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4. Given in rectangle $A B C D$, $A B=72, A D=$ 56. If $A B$ is divided into 72 equal parts, and parallel lines to $A D$ are drawn through each division point; and if $A D$ is divided into 56 equal parts, and parallel lines to $A B$ are drawn through each division point, then these parallel lines divide the entire rectangle into $72 \times 56$ small squares with side length 1. Thus, the number of small squares that are internally crossed by the diagonal $A C$ (the interior of the small square contains at least two points on $A C$) is ( ) .
(A) 130
(B) 129
(C) 121
(D) 120
|
4. D.
According to the problem, establish a Cartesian coordinate system such that $A(0,0)$, $B(72,0)$, and $D(0,56)$. Then, $C(72,56)$.
Since $AC$ intersects with every horizontal line (including $AB$ and $DC$) and every vertical line (including $AD$ and $BC$), there are $57 + 73 = 130$ intersection points (including overlapping points).
The linear function representing line $AC$ is $y = \frac{7}{9} x$. Therefore, the coordinates of the overlapping points are $(9k, 7k) (k=0,1, \cdots, 8)$, which means there are 9 overlapping points.
Thus, there are $(130 - 9 =) 121$ distinct intersection points, which divide diagonal $AC$ into 120 segments, each passing through 1 small square.
Therefore, $AC$ passes through 120 small squares.
|
120
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Calculate $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}}(n \geqslant 2$, $n \in \mathbf{N})$ 的值为
The value of $\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}}(n \geqslant 2$, $n \in \mathbf{N})$ is
|
7. 100 .
$$
\begin{array}{l}
\sqrt[n]{\underbrace{99 \cdots 9}_{n \uparrow} \times \underbrace{99 \cdots 9}_{n \uparrow}+1 \underbrace{99 \cdots 9}_{n \uparrow}} \\
=\sqrt[n]{\left(10^{n}-1\right)^{2}+2 \times 10^{n}-1} \\
=\sqrt[n]{10^{2 n}}=100 .
\end{array}
$$
|
100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. If $p$ is a prime number, and $p+3$ divides $5p$, then the last digit of $p^{2009}$ is $\qquad$ .
|
8. 2 .
Since the positive divisors of $5p$ are $1, 5, p, 5p$, and $(p+3)$ divides $5p$, therefore, $p=2$.
Thus, $p^{2009}=2^{2009}=\left(2^{4}\right)^{502} \times 2=2 \times 16^{502}$.
Given that the last digit of $16^{502}$ is 6, it follows that the last digit of $p^{2009}$ is 2.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. As shown in Figure 2, in quadrilateral $A B C D$, $\angle A C B=$ $\angle B A D=105^{\circ}, \angle A B C=\angle A D C=45^{\circ}$. If $A B$ $=2$, then the length of $C D$ is $\qquad$.
|
9. 2 .
As shown in Figure 6, draw $EA \perp AB$ intersecting the extension of $BC$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle AEB=45^{\circ} \\
\quad=\angle ADC, \\
AE=AB=2 . \\
\text { Also } \angle DAC=\angle DAB-\angle CAB \\
=\angle DAB-\left(180^{\circ}-\angle ABC-\angle ACB\right) \\
=75^{\circ}=180^{\circ}-\angle ACB=\angle ACE,
\end{array}
$$
Therefore, $\triangle ACE \cong \triangle CAD \Rightarrow CD=AE=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. As shown in Figure 3, fill in the 10 spaces in the annulus with the numbers $1,2, \cdots, 10$. Add up the absolute values of the differences between all adjacent cells (cells that share a common edge). If this sum is to be maximized, then the maximum value is $\qquad$
|
10.50.
Let the two numbers in adjacent cells be $a$ and $b$ ($a > b$). Then $|a-b| = a-b$, and there are 10 differences. To maximize the sum of these 10 differences, the 10 minuends $a$ should be as large as possible, and the 10 subtrahends $b$ should be as small as possible. Since each number is adjacent to two other numbers, we can set $10, 9, 8, 7, 6$ as the minuends twice, and $5, 4, 3, 2, 1$ as the subtrahends twice, which satisfies the condition. Therefore, the maximum value we seek is
$$
\begin{array}{l}
2(10+9+8+7+6)- \\
2(5+4+3+2+1) \\
=50 .
\end{array}
$$
Figure 7 is one way to fill in the numbers that meets the condition.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given $\frac{x y}{x+y}=2, \frac{x z}{x+z}=3, \frac{y z}{y+z}=4$. Find the value of $7 x+5 y-2 z$.
|
Three, 11. Given
$$
\begin{array}{l}
\frac{1}{2}=\frac{1}{x}+\frac{1}{y}, \frac{1}{3}=\frac{1}{x}+\frac{1}{z}, \\
\frac{1}{4}=\frac{1}{y}+\frac{1}{z} .
\end{array}
$$
Solving simultaneously, we get $x=\frac{24}{7}, y=\frac{24}{5}, z=24$.
Therefore, $7 x+5 y-2 z=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The smallest positive integer $a$ that makes the inequality
$$
\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}<a-2007 \frac{1}{3}
$$
hold for all positive integers $n$ is $\qquad$
|
4. 2009 .
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2 n+1}$.
Obviously, $f(n)$ is monotonically decreasing. From the maximum value of $f(n)$, $f(1)<a-2007 \frac{1}{3}$, we get $a=2009$.
|
2009
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A station has exactly one bus arriving between $8:00 \sim 9:00$ and $9:00 \sim 10:00$ every morning, but the arrival times are random, and the two arrival times are independent of each other, as shown in Table 1. A passenger arrives at the station at $8:20$. What is the expected waiting time for the passenger (rounded to the nearest minute)?
Table 1
\begin{tabular}{|l|c|c|c|}
\hline Arrival time $8:10 \sim 9:10$ & $8:30 \sim 9:30$ & $8:50 \sim 9:50$ \\
\hline Probability & $\frac{1}{6}$ & $\frac{1}{2}$ & $\frac{1}{3}$ \\
\hline
\end{tabular}
|
8. 27 .
The distribution of passenger waiting times is shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|c|c|}
\hline \begin{tabular}{c}
Waiting Time \\
(min)
\end{tabular} & 10 & 30 & 50 & 70 & 90 \\
\hline Probability & $\frac{1}{2}$ & $\frac{1}{3}$ & $\frac{1}{6} \times \frac{1}{6}$ & $\frac{1}{2} \times \frac{1}{6}$ & $\frac{1}{3} \times \frac{1}{6}$ \\
\hline
\end{tabular}
The mathematical expectation of the waiting time is
$$
\begin{array}{l}
10 \times \frac{1}{2}+30 \times \frac{1}{3}+50 \times \frac{1}{36}+ \\
70 \times \frac{1}{12}+90 \times \frac{1}{18} \doteq 27
\end{array}
$$
|
27
|
Other
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (14 points) Let the line $l: y=k x+m(k, m \in$ Z) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Question: Does there exist a line $l$ such that the vector $\overrightarrow{A C}+\overrightarrow{B D}=\mathbf{0}$? If it exists, indicate how many such lines there are; if not, explain the reason.
|
Given the system of equations:
$$
\left\{\begin{array}{l}
y=k x+m, \\
\frac{x^{2}}{16}+\frac{y^{2}}{12}=1,
\end{array}\right.
$$
eliminating \( y \) and simplifying, we get:
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0.
$$
Let \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). Then:
$$
\begin{array}{l}
x_{1}+x_{2}=-\frac{8 k m}{3+4 k^{2}}. \\
\Delta_{1}=(8 k m)^{2}-4\left(3+4 k^{2}\right)\left(4 m^{2}-48\right)>0.
\end{array}
$$
From the system:
$$
\left\{\begin{array}{l}
y=k x+m, \\
\frac{x^{2}}{4}-\frac{y^{2}}{12}=1,
\end{array}\right.
$$
eliminating \( y \) and simplifying, we get:
$$
\left(3-k^{2}\right) x^{2}-2 k m x-m^{2}-12=0.
$$
Let \( C\left(x_{3}, y_{3}\right) \) and \( D\left(x_{4}, y_{4}\right) \). Then:
$$
x_{3}+x_{4}=\frac{2 k m}{3-k^{2}}.
$$
$$
\Delta_{2}=(-2 k m)^{2}+4\left(3-k^{2}\right)\left(m^{2}+12\right)>0.
$$
Since \( \overrightarrow{A C}+\overrightarrow{B D}=0 \), we have:
$$
\left(x_{4}-x_{2}\right)+\left(x_{3}-x_{1}\right)=0.
$$
Thus, \( \left(y_{4}-y_{2}\right)+\left(y_{3}-y_{1}\right)=0 \).
From \( x_{1}+x_{2}=x_{3}+x_{4} \), we get:
$$
-\frac{8 k m}{3+4 k^{2}}=\frac{2 k m}{3-k^{2}}.
$$
Therefore, \( 2 k m=0 \) or \( -\frac{4}{3+4 k^{2}}=\frac{1}{3-k^{2}} \).
Solving the above, we get \( k=0 \) or \( m=0 \).
When \( k=0 \), from equations (1) and (2), we get:
$$
-2 \sqrt{3}<m<2 \sqrt{3}.
$$
Since \( m \) is an integer, the values of \( m \) are:
$$
-3,-2,-1,0,1,2,3.
$$
When \( m=0 \), from equations (1) and (2), we get \( -\sqrt{3}<k<\sqrt{3} \). Since \( k \) is an integer, the values of \( k \) are:
$$
-1,0,1.
$$
Thus, there are 9 lines that satisfy the conditions.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (15 points) Find the maximum and minimum values of the function
$$
y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}
$$
|
3. The domain of the function is $[0,13]$.
$$
\begin{array}{l}
\text { Given } y=\sqrt{x}+\sqrt{x+27}+\sqrt{13-x} \\
=\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \\
\geqslant \sqrt{27}+\sqrt{13}=3 \sqrt{3}+\sqrt{13},
\end{array}
$$
we know that the equality holds when $x=0$.
Thus, the minimum value of $y$ is $3 \sqrt{3}+\sqrt{13}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
y^{2}=(\sqrt{x}+\sqrt{x+27}+\sqrt{13-x})^{2} \\
\leqslant\left(\frac{1}{2}+1+\frac{1}{3}\right)[2 x+(x+27)+3(13-x)] \\
=121 .
\end{array}
$$
Thus, $y \leqslant 11$.
By the condition for equality in the Cauchy-Schwarz inequality, we get
$$
4 x=9(13-x)=x+27 \Rightarrow x=9 \text {. }
$$
Thus, when $x=9$, the equality in (1) holds.
Therefore, the maximum value of $y$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.3. The function $f(x)=\prod_{i=1}^{2000} \cos \frac{x}{i}$ changes sign how many times in the interval $\left[0, \frac{2009 \pi}{2}\right]$?
|
10.3. 75 times.
Let $n=2009$. Consider the function $\cos \frac{x}{k}$. It changes sign at $x=\frac{k(2 m+1) \pi}{2}$. This indicates that the zeros of $f(x)$ are $x_{i}=\frac{i \pi}{2}(1 \leqslant i \leqslant n)$.
We only need to consider the sign change of $f(x)$ at $x_{i}(i=1,2, \cdots, n-1)$.
$\cos \frac{x}{k}$ changes sign at $x_{i}$ if and only if $i=k(2 m+1)$.
The number of functions $\cos x, \cos \frac{x}{2}, \cdots, \cos \frac{x}{n}$ that change sign at $x_{i}$ is equal to the number of odd divisors of $i$. Therefore, $f(x)$ changes sign at $x_{i}$ if and only if $i$ has an odd number of odd divisors.
Let $i=2^{l} j$ (where $j$ is an odd number). Then $i$ and $j$ have the same number of odd divisors. An odd number has an odd number of odd divisors if and only if
it is a perfect square, $j$ is a perfect square if and only if $2^{l} j$ is a perfect square or twice a perfect square (depending on the parity of $l$), there are $[\sqrt{n-1}]$ perfect squares and $\left[\sqrt{\frac{n-1}{2}}\right]$ double perfect squares in the range $1 \sim n-1$.
Thus, the number of sign changes is
$$
[\sqrt{n-1}]+\left[\sqrt{\frac{n-1}{2}}\right]=44+31=75 .
$$
|
75
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. 4. In a regular 2009-gon, a non-negative integer not exceeding 100 is placed at each vertex. Adding 1 to the numbers at two adjacent vertices is called an operation on these two adjacent vertices. For any given two adjacent vertices, the operation can be performed at most $k$ times. Find the minimum value of $k$ such that it is always possible to make all the numbers at the vertices equal.
|
10. 4. $k_{\min }=100400$.
Let the numbers at each vertex be $a_{1}, a_{2}, \cdots, a_{2009}$. Let $N=100400$.
$$
\begin{array}{l}
\text { (1) Let } a_{2}=a_{4}=\cdots=a_{2008}=100, \\
a_{1}=a_{3}=\cdots=a_{2008}=0, \\
S=\left(a_{2}-a_{3}\right)+\left(a_{4}-a_{5}\right)+\cdots+\left(a_{2008}-a_{2009}\right) .
\end{array}
$$
For adjacent vertices $\left(a_{1}, a_{2}\right)$, $S$ increases by 1 after one operation; for adjacent vertices $\left(a_{1}, a_{2009}\right)$, $S$ decreases by 1 after one operation; for other adjacent vertices, $S$ remains unchanged.
Initially, $S=N$, and finally, $S=0$.
Therefore, at least $N$ operations must be performed on adjacent vertices $\left(a_{1}, a_{2009}\right)$. Thus, $k \geqslant N$.
(2) Below, we prove that all numbers can be made equal through at most $N$ operations.
For adjacent vertices $\left(a_{i}, a_{i+1}\right)$, perform the operation
$$
s_{i}=a_{i+2}+a_{i+4}+\cdots+a_{i+2 \text { 008 }} \text { (times). }
$$
Then $a_{i}$ will eventually become
$$
a_{i}+s_{i-1}+s_{i}=a_{1}+a_{2}+\cdots+a_{2000} .
$$
And the number of operations on adjacent vertices $\left(a_{i}, a_{i+1}\right)$
$$
s_{i} \leqslant 1004 \times 100=N .
$$
|
100400
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.5. Let $k$ be a positive integer. An infinite strictly increasing sequence of positive integers $\left\{a_{n}\right\}$ satisfies: for any positive integer $n$, we have $a_{n+1}-a_{n} \leqslant k$, and $a_{n}$ is a multiple of 1005 or 1006, but not a multiple of 97. Find the smallest possible value of $k$.
|
10.5. $k_{\text {min }}=2010$.
Given a positive integer $N$, such that
$$
a_{1}<1005 \times 1006 \times 97 \times N=D \text {. }
$$
Since $D$ is a multiple of 97, it is not any term in the sequence $a_{n}$, hence there exists a positive integer $n$, such that $a_{n}<D<a_{n+1}$.
The largest multiple of 1005 or 1006 less than $D$ is $D-1005$, and the smallest multiple of 1005 or 1006 greater than $D$ is $D+1005$, which implies
$$
\begin{array}{l}
a_{n} \leqslant D-1005, a_{n+1} \geqslant D+1005 . \\
\text { Hence } k \geqslant a_{n+1}-a_{n} \\
\geqslant(D+1005)-(D-1005)=2010 .
\end{array}
$$
Let $\left\{a_{n}\right\}$ be the increasing sequence of all multiples of 1005 that are not multiples of 97. Then
$$
\begin{array}{l}
a_{n+1}-a_{n} \leqslant 2010 . \\
\text { Hence } k_{\min }=2010 .
\end{array}
$$
|
2010
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given three non-zero real numbers $a, b, c$, the set $A=$ $\left\{\frac{a+b}{c}, \frac{b+c}{a}, \frac{c+a}{b}\right\}$. Let $x$ be the sum of all elements in set $A$, and $y$ be the product of all elements in set $A$. If $x=2 y$, then the value of $x+y$ is $\qquad$
|
$-1 .-6$.
From the problem, we have
$$
x=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}
$$
$$
\begin{aligned}
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-3, \\
y & =\frac{a+b}{c} \cdot \frac{b+c}{a} \cdot \frac{c+a}{b} \\
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-1 .
\end{aligned}
$$
Then $x=y-2$.
Also, $x=2y$, solving this gives $y=-2$.
Thus, $x+y=2y-2=-6$.
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The total score of an exam consists of six 6-point questions, six 9-point questions, and twelve 5-point questions. Therefore, the number of different scores that can be formed by this test paper is $\qquad$
|
8. 136.
Since there are scores, and the scores can only be generated from the 150 positive integers from $1 \sim 150$, and the sum of the scores and the deducted points is also 150, we only need to consider the number of positive integers from $1 \sim 75$ that can be represented by $6x + 9y + 5z$ (where $x, y, z$ are integers, not all zero, and $0 \leqslant x \leqslant 6, 0 \leqslant y \leqslant 6, 0 \leqslant z \leqslant 12$). Clearly, the numbers $1, 2, 3, 4, 7, 8, 13$ do not meet the conditions.
Notice that
$$
\begin{array}{l}
\text { (1) } 14=5+9, 19=2 \times 5+9, \cdots \cdots \\
69=12 \times 5+9, 74=4 \times 6+10 \times 5 ; \\
\text { (2) } 15=6+9, 20=6+9+5, \cdots \cdots \\
75=6+9+12 \times 5 \text {; }
\end{array}
$$
$$
\begin{array}{l}
\text { (3) } 16=6+2 \times 5, 21=6+3 \times 5, \cdots \cdots \\
66=6+12 \times 5, 71=2 \times 6+9+10 \times 5 ; \\
\text { (4) } 17=2 \times 6+5, 22=2 \times 6+2 \times 5, \\
\cdots \cdots 72=2 \times 6+12 \times 5 ; \\
\text { (5) } 18=2 \times 9, 23=2 \times 9+5, \cdots \cdots \\
73=2 \times 9+11 \times 5 .
\end{array}
$$
And 5, 6, 9, 10, 11, 12 also meet the conditions. Therefore, there are 68 numbers that meet the conditions in the range 1 to 75. Correspondingly, there are 67 numbers that meet the conditions in the range 76 to 149, plus 150. Hence, the total number of positive integers that meet the conditions is 136, meaning the test paper can form 136 different score combinations.
|
136
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (14 points) Let $n$ be a positive integer greater than 16, and let $A(n, k)$ and $B(n, k) (k \geqslant 3)$ be the number of $k$-term monotonic arithmetic sequences and geometric sequences with integer common ratios, respectively, in the set $\{1,2, \cdots, n\}$.
(1) Find the value of $A(2009,50)$;
(2) Prove: $n-4 \sqrt{n}+2<B(n, 3)<2 n-2 \sqrt{n}$.
|
1. (1) First, find the general $A(n, \dot{k})$.
Let the set $\{1,2, \cdots, n\}$ have a monotonically increasing arithmetic sequence $\left\{a_{i}\right\}$, with common difference $d$, and $a_{k}=a_{1}+(k-1) d$. Then
$$
\begin{array}{l}
1+(k-1) d \leqslant a_{1}+(k-1) d \leqslant n \\
\Rightarrow d \leqslant \frac{n-1}{k-1} .
\end{array}
$$
Let $p=\left[\frac{n-1}{k-1}\right]$.
Now, let $d \in\{1,2, \cdots, p\}$, and $a_{d} \in\{1,2, \cdots, n\}$ be the largest first term with common difference $d$. Thus,
$$
a_{d}+(k-1) d \leqslant n \text {. }
$$
Therefore, $a_{d}$ is the number of monotonically increasing arithmetic sequences with common difference $d$ that satisfy the given conditions. Then
$$
\begin{array}{l}
A(n, k)=2 \sum_{d=1}^{p} a_{d}=2 \sum_{d=1}^{p}[n-(k-1) d] \\
=2 n p-p(p+1)(k-1) .
\end{array}
$$
For $n=2009, k=50$, we have
$$
p=40, A(2009,50)=80360 \text {. }
$$
(2) Similarly, let the set $\{1,2, \cdots, n\}$ have a monotonically increasing geometric sequence $\left\{b_{i}\right\}$, with common ratio $r(r \geqslant 2, r \in \mathbf{N})$, and $b_{k}=b_{1} r^{k-1}$. Then
$$
r^{k-1} \leqslant b_{1} r^{k-1} \leqslant n \Rightarrow r \leqslant \sqrt[k-1]{n} .
$$
Let $q=[\sqrt[k-1]{n}]$, and let $r \in\{2,3, \cdots, q\}$, and $b_{r} \in\{1,2, \cdots, n\}$ be the largest first term with common ratio $r$. Thus, $b_{r} r^{k-1} \leqslant n$. Therefore, $b_{r}$ is the number of monotonically increasing geometric sequences with common ratio $r$ that satisfy the given conditions. Then
$$
B(n, k)=2 \sum_{r=2}^{q} b_{r}=2 \sum_{r=2}^{q}\left[\frac{n}{r^{k-1}}\right] \text {. }
$$
When $k=3$, we have
$$
\sqrt{n}-12 \sum_{r=2}^{q}\left(\frac{n}{r^{2}}-1\right) \\
> 2 \sum_{r=2}^{q} \frac{n}{r(r+1)}-2(q-1) \\
= 2 n\left(\frac{1}{2}-\frac{1}{q+1}\right)-2 q+2 \\
> n-4 \sqrt{n}+2 .
\end{array}
$$
Therefore, the conclusion holds.
|
80360
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Let $A=\{i \in \mathbf{N} \mid 1 \leqslant i \leqslant$ $2880\}, B \subseteq A,|B| \geqslant 9$. If all elements in set $A$ can be represented by the sum of no more than 9 different elements from $B$, find $\min |B|$, and construct a set corresponding to the minimum $|B|$.
|
Let $|B|=k \geqslant 9$.
According to the problem, we should have $S_{k}=\sum_{i=1}^{9} \mathrm{C}_{k}^{i} \geqslant 2880$.
Notice that $S_{9}=2^{9}-1, S_{10}=2^{10}-2$,
$$
\begin{array}{l}
S_{11}=2^{11}-2-\mathrm{C}_{11}^{10}=2035, \\
S_{12}=2^{12}-2-\mathrm{C}_{12}^{10}-\mathrm{C}_{12}^{11}=4016>2880 .
\end{array}
$$
Thus, $k \geqslant 12$.
Next, we prove:
$$
\begin{aligned}
B= & \left\{1,2,2^{2}, 2^{3}, 2^{4}, 2^{5}-1,2^{6}-2,2^{7}-4,\right. \\
& \left.2^{8}-8,2^{9}-16,2^{10}-32,2^{11}-80\right\} \\
= & \left\{b_{i} \mid i=0,1, \cdots, 11\right\}
\end{aligned}
$$
satisfies the conditions.
(1) First, we prove by mathematical induction: For any $n \in\left[1, b_{i}-1\right](i=5,6, \cdots, 10)$, $n$ can be expressed as the sum of at most $i-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, i-1\right\}$.
When $i=5$, for any $n \in\left[1, b_{5}-1\right]$, by binary knowledge,
$$
n=\sum_{j=0}^{4} \varepsilon_{j} j^{j}=\sum_{j=0}^{4} \varepsilon_{j} b_{j},
$$
where $\varepsilon_{j}=0$ or $1$, $\varepsilon_{j}$ are not all $1$, $j=0,1, \cdots, 4$.
That is, $n$ can be expressed as the sum of at most 4 different elements from $\left\{b_{j} \mid j=0,1, \cdots, 4\right\}$.
Assume the proposition holds for $i=k \geqslant 5$.
When $i=k+1$, by the induction hypothesis, it is easy to see that the proposition holds for $n \in \left[1, b_{k}\right]$; for $n \in\left[b_{k}+1, b_{k+1}-1\right]$, $n-b_{k} \in\left[1, b_{k}-1\right]$.
By the induction hypothesis, $n-b_{k}$ can be expressed as the sum of at most $k-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}$, hence $n$ can be expressed as the sum of at most $k$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k\right\}$.
(2) For $n \in\left[b_{10}+1, b_{11}-1\right]$, take $k \in \mathbf{N}$ such that
$$
\sum_{j=k}^{10} b_{j}>n \geqslant \sum_{j=k+1}^{10} b_{j} .
$$
If $k \leqslant 3$, then $n \geqslant \sum_{j=4}^{10} b_{j}=b_{11}+1$, which is a contradiction.
If $k=4$, then $n-\sum_{j=5}^{10} b_{j} \in\left[0, b_{4}-2\right]$, by (1) we know $n-\sum_{j=5}^{10} b_{j}$ can be expressed as the sum of at most 3 different elements from $\left\{b_{j} \mid j=0,1,2,3\right\}$. Hence $n$ can be expressed as the sum of at most 9 different elements from $\left\{b_{j} \mid j=0,1, \cdots, 10\}$.
If $k \geqslant 4$, then $n-\sum_{j=k+1}^{10} b_{j} \in\left[0, b_{k}-1\right]$, by (1) we know $n-\sum_{j=k+1}^{10} b_{j}$ can be expressed as the sum of at most $k-1$ different elements from $\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}$. Hence $n$ can be expressed as the sum of at most
$$
(10-k)+(k-1)=9
$$
different elements from $\left\{b_{j} \mid j=0,1, \cdots, 10\}$.
(3) For $n \in\left[b_{11}+1,2880\right]$, then
$n-b_{11} \in\left[1, b_{10}-1\right]$.
Take $k \in \mathbf{N}$ such that $\sum_{j=k}^{9} b_{j}>n-b_{11} \geqslant \sum_{j=k+1}^{9} b_{j}$.
Thus, $n-b_{11}-\sum_{j=k+1}^{9} b_{j} \in\left[0, b_{k}-1\right]$.
By (1) we know $n-b_{11}-\sum_{j=k+1}^{9} b_{j}$ can be expressed as the sum of
$$
\left\{b_{j} \mid j=0,1, \cdots, k-1\right\}
$$
of at most $k-1$ different elements.
Hence $n$ can be expressed as the sum of at most
$$
1+(9-k)+(k-1)=9
$$
different elements from $\left\{b_{j} \mid j=0,1, \cdots, 11\}$.
In summary, $\min |B|=12, B=\{1,2,4,8,16, 31,62,124,248,496,992,1968\}$ is a set that satisfies the conditions.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initially 261 the perimeter of an integer-sided triangle is 75, and squares are constructed on each side. The sum of the areas of the three squares is 2009. Find the difference between the longest and shortest sides of this triangle.
|
Solution: Given that the sum of three numbers is 75, their average is 25.
Let $2009=(25+a)^{2}+(25+b)^{2}+(25+c)^{2}$, where $a+b+c=0$. It is easy to see that
$a^{2}+b^{2}+c^{2}=134$.
Decomposing 134 into the sum of squares of three positive integers, we find the following four cases:
(1) $a^{2}+b^{2}+c^{2}=134=121+9+4$;
(2) $a^{2}+b^{2}+c^{2}=134=100+25+9$;
(3) $a^{2}+b^{2}+c^{2}=134=81+49+4$;
(4) $a^{2}+b^{2}+c^{2}=134=49+49+36$.
For (3), let $a=9, b=-2, c=-7$. The lengths of the three sides of the triangle are
$25+9=34, 25-2=23, 25-7=18$.
Thus, the difference between the longest side and the shortest side is $34-18=16$;
Let $a=7, b=2, c=-9$. The lengths of the three sides of the triangle are
$$
25+7=32, 25+2=27, 25-9=16.
$$
Thus, the difference between the longest side and the shortest side is $32-16=16$.
However, cases (1), (2), and (4) have no solutions.
In conclusion, the difference between the longest side and the shortest side of this triangle is 16.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let point $O$ be outside $\triangle A B C$, and
$$
\overrightarrow{O A}-2 \overrightarrow{O B}-3 \overrightarrow{O C}=0 \text {. }
$$
Then $S_{\triangle A B C}: S_{\triangle O B C}=$
|
$-1.4$.
As shown in Figure 4, let $D$ and $E$ be the midpoints of sides $AB$ and $BC$, respectively, and connect $CD$. Then
$$
\begin{array}{l}
\overrightarrow{OA}+\overrightarrow{OB}=2 \overrightarrow{OD}, \\
\overrightarrow{OB}+\overrightarrow{OC}=2 \overrightarrow{OE} .
\end{array}
$$
(1) - (2) $\times 3$ gives
$$
\begin{array}{l}
0=\overrightarrow{OA}-2 \overrightarrow{OB}-3 \overrightarrow{OC} \\
=2 \overrightarrow{OD}-6 \overrightarrow{OE} . \\
\text { Therefore, } \overrightarrow{OD}=3 \overrightarrow{OE} .
\end{array}
$$
Thus, $\overrightarrow{OD}$ and $\overrightarrow{OE}$ are collinear, and $|\overrightarrow{OD}|=3|\overrightarrow{OE}|$.
Hence, $|\overrightarrow{DE}|=2|\overrightarrow{OE}|$.
$$
\text { Therefore, } \frac{S_{\triangle BCD}}{S_{\triangle OBC}}=\frac{2}{1}=2, \frac{S_{\triangle ABC}}{S_{\triangle OBC}}=\frac{2 S_{\triangle BCD}}{S_{\triangle OBC}}=4 \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (14 points) In the Cartesian coordinate system $x O y$, points with both integer coordinates are called integer points. Given $O(0,0), A(2,1)$, and $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$. If $S_{\triangle O M M}=3$, find the number of integer points $M$ that satisfy this condition.
|
9. Connect $O A$. It is easy to know that there are two integer points $M_{1}(-6,0)$ and $M_{2}(6,0)$ on the $x$-axis inside the ellipse that satisfy the problem.
Draw two lines $l_{1}$ and $l_{2}$ parallel to the line $O A$ through points $M_{1}$ and $M_{2}$, respectively.
According to the principle that triangles with equal bases and heights have equal areas, all integer points $M$ that meet the conditions lie on the lines $l_{1}$ and $l_{2}$. It is easy to know
$$
k_{O A}=\frac{1-0}{2-0}=\frac{1}{2} \text {. }
$$
Therefore, the equations of the lines $l_{1}$ and $l_{2}$ are
$$
y=\frac{1}{2}(x+6), y=\frac{1}{2}(x-6) \text {. }
$$
Given that $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$, we have
$$
\frac{x^{2}}{200}+\frac{y^{2}}{8}<1 \text {. }
$$
Solve $\left\{\begin{array}{l}\frac{x^{2}}{200}+\frac{y^{2}}{8}<1, \\ y=\frac{1}{2}(x+6)\end{array}\right.$ and $\left\{\begin{array}{l}\frac{x^{2}}{200}+\frac{y^{2}}{8}<1, \\ y=\frac{1}{2}(x-6),\end{array}\right.$ to get $-10<x<-\frac{10}{29}, \frac{10}{29}<x<10$.
Since $M$ is an integer point and lies on the lines $l_{1}$ and $l_{2}$, $x$ must be an even number.
Therefore, in $-10<x<-\frac{10}{29}$ and $\frac{10}{29}<x<10$, $x$ has 4 even numbers each.
Thus, the number of integer points that meet the conditions is 8.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Two, (50 points) The four-digit numbers $m$ and $n$ are reverse positive integers of each other, and $m+n=18 k+9\left(k \in \mathbf{N}_{+}\right), m$ and $n$ have 16 and 12 positive divisors (including 1 and themselves), respectively. The prime factors of $n$ are also prime factors of $m$, but $n$ has one fewer prime factor than $m$. Find all possible values of $m$.
|
Let $m=\overline{a b c d}, a d \neq 0$. Then $n=\overline{d c b a}$.
Given $m+n=9(2 k+1)$, then $9 \mid (m+n)$.
Thus, $9 \mid [(1000 a+100 b+10 c+d)+$
$$
\begin{array}{c}
(1000 d+100 c+10 b+a)], \\
9 \mid 2(a+b+c+d), 9 \mid (a+b+c+d).
\end{array}
$$
Therefore, $9 \mid m, 9 \mid n$.
Since $m+n$ is odd, it follows that $m$ and $n$ are of opposite parity.
If $n$ is even, i.e., $2 \mid n$, then $2 \mid m$, making $m$ even, which is a contradiction.
Thus, $m$ is even, and $n$ is odd.
Let the number of 3's in the prime factorization of $m$ be $\alpha_{1}$, and the number of 2's be $\alpha_{2}$. Then $\alpha_{1} \geqslant 2, \alpha_{2} \geqslant 1$.
By the divisor count theorem,
$\left[\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\right] \mid 16$.
Thus, $\left(\alpha_{1}+1\right) \mid 8, \alpha_{1}+1 \geqslant 3$.
So, $\alpha_{1}+1=4$ or $8, \alpha_{1}=3$ or 7.
Hence, $m$ has at most three prime factors.
Thus, $n$ has at most two prime factors, and 3 is one of the prime factors of $n$.
If $n$ has only one prime factor, then this prime factor is 3. Thus, $n=3^{11}>10000$, which contradicts the fact that $n$ is a four-digit number.
Therefore, $n$ has two prime factors.
Let the other prime factor of $n$ be $p$.
Since $9 \mid n$, we have
$n=3^{2} p^{3}$ or $3^{3} p^{2}$ or $3^{5} p(p>3)$.
Thus, $m=2^{\alpha_{2}} 3^{\alpha_{1}} p^{\alpha_{3}}\left(\alpha_{1}, \alpha_{2}, \alpha_{3} \in \mathbf{N}_{+}\right)$.
Also, $\left(\alpha_{2}+1\right)\left(\alpha_{1}+1\right)\left(\alpha_{3}+1\right)=16$, so $\alpha_{1}=3, \alpha_{2}=\alpha_{3}=1$, i.e., $m=54 p$.
Since $m \geqslant 1000$, we have $p \geqslant 19$.
At this point, $3^{2} p^{3}$ is greater than 9999.
When $p=19$, $3^{3} p^{2}=9747$.
But $m=54 p=1026$ are not reverse numbers, so $p \geqslant 23$. At this point, $3^{3} p^{2}>9999$.
Therefore, $n=3^{5} p$. Thus,
$\frac{m}{n}=\frac{2}{9} \Rightarrow 9 m=2 n$,
$9(1000 a+100 b+10 c+d)$
$=2(1000 d+100 c+10 b+a)$,
$818 a+80 b=181 d+10 c$.
$818 a \leqslant 181 \times 9+10 \times 9=1719<3 \times 818$.
Thus, $a<3$.
Since $n$ is odd, $a$ must be odd. Hence, $a=1$.
From equation (1),
$d=\frac{818+80 b-10 c}{181} \geqslant \frac{818-10 \times 9}{181}=4 \frac{4}{181}$.
Since $m$ is even, $d$ must be even.
Thus, $d=6$ or 8.
When $d=6$, from equation (1),
$880 b-110 c=2948$.
Since $51 \mid (80 b-10 c)$, we have $d=8$.
Thus, $8 b-c=63, 8 b=63+c \geqslant 63, b \geqslant 7 \frac{7}{8}$.
Thus, $b=8$ or 9.
When $b=8$, $c=1$;
When $b=9$, $c=9$.
Thus, $m=1818$ or 1998.
Since $271 \mid m$, $m \neq 1818$.
Also, $m=1998=2 \times 3^{3} \times 37, n=8991=$ $3^{5} \times 37$ satisfies the conditions.
Therefore, $m=1998$.
|
1998
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Does there exist 4098 sets
\[
\begin{aligned}
B_{i} & =\left\{a_{i 1}, a_{i 2}, \cdots, a_{i n}\right\}(i=1,2, \cdots, 4098, \\
a_{i j} \in A_{j} & =\{3 j-2,3 j-1,3 j\}, j=1,2, \cdots, 12)
\end{aligned}
\]
satisfying the following three conditions:
(1) \( B_{i} \cap B_{i+1}=\varnothing(i=1,2, \cdots, 4098\), with the convention \( B_{4099}=B_{1} \));
(2) When \( i \neq j \), \( B_{i} \neq B_{j} \);
(3) When \( 2 \leqslant|i-j| \leqslant 4096 \), \( B_{i} \cap B_{j} \neq \varnothing \)?
|
Let $(s, t)$ denote a set circle with $s$ sets $A_{1}, A_{2}, \cdots, A_{s}$, and $t$ sets $B_{1}, B_{2}, \cdots, B_{t}$ that satisfy the given conditions. Use $\left(B_{i}, p\right)$ to denote a set formed by all elements of $B_{i}$ and the $p$-th element of $A_{i+1}$.
If $B_{1}, B_{2}, \cdots, B_{t}$ satisfy the given conditions, then
(i) when $t$ is odd,
$$
\begin{array}{l}
\quad\left(B_{1}, 1\right),\left(B_{2}, 2\right),\left(B_{3}, 1\right),\left(B_{4}, 2\right), \cdots, \\
\left(B_{t}, 1\right),\left(B_{1}, 2\right),\left(B_{2}, 1\right),\left(B_{3}, 2\right), \cdots, \\
\left(B_{t}, 2\right)
\end{array}
$$
is a $(s+1,2 t)$ set circle (the second component alternates between 1 and 2);
(ii) when $t$ is even,
$$
\begin{array}{l}
\left(B_{1}, 3\right),\left(B_{2}, 1\right),\left(B_{3}, 2\right),\left(B_{4}, 1\right), \\
\left(B_{5}, 2\right), \cdots,\left(B_{k-2}, 1\right),\left(B_{k-1}, 2\right), \\
\left(B_{k}, 3\right),\left(B_{k-1}, 1\right),\left(B_{k-2}, 2\right), \cdots, \\
\left(B_{5}, 1\right),\left(B_{4}, 2\right),\left(B_{3}, 1\right),\left(B_{2}, 2\right)
\end{array}
$$
is a $(s+1,2 k-2)$ set circle, where $k$ is the largest even number not greater than $t$, and except for the second component of $B_{1}$ and $B_{k}$ being 3, the rest of the second components alternate between 1 and 2.
Clearly, there is a $(1,3)$ set circle.
According to the method of constructing new set circles, we can sequentially obtain the set circles:
$$
\begin{array}{l}
(2,6),(3,10),(4,18),(5,34), \\
(6,66),(7,130),(8,258),(9,514), \\
(10,1026),(11,2050),(12,4098) .
\end{array}
$$
Therefore, there exist 12 sets $A_{1}, A_{2}, \cdots, A_{12}$ and 4098 sets $B_{1}, B_{2}, \cdots, B_{4098}$ forming a set circle.
|
4098
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. Calculate $\frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9$ The value equals $\qquad$ .
|
$\begin{array}{l} \text { II.1.3 481. } \\ \frac{45.1^{3}-13.9^{3}}{31.2}+45.1 \times 13.9 \\ = \frac{(45.1-13.9)\left(45.1^{2}+45.1 \times 13.9+13.9^{2}\right)}{45.1-13.9}+ \\ 45.1 \times 13.9 \\ = 45.1^{2}+2 \times 45.1 \times 13.9+13.9^{2} \\ =(45.1+13.9)^{2}=59^{2}=3481 .\end{array}$
|
3481
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]$ equals $\qquad$ (where [x] denotes the greatest integer not exceeding the real number $x$).
|
2. 6 .
Given $0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$, then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]=0$ or 1. From the problem, we know that 18 of them are equal to 1. Therefore,
$$
\begin{array}{l}
{\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\left[a+\frac{11}{30}\right]=0,} \\
{\left[a+\frac{12}{30}\right]=\left[a+\frac{13}{30}\right]=\cdots=\left[a+\frac{29}{30}\right]=1}
\end{array}
$$
Thus, $0<a+\frac{11}{30}<1,1 \leqslant a+\frac{12}{30}<2$
$\Rightarrow 18 \leqslant 30 a<19 \Rightarrow 6 \leqslant 10 a<\frac{19}{3}$
$\Rightarrow[10 a]=6$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 1, in $\square A B C D$, $A D=a, C D=b$, draw the heights $h_{a}, h_{b}$ from point $B$ to sides $A D$ and $C D$ respectively. Given $h_{a} \geqslant a, h_{b} \geqslant b$, and the diagonal $A C=20$. Then the area of $\square A B C D$ is $\qquad$
|
3. 200 .
Given $h_{a} \geqslant a, h_{b} \geqslant b$, in addition, $a \geqslant h_{b}, b \geqslant h_{a}$, then $h_{a} \geqslant a \geqslant h_{b} \geqslant b \geqslant h_{a}$, i.e., $h_{a}=a=h_{b}=b$, which means that quadrilateral ABCD is a square.
Since the diagonal of this square $AC=20$, the area of $\square ABCD$ is 200.
|
200
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}=
$$
$\qquad$ .
|
5. 2 .
Notice that
$$
\begin{array}{l}
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \\
= \sqrt{(x-1)+2 \sqrt{x-1}+1}+ \\
\sqrt{(x-1)-2 \sqrt{x-1}+1} \\
= \sqrt{(\sqrt{x-1}+1)^{2}}+\sqrt{(\sqrt{x-1}-1)^{2}} \\
=|\sqrt{x-1}+1|+|\sqrt{x-1}-1| .
\end{array}
$$
Since $1 \leqslant x \leqslant 2$, we have $\sqrt{x-1}-1 \leqslant 0$.
$$
\begin{array}{l}
\text { Therefore, } \sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \\
=\sqrt{x-1}+1-(\sqrt{x-1}-1)=2 \text {. }
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given an integer $n \geqslant 3$. Find the smallest positive integer $k$, such that there exists a $k$-element set $A$ and $n$ pairwise distinct real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1}+x_{2}, x_{2}+x_{3}, \cdots$, $x_{n-1}+x_{n}, x_{n}+x_{1}$ all belong to $A$. (Xiong Bin provided)
|
2. Let $x_{1}+x_{2}=m_{1}, x_{2}+x_{3}=m_{2}, \cdots \cdots$
$x_{n-1}+x_{n}=m_{n-1}, x_{n}+x_{1}=m_{n}$.
First, $m_{1} \neq m_{2}$, otherwise, $x_{1}=x_{3}$, which is a contradiction.
Similarly, $m_{i} \neq m_{i+1}\left(i=1,2, \cdots, n, m_{n+1}=m_{1}\right)$.
Thus, $k \geqslant 2$.
If $k=2$, let $A=\{a, b\}(a \neq b)$, such that
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{2}+x_{3}=b, \\
\cdots \ldots . \\
x_{n-1}+x_{n}=b, \\
x_{n}+x_{1}=a
\end{array} \text { ( } n\right. \text { is odd), } \\
\text { or }\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{2}+x_{3}=b, \\
\cdots \cdots \\
x_{n-1}+x_{n}=a, \\
x_{n}+x_{1}=b
\end{array} \quad(n \text { is even). }\right.
\end{array}
$$
For the system (1), $x_{n}=x_{2}$, which is a contradiction.
For the system (2),
$$
\begin{array}{l}
\frac{n}{2} a=\left(x_{1}+x_{2}\right)+\left(x_{3}+x_{4}\right)+\cdots+\left(x_{n-1}+x_{n}\right) \\
=\left(x_{2}+x_{3}\right)+\left(x_{4}+x_{5}\right)+\cdots+\left(x_{n}+x_{1}\right) \\
=\frac{n}{2} b,
\end{array}
$$
Thus, $a=b$, which is a contradiction.
For $k=3$, an example can be constructed as follows:
$$
\text { Let } x_{2 k-1}=k, x_{2 k}=n+1-k(k=1,2, \cdots) \text {. }
$$
Then, when $n$ is even,
$$
x_{\mathrm{i}}+x_{\mathrm{i}+1}=\left\{\begin{array}{ll}
n+1, & i \text { is odd; } \\
n+2, & i \text { is even and } i<n ; \\
\frac{n}{2}+2, & i=n\left(x_{n+1}=x_{1}\right) .
\end{array}\right.
$$
When $n$ is odd,
$$
x_{i}+x_{i+1}=\left\{\begin{array}{ll}
n+1, & i \text { is odd and } i<n ; \\
n+2, & i \text { is even; } \\
\frac{n-1}{2}+2, & i=n\left(x_{n+1}=x_{1}\right) .
\end{array}\right.
$$
In summary, the minimum value of $k$ is 3.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. There are $n (n>12)$ people participating in a mathematics invitational competition. The test consists of fifteen fill-in-the-blank questions, with each correct answer worth 1 point and no answer or a wrong answer worth 0 points. Analyzing every possible score situation, it is found that as long as the sum of the scores of any 12 people is no less than 36 points, then among these $n$ people, at least 3 people have answered at least three of the same questions correctly. Find the minimum possible value of $n$.
（Liu Shixiong provided）
|
7. The minimum possible value of $n$ is 911.
(1) First, prove: 911 satisfies the condition.
If each student answers at least three questions correctly, since the number of different ways a student can answer three questions correctly is $\mathrm{C}_{15}^{3}=455$, then if there are 911 students participating, by the pigeonhole principle, at least 3 students must have answered the same three questions correctly.
If a student answers no more than 2 questions correctly, then among the remaining students, the number of students who answer no more than three questions correctly cannot exceed 10 (otherwise, the total score of these students and the first student would be less than 36 points). For the remaining $911-11=900$ students, each student answers at least 4 questions correctly.
Since $\mathrm{C}_{4}^{3} \times 900 > 455 \times 2$, there must be at least 3 students who answered the same three questions correctly.
(2) If there are 910 students participating, these students can be divided into 455 groups, with each group consisting of two students who answered the same three questions correctly. In this case, the condition of the problem is not satisfied.
In conclusion, $n_{\min }=911$.
|
911
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Place 3 identical white balls, 4 identical red balls, and 5 identical yellow balls into three different boxes, allowing some boxes to contain balls of different colors. The total number of different ways to do this is (answer in numbers).
允许有的盒子中球的颜色不全的不同放法共有种 (要求用数字做答).
Allowing some boxes to contain balls of different colors, the total number of different ways to do this is (answer in numbers).
|
$\begin{array}{l}\text { 9. } 3150 \text {. } \\ \mathrm{C}_{5}^{2} \cdot \mathrm{C}_{6}^{2} \cdot \mathrm{C}_{7}^{2}=3150 \text {. }\end{array}$
|
3150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$
|
11. 9 .
Since $(\sqrt{2}+\sqrt{3})^{2010}+(\sqrt{2}-\sqrt{3})^{2010}$ is an integer, therefore, the fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is
$$
\begin{array}{l}
1-(\sqrt{2}-\sqrt{3})^{2010} . \\
\text { Also } 0<(\sqrt{2}-\sqrt{3})^{2010}<0.2^{1005}<(0.008)^{300} \text {, then } \\
0.9<1-(\sqrt{2}-\sqrt{3})^{2010}<1 .
\end{array}
$$
It follows that the first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is 9 .
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11.6. A $10 \times 10$ chessboard has $k$ rooks. A square on the board that can be attacked by a rook is called "dangerous" (the square occupied by the rook itself is also considered dangerous). If removing any rook results in at least one dangerous square becoming safe, find the maximum possible value of $k$.
|
11. 6. $k_{\max }=16$.
Consider a chessboard with $k$ rooks satisfying the problem's conditions. There are two cases to consider.
(1) Each row (column) has a rook.
In this case, all squares are dangerous. If there is a row (column) with at least two rooks, removing one of these rooks will still leave all squares dangerous. This is a contradiction.
Therefore, each row (column) has only one rook. Thus, $k=10$.
(2) At least one row and one column have no rooks.
For each rook, in its row (or column), it is the only rook (otherwise, removing this rook does not reduce the number of dangerous squares). Associate this rook with this row (or column). If all rooks are associated with 9 rows (or columns), then $k=9$; if the total number of rows and columns associated with all rooks is no more than 8, then $k \leqslant 8+8=16$.
An example for $k=16$.
The 16 rooks are located at
$(1, i) 、(i, 1)(i=3,4, \cdots, 10)$.
Removing the rook from square $(1, i)$ makes square $(2, i)$ change from dangerous to safe; removing the rook from square $(i, 1)$ makes square $(i, 2)$ change from dangerous to safe.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The first 24 digits of $\pi$ are
3. 14159265358979323846264 .
Let $a_{1}, a_{2}, \cdots, a_{24}$ be any permutation of these 24 digits. Then
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\left(a_{23}-a_{24}\right) \\
\equiv \quad(\bmod 2) .
\end{array}
$$
$(\bmod 2)$.
|
3. 0 .
In the first 24 digits of $\pi$, there are 13 odd numbers. When these 13 odd numbers are placed into 12 parentheses, by the pigeonhole principle, there must be two in the same parenthesis, making the difference in this parenthesis even. Thus, the product $\left(a_{1}-a_{2}\right)\left(a_{3}-a_{4}\right) \cdots\left(a_{23}-a_{24}\right)$ is even, and an even number has a remainder of 0 when divided by 2.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If a positive integer is written on each face of a cube, and then a number is written at each vertex, which is equal to the product of the two integers on the faces passing through that vertex, then, when the sum of the numbers at the vertices of the cube is 290, the sum of the numbers on the faces of the cube is
|
3. 36 .
Let the numbers on each face of the cube be $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, $x_{5}$, $x_{6}$, and the numbers written at the vertices be
$$
\begin{array}{l}
x_{1} x_{2} x_{5}, x_{2} x_{3} x_{5}, x_{3} x_{4} x_{5}, x_{4} x_{1} x_{5}, x_{1} x_{2} x_{6}, \\
x_{2} x_{3} x_{6}, x_{3} x_{4} x_{6}, x_{4} x_{1} x_{6} .
\end{array}
$$
$$
\begin{array}{l}
\text { Then } 290=x_{1} x_{2} x_{5}+x_{2} x_{3} x_{5}+x_{3} x_{4} x_{5}+x_{4} x_{1} x_{5}+ \\
\quad x_{1} x_{2} x_{6}+x_{2} x_{3} x_{6}+x_{3} x_{4} x_{6}+x_{4} x_{1} x_{6} \\
=\left(x_{5}+x_{6}\right)\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{1}\right) \\
=\left(x_{1}+x_{3}\right)\left(x_{2}+x_{4}\right)\left(x_{5}+x_{6}\right) .
\end{array}
$$
Since $290=2 \times 5 \times 29$, it is clear that in the product $\left(x_{1}+x_{3}\right)\left(x_{2}+x_{4}\right)\left(x_{5}+x_{6}\right)$, one factor is 2, one factor is 5, and the other factor is 29.
Therefore, $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=2+5+29=36$.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Given an isosceles triangle with a vertex angle less than $60^{\circ}$, the lengths of its three sides are all positive integers. Construct a square outward on each side, such that the sum of the areas of the three squares is 2009. Find the perimeter of this isosceles triangle.
|
Let the length of the legs of the isosceles triangle be $a$, and the length of the base be $b$. According to the problem, we have
$$
2 a^{2}+b^{2}=2009 \text{. }
$$
Therefore, $a^{2}=\frac{2009-b^{2}}{2}\frac{2009}{3}>669$.
Thus, $669<a^{2}<1005 \Rightarrow 25<a<32$.
Since $a$ and $b$ are positive integers, it is verified that:
When $a=28$,
$$
b^{2}=2009-2 \times 28^{2}=441=21^{2} \Rightarrow b=21 \text{; }
$$
When $a=26$, $b^{2}=657$ is not a perfect square,
so, $b$ has no integer solution.
Similarly, when $a=27, 29, 30, 31$, $b$ has no integer solution.
In summary, the length of the legs of the required isosceles triangle is 28, the length of the base is 21, and the perimeter is $28+28+21=77$.
|
77
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $a, b, c \in \mathbf{R}$, and $a+b+c=3$. Then the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is $\qquad$
|
3. 9 .
Since the function $y=3^{x}$ is an increasing function on $(-\infty,+\infty)$, by the property of increasing functions, for any $x_{1}$, $x_{2}$, we have $\left(x_{1}-x_{2}\right)\left(3^{x_{1}}-3^{x_{2}}\right) \geqslant 0$, so,
$$
(a-1)\left(3^{a}-3^{1}\right) \geqslant 0,
$$
which means $3^{a} a-3^{\circ} \geqslant 3 a-3$.
Similarly, $3^{b} b-3^{b} \geqslant 3 b-3$, $3^{c} c-3^{c} \geqslant 3 c-3$.
Adding the three inequalities and transforming them, we get
$$
\begin{array}{l}
3^{a} a+3^{b} b+3^{c} c \\
\geqslant 3^{a}+3^{b}+3^{c}+3(a+b+c)-9 \\
=3^{a}+3^{b}+3^{c}+3 \times 3-9=3^{a}+3^{b}+3^{c} . \\
\text { Also, } 3^{a}+3^{b}+3^{c} \geqslant 3 \sqrt[3]{3^{c} \times 3^{b} \times 3^{c}}=3^{2}=9,
\end{array}
$$
with equality holding if and only if $a=b=c=1$.
Therefore, the minimum value of $3^{a} a+3^{b} b+3^{c} c$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 The number of positive integer solutions to the system of equations $\left\{\begin{array}{l}x y+y z=63, \\ x z+y z=23\end{array}\right.$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
|
From the second equation, we get $z(x+y)=23$, and since 23 is a prime number, we can obtain $\left\{\begin{array}{l}z=1, \\ x+y=23\end{array}\right.$ or $\left\{\begin{array}{l}z=23, \\ x+y=1 \text {. }\end{array}\right.$ According to the problem, $x+y=1$ is not valid, so we discard it. Substituting $z=1$ into the given equations, we get
$$
\left\{\begin{array}{l}
x y+y=63, \\
x+y=23 .
\end{array}\right.
$$
Solving this, we get $(x, y)=(20,3),(2,21)$.
Therefore, the solutions to the original system of equations are
$$
(x, y, z)=(20,3,1),(2,21,1) .
$$
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given $[x]$ represents the greatest integer not exceeding the real number $x$. If $P=\sum_{i=1}^{2010}\left[\frac{6^{i}}{7}\right]$, then the remainder when $P$ is divided by 35 is
|
6. 20 .
First, consider $S=\frac{6}{7}+\frac{6^{2}}{7}+\cdots+\frac{6^{2010}}{7}$.
In equation (1), no term is an integer, but the sum of any two adjacent terms is an integer (since $\frac{6^{k}}{7}+\frac{6^{k+1}}{7}=6^{k}$ $(k \in \mathbf{Z})$ is an integer).
If the sum of two non-integer numbers is an integer, then the sum of their integer parts is 1 less than the sum of the numbers themselves. Since there are 2010 terms in equation (1), the value of the desired sum $P$ is 1005 less than $S$.
Using the formula for the sum of a geometric series, we get
$$
\begin{array}{l}
S=\frac{6\left(6^{2010}-1\right)}{35} . \\
\text { Therefore, } 35 P=35(S-1005) \\
=6^{2011}-6-35 \times 1005 . \\
\text { Also, } 6^{2011}=6(35+1)^{1005} \\
=6 \times 35^{2} M+6 \times 35 \times 1005+6\left(M \in \mathbf{N}_{+}\right) \\
\Rightarrow 35 P=6 \times 35^{2} M+5 \times 35 \times 1005 \\
\Rightarrow P=6 \times 35 M+5 \times 1005 .
\end{array}
$$
Therefore, the remainder when $P$ is divided by 35 is equal to
$$
5 \times 1005=5025
$$
divided by 35.
Thus, the remainder when $P$ is divided by 35 is 20.
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $P_{n}(k)$ denote the number of permutations of $\{1,2, \cdots, n\}$ with $k$ fixed points. Let $a_{t}=\sum_{k=0}^{n} k^{t} P_{n}(k)$. Then
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=
\end{array}
$$
|
8. 0 .
Notice
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1} \\
=\sum_{k=0}^{n}\left(k^{5}-10 k^{4}+35 k^{3}-50 k^{2}+25 k\right) P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] . \\
\frac{n!}{(n-k)!k!} P_{n-k}(0) \\
=\sum_{k=0}^{n}\left[\frac{n!}{(n-5)!} \cdot \frac{(n-5)!}{(k-5)!(n-k)!} P_{(n-5)-(k-5)}(0)+\right. \\
\left.n \cdot \frac{(n-1)!}{(k-1)!(n-k)!} P_{(n-1)-(k-1)}(0)\right] \\
=\frac{n!}{(n-5)!} \sum_{k=0}^{n-5} P_{n-5}(k)+n \sum_{k=0}^{n-1} P_{n-1}(k) .
\end{array}
$$
Since $a_{0}=\sum_{k=0}^{n} P_{n}(k)=n!$, therefore,
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=\frac{n!}{(n-5)!}(n-5)!+n \cdot(n-1)!-2 \cdot n! \\
=0 .
\end{array}
$$
|
0
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (15 points) Given seven different points on a circle, vectors are drawn from any one point to another (for points $A$ and $B$, if vector $\overrightarrow{A B}$ is drawn, then vector $\overrightarrow{B A}$ is not drawn). If the four sides of a convex quadrilateral determined by any four points are four consecutive vectors, it is called a "zero quadrilateral". Find the maximum number of zero quadrilaterals among the convex quadrilaterals formed by any four of these seven points.
|
11. Let the seven points on the circumference be $P_{1}, P_{2}, \cdots, P_{7}$. The number of vectors starting from point $P_{i} (i=1,2, \cdots, 7)$ is $x_{i} (i=1,2, \cdots, 7)$, then $0 \leqslant x_{i} \leqslant 6$, and
$$
\sum_{i=1}^{7} x_{i}=\mathrm{C}_{7}^{2}=21 \text {. }
$$
First, find the minimum number of "non-zero quadrilaterals" among the $\mathrm{C}_{7}^{4}=35$ convex quadrilaterals.
It is easy to know that non-zero quadrilaterals only have the following two scenarios:
The first type: A non-zero quadrilateral with exactly one vertex as the starting point of two vectors (as shown in Figure 3).
The second type: A non-zero quadrilateral with exactly two vertices as the starting points of two vectors (as shown in Figure 4).
(1) When the entire figure contains the second type of non-zero quadrilateral, it is easy to know that on the basis of the four vertices of the second type of non-zero quadrilateral, each additional point will increase four convex quadrilaterals compared to the original quadrilateral, and it is easy to know from the drawing that these four convex quadrilaterals are all non-zero quadrilaterals.
Thus, in the 35 convex quadrilaterals in the entire seven-point figure, if there is one second type of non-zero quadrilateral, then on the basis of this non-zero quadrilateral, adding three points will increase at least $(4 \times 3=) 12$ non-zero quadrilaterals. Therefore, at this time, there are at least 13 non-zero quadrilaterals in the figure.
(2) When the seven-point figure does not contain the second type of non-zero quadrilateral, i.e., it only contains the first type of non-zero quadrilateral and zero quadrilateral, the problem can be converted to just finding the minimum number of the first type of non-zero quadrilateral.
And the four vertices of the first type of non-zero quadrilateral have exactly one vertex as the starting point of two vectors of this non-zero quadrilateral, and these two vectors form a diagonal vector. When these three vectors are determined, then a non-zero quadrilateral is determined. Therefore, when the figure does not contain the second type of non-zero quadrilateral, the number of the first type of non-zero quadrilateral in the figure
$$
M \geqslant \sum_{i=1}^{7} \mathrm{C}_{x_{i}}^{3} .
$$
Since $x_{1}+x_{2}+\cdots+x_{7}=21$, $M$ has a minimum value.
If there is some $x_{i}$ in $x_{1}, x_{2}, \cdots, x_{7}$ that is less than 3, then there must be some $x_{j}$ that is greater than or equal to 4.
And $C_{4}^{3}=4, C_{5}^{3}=10, C_{6}^{3}=20$, and $C_{3}^{3}=1$, then $M \geqslant 7$.
If $x_{1}, x_{2}, \cdots, x_{7}$ are all not less than 3, by
$$
x_{1}+x_{2}+\cdots+x_{7}=21
$$
we know, only $x_{1}=x_{2}=\cdots=x_{7}=3, M=7 \mathrm{C}_{3}^{3}=7$.
Below, through drawing, it is shown that the figure with $M=7$ exists (as shown in Figure 5).
Figure 5 contains seven non-zero quadrilaterals.
(1) $P_{1} P_{2} P_{3} P_{4}$, (2) $P_{2} P_{3} P_{4} P_{5}$,
(3) $P_{3} P_{4} P_{5} P_{6}$, (4) $P_{4} P_{5} P_{6} P_{7}$,
(5) $P_{5} P_{6} P_{7} P_{1}$, (6) $P_{6} P_{7} P_{1} P_{2}$,
(7) $P_{7} P_{1} P_{2} P_{3}$.
Therefore, the minimum value of $M$ is 7.
Thus, the maximum number of zero quadrilaterals is $C_{7}^{4}-7=28$.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Try to find the smallest positive integer $M$, such that the sum of all positive divisors of $M$ is 4896.
|
Let $M=\prod_{i=1}^{n} p_{i}^{\alpha_{i}}\left(p_{i}\right.$ be a prime, $\alpha_{i} \in \mathbf{N}_{+}, i=1,2, \cdots, n)$. Denote
$$
f(p, \alpha)=\sum_{k=0}^{\alpha} p^{k}\left(p\right.$ be a prime, $\left.\alpha \in \mathbf{N}_{+}\right)$.
From the problem, we know
$$
\prod_{i=1}^{n} f\left(p_{i}, \alpha_{i}\right)=4896=2^{5} \times 3^{2} \times 17.
$$
If $n=1$, then $f\left(p_{1}, \alpha_{1}\right)=4896$, which implies
$$
p_{1} \mid 4895.
$$
Since $4895=5 \times 11 \times 89 \Rightarrow p_{1} \in\{5,11, 89\}$.
$$
\begin{array}{l}
\text { But } f(5,5)=1.
\end{array}
$$
Next, we divide the positive divisors of 4896 into the following two sets:
$$
\begin{aligned}
A= & \{1,2,9,16,17,34,36,51,96,136, \\
& 144,153,204,272,288,306,408, \\
& 544,612,816,1632,4896\}, \\
B= & \{3,4,6,8,12,18,24,32,48,68,72, \\
& 102,1224,2448\}.
\end{aligned}
$$
Similarly, we know: when $t \in A$, $f(p, \alpha)=t$ has no prime solution $p$; when $t \in B$, $f(p, \alpha)=t$ has only the solution
$$
(p, \alpha)=(t-1,1).
$$
From equation (1), assume $17 \mid f\left(p_{1}, \alpha_{1}\right)$. Then
$$
f\left(p_{1}, \alpha_{1}\right) \in\{68,102,1224,2448\}.
$$
(1) If $f\left(p_{1}, \alpha_{1}\right)=2448$, then
$$
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=2,
$$
which has no solution.
(2) If $f\left(p_{1}, \alpha_{1}\right)=1224$, then
$$
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=4.
$$
Thus, $M_{1}=1223 \times 3=3669$.
(3) If $f\left(p_{1}, \alpha_{1}\right)=102$, then
$$
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=48=4 \times 12=6 \times 8.
$$
Thus, $M_{2}=101 \times 48=4747$,
$$
\begin{array}{l}
M_{3}=101 \times 3 \times 11=3333, \\
M_{4}=101 \times 5 \times 7=3535 ;
\end{array}
$$
(4) If $f\left(p_{1}, \alpha_{1}\right)=68$, then
$$
\begin{array}{l}
\prod_{i=2}^{n} f\left(p_{i}, \alpha_{i}\right)=72=3 \times 24 \\
=4 \times 18=6 \times 12=3 \times 4 \times 6.
\end{array}
$$
Thus, $M_{5}=67 \times 71=4757$,
$$
\begin{array}{l}
M_{6}=67 \times 2 \times 23=3082, \\
M_{7}=67 \times 3 \times 17=3417, \\
M_{8}=67 \times 5 \times 11=3685, \\
M_{9}=67 \times 2 \times 3 \times 5=2010.
\end{array}
$$
In summary, $M_{\min }=2010$.
|
2010
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Let $n$ be a positive integer, and let $S_{n}$ be a subset of the set $A_{n}=\left\{m \mid m \in \mathbf{N}_{+}\right.$, and $\left.m \leqslant n\right\}$. Moreover, the difference between any two numbers in $S_{n}$ is not equal to 4 or 7. If the maximum number of elements in $S_{n}$ is denoted by $M_{n}$ (for example, $M_{1}=1, M_{2}$ =2), find $\sum_{n=1}^{2012} M_{n}$.
|
It is known that the difference between any two numbers among $1,4,6,7,9$ is not 4 or 7. Adding 11 to each of these numbers gives $12, 15, 17, 18, 20$, which clearly also have the same property, and the difference between any of these numbers and any of the first five numbers is also not 4 or 7. By this reasoning, for any fixed positive integer $n$, in each set of 11 consecutive numbers, five numbers can be chosen, and from the remaining $i=n-11 \times \left[\frac{n}{11}\right]$ numbers, $R_{i}$ numbers can be chosen, forming the set $S_{n}$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
We now prove that $S_{n}$ cannot contain more numbers.
If this were not the case, then in the set $S_{n}$, at least one of the groups of 11 numbers could have six numbers chosen such that the difference between any two of them is not 4 or 7.
Consider the numbers $1,2, \cdots, 11$ and divide them into five groups:
$$
(1,8),(2,6),(3,10),(5,9),(4,7,11),
$$
where at least one group must have two numbers chosen. Clearly, from the first four groups, no more than one number can be chosen from each, so the only possibility is to choose 4 and 7 from the fifth group $(4,7,11)$.
Thus, from $(1,8)$, only 1 can be chosen, from $(5,9)$, only 9 can be chosen, and from $(2,6)$, only 6 can be chosen.
This means that from $(3,10)$, neither 3 nor 10 can be chosen, making it impossible to select a sixth number. Therefore,
$$
M_{n}=5 \times \left[\frac{n}{11}\right]+R_{i}.
$$
When $i=0, 1, 2, 3$, $R_{i}=i$;
When $i=4, 5, 6, 7, 8$, for example, if $i=8$, consider the remaining numbers $1,2, \cdots, 8$ and divide them into four groups:
$$
(1,5),(2,6),(3,7),(4,8),
$$
one number can be chosen from each group, totaling 4 numbers, so $R_{i}=4$;
When $i=9, 10$, for example, if $i=9$, consider the remaining numbers $1,2, \cdots, 9$ and divide them into five groups:
$$
(1,5),(2,6),(3,7),(4,8),(9),
$$
the numbers 1, 3, 4, 6, 9 can be chosen, so $R_{i}=5$.
Since $2012=11 \times 182+10$, we have
$$
\begin{array}{l}
\sum_{n=1}^{2012} M_{n}=\sum_{k=0}^{182} \sum_{i=0}^{10} M_{11 k+i} \\
=\sum_{k=0}^{182}(11 \times 5 k+1+2+3+4 \times 5+5 \times 2) \\
=55 \times \frac{182 \times 183}{2}+183 \times 36 \\
=922503 .
\end{array}
$$
|
922503
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
265 It is known that for all positive integers $n$,
$$
\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right) \geqslant \frac{k}{2} \sqrt[3]{19 n+8}
$$
always holds. Try to find the maximum value of $k$.
|
Let $T_{n}=\prod_{i=1}^{n}\left(1+\frac{1}{3 i-1}\right)$. Then
$$
\frac{T_{n+1}}{T_{n}}=1+\frac{1}{3 n+2}=\frac{3 n+3}{3 n+2} \text {. }
$$
Given $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$, we have $k \leqslant \frac{2 T_{n}}{\sqrt[3]{19 n+8}}$.
Let $f(n)=\frac{2 T_{n}}{\sqrt[3]{19 n+8}}$. Then
$$
\begin{array}{l}
f(n+1)=\frac{2 T_{n+1}}{\sqrt[3]{19 n+27}}, \\
\frac{f(n+1)}{f(n)}=\frac{2 T_{n+1}}{\sqrt[3]{19 n+27}} \cdot \frac{\sqrt[3]{19 n+8}}{2 T_{n}} \\
=\frac{T_{n+1}}{T_{n}} \cdot \frac{\sqrt[3]{19 n+8}}{\sqrt[3]{19 n+27}} \\
=\frac{3 n+3}{3 n+2} \sqrt[3]{\frac{19 n+8}{19 n+27}} \\
=\sqrt[3]{\frac{(3 n+3)^{3}(19 n+8)}{(3 n+2)^{3}(19 n+27)}} .
\end{array}
$$
Notice that
$$
\begin{aligned}
& (3 n+3)^{3}(19 n+8)-(3 n+2)^{3}(19 n+27) \\
= & (19 n+8)\left[(3 n+3)^{3}-(3 n+2)^{3}\right]- \\
& 19(3 n+2)^{3} \\
= & (19 n+8)\left(27 n^{2}+45 n+19\right)- \\
& 19\left(27 n^{3}+54 n^{2}+36 n+8\right) \\
= & 45 n^{2}+37 n>0 .
\end{aligned}
$$
Then $(3 n+3)^{3}(19 n+8)$
$$
\begin{array}{l}
>(3 n+2)^{3}(19 n+27) . \\
\text { Hence } \sqrt[3]{\frac{(3 n+3)^{3}(19 n+8)}{(3 n+2)^{3}(19 n+27)}}>1 \\
\Rightarrow \frac{f(n+1)}{f(n)}>1 \Rightarrow f(n)<f(n+1) .
\end{array}
$$
Therefore, $f(n)$ is a monotonically increasing function with respect to $n$.
Since $T_{n} \geqslant \frac{k}{2} \sqrt[3]{19 n+8}$ holds for all positive integers $n$,
$$
\begin{array}{l}
k \leqslant \min \{f(n) \mid n=1,2, \cdots\}=f(1) \\
=\frac{2 T_{1}}{\sqrt[3]{19+8}}=\frac{2\left(1+\frac{1}{2}\right)}{\sqrt[3]{27}}=1 .
\end{array}
$$
Thus, the maximum value of $k$ is 1.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given real numbers $a, b \neq 0$, let
$$
x=\frac{a}{|a|}+\frac{b}{|b|}+\frac{a b}{|a b|} \text {. }
$$
Then the sum of the maximum and minimum values of $x$ is $\qquad$ [1]
|
Question 1 Original Solution ${ }^{[1]}$ According to the number of negative numbers in $a$ and $b$, there are three cases:
(1) If $a$ and $b$ are both positive, then $x=3$;
(2) If $a$ and $b$ are one positive and one negative, then $x=-1$;
(3) If $a$ and $b$ are both negative, then $x=-1$.
In summary, the maximum value of $x$ is 3, and the minimum value is -1.
Therefore, the required answer is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $a b c<0$, let
$$
P=\frac{a}{|a|}+\frac{|b|}{b}+\frac{c}{|c|}+\frac{|a b|}{a b}+\frac{a c}{|a c|}+\frac{|b c|}{b c} \text {. }
$$
Find the value of $a P^{3}+b P^{2}+c P+2009$.
|
Given $a b c<0$, we get $\frac{a b c}{|a b c|}=-1$, and at least one of $a, b, c$ is negative. According to the corollary, we can change the position of the absolute value symbol, then
$$
\begin{aligned}
P & =\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{a b}{|a b|}+\frac{a c}{|a c|}+\frac{b c}{|b c|} \\
& =\left(1+\frac{a}{|a|}\right)\left(1+\frac{b}{|b|}\right)\left(1+\frac{c}{|c|}\right)-1-\frac{a b c}{|a b c|} \\
& =0-1+1=0 .
\end{aligned}
$$
Therefore, $a P^{3}+b P^{2}+c P+2009=2009$.
|
2009
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $a, b, c$ satisfy
$$
\frac{a b}{a+b}=\frac{1}{3}, \frac{b c}{b+c}=\frac{1}{4}, \frac{c a}{c+a}=\frac{1}{5} \text {. }
$$
then $a b+b c+c a=$ $\qquad$
|
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=3, \frac{1}{b}+\frac{1}{c}=4, \frac{1}{c}+\frac{1}{a}=5 \\
\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6 \\
\Rightarrow \frac{1}{c}=3, \frac{1}{a}=2, \frac{1}{b}=1 \\
\Rightarrow a=\frac{1}{2}, b=1, c=\frac{1}{3} \\
\Rightarrow a b+b c+c a=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that the altitude to the hypotenuse of right $\triangle A B C$ is 4. Then the minimum value of the area of $\triangle A B C$ is $\qquad$ .
|
2. 16 .
Let the lengths of the two legs of the right triangle $\triangle ABC$ be $a$ and $b$, and the length of the hypotenuse be $c$. From the area relationship, we have $ab = 4c$.
By the Pythagorean theorem, we know $c^2 = a^2 + b^2 \geq 2ab = 8c$.
Thus, $c \geq 8$. Therefore, $S_{\triangle ABC} = 2c \geq 16$.
When $a = b = 4\sqrt{2}$, the minimum value is achieved.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In rectangle $A B C D$, $A B=12, A D=3, E$ and $F$ are points on $A B$ and $D C$ respectively. Then the minimum length of the broken line $A F E C$ is $\qquad$ .
|
3. 15 .
As shown in Figure 5, construct the symmetric points $A_{1}$ and $C_{1}$ of $A$ and $C$ with respect to $DC$ and $AB$, respectively. Connect $A_{1}C_{1}$ to intersect $AB$ and $DC$ at points $E_{1}$ and $F_{1}$, respectively, and connect $A_{1}F$ and $C_{1}E$. Draw a perpendicular from $A_{1}$ to the extension of $BC$, with the foot of the perpendicular being $G$.
Given that $A_{1}G = AB = 12$ and $C_{1}G = 3AD = 9$, by the Pythagorean theorem, we have
$$
\begin{array}{l}
A_{1}C_{1} = \sqrt{A_{1}G^{2} + C_{1}G^{2}} = 15 . \\
\text{Therefore, } AF + FE + EC = A_{1}F + FE + EC_{1} \\
\geqslant A_{1}C_{1} = 15 .
\end{array}
$$
When points $E$ and $F$ coincide with $E_{1}$ and $F_{1}$, respectively, the minimum value is achieved.
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ are six different positive integers, taking values from $1, 2, 3, 4, 5, 6$. Let
$$
\begin{aligned}
S= & \left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\left|x_{3}-x_{4}\right|+ \\
& \left|x_{4}-x_{5}\right|+\left|x_{5}-x_{6}\right|+\left|x_{6}-x_{1}\right| .
\end{aligned}
$$
Then the minimum value of $S$ is $\qquad$
|
4. 10 .
Since equation (1) is a cyclic expression about $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{array}{l}
S \geqslant\left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
\quad\left|\left(x_{j+1}-1\right)+\left(x_{j+2}-x_{j+1}\right)+\cdots+\left(6-x_{6}\right)\right| \\
= 10 . \\
\text { Also, when } x_{i}=7-i(i=1,2, \cdots, 6) \text {, } S=10 .
\end{array}
$$
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Find the value of $\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}+15^{1998}}{7^{1998}+35^{1998}}}$.
|
$\begin{array}{l}\text { Solve the original expression }=\left(\frac{7}{3}\right)^{999} \sqrt{\frac{3^{1998}\left(1+5^{1998}\right)}{7^{1998}\left(1+5^{1998}\right)}} \\ =\left(\frac{7}{3}\right)^{999} \times\left(\frac{3}{7}\right)^{999}=\left(\frac{7}{3} \times \frac{3}{7}\right)^{999}=1 .\end{array}$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A writes down the positive integers $1, 2, \cdots$, 2009 on the blackboard, then turns away from the blackboard, and asks Person B to erase some of these numbers and then add the remainder of the sum of the erased numbers when divided by 7. After several such operations, only two numbers remain on the blackboard, one of which is a single-digit number. A asks B: “What is the larger of the two remaining numbers?” B answers: “100.” Then the single-digit number is
|
3.5.
Since $1+2+\cdots+2009 \equiv 0(\bmod 7)$, therefore, the single digit $a$ satisfies $100+a \equiv 0(\bmod 7)$. Hence $a \equiv 5(\bmod 7)$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The number of solutions to the equation $\pi^{x-1} x^{2}+\pi^{x^{2}} x-\pi^{x^{2}}=x^{2}+x-1$ is $\qquad$ ( $\pi$ is the ratio of a circle's circumference to its diameter).
|
4. 2 .
The original equation is transformed into
$$
x^{2}\left(\pi^{x-1}-1\right)+(x-1)\left(\pi^{x^{2}}-1\right)=0 \text {. }
$$
When $x \neq 0,1$, $x^{2}\left(\pi^{x-1}-1\right)$ and $(x-1)\left(\pi^{x^{2}}-1\right)$ have the same sign.
Therefore, $x=0,1$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given a regular 200-gon $A_{1} A_{2} \cdots A_{200}$, connect the diagonals $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, where $A_{i+200}=$ $A_{i}(i=1,2, \cdots, 9)$. Then these 200 diagonals have $\qquad$ different intersection points inside the regular 200-gon.
|
8. 1600.
For each diagonal $A_{i} A_{i+9}(i=1,2, \cdots, 200)$, $A_{i+1}, A_{i+2}, \cdots, A_{i+8}$ each draw two diagonals intersecting with $A_{i} A_{i+9}$. And each such intersection point is counted twice, so the number of different intersection points of these 200 diagonals inside the regular 200-gon does not exceed $\frac{2 \times 8 \times 200}{2}=1600$.
Next, we prove: it is impossible for three such diagonals to intersect at one point.
Since each diagonal is of equal length, these 200 diagonals are all tangent to a fixed circle.
Because only two tangents can be drawn from an intersection point, it is impossible for three such diagonals to intersect at one point.
|
1600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (15 points) Given a sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ that satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}(n=2,3, \cdots, m$ is a real parameter $)$. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$.
|
10. Since $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}$,
$$
\begin{array}{l}
a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}, a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, \\
a_{6}=\frac{\frac{a_{0}}{m a_{1}}}{m \frac{1}{m^{2} a_{1}}}=a_{0}, a_{7}=\frac{a_{0}}{m \frac{a_{0}}{m a_{1}}}=a_{1},
\end{array}
$$
Therefore, the sequence $\left\{a_{n}\right\}$ is a periodic sequence with a period of 6.
Thus, $\frac{a_{0}}{a_{1}}=a_{2009}=a_{5}=\frac{a_{0}}{m a_{1}} \Rightarrow m=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Let $x=2 \frac{1}{5}, y=-\frac{5}{11}, z=-2 \frac{1}{5}$. Then $x^{2}+x z+2 y z+3 x+3 z+4 x y+5=$
|
Solve: From the given, we have
$$
x y=-1, x+z=0, y z=1 \text {. }
$$
Therefore, the expression to be found is
$$
\begin{array}{l}
=x(x+z)+2 y z+3(x+z)+4 x y+5 \\
=2 \frac{1}{5} \times 0+2 \times 1+3 \times 0+4 \times(-1)+5=3 .
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (15 points) Divide the sides of an equilateral $\triangle ABC$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points where the sides of $\triangle ABC$ and these parallel lines intersect are called grid points. If $n$ grid points are chosen from these 10 grid points, there will always be three grid points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.
|
11. $n_{\min }=5$.
Let the two equal division points on side $AB$ from point $A$ to $B$ be $D$ and $E$, the two equal division points on side $BC$ from point $B$ to $C$ be $F$ and $G$, and the two equal division points on side $CA$ from point $C$ to $A$ be $H$ and $I$, with the central grid point being $K$.
If the minimum value of $n$ is 4, taking the grid points $A, D, E, B$, then there do not exist three grid points that can form an isosceles triangle.
Therefore, $n \geqslant 5$.
We now prove: Any selection of five grid points will definitely contain three grid points that can form an isosceles triangle.
Assume the selected points are red points.
We only need to prove: There must exist an isosceles triangle formed by red points.
If these five red points include grid point $K$, divide the other nine grid points into three sets:
$$
L=\{D, E, F\}, M=\{G, H, I\}, N=\{A, B, C\}.
$$
By the pigeonhole principle, there must be at least one set containing at least two red points. Regardless of which set and which two points are red, they will form an isosceles triangle with red point $K$.
If these five red points do not include grid point $K$, when grid point $A$ is a red point, in
$$
\begin{array}{l}
U=\{D, I\}, V=\{E, H\}, \\
W=\{F, G\}, T=\{B, C\}
\end{array}
$$
if one of these sets contains two red points, the conclusion holds; otherwise, each set contains exactly one red point.
Assume $D$ is a red point, then $I$ is not a red point.
If $B$ is a red point, then $G, C$ are not red points. Thus, $F$ is a red point, and regardless of whether $E$ or $H$ is a red point, they will form an isosceles triangle with $D$ and $F$.
If $B$ is not a red point, then $C$ is a red point, thus $E$ is not a red point, and $H$ is a red point. Regardless of whether $F$ or $G$ is a red point, they will form an isosceles triangle with $D$ and $H$ or $H$ and $C$.
Similarly, when grid point $B$ or $C$ is a red point, the conclusion still holds.
If $K, A, B, C$ are not red points, then among $D, E, F, G, H, I$, there are five red points, and the conclusion is clearly true.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let a tangent line of the circle $x^{2}+y^{2}=1$ intersect the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then the minimum value of $|AB|$ is $\qquad$ .
|
3. 2 .
By symmetry, without loss of generality, assume the point of tangency is
$$
P(\cos \theta, \sin \theta)\left(0<\theta<\frac{\pi}{2}\right) \text {. }
$$
Then $|P A|=\tan \theta,|P B|=\cot \theta$.
Thus, $|A B|=|P A|+|P B|=\tan \theta+\cot \theta \geqslant 2$.
Equality holds if and only if $\tan \theta=\cot \theta$, i.e., $\theta=\frac{\pi}{4}$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $f(x)$ be an odd function defined on $\mathbf{R}$, $f(1)=2$, when $x>0$, $f(x)$ is an increasing function, and for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$. Then the maximum value of the function $f(x)$ on the interval $[-3,-2]$ is
|
5. -4 .
Since $f(x)$ is an odd function and is increasing on $(0,+\infty)$, therefore, $f(x)$ is also increasing on $(-\infty, 0)$. Thus, $f(-3) \leqslant f(x) \leqslant f(-2)$.
$$
\begin{array}{l}
\text { Also, } f(2)=f(1)+f(1)=4, \text { then } \\
f(-2)=-f(2)=-4 .
\end{array}
$$
Therefore, the maximum value of the function $f(x)$ on $[-3,-2]$ is -4.
|
-4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. As shown in Figure $2, A B$ is the diameter of the semicircle $\odot O$, and $C, D$ are two moving points on the semicircle, with $C D / / A B$. If the radius of the semicircle $\odot O$ is 1, then the maximum value of the perimeter of trapezoid $A B C D$ is $\qquad$ .
|
7.5.
In Figure 2, connect $A C$, and draw $C H \perp A B$ at point $H$. Let $\angle A B C=\theta\left(0<\theta<\frac{\pi}{2}\right)$. Then $A D=B C=A B \cos \theta=2 \cos \theta$.
Thus, $B H=B C \cos \theta=2 \cos ^{2} \theta$.
Therefore, $C D=A B-2 B H=2-4 \cos ^{2} \theta$.
Hence, the perimeter of trapezoid $A B C D$ is
$$
\begin{aligned}
l & =A B+B C+C D+D A \\
& =4+4 \cos \theta-4 \cos ^{2} \theta \\
& =5-4\left(\cos \theta-\frac{1}{2}\right)^{2} .
\end{aligned}
$$
Therefore, when $\cos \theta=\frac{1}{2}$, i.e., $\theta=\frac{\pi}{3}$, $l_{\max }=5$.
|
5
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 3, in $\triangle A B C$, $A B=3, A C=5$. If $O$ is the circumcenter of $\triangle A B C$, then the value of $\overrightarrow{A O} \cdot \overrightarrow{B C}$ is
|
8. 8 .
Let $D$ be the midpoint of side $B C$, and connect $O D, A D$. Then $O D \perp B C$. Therefore,
$$
\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{B C}=(\overrightarrow{A D}+\overrightarrow{D O}) \cdot \overrightarrow{B C} \\
=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{D O} \cdot \overrightarrow{B C}=\overrightarrow{A D} \cdot \overrightarrow{B C} \\
=\frac{1}{2}(\overrightarrow{A C}+\overrightarrow{A B}) \cdot(\overrightarrow{A C}-\overrightarrow{A B}) \\
=\frac{1}{2}\left(|\overrightarrow{A C}|^{2}-|\overrightarrow{A B}|^{2}\right)=8
\end{array}
$$
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Find the sum:
$$
\begin{array}{l}
\left(\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{60}\right)+\left(\frac{2}{3}+\frac{2}{4}+\cdots+\frac{2}{60}\right)+ \\
\left(\frac{3}{4}+\frac{3}{5}+\cdots+\frac{3}{60}\right)+\cdots+\left(\frac{58}{59}+\frac{59}{60}\right)
\end{array}
$$
|
$$
\begin{aligned}
= & \frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\cdots+ \\
& \left(\frac{1}{60}+\frac{2}{60}+\cdots+\frac{59}{60}\right) \\
= & \frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\cdots+\frac{59}{2} \\
= & \frac{1}{2}(1+2+\cdots+59) \\
= & \frac{1}{2} \times \frac{59(1+59)}{2}=885 .
\end{aligned}
$$
|
885
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the set
$$
M=\left\{x|5-| 2 x-3 \mid \in \mathbf{N}_{+}\right\} \text {. }
$$
Then the number of all non-empty proper subsets of $M$ is ( ).
(A) 254
(B) 255
(C) 510
(D) 511
|
- 1. C.
Given $5-|2 x-3| \in \mathbf{N}_{+}$, i.e., $|2 x-3|=0,1, 2,3,4$, we have
$$
M=\left\{-\frac{1}{2}, 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3, \frac{7}{2}\right\} .
$$
Since $M$ contains 9 elements, the number of all non-empty proper subsets of $M$ is $2^{9}-2=510$.
|
510
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Calculate:
$$
\frac{2009^{2}-2008^{2}}{19492009^{2}-19492008 \times 19492010+2 \times 2008} .
$$
|
Let $19492009=a, 2008=b$. Then the original expression $=\frac{(b+1)^{2}-b^{2}}{a^{2}-(a-1)(a+1)+2 b}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (40 points) Let the number of ways to choose $k$ pairwise coprime numbers from the set $\{1,2, \cdots, 28\}$ be $T(k)$. Find $T(2)+T(3)+\cdots+T(12)$.
|
Obviously, the prime numbers in $\{1,2, \cdots, 28\}$ are 2, 3, $5,7,11,13,17,19,23$, a total of 9.
When $k \geqslant 11$, the $k$ numbers taken out, except for 1, must include at least 10 numbers. By the pigeonhole principle, there must be two numbers divisible by the same prime number, which means these two numbers are not coprime, leading to a contradiction.
Therefore, $T(k)=0(k \geqslant 11)$.
Thus, $T(2)+T(3)+\cdots+T(12)$ is the number of all subsets $M$ of $\{1,2, \cdots, 28\}$, where $|M| \geqslant 2$, and the numbers in $M$ are pairwise coprime.
Below, we calculate $T$ by considering the prime factorization of the numbers in $M$.
(1) All numbers in $M$ have exactly one prime factor, then $M$ is formed by taking one number from each of the following sets:
$$
\begin{array}{l}
\left\{2,2^{2}, 2^{3}, 2^{4}\right\},\left\{3,3^{2}, 3^{3}\right\},\left\{5,5^{2}\right\},\{7\}, \\
\{11\},\{13\},\{17\},\{19\},\{23\}
\end{array}
$$
Similar to the calculation of the number of divisors of a number, the number of $M$ is
$$
T_{1}=(1+4)(1+3)(1+2)(1+1)^{6}-1=3839 \text {, }
$$
where there are 15 single-element subsets.
(2) $M$ has exactly one number $a$ with two prime factors, and the rest of the numbers have only one prime factor.
The prime factors of $a$ are $\{2,3\}, \{2,5\}, \{2,7\}, \{2,11\}$, $\{2,13\}, \{3,5\}$, or $\{3,7\}$. Then $a$ is one of $6i (i=1,2,3,4)$, $10i (i=1,2)$, $14i (i=1,2)$, $22$, $26$, $15$, or $21$, and the rest of the numbers are taken from the remaining sets in (1). The number of $M$ is $T_{2}=3616$, where there are 12 single-element subsets.
(3) $M$ has exactly two numbers $a, b$ each with two prime factors, and the rest of the numbers have only one prime factor.
The prime factors of $a, b$ are $\{2,5\}$ and $\{3,7\}$, $\{2,7\}$ and $\{3,5\}$, $\{2,11\}$ and $\{3,5\}$, $\{2,11\}$ and $\{3,7\}$, $\{2,13\}$ and $\{3,5\}$, $\{2,13\}$ and $\{3,7\}$. Then $(a, b)$ is one of $(10i, 21) (i=1,2)$, $(14i, 15) (i=1,2)$, $(22,15)$, $(22,21)$, $(26,15)$, $(26,21)$,
The rest of the numbers are taken from the remaining sets in (1). The number of $M$ is $T_{3}=288$.
(4) $M$ has three numbers $a, b, c$ each with two prime factors, let these six prime numbers be $p_{i} (i=1,2, \cdots, 6)$. Then
$$
p_{1} \geqslant 2, p_{2} \geqslant 3, p_{3} \geqslant 5, p_{4} \geqslant 7, p_{5} \geqslant 11, p_{6} \geqslant 13 \text {. }
$$
Since $13 \times 3 > 28$, $p_{6}$ can only be a prime factor of 2, but $p_{5} \times 3 \geqslant 11 \times 3 > 28$, leading to a contradiction.
(5) $M$ has one number $a$ with three prime factors, then $a \geqslant 2 \times 3 \times 5 > 28$, leading to a contradiction.
In summary, and considering the inclusion of "1" and the single-element subsets in (1) and (2), we have
$$
T=2\left(T_{1}+T_{2}+T_{3}\right)-27=15459 .
$$
|
15459
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (50 points) Distribute 2010 red cards and 2010 white cards arbitrarily to 2010 players participating in the game, with each person receiving two cards. All players sit in a circle facing down. The game rule is that each operation requires each player to simultaneously follow the following principle: if a player has at least one red card, he will give a red card to the player on his left; if he has no red card, he will give a white card to the player on his left. Find the maximum number of operations required for the first time that each player has exactly one red card and one white card in their hands.
|
In 2009.
Let $n=2$ 010. A wheel (divided into $n$ sections) can be placed in a circle of $n$ players. Each player, according to the game rules, when the section they face contains a white card and they hold a red card, they place the red card in the section and take back the white card; in all other cases, it is assumed that no card exchange is necessary.
The 0th operation is set as all players placing a card in the section they face (players with a red card place a red card, otherwise, they place a white card), then the wheel rotates one section clockwise.
The following proof shows: the wheel will have all red cards after at most $n-1$ rotations, meaning each player will have exactly one red and one white card.
By relativity, the wheel can be considered stationary, and the $n$ players move one section counterclockwise around the wheel, with the same card exchange rules. When they move $n-1$ times, each player will have visited all $n$ sections of the wheel, and players with red cards will have placed all of them in a section of the wheel, taking back a white card, making all the cards on the wheel red.
Additionally, if $n$ red cards are given to the $\left[\frac{n+1}{2}\right]$ players adjacent to the left (let's say players 1, 2, ..., $\left[\frac{n+1}{2}\right]$), then initially, the first $\left[\frac{n+1}{2}\right]$ sections of the wheel will have red cards, and the rest will have white cards. Player 1 must place their red card in the $n$th section, meaning the operation has been performed $n-1$ times.
(Song Qiang, compiled)
|
2009
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
The initial 267 four-digit number $w_{1}$ and the sum of its four digits is the four-digit number $w_{2}, w_{2}$ and the sum of its four digits is the four-digit number $w_{3}, w_{3}$ and the sum of its four digits is the four-digit number $w_{4}, w_{4}$ and the sum of its four digits is the four-digit number $w_{5}, w_{5}$ and the sum of its four digits is 2009. Find $w_{1}$.
|
Let $w_{5-i}=a_{i} b_{i} c_{i} d_{i}(i=0,1, \cdots, 4)$. Then $1001 a_{0}+101 b_{0}+11 c_{0}+2 d_{0}=2009$.
Obviously, $a_{0}=1$ (otherwise, if $a_{0}=2$, then $101 b_{0}+11 c_{0}+2 d_{0}=7$, which only has the solution $b_{0}=c_{0}=0, d_{0}=\frac{7}{2}$, which is not an integer).
Thus, $101 b_{0}+11 c_{0}+2 d_{0}=1008$.
If $b_{0} \leqslant 8$, then
$$
\begin{array}{l}
101 b_{0}+11 c_{0}+2 d_{0} \leqslant 808+99+18 \\
=925925$.
Therefore, $b_{1}=9$.
Thus, $11 c_{1}+2 d_{1}=80$, which only has the solution $c_{1}=6, d_{1}=7$.
So, $w_{4}=1967$.
Then $1001 a_{2}+101 b_{2}+11 c_{2}+2 d_{2}=1967$.
Obviously, $a_{2}=1$.
Thus, $101 b_{2}+11 c_{2}+2 d_{2}=966>925$.
Therefore, $b_{2}=9$.
Thus, $11 c_{2}+2 d_{2}=57$, which only has the solution $c_{2}=5, d_{2}=1$.
So, $w_{3}=1951$.
Then $1001 a_{3}+101 b_{3}+11 c_{3}+2 d_{3}=1951$.
Obviously, $a_{3}=1$.
Thus, $101 b_{3}+11 c_{3}+2 d_{3}=950>925$.
Therefore, $b_{3}=9$.
Thus, $11 c_{3}+2 d_{3}=41$, which only has the solution $c_{3}=3, d_{3}=4$.
So, $w_{2}=1934$.
Then $1001 a_{4}+101 b_{4}+11 c_{4}+2 d_{4}=1934$.
Obviously, $a_{4}=1$.
Thus, $101 b_{4}+11 c_{4}+2 d_{4}=933>925$.
Therefore, $b_{4}=9$.
Thus, $11 c_{4}+2 d_{4}=24$, which only has the solution $c_{4}=2, d_{4}=1$.
Therefore, $w_{1}=1921$.
(Tian Yonghai, Shao Hua Educational College, Heilongjiang Province, 152054)
|
1921
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given real numbers $x, y, z$ satisfy $x-y=8$, $xy+z^{2}=-16$. Then $x+y+z=$ $\qquad$ .
|
Let $x=a+b, y=a-b$. Then $x-y=2b \Rightarrow b=4$.
Since $xy+z^{2}=-16$, we have $(a+b)(a-b)+z^{2}+16=0$. Also, $-b^{2}=-16$, then $a^{2}+z^{2}=0$. By the property of non-negative numbers, we know $a=0, z=0$. Therefore, $x=4, y=-4$. Hence, $x+y+z=4-4+0=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. As shown in Figure 1, it is known that rectangle $A B C D$ can be exactly divided into seven small rectangles of the same shape and size. If the area of the small rectangle is 3, then the perimeter of rectangle $A B C D$ is
|
$-1.19$.
Let the length of the small rectangle be $a$ and the width be $b$. Then $3a = 4b \Rightarrow a = \frac{4}{3}b \Rightarrow ab = \frac{4}{3}b^2 = 3$. Solving this, we get $b = \frac{3}{2}, a = 2$.
Therefore, the perimeter of rectangle $ABCD$ is $2(4a + b) = 8a + 2b = 16 + 3 = 19$.
|
19
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an "integer point". The number of integer points on the graph of the function $y=\frac{x+12}{2 x-1}$ is $\qquad$
|
2. 6 .
Notice that $2 y=1+\frac{25}{2 x-1}$.
Since $y$ is an integer, $\frac{25}{2 x-1}$ must also be an integer. Therefore, $2 x-1= \pm 1, \pm 5, \pm 25$.
Solving for $x$ gives $x=-12,-2,0,1,3,13$.
Correspondingly, $y=0,-2,-12,13,3,1$.
Thus, there are 6 integer points.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $a$ and $b$ are integers. Then the number of ordered pairs $(a, b)$ that satisfy $a + b + ab = 2008$ is $\qquad$ groups.
|
4. 12 .
$$
\begin{array}{r}
\text { Since }(a+1)(b+1) \\
=2009=41 \times 7^{2} \text {, and given }
\end{array}
$$
that $a, b$ take integer values, so, $a+1, b+1$ have 12 factor combinations, i.e., $1 \times 2009, 7 \times 287, 41 \times 49$ and their negatives as well as swaps.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Arrange the numbers $1,2, \cdots, 10$ in a row in some order, such that the sum of any three consecutive numbers does not exceed $n$. Answer the following:
(1) When $n=10$, can it be arranged? Please explain your reasoning;
(2) When it can be arranged, what is the minimum value of $n$?
|
8. (1) Suppose $n=10$ is already arranged. Then the sum of the last nine numbers is less than or equal to 30. Thus, the first number is not less than 25, which is a contradiction. Therefore, it cannot be arranged.
(2) With the same consideration as (1).
When $n=11,12,13,14$, none can be arranged.
When $n=15$, the sum of the first nine numbers is less than or equal to 45, leading to the tenth number being 10; also, the sum of the last nine numbers is less than or equal to 45, leading to the first number being 10. However, there is only one 10, so it cannot be arranged.
When $n=16$, it can be arranged. For example,
$$
10,5,1,7,6,2,8,3,4,9
$$
or $9,4,3,7,2,6,8,1,5,10$.
From this, we know that the minimum value of $n$ for which it can be arranged is 16.
$$
x_{a}=\frac{2 S}{a+h_{a}}, x_{b}=\frac{2 S}{b+h_{b}}, x_{c}=\frac{2 S}{c+h_{c}} \text {. }
$$
Since $a+h_{a}-\left(b+h_{b}\right)=a-b+h_{a}-h_{b}$
$$
\begin{array}{l}
=a-b+\frac{b h_{b}}{a}-h_{b}=a-b+\frac{b-a}{a} h_{b} \\
=(a-b)\left(1-\frac{h_{b}}{a}\right)>0
\end{array}
$$
(Given $a>b, h_{b}$ is the height on $b$, so $h_{b}>0$.
By analogy, we have $x_{a}<x_{b}<x_{c}$.
Therefore, when a square has two vertices on the shorter side, the area of the square is maximized.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $x_{1}, x_{2}, \cdots, x_{n}$ take values 7 or -7, and satisfy
(1) $x_{1}+x_{2}+\cdots+x_{n}=0$;
(2) $x_{1}+2 x_{2}+\cdots+n x_{n}=2009$.
Determine the minimum value of $n$.
|
9. First, when $n=34$,
$$
\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{16}=x_{18}=-7, \\
x_{17}=x_{19}=x_{20}=\cdots=x_{34}=7
\end{array}
$$
satisfies the conditions of the problem.
Second, divide both sides of condition (2) by 7, and let $y_{i}=\frac{x_{i}}{7}(i=1,2, \cdots, n)$. Then
$$
\begin{array}{l}
y_{i}= \pm 1, \\
y_{1}+y_{2}+\cdots+y_{n}=0, \\
y_{1}+2 y_{2}+\cdots+n y_{n}=287 .
\end{array}
$$
From equation (1), we know that $n$ is even.
Let $n=2 k$. Then $y_{1}, y_{2}, \cdots, y_{n}$ contain $k$ -1s and $k$ 1s.
From equation (2), we get
$$
\begin{array}{r}
287 \leqslant(2 k+2 k-1+\cdots+k+1)- \\
\quad(k+k-1+\cdots+2+1)=k^{2} .
\end{array}
$$
Therefore, $k^{2} \geqslant 287 \Rightarrow k \geqslant 17 \Rightarrow n \geqslant 34$.
Thus, the minimum value of $n$ is 34.
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (15 points) Fill seven different perfect squares into seven consecutive boxes so that the sum of any three adjacent boxes is greater than 100. Find the minimum possible value of the sum of these seven perfect squares.
|
3. Since the three perfect squares at positions $1, 4$, and $7$ are all different, considering the smallest possible values, one of these numbers must be at least 9. After positioning 9, the remaining six positions can be divided into two groups, each with three consecutive numbers whose sum is at least 101. Therefore, the sum of these seven numbers is at least $101 \times 2 + 9 = 211$. Thus, the smallest possible value is 211.
The arrangement $1, 36, 64, 9, 81, 16, 4$ satisfies the conditions, with a sum of 211.
|
211
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let convex quadrilateral $ABCD$ satisfy $AB=AD=1$, $\angle A=160^{\circ}, \angle C=100^{\circ}$. Then the range of the length of diagonal $AC$ is $\qquad$ .
|
2. $\{1\}$.
Since $\angle C=180^{\circ}-\frac{\angle A}{2}$, therefore, points $B$, $C$, and $D$ lie on a circle with $A$ as the center and a radius of 1.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If the three-digit decimal number $n=abc$ satisfies that $a$, $b$, $c$ form an arithmetic sequence, then the maximum possible value of a prime factor of $n$ is $\qquad$
|
3. 317 .
$$
\begin{array}{l}
31 n, 999=27 \times 37, 987=3 \times 7 \times 47, \\
963=9 \times 107, 951=3 \times 317 .
\end{array}
$$
|
317
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The sum of the radii of all circles passing through point $A(1505,1008)$ and tangent to the lines $l_{1}: y=0$ and $l_{2}: y=\frac{4}{3} x$ is $\qquad$
|
5. 2009 .
From $\tan 2 \theta=\frac{4}{3} \Rightarrow \tan \theta=\frac{1}{2}$, the center of the circle lies on the line $y=\frac{x}{2}$, and the center is $(2 r, r)$ (where $r$ is the radius of the circle).
$$
\begin{array}{l}
\text { Hence }(1505-2 r)^{2}+(1008-r)^{2}=r^{2} \\
\Rightarrow 4 r^{2}-8036 r+1505^{2}+1008^{2}=0 \\
\Rightarrow r_{1}+r_{2}=2009 .
\end{array}
$$
|
2009
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let the lines $l_{1} / / l_{2}$, and take 10 points $A_{1}, A_{2}, \cdots, A_{10}$ and $B_{1}, B_{2}, \cdots, B_{10}$ on $l_{1}$ and $l_{2}$ respectively. Then the line segments $A_{1} B_{1}, A_{2} B_{2}, \cdots, A_{10} B_{10}$ can divide the strip region between $l_{1}$ and $l_{2}$ into at most $\qquad$ non-overlapping parts.
|
8. 56 .
A line segment divides the original region into two parts. The $k$-th line segment can be divided into at most $k$ segments by the previous $k-1$ line segments, and it can add at most $k$ parts. Therefore, $k$ line segments can divide the region into at most $2+2+3+\cdots+k=\frac{k^{2}+k+2}{2}$ parts.
|
56
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 10 Simplify:
$$
\frac{(\sqrt{x}-\sqrt{y})^{3}+2 x \sqrt{x}+y \sqrt{y}}{x \sqrt{x}+y \sqrt{y}}+\frac{3 \sqrt{x y}-3 y}{x-y} .
$$
|
Let $\sqrt{x}=a+b, \sqrt{y}=a-b$. Then
$$
\sqrt{x y}=a^{2}-b^{2}, \sqrt{x}+\sqrt{y}=2 a, \sqrt{x}-\sqrt{y}=2 b \text {. }
$$
Therefore, the original expression is
$$
\begin{aligned}
= & \frac{(2 b)^{3}+2(a+b)^{3}+(a-b)^{3}}{(a+b)^{3}+(a-b)^{3}}+ \\
& \frac{3\left(a^{2}-b^{2}\right)-3(a-b)^{2}}{4 a b} \\
= & \frac{3\left(a^{2}+3 b^{2}\right)(a+b)}{2 a\left(a^{2}+3 b^{2}\right)}+\frac{3(a-b)}{2 a} \\
= & \frac{3 a+3 b+3 a-3 b}{2 a}=\frac{6 a}{2 a}=3 .
\end{aligned}
$$
[Note] This problem uses sum and difference substitution to transform the simplification of radicals into the simplification of fractions, thereby reducing the difficulty and improving the speed of solving the problem.
From the above examples, it can be seen that the essence of sum and difference substitution
is substitution and transformation, the key is to construct and set variables, the purpose is to introduce new variables to make complex calculations and proofs easier to handle.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (22 points) Suppose a set of planar points $S$ has the following properties:
(1) No three points are collinear;
(2) The distance between any two points is unique.
For two points $A$ and $B$ in $S$, if there exists a point $C \in S$ such that $|A C|<|A B|<|B C|$, then $A B$ is called a "middle edge" of $S$. For three points $A$, $B$, and $C$ in $S$, if $A B$, $A C$, and $B C$ are all middle edges of $S$, then $\triangle A B C$ is called a "middle edge triangle" of $S$. Find the smallest $n$ such that any $n$-element planar point set $S$ with properties (1) and (2) must contain a middle edge triangle.
|
4. Color all middle edges of $S$ red, and color other edges blue.
When $n \geqslant 6$, according to Ramsey's theorem, there must exist a monochromatic triangle, which must have a middle edge, and it must be a middle edge triangle.
The following set of five points with properties (1) and (2) does not have a middle edge triangle:
Assume five points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are arranged in a counterclockwise order on the circumference of a circle, and
$$
\begin{array}{c}
{\overparen{P_{1} P_{2}}}_{2}^{\circ}=\frac{\pi}{10},{\overparen{P_{2} P_{3}}}^{0}=\frac{3 \pi}{5},{\overparen{P_{3} P_{4}}}^{0}=\frac{3 \pi}{10}, \\
\overparen{P}_{4} P_{5}^{0}=\frac{3 \pi}{4},{\overparen{P_{5} P_{1}}}^{0}=\frac{\pi}{4} .
\end{array}
$$
Then the distances between points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are all different, and $P_{2} P_{3}, P_{3} P_{1}, P_{1} P_{5}, P_{5} P_{4}, P_{4} P_{2}$ are middle edges, but there is no middle edge triangle.
For cases with fewer than five points, simply remove some points from the previous example, and there will still be no middle edge triangle.
(Provided by An Sai)
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. For the positive integer $n$, define $a_{n}$ as the unit digit of $n^{(n+1)^{n+2}}$. Then $\sum_{n=1}^{2010} a_{n}=$ $\qquad$ .
|
4.5 829 .
When $n \equiv 0,1,5,6(\bmod 10)$, $a_{n} \equiv n^{(n+1)^{n+2}} \equiv n(\bmod 10)$;
When $n \equiv 2,4,8(\bmod 10)$,
$$
\begin{array}{l}
(n+1)^{n+2} \equiv 1(\bmod 4) \\
\Rightarrow a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k+1} \equiv n(\bmod 10) ;
\end{array}
$$
When $n \equiv 3,7,9(\bmod 10)$,
$$
\begin{array}{l}
(n+1)^{n+2} \equiv 0(\bmod 4) \\
\Rightarrow a_{n} \equiv n^{(n+1)^{n+2}}=n^{4 k} \equiv 1(\bmod 10) .
\end{array}
$$
Therefore, $a_{n}$ is sequentially $1,2,1,4,5,6,1,8,1,0,1$, $2, \cdots$ which is a periodic sequence, with the smallest positive period being 10.
Thus, $\sum_{n=1}^{2010} a_{n}=201 \times 29=5829$.
|
5829
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. The line $l: x+y=t$ intersects the circle $\odot O: x^{2}+y^{2}=20$ at points $A$ and $B$, and $S_{\triangle O A B}$ is an integer. Then the number of all positive integer values of $t$ that satisfy the condition is $\qquad$ .
|
7.2.
Let $\angle A O B=2 \alpha, O C \perp A B$, with the foot of the perpendicular being $C$. Then $O C=\sqrt{20} \cos \alpha=\frac{t}{\sqrt{1+1}}=\frac{t}{\sqrt{2}}$ $\Rightarrow \cos \alpha=\frac{t}{2 \sqrt{10}}$.
Also, $S_{\triangle O A B}=10 \sin 2 \alpha \leqslant 10$, so $\sin 2 \alpha=\frac{k}{10}(k \in\{1,2, \cdots, 10\})$.
Thus, $2 \cos \alpha \sqrt{1-\cos ^{2} \alpha}=\frac{k}{10}$.
Substituting equation (1) into the above equation and simplifying, we get $t^{2}\left(40-t^{2}\right)=4 k^{2}$.
It is easy to see that $t$ is an even number. Let $t=2 t_{0}$. Then $4 t_{0}^{2}\left(10-t_{0}^{2}\right)=k^{2} \Rightarrow t_{0}<\sqrt{10}$. Upon inspection, $t_{0}=1,3$ satisfy the conditions. Therefore, $t=2,6$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given the function $f:\{0,1, \cdots, 2010\} \rightarrow \mathbf{N}$. If for all possible integers $x$, we have
$$
\begin{array}{l}
f(4 x+2)=f(4 x+1), \\
f(5 x+3)=f(5 x+2), \\
f(7 x+5)=f(7 x+4),
\end{array}
$$
then $f(x)$ can take at most $\qquad$ different values.
|
8. 1033 .
For the function $f$, construct the function $g$, defined as:
$$
\begin{array}{l}
g(0)=1, \\
g(i+1)=\left\{\begin{array}{l}
1, f(i+1) \neq f(i) ; \\
0, f(i+1)=f(i),
\end{array}\right.
\end{array}
$$
where $i=0,1, \cdots, 2009$.
Then $g$ is a mapping from $\{0,1, \cdots, 2010\}$ to $\{0,1\}$, and for all possible $x$, we have
$$
g(4 x+2)=g(5 x+3)=g(7 x+5)=0 .
$$
Thus, $f(x)$ can take at most $\sum_{i=0}^{2010} g(i)$ different values.
Let $g(x)=h(x+2)$. Then $h$ is a mapping from $\{2,3, \cdots, 2012\}$ to $\{0,1\}$, and for all possible $x$, we have
$$
\begin{array}{l}
h(4 x)=h(5 x)=h(7 x)=0 . \\
\text { Since }\left(\sum_{i=2}^{2012} h(i)\right)_{\max } \\
=2011-\left[\frac{2012}{4}\right]-\left[\frac{2012}{5}\right]-\left[\frac{2012}{7}\right]+ \\
\quad\left[\frac{2012}{4 \times 5}\right]+\left[\frac{2012}{4 \times 7}\right]+\left[\frac{2012}{5 \times 7}\right]-\left[\frac{2012}{4 \times 5 \times 7}\right] \\
=1033,
\end{array}
$$
$\left(\right.$ at this time, $h(i)=\left\{\begin{array}{l}0,4 \mid i \text { or } 5 \mid i \text { or } 7 \mid i ; \\ 1, \text { otherwise. }\end{array}\right)$ and $\sum_{i=0}^{2010} g(i)=\sum_{i=2}^{2012} h(i)$, so $f(x)$ can take at most 1033 different values.
|
1033
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $\left(2+x-2 x^{2}\right)^{1005}=\sum_{k=0}^{2010} a_{k} x^{k}$. Then
$$
\begin{array}{l}
a_{1}+3 a_{3}+5 a_{5}+\cdots+2009 a_{2009} \\
=
\end{array}
$$
|
3. 1005.
Let $f(x)=\left(2+x-2 x^{2}\right)^{1005}$. Then $f^{\prime}(x)=1005\left(2+x-2 x^{2}\right)^{1004}(1-4 x)$. Therefore, $f^{\prime}(1)=1005 \times(-3)$, $f^{\prime}(-1)=1005 \times 5$.
Given $f(x)=\sum_{k=0}^{2010} a_{k} x^{k}$, we have
$$
\begin{array}{l}
f^{\prime}(x)=\sum_{k=1}^{2010} k a_{k} x^{k-1} . \\
\text { Hence } a_{1}+3 a_{3}+5 a_{5}+\cdots+2009 a_{2000} \\
=\frac{1}{2}\left(f^{\prime}(1)+f^{\prime}(-1)\right)=1005 .
\end{array}
$$
|
1005
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If the polynomial $f(x)=x^{3}-6 x^{2}+a x+a$ has three roots $x_{1}, x_{2}, x_{3}$ that satisfy
$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0 \text {, }
$$
then the value of the real number $a$ is .. $\qquad$
|
6. -9 .
From the problem, we have
$$
\begin{array}{l}
g(t)=f(t+3) \\
=(t+3)^{3}-6(t+3)^{2}+a(t+3)+a \\
=t^{3}+3 t^{2}+(a-9) t+4 a-27
\end{array}
$$
The three roots \( t_{1}, t_{2}, t_{3} \) satisfy \( t_{1}^{3}+t_{2}^{3}+t_{3}^{3}=0 \).
By the relationship between roots and coefficients, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
t_{1}+t_{2}+t_{3}=-3, \\
t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}=a-9, \\
t_{1} t_{2} t_{3}=-(4 a-27) .
\end{array}\right. \\
\text { Substituting into } t_{1}^{3}+t_{2}^{3}+t_{3}^{3}-3 t_{1} t_{2} t_{3} \\
=\left(t_{1}+t_{2}+t_{3}\right)\left[\left(t_{1}+t_{2}+t_{3}\right)^{2}-\right. \\
\left.3\left(t_{1} t_{2}+t_{2} t_{3}+t_{3} t_{1}\right)\right],
\end{array}
$$
we get \( 3(4 a-27)=-3[9-3(a-9)] \).
Solving for \( a \), we get \( a=-9 \).
|
-9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given an ellipse centered at the origin, with foci on the $x$-axis, the length of the major axis is twice the length of the minor axis, and it passes through the point $M(2,1)$. A line $l$ parallel to $OM$ has a $y$-intercept of $m (m<0)$, and intersects the ellipse at two distinct points $A$ and $B$. Find the $x$-coordinate of the incenter $I$ of $\triangle ABM$.
untranslated text remains the same as the source text.
|
10. Let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
Then $\left\{\begin{array}{l}a=2 b, \\ \frac{4}{a^{2}}+\frac{1}{b^{2}}=1\end{array} \Rightarrow\left\{\begin{array}{l}a^{2}=8, \\ b^{2}=2 \text {. }\end{array}\right.\right.$
Therefore, the equation of the ellipse is $\frac{x^{2}}{8}+\frac{y^{2}}{2}=1$.
As shown in Figure 1, because
the line $l$ is parallel
to $O M$, and the y-intercept
is $m$, and $k_{O S}=\frac{1}{2}$,
so, the equation of the line $l$ is $y=\frac{1}{2} x+m$.
From $\left\{\begin{array}{l}y=\frac{1}{2} x+m, \\ \frac{x^{2}}{8}+\frac{y^{2}}{2}=1\end{array} \Rightarrow x^{2}+2 m x+2 m^{2}-4=0\right.$.
Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$. Then
$x_{1}+x_{2}=-2 m, x_{1} x_{2}=2 m^{2}-4$.
Let the slopes of the lines $M A 、 M B$ be $k_{1} 、 k_{2}$. Then
$k_{1}=\frac{y_{1}-1}{x_{1}-2}, k_{2}=\frac{y_{2}-1}{x_{2}-2}$.
Thus, $k_{1}+k_{2}=\frac{y_{1}-1}{x_{1}-2}+\frac{y_{2}-1}{x_{2}-2}$
$=\frac{\left(y_{1}-1\right)\left(x_{2}-2\right)+\left(y_{2}-1\right)\left(x_{1}-2\right)}{\left(x_{1}-2\right)\left(x_{2}-2\right)}$.
The numerator of the above expression
$$
\begin{array}{l}
=\left(\frac{1}{2} x_{1}+m-1\right)\left(x_{2}-2\right)+\left(\frac{1}{2} x_{2}+m-1\right)\left(x_{1}-2\right) \\
=x_{1} x_{2}+(m-2)\left(x_{1}+x_{2}\right)-4(m-1) \\
=2 m^{2}-4+(m-2)(-2 m)-4(m-1) \\
=0 .
\end{array}
$$
Therefore, $k_{1}+k_{2}=0$.
Hence, the angle bisector $M I$ of $\triangle A B M$ is perpendicular to the x-axis. Therefore, the x-coordinate of the incenter $I$ is equal to the x-coordinate of point $M$, which is 2.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (20 points) Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
a_{1}=1, a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Try to find $\left[a_{2009}^{2}\right]$.
|
11. From $a_{n+1}=\frac{a_{n}}{n}+\frac{n}{a_{n}}$, we get $a_{n+1}^{2}=\frac{a_{n}^{2}}{n^{2}}+\frac{n^{2}}{a_{n}^{2}}+2$. Given $a_{1}=1$, we have $a_{2}^{2}=4, a_{3}^{2}=4, a_{4}^{2}=4+\frac{25}{36}$.
Next, we use mathematical induction to prove:
When $n \geqslant 4$, $n+\frac{2}{n}\frac{n+1}{n^{2}}+\frac{n^{2}}{n+1}+2 \\
=\frac{n+1}{n^{2}}+\frac{1}{n+1}+(n-1)+2 \\
>n+1+\frac{2}{n+1} .
\end{array}
$
On the other hand,
$
\begin{array}{l}
a_{n+1}^{2}=\frac{a_{n}^{2}}{n^{2}}+\frac{n^{2}}{a_{n}^{2}}+2\left(n^{2}+2\right)^{2} \\
\Leftrightarrow n^{4}>4 n^{2}+4 . \\
\text { When } n \geqslant 4 \text {, } n^{4} \geqslant 4 \times 4 n^{2}>4 n^{2}+4, \text { (3) (3) }
\end{array}
$
is true.
From equations (2) and (3), we get $a_{n+1}^{2}<n+2$.
In summary, $n+1+\frac{2}{n+1}<a_{n+1}^{2}<n+2$.
By the principle of induction, when $n \geqslant 4$, equation (1) always holds. Therefore, $\left[a_{2009}^{2}\right]=2009$.
|
2009
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}, x y z>0$,
and
$$
\begin{array}{l}
\sqrt[3]{\frac{2010}{x^{2}}+\frac{2011}{y^{2}}+\frac{2012}{z^{2}}} \\
=\sqrt[3]{2010}+\sqrt[3]{2011}+\sqrt[3]{2012} .
\end{array}
$$
Find the value of $x+y+z$.
|
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}=k$, obviously $k \neq 0$.
Then $2010=x^{3} k, 2011=y^{3} k, 2012=z^{3} k$.
From the given, we have
$$
\sqrt[3]{x k+y k+z k}=\sqrt[3]{x^{3} k}+\sqrt[3]{y^{3} k}+\sqrt[3]{z^{3} k},
$$
which means $\sqrt[3]{k} \sqrt[3]{x+y+z}=\sqrt[3]{k}(x+y+z)$.
Since $k \neq 0$, we have $\sqrt[3]{x+y+z}=x+y+z$.
Given that $x>0, y>0, z>0$.
Therefore, $x+y+z=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find all positive integers $n$, such that $2^{n-1} n+1$ is a perfect square.
(2004, Slovenia IMO National Team Selection Test
|
Let $2^{n-1} n+1=m^{2}\left(m \in \mathbf{N}_{+}\right)$. Then $2^{n-1} n=(m+1)(m-1)$.
When $n=1,2,3,4$, $2^{n-1} n+1$ is not a perfect square.
Therefore, $n \geqslant 5, 16 \mid (m+1)(m-1)$.
Since $m+1$ and $m-1$ have the same parity, both $m+1$ and $m-1$ are even, and $m$ is odd.
Let $m=2 k-1\left(k \in \mathbf{N}_{+}\right)$. Then $2^{n-1} n=2 k(2 k-2)$.
Thus, $2^{n-3} n=k(k-1)$.
Since $k$ and $k-1$ have different parities, $2^{n-3}$ can only be a divisor of one of them.
Also, $2^{n-3} n=k(k-1) \neq 0$, so
$2^{n-3} \leqslant k$.
Therefore, $n \geqslant k-1$.
Hence, $2^{n-3} \leqslant k \leqslant n+1$.
By the properties of functions or mathematical induction, when $n \geqslant 6$, $2^{n-3}>n+1$.
Thus, $n \leqslant 5$. Since $n \geqslant 5$, we have $n=5$.
At this point, $2^{n-1} n+1=81$ is a perfect square, satisfying the condition.
In summary, the required positive integer $n=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Proof: There exist positive integers $a_{i}(1 \leqslant i \leqslant 8)$, such that
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2}-1}}+\cdots+ \\
\sqrt{\sqrt{a_{8}}-\sqrt{a_{8}-1}}=2 .
\end{array}
$$
|
Prove the construction of the identity:
$$
\begin{array}{l}
\sqrt{(2 i+1)^{2}}-\sqrt{(2 i+1)^{2}-1} \\
=2 i+1-2 \sqrt{i(i+1)} \\
=(\sqrt{i+1}-\sqrt{i})^{2} .
\end{array}
$$
Take \( a_{i}=(2 i+1)^{2} \) for \( i=1,2, \cdots, 8 \), then
$$
\begin{array}{l}
\sqrt{\sqrt{a_{1}}-\sqrt{a_{1}-1}}+\sqrt{\sqrt{a_{2}}-\sqrt{a_{2}-1}}+\cdots+ \\
\sqrt{\sqrt{a_{8}}-\sqrt{a_{8}-1}} \\
=\sqrt{1+1}-\sqrt{1}+\sqrt{2+1}-\sqrt{2}+\cdots+\sqrt{8+1}-\sqrt{8} \\
=\sqrt{9}-\sqrt{1}=2 .
\end{array}
$$
|
2
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that $a$ is a root of the equation $x^{2}-3 x+1=0$. Then the value of the fraction $\frac{2 a^{6}-6 a^{4}+2 a^{5}-a^{2}-1}{3 a}$ is $\qquad$ -
|
II. 7. -1.
According to the problem, we have $a^{2}-3 a+1=0$.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 a^{3}\left(a^{2}-3 a+1\right)-\left(a^{2}+1\right)}{3 a} \\
=-\frac{a^{2}+1}{3 a}=-1 .
\end{array}
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Robots A and B simultaneously conduct a $100 \mathrm{~m}$ track test at a uniform speed, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the finish line. After calculation, this track is not standard. Then this track is $\qquad$ m longer than $100 \mathrm{~m}$.
|
8. 1 .
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙 }}$, respectively. When 甲 is $1 \mathrm{~m}$ away from the finish line, the time spent is $t$. Then $v_{\text {乙 }} t=x-2$.
Thus, $\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}$.
Let the time 甲 takes to reach the finish line be $t^{\prime}$. Then $v_{\text {甲 }} t^{\prime}=x, v_{\text {乙 }} t^{\prime}=x-1.01$.
Thus, $\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x}{x-1.01}$.
We can get the equation $\frac{x-1}{x-2}=\frac{x}{x-1.01} \Rightarrow x=101 \mathrm{~m}$.
Therefore, this track is $1 \mathrm{~m}$ longer than $100 \mathrm{~m}$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The positive numbers $b_{1}, b_{2}, \cdots, b_{60}$ are arranged in sequence and satisfy $\frac{b_{2}}{b_{1}}=\frac{b_{3}}{b_{2}}=\cdots=\frac{b_{60}}{b_{59}}$.
Determine the value of $\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right)$.
|
Given that $\left\{b_{k}\right\}$ is a geometric sequence with a common ratio of $q$, then $b_{k}=b_{1} q^{k-1}$. Therefore, $b_{k} b_{61-k}=b_{1} b_{60}$. Hence,
$$
\begin{array}{l}
\log _{b_{11} b_{50}}\left(b_{1} b_{2} \cdots b_{60}\right) \\
=\log _{b_{1} b_{60}}\left(b_{1} b_{60}\right)^{30}=30 .
\end{array}
$$
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $[x]$ represents the greatest integer not exceeding the real number $x$. If
$$
\left[\log _{3} 6\right]+\left[\log _{3} 7\right]+\cdots+\left[\log _{3} n\right]=2009 \text {, }
$$
determine the value of the positive integer $n$.
|
4. For $3^{k} \leqslant i \leqslant 3^{k+1}-1$, we have $\left[\log _{3} i\right]=k$. Therefore, $\sum_{i=3^{k}}^{3 k+1}\left[\log _{3} i\right]=2 \times 3^{k} k$.
Also, $3+\sum_{k=2}^{4} 2 \times 3^{k} k<2009<3+\sum_{k=2}^{5} 2 \times 3^{k} k$, hence $3^{5}<n<3^{6}-1$.
From $3+\sum_{k=2}^{4} 2 \times 3^{k} k+5\left(n-3^{5}+1\right)=2009$, solving for $n$ gives $n=474$.
|
474
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The graphs of the functions $f(x)=2 x^{2}-2 x-1$ and $g(x)=$ $-5 x^{2}+2 x+3$ intersect at two points. The equation of the line passing through these two points is $y=a x+b$. Find the value of $a-b$.
|
6. The graphs of the functions $f(x)$ and $g(x)$ intersect at two points, which are the solutions to the system of equations $\left\{\begin{array}{l}y=2 x^{2}-2 x-1, \\ y=-5 x^{2}+2 x+3\end{array}\right.$. Therefore,
$$
\begin{array}{l}
7 y=5\left(2 x^{2}-2 x-1\right)+2\left(-5 x^{2}+2 x+3\right) \\
=-6 x+1 .
\end{array}
$$
Thus, the two intersection points lie on the line $y=-\frac{6}{7} x+\frac{1}{7}$, i.e., $a=-\frac{6}{7}, b=\frac{1}{7}$.
Therefore, $a-b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that the 6027-digit number $\frac{a b c a b c \cdots a b c}{2000 \uparrow a b c}$ is a multiple of 91. Find the sum of the minimum and maximum values of the three-digit number $\overline{a b c}$.
|
7. From $91=7 \times 13, 1001=7 \times 11 \times 13$, we know 91 | 1001.
And $\overline{a b c a b c}=1001 \times \overline{a b c}$, thus, $91 \mid \overline{a b c a b c}$. $2009 \uparrow a b c$
$$
\overline{a_{2009 \uparrow a b c}^{a b c a b c}}=\underset{2 \times 1004 \uparrow a b c}{\overline{a b c a b c \cdots a b c}} \times 1000+\overline{a b c},
$$
Since $\overline{a b c a b c \cdots a b c}$ is divisible by 91, then $\overline{a b c}$ is divisible by 91.
$2 \times 1004 \uparrow a b$.
The smallest three-digit number divisible by 91 is 182, and the largest is 910, so the sum of the minimum and maximum values of $\overline{a b c}$ is 1092.
|
1092
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A three-digit number has 3 digits, none of which are 0, and its square is a six-digit number that has exactly 3 digits as 0. Write down one such three-digit number: $\qquad$
|
8.448 or 548 or 949
|
448
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { Three. (20 points) (1) Prove: } \\
\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=4 \sin ^{2} 2 x-3(x \in \mathbf{R}) ;
\end{array}
$$
(2) Find the value: $\prod_{t=0}^{2^{8}}\left(4 \sin ^{2} \frac{t \pi}{2^{9}}-3\right)$.
|
$$
\begin{array}{l}
=(1)\left(4 \sin ^{2} x-3\right)\left(4 \cos ^{2} x-3\right) \\
=16 \sin ^{2} x \cdot \cos ^{2} x-12\left(\sin ^{2} x+\cos ^{2} x\right)+9 \\
=4 \sin ^{2} 2 x-3
\end{array}
$$
(2) Let $A_{k}=\prod_{i=0}^{2 k-1}\left(4 \sin ^{2} \frac{t \pi}{2^{k}}-3\right)$.
When $k \geqslant 2$, by (1) we have
$$
\begin{aligned}
A_{k}= & \prod_{t=0}^{2^{k-2}-1}\left\{\left(4 \sin ^{2} \frac{t \pi}{2^{k}}-3\right)\right. \\
& {\left.\left[4 \sin ^{2} \frac{\left(2^{k-1}-t\right) \pi}{2^{k}}-3\right]\right\} } \\
& =\prod_{t=0}^{2^{k-2}-1}\left(4 \sin ^{2} \frac{2^{k-1} \pi}{2^{k}}-3\right) \\
& \left.=-\sin ^{2} \frac{t \pi}{2^{k-1}}-3\right)\left[-\left(4 \sin ^{2} \frac{2^{k-2} \pi}{2^{k-1}}-3\right)\right]
\end{aligned}
$$
Therefore, $A_{9}=A_{1}=\left(4 \sin ^{2} 0-3\right)\left(4 \sin ^{2} \frac{\pi}{2}-3\right)$ $=-3$.
|
-3
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
For a positive integer $n$, let $t_{n}=\frac{n(n+1)}{2}$. Writing down the last digits of $t_{1}=1, t_{2}=3, t_{3}=6, t_{4}=10, t_{5}=15 \cdots \cdots$ can form an infinite repeating decimal: $0.13605 \cdots$. Find the length of the repeating cycle of this decimal.
|
$$
\begin{array}{l}
t_{n+20}-t_{n}=\frac{(n+20)(n+20+1)}{2}-\frac{n(n+1)}{2} \\
=20 n+210=10(2 n+21)
\end{array}
$$
That is, the last digit of $t_{n+20}$ is the same as that of $t_{n}$.
Therefore, 20 is the length of the repeating cycle of this repeating decimal. This repeating decimal is
$$
0.13605186556815063100 .
$$
|
20
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$ (for example, $f(3)$ is the sum of the digits of $3 \times 3^{2}+3+1=31$, i.e., $f(3)=4$).
(1) Prove that for any positive integer $n, f(n) \neq 1$, and $f(n) \neq 2$;
(2) Try to find a positive integer $n$ such that $f(n)=3$.
|
(1) Since $3 n^{2}+n+1$ is an odd number greater than 3, hence $f(n) \neq 1$.
If $f(n)=2$, then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e., $3 n^{2}+n+1=10^{k}+1$ (where $k$ is an integer greater than 1).
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Since $(n, 3 n+1)=1$, we have $\left\{\begin{array}{l}n=2^{k}, \\ 3 n+1=5^{k} \text {. }\end{array}\right.$
Therefore, $3 n+1 \leqslant 4 n=4 \times 2^{k}<5^{k}$. This is a contradiction.
Hence, $f(n) \neq 2$.
(2) When $n=8$, since $3 n^{2}+n+1=201$, thus, $f(8)=3$.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given a three-digit number $x y z(1 \leqslant x \leqslant 9,0 \leqslant y, z$ $\leqslant 9)$. If $x y z=x!+y!+z!$, then the value of $x+y+z$ is
|
7. 10 .
Since $6!=720$, we have $0 \leqslant x, y, z \leqslant 5$.
And $1!=1, 2!=2, 3!=6, 4!=24, 5!=120$.
By observation, we get $x=1, y=4, z=5$. Therefore,
$$
x+y+z=10 \text {. }
$$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Two quadratic equations with unequal leading coefficients
$$
\begin{array}{l}
(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0, \\
(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0
\end{array}
$$
$\left(a 、 b \in \mathbf{N}_{+}\right)$ have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$.
|
Given the known equations, we have $a \neq 1, b \neq 1$. Therefore, $a > 1, b > 1$, and $a \neq b$. Let $x_{0}$ be the common root of the two equations. It is easy to see that $x_{0} \neq 1$. By the definition of the roots of the equation, $a$ and $b$ are the two distinct real roots of the equation
$$
\left(1-x_{0}\right) t^{2}+\left(2+x_{0}^{2}\right) t-\left(2 x_{0}+x_{0}^{2}\right)=0
$$
Then,
$$
\begin{array}{l}
a+b=\frac{2+x_{0}^{2}}{x_{0}-1}, \quad a b=\frac{2 x_{0}+x_{0}^{2}}{x_{0}-1} \\
\Rightarrow (a-1)(b-1)=3 \\
\Rightarrow \{a, b\}=\{2,4\} \\
\Rightarrow \frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=a^{b} b^{a}=2^{4} \times 4^{2}=256 .
\end{array}
$$
|
256
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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