problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
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3. In an isosceles right $\triangle ABC$, $AB=BC=5$, $P$ is a point inside $\triangle ABC$, and $PA=\sqrt{5}, PC=5$. Then $PB=\quad$ $\qquad$ | 3. $\sqrt{10}$.
As shown in Figure 3, draw $P E \perp$ $A B$ and $P F \perp B C$ intersecting $A B$ and $B C$ at points $E$ and $F$ respectively.
Let $P E=m$,
$$
P F=n \text{. }
$$
Using the Pythagorean theorem in $\triangle P A E$ and
$\triangle P C F$ respectively, we get
$$
\begin{array}{l}
m^{2}+(5-n)^{2}=5, \\
(... | \sqrt{10} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,952 |
4. Arrange several balls of two colors, red and black, in a row, requiring that both colors of balls must appear, and any two balls separated by 5 or 10 balls must be of the same color. Arrange according to this requirement, the maximum number of balls that can be placed is. | 4. 15.
Label the positions of these balls in order as $1,2, \cdots$.
According to the problem, when $|i-j|=6$ or 11, the $i$-th ball and the $j$-th ball are the same color, denoted as $i \sim j$.
Thus, $6 \sim 12 \sim 1 \sim 7 \sim 13 \sim 2 \sim 8 \sim 14 \sim 3 \sim 9$ $\sim 15 \sim 4 \sim 10,5 \sim 11$.
Therefore,... | 15 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false | 722,953 |
One. (20 points) Let positive integers $a, b, c (a \geqslant b \geqslant c)$ be the lengths of the sides of a triangle, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-a c-b c=13 \text {. }
$$
Find the number of triangles that meet the conditions and have a perimeter not exceeding 30. | Given the known equation:
$$
(a-b)^{2}+(b-c)^{2}+(a-c)^{2}=26 \text {. }
$$
Let $a-b=m, b-c=n$.
Then $a-c=m+n(m, n$ are natural numbers $)$.
Thus, equation (1) becomes
$$
m^{2}+n^{2}+m n=13 \text {. }
$$
Therefore, the pairs $(m, n)$ that satisfy equation (2) are:
$$
(m, n)=(3,1),(1,3) \text {. }
$$
(1) When $(m, n)=... | 11 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,954 |
II. (25 points) As shown in Figure 1, in isosceles $\triangle ABC$, it is known that $AB = AC$, and the angle bisector of $\angle C$ intersects side $AB$ at point $P$. $M$ is the point where the incircle $\odot I$ of $\triangle ABC$ touches side $BC$. Draw $MD \parallel AC$, intersecting $\odot I$ at point $D$. Prove:
... | As shown in Figure 4, draw the tangent line $PQ$ (with the point of tangency at $Q$) from point $P$ to $\odot I$, and extend it to intersect $BC$ at point $N$.
It is easy to see that
$\angle ACP$
$=\angle BCP$.
Since $PA, PQ$ are both tangents to $\odot I$,
we have
$\angle APC = \angle NPC$.
Also, $CP$ is a common side... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,955 |
Three. (25 points) Given the quadratic function $y=x^{2}+b x-c$ whose graph passes through two points $P(1, a)$ and $Q(2,10 a)$.
(1) If $a$, $b$, and $c$ are all integers, and $c<b<8 a$, find the values of $a$, $b$, and $c$;
(2) Let the graph of the quadratic function $y=x^{2}+b x-c$ intersect the $x$-axis at points $A... | Three, given that points $P(1, a)$ and $Q(2, 10a)$ are on the graph of the quadratic function $y = x^2 + bx - c$, we have
$$
1 + b - c = a, \quad 4 + 2b - c = 10a.
$$
Solving these equations, we get $b = 9a - 3$ and $c = 8a - 2$.
(1) From $c < b < 8a$, we know
$$
8a - 2 < 9a - 3 < 8a \Rightarrow 1 < a < 3.
$$
Since $a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,956 |
Three. (25 points) Let $p$ be a prime number greater than 2, and $k$ be a positive integer. If the graph of the function $y=x^{2}+p x+(k+1) p-4$ intersects the $x$-axis at two points, at least one of which has an integer coordinate, find the value of $k$.
---
The function is given by:
\[ y = x^2 + px + (k+1)p - 4 \]
... | From the problem, we know that the equation
$$
x^{2}+p x+(k+1) p-4=0
$$
has at least one integer root among its two roots $x_{1}, x_{2}$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-p, x_{1} x_{2}=(k+1) p-4 . \\
\text { Hence }\left(x_{1}+2\right)\left(x_{2}+2\right) \\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,957 |
1. Set
$$
A=\{1,3,5,7\}, B=\{2,4,6,8,20\} \text {. }
$$
If $C=\{s \mid s=a+b, a \in A, b \in B\}$, then the number of elements in set $C$ is ( ).
(A) 9
(B) 11
(C) 13
(D) 20 | -1. B.
Note that $s$ is odd, with a minimum value of 3 and a maximum value of 27, and it is missing 17 and 19. Therefore, set $C$ has a total of 11 elements. | B | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,958 |
2. “Function $f(x)$ is monotonic on $[a, b]$” is “Function $f(x)$ has a maximum value and a minimum value on $[a, b]$” ( ).
(A) a sufficient condition but not a necessary condition
(B) a necessary condition but not a sufficient condition
(C) a sufficient and necessary condition
(D) neither a sufficient condition nor a ... | 2. A.
If $f(x)$ is monotonic on $[a, b]$, then $f(a)$ and $f(b)$ are one the maximum value and the other the minimum value. Therefore, it is a sufficient condition;
$\sin x$ on $[0, \pi]$ has a maximum value of 1 and a minimum value of 0, but it is not monotonic, hence it is not a necessary condition. | A | Calculus | MCQ | Yes | Yes | cn_contest | false | 722,959 |
Example 6 Let $a, b$ be positive integers, and satisfy $2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{15}{b}}\right)$ is an integer. Then the number of such ordered pairs $(a, b)$ is $\qquad$ pairs.
$(2009$, National Junior High School Mathematics League) | First, guess that $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
The following is an attempt to prove this.
Let $\frac{15}{a}=A, \frac{15}{b}=B$. According to the problem, we can assume
$\sqrt{A}+\sqrt{B}=C\left(A, B, C \in \mathbf{Q}_{+}\right)$.
Thus, $\sqrt{A}=C-\sqrt{B} \Rightarrow A=(C-... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,960 |
3. In the afternoon, 5 classes need to be scheduled: Physics, Chemistry, Biology, and two self-study periods. If the first class cannot be Biology and the last class cannot be Physics, then the number of different scheduling methods is ( ) kinds.
(A) 36
(B) 39
(C) 60
(D) 78 | 3. B.
The total number of unrestricted arrangements is $\frac{A_{5}^{5}}{2}=60$; the number of arrangements with biology in the first period or physics in the last period is $\frac{\mathrm{A}_{4}^{4}}{2}=12$ each, and among these, the number of arrangements with biology in the first period and physics in the last peri... | B | Combinatorics | MCQ | Yes | Yes | cn_contest | false | 722,961 |
4. The sequence $\frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \cdots$, $\frac{1}{m+1}, \frac{2}{m+1}, \cdots, \frac{m}{m+1}, \cdots$ has the sum of its first 40 terms as ( ).
(A) $23 \frac{1}{2}$
(B) $19 \frac{1}{9}$
(C) 19
(D) 18 | 4. C.
Let the denominator of the 40th term $a_{40}$ be $m+1$. Then
$$
\frac{m(m-1)}{2}<40 \leqslant \frac{m(m+1)}{2} \text {. }
$$
Solving this, we get $m=9 \Rightarrow \frac{m(m-1)}{2}=36$.
Thus, $a_{40}=\frac{4}{10}$,
$$
\begin{array}{l}
S_{40}=\sum_{k=1}^{8} \sum_{i=1}^{k} \frac{i}{k+1}+\frac{1}{10}(1+2+3+4) \\
=\... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,962 |
5. Let
$$
A=\{(m, n) \mid 0<m<2,0<n<2\} \text {. }
$$
Then for any $(m, n) \in A$, the equation in terms of $x$
$$
\frac{m}{4} x^{2}+x+n=0
$$
has real roots with a probability of ( ).
(A) $\frac{1+2 \ln 2}{4}$
(B) $\frac{1+\ln 2}{2}$
(C) $\frac{3-2 \ln 2}{4}$
(D) $\frac{1-\ln 2}{2}$ | 5. A.
Given that the equation $\frac{m}{4} x^{2}+x+n=0$ has real solutions, we know $\Delta=1-m n \geqslant 0 \Rightarrow n \leqslant \frac{1}{m}$.
Therefore, the required probability is
$$
P=\frac{1}{4}\left(1+\int_{\frac{1}{2}}^{2} \frac{1}{x} \mathrm{~d} x\right)=\frac{1+2 \ln 2}{4} .
$$ | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,963 |
6. If the edge length of a cube is $a$, then the area of the largest section perpendicular to the cube's diagonal is ( ).
(A) $\frac{3 \sqrt{2}}{4} a^{2}$
(B) $\sqrt{2} a^{2}$
(C) $\frac{\sqrt{3}}{2} a^{2}$
(D) $\frac{3 \sqrt{3}}{4} a^{2}$ | 6. D.
The maximum cross-section through the center of the cube is a regular hexagon with a side length of $\frac{\sqrt{2}}{2} a$, and the area is
$$
\frac{\sqrt{3}}{4} \times \frac{1}{2} a^{2} \times 6=\frac{3 \sqrt{3}}{4} a^{2} .
$$ | D | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,964 |
$$
\begin{array}{l}
\text { 7. Let }\left(3+x+2 x^{2}\right)^{n} \\
=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n}\left(n \in \mathbf{N}_{+}\right)
\end{array}
$$
hold for $x \in \mathbf{R}$. Then
$$
a_{1}+a_{2}+\cdots+a_{2 n-1}=
$$
$\qquad$ | Let $x=1$. Then $a_{0}+a_{1}+\cdots+a_{2 n}=6^{n}$.
It is easy to see that $a_{0}=3^{n}, a_{2 n}=2^{n}$.
Therefore, $a_{1}+a_{2}+\cdots+a_{2 n-1}=6^{n}-3^{n}-2^{n}$. | 6^{n}-3^{n}-2^{n} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,965 |
8. The range of the function $f(x)=2|\sin x|+3|\cos x|$ is $\qquad$ . | 8. $[2, \sqrt{13}]$.
Since $f(x)$ is an even function with period $T=\pi$, we can assume $x \in\left[0, \frac{\pi}{2}\right]$. Therefore,
$$
f(x)=2 \sin x+3 \cos x=\sqrt{13} \sin (x+\varphi),
$$
where $\sin \varphi=\frac{3}{\sqrt{13}}, \cos \varphi=\frac{2}{\sqrt{13}}, \varphi$ is an acute angle. $x+\varphi \in\left[... | [2, \sqrt{13}] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,966 |
9. The highest point of the ellipse obtained by rotating the ellipse $\frac{x^{2}}{2}+y^{2}=1$ counterclockwise by $45^{\circ}$ around the coordinate origin is $\qquad$ .
Translating the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 9. $\frac{\sqrt{15}}{3}$.
Draw the tangent line to the ellipse at its highest point, then the slope of the tangent line is 0. After rotating the ellipse $45^{\circ}$ clockwise to restore it, the slope of the tangent line is -1.
Let the point of tangency be $\left(x_{0}, y_{0}\right)$, and the tangent line be $\frac{x... | \frac{\sqrt{15}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,967 |
10. The number of integer solutions to the inequality $\log _{6}(1+\sqrt{x})>\log _{25} x$ is $\qquad$ . | 10. 24 .
Let $\log _{25} x=t$. Then $x=25^{t}$.
Thus the original inequality $\Leftrightarrow \log _{6}\left(1+5^{t}\right)>t$
$$
\Leftrightarrow 1+5^{t}>6^{t} \Leftrightarrow\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}>1 \text {. }
$$
Define $f(t)=\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t... | 24 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 722,968 |
11. If the equation about $x$
$$
x^{2}-\left(a^{2}+b^{2}-6 b\right) x+a^{2}+b^{2}+2 a-4 b+1=0
$$
has two real roots $x_{1}, x_{2}$ satisfying $x_{1} \leqslant 0 \leqslant x_{2} \leqslant 1$, then the sum of the minimum and maximum values of $a^{2}+b^{2}+4 a+4$ is $\qquad$ | $11.9 \frac{1}{2}+4 \sqrt{5}$.
Let $f(x)=x^{2}-\left(a^{2}+b^{2}-6 b\right) x+a^{2}+b^{2}+2 a-4 b+1$.
From $f(x)=0$ with roots $x_{1} \leqslant 0 \leqslant x_{2} \leqslant 1$, we get $f(0) \leqslant 0, f(1) \geqslant 0$.
Simplifying, we obtain $(a+1)^{2}+(b-2)^{2} \leqslant 4$, and
$$
a+b+1 \geqslant 0 \text {. }
$$
I... | 9 \frac{1}{2}+4 \sqrt{5} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,969 |
12. Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $F_{1} 、 F_{2}$ are its left and right foci, $Q$ is any point on the ellipse $C$, and the centroid and incenter of $\triangle F_{1} Q F_{2}$ are $G 、 I$, respectively. The line $I G$ is parallel to the $x$-axis. Then the eccentricity of the el... | 12. $\frac{1}{2}$.
Let $Q\left(x_{0}, y_{0}\right)$. From the problem, the y-coordinates of points $G$ and $I$ are both $\frac{1}{3}\left|y_{0}\right|$. Therefore, the inradius of $\triangle F_{1} Q F_{2}$ is
$$
r=\frac{1}{3}\left|y_{0}\right| .
$$
Then, $S_{\triangle F_{1} Q F_{2}}=\frac{1}{2}\left|F_{1} F_{2}\right... | \frac{1}{2} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,970 |
Example 7 Given that $a, b, c$ are positive integers, and $\frac{\sqrt{3} a+b}{\sqrt{3} b+c}$ is a rational number. Prove: $\frac{a^{2}+b^{2}+c^{2}}{a+b+c}$ is an integer.
(2010, National Junior High School Mathematics League, Jiangxi Province Preliminary) | Given $\frac{\sqrt{3} a+b}{\sqrt{3} b+c}=q$ (where $q$ is a rational number). Then $\sqrt{3} a+b=\sqrt{3} b q+c q$, which means
$$
\sqrt{3}(a-b q)=c q-b \text{. }
$$
If $a-b q \neq 0$, then $\sqrt{3}=\frac{c q-b}{a-b q}$, which is a contradiction.
Hence $a-b q=0 \Rightarrow c q-b=0$
$$
\begin{array}{l}
\Rightarrow b=c... | a-b+c | Number Theory | proof | Yes | Yes | cn_contest | false | 722,971 |
13. Given $\frac{1}{3} \leqslant a \leqslant 1$. If $f(x)=a x^{2}-2 x+1$ has a maximum value $M(a)$ and a minimum value $N(a)$ on $[1,3]$, let $g(a)=M(a)-N(a)$.
(1) Find the function expression for $g(a)$;
(2) Prove that $g(a) \geqslant \frac{1}{2}$ always holds. | (1) Notice that
$$
f(x)=a\left(x-\frac{1}{a}\right)^{2}+\left(1-\frac{1}{a}\right) \text {. }
$$
Then $N(a)=f\left(\frac{1}{a}\right)=1-\frac{1}{a}$.
When $1 \leqslant \frac{1}{a} \leqslant 2$, i.e., $\frac{1}{2} \leqslant a \leqslant 1$,
$$
M(a)=f(3)=9 a-5 \text {. }
$$
Thus, $g(a)=9 a-6+\frac{1}{a}$.
When $2 \leqsl... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,972 |
14. Given $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on the ellipse. In $\triangle F_{1} P F_{2}$, the external angle bisector of $\angle F_{1} P F_{2}$ is $l$, and the symmetric point of $F_{2}$ with respect to $l$ is $Q$. $F... | 14. (1) Let the moving point $R(x, y)$ and $Q\left(x_{1}, y_{1}\right)$. Then
$$
\left\{\begin{array}{l}
x = \frac{x_{1} + c}{2}, \\
y = \frac{y_{1}}{2}
\end{array} \Rightarrow \left\{\begin{array}{l}
x_{1} = 2x - c, \\
y_{1} = 2y
\end{array}\right.\right.
$$
Since point $P$ is on the ellipse, we have
$$
\left|P F_{1}... | k = \pm \frac{\sqrt{3}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,973 |
15. As shown in Figure $1, \odot O_{1}$ is internally tangent to $\odot O_{2}$ at point $P, \odot O_{1}$'s chord $A B$ is tangent to $\odot O_{2}$ at point $C$, extending $P C$ intersects $\odot O_{1}$ at point $G$, $P A$ and $P B$ intersect $\odot O_{2}$ at points $E$ and $F$ respectively, $E F$ intersects $P C$ at po... | 15. (1) As shown in Figure 2, draw a tangent line from point $P$ to $\odot O_{1}$, and take a point $Q$ on this tangent line to the left of $P$, then
$$
\begin{array}{c}
\angle Q P C \\
=\angle A C P . \\
\text { Therefore, } \angle Q P A+\angle A P G \\
=\angle G P B+\angle P B C .
\end{array}
$$
Since $\angle Q P A=... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,974 |
16. Given a sequence $\left\{a_{n}\right\}$ where all terms are non-zero, the sum of its first $n$ terms is $S_{n}$, and for any $n \in \mathbf{N}$, we have
$(1-p) S_{n}=p-p a_{n}$ (where $p$ is a constant greater than 1).
Let $f(n)=\frac{1+\mathrm{C}_{n}^{1} a_{1}+\mathrm{C}_{n}^{2} a_{2}+\cdots+\mathrm{C}_{n}^{n} a_{... | 16. (1) Notice
$$
(1-p) S_{n}=p-p a_{n} \text {. }
$$
Then $(1-p) S_{n+1}=p-p a_{n+1}$.
(2)
(2) - (1) gives
$$
\text { (1-p) } a_{n+1}=-p a_{n+1}+p a_{n} \text {, }
$$
which means $a_{n+1}=p a_{n}$.
Substituting $n=1$ into equation (1), we get $a_{1}=p$.
Thus, $a_{n}=p^{n}$.
Therefore, $S_{n}=\frac{p\left(p^{n}-1\rig... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 722,975 |
11.1. Do there exist non-zero real numbers $a_{1}, a_{2}, \cdots, a_{10}$, such that
$$
\prod_{i=1}^{10}\left(a_{i}+\frac{1}{a_{i}}\right)=\prod_{i=1}^{10}\left(a_{i}-\frac{1}{a_{i}}\right) ?
$$ | 11. 1. does not exist.
Notice that for $k=1,2, \cdots, 10$, we have
$$
\begin{array}{l}
\left|a_{k}+\frac{1}{a_{k}}\right|=\left|a_{k}\right|+\left|\frac{1}{a_{k}}\right| \\
>\max \left\{\left|a_{k}\right|, \frac{1}{\left|a_{k}\right|}\right\} \geqslant\left|a_{k}-\frac{1}{a_{k}}\right| \geqslant 0 .
\end{array}
$$
... | proof | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,976 |
11. 2. In an $n \times n(n \geqslant 4)$ grid, each cell on one of the diagonals contains a “+” sign, and each of the other cells contains a “_” sign. Changing all the signs in any row or column is called one operation. Prove that after any finite number of operations, the grid will still contain at least $n$ “+” signs... | 11. 2. Let $(i, j)$ denote the square in the $i$-th row and $j$-th column. Suppose that $(i, i) (i=1,2, \cdots, n)$ contains a “+” sign. Clearly, the operation cannot change the parity of the number of “+” signs in the four corners (referred to as quadruple corners) of any rectangle in the grid.
Consider the following ... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,977 |
11.3. For the cyclic quadrilateral $ABCD$ inscribed in circle $\omega$, the diagonals intersect at point $K$. $M_{1}, M_{2}, M_{3}, M_{4}$ are the midpoints of the arcs $\overparen{AB}, \overparen{BC}, \overparen{CD}, \overparen{DA}$ (excluding the other vertices), and $I_{1}, I_{2}, I_{3}, I_{4}$ are the incenters of ... | 11.3. Note that $I_{1}, I_{2}, I_{3}, I_{4}$ are the intersections of $A M_{2}$ with $B M_{4}$, $B M_{3}$ with $C M_{1}$, $C M_{4}$ with $D M_{2}$, and $D M_{1}$ with $A M_{3}$, respectively.
Since $\overparen{A M}_{1} + \overparen{C M}_{3} = \overparen{B M}_{1} + \overparen{D M}_{3}$, the line $M_{1} M_{3}$ forms equ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,978 |
11. 4. Given a positive integer $n(n \geqslant 3)$. Find the smallest positive integer $k$ such that the following conclusion always holds.
For any $n$ points $A_{i}=\left(x_{i}, y_{i}\right)$ in the plane, no three of which are collinear, and any $n$ real numbers $c_{i}(1 \leqslant i \leqslant n)$, there exists a biv... | 11.4. $k=\left[\frac{n}{2}\right]$.
First, prove a lemma.
Lemma For any $n$ given points $A_{i}=\left(x_{i}, y_{i}\right)(1 \leqslant i \leqslant n)$ in the plane, no three of which are collinear, there exists a polynomial $P(x, y)$ of degree not greater than $\left[\frac{n}{2}\right]$, satisfying
$$
P\left(x_{n}, y_{... | k=\left[\frac{n}{2}\right] | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,979 |
11.5. Given a positive integer $n(n>1)$. Prove: There exist $n$ consecutive positive integers, such that their product is a multiple of all prime numbers not exceeding $2 n+1$, but not a multiple of any other prime number. | 11.5. If $n+1$ is a composite number, then $n+2, n+3$, $\cdots, 2 n+1$ these $n$ consecutive positive integers satisfy the condition.
Obviously, it is not a multiple of any prime number greater than $2 n+1$, it is a multiple of all prime numbers between $n+2$ and $2 n+1$.
Moreover, the product of any $n$ consecutive ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 722,980 |
11. 6. Can the incenter of the four faces of a tetrahedron be coplanar? | 11.6. Not Coplanar.
Let $A, B, C, D$ be the four vertices of a tetrahedron, and $I_{A}$, $I_{B}$, $I_{C}$, $I_{D}$ be the incenters of $\triangle BCD$, $\triangle ACD$, $\triangle ABD$, and $\triangle ABC$, respectively. Suppose they are coplanar. Clearly, any three of them are not collinear, so they form the vertices... | Not Coplanar | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,981 |
Example 8 Given real numbers $x, y, z$ satisfy
$$
\sqrt[5]{x-y}+\sqrt[5]{y-z}=3 \text{, and } x-z=33 \text{. }
$$
Find the value of the algebraic expression $x-2 y+z$. | Let $\sqrt[5]{x-y}=x_{1}, \sqrt[5]{y-z}=x_{2}$. Then $x_{1}+x_{2}=3, x_{1}^{5}+x_{2}^{5}=33$.
Let $x_{1} x_{2}=\lambda$. Then $x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+\lambda=0$.
Let $x_{1}^{n}+x_{2}^{n}=a_{n}$ ( $n$ is a positive integer $)$. Then
$$
a_{n}-3 a_{n-1}+\lambda a_{n-2}=0 \text {, }
$$
i.e.,
$$
a_{... | 31 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,982 |
11.8. In a boarding school, there are 512 students living in 256 dormitories, with each pair of roommates sharing a dormitory. It is known that these students have chosen 9 courses, and no two students have chosen exactly the same set of courses. Prove: All students can be arranged in a circle such that:
(1) Any two ro... | 11.8. Use mathematical induction to prove the conclusion for the general case ( $n$ courses, $2^{n}$ students, $2^{n-1}$ dormitories).
It is not difficult to discuss that the conclusion holds when $n=2$.
Assume $n>2$.
Without loss of generality, let physics be an elective course and a pair of roommates, one of whom has... | proof | Combinatorics | proof | Yes | Yes | cn_contest | false | 722,984 |
1. Given real numbers $a, b, c$ satisfy
$$
\sqrt[3]{a-b}+\sqrt[3]{b-c}=2, a-c=26 \text {. }
$$
Then the value of $a-2 b+c$ is ( ).
(A) 26
(B) 28
(C) -28
(D) $\pm 28$ | -1. D.
Let $m=\sqrt[3]{a-b}, n=\sqrt[3]{b-c}$. Thus, $m+n=2, m^{3}+n^{3}=a-c=26$.
Solving this, we get $(m, n)=(3,-1)$ or $(-1,3)$.
Therefore, $a-2 b+c=m^{3}-n^{3}= \pm 28$. | D | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,985 |
2. As shown in Figure 1, let $M$ and $N$ be two points on the diagonal $BD$ of square $ABCD$, with $BM=3$, $DN=4$, and $BD=12$. Then $\angle MAB + \angle NCD$ equals ( ).
(A) $40^{\circ}$
(B) $45^{\circ}$
(C) $48^{\circ}$
(D) $60^{\circ}$ | 2. B.
As shown in Figure 6, construct the congruent triangle $\triangle A B^{\prime} M$ of $\triangle A B M$, and let the angle bisector of $\angle B^{\prime} A D$ intersect $B D$ at point $P$. Thus,
$$
\triangle A D P \cong \triangle A B^{\prime} P \text {. }
$$
Then $B M=B^{\prime} M$,
$P D=P B^{\prime}$,
$\angle P... | B | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,986 |
3. Given that $a$ and $b$ are constants, the solution set of the inequality $a b(x-1)>a x+b$ with respect to $x$ is $x<2$. Then the value of the algebraic expression
$$
|a-1|+|b-2|-|a+b-3|
$$
is ( ).
(A) 0
(B) $2 a-2$
(C) $2-2 a$
(D) $2 b-6$ | 3. A.
From the given, we know that $ab = 2a + b$.
Clearly, $b \neq 2$, and $a = \frac{b}{b-2}$.
Thus, $a-1 = \frac{2}{b-2}$.
Then, $a-1$ and $b-2$ have the same sign.
Therefore, $|a-1| + |b-2| - |a+b-3| = 0$. | A | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,987 |
4. As shown in Figure $2, D, E$ are points on the sides $BC, AD$ of the equilateral $\triangle ABC$ respectively, $\angle BEC$ $=90^{\circ}, \angle CED=30^{\circ}$. Then the value of $BD: CD$ is ( ).
(A) $2: 1$
(B) 3: 1
(C) $5: 2$
(D) $20: 9$ | 4. A.
As shown in Figure 7, extend $E D$ to point $F$ such that $E F = B E$, and connect $B F$ and $C F$.
It is easy to see that $\triangle B E F$ is an equilateral triangle.
$$
\begin{array}{c}
\Rightarrow \angle B F E = 60^{\circ} \\
= \angle A C B \\
\Rightarrow A, B, F, C \text{ are four }
\end{array}
$$
points ... | A | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,988 |
5. Given that $x$ and $y$ are any real numbers, and $M$ is the maximum value of the algebraic expressions
$$
x^{2}+2 x y+y^{2} 、 x^{2}-2 x y+y^{2} 、 x^{2}+4 x+4
$$
Then the minimum value of $M$ is ( ).
(A) $\frac{1}{4}$
(B) $\frac{1}{2}$
(C) 1
(D) $\frac{3}{2}$ | 5. C.
From the problem, we have
$$
M \geqslant x^{2}+2 x y+y^{2} \Rightarrow \sqrt{M} \geqslant|x+y| \text{. }
$$
Similarly, $\sqrt{M} \geqslant|x-y|, \sqrt{M} \geqslant|x+2|$.
Thus, $4 \sqrt{M} \geqslant|x+y|+|x-y|+|-2 x-4| \geqslant 4$
$$
\Rightarrow \sqrt{M} \geqslant 1 \Rightarrow M \geqslant 1 \text{. }
$$
When... | C | Algebra | MCQ | Yes | Yes | cn_contest | false | 722,989 |
6. As shown in Figure 3, given that $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, $OI \perp AI$, $AB=5, AC=9$. Then the length of side $BC$ is ( ).
(A) 8
(B) 7.5
(C) 7
(D) 6.5 | 6. C.
As shown in Figure 8, extend $AI$ to intersect the circumcircle $\odot O$ of $\triangle ABC$ at point $D$, connect $IC$, $CD$, $OD$, and let $OD$ intersect $BC$ at point $E$. Draw $IF \perp AB$ at point $F$.
Since $I$ is the incenter of $\triangle ABC$, we have
$$
\begin{array}{l}
AF=\frac{1}{2}(AB+AC-BC)=7-\fr... | C | Geometry | MCQ | Yes | Yes | cn_contest | false | 722,990 |
1. If real numbers $a, b$ satisfy
$$
\left(a-\sqrt{a^{2}+2010}\right)\left(b+\sqrt{b^{2}+2010}\right)+2010=0 \text {, }
$$
then $a \sqrt{b^{2}+2011}-b \sqrt{a^{2}+2011}=$ $\qquad$ | From the given, we have
$$
\begin{array}{l}
a-\sqrt{a^{2}+2010}=-\frac{2010}{b+\sqrt{b^{2}+2010}} \\
=b-\sqrt{b^{2}+2010} .
\end{array}
$$
Similarly,
$$
b+\sqrt{b^{2}+2010}=a+\sqrt{a^{2}+2010} \text {. }
$$
Subtracting (2) from (1) and simplifying, we get $a=b$.
Therefore, the answer is 0. | 0 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,991 |
2. Given that $P$ is a point inside the circle $\odot O$ with radius 15, among all the chords passing through point $P$, 24 chords have integer lengths. Then $O P=$ $\qquad$ . | 2. 12 .
Among the chords of circle $\odot O$ passing through point $P$, the diameter is the longest chord, and there is only one; the shortest chord is the one perpendicular to $OP$, and there is also only one. Therefore, there is only one chord each with lengths of $30$ and $18$, and two chords each with lengths of $... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,992 |
Example 9 Prove: $\frac{(2+\sqrt{3})^{2010}+(2-\sqrt{3})^{2010}}{2}$ is an integer, and find its remainder when divided by 4. | Notice
$$
\begin{array}{l}
(2+\sqrt{3})+(2-\sqrt{3})=4, \\
(2+\sqrt{3})(2-\sqrt{3})=1 .
\end{array}
$$
Therefore, $2+\sqrt{3}$ and $2-\sqrt{3}$ are the two roots of $x^{2}-4 x+1=0$.
Let $a_{n}=\frac{1}{2}(2+\sqrt{3})^{n}+\frac{1}{2}(2-\sqrt{3})^{n}$. Then
$$
a_{n}-4 a_{n-1}+a_{n-2}=0
$$
That is, $a_{n}=4 a_{n-1}-a_{n... | 3 | Number Theory | proof | Yes | Yes | cn_contest | false | 722,993 |
3. $[x]$ represents the greatest integer not exceeding the real number $x$ (for example, $[\pi]=3,[-\pi]=-4,[-4]=-4$). Let $M=[x]+[2x]+[3x]$. Positive integers that cannot be expressed in the form of $M$ are called "invisible numbers". If the invisible numbers are arranged in ascending order, the 2009th invisible numbe... | 3.6028 .
Let $n$ be a natural number.
When $n \leqslant x < n+\frac{1}{3}$,
$2 n \leqslant 2 x < 2 n+\frac{2}{3}$,
$3 n \leqslant 3 x < 3 n+1$.
Thus, $M=n+2 n+3 n=6 n$.
When $n+\frac{1}{3} \leqslant x < n+\frac{1}{2}$,
$2 n+\frac{2}{3} \leqslant 2 x < 2 n+1$,
$3 n+1 \leqslant 3 x < 3 n+\frac{3}{2}$.
Thus, $M=n+2 n+(3 ... | 6028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,994 |
4. As shown in Figure 4, in quadrilateral $ABCD$, $\angle BAD$ and $\angle CAD$ are complementary, $AB=AC$, $AD=5\sqrt{2}$, $BD=8$, $CD=6$. Then the degree measure of $\angle ABD$ is $\qquad$ | $4.45^{\circ}$.
As shown in Figure 9, rotate $AD$ clockwise around point $A$ by $90^{\circ}$, and connect $AD^{\prime}$, $BD^{\prime}$, and $DD^{\prime}$. Then
$$
\begin{array}{l}
AD^{\prime}=AD=5 \sqrt{2}, \\
\angle DAD^{\prime}=90^{\circ}.
\end{array}
$$
From $\angle BAD$ and $\angle CAD$ being complementary, we kno... | 45^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 722,995 |
One, (20 points) A telecommunications company's mobile phone charging standards are: a monthly rental fee of 20 yuan, and a call fee of 0.2 yuan per minute (including both outgoing and incoming calls). The company has recently launched two promotional projects.
Project $A$: A monthly charge of only 1 yuan for incoming... | (1) Pay according to project $A$
$$
20+1+0.2 \times 300=81 \text { (Yuan). }
$$
Also, $20+0.2(300+280)=136$ Yuan, thus, according to (1) of project $B$, paying 60 Yuan for 100 Yuan should result in
$$
136-(100-60)=96 \text { (Yuan). }
$$
Therefore, according to project $A$, Xiao Song should pay 81 Yuan, and according... | null | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 722,996 |
II. (25 points) As shown in Figure 5, given that $O$, $I$, and $H$ are the circumcenter, incenter, and orthocenter of $\triangle ABC$ respectively, $\angle ABC=40^{\circ}$, $\angle ACB=80^{\circ}$. Prove:
(1) $OI=IH=HC$;
(2) $AO=AH$.
保留了原文的换行和格式。 | (1) From the given information,
$\angle B A C=60^{\circ}$.
Thus, $\angle B O C=2 \angle B A C=120^{\circ}$,
$\angle B I C=90^{\circ}+\frac{1}{2} \angle B A C=120^{\circ}$,
$\angle B H C=180^{\circ}-\angle B A C=120^{\circ}$.
Therefore, points $B 、 C 、 O 、 I 、 H$ are concyclic.
It is easy to find that $\angle O B I=\ang... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 722,997 |
Three. (25 points) Given a positive integer $M$ has $k$ positive divisors, among which, only two divisors are prime, and the sum of the reciprocals of these $k$ positive divisors is $\frac{342}{287}$. Find all values of $M$.
---
The text has been translated while preserving the original formatting and line breaks. | $$
\begin{array}{l}
\text{Because } 287=7 \times 41, \text{ so, 7 and 41 are exactly the two prime factors of } M. \\
\text{Let } M=7^{m} \times 41^{n}\left(m, n \in \mathbf{N}_{+}\right), d_{1}, d_{2}, \cdots, d_{k} \text{ be all the positive divisors of } M \text{ arranged in ascending order. Thus, } d_{1}=1, d_{k}=M... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 722,998 |
1. Using the vertices of a regular dodecagon as the vertices of triangles, the total number of acute and obtuse triangles that can be formed is $\qquad$.
untranslated part: $\qquad$ (This part is typically left as a blank space for the answer to be filled in.) | -1.160 .
The circumference of a circle with 12 points can form $\mathrm{C}_{12}^{3}=220$ triangles; among the chords of a circle equally divided into 12 parts, there are 6 diameters, and each diameter corresponds to 10 right-angled triangles, thus yielding 60 right-angled triangles. Therefore, acute and obtuse triangle... | 160 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 722,999 |
2. Let the polynomial $f(x)$ satisfy
$$
2 f(x+1)+3 f(x-1)=10 x^{2}+11 x+32 \text {. }
$$
Then $f(x)=$ | 2. $2 x^{2}+3 x+5$.
Notice that the degrees of $f(x+1)$ and $f(x-1)$ are the same, and the right side is quadratic, so let $f(x)=a x^{2}+b x+c$.
Substitute into the given equation and compare the coefficients on both sides to get
$$
a=2, b=3, c=5 \text {. }
$$
Therefore, $f(x)=2 x^{2}+3 x+5$. | 2 x^{2}+3 x+5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,000 |
4. $F$ is the right focus of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and $P$ is a moving point on the ellipse. For the fixed point $A(-2, \sqrt{3}),|P A|+$ $2|P F|$ the minimum value is $\qquad$ . | 4. 10 .
It is known that the eccentricity of the ellipse is $\frac{1}{2}$, and the equation of the right directrix $l$ is $x=8$. The distance from point $A$ to $l$ is 10.
Let the projection of point $P$ onto $l$ be $H$. Then
$\frac{|P F|}{|P H|}=\frac{1}{2}$.
Thus, $|P A|+2|P F|=|P A|+|P H| \geqslant 10$.
The minimum ... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,002 |
$$
\begin{array}{l}
\text { 5. Let } \frac{\sin ^{4} \theta}{a}+\frac{\cos ^{4} \theta}{b}=\frac{1}{a+b}(a, b>0) \text {. } \\
\text { Then } \frac{\sin ^{8} \theta}{a^{3}}+\frac{\cos ^{8} \theta}{b^{3}}=
\end{array}
$$ | 5. $\frac{1}{(a+b)^{3}}$.
Let $\sin ^{2} \theta=x$. Then the given equation becomes $\frac{x^{2}}{a}+\frac{(1-x)^{2}}{b}=\frac{1}{a+b}$.
Solving for $x$, we get $x=\frac{a}{a+b}, 1-x=\frac{b}{a+b}$.
Thus, $\frac{\sin ^{8} \theta}{a^{3}}+\frac{\cos ^{8} \theta}{b^{3}}=\frac{x^{4}}{a^{3}}+\frac{(1-x)^{4}}{b^{3}}$
$=\fra... | \frac{1}{(a+b)^{3}} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,003 |
1. Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+$ $\cdots+\sqrt{1+\frac{1}{1999^{2}}+\frac{1}{2000^{2}}}$. Find the greatest integer not exceeding $S$.
(2000, Taiyuan Junior High School Mathematics Competition) | Prompt: Following Example 2, the largest integer not exceeding $S$ is
$$
1999 .
$$ | 1999 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,004 |
6. In tetrahedron $ABCD$, one edge has a length of 3, and the other five edges have a length of 2. Then the radius of its circumscribed sphere is | 6. $\frac{\sqrt{21}}{3}$.
As shown in Figure 2, let
$$
\begin{array}{l}
B C=3, \\
A B=A C=A D \\
=B D=C D \\
=2,
\end{array}
$$
$E$ and $F$ are the midpoints of $B C$ and $A D$, respectively. The projection of point $D$ on plane $A B C$ is $H$, which is the circumcenter of $\triangle A B C$, and the center of the circ... | \frac{\sqrt{21}}{3} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,005 |
7. If in any $n$ consecutive positive integers, there must exist a number whose sum of digits is a multiple of 7, then the smallest $n=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 7. 13 .
Firstly, note that in 12 consecutive positive integers (such as 994, $995, \cdots, 1005$), the sum of the digits of any number is not a multiple of 7, hence $n \geqslant 13$.
Secondly, we need to show that in any 13 consecutive positive integers, there must be one number whose sum of digits is a multiple of 7... | null | Calculus | math-word-problem | Yes | Yes | cn_contest | false | 723,006 |
8. The set of positive real numbers $X$ satisfies: $x \in X$ if and only if $x+\{x\}^{2}$ is an integer (where, $\{x\}$ denotes the fractional part of $x$). If the numbers in $X$ are arranged in strictly increasing order, then the sum of the first 100 terms is $\qquad$ . | 8. $2475+25 \sqrt{5}$.
Obviously, every positive integer $x$ belongs to $X$, and we denote them in strictly increasing order as $a_{1}, a_{2}, \cdots$.
Next, consider the elements in $X$ that are not positive integers, and we denote them in strictly increasing order as $b_{1}, b_{2}, \cdots$.
Since $x+\{x\}^{2}$ is ... | 2475+25 \sqrt{5} | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,007 |
9. (16 points) Given a tetrahedron, if there exists a face (lying in the plane $\alpha$) such that when the other three faces are rotated outward around their edges lying on the plane $\alpha$ until they lie on the plane $\alpha$, the four faces together form a triangle on the plane $\alpha$, then the tetrahedron is ca... | 9. As shown in Figure 3, if the tetrahedron PDEF is a flat tetrahedron, and the triangle after being flattened along the plane of $\triangle DEF$ is $\triangle ABC$, then $DE$, $EF$, and $FD$ are the midlines of $\triangle ABC$. The four congruent triangles formed by the three midlines of $\triangle ABC$ constitute the... | \frac{1}{12} \sqrt{2\left(b^{2} + c^{2} - a^{2}\right)\left(c^{2} + a^{2} - b^{2}\right)\left(a^{2} + b^{2} - c^{2}\right)} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,008 |
10. (20 points) $P$ is any fixed point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. A circle is drawn with $P$ as the center, intersecting the major axis $A B$ of the ellipse at points $M$ and $N$. Let $P M$ and $P N$ intersect the ellipse at points $E$ and $F$. Prove: The slope of the line $E F$ ... | 10. As shown in Figure 4, construct $\odot O: x^{2}+y^{2}=a^{2}$, and draw perpendiculars from points $P$, $E$, and $F$ to the $x$-axis, with the feet of the perpendiculars being $H$, $L$, and $T$, respectively, intersecting $\odot O$ at points $C$, $D$, $G$, and $Q$. The coordinates of the points on the ellipse are gi... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,009 |
11. (20 points) Given two sequences, satisfying
$$
\begin{array}{l}
\left\{a_{n}\right\}: a_{0}=1, a_{1}=13, \\
a_{n+1}=\frac{a_{n}^{2}+64}{a_{n-1}}\left(n \in \mathbf{N}_{+}\right) ; \\
\left\{b_{n}\right\}: b_{0}=1, \\
b_{n+1}=9 b_{n}+\sqrt{80 b_{n}^{2}-64}(n \in \mathbf{N}) .
\end{array}
$$
Prove: For any $n \in \m... | 11. For the sequence $\left\{a_{n}\right\}$, we have
$$
\begin{array}{l}
a_{0}=1, a_{1}=13, a_{2}=233 . \\
\begin{array}{l}
\text { By } \frac{a_{n}+a_{n+2}}{a_{n+1}}=\frac{a_{n} a_{n+2}+a_{n+2}^{2}}{a_{n+1} a_{n+2}} \\
=\frac{\left(a_{n+1}^{2}+64\right)+a_{n+2}^{2}}{a_{n+1} a_{n+2}} \\
=\frac{a_{n+1}^{2}+\left(a_{n+2... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,010 |
One. (40 points) As shown in Figure 1, quadrilateral $ABCD$ is inscribed in a circle, $E$ is any point on arc $\overparen{CD}$, and the reflections of point $D$ over sides $BC$, $CA$, and $AB$ are $D_{1}$, $D_{2}$, and $D_{3}$, respectively. Connecting $E D_{1}$, $E D_{2}$, and $E D_{3}$ intersect the lines $BC$, $CA$,... | (1) Let $D D_{1}$, $D D_{2}$, $D D_{3}$ intersect $B C$, $C A$, $A B$ at points $F_{1}$, $F_{2}$, $F_{3}$, respectively.
By the Simson line theorem, $F_{1}$, $F_{2}$, $F_{3}$ are collinear.
Since $D_{1} D_{2} \parallel F_{1} F_{2}$, $D_{1} D_{3} \parallel F_{1} F_{3}$, then $D_{1}$, $D_{2}$, $D_{3}$ are collinear.
(2) ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,011 |
II. (40 points) Prove: The set $E$ of real numbers $x$ that satisfy the inequality
$$
\frac{1}{x-1}+\frac{2}{x-2}+\cdots+\frac{200}{x-200}>10
$$
can be expressed as the union of some mutually disjoint open intervals. Find the total length of these intervals. | Consider the function
$$
f(x)=\frac{1}{x-1}+\frac{2}{x-2}+\cdots+\frac{200}{x-200}-10 \text {. }
$$
For any real number $x$.
For any $k$ in the set $\{1,2, \cdots, 200\}$, when $x \rightarrow k-0$, $f(x) \rightarrow-\infty$, and when $x \rightarrow k+0$, $f(x) \rightarrow+\infty$. Also, when $x \rightarrow+\infty$, $... | 2010 | Inequalities | proof | Yes | Yes | cn_contest | false | 723,012 |
Three, (50 points) A positive integer is called "simple" if it does not have any square factors greater than 1. Determine how many numbers in $1,2, \cdots, 2010$ are simple.
| Three, notice that $2010 < 45^2$.
Thus, if a number $n$ in the set $M = \{1, 2, \cdots, 2010\}$ is not simple, it must contain one or more square factors from the set of primes
$$
N = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43\}.
$$
Let
$$
\left[\frac{2010}{p^2}\right] = s(p), \left[\frac{2010}{p^2 q^2}\righ... | 1221 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,013 |
For the set $M_{n}=\{1,2, \cdots, 2 n\}$ $\left(n \in \mathbf{N}_{+}\right)$, if there exist two sequences
$$
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right), B=\left(b_{1}, b_{2}, \cdots, b_{n}\right)
$$
satisfying
(i) $\left\{a_{k}, b_{k} \mid k=1,2, \cdots, n\right\}=M_{n}$,
(ii) $a_{k}-b_{k}=k(k=1,2, \cdots, n)$,
then... | (1) For any friendship arrangement $(A, B)$ of $M_{n}$, construct the dual arrangement $(\bar{A}, \bar{B})$, where
$$
\begin{array}{l}
\bar{A}=\left(\bar{a}_{1}, \bar{a}_{2}, \cdots, \bar{a}_{n}\right), \bar{B}=\left(\bar{b}_{1}, \bar{b}_{2}, \cdots, \bar{b}_{n}\right), \\
\bar{a}_{i}=2 n+1-b_{i}, \bar{b}_{i}=2 n+1-a_{... | proof | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,014 |
Example 1 Let $\alpha=\frac{\sqrt{5}+1}{2}$. Then $\left[\alpha^{16}\right]=$ $\qquad$ .
(2008, "Five Sheep Cup" Mathematics Competition (Junior High School))
[Analysis] Note that $\alpha=\frac{\sqrt{5}+1}{2}$ and $\beta=\frac{\sqrt{5}-1}{2}$ $(0<\beta<1)$ can make $\alpha+\beta=\sqrt{5}$ and $\alpha \beta=1$. Therefor... | Let $\beta=\frac{\sqrt{5}-1}{2}$. Then
$$
\begin{array}{l}
\alpha+\beta=\sqrt{5}, \alpha \beta=1 . \\
\text { Therefore, } \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3, \\
\alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2}=7, \\
\alpha^{8}+\beta^{8}=\left(\alpha^{4}+\beta^{4}\rig... | 2206 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,015 |
Example 2 Calculate $[\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}]$ (2008 appears 2008 times).
(2008, International Youth Math Invitational Competition)
【Analysis】Although there are 2008 square root operations, as long as you patiently estimate from the inside out, the pattern will naturally become apparent. | Let $a_{n}=\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}$ (with $n$ 2008s, $n=1,2, \cdots$ ).
Given $44<\sqrt{2008}<45$, we have
$$
\begin{array}{l}
45^{2}<2008+44<2008+\sqrt{2008} \\
<2008+45<46^{2} .
\end{array}
$$
Thus, $45<\sqrt{2008+\sqrt{2008}}<46$, which means $45<a_{2}<46$.
Therefore, $45^{2}<2008+45<2008+a_{2}$
... | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,016 |
Example 11 Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a$ ] equals $\qquad$
$(2009$, Beijing Mathematical Competition (Grade 8)) | Notice that
$0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$.
Then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]$ equals
0 or 1.
By the problem statement, 18 of these are equal to 1. Therefore,
$\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\lef... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,017 |
Example 12 Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$. Then the number of such $n$ is $\qquad$.
(2006, National Junior High School Mathematics Competition) | From the properties of the Gaussian function, we know that $\left[\frac{n}{3}\right] \leqslant \frac{n}{3}$, with equality holding if and only if $\frac{n}{3}$ is an integer; $\left[\frac{n}{6}\right] \leqslant \frac{n}{6}$, with equality holding if and only if $\frac{n}{6}$ is an integer.
Therefore, $\left[\frac{n}{3}... | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,018 |
2. The algebraic expression $\left[(\sqrt{6}+\sqrt{5})^{6}\right]=$ $\qquad$
(2004, "Five Sheep Cup" Mathematics Competition (Grade 9)) | Prompt: Example 1. Answer: 10581. | 10581 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,020 |
3. The sum of all real numbers $x$ that satisfy $25\{x\}+[x]=125$ is $\qquad$
(2007, International Invitational Competition for Young Mathematicians in Cities) | Hint: $25\{x\}+[x]=125$ can be transformed into $25 x-24[x]=125$.
Example 6. Answer: 2837. | 2837 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,021 |
4. The real solutions of the equation $[2 x]+[3 x]=8 x-\frac{7}{2}$ are $\qquad$
$(2006,($ Xin Zhi Cup $)$ Shanghai Junior High School Mathematics Competition) | Prompt: Example 7. Answer: $\frac{13}{16}, \frac{17}{16}$. | \frac{13}{16}, \frac{17}{16} | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,022 |
5. The positive integer $n$ is less than 100 and satisfies the equation $\left[\frac{n}{2}\right]+\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=n$.
How many such positive integers $n$ are there?
(A) 2
(B) 3
(C) 12
(D) 16
$(2000$, National Junior High School Mathematics League) | Prompt: Example 12. Answer: (D).
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided text is already in English, so no translation is needed. If you intended to translate a different text, please provide t... | null | Number Theory | MCQ | Yes | Yes | cn_contest | false | 723,023 |
Example 1 Proof: If $(a, b)=1, p$ is a prime, and $p \mid\left(a^{2}+b^{2}\right)$, then $p \neq \equiv-1(\bmod 4)$. | Proof by contradiction.
Assume $p=4 k-1\left(k \in \mathbf{N}_{+}\right)$.
Given $(a, b)=1$, then $a$ and $b$ cannot both be divisible by $p$.
(1) When exactly one of $a$ and $b$ is divisible by $p$, assume without loss of generality that $p \mid a, p \nmid b$, then $p \nmid b^{2} \Rightarrow p \nmid\left(a^{2}+b^{2}\r... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,024 |
Example 2 In the set of integers, find the solutions to
$$
x^{2010}-2006=4 y^{2009}+4 y^{2008}+2007 y
$$
$^{[1]}$
(2009, Macedonian Mathematical Olympiad) | Solve the original equation, which can be transformed into
$$
\begin{array}{l}
x^{2010}+1=4 y^{2009}+4 y^{2008}+2007 y+2007 \\
=\left(4 y^{2008}+2007\right)(y+1) .
\end{array}
$$
Since $4 y^{2008}+2007 \equiv-1(\bmod 4)$, then $4 y^{2008}+2007$ must have a prime factor of the form $4 k-1$.
However, $(x, 1)=1$, so $x^... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,025 |
Example 3 Proof: The equation $x^{2}+5=y^{3}$ has no integer solutions. (1979, Bulgarian Mathematical Olympiad) | Proof Assume the equation has integer solutions $(x, y)$.
(1) When $x$ is odd, then $x^{2}+5 \equiv 2(\bmod 4)$.
Thus, $y^{3} \equiv 2(\bmod 4)$.
Therefore, $y$ is even, i.e., $y \equiv 0(\bmod 2)$.
Hence, $y^{3} \equiv 0(\bmod 4)$, which is a contradiction.
(2) When $x$ is even, let $x=2 n$.
By $y^{3}=x^{2}+5 \equiv ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,026 |
Example 3 Let
$$
\begin{aligned}
S & =\frac{1}{\left[\frac{(10 \times 11-1)^{2}}{10 \times 11}\right]}+\frac{1}{\left[\frac{(11 \times 12-1)^{2}}{11 \times 12}\right]}+\cdots+ \\
& \frac{1}{\left[\frac{(49 \times 50-1)^{2}}{49 \times 50}\right]}
\end{aligned}
$$
Then $[30 S]=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 0
(2002, "F... | When $n \geqslant 10$, and $n$ is an integer,
$$
\begin{array}{l}
{\left[\frac{[n(n+1)-1]^{2}}{n(n+1)}\right]=\left[n(n+1)-2+\frac{1}{n(n+1)}\right]} \\
=n(n+1)-2=(n-1)(n+2) .
\end{array}
$$
Thus, $S=\frac{1}{9 \times 12}+\frac{1}{10 \times 13}+\cdots+\frac{1}{48 \times 51}$.
Notice that $\frac{1}{k(k+3)}=\frac{1}{3}\... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,027 |
Example 4 Proof: In the sequence $\left\{2^{n}-3\right\}$, there must be an infinite subsequence where any two terms are coprime.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Proof: In the sequence $\left\{2^{n}-3\right\}$, there are already $k$ terms that are pairwise coprime, denoted as $u_{1}, u_{2}, \cdots, u_{k}$. We define
$$
u_{k+1}=2^{\varphi\left(u_{1} w_{2} \cdots u_{k}\right)+1}-3 \text {, }
$$
where $\varphi(x)$ is the Euler's totient function.
By Euler's theorem, we have
$$
\b... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,028 |
Example 6 Find all pairs of primes $(p, q)$ such that
$$
p q \mid\left(5^{p}+5^{q}\right) .{ }^{[2]}
$$
(2009, China Mathematical Olympiad) | (1) If $\min \{p, q\}=2$, without loss of generality, let $p=2$, then to make $2 q l\left(5^{2}+5^{q}\right)=25+5^{q}$, it is necessary that
$$
q \mid\left(25+5^{q}\right) \text {. }
$$
Since $5^{q}+25=5^{q}-5+30$, by Fermat's Little Theorem, we have
$$
q\left|\left(5^{q}-5\right) \Rightarrow q\right| 30 \text {. }
$$... | (2,3),(3,2),(2,5),(5,2),(5,5),(5,313),(313,5) | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,029 |
Example 8 Let $m$ be a positive integer. If $2^{m+1}+1$ divides $3^{2 m}+1$, prove: $2^{m+1}+1$ is a prime number.
(2003, Korean Mathematical Olympiad) | Let $q=2^{m+1}+1$. Given $q \mid\left(3^{2 m}+1\right)$, then $3^{2 m} \equiv-1(\bmod q)$.
Hence $(3, q)=1$.
From equation (1), we get $\left(3^{2 m}\right)^{2} \equiv(-1)^{2}(\bmod q)$, which means
$$
3^{2^{m+1}} \equiv 1(\bmod q) \text {. }
$$
From equations (1) and (2), it is easy to see that the order of 3 modulo ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,031 |
3. Let positive integers $a, b$ be such that $15a + 16b$ and $16a - 15b$ are both squares of positive integers. Find the smallest value that the smaller of these two squares can take. ${ }^{[3]}$
(1996, China Mathematical Olympiad) | Given: $15a + 16b = r^2, 16a - 15b = s^2$ $\left(r, s \in \mathbf{N}_{+}\right)$.
Eliminating $a$ gives:
$481b = 16r^2 - 15s^2$
$\Rightarrow 481 \mid (16r^2 - 15s^2)$.
From equation (1):
$16r^2 \equiv 15s^2 \pmod{13}$.
If $13 \times r, 13 \times s$, then by Fermat's Little Theorem, $s^{12} \equiv 1 \pmod{13}$.
From equ... | 481^2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,032 |
4. Find all positive integers $n$ such that every positive integer whose decimal representation consists of $n-1$ digits 1 and one digit 7 is a prime number.
(31st IMO Shortlist) | Obviously, $n=1,2$ satisfy the condition.
Below is the proof that $n \geqslant 3$ does not satisfy the condition.
Every positive integer $N$ formed by $n-1$ ones and one 7 can be expressed as
$$
N=A_{n}+6 \times 10^{k}(0 \leqslant k \leqslant n) \text {. }
$$
Here, $A_{n}=\sum_{i=0}^{n-1} 10^{i}$.
Clearly, when $3 \mi... | n=1,2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,033 |
5. If the positive integer $n(n>1)$, satisfies $n^{2} \mid\left(2^{n}+1\right)$, prove: $31 n$.
The statement to prove should be $31 \mid n$. Here is the translation with the correction:
5. If the positive integer $n(n>1)$, satisfies $n^{2} \mid\left(2^{n}+1\right)$, prove: $31 \mid n$. | Let $p$ be the smallest prime factor of $n$. Then, since $2^{n}+1$ is odd, we have $p \geqslant 3$.
Let the order of 2 modulo $p$ be $l$. Then
$2^{l} \equiv 1(\bmod p)$.
Also, $n^{2} \mid\left(2^{n}+1\right)$, and $p \mid n$, so
$2^{n} \equiv-1(\bmod p), 2^{2 n} \equiv 1(\bmod p)$.
By Fermat's Little Theorem, we have
$... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,034 |
Example 1 In the right triangle $ABC$, it is known that $\angle B=90^{\circ}$, $AB > BC$, a semicircle $\Gamma$ with $AB$ as its diameter is on the same side of $AB$ as point $C$, $P$ is a point on the semicircle $\Gamma$, and satisfies $BP = BC$, $Q$ is a point on $AB$ and satisfies $AP = AQ$. Prove: The midpoint of $... | Prove: As shown in Figure 1, with $B$ as the origin, the line $BA$ as the $x$-axis, and the line $BC$ as the $y$-axis, establish a Cartesian coordinate system.
Let $BC = b$, $BA = a$.
Then the polar equation of the semicircle $\Gamma$ is
$\rho = a \cos \theta$.
Also, $AQ = AP = \sqrt{AB^2 - BP^2} = \sqrt{a^2 - b^2}$, s... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,035 |
Example 2 Let $A D$ be a median of $\triangle A B C$, $d$ a line perpendicular to $A D$, $M$ a point on $d$, $E, F$ the midpoints of $M B, M C$ respectively, and the lines through $E, F$ perpendicular to $d$ intersect $A B, A C$ at points $P, Q$ respectively. Let $d^{\prime}$ be the line through $M$ perpendicular to $P... | Proof As shown in Figure 2, establish a Cartesian coordinate system with the line $AD$ as the $y$-axis and the line $d$ as the $x$-axis. Let the points be
$$
\begin{array}{l}
A(0, a) \text {, } \\
D(0, -d) \text {, } \\
M(m, 0) \text {, } \\
B(-b, -d+c) \text {, } C(b, -d-c) \text { . }
\end{array}
$$
Thus, the equati... | \left(\frac{-b c}{a+d}, \frac{-b^{2}}{a+d}\right) | Geometry | proof | Yes | Yes | cn_contest | false | 723,036 |
Example 4 Algebraic Expression
$[\sqrt[3]{1 \times 2 \times 3}]+[\sqrt[3]{2 \times 3 \times 4}]+\cdots+$ $[\sqrt[3]{2000 \times 2001 \times 2002}]=(\quad)$.
(A) 2000000
(B) 2001000
(C) 2002000
(D) 2003001
(2000, "Five Sheep Cup" Mathematics Competition (Grade 9))
[Analysis] Notice that each term in the sum can be expre... | Since $k^{3}<k(k+1)(k+2)<(k+1)^{3}$,
thus, $k<\sqrt[3]{k(k+1)(k+2)}<k+1$.
Therefore, $[\sqrt[3]{k(k+1)(k+2)}]=k$.
Hence the original expression $=1+2+\cdots+2000=2001000$. | 2001000 | Algebra | MCQ | Yes | Yes | cn_contest | false | 723,037 |
Example 3 Let the diagonals of convex quadrilateral $ABCD$ intersect at point $O$, and the circumcircles of $\triangle OAD$ and $\triangle OBC$ intersect at points $O$ and $M$. Line $OM$ intersects the circumcircles of $\triangle OAB$ and $\triangle OCD$ at points $T$ and $S$, respectively. Prove that $M$ is the midpoi... | Proof: Let the circumcenters of $\triangle A O D, \triangle B O C, \triangle A O B$, and $\triangle C O D$ be $O_{1}, O_{2}, O_{3}, O_{4}$, respectively.
Then $O_{1} O_{3} \perp A O, O_{2} O_{4} \perp \mathrm{CO}$.
Therefore, $O_{1} O_{3} / / O_{2} O_{4}$.
Similarly, $\mathrm{O}_{2} \mathrm{O}_{3} / / \mathrm{O}_{1} \m... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,038 |
Example 4 Two fixed circles $\odot O_{1}$ and $\odot O_{2}$ are tangent at point $\mathrm{P},$ the radius of $\odot O_{2}$ is greater than that of $\odot O_{1},$ and $A$ is a point on $\odot O_{2}.$ The tangent line to $\odot O_{2}$ at $A$ is $l,$ and $A B, A C$ are the tangents from point $A$ to $\odot O_{1},$ with po... | Prove that with $P$ as the origin, the line $O_{1} O_{2}$ as the $x$-axis, and the common tangent of the two circles as the $y$-axis, we establish a Cartesian coordinate system.
Let $\odot O_{1}:(x-a)^{2}+y^{2}=a^{2}$,
$\odot O_{2}:(x-b)^{2}+y^{2}=b^{2}$.
Let point $A\left(x_{0}, y_{0}\right)$, and let $B C$ intersect ... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,039 |
Let $\alpha$ be a real number greater than 1, and the sequence $\left\{s_{n}\right\}_{n>1}$ is defined as follows: $s_{1}=1, s_{2}=\alpha$. If for some $n$ $(n \geqslant 1), s_{1}, s_{2}, \cdots, s_{2 n}$ have been defined, then
$$
2^{n}+1 \leqslant j \leqslant 2^{n+1}, s_{j}=\alpha s_{j-2}
$$
(The first $n$ terms are ... | Prove by mathematical induction on $k$.
When $k=0$, $n=2^{0}$, it is only necessary to prove:
$$
c_{20}=(1+\alpha)^{\epsilon_{0}} \text {. }
$$
By the problem, $c_{2^{0}}=s_{1}=1$.
When $e_{0} \geqslant 1$,
$$
\begin{array}{l}
c_{20}=c_{200-1}+\sum_{i=2^{* 0-1}}^{200} s_{i} \\
=c_{2 \theta-1}+\alpha \sum_{i=1}^{2 \inf... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,040 |
2. If a positive integer $N$ satisfies $N=1$ or $N$ can be written as the product of an even number of prime numbers (not necessarily distinct), then $N$ is called "balanced." Given positive integers $a, b$, define the polynomial $P(x)=(x+a)(x+b)$. Prove:
(1) There exist different positive integers $a, b$ such that all... | 2. (1) Define the function $f: \mathbf{N}_{+} \rightarrow\{0,1\}$.
If $n$ is balanced, then $f(n)=0$;
If $n$ is not balanced, then $f(n)=1$.
Thus, for all positive integers $n, m$, we have $f(n m) \equiv f(n)+f(m)(\bmod 2)$.
For each positive integer $n$, consider the $0-1$ sequence
$$
\{f(n+1), f(n+2), \cdots, f(n+50... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,041 |
3. Let $f$ be a non-constant function from the set of positive integers to the set of positive integers, such that for all distinct positive integers $a$ and $b$, $a-b$ divides $f(a)-f(b)$. Prove that there exist infinitely many primes $p$, such that for each $p$, there exists a positive integer $c$ such that $p \mid f... | 3. Let $v_{p}(a)$ denote the exponent of the prime $p$ in the prime factorization of $a$. Assume that all elements in the range of $f$ have only a finite number of distinct prime factors, denoted as $p_{1}, p_{2}, \cdots, p_{m}$.
Clearly, there are infinitely many positive integers $a$ such that
$v_{p_{i}}(a)>v_{p_{i}}... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,042 |
4. Find all positive integers $n$, such that there exists a sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}$, for every positive integer $k(2 \leqslant$ $k \leqslant n-1)$, we have
$$
a_{k+1}=\frac{a_{k}^{2}+1}{a_{k-1}+1}-1 .
$$ | 4. When $n=1,2,3,4$, such sequences exist.
In fact, if for some $n$, such a sequence exists, then for all sequences with fewer terms than $n$, they also exist.
An example for $n=4$:
$a_{1}=4, a_{2}=33, a_{3}=217, a_{4}=1384$.
Below is the proof: When $n \geqslant 5$, such sequences do not exist.
In fact, it is suffici... | proof | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,043 |
5. Let $P(x)$ be a non-constant polynomial with integer coefficients. Prove that there does not exist a function $T: \mathbf{Z} \rightarrow \mathrm{Z}$, such that for each integer $n(n \geqslant 1)$, the number of integers $x$ satisfying $T^{n}(x)=x$ is equal to $P(n)$, where $T^{n}$ denotes the $n$-th iteration of $T$... | 5. Suppose there exists a polynomial $P(x)$ of degree at least 1 and a function $T$ that satisfies the given conditions. Let $A(n)$ denote the set of all integers $x$ such that $T^{n}(x)=x$, and $B(n)$ denote the set of all integers $x$ such that $T^{n}(x)=x$, and for all $k(1 \leqslant k < n)$, $T^{k}(x) \neq x$. Then... | proof | Algebra | proof | Yes | Yes | cn_contest | false | 723,044 |
6. Given that $t$ is a positive integer. Prove: If there exists an integer sequence $a_{0}, a_{1}, \cdots$ satisfying
$$
a_{n}=\frac{a_{n-1}+n^{t}}{n}(n=1,2, \cdots),
$$
then $t-2$ is divisible by 3. | 6. First, prove that for each positive integer $k$, there exists an integer-coefficient polynomial $P_{k}$ of degree $k-1$ and an integer $q_{k}$ such that
$$
x P_{k}(x)=x^{k}+P_{k}(x-1)+q_{k} \text {. }
$$
In fact, for a given $x$, let
$$
P_{k}(x)=b_{k-1} x^{k-1}+\cdots+b_{1} x+b_{0} \text {. }
$$
Clearly, $b_{k-1}=... | 3 \mid (t-2) | Number Theory | proof | Yes | Yes | cn_contest | false | 723,045 |
7. Let $a, b$ be distinct integers greater than 1. Prove: There exists a positive integer $n$, such that $\left(a^{n}-1\right)\left(b^{n}-1\right)$ is not a perfect square. | 7. Let $a^{2}=A, b^{2}=B$,
$$
z_{n}=\sqrt{\left(A^{n}-1\right)\left(B^{n}-1\right)} \text {. }
$$
Assume that for $n=1,2, \cdots, z_{n}$ is an integer. Without loss of generality, assume $b>0$. Then,
$$
\begin{array}{l}
A^{n} B^{n}-A^{n}-B^{n}+1=z_{n}^{2} \\
=\left(\delta_{0}(a b)^{n}-\delta_{1}\left(\frac{a}{b}\righ... | proof | Number Theory | proof | Yes | Yes | cn_contest | false | 723,046 |
1. Let $n$ ($n<100$) be a positive integer, and there exists a positive integer $k$, such that $1 \leqslant k \leqslant n-1$, satisfying
$$
\frac{4 k+1}{2 n}+\frac{1-2 k^{2}}{n^{2}}=\frac{1}{2} \text {. }
$$
How many values of $n$ satisfy the condition? Prove your conclusion. | 1. From equation (1), we get $(2 k-n)^{2}=n+2$.
Solving, we get $k=\frac{1}{2}(n \pm \sqrt{n+2})$.
Since $k$ is an integer, therefore, $n=m^{2}-2(m \in \mathbf{N})$.
Thus, $k=\frac{1}{2}\left(m^{2} \pm m-2\right)$.
Also, $1 \leqslant k \leqslant n-1$, so
$$
1 \leqslant \frac{m^{2} \pm m-2}{2} \leqslant m^{2}-3 \text {... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,047 |
Example 5 Calculate $\left[\frac{23 \times 1}{101}\right]+\left[\frac{23 \times 2}{101}\right]+\cdots+$ $\left[\frac{23 \times 100}{101}\right]$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Since 23 and 101 are coprime, for $k=1$, $2, \cdots, 100$, $\frac{23 k}{101}$ is not an integer. Therefore,
$$
\begin{array}{l}
\frac{23 k}{101}-1<\left[\frac{23 k}{101}\right]<\frac{23 k}{101}, \\
\frac{23(101-k)}{101}-1<\left[\frac{23(101-k)}{101}\right] \\
<\frac{23(101-k)}{101}. \\
\text{ Hence }\left(\frac{23 k}{1... | 1100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false | 723,048 |
2. As shown in Figure 1, let $A D$ be the median of $\triangle A B C$, and the circumcenters of $\triangle A B D$ and $\triangle A D C$ be $E$ and $F$, respectively. The lines $B E$ and $C F$ intersect at point $G$. If $D G = \frac{1}{2} B C$, prove:
$$
\angle A D G = 2 \angle A C G .
$$ | 2. As shown in Figure 3, connect $DE$ and $DF$.
It is easy to prove that $\triangle BGC$ is a right triangle.
Since $E$ and $F$ are the circumcenters of $\triangle ABD$ and $\triangle ADC$ respectively, then
$$
\begin{array}{l}
\angle GBC=90^{\circ}-\angle BAD, \\
\angle GCB=90^{\circ}-\angle DAC.
\end{array}
$$
The... | proof | Geometry | proof | Yes | Yes | cn_contest | false | 723,049 |
3. Find a prime number $p$ greater than 2, such that the parabola
$$
y=\left(x-\frac{1}{p}\right)\left(x-\frac{p}{2}\right)
$$
has a point $\left(x_{0}, y_{0}\right)$ where $x_{0}$ is a positive integer and $y_{0}$ is the square of a prime number. | 3. From the problem, we have
$$
\begin{array}{l}
y_{0}=\left(x_{0}-\frac{1}{p}\right)\left(x_{0}-\frac{p}{2}\right)=\frac{x_{0} p-1}{2} \cdot \frac{2 x_{0}-p}{p} . \\
\text { From } \frac{x_{0} p-1}{2}>\frac{2 x_{0}-p}{p}, \text { we know } \frac{2 x_{0}-p}{p}=1 .
\end{array}
$$
Therefore, $x_{0}=p, y_{0}=\frac{p^{2}-... | p=3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,050 |
1. Given $a^{2}(b+c)=b^{2}(a+c)=2010$, and $a \neq b$. Then $c^{2}(a+b)=$ | 1. 2010 .
Notice
$$
\begin{array}{l}
a^{2}(b+c)-b^{2}(a+c) \\
=(a-b)(a c+b c+a b)=0 .
\end{array}
$$
Since $a \neq b$, it follows that $a c+b c+a b=0$.
$$
\begin{array}{l}
\text { Then } c^{2}(a+b)-b^{2}(a+c) \\
=(c-b)(a c+b c+a b)=0 .
\end{array}
$$
Therefore, $c^{2}(a+b)=b^{2}(a+c)=2010$. | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false | 723,051 |
2. If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-$ $xy+y^{2}$ is $\qquad$ | 2. 1 .
Notice that
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also }|x \pm y| \leqslant|x|+|y| \leqslant 1 \text {, then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ take 0 and 1 respectively, the equality holds. | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false | 723,052 |
3. The difference between the maximum area and the minimum area of right-angled triangles with integer side lengths and where the area value equals the perimeter value 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. 6 .
Let the lengths of the three sides of a triangle be $a$, $b$, and $c$. According to the problem, we have
$$
\begin{array}{l}
a^{2}+b^{2}=c^{2}, \\
a b=2(a+b+c) .
\end{array}
$$
Then $a b-4(a+b)+8=0$.
Solving this, we get $(a, b)=(5,12)$ or $(6,8)$.
Thus, $S_{\max }=30, S_{\text {min }}=24$.
Therefore, $S_{\max... | null | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false | 723,053 |
4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively. Given that $\angle C=2 \angle B, a b=24$. Then the range of values for $c$ is $\qquad$ . | 4. $(4 \sqrt{2},+\infty)$.
As shown in Figure 4, draw the angle bisector $CD$. Then
$$
\begin{array}{l}
b^{2}a \Rightarrow c>\frac{a+b}{2} \text{. }
\end{array}
$$
Therefore, $c^{2}=a b+b^{2}>\frac{1}{4}(a+b)^{2}$, which means
$$
3 b^{2}+2 a b-a^{2}>0 \Rightarrow a8$.
Thus, $c^{2}=24+b^{2}>32 \Rightarrow c>4 \sqrt{2}... | c \in (4 \sqrt{2}, +\infty) | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,054 |
5. As shown in Figure 2, in $\triangle A B C$, $A C=B C, \angle C=20^{\circ}$, and points $M, N$ are on sides $A C, B C$ respectively, such that $\angle B A N=50^{\circ}, \angle A B M=$ $60^{\circ}$. Then $\angle N M B=$ $\qquad$ | 5. $30^{\circ}$.
It is easy to know that $A B=B N$.
As shown in Figure 5, take a point $D$ on side $A M$ such that $B D=B A$.
Then $B D=B N$, and
$\angle A B D=20^{\circ}$.
Thus, $\angle B D N=60^{\circ}$,
$\angle M D N=40^{\circ}$.
Therefore, $\angle D M N=70^{\circ}$,
$\angle N M B=30^{\circ}$. | 30^{\circ} | Geometry | math-word-problem | Yes | Yes | cn_contest | false | 723,055 |
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