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Example 3 㷵 On a board, there is a convex 2011-gon, and Betya draws its diagonals one by one. It is known that each diagonal drawn intersects at most one of the previously drawn diagonals at an interior point. Question: What is the maximum number of diagonals Betya can draw? [3] | To prove by induction: For a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex polygon. The following $2n-6$ diagonals can be drawn sequentially:
$A_{2} A_{4}, A_{3} A_{5}, \cdots, A_{n-2} A_{n}, A_{1} A_{3}, A_{1} A_{4}, \cdots, A_{1} A_{n-1}$.
When $n=3$, the... | 4016 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
\text { 11. If } a+b-2 \sqrt{a-1}-4 \sqrt{b-2} \\
=3 \sqrt{c-3}-\frac{1}{2} c-5 \text {, }
\end{array}
$$
then $a+b+c=$ . $\qquad$ | Ni, 11.20.
Notice,
$$
(\sqrt{a-1}-1)^{2}+(\sqrt{b-2}-2)^{2}+\frac{1}{2}(\sqrt{c-3}-3)^{2}=0 \text {. }
$$
Therefore, $a=2, b=6, c=12$.
So $a+b+c=20$. | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the function $y=f(x)$ have the domain $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
2 f\left(x^{2}+x\right)+f\left(x^{2}-3 x+2\right) \\
=9 x^{2}-3 x-6 .
\end{array}
$$
Then the value of $f(60)$ is . $\qquad$ | 2. 176.
Substituting $1-x$ for $x$ in equation (1) yields
$$
\begin{array}{l}
2 f\left(x^{2}-3 x+2\right)+f\left(x^{2}+x\right) \\
=9 x^{2}-15 x .
\end{array}
$$
Combining equations (1) and (2) gives
$$
\begin{array}{l}
f\left(x^{2}+x\right)=3 x^{2}+3 x-4 \\
=3\left(x^{2}+x\right)-4 .
\end{array}
$$
Therefore, $f(60... | 176 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let positive real numbers $a, b, c, d, e$ satisfy $a<b<c<d$ $<e$, and the smallest three of the 10 products of any two numbers are $28, 32, 56$, and the largest two are 128, 240. Then $e=$ $\qquad$ | 2.16.
From the problem, we know
$$
a b=28, a c=32, c e=128, d e=240 \text {. }
$$
Then $c=\frac{8}{7} b, d=\frac{15}{8} c=\frac{15}{7} b$.
Thus, $a d=\frac{15}{7} a b=60>56$.
Therefore, $b c=56 \Rightarrow \frac{8}{7} b^{2}=56 \Rightarrow b=7$.
Hence, $a=4, c=8, d=15, e=16$. | 16 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) If placing the positive integer $N$ to the left of the positive integer $n$ results in a new number that is divisible by 7, then $N$ is called a "magic number" of $n$. $M$ is a set of positive integers such that for any positive integer $n$, there exists a positive integer in set $M$ that is a magic ... | For $n=1,2, \cdots, 7$.
If $|M| \leqslant 6$, then by the pigeonhole principle, there must be a positive integer $N$ in set $M$ that is a common magic number of $i, j(1 \leqslant i<j \leqslant 7)$, i.e.,
$71(10 N+i), 7 I(10 N+j)$.
Then $7!(j-i)$, but $0<j-i \leqslant 6$, a contradiction.
Therefore, $|M| \geqslant 7$.
T... | 28 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the three sides of $\triangle A B C$ are $a, b, c$. If $a+b+c=16$, then
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
$=$ | 6. 64 .
Let the circumradius of $\triangle ABC$ be $R$. Then
$$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}... | 64 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. An $8 \times 8$ chessboard is colored in the usual way, with 32 black squares and 32 white squares. A "path" consists of 8 white squares, one in each row, and adjacent white squares share a common vertex. The number of such paths is $\qquad$. | 8. 296
We can use the number labeling method:
\begin{tabular}{cccccccc}
1 & Black & 1 & Black & 1 & Black & 1 & Black \\
Black & 2 & Black & 2 & Black & 2 & Black & 1 \\
2 & Black & 4 & Black & 4 & Black & 3 & Black \\
Black & 6 & Black & 8 & Black & 7 & Black & 3 \\
6 & Black & 14 & Black & 15 & Black & 10 & Black \\... | 296 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) There are 12 points on a plane, and no three points are collinear. Using any one of these points as the starting point and another point as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of zero t... | Let the 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $C_{12}^{3}$. Let the number of vectors starting from $P_{i}(i=1,2, \cdots, 12)$ be $x_{i}$ $\left(0 \leqslant x_{i} \leqslant 11\right)$.
If a triangle with three points as vertices is a non-zero triangle, th... | 70 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
316 Given that the function $f: \mathbf{N}_{+} \rightarrow \mathbf{N}_{+}$ is a monotonically increasing function. If $f(f(n))=3 n$, find $f(2011)$. | Let $f(1)=m$. Then $f(m)=f(f(1))=3$.
Thus, $m>1 \Rightarrow f(m)>f(1) \Rightarrow 3>m$.
Therefore, $f(1)=m=2 \Rightarrow f(2)=f(f(1))=3$.
Similarly, $f(3)=6, f(6)=9$.
From $f(3)<f(4)<f(5)<f(6)$, we get
$6<f(4)<f(5)<9$.
Hence, $f(4)=7, f(5)=8$.
Furthermore, $f(7)=f(f(4))=12$,
$f(8)=f(f(5))=15$,
$f(9)=f(f(6))=18$,
$f(12)... | 3846 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 If the two roots of the equation
$$
x^{2}-3 x-1=0
$$
are also roots of the equation
$$
x^{4}+a x^{2}+b x+c=0
$$
then the value of $a+b-2 c$ is ( ).
(A) -13
(B) -9
(C) 6
(D) 0 | $$
\begin{array}{l}
\text { From the problem, we know that } x^{4}+a x^{2}+b x+c \text { can definitely be divided by } x^{2}-3 x-1. \\
=\left(x^{2}-3 x-1\right)\left(x^{2}+3 x+a+10\right)+ \\
{[(3 a+b+33) x+(a+c+10)], } \\
\text { then }\left\{\begin{array}{l}
3 a+b+33=0, \\
a+c+10=0
\end{array}\right. \\
\Rightarrow... | -13 | Algebra | MCQ | Yes | Yes | cn_contest | false |
6. Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a$, $b$, and $c$ in $S$ such that $a+b$, $b+c$, and $c+a$ are all in the set $A$. | 6. Let $a<b<c$. Let
$x=a+b, y=a+c, z=b+c$.
Then $x\langle y\langle z, x+y\rangle z$, and $x+y+z$ is even. (1)
Conversely, if there exist $x, y, z \in A$ satisfying property (1), then take
$$
a=\frac{x+y-z}{2}, b=\frac{x+z-y}{2}, c=\frac{y+z-x}{2},
$$
we have $a, b, c \in \mathbf{Z}, 1 \leqslant a<b<c \leqslant 2012$, ... | 1008 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Two quadratic equations with unequal leading coefficients $(a-1) x^{2}-\left(a^{2}+2\right) x+\left(a^{2}+2 a\right)=0$, (1) and $(b-1) x^{2}-\left(b^{2}+2\right) x+\left(b^{2}+2 b\right)=0$ ( $a, b$ are positive integers) have a common root. Find the value of $\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}$. | Given the conditions $a>1, b>1, a \neq b$.
Assume the common root of equations (1) and (2) is $x_{0}$. Then
$$
\begin{array}{l}
(a-1) x_{0}^{2}-\left(a^{2}+2\right) x_{0}+\left(a^{2}+2 a\right)=0, \\
(b-1) x_{0}^{2}-\left(b^{2}+2\right) x_{0}+\left(b^{2}+2 b\right)=0 .
\end{array}
$$
(3) $\times(b-1)$ - (4) $\times(a-1... | 256 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. When $s$ and $t$ take all real values,
$$
(s+7-|\cos t|)^{2}+(s-2|\sin t|)^{2}
$$
the minimum value is $\qquad$ | 6. 18
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the text "6. 18" is already in a numerical form that is universal, it does not require translation. Here is the retained format:
6. 18 | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (20 points) It is known that $\left\{a_{n}\right\}$ is an arithmetic sequence with the first term 9 and common difference 7.
(1) Prove: The sequence $\left\{a_{n}\right\}$ contains infinitely many perfect squares;
(2) The 100th perfect square in the sequence $\left\{a_{n}\right\}$ is the nth term? | (1) The general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=9+7(n-1) \text {. }
$$
When $n=7 k^{2} \pm 6 k+1(k=0,1, \cdots)$, we have
$$
\begin{array}{l}
a_{n}=9+7(n-1) \\
=9+7\left(7 k^{2} \pm 6 k\right)=(7 k \pm 3)^{2} .
\end{array}
$$
Therefore, the sequence $\left\{a_{n}\right\}$ has infinitel... | 17201 | Number Theory | proof | Yes | Yes | cn_contest | false |
4. If $4n+1$ and $6n+1$ are both perfect squares, then the smallest positive integer $n$ is $\qquad$ | 4. 20 .
Obviously, $4 n+1, 6 n+1$ are both odd square numbers.
Let $6 n+1=(2 m+1)^{2}=4 m(m+1)+1$.
Then $3 n=2 m(m+1)$.
Since $m(m+1)$ is even, $4 \mid n$.
Let $n=4 k$. Then
$4 n+1=16 k+1, 6 n+1=24 k+1$.
When $k=1,2,3,4$, $4 n+1, 6 n+1$ are not both square numbers, but when $k=5$, i.e., $n=20$, $4 n+1=81$, $6 n+1=121$... | 20 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. If the four digits of the four-digit number $\overline{a b c d}$ satisfy $a+b=c+d$, then it is called a "good number" (for example, 2011 is a good number). Then, the number of good numbers is $\qquad$ | 8. 615 .
Let $k=a+b=c+d$.
Since $1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant 9$, then
$1 \leqslant k \leqslant 18$.
When $1 \leqslant k \leqslant 9$, in the above equation, $a$ can take any value in $\{1,2$, $\cdots, k\}$, $c$ can take any value in $\{0,1, \cdots, k\}$, and once $a 、 c$ are determined, $... | 615 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: There are $n$ people registered to participate in four sports events: A, B, C, and D. It is stipulated that each person must participate in at least one event and at most two events, but events B and C cannot be registered for simultaneously. If in all different registration methods, there must be at least o... | Let the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{T}}\right)$ represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is 1 (for example, if participating in event 甲, then $a_{\text {甲 }}=1... | 172 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
7. Let $M=\{1,2,3,4,5\}$. Then the number of mappings $f: M \rightarrow M$ such that
$$
f(f(x))=f(x)
$$
is $\qquad$ | 7. 196.
For the case where $M$ contains $n$ elements, from
$$
f(f(x))=f(x) \text {, }
$$
we know that for any $a \in M$ (let $f(c)=a$), we have
$$
f(f(c))=f(a)=f(c)=a .
$$
If the range of $f$ contains $k$ elements, then the $n-k$ elements of set $M$ that are not in the range have $k^{n-k}$ possible mappings. Therefo... | 196 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Find
$$
f(x)=|x-1|+2|x-2|+\cdots+2011|x-2011|
$$
the minimum value. | 10. Obviously, when $x=2011$, $f(x)$ has no minimum value.
The following assumes $x \in[1,2011]$.
When $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2010)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2011} i(i-x) \\
=\left(k^{2}+k-2011 \times 1006\right) x+ \\
\quad \frac{2011 \times 2012 \time... | 794598996 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\},
$$
where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedron. Question: What is the maximum number of triangles in graph $G$? | Four, first prove a lemma.
Lemma If in a space graph with $n$ points there is no triangle, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$ ( $[x]$ represents the largest integer not exceeding the real number $x$).
Proof Let the $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, among which, the numbe... | 27 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given a set of complex numbers $D$, a complex number $z \in D$ if and only if there exists a complex number $z_{1}$ with modulus 1, such that
$$
|z-2005-2006 \mathrm{i}|=\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \text {. }
$$
Then the number of complex numbers in $D$ whose real and imaginary parts are both intege... | Solve: Establish a complex plane, let
$$
z=x+y \mathrm{i}(x, y \in \mathbf{R}) \text {. }
$$
From $\left|z_{1}\right|=1$, we know $z_{1}^{2}$ is also on the unit circle $\odot 0$.
$$
\begin{array}{l}
\text { Then }\left|z_{1}^{4}+1-2 z_{1}^{2}\right| \\
=\left|z_{1}^{2}-1\right|^{2}
\end{array}
$$
This represents the... | 49 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For example, 32006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find
$$
a_{1}+a_{2}+\cdots+a_{2000}
$$
the minimum value. ${ }^{[1]}$ | First, we prove: For any 119 positive integers \( b_{1}, b_{2}, \cdots, b_{119} \), there must exist a subset (at least one, or all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
\[ b_{1}, b_{1}+b_{2}, b_{1}+b_{2}+b_{3}, \cdots, b_{1}+b_{2}+\cdots+b_{119}. \]
If one of them is a... | 3910 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 In the positive integers not exceeding 2012, take $n$ numbers arbitrarily, so that there are definitely two numbers whose ratio is within the interval $\left[\frac{2}{3} ; \frac{3}{2}\right]$. Try to find the minimum value of $n$.
| Solve for
$$
\begin{array}{l}
\{1,2,4,7,11,17,26,40,61,92,139, \\
209,314,472,709,1064,1597\}
\end{array}
$$
When any two numbers are taken, their ratio is either greater than $\frac{3}{2}$ or less than $\frac{2}{3}$.
Therefore, $n \geqslant 18$.
Next, we prove: When 18 numbers are taken, there exist two numbers whose... | 18 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Arrange the eight numbers $-7, -5, -3, -2, 2, 4, 6, 13$ as $a, b, c, d, e, f, g, h$, such that
$$
(a+b+c+d)^{2}+(e+f+g+h)^{2}
$$
is minimized. Then this minimum value is $\qquad$ | 4.34.
Let $x=a+b+c+d$. Then
$$
\begin{array}{l}
e+f+g+h=8-x, \\
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
=x^{2}+(8-x)^{2}=2(x-4)^{2}+32 .
\end{array}
$$
From the known eight numbers, the sum of any four numbers $x$ is an integer and cannot be 4, so we get
$$
\begin{array}{l}
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
\geqslant 2+32=34,
\en... | 34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. There are 10 chess players participating in a round-robin tournament (i.e., each pair of players competes in one match). The rules state that a win earns 2 points, a draw earns 1 point for each player, and a loss earns 0 points. After the tournament, it is found that each player's score is unique, and the second-pla... | 7. 16 .
The last five contestants have to play 10 matches, so the total score of the last five contestants is greater than or equal to $2 \times 10=20$.
Thus, the score of the second-place contestant is greater than or equal to
$$
\frac{4}{5} \times 20=16 \text {. }
$$
The score of the first-place contestant is less ... | 16 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $a b c d$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$ . | 8. 22 .
$$
\begin{array}{l}
\text { Let } a b c d=k+(k+1)+\cdots+(k+34) \\
=35(k+17),
\end{array}
$$
where $k$ is a positive integer. Then the prime numbers $a, b, c, d$ must include one 5 and one 7.
Assume $a=5, b=7$. Then $c d=k+17$.
When $k \geqslant 8$,
$$
c+d \geqslant 2 \sqrt{c d}=2 \sqrt{k+17} \geqslant 2 \sqrt... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Given a positive integer $n$ that satisfies the following condition: In any $n$ integers greater than 1 and not exceeding 2009 that are pairwise coprime, at least one is a prime number. Find the minimum value of $n$. ${ }^{[2]}$ | Since $44<\sqrt{2009}<45$, any composite number greater than 1 and not exceeding 2009 must have a prime factor not exceeding 44.
There are 14 prime numbers not exceeding 44, as follows:
$2,3,5,7,11,13,17$,
$19,23,29,31,37,41,43$.
On one hand,
$2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}$,
$19^{2}, 23^{2}, 29^{2}... | 15 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The quadratic trinomial
$$
x^{2}+a x+b(a 、 b \in \mathbf{N})
$$
has real roots, and $a b=2^{2011}$. Then the number of such quadratic trinomials is $\qquad$. | $-1.1341$.
It is known that the numbers $a$ and $b$ are powers of 2 with non-negative integer exponents, i.e., $a=2^{k}, b=2^{2011-k}$. Therefore, we have
$$
\begin{array}{l}
\Delta=a^{2}-4 b \geqslant 0 \\
\Rightarrow 2 k \geqslant 2013-k \Rightarrow k \geqslant \frac{2013}{3}=671 .
\end{array}
$$
But $k \leqslant 20... | 1341 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (15 points) On a plane, $n$ points are called a "standard $n$-point set" if among any three of these points, there are always two points whose distance is no more than 1. To ensure that a circular paper with a radius of 1 can cover at least 25 points of any standard $n$-point set, find the minimum value of $n$.
| First, prove: $n_{\text {min }}>48$.
Draw a line segment $AB$ of length 5 on the plane, and construct two circles with radii of 0.5 centered at $A$ and $B$, respectively. Take 24 points in each circle. Then there are 48 points on the plane that satisfy the problem's condition (any three points must have at least two po... | 49 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Five. (15 points) Given a function $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y, z$ we have
$$
\begin{array}{l}
\frac{1}{2} f(x y)+\frac{1}{2} f(x z)-f(x) f(y z) \geqslant \frac{1}{4} . \\
\text { Find }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)]
\end{array}
$$
where $[a]$ denotes th... | $$
f(1)=\frac{1}{2} \text {. }
$$
Let $y=z=0$. Then
$$
-\frac{1}{2} f(0)+\frac{1}{2} f(0)-f(x) f(0) \geqslant \frac{1}{4} \text {. }
$$
Substituting $f(0)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \leqslant \frac{1}{2} \text {. }
$$
Now let $y=z=1$. Then
$$
\frac{1}{2} f(x)+\frac{1}{2} f(x)-f(x) f(1... | 1011030 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $A$ and $B$ be two sets, and call $(A, B)$ a "pair". When $A \neq B$, consider $(A, B)$ and $(B, A)$ as different pairs. Then the number of different pairs $(A, B)$ that satisfy the condition
$$
A \cup B=\{1,2,3,4\}
$$
is $\qquad$ | 2. 81.
When set $A$ has no elements, i.e., $A=\varnothing$, set $B$ has 4 elements, there is 1 case; when set $A$ contains $k(k=1,2,3,4)$ elements, set $B$ contains the other $4-k$ elements besides these $k$ elements, the elements in set $A$ may or may not be in set $B$, there are $\mathrm{C}_{4}^{k} \times 2^{k}$ cas... | 81 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Given three points $A, B, C$ in a plane satisfying $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=4,|\overrightarrow{C A}|=5$.
Then $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}=$ $\qquad$ | 8. -25 .
Given that $\overrightarrow{A B} \perp \overrightarrow{B C}$. Then
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B} \\
=\overrightarrow{C A} \cdot(\overrightarrow{A B}+\overrightarrow{B C})=-\ove... | -25 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Fill in each small square of a $9 \times 9$ grid with a number, with no more than four different numbers in each row and each column. What is the maximum number of different numbers that can be in this grid? ${ }^{[4]}$ | If there are 29 different numbers in this grid, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is the first row).
The remaining 25 numbers are in rows $2 \sim 9$. Similarly, by the pigeonhole principle, there must be a row with four different numbers (let's assume it is t... | 28 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. Let the sum of the first $n$ terms of the real geometric sequence $\left\{a_{n}\right\}$ be
$S_{n}$. If $S_{10}=10, S_{30}=70$, then $S_{40}=$ $\qquad$ . | 9. 150 .
$$
\begin{array}{l}
\text { Let } b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, \\
b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30} .
\end{array}
$$
Let $q$ be the common ratio of $\left\{a_{n}\right\}$. Then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with common ratio $r=q^{10}$. Therefore,
$$
\begin{array}{l}
70=S... | 150 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{aligned}
M= & |2012 x-1|+|2012 x-2|+\cdots+ \\
& |2012 x-2012|
\end{aligned}
$$
The minimum value of the algebraic expression is . $\qquad$ | 2. 1012036 .
By the geometric meaning of absolute value, we know that when
$$
1006 \leqslant 2012 x \leqslant 1007
$$
$M$ has the minimum value.
$$
\begin{array}{l}
\text { Then } M_{\text {min }}=(-1-2-\cdots-1006)+ \\
(1007+1008+\cdots+2012) \\
=1006 \times 1006=1012036 .
\end{array}
$$ | 1012036 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. If real numbers $m, n, p, q$ satisfy the conditions
$$
\begin{array}{l}
m+n+p+q=22, \\
m p=n q=100,
\end{array}
$$
then the value of $\sqrt{(m+n)(n+p)(p+q)(q+m)}$ is
$\qquad$ | 3. 220 .
From the given, we have
$$
\begin{aligned}
& (m+n)(n+p)(p+q)(q+m) \\
= & {[(m+n)(p+q)][(n+p)(q+m)] } \\
= & (200+m q+n p)(200+m n+p q) \\
= & 200^{2}+100\left(m^{2}+n^{2}+p^{2}+q^{2}+2 m n+\right. \\
& 2 m q+2 n p+2 p q) \\
= & 200^{2}+100\left[(m+n+p+q)^{2}-400\right] \\
= & {[10(m+n+p+q)]^{2} . }
\end{align... | 220 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Select $k$ numbers from 1 to 2012, such that among the selected $k$ numbers, there are definitely three numbers that can form the lengths of the sides of a triangle (the lengths of the three sides of the triangle must be distinct). What is the minimum value of $k$ that satisfies the condition? | Three, the problem is equivalent to:
Select $k-1$ numbers from $1,2, \cdots, 2012$, such that no three of these numbers can form the sides of a triangle with unequal sides. What is the maximum value of $k$ that satisfies this condition?
Now consider the arrays that meet the above conditions.
When $k=4$, the smallest th... | 17 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. A five-digit number $\overline{a b c d e}$ satisfies
$$
ac>d, dd, b>e
$$
(for example, 37201, 45412), if its digits change with the position in a manner similar to the monotonicity of a sine function over one period, then this five-digit number is said to conform to the "sine rule". Therefore, there are $\qquad$ fiv... | 6.2892 .
From the problem, we know that $b$ and $d$ are the maximum and minimum numbers among $a, b, c, d, e$. It is easy to see that $2 \leqslant b-d \leqslant 9$.
Let $b-d=k$. In this case, there are $10-k$ ways to choose $(b, d)$, and $a, c, e$ each have $k-1$ ways to be chosen, i.e., $(a, c, e)$ has $(k-1)^{3}$ g... | 2892 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
320 Suppose there are $2 k+1$ consecutive natural numbers,
where the sum of the squares of the $k$ larger natural numbers is equal to the sum of the squares of the remaining $k+1$ smaller natural numbers (such as $5^{2}=4^{2}+3^{2}$ $\left.(k=1), 14^{2}+13^{2}=12^{2}+11^{2}+10^{2}(k=2)\right)$. For convenience, such an... | Let
\[
\begin{array}{l}
(a+k)^{2}+(a+k-1)^{2}+\cdots+(a+1)^{2} \\
=a^{2}+(a-1)^{2}+\cdots+(a-k)^{2}
\end{array}
\]
be the $k$-th transformation method,
\[
w_{k}=(a+k, a-k)
\]
indicates that the largest number is $a+k$ and the smallest number is $a-k$.
Clearly, $a^{2}=4 \times(1+2+\cdots+k) a$, or $a=4 \times(1+2+\cdo... | 2012 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
B. Let $n$ be an integer, and $1 \leqslant n \leqslant 2012$. If $\left(n^{2}-n+3\right)\left(n^{2}+n+3\right)$ is divisible by 5, then the number of all $n$ is $\qquad$. | B. 1610.
Notice,
$$
\begin{array}{l}
\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \\
=n^{4}+5 n^{2}+9 \\
=(n-1)(n+1)\left(n^{2}+1\right)+5 n^{2}+10 .
\end{array}
$$
When $n$ is divided by 5, the remainder is 1 or 4, then $n-1$ or $n+1$ is divisible by 5, so
$$
51\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {; }... | 1610 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
13. A. Given integers $a, b$ satisfy $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$. | 13. A. Let $a-b=m$ (where $m$ is a prime number), and $ab=n^2$ (where $n$ is a non-negative integer).
When $b \neq 0$,
$$
\begin{array}{l}
\text { From }(a+b)^{2}-4ab=(a-b)^{2} \\
\Rightarrow(2a-m)^{2}-4n^{2}=m^{2} \\
\Rightarrow(2a-m+2n)(2a-m-2n)=m^{2} .
\end{array}
$$
Since $2a-m+2n$ and $2a-m-2n$ are both positive ... | 2017 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, a_{2}=4, a_{3}=9, \\
a_{n}=a_{n-1}+a_{n-2}-a_{n-3}(n=4,5, \cdots) .
\end{array}
$$
Then $a_{2011}=$ | $-、 1.8041$.
From the problem, we have
$$
a_{2}-a_{1}=3, a_{3}-a_{2}=5 \text {, }
$$
and $a_{n}-a_{n-1}=a_{n-2}-a_{n-3}(n \geqslant 4)$.
Thus, $a_{2 n}-a_{2 n-1}=3, a_{2 n+1}-a_{2 n}=5\left(n \in \mathbf{N}_{+}\right)$.
Therefore, $a_{2 n+1}-a_{2 n-1}=8$.
Hence, $a_{2011}=\sum_{k=1}^{1005}\left(a_{2 k+1}-a_{2 k-1}\rig... | 8041 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Given a function $f(n)$ defined on the set of positive integers satisfies the conditions:
(1) $f(m+n)=f(m)+f(n)+m n\left(m, n \in \mathbf{N}_{+}\right)$;
(2) $f(3)=6$.
Then $f(2011)=$ . $\qquad$ | 3.2023066.
In condition (1), let $n=1$ to get
$$
f(m+1)=f(m)+f(1)+m \text {. }
$$
Let $m=n=1$ to get
$$
f(2)=2 f(1)+1 \text {. }
$$
Let $m=2, n=1$, and use condition (2) to get
$$
6=f(3)=f(2)+f(1)+2 \text {. }
$$
From equations (3) and (2), we get
$$
f(1)=1, f(2)=3 \text {. }
$$
Substitute into equation (1) to get... | 2023066 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. 10 students stand in a row, and a red, yellow, or blue hat is to be given to each student. It is required that each color of hat must be present, and the hats of adjacent students must be of different colors. Then the number of ways to distribute the hats that meet the requirements is $\qquad$ kinds. | 8. 1530.
Generalize to the general case.
Let the number of ways to arrange $n$ students according to the given conditions be $a_{n}$. Then
$$
\begin{array}{l}
a_{3}=6, a_{4}=18, a_{n+1}=2 a_{n}+6(n \geqslant 3) . \\
\text { Hence } a_{n+1}+6=2\left(a_{n}+6\right) \\
\Rightarrow a_{n}=\left(a_{3}+6\right) \times 2^{n-3... | 1530 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 In $\triangle A B C$, $A C=B C, \angle A C B=$ $90^{\circ}, D, E$ are two points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$
(2006, Beijing Middle School Mathematics Competition (Grade 8)) | Solve as shown in Figure 1, construct square $C A H B$, and extend $C D$, $C E$ to intersect $A H$, $B H$ at points $G$, $F$ respectively.
Let $D E=x$.
Since $A C / / B F$
$$
\Rightarrow \frac{3+x}{4}=\frac{A C}{B F} \text {, }
$$
$B C / / A G$
$$
\begin{array}{c}
\Rightarrow \frac{4+x}{3}=\frac{B C}{A G} . \\
\text { ... | 36 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 In $\triangle A B C$, it is known that $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=2, C D=3$, then $S_{\triangle A B C}$ $=$ $\qquad$
(2007, Shandong Province Junior High School Mathematics Competition) | Solve as shown in Figure 5, with $AB$ as the axis of symmetry, construct the symmetric figure of $\triangle ADB$ as $\triangle AGB$, and with $AC$ as the axis of symmetry, construct the symmetric figure of $\triangle ADC$ as $\triangle AFC$, and extend $GB$ and $FC$ to intersect at point $E$. Then it is easy to know th... | 15 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Given four distinct positive real numbers $a, b, c, d$ satisfy
$$
\begin{array}{l}
\left(a^{2012}-c^{2012}\right)\left(a^{2012}-d^{2012}\right)=2012, \\
\left(b^{2012}-c^{2012}\right)\left(b^{2012}-d^{2012}\right)=2012 . \\
\text { Then }(a b)^{2012}-(c d)^{2012}=(\quad) .
\end{array}
$$
(A) -2012
(B) -2011
(C) 2012... | - 1. A.
From the problem, we know that \(a^{2012}\) and \(b^{2012}\) are the two distinct real roots of the quadratic equation in \(x\):
\[
\left(x-c^{2012}\right)\left(x-d^{2012}\right)=2012,
\]
which is \(x^{2}-\left(c^{2012}+d^{2012}\right) x+(c d)^{2012}-2012=0\).
Thus, \(a^{2012} b^{2012}=(c d)^{2012}-2012\).
The... | -2012 | Algebra | MCQ | Yes | Yes | cn_contest | false |
5. There are $n$ people registered to participate in four sports competitions: A, B, C, and D. It is stipulated that each person must participate in at least one competition and at most two competitions, but competitions B and C cannot be registered for simultaneously. If in all different registration methods, there mu... | 5. B.
Use the ordered array $\left(a_{\text {甲 }}, b_{\text {乙 }}, c_{\text {丙 }}, d_{\mathrm{J}}\right)$ to represent each person's registration for the four sports events 甲, 乙, 丙, and 丁. If a person participates in a certain event, the corresponding number is recorded as 1 (for example, if participating in event 甲, ... | 172 | Combinatorics | MCQ | Yes | Yes | cn_contest | false |
11. (16 points) For an integer $k$, define the set
$$
S_{k}=\{n \mid 50 k \leqslant n<50(k+1), n \in \mathbf{Z}\} \text {. }
$$
How many of the 600 sets $S_{0}, S_{1}, \cdots, S_{599}$ do not contain any perfect squares? | 11. Notice,
$$
\begin{array}{l}
(x+1)^{2}-x^{2}=2 x+1 \leqslant 50(x \in \mathbf{N}) \\
\Leftrightarrow x \leqslant 24(x \in \mathbf{N}) .
\end{array}
$$
While $(24+1)^{2}=625 \in S_{12}$, thus, the square numbers in $S_{0}, S_{1}$, $\cdots, S_{12}$ do not exceed $25^{2}$, and each set consists of 50 consecutive non-n... | 439 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given $a, b, c \in \mathbf{R}_{+}$, and $abc=4$. Then the minimum value of the algebraic expression $a^{a+b} b^{3b} c^{c+b}$ 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. 64.
It is easy to know that $a^{a+b} b^{3 b} c^{c+b}=a^{a} b^{2 b} c^{c} \cdot 4^{b}=a^{a}(2 b)^{2 b} c^{c}$.
And $a b c=4$ can be transformed into $a \cdot 2 b \cdot c=8$, considering $2 b$ as a whole.
Since the function $f(x)=\ln x$ is increasing on $(0,+\infty)$, for any $a, b \in(0,+\infty)$, we always have $(... | 64 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
9. (16 points) For a given positive integer $M$, define $f_{1}(M)$ as the square of the sum of the digits of $M$. When $n>1$ and $n \in \mathbf{N}$, $f_{n}\left(f_{n-1}(M)\right)$ represents the $r_{n}$-th power of the sum of the digits of $f_{n-1}(M)$, where, when $n$ is odd, $r_{n}=2$; when $n$ is even, $r_{n}=3$. Fi... | Let the positive integer $M=a_{1} a_{2} \cdots a_{m}$, where $a_{i} \in \mathbf{N}, a_{1}>0 (i=1,2, \cdots, m)$. Then
$$
f_{1}(M)=\left(a_{1}+a_{2}+\cdots+a_{m}\right)^{2} \equiv M^{2}(\bmod 9) .
$$
When $M=3^{2010}$, $f_{1}(M) \equiv M^{2} \equiv 0(\bmod 9)$.
By mathematical induction, it is easy to see that for $n \... | 729 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $a$ is a prime number less than 2012, and 2012 $-a$ is coprime with $a$. Then the number of $a$ that satisfies the condition is ( ).
(A) 1003
(B) 1004
(C) 1005
(D) 1006 | 5. B.
If $a$ is even, then $2012-a$ is also even. In this case, $a$ and $2012-a$ have a common divisor of 2, so they are not coprime. Therefore, $a$ cannot be even and must be odd.
Since $2012=503 \times 4$, and 503 is a prime number, we have
$$
a \neq 503 \times 1 \text { and } 503 \times 3 \text {. }
$$
There are 1... | 1004 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
3. If $a-b=2, \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4$, then $a^{5}-b^{5}=$ | 3. 82 .
$$
\begin{aligned}
\text { Given } & \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4 \\
\Rightarrow & a(1-a)^{2}-b(1+b)^{2}=4 a b \\
\Rightarrow & a-2 a^{2}+a^{3}-b-2 b^{2}-b^{3}=4 a b \\
\Rightarrow & (a-b)-2\left(a^{2}+b^{2}\right)+\left(a^{3}-b^{3}\right)=4 a b \\
\Rightarrow & (a-b)-2\left[(a-b)^{2}+2 a b\right]+... | 82 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) Find the integer part of $\left(\frac{1+\sqrt{5}}{2}\right)^{19}$. | Let $a=\frac{1+\sqrt{5}}{2}, b=\frac{1-\sqrt{5}}{2}$. Then $a+b=1, a^{2}=a+1, b^{2}=b+1$.
Thus $a^{2}+b^{2}=(a+1)+(b+1)$ $=(a+b)+2=3$, $a^{3}+b^{3}=\left(a^{2}+a\right)+\left(b^{2}+b\right)$ $=\left(a^{2}+b^{2}\right)+(a+b)=1+3=4$, $a^{4}+b^{4}=\left(a^{3}+a^{2}\right)+\left(b^{3}+b^{2}\right)$ $=\left(a^{3}+b^{3}\righ... | 9349 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Given a positive integer $k$ that satisfies for any positive integer $n$, the smallest prime factor of $n^{2}+n-k$ is no less than 11. Then, $k_{\text {min }}$ $=$ . $\qquad$ | 6.43 .
Notice,
$$
\begin{array}{l}
n^{2}+n \equiv 0(\bmod 2), \\
n^{2}+n \equiv 0,2(\bmod 3),
\end{array}
$$
$$
\begin{array}{l}
n^{2}+n \equiv 0,2,1(\bmod 5), \\
n^{2}+n \equiv 0,2,6,5(\bmod 7),
\end{array}
$$
and the smallest prime factor of $n^{2}+n-k$ is not less than 11, then
$$
\begin{aligned}
k & \equiv 1(\bmo... | 43 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given positive integers $a, b$ satisfy
$$
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \text {. }
$$
Then $|10(a-5)(b-15)|+2=$ | 7.2012.
Notice,
$$
\begin{array}{l}
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \\
\Leftrightarrow 16 a b^{3}=\left(2 b^{2}-a\right)\left(a^{2}+4 a b+4 b^{2}\right) \\
\Leftrightarrow a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right) \\
\Leftrightarrow a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2} .
\end{... | 2012 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) How many nine-digit numbers $\overline{a_{1} a_{2} \cdots a_{9}}$ satisfy $a_{1} \neq 0$, all digits are distinct, and
$$
a_{1}+a_{3}+a_{5}+a_{9}=a_{2}+a_{4}+a_{6}+a_{8} .
$$ | 10. Let the missing digit be $a$, and
$$
\sum_{i=\pi} a_{i}=\sum_{i=m} a_{i}=t .
$$
Then $2 t=\sum_{i=1}^{9} a_{i}=45-a$.
Thus, $a \in\{1,3,5,7,9\}$.
If $a=1$, then $t=22$, we have
$$
\begin{array}{l}
22=2+3+8+9=2+4+7+9 \\
=2+5+6+9=2+5+7+8 \\
=3+4+6+9=3+4+7+8 \\
=3+5+6+8=4+5+6+7 \\
=0+5+8+9=0+6+7+9 ;
\end{array}
$$
I... | 120384 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Let $n$ be a positive integer, $[x]$ be the greatest integer not exceeding the real number $x$, and $\{x\}=x-[x]$.
(1) Find all positive integers $n$ that satisfy
$$
\sum_{k=1}^{2013}\left[\frac{k n}{2013}\right]=2013+n
$$
(2) Find all positive integers $n$ that maximize $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right... | (1) Let $m=2013, d=(m, n), m=d m_{1}, n=d n_{1}$.
For $1 \leqslant k \leqslant m-1$, we have
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=\left[\frac{k n_{1}}{m_{1}}\right]+\left[\frac{(m-k) n_{1}}{m_{1}}\right] \text{. }
$$
When $m_{1} \mid k$,
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\rig... | 1006 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1} \leqslant \frac{a_{n+2}+a_{n}}{2}, a_{1}=1, a_{403}=2011 \text {. }
$$
Then the maximum value of $a_{5}$ is $\qquad$ | - 1.21.
Obviously, the sequence of points $\left(n, a_{n}\right)$ is arranged in a convex function. When the sequence of points is distributed on the line determined by the points $(1,1)$ and $(403,2011)$, $a_{5}$ takes the maximum value 21. | 21 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Six college graduates apply to three employers. If each employer hires at least one of them, the number of different hiring scenarios is $\qquad$ .
| 7. 2100 .
The number of ways for three people to be hired is $\mathrm{A}_{6}^{3}=120$; the number of ways for four people to be hired is $\mathrm{C}_{6}^{4} \mathrm{C}_{4}^{2} \mathrm{~A}_{3}^{3}=15 \times 6 \times 6=540$ (ways); the number of ways for five people to be hired is $C_{6}^{5}\left(C_{5}^{3} A_{3}^{3}+\fr... | 2100 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
15. Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than
1. Then
$$
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=
$$
$\qquad$ | 15.2011.
When the common ratio is 1, $\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011$.
When the common ratio is $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q} ... | 2011 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
16. Figure 3 is a defective $3 \times 3$ magic square, in which the sum of the three numbers in each row, each column, and each diagonal is equal. Then the value of $x$ is . $\qquad$ | $$
\begin{array}{l}
\text { Magic Square } \\
4017+2012 \\
=x-2003+x \\
\Rightarrow x=4016 \text {. } \\
\end{array}
$$ | 4016 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
$$
\begin{array}{l}
1 \times 2 - 3 \times 4 + 5 \times 6 - 7 \times 8 + \cdots + \\
2009 \times 2010 - 2011 \times 2012 \\
= \quad .
\end{array}
$$ | $$
\text { II, 1. }-2025078 \text {. }
$$
Notice that,
$$
\begin{array}{l}
(n+2)(n+3)-n(n+1)=4 n+6 \\
=n+(n+1)+(n+2)+(n+3) .
\end{array}
$$
Therefore, the original expression is
$$
=-(1+2+\cdots+2012)=-2025078 .
$$ | -2025078 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (15 points) It is known that a positive integer $n$ can be expressed as the sum of 2011 identical natural numbers, and also as the sum of 2012 identical natural numbers. Determine the minimum value of $n$.
| Let $a_{1}, a_{2}, \cdots, a_{2011}$ be 2011 natural numbers with the same digit sum, and $b_{1}, b_{2}, \cdots, b_{2012}$ be 2012 natural numbers with the same digit sum, and they satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=n \\
=b_{1}+b_{2}+\cdots+b_{2012} .
\end{array}
$$
Since each of the numbers $a_{1... | 10055 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. The number of positive integers not exceeding 2012 and having exactly three positive divisors is $\qquad$ . | 10. 14.
Let $1 \leqslant a \leqslant 2012$ and $a$ has only three positive divisors. Then $a$ is the square of a prime number, i.e., $a=p^{2} \leqslant 2012$.
Thus, $2 \leqslant p \leqslant 43$.
$$
\begin{array}{c}
\text { Hence } p=2,3,5,7,11,13,17,19,23, \\
29,31,37,41,43 .
\end{array}
$$ | 14 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. For a certain game activity,
the rewards are divided into first, second, and third prizes (all participants in the game activity will receive a prize), and the corresponding winning probabilities form a geometric sequence with the first term $a$ and a common ratio of 2. The corresponding prize money forms an arithm... | 7.500 .
From the problem, we know the probabilities of winning the first, second, and third prizes are
$$
P_{1}=a, P_{2}=2 a, P_{3}=4 a .
$$
From $P_{1}+P_{2}+P_{3}=1$, we get $a=\frac{1}{7}$.
Thus, $P_{1}=\frac{1}{7}, P_{2}=\frac{2}{7}, P_{3}=\frac{4}{7}$.
The prizes for winning the first, second, and third prizes a... | 500 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
10. Transporting utility poles from a construction site by the roadside along a straight road in the same direction to plant them 500 m away on the roadside, plant one at the 500 m mark, and then plant one every 50 m along the roadside. Knowing that the transport vehicle can carry a maximum of 3 poles at a time, to com... | 10. 14000 .
Assuming the completion of transporting and planting 21 utility poles, 3 poles each time, let the round trip distance for the $k(k=1,2, \cdots, 7)$-th time be $a_{k}$. Then
$$
a_{1}=2 \times 600=1200,
$$
and $a_{k+1}=a_{k}+2 \times 150(k=1,2, \cdots, 6)$.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic... | 14000 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Four. (30 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{2}, a_{n}=2 a_{n} a_{n+1}+3 a_{n+1}\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) If the sequence $\left\{b_{n}\right\}$ satisfies $b_{n}=1+\frac{1}{a_{n... | (1) From
$$
\begin{array}{l}
a_{n}=2 a_{n} a_{n+1}+3 a_{n+1} \\
\Rightarrow a_{n+1}=\frac{a_{n}}{2 a_{n}+3} \Rightarrow a_{n}>0 . \\
\text { Hence } \frac{1}{a_{n+1}}-\frac{3}{a_{n}}=2 \Rightarrow \frac{1}{a_{n+1}}+1=3\left(\frac{1}{a_{n}}+1\right) .
\end{array}
$$
Therefore, $\left\{\frac{1}{a_{n}}+1\right\}$ is a ge... | 13 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Given that among the vertices of a regular 2012-gon, there exist $k$ vertices such that the convex $k$-gon formed by these $k$ vertices has no two parallel sides. Find the maximum value of $k$.
| 2. The maximum value of $k$ is 1509.
Let $A_{1}, A_{2}, \cdots, A_{2012}$ be the set of vertices of a regular polygon.
Consider the set of four vertices
$$
\begin{array}{l}
\left(A_{1}, A_{2}, A_{1000}, A_{1008}\right),\left(A_{3}, A_{4}, A_{100}, A_{1010}\right), \\
\cdots,\left(A_{1005}, A_{1000}, A_{2011}, A_{2012}... | 1509 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Let $n \geqslant 2$,
$$
A_{n}=\left\{x \in \mathbf{R} \left\lvert\, x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right]\right.\right\} .
$$ | Prove: $A=\underset{n \geqslant 2}{\cup} A_{n}$ is a finite set, and find the maximum and minimum elements of $A$.
(2010, Romanian Mathematical Olympiad (Final)) [Analysis] Start with simple cases in mathematical experiments.
Elements in $A_{2}$ satisfy $x=\left[\frac{x}{2}\right]$, and $x \in \mathbf{Z}$. Hence, $\fra... | 23 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. The largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is ( ).
(A) 209520
(B)209 519
(C)209 518
(D) 209517 | 4. B.
Let $3 \sqrt{2}+2 \sqrt{3}=a, 3 \sqrt{2}-2 \sqrt{3}=b$.
Then $a+b=6 \sqrt{2}, ab=6$
$$
\begin{array}{l}
\Rightarrow a^{2}+b^{2}=(a+b)^{2}-2ab=60 \\
\Rightarrow a^{6}+b^{6}=\left(a^{2}+b^{2}\right)^{3}-3(ab)^{2}\left(a^{2}+b^{2}\right) \\
\quad=209520 .
\end{array}
$$
Since $0<b<1$, we have
$$
209519<a^{6}<20952... | 209519 | Algebra | MCQ | Yes | Yes | cn_contest | false |
4. As shown in Figure 5, an equilateral $\triangle ABC$ with side length 26 is inscribed in a circle, and chord $DE \parallel BC$. The extension of $DE$ intersects the extensions of $AB$ and $AC$ at points $F$ and $G$, respectively. If the lengths of segments $AF$ and $DF$ are both positive integers, then the chord $DE... | 4. 16 .
Let $A F=x, D F=y$. Then
$$
B F=x-26, D E=x-2 y, E F=x-y \text {. }
$$
By the secant theorem, we have
$$
\begin{array}{l}
A F \cdot B F=D F \cdot E F \\
\Rightarrow x(x-26)=y(x-y) \\
\Rightarrow x^{2}-(26+y) x+y^{2}=0 .
\end{array}
$$
Since $A F$ and $D F$ are both positive integers, we have
$$
\Delta=(26+y)... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Write out a sequence of consecutive positive integers starting from 1, then erase one of the numbers, so that the average of the remaining numbers is $43 \frac{14}{17}$. What is the number that was erased? | Three, suppose we have written $n$ consecutive positive integers 1, 2, ..., $n$. If the number $k$ is erased, then
$$
\begin{array}{l}
\frac{(1+2+\cdots+n)-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2}+\frac{n-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2} \leqslant 43 \frac{14}{17} \leqslant \frac{n}{2}+1 \... | 16 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Given that $f(x)$ is a function defined on $\mathbf{R}$. If $f(0)=0$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+4)-f(x) \leqslant x^{2}, \\
f(x+16)-f(x) \geqslant 4 x^{2}+48 x+224,
\end{array}
$$
then $f(64)=$ $\qquad$ | 5. 19840.
Notice that,
$$
\begin{array}{l}
f(x+4)-f(x) \\
=(f(x+16)-f(x))-(f(x+16)- \\
f(x+12))-(f(x+12)-f(x+8))- \\
(f(x+8)-f(x+4)) \\
\geqslant\left(4 x^{2}+48 x+224\right)-(x+12)^{2}- \\
(x+8)^{2}-(x+4)^{2} \\
= x^{2} .
\end{array}
$$
Since $f(x+4)-f(x) \leqslant x^{2}$, we have,
$$
\begin{array}{l}
f(x+4)-f(x)=x... | 19840 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4: From the numbers $1, 2, \cdots, 2012$, select a set of numbers such that the sum of any two numbers cannot be divisible by their difference. How many such numbers can be selected at most?
(2012, Joint Autonomous Admission Examination of Peking University and Other Universities) | Solve: Divide $1,2, \cdots, 2012$ into
$$
\begin{array}{l}
(1,2,3),(4,5,6), \cdots, \\
(2008,2009,2010),(2011,2012)
\end{array}
$$
these 671 groups.
If at least 672 numbers are taken, then by the pigeonhole principle, there must be two numbers in the same group, let's say $a$ and $b$ ($a>b$).
Thus, $a-b=1$ or 2.
When ... | 671 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Sets $S_{1}, S_{2}, \cdots, S_{n}$ are pairwise distinct and satisfy the following conditions:
(1) $\left|S_{i} \cup S_{j}\right| \leqslant 2004(1 \leqslant i, j \leqslant n, i, j \in \mathbf{N}_{+})$;
(2) $S_{i} \cup S_{j} \cup S_{k}=\{1,2, \cdots, 2008\}(1 \leqslant i < j < k \leqslant n, i, j, k \in \mathb... | Let the complement of $S_{i}$ in $Y=\{1,2, \cdots, 2008\}$ be $A_{i}(1 \leqslant i \leqslant n)$. Then
$\left|A_{i} \cap A_{j}\right| \geqslant 4, A_{i} \cap A_{j} \cap A_{k}=\varnothing$.
Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$.
Consider the bipartite graph $X+Y$, where a vertex $A_{i} \in X$ is adjacent to... | 32 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 There are 10 sets of test papers, each set containing 4 questions, and at most one question is the same between any two sets. Among these test papers, what is the minimum number of different questions?
(2005, Taiwan Mathematical Olympiad) | Consider each test paper as a vertex, these vertices form a set $X$; consider each question as a vertex, these vertices form a set $Y$.
If a test paper $x \in X$ contains a question $y \in Y$, connect a line between the corresponding vertices $x$ and $y$. This results in a bipartite graph $X+Y$.
Let $Y$ have $n$ vert... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given any 2012 points in the plane. Prove: the distances between each pair of them take at least 32 different values. | Prove that with these 2012 points as vertices, connecting edges between points of the same distance, we obtain a graph $G$.
Since there is at most one point that is equidistant from three given points, the graph $G$ does not contain $K_{2,3}$.
By Theorem 2, the number of edges in graph $G$ is
$$
m \leqslant \frac{2012... | 32 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Example 8 Let $X$ be a set with 56 elements. Find the smallest positive integer $n$, such that for any 15 subsets of $X$, if the union of any seven of these subsets has at least $n$ elements, then there must exist three of these 15 subsets whose intersection is non-empty. ${ }^{[2]}$
(2006, China Mathematical Olympiad) | First, we prove by contradiction that \( n = 41 \) satisfies the requirement.
Assume there exist 15 subsets \( A_{1}, A_{2}, \cdots, A_{15} \), such that the union of any seven subsets has at least 41 elements, but the intersection of any three subsets is empty.
At this point, let \( Y = \{A_{1}, A_{2}, \cdots, A_{15}... | 41 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given the set of integers
$M=\left\{m \mid x^{2}+m x-36=0\right.$ has integer solutions $\}$, set $A$ satisfies the conditions:
(1) $\varnothing \subset A \subseteq M$;
(2) If $a \in A$, then $-a \in A$.
The number of all such sets $A$ is ( ).
(A) 15
(B) 16
(C) 31
(D) $32^{[1]}$
(2010, National High School Ma... | Let $\alpha, \beta$ be the integer roots of the equation
$$
x^{2}+m x-36=0
$$
Assume $|\alpha| \geqslant|\beta|$. Then
$$
\begin{array}{l}
\alpha \beta=-36 \\
\Rightarrow(|\alpha|,|\beta|) \\
=(1,36)(2,18),(3,12),(4,9),(6,6) \\
\Rightarrow m= \pm 35, \pm 16, \pm 9, \pm 5,0 \\
\Rightarrow M=\{0\} \cup\{-5,5\} \cup\{-9,... | 31 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 6 Let $M \subseteq\{1,2, \cdots, 2011\}$ satisfy: in any three elements of $M$, there can always be found two elements $a, b$ such that $a \mid b$ or $b \mid a$. Find the maximum value of $|M|$. ${ }^{[2]}$
(2011, China Western Mathematical Olympiad) | Solve for when
$$
M=\left\{2^{k}, 3 \times 2^{l} \mid k=0,1, \cdots, 10 ; l=0,1, \cdots, 9\right\}
$$
the condition is satisfied, at this time, $|M|=21$.
Assume $|M| \geqslant 22$, let the elements of $M$ be
$$
a_{1}2011$, a contradiction.
In summary, $|M|_{\max }=21$. | 21 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Let the set
$$
A=\left\{y \left\lvert\, y=\sin \frac{k \pi}{4}(k=0, \pm 1, \pm 2, \cdots)\right.\right\} \text {. }
$$
Then the number of proper subsets of set $A$ is $\qquad$ (2009, Tongji University Independent Admission Examination) | Hint: Calculate $A=\left\{0, \pm 1, \pm \frac{\sqrt{2}}{2}\right\}$. The number of its proper subsets is $2^{5}-1=31$. | 31 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. On a square table composed of unit squares of size $2011 \times 2011$, a finite number of napkins are placed, each covering a $52 \times 52$ square. In each unit square, write the number of napkins covering it, and let the maximum number of unit squares with the same number be $k$. For all possible configurations of... | 7. The maximum value of $k$ is
$2011^{2}-\left[\left(52^{2}-35^{2}\right) \times 39-17^{2}\right]$
$=4044121-57392=3986729$.
Let $m=39$. Then $2011=52 m-17$.
Below is an example where there are 3986729 unit squares with the same number written in them.
Let the column numbers from left to right, and the row numbers from... | 3986729 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. (50 points) Find the smallest prime $p$ that satisfies $(p, N)=1$, where $N$ is the number of all $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ that meet the following conditions:
(1) $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ is a permutation of $0,1, \cdots, 2012$;
(2) For any positive divisor $m$ of 2013 and a... | 3. From (2), we know that for any $i, j (0 \leqslant i < j \leqslant 2012)$ and $m=3^{\alpha} \times 11^{\beta} \times 61^{\gamma} (\alpha, \beta, \gamma \in \{0,1\})$, we have
$$
a_{i} \equiv a_{j}(\bmod m) \Leftrightarrow i \equiv j(\bmod m).
$$
Let
$$
\begin{array}{l}
A_{i}=\{x \in \mathrm{N} \mid 0 \leqslant x \le... | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
8. How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? | 8 . Note Quality.
First, consider whether the combination number $\mathrm{C}_{2012}^{k}$ is a multiple of $p=503$.
If $p \nmid k$, then
$\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1}$
$\Rightarrow p\left|k \mathrm{C}_{4 p}^{k} \Rightarrow p\right| \math... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given positive integers $m, n$ can be written as
$$
a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}
$$
where $a_{i} (i=0,1,2,3)$ are positive integers from 1 to 7, and
$$
m+n=2012(m>n) .
$$
Then the number of pairs $(m, n)$ that satisfy the condition is $(\quad)$.
(A) 606
(B) 608
(C) 610
(D) 612 | 4. A.
There are $7^{4}=2401$ positive integers of the given form, the largest of which is
$$
7 \times 7^{3}+7 \times 7^{2}+7 \times 7+7=2800,
$$
and the smallest is
$$
1 \times 7^{3}+1 \times 7^{2}+1 \times 7+1=400 .
$$
Since $m+n=2012(m>n)$, we have
$$
1007 \leqslant m \leqslant 2012-400=1612 \text {. }
$$
Therefo... | 606 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
5. Divide the natural numbers from 1 to 30 into two groups, such that the product of all numbers in the first group $A$ is divisible by the product of all numbers in the second group $B$. Then the minimum value of $\frac{A}{B}$ is ( ).
(A) 1077205
(B) 1077207
(C) 1077209
(D) 1077211 | 5. A.
$$
\begin{array}{l}
\text { Given } A B=30 \times 29 \times \cdots \times 1 . \\
=2^{26} \times 3^{14} \times 5^{7} \times 7^{4} \times 11^{2} \times 13^{2} \times \\
17 \times 19 \times 23 \times 29 .
\end{array}
$$
Let $C=2^{13} \times 3^{7} \times 5^{4} \times 7^{2} \times 11 \times 13 \times 17 \times 19 \ti... | 1077205 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 8 Given that $P(x)$ is a polynomial with integer coefficients, satisfying $P(17)=10, P(24)=17$. If the equation $P(n)=n+3$ has two distinct integer solutions $n_{1}, n_{2}$, find the value of $n_{1} n_{2}$. ${ }^{[7]}$
(2005, American Invitational Mathematics Examination) | 【Analysis】From the conditions of the problem, we cannot determine the zeros of the integer-coefficient polynomial $P(x)$. Let's construct another polynomial function $T(x)$ such that 17 and 24 are zeros of $T(x)$, and solve the problem using the zero-product property.
Solution Let $S(x)=P(x)-x-3$. Then
$S(17)=-10, S(24... | 418 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let positive integers $k_{1} \geqslant k_{2} \geqslant \cdots \geqslant k_{n}\left(n \in \mathbf{N}_{+}\right)$, and $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=2012$.
Then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is $\qquad$ | 2. 49 .
Notice that,
$$
2012=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{4}+2^{3}+2^{2} \text {. }
$$
Also, $2^{k+1}=2^{k}+2^{k}$, and $k+1 \leqslant 2 k(k \geqslant 1$ when $)$, then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is
$$
10+9+8+7+6+4+3+2=49 \text {. }
$$ | 49 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Group all positive integers that are coprime with 2012 in ascending order, with the $n$-th group containing $2n-1$ numbers:
$$
\{1\},\{3,5,7\},\{9,11,13,15,17\}, \cdots \text {. }
$$
Then 2013 is in the $\qquad$ group. | 4.32.
Notice that, $2012=2^{2} \times 503$, where $2$ and $503$ are both prime numbers.
Among the positive integers not greater than 2012, there are 1006 multiples of 2, 4 multiples of 503, and 2 multiples of $2 \times 503$. Therefore, the numbers that are not coprime with 2012 are
$$
1006+4-2=1008 \text { (numbers).... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $n=\sum_{a_{1}=0}^{2} \sum_{a_{2}=0}^{a_{1}} \cdots \sum_{a_{2} 012=0}^{a_{2} 011}\left(\prod_{i=1}^{2012} a_{i}\right)$. Then the remainder when $n$ is divided by 1000 is . $\qquad$ | 6. 191 .
It is evident that from $a_{1}$ to $a_{2012}$ forms a non-increasing sequence, and the maximum element does not exceed 2. Therefore, their product is a power of 2 or 0. Since each power of 2 can only be represented in one way (the sequence being non-increasing), we have
$$
\begin{array}{l}
n=1+2+4+\cdots+2^{2... | 191 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $X$ be the set of irreducible proper fractions with a denominator of 800, and $Y$ be the set of irreducible proper fractions with a denominator of 900, and let $A=\{x+y \mid x \in X, y \in Y\}$. Find the smallest denominator of the irreducible fractions in $A$. | 【Analysis】This problem is adapted from the 35th Russian Mathematical Olympiad question ${ }^{[1]}$.
Let $x=\frac{a}{800} \in X, y=\frac{b}{900} \in Y$, where,
$$
\begin{array}{l}
1 \leqslant a \leqslant 799, (a, 800)=1, \\
1 \leqslant b \leqslant 899, (b, 900)=1 .
\end{array}
$$
Then $x+y=\frac{9 a+8 b}{7200}$.
Since ... | 288 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. ${ }^{[3]}$
(2011, Xin Zhi Cup Shanghai Junior High School Mathematics Competition) | 【Analysis】According to the problem, we set
$$
\begin{array}{l}
a b c d=k+(k+1)+\cdots+(k+34)\left(k \in \mathbf{N}_{+}\right) \\
\Rightarrow \frac{(2 k+34) \times 35}{2}=a b c d \\
\Rightarrow(k+17) \times 5 \times 7=a b c d .
\end{array}
$$
By symmetry, without loss of generality, let $c=5, d=7, a \leqslant b$.
We on... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, on a plane there are $n(n \geqslant 4)$ lines. For lines $a$ and $b$, among the remaining $n-2$ lines, if at least two lines intersect with both lines $a$ and $b$, then lines $a$ and $b$ are called a "congruent line pair"; otherwise, they are called a "separated line pair". If the number of congruent line pairs a... | (1) Among these $n$ lines, if there exist four lines that are pairwise non-parallel, then any two lines are coincident line pairs. However, $\mathrm{C}_{n}^{2}=2012$ has no integer solution, so there does not exist an $n$ that satisfies the condition.
(2) If the $n$ lines have only three different directions, let the n... | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given positive integers $a, b$ satisfy that $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$.
保留源文本的换行和格式，直接输出翻译结果。 | Given the problem, let's set
$a-b=p(p$ is a prime number $), ab=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $a(a-p)=k^{2} \Rightarrow a^{2}-k^{2}=ap$
$$
\Rightarrow(a+k)(a-k)=ap \text {. }
$$
Since $a+k, a, p$ are all positive integers, we have
$$
a-k>0 \text {. }
$$
Given that $p$ is a prime number, from equation ... | 2025 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. If a non-negative integer $m$ and the sum of its digits are both multiples of 6, then $m$ is called a "Lucky Six Number". Find the number of Lucky Six Numbers among the non-negative integers less than 2012. | 5. Solution 1 It is easy to know that a non-negative integer is a hexagonal number if and only if its last digit is even and the sum of its digits is a multiple of 6.
For convenience, let
$$
M=\{0,1, \cdots, 2011\}
$$
write each number in $M$ as a four-digit number $\overline{a b c d}$ (when it is less than four digit... | 168 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Find the smallest positive integer $n$ such that
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}} \\
<\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$ | 6. From the known, we must have $n \geqslant 2$ 013. At this time,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}4023, \\
\sqrt[3]{\frac{n-2013}{2011}} \geqslant \sqrt[3]{\frac{n-2011}{2013}} \\
\Leftrightarrow 2013(n-2013) \geqslant 2011(n-2011) \\
\Leftrightarrow n \geqslant 4024 .
\end{array}
$$
From equations (1) ... | 4024 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a, b$ are both integers, the equation
$$
a x^{2}+b x-2008=0
$$
has two distinct roots that are prime numbers, then $3 a+b=$ $\qquad$ (2008, Taiyuan Junior High School Mathematics Competition) | Let the two prime roots of the equation be \(x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)\). From the problem, we have
\[
x_{1} x_{2}=\frac{-2008}{a} \Rightarrow a x_{1} x_{2}=-2008 \text{. }
\]
It is easy to see that, \(2008=2^{3} \times 251\) (251 is a prime number).
Thus, \(x_{1}=2, x_{2}=251\).
Therefore, \(3 a+b=1000\). | 1000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the real-coefficient equation $a x^{3}-x^{2}+b x-1=0$ has three positive real roots. Then
$$
P=\frac{5 a^{2}-6 a b+3}{a^{3}(b-a)}
$$
the minimum value of $P$ is | 7. 108.
Let the three positive real roots of $a x^{3}-x^{2}+b x-1=0$ be $v_{1}, v_{2}, v_{3}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
v_{1}+v_{2}+v_{3}=\frac{1}{a}, \\
v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}=\frac{b}{a}, \\
v_{1} v_{2} v_{3}=\frac{1}{a} .
\end{array}
$$
From (1) and (2), we get $a>0, b>0$.
From... | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 3, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value. | Solve as shown in Figure 4, construct the symmetric point $B'$ of point $B$ with respect to $AC$, and connect $AB'$. Construct $B'N \perp AB$, intersecting $AC$ at point $M$, and connect $BM$. Then $BM=B'M$.
Therefore, $BM+MN=B'M+MN$.
By the shortest distance of a perpendicular segment, we know that $B'M+MN$ is minimiz... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
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