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3. If real numbers $m, n, p, q$ satisfy the conditions
$$
\begin{array}{l}
m+n+p+q=22, \\
m p=n q=100,
\end{array}
$$
then the value of $\sqrt{(m+n)(n+p)(p+q)(q+m)}$ is
$\qquad$
|
3. 220 .
From the given, we have
$$
\begin{aligned}
& (m+n)(n+p)(p+q)(q+m) \\
= & {[(m+n)(p+q)][(n+p)(q+m)] } \\
= & (200+m q+n p)(200+m n+p q) \\
= & 200^{2}+100\left(m^{2}+n^{2}+p^{2}+q^{2}+2 m n+\right. \\
& 2 m q+2 n p+2 p q) \\
= & 200^{2}+100\left[(m+n+p+q)^{2}-400\right] \\
= & {[10(m+n+p+q)]^{2} . }
\end{aligned}
$$
Therefore, the original expression $=10(m+n+p+q)=220$.
|
220
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) If the two sides of a right triangle, $x, y$, are both prime numbers, and make the algebraic expressions $\frac{2 x-1}{y}$ and $\frac{2 y+3}{x}$ both positive integers, find the inradius $r$ of this right triangle.
|
(1) If $x>y$, then
$$
1 \leqslant \frac{2 y+3}{x}<\frac{2 x+3}{x}<4 \text {. }
$$
It is easy to see that $\frac{2 y+3}{x}=1$ or 2.
$$
\begin{array}{l}
\text { (i) From } \frac{2 y+3}{x}=1 \\
\Rightarrow x=2 y+3 \\
\Rightarrow \frac{2 x-1}{y}=\frac{2(2 y+3)-1}{y}=4+\frac{5}{y} \\
\Rightarrow y=5, x=13 ;
\end{array}
$$
(ii) From $\frac{2 y+3}{x}=2 \Rightarrow 2 x=2 y+3$, the equation is obviously not valid, leading to a contradiction.
(2) If $x=y$, then $\frac{2 x-1}{y}$ is not a positive integer, which contradicts the given condition.
(3) If $x<y$, then
$$
\begin{array}{l}
1 \leqslant \frac{2 x-1}{y}<2 \Rightarrow \frac{2 x-1}{y}=1 \Rightarrow y=2 x-1 \\
\Rightarrow \frac{2 y+3}{x}=\frac{2(2 x-1)+3}{x}=4+\frac{1}{x}
\end{array}
$$
It cannot be a positive integer, which contradicts the given condition.
In summary, only $y=5, x=13$ meets the conditions of the problem. The third side of the right triangle is 12 or $\sqrt{194}$.
$$
\text { Hence } r=\frac{5+12-13}{2}=2
$$
or $r=\frac{5+13-\sqrt{194}}{2}=\frac{18-\sqrt{194}}{2}$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the function
$$
f(x)=a x+\sin x
$$
has perpendicular tangents on its graph, then the real number $a$ is
$\qquad$ .
|
4.0.
Notice that, $f^{\prime}(x)=a+\cos x$.
If the function $f(x)$ has two perpendicular tangents, then there exist $x_{1}, x_{2} \in \mathbf{R}$, such that
$$
\begin{array}{l}
f^{\prime}\left(x_{1}\right) f^{\prime}\left(x_{2}\right)=-1 \\
\Leftrightarrow\left(a+\cos x_{1}\right)\left(a+\cos x_{2}\right)=-1 \\
\Leftrightarrow a^{2}+a\left(\cos x_{1}+\cos x_{2}\right)+\cos x_{1} \cdot \cos x_{2}+1=0 \\
\Leftrightarrow\left(a+\frac{\cos x_{1}+\cos x_{2}}{2}\right)^{2}+1- \\
\quad\left(\frac{\cos x_{1}-\cos x_{2}}{2}\right)^{2}=0 \\
\Leftrightarrow \cos x_{1}=-\cos x_{2}= \pm 1, a=0 .
\end{array}
$$
|
0
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 The equation $2 x^{2}+5 x y+2 y^{2}=2007$ has $\qquad$ different integer solutions.
(2007, National Junior High School Mathematics League Sichuan Preliminary Competition)
|
The original equation can be transformed into
$$
(2 x+y)(x+2 y)=2007 \text {. }
$$
Since $x$ and $y$ are integers, without loss of generality, assume $x \leqslant y$, so,
$$
2 x+y \leqslant x+2 y \text {. }
$$
Notice that, $31[2 x+y)+(x+2 y)]$.
Thus, from equation (1), we get the system of equations
$$
\left\{\begin{array} { l }
{ 2 x + y = 3 , } \\
{ x + 2 y = 6 6 9 ; }
\end{array} \left\{\begin{array}{l}
2 x+y=-669, \\
x+2 y=-3 .
\end{array}\right.\right.
$$
The integer solutions are
When $x \geqslant y$, the original equation has two more sets of integer solutions
$$
\left\{\begin{array} { l }
{ x = 4 4 5 , } \\
{ y = - 2 2 1 , }
\end{array} \left\{\begin{array}{l}
x=221, \\
y=-445 .
\end{array}\right.\right.
$$
In summary, the original equation has 4 sets of integer solutions.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Select $k$ numbers from 1 to 2012, such that among the selected $k$ numbers, there are definitely three numbers that can form the lengths of the sides of a triangle (the lengths of the three sides of the triangle must be distinct). What is the minimum value of $k$ that satisfies the condition?
|
Three, the problem is equivalent to:
Select $k-1$ numbers from $1,2, \cdots, 2012$, such that no three of these numbers can form the sides of a triangle with unequal sides. What is the maximum value of $k$ that satisfies this condition?
Now consider the arrays that meet the above conditions.
When $k=4$, the smallest three numbers are $1, 2, 3$. This array can be expanded continuously, as long as the added number is greater than or equal to the sum of the two largest numbers in the existing array. To maximize $k$, the added number should equal the sum of the two largest numbers in the existing array. Thus, we get
\[
\begin{array}{l}
1,2,3,5,8,13,21,34,55,89,144, \\
233,377,610,987,1597,
\end{array}
\]
a total of 16 numbers.
For any array $a_{1}, a_{2}, \cdots, a_{n}$ that meets the above conditions, it is clear that $a_{i}$ is always greater than or equal to the $i$-th number in (1).
Therefore, $n \leqslant 16 \leqslant k-1$.
Thus, the minimum value of $k$ is 17.
(Chang Yongsheng, Nankai High School, Tianjin, 300100)
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A five-digit number $\overline{a b c d e}$ satisfies
$$
ac>d, dd, b>e
$$
(for example, 37201, 45412), if its digits change with the position in a manner similar to the monotonicity of a sine function over one period, then this five-digit number is said to conform to the "sine rule". Therefore, there are $\qquad$ five-digit numbers that conform to the sine rule.
|
6.2892 .
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers among $a, b, c, d, e$. It is easy to see that $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$. In this case, there are $10-k$ ways to choose $(b, d)$, and $a, c, e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ groups. Therefore, there are $(k-1)^{3}(10-k)$ numbers that conform to the sine rule.
Thus, the total number of numbers that conform to the sine rule is
$$
\sum_{k=2}^{9}(k-1)^{3}(10-k)=2892 \text { (numbers). }
$$
|
2892
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
320 Suppose there are $2 k+1$ consecutive natural numbers,
where the sum of the squares of the $k$ larger natural numbers is equal to the sum of the squares of the remaining $k+1$ smaller natural numbers (such as $5^{2}=4^{2}+3^{2}$ $\left.(k=1), 14^{2}+13^{2}=12^{2}+11^{2}+10^{2}(k=2)\right)$. For convenience, such an equation is called a "modified way". Clearly, the natural numbers $3, 4, 5, 10, 11, 12, 13, 14$ are all numbers in the modified way. The question is: Is the four-digit number 2012 a number in the modified way?
|
Let
\[
\begin{array}{l}
(a+k)^{2}+(a+k-1)^{2}+\cdots+(a+1)^{2} \\
=a^{2}+(a-1)^{2}+\cdots+(a-k)^{2}
\end{array}
\]
be the $k$-th transformation method,
\[
w_{k}=(a+k, a-k)
\]
indicates that the largest number is $a+k$ and the smallest number is $a-k$.
Clearly, $a^{2}=4 \times(1+2+\cdots+k) a$, or $a=4 \times(1+2+\cdots+k)=2 k(k+1)$.
From $2 k(k+1)-k \leqslant 2012 \leqslant 2 k(k+1)+k$, solving gives $k=31$.
It is easy to see that $w_{31}=(2015,1953)$.
Thus, 2012 is the 31st transformation method number.
(Tian Yonghai, Suihua Educational Institute, Heilongjiang, 152054)
|
2012
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Suppose the lengths of the two legs of a right triangle are $a$ and $b$, and the length of the hypotenuse is $c$. If $a$, $b$, and $c$ are all positive integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition.
(2010, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
Solve: By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.
Also, $c=\frac{1}{3} a b-(a+b)$, so
$$
c^{2}=\left[\frac{1}{3} a b-(a+b)\right]^{2} \text {. }
$$
Rearranging gives $a b-6(a+b)+18=0$.
Thus, $(a-6)(b-6)=18$.
Since $a$ and $b$ are both positive integers, and without loss of generality, let $a<b$, then,
$$
\left\{\begin{array} { l }
{ a - 6 = 1 , } \\
{ b - 6 = 1 8 ; }
\end{array} \left\{\begin{array} { l }
{ a - 6 = 2 , } \\
{ b - 6 = 9 ; }
\end{array} \left\{\begin{array}{l}
a-6=3, \\
b-6=6 .
\end{array}\right.\right.\right.
$$
Solving gives $\left\{\begin{array}{l}a=7, \\ b=24 ;\end{array}\left\{\begin{array}{l}a=8, \\ b=15 ;\end{array}\left\{\begin{array}{l}a=9, \\ b=12 .\end{array}\right.\right.\right.$
The corresponding $c=25,17,15$.
Therefore, there are 3 right-angled triangles that satisfy the conditions.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 It is known that for any $x$,
$$
a \cos x + b \cos 2x \geqslant -1
$$
always holds. Find the minimum value of $a + b$.
(2009, Peking University Independent Admission Examination)
|
When $x=0$, $a+b \geqslant-1$. Taking $a=-\frac{4}{5}, b=-\frac{1}{5}$, then
$$
\begin{array}{l}
a \cos x+b \cos 2 x \\
=-\frac{2}{5}(\cos x+1)^{2}+\frac{3}{5} \\
\geqslant-\frac{2}{5}(1+1)^{2}+\frac{3}{5} \\
=-1 .
\end{array}
$$
Therefore, the minimum value of $a+b$ is -1.
|
-1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. If $a, b, c$ are positive numbers, and satisfy
$$
\begin{array}{c}
a+b+c=9, \\
\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{10}{9}, \\
\text { then } \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=
\end{array}
$$
|
B. 7 .
From the given information, we have
$$
\begin{array}{l}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\
=\frac{9-b-c}{b+c}+\frac{9-c-a}{c+a}+\frac{9-a-b}{a+b} \\
=\frac{9}{b+c}+\frac{9}{c+a}+\frac{9}{a+b}-3 \\
=9 \times \frac{10}{9}-3=7 .
\end{array}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. A. As shown in Figure 4, the side length of square $ABCD$ is $2 \sqrt{15}, E, F$ are the midpoints of sides $AB, BC$ respectively, $AF$ intersects $DE, DB$ at points $M, N$. Then the area of $\triangle DMN$ is $\qquad$.
|
7. A. 8 .
Connect $D F$. Let the side length of the square $A B C D$ be $2 a$. From the problem, we easily know
$$
\begin{array}{l}
\triangle B F N \backsim \triangle D A N \\
\Rightarrow \frac{A D}{B F}=\frac{A N}{N F}=\frac{D N}{B N}=\frac{2}{1} \\
\Rightarrow A N=2 N F \Rightarrow A N=\frac{2}{3} A F .
\end{array}
$$
In the right triangle $\triangle A B F$, given $A B=2 a, B F=a$, we have
$$
A F=\sqrt{A B^{2}+B F^{2}}=\sqrt{5} a \text {. }
$$
Thus, $\cos \angle B A F=\frac{A B}{A F}=\frac{2 \sqrt{5}}{5}$.
From the problem, we know
$$
\begin{array}{c}
\triangle A D E \cong \triangle B A F \Rightarrow \angle A E D=\angle A F B \\
\Rightarrow \angle A M E=180^{\circ}-\angle B A F-\angle A E D \\
=180^{\circ}-\angle B A F-\angle A F B=90^{\circ} .
\end{array}
$$
Therefore, $A M=A E \cos \angle B A F=\frac{2 \sqrt{5}}{5} a$,
$$
M N=A N-A M=\frac{2}{3} A F-A M=\frac{4 \sqrt{5}}{15} a \text {. }
$$
Thus, $\frac{S_{\triangle M N D}}{S_{\triangle A F D}}=\frac{M N}{A F}=\frac{4}{15}$.
Since $S_{\triangle A F D}=\frac{1}{2}(2 a)(2 a)=2 a^{2}$, then
$$
S_{\triangle M N D}=\frac{4}{15} S_{\triangle A F D}=\frac{8}{15} a^{2} .
$$
Given $a=\sqrt{15}$, we have $S_{\triangle M N D}=8$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. Let $n$ be an integer, and $1 \leqslant n \leqslant 2012$. If $\left(n^{2}-n+3\right)\left(n^{2}+n+3\right)$ is divisible by 5, then the number of all $n$ is $\qquad$.
|
B. 1610.
Notice,
$$
\begin{array}{l}
\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \\
=n^{4}+5 n^{2}+9 \\
=(n-1)(n+1)\left(n^{2}+1\right)+5 n^{2}+10 .
\end{array}
$$
When $n$ is divided by 5, the remainder is 1 or 4, then $n-1$ or $n+1$ is divisible by 5, so
$$
51\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {; }
$$
When $n$ is divided by 5, the remainder is 2 or 3, then $n^{2}+1$ is divisible by 5, so
$$
5 I\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {; }
$$
When $n$ is divided by 5, the remainder is 0,
$$
5 \nmid\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {. }
$$
Therefore, the number of all $n$ that meet the requirements of the problem is
$$
\frac{2010}{10} \times 8+2=1610 .
$$
|
1610
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. A. 2 eighth-grade students and $m$ ninth-grade students participate in a single round-robin chess tournament, where each participant plays against every other participant exactly once. The scoring rule is: the winner of each match gets 3 points, the loser gets 0 points, and in the case of a draw, both players get 1 point. After the tournament, the total score of all students is 130 points, and the number of draws does not exceed half of the total number of matches. Then the value of $m$ is $\qquad$ .
|
9. A. 8 .
Let the number of draws be $a$, and the number of wins (losses) be $b$.
From the problem, we know
$$
2 a+3 b=130 \text {. }
$$
This gives $0 \leqslant b \leqslant 43$.
Also, $a+b=\frac{(m+1)(m+2)}{2}$
$$
\Rightarrow 2 a+2 b=(m+1)(m+2) \text {. }
$$
Thus, $0 \leqslant b=130-(m+1)(m+2) \leqslant 43$.
Therefore, $87 \leqslant(m+1)(m+2) \leqslant 130$.
From this, we get $m=8$ or 9 .
When $m=8$, $b=40, a=5$;
When $m=9$, $b=20, a=35, a>\frac{55}{2}$, which does not meet the problem's conditions.
Hence, $m=8$.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. Given that $n$ is even, and $1 \leqslant n \leqslant 100$. If there is a unique pair of positive integers $(a, b)$ such that $a^{2}=b^{2}+n$ holds, then the number of such $n$ is
|
B. 12.
From the given, we have $(a-b)(a+b)=n$, and $n$ is even, so $a-b$ and $a+b$ are both even.
Therefore, $n$ is a multiple of 4.
Let $n=4m$. Then $1 \leqslant m \leqslant 25$.
(1) If $m=1$, we get $b=0$, which contradicts that $b$ is a positive integer.
(2) If $m$ has at least two different prime factors, then there are at least two pairs of positive integers $(a, b)$ that satisfy
$$
\frac{a-b}{2} \cdot \frac{a+b}{2}=m.
$$
If $m$ is exactly a power of a prime, and this power is not less than 3, then there are at least two pairs of positive integers $(a, b)$ that satisfy equation (1).
(3) If $m$ is a prime number, or $m$ is exactly a power of a prime, and this power is 2, then there is a unique pair of positive integers $(a, b)$ that satisfies equation (1).
Since there is a unique pair of positive integers $(a, b)$, the possible values of $m$ are
$$
2,3,4,5,7,9,11,13,17,19,23,25,
$$
a total of 12.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. A. Given integers $a, b$ satisfy $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$.
|
13. A. Let $a-b=m$ (where $m$ is a prime number), and $ab=n^2$ (where $n$ is a non-negative integer).
When $b \neq 0$,
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} .
\end{array}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive integers, and $2a-m+2n > 2a-m-2n$ (since $m$ is a prime number),
we have $\left\{\begin{array}{l}2a-m+2n=m^{2} \\ 2a-m-2n=1 .\end{array}\right.$,
Solving these, we get $a=\frac{(m+1)^{2}}{4}, n=\frac{m^{2}-1}{4}$.
Thus, $b=a-m=\frac{(m-1)^{2}}{4}$.
Also, $a \geqslant 2012$, i.e., $\frac{(m+1)^{2}}{4} \geqslant 2012$.
Since $m$ is a prime number, solving this gives $m \geqslant 89$.
At this point, $a \geqslant \frac{(89+1)^{2}}{4}=2025$.
When $b=0$,
from $a-b=a=m$, we know that $a=m$ is a prime number.
Also, $a \geqslant 2012$, so the smallest prime number is 2017.
Therefore, the minimum value of $a$ is 2017.
|
2017
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. In a convex $n$-sided polygon, what is the maximum number of interior angles that can be $150^{\circ}$? Explain your reasoning.
|
B. Suppose in a convex $n$-sided polygon, there are $k$ interior angles equal to $150^{\circ}$. Then there are $n-k$ interior angles not equal to $150^{\circ}$.
(1) If $k=n$, then
$$
n \times 150^{\circ}=(n-2) \times 180^{\circ},
$$
we get $n=12$. Thus, in a regular dodecagon, all 12 interior angles are $150^{\circ}$.
(2) If $k(n-2) \times 180^{\circ}$,
$$
we get $k<12$, i.e., $k \leqslant 11$.
When $k=11$, there exists a convex $n$-sided polygon with 11 interior angles of $150^{\circ}$, and the remaining $n-k$ angles are
$$
\begin{aligned}
\alpha & =\frac{(n-2) \times 180^{\circ}-11 \times 150^{\circ}}{n-11} \\
& =\left(6-\frac{1}{n-11}\right) \times 30^{\circ} \in\left[165^{\circ}, 180^{\circ}\right) .
\end{aligned}
$$
(3) If $k<n$, and $8 \leqslant n \leqslant 11$, when $k=n-1$, there exists a convex $n$-sided polygon with $n-1$ interior angles of $150^{\circ}$, and the other angle is
$$
\begin{array}{l}
\alpha=(n-2) \times 180^{\circ}-(n-1) \times 150^{\circ} \\
=(n-7) \times 30^{\circ} \in\left[30^{\circ}, 120^{\circ}\right] .
\end{array}
$$
(4) If $k<n$, and $3 \leqslant n \leqslant 7$, from (3) we know $k \leqslant n-2$, when $k=n-2$, there exists a convex $n$-sided polygon with $n-2$ interior angles of $150^{\circ}$, and the other two angles are both $(n-2) \times 15^{\circ}$.
In summary, when $n=12$, the maximum value of $k$ is 12;
when $n \geqslant 13$, the maximum value of $k$ is 11;
when $8 \leqslant n \leqslant 11$, the maximum value of $k$ is $n-1$;
when $3 \leqslant n \leqslant 7$, the maximum value of $k$ is $n-2$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, a_{2}=4, a_{3}=9, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}(n=4,5, \cdots) .
\end{array}
$$
Then $a_{2011}=$
|
$-、 1.8041$.
From the problem, we have
$$
a_{2}-a_{1}=3, a_{3}-a_{2}=5 \text {, }
$$
and $a_{n}-a_{n-1}=a_{n-2}-a_{n-3}(n \geqslant 4)$.
Thus, $a_{2 n}-a_{2 n-1}=3, a_{2 n+1}-a_{2 n}=5\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $a_{2 n+1}-a_{2 n-1}=8$.
Hence, $a_{2011}=\sum_{k=1}^{1005}\left(a_{2 k+1}-a_{2 k-1}\right)+a_{1}$
$$
=1005 \times 8+1=8041 \text {. }
$$
|
8041
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given a function $f(n)$ defined on the set of positive integers satisfies the conditions:
(1) $f(m+n)=f(m)+f(n)+m n\left(m, n \in \mathbf{N}_{+}\right)$;
(2) $f(3)=6$.
Then $f(2011)=$ . $\qquad$
|
3.2023066.
In condition (1), let $n=1$ to get
$$
f(m+1)=f(m)+f(1)+m \text {. }
$$
Let $m=n=1$ to get
$$
f(2)=2 f(1)+1 \text {. }
$$
Let $m=2, n=1$, and use condition (2) to get
$$
6=f(3)=f(2)+f(1)+2 \text {. }
$$
From equations (3) and (2), we get
$$
f(1)=1, f(2)=3 \text {. }
$$
Substitute into equation (1) to get
$$
\begin{array}{l}
f(m+1)-f(m)=m+1 . \\
\text { Hence } f(2011)=\sum_{k=1}^{2010}[f(k+1)-f(k)]+f(1) \\
=\sum_{k=1}^{2010}(k+1)+1=\frac{2011 \times 2012}{2} \\
=2023066 .
\end{array}
$$
|
2023066
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Equation
$$
|| \cdots|||x|-1|-2| \cdots|-2011|=2011
$$
has $\qquad$ solutions.
|
4. 4.
The solutions to the equation $||x|-1|=1$ are $x=0$ or $\pm 2$;
The solutions to the equation $|||x|-1|-2|=2$ are $x= \pm 1$ or $\pm 5$;
The solutions to the equation $||||x|-1|-2|-3|=3$ are $x= \pm 3$ or $\pm 9$;
In general, the solutions to the equation
$$
|1 \cdots||| x|-1|-2|\cdots|-n \mid=n(n \geqslant 2)
$$
are
$$
x= \pm \frac{n(n-1)}{2} \text { or } \pm \frac{n(n+3)}{2} \text {. }
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. 10 students stand in a row, and a red, yellow, or blue hat is to be given to each student. It is required that each color of hat must be present, and the hats of adjacent students must be of different colors. Then the number of ways to distribute the hats that meet the requirements is $\qquad$ kinds.
|
8. 1530.
Generalize to the general case.
Let the number of ways to arrange $n$ students according to the given conditions be $a_{n}$. Then
$$
\begin{array}{l}
a_{3}=6, a_{4}=18, a_{n+1}=2 a_{n}+6(n \geqslant 3) . \\
\text { Hence } a_{n+1}+6=2\left(a_{n}+6\right) \\
\Rightarrow a_{n}=\left(a_{3}+6\right) \times 2^{n-3}-6 \\
\Rightarrow a_{10}=12 \times 2^{7}-6=1530 .
\end{array}
$$
|
1530
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the parabola
$$
y=x^{2}+(k+1) x+1
$$
intersects the $x$-axis at two points $A$ and $B$, not both on the left side of the origin. The vertex of the parabola is $C$. To make $\triangle A B C$ an equilateral triangle, the value of $k$ is $\qquad$
|
$=1 .-5$.
From the problem, we know that points $A$ and $B$ are to the right of the origin, and
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\sqrt{(k+1)^{2}-4} \text {. } \\
\text { Then } \frac{\sqrt{3}}{2} \sqrt{(k+1)^{2}-4}=\left|1-\left(\frac{k+1}{2}\right)^{2}\right| \text {. }
\end{array}
$$
Solving this, we get $k=-5$.
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $A M$ is the median of $\triangle A B C$ on side $B C$, $P$ is the centroid of $\triangle A B C$, and a line $E F$ through point $P$ intersects sides $A B$ and $A C$ at points $E$ and $F$ respectively. Then $\frac{B E}{A E}+\frac{C F}{A F}=$ $\qquad$
|
3. 1 .
Draw $B G$ and $C K$ parallel to $A M$ through points $B$ and $C$ respectively, intersecting line $E F$ at points $G$ and $K$. Then
$$
\frac{B E}{A E}=\frac{B G}{A P}, \frac{C F}{A F}=\frac{C K}{A P} \text {. }
$$
Adding the two equations gives
$$
\frac{B E}{A E}+\frac{C F}{A F}=\frac{B G+C K}{A P} .
$$
In trapezoid BCKG, we have
$$
P M=\frac{1}{2}(B G+C K) \text {. }
$$
Since $P$ is the centroid, $A P=2 P M$.
$$
\text { Therefore, } \frac{B E}{A E}+\frac{C F}{A F}=\frac{2 P M}{2 P M}=1 \text {. }
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 In $\triangle A B C$, $A C=B C, \angle A C B=$ $90^{\circ}, D, E$ are two points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$
(2006, Beijing Middle School Mathematics Competition (Grade 8))
|
Solve as shown in Figure 1, construct square $C A H B$, and extend $C D$, $C E$ to intersect $A H$, $B H$ at points $G$, $F$ respectively.
Let $D E=x$.
Since $A C / / B F$
$$
\Rightarrow \frac{3+x}{4}=\frac{A C}{B F} \text {, }
$$
$B C / / A G$
$$
\begin{array}{c}
\Rightarrow \frac{4+x}{3}=\frac{B C}{A G} . \\
\text { Then } \frac{(3+x)(4+x)}{12}=\frac{A C^{2}}{B F \cdot A G} .
\end{array}
$$
Also, $\angle D C E=\angle F B E=\angle G A D=45^{\circ}$, hence $\triangle A D G \backsim \triangle C D E \backsim \triangle B F E$
$$
\Rightarrow \frac{B F}{A D}=\frac{B E}{A G} \Rightarrow B F \cdot A G=12 \text {. }
$$
Notice that,
$$
A C^{2}=B C^{2}=\frac{1}{2} A B^{2}=\frac{1}{2}(3+x+4)^{2} \text {. }
$$
Substitute equations (2) and (3) into equation (1), simplify and rearrange to get
$$
x^{2}=3^{2}+4^{2}=25 \Rightarrow x=5 \text {. }
$$
Therefore, $S_{\triangle A B C}=\frac{1}{4} A B^{2}=\frac{1}{4}(3+5+4)^{2}=36$.
|
36
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 In $\triangle A B C$, it is known that $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=2, C D=3$, then $S_{\triangle A B C}$ $=$ $\qquad$
(2007, Shandong Province Junior High School Mathematics Competition)
|
Solve as shown in Figure 5, with $AB$ as the axis of symmetry, construct the symmetric figure of $\triangle ADB$ as $\triangle AGB$, and with $AC$ as the axis of symmetry, construct the symmetric figure of $\triangle ADC$ as $\triangle AFC$, and extend $GB$ and $FC$ to intersect at point $E$. Then it is easy to know that quadrilateral $AGEF$ is a square.
Assume $AD=h$. Then
$$
\begin{array}{l}
BE=h-2, CE=h-3 . \\
\text { By } BC^{2}=BE^{2}+CE^{2} \\
\Rightarrow(h-2)^{2}+(h-3)^{2}=5^{2} \\
\Rightarrow h^{2}-5 h-6=0 \\
\Rightarrow h=6 \\
\Rightarrow S_{\triangle ABC}=\frac{1}{2} BC \cdot AD=\frac{1}{2} \times 5 \times 6=15 .
\end{array}
$$
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given four distinct positive real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2012}-c^{2012}\right)\left(a^{2012}-d^{2012}\right)=2012, \\
\left(b^{2012}-c^{2012}\right)\left(b^{2012}-d^{2012}\right)=2012 . \\
\text { Then }(a b)^{2012}-(c d)^{2012}=(\quad) .
\end{array}
$$
(A) -2012
(B) -2011
(C) 2012
(D) 2011
|
- 1. A.
From the problem, we know that \(a^{2012}\) and \(b^{2012}\) are the two distinct real roots of the quadratic equation in \(x\):
\[
\left(x-c^{2012}\right)\left(x-d^{2012}\right)=2012,
\]
which is \(x^{2}-\left(c^{2012}+d^{2012}\right) x+(c d)^{2012}-2012=0\).
Thus, \(a^{2012} b^{2012}=(c d)^{2012}-2012\).
Therefore, \((a b)^{2012}-(c d)^{2012}=-2012\).
|
-2012
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. There are $n$ people registered to participate in four sports competitions: A, B, C, and D. It is stipulated that each person must participate in at least one competition and at most two competitions, but competitions B and C cannot be registered for simultaneously. If in all different registration methods, there must exist one method where at least 20 people register, then the minimum value of $n$ is $(\quad$.
(A) 171
(B) 172
(C) 180
(D) 181
|
5. B.
Use the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{J}}\right)$ to represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is recorded as 1 (for example, if participating in event 甲, record $a_{\text {甲 }}=1$); if not participating in a certain event, the corresponding number is recorded as 0 (for example, if not participating in event 甲, record $a_{\text {甲 }}=0$).
Thus, each person's registration method for the events can be one of the following nine possibilities:
$$
\begin{array}{l}
(1,1,0,0),(1,0,1,0),(1,0,0,1), \\
(0,1,0,1),(0,0,1,1),(1,0,0,0), \\
(0,1,0,0),(0,0,1,0),(0,0,0,1) .
\end{array}
$$
Using these nine possibilities as the nine drawers, by the known pigeonhole principle, we have $n=19 \times 9+r$.
When $r=1$, the minimum value of $n$ is 172.
|
172
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the circumradius of $\triangle A B C$ is $1, \angle A$, $\angle B$, and $\angle C$'s angle bisectors intersect the circumcircle of $\triangle A B C$ at points $A_{1}$, $B_{1}$, and $C_{1}$, respectively. Then
$$
\frac{A A_{1} \cos \frac{A}{2}+B B_{1} \cos \frac{B}{2}+C C_{1} \cos \frac{C}{2}}{\sin A+\sin B+\sin C}=
$$
$\qquad$
|
4. 2 .
Connect $B A_{1}$. By the Law of Sines, we have
$$
\begin{array}{l}
A A_{1} \cos \frac{A}{2}=2 \sin \left(B+\frac{A}{2}\right) \cdot \cos \frac{A}{2} \\
=\sin C+\sin B .
\end{array}
$$
Similarly, $B B_{1} \cos \frac{B}{2}=\sin C+\sin A$,
$$
C C_{1} \cos \frac{C}{2}=\sin A+\sin B \text {. }
$$
Therefore, the original expression $=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $f(x)=a \sin [(x+1) \pi]+b \sqrt[3]{x-1}+2$, where $a$ and $b$ are real constants. If $f(\lg 5)=5$, then $f(\lg 20)=$
|
5. -1 .
Obviously,
$$
f(x)=a \sin [(x-1) \pi]+b \sqrt[3]{x-1}+2 .
$$
Let $t=x-1$. Then
$$
f(t+1)=a \sin \pi t+b \sqrt[3]{t}+2=g(t)+2,
$$
where, $g(t)=a \sin \pi t+b \sqrt[3]{t}$ is an odd function.
According to the problem,
$$
\begin{aligned}
5 & =f(\lg 5)=f(1-\lg 2) \\
& =g(-\lg 2)+2=-g(\lg 2)+2 .
\end{aligned}
$$
Thus, $g(\lg 2)=-3$.
$$
\text { Therefore, } f(\lg 20)=f(1+\lg 2)=g(\lg 2)+2=-1 \text {. }
$$
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the Cartesian coordinate system, $O$ is the origin, points $A(3, a) 、 B(3, b)$ make $\angle A O B=45^{\circ}$, where $a 、 b$ are integers, and $a>b$. Then the number of pairs $(a, b)$ that satisfy the condition is.
|
6.6.
Let $\angle A O X=\alpha, \angle B O X=\beta$. Then $\tan \alpha=\frac{a}{3}, \tan \beta=\frac{b}{3}$.
Given $a>b$, we have
$$
\begin{array}{l}
1=\tan 45^{\circ}=\tan (\alpha-\beta) \\
=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \cdot \tan \beta}=\frac{3(a-b)}{9+a b} .
\end{array}
$$
Rearranging gives $(a+3)(b-3)=-18$.
Since $a>b$, and both are integers, we have
$$
\begin{array}{l}
(a+3, b-3) \\
=(18,-1),(9,-2),(6,-3), \\
\quad(3,-6),(2,-9),(1,-18) .
\end{array}
$$
Therefore, there are 6 pairs of integers $(a, b)$ that satisfy the condition.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (16 points) For an integer $k$, define the set
$$
S_{k}=\{n \mid 50 k \leqslant n<50(k+1), n \in \mathbf{Z}\} \text {. }
$$
How many of the 600 sets $S_{0}, S_{1}, \cdots, S_{599}$ do not contain any perfect squares?
|
11. Notice,
$$
\begin{array}{l}
(x+1)^{2}-x^{2}=2 x+1 \leqslant 50(x \in \mathbf{N}) \\
\Leftrightarrow x \leqslant 24(x \in \mathbf{N}) .
\end{array}
$$
While $(24+1)^{2}=625 \in S_{12}$, thus, the square numbers in $S_{0}, S_{1}$, $\cdots, S_{12}$ do not exceed $25^{2}$, and each set consists of 50 consecutive non-negative integers. Therefore, each set contains at least 1 square number.
In the sets $S_{13}, S_{14}, \cdots, S_{599}$, if they contain square numbers, they are no less than $26^{2}$.
When $x \geqslant 26$, $2 x+1 \geqslant 53$, thus, in $S_{13}, S_{14}$, $\cdots, S_{599}$, each set contains at most 1 square number.
On the other hand, the largest number in $S_{599}$ is
$$
600 \times 50-1=29999 \text {, }
$$
and $173^{2}<29999<174^{2}$, so the square numbers in $S_{13}, S_{14}, \cdots, S_{599}$ do not exceed $173^{2}$.
Therefore, there are exactly $173-$ $25=148$ sets in $S_{13}, S_{14}, \cdots, S_{599}$ that contain square numbers.
In summary, in $S_{0}, S_{1}, \cdots, S_{599}$, there are
$$
600-13-148=439
$$
sets that do not contain square numbers.
|
439
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The sequence satisfies $a_{0}=\frac{1}{4}$, and for natural number $n$, $a_{n+1}=a_{n}^{2}+a_{n}$. Then the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is
|
6.3.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{a_{n+1}}=\frac{1}{a_{n}\left(a_{n}+1\right)}=\frac{1}{a_{n}}-\frac{1}{a_{n}+1} \\
\Rightarrow \frac{1}{a_{n}+1}=\frac{1}{a_{n}}-\frac{1}{a_{n+1}} \\
\Rightarrow \sum_{n=0}^{2011} \frac{1}{a_{n}+1}=\sum_{n=0}^{2011}\left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right) \\
\quad=\frac{1}{a_{0}}-\frac{1}{a_{2011}}=4-\frac{1}{a_{2011}} .
\end{array}
$$
Since $a_{2011}>1$, the integer part of $\sum_{n=0}^{2011} \frac{1}{a_{n}+1}$ is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. $[x]$ represents the greatest integer not exceeding the real number $x$. The area of the figure formed by points satisfying $[x]^{2}+[y]^{2}=50$ on the plane is $\qquad$ .
|
8. 12 .
First, consider the case in the first quadrant.
When $x>0, y>0$, from $[x]^{2}+[y]^{2}=50$, we get
$$
\begin{array}{l}
\Rightarrow\left\{\begin{array} { l }
{ 7 \leqslant x < 8 , } \\
{ 1 \leqslant y < 2 }
\end{array} \left\{\begin{array} { l }
{ 5 \leqslant x < 6 , } \\
{ 5 \leqslant y < 6 ; }
\end{array} \left\{\begin{array}{l}
1 \leqslant x<2, \\
7 \leqslant y<8 .
\end{array}\right.\right.\right. \\
\end{array}
$$
Thus, the area of the figure formed in the first quadrant is 3.
Therefore, the area of the figure given in the problem is 12.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $\sqrt{2 \sqrt{3}-3}=\sqrt{\sqrt{3} x}-\sqrt{\sqrt{3} y}(x, y$ are rational numbers). Then $x-y=$ $\qquad$ .
|
3. 1 .
From the given, $\sqrt{2-\sqrt{3}}=\sqrt{x}-\sqrt{y}$.
Then $\sqrt{x}-\sqrt{y}=\sqrt{\frac{4-2 \sqrt{3}}{2}}=\frac{\sqrt{(\sqrt{3}-1)^{2}}}{\sqrt{2}}$ $=\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.
Thus, $x+y-2 \sqrt{x y}=\frac{3}{2}+\frac{1}{2}-2 \sqrt{\frac{3}{2} \times \frac{1}{2}}$.
Since $x, y$ are rational numbers, we have $x y=\frac{3}{2} \times \frac{1}{2}, x+y=\frac{3}{2}+\frac{1}{2}$.
Also, $x>y$, so $x=\frac{3}{2}, y=\frac{1}{2} \Rightarrow x-y=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { II. (25 points) (1) Given the parabola } \\
y=x^{2}-2 m x+4 m-8
\end{array}
$$
with vertex $A$, construct an inscribed equilateral $\triangle A M N$ (points $M, N$ are on the parabola). Find the area of $\triangle A M N$;
(2) If the parabola $y=x^{2}-2 m x+4 m-8$ intersects the $x$-axis at points with integer coordinates, find the integer value of $m$.
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|
(1) From the symmetry of the parabola and the equilateral triangle, we know that $M N \perp y$-axis.
As shown in Figure 8, let the axis of symmetry of the parabola intersect $M N$ at point $B$. Then
$$
A B=\sqrt{3} B M .
$$
Let $M(a, b)$ $(m<a)$. Then
$$
\begin{aligned}
& B M=a-m . \\
& \text { Also, } A B=y_{B}-y_{A} \\
= & \left(a^{2}-m^{2}\right)-2 m(a-m) \\
= & (a-m)^{2},
\end{aligned}
$$
Thus, $B M^{2}=A B=\sqrt{3} B M$, which means $B M=\sqrt{3}$.
Therefore, $S_{\triangle A M N}=\sqrt{3} B M^{2}=3 \sqrt{3}$.
(2) According to the problem, $\Delta=4\left[(m-2)^{2}+4\right]$ is a perfect square.
Let $(m-2)^{2}+4=n^{2}$, then
$$
(n+m-2)(n-m+2)=4 \text {. }
$$
Since $n+m-2$ and $n-m+2$ have the same parity, we have
$$
\left\{\begin{array} { l }
{ n + m - 2 = 2 , } \\
{ n - m + 2 = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
n+m-2=-2, \\
n-m+2=-2 .
\end{array}\right.\right.
$$
Solving these, we get $\left\{\begin{array}{l}m=2, \\ n=2\end{array}\right.$ or $\left\{\begin{array}{l}m=2, \\ n=-2 .\end{array}\right.$
In summary, $m=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $a, b, c \in \mathbf{R}_{+}$, and $abc=4$. Then the minimum value of the algebraic expression $a^{a+b} b^{3b} c^{c+b}$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3. 64.
It is easy to know that $a^{a+b} b^{3 b} c^{c+b}=a^{a} b^{2 b} c^{c} \cdot 4^{b}=a^{a}(2 b)^{2 b} c^{c}$.
And $a b c=4$ can be transformed into $a \cdot 2 b \cdot c=8$, considering $2 b$ as a whole.
Since the function $f(x)=\ln x$ is increasing on $(0,+\infty)$, for any $a, b \in(0,+\infty)$, we always have $(a-b)(f(a)-f(b)) \geqslant 0$, i.e., $a \ln a+b \ln b \geqslant a \ln b+b \ln a$.
Substituting $2 b$ for $b$ in the above inequality, we get $a \ln a+2 b \ln 2 b \geqslant a \ln 2 b+2 b \ln a$.
Similarly, $2 b \ln 2 b+c \ln c \geqslant 2 b \ln c+c \ln 2 b$, (2) $c \ln c+a \ln a \geqslant c \ln a+a \ln c$.
(1) + (2) + (3) and rearranging gives
$3 \ln \left[a^{a}(2 b)^{2 b} c^{c}\right] \geqslant(a+2 b+c) \ln (a \cdot 2 b \cdot c)$,
i.e., $a^{a}(2 b)^{2 b} c^{c} \geqslant(a \cdot 2 b \cdot c)^{\frac{a+2 b+c}{3}} \geqslant 8 \sqrt[3 / 2 \cdot 2 b \cdot c]{ }=64$.
The equality holds if and only if $a=c=2, b=1$. Therefore, the minimum value of $a^{a+b} b^{3 b} c^{c+b}$ is 64.
|
64
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given vectors
$$
a=(x-y+1, x-y), b=\left(x-y+1,10^{x}\right) \text {. }
$$
Then the number of all integer pairs $(x, y)$ that satisfy $a \cdot b=2012$ is $\qquad$
|
5.0.
From the problem, we have
$$
\begin{array}{l}
(x-y+1)^{2}+10^{x}(x-y)=2012 \\
\Rightarrow(x-y)\left(x-y+2+10^{x}\right)=2011 .
\end{array}
$$
Obviously, $x \neq 0$, otherwise, $y(y-3)=2011$, this equation has no integer solutions.
If $x>0$, then $10^{x}$ is a positive integer.
Thus, $x-y+2+10^{x}>x-y$.
We get the following two systems of equations:
$$
\begin{array}{l}
\left\{\begin{array}{l}
x-y+2+10^{x}=2011, \\
x-y=1 ;
\end{array}\right. \\
\left\{\begin{array}{l}
x-y+2+10^{x}=-1, \\
x-y=-2011 .
\end{array}\right.
\end{array}
$$
Calculations show that neither system has integer solutions.
In conclusion, the number of integer pairs $(x, y)$ is 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) For a given positive integer $M$, define $f_{1}(M)$ as the square of the sum of the digits of $M$. When $n>1$ and $n \in \mathbf{N}$, $f_{n}\left(f_{n-1}(M)\right)$ represents the $r_{n}$-th power of the sum of the digits of $f_{n-1}(M)$, where, when $n$ is odd, $r_{n}=2$; when $n$ is even, $r_{n}=3$. Find the value of $f_{2012}\left(f_{2011}\left(3^{2010}\right)\right)$.
|
Let the positive integer $M=a_{1} a_{2} \cdots a_{m}$, where $a_{i} \in \mathbf{N}, a_{1}>0 (i=1,2, \cdots, m)$. Then
$$
f_{1}(M)=\left(a_{1}+a_{2}+\cdots+a_{m}\right)^{2} \equiv M^{2}(\bmod 9) .
$$
When $M=3^{2010}$, $f_{1}(M) \equiv M^{2} \equiv 0(\bmod 9)$.
By mathematical induction, it is easy to see that for $n \in \mathbf{N}$, and $n \geqslant 2$, we always have
$$
f_{n}\left(f_{n-1}(M)\right) \equiv 0(\bmod 9) .
$$
Since $3^{2010}=9^{1005}<10^{1005}$, the sum of its digits is less than $9 \times 1005=9045$.
Thus, $f_{1}\left(3^{2010}\right)<9045^{2}<9 \times 10^{7}$.
Therefore, the sum of the digits of $f_{1}\left(3^{2010}\right)$ is less than $9 \times 8=72$.
Hence, $f_{2}\left(f_{1}\left(3^{2010}\right)\right)<72^{3}<400000$, and $f_{3}\left(f_{2}\left(3^{2010}\right)\right)<(3+9 \times 5)^{2}=48^{2}<50^{2}$. Let the sum of the digits of $f_{3}\left(f_{2}\left(3^{2010}\right)\right)$ be $a$. Then $1 \leqslant a<1+9 \times 3=28$,
and $a \equiv f_{3}\left(f_{2}\left(3^{2010}\right)\right) \equiv 0(\bmod 9)$.
Therefore, $a \in\{9,18,27\}$.
Thus, $f_{4}\left(f_{3}\left(3^{2010}\right)\right)=a^{3} \in\left\{9^{3}, 18^{3}, 27^{3}\right\}$
$$
=\{729,5832,19683\} \text {. }
$$
The sums of the digits of $729$, $5832$, and $19683$ are $18$, $18$, and $27$, respectively. Hence,
$$
f_{5}\left(f_{4}\left(3^{2010}\right)\right) \in\left\{18^{2}, 27^{2}\right\}=\{324,729\} \text {. }
$$
Furthermore,
$$
\begin{array}{l}
f_{6}\left(f_{5}\left(3^{2010}\right)\right) \in\left\{9^{3}, 18^{3}\right\}=\{729,5832\} \\
\Rightarrow f_{7}\left(f_{6}\left(3^{2010}\right)\right)=18^{2}=324 \\
\Rightarrow f_{8}\left(f_{7}\left(3^{2010}\right)\right)=9^{3}=729 \\
\Rightarrow f_{9}\left(f_{8}\left(3^{2010}\right)\right)=18^{2}=324 \\
\Rightarrow \cdots \cdots \\
\Rightarrow f_{2012}\left(f_{2011}\left(3^{2010}\right)\right)=729
\end{array}
$$
|
729
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that $a$ is a prime number less than 2012, and 2012 $-a$ is coprime with $a$. Then the number of $a$ that satisfies the condition is ( ).
(A) 1003
(B) 1004
(C) 1005
(D) 1006
|
5. B.
If $a$ is even, then $2012-a$ is also even. In this case, $a$ and $2012-a$ have a common divisor of 2, so they are not coprime. Therefore, $a$ cannot be even and must be odd.
Since $2012=503 \times 4$, and 503 is a prime number, we have
$$
a \neq 503 \times 1 \text { and } 503 \times 3 \text {. }
$$
There are 1006 odd numbers in the range $1 \sim 2012$, so the number of $a$ that meet the conditions is
$$
1006-2=1004 \text {. }
$$
|
1004
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
$\qquad$ 1. A six-digit number $\overline{a b c d e f}$, when multiplied by 4, becomes $\overline{f a b c d e}$. The number of six-digit numbers that satisfy this condition is $\qquad$.
|
$=1.6$
Let $\overline{a b c d e}=x$. Then
$$
\begin{array}{l}
4(10 x+f)=100000 f+x \\
\Rightarrow x=2564 f .
\end{array}
$$
Since $f$ is a single digit and $x$ is a five-digit number, it is easy to see that,
$$
f=4,5,6,7,8,9 \text {. }
$$
Therefore, the six-digit numbers $\overline{a b c d e f}$ that satisfy the condition are
$$
\begin{array}{l}
102564,128205,153846, \\
179487,205128,230769,
\end{array}
$$
a total of 6.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $a-b=2, \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4$, then $a^{5}-b^{5}=$
|
3. 82 .
$$
\begin{aligned}
\text { Given } & \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4 \\
\Rightarrow & a(1-a)^{2}-b(1+b)^{2}=4 a b \\
\Rightarrow & a-2 a^{2}+a^{3}-b-2 b^{2}-b^{3}=4 a b \\
\Rightarrow & (a-b)-2\left(a^{2}+b^{2}\right)+\left(a^{3}-b^{3}\right)=4 a b \\
\Rightarrow & (a-b)-2\left[(a-b)^{2}+2 a b\right]+ \\
& (a-b)\left[(a-b)^{2}+3 a b\right]=4 a b .
\end{aligned}
$$
Substituting $a-b=2$ into the equation, we get $a b=1$.
$$
\begin{array}{l}
\text { Then } a^{2}+b^{2}=(a-b)^{2}+2 a b=6, \\
a^{3}-b^{3}=(a-b)\left[(a-b)^{2}+3 a b\right]=14. \\
\text { Therefore, } a^{5}-b^{5}=\left(a^{2}+b^{2}\right)\left(a^{3}-b^{3}\right)-a^{2} b^{2}(a-b) \\
=6 \times 14-1 \times 2=82.
\end{array}
$$
|
82
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Find the integer part of $\left(\frac{1+\sqrt{5}}{2}\right)^{19}$.
|
Let $a=\frac{1+\sqrt{5}}{2}, b=\frac{1-\sqrt{5}}{2}$. Then $a+b=1, a^{2}=a+1, b^{2}=b+1$.
Thus $a^{2}+b^{2}=(a+1)+(b+1)$ $=(a+b)+2=3$, $a^{3}+b^{3}=\left(a^{2}+a\right)+\left(b^{2}+b\right)$ $=\left(a^{2}+b^{2}\right)+(a+b)=1+3=4$, $a^{4}+b^{4}=\left(a^{3}+a^{2}\right)+\left(b^{3}+b^{2}\right)$ $=\left(a^{3}+b^{3}\right)+\left(a^{2}+b^{2}\right)=3+4=7$.
By analogy,
$$
\begin{array}{l}
a^{5}+b^{5}=4+7=11, \\
a^{6}+b^{6}=7+11=18, \\
\cdots \cdots \\
a^{19}+b^{19}=9349 .
\end{array}
$$
Since $-1<\left(\frac{1-\sqrt{5}}{2}\right)^{19}<0$, it follows that $9349<\left(\frac{1+\sqrt{5}}{2}\right)^{19}<9350$.
Therefore, the integer part of $\left(\frac{1+\sqrt{5}}{2}\right)^{19}$ is 9349.
|
9349
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (25 points) Given that $D$ is a point inside $\triangle A B C$, $E$ is the midpoint of side $A C$, $A B=6, B C=10, \angle B A D=$ $\angle B C D, \angle E D C=\angle A B D$. Find the length of $D E$.
---
The above text has been translated into English, preserving the original text's line breaks and format.
|
II. As shown in Figure 4, extend \( CD \) to point \( F \) such that \( DF = CD \), and connect \( AF \) and \( BF \).
Then \( AF \parallel DE \), and \( DE = \frac{1}{2} AF \).
Thus, \( \angle AFD = \angle EDC = \angle ABD \)
\(\Rightarrow A, F, B, D\) are concyclic
\(\Rightarrow \angle BFD = \angle BAD = \angle BCD \)
\(\Rightarrow BF = BC = 10 \).
Since \( D \) is the midpoint of \( FC \), it follows that \( BD \perp FC \).
Therefore, \( \angle FAB = \angle FDB = 90^\circ \).
In the right triangle \( \triangle FAB \),
\( FA = \sqrt{BF^2 - AB^2} = \sqrt{10^2 - 6^2} = 8 \).
Hence, \( DE = \frac{1}{2} FA = 4 \).
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Place 27 balls numbered $1 \sim 27$ into three bowls, Jia, Yi, and Bing, such that the average values of the ball numbers in bowls Jia, Yi, and Bing are $15$, $3$, and $18$, respectively, and each bowl must contain no fewer than 4 balls. Then the maximum value of the smallest ball number in bowl Jia is $\qquad$
|
4. 10 .
Let there be $a$, $b$, and $c$ balls in bowls 甲, 乙, and 丙, respectively. Then,
$$
\begin{aligned}
a+b+c & =27, \\
15 a+3 b+18 c & =\frac{27 \times 28}{2} .
\end{aligned}
$$
From equation (2), we get
$$
5 a+b+6 c=126 \text {. }
$$
If $b \geqslant 6$, then the average value of the ball numbers in bowl 乙 is not less than
$$
\frac{1+2+\cdots+b}{b}=\frac{b+1}{2}>3,
$$
which is a contradiction.
Therefore, $b \leqslant 5 \Rightarrow b=4$ or 5.
$6 \times$ (1) - (3) gives $a+5 b=36$.
Solving this, we get $(a, b)=(11,5),(16,4)$.
We discuss two cases.
(1) $(a, b, c)=(11,5,11)$.
In this case, bowl 乙 must contain balls numbered 1 to 5. When the ball numbers in bowl 甲 are $10 \sim 20$, the minimum ball number reaches its maximum value of 10.
(2) $(a, b, c)=(16,4,7)$.
In this case, bowl 乙 must contain balls numbered $(1,2,4,5)$ or $(1,2,3,6)$. When the ball numbers in bowl 甲 are $7 \sim 14,16 \sim 23$, the minimum ball number reaches its maximum value of 7.
In summary, the maximum value of the minimum ball number in bowl 甲 is 10.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for all non-negative integers $m, n (m \geqslant n)$, we have
$$
a_{m+n}+a_{m-n}+m-n-1=\frac{1}{2}\left(a_{2 m}+a_{2 n}\right) \text {. }
$$
Then the remainder when $a_{2012}$ is divided by 2012 is ـ. $\qquad$
|
5.1.
In equation (1), let $n=0$, we get
$$
a_{2 m}=4 a_{m}+2 m-2-a_{0} \text {. }
$$
In equation (2), let $m=0,1$, we get
$$
a_{0}=1, a_{2}=3 \text {. }
$$
In equation (1), let $n=1$, we get
$$
\begin{array}{l}
a_{m+1}+a_{m-1}+m-2=\frac{1}{2}\left(a_{2 m}+a_{2}\right) \\
=\frac{1}{2}\left(4 a_{m}+2 m-3+3\right)=2 a_{m}+m \\
\Rightarrow a_{m+1}-a_{m}=a_{m}-a_{m-1}+2 \\
\Rightarrow a_{m}-a_{m-1}=a_{1}-a_{0}+2(m-1)=2(m-1) \\
\Rightarrow a_{m}=a_{0}+\sum_{i=1}^{m}\left(a_{i}-a_{i-1}\right)=1+m(m-1) \\
\Rightarrow a_{m} \equiv 1(\bmod m) \\
\Rightarrow a_{2012} \equiv 1(\bmod 2012) .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given a positive integer $k$ that satisfies for any positive integer $n$, the smallest prime factor of $n^{2}+n-k$ is no less than 11. Then, $k_{\text {min }}$ $=$ . $\qquad$
|
6.43 .
Notice,
$$
\begin{array}{l}
n^{2}+n \equiv 0(\bmod 2), \\
n^{2}+n \equiv 0,2(\bmod 3),
\end{array}
$$
$$
\begin{array}{l}
n^{2}+n \equiv 0,2,1(\bmod 5), \\
n^{2}+n \equiv 0,2,6,5(\bmod 7),
\end{array}
$$
and the smallest prime factor of $n^{2}+n-k$ is not less than 11, then
$$
\begin{aligned}
k & \equiv 1(\bmod 2), \quad k \equiv 1(\bmod 3), \\
k & \equiv 3,4(\bmod 5), k \equiv 1,3,4(\bmod 7) .
\end{aligned}
$$
Therefore, $k \equiv 43,73,109,169,193,199(\bmod 210)$.
Thus, $k_{\min }=43$.
|
43
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given positive integers $a, b$ satisfy
$$
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \text {. }
$$
Then $|10(a-5)(b-15)|+2=$
|
7.2012.
Notice,
$$
\begin{array}{l}
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \\
\Leftrightarrow 16 a b^{3}=\left(2 b^{2}-a\right)\left(a^{2}+4 a b+4 b^{2}\right) \\
\Leftrightarrow a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right) \\
\Leftrightarrow a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2} .
\end{array}
$$
Let $d=(a, b), a=a_{0} d, b=b_{0} d,\left(a_{0}, b_{0}\right)=1$.
Then equation (1) $\Leftrightarrow a_{0}\left(a_{0}+2 b_{0}\right)^{2}=2 d b_{0}^{2}\left(a_{0}-2 b_{0}\right)^{2}$.
Since $b_{0} \mid a_{0}\left(a_{0}+2 b_{0}\right)^{2}$, we get
$$
b_{0} \mid a_{0}^{3} \Rightarrow b_{0}=1 \text {. }
$$
Then $a_{0}\left(a_{0}+2\right)^{2}=2 d\left(a_{0}-2\right)^{2}$
$$
\Rightarrow 2 d=a_{0}+\frac{8 a_{0}^{2}}{\left(a_{0}-2\right)^{2}}=a_{0}+2\left(\frac{2 a_{0}}{a_{0}-2}\right)^{2} \text {. }
$$
Thus $\left(a_{0}-2\right)\left|2 a_{0} \Rightarrow\left(a_{0}-2\right)\right| 4$
$$
\Rightarrow a_{0}-2= \pm 1,2,4 \text {. }
$$
From equation (2), we know $2 \mid a_{0}$. Therefore, $a_{0}=4,6$.
Substituting into equation (2) gives $d=18,12$.
Accordingly, $(a, b)=(72,18),(72,12)$.
Thus, $|10(a-5)(b-15)|+2=2012$.
|
2012
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given a cyclic quadrilateral $A_{1} A_{2} A_{3} A_{4}$ with an inscribed circle $\odot I$ that is tangent to the sides $A_{1} A_{2}$, $A_{2} A_{3}$, $A_{3} A_{4}$, and $A_{4} A_{1}$ at points $B_{1}$, $B_{2}$, $B_{3}$, and $B_{4}$ respectively, then
$$
\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{2} A_{3}}{B_{2} B_{3}}\right)^{2}+\left(\frac{A_{3} A_{4}}{B_{3} B_{4}}\right)^{2}+\left(\frac{A_{4} A_{1}}{B_{4} B_{1}}\right)^{2}
$$
the minimum value is $\qquad$
|
8. 8 .
As shown in Figure 5, let the radius of $\odot I$ be $r$,
$$
\begin{array}{l}
A_{i} B_{i}=A_{i} B_{i-1} \\
=a_{i},
\end{array}
$$
where $i=1,2,3,4, B_{0}=B_{4}$.
Then $A_{i} A_{i+1}=a_{i}+a_{i+1}$,
$$
B_{i} B_{i+1}=\frac{2 A_{i+1} B_{i} \cdot B_{i} I}{\sqrt{A_{i+1} B_{i}^{2}+B_{i} I^{2}}}=2\left(a_{i+1}^{-2}+r^{-2}\right)^{-\frac{1}{2}} \text {. }
$$
From the right triangle $\triangle A_{i} B_{i} I \subset$ right triangle $\triangle I B_{i+2} A_{i+2}$
$$
\Rightarrow a_{i} a_{i+2}=r^{2}(i=1,2) \text {. }
$$
$$
\begin{array}{l}
\text { Then }\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{2} A_{3}}{B_{2} B_{3}}\right)^{2}+\left(\frac{A_{3} A_{4}}{B_{3} B_{4}}\right)^{2}+\left(\frac{A_{4} A_{1}}{B_{4} B_{1}}\right)^{2} \\
=\frac{1}{4} \sum_{i=1}^{4}\left(a_{i}+a_{i+1}\right)^{2}\left(a_{i+1}^{-2}+r^{-2}\right) . \\
\text { Also } \sum_{i=1}^{4} \frac{\left(a_{i}+a_{i+1}\right)^{2}}{4 a_{i+1}^{2}} \\
=\frac{1}{4} \sum_{i=1}^{4}\left(1+\frac{2 a_{i}}{a_{i+1}}+\frac{a_{i}^{2}}{a_{i+1}^{2}}\right) \geqslant 4, \\
\sum_{i=1}^{4} \frac{\left(a_{i}+a_{i+1}\right)^{2}}{4 r^{2}} \geqslant \sum_{i=1}^{4} \frac{a_{i} a_{i+1}}{r^{2}} \\
\geqslant \frac{4}{r^{2}} \sqrt{\prod_{i=1}^{4} a_{i} a_{i+1}}=4, \\
\text { Therefore }\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}\right)^{2}+\left(\frac{A_{2} A_{3}}{B_{2} B_{3}}\right)^{2}+\left(\frac{A_{3} A_{4}}{B_{3} B_{4}}\right)^{2}+\left(\frac{A_{4} A_{1}}{B_{4} B_{1}}\right)^{2} \geqslant 8 .
\end{array}
$$
Equality holds if and only if $a_{1}=a_{2}=a_{3}=a_{4}$.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) How many nine-digit numbers $\overline{a_{1} a_{2} \cdots a_{9}}$ satisfy $a_{1} \neq 0$, all digits are distinct, and
$$
a_{1}+a_{3}+a_{5}+a_{9}=a_{2}+a_{4}+a_{6}+a_{8} .
$$
|
10. Let the missing digit be $a$, and
$$
\sum_{i=\pi} a_{i}=\sum_{i=m} a_{i}=t .
$$
Then $2 t=\sum_{i=1}^{9} a_{i}=45-a$.
Thus, $a \in\{1,3,5,7,9\}$.
If $a=1$, then $t=22$, we have
$$
\begin{array}{l}
22=2+3+8+9=2+4+7+9 \\
=2+5+6+9=2+5+7+8 \\
=3+4+6+9=3+4+7+8 \\
=3+5+6+8=4+5+6+7 \\
=0+5+8+9=0+6+7+9 ;
\end{array}
$$
If $a=3$, then $t=21$, we have
$$
\begin{array}{l}
21=1+4+7+9=1+5+6+9 \\
=1+5+7+8=2+4+6+9 \\
=2+4+7+8=2+5+6+8 \\
=0+4+8+9=0+5+7+9 \\
=0+6+7+8 ;
\end{array}
$$
If $a=5$, then $t=20$, we have
$$
\begin{array}{l}
20=1+2+8+9=1+3+7+9 \\
=1+4+6+9=1+4+7+8 \\
=2+3+6+9=2+3+7+8 \\
=2+4+6+8=3+4+6+7 \\
=0+3+8+9=0+4+7+9 ;
\end{array}
$$
If $a=7$, then $t=19$, we have
$$
\begin{array}{l}
19=1+3+6+9=1+4+5+9 \\
=1+4+6+8=2+3+5+9 \\
=2+3+6+8=2+4+5+8 \\
=0+2+8+9=0+4+6+9 \\
=0+5+6+8 ;
\end{array}
$$
If $a=9$, then $t=18$, we have
$$
\begin{array}{l}
18=1+2+7+8=1+3+6+8 \\
=1+4+5+8=1+4+6+7 \\
=2+3+5+8=2+3+6+7 \\
=2+4+5+7=3+4+5+6 \\
=0+3+7+8=0+4+6+8 \\
=0+5+6+7 .
\end{array}
$$
In summary, the number of nine-digit numbers that satisfy the conditions is
$$
49 A_{4}^{4} A_{5}^{5}-36 A_{4}^{4} A_{4}^{4}=120384 \text { (numbers). }
$$
|
120384
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Write down all positive integers from 1 to 10000 from left to right, then remove those numbers that are divisible by 5 or 7, and concatenate the remaining numbers to form a new number. Try to find:
(1) The number of digits in the new number;
(2) The remainder when the new number is divided by 11.
|
Three, (1) Obviously, the remainder of the remaining numbers when divided by 35 is
$$
\begin{array}{l}
1,2,3,4,6,8,9,11,12,13,16,17,18, \\
19,22,23,24,26,27,29,31,32,33,34,
\end{array}
$$
denoted as $a_{i}(i=1,2, \cdots, 24)$ in sequence.
Among the numbers from $1 \sim 9$, 7 numbers remain, denoted as $b_{i}$ $(i=1,2, \cdots, 7)$ in sequence.
Among the numbers from $10 \sim 99$, since $99=35 \times 2+29$, there are $24 \times 2+20-7=61$ remaining numbers, denoted as $b_{i}(i=$ $8,9, \cdots, 68)$ in sequence.
Among the numbers from $100 \sim 999$, since $999=35 \times 28+19$, there are $24 \times 28+14-68=618$ remaining numbers, denoted as $b_{i}(i=69,70, \cdots, 686)$ in sequence.
Among the numbers from $1000 \sim 9999$, since
$$
9999=35 \times 285+24 \text {, }
$$
there are $24 \times 285+17-686=6171$ remaining numbers, denoted as $b_{i}(i=687,688, \cdots, 6857)$ in sequence.
In summary, the number of digits in the new number is
$$
7+61 \times 2+618 \times 3+6171 \times 4=26667 \text {. }
$$
(2) First, there are two conclusions:
(i) If $c_{i}(i=1,2, \cdots, k)$ are all $n$-digit positive integers, then
$$
\overline{c_{1} c_{2} \cdots c_{k}} \equiv \sum_{i=1}^{k} c_{i}(-1)^{(k-i) n}(\bmod 11) .
$$
(ii) The sum of any 11 consecutive terms in an arithmetic sequence composed of positive integers is divisible by 11.
Obviously, $b_{24 k+j}=35 k+a_{j}(j=1,2, \cdots, 24)$,
$$
\begin{array}{l}
\sum_{j=1}^{24} b_{24 k+j}=35 k \times 24+\sum_{j=1}^{24} a_{j} \\
\equiv 4 k+2(\bmod 11), \\
\sum_{j=1}^{24}(-1)^{j} b_{24 k+j}=\sum_{j=1}^{24}(-1)^{j} a_{j} \\
\equiv 5(\bmod 11) .
\end{array}
$$
Let the new number be $A=\overline{A_{1} A_{2} A_{3} A_{4}}$, where,
$$
\begin{array}{l}
A_{1}=\overline{1234689} \equiv 1-2+3-4+6-8+9 \\
\equiv 5(\bmod 11) \text {, } \\
A_{2}=\overline{b_{8} b_{9} \cdots b_{68}} \equiv \sum_{i=8}^{68} b_{i} \\
\equiv \sum_{k=0}^{2} \sum_{j=1}^{24} b_{24 k+j}-\sum_{j=1}^{7} b_{j}-\sum_{j=21}^{24} b_{48+j} \\
\equiv \sum_{k=0}^{2}(4 k+2)-0-\sum_{j=21}^{24}\left(70+a_{j}\right) \\
\equiv 18-4 \times 70-\sum_{j=11}^{24} a_{j} \\
\equiv 2-31 \equiv 4(\bmod 11) \text {, } \\
A_{3}=\overline{b_{69} b_{70} \cdots b_{686}} \equiv \sum_{i=69}^{686}(-1)^{i} b_{i} \\
\equiv \sum_{j=21}^{24}(-1)^{j} b_{48+j}+\sum_{k=3}^{27} \sum_{j=1}^{24}(-1)^{j} b_{2 k+j}+ \\
\sum_{j=1}^{14}(-1)^{j} b_{24 \times 28+j} \\
\equiv \sum_{j=21}^{24}(-1)^{j} a_{j}+5 \times 25+\sum_{j=1}^{14}(-1)^{j} a_{j} \\
\equiv 5 \times 26-\sum_{j=15}^{20}(-1)^{j} a_{j} \\
\equiv 5 \times 26-5 \equiv 4(\bmod 11) \text {, } \\
A_{4}=\overline{b_{687} b_{688} \cdots b_{6857}} \equiv \sum_{i=687}^{6857} b_{i} \equiv \sum_{i=687}^{785} b_{i} \\
\equiv \sum_{j=15}^{24} b_{24 \times 28+j}+\sum_{k=29}^{31} \sum_{j=1}^{24} b_{24 k+j}+\sum_{j=1}^{17} b_{24 \times 32+j} \\
=\sum_{j=15}^{24}\left(35 \times 28+a_{j}\right)+\sum_{k=2}^{31}(4 k+2)+\sum_{j=1}^{17}\left(35 \times 32+a_{j}\right) \\
=1+\sum_{j=1}^{24} a_{j}+\sum_{j=15}^{17} a_{j} \\
\equiv 1+2+3 \equiv 6(\bmod 11) \text {. } \\
\end{array}
$$
Therefore, $A \equiv \sum_{i=1}^{4} A_{i} \equiv 8(\bmod 11)$.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Find all real roots of the equation
$$
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right)
$$
|
Solve:
$$
\begin{array}{l}
x^{6}-x^{3}+1=\left(x^{6}+x^{3}+1\right)\left(x^{2}+2 x+4\right) \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}=x^{2}+2 x+4 \\
\Rightarrow \frac{x^{6}-x^{3}+1}{x^{6}+x^{3}+1}-3=x^{2}+2 x+1 \\
\Rightarrow \frac{-2\left(x^{3}+1\right)^{2}}{x^{6}+x^{3}+1}=(x+1)^{2} \\
\Rightarrow(x+1)^{2}\left[1+\frac{2\left(x^{2}-x+1\right)^{2}}{x^{6}+x^{3}+1}\right]=0 .
\end{array}
$$
Notice that, $x^{6}+x^{3}+1=\left(x^{3}+\frac{1}{2}\right)^{2}+\frac{3}{4}>0$.
Therefore, the original equation has a unique real root -1.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $n$ be a positive integer, $[x]$ be the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$.
(1) Find all positive integers $n$ that satisfy
$$
\sum_{k=1}^{2013}\left[\frac{k n}{2013}\right]=2013+n
$$
(2) Find all positive integers $n$ that maximize $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right\}$, and determine this maximum value.
|
(1) Let $m=2013, d=(m, n), m=d m_{1}, n=d n_{1}$.
For $1 \leqslant k \leqslant m-1$, we have
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=\left[\frac{k n_{1}}{m_{1}}\right]+\left[\frac{(m-k) n_{1}}{m_{1}}\right] \text{. }
$$
When $m_{1} \mid k$,
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=n,
$$
and there are $d-1$ values of $k$ satisfying $1 \leqslant k \leqslant m-1$;
When $m_{1} \nmid k$,
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=n-1 \text{, }
$$
and there are
$$
\begin{array}{l}
m-1-(d-1)=m-d \text{ (values) }. \\
\text{ Then } \sum_{k=1}^{m}\left[\frac{k n}{m}\right]=n+\sum_{k=1}^{m-1}\left[\frac{k n}{m}\right] \\
=n+\frac{1}{2} \sum_{k=1}^{m-1}\left(\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]\right) \\
=n+\frac{1}{2}[n(d-1)+(n-1)(m-d)] \\
=\frac{1}{2}(m n+n-m+d) .
\end{array}
$$
Thus, when $m=2013$, we have
$$
\frac{1}{2}(2013 n+n-2013+d)=2013+n \text{, }
$$
which implies $3 \times 2013=2012 n+(2013, n)>2012 n$.
Therefore, $n \leqslant 3$.
Upon verification, $n=3$.
(2) From (1), we get
$$
\begin{array}{l}
\sum_{k=1}^{m}\left\{\frac{k n}{m}\right\}=\sum_{k=1}^{m}\left(\frac{k n}{m}-\left[\frac{k n}{m}\right]\right) \\
=\frac{n}{m} \cdot \frac{m(m+1)}{2}-\sum_{k=1}^{m}\left[\frac{k n}{m}\right] \\
=\frac{n(m+1)}{2}-\frac{1}{2}(m n+n-m+d) \\
=\frac{m-d}{2} \leqslant \frac{m-1}{2} .
\end{array}
$$
In particular, for $m=2013$, when $d=(2013, n)=1$, i.e., when $n$ is not divisible by $3, 11, 61$, the maximum value of $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right\}$ is
$$
\frac{2013-1}{2}=1006 .
$$
|
1006
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If $\frac{1}{a}+\frac{1}{b}=\frac{5}{a+b}$, then $\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=$
|
2. 7 .
$$
\begin{array}{l}
\text { Given } \frac{1}{a}+\frac{1}{b}=\frac{5}{a+b} \Rightarrow \frac{b}{a}+\frac{a}{b}=3 \\
\Rightarrow \frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}=\left(\frac{b}{a}+\frac{a}{b}\right)^{2}-2=7 .
\end{array}
$$
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $x-1$ is a factor of $x^{3}+a x^{2}+1$, then the value of $a$ is $\qquad$ .
|
3. -2 .
Let $x^{3}+a x^{2}+1=(x-1)\left(x^{2}-m x-1\right)$, that is
$$
\begin{array}{l}
x^{3}+a x^{2}+1 \\
=x^{3}-(m+1) x^{2}+(m-1) x+1 .
\end{array}
$$
By comparing, we get $m-1=0, a=-(m+1)$.
Thus, $m=1, a=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (25 points) Let the two intersection points of the functions $y=2x$ and $y=\frac{4}{x}$ be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}>x_{2}\right)$, and point $C(\sqrt{2},-2 \sqrt{2})$. Find the area of $\triangle ABC$.
|
From $\left\{\begin{array}{l}y=2 x, \\ y=\frac{4}{x},\end{array}\right.$ eliminating $y$ we get
$$
2 x=\frac{4}{x} \Rightarrow x^{2}=2 \Rightarrow x= \pm \sqrt{2} \text {. }
$$
Therefore, $A(\sqrt{2}, 2 \sqrt{2}), B(-\sqrt{2},-2 \sqrt{2})$.
Since point $C(\sqrt{2},-2 \sqrt{2})$, $\triangle A B C$ is a right triangle, and $\angle A C B=90^{\circ}$.
Obviously, $A C=4 \sqrt{2}, B C=2 \sqrt{2}$.
Thus, $S_{\triangle A C B}=\frac{1}{2} A C \cdot B C=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1} \leqslant \frac{a_{n+2}+a_{n}}{2}, a_{1}=1, a_{403}=2011 \text {. }
$$
Then the maximum value of $a_{5}$ is $\qquad$
|
- 1.21.
Obviously, the sequence of points $\left(n, a_{n}\right)$ is arranged in a convex function. When the sequence of points is distributed on the line determined by the points $(1,1)$ and $(403,2011)$, $a_{5}$ takes the maximum value 21.
|
21
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $a_{1}=1, a_{2}=3$,
$$
a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n} \text {, }
$$
when $m \geqslant n$, $a_{m}$ is divisible by 9. Then the minimum value of $n$ is $\qquad$ .
|
5.5.
Notice,
$$
\begin{array}{l}
a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right) \\
=\cdots=(n+2)(n+1) n \cdots \cdots \cdot 3\left(a_{2}-a_{1}\right) \\
=(n+2)!.
\end{array}
$$
Thus, $a_{n}=a_{1}+\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)=\sum_{i=1}^{n} i!$.
Given $a_{1}=1, a_{2}=3$, we have
$$
a_{3}=9, a_{4}=33, a_{5}=153 \text {, }
$$
At this point, 153 is divisible by 9.
When $m>5$, $a_{m}=a_{5}+\sum_{k=6}^{m} k!$, and for $k \geqslant 6$, $k!$ is divisible by 9, so when $m \geqslant 5$, $a_{m}$ is divisible by 9.
Therefore, the smallest value of $n$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Six college graduates apply to three employers. If each employer hires at least one of them, the number of different hiring scenarios is $\qquad$ .
|
7. 2100 .
The number of ways for three people to be hired is $\mathrm{A}_{6}^{3}=120$; the number of ways for four people to be hired is $\mathrm{C}_{6}^{4} \mathrm{C}_{4}^{2} \mathrm{~A}_{3}^{3}=15 \times 6 \times 6=540$ (ways); the number of ways for five people to be hired is $C_{6}^{5}\left(C_{5}^{3} A_{3}^{3}+\frac{C_{5}^{2} C_{3}^{2}}{2!} A_{3}^{3}\right)=900$ (ways); the number of ways for all six people to be hired is $3^{6}-C_{3}^{1} \times 2^{6}+C_{3}^{2} \times 1^{6}=540$ (ways). Therefore, the total number of ways is $120+540+900+540=2100$.
|
2100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (12 points) Given
$$
\begin{array}{l}
f(x, y) \\
=x^{3}+y^{3}+x^{2} y+x y^{2}-3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{array}
$$
and $x, y \geqslant \frac{1}{2}$. Find the minimum value of $f(x, y)$.
|
12. When $x \neq y$, multiplying both sides of the function by $x-y$ gives
$$
\begin{array}{l}
(x-y) f(x, y) \\
=\left(x^{4}-y^{4}\right)-3\left(x^{3}-y^{3}\right)+3\left(x^{2}-y^{2}\right) . \\
\text { Let } g(x)=x^{4}-3 x^{3}+3 x^{2} .
\end{array}
$$
Then $f(x, y)=\frac{g(x)-g(y)}{x-y}$ is the slope of the line segment connecting two points on the graph of $g(x)$.
When $x=y$, $f(x, y)=4 x^{3}-9 x^{2}+6 x$.
Thus, we only need to find the minimum value of the derivative of $g(x)$ for $x \geqslant \frac{1}{2}$,
$$
h(x)=4 x^{3}-9 x^{2}+6 x
$$
It is easy to find that when $x \geqslant \frac{1}{2}$,
$$
h(x)=4 x^{3}-9 x^{2}+6 x
$$
the minimum value is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than
1. Then
$$
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=
$$
$\qquad$
|
15.2011.
When the common ratio is 1, $\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011$.
When the common ratio is $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q} \sum_{i=1}^{2011}\left(\frac{1}{\lg a_{i}}-\frac{1}{\lg a_{i+1}}\right) \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q}\left(\frac{1}{\lg a_{1}}-\frac{1}{\lg a_{2012}}\right)=2011 .
\end{array}
$$
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Figure 3 is a defective $3 \times 3$ magic square, in which the sum of the three numbers in each row, each column, and each diagonal is equal. Then the value of $x$ is . $\qquad$
|
$$
\begin{array}{l}
\text { Magic Square } \\
4017+2012 \\
=x-2003+x \\
\Rightarrow x=4016 \text {. } \\
\end{array}
$$
|
4016
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
1 \times 2 - 3 \times 4 + 5 \times 6 - 7 \times 8 + \cdots + \\
2009 \times 2010 - 2011 \times 2012 \\
= \quad .
\end{array}
$$
|
$$
\text { II, 1. }-2025078 \text {. }
$$
Notice that,
$$
\begin{array}{l}
(n+2)(n+3)-n(n+1)=4 n+6 \\
=n+(n+1)+(n+2)+(n+3) .
\end{array}
$$
Therefore, the original expression is
$$
=-(1+2+\cdots+2012)=-2025078 .
$$
|
-2025078
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The ten positive integers from $1 \sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\cdots, S_{10}=a_{1}+a_{2}+\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \cdots, S_{10}$, the maximum number of primes that can occur is .
|
2. 7 .
Adding an odd number changes the sum to the opposite parity, and among even numbers, only 2 is a prime. Let $b_{i}$ be the $i$-th ($i=1,2,3,4,5$) odd number in this row. Then, when adding $b_{2}$ and $b_{4}$, the sums $S_{k}$ and $S_{n}$ are even numbers greater than 2. Therefore, $S_{k} \backslash S_{n}$ and $S_{10}=55$ must be composite numbers, meaning that in $S_{1}, S_{2}, \cdots, S_{10}$, the primes are no greater than 7. The example in Table 1 shows that there can be 7 primes.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$i$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$a_{i}$ & 7 & 6 & 4 & 2 & 9 & 1 & 3 & 5 & 10 & 8 \\
\hline$S_{i}$ & 7 & 13 & 17 & 19 & 28 & 29 & 32 & 37 & 47 & 55 \\
\hline
\end{tabular}
Therefore, in $S_{1}, S_{2}, \cdots, S_{10}$, there can be at most 7 primes.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (10 points) Prove: For any three distinct numbers $a, b, c$, we have
$$
\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}
$$
is a constant.
|
$$
\begin{array}{l}
S=\frac{(a+b-c)^{2}}{(a-c)(b-c)}+\frac{(b+c-a)^{2}}{(b-a)(c-a)}+\frac{(c+a-b)^{2}}{(c-b)(a-b)}, \\
p=a^{2}+b^{2}+c^{2}-2 a b-2 b c-2 c a . \\
\text { Then } (a+b-c)^{2}=p+4 a b, \\
(b+c-a)^{2}=p+4 b c, \\
(c+a-b)^{2}=p+4 c a . \\
\text { Hence } S=\frac{p+4 a b}{(a-c)(b-c)}+\frac{p+4 b c}{(b-a)(c-a)}+ \\
\quad \frac{p+4 c a}{(c-b)(a-b)} \\
=p q+4 r,
\end{array}
$$
where,
$$
\begin{aligned}
q & =\frac{1}{(a-c)(b-c)}+\frac{1}{(b-a)(c-a)}+\frac{1}{(c-b)(a-b)} \\
& =\frac{(a-b)+(b-c)-(a-c)}{(a-b)(b-c)(a-c)}=0, \\
r & =\frac{a b}{(a-c)(b-c)}+\frac{b c}{(b-a)(c-a)}+\frac{c a}{(c-b)(a-b)} \\
& =\frac{a b(a-b)+b c(b-c)-a c(a-c)}{(a-b)(b-c)(a-c)} \\
& =\frac{\left(a^{2}-c^{2}\right) b+(c-a) b^{2}-a c(a-c)}{(a-b)(b-c)(a-c)} \\
& =\frac{(a+c) b-b^{2}-a c}{(a-b)(b-c)}=1 .
\end{aligned}
$$
Therefore, $S=p q+4 r=4$.
|
4
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (15 points) It is known that a positive integer $n$ can be expressed as the sum of 2011 identical natural numbers, and also as the sum of 2012 identical natural numbers. Determine the minimum value of $n$.
|
Let $a_{1}, a_{2}, \cdots, a_{2011}$ be 2011 natural numbers with the same digit sum, and $b_{1}, b_{2}, \cdots, b_{2012}$ be 2012 natural numbers with the same digit sum, and they satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=n \\
=b_{1}+b_{2}+\cdots+b_{2012} .
\end{array}
$$
Since each of the numbers $a_{1}, a_{2}, \cdots, a_{2011}$ has the same digit sum, each of them has the same remainder when divided by 9, denoted by $r$ $(0 \leqslant r \leqslant 8)$. Similarly, for the numbers $b_{1}, b_{2}, \cdots, b_{2012}$, each has the same digit sum, and thus the same remainder when divided by 9, denoted by $s$ $(0 \leqslant s \leqslant 8)$. Therefore, the numbers $n-2011 r$ and $n-2012 s$ are both multiples of 9. This implies,
$$
\begin{array}{l}
(n-2011 r)-(n-2012 s) \\
=2012 s-2011 r \\
=2012(r+s)-4023 r
\end{array}
$$
is a multiple of 9.
Since 4023 is a multiple of 9 and 2012 is coprime with 9, the number $r+s$ must be a multiple of 9.
If $r=s=0$, then $n \geqslant 9 \times 2012$ (since $b_{1}, b_{2}, \cdots, b_{2012}$ are divisible by 9).
If $r \neq 0$, then $r+s=9$, so at least one of $r \geqslant 5$ or $s \geqslant 5$ holds.
For the number $n$, we get
$$
\begin{array}{l}
n \geqslant 5 \times 2011 \text { and } n \geqslant 5 \times 2012 . \\
\text { Also, } 10055=5 \times 2011 \\
=4 \times 2011+2011 \times 1
\end{array}
$$
and the numbers 4 and 2011 have the same digit sum, so the number 10055 is the desired number.
|
10055
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The constant term in the expansion of $\left(x^{2}+x-\frac{1}{x}\right)^{6}$ is $\qquad$ (answer with a specific number).
|
8. -5 .
From the conditions, the constant term is
$$
C_{6}^{2}(-1)^{4}+C_{6}^{3}(-1)^{3}=-5 .
$$
|
-5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the sequence $\left\{a_{n}\right\}$ be a geometric sequence with the sum of the first $n$ terms denoted as $S_{n}$, satisfying $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$. Then the value of $S_{20}$ is $\qquad$.
|
9. 0 .
From the condition, we know $a_{1}=\frac{\left(a_{1}+1\right)^{2}}{4}$, solving this gives $a_{1}=1$.
When $n \geqslant 2$, from $S_{n}=\frac{\left(a_{n}+1\right)^{2}}{4}$, we know
$$
S_{n-1}=\frac{\left(a_{n-1}+1\right)^{2}}{4} \text {. }
$$
Then $a_{n}=\frac{1}{4}\left(a_{n}+1\right)^{2}-\frac{1}{4}\left(a_{n-1}+1\right)^{2}$
$$
\Rightarrow\left(a_{n}-1\right)^{2}=\left(a_{n-1}+1\right)^{2}
$$
$\Rightarrow a_{n}=a_{n-1}+2$ (discard) or $a_{n}=-a_{n-1}$.
Thus, $\left\{a_{n}\right\}$ is a geometric sequence with the first term 1 and the common ratio -1.
$$
\text { Hence } S_{20}=\frac{1 \times\left[1-(-1)^{20}\right]}{1-(-1)}=0 \text {. }
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The number of positive integers not exceeding 2012 and having exactly three positive divisors is $\qquad$ .
|
10. 14.
Let $1 \leqslant a \leqslant 2012$ and $a$ has only three positive divisors. Then $a$ is the square of a prime number, i.e., $a=p^{2} \leqslant 2012$.
Thus, $2 \leqslant p \leqslant 43$.
$$
\begin{array}{c}
\text { Hence } p=2,3,5,7,11,13,17,19,23, \\
29,31,37,41,43 .
\end{array}
$$
|
14
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Let the function $f(x)=\sin x+\sqrt{3} \cos x+1$.
(1) Find the maximum and minimum values of the function $f(x)$ on $\left[0, \frac{\pi}{2}\right]$;
(2) If real numbers $a$, $b$, and $c$ satisfy
$$
a f(x)+b f(x-c)=1
$$
for any $x \in \mathbf{R}$, find the value of $\frac{b \cos c}{a}$.
|
Three, 13. (1) From the given conditions,
$$
f(x)=2 \sin \left(x+\frac{\pi}{3}\right)+1 \text {. }
$$
From $0 \leqslant x \leqslant \frac{\pi}{2} \Rightarrow \frac{\pi}{3} \leqslant x+\frac{\pi}{3} \leqslant \frac{5 \pi}{6}$.
Thus, $\frac{1}{2} \leqslant \sin \left(x+\frac{\pi}{3}\right) \leqslant 1$.
Therefore, when $x=\frac{\pi}{2}$, $f(x)$ has a minimum value of 2;
when $x=\frac{\pi}{6}$, $f(x)$ has a maximum value of 3.
(2) From the given conditions, for any $x \in \mathbf{R}$,
$$
\begin{array}{l}
2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}-c\right)+a+b=1 \\
\Rightarrow 2 a \sin \left(x+\frac{\pi}{3}\right)+2 b \sin \left(x+\frac{\pi}{3}\right) \cdot \cos c- \\
2 b \cos \left(x+\frac{\pi}{3}\right) \cdot \sin c+(a+b-1)=0 \\
\Rightarrow 2(a+b \cos c) \cdot \sin \left(x+\frac{\pi}{3}\right)- \\
2 b \sin c \cdot \cos \left(x+\frac{\pi}{3}\right)+(a+b-1)=0 . \\
\text { Hence }\left\{\begin{array}{l}
a+b \cos c=0, \\
\sin c=0, \\
a+b-1=0 .
\end{array}\right.
\end{array}
$$
From $b \sin c=0$, we know $b=0$ or $\sin c=0$.
If $b=0$, then from $a+b \cos c=0$, we know $a=0$, which contradicts $a+b-1=0$.
If $\sin c=0$, then $\cos c=1$ (discard), $\cos c=-1$.
Solving gives $a=b=\frac{1}{2}, c=(2 k+1) \pi$.
Thus, $\frac{b \cos c}{a}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $\triangle A B C$ is an isosceles right triangle, $\angle A$ $=90^{\circ}$, and $\overrightarrow{A B}=a+b, \overrightarrow{A C}=a-b$.
If $a=(\cos \theta, \sin \theta)(\theta \in \mathbf{R})$, then $S_{\triangle A B C}$ $=$ . $\qquad$
|
4. 1.
From the problem, we know that $A B \perp A C,|A B|=|A C|$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
(a+b) \cdot(a-b)=0, \\
|a+b|=|a-b|
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
|a|=|b|, \\
a \cdot b=0 .
\end{array}\right.
\end{array}
$$
Since $|a|=1$, we have $|b|=1$.
According to the geometric meaning of vector addition and subtraction,
$$
|a+b|=|a-b|=\sqrt{2} \text {. }
$$
Therefore, $S_{\triangle A B C}=\frac{1}{2}|a+b| \cdot|a-b|=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In a regular tetrahedron $ABCD$, $AO \perp$ plane $BCD$, with the foot of the perpendicular being $O$. Let $M$ be a point on the line segment $AO$ such that $\angle BMC=90^{\circ}$. Then $\frac{AM}{MO}=$ $\qquad$
|
5. 1.
As shown in Figure 3, connect $O B$. Let the edge length of the regular tetrahedron $A B C D$ be $a$. Then
$$
\begin{array}{c}
O B=\frac{\sqrt{3}}{3} a, \\
M B=\frac{\sqrt{2}}{2} a . \\
\text { Therefore, } M O=\sqrt{M B^{2}-O B^{2}} \\
=\frac{\sqrt{6}}{6} a=\frac{1}{2} A O=A M . \\
\text { Hence, } \frac{A M}{M O}=1 .
\end{array}
$$
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. For a certain game activity,
the rewards are divided into first, second, and third prizes (all participants in the game activity will receive a prize), and the corresponding winning probabilities form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prize money forms an arithmetic sequence with the first term of 700 yuan and a common difference of -140. Then the expected prize money for participating in this game activity is $\qquad$ yuan.
|
7.500 .
From the problem, we know the probabilities of winning the first, second, and third prizes are
$$
P_{1}=a, P_{2}=2 a, P_{3}=4 a .
$$
From $P_{1}+P_{2}+P_{3}=1$, we get $a=\frac{1}{7}$.
Thus, $P_{1}=\frac{1}{7}, P_{2}=\frac{2}{7}, P_{3}=\frac{4}{7}$.
The prizes for winning the first, second, and third prizes are
$$
\begin{array}{l}
\quad \xi_{1}=700, \xi_{2}=560, \xi_{3}=420 . \\
\quad \text { Therefore, } E \xi=700 \times \frac{1}{7}+560 \times \frac{2}{7}+420 \times \frac{4}{7} \\
=500 \text { (yuan). }
\end{array}
$$
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $p$ and $q$ be two different prime numbers. Then the remainder when $p^{q-1}+q^{p-1}$ is divided by $p q$ is $\qquad$
|
8. 1 .
Since $p$ and $q$ are different prime numbers, by Fermat's Little Theorem, we have
$$
p^{q-1} \equiv 1(\bmod q) \text {. }
$$
Also, $q^{p-1} \equiv 0(\bmod q)$, thus
$$
p^{q-1}+q^{p-1} \equiv(\bmod q) \text {. }
$$
Similarly, $p^{q-1}+q^{p-1} \equiv 1(\bmod p)$.
Therefore, $p^{q-1}+q^{p-1} \equiv 1(\bmod p q)$.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Transporting utility poles from a construction site by the roadside along a straight road in the same direction to plant them 500 m away on the roadside, plant one at the 500 m mark, and then plant one every 50 m along the roadside. Knowing that the transport vehicle can carry a maximum of 3 poles at a time, to complete the task of transporting and planting 20 poles, and returning to the construction site, the minimum total distance the transport vehicle must travel is $\qquad$ m.
|
10. 14000 .
Assuming the completion of transporting and planting 21 utility poles, 3 poles each time, let the round trip distance for the $k(k=1,2, \cdots, 7)$-th time be $a_{k}$. Then
$$
a_{1}=2 \times 600=1200,
$$
and $a_{k+1}=a_{k}+2 \times 150(k=1,2, \cdots, 6)$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 1200 and a common difference of 300.
Thus, $S_{7}=7 \times 1200+\frac{7 \times 6}{2} \times 300=14700(\mathrm{~m})$.
However, in reality, only 20 poles are transported, so there must be one time when 2 poles are transported, and the other 6 times, 3 poles each.
If the transportation of 2 poles is arranged on the $k(k=1,2, \cdots, 7)$-th time, then $a_{1}, a_{2}, \cdots, a_{k-1}$ remain unchanged, and $a_{k}, a_{k+1}, \cdots, a_{7}$ each decrease by $100 \mathrm{~m}$.
$$
\begin{array}{l}
\text { Hence } S_{7}(k)=S_{7}-100(8-k) \\
=13900+100 k .
\end{array}
$$
Obviously, when $k=1$, that is, the first time 2 poles are transported, and the rest each time 3 poles, the total distance is minimized at $14000 \mathrm{~m}$.
|
14000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (30 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=1+\frac{1}{a_{n}}\left(n \in \mathbf{N}_{+}\right)$, and for any positive integer $n(n \geqslant 2)$, the inequality
$$
\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24}
$$
always holds, find the maximum value of the integer $m$.
|
(1) From
$$
\begin{array}{l}
a_{n}=2 a_{n} a_{n+1}+3 a_{n+1} \\
\Rightarrow a_{n+1}=\frac{a_{n}}{2 a_{n}+3} \Rightarrow a_{n}>0 . \\
\text { Hence } \frac{1}{a_{n+1}}-\frac{3}{a_{n}}=2 \Rightarrow \frac{1}{a_{n+1}}+1=3\left(\frac{1}{a_{n}}+1\right) .
\end{array}
$$
Therefore, $\left\{\frac{1}{a_{n}}+1\right\}$ is a geometric sequence with the first term 3 and common ratio 3.
Thus, $\frac{1}{a_{n}}+1=3^{n} \Rightarrow a_{n}=\frac{1}{3^{n}-1}\left(n \in \mathbf{N}_{+}\right)$.
(2) From (1), we get $b_{n}=1+\frac{1}{a_{n}}=3^{n}$.
Therefore, $\log _{3} b_{k}=k(k=1,2, \cdots)$.
Hence, $\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24}$
$\Leftrightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}>\frac{m}{24}$.
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}$. Then
$$
\begin{array}{l}
f(n+1)-f(n) \\
=\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1} \\
=\frac{1}{2 n+1}-\frac{1}{2 n+2}>0 .
\end{array}
$$
Therefore, $f(n)$ is monotonically increasing.
Hence, $f(n)_{\min }=f(2)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$.
Thus, $\frac{7}{12}>\frac{m}{24} \Rightarrow m<14$.
Therefore, the maximum integer value of $m$ is 13.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that among the vertices of a regular 2012-gon, there exist $k$ vertices such that the convex $k$-gon formed by these $k$ vertices has no two parallel sides. Find the maximum value of $k$.
|
2. The maximum value of $k$ is 1509.
Let $A_{1}, A_{2}, \cdots, A_{2012}$ be the set of vertices of a regular polygon.
Consider the set of four vertices
$$
\begin{array}{l}
\left(A_{1}, A_{2}, A_{1000}, A_{1008}\right),\left(A_{3}, A_{4}, A_{100}, A_{1010}\right), \\
\cdots,\left(A_{1005}, A_{1000}, A_{2011}, A_{2012}\right) .
\end{array}
$$
If $k$ points contain a set of four points
$$
\left(A_{2 i-1}, A_{2 i}, A_{2 i+1005}, A_{2 i+1006}\right),
$$
then the convex $k$-gon obtained has two parallel sides $A_{2 i-1} A_{2 i}$ and $A_{2 i+1005} A_{2 i+1006}$, which is a contradiction.
This indicates that at most three points can be taken from each set of four points, hence
$$
k \leqslant 503 \times 3=1509 \text {. }
$$
Below is an example for $k=1509$.
It is easy to see that in the convex polygon $A_{1} A_{2} \cdots A_{1006} A_{1008} A_{1010} \cdots A_{2012}$, no two sides are parallel.
|
1509
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $n \geqslant 2$,
$$
A_{n}=\left\{x \in \mathbf{R} \left\lvert\, x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right]\right.\right\} .
$$
|
Prove: $A=\underset{n \geqslant 2}{\cup} A_{n}$ is a finite set, and find the maximum and minimum elements of $A$.
(2010, Romanian Mathematical Olympiad (Final)) [Analysis] Start with simple cases in mathematical experiments.
Elements in $A_{2}$ satisfy $x=\left[\frac{x}{2}\right]$, and $x \in \mathbf{Z}$. Hence, $\frac{x-1}{2} \leqslant x=\left[\frac{x}{2}\right] \leqslant \frac{x}{2}$, which means $-1 \leqslant x \leqslant 0$. After verification, $x=0,-1$ are elements of $A_{2}$. Therefore, $A_{2}=\{-1,0\}$.
Elements in $A_{3}$ satisfy
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right] \text {, and } x \in \mathbf{Z} . \\
\text { Hence } \frac{x-1}{2}+\frac{x-2}{3} \leqslant x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right] \\
\leqslant \frac{x}{2}+\frac{x}{3},
\end{array}
$$
which means $-7 \leqslant x \leqslant 0$.
After verification, $x=-7,-5,-4,-3,-2,0$ are elements of $A_{3}$, i.e.,
$$
A_{3}=\{-7,-5,-4,-3,-2,0\} \text {. }
$$
Elements in $A_{4}$ satisfy
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right], \text { and } x \in \mathbf{Z} . \\
\text { Hence } \frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4} \leqslant x \\
=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right] \leqslant \frac{x}{2}+\frac{x}{3}+\frac{x}{4} .
\end{array}
$$
Solving this, we get $0 \leqslant x \leqslant 23$.
Therefore, $A_{4} \subseteq\{0,1, \cdots, 23\}$.
When $x=23$,
$$
\left[\frac{x}{2}\right]=11,\left[\frac{x}{3}\right]=7,\left[\frac{x}{4}\right]=5,
$$
satisfying $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right]=23$.
Thus, $23 \in A_{4}$.
When $n \geqslant 5$,
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right], \text { and } x \in \mathbf{Z} . \\
\text { Hence } x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right] \\
\leqslant \frac{x}{2}+\frac{x}{3}+\cdots+\frac{x}{n},
\end{array}
$$
which means $x \geqslant 0$.
On the other hand,
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right] \\
\geqslant\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right] \\
\geqslant \frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4},
\end{array}
$$
which gives $x \leqslant 23$.
In summary, $A=\cup_{n \geqslant 2} A_{n} \subseteq\{-7,-6, \cdots, 23\}$ is a finite set, with the maximum element being $23 \in A_{4}$ and the minimum element being $-7 \in A_{3}$.
|
23
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is ( ).
(A) 209520
(B)209 519
(C)209 518
(D) 209517
|
4. B.
Let $3 \sqrt{2}+2 \sqrt{3}=a, 3 \sqrt{2}-2 \sqrt{3}=b$.
Then $a+b=6 \sqrt{2}, ab=6$
$$
\begin{array}{l}
\Rightarrow a^{2}+b^{2}=(a+b)^{2}-2ab=60 \\
\Rightarrow a^{6}+b^{6}=\left(a^{2}+b^{2}\right)^{3}-3(ab)^{2}\left(a^{2}+b^{2}\right) \\
\quad=209520 .
\end{array}
$$
Since $0<b<1$, we have
$$
209519<a^{6}<209520 \text {. }
$$
Therefore, the largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is
209519.
|
209519
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 4, in the "dart-shaped" quadrilateral $ABCD$, $AB=4\sqrt{3}$, $BC=8$, $\angle A=\angle B=\angle C=30^{\circ}$. Then the distance from point $D$ to $AB$ is $\qquad$
|
3. 1.
As shown in Figure 9, extend $A D$ to intersect $B C$ at point $E$. Then $A E=B E$.
Draw $E F \perp A B$ at point $F$.
It is easy to see that $A F=B F=2 \sqrt{3}$,
$$
E F=2, A E=4 \text {. }
$$
Thus, $C E=4$.
Also, $\angle A D C=90^{\circ}$
$\Rightarrow D E=2$
$\Rightarrow D$ is the midpoint of $A E$
$\Rightarrow$ The distance from $D$ to $A B$ is equal to $\frac{1}{2} E F=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 5, an equilateral $\triangle ABC$ with side length 26 is inscribed in a circle, and chord $DE \parallel BC$. The extension of $DE$ intersects the extensions of $AB$ and $AC$ at points $F$ and $G$, respectively. If the lengths of segments $AF$ and $DF$ are both positive integers, then the chord $DE=$ $\qquad$
|
4. 16 .
Let $A F=x, D F=y$. Then
$$
B F=x-26, D E=x-2 y, E F=x-y \text {. }
$$
By the secant theorem, we have
$$
\begin{array}{l}
A F \cdot B F=D F \cdot E F \\
\Rightarrow x(x-26)=y(x-y) \\
\Rightarrow x^{2}-(26+y) x+y^{2}=0 .
\end{array}
$$
Since $A F$ and $D F$ are both positive integers, we have
$$
\Delta=(26+y)^{2}-4 y^{2}>0 \Rightarrow y<26 \text {. }
$$
Upon verification, only $y=8$ satisfies the condition. Then $x=32$. Therefore, $D E=32-2 \times 8=16$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Write out a sequence of consecutive positive integers starting from 1, then erase one of the numbers, so that the average of the remaining numbers is $43 \frac{14}{17}$. What is the number that was erased?
|
Three, suppose we have written $n$ consecutive positive integers 1, 2, ..., $n$. If the number $k$ is erased, then
$$
\begin{array}{l}
\frac{(1+2+\cdots+n)-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2}+\frac{n-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2} \leqslant 43 \frac{14}{17} \leqslant \frac{n}{2}+1 \\
\Rightarrow 85 \frac{11}{17} \leqslant n \leqslant 87 \frac{11}{17} \Rightarrow n=86,87 .
\end{array}
$$
From equation (1), we know that $n-1$ must be a multiple of 17.
Thus, $n=86 \Rightarrow k=16$.
Therefore, the number erased is 16.
(Chen Qian, Yujing Middle School, Xishui County, Huanggang City, Hubei Province, 438200 Ma Xuemin, Xishui County Special Education School, Huanggang City, Hubei Province, 438200)
|
16
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that $f(x)$ is a function defined on $\mathbf{R}$. If $f(0)=0$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant x^{2}, \\
f(x+16)-f(x) \geqslant 4 x^{2}+48 x+224,
\end{array}
$$
then $f(64)=$ $\qquad$
|
5. 19840.
Notice that,
$$
\begin{array}{l}
f(x+4)-f(x) \\
=(f(x+16)-f(x))-(f(x+16)- \\
f(x+12))-(f(x+12)-f(x+8))- \\
(f(x+8)-f(x+4)) \\
\geqslant\left(4 x^{2}+48 x+224\right)-(x+12)^{2}- \\
(x+8)^{2}-(x+4)^{2} \\
= x^{2} .
\end{array}
$$
Since $f(x+4)-f(x) \leqslant x^{2}$, we have,
$$
\begin{array}{l}
f(x+4)-f(x)=x^{2} \\
\Rightarrow f(4 k+4)-f(4 k)=(4 k)^{2} \\
\Rightarrow f(64)=\sum_{k=0}^{15}(f(4(k+1))-f(4 k)) \\
\quad=\sum_{k=0}^{15} 16 k^{2}=19840 .
\end{array}
$$
|
19840
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Given that $f(x)$ is a function defined on the set of real numbers $\mathbf{R}$, $f(0)=2$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(5+2 x)=f(-5-4 x), \\
f(3 x-2)=f(5-6 x) .
\end{array}
$$
Find the value of $f(2012)$.
|
In equation (1), let $x=\frac{1}{2} y-\frac{3}{2}(y \in \mathbf{R})$, we get $f(2+y)=f(1-2 y)$.
In equation (2), let $x=\frac{1}{3} y+\frac{2}{3}(y \in \mathbf{R})$, we get $f(y)=f(1-2 y)$.
Thus, $f(2+y)=f(y)$, which means $f(x)$ is a periodic function with a period of 2.
Therefore, $f(2012)=f(2 \times 1006+0)=f(0)=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Let $x, y$ be real numbers. Then the minimum value of the algebraic expression
$$
2 x^{2}+4 x y+5 y^{2}-4 x+2 y-5
$$
is $\qquad$ [1]
(2005, National Junior High School Mathematics League Wuhan CASIO Cup Selection Competition)
|
$$
\begin{array}{l}
\text { Original expression }= x^{2}+4 x y+4 y^{2}+x^{2}-4 x+4+ \\
y^{2}+2 y+1-10 \\
=(x+2 y)^{2}+(x-2)^{2}+(y+1)^{2}-10 .
\end{array}
$$
Therefore, when $x=2, y=-1$, the minimum value of the required algebraic expression is -10.
|
-10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $m>0$. If the function
$$
f(x)=x+\sqrt{100-m x}
$$
has a maximum value of $g(m)$, find the minimum value of $g(m)$.
(2011, National High School Mathematics League Sichuan Province Preliminary Contest)
|
First, use the discriminant method to find $g(m)$.
Notice that the original function is $y-x=\sqrt{100-m x}$. Squaring both sides and rearranging, we get
$$
x^{2}+(m-2 y) x+y^{2}-100=0 \text {. }
$$
By $\Delta \geqslant 0$, we have $y \leqslant \frac{m}{4}+\frac{100}{m}$.
Thus, $g(m)=\frac{m}{4}+\frac{100}{m}$
$$
\geqslant 2 \sqrt{\frac{m}{4} \times \frac{100}{m}}=10 \text {. }
$$
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4: From the numbers $1, 2, \cdots, 2012$, select a set of numbers such that the sum of any two numbers cannot be divisible by their difference. How many such numbers can be selected at most?
(2012, Joint Autonomous Admission Examination of Peking University and Other Universities)
|
Solve: Divide $1,2, \cdots, 2012$ into
$$
\begin{array}{l}
(1,2,3),(4,5,6), \cdots, \\
(2008,2009,2010),(2011,2012)
\end{array}
$$
these 671 groups.
If at least 672 numbers are taken, then by the pigeonhole principle, there must be two numbers in the same group, let's say $a$ and $b$ ($a>b$).
Thus, $a-b=1$ or 2.
When $a-b=1$, $(a-b) \mid (a+b)$, which is a contradiction.
When $a-b=2$, $a$ and $b$ have the same parity, thus, $a+b$ is even. Therefore, $(a-b) \mid (a+b)$, which is a contradiction.
Therefore, the maximum number of numbers that can be taken is 671.
For example, taking $1,4,6, \cdots, 2011$ these 671 numbers satisfies the condition.
|
671
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given that $a$ and $b$ are real numbers, and
$$
a^{2}+a b+b^{2}=3 \text {. }
$$
If the maximum value of $a^{2}-a b+b^{2}$ is $m$, and the minimum value is $n$, find the value of $m+n$. ${ }^{[3]}$
$(2008$, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
Let $a^{2}-a b+b^{2}=k$.
From $\left\{\begin{array}{l}a^{2}+a b+b^{2}=3 \\ a^{2}-a b+b^{2}=k,\end{array}\right.$, we get $a b=\frac{3-k}{2}$.
Thus, $(a+b)^{2}=\left(a^{2}+a b+b^{2}\right)+a b$
$=3+\frac{3-k}{2}=\frac{9-k}{2}$.
Since $(a+b)^{2} \geqslant 0$, then $\frac{9-k}{2} \geqslant 0$, which means $k \leqslant 9$.
Therefore, $a+b= \pm \sqrt{\frac{9-k}{2}}, a b=\frac{3-k}{2}$.
Hence, the real numbers $a, b$ can be considered as the two roots of the quadratic equation
$$
x^{2} \mp \sqrt{\frac{9-k}{2}}+\frac{3-k}{2}=0
$$
Thus, $\Delta=\left(\sqrt{\frac{9-\bar{k}}{2}}\right)^{2}-4 \times \frac{3-k}{2}=\frac{3 k-3}{2} \geqslant 0$.
Solving this, we get $k \geqslant 1$.
Therefore, $1 \leqslant k \leqslant 9$.
So, the maximum value of $a^{2}-a b+b^{2}$ is $m=9$, and the minimum value is $n=1$.
Thus, $m+n=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-x y+y^{2}$ is $\qquad$ [4]
(2010, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
|
Solve: Completing the square for $x^{2}-x y+y^{2}$ yields
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also, }|x \pm y| \leqslant|x|+|y| \leqslant 1, \text { then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ are such that one is 0 and the other is 1, the equality holds.
Therefore, the maximum value of $x^{2}-x y+y^{2}$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{6}$ are six different positive integers, taking values from $1,2, \cdots, 6$. 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 $\qquad$ [5]
(2009, National Junior High School Mathematics League Sichuan Preliminary)
|
Since equation (1) is a cyclic expression about $x_{1}, x_{2}, \cdots, x_{6}$, we can assume $x_{1}=6, x_{j}=1(j \neq 1)$. Then
$$
\begin{aligned}
S \geqslant & \left|\left(6-x_{2}\right)+\left(x_{2}-x_{3}\right)+\cdots+\left(x_{j-1}-1\right)\right|+ \\
& \left|\left(x_{j+1}-1\right)+\left(x_{j+2}-x_{j+1}\right)+\cdots+\left(6-x_{6}\right)\right| \\
= & 10 .
\end{aligned}
$$
When $x_{i}=7-i(i=1,2, \cdots, 6)$, $S$ achieves its minimum value of 10.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the function $S=|x-2|+|x-4|$.
(1) Find the minimum value of $S$;
(2) If for any real numbers $x, y$ the inequality
$$
S \geqslant m\left(-y^{2}+2 y\right)
$$
holds, find the maximum value of the real number $m$.
|
(1) Using the absolute value inequality, the answer is 2.
(2) From the problem, we know that for any real number $y$,
$$
m\left(-y^{2}+2 y\right) \leqslant 2
$$
holds. It is easy to find that the maximum value of $-y^{2}+2 y$ is 1. Therefore, $0 \leqslant m \leqslant 2$, and the maximum value of $m$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) A fly and $k$ spiders are placed at some intersections of a $2012 \times 2012$ grid. An operation consists of the following steps: first, the fly moves to an adjacent intersection or stays in place, then each spider moves to an adjacent intersection or stays in place (multiple spiders can occupy the same intersection). Assume that each spider and the fly always know the positions of the other spiders and the fly.
(1) Find the smallest positive integer $k$ such that the spiders can catch the fly in a finite number of operations, regardless of their initial positions;
(2) In a $2012 \times 2012 \times 2012$ three-dimensional grid, what is the conclusion for (1)?
[Note] In the problem, adjacent means that one intersection has only one coordinate different from another intersection's corresponding coordinate, and the difference is 1; catching means the spider and the fly are at the same intersection.
|
The minimum value of $k$ is 2.
(1) First, prove that a single spider cannot catch the fly.
Establish a Cartesian coordinate system, then the range of the spider and fly's movement is
$$
\{(x, y) \mid 0 \leqslant x, y \leqslant 2012, x, y \in \mathbf{N}\}.
$$
For each point in the above set, there are at least two points in the set that are adjacent to it. Therefore, if the spider is not adjacent to the fly after a move, the fly can remain still; if the spider is adjacent to the fly after a move, the fly can move to another adjacent point in the set. This way, the spider cannot catch the fly.
Second, if there are two spiders $S_{1}$ and $S_{2}$, let the distance between $S_{i}$ and the fly $F$ after $n$ moves be $a_{i, n} (i=1,2)$, where the distance between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is defined as
$$
d=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|.
$$
First, it can be achieved through several operations that
$$
S_{1}(0,0), S_{2}(2012,2012).
$$
Next, we prove:
(i) After each move, $S_{1}$ and $S_{2}$ can be appropriately moved so that the fly $F$ is inside or on the boundary of the rectangle with $S_{1}$ and $S_{2}$ as the diagonal (the line connecting $S_{1}$ and $S_{2}$ is not parallel to the coordinate axes);
(ii) Keep $a_{1, n}+a_{2, n}$ non-increasing, and it will strictly decrease after several moves.
In fact, let the unit vectors in the positive directions of the $x$ and $y$ axes be $\boldsymbol{i}$ and $\boldsymbol{j}$. The strategy is as follows:
If the fly's coordinate value is not inside or on the boundary of the rectangle with $S_{1}$ and $S_{2}$ as the diagonal after a move, then $S_{1}$ and $S_{2}$ move in that direction towards the grid where $\boldsymbol{F}$ is located; otherwise, $S_{1}$ can move $\boldsymbol{i}$ or $\boldsymbol{j}$, and $S_{2}$ can move $\boldsymbol{i}$ or $-\boldsymbol{j}$, appropriately chosen to ensure that the line connecting $S_{1}$ and $S_{2}$ is not parallel to the coordinate axes, and the fly's position is inside or on the boundary of the rectangle. By induction, (i) is easily proven.
Next, we prove (ii).
In fact, if the fly is strictly inside the rectangle with $S_{1}$ and $S_{2}$ as the diagonal before the $n$-th operation, then $a_{1, n}+a_{2, n}$ must decrease after that round; otherwise, the fly must be on one of the rectangle's sides and must move outside the rectangle, and by the rule, $a_{1, n}+a_{2, n}$ does not increase, but the grid is bounded, so after at most 4024 steps, the fly can no longer move outside the rectangle. At this point, $a_{1, n}+a_{2, n}$ strictly decreases.
Since the initial value of $a_{1,0}+a_{2,0}$ is 4024, it cannot decrease indefinitely, so after a finite number of operations, the spider can catch the fly.
(2) The conclusion remains unchanged, and the minimum value of $k$ is still 2.
Establish a spatial Cartesian coordinate system, then the range of $S_{i}$ and $F$ is $\{(x, y, z) \mid 0 \leqslant x, y, z \leqslant 2012, x, y, z \in \mathbf{N}\}$.
When $k=1$, the fly can adopt a similar strategy to ensure it is not caught;
When $k=2$, let $S_{1}$ and $S_{2}$ move to $(0,0,0)$ and $(2012,2012,2012)$ through several operations, and each operation keeps the fly inside or on the boundary (face, edge, vertex) of the cube with $S_{1}$ and $S_{2}$ as the body diagonal, and the line connecting $S_{1}$ and $S_{2}$ is not parallel to the planes $x O y$, $y O z$, and $z O x$. It can be completely similarly proven that after each operation, $a_{1, n}+a_{2, n}$ does not increase, and after at most $3 \times 2012$ steps, it will strictly decrease.
Since the initial value of $a_{1,0}+a_{2,0}$ is 6036, after a finite number of operations, the spider will catch the fly.
|
2
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Sets $S_{1}, S_{2}, \cdots, S_{n}$ are pairwise distinct and satisfy the following conditions:
(1) $\left|S_{i} \cup S_{j}\right| \leqslant 2004(1 \leqslant i, j \leqslant n, i, j \in \mathbf{N}_{+})$;
(2) $S_{i} \cup S_{j} \cup S_{k}=\{1,2, \cdots, 2008\}(1 \leqslant i < j < k \leqslant n, i, j, k \in \mathbf{N}_{+})$.
Find the maximum possible value of $n$. ${ }^{[1]}$
$(2009$, Serbian Mathematical Olympiad)
|
Let the complement of $S_{i}$ in $Y=\{1,2, \cdots, 2008\}$ be $A_{i}(1 \leqslant i \leqslant n)$. Then
$\left|A_{i} \cap A_{j}\right| \geqslant 4, A_{i} \cap A_{j} \cap A_{k}=\varnothing$.
Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$.
Consider the bipartite graph $X+Y$, where a vertex $A_{i} \in X$ is adjacent to $y \in Y$ if and only if $y$ is an element of $A_{i}$.
Since $\left|A_{i} \cap A_{j}\right| \geqslant 4$, for any two vertices in $X$, their common neighbors (referring to points adjacent to both, the same below) are at least four. In other words, for any two vertices in $X$, there are at least four vertices in $Y$ that form an angle with them. Therefore, the number of angles formed by vertices in $Y$ is greater than or equal to $4 \mathrm{C}_{n}^{2}$.
On the other hand, since $A_{i} \cap A_{j} \cap A_{k}=\varnothing$, the degree of each vertex in $Y$ is less than or equal to 2, meaning that at most one angle is formed at each vertex in $Y$. Therefore, the number of angles formed by vertices in $Y$ is less than or equal to $|Y|=2008$.
In summary, $4 \mathrm{C}_{n}^{2} \leqslant 2008(n \leqslant 32)$.
Using the bipartite graph $X+Y$, it is not difficult to construct an example where $n=32$.
Therefore, the largest positive integer sought is 32.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 There are 10 sets of test papers, each set containing 4 questions, and at most one question is the same between any two sets. Among these test papers, what is the minimum number of different questions?
(2005, Taiwan Mathematical Olympiad)
|
Consider each test paper as a vertex, these vertices form a set $X$; consider each question as a vertex, these vertices form a set $Y$.
If a test paper $x \in X$ contains a question $y \in Y$, connect a line between the corresponding vertices $x$ and $y$. This results in a bipartite graph $X+Y$.
Let $Y$ have $n$ vertices, with degrees $x_{1}, x_{2}, \cdots, x_{n}$.
Since each test paper has 4 questions, each vertex in $X$ has a degree of 4. Therefore,
$x_{1}+x_{2}+\cdots+x_{n}=4 \times 10=40$.
Furthermore, since at most one question is the same between any two test papers, for any two vertices in $X$, there is at most one vertex in $Y$ that forms an angle with them, meaning the number of such angles is less than or equal to $\mathrm{C}_{10}^{2}$.
In $Y$, the number of angles at the $i$-th vertex is $\mathrm{C}_{x_{i}}^{2}$, so the total number of angles is
$\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2}$.
Thus, $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \leqslant \mathrm{C}_{10}^{2}$.
Combining this with the previous equation, we can simplify to
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \leqslant 2 \mathrm{C}_{10}^{2}+40=130 \text {. }
$$
Using a common inequality, we get
$$
\begin{array}{l}
\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \leqslant n\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \\
\Rightarrow 40^{2} \leqslant 130 n \Rightarrow n \geqslant 13 .
\end{array}
$$
Using the aforementioned bipartite graph, it is not difficult to construct an example with $n=13$ (take $x_{1}=x_{2}=4, x_{3}=x_{4}=\cdots=x_{12}=3$, $x_{13}=2$).
Therefore, there are at least 13 different questions.
The idea in Example 6 can be used to prove a general conclusion.
Theorem 2 Let a simple graph $G$ have $n$ vertices and $m$ edges. If graph $G$ does not contain $K_{2, r+1}$, then
$$
m \leqslant \frac{n}{4}[1+\sqrt{4 r(n-1)+1}] .
$$
Proof Since graph $G$ does not contain $K_{2, r+1}$, for any two vertices, the number of angles they form is less than or equal to $r$.
On the other hand, let the degrees of the vertices in the graph be $x_{1}$, $x_{2}, \cdots, x_{n}$, then the number of angles in the graph is
$$
\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \text {. }
$$
Therefore, $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \leqslant r \mathrm{C}_{n}^{2}$.
Combining this with $x_{1}+x_{2}+\cdots+x_{n}=2 m$, we get
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=r n(n-1)+2 m \text {. }
$$
As in Example 6, we can derive the inequality
$$
(2 m)^{2} \leqslant n[m(n-1)+2 m] \text {. }
$$
Thus, the theorem is proved.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given any 2012 points in the plane. Prove: the distances between each pair of them take at least 32 different values.
|
Prove that with these 2012 points as vertices, connecting edges between points of the same distance, we obtain a graph $G$.
Since there is at most one point that is equidistant from three given points, the graph $G$ does not contain $K_{2,3}$.
By Theorem 2, the number of edges in graph $G$ is
$$
m \leqslant \frac{2012}{4}(1+\sqrt{4 \times 2 \times 2011+1}) \text {, }
$$
i.e., $m \leqslant 64304$.
Therefore, the number of different distances is at least $\frac{\mathrm{C}_{2012}^{2}}{64304}$.
Thus, the number of different distances is at least 32.
[Note] This approach originates from P$\cdot$Erdös. In fact, using other methods, 32 can be improved to 45.
|
32
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Let $X$ be a set with 56 elements. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any seven of these subsets has at least $n$ elements, then there must exist three of these 15 subsets whose intersection is non-empty. ${ }^{[2]}$
(2006, China Mathematical Olympiad)
|
First, we prove by contradiction that \( n = 41 \) satisfies the requirement.
Assume there exist 15 subsets \( A_{1}, A_{2}, \cdots, A_{15} \), such that the union of any seven subsets has at least 41 elements, but the intersection of any three subsets is empty.
At this point, let \( Y = \{A_{1}, A_{2}, \cdots, A_{15}\} \), then \( X + Y \) forms a bipartite graph, where \( x \in X \) is adjacent to \( A_{j} \in Y \) if and only if \( x \) is an element of \( A_{j} \).
Since the intersection of any three subsets is empty, each element in \( X \) appears in at most two subsets, i.e., the degree of each vertex in \( X \) is less than or equal to 2.
Without loss of generality, assume each vertex in \( X \) has a degree of 2 (if a vertex has a degree \( d < 2 \), we can arbitrarily add \( 2 - d \) neighbors).
Thus, the number of edges in the graph is \( 2 \times 56 = 112 \), and the graph has exactly 56 (vertices in \( X \)) angles.
For any seven vertices in \( Y \), their number of neighbors is at least 41, which means the number of angles related to these seven vertices is at least 41. Thus, for the remaining eight vertices, the number of angles formed by any two of them is at most \( 56 - 41 = 15 \).
Next, fix 14 vertices in \( Y \), among which there are \( \binom{14}{8} \) "eight-point groups," and each eight-point group has at most 15 angles, while each angle appears in \( \binom{12}{6} \) eight-point groups. Therefore, the number of angles formed by these 14 vertices is less than or equal to \( 15 \times \frac{\binom{14}{8}}{\binom{12}{6}} < 49 \).
This indicates that the number of angles related to the remaining vertex in \( Y \) is greater than 7, i.e., the degree of this vertex is at least 8.
Thus, the degree of any vertex in \( Y \) is at least 8, and the number of edges in the graph is at least \( 8 \times 15 = 120 \). This contradicts the fact that the number of edges is 112.
In conclusion, \( n = 41 \) satisfies the given requirements.
Next, we show that when \( n = 40 \), there exist 15 subsets \( A_{1}, A_{2}, \cdots, A_{15} \), such that the union of any seven subsets has at least 40 elements, but the intersection of any three subsets is empty.
In fact, consider such a bipartite graph \( X + Y \), where \( Y = \{A_{1}, A_{2}, \cdots, A_{15}\} \), and for each \( 1 \leq i \leq 7 \) and \( 8 \leq j \leq 15 \), there is exactly one vertex in \( X \) adjacent to both \( A_{i} \) and \( A_{j} \) (i.e., it forms an angle with them).
It is easy to verify that any eight vertices in \( Y \) form at most \( x(8 - x) \leq 16 \) angles, thus, the number of angles related to the remaining seven vertices is at least 40, i.e., their number of neighbors is at least 40.
In conclusion, the smallest positive integer that satisfies the requirements is 41.
|
41
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Given the sets
$$
\begin{array}{l}
M=\{(x, y) \mid x(x-1) \leqslant y(1-y)\}, \\
N=\left\{(x, y) \mid x^{2}+y^{2} \leqslant k\right\} .
\end{array}
$$
If $M \subset N$, then the minimum value of $k$ is $\qquad$ .
(2007, Shanghai Jiao Tong University Independent Admission Examination)
|
Notice that,
$$
M=\left\{(x, y) \left\lvert\,\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant \frac{1}{2}\right.\right\}
$$
represents a disk with center $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{\sqrt{2}}{2}$.
By $M \subset N \Rightarrow \sqrt{k} \geqslant \sqrt{2} \times \frac{1}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$
$$
\Rightarrow k_{\min }=2 \text {. }
$$
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given the set of integers
$M=\left\{m \mid x^{2}+m x-36=0\right.$ has integer solutions $\}$, set $A$ satisfies the conditions:
(1) $\varnothing \subset A \subseteq M$;
(2) If $a \in A$, then $-a \in A$.
The number of all such sets $A$ is ( ).
(A) 15
(B) 16
(C) 31
(D) $32^{[1]}$
(2010, National High School Mathematics League Shandong Province Preliminary Contest)
|
Let $\alpha, \beta$ be the integer roots of the equation
$$
x^{2}+m x-36=0
$$
Assume $|\alpha| \geqslant|\beta|$. Then
$$
\begin{array}{l}
\alpha \beta=-36 \\
\Rightarrow(|\alpha|,|\beta|) \\
=(1,36)(2,18),(3,12),(4,9),(6,6) \\
\Rightarrow m= \pm 35, \pm 16, \pm 9, \pm 5,0 \\
\Rightarrow M=\{0\} \cup\{-5,5\} \cup\{-9,9\} \cup \\
\{-16,16\} \cup\{-35,35\} \text {. } \\
\end{array}
$$
By the problem statement, $A \neq \varnothing$, and the above five subsets of $M$ are either subsets of $A$ or disjoint from $A$.
Thus, there are $2^{5}-1=31$ different non-empty sets $A$.
|
31
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can always be found two elements $a, b$ such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$. ${ }^{[2]}$
(2011, China Western Mathematical Olympiad)
|
Solve for when
$$
M=\left\{2^{k}, 3 \times 2^{l} \mid k=0,1, \cdots, 10 ; l=0,1, \cdots, 9\right\}
$$
the condition is satisfied, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011$, a contradiction.
In summary, $|M|_{\max }=21$.
|
21
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let the set
$$
A=\left\{y \left\lvert\, y=\sin \frac{k \pi}{4}(k=0, \pm 1, \pm 2, \cdots)\right.\right\} \text {. }
$$
Then the number of proper subsets of set $A$ is $\qquad$ (2009, Tongji University Independent Admission Examination)
|
Hint: Calculate $A=\left\{0, \pm 1, \pm \frac{\sqrt{2}}{2}\right\}$. The number of its proper subsets is $2^{5}-1=31$.
|
31
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Find the largest positive integer $k$ such that the set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \cdots, A_{k}$, so that for all integers $n(n \geqslant 15)$ and all $i \in\{1,2, \cdots, k\}$, there exist two distinct elements in $A_{i}$ whose sum is $n$.
|
4. The largest positive integer $k$ is 3.
When $k=3$, let
$A_{1}=\{1,2,3\} \cup\{3 m \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{3}=\{7,8,9\} \cup\{3 m-21 m \in \mathbf{Z}$, and $m \geqslant 4\}$.
Then the sum of two different elements in $A_{1}$ can represent all
positive integers $n$ greater than or equal to $1+12=13$; the sum of two different elements in $A_{2}$ can represent all positive integers $n$ greater than or equal to $4+11=15$; the sum of two different elements in $A_{3}$ can represent all positive integers $n$ greater than or equal to $7+10=17$ and $7+8=15, 7+9=16$.
If $k \geqslant 4$, and there exist sets $A_{1}, A_{2}, \cdots, A_{k}$ that satisfy the conditions, then the sets $A_{1}, A_{2}, A_{3}, A_{4} \cup A_{5} \cup \cdots \cup A_{k}$ also satisfy the conditions.
Thus, we can assume $k=4$.
Let $B_{i}=A_{i} \cap\{1,2, \cdots, 23\}(i=1,2,3,4)$.
For each $i=1,2,3,4$, the integers $15,16, \cdots, 24$ can be written as the sum of two different elements in $B_{i}$. Therefore, $B_{i}$ must have at least five elements.
Since $\left|B_{1}\right|+\left|B_{2}\right|+\left|B_{3}\right|+\left|B_{4}\right|=23$, there exists $j \in\{1,2,3,4\}$ such that $\left|B_{j}\right|=5$.
Let $B_{j}=\left\{x_{1}, x_{2}, \cdots, x_{5}\right\}$. Then the sums of two different elements in $A_{j}$ that represent the integers $15,16, \cdots, 24$ are exactly the sums of all two different elements in $B_{j}$.
$$
\begin{array}{l}
\text { Hence } 4\left(x_{1}+x_{2}+\cdots+x_{5}\right) \\
=15+16+\cdots+24=195 .
\end{array}
$$
Since 195 is not divisible by 4, the above equation does not hold.
Thus, a contradiction arises.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. On a square table composed of unit squares of size $2011 \times 2011$, a finite number of napkins are placed, each covering a $52 \times 52$ square. In each unit square, write the number of napkins covering it, and let the maximum number of unit squares with the same number be $k$. For all possible configurations of napkins, find the maximum value of $k$.
|
7. The maximum value of $k$ is
$2011^{2}-\left[\left(52^{2}-35^{2}\right) \times 39-17^{2}\right]$
$=4044121-57392=3986729$.
Let $m=39$. Then $2011=52 m-17$.
Below is an example where there are 3986729 unit squares with the same number written in them.
Let the column numbers from left to right, and the row numbers from bottom to top be
$1,2, \cdots, 2011$.
Each napkin is represented by the coordinates of its lower left corner, and the napkins are divided into four categories:
The coordinates of the first category are
$(52 i+36,52 j+1)(0 \leqslant j \leqslant i \leqslant m-2)$;
The coordinates of the second category are
$(52 i+1,52 j+36)(0 \leqslant i \leqslant j \leqslant m-2)$;
The coordinates of the third category are
$(52 i+36,52 i+36)(0 \leqslant i \leqslant m-2)$;
The coordinates of the fourth category are $(1,1)$.
In Figure 2, different shading indicates different categories of napkins.
Except for the unit squares with at least two different types of shading, the number written in all other unit squares is 1. It is easy to calculate that the number of unit squares with at least two different types of shading is
$\left(52^{2}-35^{2}\right) m-17^{2}=57392$.
Below is the proof: the number of unit squares with the same number written in them does not exceed 3986729.
For any configuration of napkins and any positive integer $M$, assume there are $g$ unit squares with numbers different from $M$. We need to prove that $g \geqslant 57392$.
A line is defined as either a row or a column.
Consider any line $l$, and let the numbers written in the adjacent unit squares be $a_{1}, a_{2}, \cdots, a_{52 m-17}$.
For $i=1,2, \cdots, 52$, let $s_{i}=\sum_{t=i(\bmod 52)} a_{t}$, where $s_{1}, s_{2}, \cdots, s_{35}$ each have $m$ terms, and $s_{36}, s_{37}, \cdots, s_{52}$ each have $m-1$ terms. Each napkin intersecting line $l$ contributes exactly 1 to each $s_{i}$.
Thus, the number of napkins intersecting $l$, denoted as $s$, satisfies
$s_{1}=s_{2}=\cdots=s_{52}=s$.
If $s>(m-1) M$, then line $l$ is called "rich"; otherwise, it is called "poor".
Assume $l$ is rich. Then each of $s_{36}, s_{37}, \cdots, s_{52}$ has at least one term greater than $M$. Consider all the unit squares corresponding to these terms, and call them "rich bad squares" for this line. Thus, each rich line has at least 17 unit squares that are rich bad squares for this line.
On the other hand, if $l$ is poor, then
$s \leqslant(m-1) M<m M$.
Thus, each of $s_{1}, s_{2}, \cdots, s_{35}$ has at least one term less than $M$.
Consider all the unit squares corresponding to these terms, and call them "poor bad squares" for this line. Thus, each poor line has at least 35 unit squares that are poor bad squares for this line.
Call the labels "small" if they are congruent to $1,2, \cdots, 35$ modulo 52, and "large" otherwise, i.e., congruent to $36,37, \cdots, 52$ modulo 52.
Considering the column numbers from left to right and the row numbers from bottom to top are $1,2, \cdots, 52 m-17$, a line is called "large" or "small" depending on whether its label is large or small.
By definition, the rich bad squares of a row belong to large columns, and the poor bad squares of a row belong to small columns, and vice versa.
On each line, place a strawberry in each bad square for that line, and additionally, place an extra strawberry in each rich bad square for each "small rich" line. A unit square can independently receive strawberries from its row and column.
Note that a unit square with one strawberry has a number different from $M$. If this unit square receives an extra strawberry, then the number in this unit square is greater than $M$. Thus, it is either in a small row or a large column, and vice versa. Assume it is in a small row, then it is not bad for this column. Thus, in this case, it has no more than two strawberries. If the extra rule does not apply to this unit square, then it also has no more than two strawberries. Thus, the total number of strawberries $N$ is at most $2 g$.
Below, we estimate $N$ using a different method.
For the $2 \times 35 \mathrm{~m}$ small lines, if it is rich, there are at least 34 strawberries; if it is poor, there are at least 35 strawberries. Therefore, in either case, there are at least 34 strawberries.
Similarly, for the $2 \times 17(m-1)$ large lines, there are at least $\min \{17,35\}=17$ strawberries.
Summing over all lines, we get
$$
\begin{array}{l}
2 g \geqslant N \geqslant 2[35 m \times 34+17(m-1) \times 17] \\
=2(1479 m-289)=2 \times 57392, \\
g \geqslant 57392 .
\end{array}
$$
Thus, $g \geqslant 57392$.
|
3986729
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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