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1. Given $S=1^{2}-2^{2}+3^{2}-4^{2}+\cdots-100^{2}+$ $101^{2}$. Then the remainder when $S$ is divided by 103 is | $=.1 .1$.
Notice that
$$
\begin{array}{l}
S=1+\left(3^{2}-2^{2}\right)+\left(5^{2}-4^{2}\right)+\cdots+\left(101^{2}-100^{2}\right) \\
=1+2+3+\cdots+100+101 \\
=\frac{101 \times 102}{2}=5151=50 \times 103+1 .
\end{array}
$$
Therefore, the required remainder is 1. | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. The parabola $y=a x^{2}+b x+c$ intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If $\triangle A B C$ is a right triangle, then $a c=$ $\qquad$ | 2. -1 .
Let $A\left(x_{1}, 0\right)$ and $B\left(x_{2}, 0\right)$. Since $\triangle ABC$ is a right triangle, it is known that $x_{1}$ and $x_{2}$ must have opposite signs, so $x_{1} x_{2}=\frac{c}{a}<0$. By the projection theorem, we know that $|O C|^{2}=|A O| \cdot|B O|$, which means $c^{2}=\left|x_{1}\right| \cdot\... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $x, y, z$ satisfy
$$
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \leqslant 60 \text {. }
$$
Then the value of $x+y-z$ is ( ).
(A) 3
(B) 2
(C) 1
(D) 0 | -.1.D.
From the given equation, we have
$$
\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \leqslant 60 \text {. }
$$
Therefore, $x=-2, y=5, z=3$.
Thus, $x+y-z=-2+5-3=0$. | 0 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Given integers $a, b, c, d$ satisfy
$$
27\left(3^{a}+3^{b}+3^{c}+3^{d}\right)=6888 \text {. }
$$
Then the value of $a+b+c+d$ is ( ).
(A) 4
(B) 5
(C) 6
(D) 7 | 4.C.
Assume $a \leqslant b \leqslant c \leqslant d$. From the given equation, we have
$$
3^{a+3}\left(1+3^{b-a}+3^{c-a}+3^{d-a}\right)=3 \times 2296 \text {. }
$$
(1) If $3^{a+3}=1$, i.e., $a=-3$, then
$$
31\left(1+3^{b+3}+3^{c+3}+3^{d+3}\right)=3 \times 2296 \text {. }
$$
Since $3^{b+3} \leqslant 3^{c+3} \leqslant 3... | 6 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. Given that $x$ is a real number. Then the maximum value of $\sqrt{2000-x}+$ $\sqrt{x-2000}$ is $\qquad$ . | 4.4.
$$
\begin{array}{l}
\text { Let } t_{1}=\sqrt{2008-x}+\sqrt{x-2000} \text {. Then } \\
t_{1}^{2}=8+2 \sqrt{(2008-x)(x-2000)} \\
\leqslant 8+8=16 .
\end{array}
$$
Therefore, $t \leqslant 4$, i.e., $t_{\max }=4$. At this point, $x=2004$. | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 2, each segment of the broken line $A-B-C-D$ is parallel to the sides of the rectangle, and it divides the rectangle into two equal areas. Point $E$ is on the side of the rectangle such that segment $A E$ also bisects the area of the rectangle.
Given that segment $A B=30, B C=$
$24, C D=10$. Then ... | 8.12.
Solution 1: Let line segment $A E$ intersect $B C$ at point $M$, and draw $E P \perp B C$, with the foot of the perpendicular being $P$.
Let $E D=x, B M=y$. Then $M P=24-x-y$.
By the problem, $S_{\triangle B B Y}=S_{\text {quadrilateral } U C D E}$, then
$15 y=120-5 y+5 x$.
Thus, $4 y=24+x$.
Also, $\triangle A M... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. (40 points) A rectangular box with dimensions $a_{1} \times b_{1} \times c_{1}$ can fit into another rectangular box with dimensions $a_{2} \times b_{2} \times c_{2}$ if and only if $a_{1} \leqslant a_{2} 、 b_{1} \leqslant b_{2} 、 c_{1} \leqslant c_{2}$. Therefore, among the rectangular boxes with dimensions $a \tim... | 4. Among the boxes that meet the conditions, there are 5 types of boxes that are cubes, and 20 types of boxes that are square-based but not cubic prisms, because they have five different heights and four different choices for the square side length.
In addition, there are 10 types of boxes where the length, width, and... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7.(40 points) As shown in Figure 7, quadrilateral $ABCD$ is inscribed in a circle, $AB=AD$, and its diagonals intersect at point $E$. Point $F$ lies on segment $AC$ such that $\angle BFC = \angle BAD$. If $\angle BAD = 2 \angle DFC$, find the value of $\frac{BE}{DE}$.
保留源文本的换行和格式，直接输出翻译结果。 | 7. From $AB=AD$, we know $\angle ABD=\angle ADB=\theta$. By the property of equal arcs subtending equal angles at the circumference, we have
$$
\angle ACD=\angle ACB=\theta.
$$
Let $\angle DFC=\varphi$. Then
$$
\angle BAD=\angle BFC=2\varphi.
$$
Therefore, $\angle ABD + \angle ADB + \angle BAD$
$$
=\theta + \theta + ... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. A sequence of numbers, the first three numbers are $1, 9, 9$, and each subsequent number is the remainder of the sum of the three preceding numbers divided by 3. What is the 1999th number in this sequence? | (Tip: Apart from the first three numbers $1,9,9$, this sequence repeats every 13 terms (i.e., $1,1,2,1,1,1,0,2,0,2,1,0,0$). Since $1999-3=13 \times 153+7$, the 1999th number is the 7th number in the 154th cycle, which is exactly 0.) | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Several 1s and 2s are arranged in a row
$$
1,2,1,2,2,1,2,2,2,1,2, \cdots
$$
The rule is: the 1st number is 1, the 2nd number is 2, the 3rd number is 1, ... Generally, first write a row of 1s, then insert $k$ 2s between the $k$th 1 and the $(k+1)$th 1 ($k=1$, $2, \cdots$). Try to answer:
(1) Is the 2005th num... | Explanation: Clearly, the position of 1 is somewhat special. For the convenience of calculation, we might as well divide this sequence of numbers into $n$ groups: the 1st group has 1 number, the 2nd group has 2 numbers, the 3rd group has 3 numbers, ..., the $n$th group has $n$ numbers, and the last number of each group... | 2 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure $2, \odot O$ is internally tangent to $\odot O^{\prime}$ at point $P, \odot O$'s chord $A B$ is tangent to $\odot O^{\prime}$ at point $C$, and $A B / / O O^{\prime}$. If the area of the shaded part is $4 \pi$, then the length of $A B$ is $\qquad$ | 4.4.
As shown in Figure 10, connect $O^{\prime} C$ and $O A$. Draw $O D \perp A B$ at $D$. Then quadrilateral $O O^{\prime} C D$ is a rectangle.
Therefore, $O^{\prime} C=O D$.
Thus, $S_{\text {fill }}$
$=\pi O A^{2}-\pi O^{\prime} C^{2}$ $=\pi\left(O A^{2}-O D^{2}\right)=\pi A D^{2}=4 \pi$.
So, $A D=2$.
By the Perpend... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. The digit at the 2007th position after the decimal point of the irrational number $0.2342343423434342343434342 \cdots$ is $\qquad$ . | (Observation: Note the position of the digit 2 after the decimal point. The $n$th 2 is at the $n^{2}$th position after the decimal point. Since $44^{2}<2007<45^{2}$, the digit at the 2007th position after the decimal point is between the 44th and 45th 2, with 44 instances of 34 in between. The digit 3 is in the odd pos... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Find four distinct natural numbers such that the sum of any two of them can be divided by their difference. If the sum of the largest and smallest of these four numbers is to be minimized, what is the sum of the middle two numbers?
(3rd Hua Luogeng Cup) | Analysis: Let $a_{1}, a_{2}, a_{3}, a_{4}$ be four numbers that meet the conditions, and $a_{1}8$, which contradicts the condition that $a_{1}+a_{4}$ is the smallest.
In summary, $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(2,3,4,6)$, the sum of the middle two numbers is 7. . | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 If a natural number $N$ is appended to the right of any natural number, the resulting number can be divided by $N$ (for example, 2 appended to 35 results in 352, which is divisible by 2), then $N$ is called a "magic number". Among the natural numbers less than 130, how many magic numbers are there? | Analysis: To calculate how many magic numbers there are, we first need to clarify what constitutes a magic number. Although the problem provides the definition of a magic number, it is not convenient to directly use this definition to determine whether a number is a magic number. Therefore, we need to explore a simplif... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
244 Given real numbers $a, b, c, d$ satisfy
$$
a+b+c+d=ab+ac+ad+bc+bd+cd=3 \text{. }
$$
Find the maximum real number $k$, such that the inequality
$$
a+b+c+2ab+2bc+2ca \geqslant k d
$$
always holds. | Given:
$$
\begin{aligned}
& (a+b+c+d)^{2}=3^{2} \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}+2 a b+2 a c+2 a d+ \\
& 2 b c+2 b d+2 c d=9 \\
\Rightarrow & a^{2}+b^{2}+c^{2}+d^{2}=3 .
\end{aligned}
$$
Let \( y=(x-a)^{2}+(x-b)^{2}+(x-c)^{2} \).
Then \( y=3 x^{2}-2(a+b+c) x+\left(a^{2}+b^{2}+c^{2}\right) \).
Since \( y \geqsl... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given the function $f(x)=x+\frac{t}{x}(t>0)$ and the point $P(1,0)$, draw two tangent lines $P M$ and $P N$ from $P$ to the curve $y=f(x)$, with the points of tangency being $M$ and $N$.
(1) Let $|M N|=g(t)$, find the expression for the function $g(t)$.
(2) Does there exist a $t$ such that $M$, $N$, and $A(0,... | Solution: (1) The equation of the chord of tangents $MN$ passing through the external point $P(1,0)$ of the curve $y=x+\frac{t}{x}(t>0)$ is
$$
y=2 x+2 t \text {. }
$$
Substituting into $y=x+\frac{t}{x}(t>0)$, we get $x^{2}+2 t x-t=0$.
Thus, $x_{1}+x_{2}=-2 t, x_{1} x_{2}=-t \ldots$
Therefore, $g(t)=\sqrt{\left(x_{1}-x... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Choose $n$ numbers from $1,2, \cdots, 9$, among which there must be several numbers (at least one, or all), the sum of which can be divisible by 10. Find the minimum value of $n$.
(2008, National Junior High School Mathematics Competition) | Solution: When $n \leqslant 4$, since no combination of $n$ numbers from $1,3,5,8$ can sum to a multiple of 10, we have $n \geqslant 5$.
Below, we prove that $n=5$ meets the requirement, i.e., we prove that from $1,2, \cdots, 9$, any 5 numbers chosen will always include some numbers whose sum is a multiple of 10.
We u... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. As shown in Figure 5, in quadrilateral $A B C D$, $\angle A=$ $\angle B C D=90^{\circ}, B C=$ $C D, E$ is a point on the extension of $A D$. If $D E=A B$ $=3, C E=4 \sqrt{2}$, then the length of $A D$ is | 11.5.
In Figure 5, connect $A C$. It is easy to prove that
$\triangle C D E \cong \triangle C B A, \angle A C E=90^{\circ}$.
Since $C A=C E=4 \sqrt{2}$, therefore, $A E=8$.
Thus, $A D=5$. | 5 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
13. Xiao Wang walks along the street at a constant speed, and he notices that a No. 18 bus passes him from behind every $6 \mathrm{~min}$, and a No. 18 bus comes towards him every $3 \mathrm{~min}$. Assuming that each No. 18 bus travels at the same speed, and the No. 18 bus terminal dispatches a bus at fixed intervals,... | 13.4 .
Let the speed of bus No. 18 be $x \mathrm{~m} / \mathrm{min}$, the walking speed of Xiao Wang be $y \mathrm{~m} / \mathrm{min}$, and the distance between two consecutive buses traveling in the same direction be $s \mathrm{~m}$. From the problem, we have
$$
\left\{\begin{array}{l}
6 x-6 y=s, \\
3 x+3 y=s .
\end{... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $n$ is an integer, and the quadratic equation in $x$
$$
(n-1)^{2} x^{2}-5 n(n-1) x+\left(6 n^{2}-n-1\right)=0
$$
has at least one integer root. Then the sum of all possible values of $n$ is $\qquad$ | $$
[(n-1) x-(2 n-1)][(n-1) x-(3 n+1)]=0 .
$$
Since $n \neq 1$, then
$$
\begin{array}{l}
x_{1}=\frac{2 n-1}{n-1}=2+\frac{1}{n-1}, \\
x_{2}=\frac{3 n+1}{n-1}=3+\frac{4}{n-1} .
\end{array}
$$
Because the original equation has at least one integer root, and $n$ is an integer, so, $n-1= \pm 1, \pm 2, \pm 4$.
Solving gives... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9}$, $x \in[0,1]$. Provide the following conclusions:
(1) $f(x)>0$;
(2) $f(x)<0$;
(3) There exists $x_{0} \in[0,1)$, such that $f\left(x_{0}\right)=0$;
(4) There exists $x_{0} \in[0,1]$, such that $f\left(x_{0}\right)<0$.
Among them, the correct conclusion numbers are... | 2. (1)
$$
\begin{array}{l}
f(x)=-x^{4}+4 x^{3}-2 x^{2}-2 x+\frac{13}{9} \\
=x^{3}(1-x)+3 x^{3}-3 x^{2}+(x-1)^{2}+\frac{4}{9} \\
=x^{3}(1-x)+(x-1)^{2}+\frac{3}{2} x^{3}+\frac{3}{2} x^{3}+\frac{4}{9}-3 x^{2} \\
\geqslant x^{3}(1-x)+(x-1)^{2}+3 \sqrt[3]{\frac{3}{2} x^{3} \cdot \frac{3}{2} x^{3} \cdot \frac{4}{9}}-3 x^{2} ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $f(x)=x^{2}+2 x+1$, there exists a real number $t$ such that when $x \in[1, m]$, $f(x+t) \leqslant x$ always holds, then the maximum value of $m$ is $\qquad$ . | 2.4.
Translate the graph of $f(x)$ to the right by $-t$ units, and from the graphical analysis, the maximum value of $m$ is the larger of the x-coordinates of the two intersection points of $\left\{\begin{array}{l}y=x, \\ y=f(x+t)\end{array}\right.$
From $f(1+t)=1$, we get $t=-1, t=-3$.
Then, from $f(x-3)=x$, we have
... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $n$ be a positive integer. If
$$
A(n)=1986^{n}+1987^{n}+1988^{n}+1989^{n}
$$
is not divisible by 5, find $n$. | (Tip: Only consider the unit digit $G(A(n))$ of $A(n)$. When $n=4 k\left(k \in \mathbf{N}_{+}\right)$,
$$
\begin{array}{l}
G(A(n))=G\left(6^{n}+7^{n}+8^{n}+9^{n}\right) \\
=G\left(6^{4}+7^{4}+8^{4}+9^{4}\right) \\
=G(6+1+6+1)=4,
\end{array}
$$
at this time, 5 does not divide $A(n)$;
$$
\begin{array}{l}
\text { When } ... | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. There are 10 positive integers arranged from smallest to largest: $1, 4, 8$, $10, 16, 19, 21, 25, 30, 43$. How many groups of consecutive numbers have a sum that is divisible by 11? | Let 10 positive integers be $a_{1}, a_{2}$, $\cdots, a_{10}$, and let $S_{n}=a_{1}+a_{2}+\cdots+a_{n}$. Then
$$
a_{i+1}+a_{i+2}+\cdots+a_{j}=S_{j}-S_{i} .
$$
The sequence $a_{n}(n=1,2, \cdots, 10)$ has remainders modulo 11 of $1,4,-3,-1,5,-3,-1,3,-3,-1$; the sum of the first $n$ terms $S_{n}$ has remainders modulo 11 ... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. If
$$
\frac{4^{5}+4^{5}+4^{5}+4^{5}}{3^{5}+3^{5}+3^{5}} \times \frac{6^{5}+6^{5}+6^{5}+6^{5}+6^{5}+6^{5}}{2^{5}+2^{5}}=2^{n},
$$
then, $n=$ . $\qquad$ | $-1.12$
The left side of the original equation can be transformed into
$$
\frac{4 \times 4^{5}}{3 \times 3^{5}} \times \frac{6 \times 6^{5}}{2 \times 2^{5}}=\frac{4^{6}}{3^{6}} \times \frac{6^{6}}{2^{6}}=4^{6}=2^{12} .
$$
Thus, $2^{12}=2^{n}$. Therefore, $n=12$. | 12 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let the integer $a$ divided by 7 leave a remainder of 3, and the integer $b$ divided by 7 leave a remainder of 5. If $a^{2}>4 b$, find the remainder when $a^{2}-4 b$ is divided by 7.
(1994, Tianjin City Junior High School Mathematics Competition) | Solution: Let $a=7 m+3, b=7 n+5$. Then
$$
\begin{array}{l}
a^{2}-4 b=(7 m+3)^{2}-4(7 n+5) \\
=49 m^{2}+42 m+9-28 n-20 \\
=7\left(7 m^{2}+6 m-4 n-2\right)+3 .
\end{array}
$$
Therefore, the remainder when $a^{2}-4 b$ is divided by 7 is 3. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $a, b, c$ simultaneously satisfy $a-7b+8c=4$ and $8a+4b-c=7$. Then, $a^{2}-b^{2}+c^{2}=$ $\qquad$ . | 4.1.
According to the conditions $a+8c=4+7b, 8a-c=7-4b$. Squaring both sides of the two equations and then adding them yields
$$
\begin{array}{l}
(a+8c)^{2}+(8a-c)^{2} \\
=(7+4b)^{2}+(7-4b)^{2} .
\end{array}
$$
Simplifying and organizing, we get $65\left(a^{2}+c^{2}\right)=65\left(1+b^{2}\right)$.
Therefore, $a^{2}-b... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. In an equilateral triangle $\triangle A B C$ with side length $2 \sqrt{7}$, $M$ is the midpoint of side $B C$, and $P$ is a point on $A C$.
(1) For what value of $P C$ is $B P+P M$ minimized?
(2) Find the minimum value of $B P+P M$. | 7. As shown in Figure 8, connect $A M$, and flip $\triangle A B C$ along the axis $A C$ by $180^{\circ}$ to get $\triangle A D C$. Let $N$ be the symmetric point of $M$ (about $A C$). At this point, we always have $P M = P N$. Since $B$ and $N$ are fixed points, $B P + P N \geqslant B N$, with equality holding if and o... | 7 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 There is a sequence of numbers: $1,3,4,7,11,18, \cdots$, starting from the third number, each number is the sum of the two preceding numbers.
(1) What is the remainder when the 1991st number is divided by 6?
(2) Group the above sequence as follows:
$(1),(3,4),(7,11,18), \cdots$,
where the $n$-th group has ex... | Solution: Let the $n$-th number of the sequence be $a_{n}$. Then
$$
a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3) .
$$
Through experimentation, it is known that the remainders of $a_{1}, a_{2}, \cdots, a_{26}$ when divided by 6 are
$$
\begin{array}{l}
1,3,4,1,5,0,5,5,4,3,1,4,5, \\
3,2,5,1,0,1,1,2,3,5,2,1,3 .
\end{array}
$$
No... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (15 points) The function $f(x)$ defined on the interval $[0,1]$ satisfies $f(0)=f(1)=0$, and for any $x_{1}, x_{2} \in [0,1]$, we have
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant f\left(x_{1}\right)+f\left(x_{2}\right) \text {. }
$$
(1) Prove: For any $x \in [0,1]$, we have
$$
f(x) \geqslant 0 \text {; }
$$
... | (1) Take $x_{1}=x_{2}=x \in[0,1]$, then
$$
f\left(\frac{2 x}{2}\right) \leqslant f(x)+f(x),
$$
i.e., $f(x) \leqslant 2 f(x)$.
Thus, $f(x) \geqslant 0$.
Therefore, for any $x \in[0,1]$, $f(x) \geqslant 0$.
(2) From $f(0)=f(1)=0$, we get
$$
f\left(\frac{0+1}{2}\right) \leqslant f(0)+f(1)=0+0=0.
$$
Thus, $f\left(\frac{1... | 0 | Algebra | proof | Yes | Yes | cn_contest | false |
Three. (20 points) For a set $S\left(S \subseteq \mathbf{N}_{+}\right)$, if for any $x \in S, \dot{x}$ cannot divide the sum of elements of any non-empty subset of $S \backslash\{x\}$, then $S$ is called a "good set" ($S \backslash\{x\}$ represents the set $S$ after removing the element $x$).
(1) If $\{3,4, n\}$ is a g... | Three, (1) Obviously, $n>4$.
If $n=5$, then $4 \mid (3+5)$; if $n=6$, then $3 \mid 6$; if $n=7$, then $7 \mid (3+4)$; if $n=8$, then $4 \mid 8$; if $n=9$, then $3 \mid 9$.
When $n=10$, it is verified that $\{3,4,10\}$ is a good set.
Therefore, the minimum value of $n$ is $n_{0}=10$.
(2) If $\{3,4,10, m\}$ is a good set... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 10 Given Theorem: "If three prime numbers $a, b, c$ greater than 3 satisfy the equation $2a + 5b = c$, then $a + b + c$ is a multiple of the integer $n$." What is the maximum possible value of the integer $n$ in the theorem? Prove your conclusion.
(1997, National Junior High School Mathematics League) | Solution: Let $a=3 k_{1}+r_{1}, b=3 k_{2}+r_{2}$. Then
$$
\begin{array}{l}
a+b+c=3(a+2 b) \\
=3\left(3 k_{1}+r_{1}+6 k_{2}+2 r_{2}\right) \\
=9\left(k_{1}+2 k_{2}\right)+3\left(r_{1}+2 r_{2}\right) .
\end{array}
$$
Since $a, b$ are both primes greater than 3, we have
$$
r_{1} r_{2} \neq 0.
$$
If $r_{1} \neq r_{2}$, t... | 9 | Number Theory | proof | Yes | Yes | cn_contest | false |
3. Find the remainder when $47^{37^{2}}$ is divided by 7.
Try to find the remainder of $47^{37^{2}}$ when divided by 7. | (Note that $47^{6} \equiv(-2)^{6} \equiv 2^{6} \equiv 8^{2} \equiv 1^{2}$ $\equiv 1(\bmod 7)$. Also, $37^{23} \equiv 1^{23} \equiv 1(\bmod 6)$, let $37^{23}$ $=6k+1$. Then $47^{37^{23}}=47^{k+1}=\left(47^{6}\right)^{k} \times 47 \equiv$ $\left.1^{k} \times 47 \equiv 47 \equiv 5(\bmod 7).\right)$ | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{2009}\right]$ when divided by 7 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 7.6.
Let $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{1-\sqrt{5}}{2}$. Then
$$
-1<\beta<0,1<\alpha<2 \text {, }
$$
and $\alpha+\beta=1, \alpha \beta=-1$.
Therefore, $\alpha, \beta$ are the two distinct real roots of $x^{2}-x-1=0$.
Let $A_{n}=\alpha^{n}+\beta^{n}$. Then $A_{n+2}=A_{n+1}+A_{n}$.
By $A_{1}=\alpha+\beta=1$,... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that the three non-zero real roots of the equation $x^{3}+a x^{2}+b x+c$ $=0$ form a geometric progression. Then $a^{3} c-b^{3}$ $=$ . $\qquad$ | 8.0.
Let the three roots be $d$, $d q$, and $d q^{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
d+d q+d q^{2}=-a, \\
d^{2} q+d^{2} q^{2}+d^{2} q^{3}=b, \\
d^{3} q^{3}=-c .
\end{array}\right.
$$
Dividing (2) by (1) gives $d q=-\frac{b}{a}$.
Substituting into (3) yields $\left(-\frac{b}{a}\right)^{3}=-c$... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (15 points) Given two lines
$$
\begin{array}{l}
l_{1}: 3 x+4 y-25=0, \\
l_{2}: 117 x-44 y-175=0,
\end{array}
$$
Point $A$ has projections $B$ and $C$ on $l_{1}$ and $l_{2}$, respectively.
(1) Find the locus curve $\Gamma$ of point $A$ such that $S_{\triangle A B C}=\frac{1728}{625}$;
(2) If $\odot T:\left(x-\frac{... | 10. (1) Let $A(x, y)$. It is easy to know
$$
\begin{array}{l}
A B=\frac{|3 x+4 y-25|}{\sqrt{3^{2}+4^{2}}}=\frac{|3(x-3)+4(y-4)|}{5}, \\
A C=\frac{|117 x-44 y-175|}{\sqrt{117^{2}+44^{2}}} \\
=\frac{|117(x-3)-44(y-4)|}{125} .
\end{array}
$$
Let the angles of inclination of $l_{1}$ and $l_{2}$ be $\dot{\alpha}_{1}$ and $... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange $1, 2, 3$ in a cyclic order from left to right, forming a 2009-digit number: $x=123123 \cdots 12312$. Find the remainder when $x$ is divided by 101. | (Tip: Since $100 \equiv -1 \pmod{101}$, we have
$$
\begin{array}{l}
123123 = 123 \times 100 + 123 \\
\equiv -123 + 123 \equiv 0 \pmod{101}.
\end{array}
$$
Therefore, $123123 \cdots 12312 = 123123 \cdots 12300 + 12$
$$
\equiv 0 + 12 \equiv 12 \pmod{101}.
$$
) | 12 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Find the unit digit of the natural number $2^{100}+3^{101}+4^{100}$. | Solution: Notice that
$$
\begin{array}{l}
2^{100} \equiv 2^{1 \times 25} \equiv\left(2^{4}\right)^{25} \equiv 16^{25} \\
\equiv 6^{25} \equiv 6(\bmod 10), \\
3^{101} \equiv 3^{4 \times 25+1} \equiv\left(3^{4}\right)^{25} \times 3^{1} \\
\equiv 1^{25} \times 3^{1} \equiv 3(\bmod 10), \\
4^{102} \equiv 4^{2 \times 51} \e... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The inequality $x^{2}+|2 x-6| \geqslant a$ holds for all real numbers $x$. Then the maximum value of the real number $a$ is $\qquad$ | 2.5.
When $x \geqslant 3$,
left side $=x^{2}+2 x-6=(x+1)^{2}-7 \geqslant 9$;
When $x<3$,
left side $=x^{2}-2 x+6=(x-1)^{2}+5 \geqslant 5$.
Therefore, for the given inequality to hold for all real numbers $x$, the maximum value of $a$ should be 5. | 5 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Find the largest positive number $\lambda$, such that for any real numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$, the inequality
$|\lambda x y+y z| \leqslant \frac{\sqrt{5}}{2}$ holds.
(Zhang Zhengjie) | 5. Notice that
$$
\begin{array}{l}
1=x^{2}+y^{2}+z^{2}=x^{2}+\frac{\lambda^{2}}{1+\lambda^{2}} y^{2}+\frac{1}{1+\lambda^{2}} y^{2}+z^{2} \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(\lambda|x y|+|y z|) \\
\geqslant \frac{2}{\sqrt{1+\lambda^{2}}}(|\lambda x y+y z|),
\end{array}
$$
and when $y=\frac{\sqrt{2}}{2}, x=\frac... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $n$ be a positive integer, and let $f(n)$ denote the number of $n$-digit numbers (called wave numbers) $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfy the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and $a_{i} \neq$ $a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, the signs of $a_{i}-a_{i... | And the $n$-digit waveform number $\overline{a_{1} a_{2} \cdots a_{n}}$ that satisfies $a_{1}>a_{2}$ is called a "B-type number". According to symmetry, when $n \geqslant 2$, the number of B-type numbers is also $g(n)$. Therefore, $f(n)=2 g(n)$.
Next, we find $g(n)$: Let $m_{k}(i)$ represent the number of $k$-digit A-... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Given that positive integers $p, q$ are both prime numbers, and $7p+q, pq+11$ are also prime numbers. Find $p+q$.
保留源文本的换行和格式，直接输出翻译结果。 | When $p \equiv q \equiv 1(\bmod 2)$,
$$
7 p+q \equiv 7+1 \equiv 0(\bmod 2) \text{. }
$$
But $7 p+q \geqslant 7>2$, which contradicts the fact that $7 p+q$ is a prime number. Therefore, one of $p, q$ must be even.
Since $p, q$ are both primes, then $p=2$ or $q=2$.
(1) When $p=2$, $14+q$ and $2 q+11$ are both primes.
If... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. If the equation $z^{2009}+z^{2008}+1=0$ has roots of modulus 1, then the sum of all roots of modulus 1 is $\qquad$ . | 8. -1 .
Let $z$ be a root satisfying the condition. Then the original equation is equivalent to
$$
z^{2008}(z+1)=-1 \text {. }
$$
Taking the modulus on both sides, we get $|z+1|=1$.
Since $|z|=1$, all roots with modulus 1 can only be the complex numbers corresponding to the intersection points of the circle I: with c... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
15. There are 10 players $A_{1}, A_{2}, \cdots, A_{10}$, whose points are $9,8,7,6,5,4,3,2,1,0$, and their rankings are 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th. Now a round-robin tournament (i.e., each pair of players plays exactly one match) is held, and each match must have a winner. If the player with a hi... | 15. The minimum cumulative score of the new champion is 12.
If the new champion's score does not exceed 11 points, then
$A_{1}$ can win at most 2 games; $A_{2}$ can win at most 3 games;
$A_{3}$ can win at most 4 games; $A_{4}$ can win at most 5 games.
$A_{5}$ can increase by at most 6 points, but there are only 5 play... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $x$ and $y$ are integers, $y=\sqrt{x+2003}-$ $\sqrt{x-2009}$. Then the minimum value of $y$ is $\qquad$ . | 3.2 .
Let $\sqrt{x+2003}=a, \sqrt{x-2009}=b$.
Since $x, y$ are integers,
$y=a-b=\frac{a^{2}-b^{2}}{a+b}=\frac{4012}{a+b} \in \mathbf{Z}_{+}$.
Thus, $a-b \in \mathbf{Q}, a+b \in \mathbf{Q}$
$\Rightarrow a \in \mathbf{Q}, b \in \mathbf{Q} \Rightarrow a, b \in \mathbf{Z}$.
Then $(a+b)(a-b)=4012=2006 \times 2$.
Since $a+b... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, on the hypotenuse $A B$ respectively intercept $A D=$
$$
A C, B E=B C, D E=6 \text{, }
$$
$O$ is the circumcenter of $\triangle C D E$, as shown in Figure 2. Then the sum of the distances from $O$ to the three sides of $\triangle A B C$ is . $\qquad$ | 4.9.
As shown in Figure 8, connect
$$
O A, O B, O C, O D \text {, and }
$$
$O E$.
Since $O C=O D$,
we know that point $O$ lies on the perpendicular bisector of segment $C D$.
Also, $A D=A C$, so $O A$ bisects $\angle C A D$.
Similarly, $O B$ bisects $\angle C B D$.
Therefore, $O$ is the incenter of $\triangle A B C$.... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sum of $2 n+1\left(n \in \mathbf{N}_{+}\right)$ consecutive positive integers is $a$, and the difference between the sum of the squares of the last $n$ numbers and the sum of the squares of the first $n$ numbers is $b$. If $\frac{a}{b}=\frac{11}{60}$, then the value of $n$ is | 2.5
Let the $(n+1)$-th number be $m$. Then
$$
\begin{array}{l}
a=(2 n+1) m, \\
b=\sum_{i=1}^{n}(m+i)^{2}-\sum_{i=1}^{n}(m-i)^{2} \\
=2 m \sum_{i=1}^{n} 2 i=2 m n(n+1) . \\
\text { Hence } \frac{a}{b}=\frac{2 n+1}{2 n(n+1)}=\frac{11}{60} .
\end{array}
$$
Solving gives $n=5$ or $-\frac{6}{11}$ (discard).
| 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. From the arithmetic sequence $2,5,8,11, \cdots$, take $k$ terms such that the sum of their reciprocals is 1. Then the minimum value of $k$ is $\qquad$ | 2.8 .
First, take $2,5,8,11,20,41,110,1640$, it is easy to see that the sum of their reciprocals is 1, i.e., $k=8$ satisfies the requirement.
Second, suppose we take $x_{1}, x_{2}, \cdots, x_{k}$ from the sequence, such that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{k}}=1$.
Let $y_{i}=\frac{x_{1} x_{2} \cdot... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$ are any two points (which can coincide) on the graph of the function
$$
f(x)=\left\{\begin{array}{ll}
\frac{2 x}{1-2 x}, & x \neq \frac{1}{2} \\
-1, & x=\frac{1}{2}
\end{array}\right.
$$
Point $M$ lies on the line $x=\frac{1}{2}$, and $\overrightarrow{... | 4. -2 .
Given that point $M$ is on the line $x=\frac{1}{2}$, let $M\left(\frac{1}{2}, y_{M}\right)$.
Also, $\overrightarrow{A M}=\overrightarrow{M B}$, that is,
$$
\begin{array}{l}
\overrightarrow{A M}=\left(\frac{1}{2}-x_{1}, y_{M}-y_{1}\right), \\
\overrightarrow{M B}=\left(x_{2}-\frac{1}{2}, y_{2}-y_{M}\right) .
\e... | -2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. If $n \in \mathbf{N}$, and $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$ is a positive integer, then $n=$ $\qquad$ | 6.5.
Notice that $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$
$$
=\frac{33}{\sqrt{n^{2}+24}+\sqrt{n^{2}-9}},
$$
thus $\sqrt{n^{2}+24}+\sqrt{n^{2}-9}$ could be $1,3,11,33$. Therefore, the solution is $n=5$. | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a, b \in \mathbf{N}_{+}$, when $a^{2}+b^{2}$ is divided by $a+b$, the quotient is $q$, and the remainder is $r$. Then the number of pairs $(a, b)$ that satisfy $q^{2}+r=2009$ is $\qquad$ pairs. | -1.0 .
From $2\left(a^{2}+b^{2}\right) \geqslant(a+b)^{2}$, we get $\frac{a^{2}+b^{2}}{a+b} \geqslant \frac{a+b}{2}$.
Then $q+1>q+\frac{r}{a+b}=\frac{a^{2}+b^{2}}{a+b}$ $\geqslant \frac{a+b}{2} \geqslant \frac{r+1}{2}$.
Thus, $r<2 q+1$.
From $q^{2}+r=2009<(q+1)^{2}$, we get $q \geqslant 44$. From $q^{2}+r=2009 \geqslan... | 0 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $a, b, c, d \in \mathbf{Z}$,
$$
f(x)=a x^{3}+b x^{2}+c x+d \text {. }
$$
Then among $A(1,1)$, $B(2009,1)$, $C(-6020,2)$, $D(2,-6020)$, at most $\qquad$ of them can be on the curve $y=f(x)$. | 4.3.
If $A$ and $C$ are both on $y=f(x)$, then $f(1)=1, f(-6020)=2$.
It is easy to see that $\left(x_{1}-x_{2}\right) \mid \left(f\left(x_{1}\right)-f\left(x_{2}\right)\right)$. Therefore, $[1-(-6020)] \mid (f(1)-f(-6020))=-1$, which is a contradiction.
Thus, $A$ and $C$ cannot both be on $y=f(x)$.
Similarly, $B$ and ... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. (16 points) On the Cartesian plane, a point whose both coordinates are rational numbers is called a rational point. Find the smallest positive integer $k$ such that: for every circle that contains $k$ rational points on its circumference, the circle must contain infinitely many rational points on its circumference. | 12. First, prove: If a circle's circumference contains 3 rational points, then the circumference must contain infinitely many rational points.
Let $\odot C_{0}$ in the plane have 3 rational points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ on its circumference, with the center $C_{0}\left(x_{0}, y_{0}\right)$.
Since t... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $a, b, c \geqslant 0$, and $a b+b c+c a=\frac{1}{3}$. Prove:
$$
S=\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-a c+1}+\frac{1}{c^{2}-a b+1} \leqslant 3 .
$$ | Explanation: When $a=b=c=\frac{1}{3}$, $S=3$.
Also, when $a=0, b=\frac{\sqrt{3}}{3}, c=\frac{\sqrt{3}}{3}$, $S=3$ as well.
Therefore, the maximum value 3 is both conventional and unconventional, which is quite rare. Moreover, $S$ has no other extremum (the lower bound of $S$ is $\frac{5}{2}$, $S>\frac{5}{2}$ but cannot... | 3 | Inequalities | proof | Yes | Yes | cn_contest | false |
(1) Find the minimum value of the smallest side length of a peculiar triangle;
(2) Prove that there are infinitely many isosceles peculiar triangles;
(3) How many non-isosceles peculiar triangles are there? | (1) Let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of a peculiar triangle. Then, by Heron's formula, we have
$$
\begin{aligned}
16 \Delta^{2}= & (a+b+c)(a+b-c) . \\
& (a-b+c)(-a+b+c) .
\end{aligned}
$$
Since $(a, b, c)=1$, at least one of $a, b, c$ must be odd. If there is an odd number of odd numbers a... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. In trapezoid $A B C D$, it is known that $A D / / B C(B C>$ $A D), \angle D=90^{\circ}, B C=C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$
(2004, "Xinli Cup" National Junior High School Mathematics Competition) | (提示: Extend $D A$ to point $F$, complete the trapezoid into a square $C B F D$, then rotate $\triangle A B F$ $90^{\circ}$ around point $B$ to the position of $\triangle B C G$. It is easy to see that $\triangle A B E \cong \triangle G B E$. Therefore, $A E$ $=E G=E C+A F=10$. Let $E C=x$, then $A F=10-$ $x, D E=12-x, ... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. In trapezoid $A B C D$, it is known that $A D / / B C, A D \perp$ $C D, B C=C D=2 A D, E$ is a point on side $C D$, $\angle A B E=45^{\circ}$. Then $\tan \angle A E B=$ $\qquad$
(2007, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | (提示: Extend $D A$ to point $F$, such that $A F=A D$, to get the square $B C D F$. Then extend $A F$ to point $G$, such that $F G=C E$. Then $\triangle B F G \cong \triangle B C E$. It is also easy to prove that $\triangle A B G \cong \triangle A B E$. Let $C E=x, B C=2 a$. Then $F G=x, A F=A D=a, A E=A G=x+a, D E=2 a-x... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
(2008, National High School Mathematics Competition) | Explanation: By the symmetry of the parabola, we can assume
$$
P\left(2 t^{2}, 2 t\right)(t>0) \text {. }
$$
Since the equation of the circle is $x^{2}+y^{2}-2 x=0$, the equation of the chord of contact $M N$ is
$$
2 t^{2} x+2 t y-\left(x+2 t^{2}\right)=0,
$$
which simplifies to $\left(2 t^{2}-1\right) x+2 t y-2 t^{2... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Quadratic Function
$$
f(x)=a x^{2}+b x+c(a, b \in \mathbf{R} \text {, and } a \neq 0)
$$
satisfies the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and
$$
f(x) \geqslant x \text {; }
$$
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\math... | Explanation: First, derive the analytical expression of $f(x)$ from the given conditions, then classify and discuss $m$ and $t$ to determine the value of $m$.
Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about the line $x=-1$. Therefore,
$$
-\frac{b}{2 a}=-1 \Rightarrow b=2 a \text {. }
$$
From condi... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 A $6 \times 6$ chessboard has 36 squares. Placing chess pieces in the squares, if there are 4 pieces of the same color in a straight line (horizontal, vertical, or at a $45^{\circ}$ diagonal), it is called a "four-in-a-row". Player A places white pieces, and Player B places black pieces. If A goes first, what... | Solution: At least 10 chess pieces need to be placed.
This is because, to prevent the four-in-a-row formation by the black pieces placed subsequently by player B, player A must place at least one white piece in each $1 \times 4$ rectangle, meaning that at least 8 white pieces must be placed in the $2 \times 4$ rectangl... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 4, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5... | Analysis: The difficulty ratio of this problem has significantly increased from 1, mainly due to its "apparent" asymmetry (the number of rows and columns are inconsistent). Therefore, the key to solving the problem is to find the symmetry hidden behind this asymmetry.
Inspired by Example 1, we will also start from the ... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five distinct integers satisfying the condition
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=9
$$
If $b$ is an integer root of the equation
$$
\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\left(x-a_{4}\right)\left(x-a_{5}\right)=2009
$$
then the value of $b$ is... | 8. 10 .
Notice
$$
\left(b-a_{1}\right)\left(b-a_{2}\right)\left(b-a_{3}\right)\left(b-a_{4}\right)\left(b-a_{5}\right)=2009,
$$
and $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five different integers, so all $b-a_{1}, b-a_{2}, b-a_{3}, b-a_{4}, b-a_{5}$ are also five different integers.
$$
\begin{array}{l}
\text { Also, ... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A. $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following conditions:
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. | 14. A. Let in $a_{1}, a_{2}, \cdots, a_{n}$, after removing $a_{i}(i=1$, $2, \cdots, n)$, the arithmetic mean of the remaining $n-1$ numbers is a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Therefore, for any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
b_{i... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x_{1}, x_{2}, \cdots, x_{6}$ be positive integers, and they satisfy the relation
$$
\begin{aligned}
x_{6} & =2288, \\
x_{n+3} & =x_{n+2}\left(x_{n+1}+2 x_{n}\right)(n=1,2,3) .
\end{aligned}
$$
Then $x_{1}+x_{2}+x_{3}=$ | 3. 8 .
Substituting $n=1,2,3$ into the relation
$$
\begin{array}{l}
x_{n+3}=x_{n+2}\left(x_{n+1}+2 x_{n}\right) . \\
\text { we get } x_{4}=x_{3}\left(x_{2}+2 x_{1}\right), \\
x_{5}=x_{4}\left(x_{3}+2 x_{2}\right)=x_{3}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right), \\
x_{6}=x_{3}^{2}\left(x_{2}+2 x_{1}\right)\l... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
(2003, National Junior High School Mathematics Competition) | Solution: (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$.
From the problem, we have $a>0$, and $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$
Therefore,
$$
\begin{array}{l}
\Delta=[-(2... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. For any real numbers $x, y$, the inequality
$$
|x-1|+|x-3|+|x-5| \geqslant k(2-|y-9|)
$$
always holds. Then the maximum value of the real number $k$ is $\qquad$ | 3. 2 .
From the geometric meaning of absolute value, we know that the minimum value of $|x-1|+|x-3|+|x-5|$ is 4 (at this time $x=3$), and the maximum value of $2-|y-9|$ is 2 (at this time $y=9$). According to the condition, we get $4 \geqslant k \cdot 2$, so, $k \leqslant 2$. Therefore, the maximum value of $k$ is 2. | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
7. $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+p_{5}^{2}=p_{6}^{2}$ has $\qquad$ groups of positive prime solutions $\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)$. | 7.5.
Obviously, $p_{6} \neq 2$.
Since $p^{2} \equiv 1$ or $4(\bmod 8)$ (where $p$ is a prime), and $p^{2} \equiv 4(\bmod 8)$ if and only if $p=2$, it follows that among $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$, there must be 4 twos (let's assume they are $p_{1}$, $p_{2}$, $p_{3}$, and $p_{4}$), and the other on... | 5 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If $a$, $b$, $c$ are all real numbers, and $a+b+c=0$, $abc=2$, then the minimum value that $|a|+|b|+|c|$ can reach is $\qquad$ . | (Tip: Refer to Example 3. Answer: 4. ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Find the maximum value of the function $y=\sqrt{2 x^{2}+3 x+1}+\sqrt{7-2 x^{2}-3 x}$. | ( Hint: Refer to Example 8. Answer: The maximum value is 4 . ) | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Each point in the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such th... | Explanation: If $n=4$, then a colored plane can be constructed such that: there is exactly one point on a circle with three colors, and the rest of the points are of another color, satisfying the problem's conditions and ensuring no four points of different colors are concyclic. Therefore, $n \geqslant 5$.
When $n=5$,... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find the minimum value of $y=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$. | Solution: As shown in Figure 2, construct Rt $\triangle P A C$ and Rt $\triangle P B D$ such that $A C=1, B D=2, C P=x, P D=4-x$. Thus, the original problem is transformed into: finding a point $P$ on line $l$ such that the value of $P A + P B$ is minimized.
Since $y = P A + P B \geqslant A B$, the minimum value of $y... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Consider rectangles in the plane with sides parallel to the coordinate axes (both length and width are greater than 0), and call such a rectangle a "box". If two boxes have a common point (including common points on the interior or boundary of the boxes), then the two boxes are said to "intersect". Find the largest ... | 1. The maximum value that satisfies the condition is 6.
An example is shown in Figure 1.
Below is the proof: 6 is the maximum value.
Assume the boxes $B_{1}$, $B_{2}, \cdots, B_{n}$ satisfy the condition. Let the closed intervals corresponding to the projections of $B_{k}$ on the $x$-axis and $y$-axis be $I_{k}$ and ... | 6 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$ is a root of the equation $x^{2}-5 x+1=0$. Then the unit digit of $a^{4}+a^{-4}$ is $\qquad$ . | ニ、1.7.
Obviously, $a^{-1}$ is another root of the equation. Then $a+a^{-1}=5$.
Thus, $a^{2}+a^{-2}=\left(a+a^{-1}\right)^{2}-2=23$, $a^{4}+a^{-4}=\left(a^{2}+a^{-2}\right)^{2}-2=527$. | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the maximum value of the function $y=\sqrt{5-2 x}+\sqrt{3+2 x}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This part is not translated as it contains instructions for the translation task itself. Here is the requested translation above. | Solution: From the given condition, we know $y>0$.
Notice that the variance of the two numbers $\sqrt{5-2 x}$ and $\sqrt{3+2 x}$ is
$$
\begin{array}{l}
S^{2}=\frac{1}{2}\left[(\sqrt{5-2 x})^{2}+(\sqrt{3+2 x})^{2}-2\left(\frac{y}{2}\right)^{2}\right] \\
\geqslant 0 .
\end{array}
$$
Solving this, we get $16-y^{2} \geqsl... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. As shown in Figure 1, given that the circumradius $R$ of the acute triangle $\triangle ABC$ is $1$, $\angle BAC = 60^{\circ}$, and the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively. The line segment $OH$ intersects the extension of $BC$ at point $P$. Find:
(1) The area of the concave ... | 14. (1) As shown in Figure 2, connect $A H$, and draw $O D \perp B C$ at point D.
Since $O$ is the
circumcenter of
$\triangle A B C$ and $\angle B A C$ $=60^{\circ}$, we have
$$
\begin{array}{l}
\angle B O C \\
=2 \angle B A C \\
=120^{\circ}, \\
O D=O C \cos 60^{\circ}=\frac{1}{2} .
\end{array}
$$
By the property of ... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
15. Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$. | 15. When $a=b=c=1$, we can get $k \geqslant 2$.
Below is the proof: the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. Given
$$
x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}=1 \text {. }
$$
then the value of $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}$ is | $$
\text { II, 1. }-1 \text {. }
$$
Notice that
$$
\begin{array}{l}
x \sqrt{x^{2}+3 x+18}+x \sqrt{x^{2}-6 x+18} \\
=\frac{x^{2}\left(x^{2}+3 x+18\right)-x^{2}\left(x^{2}-6 x+18\right)}{x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}} \\
=\frac{x^{2} \cdot 9 x}{1}=9 x^{3} .
\end{array}
$$
Subtracting the above equation fr... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that the three vertices $A$, $B$, and $C$ of the right triangle $\triangle ABC$ are all on the parabola $y=x^{2}$, and the hypotenuse $AB$ is parallel to the $x$-axis. Then the height $h$ from the hypotenuse is $\qquad$. | 3. 1 .
Let point $A\left(a, a^{2}\right)$ and $C\left(c, c^{2}\right)(|c|<a)$. Since $|AB|=|AC|$, we have $a^{2}-c^{2}=1$.
Therefore, the altitude from $C$ to the hypotenuse $AB$ is $a^{2}-c^{2}=1$. | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 2, in $\triangle A B C$, it is known that $A B=9, B C$ $=8, C A=7, A D$ is the angle bisector, and a circle is drawn with $A D$ as a chord,
touching $B C$ and intersecting $A B$ and $A C$ at points $M$ and $N$, respectively. Then $M N=$
$\qquad$ | 4. 6 .
As shown in Figure 4, connect $D M$.
$$
\begin{array}{l}
\text { By } \angle B D M \\
=\angle B A D \\
=\angle C A D \\
=\angle D M N,
\end{array}
$$
we get $M N \parallel B C$.
Thus, $\triangle A M N \sim \triangle A B C$.
It is easy to know $B D=\frac{9}{9+7} \times 8=\frac{9}{2}, B M \cdot B A=B D^{2}$
$\Ri... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. If the function $f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x$ has a range of $[n, M]$ on the interval $[-k, k](k>0)$, then $M+n$ $=$ $\qquad$ | 2. 2 .
Notice that
$$
f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x=1+\frac{2^{x}-1}{2^{x}+1}+\sin x \text {. }
$$
Let $g(x)=f(x)-1$. Then
$$
g(-x)=\frac{2^{-x}-1}{2^{-x}+1}+\sin (-x)=\frac{1-2^{x}}{1+2^{x}}-\sin x=-g(x)
$$
is an odd function.
Let the maximum value of $g(x)$ be $g\left(x_{0}\right)$. By the given information,... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The 29th Summer Olympic Games were held in Beijing on August 8, 2008, forming a memorable number 20080808. The number of different positive divisors of 20080808 divided by 8 is $\qquad$ | 3. 8 .
From $20080808=2^{3} \times 11 \times 17 \times 31 \times 433$, we know that the number of different positive divisors of 20080808 is
$$
N=(3+1)(1+1)^{4}=8^{2} \text { (divisors). }
$$
Dividing by 8 gives 8. | 8 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. A line $l$ passing through the right focus $F$ of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$, and there are exactly 3 such lines, find $\lambda$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 | (The polar equation of the hyperbola is
$$
\rho=\frac{2}{1-\sqrt{3} \cos \theta} \text {. }
$$
Let $A B$ be a chord passing through the right focus and intersecting only the right branch.
Then $|A B|=\rho_{1}+\rho_{2}$
$$
\begin{array}{l}
=\frac{2}{1-\sqrt{3} \cos \theta}+\frac{2}{1-\sqrt{3} \cos (\theta+\pi)} \\
=\fr... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. As shown in Figure 2, given that $A, B, C, D$ are four points on a plane that are not concyclic,
$$
\begin{array}{l}
\triangle A B D, \triangle A D C \text {, } \\
\triangle B C D, \triangle A B C
\end{array}
$$
have circumcenters
$$
\text { as } E, F, G, H \text {, }
$$
respectively. The line segments $E G, F H$
... | 8.4.
Since points $E$ and $G$ lie on the perpendicular bisector of $BD$, $EG$ is the perpendicular bisector of segment $BD$. Similarly, $FH$ is the perpendicular bisector of segment $AC$. Since $I$ is the intersection of $EG$ and $FH$, it follows that $CI = AI = 4$. | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. (40 points) There is a stack of cards numbered $1 \sim 15$. After arranging them in a certain order, the following two operations are performed: Place the top card on the table, then place the second card at the bottom of the stack. These two operations are repeated until all 15 cards are sequentially placed on the ... | 1. Assuming the cards are arranged in descending order from top to bottom as shown from left to right in Table 1. The first card originally becomes the bottom card, so it must be 15. The third card must be 14. And so on.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & $\alpha$ & 14 & $\beta$ & 13 &... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. In a $10 \times 10$ grid, there is a shape composed of $4 n$ $1 \times 1$ small squares, which can be covered by $n$ "田" shaped figures, or by $n$ "世" or "円" shaped figures (which can be rotated). Find the minimum value of the positive integer $n$.
(Zhu Huaiwei, problem contributor) | 7. Let the given shapes be denoted as type $A$ and type $B$.
First, we prove that $n$ is even.
Color the $10 \times 10$ grid as shown in Figure 6.
Regardless of which 4 squares type $A$ covers, the number of black squares must be even, while for type $B$ it is odd. If $n$ is odd, the number of black squares covered b... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given $41^{x}=2009,7^{y}=2009$. Then the value of $\frac{1}{x}+\frac{2}{y}$ is . $\qquad$ | 4. 1 .
From $41^{x}=2009$, we get $41^{x y}=2009^{y}$.
$$
\begin{array}{l}
\text { Also } 7^{y}=2009 \Rightarrow 7^{2 y}=2009^{2} \\
\Rightarrow 49^{y}=2009^{2} \Rightarrow 49^{x y}=2009^{2 x} .
\end{array}
$$
$$
\text { Also } 41 \times 49=2009 \Rightarrow 41^{x y} \times 49^{x y}=2009^{x y} \text {. }
$$
Therefore ... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
9. (14 points) Let positive real numbers $x, y, z$ satisfy $xyz=1$. Try to find the maximum value of
$$
f(x, y, z)=(1-yz+z)(1-xz+x)(1-xy+y)
$$
and the values of $x, y, z$ at that time. | $$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 - y z + z < 0 , } \\
{ 1 - x z + x < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+x z<1, \\
y+x y<1
\end{array}\right.\right. \\
\Rightarrow(x+x z)(y+x y)<1 \\
\Leftrightarrow x+x y+x^{2} y<0,
\end{array}
$$
Contradiction. Therefore, among $1-y z+z$, $1-... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. If $p$ is a prime number, and $p+3$ divides $5p$, then the last digit of $p^{2009}$ is $\qquad$ . | 8. 2 .
Since the positive divisors of $5p$ are $1, 5, p, 5p$, and $(p+3)$ divides $5p$, therefore, $p=2$.
Thus, $p^{2009}=2^{2009}=\left(2^{4}\right)^{502} \times 2=2 \times 16^{502}$.
Given that the last digit of $16^{502}$ is 6, it follows that the last digit of $p^{2009}$ is 2. | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
9. As shown in Figure 2, in quadrilateral $A B C D$, $\angle A C B=$ $\angle B A D=105^{\circ}, \angle A B C=\angle A D C=45^{\circ}$. If $A B$ $=2$, then the length of $C D$ is $\qquad$. | 9. 2 .
As shown in Figure 6, draw $EA \perp AB$ intersecting the extension of $BC$ at point $E$.
$$
\begin{array}{l}
\text { Then } \angle AEB=45^{\circ} \\
\quad=\angle ADC, \\
AE=AB=2 . \\
\text { Also } \angle DAC=\angle DAB-\angle CAB \\
=\angle DAB-\left(180^{\circ}-\angle ABC-\angle ACB\right) \\
=75^{\circ}=180... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. Given $\frac{x y}{x+y}=2, \frac{x z}{x+z}=3, \frac{y z}{y+z}=4$. Find the value of $7 x+5 y-2 z$. | Three, 11. Given
$$
\begin{array}{l}
\frac{1}{2}=\frac{1}{x}+\frac{1}{y}, \frac{1}{3}=\frac{1}{x}+\frac{1}{z}, \\
\frac{1}{4}=\frac{1}{y}+\frac{1}{z} .
\end{array}
$$
Solving simultaneously, we get $x=\frac{24}{7}, y=\frac{24}{5}, z=24$.
Therefore, $7 x+5 y-2 z=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. (14 points) Let the line $l: y=k x+m(k, m \in$ Z) intersect the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$ at two distinct points $A, B$, and intersect the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{12}=1$ at two distinct points $C, D$. Question: Does there exist a line $l$ such that the vector $\overrightarrow{A C}... | Given the system of equations:
$$
\left\{\begin{array}{l}
y=k x+m, \\
\frac{x^{2}}{16}+\frac{y^{2}}{12}=1,
\end{array}\right.
$$
eliminating \( y \) and simplifying, we get:
$$
\left(3+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-48=0.
$$
Let \( A\left(x_{1}, y_{1}\right) \) and \( B\left(x_{2}, y_{2}\right) \). Then:
$$
\beg... | 9 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. (15 points) Find the maximum and minimum values of the function
$$
y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x}
$$ | 3. The domain of the function is $[0,13]$.
$$
\begin{array}{l}
\text { Given } y=\sqrt{x}+\sqrt{x+27}+\sqrt{13-x} \\
=\sqrt{x+27}+\sqrt{13+2 \sqrt{x(13-x)}} \\
\geqslant \sqrt{27}+\sqrt{13}=3 \sqrt{3}+\sqrt{13},
\end{array}
$$
we know that the equality holds when $x=0$.
Thus, the minimum value of $y$ is $3 \sqrt{3}+\s... | 11 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given three non-zero real numbers $a, b, c$, the set $A=$ $\left\{\frac{a+b}{c}, \frac{b+c}{a}, \frac{c+a}{b}\right\}$. Let $x$ be the sum of all elements in set $A$, and $y$ be the product of all elements in set $A$. If $x=2 y$, then the value of $x+y$ is $\qquad$ | $-1 .-6$.
From the problem, we have
$$
x=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}
$$
$$
\begin{aligned}
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-3, \\
y & =\frac{a+b}{c} \cdot \frac{b+c}{a} \cdot \frac{c+a}{b} \\
& =\frac{(a+b+c)(ab+bc+ca)}{abc}-1 .
\end{aligned}
$$
Then $x=y-2$.
Also, $x=2y$, solving this gives $y=-2$.
Thus... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Let $A=\{i \in \mathbf{N} \mid 1 \leqslant i \leqslant$ $2880\}, B \subseteq A,|B| \geqslant 9$. If all elements in set $A$ can be represented by the sum of no more than 9 different elements from $B$, find $\min |B|$, and construct a set corresponding to the minimum $|B|$.
| Let $|B|=k \geqslant 9$.
According to the problem, we should have $S_{k}=\sum_{i=1}^{9} \mathrm{C}_{k}^{i} \geqslant 2880$.
Notice that $S_{9}=2^{9}-1, S_{10}=2^{10}-2$,
$$
\begin{array}{l}
S_{11}=2^{11}-2-\mathrm{C}_{11}^{10}=2035, \\
S_{12}=2^{12}-2-\mathrm{C}_{12}^{10}-\mathrm{C}_{12}^{11}=4016>2880 .
\end{array}
$$... | 12 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Let point $O$ be outside $\triangle A B C$, and
$$
\overrightarrow{O A}-2 \overrightarrow{O B}-3 \overrightarrow{O C}=0 \text {. }
$$
Then $S_{\triangle A B C}: S_{\triangle O B C}=$ | $-1.4$.
As shown in Figure 4, let $D$ and $E$ be the midpoints of sides $AB$ and $BC$, respectively, and connect $CD$. Then
$$
\begin{array}{l}
\overrightarrow{OA}+\overrightarrow{OB}=2 \overrightarrow{OD}, \\
\overrightarrow{OB}+\overrightarrow{OC}=2 \overrightarrow{OE} .
\end{array}
$$
(1) - (2) $\times 3$ gives
$$
\... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
9. (14 points) In the Cartesian coordinate system $x O y$, points with both integer coordinates are called integer points. Given $O(0,0), A(2,1)$, and $M$ is an integer point inside the ellipse $\frac{x^{2}}{200}+\frac{y^{2}}{8}=1$. If $S_{\triangle O M M}=3$, find the number of integer points $M$ that satisfy this con... | 9. Connect $O A$. It is easy to know that there are two integer points $M_{1}(-6,0)$ and $M_{2}(6,0)$ on the $x$-axis inside the ellipse that satisfy the problem.
Draw two lines $l_{1}$ and $l_{2}$ parallel to the line $O A$ through points $M_{1}$ and $M_{2}$, respectively.
According to the principle that triangles w... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a]$ equals $\qquad$ (where [x] denotes the greatest integer not exceeding the real number $x$). | 2. 6 .
Given $0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$, then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]=0$ or 1. From the problem, we know that 18 of them are equal to 1. Therefore,
$$
\begin{array}{l}
{\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. When $1 \leqslant x \leqslant 2$, simplify
$$
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}}=
$$
$\qquad$ . | 5. 2 .
Notice that
$$
\begin{array}{l}
\sqrt{x+2 \sqrt{x-1}}+\sqrt{x-2 \sqrt{x-1}} \\
= \sqrt{(x-1)+2 \sqrt{x-1}+1}+ \\
\sqrt{(x-1)-2 \sqrt{x-1}+1} \\
= \sqrt{(\sqrt{x-1}+1)^{2}}+\sqrt{(\sqrt{x-1}-1)^{2}} \\
=|\sqrt{x-1}+1|+|\sqrt{x-1}-1| .
\end{array}
$$
Since $1 \leqslant x \leqslant 2$, we have $\sqrt{x-1}-1 \leq... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given an integer $n \geqslant 3$. Find the smallest positive integer $k$, such that there exists a $k$-element set $A$ and $n$ pairwise distinct real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1}+x_{2}, x_{2}+x_{3}, \cdots$, $x_{n-1}+x_{n}, x_{n}+x_{1}$ all belong to $A$. (Xiong Bin provided) | 2. Let $x_{1}+x_{2}=m_{1}, x_{2}+x_{3}=m_{2}, \cdots \cdots$
$x_{n-1}+x_{n}=m_{n-1}, x_{n}+x_{1}=m_{n}$.
First, $m_{1} \neq m_{2}$, otherwise, $x_{1}=x_{3}$, which is a contradiction.
Similarly, $m_{i} \neq m_{i+1}\left(i=1,2, \cdots, n, m_{n+1}=m_{1}\right)$.
Thus, $k \geqslant 2$.
If $k=2$, let $A=\{a, b\}(a \neq b)$... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. The first digit after the decimal point of $(\sqrt{2}+\sqrt{3})^{2010}$ is $\qquad$ | 11. 9 .
Since $(\sqrt{2}+\sqrt{3})^{2010}+(\sqrt{2}-\sqrt{3})^{2010}$ is an integer, therefore, the fractional part of $(\sqrt{2}+\sqrt{3})^{2010}$ is
$$
\begin{array}{l}
1-(\sqrt{2}-\sqrt{3})^{2010} . \\
\text { Also } 0<(\sqrt{2}-\sqrt{3})^{2010}<0.2^{1005}<(0.008)^{300} \text {, then } \\
0.9<1-(\sqrt{2}-\sqrt{3})^... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
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