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2. Let $S$ be a subset of $\{1,2, \cdots, 9\}$ such that the sum of any two distinct elements of $S$ is unique. How many elements can $S$ have at most?
(2002, Canadian Mathematical Olympiad)
|
(When $S=\{1,2,3,5,8\}$, $S$ meets the requirements of the problem. If $T \subseteq\{1,2, \cdots, 9\},|T| \geqslant 6$, then since the sum of any two different numbers in $T$ is between 3 and 17, at most 15 different sum numbers can be formed. And choosing any two numbers from $T$, there are at least $\mathrm{C}_{6}^{2}=15$ ways. If $T$ meets the condition, then the sum numbers must include both 3 and 17, meaning that $1,2,8,9$ must all appear in $T$, but in this case $1+9=2+8$, making $T$ not meet the requirement. Therefore, $S$ can have at most 5 elements.)
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x=\frac{1}{\sqrt{2}-1}, a$ be the fractional part of $x$, and $b$ be the fractional part of $-x$. Then $a^{3}+b^{3}+3 a b=$ $\qquad$ .
|
ニ、1. 1 .
Since $x=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$, and $2<\sqrt{2}+1<3$, therefore,
$$
\begin{array}{l}
a=x-2=\sqrt{2}-1 . \\
\text { Also }-x=-\sqrt{2}-1, \text { and }-3<-\sqrt{2}-1<-2 \text {, so, } \\
b=-x-(-3)=2-\sqrt{2} .
\end{array}
$$
Then $a+b=1$.
Thus $a^{3}+b^{3}+3 a b=(a+b)\left(a^{2}-a b+b^{2}\right)+3 a b$ $=a^{2}-a b+b^{2}+3 a b=(a+b)^{2}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given a right trapezoid $A B C D$ with side lengths $A B=2, B C=C D=10, A D=6$, a circle is drawn through points $B$ and $D$, intersecting the extension of $B A$ at point $E$ and the extension of $C B$ at point $F$. Then the value of $B E-B F$ is $\qquad$
|
3.4 .
As shown in Figure 3, extend $C D$ to intersect $\odot O$ at point $G$. Let the midpoints of $B E$ and $D G$ be $M$ and $N$, respectively. It is easy to see that $A M = D N$. Since $B C = C D = 10$, by the secant theorem, it is easy to prove that
$$
\begin{aligned}
B F = & D G. \text{ Therefore, } \\
& B E - B F = B E - D G = 2(B M - D N) \\
& = 2(B M - A M) = 2 A B = 4 .
\end{aligned}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let positive integers $a, b, k$ satisfy $\frac{a^{2}+b^{2}}{a b-1}=k$. Prove: $k=5$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
For every $k$ that satisfies the above equation, let $a_{0}, b_{0}$ satisfy $\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=k$, and $a_{0}+b_{0}$ is the smallest pair. Without loss of generality, assume $a_{0} \geqslant b_{0}$.
(1) If $a_{0}=b_{0}$, then
$$
k=\frac{a_{0}^{2}+b_{0}^{2}}{a_{0} b_{0}-1}=2+\frac{2}{a_{0}^{2}-1}.
$$
Since $a_{0}^{2}-1 \neq 1,2$, $k$ is not an integer.
(2) If $a_{0}>b_{0}$, the quadratic equation in $a$
$$
a^{2}-b_{0} k a+b_{0}^{2}+k=0
$$
has one root $a_{0}$. By Vieta's formulas, the other root $a^{\prime}$ is a positive integer.
Thus, by the minimality of $a_{0}+b_{0}$, we have $a^{\prime} \geqslant a_{0}$.
Therefore, $a_{0}^{2} \leqslant a_{0} a^{\prime}=b_{0}^{2}+k$.
Hence, $k \geqslant a_{0}^{2}-b_{0}^{2} \geqslant a_{0}^{2}-\left(a_{0}-1\right)^{2}=2 a_{0}-1$.
When $b_{0}=1$,
$$
k=\frac{a_{0}^{2}+1}{a_{0}-1}=a_{0}+1+\frac{2}{a_{0}-1}.
$$
Thus, $a_{0}-1=1$ or 2. Therefore, $k=5$.
When $b_{0} \geqslant 2$, we have
$$
\begin{array}{l}
2 a_{0}^{2}>a_{0}^{2}+b_{0}^{2}=k\left(a_{0} b_{0}-1\right) \\
\geqslant\left(2 a_{0}-1\right)^{2}=4 a_{0}^{2}-4 a_{0}+1 .
\end{array}
$$
Thus, $2\left(a_{0}-1\right)^{2}<1$, which is impossible.
In conclusion, $k=5$.
|
5
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $100a+64$ and $201a+64$ are both four-digit numbers, and both are perfect squares, then the value of the integer $a$ is
|
4.17.
Let $100 a+64=m^{2}, 201 a+64=n^{2}$, then
$$
32 \leqslant m, n<100 \text {. }
$$
Subtracting the two equations gives
$$
101 a=n^{2}-m^{2}=(n+m)(n-m) \text {. }
$$
Since 101 is a prime number, and $-101<n-m<101, 0<n$ $+m<200$, so, $n+m=101$.
Thus, $a=n-m=2 n-101$.
Substituting $a=2 n-101$ into $201 a+64=n^{2}$, and simplifying gives
$$
n^{2}-402 n+20237=0 \text {. }
$$
Solving for $n$ yields $n=59$ or $n=343$ (discard $n=343$).
Therefore, $a=2 n-101=17$.
|
17
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $a$ be a positive integer, the quadratic function $y=x^{2}+(a+17) x+38-a$, and the reciprocal function $y=\frac{56}{x}$. If the intersection points of the two functions are all integer points (points with both coordinates as integers), find the value of $a$.
---
The above text has been translated into English, preserving the original text's line breaks and format.
|
Three, by eliminating $y$ from the two equations, we get
$$
x^{2}+(a+17) x+38-a=\frac{56}{x} \text {, }
$$
which simplifies to $x^{3}+(a+17) x^{2}+(38-a) x-56=0$.
Factoring, we get
$$
(x-1)\left[x^{2}+(a+18) x+56\right]=0 \text {. }
$$
Clearly, $x_{1}=1$ is a root of equation (1), and $(1,56)$ is one of the intersection points of the two function graphs.
Since $a$ is a positive integer, the discriminant of the equation
$$
x^{2}+(a+18) x+56=0
$$
is $\Delta=(a+18)^{2}-224>0$, so it must have two distinct real roots.
Since the intersection points of the two function graphs are all integer points, the roots of equation (2) must all be integers. Therefore, its discriminant $\Delta=(a+18)^{2}-224$ should be a perfect square.
Let $(a+18)^{2}-224=k^{2}$ (where $k$ is a non-negative integer), then $(a+18)^{2}-k^{2}=224$,
which simplifies to $(a+18+k)(a+18-k)=224$.
Following the solution method of the third problem in the A paper, we get
$$
\left\{\begin{array} { l }
{ a = 3 9 , } \\
{ k = 5 5 }
\end{array} \text { or } \left\{\begin{array}{l}
a=12, \\
k=26 .
\end{array}\right.\right.
$$
When $a=39$, equation (2) becomes $x^{2}+57 x+56=0$, whose roots are -1 and -56. In this case, the two function graphs have two additional intersection points $(-1,-56)$ and $(-56,-1)$.
When $a=12$, equation (2) becomes $x^{2}+30 x+56=0$, whose roots are -2 and -28. In this case, the two function graphs have two additional intersection points $(-2,-28)$ and $(-28,-2)$.
|
12
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given that for all real numbers $x$, we have
$$
|x+1|+\sqrt{x-1} \geqslant m-|x-2|
$$
always holds. Then the maximum value that $m$ can take is $\qquad$ .
|
When $-1 \leqslant x \leqslant 2$, the minimum value of $|x+1|+|x-2|$ is
3. Since $\sqrt{x-1} \geqslant 0$, when $x=1$,
$$
|x+1|+\sqrt{x-1}+|x-2|
$$
the minimum value is 3, so, $3 \geqslant m$, hence the maximum value of $m$ is 3.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. The military training base purchases apples to comfort the trainees. It is known that the total number of apples, when represented in octal (base 8), is $\overline{a b c}$, and when represented in septenary (base 7), it is $\overline{c b a}$. Therefore, the total number of apples, when represented in decimal (base 10), is $\qquad$
|
13.220.
$$
\begin{array}{l}
\text { Given } 1 \leqslant a, b, c \leqslant 6, \\
a \times 8^{2}+b \times 8+c \\
=c \times 7^{2}+b \times 7+a, \\
63 a+b-48 c=0, \\
b=3(16 c-21 a),
\end{array}
$$
Therefore, $b=0,3,6$.
Upon verification, $b=3$ meets the condition. Hence, $b=3, c=4, a=3$. Thus, $3 \times 8^{2}+3 \times 8+4=220$.
|
220
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x$ and $y$ are real numbers, and $x^{2}+x y+y^{2}=3$. Let the maximum and minimum values of $x^{2}-x y+y^{2}$ be $m$ and $n$, respectively. Then the value of $m+n$ is $\qquad$
|
$=, 1.10$.
Given the equation is symmetric about $x, y$, let $x=a+b, y=a-b$, then $3=x^{2}+xy+y^{2}=3a^{2}+b^{2}$.
Therefore, $3a^{2}=3-b^{2}$.
So, $0 \leqslant b^{2} \leqslant 3$.
$$
\begin{array}{l}
\text { Also, } x^{2}-xy+y^{2}=a^{2}+3b^{2}=\frac{1}{3}\left(3a^{2}+b^{2}\right)+\frac{8}{3}b^{2} \\
=\frac{1}{3} \times 3+\frac{8}{3}b^{2}=1+\frac{8}{3}b^{2},
\end{array}
$$
Then $1 \leqslant 1+\frac{8}{3}b^{2} \leqslant 1+\frac{8}{3} \times 3=9$.
Thus, $m=9, n=1$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 3, on a $2 \times 3$ rectangular grid paper, the vertices of each small square are called grid points. Then the number of isosceles right triangles with grid points as vertices is ( ).
(A) 24
(B) 38
(C) 46
(D) 50
This problem can be solved using the labeling method, making the solution clear at a glance.
|
Solution: As shown in Figure 4, at each grid point, mark the number of isosceles right triangles with it as the right-angle vertex. For example, $A_{2+0}$ indicates that there are 2 upright and 0 slanted isosceles right triangles with $A$ as the right-angle vertex; $B_{3+2}$ indicates that there are 3 upright and 2 slanted isosceles right triangles with $B$ as the right-angle vertex, among which 1 slanted one (indicated by a dashed line) is not easy to find.
By the symmetry of the figure, it is only necessary to mark the numbers for the grid points within $\frac{1}{4}$ of the area. Note that there are 4 points of type $A$ and 4 points of type $B$, and 2 points of type $C$ and 2 points of type $D$. Therefore, there are a total of $4(2+5)+2(3+8)=50$ isosceles right triangles.
The paper [4] provides the calculation formulas for $f(1, n)$ and $f(2, n)$:
$$
\begin{array}{l}
f(1, n)=6 n-2, \\
f(2, n)=\left\{\begin{array}{ll}
10, & n=1 ; \\
28, & n=2 ; \\
24 n-22, & n \geqslant 3 .
\end{array}\right.
\end{array}
$$
where $f(m, n)$ represents the number of isosceles right triangles with grid points as vertices in an $m \times n$ rectangular grid.
Using the marking method, the derivation of these formulas is straightforward.
For a $1 \times n$ rectangular grid (Figure 5), except for the 4 vertices of the rectangle, the marking at the remaining grid points is $2+1$, and the marking at the 4 vertices is $1+0$. Therefore,
$$
\begin{array}{l}
f(1, n)=4 \times 1+3(2 n-2) \\
=4+6 n-6=6 n-2 .
\end{array}
$$
For a $2 \times n$ rectangular grid, when $n=1$, it is obvious that
$$
f(2,1)=f(1,2)=10 \text {. }
$$
When $n=2$, there are 4 points of type $A$ and 4 points of type $B$, and 1 point of type $C$, with the markings at each type of point as shown in Figure 6.
Thus, $f(2,2)=4 \times 2+4 \times 3+1 \times 8=28$.
When $n \geqslant 3$, there are 4 points of type $A$ and 4 points of type $B$, $2(n+1-4)=2 n-6$ points of type $C$, 2 points of type $D$, and $n+1-2=n-1$ points of type $E$, with the markings at each type of point as shown in Figure 7. Some grid points have only one leg of the slanted isosceles right triangle marked (dashed line), and the other leg can be obtained by rotating the marked leg $90^{\circ}$ clockwise (the same applies below). Thus,
$$
\begin{array}{l}
f(2, n) \\
=4 \times 2+4 \times 5+8(2 n-6)+2 \times 3+8(n-1) \\
=24 n-22 .
\end{array}
$$
Using a similar method, it is not difficult to derive the formula for $f(3, n)$.
First, it is obvious that
$$
\begin{array}{l}
f(3,1)=f(1,3)=16, \\
f(3,2)=f(2,3)=50 . \\
\text { Second, when } n=3 \text {, }
\end{array}
$$
there are 4 points of type $A$ and 4 points of type $C$, and 8 points of type $B$, with the markings at each type of point as shown in Figure 8. Thus,
$$
\begin{array}{l}
f(3,3) \\
=4 \times 3+4 \times 11+8 \times 5=96 .
\end{array}
$$
When $n=4$, there are 4 points of type $A$, 4 points of type $B$, 4 points of type $D$, and 4 points of type $E$, and 2 points of type $C$ and 2 points of type $F$, with the markings at each type of point as shown in Figures 9 and 10.
Thus, $f(3,4)=4(3+7+5+11)+2(8+15)$
$=150$.
When $n \geqslant 5$, there are 4 points of type $A$, 4 points of type $B$, 4 points of type $C$, 4 points of type $E$, 4 points of type $F$, and 4 points of type $G$, and $2(n+1-6)=2 n-10$ points of type $D$ and $2 n-10$ points of type $H$, with the markings at each type of point as shown in Figures 11 and 12.
Thus, $f(3, n)$ $=4(3+7+11+5+11+15)+(15+15)(2 n-10)$
$$
\begin{array}{l}
=60 n-92 \text {. } \\
\text { In summary, } f(3, n)=\left\{\begin{array}{ll}
16, & n=1 ; \\
50, & n=2 ; \\
96, & n=3 ; \\
150, & n=4 ; \\
60 n-92, & n \geqslant 5 .
\end{array}\right. \\
\end{array}
$$
How to derive the formula for $f(m, n)$? This is a rather complex problem, and it seems difficult to solve using the marking method, so it is necessary to improve the method.
Below are two more examples of counting using the marking method.
|
50
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $a$ and $b$ be integers, and one root of the equation $x^{2} + a x + b = 0$ is $\sqrt{4 - 2 \sqrt{3}}$. Then the value of $a + b$ is ( ).
(A) -1
(B) 0
(C) 1
(D) 2
|
4.B.
Notice that $\sqrt{4-2 \sqrt{3}}=\sqrt{3}-1$.
According to the problem, we have $(\sqrt{3}-1)^{2}+a(\sqrt{3}-1)+b=0$, which simplifies to $(a-2) \sqrt{3}+4-a+b=0$.
Thus, $a-2=0$ and $4-a+b=0$.
Solving these, we get $a=2, b=-2$. Therefore, $a+b=0$.
|
0
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 1, in the right trapezoid $A B C D$, $A B=B C=4$, $M$ is a point on the leg $B C$, and $\triangle A D M$ is an equilateral triangle. Then $S_{\triangle C D M}$ : $S_{\triangle A B M}=$ $\qquad$ .
|
2.2.
As shown in Figure 6, draw $A E \perp$ $C D$ intersecting the extension of $C D$ at point $E$, then quadrilateral $A B C E$ is a square.
It is easy to prove
$\mathrm{Rt} \triangle A B M \cong \mathrm{Rt} \triangle A E D$.
Therefore, $B M=D E$.
Thus, $C M=C D$.
Let this value be $x$, then
$$
\begin{array}{l}
x^{2}+x^{2}=4^{2}+(4-x)^{2} \Rightarrow x^{2}=32-8 x . \\
\text { Hence, } \frac{S_{\triangle C D M}}{S_{\triangle B M}}=\frac{\frac{1}{2} x^{2}}{\frac{1}{2} \times 4(4-x)}=\frac{x^{2}}{16-4 x}=2 .
\end{array}
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The quality requirements of a product are divided into four different levels from low to high, labeled as $1,2,3,4$. If the working hours remain unchanged, the workshop can produce 40 units of the lowest level (i.e., level 1) product per day, with a profit of 16 yuan per unit; if the level is increased by one, the profit per unit increases by 1 yuan, but the daily production decreases by 2 units. Now the workshop plans to produce only one level of product. To maximize profit, the workshop should produce level $\qquad$ product.
|
3.3.
Let the profit obtained from producing products of the $x$-th grade in the workshop be $y$. According to the problem, we have
$$
\begin{array}{l}
y=[40-2(x-1)][16+(x-1)] \\
=-2 x^{2}+12 x+630=-2(x-3)^{2}+648 .
\end{array}
$$
Therefore, when $x=3$, the profit $y$ is maximized, at 648 yuan.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The number of all different integer solutions to the equation $2 x^{2}+5 x y+2 y^{2}=2007$ is $\qquad$ groups.
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。
Note: The note above is not part of the translation but is provided to clarify the instruction. The actual translation is above this note.
|
4.4 .
Let's first assume $x \geqslant y$, the original equation can be transformed into $(2 x+y)(x+2 y)=2007$.
Since $2007=2007 \times 1=669 \times 3=223 \times 9$
$$
\begin{array}{l}
=(-1) \times 2007=(-3) \times(-669) \\
=(-9) \times(-223),
\end{array}
$$
Therefore, $\left\{\begin{array}{l}2 x+y=2007, \\ x+2 y=1 .\end{array}\right.$,
Solving gives $3(x+y)=2008$, which has no integer solutions.
Similarly, $\left\{\begin{array}{l}2 x+y=223, \quad-1,-9 \\ x+2 y=9,-2007,-223,\end{array}\right.$ all have no integer solutions;
$\left\{\begin{array}{l}2 x+y=-3, \\ x+2 y=-669\end{array}\right.$ has integer solutions $\left\{\begin{array}{l}x=221, \\ y=-445 .\end{array}\right.$
If $y \geqslant x$, there are two more sets of integer solutions
$$
\left\{\begin{array} { l }
{ x = - 2 2 1 , } \\
{ y = 4 4 5 , }
\end{array} \left\{\begin{array}{l}
x=-445, \\
y=221 .
\end{array}\right.\right.
$$
Therefore, there are a total of 4 sets of integer solutions.
|
4
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(1) Find $f(2006)$;
(2) If $f(n) \geqslant 30$, find the minimum value of $n$.
For a positive integer $n$, let $f(n)$ be the sum of the digits of $n^2$.
|
(1) Since $2006^{2}=4024036$, we have
$$
f(2006)=19.
$$
(2) Since $f(n) \geqslant 30$, then $n^{2} \geqslant 3999$.
Thus, $n>63$.
$$
\begin{array}{l}
\text{And } f(64)=19, f(65)=13, f(66)=18, \\
f(67)=25, f(68)=16, f(69)=18, \\
f(70)=13, f(71)=10, f(72)=18, \\
f(73)=19, f(74)=22, f(75)=18, \\
f(76)=25, f(77)=25, f(78)=18, \\
f(79)=13, f(80)=10, f(81)=18, \\
f(82)=19, f(83)=31,
\end{array}
$$
Therefore, the smallest positive integer $n$ such that $f(n) \geqslant 30$ is 83.
Note: Using a CASIO $\mathrm{fx}-82 \mathrm{MS}$ calculator, you can press the following keys:
$$
63=\text{ Ans }^{2}: \sqrt{\text{ (Ans) }}+1 \text{, }
$$
Then, repeatedly press the $“=$" key to get the values of $64^{2}, 65^{2}, 66^{2}, \cdots$.
Using a TI-83Plus or T1-92Plus calculator, you can press the following keys:
$$
63 \rightarrow \mathrm{N} \text{ ENTER } 1+\mathrm{N} \rightarrow \mathrm{N}:\left\{\mathrm{N}, \mathrm{N}^{2}\right\},
$$
Then, repeatedly press the “ENTER” key to get
$$
\{64,4096\},\{65,4225\}, \cdots .
$$
|
83
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) How many real roots does the equation $\sqrt{x} \sin x^{2}-2=0$ have in the interval $[0,20]$?
|
Three, Solution 1: Let $y=\sqrt{x} \sin x^{2}$.
The solutions to the equation $y=0$ are $x=\sqrt{k \pi} \in[0,20], k \in \mathbf{Z}$. Therefore, $x=\sqrt{k \pi}(k=0,1, \cdots, 127)$.
When $\sqrt{x} \geqslant 2$, the graph of $y=\sqrt{x} \sin x^{2}$ may intersect with the graph of $y=2$, so $x \geqslant 4$.
Also, $\sqrt{5 \pi}4$, when $x \in[\sqrt{6 \pi}, \sqrt{7 \pi}]$, $\sqrt{x} \sin x^{2} \geqslant 0$, the function $y=\sqrt{x} \sin x^{2}$ increases first and then decreases on $[\sqrt{6 \pi}, \sqrt{7 \pi}]$, and when $x=\sqrt{6 \pi+\frac{\pi}{2}}$,
$$
\begin{array}{l}
y_{\max } \geqslant \sqrt[4]{6 \pi+\frac{\pi}{2}} \sin \left(\sqrt{6 \pi+\frac{\pi}{2}}\right)^{2} \\
=\sqrt[4]{6 \pi+\frac{\pi}{2}}>2 .
\end{array}
$$
Therefore, within the interval $[\sqrt{6 \pi}, \sqrt{7 \pi}]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has two real roots.
Similarly, within the intervals $[\sqrt{8 \pi}, \sqrt{9 \pi}],[\sqrt{10 \pi}, \sqrt{11 \pi}], \cdots$, $[\sqrt{126 \pi}, \sqrt{127 \pi}]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has two real roots each.
Thus, there are a total of 122 real roots.
In summary, within the interval $[0,20]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has a total of 122 real roots.
Solution 2: Use a graphing calculator to plot the graph of the function $y=\sqrt{x} \sin x^{2}$. To clearly see the number of roots, consider observing several intervals.
For example, when $x \in[0,10]$, i.e., $x_{\text {min }}=0, x_{\max }=10$, the equation has 26 real roots.
Similarly, when $x \in(10,13]$, the equation has 22 real roots;
when $x \in(13,17]$, the equation has 38 real roots;
when $x \in(17,20]$, the equation has 36 real roots.
Therefore, within the interval $[0,20]$, the equation $\sqrt{x} \sin x^{2}-2=0$ has a total of 122 real roots.
|
122
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given the set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$
Then the number of complex numbers in $D$ whose real and imaginary parts are both integers is $\qquad$
|
8.49 .
Given $\left|z_{1}\right|=1$, let $z_{1}=\cos \theta+\mathrm{i} \sin \theta$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right|=\left|\left(z_{1}^{2}-1\right)^{2}\right|=\left|z_{1}^{2}-1\right|^{2} \\
=|(\cos 2 \theta-1)+\mathrm{i} \sin 2 \theta|^{2} \\
=(\cos 2 \theta-1)^{2}+\sin ^{2} 2 \theta=2-2 \cos 2 \theta=4 \sin ^{2} \theta . \\
\text { Therefore, } z=x+y \mathrm{i} \in D \\
\Leftrightarrow|z-2005-2006 \mathrm{i}| \leqslant 4 \\
\Leftrightarrow(x-2005)^{2}+(y-2006)^{2} \leqslant 16 .
\end{array}
$$
Let $s=x-2005, t=y-2006$, then
$$
z \in D \Leftrightarrow s^{2}+t^{2} \leqslant 16 \text {. }
$$
The latter has a total of 49 solutions.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. For any 4 vertices $A, B, C, D$ of a cube that do not lie on the same plane, the number of cosine values of the dihedral angle $A-BC-D$ that are less than $\frac{1}{2}$ is
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
9.4 .
Among the cosines of the dihedral angles with the body diagonal as the edge,
$$
\begin{array}{l}
\cos \angle A_{1} O_{1} C_{1} \\
=\cos 120^{\circ}=-\frac{1}{2} .
\end{array}
$$
Among the cosines of the dihedral angles with the edge as the edge, only 0 is less than $\frac{1}{2}$.
Among the cosines of the dihedral angles with the face diagonal as the edge,
$$
\begin{array}{l}
\cos \angle A O C_{1} \\
=\frac{\left(\frac{\sqrt{2}}{2}\right)^{2}+\left(1+\frac{1}{2}\right)-(\sqrt{3})^{2}}{2 \times \frac{\sqrt{2}}{2} \times \sqrt{1+\frac{1}{2}}}=-\frac{\sqrt{3}}{3}<\frac{1}{2} \\
\cos \angle A_{1} O C_{1}=\frac{\frac{3}{2}+\frac{3}{2}-2}{2 \sqrt{\frac{3}{2}} \times \sqrt{\frac{3}{2}}}=\frac{1}{3}<\frac{1}{2} .
\end{array}
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $x_{1}, x_{2}, \cdots, x_{n}$ be numbers that can take one of the values $-3, 0, 1$, and
$$
\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=279, \\
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=-585 .
\end{array}
$$
Then the value of $x_{1}^{4}+x_{2}^{4}+\cdots+x_{n}^{4}$ is ( ).
(A) 2005
(B) 2006
(C) 2007
(D) 2008
|
- 1.C.
Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ numbers, where $-3$ appears $b$ times, $1$ appears $b$ times, and $0$ appears $c$ times. Then
$$
\left\{\begin{array} { l }
{ 9 a + b = 2 7 9 , } \\
{ - 2 7 a + b = - 5 8 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=24, \\
b=63 .
\end{array}\right.\right.
$$
Thus, the original expression $=24 \times(-3)^{4}+63 \times 1=2007$.
|
2007
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
3. Divide the sides of the equilateral $\triangle A B C$ into four equal parts, as shown in Figure 1. Then the number of pairs of congruent equilateral triangles in the figure is ( ).
(A) 100
(B) 121
(C) 144
(D) 169
|
3. C.
Let the side length of the equilateral triangle be 4. From Figure 1, we can see:
(1) There are 16 equilateral triangles with a side length of 1, all of which are congruent, and the number of pairs is $15+14+\cdots+1=120$;
(2) There are 7 equilateral triangles with a side length of 2, all of which are congruent, and the number of pairs is $6+5+4+3+2+1=21$;
(3) There are 3 equilateral triangles with a side length of 3, and the number of pairs is $2+1=3$.
In summary, there are $120+21+3=144$ pairs of congruent triangles.
|
144
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given that $m$ and $n$ ($m>n$) are positive integers, and the last three digits of $3^{m}$ and $3^{n}$ are the same. Find the minimum value of $m-n$.
保留源文本的换行和格式，翻译结果如下：
```
Three, (25 points) Given that $m$ and $n$ ($m>n$) are positive integers, and the last three digits of $3^{m}$ and $3^{n}$ are the same. Find the minimum value of $m-n$.
```
|
Three, from the given information, $3^{m}-3^{n}$ is a multiple of 1000, i.e.,
$$
3^{m}-3^{n}=3^{n}\left(3^{m-n}-1\right)
$$
is a multiple of 1000.
Also, $\left(3^{n}, 1000\right)=1$, hence $3^{m-n}-1$ is a multiple of 1000.
Let $s=m-n$, then $3^{s}-1$ is a multiple of 1000.
We only need to find the smallest $s$ such that the last three digits of $3^{s}$ are 001.
From Table 1, we know that $s$ is a multiple of 4.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$s$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & $\cdots$ \\
\hline The last digit of $3^{s}$ & 3 & 9 & 7 & 1 & 3 & 9 & 7 & 1 & $\cdots$ \\
\hline
\end{tabular}
From Table 2, we know that $s$ is a multiple of 20.
Table 2
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$s$ & 4 & 8 & 12 & 16 & 20 & 24 & 28 & 32 & 36 & 40 & $\cdots$ \\
\hline \begin{tabular}{l}
The last two \\
digits of $3^{s}$
\end{tabular} & 81 & 61 & 41 & 21 & 01 & 81 & 61 & 41 & 21 & 01 & $\cdots$ \\
\hline
\end{tabular}
From Table 3, we know that the smallest value of $s=m-n$ is 100.
Table 3
\begin{tabular}{|c|r|r|r|r|r|r|}
\hline$s$ & 20 & 40 & 60 & 80 & 100 & $\cdots$ \\
\hline The last three digits of $3^{s}$ & 401 & 801 & 201 & 601 & 001 & $\cdots$ \\
\hline
\end{tabular}
|
100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a, b, c$ be three distinct positive integers. If
$$
\begin{array}{l}
\{a+b, b+c, c+a\} \\
=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\}\left(n \in \mathbf{N}_{+}\right),
\end{array}
$$
then the minimum value of $a^{2}+b^{2}+c^{2}$ is ( ).
(A) 2007
(B) 1949
(C) 1297
(D) 1000
|
- 1.C.
Assume $a>b>c$, then $a+b>c+a>b+c$.
Since $(a+b)+(b+c)+(c+a)=2(a+b+c)$ is an even number, $n^{2} 、(n+1)^{2} 、(n+2)^{2}$ must be two odd and one even. Therefore, $n$ must be an odd number.
Also, since $b+c>1$, $n$ must be an odd number no less than 3.
If $n=3$, then $\{a+b, b+c, c+a\}=\left\{3^{2}, 4^{2}, 5^{2}\right\}$.
Thus, $a+b+c=\frac{1}{2}\left(3^{2}+4^{2}+5^{2}\right)=5^{2}$, and $a+b=5^{2}$.
Therefore, $c=0$, which does not meet the requirement.
If $n=5$, then $\{a+b, b+c, c+a\}=\left\{5^{2}, 6^{2}, 7^{2}\right\}$.
Thus, $\left\{\begin{array}{l}a+b=7^{2}, \\ c+a=6^{2} \\ b+c=5^{2} .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}a=30, \\ b=19, \\ c=6 .\end{array}\right.$
At this point, $a^{2}+b^{2}+c^{2}=1297$.
|
1297
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $m, n$ be positive integers. If there exists a positive integer $k$ such that $\log _{2} m, \log _{2} n$ and $\log _{2} k$ can be the lengths of the three sides of a triangle, then $k$ is called a “good number”. It is known that there are exactly 100 good numbers $k$. Then the maximum possible value of $m n$ is ( ).
(A) 100
(B) 101
(C) 134
(D) 2007
|
6.C.
Assume $m \geqslant n$. From the problem, we have
$$
\begin{array}{l}
\log _{2} m-\log _{2} n\frac{m}{n}, m n-101 \leqslant \frac{m}{n},
\end{array}
$$
which implies $\frac{100}{1-\frac{1}{n^{2}}}<m n \leqslant \frac{101}{1-\frac{1}{n^{2}}}$.
It is easy to see that when $n=2$, $\frac{101}{1-\frac{1}{n^{2}}}$ reaches its maximum value $\frac{404}{3}$, at this time,
$$
\frac{400}{3}<m n \leqslant \frac{404}{3} \text {. }
$$
Therefore, when $m=67, n=2$, $m n$ reaches its maximum value 134.
|
134
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Figure 1 is a shape composed of 4 unit squares, with each vertex of the unit squares called a grid point. By taking any three grid points as vertices, how many different isosceles right triangles can be formed?
Regarding this problem, the answer given in [1] is 24, [2] supplements it to 30, and [3] provides the correct answer as 32. This shows that although the problem seems simple, a slight oversight can lead to errors in solving it.
This article introduces an effective method for solving such counting problems—the labeling method. It can effectively avoid the aforementioned mistakes.
The basic approach is to label each grid point with a number, and the number labeled at grid point $A$ represents the number of counting objects (such as isosceles right triangles) with $A$ as the special element.
For Example 1, since each isosceles right triangle has a unique right-angle vertex, we can consider the number of isosceles right triangles with each grid point as the right-angle vertex.
Of course, once the right-angle vertex is determined, the isosceles right triangles can be divided into two major categories:
(1) The legs are parallel to the grid lines (called upright);
(2) The legs are not parallel to the grid lines (called slanted).
For each major category, they can be further classified by different directions. Once the direction is determined, they can be classified by the length of the legs.
|
Solution: Through the above classification, mark the number of isosceles right triangles with each grid point as the right-angle vertex (Figure 2). For example: $B_{3+1}$ indicates that there are 3 upright and 1 slanted isosceles right triangles with $B$ as the right-angle vertex; $E_{1+2}$ indicates that there are 1 upright and 2 slanted isosceles right triangles with $E$ as the right-angle vertex, among which 1 slanted one (indicated by a dashed line) is not easily noticed.
Noticing the symmetry of the figure, it is only necessary to number half of the grid points. Since each type of grid point appears twice, there are a total of $2(1+4+2+6+3)=32$ isosceles right triangles.
Interestingly, a similar problem appeared in the 2004 National Junior High School Mathematics Competition.
|
32
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. There are 5 rooms $A, B, C, D, E$ arranged in a circular pattern, with the number of people living in them being $17, 9, 14, 16, 4$ respectively. Now, adjustments are to be made so that the number of people in each room is the same, and it is stipulated that people can only move to the adjacent left or right room. How many people should each room move to the left or right so that the total number of people moved is minimized?
|
(Tip: Let the number of people moved from room $A$ to room $B$ be $x_{B}$, and so on, then we have
$$
\begin{array}{l}
9+x_{B}-x_{C}=14+x_{C}-x_{D}=16+x_{D}-x_{E} \\
=4+x_{E}-x_{A}=17+x_{A}-x_{B} \\
=\frac{1}{5}(17+9+14+16+4)=12 .
\end{array}
$$
This is transformed into finding
$$
\begin{array}{l}
y=\left|x_{B}-5\right|+\left|x_{B}\right|+\left|x_{B}-3\right|+ \\
\left|x_{B}-1\right|+\left|x_{B}+3\right|
\end{array}
$$
The minimum value of $y$. By the shortest distance between two points is a straight line, we get $x_{B}=1$. Thus, $x_{A}=$ $-4, x_{B}=1, x_{C}=-2, x_{D}=0, x_{E}=4$. The minimum total number of people moved is $4+1+2+0+4=11$.)
|
11
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. Given that the weights of $A$, $B$, $C$, and $D$ are all integers in kilograms, where $A$ is the lightest, followed by $B$, $C$, and $D$, the weights of each pair of them are as follows (unit: kg):
$45,49,54,55,60,64$.
Then the weight of $D$ is $\qquad$ kg.
|
14.35.
Since $A+B=45, A+C=49, B+D=60, C+D$ $=64$, therefore, $C-B=4$.
Then $B+C=B+(B+4)=2B+4$ is an even number.
Among $54 \text{~kg}$ and $55 \text{~kg}$, only 54 is an even number, so $B=25$. Then $D=35$.
|
35
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. (12 points) There are $a$ matchsticks of the same length. When placed as shown in Figure 7, they can form $m$ squares; when placed as shown in Figure 8, they can form $2 n$ squares.
(1) Express $m$ in terms of $n$;
(2) When these $a$ matchsticks can also be arranged as shown in Figure 9, find the minimum value of $a$.
|
16. (1) The total number of matchsticks in Figure 7 is $3 m+1$, and the total number of matchsticks in Figure 8 is $5 n+2$.
Since the total number of matchsticks is the same, we have $3 m+1=5 n+2$.
Solving for $m$ gives $m=\frac{5 n+1}{3}$.
(2) Suppose there are $3 p$ squares in Figure 9, then the total number of matchsticks is $7 p+3$.
By the problem, we have $a=3 m+1=5 n+2=7 p+3$.
Thus, $p=\frac{3 m-2}{7}=\frac{5 n-1}{7}$.
Since $m$, $n$, and $p$ are all positive integers, when $m=17$, $n=10$, then $p=7$.
At this point, the minimum value of $a$ is
$$
a=3 \times 17+1=5 \times 10+2=7 \times 7+3=52 \text{. }
$$
|
52
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The number of triangles with integer side lengths and a perimeter of 20 is $\qquad$ .
|
2.8.
Let the three sides of a triangle be $a, b, c$, and $a \geqslant b \geqslant c$, $a+b+c=20$, then $a \geqslant 7$.
Also, from $b+c>a$, we get $2 a<a+b+c=20 \Rightarrow a<10$. Therefore, $7 \leqslant a \leqslant 9$. We can list
$$
\begin{array}{l}
(a, b, c)=(9,9,2),(9,8,3),(9,7,4),(9,6,5), \\
(8,8,4),(8,7,5),(8,6,6),(7,7,6) .
\end{array}
$$
There are 8 sets in total.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange the four numbers $1, 2, 3, 4$ to form a four-digit number, such that this number is a multiple of 11. Then the number of such four-digit numbers is $\qquad$.
|
4.8.
Since $1+4=2+3$, we can place 1 and 4 in the even positions, and 2 and 3 in the odd positions, which gives us four arrangements; placing 2 and 3 in the even positions, and 1 and 4 in the odd positions also gives us four arrangements.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) If a number $a$ can be expressed as the sum of the squares of two natural numbers (allowing the same), then $a$ is called a "good number". Determine how many good numbers there are among the first 200 positive integers $1,2, \cdots, 200$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Three, the squares not exceeding 200 are $0^{2}, 1^{2}, \cdots, 14^{2}$.
Obviously, each number $k^{2}$ in $1^{2}, 2^{2}, \cdots, 14^{2}$ can be expressed in the form of $k^{2}+$ $0^{2}$, and there are 14 such numbers.
Each pair of numbers in $1^{2}, 2^{2}, \cdots, 10^{2}$ (which can be the same) has a sum not exceeding 200, and there are $\frac{10 \times 9}{2}+10=55$ such numbers (among which, numbers in the form of $x^{2}+x^{2}$ are 10, and numbers in the form of $x^{2}+y^{2}(x \neq y)$ are $\frac{10 \times 9}{2}=45$).
Next, numbers in the form of $11^{2}+x^{2}(x=1,2, \cdots, 8)$ are 8; numbers in the form of $12^{2}+x^{2}(x=1,2, \cdots, 7)$ are 7; numbers in the form of $13^{2}+x^{2}(x=1,2, \cdots, 5)$ are 5; and numbers in the form of $14^{2}+x^{2}(x=1,2)$ are 2, totaling 22.
Considering the repeated cases, use the following fact:
$$
\begin{array}{l}
\text { If } x=a^{2}+b^{2}, y=c^{2}+d^{2}(a \neq b, c \neq d), \text { then } \\
x y=(a c+b d)^{2}+(a d-b c)^{2} \\
=(a c-b d)^{2}+(a d+b c)^{2} .
\end{array}
$$
Numbers not exceeding 40 and can be expressed as the sum of squares of two different positive integers are $5,10,13,17,20,25,26,29,34,37,40$. Each number in this group, when multiplied by 5, and $13^{2}$, does not exceed 200 and can be expressed as the sum of squares in two ways, so each is counted twice, totaling 12 repeated counts $(10,13,17,20$ when multiplied by 10 are already included in the above product group).
Therefore, the total number of numbers that meet the condition is $14+55+22-12=79$.
(Supplied by Tao Pingsheng)
|
79
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given the sequence
$$
\begin{array}{l}
a_{1}=\frac{1}{2}, a_{2}=\frac{1}{3}+\frac{2}{3}, \cdots, \\
a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}, \cdots .
\end{array}
$$
Let $S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}}$. Then the integer closest to $S_{2006}$ is $(\quad)$.
(A) 2
(B) 3
(C) 4
(D) 5
|
3.C.
Since $a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+\cdots+\frac{n}{n+1}=\frac{n}{2}$, we have
$$
\begin{array}{l}
S_{n}=\frac{1}{a_{1} a_{2}}+\frac{1}{a_{2} a_{3}}+\cdots+\frac{1}{a_{n} a_{n+1}} \\
=\frac{4}{1 \times 2}+\frac{4}{2 \times 3}+\cdots+\frac{4}{n(n+1)} \\
=4\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)\right] \\
=4\left(1-\frac{1}{n+1}\right) .
\end{array}
$$
Therefore, $S_{2000}=4\left(1-\frac{1}{2007}\right)$ is closest to the integer 4.
|
4
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$. Then
$$
f(1)+f(2)+\cdots+f(2006)=
$$
$\qquad$
|
2.0.
Given that the graph of $y=f(x)$ is symmetric about the line $x=\frac{1}{2}$, we know that $f(x)=f(1-x)$.
Also, since $f(x)$ is an odd function defined on $\mathbf{R}$, it follows that $f(1-x)=-f(x-1)$.
Therefore, $f(x)+f(x-1)=0$.
Thus, $f(1)+f(2)+\cdots+f(2006)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let the increasing sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=6, n \in \mathbf{N}_{+}$, and when $n \geqslant 2$, $a_{n}+a_{n-1}=\frac{9}{a_{n}-a_{n-1}}+8$. Then $a_{x}=$ $\qquad$
|
4.29.
$$
\begin{array}{l}
\text { Given } a_{k}+a_{k-1}=\frac{9}{a_{k}-a_{k-1}}+8 \\
\Rightarrow a_{k}^{2}-a_{k-1}^{2}=8\left(a_{k}-a_{k-1}\right)+9 .
\end{array}
$$
Taking $k=2,3, \cdots, n$, summing up we get
$$
a_{n}^{2}-a_{1}^{2}=8\left(a_{n}-a_{1}\right)+9(n-1) \text {. }
$$
Thus, $a_{10}^{2}-6^{2}=8\left(a_{10}-6\right)+9 \times 69$, which means
$$
a_{70}^{2}-8 a_{70}-609=0 \text {. }
$$
Therefore, $a_{70}=29$ (negative value is discarded).
|
29
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $\triangle A B C$ with three sides $A B=\sqrt{34}, B C$ $=5 \sqrt{10}, C A=2 \sqrt{26}$. Then the area of $\triangle A B C$ is $\qquad$
|
5.10.
As shown in Figure 4, with $BC$ as the hypotenuse, construct a right triangle $\triangle BCD$ on one side of $\triangle ABC$ such that $\angle BDC = 90^\circ$, $BD = 5$, and $CD = 15$.
Then construct a rectangle $DEA'F$ such that $DE = 2$ and $DF = 5$. Thus, $BE = 3$ and $CF = 10$. At this point,
$$
\begin{array}{l}
A'C = \sqrt{104} = AC, \\
A'B = \sqrt{34} = AB.
\end{array}
$$
Therefore, point $A$ coincides with point $A'$.
Hence,
$$
\begin{array}{l}
S_{\triangle ABC} = S_{\triangle BCD} - S_{\triangle ABE} - \\
\quad S_{\triangle ACF} - S_{\text{rect } EAF} \\
= \frac{1}{2} \times 5 \times 15 - \frac{1}{2} \times 3 \times 5 - \frac{1}{2} \times 2 \times 10 - 2 \times 5 \\
= 10.
\end{array}
$$
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=$ $9 \sqrt[3]{a_{n}}(n \geqslant 1)$. Then $\lim _{n \rightarrow \infty} a_{n}=$ $\qquad$ .
|
8.27.
Given $\log _{3} a_{n+1}=2+\frac{1}{3} \log _{3} a_{n}$, let $b_{n}=\log _{3} a_{n}$, then
$$
\begin{array}{l}
b_{n+1}=\frac{1}{3} b_{n}+2, b_{1}=1 \\
\Rightarrow b_{n+1}-3=\frac{1}{3}\left(b_{n}-3\right) \\
\Rightarrow b_{n}-3=\left(b_{1}-3\right)\left(\frac{1}{3}\right)^{n-1}=-2 \times\left(\frac{1}{3}\right)^{n-1} \\
\Rightarrow b_{n}=-\frac{2}{3^{n-1}}+3 \Rightarrow a_{n}=3^{-\frac{2}{3^{n-1}}+3} . \\
\text { Also } \lim _{n \rightarrow \infty}\left(-\frac{2}{3^{n-1}}+3\right)=3, \text { so } \lim _{n \rightarrow \infty} a_{n}=3^{3}=27 .
\end{array}
$$
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If $(2 x-1)^{8}=a_{8} x^{8}+a_{7} x^{7}+\cdots+a_{1} x$ $+a_{0}$, then $a_{8}+a_{6}+a_{4}+a_{2}=$ $\qquad$
|
9.3280 .
Let $f(x)=(2 x-1)^{8}$, then
$$
\begin{array}{l}
a_{0}=f(0)=1, \\
a_{8}+a_{7}+\cdots+a_{0}=f(1)=1, \\
a_{8}-a_{7}+a_{6}-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0} \\
=f(-1)=(-3)^{8}=6561 .
\end{array}
$$
Therefore, $a_{8}+a_{6}+a_{4}+a_{2}$
$$
=\frac{1}{2}(f(1)+f(-1))-f(0)=3280 \text {. }
$$
|
3280
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Let $n$ be a positive integer, and let $f(n)$ denote the number of positive integers not exceeding $\sqrt{n}$ (for example, $f(3)=1$, $f(9)=3$).
(1) Find $f(2007)$;
(2) Find the positive integer $n$ such that it satisfies
$$
f(1)+f(2)+\cdots+f(n)=2009 .
$$
|
(1) From $44<\sqrt{2007}<45$, we have $f(2007)=44$.
(2) When $k=1,2,3$, $f(k)=1$, then
$$
f(1)+f(2)+f(3)=1 \times 3 \text {. }
$$
When $k=4,5, \cdots, 8$, $f(k)=2$, then
$$
f(4)+f(5)+\cdots+f(8)=2 \times 5 \text {. }
$$
When $k=9,10, \cdots, 15$, $f(k)=3$, then
$$
f(9)+f(10)+\cdots+f(15)=3 \times 7 \text {. }
$$
When $k=16,17, \cdots, 24$, $f(k)=4$, then
$$
f(16)+f(17)+\cdots+f(24)=4 \times 9 \text {. }
$$
When $k=169,170, \cdots, 195$, $f(k)=13$, then
$$
f(169)+f(170)+\cdots+f(195)=13 \times 27 \text {. }
$$
When $k=196,197, \cdots, 224$, $f(k)=14$.
From the original equation, we have
$$
\begin{array}{l}
1 \times 3+2 \times 5+3 \times 7+4 \times 9+\cdots+13 \times 27+14 x \\
=2009,
\end{array}
$$
Solving for $x$ gives $x=20$.
Thus, $n=195+20=215$, which means
$$
f(1)+f(2)+\cdots+f(215)=2009 \text {. }
$$
Therefore, the natural number $n=215$.
|
215
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Given three natural numbers $a, b, c$ where at least $a$ is a prime number, and they satisfy
$$
\left\{\begin{array}{l}
(4 a+2 b-4 c)^{2}=443(2 a-442 b+884 c), \\
\sqrt{4 a+2 b-4 c+886}-\sqrt{42 b-2 a+2 c-443}=\sqrt{443} .
\end{array}\right.
$$
Find the value of $a b c$.
|
Let $x=\frac{4 a+2 b-4 c}{443}, y=\frac{2 a-442 b+884 c}{443}$. Then
$4 a+2 b-4 c=443 x$,
$2 a-442 b+884 c=443 y$.
Hence $x^{2}=y$.
From equation (1), we know that $(4 a+2 b-4 c)^{2}$ is divisible by 443. Since 443 is a prime number, 443 divides $(4 a+2 b-4 c)$. Therefore, $x$ is an integer.
From equation (5), we know that $y$ is an integer.
(3) $\times 221+$ (4) gives
$$
4 \times 221 a+2 a=443 \times 221 x+443 y \text {, }
$$
which simplifies to $2 a=221 x+y$.
(5) - (6) gives $x^{2}+221 x-2 a=0$.
Then $\Delta=221^{2}-4 \times 1 \times(-2 a)=221^{2}+8 a$.
Since $x$ is an integer, $\Delta$ must be a perfect square.
Let $\Delta=t^{2}(t \in \mathbf{N})$. Then $t^{2}=221^{2}+8 a$, which implies $(t+221)(t-221)=8 a$.
Since $t+221$ and $t-221$ have the same parity and $t+221>221$, we have $\left\{\begin{array}{l}t+221=4 a, \\ t-221=2\end{array}\right.$ or $\left\{\begin{array}{l}t+221=2 a, \\ t-221=4 .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}t=223, \\ a=111\end{array}\right.$ (discard) or $\left\{\begin{array}{l}t=225, \\ a=223 .\end{array}\right.$
Therefore, $x^{2}+221 x-2 \times 223=0$.
Thus, $x=2$ or $x=-223$.
From equation (1), we know that $x$ is even, so $x=2$.
From equation (3), we know that $b=2 c-3$.
Substituting $a=223, b=2 c-3$ into equation (2) gives
$$
c=3, b=3 \text {. }
$$
Therefore, $a b c=223 \times 3 \times 3=2007$.
(Xie Wenxiao, Huanggang High School, Hubei Province, 438000)
|
2007
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The area of the figure enclosed by the two curves $y=x^{3}$ and $x=y^{3}$ on the Cartesian plane is $\qquad$ .
|
5.1.
Since the two curves are symmetric with respect to the origin, it is only necessary to calculate the area $A$ of the figure enclosed by the two curves in the first quadrant.
When $x>1$, $x^{3}>\sqrt[3]{x}$;
When $0<x<1$, $x^{3}<\sqrt[3]{x}$.
Therefore, the two curves have a unique intersection point $(1,1)$ in the first quadrant.
$$
\begin{array}{l}
\text { Also, } A=\int_{0}^{1}\left(\sqrt[3]{x}-x^{3}\right) \mathrm{d} x \\
=\left.\left(\frac{3}{4} x^{\frac{4}{3}}-\frac{x^{4}}{4}\right)\right|_{0} ^{1}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2},
\end{array}
$$
Therefore, the area of the figure enclosed by the two curves is $2 A=1$.
|
1
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8.2. Write the 100 natural numbers from 1 to 100 in a $10 \times 10$ grid, with one number in each cell. In each operation, you can swap the positions of any two numbers. Prove that it is possible to ensure that the sum of any two numbers in cells that share a common edge is a composite number after just 35 operations.
|
8.2. A vertical straight line $m$ divides the grid into two halves. In one of the halves, there are no more than 25 even numbers, let's assume it is the right half. Thus, in the left half, there is an equal number of odd numbers. By swapping the even numbers in the right half with the odd numbers in the left half one by one, after no more than 25 operations, the right half will be filled with odd numbers, and the left half will be filled with even numbers. At this point, the sum of any two adjacent cells in the same half is even, and thus composite.
From the above, only the sum of the numbers $l_{j}$ and $r_{j}$ in the cells on either side of the line $m$ in the same row might be prime.
Next, we show that by operating only in the right half, we can ensure that the sum of each pair $\left(l_{j}, r_{j}\right)(j=1,2, \cdots, 10)$ is a multiple of 3. Note that in the right half, all odd numbers from $1 \sim 99$ are written, and there are at least 16 numbers each with remainders of 0, 1, and 2 when divided by 3. To make each sum $l_{j}+r_{j}$ a multiple of 3, we need at most 10 of these numbers, so it is definitely possible. In this way, a total of no more than $25+10=35$ operations are required.
|
35
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8.3. On the side $BC$ of the rhombus $ABCD$, take a point $M$. Draw perpendiculars from $M$ to the diagonals $BD$ and $AC$, intersecting the line $AD$ at points $P$ and $Q$. If the lines $PB$ and $QC$ intersect $AM$ at the same point, find the ratio $\frac{BM}{MC}$.
|
8.3. As shown in Figure 1, let the intersection of lines $P B$, $Q C$, and $A M$ be $R$.
From the given conditions, $P M \parallel A C$ and $M Q \parallel B D$, thus quadrilaterals $P M C A$ and $Q M B D$ are both parallelograms. Therefore,
$$
\begin{aligned}
M C &= P A, B M = D Q, \text{ and } \\
P Q &= P A + A D + D Q = M C + A D + B M = 2 B C .
\end{aligned}
$$
Since $B C \parallel P Q$ and $B C = \frac{1}{2} P Q$, $B C$ is the midline of $\triangle P R Q$.
This implies that $B M$ is the midline of $\triangle A R P$.
$$
\text{Thus, } M C = P A = 2 B M \text{. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8.7. For a natural number $n(n>3)$, we use “$n$ ?" to denote the product of all prime numbers less than $n$. Solve the equation
$$
n ?=2 n+16
$$
|
8.7. The given equation is
$$
n ?-32=2(n-8) \text {. }
$$
Since $n$ ? cannot be divisible by 4, it follows from equation (1) that $n-8$ is odd.
Assume $n>9$, then $n-8$ has an odd prime factor $p$.
Also, $p2 \times 9+16$.
When $n=7$, it is clearly a root of the equation:
However, when $n=5$, we have $n ?=6<16$.
Therefore, the equation has a unique root $n=7$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.8. Ji Ma calculated the reciprocal of the factorial of each integer from $80 \sim 99$, and printed the resulting decimal fractions on 20 infinitely long strips of paper (for example, Sasha cut a segment from one of the strips, which had exactly $n$ digits without a decimal point. If Sasha does not want Ji Ma to guess which strip the $n$ digits were cut from, what is the maximum value of $n$?
|
9.8. The maximum value of $n$ is 155.
Assuming that on the slips of paper, $\frac{1}{k!}$ and $\frac{1}{l!}$ (where $k < l$) are written, the first 156 digits of the decimal expansion of $\frac{1}{k!} - \frac{1}{l!}$ are greater than $\frac{1}{10^{16}}$. Therefore, $n < 156$.
Thus, for any segment of 156 digits, we can determine its origin. However, when $n = 155$, on the slips of paper with $\frac{1}{98!}$ and $\frac{1}{99!}$, there is a completely identical segment of 155 digits:
$$
\begin{array}{l}
=0 . \underbrace{00 \cdots 0106 \cdots}_{153 \uparrow} \text {, } \\
\end{array}
$$
|
155
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10.4. A magician and his assistant perform the following act: First, the assistant asks the audience to write down $N$ numbers in a row on a blackboard, then the assistant covers up two adjacent numbers. After that, the magician comes on stage and guesses the two adjacent covered numbers (including their order). To ensure that the magician can guess the result according to the prior agreement with the assistant, find the minimum value of $N$.
|
10.4. $N=101$.
For convenience, a sequence of $m$ digits is called an “$m$-digit number”. Suppose for some value of $N$, the magician can guess the result, so the magician can restore any two-digit number to the original $N$-digit number (the number of $N$-digit numbers that can be restored is denoted as $k_{1}$). This means that for any two-digit number, the magician can map it to an $N$-digit number (the number of such $N$-digit numbers is denoted as $k_{2}$). Therefore, $k_{1} \geqslant k_{2}$. It is easy to see that $k_{1}=(N-1) \times 10^{N-2}$ (the given two-digit number has $N-1$ ways to choose its position, and each of the remaining $N-2$ digits has 10 different ways to place the digits).
It is not difficult to see that $k_{2}=10^{N}$.
Thus, by $k_{1} \geqslant k_{2} \Rightarrow N-1 \geqslant 100 \Rightarrow N \geqslant 101$.
Below, we show that for $N=101$, the magician can guess the result.
Number the 101 digits from left to right as 0 to 100. Let the sum of the digits in all odd positions modulo 10 be $s$, and the sum of the digits in all even positions modulo 10 be $t$. Let $p=10 s+t$.
The magician and the assistant agree to cover the digits at the $p$-th and $(p+1)$-th positions. Thus, as soon as the magician sees the positions of the covered digits, he immediately knows the value of $p$. Therefore, he can determine $s$ and $t$.
Since the $p$-th and $(p+1)$-th positions are one odd and one even, once $s$ is known, the magician can calculate the digit at the covered odd position based on the sum of the digits at the uncovered odd positions.
Similarly, the magician can also calculate the digit at the covered even position.
|
101
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $N=99999$. Then $N^{3}=$
|
$$
\text { Two, 1.999970000 } 299999 \text {. }
$$
Since $N=10^{5}-1$, then
$$
\begin{array}{l}
N^{3}=\left(10^{5}-1\right)^{3}=10^{15}-3 \times 10^{10}+3 \times 10^{5}-1 \\
=10^{10}\left(10^{5}-3\right)+3 \times 10^{5}-1 \\
=999970000299999 .
\end{array}
$$
|
999970000299999
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In the arithmetic sequence $\left\{a_{n}\right\}$, if $a_{2}+a_{4}+a_{6}+$ $a_{8}+a_{10}=80$, then $a_{7}-\frac{1}{2} a_{8}=(\quad)$.
(A) 4
(B) 6
(C) 8
(D) 10
|
4.C.
Since $a_{2}+a_{4}+a_{6}+a_{8}+a_{10}=5 a_{6}=80$, therefore, $a_{6}=16$.
Thus, $a_{7}-\frac{1}{2} a_{8}=a_{6}+d-\frac{1}{2}\left(a_{6}+2 d\right)=\frac{1}{2} a_{6}=8$.
|
8
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. If the three medians $A D$, $B E$, $C F$ of $\triangle A B C$ intersect at point $M$, then $M A+M B+M C=$
|
Ni, 11.0.
Diagram (omitted). Let the midpoint of $A B$ be $D$. By the parallelogram rule, we have
$$
\begin{array}{l}
M A+M B=2 M D=-M C \text {. } \\
\text { Therefore, } M A+M B+M C=0 \text {. } \\
\end{array}
$$
|
0
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In the sequence $\left\{a_{n}\right\}$, $a_{1}=2, a_{n}+a_{n+1}=1$ $\left(n \in \mathbf{N}_{+}\right)$, let $S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$. Then
$$
\mathrm{S}_{2007}-2 \mathrm{~S}_{2006}+\mathrm{S}_{2000}=
$$
$\qquad$
|
13.3.
When $n$ is even, we have
$$
a_{1}+a_{2}=a_{3}+a_{4}=\cdots=a_{n-1}+a_{n}=1 .
$$
Thus, $S_{n}=\frac{n}{2}$.
When $n$ is odd, we have
$$
a_{1}=2, a_{2}+a_{3}=a_{4}+a_{5}=\cdots=a_{n-1}+a_{n}=1 \text {. }
$$
Thus, $S_{n}=2+\frac{n-1}{2}=\frac{n+3}{2}$.
Therefore, $S_{2007}-2 S_{2006}+S_{2005}$
$$
=1005-2 \times 1003+1004=3 \text {. }
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. In $\triangle A B C$, it is known that $A B=\sqrt{30}, A C=$ $\sqrt{6}, B C=\sqrt{15}$, point $D$ makes $A D$ bisect $B C$, and $\angle A D B=90^{\circ}$, the ratio $\frac{S_{\triangle A D B}}{S_{\triangle A B C}}$ can be written in the form $\frac{m}{n}$, where $m, n$ are coprime positive integers. Then $m+n=$
|
14.65.
Let the midpoint of $BC$ be $E$, and $AD=\frac{x}{2}$. By the median formula, we get $AE=\frac{\sqrt{57}}{2}$.
Thus, $(\sqrt{30})^{2}-\left(\frac{x}{2}\right)^{2}$
$$
=\left(\frac{\sqrt{15}}{2}\right)^{2}-\left(\frac{x}{2}-\frac{\sqrt{57}}{2}\right)^{2} \text {. }
$$
Solving for $x$ gives $x=\frac{81}{\sqrt{57}}$.
Therefore, $\frac{m}{n}=\frac{S_{\triangle A D B}}{2 S_{\triangle E B}}=\frac{AD}{2 AE}=\frac{27}{38}$.
Thus, $m+n=27+38=65$.
|
65
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. (12 points) In $\triangle A B C$, it is known that $\sin A \cdot \cos ^{2} \frac{C}{2}+\sin C \cdot \cos ^{2} \frac{A}{2}=\frac{3}{2} \sin B$. Find the value of $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}$.
|
Three, 15. From the given,
$\sin A \cdot \frac{1+\cos C}{2}+\sin C \cdot \frac{1+\cos A}{2}=\frac{3}{2} \sin B$.
Then, $\sin A+\sin C+\sin A \cdot \cos C+\cos A \cdot \sin C$ $=3 \sin B$.
Thus, $\sin A+\sin C+\sin (A+C)=3 \sin B$, which means $\sin A+\sin C=2 \sin B$.
Therefore, $2 \sin \frac{A+C}{2} \cdot \cos \frac{A-C}{2}=4 \sin \frac{B}{2} \cdot \cos \frac{B}{2}$.
So, $\cos \frac{A-C}{2}=2 \sin \frac{B}{2}$, which is $\cos \frac{A-C}{2}-2 \sin \frac{B}{2}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Three people, A, B, and C, are playing a game. Each of them writes 100 different Chinese characters, and then they compare the characters written by each person. If a character is written by at least two people, it is deleted until there are no more identical characters. A, B, and C deleted 31, 27, and 39 characters, respectively. If from the 300 characters written by these three people, the maximum number of different characters that can be selected is $\mathrm{m}$, then the minimum value of $\mathrm{m}$ is ( ).
(A) 228
(B) 240
(C) 242
(D) 243
|
5.C.
Suppose there are $a$ Chinese characters that are the same for all three people, $b$ Chinese characters that are the same only for person A and B, $c$ Chinese characters that are the same only for person A and C, and $d$ Chinese characters that are the same only for person B and C. From the problem, we have:
$$
\left\{\begin{array}{l}
a+b+c=31, \\
a+b+d=27, \\
a+c+d=39 .
\end{array}\right.
$$
(1) - (2) gives $d=c-4$.
(2) - (3) gives $b=c-12$.
Thus, $a=43-2c$.
To satisfy $\left\{\begin{array}{l}a \geqslant 0, \\ b \geqslant 0, \\ c \geqslant 0, \\ d \geqslant 0,\end{array}\right.$ we have $\left\{\begin{array}{l}43-2c \geqslant 0, \\ c-12 \geqslant 0, \\ c \geqslant 0, \\ c-4 \geqslant 0 .\end{array}\right.$
Solving this, we get $12 \leqslant c \leqslant 21.5$.
Thus, the minimum value of $c$ is 12, and the maximum value is 21.
Therefore, the maximum number of different Chinese characters that can be selected is
$$
\begin{array}{l}
m=300-2a-b-c-d \\
=300-2(43-2c)-(c-12)-c-(c-4) \\
=230+c \geqslant 242 .
\end{array}
$$
|
242
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. There are weights of $11 \mathrm{~g}$ and $17 \mathrm{~g}$ available in sufficient quantity. To weigh an object of mass $3 \mathrm{~g}$ on a balance, at least $\qquad$ such weights are needed.
|
Let the weights of 11 g and 17 g be used $x$ times and $y$ times, respectively. Then we have $11 x - 17 y = 3$ or $17 y - 11 x = 3$.
The solutions are $\left\{\begin{array}{l}x=8+17 t, \\ y=5+11 t\end{array}\right.$ or $\left\{\begin{array}{l}x=9+17 t, \\ y=6+11 t .\end{array}\right.$
where $t$ is an integer.
Since $x$ and $y$ are positive integers, the minimum values of $x$ and $y$ can be
$$
\left\{\begin{array}{l}
x=8, \\
y=5
\end{array} \text { or } \left\{\begin{array}{l}
x=9, \\
y=6 .
\end{array}\right.\right.
$$
Therefore, the minimum value of $x+y$ is 13.
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, $\triangle ABC$ is an equilateral triangle with side length 8, $M$ is a point on side $AB$, $MP \perp AC$ at point $P$, $MQ \perp BC$ at point $Q$, and connect $PQ$.
(1) Find the minimum length of $PQ$;
(2) Find the maximum area of $\triangle CPQ$.
|
Solution: (1) Let the height of $\triangle ABC$ be $h$, then $h=4 \sqrt{3}$.
From $S_{\triangle C M}+S_{\triangle B C M}=S_{\triangle B B C}$, we get
$$
M P+M O=h=4 \sqrt{3} \text {. }
$$
As shown in Figure 11, draw perpendiculars from points $P$ and $Q$ to side $AB$, with the feet of the perpendiculars being $P_{1}$ and $Q_{1}$, respectively. Since
$$
\begin{array}{l}
\angle P M A=\angle Q M B \\
=30^{\circ},
\end{array}
$$
we have
$$
\begin{array}{l}
P_{1} M=P M \cos 30^{\circ}=\frac{\sqrt{3}}{2} P M, \\
Q_{1} M=Q M \cos 30^{\circ}=\frac{\sqrt{3}}{2} Q M, \\
P Q \geqslant P_{1} Q_{1}=P_{1} M+M Q_{1} \\
=\frac{\sqrt{3}}{2}(P M+Q M)=6 .
\end{array}
$$
When $M$ is the midpoint of $AB$, the equality holds. Therefore, the minimum value of $P Q$ is 6.
(2) Since $\angle P M A=\angle Q M B=30^{\circ}$, we have
$$
\begin{array}{l}
A P+B Q=\frac{1}{2} A M+\frac{1}{2} B M=\frac{1}{2} A B=4, \\
C P+C Q=16-(A P+B Q)=12 .
\end{array}
$$
Thus, $S_{\triangle C P Q}=\frac{1}{2} C P \cdot C Q \sin C=\frac{\sqrt{3}}{4} C P \cdot C Q$
$$
\leqslant \frac{\sqrt{3}}{4} \cdot \frac{(C P+C Q)^{2}}{4}=9 \sqrt{3} \text {. }
$$
When $M$ is the midpoint of $AB$, the equality holds. Therefore, the maximum area of $\triangle C P Q$ is $9 \sqrt{3}$.
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2-1 Given that $x, y, z$ are positive numbers, and $xyz(x+y+z)=1$.
Find the minimum value of $(x+y)(y+z)$.
(1989, All-Soviet Union Mathematical Competition)
|
Solution 1: $(x+y)(y+z)$
$$
\begin{array}{l}
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2 .
\end{array}
$$
When $x=z=1, y=\sqrt{2}-1$, $y(x+y+z) = xz$, $(x+y)(y+z)$ takes the minimum value 2.
Solution 2: As shown in Figure 1, construct
$\triangle ABC$, with side lengths
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$
Then the area of the triangle is
$$
\begin{array}{l}
S=\sqrt{p(p-a)(p-b)(p-c)} \\
=\sqrt{x y z(x+y+z)}=1 \text { (constant). } \\
\text { Also, }(x+y)(y+z)=a b=\frac{2 S}{\sin C} \geqslant 2, \text { when and }
\end{array}
$$
only when $\sin C=1\left(\angle C=90^{\circ}\right)$, $(x+y)(y+z)$ takes the minimum value 2.
This problem's integration of algebra and geometry has sparked our interest in exploration, providing some insights.
(1) Figure 1 and the corresponding transformation (1) serve as a bridge between the positive numbers $x, y, z$ and $\triangle ABC$, representing a stable approach to integrating algebra and geometry;
(2) Solution 2 tells us: Among triangles with a constant area, the product of two sides is minimized when their included angle is a right angle (right-angled triangle);
(3) Solution 1 tells us: $(x+y)(y+z)$ takes the minimum value under the condition
$$
\begin{array}{l}
x z=y(x+y+z) \\
\Leftrightarrow(z+x)^{2}=(x+y)^{2}+(y+z)^{2} .
\end{array}
$$
Combining this with the converse of the Pythagorean theorem, we see that $\triangle ABC$ is a right-angled triangle, which is consistent with the geometric observation. However, it also provides new insights, namely that from the constant area
$$
\begin{array}{c}
\sqrt{x y z(x+y+z)}=S \text { and equation (2), we get } \\
x z=y(x+y+z)=S .
\end{array}
$$
In this right-angled triangle, $y$ equals the radius of the inscribed circle, so $y(x+y+z)=S$ means that "the area of a right-angled triangle is equal to the product of the semiperimeter and the radius of the inscribed circle."
And $x z=S$ is the following conclusion.
Conclusion: The area of a right-angled triangle is equal to the product of the segments into which the hypotenuse is divided by the points of tangency of the inscribed circle.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2-3 Let quadrilateral $A B C D$ be a rectangle with an area of 2, $P$ a point on side $C D$, and $Q$ the point where the incircle of $\triangle P A B$ touches side $A B$. The product $P A \cdot P B$ varies with the changes in rectangle $A B C D$ and point $P$. When $P A \cdot P B$ is minimized,
(1) Prove: $A B \geqslant 2 B C$;
(2) Find the value of $A Q \cdot B Q$.
(2001, China Western Mathematical Olympiad)
|
Solution 1: (1) From the given information,
$S_{\triangle P A B}=\frac{1}{2} S_{\text {rectangle } A B C D}=1$.
Thus, $P A \cdot P B=\frac{2 S_{\triangle P A B}}{\sin \angle A P B} \geqslant 2$.
Equality holds if and only if $\angle A P B=90^{\circ}$. At this time, point $P$ lies on the circle with $A B$ as its diameter, meaning this circle intersects $C D$ at point $P$.
By the property that "in a right triangle, the median to the hypotenuse is not less than the altitude to the hypotenuse," when $P A \cdot P B$ reaches its minimum value, we have
$$
B C \leqslant \frac{A B}{2} \Rightarrow A B \geqslant 2 B C \text {. }
$$
(2) In the right triangle $\triangle P A B$, by the property that "the lengths of the tangents from a point outside a circle to the circle are equal," we get
$$
|P A-P B|=|A Q-B Q| .
$$
Squaring both sides, we obtain
$$
P A^{2}+P B^{2}-2 P A \cdot P B=A B^{2}-4 A Q \cdot B Q \text {. }
$$
Thus, $A Q \cdot B Q=\frac{1}{2} P A \cdot P B=1$.
Solution 2: As shown in Figure 3, let the incircle of $\triangle P A B$ touch $P A$ and $P B$ at points $M$ and $N$, respectively. Let
$$
\begin{array}{l}
A Q=A M=z, \\
B Q=B N=x, \\
P N=P M=y .
\end{array}
$$
Then $S_{\triangle P A B}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{x y z(x+y+z)} \\
=\frac{1}{2} S_{\text {rectangle } A B C D}=1 .
\end{array}
$$
Thus, $x y z(x+y+z)=1$.
$$
\begin{array}{l}
\text { Also, } P A \cdot P B=(y+z)(x+y) \\
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2,
\end{array}
$$
Therefore, when $x z=y(x+y+z)$, $P A \cdot P B$ reaches its minimum value of 2.
(1) Multiplying both sides of $x z=y(x+y+z)$ by 2 and adding $x^{2}+z^{2}$, we get
$$
(x+z)^{2}=(y+z)^{2}+(x+y)^{2},
$$
which means $P A^{2}+P B^{2}=A B^{2}$.
Thus, $\triangle P A B$ is a right triangle, and $B C$ is the altitude to the hypotenuse of the right triangle. Hence,
$$
2 B C=\frac{2 P A \cdot P B}{A B} \leqslant \frac{P A^{2}+P B^{2}}{A B}=A B .
$$
(2) Adding $x z$ to both sides of $x z=y(x+y+z)$, we get
$$
2 x z=(x+y)(y+z) \text {, }
$$
which means $2 A Q \cdot B Q=P A \cdot P B=2$.
Thus, $A Q \cdot B Q=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $p$ be a positive odd number. Then the remainder of $p^{2}$ divided by 8 is $\qquad$ .
|
Because $p$ is an odd positive integer, let $p=2k-1\left(k \in \mathbf{N}_{+}\right)$, so
$$
p^{2}=(2k-1)^{2}=4k^{2}-4k+1=4(k-1)k+1 \text{. }
$$
Since $(k-1)k$ is even, therefore, $p^{2}$ leaves a remainder of 1 when divided by 8.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given an isosceles triangle $\triangle A B C$ with side lengths $a$, $b$, and $c$ all being integers, and satisfying $a+b c+b+c a=24$. Then the number of such triangles is $\qquad$.
|
4.3.
$$
\begin{array}{l}
\text { Since } a+b c+b+a a \\
=(a+b)(c+1)=24=12 \times 2=8 \times 3=6 \times 4, \text { and }
\end{array}
$$
$\triangle A B C$ is an isosceles triangle, so the length of the base can only be $c$.
Thus, there are 3 triangles that satisfy the conditions:
$$
c=1, a=b=6 ; c=2, a=b=4 ; c=3, a=b=3 \text {. }
$$
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (20 points) Given that the graph of the linear function $y=a x+b$ passes through the points $A(\sqrt{3}, \sqrt{3}+2), B(-1, \sqrt{3})$, and $C(c, 2-c)$. Find the value of $a-b+c$.
---
The above text has been translated into English, preserving the original text's line breaks and format.
|
Three, from $\left\{\begin{array}{l}\sqrt{3}+2=\sqrt{3} a+b, \\ \sqrt{3}=-a+b\end{array} \Rightarrow\left\{\begin{array}{l}a=\sqrt{3}-1, \\ b=2 \sqrt{3}-1 .\end{array}\right.\right.$
Therefore, $2-c=a c+b=(\sqrt{3}-1) c+(2 \sqrt{3}-1)$.
Solving for $c$ gives $c=\sqrt{3}-2$.
Thus, $a-b+c=-2$.
|
-2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (50 points)
As shown in Figure 3, in a $7 \times 8$
rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected." Now, some of the 56 chess pieces are to be removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally) in sequence. What is the minimum number of chess pieces that need to be removed to meet this requirement? Explain your reasoning.
|
Second, at least 11 chess pieces must be taken out to possibly meet the requirement.
The reason is as follows:
If a square is in the $i$th row and the $j$th column, then this square is denoted as $(i, j)$.
Step 1 Proof: If any 10 chess pieces are taken, then the remaining chess pieces must have a five-in-a-row, i.e., five chess pieces are consecutively connected in a straight line (horizontally, vertically, or diagonally).
Proof by contradiction.
Assume that 10 chess pieces can be taken such that the remaining chess pieces do not have a five-in-a-row. As shown in Figure 8, one chess piece must be taken from each of the first five squares in every row, and one chess piece must be taken from each of the first five squares in the last three columns. Thus, the 10 taken chess pieces will not be distributed in the shaded area in the lower right corner of Figure 8. By symmetry, they will not be distributed in the shaded areas in the other corners.
The 1st and 2nd rows must each have 1 chess piece taken, and these can only be distributed in the squares $(1,4)$, $(1,5)$, $(2,4)$, and $(2,5)$.
Similarly, at least 2 chess pieces must be taken from the squares $(6,4)$, $(6,5)$, $(7,4)$, and $(7,5)$.
In the 1st, 2nd, and 3rd columns, at least 1 chess piece must be taken from each column, distributed in the squares $(3,1)$, $(3,2)$, $(3,3)$, $(4,1)$, $(4,2)$, $(4,3)$, $(5,1)$, $(5,2)$, and $(5,3)$.
Similarly, at least 3 chess pieces must be taken from the squares $(3,6)$, $(3,7)$, $(3,8)$, $(4,6)$, $(4,7)$, $(4,8)$, $(5,6)$, $(5,7)$, and $(5,8)$.
Thus, at least 10 chess pieces must be taken from these areas.
Therefore, no chess pieces can be taken from the central shaded area.
Since at most 2 of the 4 chess pieces (1), (2), (3), and (4) can be taken, there will inevitably be a five-in-a-row from the diagonal direction. This is a contradiction.
Step 2 Construction of a method:
Take a total of 11
chess pieces, and the
remaining chess pieces
will not have a five-in-a-row.
As shown in Figure 9, as long as the chess pieces in the marked positions are taken,
the remaining chess pieces will not
have a five-in-a-row.
In summary, at least 11 chess pieces must be taken to possibly meet the requirement.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that $\alpha, \beta$ are the roots of the equation $x^{2}+x-1=0$. Then the value of $\alpha^{16}-987 \beta+411$ is ( ).
(A) 2006
(B) 2007
(C) 2008
(D) 2009
|
5.C.
From the given, we have $\alpha^{2}=1-\alpha$ and $\alpha+\beta=-1$. Therefore,
$$
\begin{array}{l}
\alpha^{4}=(1-\alpha)^{2}=1-2 \alpha+\alpha^{2}=2-3 \alpha, \\
\alpha^{8}=(2-3 \alpha)^{2}=4-12 \alpha+9 \alpha^{2}=13-21 \alpha, \\
\alpha^{16}=(13-21 \alpha)^{2}=169-546 \alpha+441 \alpha^{2}=610-987 \alpha .
\end{array}
$$
Thus, $\alpha^{16}-987 \beta+411=(610-987 \alpha)-987 \beta+411$
$$
=1021-987(\alpha+\beta)=1021+987=2008 .
$$
|
2008
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) Let $x, y, a, m, n$ be positive integers, and $x+y=a^{m}, x^{2}+y^{2}=a^{n}$. Find how many digits $a^{30}$ has.
保留源文本的换行和格式，直接输出翻译结果。
|
Three, from the known we get
$$
a^{2 m}=x^{2}+y^{2}+2 x y=a^{n}+2 x y \text {. }
$$
From the problem and equation (1), we know that $a^{2 m}>a^{n}$. Therefore, $2 m>n$.
Dividing both sides of equation (1) by $a^{n}$, we get
$$
a^{2 m-n}=1+\frac{2 x y}{a^{n}} \text {. }
$$
Since the left side of equation (2) is a positive integer, $a^{n}$ must divide $2 x y$.
Thus, $2 x y \geqslant a^{n}=x^{2}+y^{2}$, which means $(x-y)^{2} \leqslant 0$.
Solving this, we get $x=y$.
Therefore, $2 x=a^{m}, 2 x^{2}=a^{n}$. So, $a^{2 m-n}=2$.
Considering that $a$ and $2 m-n$ are both positive integers, we have $a=2$, at this time,
$$
2 m-n=1 \text {. }
$$
Next, we need to find how many digits $2^{30}$ has.
Since $2^{30}=\left(2^{10}\right)^{3} = 1024^{3} > \left(10^{3}\right)^{3} = 10^{9}$, so, $2^{30}$ is a number with no fewer than 10 digits.
$$
\begin{array}{l}
\text { Also, } \frac{2^{30}}{10^{9}}=\frac{1}{10} \times \frac{1024^{3}}{1000^{3}}<\frac{1}{10} \times\left[\frac{1025}{1000}\right]^{3} \\
=\frac{1}{10} \times\left[\frac{41}{40}\right]^{3}<\frac{1}{10} \times\left[\frac{41}{40} \times \frac{40}{39} \times \cdots \times \frac{12}{11}\right] \\
=\frac{1}{10} \times \frac{41}{11}=\frac{41}{110}<1,
\end{array}
$$
Therefore, $2^{30}<10^{9}$.
Hence, $2^{30}$ is a number with fewer than 10 digits.
Thus, we can conclude that $2^{30}$ is a 9-digit number.
(Li Ming, Wang Chengyong, Wuhu County No.3 Middle School, Anhui Province, 233300)
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In a convex quadrilateral $A B C D$, $A B+A C+C D=16$. When the diagonals $A C$ and $B D$ are what values, will the area of quadrilateral $A B C D$ be maximized? What is the maximum area?
|
(Hint: Let $A B=x, A C=y$, then $C D=16-x-y$. And
$$
\begin{array}{l}
S_{\text {quadrilateral } A B C D}=S_{\triangle A B C}+S_{\triangle C D} \\
\leqslant \frac{1}{2} x y+\frac{1}{2} y(16-x-y) \\
=-\frac{1}{2}(y-8)^{2}+32 .
\end{array}
$$
Therefore, when $\angle B A C=\angle A C D=90^{\circ}, A C=8, B D=8 \sqrt{2}$, the maximum area of quadrilateral $A B C D$ is 32. )
|
32
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $A=\{0,1,2,3,4,5,6,7\}, f: A \rightarrow$ $A$. If $i+j=7$, then $f(i) f(j)=i j$. Then, the number of mappings $f$ is $\qquad$
|
For $0$ and $7$ elements, since $f(i) f(j)=0$, at least one of the images of $0$ and $7$ must be $0$. There are a total of $2 \times 8-1=15$ cases.
For the images of $1$ and $6$ elements, in this case, $f(i) f(j)=6=1 \times 6=2 \times 3$, so the images of $1$ and $6$ can be: $1, 6, 6, 1, 2, 3, 3, 2$, totaling 4 cases.
Similarly, the images of $2$ and $5$ elements have 2 cases, and the images of $3$ and $4$ elements have 4 cases.
Therefore, the number of mappings $f$ is $15 \times 4 \times 2 \times 4=480$.
|
480
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A five-digit number consists of only the three different digits $1, 3, 5$. Then the number of such five-digit numbers is $\qquad$.
|
3.150.
Considering overall, there are
$$
3^{5}-C_{3}^{2}\left(2^{5}-2\right)-3=150 \text { (cases). }
$$
|
150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In a $4 \times 4$ grid, eighteen lines can be drawn, including eight lines (four horizontal and four vertical), and five diagonal lines from the top left to the bottom right and from the top right to the bottom left. These diagonal lines may pass through 2, 3, or 4 small squares. Ten chips are to be placed in the grid, with a maximum of one chip per small square. If a line has an even number of chips, 1 point is scored. What is the maximum number of points that can be scored?
|
2. As shown in Figure 11, the maximum score is 17.
The following proves that 18 points cannot be achieved.
Notice that, in Figure 11, there are always two opposite corner cells not covered by the same direction of five diagonal lines, and these two cells must be either both empty or both full. At the same time, there must be one row and one column that are full. Consider three scenarios.
(1) All four corner cells are empty.
By symmetry, assume without loss of generality that the second row and the second column are full. In this case, another cell in the first row, fourth row, first column, and fourth column must be filled. This requires at least 11 chips in total.
(2) Exactly two diagonal cells are empty.
By symmetry, assume without loss of generality that one of the cells is in the first row and first column, and the other is in the fourth row and fourth column. Then, one chip must be placed in the first row, first column, fourth row, and fourth column. This indicates that the four internal cells must all contain chips.
By symmetry, assume without loss of generality that the full row is the second row. In this case, the length-2 diagonals intersecting the second row cannot all score.
(3) All four corner cells are full.
Claim: The full row is either the first row or the fourth row.
If the second row is full, then the first column and the fourth column must also be filled, which already uses 10 chips, but multiple diagonals cannot score. This confirms the claim.
By symmetry, assume without loss of generality that the first row and the first column are full. In this case, no matter how the remaining 2 chips are placed, some diagonal will not score.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Mary found that when a certain three-digit number is squared, the last three digits of the product are the same as the original three-digit number. What is the sum of all different three-digit numbers that satisfy this property?
|
4. Let the three-digit number be $\overline{a b c}$, then
$$
\overline{a b c}^{2}=1000 k+\overline{a b c} \text {, }
$$
i.e., $\overline{a b c}(\overline{a b c}-1)=2^{3} \times 5^{3} k$.
Since $(\overline{a b c}, \overline{a b c}-1)=1$, therefore,
$2^{3} \mid \overline{a b c}$, and $5^{3} \mid(\overline{a b c}-1)$,
or $2^{3} \mid (\overline{a b c}-1)$, and $5^{3} \mid \overline{a b c}$.
(1) If $2^{3} \mid \overline{a b c}$, and $5^{3} \mid(\overline{a b c}-1)$, then
$$
\overline{a b c}-1=125,375,625,875 \text {. }
$$
Only $\overline{a b c}=376$ satisfies $2^{3} \mid \overline{a b c}$, meeting the condition.
(2) If $2^{3} \mid (\overline{a b c}-1)$, and $5^{3} \mid \overline{a b c}$, then
$$
\overline{a b c}=125,375,625,875 \text {. }
$$
Only $\overline{a b c}=625$ satisfies $2^{3} \mid (\overline{a b c}-1)$, meeting the condition.
Therefore, the sum is $376+625=1001$.
|
1001
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Arrange the $n$ positive integers from $1 \sim n (n>1)$ in a row so that the sum of any two adjacent numbers is a perfect square. What is the minimum value of $n$?
|
9. The smallest positive integer $n$ is 15.
Since $n>1$, it includes 2. The smallest positive integer that, when added to 2, results in a perfect square is 7, represented as $2+7=9$ (the same notation applies below), so $n \geqslant 7$.
If $n=7$, we can get three non-adjacent segments: $(1,3,6),(2,7),(4,5)$.
Adding 8 can only change the first segment to $(8,1,3,6)$, and adding 9 changes the second segment to $(2,7,9)$. Therefore, $n \geqslant 10$.
Since $8+1=9,9+7=16,10+6=16$, 8, 9, and 10 must all be at the end of the array.
$$
\text { Also, } 8+17=25,9+16=25,10+15=25 \text {, }
$$
Thus, $n \geqslant 15$.
When $n=15$, the 15 numbers can be arranged as:
$8,1,15,10,6,3,13,12,4,5,11,14,2,7,9$.
Therefore, the smallest positive integer $n$ is 15.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among the 2007 positive integers $1,2, \cdots, 2007$, what is the maximum number of integers that can be selected such that each selected number is coprime with 2007, and the sum of any three selected numbers is not a multiple of 7?
|
3. When $1,2, \cdots, 2007$ are divided by 7, the remainders 1, 2, 3, 4, 5 each have $286+1=287$ numbers; the remainders 6, 0 each have 286 numbers.
Among $1,2, \cdots, 2007$, the numbers that are not coprime with 2007 are $3,2 \times 3,3 \times 3, \cdots, 669 \times 3$ and $223,2 \times 223,4 \times 223,5 \times 223,7 \times 223,8 \times 223$.
When these numbers that are not coprime with 2007 are divided by 7, the remainders are $3,6,2,5,1,4,0,3,6,2,5,1,4,0$, $\cdots, 3,6,2,5$ and $6,5,3,2,0,6$.
Thus, among these numbers that are not coprime with 2007, the remainders 1, 2, 3, 4, 5, 6, 0 have $95,97,97,95,97,98,96$ numbers respectively.
Among $1,2, \cdots, 2007$ and coprime with 2007, the remainders 1, 2, 3, 4, 5, 6, 0 have 192, 190, 190, 192, 190, 188, 190 numbers respectively.
To ensure that the sum of any three of the selected numbers is not a multiple of 7, at most 2 numbers with a remainder of 0 can be taken. Since the sum of the remainders (1, 3, 3), (3, 2, 2), (2, 6, 6), (6, 4, 4), (4, 5, 5), (5, 1, 1) and (1, 2, 4), (3, 6, 5) are all multiples of 7, at most 2 groups of the other remainders that are not adjacent in Figure 2 can be taken.
It has been verified that taking all the numbers with remainders 1, 4 from 2 groups, and 2 numbers with a remainder of 0, meets the requirements of the problem, and the number of selected numbers reaches the maximum. Therefore, the maximum number of numbers that can be selected, each of which is coprime with 2007, and the sum of any three of which is not a multiple of 7, is $192+192+2=386$.
|
386
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) As shown in Figure 5, there are 12 equally spaced points on the circumference of a clock face, marked with the numbers $1, 2, \cdots, 12$ in sequence. Please use these equally spaced points as vertices to form 4 triangles (dividing these 12 equally spaced points into 4 groups) such that the following conditions are met:
(1) No two triangles share a common vertex;
(2) In each triangle, one of the numbers marked on the vertices is equal to the sum of the other two numbers.
Find all different grouping schemes.
(A triangle with vertices $a, b, c$ can be represented as $(a, b, c)$.)
|
Three, let 4 triangles be $\left(a_{i}, b_{i}, c_{i}\right), i=1,2, 3,4$, where $a_{i}=b_{i}+c_{i}, b_{i}>c_{i}$.
Let $a_{1}=27$.
Thus, $10 \leqslant a_{3} \leqslant 11$.
If $a_{3}=10$, then from $a_{1}+a_{2}=17, a_{2}a_{1}+a_{2}=17$, we get $a_{2}=9, a_{1}=8$, that is
$$
\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(8,9,10,12) \text {. }
$$
From $8=b_{1}+c_{1}, 9=b_{2}+c_{2}, 10=b_{3}+c_{3}, 12=b_{4}+c_{4}$, it must be that $b_{4}=11, c_{4}=1$, yielding two cases:
$$
\begin{array}{l}
12=11+1,10=7+3,9=5+4,8=6+2 ; \\
12=11+1,10=6+4,9=7+2,8=5+3 .
\end{array}
$$
Corresponding to two divisions:
$$
\begin{array}{l}
(12,11,1),(10,7,3),(9,5,4),(8,6,2) ; \\
(12,11,1),(10,6,4),(9,7,2),(8,5,3) .
\end{array}
$$
If $a_{3}=11$, then $a_{1}+a_{2}=16$.
Thus, $8<a_{2}<11$, yielding respectively
$$
\left(a_{1}, a_{2}\right)=(6,10),(7,9) \text {. }
$$
For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(6,10,11,12)$, we get three divisions:
$(12,8,4),(11,9,2),(10,7,3),(6,5,1)$; $(12,9,3),(11,7,4),(10,8,2),(6,5,1)$; $(12,7,5),(11,8,3),(10,9,1),(6,4,2)$.
For $\left(a_{1}, a_{2}, a_{3}, a_{4}\right)=(7,9,11,12)$, we also get three divisions:
$(12,8,4),(11,10,1),(9,6,3),(7,5,2)$; $(12,10,2),(11,6,5),(9,8,1),(7,4,3)$; $(12,10,2),(11,8,3),(9,5,4),(7,6,1)$.
Therefore, there are eight grouping schemes in total.
(Tao Pingsheng, Department of Mathematics and Computer Science, Jiangxi Science and Technology Normal University, 330013)
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $a, b, c$ are 3 different positive integers, and $a b c=16$, then the maximum possible value of $a^{b}-b^{c}+c^{a}$ is ( ).
(A) 249
(B) 253
(C) 263
(D) 264
|
$-1 . C$.
From the given, we easily know that $\{a, b, c\}=\{1,2,8\}$.
And when $b=1, c=2, a=8$, $a^{b}-b^{c}+c^{a}$ reaches the maximum value $8-1+2^{8}=263$.
|
263
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Arrange positive integers starting from 1 in a number array according to the pattern shown in Figure 2, where 2 is at the 1st corner, 3 is at the 2nd corner, 5 is at the 3rd corner, 7 is at the 4th corner, $\cdots \cdots$. Then, the number at the 2007th corner is $\qquad$
Figure 2
|
$$
\begin{array}{l}
a_{1}=2, a_{2 i}=a_{2 i-1}+i, \\
a_{2 i+1}=a_{2 i}+(i+1) .
\end{array}
$$
Since $2007=2 \times 1003+1$, we have,
$$
\begin{array}{l}
a_{2007}=1+2(1+2+\cdots+1003)+1004 \\
=1004^{2}+1=1008017 .
\end{array}
$$
|
1008017
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a $3 \times 3$ grid, the numbers $1,2,3,4,5,6,7,8,9$ are filled in, with one number per cell. Now, the cells containing the maximum number in each row are colored red, and the cells containing the minimum number in each row are colored green. Let $M$ be the smallest number in the red cells, and $m$ be the largest number in the green cells. Then $M-m$ can take on different values.
|
2.8
Obviously, $3 \leqslant m, M \leqslant 7$, and $m \neq M$. Therefore,
$$
M-m \in\{-4,-3,-2,-1,1,2,3,4\} \text {. }
$$
As shown in Figure 7, all 8 values can be obtained
\begin{tabular}{|l|l|l|}
\hline 7 & 9 & 8 \\
\hline 6 & 5 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 6 & 9 & 8 \\
\hline 7 & 5 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 5 & 9 & 8 \\
\hline 7 & 6 & 4 \\
\hline 3 & 2 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 5 & 9 & 8 \\
\hline 7 & 6 & 1 \\
\hline 4 & 3 & 2 \\
\hline
\end{tabular}
$M=3, m=7$
$M=3, m=6$
$M=3, m=5$
$M=4, m=5$
\begin{tabular}{|l|l|l|} \hline 9 & 8 & 5 \\ \hline 7 & 4 & 2 \\ \hline 6 & 3 & 1 \\ \hline \end{tabular}
\begin{tabular}{|l|l|l|} \hline 9 & 6 & 5 \\ \hline 8 & 4 & 2 \\ \hline 7 & 3 & 1 \\ \hline \end{tabular}
\begin{tabular}{|l|l|l|}
\hline 9 & 6 & 4 \\
\hline 8 & 5 & 2 \\
\hline 7 & 3 & 1 \\
\hline
\end{tabular}
\begin{tabular}{|l|l|l|}
\hline 9 & 6 & 3 \\
\hline 8 & 4 & 2 \\
\hline 7 & 5 & 1 \\
\hline
\end{tabular}
$M=6, m=5$
$M=7, m=5$
$M=7, m=4$
$M=7, m=3$
Figure 7
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (15 points) If for any $n$ consecutive positive integers, there always exists a number whose sum of digits is a multiple of 8. Determine the minimum value of $n$. And explain the reason.
---
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Five, first prove that when $n \leqslant 14$, the property in the problem does not hold.
When $n=14$, for
$9999993, 9999994, \cdots, 10000006$
these 14 consecutive integers, the sum of the digits of any number cannot be divisible by 8.
Therefore, when $n \leqslant 14$, the property in the problem does not hold.
Thus, to make the property in the problem hold, we should have
$n \geqslant 15$.
Next, prove that when $n=15$, the property in the problem holds.
Let $a_{1}, a_{2}, \cdots, a_{15}$ be any 15 consecutive positive integers. Then, among these 15 positive integers, the number of integers with a units digit of 0 is at most two and at least one, which can be divided into:
(1) When $a_{1}, a_{2}, \cdots, a_{15}$ have two integers with a units digit of 0, let $a_{i}<a_{j}$, and $a_{i}, a_{j}$ have a units digit of 0, then $a_{i}, a_{i}+1, \cdots, a_{i}+9, a_{j}$ are 11 consecutive integers, among which $a_{i}, a_{i}+1, \cdots, a_{i}+9$ have no carry.
Let $n_{i}$ represent the sum of the digits of $a_{i}$, then the sum of the digits of the first 10 numbers are $n_{i}, n_{i}+1, \cdots, n_{i}+9$. Therefore, among these 10 consecutive numbers, at least one is divisible by 8.
(2) When $a_{1}, a_{2}, \cdots, a_{15}$ have only one integer with a units digit of 0 (denoted as $a_{i}$):
(i) If the integer $i$ satisfies $1 \leqslant i \leqslant 8$, then there are at least 7 consecutive integers after $a_{i}$. Thus, $a_{i}, a_{i}+1, \cdots, a_{i}+7$ are 8 consecutive integers, and the sum of the digits of these 8 consecutive integers are also 8 consecutive integers. Therefore, there must be one number that is divisible by 8.
(ii) If the integer $i$ satisfies $9 \leqslant i \leqslant 15$, then there are at least 8 consecutive integers before $a_{i}$, which can be denoted as $a_{i-8}$, $a_{i-7}, \cdots, a_{i-1}$. The sum of the digits of these 8 consecutive integers are also 8 consecutive integers. Therefore, there must be one number that is divisible by 8.
In summary, for any 15 consecutive integers, there must be one number whose sum of digits is a multiple of 8.
And for any fewer than 15 consecutive integers, this property does not hold.
Therefore, the minimum value of $n$ is 15.
|
15
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of positive integers $m$ that make $m^{2}+m+7$ a perfect square is $\qquad$ .
|
3.2.
It has been verified that when $m=1$, $m^{2}+m+7=9$ is a perfect square;
when $m=2,3,4,5$, $m^{2}+m+7$ are not perfect squares;
when $m=6$, $m^{2}+m+7=49$ is a perfect square.
When $m>6$, $m^{2}+m+7$ are not perfect squares.
Therefore, there are only 2 positive integers $m$ that meet the condition.
|
2
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (20 points) For any real numbers $x, y$, we have
$$
|x-2|+|x-4| \geqslant m\left(-y^{2}+2 y\right)
$$
Determine the maximum value of the real number $m$.
|
Three, by the geometric meaning of absolute value, $|x-2|+|x-4|$ has a minimum value of 2 when $x \in [2,4]$.
And $-y^{2}+2y=-(y-1)^{2}+1$ has a maximum value of 1 when $y=1$.
From the condition, $2 \geqslant m \times 1$, then $m \leqslant 2$.
Therefore, the maximum value of $m$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (25 points) Given the system of equations in $x$ and $y$
$$
\left\{\begin{array}{l}
x^{2}-y^{2}=p, \\
3 x y+p(x-y)=p^{2}
\end{array}\right.
$$
has integer solutions $(x, y)$. Find the prime number $p$ that satisfies the condition.
|
Five, from $p=x^{2}-y^{2}=(x-y)(x+y)$ and $p$ being a prime number, we have
$\left\{\begin{array}{l}x+y=p, \\ x-y=1\end{array}\right.$ or $\quad\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$
or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p .\end{array}\right.$
Substituting these into $3 x y+p(x-y)=p^{2}$, we get $\frac{3}{4}\left(p^{2}-1\right)+p=p^{2}$,
which simplifies to $p^{2}-4 p+3=0$.
Solving this, we get $p=3$ or $p=1$ (discard).
(2) When $\left\{\begin{array}{l}x+y=-p, \\ x-y=-1\end{array}\right.$ or $\left\{\begin{array}{l}x+y=1, \\ x-y=p\end{array}\right.$ or $\left\{\begin{array}{l}x+y=-1, \\ x-y=-p\end{array}\right.$, it is found through calculation that there are no prime numbers $p$ that satisfy the conditions.
Therefore, the prime number $p$ that satisfies the conditions is $p=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In the pentagram $A B C D E$, the intersection points of each line segment are shown in Figure 3, $A G=G F=F D, B G=$ $3 E G, F C=2 F E$. If the area of pentagon $F G H I J$ is 91, try to find the area of the pentagram $A B C D E$.
|
Solution: As shown in Figure 3, connect $F H, F I, G J, H J, A E,$ and $B C$.
From $\frac{S_{\triangle A E C}}{S_{\triangle A C F}} \cdot \frac{S_{\triangle C F A}}{S_{\triangle C A G}}$.
$\frac{S_{\triangle C A G}}{S_{\triangle C A E}}=1$, we get
$$
\frac{E C}{C F} \cdot \frac{F A}{A G} \cdot \frac{G H}{H E}=1,
$$
which means $\frac{3}{2} \cdot \frac{2}{1} \cdot \frac{G H}{H E}=1$.
Thus, $\frac{G H}{H E}=\frac{1}{3}$.
Also, $B G=3 E G$, so
$$
E G: G H: H B=2: 1: 5.
$$
Similarly, $E F: F J: J C=5: 3: 7$.
Notice that
$\frac{S_{\triangle A E F}}{S_{\triangle A F C}} \cdot \frac{S_{\triangle F C A}}{S_{\triangle F A H}} \cdot \frac{S_{\triangle F A H}}{S_{\triangle F A E}}=1$,
$\frac{S_{\triangle B C J}}{S_{\triangle B E E}} \cdot \frac{S_{\triangle \triangle E B}}{S_{\triangle B H}} \cdot \frac{S_{\triangle B B}}{S_{\triangle A B C}}=1$,
so $\frac{E F}{F C} \cdot \frac{C A}{A H} \cdot \frac{H G}{G E}=1, \frac{C J}{J E} \cdot \frac{E B}{B H} \cdot \frac{H I}{I C}=1$,
which means $\frac{1}{2} \cdot \frac{C A}{A H} \cdot \frac{1}{2}=1, \frac{7}{8} \cdot \frac{8}{5} \cdot \frac{H I}{I C}=1$.
Therefore, $A H: H I: I C=4: 5: 7$.
Similarly, $D J: J I: I B=8: 7: 25$.
Let $S_{\triangle A G H}=x$. Noting that "the ratio of the areas of triangles with the same height is equal to the ratio of the corresponding bases," and using the line segment ratios obtained earlier, we have
$$
\begin{array}{c}
S_{\triangle F G H}=x, S_{\triangle E F G}=2 x, S_{\triangle E F H}=3 x, \\
S_{\triangle F H J}=\frac{3}{5} S_{\triangle E F H}=\frac{9}{5} x, \\
S_{\triangle F G J}=\frac{3}{5} S_{\triangle E F G}=\frac{6}{5} x, \\
S_{\triangle D F J}=S_{\triangle F G J}=\frac{6}{5} x, \\
S_{\triangle F I J}=\frac{7}{8} S_{\triangle D F J}=\frac{21}{20} x, \\
S_{\triangle C I J}=\frac{7}{3} S_{\triangle F I J}=\frac{49}{20} x, \\
S_{\triangle H H J}=\frac{5}{7} S_{\triangle C I J}=\frac{7}{4} x, \\
S_{\triangle B H I}=\frac{25}{7} S_{\triangle H I J}=\frac{25}{4} x .
\end{array}
$$
Notice that the area of pentagon $F G H I J$ is the sum of the areas of $\triangle F G H$, $\triangle F H J$, and $\triangle H I J$, so
$$
x+\frac{9}{5} x+\frac{7}{4} x=91 .
$$
Solving for $x$, we get $x=20$.
Therefore, $S_{\text {pentagram } A B C D E}$
$$
\begin{array}{l}
=S_{\triangle A G H}+S_{\triangle E F G}+S_{\triangle D F J}+S_{\triangle C I J}+S_{\triangle B H I}+91 \\
=x+2 x+\frac{6}{5} x+\frac{49}{20} x+\frac{25}{4} x+91=349 .
\end{array}
$$
|
349
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. From $1,2, \cdots, 2006$, at least
$\qquad$ odd numbers must be taken to ensure that there are definitely two numbers whose sum is 2008.
|
4.503.
From $1,2, \cdots, 2006$, selecting two odd numbers whose sum is 2008, there are a total of 501 pairs as follows:
$$
3+2005,5+2003, \cdots, 1003+1005 \text {. }
$$
Since 1 added to any of these odd numbers will not equal 2008, therefore, at least 503 odd numbers must be selected to ensure that there are definitely two numbers whose sum is 2008.
|
503
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Find the maximum and minimum values of $x+y+z+w$.
|
Let $z=w+a, y=w+a+b, x=w+$ $a+b+c$. Then $a, b, c \geqslant 0$, and
$$
x+y+z+w=4 w+3 a+2 b+c .
$$
Therefore, $100=5(w+a+b+c)+4(w+a+$
$$
\begin{array}{l}
\quad b)+3(w+a)+6 w \\
=18 w+12 a+9 b+5 c \\
=4(4 w+3 a+2 b+c)+(2 w+b+c) \\
\geqslant 4(x+y+z+w) .
\end{array}
$$
Thus, $x+y+z+w \leqslant 25$.
When $x=y=z=\frac{25}{3}, w=0$, the equality holds. Therefore, the maximum value of $x+y+z+w$ is 25.
$$
\begin{array}{l}
\text { Also, } 100=18 w+12 a+9 b+5 c \\
=5(4 w+3 a+2 b+c)-(2 w+3 a+b) \\
\leqslant 5(x+y+z+w),
\end{array}
$$
then $x+y+z+w \geqslant 20$.
When $x=20, y=z=w=0$, the equality holds. Therefore, the minimum value of $x+y+z+w$ is 20.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{r}
\text { Three. (25 points) In }\left[\frac{1^{2}}{2008}\right],\left[\frac{2^{2}}{2008}\right], \cdots, \\
{\left[\frac{2008^{2}}{2008}\right] \text {, how many different integers are there (where }[x]}
\end{array}
$$
indicates the greatest integer not greater than $x$)?
|
Three, let $f(n)=\frac{n^{2}}{2008}$.
When $n=2,3, \cdots, 1004$, we have
$$
\begin{array}{l}
f(n)-f(n-1) \\
=\frac{n^{2}}{2008}-\frac{(n-1)^{2}}{2008}=\frac{2 n-1}{2008}1 .
\end{array}
$$
And $f(1005)=\frac{1005^{2}}{2008}=\frac{(1004+1)^{2}}{2008}$
$$
=502+1+\frac{1}{2008}>503,
$$
Therefore, $\left[\frac{1005^{2}}{2008}\right],\left[\frac{1006^{2}}{2008}\right], \cdots,\left[\frac{2008^{2}}{2008}\right]$ are distinct integers.
Thus, in $\left[\frac{1^{2}}{2008}\right],\left[\frac{2^{2}}{2008}\right], \cdots,\left[\frac{2008^{2}}{2008}\right]$, there are a total of $503+1004=1507$ different integers.
|
1507
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Add the same integer $a(a>0)$ to the numerator and denominator of $\frac{2008}{3}$, making the fraction an integer. Then the integer $a$ added has $\qquad$ solutions.
|
$=1.3$.
From the problem, we know that $\frac{2008+a}{3+a}$ should be an integer, which means $\frac{2008+a}{3+a}=\frac{2005}{3+a}+1$ should be an integer. Therefore, $(3+a) \mid 2005=5 \times 401$.
Since 2005 has 4 divisors, at this point, $a$ can take the values $2002$, $398$, $2$, and one divisor corresponds to a value of $a$ that is less than 0. Therefore, there are 3 values of $a$ that meet the problem's requirements.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In rectangle $A B C D$, $A B=56, A D=35$. Using lines parallel to $A B$ and $A D$, it is divided into $35 \times$ 56 unit squares. The number of unit squares that are internally crossed by diagonal $A C$ is ( ) .
(A) 84
(B) 85
(C) 91
(D) 92
|
5.A.
Take $A$ as the origin, the lines $AB$ and $AD$ as the $x$-axis and $y$-axis, respectively. Apart from the coordinate axes, there are 35 horizontal lines and 56 vertical lines. The diagonal $AC$ intersects with each of these lines, resulting in $35+56=91$ intersection points (including coincident points).
Next, calculate the repeated points. Since $C(56,35)$, the greatest common divisor $(56,35)=7$, the intersections of horizontal and vertical lines $C_{k}(8k, 5k) (k=1,2, \cdots, 7)$ lie on $AC$. Therefore, the different intersection points of the grid lines on $AC$ are $91-7+1=85$ (including the origin $A$). These points divide $AC$ into 84 segments, each passing through a small square, totaling 84 small squares.
|
84
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. A bicycle tire, if installed on the front wheel, will wear out after traveling $5000 \mathrm{~km}$; if installed on the back wheel, it will wear out after traveling $3000 \mathrm{~km}$. If the front and back tires are swapped after traveling a certain distance, so that a pair of new tires wear out simultaneously, then the maximum distance that can be traveled is $\mathrm{km}$.
|
II. 1.3750.
Assume the total wear of each new tire when it is scrapped is $k$, then the wear per kilometer for a tire installed on the front wheel is $\frac{k}{5000}$, and the wear per kilometer for a tire installed on the rear wheel is $\frac{k}{3000}$. Let a pair of new tires travel $x \mathrm{~km}$ before swapping positions, and $y \mathrm{~km}$ after swapping positions. By setting up equations based on the total wear of a tire, we have
$$
\left\{\begin{array}{l}
\frac{k x}{5000}+\frac{k y}{3000}=k, \\
\frac{k y}{5000}+\frac{k x}{3000}=k .
\end{array}\right.
$$
Adding the two equations gives
$$
\frac{k(x+y)}{5000}+\frac{k(x+y)}{3000}=2 k \text {. }
$$
Thus, $x+y=\frac{2}{\frac{1}{5000}+\frac{1}{3000}}=3750(\mathrm{~km})$.
|
3750
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $x$ is a real number,
$$
\sqrt{x^{3}+2020}-\sqrt{2030-x^{3}}=54 \text {. }
$$
Then $28 \sqrt{x^{3}+2020}+27 \sqrt{2030-x^{3}}=$
|
II. 1.2007.
Let $x^{3}+2020=a, 2030-x^{3}=b$. Then $a+b=4050$.
From the problem, we have
$$
\begin{array}{l}
\sqrt{a}-\sqrt{b}=54 \\
\Rightarrow a+b-2 \sqrt{a b}=54^{2}=2916 \\
\Rightarrow 2 \sqrt{a b}=(a+b)-2916 \\
\quad=4050-2916=1134 \\
\Rightarrow(\sqrt{a}+\sqrt{b})^{2}=(a+b)+2 \sqrt{a b} \\
\quad=4050+1134=5184 \\
\Rightarrow \sqrt{a}+\sqrt{b}=72 .
\end{array}
$$
From equations (1) and (2), we get $\sqrt{a}=63, \sqrt{b}=9$.
Therefore, $28 \sqrt{x^{3}+2020}+27 \sqrt{2030-x^{3}}$ $=2007$.
|
2007
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $|a|>1$, simplify
$$
\left(a+\sqrt{a^{2}-1}\right)^{4}+2\left(1-2 a^{2}\right)\left(a+\sqrt{a^{2}-1}\right)^{2}+3
$$
the result is $\qquad$ .
|
Let $x_{0}=a+\sqrt{a^{2}-1}$. Clearly, $x_{0}$ is a root of the equation $x^{2}-2 a x+1=0$, and thus it is also a root of the equation
$$
\left(x^{2}-2 a x+1\right)\left(x^{2}+2 a x+1\right)=0
$$
Then $\left(x_{0}^{2}+1\right)^{2}-4 a^{2} x_{0}^{2}=0$, which means
$$
x_{0}^{4}+2\left(1-2 a^{2}\right) x_{0}^{2}+1=0 \text {. }
$$
Therefore, the original expression $=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $n$ is a natural number, $9 n^{2}-10 n+2009$ can be expressed as the product of two consecutive natural numbers. Then the maximum value of $n$ is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
3.2007 .
Let $9 n^{2}-10 n+2009=m(m+1)$, where $m$ is a natural number. Then
$$
9 n^{2}-10 n+\left(2009-m^{2}-m\right)=0 \text {. }
$$
Considering equation (1) as a quadratic equation in the natural number $n$, its discriminant should be the square of a natural number. Let's assume $\Delta=t^{2}(t \in \mathbf{N})$, then
$$
(-10)^{2}-4 \times 9\left(2009-m^{2}-m\right)=t^{2} \text {. }
$$
Simplifying and rearranging, we get
$$
(6 m+3)^{2}-t^{2}=72233,
$$
which is $(6 m+3+t)(6 m+3-t)=72233$.
Let $6 m+3+t=a, 6 m+3-t=b$. Then $t=\frac{1}{2}(a-b)$.
When $a=72233, b=1$, $t$ has its maximum value.
At this point, the maximum value of $t$ is 36116.
$$
\text { Also, } n=\frac{-(-10) \pm \sqrt{\Delta}}{2 \times 9}=\frac{10 \pm \sqrt{t^{2}}}{18}=\frac{10 \pm t}{18} \text {, }
$$
When $t$ takes its maximum value of 36116, $n$ takes its maximum value, which is
$$
\frac{10+36116}{18}=2007 .
$$
|
2007
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) During the New Year, the grandmother gave three grandsons a total of 400 yuan as New Year's money. There are 50 yuan, 20 yuan, and 10 yuan banknotes available in various quantities for the three grandsons to choose from, but each can only take banknotes of the same denomination. One of them takes a number of banknotes that is exactly equal to the product of the number of banknotes taken by the other two. How many ways are there to choose the denominations and the number of banknotes?
|
Three people take money in the amounts of $x y$, $x$, and $y$, with denominations of $a$ yuan, $b$ yuan, and $c$ yuan respectively, where $x \geqslant 1, y \geqslant 1$. Then $a x y + b x + c y = 400$.
(1) When $a = b = c$, i.e., the denominations chosen by the three people are the same, then
$$
x y + x + y = \frac{400}{a},
$$
which means $(x+1)(y+1) = \frac{400}{a} + 1$.
(i) If $a = 50$, then
$$
(x+1)(y+1) = 9 = 3 \times 3.
$$
Solving this, we get $x = 2, y = 2$.
Therefore, each of the three people selects 50 yuan in the amounts of 2, 2, and 4 respectively. In this case, there is one way to choose.
(ii) If $a = 20$, then
$$
(x+1)(y+1) = 21 = 3 \times 7.
$$
Solving this, we get $x = 2, y = 6$.
Therefore, there is one way to choose 20 yuan.
(iii) If $a = 10$, then $(x+1)(y+1) = 41$. In this case, $x$ and $y$ have no positive integer solutions.
(2) When two of $a$, $b$, and $c$ are equal, based on the lack of order in selection, it can be divided into two cases: $a = b \neq c$ and $b = c \neq a$.
$$
\begin{array}{l}
\text{When } a = b \neq c, \\
a x y + a x + c y = 400,
\end{array}
$$
which means $(a x + c)(y + 1) = 400 + c(y + 1 \geqslant 2)$.
(i) If $a = 50, c = 20$, then
$$
(50 x + 20)(y + 1) = 420,
$$
which means $(5 x + 2)(y + 1) = 42 = 7 \times 6$.
Solving this, we get $x = 1, y = 5$.
In this case, there is one way to choose.
(ii) If $a = 50, c = 10$, then
$(50 x + 10)(y + 1) = 410$,
which means $(5 x + 1)(y + 1) = 41$ (a prime number).
The above equation has no positive integer solutions.
(iii) If $a = 20, c = 50$, then
$(20 x + 50)(y + 1) = 450$,
which means $(2 x + 5)(y + 1) = 45 = 9 \times 5 = 15 \times 3$.
Solving this, we get $x = 2, y = 4$ or $x = 5, y = 2$.
Therefore, there are two ways to choose the denominations and amounts.
Similarly, we get:
(iv) If $a = 20, c = 10$, there are no positive integer solutions.
(v) If $a = 10, c = 50$, there are two ways to choose the denominations and amounts.
(vi) If $a = 10, c = 20$, there are five ways to choose the denominations and amounts.
(3) When $a$, $b$, and $c$ are all different, similarly, we get:
(i) If $a = 10, b = 20, c = 50$, there is one way to choose the denominations and amounts.
(ii) If $a = 20, b = 10, c = 50$, there is one way to choose the denominations and amounts.
(iii) If $a = 50, b = 10, c = 20$, there are no positive integer solutions.
In summary, there are 14 ways to choose the denominations and amounts.
(Xie Wenxiao, Huanggang High School, Hubei, 438000)
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Within the range of the independent variable $x$, $59 \leqslant x \leqslant 60$, the number of integer values of the quadratic function $y=x^{2}+x+\frac{1}{2}$ is ( ).
(A) 59
(B) 120
(C) 118
(D) 60
|
5.B.
Notice that $y=x^{2}+x+\frac{1}{2}=\left(x+\frac{1}{2}\right)^{2}+\frac{1}{4}$. When $x>-\frac{1}{2}$, $y$ increases as $x$ increases.
Since $59 \leqslant x \leqslant 60$, then
$$
59^{2}+59+\frac{1}{2} \leqslant y \leqslant 60^{2}+60+\frac{1}{2},
$$
which means $3540 \frac{1}{2} \leqslant y \leqslant 3660 \frac{1}{2}$.
This range contains $3660-3541+1=$ 120 integers.
|
120
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. $a_{1}, a_{2}, \cdots, a_{10}$ represent the ten digits $1,2,3,4,5,6$, $7,8,9,0$, respectively, to form two five-digit numbers
$$
m=\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}, n=\overline{a_{6} a_{7} a_{8} a_{9} a_{10}}(m>n) .
$$
Then the minimum value of $m-n$ is
|
2.247.
Since $m>n$, we have $a_{1}>a_{6}$.
To make $m-n$ as small as possible, we should take $a_{1}-a_{6}=1$, the last four digits of $m$ should be the smallest, and the last four digits of $n$ should be the largest. The largest four-digit number that can be formed from the different digits 1,2,3,4,5,6,7,8,9,0 is 9876, and the smallest four-digit number is 0123. The remaining two numbers are 4,5. Then, taking $a_{1}=5, a_{6}=4$, we get $m=50123, n=49876$. Therefore, $m-n=247$.
|
247
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given
$$
\frac{y+z-x}{x+y+z}=\frac{z+x-y}{y+z-x}=\frac{x+y-z}{z+x-y}=p \text {. }
$$
Then $p^{3}+p^{2}+p=$ $\qquad$ .
|
Notice
$$
\begin{array}{l}
p^{2}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x}=\frac{z+x-y}{x+y+z}, \\
p^{3}=\frac{y+z-x}{x+y+z} \cdot \frac{z+x-y}{y+z-x} \cdot \frac{x+y-z}{z+x-y} \\
=\frac{x+y-z}{x+y+z}, \\
p^{3}+p^{2}+p \\
=\frac{x+y-z}{x+y+z}+\frac{z+x-y}{x+y+z}+\frac{y+z-x}{x+y+z} \\
=\frac{x+y+z}{x+y+z}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) A company spent 4.8 million yuan to purchase the production technology of a certain product, and then invested another 15.2 million yuan to buy production equipment to process and produce the product. It is known that the cost of producing each piece of this product is 40 yuan. Market research has found that: the reasonable range for the unit selling price of the product is between 100 yuan and 300 yuan. When the unit selling price is set at 100 yuan, the annual sales volume is 200,000 pieces; when the unit selling price exceeds 100 yuan but does not exceed 200 yuan, for every 10 yuan increase in the unit selling price, the annual sales volume will decrease by 0.8 ten thousand pieces; when the unit selling price exceeds 200 yuan but does not exceed 300 yuan, for every 10 yuan increase in the unit selling price, the annual sales volume will decrease by 1 ten thousand pieces. Let the unit selling price be \( x \) (yuan), the annual sales volume be \( y \) (ten thousand pieces), and the annual profit be \( w \) (ten thousand yuan).
(1) Directly write the function relationship between \( y \) and \( x \).
(2) Find the function relationship between the first year's annual profit \( w \) and \( x \), and explain whether the company is profitable or at a loss in the first year of investment. If profitable, what is the maximum profit? If at a loss, what is the minimum loss?
(3) The company hopes that by the end of the second year, the total profit over two years will be no less than 18.42 million yuan. Please determine the range of the unit selling price at this time. Under this condition, to maximize the product sales volume, what should the unit selling price be set to?
|
$-、(1) y=\left\{\begin{array}{ll}-\frac{2}{25} x+28, & 100 \leqslant x \leqslant 200 \\ -\frac{1}{10} x+32, & 200<x \leqslant 300\end{array}\right.$
(2) When $100 \leqslant x \leqslant 200$, $w=x y-40 y-(1520+480)$.
Substituting $y=-\frac{2}{25} x+28$ into equation (1) gives $w=x\left(-\frac{2}{25} x+28\right)-40\left(-\frac{2}{25} x+28\right)-2000$.
Simplifying, we get $w=-\frac{2}{25}(x-195)^{2}-78$.
When $200<x \leqslant 300$, similarly, we get $w=-\frac{1}{10}(x-180)^{2}-40$.
Thus, $w=\left\{\begin{array}{ll}-\frac{2}{25}(x-195)^{2}-78, & 100 \leqslant x \leqslant 200 ; \\ -\frac{1}{10}(x-180)^{2}-40, & 200<x \leqslant 300 .\end{array}\right.$
If $100 \leqslant x \leqslant 200$, when $x=195$, $w_{\max }=-78$;
If $200<x \leqslant 300$, $w_{\max }<-80$.
Therefore, the company is in a loss in the first year of investment, with the minimum loss being 780,000 yuan.
(3) According to the problem, the function relationship between $w$ and $x$ in the second year is
$$
w=\left\{\begin{array}{ll}
\left(-\frac{2}{25} x+28\right)(x-40), & 100 \leqslant x \leqslant 200 ; \\
\left(-\frac{1}{10} x+32\right)(x-40), & 200<x \leqslant 300 .
\end{array}\right.
$$
When the total profit over two years is exactly 18,420,000 yuan, according to the problem, we have
$$
\left\{\begin{array}{l}
\left(-\frac{2}{25} x+28\right)(x-40)-78=1842 \\
100 \leqslant x \leqslant 200
\end{array}\right.
$$
or $\left\{\begin{array}{l}\left(-\frac{1}{10} x+32\right)(x-40)-78=1842, \\ 200<x \leqslant 300 .\end{array}\right.$
Solving, we get $x_{1}=190, x_{2}=200$.
Therefore, when $190 \leqslant x \leqslant 200$, the total profit is no less than 18,420,000 yuan.
From $y=-\frac{2}{25} x+28(100 \leqslant x \leqslant 200)$, we know that when the selling price is set to 190 yuan, the sales volume is maximized.
|
190
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the two roots of the equation $x^{2}+x-1=0$ are $\alpha, \beta$. Then the value of $\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}$ is $\qquad$
|
Let $A=\frac{\alpha^{3}}{\beta}+\frac{\beta^{3}}{\alpha}, B=\frac{\alpha^{3}}{\alpha}+\frac{\beta^{3}}{\beta}=\alpha^{2}+\beta^{2}$.
From the given information,
$$
\begin{array}{l}
\alpha+\beta=-1, \alpha \beta=-1 . \\
\text { Hence } B=(\alpha+\beta)^{2}-2 \alpha \beta=1+2=3 . \\
\text { Also, } \alpha^{3}+\beta^{3}=(\alpha+\beta)^{3}-3 \alpha \beta(\alpha+\beta)=-4, \\
\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-1}{-1}=1,
\end{array}
$$
Then $A+B=\alpha^{3}\left(\frac{1}{\beta}+\frac{1}{\alpha}\right)+\beta^{3}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$
$$
=\left(\alpha^{3}+\beta^{3}\right)\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=-4 \text {. }
$$
From equations (1) and (2), we get $A=-4-3=-7$.
|
-7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Divide the positive integers $1,2, \cdots, 2008$ into 1004 groups: $a_{1}, b_{1} ; a_{2}, b_{2} ; \cdots ; a_{1004}, b_{1004}$, and satisfy
$$
a_{1}+b_{1}=a_{2}+b_{2}=\cdots=a_{1004}+b_{1004} \text {. }
$$
For all $i(i=1,2, \cdots, 1004), a_{i} b_{i}$ the maximum value is
|
2.1009020 .
Notice that
$$
\begin{array}{l}
a_{i} b_{i}=\frac{1}{4}\left[\left(a_{i}+b_{i}\right)^{2}-\left(a_{i}-b_{i}\right)^{2}\right], \\
a_{i}+b_{i}=\frac{(1+2008) \times 1004}{1004}=2009 .
\end{array}
$$
To maximize the value of $a_{i} b_{i}$, the value of $a_{i}-b_{i}$ must be minimized, and the minimum value of $a_{i}-b_{i}$ is 1. At this point, $a_{i}+b_{i}=2009, a_{i}-b_{i}=1$. Thus, $a_{i}=1005, b_{i}=1004$, and the maximum value of $a_{i} b_{i}$ is
$$
1005 \times 1004=1009020 \text {. }
$$
|
1009020
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Among the following four propositions:
(1) A quadrilateral with one pair of opposite sides equal and one pair of opposite angles equal is a parallelogram;
(2) A quadrilateral with one pair of opposite sides equal and one diagonal bisecting the other diagonal is a parallelogram;
(3) A quadrilateral with one pair of opposite angles equal and the diagonal connecting the vertices of this pair of opposite angles being bisected by the other diagonal is a parallelogram;
(4) A quadrilateral with one pair of opposite angles equal and the diagonal connecting the vertices of this pair of opposite angles bisecting the other diagonal is a parallelogram.
Among them, the correct proposition numbers are $\qquad$
|
4. (4).
Propositions (1), (2), and (3) can be refuted with the following counterexamples:
Proposition (1): The quadrilateral $ABCD$ in Figure 5(a), where $\triangle ABD \cong \triangle CDE$.
Proposition (2): As shown in Figure 5(b), construct an isosceles $\triangle ADE$, extend the base $ED$ to any point $O$, and use $O$ as the intersection of the diagonals to construct $\square ABCE$. In this case, the quadrilateral $ABCD$ satisfies the conditions $AD = (AE =) BC$ and $AO = CO$, but it is not a parallelogram.
Proposition (3): The quadrilateral $ABCD$ in Figure 5(c), where $A$ and $C$ are any two points on the perpendicular bisector of $BD$.
The following is a proof that Proposition (4) is correct.
As shown in Figure 5(d), given $\angle BAD = \angle DCB$ and $OB = OD$.
With point $O$ as the center, rotate $\triangle ABD$ counterclockwise by $180^\circ$. Since $OB = OD$, point $D$ coincides with $B$, and point $B$ coincides with $D$, while point $A$ coincides with some point $A_1$ on ray $OC$. If $A_1$ is not $C$, then $\angle BA_1D > \angle BCD$ (if $A_1$ is inside segment $OC$) or $\angle BA_1D < \angle BCD$ (if $A_1$ is on the extension of $OC$), both of which contradict $\angle BA_1D = \angle BAD = \angle BCD$. Therefore, $A_1$ must be $C$, i.e., $OA = OA_1 = OC$. Thus, quadrilateral $ABCD$ is a parallelogram.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) In $\triangle A B C$, it is known that $\angle A>\angle B>$ $\angle C$, and $\angle A=2 \angle B$. If the three sides of the triangle are integers, and the area is also an integer, find the minimum value of the area of $\triangle A B C$.
|
Let $B C=a, C A=b, A B=c$.
As shown in Figure 6, construct the angle bisector $A D$ of $\angle B A C$, then
$$
\begin{array}{l}
\angle B A D=\angle D A C \\
= \angle B, \\
\angle A D C=\angle B+\angle B A D \\
= 2 \angle B .
\end{array}
$$
Figure 6
Therefore, $\triangle A C D \backsim \triangle B C A$. Thus,
$$
\frac{b}{a}=\frac{C D}{b} \text {. }
$$
By the Angle Bisector Theorem, we have $\frac{b}{c}=\frac{C D}{B D}$. Hence,
$$
\frac{b}{b+c}=\frac{C D}{C D+B D}=\frac{C D}{a} \text {. }
$$
From equations (1) and (2), we get
$$
\frac{b+c}{a}=\frac{a}{b} \text {. }
$$
Thus, $a^{2}=b(b+c)$.
If $(b, c)=d$, then from equation (1) we know $d \mid a$, so we can assume $(b, c)=1$. Therefore, we can let
$$
b=m^{2}, b+c=n^{2} \text {. }
$$
Then $a=m n, c=n^{2}-m^{2}$.
Since $\angle A>\angle B>\angle C$, we have $a>b>c$, i.e.,
$$
m n>m^{2}>n^{2}-m^{2} \text {. }
$$
Thus, $m1$, i.e.,
$$
m>\sqrt{2}+1 \text {. }
$$
Let the area of $\triangle A B C$ be $S$. By Heron's formula, we have
$$
\begin{aligned}
S & =\frac{1}{4} \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)} \\
= & \frac{1}{4} \sqrt{\left(m n+n^{2}\right)\left(n^{2}-m n\right)\left(n^{2}+m n-2 m^{2}\right)\left(2 m^{2}+m n-n^{2}\right)} \\
= & \frac{1}{4} n(n+m)(n-m) . \\
& \sqrt{(2 m+n)(2 m-n)} .
\end{aligned}
$$
From equation (4), we know $m \geqslant 3$. It is easy to verify from equation (3):
When $3 \leqslant m \leqslant 7$, only when $m=5$, $n=6$, $\sqrt{(2 m+n)(2 m-n)}=8$ (a rational number), at this time,
$$
S=\frac{1}{4} \times 6 \times 11 \times 1 \times 8=132 \text {. }
$$
Next, we prove that when $m \geqslant 8, n \geqslant 9$, $S>162$.
From equations (3) and (4), we know
$$
\begin{array}{l}
(2 m+n)(2 m-n) \\
>3 m(2 m-\sqrt{2} m)=(6-3 \sqrt{2}) m^{2} \\
>(6-4 \sqrt{2}) m^{2}=(2-\sqrt{2})^{2} m^{2}, \\
n(n+m)(n-m) \\
>n\left(1+\frac{\sqrt{2}}{2} n\right) \times 1=\frac{1}{2}(2+\sqrt{2}) n^{2} .
\end{array}
$$
From equation (5), we have
$S>\frac{1}{4} \times \frac{1}{2}(2+\sqrt{2}) n^{2}(2-\sqrt{2}) m=\frac{1}{4} n^{2} m$.
Thus, when $m \geqslant 8, n \geqslant 9$, we have $S>162$.
Therefore, the minimum value of $S$ is 132, at this time, $m=5, n=6$.
So, when $a=30, b=25, c=11$, the area of $\triangle A B C$ is minimized, and the minimum value is 132.
|
132
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If 6 pieces of $1 \times 2$ paper are used to cover a $3 \times 4$ grid, the number of different ways to cover it is.
|
4.11.
As shown in Figure 8, the cells of the grid are numbered. Let $M(a, b)$ denote the number of ways to cover the entire grid when the cells numbered $a$ and $b$ (which are adjacent) are covered by the same piece of paper. We focus on the covering of cell 8. It is given that $M(8,5) = 2, M(8,11) = 3, M(8,7) = M(8,9) = 3$.
Therefore, the total number of different ways to cover the entire grid is 11.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (50 points) There are 2008 students participating in a large public welfare activity. If two students know each other, then these two students are considered as a cooperative group.
(1) Find the minimum number of cooperative groups $m$, such that no matter how the students know each other, there exist three students who are pairwise in a cooperative group;
(2) If the number of cooperative groups is $\left[\frac{m}{22}\right]$, prove: there exist four students $A$, $B$, $C$, $D$, such that $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$ are each in a cooperative group.
|
(1) Let $n=1004$.
The following proof: $m=n^{2}+1$.
Divide the students into two large groups, each with $n$ students, and no students in the same large group know each other, while each student knows every student in the other large group, thus forming $n^{2}$ cooperative groups, but there do not exist three students who are all in the same cooperative group.
If there are $n^{2}+1$ cooperative groups, let student $X$ know the most students, and let them be $Y_{1}$, $Y_{2}, \cdots, Y_{k}$.
If there exist $i, j \in\{1,2, \cdots, k\}(i \neq j)$ such that $Y_{i}$ and $Y_{j}$ know each other, then $X, Y_{i}, Y_{j}$ satisfy the condition;
If $Y_{1}, Y_{2}, \cdots, Y_{k}$ do not form any cooperative group, then the number of cooperative groups does not exceed
$$
k+(2 n-k-1) k=k(2 n-k) \leqslant n^{2},
$$
which is a contradiction.
Therefore, $m=1004^{2}+1=1008017$.
(2) Let $n=1004, l=\left[\frac{m}{22}\right]=45818$, and the $2 n$ students are $X_{1}, X_{2}, \cdots, X_{2 n}$, with the number of students they know being $a_{1}, a_{2}, \cdots, a_{2 n}$, then
$$
a_{1}+a_{2}+\cdots+a_{2 n}=2 l \text {. }
$$
Consider all possible pairs of groups among the students each student knows, the total number is
$$
\begin{array}{l}
\mathrm{C}_{a_{1}}^{2}+\mathrm{C}_{a_{2}}^{2}+\cdots+\mathrm{C}_{a_{2 n}}^{2} \\
=\frac{1}{2}\left(\sum_{i=1}^{2 n} a_{i}^{2}-\sum_{i=1}^{2 n} a_{i}\right) \\
\geqslant \frac{1}{2}\left[\frac{1}{2 n}\left(\sum_{i=1}^{2 n} a_{i}\right)^{2}-\sum_{i=1}^{2 n} a_{i}\right] \\
=\frac{1}{2} \sum_{i=1}^{2 n} a_{i}\left(\frac{1}{2 n} \sum_{i=1}^{2 n} a_{i}-1\right) \\
=\frac{2 l}{2}\left(\frac{2 l}{2 n}-1\right)=l\left(\frac{l}{n}-1\right) \\
=45818\left(\frac{45818}{1004}-1\right) \\
>C_{2088}^{2} .
\end{array}
$$
Therefore, there exists a pair of students $A$ and $C$ who both know $B$ and $D$.
(Li Jianquan, Institute of Mathematics Education and Mathematical Olympiad, Tianjin Normal University, 300387)
|
1008017
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Second question As shown in Figure 8, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center point of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally) in sequence. How many chess pieces must be removed at minimum to meet this requirement? And explain the reasoning.
|
Solution: The minimum number of chess pieces to be removed is 11.
Define the square at the $x$-th column and $y$-th row as $(x, y)$ $(1 \leqslant x \leqslant 8,1 \leqslant y \leqslant 7, x, y \in \mathbf{Z})$.
A group of 5 consecutive chess pieces is called a "good group". The problem requires that no chess pieces form a "good group".
Let the number of removed chess pieces be $n$.
Lemma 1: There must be a removed chess piece in each row and each column. If no chess piece is removed in a certain row (column), then there will be more than one good group.
Lemma 2: If only 1 chess piece is removed in a certain column, then this chess piece must be in the 3rd, 4th, or 5th row of that column.
Lemma 3: If only 1 chess piece is removed in a certain row, then this chess piece must be in the 4th or 5th column of that row.
Proof of lemmas is omitted.
When $n \leqslant 7$, by the pigeonhole principle, there must be 1 column without any removed chess pieces. By Lemma 1, there must be a good group, so $n \leqslant 7$ does not meet the requirement.
When $n=8$, by Lemma 1, exactly 1 chess piece is removed from each column. By Lemma 2, these 8 chess pieces are all in the 3rd, 4th, or 5th rows. At this point, the 1st, 2nd, 6th, and 7th rows have no removed chess pieces, so there exists a good group. Therefore, $n=8$ does not meet the requirement.
When $n=9$, if 3 or more chess pieces are removed in a certain column, then in the remaining 7 columns, fewer than or equal to 6 chess pieces are removed, which means there must be a column without any removed chess pieces, and thus a good group exists. Therefore, at most 2 chess pieces can be removed in one column. By Lemma 1, a chess piece must be removed from each column. Since $9=1 \times 7+2$, it is only possible that 7 columns each have 1 chess piece removed, and the remaining 1 column has 2 chess pieces removed.
By Lemma 2, the 7 chess pieces in these 7 columns are all in the 3rd, 4th, or 5th rows, and there are 2 remaining chess pieces in the 1st, 2nd, 6th, and 7th rows. Therefore, there must be a row without any removed chess pieces, and thus a good group exists. Therefore, $n=9$ does not meet the requirement.
When $n=10$, if 4 or more chess pieces are removed in a certain column, then in the remaining 7 columns, fewer than or equal to 6 chess pieces are removed, which means there must be a column without any removed chess pieces, and thus a good group exists; if 3 chess pieces are removed in a certain column, then in the remaining 7 columns, 7 chess pieces are removed. By Lemma 1 and Lemma 2, exactly 1 chess piece is removed from each of these 7 columns, and these 7 chess pieces are distributed in the 3rd, 4th, and 5th rows. At this point, the 1st, 2nd, 6th, and 7th rows have no removed chess pieces, and there are 3 remaining chess pieces, which means there must be a row without any removed chess pieces, and thus a good group exists; by Lemma 1 and $10=1 \times 6+2 \times 2$, it is only possible that 6 columns each have 1 chess piece removed, and 2 columns each have 2 chess pieces removed. By Lemma 2, the 6 chess pieces in these 6 columns are distributed in the 3rd, 4th, and 5th rows, and the remaining 4 chess pieces are distributed in the 1st, 2nd, 6th, and 7th rows. By Lemma 1, these 4 chess pieces correspond to the 1st, 2nd, 6th, and 7th rows, one-to-one, without overlap. By Lemma 3, the two columns with 2 chess pieces removed are the 4th and 5th columns. Since the two chess pieces in the same column are distributed in the 1st, 2nd, 6th, and 7th rows, the possible combinations of removed chess pieces are:
$$
\begin{array}{l}
(4,1) \text { and }(4,6),(4,2) \text { and }(4,7), \\
(4,1) \text { and }(4,7),(4,2) \text { and }(4,6),
\end{array}
$$
Among them, $(4,1)$ and $(4,7)$ do not meet the requirement (because $(4,2)$, $(4,3), \cdots,(4,6)$ form a good group).
Similarly, $(5,1)$ and $(5,7)$ do not meet the requirement.
Also, because only 1 chess piece is removed in each row, $(4,2)$ and $(4,6)$ do not meet the requirement.
By symmetry, assume the removed chess pieces are
$(4,1),(4,6),(5,2),(5,7)$.
At this point, as shown in Figure 9 ("○" represents a chess piece, "×" represents a removed chess piece, and blank squares represent undetermined positions).
Since only 1 chess piece is removed in the 3rd column, at most 1 "×" is in $(3,3),(3,5)$, so in $(1,1),(2,2),(3,3),(4,4)$, $(5,5)$ and $(1,7),(2,6),(3,5),(4,4),(5,3)$, there must be 1 good group.
Therefore, $n=10$ does not meet the requirement.
When $n=11$, an example can be constructed as shown in Figure 10.
In summary, the minimum number of chess pieces to be removed is 11.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If the system of equations concerning $x$ and $y$ $\left\{\begin{array}{l}a x+b y=1, \\ x^{2}+y^{2}=10\end{array}\right.$ has solutions, and all solutions are integers, then the number of ordered pairs $(a, b)$ is $\qquad$
|
$=7.32$.
Since the integer solutions of $x^{2}+y^{2}=10$ are
$$
\begin{array}{l}
(1,3),(3,1),(1,-3),(-3,1), \\
(-1,3),(3,-1),(-1,-3),(-3,-1),
\end{array}
$$
Therefore, the number of lines connecting these eight points that do not pass through the origin is 24, and the number of tangents passing through these eight points is 8. Each line determines a unique ordered pair $(a, b)$. Hence, the number of ordered pairs $(a, b)$ is 32.
|
32
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Equation
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+ \\
\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}+\frac{x^{3}+4^{3}}{(x+4)^{3}}+\frac{x^{3}+9^{3}}{(x+9)^{3}}\right]=1
\end{array}
$$
The number of distinct non-zero integer solutions is
|
10.4 .
Using $a^{3}+b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)$, the original equation is
$$
\begin{array}{l}
\frac{(x-1)(x-4)(x-9)}{(x+1)(x+4)(x+9)}+1+\frac{2}{3}\left[\frac{x^{3}+1}{(x+1)^{3}}-\right. \\
\left.1+\frac{x^{3}+4^{3}}{(x+4)^{3}}-1+\frac{x^{3}+9^{3}}{(x+9)^{3}}-1\right]=0 \\
\Leftrightarrow \frac{x^{3}+49 x}{(x+1)(x+4)(x+9)}- \\
\quad\left[\frac{x}{(x+1)^{2}}+\frac{4 x}{(x+4)^{2}}+\frac{9 x}{(x+9)^{2}}\right]=0 .
\end{array}
$$
Dividing both sides of the equation by $x$, and rearranging, we get
$$
x\left(x^{4}-98 x^{2}-288 x+385\right)=0 \text {. }
$$
Dividing by $x$ again, we have
$$
\left(x^{2}-31\right)^{2}-(6 x+24)^{2}=0 \text {, }
$$
which is $\left(x^{2}+6 x-7\right)\left(x^{2}-6 x-55\right)=0$.
Thus, $(x+7)(x-1)(x+5)(x-11)=0$.
Upon verification, $x_{1}=-7, x_{2}=1, x_{3}=-5, x_{4}=11$ are all roots of the original equation. Therefore, the original equation has 4 integer roots.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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