallest positive integer $n$ that satisfies the condition is
49.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given $0<a<b<c<d<500$, and $a+d=b+c$. Also, $bc-ad=93$.
Then the number of ordered quadruples of integers ( $\dot{a}, b, c$,
$d)$ that satisfy the conditions is . $\qquad$
|
7.870.
Since $a+d=b+c$, we set
$$
(a, b, c, d)=(a, a+x, a+y, a+x+y) \text {, }
$$
where $x$ and $y$ are integers, and $0<x<y$.
Then $93=b c-a d$
$$
=(a+x)(a+y)-a(a+x+y)=x y \text {. }
$$
Therefore, $(x, y)=(1,93)$ or $(3,31)$.
First case
$$
(a, b, c, d)=(a, a+1, a+93, a+94) \text {, }
$$
where $a=1,2, \cdots, 405$;
Second case
$$
(a, b, c, d)=(a, a+3, a+31, a+34) \text {, }
$$
where $a=1,2, \cdots, 465$.
Thus, the total number of integer quadruples that satisfy the condition is
$$
405+465=870 \text { (sets). }
$$
|
870
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Each cell of a $2011 \times 2011$ grid is labeled with an integer from $1,2, \cdots, 2011^{2}$, such that each number is used exactly once. Now, the left and right boundaries, as well as the top and bottom boundaries of the grid, are considered the same, forming a torus (which can be viewed as the surface of a "doughnut"). Find the largest positive integer $M$ such that for any labeling method, there exist two adjacent cells (cells sharing a common edge) whose numbers differ (the larger minus the smaller) by at least $M$.
【Note】Using coordinates, a cell $(x, y)$ and $\left(x^{\prime}, y^{\prime}\right)$ are adjacent if:
$$
\begin{aligned}
x=x^{\prime}, y-y^{\prime} \equiv \pm 1(\bmod 2011) \\
\text { or } \quad y=y^{\prime}, x-x^{\prime} \equiv \pm 1(\bmod 2011) .
\end{aligned}
$$
|
6. Let $N=2011$.
Consider a general $N \times N$ table.
When $N=2$, the conclusion is obvious, and the required $M=2$. An example is shown in Table 1.
Table 1
\begin{tabular}{|l|l|}
\hline 1 & 2 \\
\hline 3 & 4 \\
\hline
\end{tabular}
When $N \geqslant 3$, first prove:
$M \geqslant 2 N-1$.
Starting from a state where each small square in the table is white, write the numbers $1,2, \cdots$ in the table while coloring the marked squares black. Stop the operation when the following condition is first met: every row or every column has at least two black squares. Let the last number written be $k$.
Before marking $k$, there must be one row and one column with at most one black square.
Assume, when marking $k$, every row has two black squares. At this point, there is at most one row with all black squares. This is because if there are two rows with all black squares, then if $k$ is marked in one of these two rows, each row already had two black squares before (using $N \geqslant 3$); if $k$ is marked in another row, then each column already had two black squares.
Color a black square red if it has an adjacent white square. Since, except for the possible all-black row, each row has two black squares and one white square, each of these rows has at least two red squares. Furthermore, the row adjacent to the possible all-black row must have at least one white square. Therefore, at least one black square in the all-black row is colored red. Thus, the number of red squares is at least $2(N-1)+1=2 N-1$.
Therefore, the smallest number in all red squares is at most $k+1-(2 N-1)$.
When the adjacent white square of this red square is marked (the number marked is at least $k+1$), the difference between these two adjacent squares is at least $2 N-1$.
Since $N=2011$, it is sufficient to construct an example for $N=2 n+1(\geqslant 2)$.
Table 2 provides an example where $M=2 N-1$.
Therefore, the required $M=4021$.
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline$(2 n+1)^{2}-2$ & $(2 n+1)^{2}-9$ & $\cdots$ & $\cdots$ & $n(2 n-1)+1$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-10$ & $(2 n+1)^{2}-3$ \\
\hline$(2 n+1)^{2}-8$ & $\cdots$ & $\cdots$ & $n(2 n-1)+2$ & $\cdots$ & $n(2 n-1)$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-11$ \\
\hline$\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ \\
\hline$\cdots$ & $2 n^{2}$ & $\cdots$ & 8 & 2 & 6 & $\cdots$ & $2 n(n-1)+2$ & $2 n(n+1)+2$ \\
\hline $2 n^{2}+1$ & $\cdots$ & $\cdots$ & 3 & 1 & 5 & $\cdots$ & $\cdots$ & $2 n(n+1)+1$ \\
\hline$\cdots$ & $2 n^{2}+2$ & $\cdots$ & 10 & 4 & 12 & $\cdots$ & $2 n(n+1)$ & $\cdots$ \\
\hline$\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\ddots$ & $\vdots$ & $\vdots$ \\
\hline$(2 n+1)^{2}-7$ & $\cdots$ & $\cdots$ & $n(2 n+1)$ & $\cdots$ & $n(2 n+1)+2$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-4$ \\
\hline$(2 n+1)^{2}-1$ & $(2 n+1)^{2}-6$ & $\cdots$ & $\cdots$ & $n(2 n+1)+1$ & $\cdots$ & $\cdots$ & $(2 n+1)^{2}-5$ & $(2 n+1)^{2}$ \\
\hline
\end{tabular}
|
4021
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.1. A table of numbers consisting of $n$ rows and 10 columns, with each element being an integer from $0 \sim 9$, satisfies the following condition: for any row $A$ and any two columns $B$ and $C$, there exists another row $D$ such that $D$ differs from $A$ only in the numbers in columns $B$ and $C$. Prove: $n \geqslant 512$.
|
10.1. Let $R_{0}$ be the first row. Arbitrarily select $2 m$ columns $C_{1}$, $C_{2}, \cdots, C_{2 m}$. By the given condition, there exists a row $R_{1}$, which differs from $R_{0}$ only in $C_{1}$ and $C_{2}$; further, there exists a row $R_{2}$, which differs from $R_{1}$ only in $C_{3}$ and $C_{4}$; $\cdots \cdots$ there exists a row $R_{m}$, which differs from $R_{m-1}$ only in $C_{2 m-1}$ and $C_{2 m}$ (if $m=0$, then $R_{m}=R_{0}$).
Obviously, $R_{m}$ differs from $R_{0}$ only in $C_{1}, C_{2}, \cdots, C_{2 m}$. Therefore, the corresponding $R_{m}$ for different column selections are also different. Since the total number of different even column selections is $\mathrm{C}_{10}^{0}+\mathrm{C}_{10}^{2}+\mathrm{C}_{10}^{4}+\mathrm{C}_{10}^{6}+\mathrm{C}_{10}^{8}+\mathrm{C}_{10}^{10}=512$ (ways), thus, $n \geqslant 512$.
|
512
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. If the length, width, and height of a rectangular prism are all prime numbers, and the sum of the areas of two adjacent sides is 341, then the volume of this rectangular prism $V=$ $\qquad$ .
|
3. 638.
Let the length, width, and height of the rectangular prism be $x, y, z$. From the problem, we have
$$
\begin{array}{l}
x(y+z)=341=11 \times 31 \\
\Rightarrow(x, y+z)=(11,31),(31,11) .
\end{array}
$$
Since $y+z$ is odd, one of $y, z$ must be 2 (let's assume $z=2$).
Also, $11-2=9$ is not a prime number, so
$$
(x, y, z)=(11,29,2) \text {. }
$$
Therefore, $V=x y z=638$.
|
638
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let
$$
\begin{array}{l}
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \\
\text { then } f(1)+f(2)+\cdots+f(50)=
\end{array}
$$
|
$$
\begin{array}{l}
x^{2}-53 x+196=(x-4)(x-49) . \\
\text { Therefore, when } 4 \leqslant x \leqslant 49, f(x)=0 . \\
\text { Then } f(1)+f(2)+\cdots+f(50) \\
=f(1)+f(2)+f(3)+f(50)=660 \text {. }
\end{array}
$$
|
660
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Two boxes are filled with black and white balls. The total number of balls in both boxes is 25. Each time, a ball is randomly drawn from each box. The probability that both balls are black is $\frac{27}{50}$, and the probability that both balls are white is $\frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)=1\right)$. Then $m+n=$ $\qquad$ .
|
5.26.
Let the first box contain $x$ balls, of which $p$ are black, and the second box contain $25-x$ balls, of which $q$ are black. Then
$$
\begin{array}{l}
\frac{p}{x} \cdot \frac{q}{25-x}=\frac{27}{50}, \\
50 p q=27 x(25-x) .
\end{array}
$$
Thus, $x$ is a multiple of 5.
Substituting $x=5$ and $x=10$ into the two equations, we get
$$
\begin{array}{l}
x=5, p=3, q=18 ; \\
x=10, p=9, q=9 .
\end{array}
$$
For both cases, $\frac{m}{r_{0}}=\frac{1}{25}$. Therefore, $m+n=26$.
|
26
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The equation $x+y+z=2011$ satisfies $x<y<z$ to prove: the number of solutions $(x, y, z)$ is $\qquad$ groups.
|
5. 336005 .
The equation $x+y+z=2011$ has $\mathrm{C}_{2010}^{2}$ sets of positive integer solutions. In each solution, $x, y, z$ cannot all be equal, and there are $3 \times 1005$ sets of solutions where two of $x, y, z$ are equal. Therefore, the number of solutions that meet the requirements is
$$
\frac{C_{2010}^{2}-3 \times 1005}{3!}=336005 \text { (sets). }
$$
|
336005
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given positive integers $a, b, c$ satisfy
$$
\left\{\begin{array}{l}
a b+b c+c a+2(a+b+c)=8045, \\
a b c-a-b-c=-2 .
\end{array}\right.
$$
then $a+b+c=$ $\qquad$
|
$$
-1.2012 .
$$
Note that
$$
\begin{array}{l}
(a+1)(b+1)(c+1) \\
=a b c+a b+b c+c a+a+b+c+1 \\
=8045+(-2)+1=8044 .
\end{array}
$$
Since $a, b, c$ are positive integers, we have
$$
a+1 \geqslant 2, b+1 \geqslant 2, c+1 \geqslant 2 \text{. }
$$
Therefore, 8044 can only be factored as $2 \times 2 \times 2011$. Hence $a+b+c=2010+1+1=2012$.
|
2012
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a class, there are two types of students: one type always lies, and the other type never lies. Each student knows what type the other students are. During a gathering today, each student has to state what type the other students are, and all students together said "liar" 240 times. At a similar gathering yesterday, one student was absent, but all students together said "liar" 216 times. Then the total number of students who participated in today's gathering is.
|
2. 22 .
Consider four possible scenarios:
(1) If student $A$ is a liar and student $B$ is not a liar, then $A$ will say $B$ is a liar;
(2) If student $A$ is a liar and student $B$ is also a liar, then $A$ will say $B$ is not a liar;
(3) If student $A$ is not a liar and student $B$ is also not a liar, then $A$ will say $B$ is not a liar;
(4) If student $A$ is not a liar and student $B$ is a liar, then $A$ will say $B$ is a liar.
Thus, if there are $p$ liars and $q$ non-liars in the class, the number of times "liar" is mentioned is $2pq$.
Combining the given conditions, we have
$pq = 120$,
$(p-1)q = 108$ or $p(q-1) = 108$.
Solving these, we get $(p, q) = (10, 12)$ or $(12, 10)$.
Therefore, the total number of students attending today's gathering is 22.
|
22
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that $n+1$ is four times the sum of the other two divisors. Then $n=$
|
5.95.
Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p \neq q$).
In the first case, all positive divisors are $1, p, p^{2}, p^{3}$, then $1+p^{3}=4\left(p+p^{2}\right)$, but $p \nmid \left(1+p^{3}\right)$, which is a contradiction.
In the second case, all positive divisors are $1, p, q, p q$, then
$$
\begin{array}{l}
1+p q=4(p+q) \\
\Rightarrow(p-4)(q-4)=15 .
\end{array}
$$
Thus, $q$ is a prime and $q-4$ is a divisor of 15. Hence, $q \in\{5,7,19\}$.
Since $p$ is also a prime, the only solution is $\{p, q\}=\{5,19\}$. Therefore, $n=95$.
|
95
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given a positive integer $n$ that satisfies the following conditions:
(1) It is an eight-digit number, and all its digits are 0 or 1;
(2) Its first digit is 1;
(3) The sum of the digits in the even positions equals the sum of the digits in the odd positions.
How many such $n$ are there?
|
10.35.
From the fact that the sum of the digits in the even positions equals the sum of the digits in the odd positions, we know that the number of 1s in the even positions equals the number of 1s in the odd positions.
Since the first digit is fixed, there are only three positions in the odd positions that can be freely changed.
Therefore, the total is $\sum_{k=1}^{4} \mathrm{C}_{3}^{k-1} \mathrm{C}_{4}^{k}=4+18+12+1=35$ possible integers.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Given that 15 rays share a common endpoint. Question: What is the maximum number of obtuse angles (considering the angle between any two rays to be the one not greater than $180^{\circ}$) that these 15 rays can form?
|
15. First, it is explained that constructing 75 obtuse angles is achievable.
The position of the rays is represented by their inclination angles, with 15 rays placed near the positions of $0^{\circ}$, $120^{\circ}$, and $240^{\circ}$. Within each group, the five rays are sufficiently close to each other (as shown in Figure 10). Thus, the total number of obtuse angles formed is
$$
\mathrm{C}_{15}^{2}-3 \mathrm{C}_{5}^{2}=75
$$
Next, we prove that it is impossible to obtain more than 75 obtuse angles.
For any arrangement, if there do not exist three rays such that the angles between them are all obtuse, then the total number of obtuse angles does not exceed $\frac{\mathrm{C}_{15}^{2} \times 2}{3}=70$; if there exist three rays such that the angles between each pair are all obtuse, then any other ray can form at most two obtuse angles with these three rays. Removing these initial three rays, the total number of obtuse angles is reduced by at most
$$
3+2 \times 12=27 \text{.}
$$
In the remaining 12 rays, continue to search for such three rays (if such three rays do not exist, then the initial total number of obtuse angles does not exceed $27+44=71$), and remove these three rays. Thus, the number of obtuse angles is reduced by at most $3+2 \times 9=21$. Continue this process, and the number of obtuse angles is reduced by at most $3+2 \times 6=15$ and $3+2 \times 3=9$. In the end, only three rays remain.
Therefore, the initial number of obtuse angles is at most
$$
27+21+15+9+3=75
$$
|
75
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Now arrange for 7 students to participate in 5 sports events, requiring that students A and B cannot participate in the same event, each event must have participants, and each person can only participate in one event. The number of different arrangements that meet the above requirements is $\qquad$ (answer in numbers).
安排 7 名同学去参加 5 个运动项目 -> Arrange for 7 students to participate in 5 sports events
要求甲、乙两同学不能参加同一个项目 -> Requiring that students A and B cannot participate in the same event
每个项目都有人参加 -> Each event must have participants
每人只参加一个项目 -> Each person can only participate in one event
则满足上述要求的不同安排方案数为 $\qquad$ (用数字作答) -> The number of different arrangements that meet the above requirements is $\qquad$ (answer in numbers)
|
5. 15000 .
According to the problem, there are two scenarios that meet the conditions:
(1) One project has 3 participants, with a total of
$$
C_{7}^{3} \times 5! - C_{5}^{1} \times 5! = 3600
$$
schemes;
(2) Two projects each have 2 participants, with a total of
$$
\frac{1}{2}\left(\mathrm{C}_{7}^{2} \mathrm{C}_{5}^{2}\right) \times 5! - \mathrm{C}_{5}^{2} \times 5! = 11400
$$
schemes.
Therefore, the total number of schemes is
$$
3600 + 11400 = 15000 .
$$
|
15000
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given
$$
a_{n}=\mathrm{C}_{200}^{n}(\sqrt[3]{6})^{200-n}\left(\frac{1}{\sqrt{2}}\right)^{n}(n=1,2, \cdots, 95) \text {. }
$$
The number of integer terms in the sequence $\left\{a_{n}\right\}$ is $\qquad$
|
8. 15 .
Notice that $a_{n}=\mathrm{C}_{200}^{n} \times 3^{\frac{200-n}{3}} \times 2^{\frac{200-5 n}{6}}$.
To make $a_{n}(1 \leqslant n \leqslant 95)$ an integer, it must be that $\frac{200-n}{3}$ and $\frac{400-5 n}{6}$ are both integers, i.e., $61(n+4)$.
When $n=6 k+2(k=0,1, \cdots, 13)$, $\frac{200-n}{3}, \frac{400-5 n}{6}$ are both non-negative integers. Therefore, $a_{n}$ is an integer, with a total of 14.
When $n=86$, $a_{86}=\mathrm{C}_{200}^{86} \times 3^{38} \times 2^{-5}$, in $\mathrm{C}_{200}^{86}=\frac{200!}{86!\times 114!}$, the number of factor 2 in $200!$ is $\left[\frac{200}{2}\right]+\left[\frac{200}{2^{2}}\right]+\cdots+\left[\frac{200}{2^{7}}\right]=197$.
Similarly, the number of factor 2 in 86! is 82, and the number of factor 2 in 114! is 110.
Thus, the number of factor 2 in $\mathrm{C}_{200}^{86}$ is
$197-82-110=5$.
Therefore, $a_{86}$ is an integer.
When $n=92$, $a_{92}=\mathrm{C}_{200}^{92} \times 3^{36} \times 2^{-10}$.
Similarly, the number of factor 2 in $\mathrm{C}_{200}^{92}$ is less than 10.
Therefore, $a_{92}$ is not an integer.
Thus, the number of integer terms is $14+1=15$.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Let $A$ be a $3 \times 9$ grid, with each small cell filled with a positive integer. If the sum of all numbers in an $m \times n (1 \leqslant m \leqslant 3, 1 \leqslant n \leqslant 9)$ subgrid of $A$ is a multiple of 10, then it is called a "good rectangle"; if a $1 \times 1$ cell in $A$ is not contained in any good rectangle, then it is called a "bad cell". Find the maximum number of bad cells in $A$.
|
First, we prove by contradiction that there are no more than 25 bad cells in $A$.
Assume the conclusion is not true. Then, in the grid $A$, there is at most 1 cell that is not a bad cell. By the symmetry of the grid, we can assume that all cells in the first row are bad cells.
Let the numbers filled in the $i$-th column from top to bottom be $a_{i}, b_{i}, c_{i} (i=1,2, \cdots, 9)$. Define
$$
\begin{array}{l}
S_{k}=\sum_{i=1}^{k} a_{i}, \\
T_{k}=\sum_{i=1}^{k}\left(b_{i}+c_{i}\right)(k=0,1, \cdots, 9),
\end{array}
$$
where $S_{0}=T_{0}=0$.
We will prove that the three sets of numbers $S_{0}, S_{1}, \cdots, S_{9}; T_{0}, T_{1}, \cdots, T_{9}$, and $S_{0}+T_{0}, S_{1}+T_{1}, \cdots, S_{9}+T_{9}$ are all complete residue systems modulo 10.
In fact, suppose there exist $m, n (0 \leqslant m < n \leqslant 9)$ such that $S_{m} \equiv S_{n} (\bmod 10)$. Then
$$
\sum_{i=m+1}^{n} a_{i} = S_{n} - S_{m} \equiv 0 (\bmod 10),
$$
which means the cells from the $(m+1)$-th column to the $n$-th column in the first row form a good rectangle, contradicting the assumption that all cells in the first row are bad cells.
Similarly, suppose there exist $m, n (0 \leqslant m < n \leqslant 9)$ such that
$$
\begin{array}{l}
T_{m} \equiv T_{n} (\bmod 10). \\
\text{Then } \sum_{i=m+1}^{n}\left(b_{i}+c_{i}\right) = T_{n} - T_{m} \equiv 0 (\bmod 10),
\end{array}
$$
which means the cells from the $(m+1)$-th column to the $n$-th column in the second and third rows form a good rectangle.
Thus, there are at least 2 cells that are not bad cells, which is a contradiction.
Similarly, there do not exist $m, n (0 \leqslant m < n \leqslant 9)$ such that
$$
\begin{array}{l}
S_{m} + T_{m} \equiv S_{n} + T_{n} (\bmod 10). \\
\text{Hence } \sum_{k=0}^{9} S_{k} \equiv \sum_{k=0}^{9} T_{k} \equiv \sum_{k=0}^{2}\left(S_{k} + T_{k}\right) \\
\equiv 0 + 1 + \cdots + 9 \equiv 5 (\bmod 10). \\
\text{Then } \sum_{k=0}^{2}\left(S_{k} + T_{k}\right) \equiv \sum_{k=0}^{2} S_{k} + \sum_{k=0}^{2} T_{k} \\
\equiv 5 + 5 \equiv 0 (\bmod 10),
\end{array}
$$
which is a contradiction.
Therefore, the assumption is false, meaning that the number of bad cells cannot exceed 25.
Next, we construct a $3 \times 9$ grid (Table 1), and it can be verified that each cell not filled with 10 is a bad cell. In this case, there are 25 bad cells.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline 1 & 1 & 1 & 2 & 1 & 1 & 1 & 1 & 10 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 1 & 1 & 1 & 10 & 1 & 1 & 1 & 1 & 2 \\
\hline
\end{tabular}
In conclusion, the maximum number of bad cells is 25.
(Ding Longyun provided)
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given a sequence of numbers
$$
\frac{1}{1}, \frac{2}{1}, \frac{1}{2}, \frac{3}{1}, \frac{2}{2}, \frac{1}{3}, \cdots, \frac{k}{1}, \frac{k-1}{2}, \cdots, \frac{1}{k} \text {. }
$$
In this sequence, the index of the 40th term that equals 1 is ( ).
(A) 3120
(B) 3121
(C) 3200
(D) 3201
|
- 1. B.
For the terms where the sum of the numerator and denominator is $k+1$, we denote them as the $k$-th group. According to the arrangement rule, the 40th term with a value of 1 should be the 40th number in the $2 \times 40-1=79$ group, with the sequence number being
$$
\begin{array}{l}
(1+2+\cdots+78)+40 \\
=\frac{(1+78) \times 78}{2}+40=3121 .
\end{array}
$$
|
3121
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{199}-\frac{1}{200}}{\frac{1}{201^{2}-1^{2}}+\frac{1}{202^{2}-2^{2}}+\cdots+\frac{1}{300^{2}-100^{2}}} \\
= \\
\end{array}
$$
|
3. 400 .
Original expression
$$
\begin{array}{l}
=\frac{\left(1+\frac{1}{2}+\cdots+\frac{1}{200}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{200}\right)}{\frac{1}{202 \times 200}+\frac{1}{204 \times 200}+\cdots+\frac{1}{400 \times 200}} \\
=\frac{\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}}{\frac{1}{400}\left(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}\right)}=400 .
\end{array}
$$
|
400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. A real-coefficient polynomial $P(x)$ of degree not exceeding 2011 takes integer values for any integer $x$, and the remainders when $P(x)$ is divided by $x-1, x-2, \cdots, x-2011$ are $1, 2, \cdots, 2011$ respectively. Then $\max _{x \in \{-1, -2, \cdots, -2011\}}|P(x)|$ has the minimum value of $\qquad$
|
$-1.2011$
First, $P(x)-x$. can be divided by $x-1, x-2, \cdots$, $x-2011$ respectively, so we can assume
$$
P(x)-x=q(x-1)(x-2) \cdots(x-2011) \text {, }
$$
where, $q \in \mathbf{Q}$.
If $q \neq 0$, then
$$
\begin{array}{l}
|P(-2011)|=\left|-2011-q \cdot \frac{4022!}{2011!}\right| \\
\geqslant\left|q \cdot \frac{4022!}{2011!}\right|-2011 \\
>4024|q \cdot 2012!|-2011 \\
\geqslant 4024-2011=2013 .
\end{array}
$$
If $q=0$, then
$$
\begin{array}{l}
P(x)=x, \\
\max _{x \in 1-1,-2, \cdots,-20111}|P(x)|=2011 .
\end{array}
$$
In summary, $\max _{x \in 1-1,-2, \cdots,-20111}|P(x)|$'s minimum value is 2011 .
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $x$ is a positive integer, and $2011-x$ is a perfect cube. Then the minimum value of $x$ is $\qquad$ .
|
1. 283.
From $12^{3}=1728<2011<13^{3}=2197$, we know the minimum value of $x$ is
$$
2011-1728=283 \text {. }
$$
|
283
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Divide a cube with an edge length of a positive integer into 99 smaller cubes, among which, 98 smaller cubes are unit cubes. Find the surface area of the original cube.
|
Let the side length of the original cube be $x$, and the side lengths of the 99 smaller cubes be 1 and $y(x>y>1)$. Then
$$
\begin{array}{l}
x^{3}-y^{3}=98 \\
\Rightarrow(x-y)\left[(x-y)^{2}+3 x y\right]=98 \\
\Rightarrow(x-y) \mid 98=7^{2} \times 2 \\
\Rightarrow x-y=1,2,7,14,49,98 . \\
\text { By }(x-y)^{3}<x^{3}-y^{3}=98 \\
\Rightarrow x-y=1,2 .
\end{array}
$$
When $x-y=1$, we have $3 x y=97$, which is a contradiction;
When $x-y=2$, we have $x y=15$, solving gives $(x, y)=(5,3)$.
Therefore, the surface area of the original cube is $6 \times 5^{2}=150$.
|
150
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the number of ordered integer pairs $(a, b)$ such that
$$
x^{2}+a x+b=167 y
$$
has integer solutions $(x, y)$, where $1 \leqslant a, b \leqslant 2004 .^{[4]}$ (2004, Singapore Mathematical Olympiad)
|
If $x^{2}+a x+b \equiv 0(\bmod 167)$, then completing the square gives
$$
a^{2}-4 b \equiv(2 x+a)^{2}(\bmod 167) .
$$
Therefore, for a fixed $a$, $a^{2}-4 b$ is a quadratic residue modulo 167.
Hence, by the lemma, $b$ can take $\frac{167-1}{2}+1=84$ values modulo 167.
And $\frac{2004}{167}=12$, so each $a$ corresponds to $84 \times 12$ values of $b$, totaling
$$
2004 \times 84 \times 12=2020032
$$
ordered integer solutions.
The lemma can lead to the important Euler's criterion.
Theorem 3 (Euler's criterion) Let $p$ be an odd prime. Then
$$
a^{\frac{p-1}{2}} \equiv\left(\frac{a}{p}\right)(\bmod p) \text {. }
$$
Proof: Take a primitive root of $p$ to be $g$, so the residue system $\{1,2, \cdots, p-1\}$ can be represented as
$$
\left\{g, \cdots, g^{p-2}, g^{p-1}=1\right\} \text {. }
$$
By the lemma, there are exactly $\frac{p-1}{2}$ quadratic residues.
Notice that, $g^{2 k}\left(1 \leqslant k \leqslant \frac{p-1}{2}\right)$ has an even exponent and must be a quadratic residue. Therefore, exactly the even powers of $g$ are quadratic residues, and the odd powers are quadratic non-residues.
Thus, when $\left(\frac{a}{p}\right)=1$,
$a^{\frac{p-1}{2}} \equiv\left(g^{2 k}\right)^{\frac{p-1}{2}} \equiv 1(\bmod p) ;$
when $\left(\frac{a}{p}\right)=-1$,
$$
a^{\frac{p-1}{2}} \equiv\left(g^{2 k+1}\right)^{\frac{p-1}{2}} \equiv g^{\frac{p-1}{2}} \equiv-1(\bmod p) \text {. }
$$
The last equation is due to
$$
g^{p-1} \equiv 1(\bmod p) \Rightarrow g^{\frac{p-1}{2}} \equiv \pm 1(\bmod p),
$$
and since $g$ is a primitive root with order $p-1$, it must be
$$
g^{\frac{p-1}{2}} \equiv -1(\bmod p) \text {. }
$$
Since each quadratic residue modulo $p$ can be characterized by a power of a primitive root of $p$, it is easy to see that the Legendre symbol is multiplicative.
|
2020032
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Figure 2 is a rectangle composed of 6 squares. If the area of the smallest square is 1, then the area of this rectangle is $\qquad$ .
|
3. 143 .
Let the side lengths of the six squares, from smallest to largest, be
$$
1, x, x, x+1, x+2, x+3 \text{.}
$$
Then, by the equality of the top and bottom sides of the rectangle, we have
$$
\begin{array}{l}
x+x+(x+1)=(x+2)+(x+3) \\
\Rightarrow x=4 .
\end{array}
$$
Thus, the length and width of the rectangle are
$$
x+x+(x+1)=13, x+(x+3)=11 \text{.}
$$
Therefore, the area of the rectangle is $13 \times 11=143$.
|
143
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (25 points) There are several (more than enough) socks in red, yellow, blue, and white. If any two socks of the same color can make 1 pair, the question is: What is the minimum number of socks needed to ensure that 10 pairs of socks can be formed?
|
Solution 1 Since there are 4 colors, among 5 socks, there must be 1 pair.
After taking out 1 pair, 3 socks remain. By adding 2 more socks, another pair can be formed.
Following this logic, the number of pairs of socks $(x)$ and the number of socks needed $(y)$ have the following relationship:
$$
y=2 x+3 \text {. }
$$
Thus, to form 10 pairs of socks, only 23 socks are needed.
If 22 socks are taken out, 9 pairs can certainly be formed. If the remaining 4 socks are all different colors, then 22 socks cannot form 10 pairs.
Therefore, at least 23 socks are needed to ensure 10 pairs can be formed.
Solution 2 Since the maximum number of single-color socks that can remain is 4, 24 socks can certainly form 10 pairs.
When 23 socks are taken out, 9 pairs can certainly be formed, leaving 5 socks;
Among the 5 socks, 1 pair can be formed, so 23 socks can also form 10 pairs.
The rest is the same as Solution 1.
|
23
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 As shown in Figure 5, in $\triangle A B C$, $\angle C=90^{\circ}$, $I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$
(2004, Beijing Middle School Mathematics Competition (Grade 8))
|
Solve As shown in Figure 5, construct the symmetric points $F, G$ of points $E, D$ with respect to $AD, BE$ respectively. Then $F, G$ lie on $AB$. Connect $IF, IG$. It is easy to know that
$$
\angle AIB=90^{\circ}+\frac{1}{2} \angle C=135^{\circ}.
$$
By the properties of axial symmetry, we have
$$
\begin{array}{l}
IF=IE, ID=IG, \\
\angle AIE=\angle AIF=\angle BID=\angle BIG=45^{\circ}. \\
\text{Therefore, } \angle FIG=\angle AIB-\angle AIF-\angle BIG \\
=135^{\circ}-45^{\circ}-45^{\circ}=45^{\circ}=\angle BID.
\end{array}
$$
Construct $DH \perp BE$ at point $H$, and $GK \perp IF$ at point $K$.
It is easy to prove $\triangle IDH \cong \triangle IGK$.
Thus, $GK=DH$.
Therefore, $\frac{1}{2} IE \cdot DH=\frac{1}{2} IF \cdot GK$, which means
$S_{\triangle DDE}=S_{\triangle ICF}$.
Hence, $S_{\text{quadrilateral } ABDE}=2 S_{\triangle AIB}=24$.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the set $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can be found two elements $a, b$, such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$ (where $|M|$ denotes the number of elements in the set $M$). (Supplied by Feng Zhigang)
|
2. When
$$
M=\left\{1,2,2^{2}, \cdots, 2^{10}, 3,3 \times 2,3 \times 2^{2}, \cdots, 3 \times 2^{9}\right\}
$$
it satisfies the condition, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011,
\end{array}
$$
contradiction.
In summary, the maximum value of $|M|$ is 21.
|
21
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Nine positive real numbers $a_{1}, a_{2}, \cdots, a_{9}$ form a geometric sequence, and
$$
a_{1}+a_{2}=\frac{3}{4}, a_{3}+a_{4}+a_{5}+a_{6}=15 .
$$
Then $a_{7}+a_{8}+a_{9}=$ . $\qquad$
|
1. 112.
Let the common ratio be $q$. Then, from the given conditions, we have
$$
\begin{array}{l}
a_{1}(1+q)=\frac{3}{4} \\
a_{1} q^{2}\left(1+q+q^{2}+q^{3}\right)=15
\end{array}
$$
Dividing the above two equations yields $q^{2}\left(1+q^{2}\right)=20$.
Thus, $q=2, a_{1}=\frac{1}{4}$.
Therefore, $a_{7}+a_{8}+a_{9}=a_{1} q^{6}\left(1+q+q^{2}\right)=112$.
|
112
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 10, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
(1998, Beijing Junior High School Mathematics Competition)
|
As shown in Figure 10, construct the symmetric point $B^{\prime}$ of point $B$ with respect to line $A C$, and let $B B^{\prime}$ intersect $A C$ at point $E$. Draw $B^{\prime} N \perp A B$ at point $N$, and let $B^{\prime} N$ intersect $A C$ at point $M$. Then, $M$ and $N$ are the required points. The minimum value sought is $B^{\prime} N$.
From $S_{\triangle A B C}=\frac{1}{2} A B \cdot B C=\frac{1}{2} A C \cdot B E$, we get $B E=4 \sqrt{5}$.
Therefore, $B B^{\prime}=2 B E=8 \sqrt{5}$.
From $\triangle B^{\prime} N B \sim \triangle A B C$
$\Rightarrow \frac{B^{\prime} N}{A B}=\frac{B^{\prime} B}{A C} \Rightarrow B^{\prime} N=16$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the set of positive real numbers $A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}$, and the set $S=\{(a, b) \mid a \in A, b \in A, a-b \in A\}$. Then the set $S$ can have at most $\qquad$ elements.
|
9.4950 .
The number of ordered pairs of real numbers $(a, b)$ formed by the elements of set $A$ is $100^{2}=10000$.
Since $a_{i}-a_{i}=0 \notin A$, we have $\left(a_{i}, a_{i}\right) \notin S(i=1,2, \cdots, 100)$.
When $\left(a_{i}, a_{j}\right) \in S$, then $\left(a_{j}, a_{i}\right) \notin S$.
Therefore, the maximum number of elements in set $S$ is
$$
\frac{1}{2}(10000-100)=4950 \text {. }
$$
When $a_{i}=i(i=1,2, \cdots, 100)$, the number of elements in $S$ is 4950 .
|
4950
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Each vertex of the convex pentagon $A B C D E$ is colored with one of five colors, such that the two endpoints of each diagonal have different colors. The number of such coloring methods is $\qquad$ (answer with a number),
|
12. 1020 .
The number of coloring ways where all vertices have different colors is $A_{5}^{5}=$ 120. The number of coloring ways where two adjacent vertices have the same color and the rest of the vertices have different colors is
$$
A_{5}^{1} A_{5}^{4}=5(5 \times 4 \times 3 \times 2)=600 \text { ways. }
$$
The number of coloring ways where two pairs of adjacent vertices have the same color and the remaining vertex has a different color is
$$
A_{5}^{1} A_{5}^{3}=5(5 \times 4 \times 3)=300 \text { ways. }
$$
Therefore, there are a total of 1020 coloring ways.
|
1020
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given positive integers $x, y, z$ satisfying $x y z=(14-x)(14-y)(14-z)$, and $x+y+z<28$. Then the maximum value of $x^{2}+y^{2}+z^{2}$ is . $\qquad$
|
7. 219 .
From the problem, we know that $x$, $y$, and $z$ are all positive integers less than 14. On the other hand, expanding the given equation, we get
$$
2 x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x) \text {. }
$$
Thus, $71 x y z$.
Since $x$, $y$, and $z$ are all less than 14, at least one of $x$, $y$, or $z$ must be 7.
Without loss of generality, let $z=7$. Then the given equation simplifies to
$$
x y=(14-x)(14-y) \Rightarrow x+y=14 \text {. }
$$
In this case, the maximum value of $x^{2}+y^{2}$ is $1^{2}+13^{2}$.
Therefore, the maximum value of $x^{2}+y^{2}+z^{2}$ is
$$
1^{2}+13^{2}+7^{2}=219 \text {. }
$$
|
219
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. For a positive integer $n$, let $x_{n}$ be the real root of the equation
$$
n x^{3}+2 x-n=0
$$
with respect to $x$, and let
$$
a_{n}=\left[(n+1) x_{n}\right](n=2,3, \cdots),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
\frac{1}{1005}\left(a_{2}+a_{3}+\cdots+a_{2011}\right)=
$$
$\qquad$
|
9.2013.
Let $f(x)=n x^{3}+2 x-n$.
It is easy to see that when $n$ is a positive integer, $f(x)$ is an increasing function.
When $n \geqslant 2$,
$$
\begin{array}{l}
f\left(\frac{n}{n+1}\right)=n\left(\frac{n}{n+1}\right)^{3}+2 \times \frac{n}{n+1}-n \\
=\frac{n}{(n+1)^{3}}\left(-n^{2}+n+1\right)0$.
Therefore, when $n \geqslant 2$, the equation $n x^{3}+2 x-n=0$ has a unique real root $x_{n}$, and $x_{n} \in\left(\frac{n}{n+1}, 1\right)$.
Thus, $n<(n+1) x_{n}<n+1$,
$$
a_{n}=\left[(n+1) x_{n}\right]=n \text {. }
$$
Then $\frac{1}{1005} \sum_{n=2}^{2011} a_{n}=\frac{1}{1005} \sum_{n=2}^{2011} n=2013$ :
|
2013
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. In the Cartesian coordinate system, given the point set $I=\{(x, y) \mid x, y$ are integers, and $0 \leqslant x, y \leqslant 5\}$. Then the number of different squares with vertices in the set $I$ is . $\qquad$
|
10. 105.
It is easy to know that there are only two types of squares that meet the conditions: squares whose sides lie on lines perpendicular to the coordinate axes, called "standard squares," and squares whose sides lie on lines not perpendicular to the coordinate axes, called "oblique squares."
(1) In standard squares, the number of squares with side length $k(k=1,2$, $\cdots, 5)$ is $(6-k)^{2}$.
(2) Since the oblique squares with vertices in the point set $I$ are all inscribed squares of some standard square, we only need to consider the number of inscribed squares of standard squares.
Obviously, a square with side length $k(k=1,2, \cdots, 5)$ has $k-1$ inscribed squares.
In summary, the number of squares that meet the conditions is
$$
\sum_{k=1}^{5}\left[(6-k)^{2}+(6-k)^{2}(k-1)\right]=105(\text { squares }) .
$$
|
105
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. There are 4 colors of light bulbs (with enough of each color), and we need to install a light bulb at each vertex of the triangular prism $A B C-A_{1} B_{1} C_{1}$. The requirement is that the light bulbs at the two endpoints of the same edge must be of different colors, and each color of light bulb must be used at least once. The number of installation methods is $\qquad$ kinds.
|
9.216 .
We can first install $A$, $B$, and $C$, which has $\mathrm{A}_{4}^{3}$ ways; then select one vertex from $A_{1}$, $B_{1}$, and $C_{1}$ to install the fourth color of the light bulb, which has $\mathrm{C}_{3}^{1}$ ways; finally, there are 3 ways to install the remaining two vertices.
Therefore, there are 216 different installation methods.
|
216
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers is $\qquad$ .
|
10. 495 .
$$
\begin{array}{l}
\text { Let } t=l+(l+1)+\cdots+(l+8) \\
=m+(m+1)+\cdots+(m+9) \\
=n+(n+1)+\cdots+(n+10)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$
Then $l=n+2+\frac{2 n+1}{9}$,
$$
m=\frac{n}{10}+n+1 .
$$
Therefore, $2 n+1 \equiv 0(\bmod 9)$,
$$
n \equiv 0(\bmod 10) \text {. }
$$
Thus, the smallest positive integer $n$ that satisfies the condition is $n=40$, i.e.,
$$
t_{\min }=40+41+\cdots+50=495 .
$$
|
495
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=4$, the sum of the first $n$ terms is $S_{n}$, and it satisfies
$$
S_{n+1}-5 S_{n}-4 n-4=0\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Then the last four digits of $a_{2 \text { on }}$ are $\qquad$
|
7.8124.
Given $S_{n+1}-5 S_{n}-4 n-4=0$, and
$$
S_{n}-5 S_{n+1}-4 n=0 \text {, }
$$
subtracting the two equations yields
$$
\begin{array}{l}
a_{n+1}-5 a_{n}-4=0 \\
\Rightarrow a_{n+1}+1=5\left(a_{n}+1\right) \\
\Rightarrow a_{n}=5^{n}-1 .
\end{array}
$$
Notice that, when $k \in \mathbf{N}_{+}$,
$$
\begin{aligned}
5^{4 k+1} & \equiv 3125\left(\bmod 10^{4}\right), \\
5^{4 k+2} & \equiv 5625\left(\bmod 10^{4}\right), \\
5^{4 k+3} & \equiv 8125\left(\bmod 10^{4}\right), \\
5^{4 k+4} & \equiv 625\left(\bmod 10^{4}\right) .
\end{aligned}
$$
Thus, $a_{2011} \equiv 5^{4 \times 502+3}-1 \equiv 8124\left(\bmod 10^{4}\right)$.
|
8124
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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