problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 1
43
| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
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.
$$
Solving, we get $a>0, b>0$.
Since $a$ and $b$ are both integers, it follows that $a$ and $b$ are both positive integers.
Let $y=a x^{2}+b x+2$. Then this parabola opens upwards, intersects the $x$-axis at two different points, and when $x=-1$, $y>0$. Therefore,
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ \Delta = b ^ { 2 } - 8 a > 0 , } \\
{ a - b + 2 > 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
8 a \leqslant b^{2}-1, \\
a \geqslant b-1
\end{array}\right. \\
\Rightarrow b^{2}-1 \geqslant 8 b-8 \\
\Rightarrow(b-1)(b-7) \geqslant 0 .
\end{array}
$$
Since $b-1 \geqslant 0$, then $b-7 \geqslant 0 \Rightarrow b \geqslant 7$.
When $b=7$, substituting into equations (1) and (2) gives $a=6$, at which point the roots of the original equation are $-\frac{1}{2},-\frac{2}{3}$.
Therefore, the minimum value of $b$ is 7.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given that there are only three positive integers between the fractions $\frac{112}{19}$ and $\frac{112+x}{19+x}$. Find the sum of all possible integer values of $x$.
|
Notice that $50$ or $x<-19$.
Therefore, the three positive integers between $\frac{112+x}{19+x}$ and $\frac{112}{19}$ are $3,4,5$.
From $2<\frac{112+x}{19+x}<3$, we get $\frac{55}{2}<x<74$.
Since $x$ is an integer greater than 0, thus, $x=28,29, \cdots, 73$.
Therefore, the sum of these numbers is
$$
\frac{(28+73) \times 46}{2}=2323 \text {. }
$$
(2) If $\frac{112}{19}<\frac{112+x}{19+x}$, then $-19<x<0$.
Therefore, the three positive integers between $\frac{112}{19}$ and $\frac{112+x}{19+x}$ are $6,7,8$.
From $8<\frac{112+x}{19+x}<9$, we get $-\frac{59}{8}<x<-\frac{40}{7}$.
Since $-19<x<0$, thus, $x=-7,-6$.
Therefore, the sum of these numbers is -13.
Hence, the sum of all possible values of the integer $x$ is 2310.
|
2310
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
At first 272, from the eight points consisting of the vertices and midpoints of the sides of a square, how many isosceles triangles can be formed by selecting three points?
Will the above text be translated into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Considering that there are no equilateral triangles that meet the requirements in this problem, we classify the isosceles triangles by their vertices.
(1) Using the vertices of the square as the vertices of the isosceles triangle. As shown in Figure 3, if $A$ is the vertex,
then $\triangle A G H$, $\triangle A B D$, and $\triangle A E F$ all meet the requirements.
The same applies to points $B$, $C$, and $D$.
Therefore, there are
$$
3 \times 4=12 \text{ such triangles. }
$$
(2) Using the midpoints of the square's sides as the vertices of the isosceles triangle.
As shown in Figure 3, if $G$ is the vertex, then $\triangle G H F$ and $\triangle G B C$ both meet the requirements.
The same applies to points $H$, $E$, and $F$.
Therefore, there are $2 \times 4=8$ such triangles.
In summary, there are 20 triangles that meet the requirements.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. How many pairs $(n, r)$ are there in the array satisfying $0 \leqslant r \leqslant n \leqslant 63$ for which the binomial coefficient $\mathrm{C}_{n}^{r}$ is even (assuming $\left.\mathrm{C}_{0}^{0}=1\right) ?$
|
From Example 5, we know that the number of odd numbers in $\mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ is $2^{S(n)},$ where $S(n)$ is the sum of the binary digits of $n$.
Since $63=2^{6}-1=(111111)_{2}$, when $0 \leqslant n \leqslant 63$, we have $0 \leqslant S(n) \leqslant 6$.
Classifying and counting by the sum of the binary digits $S(n)$ of $n$, we know that the number of $n$ satisfying $S(n)=k \in\{0,1, \cdots, 6\}$ is $\mathrm{C}_{6}^{k}$.
Thus, the number of odd binomial coefficients $\mathrm{C}_{n}^{r}$ is
$$
\sum_{k=0}^{6} 2^{k} \mathrm{C}_{6}^{k}=(1+2)^{6}=729 \text {. }
$$
The total number of binomial coefficients is
$$
1+2+\cdots+64=2080 \text {. }
$$
Therefore, the number of the required arrays is
$$
2080-729=1351 .
$$
|
1351
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 2, fold $\triangle A B C$ along the dotted line $D E$ to get a heptagon $A D E C F G H$. If the area ratio of the heptagon to the original triangle is $2: 3$, and the area of the overlapping part after folding is 4, then the area of the original $\triangle A B C$ is
|
3. 12 .
Let the area of the non-overlapping part after folding be $x$. Then the area of the original triangle is $8+x$, and the area of the heptagon is $4+x$. From the given condition, we have
$$
\begin{array}{l}
(8+x):(4+x)=3: 2 \\
\Rightarrow 16+2 x=12+3 x \Rightarrow x=4 .
\end{array}
$$
$$
\text { Hence } S_{\triangle A B C}=12 \text {. }
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ The value is
The value of $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ is
|
4. 4 .
Let $\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}=x$.
Cubing both sides and simplifying, we get
$$
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 {. }
$$
Since $\Delta=4^{2}-4 \times 10=-24<0$, the equation $x^{2}+4 x+10=0$ has no real roots.
Hence, the equation $x^{3}-6 x-40=0$ has only one real root $x=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. Given the polynomial
$$
\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 {, }
$$
which simplifies to $(m+n)(m-n)=48$.
Since $m+n$ and $m-n$ have the same parity, the factorization of 48 can only be the following three:
$$
48=24 \times 2=12 \times 4=8 \times 6 .
$$
Thus, when $(m, n)=(13,11),(8,4),(7,1)$, there are Olympic points $B_{1}, B_{2}, B_{3}$.
Additionally, it is easy to see that the point $B_{3}$ symmetric to the line $A H$ is also an Olympic point, denoted as $B_{3}^{\prime}$.
Therefore, there are four Olympic points: $B_{1}, B_{2}, B_{3}, B_{3}^{\prime}$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $2n+1, 20n+1 \left(n \in \mathbf{N}_{+}\right)$ are powers of the same positive integer, then all possible values of $n$ are
|
6. 4 .
According to the problem, we know that $(2 n+1) \mid(20 n+1)$. Then $(2 n+1) \mid[10(2 n+1)-(20 n+1)]=9$. Therefore, $n \in\{1,4\}$. Upon verification, $n=4$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1: There is an electronic flea jumping back and forth on a number line, and it lights up a red light when it jumps to a negative number, but does not light up when it jumps to a positive number. The starting point is at the point representing the number -2 (record one red light), the first step is to jump 1 unit to the left, the second step is to jump 2 units to the right, the third step is to jump 3 units to the left, the fourth step is to jump 4 units to the right, and so on.
(1) When it jumps to the tenth step, how far is the electronic flea from the origin? How many times has the red light been lit?
(2) If the electronic flea needs to light up the red light ten times, what is the minimum number of jumps it needs to make? At this point, how far is it from the origin?
|
(1) Notice
$$
\begin{array}{l}
S=\mid-2-1+2-3+4-5+\cdots-9+101 \\
=1-2+1 \times 5 \mid=3 .
\end{array}
$$
Therefore, the negative numbers appear as $-2,-3,-1,-4,-5$, $-6,-7$, a total of seven times. So, the red light will flash seven times.
(2) In fact, after the electronic flea jumps four times, it returns to the origin, and the red light has already flashed four times (including the initial jump). After that, the red light flashes once for each odd-numbered jump. Therefore, to make the red light flash ten times, the electronic flea needs to jump $4+$ $2 \times 6-1=15$ times. At this point, the distance of the electronic flea from the origin is
$$
\begin{array}{l}
S=|-2-1+2-3+4-5+\cdots-15| \\
=|-2+1 \times 7-15|=10 .
\end{array}
$$
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Divide the numbers $1,2, \cdots, 200$ into two groups arbitrarily, each containing 100 numbers. Arrange one group in ascending order (denoted as $a_{1}<a_{2}<\cdots<a_{100}$) and the other in descending order (denoted as $b_{1}>b_{2}>\cdots>b_{100}$). Try to find
$$
\left|a_{1}-b_{1}\right|+\left|a_{2}-b_{2}\right|+\cdots+\left|a_{100}-b_{100}\right|
$$
the value of the expression.
|
First, prove: For any term in the algebraic expression
$$
\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}
$$
we get that $a_{1}, a_{2}, \cdots, a_{k}, b_{k}, b_{k+1}, \cdots, b_{100}$ are 101 numbers, all of which do not exceed 100, which is impossible;
(2) If $a_{k}>100$ and $b_{k}>100$, then by
$$
a_{100}>a_{99}>\cdots>a_{k+1}>a_{k}>100
$$
and $b_{1}>b_{2}>\cdots>b_{k}>100$
we get that $b_{1}, b_{2}, \cdots, b_{k}, a_{k}, a_{k+1}, \cdots, a_{100}$ are 101 numbers, all of which are greater than 100, which is also impossible.
Thus, in the 100 absolute values of the algebraic expression
$$
\left|a_{1}-b_{1}\right|,\left|a_{2}-b_{2}\right|, \cdots,\left|a_{100}-b_{100}\right|
$$
the smaller numbers are $1,2, \cdots, 100$, and the larger numbers are 101, $102, \cdots, 200$.
Therefore, the original expression is
$$
\begin{array}{l}
=(101+102+\cdots+200)-(1+2+\cdots+100) \\
=10000 .
\end{array}
$$
|
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^{4}, 7^{3}, 7^{2}, 7^{1}$;
(5) $11^{5}, 11^{4}, 11^{3}, 11^{2}, 11^{1}$.
On one hand, divide these 25 positive integers into five groups. Then, any selection of six numbers will definitely have two numbers in the same group. Clearly, one of these two numbers in the same group can divide the other.
On the other hand, since each group has only 5 numbers, the six selected numbers must come from at least two groups, meaning that there is no number among the six selected numbers that can be divided by the other five.
Therefore, when $n=25$, the condition is not satisfied.
For $n<25$, a similar proof can be made.
【Step 2】When $n=26$, the condition is satisfied.
If a set of numbers is all within the given 26 positive integers, denote the largest one as $a$. Among the remaining 25 numbers, there are no multiples of $a$, and all divisors of $a$ are included in this set, then this set is called a "good set" (a good set can contain only one number).
Next, we prove: there can be at most five good sets. Otherwise, there must be six good sets.
Consider the largest numbers in these six good sets, denoted as $a, b, c, d, e, f$. By the condition, among these six numbers, there must be one that can divide another, say $b \mid a$, meaning that the divisor $b$ of $a$ is not in the good set containing $a$.
This contradicts the definition of a good set, hence there can be at most five good sets.
Since there can be at most five good sets, and there are 26 given positive integers, there must be at least one good set with six numbers.
Consider the largest number in this good set. By the definition of a good set, there are at least five other numbers in this set that can divide this largest number.
In conclusion, the smallest positive integer $n=26$.
|
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 {. }
$$
From equations (2) and (3), we solve to get
$$
\begin{array}{l}
x+y=-14, x y=-38 . \\
\text { Therefore, } a x^{5}+b y^{5} \\
=\left(a x^{4}+b y^{4}\right)(x+y)-\left(a x^{3}+b y^{3}\right) x y \\
=42 \times(-14)-16 \times(-38)=20 .
\end{array}
$$
Solution 2: This problem can be handled using the perspective of a second-order linear recurrence sequence. The general term of a second-order linear recurrence sequence is
$$
a_{n}=a x^{n}+b y^{n} \text {. }
$$
Conversely, $\left\{a x^{n}+b y^{n}\right\}$ is a second-order recurrence sequence. The recurrence relation is
$$
a_{n+1}=c a_{n}+d a_{n-1},
$$
where $c=x+y, d=-x y$.
Thus,
$$
\left\{\begin{array}{l}
16=7 c+3 d, \\
42=16 c+7 d, \\
a_{5}=42 c+16 d .
\end{array}\right.
$$
Considering this as a system of equations in $1, c, d$,
$$
\left\{\begin{array}{l}
16-7 c-3 d=0, \\
42-16 c-7 d=0, \\
a_{5}-42 c-16 d=0 .
\end{array}\right.
$$
Viewing $(1, c, d)$ as a non-zero solution to the system of equations (4), the determinant of the coefficient matrix is 0, i.e.,
$$
\left|\begin{array}{lll}
16 & -7 & -3 \\
42 & -16 & -7 \\
a_{5} & -42 & -16
\end{array}\right|=0 \text {. }
$$
Solving this, we get $a_{5}=20$.
This is a clever solution, but some students might ask if there is a direct method, using "brute force" to solve for $a, b, x, y$ from the system of equations (1) and then substitute back into $a x^{5}+b y^{5}$. Of course, this is a complex approach for ordinary people, but for the Indian mathematician Ramanujan (1887-1920), it would be a piece of cake.
Ramanujan proposed and solved the following tenth-order system of equations:
$$
\begin{array}{l}
x+y+z+u+v=2, \\
p x+q y+r z+s u+t v=3, \\
p^{2} x+q^{2} y+r^{2} z+s^{2} u+t^{2} v=16, \\
p^{3} x+q^{3} y+r^{3} z+s^{3} u+t^{3} v=31, \\
p^{4} x+q^{4} y+r^{4} z+s^{4} u+t^{4} y=103, \\
p^{5} x+q^{5} y+r^{5} z+s^{5} u+t^{5} v=235, \\
p^{6} x+q^{6} y+r^{6} z+s^{6} u+t^{6} v=674, \\
p^{7} x+q^{7} y+r^{7} z+s^{7} u+t^{7} v=1669, \\
p^{8} x+q^{8} y+r^{8} z+s^{8} u+t^{8} v=4526, \\
p^{9} x+q^{9} y+r^{9} z+s^{9} u+t^{9} v=11595 .
\end{array}
$$
Ramanujan first considered the general system of equations
$$
x_{1}+x_{2}+\cdots+x_{n}=a_{1} \text {, }
$$
$$
\begin{array}{l}
x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n}=a_{2}, \\
x_{1} y_{1}^{2}+x_{2} y_{2}^{2}+\cdots+x_{n} y_{n}^{2}=a_{3}, \\
\cdots \cdots \cdots \\
x_{1} y_{1}^{2 n-1}+x_{2} y_{2}^{2 n-1}+\cdots+x_{n} y_{n}^{2 n-1}=a_{2 n} . \\
\text { Let } F(\theta)=\frac{x_{1}}{1-\theta y_{1}}+\frac{x_{2}}{1-\theta y_{2}}+\cdots+\frac{x_{n}}{1-\theta y_{n}} . \\
\text { But } \frac{x_{i}}{1-\theta y_{i}}=x_{i}\left(1+\theta y_{i}+\theta^{2} y_{i}^{2}+\theta^{3} y_{i}^{3}+\cdots\right)
\end{array}
$$
$(i=1,2, \cdots, n)$, hence
$$
\begin{array}{l}
F(\theta)=\sum_{i=1}^{n} x_{i} \cdot \sum_{k=0}^{\infty}\left(\theta y_{i}\right)^{k} \\
=\sum_{k=0}^{\infty}\left(\sum_{i=1}^{n} x_{i} y_{i}^{k}\right) \theta^{k}=\sum_{k=0}^{\infty} a_{k+1} \theta^{k}
\end{array}
$$
Converting it to a rational function with a common denominator, we get
$F(\theta)=\frac{A_{1}+A_{2} \theta+A_{3} \theta^{2}+\cdots+A_{n} \theta^{n-1}}{1+B_{1} \theta+B_{2} \theta^{2}+\cdots+B_{n} \theta^{n}}$.
Thus, $\sum_{k=0}^{\infty} a_{k+1} \theta^{k} \cdot \sum_{s=0}^{n} B_{s} \theta^{s}=\sum_{i=1}^{n} A_{t} \theta^{t-1}\left(B_{0}=1\right)$.
Hence, $A_{t}=\sum_{k=0}^{t-1} a_{k+1} B_{t-1-k}(t=1,2, \cdots, n)$, $0=\sum_{s=0}^{n} B_{s} a_{n+l-s}(t=1,2, \cdots, n)$.
Since $a_{1}, a_{2}, \cdots, a_{n}, a_{n+1}, \cdots, a_{2 n}$ are known,
we can solve for $B_{1}, B_{2}, \cdots, B_{n}$ from the last $n$ equations, and then substitute into the first $n$ equations to solve for $A_{1}, A_{2}, \cdots, A_{n}$. Knowing $A_{i} 、 B_{i}(i=1,2, \cdots, n)$, we can construct the rational function $F(\theta)$, and then expand it into partial fractions. Thus, we get
$$
F(\theta)=\frac{p_{1}}{1-q_{1} \theta}+\frac{p_{2}}{1-q_{2} \theta}+\cdots+\frac{p_{n}}{1-q_{n} \theta} .
$$
Clearly, $x_{i}=p_{i}, y_{i}=q_{i}(i=1,2, \cdots, n)$.
This is the solution to the general system of equations.
For the considered case, we have
$F(\theta)=\frac{2+\theta+3 \theta^{2}+2 \theta^{3}+\theta^{4}}{1-\theta-5 \theta^{2}+\theta^{3}+3 \theta^{4}-\theta^{5}}$.
Expanding it into partial fractions, we get the values of the unknowns:
$x=-\frac{3}{5}, p=-1$;
$y=\frac{18+\sqrt{5}}{10}, q=\frac{3+\sqrt{5}}{2} ;$
$z=\frac{18-\sqrt{5}}{10}, r=\frac{3-\sqrt{5}}{2}$;
$u=-\frac{8+\sqrt{5}}{2 \sqrt{5}}, s=\frac{\sqrt{5}-1}{2}$;
$v=\frac{8-\sqrt{5}}{2 \sqrt{5}}, t=-\frac{\sqrt{5}+1}{2}$.
Readers can apply this method to the problem mentioned at the beginning.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 If real numbers $a, b$ satisfy the conditions
$$
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}
$$
Thus, $-2 a^{2}-2 a \geqslant 0 \Rightarrow-1 \leqslant a \leqslant 0$.
Hence, $1-2 a+b \geqslant 0$.
Therefore, $1-2 a+b=-2 a^{2}-2 a$, which means
$$
1+b=-2 a^{2}=-2\left(1-b^{2}\right) \text {. }
$$
Rearranging gives $2 b^{2}-b-3=0$.
Solving this, we get $b=-1$ (the other root $b=\frac{3}{2}$ is discarded).
Substituting $b=-1$ into $1+b=-2 a^{2}$, we get $a=0$.
Thus, $a+b=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Mother's Day is here, and Xiao Hong, Xiao Li, and Xiao Ying go to a flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the above flowers. Then she should pay $\qquad$ yuan.
|
10. 20 .
Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $z$ yuan, respectively. Then
$$
\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 {. }
$$
From equations (2) and (3), we get
$$
x+y+z=10 \Rightarrow 2(x+y+z)=20 \text {. }
$$
Therefore, Xiaoying should pay 20 yuan.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition.
|
12. By the Pythagorean theorem, we have
$$
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 \\
\Rightarrow(a-6)(b-6)=18 .
\end{array}
$$
Since $a, b$ are both positive integers, without loss of generality, let $a<b$. Then
$$
\left\{\begin{array} { l }
{ a - 6 = 1 , } \\
{ b - 6 = 1 8 }
\end{array} \text { or } \left\{\begin{array} { l }
{ a - 6 = 2 , } \\
{ b - 6 = 9 }
\end{array} \text { or } \left\{\begin{array}{l}
a-6=3, \\
b-6=6 .
\end{array}\right.\right.\right.
$$
Solving, we get $(a, b, c)$
$$
=(7,24,25),(8,15,17),(9,12,15) \text {. }
$$
Therefore, there are three right triangles that satisfy the conditions.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In $\triangle A B C$, it is known that $A B=b^{2}-1, B C=$ $a^{2}, C A=2 a$, where $a$ and $b$ are both integers greater than 1. Then the value of $a-b$ is $\qquad$
|
2.0.
Since $a$ is an integer greater than 1, we have $a^{2} \geqslant 2 a$. By the triangle inequality, we get
$$
\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 \text {, so } \\
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=5 .
\end{array}
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $a, b$ be the roots of the equation $x^{2}+68 x+1=0$, and $c, d$ be the roots of the equation $x^{2}-86 x+1=0$. Then
$$
(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
$$
sets of positive integer solutions.
For each set of positive integer solutions, adding two negative signs can yield 3 sets of integer solutions with two negatives and one positive.
Therefore, the total number of integer solutions that satisfy the condition is
$$
18 \times(1+3)=72 \text { (sets). }
$$
|
72
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (15 points) Given a positive integer $n$ that satisfies the following condition: among any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$.
|
Because $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor no greater than 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17,19,23$,
$29,31,37,41,43$.
Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive integers greater than 1 and not exceeding 2009 that are pairwise coprime. If they are all composite, let $p_{i}$ be the prime factor of $a_{i}$ that does not exceed 44.
Since $a_{1}, a_{2}, \cdots, a_{n}$ are pairwise coprime, $p_{1}, p_{2}, \cdots, p_{n}$ are all distinct.
Thus, $n \leqslant 14$.
In other words, when $n \geqslant 15$, at least one of $a_{1}, a_{2}, \cdots, a_{n}$ must be a prime number.
On the other hand, $2^{2}, 3^{2}, 5^{2}, \cdots, 43^{2}$ are 14 positive integers greater than 1 and not exceeding 2009 that are pairwise coprime and contain no prime numbers.
Therefore, the minimum value of $n$ is 15.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, the minimum value of $7 y=2 x^{2}+4|x|-1$ is
$\qquad$
(2007, National Junior High School Mathematics Competition, Zhejiang Province Re-test)
|
Let $t=|x| \geqslant 0$. Then
$$
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) .
$$
Therefore, $f(x+4)=-f(x+2)=f(x)$, which means $f(x)$ is a periodic function with 4 as one of its periods.
Since $f(x)$ is an odd function defined on $\mathbf{R}$, we know $f(0)=0$.
Thus, the equation $f(x)=-\frac{1}{3}+f(0)$ simplifies to
$$
f(x)=-\frac{1}{3} \text {. }
$$
From the graph, we can see that $f(x)=-\frac{1}{3}$ has one real root in $(0,1)$ and one in $(1,2)$, and the sum of these two roots is 2; $f(x)=-\frac{1}{3}$ has one real root in $(4,5)$ and one in $(5,6)$, and the sum of these two roots is 10; $f(x)=-\frac{1}{3}$ has one real root in $(8,9)$ and one in $(9,10)$, and the sum of these two roots is 18.
Therefore, the original equation has six distinct real roots in the interval $(0,10)$, and their sum is 30.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If for any real number $x$, the function
$$
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,
$$
which is $f(x)=\left\{\begin{array}{ll}x^{2}, & x \leqslant 2 ; \\ x^{2}-4 x+8, & x>2\end{array}\right.$.
It is easy to see that, for any real number $x$, the value of $f(x)$ is always a non-negative real number.
Therefore, $a=1$ meets the requirement.
Hence, the maximum value of the real number $a$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Satisfy $0 \leqslant k_{i} \leqslant 20(i=1,2,3,4)$, and $k_{1}+k_{3}=k_{2}+k_{4}$ of the ordered integer tuples $\left(k_{1}, k_{2}, k_{3}, k_{4}\right)$ the number is $\qquad$ .
|
10. 6181.
For $0 \leqslant m \leqslant 20$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(0 \leqslant j \leqslant m),
$$
there are $m+1$ solutions;
When $20<m \leqslant 40$, the non-negative integer solutions satisfying $x+y=m$ and $0 \leqslant x, y \leqslant 20$ are
$$
(x, y)=(j, m-j)(m-20 \leqslant j \leqslant 20),
$$
there are $40-m+1$ solutions.
Therefore, the number of solutions satisfying $k_{1}+k_{3}=k_{2}+k_{4}$ is
$$
\begin{array}{l}
\sum_{m=0}^{20}(m+1)^{2}+\sum_{m=21}^{40}(40-m+1)^{2} \\
=2 \times \frac{1}{6} \times 20 \times 21 \times 41+21^{2}=6181 .
\end{array}
$$
|
6181
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Find the minimum value of the function
$$
f(x)=\max \left\{|x+1|,\left|x^{2}-5\right| \right\}
$$
and find the value of the independent variable $x$ when $f(x)$ takes the minimum value.
|
Solve As shown in Figure 1, in the same Cartesian coordinate system, draw the graphs of $f_{1}(x)=|x+1|$ and $f_{2}(x)=\left|x^{2}-5\right|$. The two graphs intersect at four points, $A$, $B$, $C$, and $D$, whose x-coordinates can be obtained by solving the equation $|x+1|=\left|x^{2}-5\right|$.
Removing the absolute value signs gives
$$
x+1=x^{2}-5 \text { or } x+1=5-x^{2} \text {. }
$$
Solving these, we get $x_{1}=3, x_{2}=-2$,
$$
x_{3}=\frac{-1+\sqrt{17}}{2}, x_{4}=\frac{-1-\sqrt{17}}{2} \text {. }
$$
From Figure 1, it is clear that the x-coordinates of points $A$, $B$, $C$, and $D$ are $\frac{-1-\sqrt{17}}{2},-2, \frac{-1+\sqrt{17}}{2}, 3$ respectively.
According to the definition of $f(x)$, its graph is shown by the solid line in the figure, and the y-coordinate of point $B$ is the minimum value of the function $f(x)$, at which $f(-2)=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Let $x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\}(i=1,2, \cdots, 2010)$. Let $S=x_{1} x_{2}+x_{3} x_{4}+\cdots+x_{2009} x_{2010}$.
(1) Can $S$ be equal to 2010? Prove your conclusion;
(2) How many different integer values can $S$ take?
|
14. (1) Since $(\sqrt{2}-1)^{2}=3-2 \sqrt{2}$,
$$
\begin{array}{l}
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, \\
(\sqrt{2}-1)(\sqrt{2}+1)=1
\end{array}
$$
Therefore, $x_{2 i-1} x_{2 i} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\}$.
Let the sum $S$ contain $a$ terms of $3+2 \sqrt{2}$, $b$ terms of $3-2 \sqrt{2}$, and $c$ terms of 1. Then $a$, $b$, and $c$ are non-negative integers, and
$$
\begin{array}{l}
a+b+c=1005 . \\
\text { Hence } S=\sum_{i=1}^{1005} x_{2 i-1} x_{2 i} \\
=(3+2 \sqrt{2}) a+(3-2 \sqrt{2}) b+c \\
=3 a+3 b+c+2 \sqrt{2}(a-b) .
\end{array}
$$
If $S=2010$, then $a=b$. At this time,
$$
S=6 a+c=6 a+(1005-a-a)=4 a+1005
$$
is an odd number.
Therefore, $S$ cannot be equal to 2010.
(2) From (1), we know that if $S$ is an integer, then $a=b, S=4 a+1005$.
By $a+b+c=2 a+c=1005,0 \leqslant a \leqslant 502$, then $S$ can take 503 different integer values.
|
503
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Real numbers $a, b$ satisfy
$$
|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 \\
\Rightarrow-12 \leqslant(a+1)(b+1) \leqslant 2 \\
\Rightarrow-13 \leqslant a b+a+b \leqslant 1 .
\end{array}
$$
Therefore, $m=1, n=-13$.
Thus, $m+n=-12$.
|
-12
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 On the blackboard, there are two positive integers, one is 2002, and the other is a number less than 2002. If the average of these two numbers $m$ is an integer, then the following operation can be performed: one of the numbers is erased and replaced by $m$. How many times can such an operation be performed at most?
|
Without loss of generality, let these two positive integers be $a, b(a>b)$.
Consider the absolute value of the difference between the two numbers.
Initially, it is $|a-b|$. After the first operation, the absolute value of the difference between the two numbers is $\left|a-\frac{a+b}{2}\right|=\left|\frac{a-b}{2}\right|$. Proceeding in this manner, after each operation, the absolute value of the difference between the two numbers becomes half of the original absolute value of the difference, and the average $m$ of the two numbers is an integer. Therefore, the number of allowed operations $n$ should satisfy
$2^{n}<2002$.
Clearly, the maximum value of $n$ is 10, meaning the operation can be performed at most 10 times.
At this point, the initial two positive integers on the blackboard are 2002 and 978.
|
10
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $a$, $b$, $x$, $y$ satisfy $x y=2008^{2008}$,
$$
\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 \\
\Rightarrow 2008^{a+b-1}=1 .
\end{array}
$$
Therefore, $2008^{a+b}=2008$.
|
2008
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Function
$$
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,
$$
which means when $x=1$, $f(x)$ takes the minimum value -53.
|
-53
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In the sequence $\left\{a_{n}\right\}$,
$$
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 {. }
$$
Then when $n=k+1$,
$$
\begin{array}{l}
a_{k+3}-a_{k+1} \\
=2 a_{k+2}-\left[a_{k+2}\right]-\left(2 a_{k}-\left[a_{k}\right]\right) \\
=2\left(a_{k+2}-a_{k}\right)-\left(\left[a_{k+2}\right]-\left[a_{k}\right]\right) \\
=2-\left(\left[a_{k}+1\right]-\left[a_{k}\right]\right)=1, \\
a_{k+1}+a_{k+2}=a_{k+1}+\left(a_{k}+1\right)=k+1 .
\end{array}
$$
Thus, when $n=k+1$, the conclusion also holds.
In summary, $a_{n+2}-a_{n}=1, a_{n}+a_{n+1}=n$ always holds. Therefore, $a_{2009}+a_{2010}=2009$.
|
2009
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If the set $A=\{x \mid x=6 n-1, n \in \mathbf{N}\}$,
$$
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
$$
has two distinct real roots in $0<x<\pi$, then the minimum value of the positive integer $n$ is $\qquad$.
|
4.4.
From the given, we have $\frac{1}{n}=-1-\frac{-1-\sin x}{2-\cos x}$.
And $\frac{-1-\sin x}{2-\cos x}$ represents the slope $k$ of the line connecting a moving point $P(\cos x, \sin x)$ on the upper half of the unit circle (excluding endpoints) and a fixed point $Q(2,-1)$.
To satisfy the problem, the line $PQ$ must intersect the upper half of the unit circle (excluding endpoints) at two distinct points, in which case, $k \in\left(-\frac{4}{3},-1\right)$.
Thus, $\frac{1}{n} \in\left(0, \frac{1}{3}\right)$, which gives $n>3$.
Therefore, the smallest positive integer value of $n$ is 4.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The sum of the x-coordinates of the points where the graph of the function $y=x^{2}-2009|x|+2010$ intersects the x-axis is $\qquad$ .
|
3. 0 .
The original problem can be transformed into finding the sum of all real roots of the equation
$$
x^{2}-2009|x|+2010=0
$$
If a real number $x_{0}$ is a root of equation (1), then its opposite number $-x_{0}$ is also a root of equation (1).
Therefore, the sum of all real roots of the equation is 0, that is, the sum of the x-coordinates of the points where the graph intersects the x-axis is 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given a positive integer $n$ such that the last two digits of $3^{n}$ form a two-digit prime number. Then the sum of all $n$ that satisfy this condition and do not exceed 2010 is $\qquad$ .
|
$-, 1.909128$.
Considering $3^{k}(k=1,2, \cdots)$ modulo 100, the remainders are sequentially $3,9,27,81,43,29,87,61,83,49,47,41$, $23,69,7,21,63,89,67,1 ; 3,9, \cdots$,
where $43,29,61,83,47,41,23,89,67$ are two-digit primes, corresponding to
$$
k \equiv 5,6,8,9,11,12,13,18,19(\bmod 20) .
$$
Thus, the sum of all $n$ that satisfy the condition is
$$
\begin{array}{l}
\sum_{i=0}^{9}(9 \times 20 i+5+6+8+9+11+12+13+18+19)+ \\
\quad 2005+2006+2008+2009 \\
=909128 .
\end{array}
$$
|
909128
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
[\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}
$$
Contradiction.
If $k=3$, let $A=m\left(m^{2}-1\right)(m \in \mathbf{Z})$.
But $A \equiv 2^{2012 n}-2^{4 n}+2 \times(-1)^{7 n}$
$$
\equiv 2 \times(-1)^{n} \equiv \pm 2(\bmod 5),
$$
This contradicts $m\left(m^{2}-1\right) \equiv 0$ or $\pm 1(\bmod 5)$.
If $k=2$, let $A=m(m+1)$. Then
$$
(2 m+1)^{2}=4 A+1 \text {. }
$$
But when $n \geqslant 1$, we have
$$
\left(2 \times 17^{100 \sigma_{n}}\right)^{2}<4 A+1<\left(2 \times 17^{100 \sigma_{n}}+1\right)^{2} \text {, }
$$
Contradiction.
Therefore, $n=0$. In this case, $A=12=3 \times 4$ satisfies the condition.
Thus, $(n, k)=(0,2)$. Hence $n+k=2$.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A photographer took some photos of eight people at a party, with any two people (there are 28 possible combinations) appearing in exactly one photo. Each photo can be a duo or a trio. How many photos did the photographer take at least?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
6. 12 .
Let the number of group photos with three people be $x$, and the number of group photos with two people be $y$. Then $3 x+y=28$.
Thus, when $x$ is maximized, the total number of photos taken is minimized.
When $x>8$, by $\frac{3 x}{8}>3$ we know that there is a person $A$ who appears 4 times, and in the 4 group photos where $A$ appears, there are a total of eight appearances by other people, so there must be a person $B$ who appears with $A$ in two group photos, which is a contradiction.
When $x=8$, $y=4$, it can be done.
Number the eight people as $0,1, \cdots, 7$, the 12 photos taken are
$$
\{012,034,056,713,745,726,146,235 \text {, }
$$
$07,15,24,36\}$.
|
12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Select 1005 numbers $a_{1}, a_{2}, \cdots, a_{1005}$ from 1 to 2010 such that their total sum is 1006779, and the sum of any two of these 1005 numbers is not equal to 2011.
(1) Prove that $\sum_{i=1}^{1005} a_{i}^{2}-4022 \sum_{i=1}^{1005} a_{i}^{3}+\sum_{i=1}^{1005} a_{i}^{4}$ is a constant;
(2) When $\sum_{i=1}^{1005} a_{i}^{3}$ is minimized, find the sum of all numbers in $\left\{a_{i}\right\}(i=1, 2, \cdots, 1005)$ that are less than 1005.
|
Divide $\{1,2, \cdots, 2010\}$ into 1005 groups:
$$
A_{i}=\{i, 2011-i\}(i=1,2, \cdots, 1005) .
$$
Since the sum of any two numbers in $\left\{a_{i}\right\}$ is not equal to 2011, exactly one number is taken from each group. First, take the even numbers from each group to form
$\left\{b_{i}\right\}$, where $b_{i} \in A_{i}(i=1,2, \cdots, 1005)$.
$$
\text { Also, } \sum_{i=1}^{1005} b_{i}=\sum_{i=1}^{1005} 2 i=1011030>\sum_{i=1}^{1005} a_{i} \text {, }
$$
Therefore, some even numbers must be replaced by the odd numbers in the same group.
Let $A_{j}$ be the group where the replacement of a number results in an increment of $\sum_{i=1}^{1005} a_{i}^{k}-\sum_{i=1}^{1005} b_{i}^{k}$ by $d_{k}(j)$, where $k=1,2,3$. Then
$$
\sum_{i=1}^{1005} d_{1}(i)=\sum_{i=1}^{1005} a_{i}-\sum_{i=1}^{1005} b_{i}=-4251 .
$$
In the $A_{2 j}$ group, if $2 j$ is replaced by $2011-2 j$, then
$$
\begin{array}{l}
d_{1}(2 j)=(2011-2 j)-2 j \\
=4(503-j)-1=e_{503-j} ;
\end{array}
$$
In the $A_{2 j-1}$ group, if $2012-2 j$ is replaced by $2 j-1$, then
and $d_{2}(j)=2011 d_{1}(j)$,
$d_{3}(j)=\frac{1}{4} d_{1}(j)\left(3 \times 2011^{2}+d_{1}(j)^{2}\right)$.
Thus, $\sum_{i=1}^{1005} a_{i}^{2}=\sum_{i=1}^{1005}(2 i)^{2}+\sum_{i=1}^{1005} d_{2}(i)$
$=\sum_{i=1}^{1005}(2 i)^{2}+2011 \sum_{i=1}^{1005} d_{1}(i)$, $\sum_{i=1}^{1005} a_{i}^{3}=\sum_{i=1}^{1005}(2 i)^{3}+\sum_{i=1}^{1005} d_{3}(i)$ $=\sum_{i=1}^{1005}(2 i)^{3}+\frac{3}{4} \times 2011^{2} \sum_{i=1}^{1005} d_{1}(i)+$ $\frac{1}{4} \sum_{i=1}^{1005} d_{1}(i)^{3}$.
(1) From equation (1), we know that $\sum_{i=1}^{1005} a_{i}^{2}$ is a constant, and
$$
\begin{array}{l}
\sum_{i=1}^{1005} a_{i}^{4}-4022 \sum_{i=1}^{1005} a_{i}^{3} \\
=\sum_{i=1}^{1005} a_{i}^{2}\left[\left(2011-a_{i}\right)^{2}-2011^{2}\right] \\
=\sum_{i=1}^{100}[(2 i)(2011-2 i)]^{2}-2011^{2} \sum_{i=1}^{1005} a_{i}^{2}
\end{array}
$$
is a constant.
(2) From equation (2), we know that $\sum_{i=1}^{1005} a_{i}^{3}$ takes its minimum value if and only if $\sum_{i=1}^{1005} d_{1}(i)^{3}$ takes its minimum value.
First, find the subsets $I$ and $J$ of $\{0,1, \cdots, 502\}$ such that
$$
\sum_{i \in 1} e_{i}-\sum_{j \in j} f_{j}=\sum_{i=1}^{1 \infty} d_{1}(i)=-4251 \text {, }
$$
and $A(I, J)=\sum_{i \in 1} e_{i}^{3}-\sum_{j \in J} f_{j}^{3}$
is minimized, where $e_{i}=4 i-1(i=1,2, \cdots, 502), f_{j}=$ $4 j+1(j=0,1, \cdots, 502)$.
Let $I_{0}=\{1,2, \cdots, n\}, J_{0}=\{m, m+1, \cdots$, $502\}$, where $1 \leqslant n \leqslant m \leqslant 502$. Then
$$
\begin{array}{l}
\sum_{i=1}^{n} e_{i}=\sum_{i=1}^{n}(4 i-1)=n(2 n+1), \\
\sum_{j=m}^{502} f_{j}=\sum_{j=m}^{502}(4 j+1)=(503-m)(1005+2 m) .
\end{array}
$$
By $4 n-1A\left(I_{0}, J_{0}\right)$.
Let $I_{1}=I_{0} \backslash V^{\prime}, I_{2}=I^{\prime} \bigcup_{0}, J_{1}=J^{\prime} \bigcup_{0}, J_{2}=J_{0} \cup^{\prime}$.
Then $\sum_{i \in J_{0}} e_{i}-\sum_{j \in J_{0}} f_{j}=\sum_{i \in I} e_{i}-\sum_{j \in J^{J}} f_{j}$
$\Rightarrow \sum_{i \in J_{1}} e_{i}-\sum_{j \in J_{2}} f_{j}=\sum_{i \in I_{2}} e_{i}-\sum_{j \in J_{1}} f_{j}$
$\Rightarrow \sum_{i \in J_{1}} e_{i}+\sum_{j \in J_{1}} f_{j}=\sum_{i \in J_{2}} e_{i}+\sum_{j \in J_{2}} f_{j}$.
Notice that each number on the left side of equation (3) is less than each number on the right side. By the adjustment method, it is easy to see that
$$
\begin{array}{l}
\sum_{i \in I_{1}} e_{i}^{3}+\sum_{j \in J_{1}} f_{j}^{3}<\sum_{i \in I_{2}} e_{i}^{3}+\sum_{j \in J_{2}} f_{j}^{3} \\
\Rightarrow \sum_{i \in I_{0}} e_{i}^{3}-\sum_{j \in J_{0}} f_{j}^{3}<\sum_{i \in I} e_{i}^{3}-\sum_{j \in J} f_{j}^{3} .
\end{array}
$$
Therefore, $I=\{1,2, \cdots, 354\}$,
$J=\{354,355, \cdots, 502\}$.
Thus, when $\sum_{i=1}^{1005} a_{i}^{3}$ takes its minimum value, we have
$$
\begin{aligned}
\left\{a_{i}\right\}= & \{2 j \mid j=1,2, \cdots, 148\} \cup \\
& \{2011-2 j \mid j=149,150, \cdots, 502\} \cup \\
& \{2 j-1 \mid j=1,2, \cdots, 149\} \cup \\
& \{2012-2 j \mid j=150,151, \cdots, 503\},
\end{aligned}
$$
where the sum of all numbers less than 1005 is
$$
\sum_{j=1}^{148} 2 j+\sum_{j=1}^{149}(2 j-1)=44253 .
$$
(Song Qiang)
|
44253
|
Algebra
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its divisors 1, $2,17,34,59,118,1003,2006$ is 3240. Find the smallest good number. ${ }^{[3]}$
(2006, Brazilian Mathematical Olympiad)
|
\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdots\left(1+\alpha_{k}\right)=8.
Therefore, when $k=1$, $\alpha_{1}=7$;
when $k=2$, $\alpha_{1}=1, \alpha_{2}=3$ or $\alpha_{1}=3, \alpha_{2}=1$;
when $k=3$, $\alpha_{1}=\alpha_{2}=\alpha_{3}=1$;
when $k \geqslant 4$, there is no solution.
(1) If $n=p^{7}$ ( $p$ is a prime number), then $\sum_{i=0}^{7} p^{i}=3240$.
(i) If $p \geqslant 3$, then
$$
\sum_{j=0}^{7} p^{j} \geqslant \sum_{j=0}^{7} y^{j}=\frac{3^{8}-1}{2}=3280>3240 \text {. }
$$
(ii) If $p=2$, then
$$
\sum_{j=0}^{7} p^{j}=2^{8}-1=511 \neq 3240 \text {. }
$$
Therefore, when $n=p^{7}$, there is no solution that satisfies the problem.
(2) If $n=p^{3} q(p, q$ are prime numbers, and $p \neq q)$, then the sum of positive divisors is
$$
\begin{array}{l}
3240=\left(1+p+p^{2}+p^{3}\right)(1+q) \\
=(1+p)\left(1+p^{2}\right)(1+q) \\
=2^{3} \times 3^{4} \times 5 .
\end{array}
$$
Since $q \geqslant 2$, we have $1+q \geqslant 3$.
Thus, $1+p+p^{2}+p^{3} \leqslant 1080$. Therefore, $p \leqslant 7$.
(i) If $p=7$, then $1+p^{2}=50 \times 3240$, which contradicts equation (1).
(ii) If $p=5$, then $1+p^{2}=26 \times 3240$, which is a contradiction.
(iii) If $p=3$, then $1+q=\frac{3240}{40}=81$, thus, $q=80$, which is a contradiction.
(iv) If $p=2$, then $1+q=\frac{3240}{15}=216$, thus, $q=215$, which is also a contradiction.
(3) If $n=p q r(p<q<r)$, then
$(1+p)(1+q)(1+r)$
$=3240=2^{3} \times 3^{4} \times 5$.
(i) If $p=2$, then
$(1+q)(1+r)=2^{3} \times 3^{3} \times 5$.
To make $p q r$ as small as possible, $q r$ should be as small as possible. Thus, from
$q r=(1+q)(1+r)-(q+1)-(r+1)+1$, we know that $(q+1)+(r+1)$ should be as large as possible.
Considering $(1+q)(1+r)=2^{3} \times 3^{3} \times 5$, we know that $q+1$ should be as small as possible, and $r+1$ should be as large as possible.
Therefore, take $q=3, r=269$.
At this point, $n_{\text {min }}=2 \times 3 \times 269=1614$.
(ii) If $p \geqslant 3$, then
$\frac{1+p}{2} \cdot \frac{1+q}{2} \cdot \frac{1+r}{2}=3^{4} \times 5$.
Due to the size relationship of $p, q, r$, there are only two cases:
$\frac{1+p}{2}=3, \frac{1+q}{2}=3^{2}, \frac{1+r}{2}=3 \times 5$;
and $\frac{1+p}{2}=3, \frac{1+q}{2}=5, \frac{1+r}{2}=3^{3}$.
It is easy to solve the corresponding
$(p, q, r)=(5,17,29)$ or $(5,9,53)$.
Since 9 is not a prime number, $(p, q, r)=(5,17,29)$.
Therefore, $n=2465$.
In summary, the $n_{\min }=1614$ that satisfies the problem.
|
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 another $5 \mathrm{~min}$, the car catches up with the bus; then after $t \min$, the truck catches up with the bus. What is the value of $t$?
|
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 have
$$
\begin{array}{l}
10(a-b)=S, 15(a-c)=2 S, \\
x(b-c)=S .
\end{array}
$$
Thus, $30(b-c)=S$.
Therefore, $x=30$.
So, $t=30-10-5=15(\mathrm{~min})$.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. For $i=2,3, \cdots, k$, the remainder when the positive integer $n$ is divided by $i$ is $i-1$. If the smallest value of $n$, $n_{0}$, satisfies $2000<n_{0}<3000$, then the smallest value of the positive integer $k$ is
|
10.9.
Since $n+1$ is a multiple of $2,3, \cdots, k$, the smallest value of $n$, $n_{0}$, satisfies
$$
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, 10]=2520,} \\
{[2,3, \cdots, 11]=27720,}
\end{array}
$$
Therefore, the smallest positive integer $k$ that satisfies $2000<n_{0}<3000$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. B. Let real numbers $a, b$ satisfy
$$
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}
$$
Thus, when $b=-1$, the minimum value of $u$ is -34.
(2) When $3 a-4 b+5=0$,
$$
\begin{array}{l}
\quad u=9 a^{2}+72 b+2=16 b^{2}+32 b+27 \\
=16(b+1)^{2}+11 .
\end{array}
$$
Thus, when $b=-1$, the minimum value of $u$ is 11.
In summary, the minimum value of $u$ is -34.
|
-34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. A. From the 2010 positive integers $1,2, \cdots, 2010$, what is the maximum number of integers that can be selected such that the sum of any three selected numbers is divisible by 33?
|
14. A. First, the following 61 numbers: $11, 11+33, 11+2 \times 33, \cdots, 11+60 \times 33$ (i.e., 1991) satisfy the conditions of the problem.
On the other hand, let $a_{1}<a_{2}<\cdots<a_{n}$ be the numbers selected from 1, 2, $\cdots, 2010$ that satisfy the conditions of the problem. For any four numbers $a_{i}, a_{j}, a_{k}, a_{m}$ among these $n$ numbers, since
$$
33 \mid \left(a_{i}+a_{k}+a_{m}\right), 33 \mid \left(a_{j}+a_{k}+a_{m}\right),
$$
it follows that $33 \mid \left(a_{j}-a_{i}\right)$.
Therefore, the difference between any two of the selected numbers is a multiple of 33.
Let $a_{i}=a_{1}+33 d_{i} (i=1,2, \cdots, n)$.
From $33 \mid \left(a_{1}+a_{2}+a_{3}\right)$, we get
$33 \mid \left(3 a_{1}+33 d_{2}+33 d_{3}\right)$.
Thus, $33 \mid 3 a_{1}$, and $11 \mid a_{1}$, which means $a_{1} \geqslant 11$.
Therefore, $d_{i}=\frac{a_{i}-a_{1}}{33} \leqslant \frac{2010-11}{33}<61$.
Hence, $d_{i} \leqslant 60$. So, $n \leqslant 61$.
In conclusion, the maximum value of $n$ is 61.
|
61
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. B. Color the five sides and five diagonals of the convex pentagon $A B C D E$, such that any two segments sharing a common vertex are of different colors. Find the minimum number of colors needed.
|
14. B. Since vertex $A$ is the common point of four segments $A B$, $A C$, $A D$, and $A E$, at least four colors are needed.
If only four colors are used, let's assume they are red, yellow, blue, and green. Then, the four segments extending from each vertex must include one each of red, yellow, blue, and green. Therefore, the number of red segments would be $\frac{5}{2}$, which is a contradiction. Thus, at least five colors are needed.
The following example shows that five colors can be used to color the ten segments to meet the conditions: color $A B$ and $C E$ with color 1; color $B C$ and $D A$ with color 2; color $C D$ and $E B$ with color 3; color $D E$ and $A C$ with color 4; color $E A$ and $B D$ with color 5. In this way, any two segments sharing a common vertex will have different colors.
In conclusion, the minimum number of colors required is 5.
(Provided by Ben Guo Min)
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If the sum of $k$ consecutive positive integers is 2010, then the maximum value of $k$ is
|
Ni. 1.60 .
Let $2010=(n+1)+(n+2)+\cdots+(n+k)$.
Then $k(2 n+k+1)=4020$.
Notice that $k<2 n+k+1$, and
$$
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 {. }
$$
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In $\pm 1 \pm 2 \pm 3 \pm 5 \pm 20$, by appropriately choosing + or -, different algebraic sums can be obtained $\qquad$.
|
4.24.
Among $1,2,3,5,20$, there are three odd numbers, so their algebraic sum must be odd.
Observing, we see that from $1,2,3,5$ we can obtain all odd numbers with absolute values not exceeding 11.
According to the problem, the expression must include 1, 2, 3, 5, and 20. Therefore, the integers that can be obtained are $\pm 20$ plus or minus $1,3,5,7,9, 11$, i.e., 12 positive odd numbers 9, $11, \cdots, 31$ and 12 negative odd numbers $-9,-11, \cdots$, -31.
Thus, the total number of integers that can be expressed is 24.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a_{1}, a_{2}, \cdots, a_{10} \in(1,+\infty)$. Then
$$
\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}
$$
Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{10}$. Therefore, the minimum value of the original expression is 100.
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $A$ be the set of all positive integers not exceeding 2009, i.e., $A=\{1,2, \cdots, 2009\}$, and let $L \subseteq A$, where the difference between any two distinct elements of $L$ is not equal to 4. Then the maximum possible number of elements in the set $L$ is
|
7. 1005.
Divide the set $A$ into the following 1005 subsets:
$$
\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 exist two elements in $L$ whose difference is 4.
Furthermore, the set $L$ formed by taking one element from each of the first 1005 subsets satisfies the condition.
Therefore, the maximum possible number of elements in the set $L$ is 1005.
|
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 four vertices of the convex decagon, and these four vertices and the intersection point of the two diagonals can form four triangles with only two vertices being vertices of the convex decagon. Therefore, the number of triangles with only two vertices being vertices of the convex decagon is $4 C_{10}^{4}$.
In summary, the total number of triangles is $C_{10}^{3}+4 C_{10}^{4}=960$
|
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$ is
$$
\begin{array}{l}
\sum_{k=1}^{5} \mathrm{C}_{5}^{k}\left(2^{k}-1\right)=\sum_{k=0}^{5} \mathrm{C}_{5}^{k}\left(2^{k}-1\right) \\
=\sum_{k=0}^{5} \mathrm{C}_{5}^{k} 2^{k}-\sum_{k=0}^{5} \mathrm{C}_{5}^{k}=3^{5}-2^{5} .
\end{array}
$$
Similarly, the number of ordered pairs $(A, B)$ satisfying $B \subset A$ is also $3^{5}-2^{5}$.
Therefore, the number of ordered pairs $(A, B)$ satisfying the conditions is
$$
2^{5}\left(2^{5}-1\right)-2\left(3^{5}-2^{5}\right)=570 \text {. }
$$
|
570
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The body diagonal of a rectangular prism is 10, and the projection of this diagonal on one of the surfaces of the prism is 8. Then the maximum volume of this rectangular prism is $\qquad$ .
|
4. 192 .
According to the problem, the height of the cuboid is $\sqrt{10^{2}-8^{2}}=6$. Let the side lengths of the base of the cuboid be $a$ and $b$. Then $a^{2}+b^{2}=64$.
Thus, the volume of the cuboid is $V=6 a b \leqslant 3\left(a^{2}+b^{2}\right)=192$.
The equality holds if and only if $a=b=4 \sqrt{2}$. Therefore, $V_{\max }=192$.
|
192
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If the equation about $x$
$$
x^{3}+a x^{2}+b x-4=0\left(a 、 b \in \mathbf{N}_{+}\right)
$$
has a positive integer solution, then $|a-b|=$
|
7.1.
Solution 1 Let $m$ be a positive integer solution of the equation.
If $m \geqslant 2$, then $a m^{2}+b m=4-m^{3}<0$, which contradicts that $a$ and $b$ are both positive integers.
Therefore, only $m=1$. Substituting it in, we get $a+b=3$.
Since $a, b \in \mathbf{N}_{+}$, $\{a, b\}=\{1,2\}$.
Thus, $|a-b|=1$.
Solution 2 It is easy to see that the positive integer solution of the equation must be a divisor of 4. The divisors of 4 are $1,2,4$. Substituting them into the original equation, we get
$$
\begin{array}{l}
a+b-3=0, \\
4 a+2 b+4=0, \\
16 a+4 b+60=0 .
\end{array}
$$
It is easy to see that the last two equations have no positive integer solutions.
Therefore, $a+b=3$.
The rest is the same as Solution 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $|x| \leqslant 1,|y| \leqslant 1$, and
$$
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{. }
$$
Solving for $t$ yields $t= \pm \frac{7}{2}$.
Since $x$ and $y$ are both positive integers, when $t=\frac{7}{2}$, we have $x+y=8, x y=15$;
When $t=-\frac{7}{2}$, we have $x+y=15, x y=8$ (discard this case). Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=34$.
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Let real numbers $x, y, z$ satisfy
$$
\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}
\text { From } x^{2}+y^{2}=(x+y)^{2}-2 x y \\
=(z-1)^{2}-2\left(z^{2}-7 z+14\right) \\
=-(z-6)^{2}+9,
\end{array}
$$
Thus, when $z=5$, the maximum value of $x^{2}+y^{2}$ is 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$, $b$, $c$, and $d$ are prime numbers, and $a b c d$ is the sum of 77 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$
|
Given 77 consecutive positive integers, with the middle one being $x$. Then $a b c d=77 x=7 \times 11 x$, and $x \geqslant 39$.
Since $a, b, c, d$ are prime numbers, thus, $x$ can be decomposed into the product of two prime numbers, and the sum of these two prime numbers is minimized.
It is easy to verify that when $x$ is $49=7 \times 7$, the sum of these two prime numbers is the smallest.
Therefore, the minimum value of $a+b+c+d$ is $7+11+7+7=32$.
|
32
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given an equilateral $\triangle A B C$ with side length $2, P$ is a point inside $\triangle A B C$, and the distances from point $P$ to the three sides $B C, A C, A B$ are $x, y, z$ respectively, and their product is $\frac{\sqrt{3}}{9}$. Then the sum of the squares of $x, y, z$ is
|
2. 1 .
As shown in Figure 4, connect
$$
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{. }
$$
Since the equality holds, we have
$$
x=y=z=\frac{\sqrt{3}}{3} \text{. }
$$
Therefore, $x^{2}+y^{2}+z^{2}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the quadratic function
$$
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 .
$$
By the Intermediate Value Theorem, $3 a x^{2}+2 b x-(a+b)=0$ has one root when $0<x<\frac{1}{2}$, and one root when $\frac{1}{2}<x<1$. Therefore, there are two intersection points.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) From the natural numbers $1, 2, \cdots, 2010$, take $n$ numbers such that the sum of any three of the taken numbers is divisible by 21. Find the maximum value of $n$.
|
$$
\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=\frac{8}{3} \text { or } 3 .
\end{array}
$$
Therefore, the length of $AE$ is $\frac{8}{3}$ or 3.
Three, let $a, b, c, d$ be any four numbers taken. By the problem, we have
$$
a+b+c=21 m, a+b+d=21 n,
$$
where $m, n$ are positive integers.
Thus, $c-d=21(m-n)$.
The above equation shows that the difference between any two of the numbers taken is a multiple of 21, meaning that the remainder when each of the numbers is divided by 21 is the same.
Let this remainder be $k$. Then,
$$
\begin{array}{l}
a=21 a_{1}+k, b=21 b_{1}+k, \\
c=21 c_{1}+k,
\end{array}
$$
where $a_{1}, b_{1}, c_{1}$ are integers, $0 \leqslant k<21$.
Then $a+b+c=21\left(a_{1}+b_{1}+c_{1}\right)+3 k$.
Since $a+b+c$ is divisible by 21, $3 k$ must be divisible by 21, which means $k$ must be divisible by 7.
Therefore, $k=0, 7$ or 14.
When $k=0$, we can take $21, 42, 63, \cdots, 1995$, a total of 95 numbers, which meets the requirement;
When $k=7$, we can take $7, 28, 49, \cdots, 2002$, a total of 96 numbers, which meets the requirement;
When $k=14$, we can take $14, 35, 56, \cdots, 2009$, a total of 96 numbers, which meets the requirement.
In summary, the maximum value of $n$ is 96.
|
96
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $A \cup B=\{1,2, \cdots, 10\},|A|=|B|$. Then the number of all possible ordered pairs of sets $(A, B)$ is
|
- 1. 9953.
If $|A|=k(k=5,6, \cdots, 10)$, then $A$ has $\mathrm{C}_{10}^{k}$ possibilities.
For each possibility of $A$, since $A \cup B=\{1,2, \cdots, 10\}$, we know that $C_{A \cup B} A \subseteq B$.
Thus, $B$ already has $10-k$ elements.
To ensure $|A|=|B|$, it is necessary to select $2 k-10$ elements from $A$ to belong to $B$, hence, $B$ has $\mathrm{C}_{k}^{2 k-10}$ possibilities. Therefore, the ordered set pair $(A, B)$ has $\mathrm{C}_{10}^{k} \mathrm{C}_{k}^{2 k-10}$ possibilities.
Noting that $k=5,6, \cdots, 10$, the total number of all possible ordered set pairs $(A, B)$ is
$$
\begin{array}{l}
\sum_{k=5}^{10} \mathrm{C}_{10}^{k} \mathrm{C}_{k}^{2 k-10}=\mathrm{C}_{10}^{5} \mathrm{C}_{5}^{0}+\mathrm{C}_{10}^{6} \mathrm{C}_{6}^{2}+\cdots+\mathrm{C}_{10}^{10} \mathrm{C}_{10}^{10} \\
=8953
\end{array}
$$
|
8953
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x, y, z \in \mathbf{R}_{+}$. Then the minimum value of $\frac{\left(x^{2}+y^{2}\right)^{3}+z^{6}}{2 x^{3} y^{3}+y^{3} z^{3}+z^{3} x^{3}}$ is $\qquad$ .
|
3.2.
Let $x=y=1, z=\sqrt[3]{2}$, we get
$$
\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)+\left(x^{6}+y^{6}+x^{4} y^{2}+x^{2} y^{4}\right)+z^{6} \\
\geqslant 4 x^{3} y^{3}+\left(x^{3}+y^{3}\right)^{2}+\left(z^{3}\right)^{2} \\
\geqslant 4 x^{3} y^{3}+2\left(x^{3}+y^{3}\right) z^{3} \\
=4 x^{3} y^{3}+2 x^{3} z^{3}+2 z^{3} y^{3} .
\end{array}
$$
Therefore, the minimum value sought is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Fill $1,2, \cdots, n^{2}$ into an $n \times n$ chessboard, with each cell containing one number, and each row forming an arithmetic sequence with a common difference of 1. If any two of the $n$ numbers on the chessboard are neither in the same row nor in the same column, then the sum of these $n$ numbers is called a “basic sum”. If a basic sum is randomly selected, the probability that this basic sum equals the arithmetic mean of all basic sums is
|
5.1.
The numbers in the $k$-th row of the number table are
$$
(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 {. }
$$
Since these numbers are from different columns, $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$.
Therefore, the sum of the $n$ numbers taken is
$$
\begin{array}{l}
\sum_{k=1}^{n}\left[(k-1) n+a_{k}\right]=n \sum_{k=1}^{n}(k-1)+\sum_{k=1}^{n} a_{k} \\
=n \frac{n(n-1)}{2}+\frac{n(n+1)}{2}=\frac{n\left(n^{2}+1\right)}{2} .
\end{array}
$$
Thus, any basic sum is equal to the arithmetic mean of all basic sums.
Therefore, the required probability is 1.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. There are 9 students participating in a math competition in the same classroom, with seats arranged in 3 rows and 3 columns, represented by a $3 \times 3$ grid, where each cell represents a seat. To prevent cheating, three types of exams, $A$, $B$, and $C$, are used, and it is required that any two adjacent seats (cells sharing a common edge) receive different types of exams. The number of ways to distribute the exams that meet the conditions is $\qquad$ kinds.
|
8. 246.
Let $a_{ij}$ denote the square at the $i$-th row and $j$-th column. First, consider the number of ways to distribute type $A$ papers at $a_{22}$.
$$
\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}
$$
Consider the types of papers on the squares in $M$. There are several scenarios:
(1) All are type $B$ papers. In this case, there is a unique way to distribute the papers in $M$, and each square in $N$ has 2 ways to distribute the papers, resulting in $2^4 = 16$ ways.
(2) All are type $C$ papers. Similarly to (1), there are $2^4 = 16$ ways.
(3) One square is a type $B$ paper, and the other three are type $C$ papers (as shown in Figure 2). In this case, there are 4 ways to choose one square in $M$ to place the type $B$ paper, and the other squares in $M$ have a unique distribution method. Each square in $N$ has $1 \times 1 \times 2 \times 2 = 4$ ways to distribute the papers, resulting in $4 \times 4 = 16$ ways.
(4) One square is a type $C$ paper, and the other three are type $B$ papers. Similarly to (3), there are $4 \times 4 = 16$ ways.
(5) Two squares are type $B$ papers, and the other two are type $C$ papers. In this case, if the two squares with type $B$ papers are in the same row (or column) (as shown in Figure 3(a)), there are 2 ways to choose the squares in $M$ to place the type $B$ papers, and the other squares in $M$ have a unique distribution method. Each square in $N$ has a unique distribution method, resulting in 2 ways. If the two squares with type $B$ papers are in different rows and columns (as shown in Figure 3(b)), there are 4 ways to choose the squares in $M$ to place the type $B$ papers, and the other squares in $M$ have a unique distribution method. Each square in $N$ has $2 \times 2 \times 1 \times 1 = 4$ ways to distribute the papers, resulting in $4 \times 4 = 16$ ways. Thus, in this scenario, there are $2 + 16 = 18$ ways.
By symmetry, the total number of ways to distribute the papers that meet the requirements is $3(4 \times 16 + 18) = 246$ ways.
|
246
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Find the smallest positive integer $k$, such that $625^{k} \equiv 1(\bmod 343)$
|
$$
\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(\bmod 7^{3}\right) .
\end{array}
$$
Thus, $7 \mid\left(2^{k}-1\right)$, i.e., $2^{k} \equiv 1(\bmod 7)$.
Given $2^{1} \equiv 2(\bmod 7), 2^{2} \equiv 4(\bmod 7), 2^{3} \equiv$ $1(\bmod 7)$, the order of 2 modulo 7 is 3, hence $3 \mid k$.
Let $k=3 s$. Substituting into equation (1), we get
$$
\begin{array}{l}
\mathrm{C}_{38}^{2}(89 \times 7)^{2} \times 2^{34-2}+\mathrm{C}_{35}^{1}(89 \times 7) \times 2^{34-1}+(7+1)^{5}-1 \\
\equiv 0\left(\bmod 7^{3}\right) .
\end{array}
$$
Then $\mathrm{C}_{3}^{2}(89 \times 7)^{2} \times 2^{3-2}+\mathrm{C}_{3}^{1}(89 \times 7) \times 2^{3-1}+$ $\mathrm{C}_{s}^{2} 7^{2}+\mathrm{C}_{s}^{1} 7 \equiv 0\left(\bmod 7^{3}\right)$.
Thus, $\mathrm{C}_{38}^{2} 89^{2} \times 7 \times 2^{3 s-2}+\mathrm{C}_{38}^{1} 89 \times 2^{34-1}+$ $\mathrm{C}_{3}^{2} 7^{2}+\mathrm{C}_{3}^{1}=0\left(\bmod 7^{2}\right)$.
Therefore, $3 s \times 89 \times 2^{3 s-1}+s \equiv 0(\bmod 7)$.
Also, $2^{3-1} \equiv 2^{3-3} \times 2^{2} \equiv\left(2^{3}\right)^{s-1} \times 2^{2} \equiv 2^{2}(\bmod 7)$, so $3 s \times 40 \times 2^{2}+s \equiv 0(\bmod 7)$, i.e., $481 s \equiv 0(\bmod 7)$.
Since $(481,7)=1$, it follows that $7 \mid s$.
Let $s=7 t$. Substituting into equation (2), we get
$$
\begin{array}{l}
\frac{21 t(21 t-1)}{2} \times 89^{2} \times 7 \times 2^{21 t-2}+21 t \times 89 \times 2^{21 t-1}+ \\
\frac{7 t(7 t-1)}{2} \times 7+7 t=0\left(\bmod 7^{2}\right) . \\
\text { Then } \frac{21 t(21 t-1)}{2} \times 89^{2} \times 2^{21 t-2}+3 t \times 89 \times 2^{21 t-1}+ \\
\frac{7 t(7 t-1)}{2}+t=0(\bmod 7) .
\end{array}
$$
Therefore, $3 t \times 89 \times 2^{21 t-1}+t \equiv 0(\bmod 7)$.
Also, $2^{21 t-1} \equiv 2^{21 t-3} \times 2^{2} \equiv\left(2^{3}\right)^{7 t-1} \times 2^{2}$ $=2^{2}(\bmod 7)$,
so $3 t \times 40 \times 2^{2}+t \equiv 0(\bmod 7)$, i.e., $481 t \equiv 0(\bmod 7)$.
Since $(481,7)=1$, it follows that $7 \mid t$.
Let $t=7 r$. Then $k=3 s=21 t=147 r \geqslant 147$. Therefore, the minimum value of $k$ is 147.
|
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 wrong answer earns 0 points, with no other point values; each problem-solving question is worth 20 points, and the step scores can only be $0, 5, 10, 15, 20$ points, with no other point values. How many different possible scores are there?
|
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 consider half of the scores. Let the number of different scores that can be obtained between 0 and 74 be $a$. Since 75 can be obtained, the total number of different scores that can be obtained is $2a+1$. Therefore, we only need to discuss up to $k=14$.
(1) When $k=0$, the score is 0;
(2) When $k=2,4,6,8,10$, the scores are all even numbers between $10 \sim 14, 20 \sim 28, 30 \sim 42, 40 \sim 56, 50 \sim 70$ respectively;
(3) When $k=12,14$, the scores are all even numbers between $60 \sim 80, 70 \sim 90$ respectively;
(4) When $k=1$, the scores are all odd numbers between $5 \sim 7$;
(5) When $k=3,5,7,9$, the scores are all odd numbers between $15 \sim 21, 25 \sim 35, 35 \sim 49, 45 \sim 63$ respectively;
(6) When $k=11,13$, the scores are all odd numbers between $55 \sim 75, 65 \sim 85$ respectively.
In summary, the odd numbers that can be obtained between $0 \sim 74$ are 31, and the even numbers are 32. Therefore, $a=63$.
Thus, $2a+1=127$.
|
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$. Then, decompose 5 in the following ways:
$$
\begin{array}{l}
5=11111=1112=1121=1211 \\
=2111=122=221=212,
\end{array}
$$
There are 8 methods in total.
Solution 2 (Recursive Method) Consider the number of ways to enter the $n$-th grid, denoted as $F_{n}$. It is easy to see that $F_{1}=1, F_{2}=1$ (Note: a person can only enter the first grid from outside the grid). For $n \geqslant 3$, the number of ways to enter the $n$-th grid, $F_{n}$, can be divided into two categories: taking one step from the $(n-1)$-th grid; taking two steps from the $(n-2)$-th grid. Therefore,
$$
F_{n}=F_{n-1}+F_{n-2} \text {, }
$$
This is a Fibonacci sequence. Extracting $F_{6}=8$ gives the answer to this problem.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 The condition for three line segments to form a triangle is: the sum of the lengths of any two line segments is greater than the length of the third line segment. There is a wire of length $144 \mathrm{~cm}$, which is to be cut into $n$ $(n>2)$ small segments, with each segment being no less than $1 \mathrm{~cm}$ in length. If any three of these segments cannot form a triangle, then the maximum value of $n$ is $\qquad$
|
Since the necessary and sufficient condition for forming a triangle is that the sum of any two sides is greater than the third side, the condition for not forming a triangle is that the sum of any two sides is less than or equal to the largest side. The shortest piece of wire is 1, so we can place 2 ones, and the third segment is 2 (to maximize $n$, the remaining wire should be as long as possible, so each segment is always the sum of the previous two segments), in sequence:
$$
1,1,2,3,5,8,13,21,34,55 \text {, }
$$
The sum of the above numbers is 143, which is 1 less than 144. Therefore, the last segment can be 56, at which point $n$ reaches its maximum of 10.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. There are 25 cards numbered $1, 3, \cdots, 49$. If a card with number $a$ is drawn, the next card to be drawn is the card with the largest odd divisor of $99-a$. This process is repeated until no more cards can be drawn. Xiao Hua drew the card numbered 1. When the operation ends, how many cards are left?
(A) 9
(B) 10
(C) 16
(D) 17
|
4. C.
The card numbers drawn are
$$
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}
$$
Then 2009 is in the 45th group.
|
45
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A certain linear function graph is parallel to the line $y=\frac{5}{4} x+\frac{95}{4}$, and intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively, and passes through the point $(-1,-25)$. Then on the line segment $AB$ (including $A$ and $B$), the number of points with both integer coordinates is $\qquad$ .
|
3.5.
Let the linear function be $y=\frac{5}{4} x+b$.
Given that it passes through the point $(-1,-25)$, we get $b=-\frac{95}{4}$.
Therefore, $y=\frac{5}{4} x-\frac{95}{4}$.
Thus, $A(19,0)$ and $B\left(0,-\frac{95}{4}\right)$.
From $y=\frac{5}{4} x-\frac{95}{4}(0 \leqslant x \leqslant 19)$, taking $x=3,7,11$,
15,19, $y$ is an integer, so there are 5 points that satisfy the condition.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the function $f(x)$ satisfies for all real numbers $x, y$,
$$
\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,
\[
\begin{array}{l}
|C A|=x_{0}+2,|D B|=\frac{8}{x_{0}}+4 . \\
\text { Therefore, } S=\frac{1}{2}\left(x_{0}+2\right)\left(\frac{8}{x_{0}}+4\right) \\
=2\left(x_{0}+\frac{4}{x_{0}}\right)+8 \geqslant 16 .
\end{array}
\]
Thus, the minimum value of the area of quadrilateral \( A B C D \) is 16.
|
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, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-x^{2}-3 y^{2}=8 x^{2}-3 .
\end{array}
$$
Therefore, $-3 \leqslant t \leqslant-\frac{1}{3}$.
Hence, $-3 \times\left(-\frac{1}{3}\right)=1$.
|
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 $B M \cdot B A=B D^{2} \Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2}$
$\Rightarrow B M=\frac{9}{4} \Rightarrow A M=\frac{27}{4}$.
Also, $\frac{M N}{B C}=\frac{A M}{A B}=\frac{\frac{27}{4}}{9}=\frac{3}{4}$, thus $M N=6$.
|
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 integer $m$ such that for any positive integer $n$, $b_{n} \geqslant b_{m}$? If it exists, find the value of $m$; if not, explain the reason.
|
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., $a_{n}=a a_{n-1}$.
Since $a_{1}=a \neq 0$, thus, $\left\{a_{n}\right\}$ is a geometric sequence with the first term and common ratio both being $a$.
Therefore, $a_{n}=a^{n}, b_{n}=a_{n} \lg \left|a_{n}\right|=n a^{n} \lg |a|$.
Also, $a=-\frac{\sqrt{7}}{3} \in(-1,0)$, then $\lg |a|<0$.
It can be seen that if there exists a positive integer $m$ that satisfies the condition, then $m$ must be even.
Notice that
$$
\begin{array}{l}
b_{2 k+2}-b_{2 k}=\left[(2 k+2) a^{2 k+2}-2 k a^{2 k}\right] \lg |a| \\
=2 a^{2 k}\left[(k+1) a^{2}-k\right] \lg |a| \\
=2 a^{2 k}\left[(k+1) \times \frac{7}{9}-k\right] \lg |a| \\
=\frac{2 a^{2 k}}{9}(7-2 k) \lg |a|\left(k \in \mathbf{N}_{+}\right) . \\
\text {Also, } \frac{a^{2}}{1-a^{2}}=\frac{7}{2}, \text { thus, }
\end{array}
$$
When $2 k>7$, $b_{2 k+2}>b_{2 k}$, i.e.,
$$
b_{8}<b_{10}<b_{12}<\cdots \text {; }
$$
When $2 k<7$, $b_{2 k+2}<b_{2 k}$, i.e.,
$$
b_{8}<b_{6}<b_{4}<b_{2} \text {. }
$$
Therefore, there exists a positive integer $m=8$, such that for any positive integer $n$, $b_{n} \geqslant b_{m}$.
|
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 both coordinates as integers are called integer points).
Question: What is the maximum number of such rectangles that can be drawn? Explain your reasoning.
|
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 coincide with the coordinate axes.
We will prove: There are at most 81 good rectangles.
In fact, draw lines $l_{1}$ and $l_{2}$ through $O$ parallel to the coordinate axes, and let $l_{1} \backslash\{O\}$ contain $m$ points chosen from the 100 integer points, and $l_{2} \backslash\{O\}$ contain $n$ points chosen from the 100 integer points. Let $P$ be one of the 100 integer points chosen, and not on $l_{1}$ and $l_{2}$. Then, at most one good rectangle can have $P$ as one of its vertices. There are at most $99-m-n$ such points, and each good rectangle must have one vertex as such a point. Therefore,
(i) If $m+n \geqslant 18$, then the number of good rectangles is at most $99-m-n \leqslant 81$;
(ii) If $m+n \leqslant 18$, consider the point pairs $(R, Q)$, where $R \in l_{1} \backslash\{O\}$ and $Q \in l_{2} \backslash\{O\}$. Each pair $(R, Q)$ can form at most one good rectangle, so the number of good rectangles is less than or equal to
$$
m n \leqslant m(18-m) \leqslant 9 \times 9=81 \text { . }
$$
In summary, for any point $O$ among the 100 integer points chosen, the number of good rectangles with $O$ as one of its vertices is at most 81. Therefore, the number of rectangles that meet the conditions is less than or equal to $\frac{81 \times 100}{4}=2025$ (here we divide by 4 because each rectangle has 4 vertices).
Below is an example of 2025 such rectangles.
Consider the point set
$$
A=\{(x, y) \mid 1 \leqslant x \leqslant 10,1 \leqslant y \leqslant 10, x y \in \mathbf{N}\} .
$$
By choosing 100 points from the point set $A$, we can exactly draw 2025 rectangles that meet the problem's conditions.
Therefore, the maximum number of such rectangles that can be drawn is 2025.
|
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 student.
Thus, the conclusion holds.
Next, we provide an example to show: There exists a distribution method such that any two students have at most three colors of pencils.
One student has 4 pencils of the second color, one student has 4 pencils of the third color, one student has 4 pencils of the fourth color, one student has 2 pencils of the first color and 2 pencils of the second color, one student has 2 pencils of the first color and 2 pencils of the third color, one student has 2 pencils of the first color and 2 pencils of the fourth color.
|
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 other is wearing a red hat. Find the minimum total number of hat changes in a day.
|
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 meet. Therefore, at least 998 dwarfs must have changed the color of their hats, and the total number of hat color changes is greater than or equal to 998.
Below is an explanation that 998 times is achievable.
Let the dwarfs be $1, 2, \cdots, 1000$. Initially, 1 wears a blue hat, and $2, 3, \cdots, 1000$ wear red hats. After 1 meets $2, 3, \cdots, 1000$, 2 changes the color of his hat to blue and then meets 3, $4, \cdots, 1000$, $\cdots \cdots$ 999 changes the color of his hat to blue and then meets 1000, for a total of 998 hat color changes.
|
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 rational number.
(1991, Shanghai Junior High School Mathematics Competition)
|
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, $\frac{O C}{C D}=\frac{k}{2 k-1}$.
(1) $\times$ (2) gives $\frac{O C}{B C}=\sqrt{\frac{k}{2 k-1}}$ is a rational number, and $(k, 2 k-1)=1$.
Thus, both $k$ and $2 k-1$ are perfect squares.
When $k=4,9,16$, $2 k-1=7,17,31$;
When $k=25$, $2 k-1=49$.
Therefore, the smallest positive integer $k$ that makes $\frac{O C}{B C}$ a rational number is $k=25$.
|
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 $\mathrm{B}$, there is a monochromatic triangle (let $\triangle x_{1} x_{2} x_{3}$ be red). The complete graph $K_{6}$ formed by the other six points, if all edges are red, the conclusion is evident.
Assume the conclusion does not hold. Suppose there is an edge $v_{4} v_{5}$ that is blue. By the same argument as in problem $\mathrm{B}$ (Figure 1), we can assume there is another red $\triangle x_{1} x_{2} x_{4}$.
The complete graph $K_{5}$ formed by the other five points $x_{5} 、 x_{6} 、 x_{7} 、 x_{8} 、 x_{9}$, if all edges are red, the conclusion is evident.
Suppose the edge $x_{8} x_{9}$ is blue (as in Figure 5).
If the conclusion does not hold, $x_{4} x_{8} 、 x_{4} x_{9}$ are not both blue.
Assume $x_{4} x_{8}$ is red (Figure 6). Then $x_{1} x_{8} 、 x_{2} x_{8}$ are blue, $x_{9} x_{1} 、 x_{9} x_{2}$ are red, $x_{9} x_{4} 、 x_{9} x_{3}$ are blue, $x_{3} x_{4}$ is red, $\triangle x_{2} x_{3} x_{4}$ and $\triangle x_{2} x_{1} x_{9}$ are two red triangles with exactly one common vertex.
Thus, the conclusion holds.
|
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" of the triangle; if all three sides of a triangle are middle sides (not necessarily the middle sides of this triangle), then the triangle is called a "middle side triangle" in the set $D$. A line $l$ that does not pass through any point $P_{i}(i=1,2, \cdots, n)$ divides the set $D$ into two subsets $D_{1}$ and $D_{2}$. If, regardless of how these $n$ points are distributed and how $l$ is chosen, there always exists a subset $D_{k}(k \in\{1,2\})$ such that $D_{k}$ contains a middle side triangle. Find the minimum value of $n$.
|
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 Ramsey's theorem, there must exist a monochromatic triangle. Since every triangle has a middle edge, this monochromatic triangle must be red. Therefore, there exists a middle edge triangle in $D_{1}$.
Below is the proof: When $n \leqslant 10$, there exists a point set $D$ and a line $l$, such that neither $D_{1}$ nor $D_{2}$ contains a middle edge triangle.
If $n=10$, consider the two subsets $D_{1}$ and $D_{2}$ on either side of the line $l$, each containing five points, and the distribution is the same.
Assume the five points in $D_{1}$ are $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$, and they are arranged in a counterclockwise order on the circumference of a circle, with
\[
\begin{array}{l}
\overparen{P_{1} P_{2}}=\frac{\pi}{10}, \overparen{P_{2} P_{3}}=\frac{3 \pi}{5}, \overparen{P_{3} P_{4}}=\frac{3 \pi}{10}, \\
\overparen{P_{4} P_{5}}=\frac{3 \pi}{4}, \overparen{P_{5} P_{1}}=\frac{\pi}{4} .
\end{array}
\]
Then the distances between $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ are all different, and $P_{2} P_{3}, P_{3} P_{1}, P_{1} P_{5}, P_{5} P_{4}, P_{4} P_{2}$ are middle edges, but there is no middle edge triangle.
If $n<10$, for the cases where $D_{1}$ and $D_{2}$ have fewer than five points, simply remove some points from the previous example, and there will still be no middle edge triangle.
In summary, the minimum value of $n$ is 11.
|
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 Mathematics League)
|
(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
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D} \\
=S_{\triangle A B C}+S_{\triangle A D C} \\
=\frac{1}{2}\left(h_{1} a+h_{2} b\right) \\
\leqslant \frac{1}{2} l(a+b) .
\end{array}
$$
Equality holds if and only if $h_{1}=h_{2}=l$, i.e., $A C \perp A B$ and $A C \perp A D$.
From the given, we have $32 \leqslant \frac{1}{2} l(a+b)$.
Also, by the problem's condition, $a+b=16-l$, so we get
$$
64 \leqslant l(16-l)=64-(l-8)^{2} \leqslant 64 \text {. }
$$
Thus, it must be that $l=8, a+b=8$, and at this time, $A C \perp A B$ and $A C \perp C D$.
Therefore, such quadrilaterals are as follows:
$$
\begin{array}{l}
a=1, b=7, l=8 ; a=2, b=6, l=8 ; \\
a=3, b=5, l=8 ; a=b=4, l=8 .
\end{array}
$$
These are all trapezoids or parallelograms with $A C$ as the height.
(2) From $A B=a, C D=8-a$, we get
$$
B C^{2}=8^{2}+a^{2}, A D^{2}=8^{2}+(8-a)^{2} \text {. }
$$
Therefore, the sum of the squares of the side lengths of such quadrilaterals is
$$
\begin{array}{l}
2 a^{2}+2(8-a)^{2}+2 \times 8^{2} \\
=4(a-4)^{2}+192 .
\end{array}
$$
Hence, when $a=b=4$, the above sum of squares is minimized, and the minimum value is 192.
|
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 rules to: the passenger specifies a time (i.e., platform number), the machine first arrives at the specified platform, then stops every 4 stations, at the 5th station. If the platform number is a positive integer power of 2, it moves back 2 stations to stop (e.g., $17 \rightarrow 22 \rightarrow 27 \rightarrow 32 \rightarrow 30 \rightarrow 35 \rightarrow \cdots$); if it stops at the first platform, it stops working. Try to answer:
(1) Can this machine get lost in the time track and fail to return to the present (i.e., not stop at the first platform)?
(2) If it can eventually return to the present, what is the maximum number of platforms the machine can stop at?
|
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) $5 k+1(k=4,5, \cdots, 51)$
$\xrightarrow{51-k \text { times }} 256 \rightarrow 254=5 \times 50+4$;
(3) $5 k+1(k=52,53, \cdots, 401)$
$\xrightarrow{401-k \text { times }} 2006 \rightarrow 2$.
(ii) (1) $2 \rightarrow 2009 \rightarrow 5$;
$$
\begin{array}{l}
\text { (2) } 5 k+2(k=1,2, \cdots, 6) \\
\xrightarrow{6-k \text { times }} 32 \rightarrow 30=5 \times 6 \text {; }
\end{array}
$$
$$
\begin{array}{l}
\text { (3) } 5 k+2(k=7,8, \cdots, 102) \\
\xrightarrow{102-k \text { times }} 512 \rightarrow 510=5 \times 102 \text {; } \\
\text { (4) } 5 k+2(k=103,104, \cdots, 401) \\
\xrightarrow{401-k \text { times }} 2007 \rightarrow 3 \text {. } \\
\end{array}
$$
$$
\begin{array}{l}
\text { (iii) (1) } 3 \rightarrow 8 \rightarrow 6=5+1 \text {; } \\
\text { (2) } 5 k+3(k=1,2, \cdots, 25) \\
\xrightarrow{25-k \text { times }} 128 \rightarrow 126=5 \times 25+1 \text {; } \\
\text { (3) } 5 k+3(k=26,27, \cdots, 401) \\
\xrightarrow{401-k \text { times }} 2008 \rightarrow 4 \text {. } \\
\end{array}
$$
(iv) (1) $4 \rightarrow 2 \rightarrow 2009 \rightarrow 5$;
(2) $5 k+4(k=1,2, \cdots, 12)$
$\xrightarrow{12-k \text { times }} 64 \rightarrow 62=5 \times 12+2$;
(3) $5 k+4(k=13,14, \cdots, 204)$
$\xrightarrow{204-k \text { times }} 1024 \rightarrow 1022=5 \times 204+2$;
(4) $5 k+4(k=205,206, \cdots, 401)$
$\xrightarrow{401-k \text { times }} 2009 \rightarrow 5$.
(v) $5 k(k=1,2, \cdots, 401) \xrightarrow{401-k \text { times }} 2005 \rightarrow 1$.
Let the passenger's designated station number be $a(1<a \leqslant 2009)$.
From (i) to (v), we know that regardless of the value of $a$, the machine can always return to the present.
(2) Let the track from station 2009 to station 1 be denoted as $A$.
Let the number of times the machine passes through $A$ be $s$, and the total number of stations it stops at be $t$, where it stops at stations with numbers that are powers of 2 $v$ times, and at other stations $u$ times. Then
$2009 s+1=a+5(u-1)-2 v=a+5 t-5-7 v$.
Thus, $t=\frac{2009 s+7 v-a+6}{5}$
We now prove that the machine will not stop at the same station twice.
Otherwise, the sequence of stations the machine stops at would form a loop. The result would be that it would either never stop at the first station (contradicting (1)), or it would have already stopped at the first station in the first loop, thus stopping and not forming a loop, which is a contradiction.
Therefore, $v \leqslant 10$.
Consider the station number the machine stops at first after each pass through $A$, which is contained in $\{1,2,3,4,5\}$.
Notice
$3 \rightarrow 8 \rightarrow 6 \xrightarrow{2 \text { times }} 16 \rightarrow 14 \xrightarrow{10 \text { times }} 64 \rightarrow 62$
$\xrightarrow{90 \text { times }} 512 \rightarrow 510 \xrightarrow{299 \text { times }} 2005 \rightarrow 1$,
$$
4 \rightarrow 2 \rightarrow 2009 \rightarrow 5 \xrightarrow{400 \text { times }} 2005 \rightarrow 1 \text {. }
$$
Thus, the machine either stops at station 3, or at stations 4, 2, 5, but it cannot stop at both.
Therefore, $s \leqslant 2$.
If $s=1$, then $t \leqslant \frac{2009+70-0+6}{5}=417$.
If $s=2$, then the second loop is
$3 \xrightarrow{407 \text { times }} 1$, or $2 \xrightarrow{403 \text { times }} 1$, or $4 \xrightarrow{404 \text { times }} 1$.
We now use (1) (i) to (v) to trace back the first loop from 3 or 2, 4:
$$
\begin{array}{l}
254 \leftarrow 256 \stackrel{26}{\leftarrow} 126 \leftarrow 128 \stackrel{23}{\longleftarrow} 13 \text {, } \\
2 \leftarrow 2006 \stackrel{349}{\leftarrow} 261 \text {, } \\
4 \leftarrow 2008 \stackrel{375 \text { times }}{\longleftarrow} 133 \text {. } \\
\text { In summary, if } a=13 \text {, then } v=7, t=812 \text {; } \\
\text { if } a=261 \text {, then } v=1, t=754 \text {; } \\
\text { if } a=133 \text {, then } v=2, t=781 \text {. } \\
\end{array}
$$
Therefore, the machine can stop at a maximum of 812 stations.
|
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 owned by at least one student. Therefore, the conclusion holds.
Next, we provide an example to show that there exists a distribution method such that any two students have at most three colors of pencils.
Two students have 4 pencils of the second color, two students have 4 pencils of the third color, two students have 4 pencils of the fourth color, one student has 4 pencils of the first color, one student has 2 pencils of the first color and 2 pencils of the second color, one student has 2 pencils of the first color and 2 pencils of the third color, and one student has 2 pencils of the first color and 2 pencils of the fourth color.
|
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 is known that for any such number grid, at least $d$ cells can be colored according to the above rule. Find the maximum value of $d$.
|
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 conclusion holds.
Suppose the numbers in the leftmost 17 cells are $a_{1}$, $a_{2}, \cdots, a_{17}$. Then among
$$
0, a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{17}
$$
there exist two numbers such that their difference is divisible by 17, i.e.,
$$
17 \mid\left(a_{i}+a_{i+1}+\cdots+a_{j}\right) .
$$
Thus, the cells from the $i$-th to the $j$-th form a good rectangle A. Removing A from the grid, we get a new $1 \times (k-(j-i+1))$ grid. By the induction hypothesis, the new grid can be expressed as the union of pairwise disjoint good rectangles, except for at most 16 cells. If none of these good rectangles contain both the $(i-1)$-th and $(j+1)$-th cells of the original grid, then they are also good rectangles in the original grid; if one good rectangle contains both the $i$-th and $j$-th cells of the original grid, then its union with A forms a good rectangle in the original grid. In either case, the conclusion holds for $n=k$.
Next, we prove: at least 9744 cells in the grid can be colored.
Consider each column as a large cell, with the number being the sum of the numbers in that column. By the lemma, except for at most 16 columns, the grid can be expressed as the union of several good rectangles of height 100. For the remaining columns, by the lemma, all but at most 16 cells in each column can be colored. Thus, the entire grid has at most $16^{2}=256$ cells that are not colored.
Below is an example where at least 256 cells are not colored.
In a $16 \times 16$ subgrid B of the grid, each cell is filled with 1, and the rest of the cells are filled with 17, then no cell in B can be colored.
In fact, no good rectangle can contain such a cell.
Let A be any good rectangle. If A contains 1, let its intersection with B be an $a \times b$ rectangle, then
$$
a b \equiv 0(\bmod 17),
$$
which contradicts $a, b \leqslant 16$.
|
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 jewelry in both boxes is ( ) ten thousand yuan.
(A) 11
(B) 12
(C) 13
(D) 14
|
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 $\left\{\begin{array}{l}\frac{x-5}{n-1}+1=\frac{x}{n}, \\ \frac{y+5}{n+1}-1=\frac{y}{n}\end{array} \Rightarrow\left\{\begin{array}{l}x=-n^{2}+6 n, \\ y=-n^{2}+4 n .\end{array}\right.\right.$
Since $x>0, y>0$, we have $0<n<4$. Therefore, $n=2$ or 3.
Thus, $x+y=-2 n^{2}+10 n=12$.
|
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
$$
a_{1}^{2} + b_{1}^{2} + c_{1}^{2} = \frac{2011}{d^{2}}, \quad 1 + m = \frac{388}{d}.
$$
Thus, $d^{2} \mid 2011$, $d \mid 388$, and $d < 45$, $388 = 2^{2} \times 97$, and $d$ is odd.
Therefore, $d = 1$.
Hence, $m = 387 = 9 \times 43$, and $a_{1}$, $b_{1}$, and $c_{1}$ are all less than 45, and they are divisors of $m$, which are $1, 3, 9, 43$.
Upon inspection, $a_{1} = 43$, $b_{1} = c_{1} = 9$.
Therefore, $a + b + c = 61$.
|
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 get $x_{1}=0, x_{2}=1$,
$$
x_{3}=\frac{-1+\sqrt{5}}{2}, x_{4}=\frac{-1-\sqrt{5}}{2} \text {. }
$$
Noting that $\left|x_{4}\right|>1$, we discard it.
Therefore, the system of equations has three solutions
$$
(0,1),(1,0),\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right) \text {. }
$$
Since set $A$ has 3 elements, set $A$ has $2^{3}=8$ subsets.
|
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 $25\%$ each subsequent year.
Based on this, the company will recover all its investments by $\qquad$ year.
|
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
$$
\begin{array}{l}
A_{n}=\sum_{k=1}^{n} 1000(1-20 \%)^{k-1} \\
=\frac{1000\left[1-\left(\frac{4}{5}\right)^{n}\right]}{1-\frac{4}{5}}=5000\left[1-\left(\frac{4}{5}\right)^{n}\right] .
\end{array}
$$
Similarly, from (2) we know
$$
B_{n}=\frac{500\left[\left(\frac{5}{4}\right)^{n}-1\right]}{\frac{5}{4}-1}=2000\left[\left(\frac{5}{4}\right)^{n}-1\right] \text {. }
$$
To recover the entire investment, we should have $B_{n} \geqslant A_{n}$, which means
$$
\begin{array}{l}
2000\left[\left(\frac{5}{4}\right)^{n}-1\right] \geqslant 5000\left[1-\left(\frac{4}{5}\right)^{n}\right] \\
\Leftrightarrow\left[5\left(\frac{4}{5}\right)^{n}-2\right]\left[\left(\frac{4}{5}\right)^{n}-1\right] \geqslant 0 \\
\Leftrightarrow 5\left(\frac{4}{5}\right)^{n}-2 \leqslant 0 \\
\Leftrightarrow\left(\frac{4}{5}\right)^{n} \leqslant \frac{2}{5} \Leftrightarrow 0.8^{n} \leqslant 0.4 .
\end{array}
$$
Given $0.8^{5}=0.32768<0.4<0.4096=0.8^{4}$, we get $n \geqslant 5$.
Therefore, the company will recover the entire investment by 2013.
|
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 $a=4$, $c=8, n=408$.
Therefore, the maximum value of $\varphi(n)$ is
$$
\begin{array}{l}
\varphi(408)=\varphi\left(2^{3} \times 3 \times 17\right) \\
=408\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{17}\right)=128 .
\end{array}
$$
|
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
$$
\begin{array}{l}
b_{n+10} \equiv b_{n+6}+b_{n+9} \\
\equiv\left(b_{n+2}+b_{n+5}\right)+\left(b_{n+5}+b_{n+8}\right) \\
\equiv b_{n+2}+b_{n+8} \\
\equiv b_{n+2}+\left(b_{n+4}+b_{n+7}\right) \\
\equiv b_{n+2}+\left(b_{n}+b_{n+3}\right)+\left(b_{n+3}+b_{n+6}\right) \\
\equiv b_{n+2}+b_{n}+b_{n+6} \\
\equiv b_{n+2}+b_{n}+\left(b_{n+2}+b_{n+5}\right) \\
\equiv b_{n}+b_{n+5}(\bmod 2) .
\end{array}
$$
Then, $b_{n+15} \equiv b_{(n+5)+10} \equiv b_{n+5}+b_{n+10}$
$$
\begin{array}{l}
\equiv b_{n+5}+\left(b_{n}+b_{n+5}\right) \\
\equiv b_{n}(\bmod 2) .
\end{array}
$$
It is evident that $\left\{b_{n}\right\}$ is a periodic sequence with a period of 15, hence
$$
\begin{array}{l}
a_{2008} \equiv b_{2008} \equiv b_{15 \times 133+13} \equiv b_{13} \equiv 0(\bmod 2), \\
a_{2009} \equiv b_{2009} \equiv b_{15 \times 133+14} \equiv b_{14} \equiv 0(\bmod 2), \\
a_{2010} \equiv b_{2010} \equiv b_{15 \times 134} \equiv b_{15} \equiv 0(\bmod 2), \\
a_{2011} \equiv b_{2011} \equiv b_{15 \times 134+1} \equiv b_{1} \equiv 1(\bmod 2) .
\end{array}
$$
Since the square of an even number is divisible by 4, and the square of an odd number is congruent to 1 modulo 4, we have
$$
S=a_{2008}^{2}+a_{2008}^{2}+a_{2010}^{2}+a_{2011}^{2} \equiv 1(\bmod 4) \text {. }
$$
Therefore, $S$ is not divisible by 4.
|
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
$$
\begin{array}{l}
(\sqrt{3}+1)\left(x^{2}+y^{2}+z^{2}\right) \geqslant 2(x y+2 y z+z x) \\
\Rightarrow x^{2}+y^{2}+z^{2} \geqslant(\sqrt{3}-1)(x y+2 y z+z x) .
\end{array}
$$
Equality holds if and only if $\frac{(\sqrt{3}+1) x^{2}}{2}=(\sqrt{3}-1) y^{2}, 2 y^{2}=2 z^{2},(\sqrt{3}-1) z^{2}=\frac{(\sqrt{3}+1) x^{2}}{2}$, i.e., $y=z=\frac{(\sqrt{3}+1) x}{2}$.
$$
\begin{array}{l}
\text { Then }(x+y+z)^{2} \\
=x^{2}+y^{2}+z^{2}+2(x y+y z+z x) \\
\geqslant(\sqrt{3}-1)(x y+2 y z+z x)+2(x y+y z+z x) \\
\geqslant(\sqrt{3}+1) x y+2 \sqrt{3} y z+(\sqrt{3}+1) z x=1 .
\end{array}
$$
$$
\text { Thus, } x+y+z \geqslant 1 \text {. }
$$
Equality holds if and only if $y=z=\frac{(\sqrt{3}+1) x}{2}$ and $(\sqrt{3}+1) x y+2 \sqrt{3} y z+(\sqrt{3}+1) z x=1$, i.e., $y=z=\frac{\sqrt{3}-1}{2}$, $x=2-\sqrt{3}$.
Therefore, when $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$, $x+y+z$ attains its minimum value of 1.
(2) By the AM-GM inequality, we have
$$
\begin{array}{l}
\frac{y}{z}+\frac{z}{y} \geqslant 2, \\
\frac{z}{x}+\frac{(2+\sqrt{3}) x}{2 z} \geqslant \sqrt{3}+1, \\
\frac{y}{x}+\frac{(2+\sqrt{3}) x}{2 y} \geqslant \sqrt{3}+1 .
\end{array}
$$
Equality holds if and only if $y=z=\frac{(\sqrt{3}+1) x}{2}$.
Introducing positive parameters $u, v, w$, we have
$$
\begin{array}{l}
u x\left(\frac{y}{z}+\frac{z}{y}\right)+v y\left[\frac{(2+\sqrt{3}) x}{2 z}+\frac{z}{x}\right]+ \\
w z\left[\frac{(2+\sqrt{3}) x}{2 y}+\frac{y}{x}\right] \\
=\left[u+\frac{(2+\sqrt{3}) v}{2}\right] \frac{x y}{z}+(v+w) \frac{y z}{x}+ \\
{\left[u+\frac{(2+\sqrt{3}) w}{2}\right] \frac{z x}{y} .} \\
\text { Also, } u x\left(\frac{y}{z}+\frac{z}{y}\right)+v y\left[\frac{(2+\sqrt{3}) x}{2 z}+\frac{z}{x}\right]+ \\
w z\left[\frac{(2+\sqrt{3}) x}{2 y}+\frac{y}{x}\right] \\
\geqslant 2 u x+(\sqrt{3}+1) v y+(\sqrt{3}+1) w z \text {, } \\
\text { Hence, }\left[u+\frac{(2+\sqrt{3}) v}{2}\right] \frac{x y}{z}+(v+w) \frac{y z}{x}+ \\
{\left[u+\frac{(2+\sqrt{3}) w}{2}\right] \frac{z x}{y}} \\
\geqslant 2 u x+(\sqrt{3}+1) v y+(\sqrt{3}+1) w z . \\
\text { Let } u=\sqrt{3}+1, v=w=2 \text {. Substituting into the above inequality, we get } \\
(3+2 \sqrt{3}) \frac{x y}{z}+4 \times \frac{y z}{x}+(3+2 \sqrt{3}) \frac{z x}{y} \\
\geqslant 2(\sqrt{3}+1) x+2(\sqrt{3}+1) y+2(\sqrt{3}+1) z \text {, } \\
\text { Then } \frac{\sqrt{3} x y}{z}+\frac{(8-4 \sqrt{3}) y z}{x}+\frac{\sqrt{3} z x}{y} \\
\geqslant 2(\sqrt{3}-1)(x+y+z) \geqslant 2(\sqrt{3}-1) \text {. } \\
\end{array}
$$
Equality holds if and only if $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$.
Therefore, when $x=2-\sqrt{3}, y=z=\frac{\sqrt{3}-1}{2}$, $\frac{\sqrt{3} x y}{z}+$ $\frac{(8-4 \sqrt{3}) y z}{x}+\frac{\sqrt{3} z x}{y}$ attains its minimum value of $2(\sqrt{3}-1)$.
|
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 $n$ is $\qquad$
(2009, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
|
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 is
$$
\begin{array}{l}
=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+\cdots+\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right) \\
=1-\frac{1}{\sqrt{n+1}} .
\end{array}
$$
From the given information,
$$
\begin{array}{l}
\frac{19}{20}\frac{1}{\sqrt{n+1}}>\frac{1}{21} \\
\Rightarrow 20<\sqrt{n+1}<21 \\
\Rightarrow 20^{2}-1<n<21^{2}-1 .
\end{array}
$$
Since $n$ is a positive integer,
$$
n_{\max }=21^{2}-2, n_{\text {min }}=20^{2} \text {. }
$$
Therefore, $21^{2}-2-20^{2}=39$ is the answer.
|
39
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.