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One. (20 points) Given that $a$ and $b$ are integers, the equation $a x^{2} + b x + 2 = 0$ has two distinct negative real roots greater than -1. Find the minimum value of $b$.
| Let the equation $a x^{2}+b x+2=0(a \neq 0)$ have two distinct negative real roots $x_{1} 、 x_{2}\left(x_{1}<x_{2}<0\right)$.
\end{array}\right.
$$
Solving, we get $a>0, b>0$.
Since $a$ and $b$ are both integers, it follows that $a$ and $b$ are both positive integers.
Let $y=a x^{2}+b x+2$. Then this parabola opens up... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Given that there are only three positive integers between the fractions $\frac{112}{19}$ and $\frac{112+x}{19+x}$. Find the sum of all possible integer values of $x$. | Notice that $50$ or $x<-19$.
Therefore, the three positive integers between $\frac{112+x}{19+x}$ and $\frac{112}{19}$ are $3,4,5$.
From $2<\frac{112+x}{19+x}<3$, we get $\frac{55}{2}<x<74$.
Since $x$ is an integer greater than 0, thus, $x=28,29, \cdots, 73$.
Therefore, the sum of these numbers is
$$
\frac{(28+73) \time... | 2310 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
At first 272, from the eight points consisting of the vertices and midpoints of the sides of a square, how many isosceles triangles can be formed by selecting three points?
Will the above text be translated into English, please retain the original text's line breaks and format, and output the translation result direct... | Considering that there are no equilateral triangles that meet the requirements in this problem, we classify the isosceles triangles by their vertices.
(1) Using the vertices of the square as the vertices of the isosceles triangle. As shown in Figure 3, if $A$ is the vertex,
then $\triangle A G H$, $\triangle A B D$, an... | 20 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. How many pairs $(n, r)$ are there in the array satisfying $0 \leqslant r \leqslant n \leqslant 63$ for which the binomial coefficient $\mathrm{C}_{n}^{r}$ is even (assuming $\left.\mathrm{C}_{0}^{0}=1\right) ?$ | From Example 5, we know that the number of odd numbers in $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ is $2^{S(n)},$ where $S(n)$ is the sum of the binary digits of $n$.
Since $63=2^{6}-1=(111111)_{2}$, when $0 \leqslant n \leqslant 63$, we have $0 \leqslant S(n) \leqslant 6$.
Classifying and... | 1351 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. As shown in Figure 2, fold $\triangle A B C$ along the dotted line $D E$ to get a heptagon $A D E C F G H$. If the area ratio of the heptagon to the original triangle is $2: 3$, and the area of the overlapping part after folding is 4, then the area of the original $\triangle A B C$ is | 3. 12 .
Let the area of the non-overlapping part after folding be $x$. Then the area of the original triangle is $8+x$, and the area of the heptagon is $4+x$. From the given condition, we have
$$
\begin{array}{l}
(8+x):(4+x)=3: 2 \\
\Rightarrow 16+2 x=12+3 x \Rightarrow x=4 .
\end{array}
$$
$$
\text { Hence } S_{\tria... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ The value is
The value of $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is | 4. 4 .
Let $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}=x$.
Cubing both sides and simplifying, we get
$$
x^{3}-6 x-40=0 \text {. }
$$
By observation, 4 is a root of the equation. Therefore,
$$
(x-4)\left(x^{2}+4 x+10\right)=0 \text {. }
$$
Since $\Delta=4^{2}-4 \times 10=-24<0$, the equation $x^{2}+4 x+10=0$ h... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
13. Given the polynomial
$$
\begin{array}{l}
(1+x)+(1+x)^{2}+\cdots+(1+x)^{n} \\
=b_{0}+b_{1} x+\cdots+b_{n} x^{n},
\end{array}
$$
and $b_{1}+b_{2}+\cdots+b_{n}=1013$.
Then a possible value of the positive integer $n$ is | 13. 9 .
Let $x=0$, we get $b_{0}=n$.
Let $x=1$, we get
$$
2+2^{2}+\cdots+2^{n}=b_{0}+b_{1}+\cdots+b_{n},
$$
which is $2\left(2^{n}-1\right)=n+1013$.
Solving this, we get $n=9$. | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. As shown in Figure 4, given $\angle A O M=60^{\circ}$, there is a point $B$ on ray $O M$ such that the lengths of $A B$ and $O B$ are both integers, thus $B$ is called an "olympic point". If $O A=8$, then the number of olympic points $B$ in Figure 4 is $\qquad$ | 4. 4 .
As shown in Figure 7, draw $A H \perp O M$ at point $H$.
Since $\angle A O M$
$=60^{\circ}$, and $O A$
$=8$, therefore,
$O H=4$,
$A H=4 \sqrt{3}$.
Let $A B=$
$m, H B=n$
($m, n$ are positive integers). Clearly, in the right triangle $\triangle A H B$, we have
$$
m^{2}-n^{2}=(4 \sqrt{3})^{2} \text {, }
$$
which ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
6. If $2n+1, 20n+1 \left(n \in \mathbf{N}_{+}\right)$ are powers of the same positive integer, then all possible values of $n$ are | 6. 4 .
According to the problem, we know that $(2 n+1) \mid(20 n+1)$. Then $(2 n+1) \mid[10(2 n+1)-(20 n+1)]=9$. Therefore, $n \in\{1,4\}$. Upon verification, $n=4$. | 4 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: There is an electronic flea jumping back and forth on a number line, and it lights up a red light when it jumps to a negative number, but does not light up when it jumps to a positive number. The starting point is at the point representing the number -2 (record one red light), the first step is to jump 1 uni... | (1) Notice
$$
\begin{array}{l}
S=\mid-2-1+2-3+4-5+\cdots-9+101 \\
=1-2+1 \times 5 \mid=3 .
\end{array}
$$
Therefore, the negative numbers appear as $-2,-3,-1,-4,-5$, $-6,-7$, a total of seven times. So, the red light will flash seven times.
(2) In fact, after the electronic flea jumps four times, it returns to the ori... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Divide the numbers $1,2, \cdots, 200$ into two groups arbitrarily, each containing 100 numbers. Arrange one group in ascending order (denoted as $a_{1}<a_{2}<\cdots<a_{100}$) and the other in descending order (denoted as $b_{1}>b_{2}>\cdots>b_{100}$). Try to find
$$
\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\... | First, prove: For any term in the algebraic expression
$$
\left|a_{k}-b_{k}\right|(k=1,2, \cdots, 100)
$$
the larger number among $a_{k}$ and $b_{k}$ must be greater than 100, and the smaller number must not exceed 100.
(1) If $a_{k} \leqslant 100$ and $b_{k} \leqslant 100$, then by
$$
a_{1}b_{k+1}>\cdots>b_{100}
$$
w... | 10000 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four, (50 points) For $n$ distinct positive integers, among any six numbers, there are at least two numbers such that one can divide the other. Find the minimum value of $n$ such that among these $n$ numbers, there must exist six numbers where one can be divided by the other five.
| The smallest positive integer $n=26$.
The proof is divided into two steps.
【Step 1】When $n \leqslant 25$, the condition is not satisfied.
Construct the following 25 positive integers:
(1) $2^{5}, 2^{4}, 2^{3}, 2^{2}, 2^{1}$;
(2) $3^{5}, 3^{4}, 3^{3}, 3^{2}, 3^{1}$;
(3) $5^{5}, 5^{4}, 5^{3}, 5^{2}, 5^{1}$;
(4) $7^{5}, 7... | 26 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Divide the 100 natural numbers $1, 2, \cdots, 100$ into 50 groups, each containing two numbers. Now, substitute the two numbers in each group (denoted as $a$ and $b$) into $\frac{1}{2}(|a-b|+a+b)$ for calculation, and obtain 50 values. Find the maximum value of the sum of these 50 values. | Since the 100 numbers from $1 \sim 100$ are all different, in each pair of numbers, there must be a larger number. Therefore, the result of the calculation is the larger of the two numbers. Hence, the maximum sum of these 50 values is
$$
51+52+\cdots+100=3775
$$ | 3775 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Let $a, b, x, y \in \mathbf{R}$, satisfy the system of equations
$$
\left\{\begin{array}{l}
a x+b y=3, \\
a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, \\
a x^{4}+b y^{4}=42 .
\end{array}\right.
$$
Find the value of $a x^{5}+b y^{5}$.
$(1990$, American Mathematical Invitational) | Solution 1:
From
$a x^{3}+b y^{3}$
we get
$$
=\left(a x^{2}+b y^{2}\right)(x+y)-(a x+b y) x y,
$$
thus
$$
\begin{array}{l}
16=7(x+y)-3 x y . \\
\text { From } a x^{4}+b y^{4} \\
=\left(a x^{3}+b y^{3}\right)(x+y)-\left(a x^{2}+b y^{2}\right) x y,
\end{array}
$$
From $a x^{4}+b y^{4}$
we get
$$
42=16(x+y)-7 x y \text {... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 If real numbers $a, b$ satisfy the conditions
$$
a^{2}+b^{2}=1,|1-2 a+b|+2 a+1=b^{2}-a^{2} \text {, }
$$
then $a+b=$ $\qquad$
(2009, National Junior High School Mathematics Joint Competition) | Given $a^{2}+b^{2}=1$, we have
$$
b^{2}=1-a^{2} \text {, and }-1 \leqslant a \leqslant 1,-1 \leqslant b \leqslant 1 \text {. }
$$
From $|1-2 a+b|+2 a+1=b^{2}-a^{2}$, we get
$$
\begin{array}{l}
|1-2 a+b|=b^{2}-a^{2}-2 a-1 \\
=\left(1-a^{2}\right)-a^{2}-2 a-1=-2 a^{2}-2 a .
\end{array}
$$
Thus, $-2 a^{2}-2 a \geqslant ... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Mother's Day is here, and Xiao Hong, Xiao Li, and Xiao Ying go to a flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the above flowers. Then she... | 10. 20 .
Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $z$ yuan, respectively. Then
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text {. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text {. }
... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
12. Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition. | 12. By the Pythagorean theorem, we have
$$
c^{2}=a^{2}+b^{2} \text {. }
$$
Also, $c=\frac{1}{3} a b-(a+b)$, substituting into equation (1) gives
$$
\begin{array}{l}
a^{2}+b^{2} \\
=\frac{1}{9}(a b)^{2}-\frac{2}{3} a b(a+b)+a^{2}+2 a b+b^{2} .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
a b-6(a+b)+18=0 \\
\... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. In $\triangle A B C$, it is known that $A B=b^{2}-1, B C=$ $a^{2}, C A=2 a$, where $a$ and $b$ are both integers greater than 1. Then the value of $a-b$ is $\qquad$ | 2.0.
Since $a$ is an integer greater than 1, we have $a^{2} \geqslant 2 a$. By the triangle inequality, we get
$$
\begin{array}{l}
\left\{\begin{array}{l}
a^{2}+2 a>b^{2}-1, \\
2 a+b^{2}-1>a^{2}
\end{array}\right. \\
\Leftrightarrow(a-1)^{2}<b^{2}<(a+1)^{2} .
\end{array}
$$
Given $a, b \in \mathbf{N}_{+}$, hence $b=a... | 0 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the equation in $x$
$$
x^{4}+2 x^{3}+(3+k) x^{2}+(2+k) x+2 k=0
$$
has real roots. If the product of all real roots is -2, then the sum of the squares of all real roots is $\qquad$ . | 4.5.
Factorizing the left side of the equation, we get
$$
\left(x^{2}+x+2\right)\left(x^{2}+x+k\right)=0 \text {. }
$$
Since $x^{2}+x+2=0$ has no real roots, the equation $x^{2}+x+k=0$ has two real roots $x_{1} 、 x_{2}$.
According to the problem, $x_{1} x_{2}=k=-2$.
$$
\begin{array}{l}
\text { Also, } x_{1}+x_{2}=-1 ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $a, b$ be the roots of the equation $x^{2}+68 x+1=0$, and $c, d$ be the roots of the equation $x^{2}-86 x+1=0$. Then
$$
(a+c)(b+c)(a-d)(b-d)
$$
the value is $\qquad$. | 6. 2772 .
$$
\begin{array}{l}
\text { Given } a b=c d=1, a+b=-68, \\
c^{2}-8 b c+1=d^{2}-86 d+1=0, \\
\text { then }(a+c)(b+c)(a-d)(b-d) \\
=\left[a b+(a+b) c+c^{2}\right]\left[a b-(a+b) d+d^{2}\right] \\
=\left(1-68 c+c^{2}\right)\left(1+68 d+d^{2}\right) \\
=18 c \cdot 154 d=2772 .
\end{array}
$$ | 2772 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 The function $y=|x+1|+|x+2|+|x+3|$. When $x=$ $\qquad$, $y$ has its minimum value, and the minimum value is $\qquad$
(2007, National Junior High School Mathematics Competition Zhejiang Regional Finals) | It is known that $|x+1|+|x+2|+|x+3|$ represents the sum of the distances (lengths of segments) from point $x$ to points $-1$, $-2$, and $-3$ on the number line. It is easy to see that when $x=-2$, this sum of distances is minimized, i.e., the value of $y$ is the smallest, at which point, $y_{\min }=2$. | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. The number of all integer solutions $(x, y, z)$ for the equation $x y z=2009$ is $\qquad$ groups. | 8. 72 .
First consider the integer solutions for $00$, it is easy to know that there are
$$
2 \mathrm{~A}_{3}^{1}+2 \mathrm{~A}_{3}^{3}=18
$$
sets of positive integer solutions.
For each set of positive integer solutions, adding two negative signs can yield 3 sets of integer solutions with two negatives and one posit... | 72 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, (15 points) Given a positive integer $n$ that satisfies the following condition: among any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$.
| Because $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor no greater than 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17,19,23$,
$29,31,37,41,43$.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive integers greater than 1 and not exce... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For example, the minimum value of $7 y=2 x^{2}+4|x|-1$ is
$\qquad$
(2007, National Junior High School Mathematics Competition, Zhejiang Province Re-test) | Let $t=|x| \geqslant 0$. Then
$$
y=2 t^{2}+4 t-1=2(t+1)^{2}-3(t \geqslant 0) \text{.}
$$
It is easy to see that when $t=|x|=0$, i.e., $x=0$, $y$ is minimized at -1. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. If the odd function $y=f(x)$ defined on $\mathbf{R}$ is symmetric about the line $x=1$, and when $0<x \leqslant 1$, $f(x)=\log _{3} x$, then the sum of all real roots of the equation $f(x)=-\frac{1}{3}+f(0)$ in the interval $(0,10)$ is . $\qquad$ | 5. 30 .
Given that the graph of the function $y=f(x)$ is symmetric about the line $x=1$, and $f(x)$ is an odd function, we have
$$
f(x+2)=f(-x)=-f(x) .
$$
Therefore, $f(x+4)=-f(x+2)=f(x)$, which means $f(x)$ is a periodic function with 4 as one of its periods.
Since $f(x)$ is an odd function defined on $\mathbf{R}$, ... | 30 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. If for any real number $x$, the function
$$
f(x)=x^{2}-2 x-|x-1-a|-|x-2|+4
$$
is always a non-negative real number, then the maximum value of the real number $a$ is | 7. 1 .
From the conditions, we have $\left\{\begin{array}{l}f(0)=-|1+a|+2 \geqslant 0, \\ f(1)=-|a|+2 \geqslant 0 .\end{array}\right.$
Solving this, we get $-2 \leqslant a \leqslant 1$.
When $a=1$, we have
$$
f(x)=x^{2}-2 x-2|x-2|+4,
$$
which is $f(x)=\left\{\begin{array}{ll}x^{2}, & x \leqslant 2 ; \\ x^{2}-4 x+8, &... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Satisfy $0 \leqslant k_{i} \leqslant 20(i=1,2,3,4)$, and $k_{1}+k_{3}=k_{2}+k_{4}$ of the ordered integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ the number is $\qquad$ . | 10. 6181.
For $0 \leqslant m \leqslant 20$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(0 \leqslant j \leqslant m),
$$
there are $m+1$ solutions;
When $20<m \leqslant 40$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \... | 6181 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Find the minimum value of the function
$$
f(x)=\max \left\{|x+1|,\left|x^{2}-5\right| \right\}
$$
and find the value of the independent variable $x$ when $f(x)$ takes the minimum value. | Solve As shown in Figure 1, in the same Cartesian coordinate system, draw the graphs of $f_{1}(x)=|x+1|$ and $f_{2}(x)=\left|x^{2}-5\right|$. The two graphs intersect at four points, $A$, $B$, $C$, and $D$, whose x-coordinates can be obtained by solving the equation $|x+1|=\left|x^{2}-5\right|$.
Removing the absolute v... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. Let $x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\}(i=1,2, \cdots, 2010)$. Let $S=x_{1} x_{2}+x_{3} x_{4}+\cdots+x_{2009} x_{2010}$.
(1) Can $S$ be equal to 2010? Prove your conclusion;
(2) How many different integer values can $S$ take? | 14. (1) Since $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
\begin{array}{l}
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, \\
(\sqrt{2}-1)(\sqrt{2}+1)=1
\end{array}
$$
Therefore, $x_{2 i-1} x_{2 i} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ terms of $3+2 \sqrt{2}$, $b$ terms of $3-2 \sqrt{2}$, and $c$ terms of 1. Then ... | 503 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Real numbers $a, b$ satisfy
$$
|a+1|+|a+3|+|b+2|+|b-5|=9 \text {. }
$$
Let the maximum and minimum values of $ab + a + b$ be $m, n$ respectively. Then the value of $m+n$ is ( ).
(A) -12 (B) -10 (C) -14 (D) -8 | -1. A.
$$
\begin{array}{l}
\text { Since } 9=|a+1|+|a+3|+|b+2|+|b-5| \\
=(|a+1|+|-a-3|)+(|b+2|+|5-b|) \\
\geqslant|1-3|+|2+5|=9 .
\end{array}
$$
Thus $-3 \leqslant a \leqslant-1$, and $-2 \leqslant b \leqslant 5$
$$
\begin{array}{l}
\Rightarrow-2 \leqslant a+1 \leqslant 0, \text { and }-1 \leqslant b+1 \leqslant 6 \\
... | -12 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 9 On the blackboard, there are two positive integers, one is 2002, and the other is a number less than 2002. If the average of these two numbers $m$ is an integer, then the following operation can be performed: one of the numbers is erased and replaced by $m$. How many times can such an operation be performed a... | Without loss of generality, let these two positive integers be $a, b(a>b)$.
Consider the absolute value of the difference between the two numbers.
Initially, it is $|a-b|$. After the first operation, the absolute value of the difference between the two numbers is $\left|a-\frac{a+b}{2}\right|=\left|\frac{a-b}{2}\right|... | 10 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given real numbers $a$, $b$, $x$, $y$ satisfy $x y=2008^{2008}$,
$$
\frac{1}{1+2008^{a} x}+\frac{1}{1+2008^{b-2009} y}=1 \text {. }
$$
Then the value of $2008^{a+b}$ is | Ni, 1.2008.
Given the equation, after removing the denominator, we get
$$
\begin{array}{l}
1+2008^{b-2009} y+1+2008^{a} x \\
=\left(1+2008^{a} x\right)\left(1+2008^{b-2009} y\right) .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
2008^{a+b-2009} x y=1 \\
\Rightarrow 2008^{a+b-2009} \times 2008^{2008}=1 \\
\Ri... | 2008 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Function
$$
f(x)=27^{x}-3^{x+3}+1
$$
The minimum value of the function on the interval $[0,3]$ is $\qquad$ | - 1. -53.
Let $t=3^{x}(x \in[0,3])$. Then $f(x)=g(t)=t^{3}-27 t+1(t \in[0,27])$. And $g^{\prime}(t)=3 t^{2}-27=3(t-3)(t+3)$, so when $t \in[1,3]$, $g^{\prime}(t)0, g(t)$ is monotonically increasing.
Therefore, when $t=3$, $g(t)$ reaches its minimum value
$$
g(t)_{\min }=g(3)=-53,
$$
which means when $x=1$, $f(x)$ tak... | -53 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right] \text {, }
$$
where, $[x]$ denotes the greatest integer not exceeding the real number $x$. Then
$$
a_{2009}+a_{2010}=
$$ | 2. 2009.
Given $a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}$.
We will prove by mathematical induction:
$$
a_{n+2}-a_{n}=1, a_{n}+a_{n+1}=n \text {. }
$$
Obviously, when $n=1$, the conclusion holds.
Assume when $n=k$, the conclusion holds, i.e.,
$$
a_{k+2}-a_{k}=1, a_{k}+a_{k+1}=k \text {. }
$$
Then when ... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. If the set $A=\{x \mid x=6 n-1, n \in \mathbf{N}\}$,
$$
B=\{x \mid x=8 n+3, n \in \mathbf{N}\},
$$
then the number of elements in $A \cap B$ that are less than 2010 is $\qquad$ | 3. 84 .
According to the problem, if $x \in A$, then $x \equiv 5(\bmod 6)$; if $x \in B$, then $x \equiv 3(\bmod 8)$. Therefore, if $x \in A \cap B$, then $x \equiv 11(\bmod 24)$, i.e., $x=24 k+11(k \in \mathbf{N})$.
Thus, there are 84 elements that satisfy the condition. | 84 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. If the equation
$$
n \sin x+(n+1) \cos x=n+2
$$
has two distinct real roots in $0<x<\pi$, then the minimum value of the positive integer $n$ is $\qquad$. | 4.4.
From the given, we have $\frac{1}{n}=-1-\frac{-1-\sin x}{2-\cos x}$.
And $\frac{-1-\sin x}{2-\cos x}$ represents the slope $k$ of the line connecting a moving point $P(\cos x, \sin x)$ on the upper half of the unit circle (excluding endpoints) and a fixed point $Q(2,-1)$.
To satisfy the problem, the line $PQ$ mu... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The sum of the x-coordinates of the points where the graph of the function $y=x^{2}-2009|x|+2010$ intersects the x-axis is $\qquad$ . | 3. 0 .
The original problem can be transformed into finding the sum of all real roots of the equation
$$
x^{2}-2009|x|+2010=0
$$
If a real number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1).
Therefore, the sum of all real roots of the equation is 0, that is, th... | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given a positive integer $n$ such that the last two digits of $3^{n}$ form a two-digit prime number. Then the sum of all $n$ that satisfy this condition and do not exceed 2010 is $\qquad$ . | $-, 1.909128$.
Considering $3^{k}(k=1,2, \cdots)$ modulo 100, the remainders are sequentially $3,9,27,81,43,29,87,61,83,49,47,41$, $23,69,7,21,63,89,67,1 ; 3,9, \cdots$,
where $43,29,61,83,47,41,23,89,67$ are two-digit primes, corresponding to
$$
k \equiv 5,6,8,9,11,12,13,18,19(\bmod 20) .
$$
Thus, the sum of all $n$ ... | 909128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
[\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}]
$$
is equal to
(there are a total of 2009 square roots). | 2. 45 .
Let $A=\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{3+\sqrt{2}}}}}$.
Then $A>\sqrt{2010+\sqrt{2009}}$
$$
\begin{array}{l}
>\sqrt{2010+44}>45, \\
A<\sqrt{2010+\sqrt{2009+\sqrt{\cdots+\sqrt{2009}}}} \\
<\sqrt{2010+46}<46 .
\end{array}
$$
Thus, $[A]=45$. | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Given
$$
A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n}(n \in \mathbf{N})
$$
can be expressed as the product of $k(k \in \mathbf{N}, k>1)$ consecutive integers. Then $n+k=$ $\qquad$ . | 4. 2 .
If $k \geqslant 4$, then $81 A$.
$$
\begin{array}{l}
\text { But } A=17^{2012 n}+4 \times 17^{4 n}+7 \times 19^{7 n} \\
\equiv 1+4-3^{7 n} \equiv 5-3^{n} \not \equiv 0(\bmod 8),
\end{array}
$$
Contradiction.
If $k=3$, let $A=m\left(m^{2}-1\right)(m \in \mathbf{Z})$.
But $A \equiv 2^{2012 n}-2^{4 n}+2 \times(-1... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A photographer took some photos of eight people at a party, with any two people (there are 28 possible combinations) appearing in exactly one photo. Each photo can be a duo or a trio. How many photos did the photographer take at least?
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。 | 6. 12 .
Let the number of group photos with three people be $x$, and the number of group photos with two people be $y$. Then $3 x+y=28$.
Thus, when $x$ is maximized, the total number of photos taken is minimized.
When $x>8$, by $\frac{3 x}{8}>3$ we know that there is a person $A$ who appears 4 times, and in the 4 grou... | 12 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Select 1005 numbers $a_{1}, a_{2}, \cdots, a_{1005}$ from 1 to 2010 such that their total sum is 1006779, and the sum of any two of these 1005 numbers is not equal to 2011.
(1) Prove that $\sum_{i=1}^{1005} a_{i}^{2}-4022 \sum_{i=1}^{1005} a_{i}^{3}+\sum_{i=1}^{1005} a_{i}^{4}$ is a constant;
(2) When... | Divide $\{1,2, \cdots, 2010\}$ into 1005 groups:
$$
A_{i}=\{i, 2011-i\}(i=1,2, \cdots, 1005) .
$$
Since the sum of any two numbers in $\left\{a_{i}\right\}$ is not equal to 2011, exactly one number is taken from each group. First, take the even numbers from each group to form
$\left\{b_{i}\right\}$, where $b_{i} \in A... | 44253 | Algebra | proof | Yes | Yes | cn_contest | false |
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its divisors 1, $2,17,34,59,118,1003,2006$ is 3240. Find the smallest good number. ${ }^{[3]}$
(... | \left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)=8.
Therefore, when $k=1$, $\alpha_{1}=7$;
when $k=2$, $\alpha_{1}=1, \alpha_{2}=3$ or $\alpha_{1}=3, \alpha_{2}=1$;
when $k=3$, $\alpha_{1}=\alpha_{2}=\alpha_{3}=1$;
when $k \geqslant 4$, there is no solution.
(1) If $n=p^{7}$ ( $p$ is a... | 1614 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. There is a bus, a truck, and a car each traveling in the same direction at a constant speed on a straight road. At a certain moment, the bus is in front, the car is at the back, and the truck is exactly in the middle between the bus and the car. After $10 \mathrm{~min}$, the car catches up with the truck; after anot... | 7. 15 .
Suppose at a certain moment, the distances between the truck and the bus, and the truck and the car are both $S \mathrm{~km}$. The speeds of the car, truck, and bus are $a, b, c(\mathrm{~km} / \mathrm{min})$, respectively, and it takes the truck $x \mathrm{~min}$ to catch up with the bus.
From the problem, we ... | 15 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. For $i=2,3, \cdots, k$, the remainder when the positive integer $n$ is divided by $i$ is $i-1$. If the smallest value of $n$, $n_{0}$, satisfies $2000<n_{0}<3000$, then the smallest value of the positive integer $k$ is | 10.9.
Since $n+1$ is a multiple of $2,3, \cdots, k$, the smallest value of $n$, $n_{0}$, satisfies
$$
n_{0}+1=[2,3, \cdots, k],
$$
where $[2,3, \cdots, k]$ represents the least common multiple of $2,3, \cdots, k$.
$$
\begin{array}{l}
\text { Since }[2,3, \cdots, 8]=840, \\
{[2,3, \cdots, 9]=2520,} \\
{[2,3, \cdots, 1... | 9 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
11. B. Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$
Find the minimum value of $u=9 a^{2}+72 b+2$. | 11. B. From
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {, }
$$
we get $(a-2 b)(3 a-4 b+5)=0$.
Therefore, $a-2 b=0$ or $3 a-4 b+5=0$.
(1) When $a-2 b=0$,
$$
\begin{array}{l}
u=9 a^{2}+72 b+2=36 b^{2}+72 b+2 \\
=36(b+1)^{2}-34 .
\end{array}
$$
Thus, when $b=-1$, the minimum value of $u$ is -34.
(2) When $3 a-4 b+5=0$,
... | -34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. A. From the 2010 positive integers $1,2, \cdots, 2010$, what is the maximum number of integers that can be selected such that the sum of any three selected numbers is divisible by 33? | 14. A. First, the following 61 numbers: $11, 11+33, 11+2 \times 33, \cdots, 11+60 \times 33$ (i.e., 1991) satisfy the conditions of the problem.
On the other hand, let $a_{1}<a_{2}<\cdots<a_{n}$ be the numbers selected from 1, 2, $\cdots, 2010$ that satisfy the conditions of the problem. For any four numbers $a_{i}, a... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
14. B. Color the five sides and five diagonals of the convex pentagon $A B C D E$, such that any two segments sharing a common vertex are of different colors. Find the minimum number of colors needed. | 14. B. Since vertex $A$ is the common point of four segments $A B$, $A C$, $A D$, and $A E$, at least four colors are needed.
If only four colors are used, let's assume they are red, yellow, blue, and green. Then, the four segments extending from each vertex must include one each of red, yellow, blue, and green. There... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If the sum of $k$ consecutive positive integers is 2010, then the maximum value of $k$ is | Ni. 1.60 .
Let $2010=(n+1)+(n+2)+\cdots+(n+k)$.
Then $k(2 n+k+1)=4020$.
Notice that $k<2 n+k+1$, and
$$
4020=2^{2} \times 3 \times 5 \times 67 \text {, }
$$
To maximize the value of $k$, 4020 should be expressed as the product of the closest pair of factors, which is $4020=60 \times 67$.
Thus, $k_{\max }=60$. | 60 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. For the cyclic quadrilateral $ABCD$, the lengths of the four sides in sequence are $AB=2, BC=7, CD=6, DA=9$. Then the area of the quadrilateral is $\qquad$ . | 3. 30 .
Since $7^{2}+6^{2}=85=9^{2}+2^{2}$, that is,
$$
B C^{2}+C D^{2}=D A^{2}+A B^{2} \text {, }
$$
thus, $\triangle B C D$ and $\triangle D A B$ are both right triangles.
Therefore, the area of the quadrilateral is
$$
S_{\triangle B C D}+S_{\triangle D A B}=\frac{1}{2}(7 \times 6+9 \times 2)=30 \text {. }
$$ | 30 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. In $\pm 1 \pm 2 \pm 3 \pm 5 \pm 20$, by appropriately choosing + or -, different algebraic sums can be obtained $\qquad$.
| 4.24.
Among $1,2,3,5,20$, there are three odd numbers, so their algebraic sum must be odd.
Observing, we see that from $1,2,3,5$ we can obtain all odd numbers with absolute values not exceeding 11.
According to the problem, the expression must include 1, 2, 3, 5, and 20. Therefore, the integers that can be obtained ... | 24 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $a_{1}, a_{2}, \cdots, a_{10} \in(1,+\infty)$. Then
$$
\frac{\log _{a_{1}} 2009+\log _{a_{2}} 2009+\cdots+\log _{a_{10}} 2009}{\log _{a_{1,2} \cdots a_{10}} 2009}
$$
the minimum value is | -1.100 .
$$
\begin{array}{l}
\text { Original expression }=\left(\sum_{i=1}^{10} \frac{\lg 2009}{\lg a_{i}}\right) \frac{\lg \left(\prod_{i=1}^{10} a_{i}\right)}{\lg 2009} \\
=\left(\sum_{i=1}^{10} \frac{1}{\lg a_{i}}\right)\left(\sum_{i=1}^{10} \lg a_{i}\right) \\
\geqslant 10^{2},
\end{array}
$$
Equality holds if an... | 100 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Let $A$ be the set of all positive integers not exceeding 2009, i.e., $A=\{1,2, \cdots, 2009\}$, and let $L \subseteq A$, where the difference between any two distinct elements of $L$ is not equal to 4. Then the maximum possible number of elements in the set $L$ is | 7. 1005.
Divide the set $A$ into the following 1005 subsets:
$$
\begin{array}{l}
A_{4 k+i}=\{8 k+i, 8 k+i+4\}(i=1,2,3,4 ; \\
k=0,1, \cdots, 250), \\
A_{1005}=\{2009\} .
\end{array}
$$
If the number of elements in $L$ is greater than 1005, then at least one of the first 1004 subsets is a subset of $L$, meaning there ... | 1005 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. In a plane, given a convex decagon and all its diagonals, in such a graph, the number of triangles that have at least two vertices as vertices of the convex decagon is $\qquad$ (answer with a number). | 8. 960 .
The number of triangles with all three vertices being vertices of a convex decagon is $\mathrm{C}_{10}^{3}$.
The number of triangles with only two vertices being vertices of the convex decagon, and the other vertex being the intersection of two diagonals, can be determined as follows: Two diagonals define fo... | 960 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (16 points) Let $A$ and $B$ be two different subsets of the set $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$, such that $A$ is not a subset of $B$, and $B$ is not a subset of $A$. Find the number of different ordered pairs $(A, B)$. | 11. The set $\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ has $2^{5}$ subsets, and the number of different ordered pairs $(A, B)$ is $2^{5}\left(2^{5}-1\right)$.
If $A \subset B$, and suppose $B$ contains $k(1 \leqslant k \leqslant 5)$ elements. Then the number of ordered pairs $(A, B)$ satisfying $A \subset B$ i... | 570 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. The body diagonal of a rectangular prism is 10, and the projection of this diagonal on one of the surfaces of the prism is 8. Then the maximum volume of this rectangular prism is $\qquad$ . | 4. 192 .
According to the problem, the height of the cuboid is $\sqrt{10^{2}-8^{2}}=6$. Let the side lengths of the base of the cuboid be $a$ and $b$. Then $a^{2}+b^{2}=64$.
Thus, the volume of the cuboid is $V=6 a b \leqslant 3\left(a^{2}+b^{2}\right)=192$.
The equality holds if and only if $a=b=4 \sqrt{2}$. Therefor... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. If the equation about $x$
$$
x^{3}+a x^{2}+b x-4=0\left(a 、 b \in \mathbf{N}_{+}\right)
$$
has a positive integer solution, then $|a-b|=$ | 7.1.
Solution 1 Let $m$ be a positive integer solution of the equation.
If $m \geqslant 2$, then $a m^{2}+b m=4-m^{3}<0$, which contradicts that $a$ and $b$ are both positive integers.
Therefore, only $m=1$. Substituting it in, we get $a+b=3$.
Since $a, b \in \mathbf{N}_{+}$, $\{a, b\}=\{1,2\}$.
Thus, $|a-b|=1$.
Solut... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given $|x| \leqslant 1,|y| \leqslant 1$, and
$$
k=|x+y|+|y+1|+|2 y-x-4| \text {. }
$$
Then the sum of the maximum and minimum values of $k$ is $\qquad$ | 2. 10 .
From the given conditions, we know
$$
-1 \leqslant x \leqslant 1, -1 \leqslant y \leqslant 1 \text{. }
$$
Thus, $y+1 \geqslant 0, 2y-x-40$ when,
$$
k=x+y+y+1-(2y-x-4)=2x+5 \text{. }
$$
Therefore, $3 \leqslant k \leqslant 7$.
Hence, the maximum value of $k$ is 7, and the minimum value is 3. | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given real numbers $x, y, z$ satisfy $x+y=5$ and $z^{2}=x y+y-9$.
Then $x+2 y+3 z=$ $\qquad$ . | Solve: From $x+y=5$, we can set $x=\frac{5}{2}+t, y=\frac{5}{2}-t$.
Substitute into $z^{2}=x y+y-9$ and simplify to get $z^{2}+\left(t+\frac{1}{2}\right)^{2}=0$.
Thus, $z=0, t=-\frac{1}{2}$.
Therefore, $x=2, y=3$.
Hence, $x+2 y+3 z=8$. | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given that $x, y$ are positive integers, and $x y+x+y=23, x^{2} y+x y^{2}=120$. Then $x^{2}+y^{2}=$ $\qquad$ | Solve: From $x y+x+y=23$, we can set $x y=\frac{23}{2}+t, x+y=\frac{23}{2}-t$.
Multiplying the two equations gives
$$
x^{2} y+x y^{2}=\left(\frac{23}{2}\right)^{2}-t^{2}=120 \text{. }
$$
Solving for $t$ yields $t= \pm \frac{7}{2}$.
Since $x$ and $y$ are both positive integers, when $t=\frac{7}{2}$, we have $x+y=8, x y... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Let real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x+y=z-1, \\
x y=z^{2}-7 z+14 .
\end{array}\right.
$$
Question: What is the maximum value of $x^{2}+y^{2}$? For what value of $z$ does $x^{2}+y^{2}$ achieve its maximum value? | Solve: From $x+y=z-1$, we can set
$$
x=\frac{z-1}{2}+t, y=\frac{z-1}{2}-t \text {. }
$$
Substituting into $x y=z^{2}-7 z+14$ and simplifying, we get
$$
3 z^{2}-26 z-55=-4 t^{2} \text {. }
$$
Therefore, $3 z^{2}-26 z-55 \leqslant 0$.
Solving this, we get $\frac{11}{3} \leqslant z \leqslant 5$.
$$
\begin{array}{l}
\tex... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, $c$, and $d$ are prime numbers, and $a b c d$ is the sum of 77 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ | Given 77 consecutive positive integers, with the middle one being $x$. Then $a b c d=77 x=7 \times 11 x$, and $x \geqslant 39$.
Since $a, b, c, d$ are prime numbers, thus, $x$ can be decomposed into the product of two prime numbers, and the sum of these two prime numbers is minimized.
It is easy to verify that when $x... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Given an equilateral $\triangle A B C$ with side length $2, P$ is a point inside $\triangle A B C$, and the distances from point $P$ to the three sides $B C, A C, A B$ are $x, y, z$ respectively, and their product is $\frac{\sqrt{3}}{9}$. Then the sum of the squares of $x, y, z$ is | 2. 1 .
As shown in Figure 4, connect
$$
P A, P B, P C \text{. }
$$
It is easy to see that, in the equilateral $\triangle A B C$, we have
$$
\begin{array}{l}
x+y+z \\
=\frac{\sqrt{3}}{2} A B=\sqrt{3} .
\end{array}
$$
Thus, $\sqrt{3}=x+y+z$
$$
\geqslant 3 \sqrt[3]{x y z}=3 \sqrt[3]{\frac{\sqrt{3}}{9}}=\sqrt{3} \text{.... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Given the quadratic function
$$
y=3 a x^{2}+2 b x-(a+b) \text {, }
$$
when $x=0$ and $x=1$, the value of $y$ is positive. Then, when $0<x<1$, the parabola intersects the $x$-axis at $\qquad$ points. | 3. 2 .
From the given, we have $\left\{\begin{array}{l}-(a+b)>0, \\ 3 a+2 b-(a+b)>0 .\end{array}\right.$
Solving, we get $a>0$.
Let $x=\frac{1}{2}$, then
$$
y=\frac{3 a}{4}+b-(a+b)=-\frac{a}{4}<0 .
$$
By the Intermediate Value Theorem, $3 a x^{2}+2 b x-(a+b)=0$ has one root when $0<x<\frac{1}{2}$, and one root when $... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) From the natural numbers $1, 2, \cdots, 2010$, take $n$ numbers such that the sum of any three of the taken numbers is divisible by 21. Find the maximum value of $n$.
| $$
\begin{array}{l}
\Rightarrow\left(x^{2}+16+\frac{64}{x^{2}}\right)-\left(8 x+\frac{64}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}\right)^{2}-8\left(x+\frac{8}{x}\right)+\frac{119}{9}=0 \\
\Rightarrow\left(x+\frac{8}{x}-\frac{17}{3}\right)\left(x+\frac{8}{x}-\frac{7}{3}\right)=0 \\
\Rightarrow x=\fra... | 96 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $A \cup B=\{1,2, \cdots, 10\},|A|=|B|$. Then the number of all possible ordered pairs of sets $(A, B)$ is | - 1. 9953.
If $|A|=k(k=5,6, \cdots, 10)$, then $A$ has $\mathrm{C}_{10}^{k}$ possibilities.
For each possibility of $A$, since $A \cup B=\{1,2, \cdots, 10\}$, we know that $C_{A \cup B} A \subseteq B$.
Thus, $B$ already has $10-k$ elements.
To ensure $|A|=|B|$, it is necessary to select $2 k-10$ elements from $A$ to ... | 8953 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $x, y, z \in \mathbf{R}_{+}$. Then the minimum value of $\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}$ is $\qquad$ . | 3.2.
Let $x=y=1, z=\sqrt[3]{2}$, we get
$$
\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}=2 \text {. }
$$
Notice that
$$
\begin{array}{l}
\left(x^{2}+y^{2}\right)^{3}+z^{6} \\
=\left(x^{6}+y^{6}+3 x^{4} y^{2}+3 x^{2} y^{4}\right)+z^{6} \\
=\left(2 x^{4} y^{2}+2 x^{2} y^{4}\right)+\le... | 2 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
5. Fill $1,2, \cdots, n^{2}$ into an $n \times n$ chessboard, with each cell containing one number, and each row forming an arithmetic sequence with a common difference of 1. If any two of the $n$ numbers on the chessboard are neither in the same row nor in the same column, then the sum of these $n$ numbers is called a... | 5.1.
The numbers in the $k$-th row of the number table are
$$
(k-1) n+1,(k-1) n+2, \cdots,(k-1) n+n \text {. }
$$
Let the number taken from the $k$-th row be
$$
(k-1) n+a_{k} \text {. }
$$
Since these numbers are from different columns, $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$.
Therefore, t... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. There are 9 students participating in a math competition in the same classroom, with seats arranged in 3 rows and 3 columns, represented by a $3 \times 3$ grid, where each cell represents a seat. To prevent cheating, three types of exams, $A$, $B$, and $C$, are used, and it is required that any two adjacent seats (c... | 8. 246.
Let $a_{ij}$ denote the square at the $i$-th row and $j$-th column. First, consider the number of ways to distribute type $A$ papers at $a_{22}$.
$$
\begin{array}{l}
\text { Let } M=\left\{a_{12}, a_{21}, a_{32}, a_{23}\right\}, \\
N=\left\{a_{11}, a_{31}, a_{33}, a_{13}\right\} .
\end{array}
$$
Consider the ... | 246 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Find the smallest positive integer $k$, such that $625^{k} \equiv 1(\bmod 343)$ | $$
\left(5^{4}\right)^{k} \equiv 1\left(\bmod 7^{3}\right),
$$
which can be transformed into $(89 \times 7+2)^{k}-1 \equiv 0\left(\bmod 7^{3}\right)$.
Expanding it, we get
$$
\begin{array}{l}
\mathrm{C}_{k}^{2}(89 \times 7)^{2} \times 2^{k-2}+\mathrm{C}_{k}^{1}(89 \times 7) \times 2^{k-1}+2^{k}-1 \\
\equiv 0\left(\bmo... | 147 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
The 277th National Junior High School Mathematics Competition consists of 14 questions (5 multiple-choice questions, 5 fill-in-the-blank questions, 4 problem-solving questions), with a full score of 150 points. Among them, each correct answer for multiple-choice and fill-in-the-blank questions earns 7 points, and a wro... | The score can be selected from 16 5s and 10 7s, let the total number of 5s and 7s selected be $k$.
Since selection and non-selection are relative, i.e., the score obtained by selecting $m$ 5s and $n$ 7s is the same as the score obtained by selecting $16-m$ 5s and $10-n$ 7s, which sums up to 150, we only need to consid... | 127 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 2 of the hopscotch game: a person can only enter the first
square from outside; in the squares, each time they can
jump forward 1 square or 2 squares.
Then, the number of ways a person can jump
from outside to the sixth square is $\qquad$. | Solution 1 (Enumeration Method) Transform the problem as follows:
Enter the 6th grid from the 1st grid, walking 5 grids. Represent the number 5 as the sum of several 1s or 2s, where different orders of 1s and 2s represent different methods (e.g., $5=1+1+1+1+1$). For convenience, abbreviate the expression to $5=11111$. ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 The condition for three line segments to form a triangle is: the sum of the lengths of any two line segments is greater than the length of the third line segment. There is a wire of length $144 \mathrm{~cm}$, which is to be cut into $n$ $(n>2)$ small segments, with each segment being no less than $1 \mathrm{~... | Since the necessary and sufficient condition for forming a triangle is that the sum of any two sides is greater than the third side, the condition for not forming a triangle is that the sum of any two sides is less than or equal to the largest side. The shortest piece of wire is 1, so we can place 2 ones, and the third... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. There are 25 cards numbered $1, 3, \cdots, 49$. If a card with number $a$ is drawn, the next card to be drawn is the card with the largest odd divisor of $99-a$. This process is repeated until no more cards can be drawn. Xiao Hua drew the card numbered 1. When the operation ends, how many cards are left?
(A) 9
(B) 1... | 4. C.
The card numbers drawn are
$$
1 \rightarrow 49 \rightarrow 25 \rightarrow 37 \rightarrow 31 \rightarrow 17 \rightarrow 41 \rightarrow 29 \rightarrow 35 \rightarrow 1 \text {. }
$$
A total of 9 cards were drawn, leaving 16 cards. | 16 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
11. Observe the sequence of arrays $(1),(3,5),(7,9,11)$,
$(13,15,17,19), \cdots$. Then 2009 is in the $\qquad$ group. | $$
\begin{array}{l}
\text { Notice that 2009 is the 1005th positive odd number. } \\
1+2+\cdots+44<1005 \\
<1+2+\cdots+45,
\end{array}
$$
Then 2009 is in the 45th group. | 45 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. A certain linear function graph is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and passes through the point $(-1,-25)$. Then on the line segment $AB$ (including $A$ and $B$), the number of points with both integer coordinates is $... | 3.5.
Let the linear function be $y=\frac{5}{4} x+b$.
Given that it passes through the point $(-1,-25)$, we get $b=-\frac{95}{4}$.
Therefore, $y=\frac{5}{4} x-\frac{95}{4}$.
Thus, $A(19,0)$ and $B\left(0,-\frac{95}{4}\right)$.
From $y=\frac{5}{4} x-\frac{95}{4}(0 \leqslant x \leqslant 19)$, taking $x=3,7,11$,
15,19, $y$... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given the function $f(x)$ satisfies for all real numbers $x, y$,
$$
\begin{array}{l}
f(x)+f(2 x+y)=f(3 x-y)+x-2010 \text {. } \\
\text { Then } f(2010)=
\end{array}
$$ | 4.0.
Let $x=2010, y=1005$, then
$$
\begin{array}{l}
f(2010)+f(5025) \\
=f(5025)+2010-2010 .
\end{array}
$$
Therefore, $f(2010)=0$. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 1, given $A(-2,0)$, $B(0,-4)$, and $P$ as any point on the hyperbola $y=\frac{8}{x}$ $(x>0)$. A perpendicular line from $P$ to the x-axis meets at point $C$, and a perpendicular line from $P$ to the y-axis meets at point $D$. The minimum value of the area of quadrilateral $A B C D$ is $\qquad$ | 5.16.
Let point \( P\left(x_{0}, y_{0}\right) \). Then \( y_{0}=\frac{8}{x_{0}}\left(x_{0}>0\right) \). Hence \( C\left(x_{0}, 0\right) \) and \( D\left(0, \frac{8}{x_{0}}\right) \).
From the given conditions,
\[
\begin{array}{l}
|C A|=x_{0}+2,|D B|=\frac{8}{x_{0}}+4 . \\
\text { Therefore, } S=\frac{1}{2}\left(x_{0}+... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$
Then the product of the maximum and minimum values of $t$ is $\qquad$ | 8. 1.
Let $a=x+y, b=x-y$. Then
$$
(x+y)^{2}+(x+y)(x-y)+(x-y)^{2}=1 \text {. }
$$
Simplifying, we get $y^{2}=1-3 x^{2}$.
Since $y^{2} \geqslant 0$, we have $0 \leqslant x^{2} \leqslant \frac{1}{3}$.
Thus, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-x^{2}-3 y^{2}=8 x^{2}-3 .
\end{array}... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
II. (12 points) As shown in Figure 3, in $\triangle ABC$, it is given that $AB=9, BC=8, AC=7$, and $AD$ is the angle bisector. A circle is drawn with $AD$ as a chord, tangent to $BC$, and intersecting $AB$ and $AC$ at points $M$ and $N$, respectively. Find the length of $MN$. | II. As shown in Figure 3, connect DM. Given
$\angle B D M=\angle B A D=\angle C A D=\angle D M N$
$\Rightarrow M N / / B C \Rightarrow \triangle A M N \backsim \triangle A B C$.
It is easy to know that $B D=\frac{9}{2}$.
Then $B M \cdot B A=B D^{2} \Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2}$
$\Rightarrow B M=... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
2. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=a(a \neq 0$, and $a \neq 1)$, the sum of the first $n$ terms is $S_{n}$, and $S_{n}=\frac{a}{1-a}\left(1-a_{n}\right)$. Let $b_{n}=a_{n} \lg \left|a_{n}\right|\left(n \in \mathbf{N}_{+}\right)$. When $a=-\frac{\sqrt{7}}{3}$, does there exist a positive integ... | 2. When $n \geqslant 2$,
$$
S_{n}=\frac{a}{1-a}\left(1-a_{n}\right), S_{n-1}=\frac{a}{1-a}\left(1-a_{n-1}\right) .
$$
Then $a_{n}=S_{n}-S_{n-1}$
$$
\begin{array}{l}
=\frac{a}{1-a}\left[\left(1-a_{n}\right)-\left(1-a_{n-1}\right)\right] \\
=\frac{a}{1-a}\left(a_{n-1}-a_{n}\right),
\end{array}
$$
i.e., $a_{n}=a a_{n-1}... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. In a Cartesian coordinate system, draw all rectangles that simultaneously satisfy the following conditions:
(1) The sides of these rectangles are parallel or coincide with the coordinate axes;
(2) All vertices of these rectangles (repeated vertices are counted only once) are exactly 100 integer points (points with b... | 4. First, prove that the number of such rectangles does not exceed 2025.
Take any 100 integer points. Let $O$ be one of the 100 integer points we have chosen. We call a rectangle "good" if $O$ is one of its vertices, the other three vertices are also chosen from these 100 integer points, and the sides are parallel or ... | 2025 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9.1. Distribute 24 pencils of four colors (6 pencils of each color) to 6 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$.
| 9.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
In fact, there are 6 pencils of each color, and each student has 4 pencils, so there exists a student who has at least two colors of pencils. Clearly, any other color must be owned by at least one ... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9.6. In a day, 1000 dwarfs wearing red or blue hats meet each other in pairs. Dwarfs wearing red hats tell lies, while those wearing blue hats tell the truth. Each dwarf may change the color of their hat several times (i.e., red to blue, blue to red). It is known that when any two dwarfs meet, they both say that the ot... | 9.6. Clearly, when two dwarfs meet, they say that the other is wearing a red hat if and only if their hat colors are different.
Thus, in a day, if three dwarfs never change the color of their hats, then two of them must have the same hat color, and they cannot both say that the other is wearing a red hat when they mee... | 998 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 6, in isosceles $\triangle ABC$, it is known that $AB = AC = k BC$ ($k$ is a natural number greater than 1), points $D$ and $E$ are on sides $AB$ and $AC$ respectively, and $DB = BC = CE$, $CD$ intersects $BE$ at point $O$. Find the smallest positive integer $k$ such that $\frac{OC}{BC}$ is a rati... | Connect $D E$. It is easy to see that quadrilateral $B C E D$ is an isosceles trapezoid.
By the given condition $\angle B D C=\angle B C D=\angle E B C$, hence
$$
O C \cdot C D=B C^{2} \text {. }
$$
On the other hand,
$$
\frac{O C}{O D}=\frac{B C}{D E}=\frac{A B}{A D}=\frac{A B}{A B-D B}=\frac{k}{k-1} .
$$
Therefore,... | 25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
A4. Find the smallest positive integer $v$, such that for any two-coloring of the edges of the complete graph $K_{v}$, there are two monochromatic triangles that share exactly one vertex. | The solution to A4 is $v=9$.
On one hand, take two red complete graphs $K_{4}$, and color all the edges between them blue. The resulting complete graph $K_{8}$ has no two monochromatic triangles with exactly one common vertex.
On the other hand, consider any two-coloring of the complete graph $K_{9}$. By problem $\mat... | 9 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. A set of $n$ points $P_{1}, P_{2}, \cdots, P_{n}$ in the plane, no three of which are collinear, is denoted as $D$. A line segment is drawn between any two points, and the lengths of all these line segments are distinct. In a triangle, the side that is neither the longest nor the shortest is called the "middle side"... | 4. The minimum value of $n$ is 11.
When $n \geqslant 11$, regardless of how $l$ is chosen, there always exists a subset, let's assume it is $D_{1}$, such that $D_{1}$ contains at least six points.
Consider all the middle edges of the triangles in $D_{1}$, and color them red, then color the other edges blue.
By Ramse... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given that the area of quadrilateral $ABCD$ is 32, the lengths of $AB$, $CD$, and $AC$ are all integers, and their sum is 16.
(1) How many such quadrilaterals are there?
(2) Find the minimum value of the sum of the squares of the side lengths of such quadrilaterals.
(2003, National Junior High School Mathemat... | (1) As shown in Figure 3, let $A B=a, C D=b, A C=l$, and let the height from $A B$ in $\triangle A B C$ be $h_{1}$, and the height from $D C$ in $\triangle A D C$ be $h_{2}$.
Then
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D} \\
=S_{\triangle A B C}+S_{\triangle A D C} \\
=\frac{1}{2}\left(h_{1} a+h_{2} b\righ... | 192 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. A scientist invented a time machine, which looks like a circular subway track. Now (2010) is the first platform, the 2nd, 3rd, ..., 2009th platforms are 2011, 2012, ..., 4018, and the 2010th platform returns to the present (the departure platform). Later, the machine had a programming error, changing its operation r... | 8. (1) The time track has 2009 stations, and the positive integer powers of 2 not exceeding 2009 are
$2,4,8,16,32,64,128,256,512,1024$.
According to the rules,
(i) (1) $5 k+1(k=1,2,3)$
$\xrightarrow{3-k \text { times }} 16 \rightarrow 14=5 \times 2+4$;
(2) $5 k+1(k=4,5, \cdots, 51)$
$\xrightarrow{51-k \text { times }} ... | 812 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. 1. Distribute 40 pencils of four colors (10 pencils of each color) to 10 students, with each student getting 4 pencils. It is known that no matter how the pencils are distributed, there will always be $n$ students such that the $4 n$ pencils they have are of four colors. Find the minimum value of $n$. | 10.1. The minimum value of $n$ is 3.
First, we prove: There are always 3 students who have pencils of 4 different colors.
Since there are 10 pencils of each color, and each student has 4 pencils, there must be a student who has at least two different colors of pencils.
Obviously, any other color of pencils must be o... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10.4. In a $100 \times 100$ grid, each cell contains a positive integer. If the sum of the numbers in the cells of a rectangle (composed of several cells) is a multiple of 17, then the rectangle is called "good". We can color the cells of some good rectangles in the grid, with each cell being colored at most once. It i... | 10.4. $d_{\max }=9744=100^{2}-16^{2}$.
First, we prove a lemma.
Lemma: In a $1 \times n$ grid, each cell contains a positive integer, then at least $n-16$ cells can be colored. Proof by induction on $n$.
When $n \leqslant 16$, the conclusion is obviously true.
Assume $n=k \geqslant 17$, and for all $n < k$, the conclu... | 9744 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. In two boxes, Jia and Yi, each contains the same number of pieces of jewelry. After transferring a piece of jewelry worth 50,000 yuan from Jia to Yi, the average value of the jewelry in Jia decreases by 10,000 yuan, and the average value of the jewelry in Yi increases by 10,000 yuan. Then the total value of the jewe... | 4. B.
Let the original number of jewels in box A be $n(n>1)$, with a total value of $x$ million yuan, and the total value in box B be $y$ million yuan. Then $\left\{\begin{array}{l}\frac{x-5}{n-1}+1=\frac{x}{n}, \\ \frac{y+5}{n+1}-1=\frac{y}{n}\end{array} \Rightarrow\left\{\begin{array}{l}x=-n^{2}+6 n, \\ y=-n^{2}+4 n... | 12 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given three positive integers $a$, $b$, and $c$ whose squares sum to 2011, and the sum of their greatest common divisor and least common multiple is 388. Then the sum of the numbers $a$, $b$, and $c$ is $\qquad$ . | Let the greatest common divisor of $a$, $b$, and $c$ be $d$.
Then $a = d a_{1}$, $b = d b_{1}$, $c = d c_{1}$.
Assume without loss of generality that $a_{1} \geqslant b_{1} \geqslant c_{1}$, and the least common multiple of $a_{1}$, $b_{1}$, and $c_{1}$ is $m$.
From the problem, we have
$$
a_{1}^{2} + b_{1}^{2} + c_{1}... | 61 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. Among the triangles with side lengths being consecutive natural numbers and a perimeter not exceeding 100, the number of acute triangles is $\qquad$
$(1987$, National Junior High School Mathematics League) | 提示: Set the three sides of the triangle to be $n-1$, $n$, and $n+1$. When $n>24$, $(n-1)^{2}+n^{2}>(n+1)^{2}$, so the triangle is an acute triangle. Therefore, the number of acute triangles that meet the requirements is 29. | 29 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Set $A=\left\{(x, y) \left\lvert\,\left\{\begin{array}{l}y=\sqrt{1-x}, \\ y=1-x^{2}\end{array}\right\}\right.\right.$ has the number of subsets as $\qquad$ | 3. 8 .
Notice that the number of elements in set $A$ is the number of solutions to the system of equations
$$
\left\{\begin{array}{l}
y=\sqrt{1-x}, \\
y=1-x^{2}
\end{array}\right.
$$
Substituting equation (1) into equation (2) and simplifying, we get
$$
x(x-1)\left(x^{2}+x-1\right)=0 \text {. }
$$
Solving this, we g... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. A company invested in a project in 2009, with both cash inputs and cash revenues every year. It is known that
(1) In 2009, the company invested 10 million yuan, and the investment will decrease by $20\%$ each subsequent year;
(2) In 2009, the company earned 5 million yuan, and the revenue will increase by $25\%$ eac... | 7.2013.
Let the total investment of the project over $n$ years from 2009 be $A_{n}$ million yuan, and the total revenue be $B_{n}$ million yuan.
From (1) we know
$$
\begin{array}{l}
A_{n}=\sum_{k=1}^{n} 1000(1-20 \%)^{k-1} \\
=\frac{1000\left[1-\left(\frac{4}{5}\right)^{n}\right]}{1-\frac{4}{5}}=5000\left[1-\left(\fra... | 2013 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. For a positive integer $n$, let $S(n)$ denote the sum of the digits of $n$, and let $\varphi(n)$ denote the number of positive integers less than $n$ that are coprime to $n$. If $n$ is a three-digit number and satisfies $n=34 S(n)$, then the maximum value of $\varphi(n)$ is $\qquad$ | 8. 128 .
Let the three-digit number be
$$
n=100 a+10 b+c,
$$
where $a, b, c \in\{0,1, \cdots, 9\}, a>0$.
Then $100 a+10 b+c=34(a+b+c)$,
$$
8 b=22 a-11 c=11(2 a-c) .
$$
But $(8,11)=1$, so $11 \mid b \Rightarrow b=0$.
Thus, $2 a-c=0$.
When $a=1$, $c=2, n=102$;
When $a=2$, $c=4, n=204$;
When $a=3$, $c=6, n=306$;
When $... | 128 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
$$
\text { Three. (50 points) Given the sequence }\left\{a_{n}\right\}: 1,3,5,7, \cdots \text {, }
$$
starting from the 5th term, $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$. Find:
$$
a_{2008}^{2}+a_{2009}^{2}+a_{2010}^{2}+a_{2011}^{2}
$$
Can it be divisible by 4? | Consider the sequence $\left\{a_{n}\right\}$ and its modulo 2 residue sequence $\left\{b_{n}\right\}$.
Given that $a_{n+4}$ is the unit digit of $a_{n}+a_{n+3}$, we have $b_{n+4} \equiv b_{n}+b_{n+3}(\bmod 2)$.
By recursion, we get
$$
\begin{array}{l}
b_{n+10} \equiv b_{n+6}+b_{n+9} \\
\equiv\left(b_{n+2}+b_{n+5}\right... | 1 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
279 Given positive real numbers $x, y, z$ satisfying $(\sqrt{3}+1) x y + 2 \sqrt{3} y z + (\sqrt{3}+1) z x = 1$.
(1) Find the minimum value of $x + y + z$;
(2) Find the minimum value of $\frac{\sqrt{3} x y}{z} + \frac{(8-4 \sqrt{3}) y z}{x} + \frac{\sqrt{3} z x}{y}$. | (1) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{(\sqrt{3}+1) x^{2}}{2}+(\sqrt{3}-1) y^{2} \geqslant 2 x y, \\
2 y^{2}+2 z^{2} \geqslant 4 y z, \\
(\sqrt{3}-1) z^{2}+\frac{(\sqrt{3}+1) x^{2}}{2} \geqslant 2 x z .
\end{array}
$$
Adding the above three inequalities, we get
$$
\begin{array}{l}
(\sqrt{3}+1)\... | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given
$$
\begin{array}{l}
\frac{1}{1 \times \sqrt{2}+2 \sqrt{1}}+\frac{1}{2 \sqrt{3}+3 \sqrt{2}}+\cdots+ \\
\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}
\end{array}
$$
is greater than $\frac{19}{20}$ and less than $\frac{20}{21}$. Then the difference between the maximum and minimum values of the positive integer $n... | Consider the general term. The $k$-th term is
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}} \\
=\frac{1}{\sqrt{k} \sqrt{k+1}(\sqrt{k}+\sqrt{k+1})} \\
=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k} \sqrt{k+1}} \\
=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}(k=1,2, \cdots, n) .
\end{array}
$$
Thus, the original expression... | 39 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
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