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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 conclusion is obviously true.
Assume the proposition holds for all values less than $n$.
For a convex $n$-sided polygon, let the last diagonal drawn be $A_{1} A_{k}$, which intersects at most one of the previously drawn diagonals (if it exists, let it be $l$) at an interior point. All the diagonals drawn, except for $A_{1} A_{k}$ and $l$, lie entirely within the $k$-sided polygon $A_{1} A_{2} \cdots A_{k}$ or the $(n+2-k)$-sided polygon $A_{k} A_{k+1} \cdots A_{n} A_{1}$.
By the induction hypothesis, there are at most
$$
(2k-6) + [2(n+2-k)-6] = 2n-8
$$
diagonals; adding $A_{1} A_{k}$ and $l$ results in at most $2n-6$ diagonals.
Therefore, for a convex $n$-sided polygon, at most $2n-6$ diagonals can be drawn.
Thus, Betja can draw at most 4016 diagonals.
|
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, $f(60)=3 \times 60-4=176$.
|
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 number of $n$. When $|M|$ is minimized, find the minimum value of the sum of all elements in set $M$.
|
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, $|M| \geqslant 7$.
Thus, the sum of all elements in set $M$ is no less than
$$
1+2+\cdots+7=28 \text {. }
$$
When $M=\{1,2, \cdots, 7\}$, for any positive integer $n$, let it be a $k\left(k \in \mathbf{N}_{+}\right)$-digit number. Then $10^{k} i+n$ $(i=1,2, \cdots, 7)$ have distinct remainders when divided by 7.
Otherwise, there exist positive integers $i, j(1 \leqslant i<j \leqslant 7)$, such that
$$
\begin{array}{l}
7 \mid\left[\left(10^{k} j+n\right)-\left(10^{k} i+n\right)\right] \\
\Rightarrow 7\left|10^{k}(j-i) \Rightarrow 7\right|(j-i),
\end{array}
$$
a contradiction.
Therefore, there must exist a positive integer $i(1 \leqslant i \leqslant 7)$, such that $71\left(10^{k} i+n\right)$, i.e., $i$ is a magic number of $n$.
Thus, when $|M|$ is minimal, the minimum value of the sum of all elements in set $M$ is 28.
|
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
$$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\right. \\
\left.2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\frac{1-\cos B}{2}+\right. \\
\left.\frac{1-\cos C}{2}+\frac{\cos A+\cos B+\cos C-1}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{A}{2} \cdot \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2} \\
= R^{2}(\sin A+\sin B+\sin C)^{2} \\
= \frac{1}{4}(a+b+c)^{2}=64 .
\end{array}
$$
|
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 \\
Black & 20 & Black & 29 & Black & 25 & Black & 10 \\
20 & Black & 49 & Black & 54 & Black & 35 & Black \\
Black & 69 & Black & 103 & Black & 89 & Black & 35
\end{tabular}
Therefore, there are $69+103+89+35=296$ (ways).
|
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 triangles that can be formed with these 12 points as vertices.
|
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, then there is and only one point that is the starting point of the two sides vectors of this triangle. Therefore, the number of non-zero triangles with $P_{i}$ as one of the vertices and as the starting point of two sides is $\mathrm{C}_{x_{i}}^{2}$. Thus, the total number of non-zero triangles with these points as vertices is $\sum_{i=1}^{12} c_{x_{i}}^{2}$.
Therefore, the number of zero triangles is $\mathrm{C}_{12}^{3}-\sum_{i=1}^{12} \mathrm{C}_{\mathrm{x}_{i}}^{2}$.
First, find the minimum value of $\sum_{i=1}^{12} C_{x_{i}}^{2}$.
Since $\sum_{i=1}^{12} x_{i}=\mathrm{C}_{12}^{2}=66$, we have
$$
\begin{array}{l}
\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}=\sum_{i=1}^{12} \frac{x_{i}\left(x_{i}-1\right)}{2} \\
=\frac{1}{2} \sum_{i=1}^{12}\left(x_{i}^{2}-x_{i}\right)=\frac{1}{2} \sum_{i=1}^{12} x_{i}^{2}-33 .
\end{array}
$$
And the non-negative integer $x_{i}$ does not exceed 11, so $\sum_{i=1}^{12} x_{i}^{2}$ has a minimum value.
By the adjustment method, when
$$
\left|x_{i}-x_{j}\right|=0,1(1 \leqslant i<j \leqslant 12),
$$
that is, when
$$
\begin{array}{l}
\left\{x_{1}, x_{2}, \cdots, x_{12}\right\} \\
=\{5,5,5,5,5,5,6,6,6,6,6,6\}
\end{array}
$$
$\sum_{i=1}^{12} x_{i}^{2}$ takes the minimum value 366.
Thus, the minimum value of $\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$ is
$$
\frac{1}{2} \times 366-33=150 \text {. }
$$
Therefore, the maximum number of zero triangles with these 12 points as vertices is 70.
|
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, $f(7)=f(f(4))=12$,
$f(8)=f(f(5))=15$,
$f(9)=f(f(6))=18$,
$f(12)=f(f(7))=21$,
$f(10)=19, f(11)=20$.
We will prove by mathematical induction:
$$
f(n)=\left\{\begin{array}{ll}
n+3^{m}, & 3^{m} \leqslant n<2 \times 3^{m} ; \\
3 n-3^{m+1}, & 2 \times 3^{m} \leqslant n<3^{m+1}
\end{array}\left(m \in N_{+}\right) .\right.
$$
When $m=0,1$, the conclusion has been verified.
Assume the conclusion holds for $m=k-1$.
When $2 \times 3^{k-1} \leqslant n<3^{k}$,
$f(n)=3 n-3^{k} \Rightarrow f\left(3 n-3^{k}\right)=3 n$.
Let $3 n-3^{k}=p \in\left[3^{k}, 2 \times 3^{k}\right)$. Then
$$
f(p)=p+3^{k} \text {. }
$$
By the monotonicity of $f$, when $3^{k} \leqslant n<2 \times 3^{k}$, we always have $f(n)=n+3^{k}$.
Then $f\left(n+3^{k}\right)=f(f(n))=3 n$.
Let $n+3^{k}=q \in\left[2 \times 3^{k}, 3^{k+1}\right)$. Then
$$
f(q)=3 q-3^{k+1} \text {. }
$$
Therefore, the conclusion holds for $n=k$.
By mathematical induction, the conclusion holds for $n \in \mathbf{N}_{+}$.
Since $2 \times 3^{6}<2011<3^{7}$, we have
$$
f(2011)=3 \times 2011-3^{7}=3846 .
$$
|
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
|
$$
\begin{array}{l}
\text { From the problem, we know that } x^{4}+a x^{2}+b x+c \text { can definitely be divided by } x^{2}-3 x-1. \\
=\left(x^{2}-3 x-1\right)\left(x^{2}+3 x+a+10\right)+ \\
{[(3 a+b+33) x+(a+c+10)], } \\
\text { then }\left\{\begin{array}{l}
3 a+b+33=0, \\
a+c+10=0
\end{array}\right. \\
\Rightarrow a+b-2 c=-13 .
\end{array}
$$
Therefore, the answer is A.
|
-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 $a, b, c \in \mathbf{Z}, 1 \leqslant a<b<c \leqslant 2012$, and
$$
x=a+b, y=a+c, z=b+c.
$$
Thus, the condition of the problem is equivalent to the fact that for any $k$-element subset $A$, there exist $x, y, z \in A$ satisfying property (1).
If $A=\{1,2,3,5,7, \cdots, 2011\}$, then $|A|=$ 1007, and the set $A$ does not contain three elements satisfying property (1). Therefore, $k \geqslant 1008$.
Next, we prove that any 1008-element subset contains three elements satisfying property (1).
We will prove a more general conclusion:
For any integer $n(n \geqslant 4)$, any $(n+2)$-element subset of the set $\{1,2, \cdots, 2 n\}$ contains three elements satisfying property (1).
We use induction on $n$.
When $n=4$, let $A$ be a six-element subset of $\{1,2, \cdots, 8\}$. Then $A \cap\{3,4, \cdots, 8\}$ has at least 4 elements.
If $A \cap\{3,4, \cdots, 8\}$ contains three even numbers, then $4, 6, 8 \in A$ and they satisfy property (1);
If $A \cap\{3,4, \cdots, 8\}$ contains exactly two even numbers, then it also contains at least two odd numbers. Taking these two odd numbers, at least two even numbers from $4, 6, 8$ can form a three-element set satisfying property (1), since there are at least two even numbers, there exist three numbers satisfying property (1);
If $A \cap\{3,4, \cdots, 8\}$ contains exactly one even number, then it contains all three odd numbers, and this even number with 5, 7 forms a three-element set satisfying property (1).
Therefore, the conclusion holds when $n=4$.
Assume the conclusion holds for $n(n \geqslant 4)$, consider the case for $n+1$.
Let $A$ be an $(n+3)$-element subset of $\{1,2, \cdots, 2 n+2\}$. If $|A \cap\{1,2, \cdots, 2 n\}| \geqslant n+2$, then by the induction hypothesis, the conclusion holds. Thus, we only need to consider the case
$$
|A \cap\{1,2, \cdots, 2 n\}| = n+1,
$$
and $2 n+1, 2 n+2 \in A$.
In this case, if $\{1,2, \cdots, 2 n\}$ contains an odd number $x$ greater than 1 in the set $A$, then $x, 2 n+1, 2 n+2$ form a three-element set satisfying property (1);
If $\{1,2, \cdots, 2 n\}$ contains no odd numbers greater than 1 in the set $A$, then
$$
A \subset\{1,2,4,6, \cdots, 2 n, 2 n+1,2 n+2\},
$$
and the latter has exactly $n+3$ elements, hence
$$
A=\{1,2,4,6, \cdots, 2 n, 2 n+1,2 n+2\},
$$
in this case, $4, 6, 8 \in A$ satisfy property (1).
In conclusion, the smallest $k$ is 1008.
|
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) $\times(a-1)$ and simplifying, we get
$$
(a-b)(a b-a-b-2)\left(x_{0}-1\right)=0 \text {. }
$$
Since $a \neq b$, we have
$x_{0}=1$ or $a b-a-b-2=0$.
If $x_{0}=1$, substituting into equation (3) yields $a=1$, which is a contradiction.
Therefore, $x_{0} \neq 1$.
Hence, $a b-a-b-2=0$
$\Rightarrow(a-1)(b-1)=3$
$\Rightarrow a=2, b=4$ or $a=4, b=2$.
Substituting into the required algebraic expression, we get
$$
\frac{a^{b}+b^{a}}{a^{-b}+b^{-a}}=\frac{2^{4}+4^{2}}{2^{-4}+4^{-2}}=256 .
$$
|
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 infinitely many terms that are perfect squares.
(2) Let $a_{n}=9+7(n-1)=m^{2}(m$ is a positive integer). Then
$$
7(n-1)=(m-3)(m+3) \text {. }
$$
Thus, $7 \mid (m-3)(m+3)$.
Therefore, $7 \mid (m-3)$ or $7 \mid (m+3)$.
Furthermore, when $7 \mid (m-3)$ or $7 \mid (m+3)$, let $m \pm 3 = 7 k$. Then
$$
n=7 k^{2} \mp 6 k+1 \text {. }
$$
From (1), we know that at this time, $a_{n}$ is a perfect square. Hence, $a_{n}$ is a perfect square if and only if $n=7 k^{2} \mp 6 k+1(k=0,1, \cdots)$.
Since $7(k+1)^{2}-6(k+1)+1$
$$
>7 k^{2}+6 k+1 \text {, }
$$
the 100th perfect square in the sequence $\left\{a_{n}\right\}$ is the 17201st term of the sequence $\left\{a_{n}\right\}$, which is $a_{17201}=347^{2}$.
|
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., $n=20$, $4 n+1=81$, $6 n+1=121$ are both square numbers.
Therefore, the smallest positive integer $n$ is 20.
|
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, $b 、 d$ are also determined. Therefore, the number of four-digit numbers $\overline{a b c d}$ that satisfy the condition is $k(k+1)$.
When $10 \leqslant k \leqslant 18$, $a 、 b 、 c 、 d$ cannot be 0.
$$
\begin{array}{l}
\text { Let } a_{1}=10-a, b_{1}=10-b, \\
c_{1}=10-c, d_{1}=10-d .
\end{array}
$$
Then $1 \leqslant a_{1} 、 b_{1} 、 c_{1} 、 d_{1} \leqslant 9$.
Let $k_{1}=a_{1}+b_{1}=c_{1}+d_{1}$. Then $2 \leqslant k_{1} \leqslant 10$, and
the four-digit number $\overline{a b c d}$ corresponds one-to-one with the four-digit number $\overline{a_{1} b_{1} c_{1} d_{1}}$.
In the above equation, $a_{1}$ and $c_{1}$ can take any value in $\left\{1,2, \cdots, k_{1}-1\right\}$, and once $a_{1} 、 c_{1}$ are determined, $b_{1} 、 d_{1}$ are also determined.
Therefore, the number of four-digit numbers $\overline{a_{1} b_{1} c_{1} d_{1}}$ that satisfy $k_{1}=a_{1}+b_{1}=c_{1}+d_{1}$ is $\left(k_{1}-1\right)^{2}$.
Thus, the total number of good numbers is
$$
\begin{array}{l}
\sum_{k=1}^{9} k(k+1)+\sum_{k_{1}=2}^{10}\left(k_{1}-1\right)^{2} \\
=330+285=615 .
\end{array}
$$
|
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 one method where at least 20 people register, then the minimum value of $n$ is ( ).
(A) 171
(B) 172
(C) 180
(D) 181
|
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 $a_{\text {甲 }}=1$), otherwise, the corresponding number is 0.
Thus, each person has 9 possible ways to register for the events:
$$
\begin{array}{l}
(1,0,0,0),(0,1,0,0),(0,0,1,0), \\
(0,0,0,1),(1,1,0,0),(1,0,1,0), \\
(1,0,0,1),(0,1,0,1),(0,0,1,1) .
\end{array}
$$
Therefore, for $n$ people, there are 9 possible ways to register for the events, which can be considered as 9 drawers.
By the pigeonhole principle, when
$$
n=19 \times 9+r(r \geqslant 1)
$$
there must be at least one way in which at least 20 people have registered.
So, when $r=1$, $n$ takes the minimum value
$$
19 \times 9+1=172 \text {. }
$$
Hence, the answer is B.
|
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. Therefore, the number of such mappings $f$ is $\mathrm{C}_{n}^{k} k^{n-k}$.
Thus, the number of mappings $f$ that satisfy the condition is
$$
\sum_{k=1}^{n} \mathrm{C}_{n}^{k} k^{n-k} \text {. }
$$
Taking $n=5$, the number of mappings $f: M \rightarrow M$ is
196.
|
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 $k \leqslant x \leqslant k+1(1 \leqslant k \leqslant 2010)$,
$$
\begin{array}{l}
f(x)=\sum_{i=1}^{k} i(x-i)+\sum_{i=k+1}^{2011} i(i-x) \\
=\left(k^{2}+k-2011 \times 1006\right) x+ \\
\quad \frac{2011 \times 2012 \times 4023}{6}- \\
\quad \frac{k(k+1)(2 k+1)}{3}
\end{array}
$$
is a linear function, and its minimum value is attained at the endpoints of the interval $[k, k+1]$.
Therefore, the minimum term of the sequence $\{f(k)\}(1 \leqslant k \leqslant 2010)$ is the desired value.
Notice that,
$$
\begin{array}{l}
f(k+1)>f(k) \\
\Leftrightarrow k^{2}+k-2011 \times 1006>0 \\
\Leftrightarrow k \geqslant 1422 . \\
\text { Hence } f(x)_{\min }=f(1422) \\
=1422\left(1422^{2}+1422-2011 \times 1006\right)+ \\
\frac{2011 \times 2012 \times 4023}{6}-\frac{1422 \times 1423 \times 2845}{3} \\
=794598996 .
\end{array}
$$
|
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 number of edges drawn from point $A_{1}$ is the most (let's say there are $k$ edges) as $A_{1} A_{n}, A_{1} A_{n-1}, \cdots, A_{1} A_{n-k+1}$.
Since there is no triangle, there are no edges between the points $A_{n}, A_{n-1}, \cdots, A_{n-k+1}$.
Thus, each edge in the space graph has at least one endpoint among the points $A_{1}, A_{2}, \cdots, A_{n-k}$, and each $A_{i}(1 \leqslant i \leqslant n-k)$ can draw at most $k$ edges.
Therefore, the total number of edges is less than or equal to $k(n-k) \leqslant\left[\frac{n^{2}}{4}\right]$.
Back to the original problem.
Next, prove: If graph $G$ already has (at least) 28 triangles, then there is at least one tetrahedron.
Otherwise, in the 9-point set $M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\}$, by the pigeonhole principle, there must be a point that is at least
$$
\left[\frac{28 \times 3}{9}\right]+1=10
$$
vertices of triangles. Thus, at least 5 edges are drawn from this point (let this point be $A_{1}$).
(1) If 5 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{6}$ are drawn from point $A_{1}$, according to the problem, since there is no tetrahedron, the subgraph formed by the 5 points $A_{2}, A_{3}, \cdots, A_{6}$ has no triangle. By the lemma, this subgraph has at most $\left[\frac{5^{2}}{4}\right]=6$ edges. Thus, the number of triangles with $A_{1}$ as a vertex is at most 6, a contradiction.
(2) If 6 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{7}$ are drawn from point $A_{1}$, similarly to (1), the number of triangles with $A_{1}$ as a vertex is at most $\left[\frac{6^{2}}{4}\right]=9$, a contradiction.
(3) If 7 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{8}$ are drawn from point $A_{1}$, since there is no tetrahedron, the subgraph formed by the 7 points $A_{2}, A_{3}, \cdots, A_{8}$ has no triangle. This subgraph has at most $\left[\frac{7^{2}}{4}\right]=12$ edges. Thus, the number of triangles with $A_{1}$ as a vertex is at most 12, and the triangles not having $A_{1}$ as a vertex must have point $A_{9}$ as a vertex. Similarly, there are at most 12 such triangles. Therefore, the total number of triangles is less than or equal to $12 \times 2=24<28$, a contradiction.
(4) If 8 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{9}$ are drawn from point $A_{1}$, the subgraph formed by the 8 points $A_{2}, A_{3}, \cdots, A_{9}$ has no triangle. This subgraph has at most $\left[\frac{8^{2}}{4}\right]=16$ edges. Hence, the graph $G$ has at most 16 triangles, a contradiction.
Thus, the number of triangles satisfying the condition is at most 27.
Divide the 9-point set $M$ into three groups
$$
\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}, A_{6}\right\},\left\{A_{7}, A_{8}, A_{9}\right\},
$$
such that no two points in the same group are connected, while any two points in different groups are connected, thus, there are $C_{3}^{1} C_{3}^{1} C_{3}^{1}=27$ triangles, of course, there is no tetrahedron.
|
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 integers is $\qquad$ [s]
|
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 square of the distance from a point on the unit circle $\odot O$ to the point $(1,0)$ (as shown in Figure 7), with a value range of $[0,4]$.
$$
\begin{array}{l}
\text { Therefore }|z-2005-2006 \mathrm{i}| \leqslant 4 \\
\Rightarrow(x-2005)^{2}+(y-2006)^{2} \leqslant 16,
\end{array}
$$
The number of integer points (both coordinates are integers) is the same as the number of integer points in
$$
x^{2}+y^{2} \leqslant 16
$$
which is 49.
|
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 multiple of 119, then the conclusion holds.
If none of them is a multiple of 119, then the remainders when they are divided by 119 can only be \( 1, 2, \cdots, 118 \), which are 118 different cases.
By the pigeonhole principle, we know that there must be two of them that have the same remainder when divided by 119. Let's assume these two are
\[ b_{1}+b_{2}+\cdots+b_{i} \]
and \( b_{1}+b_{2}+\cdots+b_{j} (1 \leqslant i < j \leqslant 119) \).
Then, \( 119 \mid \left(b_{i+1}+b_{i+2}+\cdots+b_{j}\right) \).
Next, for any 119 numbers among \( a_{1}, a_{2}, \cdots, a_{2006} \), by the previous proof, we know that there must be a subset of these numbers whose sum (let's denote it as \( x \)) is a multiple of 119.
By the problem's condition, we know that \( x \) is not equal to 119, so \( x \) is at least \( 2 \times 119 \).
Since \( 2006 = 16 \times 119 + 102 \), we have
\[ \begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2006} \\
\geqslant 16 \times 238 + 102 = 3910.
\end{array} \]
If we set \( a_{119}=a_{238}=\cdots=a_{1904}=120 \), and the rest of the numbers are 1, then the equality holds.
Therefore, the minimum value of \( a_{1}+a_{2}+\cdots+a_{2006} \) is 3910.
|
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 ratio lies within the interval $\left[\frac{2}{3}, \frac{3}{2}\right]$.
Divide $1,2, \cdots, 2012$ into the following 17 sets:
$\{1\},\{2,3\},\{4,5,6\},\{7,8,9,10\}$,
$\{11,12, \cdots, 16\},\{17,18, \cdots, 25\}$,
$\{26,27, \cdots, 39\},\{40,41, \cdots, 60\}$,
$\{61,62, \cdots, 91\},\{92,93, \cdots, 138\}$,
$\{139,140, \cdots, 208\},\{209,210, \cdots, 313\}$,
$\{314,315, \cdots, 471\},\{472,473, \cdots, 708\}$,
$\{709,710, \cdots, 1063\}$,
$\{1064,1065, \cdots, 1596\}$,
$\{1597,1598, \cdots, 2012\}$.
In each of these 17 sets, the ratio $k$ of any two numbers satisfies
$$
\frac{2}{3} \leqslant k \leqslant \frac{3}{2} \text {. }
$$
By the pigeonhole principle, we know that from $1,2, \cdots, 2012$, if 18 numbers are arbitrarily selected, there must be two numbers from the same set, and their ratio lies within $\left[\frac{2}{3}, \frac{3}{2}\right]$.
In summary, the minimum value of $n$ is 18.
|
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
$$
\begin{array}{l}
(a+b+c+d)^{2}+(e+f+g+h)^{2} \\
\geqslant 2+32=34,
\end{array}
$$
and when $x=(-3)+(-2)+4+6=5$, the equality holds.
Therefore, the minimum value sought is 34.
|
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-place player's score is $\frac{4}{5}$ of the sum of the scores of the last five players. What is the score of the second-place player? $\qquad$
|
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 than or equal to $2 \times 9=18$.
If the second-place contestant scores 17 points (i.e., 8 wins and 1 draw), then the match between the first and second-place contestants must be a draw. This way, both the first and second-place contestants would score 17 points, which contradicts the rule that each contestant has a unique score.
Therefore, the second-place contestant scores 16 points.
|
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 $k \geqslant 8$,
$$
c+d \geqslant 2 \sqrt{c d}=2 \sqrt{k+17} \geqslant 2 \sqrt{8+17}=10 \text {. }
$$
When $1 \leqslant k \leqslant 7$, there are only two cases where $k+17$ can be written as the product of two primes:
$k=4$ when, $k+17=21=3 \times 7$. In this case,
$c+d=10$;
$k=5$ when, $k+17=22=2 \times 11$. In this case,
$$
c+d=13 \text {. }
$$
In summary, the minimum value of $a+b+c+d$ is
$$
5+7+3+7=22
$$
|
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,
$2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}, 17^{2}$,
$19^{2}, 23^{2}, 29^{2}, 31^{2}, 37^{2}, 41^{2}, 43^{2}$
are 14 positive integers greater than 1 and not exceeding 2009 that are pairwise coprime, and none of them are prime.
Therefore, $n \geqslant 15$.
On the other hand, let $a_{1}, a_{2}, \cdots, a_{15}$ be 15 positive integers greater than 1 and not exceeding 2009 that are pairwise coprime.
If they are all composite, let $p_{i}$ be the prime factor of $a_{i}$ not exceeding 44 $(i=1,2, \cdots, 15)$.
By the pigeonhole principle, there exist $i, j (1 \leqslant i<j \leqslant 15)$ such that $p_{i}=p_{j}$.
Thus, $a_{i}$ and $a_{j}$ are not coprime, which is a contradiction.
Therefore, among $a_{1}, a_{2}, \cdots, a_{15}$, there exists a prime number.
In conclusion, the minimum value of $n$ is 15.
|
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 $k \leqslant 2011$, so $k$ can take
$$
2011-671+1=1341
$$
different integer values.
Each $k$ corresponds exactly to one of the required quadratic trinomials, hence there are 1341 such quadratic trinomials.
|
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 points with a distance no greater than 1).
Obviously, it is impossible to construct a circle with a radius of 1 that contains 25 of the selected points.
Therefore, $n_{\text {min }}>48$.
Next, prove: $n_{\text {min }}=49$.
If $n=49$, let $A$ be one of the points. Construct a circle $\odot A$ with a radius of 1. If all the points are within $\odot A$, then the condition of the problem is satisfied.
Otherwise, there is at least one point $B$ not in $\odot A$. Construct another circle $\odot B$ with a radius of 1. Then the distance between points $A$ and $B$ is greater than 1 (as shown in Figure 4).
Thus, for the remaining 47 points, each point $P$ forms a triplet with $A$ and $B$, and it must be true that $PA \leqslant 1$ or $PB \leqslant 1$, meaning point $P$ is either in $\odot A$ or in $\odot B$.
According to the pigeonhole principle, one of the circles must contain at least 24 of these 47 points (let's assume it is $\odot A$). Adding the center point $A$, there are at least 25 points in the circle $\odot A$ with a radius of 1 (inside or on the circumference).
Therefore, the minimum value of $n$ is 49.
|
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 the greatest integer not exceeding the real number $a$.
|
$$
f(1)=\frac{1}{2} \text {. }
$$
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
$$
\frac{1}{2} f(x)+\frac{1}{2} f(x)-f(x) f(1) \geqslant \frac{1}{4} \text {. }
$$
Substituting $f(1)=\frac{1}{2}$, we get that for any real number $x$,
$$
f(x) \geqslant \frac{1}{2} \text {. }
$$
Combining (1) and (2), for any real number $x$,
$$
f(x)=\frac{1}{2} \text {. }
$$
Upon verification, $f(x)=\frac{1}{2}$ satisfies the given conditions.
$$
\begin{array}{l}
\text { Therefore, }[1 \times f(1)]+[2 f(2)]+\cdots+[2011 f(2011)] \\
=\left[\frac{1}{2}\right]+\left[\frac{2}{2}\right]+\cdots+\left[\frac{2011}{2}\right] \\
=0+1+1+\cdots+1005+1005 \\
=2(1+2+\cdots+1005) \\
=(1+1005) \times 1005 \\
=1011030 .
\end{array}
$$
Five, let $x=y=z=0$. Then
$$
\begin{array}{l}
\frac{1}{2} f(0)+\frac{1}{2} f(0)-f^{2}(0) \geqslant \frac{1}{4} \\
\Rightarrow\left(f(0)-\frac{1}{2}\right)^{2} \leqslant 0 \\
\Rightarrow f(0)=\frac{1}{2} .
\end{array}
$$
Now let $x=y=z=1$. Then similarly we have
|
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 $\mathrm{C}_{4}^{k} \times 2^{k}$ cases.
In summary, the total number of different pairs is
$$
1+\sum_{k=1}^{4} \mathrm{C}_{4}^{k} \times 2^{k}=81
$$
|
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
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B} \\
=\overrightarrow{C A} \cdot(\overrightarrow{A B}+\overrightarrow{B C})=-\overrightarrow{C A}^{2}=-25
\end{array}
$$
|
-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 the second row).
Next, observe the numbers in each column of the grid.
Each column already has two different numbers at the top. Therefore, each column below the second row can have at most two different numbers. This way, the maximum number of different numbers is $8 + 9 \times 2 = 26$, which is a contradiction.
Therefore, the number of different numbers is less than 29.
Table 1 constructs a $9 \times 9$ grid containing 28 different numbers, with each row and column having exactly four different numbers.
|
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,
$$
\begin{array}{l}
70=S_{30}=b_{1}+b_{2}+b_{3} \\
=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right) .
\end{array}
$$
Thus, $r^{2}+r-6=0$.
Solving, we get $r=2$ or $r=-3$ (discard).
Hence, $S_{40}=10\left(1+r+r^{2}+r^{3}\right)=150$.
|
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
$$
\begin{aligned}
& (m+n)(n+p)(p+q)(q+m) \\
= & {[(m+n)(p+q)][(n+p)(q+m)] } \\
= & (200+m q+n p)(200+m n+p q) \\
= & 200^{2}+100\left(m^{2}+n^{2}+p^{2}+q^{2}+2 m n+\right. \\
& 2 m q+2 n p+2 p q) \\
= & 200^{2}+100\left[(m+n+p+q)^{2}-400\right] \\
= & {[10(m+n+p+q)]^{2} . }
\end{aligned}
$$
Therefore, the original expression $=10(m+n+p+q)=220$.
|
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 three numbers are $1, 2, 3$. This array can be expanded continuously, as long as the added number is greater than or equal to the sum of the two largest numbers in the existing array. To maximize $k$, the added number should equal the sum of the two largest numbers in the existing array. Thus, we get
\[
\begin{array}{l}
1,2,3,5,8,13,21,34,55,89,144, \\
233,377,610,987,1597,
\end{array}
\]
a total of 16 numbers.
For any array $a_{1}, a_{2}, \cdots, a_{n}$ that meets the above conditions, it is clear that $a_{i}$ is always greater than or equal to the $i$-th number in (1).
Therefore, $n \leqslant 16 \leqslant k-1$.
Thus, the minimum value of $k$ is 17.
(Chang Yongsheng, Nankai High School, Tianjin, 300100)
|
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 $\qquad$ five-digit numbers that conform to the sine rule.
|
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 $(k-1)^{3}$ groups. Therefore, there are $(k-1)^{3}(10-k)$ numbers that conform to the sine rule.
Thus, the total number of numbers that conform to the sine rule is
$$
\sum_{k=2}^{9}(k-1)^{3}(10-k)=2892 \text { (numbers). }
$$
|
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 equation is called a "modified way". Clearly, the natural numbers $3, 4, 5, 10, 11, 12, 13, 14$ are all numbers in the modified way. The question is: Is the four-digit number 2012 a number in the modified way?
|
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 $a=4 \times(1+2+\cdots+k)=2 k(k+1)$.
From $2 k(k+1)-k \leqslant 2012 \leqslant 2 k(k+1)+k$, solving gives $k=31$.
It is easy to see that $w_{31}=(2015,1953)$.
Thus, 2012 is the 31st transformation method number.
(Tian Yonghai, Suihua Educational Institute, Heilongjiang, 152054)
|
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
$$
51\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {; }
$$
When $n$ is divided by 5, the remainder is 2 or 3, then $n^{2}+1$ is divisible by 5, so
$$
5 I\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {; }
$$
When $n$ is divided by 5, the remainder is 0,
$$
5 \nmid\left(n^{2}-n+3\right)\left(n^{2}+n+3\right) \text {. }
$$
Therefore, the number of all $n$ that meet the requirements of the problem is
$$
\frac{2010}{10} \times 8+2=1610 .
$$
|
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 integers, and $2a-m+2n > 2a-m-2n$ (since $m$ is a prime number),
we have $\left\{\begin{array}{l}2a-m+2n=m^{2} \\ 2a-m-2n=1 .\end{array}\right.$,
Solving these, we get $a=\frac{(m+1)^{2}}{4}, n=\frac{m^{2}-1}{4}$.
Thus, $b=a-m=\frac{(m-1)^{2}}{4}$.
Also, $a \geqslant 2012$, i.e., $\frac{(m+1)^{2}}{4} \geqslant 2012$.
Since $m$ is a prime number, solving this gives $m \geqslant 89$.
At this point, $a \geqslant \frac{(89+1)^{2}}{4}=2025$.
When $b=0$,
from $a-b=a=m$, we know that $a=m$ is a prime number.
Also, $a \geqslant 2012$, so the smallest prime number is 2017.
Therefore, the minimum value of $a$ is 2017.
|
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, $a_{2011}=\sum_{k=1}^{1005}\left(a_{2 k+1}-a_{2 k-1}\right)+a_{1}$
$$
=1005 \times 8+1=8041 \text {. }
$$
|
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
$$
\begin{array}{l}
f(m+1)-f(m)=m+1 . \\
\text { Hence } f(2011)=\sum_{k=1}^{2010}[f(k+1)-f(k)]+f(1) \\
=\sum_{k=1}^{2010}(k+1)+1=\frac{2011 \times 2012}{2} \\
=2023066 .
\end{array}
$$
|
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
$$
\begin{array}{l}
a_{3}=6, a_{4}=18, a_{n+1}=2 a_{n}+6(n \geqslant 3) . \\
\text { Hence } a_{n+1}+6=2\left(a_{n}+6\right) \\
\Rightarrow a_{n}=\left(a_{3}+6\right) \times 2^{n-3}-6 \\
\Rightarrow a_{10}=12 \times 2^{7}-6=1530 .
\end{array}
$$
|
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 $A C / / B F$
$$
\Rightarrow \frac{3+x}{4}=\frac{A C}{B F} \text {, }
$$
$B C / / A G$
$$
\begin{array}{c}
\Rightarrow \frac{4+x}{3}=\frac{B C}{A G} . \\
\text { Then } \frac{(3+x)(4+x)}{12}=\frac{A C^{2}}{B F \cdot A G} .
\end{array}
$$
Also, $\angle D C E=\angle F B E=\angle G A D=45^{\circ}$, hence $\triangle A D G \backsim \triangle C D E \backsim \triangle B F E$
$$
\Rightarrow \frac{B F}{A D}=\frac{B E}{A G} \Rightarrow B F \cdot A G=12 \text {. }
$$
Notice that,
$$
A C^{2}=B C^{2}=\frac{1}{2} A B^{2}=\frac{1}{2}(3+x+4)^{2} \text {. }
$$
Substitute equations (2) and (3) into equation (1), simplify and rearrange to get
$$
x^{2}=3^{2}+4^{2}=25 \Rightarrow x=5 \text {. }
$$
Therefore, $S_{\triangle A B C}=\frac{1}{4} A B^{2}=\frac{1}{4}(3+5+4)^{2}=36$.
|
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 that quadrilateral $AGEF$ is a square.
Assume $AD=h$. Then
$$
\begin{array}{l}
BE=h-2, CE=h-3 . \\
\text { By } BC^{2}=BE^{2}+CE^{2} \\
\Rightarrow(h-2)^{2}+(h-3)^{2}=5^{2} \\
\Rightarrow h^{2}-5 h-6=0 \\
\Rightarrow h=6 \\
\Rightarrow S_{\triangle ABC}=\frac{1}{2} BC \cdot AD=\frac{1}{2} \times 5 \times 6=15 .
\end{array}
$$
|
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
(D) 2011
|
- 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, \(a^{2012} b^{2012}=(c d)^{2012}-2012\).
Therefore, \((a b)^{2012}-(c d)^{2012}=-2012\).
|
-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 must exist one method where at least 20 people register, then the minimum value of $n$ is $(\quad$.
(A) 171
(B) 172
(C) 180
(D) 181
|
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, record $a_{\text {甲 }}=1$); if not participating in a certain event, the corresponding number is recorded as 0 (for example, if not participating in event 甲, record $a_{\text {甲 }}=0$).
Thus, each person's registration method for the events can be one of the following nine possibilities:
$$
\begin{array}{l}
(1,1,0,0),(1,0,1,0),(1,0,0,1), \\
(0,1,0,1),(0,0,1,1),(1,0,0,0), \\
(0,1,0,0),(0,0,1,0),(0,0,0,1) .
\end{array}
$$
Using these nine possibilities as the nine drawers, by the known pigeonhole principle, we have $n=19 \times 9+r$.
When $r=1$, the minimum value of $n$ is 172.
|
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-negative integers. Therefore, each set contains at least 1 square number.
In the sets $S_{13}, S_{14}, \cdots, S_{599}$, if they contain square numbers, they are no less than $26^{2}$.
When $x \geqslant 26$, $2 x+1 \geqslant 53$, thus, in $S_{13}, S_{14}$, $\cdots, S_{599}$, each set contains at most 1 square number.
On the other hand, the largest number in $S_{599}$ is
$$
600 \times 50-1=29999 \text {, }
$$
and $173^{2}<29999<174^{2}$, so the square numbers in $S_{13}, S_{14}, \cdots, S_{599}$ do not exceed $173^{2}$.
Therefore, there are exactly $173-$ $25=148$ sets in $S_{13}, S_{14}, \cdots, S_{599}$ that contain square numbers.
In summary, in $S_{0}, S_{1}, \cdots, S_{599}$, there are
$$
600-13-148=439
$$
sets that do not contain square numbers.
|
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 $(a-b)(f(a)-f(b)) \geqslant 0$, i.e., $a \ln a+b \ln b \geqslant a \ln b+b \ln a$.
Substituting $2 b$ for $b$ in the above inequality, we get $a \ln a+2 b \ln 2 b \geqslant a \ln 2 b+2 b \ln a$.
Similarly, $2 b \ln 2 b+c \ln c \geqslant 2 b \ln c+c \ln 2 b$, (2) $c \ln c+a \ln a \geqslant c \ln a+a \ln c$.
(1) + (2) + (3) and rearranging gives
$3 \ln \left[a^{a}(2 b)^{2 b} c^{c}\right] \geqslant(a+2 b+c) \ln (a \cdot 2 b \cdot c)$,
i.e., $a^{a}(2 b)^{2 b} c^{c} \geqslant(a \cdot 2 b \cdot c)^{\frac{a+2 b+c}{3}} \geqslant 8 \sqrt[3 / 2 \cdot 2 b \cdot c]{ }=64$.
The equality holds if and only if $a=c=2, b=1$. Therefore, the minimum value of $a^{a+b} b^{3 b} c^{c+b}$ is 64.
|
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, $r_{n}=3$. Find the value of $f_{2012}\left(f_{2011}\left(3^{2010}\right)\right)$.
|
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 $n \in \mathbf{N}$, and $n \geqslant 2$, we always have
$$
f_{n}\left(f_{n-1}(M)\right) \equiv 0(\bmod 9) .
$$
Since $3^{2010}=9^{1005}<10^{1005}$, the sum of its digits is less than $9 \times 1005=9045$.
Thus, $f_{1}\left(3^{2010}\right)<9045^{2}<9 \times 10^{7}$.
Therefore, the sum of the digits of $f_{1}\left(3^{2010}\right)$ is less than $9 \times 8=72$.
Hence, $f_{2}\left(f_{1}\left(3^{2010}\right)\right)<72^{3}<400000$, and $f_{3}\left(f_{2}\left(3^{2010}\right)\right)<(3+9 \times 5)^{2}=48^{2}<50^{2}$. Let the sum of the digits of $f_{3}\left(f_{2}\left(3^{2010}\right)\right)$ be $a$. Then $1 \leqslant a<1+9 \times 3=28$,
and $a \equiv f_{3}\left(f_{2}\left(3^{2010}\right)\right) \equiv 0(\bmod 9)$.
Therefore, $a \in\{9,18,27\}$.
Thus, $f_{4}\left(f_{3}\left(3^{2010}\right)\right)=a^{3} \in\left\{9^{3}, 18^{3}, 27^{3}\right\}$
$$
=\{729,5832,19683\} \text {. }
$$
The sums of the digits of $729$, $5832$, and $19683$ are $18$, $18$, and $27$, respectively. Hence,
$$
f_{5}\left(f_{4}\left(3^{2010}\right)\right) \in\left\{18^{2}, 27^{2}\right\}=\{324,729\} \text {. }
$$
Furthermore,
$$
\begin{array}{l}
f_{6}\left(f_{5}\left(3^{2010}\right)\right) \in\left\{9^{3}, 18^{3}\right\}=\{729,5832\} \\
\Rightarrow f_{7}\left(f_{6}\left(3^{2010}\right)\right)=18^{2}=324 \\
\Rightarrow f_{8}\left(f_{7}\left(3^{2010}\right)\right)=9^{3}=729 \\
\Rightarrow f_{9}\left(f_{8}\left(3^{2010}\right)\right)=18^{2}=324 \\
\Rightarrow \cdots \cdots \\
\Rightarrow f_{2012}\left(f_{2011}\left(3^{2010}\right)\right)=729
\end{array}
$$
|
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 1006 odd numbers in the range $1 \sim 2012$, so the number of $a$ that meet the conditions is
$$
1006-2=1004 \text {. }
$$
|
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 .
$$
\begin{aligned}
\text { Given } & \frac{(1-a)^{2}}{b}-\frac{(1+b)^{2}}{a}=4 \\
\Rightarrow & a(1-a)^{2}-b(1+b)^{2}=4 a b \\
\Rightarrow & a-2 a^{2}+a^{3}-b-2 b^{2}-b^{3}=4 a b \\
\Rightarrow & (a-b)-2\left(a^{2}+b^{2}\right)+\left(a^{3}-b^{3}\right)=4 a b \\
\Rightarrow & (a-b)-2\left[(a-b)^{2}+2 a b\right]+ \\
& (a-b)\left[(a-b)^{2}+3 a b\right]=4 a b .
\end{aligned}
$$
Substituting $a-b=2$ into the equation, we get $a b=1$.
$$
\begin{array}{l}
\text { Then } a^{2}+b^{2}=(a-b)^{2}+2 a b=6, \\
a^{3}-b^{3}=(a-b)\left[(a-b)^{2}+3 a b\right]=14. \\
\text { Therefore, } a^{5}-b^{5}=\left(a^{2}+b^{2}\right)\left(a^{3}-b^{3}\right)-a^{2} b^{2}(a-b) \\
=6 \times 14-1 \times 2=82.
\end{array}
$$
|
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 $a^{2}+b^{2}=(a+1)+(b+1)$ $=(a+b)+2=3$, $a^{3}+b^{3}=\left(a^{2}+a\right)+\left(b^{2}+b\right)$ $=\left(a^{2}+b^{2}\right)+(a+b)=1+3=4$, $a^{4}+b^{4}=\left(a^{3}+a^{2}\right)+\left(b^{3}+b^{2}\right)$ $=\left(a^{3}+b^{3}\right)+\left(a^{2}+b^{2}\right)=3+4=7$.
By analogy,
$$
\begin{array}{l}
a^{5}+b^{5}=4+7=11, \\
a^{6}+b^{6}=7+11=18, \\
\cdots \cdots \\
a^{19}+b^{19}=9349 .
\end{array}
$$
Since $-1<\left(\frac{1-\sqrt{5}}{2}\right)^{19}<0$, it follows that $9349<\left(\frac{1+\sqrt{5}}{2}\right)^{19}<9350$.
Therefore, the integer part of $\left(\frac{1+\sqrt{5}}{2}\right)^{19}$ is 9349.
|
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
$$
\begin{aligned}
k & \equiv 1(\bmod 2), \quad k \equiv 1(\bmod 3), \\
k & \equiv 3,4(\bmod 5), k \equiv 1,3,4(\bmod 7) .
\end{aligned}
$$
Therefore, $k \equiv 43,73,109,169,193,199(\bmod 210)$.
Thus, $k_{\min }=43$.
|
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,
$$
\begin{array}{l}
\sqrt{\frac{a b}{2 b^{2}-a}}=\frac{a+2 b}{4 b} \\
\Leftrightarrow 16 a b^{3}=\left(2 b^{2}-a\right)\left(a^{2}+4 a b+4 b^{2}\right) \\
\Leftrightarrow a\left(a^{2}+4 a b+4 b^{2}\right)=2 b^{2}\left(a^{2}-4 a b+4 b^{2}\right) \\
\Leftrightarrow a(a+2 b)^{2}=2 b^{2}(a-2 b)^{2} .
\end{array}
$$
Let $d=(a, b), a=a_{0} d, b=b_{0} d,\left(a_{0}, b_{0}\right)=1$.
Then equation (1) $\Leftrightarrow a_{0}\left(a_{0}+2 b_{0}\right)^{2}=2 d b_{0}^{2}\left(a_{0}-2 b_{0}\right)^{2}$.
Since $b_{0} \mid a_{0}\left(a_{0}+2 b_{0}\right)^{2}$, we get
$$
b_{0} \mid a_{0}^{3} \Rightarrow b_{0}=1 \text {. }
$$
Then $a_{0}\left(a_{0}+2\right)^{2}=2 d\left(a_{0}-2\right)^{2}$
$$
\Rightarrow 2 d=a_{0}+\frac{8 a_{0}^{2}}{\left(a_{0}-2\right)^{2}}=a_{0}+2\left(\frac{2 a_{0}}{a_{0}-2}\right)^{2} \text {. }
$$
Thus $\left(a_{0}-2\right)\left|2 a_{0} \Rightarrow\left(a_{0}-2\right)\right| 4$
$$
\Rightarrow a_{0}-2= \pm 1,2,4 \text {. }
$$
From equation (2), we know $2 \mid a_{0}$. Therefore, $a_{0}=4,6$.
Substituting into equation (2) gives $d=18,12$.
Accordingly, $(a, b)=(72,18),(72,12)$.
Thus, $|10(a-5)(b-15)|+2=2012$.
|
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
$$
\begin{array}{l}
22=2+3+8+9=2+4+7+9 \\
=2+5+6+9=2+5+7+8 \\
=3+4+6+9=3+4+7+8 \\
=3+5+6+8=4+5+6+7 \\
=0+5+8+9=0+6+7+9 ;
\end{array}
$$
If $a=3$, then $t=21$, we have
$$
\begin{array}{l}
21=1+4+7+9=1+5+6+9 \\
=1+5+7+8=2+4+6+9 \\
=2+4+7+8=2+5+6+8 \\
=0+4+8+9=0+5+7+9 \\
=0+6+7+8 ;
\end{array}
$$
If $a=5$, then $t=20$, we have
$$
\begin{array}{l}
20=1+2+8+9=1+3+7+9 \\
=1+4+6+9=1+4+7+8 \\
=2+3+6+9=2+3+7+8 \\
=2+4+6+8=3+4+6+7 \\
=0+3+8+9=0+4+7+9 ;
\end{array}
$$
If $a=7$, then $t=19$, we have
$$
\begin{array}{l}
19=1+3+6+9=1+4+5+9 \\
=1+4+6+8=2+3+5+9 \\
=2+3+6+8=2+4+5+8 \\
=0+2+8+9=0+4+6+9 \\
=0+5+6+8 ;
\end{array}
$$
If $a=9$, then $t=18$, we have
$$
\begin{array}{l}
18=1+2+7+8=1+3+6+8 \\
=1+4+5+8=1+4+6+7 \\
=2+3+5+8=2+3+6+7 \\
=2+4+5+7=3+4+5+6 \\
=0+3+7+8=0+4+6+8 \\
=0+5+6+7 .
\end{array}
$$
In summary, the number of nine-digit numbers that satisfy the conditions is
$$
49 A_{4}^{4} A_{5}^{5}-36 A_{4}^{4} A_{4}^{4}=120384 \text { (numbers). }
$$
|
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 $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right\}$, and determine this maximum value.
|
(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 $m_{1} \mid k$,
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=n,
$$
and there are $d-1$ values of $k$ satisfying $1 \leqslant k \leqslant m-1$;
When $m_{1} \nmid k$,
$$
\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]=n-1 \text{, }
$$
and there are
$$
\begin{array}{l}
m-1-(d-1)=m-d \text{ (values) }. \\
\text{ Then } \sum_{k=1}^{m}\left[\frac{k n}{m}\right]=n+\sum_{k=1}^{m-1}\left[\frac{k n}{m}\right] \\
=n+\frac{1}{2} \sum_{k=1}^{m-1}\left(\left[\frac{k n}{m}\right]+\left[\frac{(m-k) n}{m}\right]\right) \\
=n+\frac{1}{2}[n(d-1)+(n-1)(m-d)] \\
=\frac{1}{2}(m n+n-m+d) .
\end{array}
$$
Thus, when $m=2013$, we have
$$
\frac{1}{2}(2013 n+n-2013+d)=2013+n \text{, }
$$
which implies $3 \times 2013=2012 n+(2013, n)>2012 n$.
Therefore, $n \leqslant 3$.
Upon verification, $n=3$.
(2) From (1), we get
$$
\begin{array}{l}
\sum_{k=1}^{m}\left\{\frac{k n}{m}\right\}=\sum_{k=1}^{m}\left(\frac{k n}{m}-\left[\frac{k n}{m}\right]\right) \\
=\frac{n}{m} \cdot \frac{m(m+1)}{2}-\sum_{k=1}^{m}\left[\frac{k n}{m}\right] \\
=\frac{n(m+1)}{2}-\frac{1}{2}(m n+n-m+d) \\
=\frac{m-d}{2} \leqslant \frac{m-1}{2} .
\end{array}
$$
In particular, for $m=2013$, when $d=(2013, n)=1$, i.e., when $n$ is not divisible by $3, 11, 61$, the maximum value of $\sum_{k=1}^{2013}\left\{\frac{k n}{2013}\right\}$ is
$$
\frac{2013-1}{2}=1006 .
$$
|
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 $C_{6}^{5}\left(C_{5}^{3} A_{3}^{3}+\frac{C_{5}^{2} C_{3}^{2}}{2!} A_{3}^{3}\right)=900$ (ways); the number of ways for all six people to be hired is $3^{6}-C_{3}^{1} \times 2^{6}+C_{3}^{2} \times 1^{6}=540$ (ways). Therefore, the total number of ways is $120+540+900+540=2100$.
|
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 $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q} \sum_{i=1}^{2011}\left(\frac{1}{\lg a_{i}}-\frac{1}{\lg a_{i+1}}\right) \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q}\left(\frac{1}{\lg a_{1}}-\frac{1}{\lg a_{2012}}\right)=2011 .
\end{array}
$$
|
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 $a_{1}, a_{2}, \cdots, a_{2011}$ has the same digit sum, each of them has the same remainder when divided by 9, denoted by $r$ $(0 \leqslant r \leqslant 8)$. Similarly, for the numbers $b_{1}, b_{2}, \cdots, b_{2012}$, each has the same digit sum, and thus the same remainder when divided by 9, denoted by $s$ $(0 \leqslant s \leqslant 8)$. Therefore, the numbers $n-2011 r$ and $n-2012 s$ are both multiples of 9. This implies,
$$
\begin{array}{l}
(n-2011 r)-(n-2012 s) \\
=2012 s-2011 r \\
=2012(r+s)-4023 r
\end{array}
$$
is a multiple of 9.
Since 4023 is a multiple of 9 and 2012 is coprime with 9, the number $r+s$ must be a multiple of 9.
If $r=s=0$, then $n \geqslant 9 \times 2012$ (since $b_{1}, b_{2}, \cdots, b_{2012}$ are divisible by 9).
If $r \neq 0$, then $r+s=9$, so at least one of $r \geqslant 5$ or $s \geqslant 5$ holds.
For the number $n$, we get
$$
\begin{array}{l}
n \geqslant 5 \times 2011 \text { and } n \geqslant 5 \times 2012 . \\
\text { Also, } 10055=5 \times 2011 \\
=4 \times 2011+2011 \times 1
\end{array}
$$
and the numbers 4 and 2011 have the same digit sum, so the number 10055 is the desired number.
|
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 arithmetic sequence with the first term of 700 yuan and a common difference of -140. Then the expected prize money for participating in this game activity is $\qquad$ yuan.
|
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 are
$$
\begin{array}{l}
\quad \xi_{1}=700, \xi_{2}=560, \xi_{3}=420 . \\
\quad \text { Therefore, } E \xi=700 \times \frac{1}{7}+560 \times \frac{2}{7}+420 \times \frac{4}{7} \\
=500 \text { (yuan). }
\end{array}
$$
|
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 complete the task of transporting and planting 20 poles, and returning to the construction site, the minimum total distance the transport vehicle must travel is $\qquad$ m.
|
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 sequence with the first term 1200 and a common difference of 300.
Thus, $S_{7}=7 \times 1200+\frac{7 \times 6}{2} \times 300=14700(\mathrm{~m})$.
However, in reality, only 20 poles are transported, so there must be one time when 2 poles are transported, and the other 6 times, 3 poles each.
If the transportation of 2 poles is arranged on the $k(k=1,2, \cdots, 7)$-th time, then $a_{1}, a_{2}, \cdots, a_{k-1}$ remain unchanged, and $a_{k}, a_{k+1}, \cdots, a_{7}$ each decrease by $100 \mathrm{~m}$.
$$
\begin{array}{l}
\text { Hence } S_{7}(k)=S_{7}-100(8-k) \\
=13900+100 k .
\end{array}
$$
Obviously, when $k=1$, that is, the first time 2 poles are transported, and the rest each time 3 poles, the total distance is minimized at $14000 \mathrm{~m}$.
|
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 $b_{n}=1+\frac{1}{a_{n}}\left(n \in \mathbf{N}_{+}\right)$, and for any positive integer $n(n \geqslant 2)$, the inequality
$$
\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24}
$$
always holds, find the maximum value of the integer $m$.
|
(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 geometric sequence with the first term 3 and common ratio 3.
Thus, $\frac{1}{a_{n}}+1=3^{n} \Rightarrow a_{n}=\frac{1}{3^{n}-1}\left(n \in \mathbf{N}_{+}\right)$.
(2) From (1), we get $b_{n}=1+\frac{1}{a_{n}}=3^{n}$.
Therefore, $\log _{3} b_{k}=k(k=1,2, \cdots)$.
Hence, $\sum_{k=1}^{n} \frac{1}{n+\log _{3} b_{k}}>\frac{m}{24}$
$\Leftrightarrow \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}>\frac{m}{24}$.
Let $f(n)=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}$. Then
$$
\begin{array}{l}
f(n+1)-f(n) \\
=\frac{1}{2 n+1}+\frac{1}{2 n+2}-\frac{1}{n+1} \\
=\frac{1}{2 n+1}-\frac{1}{2 n+2}>0 .
\end{array}
$$
Therefore, $f(n)$ is monotonically increasing.
Hence, $f(n)_{\min }=f(2)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$.
Thus, $\frac{7}{12}>\frac{m}{24} \Rightarrow m<14$.
Therefore, the maximum integer value of $m$ is 13.
|
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
$$
\begin{array}{l}
\left(A_{1}, A_{2}, A_{1000}, A_{1008}\right),\left(A_{3}, A_{4}, A_{100}, A_{1010}\right), \\
\cdots,\left(A_{1005}, A_{1000}, A_{2011}, A_{2012}\right) .
\end{array}
$$
If $k$ points contain a set of four points
$$
\left(A_{2 i-1}, A_{2 i}, A_{2 i+1005}, A_{2 i+1006}\right),
$$
then the convex $k$-gon obtained has two parallel sides $A_{2 i-1} A_{2 i}$ and $A_{2 i+1005} A_{2 i+1006}$, which is a contradiction.
This indicates that at most three points can be taken from each set of four points, hence
$$
k \leqslant 503 \times 3=1509 \text {. }
$$
Below is an example for $k=1509$.
It is easy to see that in the convex polygon $A_{1} A_{2} \cdots A_{1006} A_{1008} A_{1010} \cdots A_{2012}$, no two sides are parallel.
|
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, $\frac{x-1}{2} \leqslant x=\left[\frac{x}{2}\right] \leqslant \frac{x}{2}$, which means $-1 \leqslant x \leqslant 0$. After verification, $x=0,-1$ are elements of $A_{2}$. Therefore, $A_{2}=\{-1,0\}$.
Elements in $A_{3}$ satisfy
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right] \text {, and } x \in \mathbf{Z} . \\
\text { Hence } \frac{x-1}{2}+\frac{x-2}{3} \leqslant x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right] \\
\leqslant \frac{x}{2}+\frac{x}{3},
\end{array}
$$
which means $-7 \leqslant x \leqslant 0$.
After verification, $x=-7,-5,-4,-3,-2,0$ are elements of $A_{3}$, i.e.,
$$
A_{3}=\{-7,-5,-4,-3,-2,0\} \text {. }
$$
Elements in $A_{4}$ satisfy
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right], \text { and } x \in \mathbf{Z} . \\
\text { Hence } \frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4} \leqslant x \\
=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right] \leqslant \frac{x}{2}+\frac{x}{3}+\frac{x}{4} .
\end{array}
$$
Solving this, we get $0 \leqslant x \leqslant 23$.
Therefore, $A_{4} \subseteq\{0,1, \cdots, 23\}$.
When $x=23$,
$$
\left[\frac{x}{2}\right]=11,\left[\frac{x}{3}\right]=7,\left[\frac{x}{4}\right]=5,
$$
satisfying $\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right]=23$.
Thus, $23 \in A_{4}$.
When $n \geqslant 5$,
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right], \text { and } x \in \mathbf{Z} . \\
\text { Hence } x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right] \\
\leqslant \frac{x}{2}+\frac{x}{3}+\cdots+\frac{x}{n},
\end{array}
$$
which means $x \geqslant 0$.
On the other hand,
$$
\begin{array}{l}
x=\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\cdots+\left[\frac{x}{n}\right] \\
\geqslant\left[\frac{x}{2}\right]+\left[\frac{x}{3}\right]+\left[\frac{x}{4}\right] \\
\geqslant \frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4},
\end{array}
$$
which gives $x \leqslant 23$.
In summary, $A=\cup_{n \geqslant 2} A_{n} \subseteq\{-7,-6, \cdots, 23\}$ is a finite set, with the maximum element being $23 \in A_{4}$ and the minimum element being $-7 \in A_{3}$.
|
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
$$
209519<a^{6}<209520 \text {. }
$$
Therefore, the largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is
209519.
|
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 $DE=$ $\qquad$
|
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
$$
\Delta=(26+y)^{2}-4 y^{2}>0 \Rightarrow y<26 \text {. }
$$
Upon verification, only $y=8$ satisfies the condition. Then $x=32$. Therefore, $D E=32-2 \times 8=16$.
|
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
$$
\begin{array}{l}
\frac{(1+2+\cdots+n)-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2}+\frac{n-k}{n-1}=43 \frac{14}{17} \\
\Rightarrow \frac{n}{2} \leqslant 43 \frac{14}{17} \leqslant \frac{n}{2}+1 \\
\Rightarrow 85 \frac{11}{17} \leqslant n \leqslant 87 \frac{11}{17} \Rightarrow n=86,87 .
\end{array}
$$
From equation (1), we know that $n-1$ must be a multiple of 17.
Thus, $n=86 \Rightarrow k=16$.
Therefore, the number erased is 16.
(Chen Qian, Yujing Middle School, Xishui County, Huanggang City, Hubei Province, 438200 Ma Xuemin, Xishui County Special Education School, Huanggang City, Hubei Province, 438200)
|
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,
$$
\begin{array}{l}
f(x+4)-f(x)=x^{2} \\
\Rightarrow f(4 k+4)-f(4 k)=(4 k)^{2} \\
\Rightarrow f(64)=\sum_{k=0}^{15}(f(4(k+1))-f(4 k)) \\
\quad=\sum_{k=0}^{15} 16 k^{2}=19840 .
\end{array}
$$
|
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 $a-b=1$, $(a-b) \mid (a+b)$, which is a contradiction.
When $a-b=2$, $a$ and $b$ have the same parity, thus, $a+b$ is even. Therefore, $(a-b) \mid (a+b)$, which is a contradiction.
Therefore, the maximum number of numbers that can be taken is 671.
For example, taking $1,4,6, \cdots, 2011$ these 671 numbers satisfies the condition.
|
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) $S_{i} \cup S_{j} \cup S_{k}=\{1,2, \cdots, 2008\}(1 \leqslant i < j < k \leqslant n, i, j, k \in \mathbf{N}_{+})$.
Find the maximum possible value of $n$. ${ }^{[1]}$
$(2009$, Serbian Mathematical Olympiad)
|
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 $y \in Y$ if and only if $y$ is an element of $A_{i}$.
Since $\left|A_{i} \cap A_{j}\right| \geqslant 4$, for any two vertices in $X$, their common neighbors (referring to points adjacent to both, the same below) are at least four. In other words, for any two vertices in $X$, there are at least four vertices in $Y$ that form an angle with them. Therefore, the number of angles formed by vertices in $Y$ is greater than or equal to $4 \mathrm{C}_{n}^{2}$.
On the other hand, since $A_{i} \cap A_{j} \cap A_{k}=\varnothing$, the degree of each vertex in $Y$ is less than or equal to 2, meaning that at most one angle is formed at each vertex in $Y$. Therefore, the number of angles formed by vertices in $Y$ is less than or equal to $|Y|=2008$.
In summary, $4 \mathrm{C}_{n}^{2} \leqslant 2008(n \leqslant 32)$.
Using the bipartite graph $X+Y$, it is not difficult to construct an example where $n=32$.
Therefore, the largest positive integer sought is 32.
|
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 $n$ vertices, with degrees $x_{1}, x_{2}, \cdots, x_{n}$.
Since each test paper has 4 questions, each vertex in $X$ has a degree of 4. Therefore,
$x_{1}+x_{2}+\cdots+x_{n}=4 \times 10=40$.
Furthermore, since at most one question is the same between any two test papers, for any two vertices in $X$, there is at most one vertex in $Y$ that forms an angle with them, meaning the number of such angles is less than or equal to $\mathrm{C}_{10}^{2}$.
In $Y$, the number of angles at the $i$-th vertex is $\mathrm{C}_{x_{i}}^{2}$, so the total number of angles is
$\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2}$.
Thus, $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \leqslant \mathrm{C}_{10}^{2}$.
Combining this with the previous equation, we can simplify to
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2} \leqslant 2 \mathrm{C}_{10}^{2}+40=130 \text {. }
$$
Using a common inequality, we get
$$
\begin{array}{l}
\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \leqslant n\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \\
\Rightarrow 40^{2} \leqslant 130 n \Rightarrow n \geqslant 13 .
\end{array}
$$
Using the aforementioned bipartite graph, it is not difficult to construct an example with $n=13$ (take $x_{1}=x_{2}=4, x_{3}=x_{4}=\cdots=x_{12}=3$, $x_{13}=2$).
Therefore, there are at least 13 different questions.
The idea in Example 6 can be used to prove a general conclusion.
Theorem 2 Let a simple graph $G$ have $n$ vertices and $m$ edges. If graph $G$ does not contain $K_{2, r+1}$, then
$$
m \leqslant \frac{n}{4}[1+\sqrt{4 r(n-1)+1}] .
$$
Proof Since graph $G$ does not contain $K_{2, r+1}$, for any two vertices, the number of angles they form is less than or equal to $r$.
On the other hand, let the degrees of the vertices in the graph be $x_{1}$, $x_{2}, \cdots, x_{n}$, then the number of angles in the graph is
$$
\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \text {. }
$$
Therefore, $\mathrm{C}_{x_{1}}^{2}+\mathrm{C}_{x_{2}}^{2}+\cdots+\mathrm{C}_{x_{n}}^{2} \leqslant r \mathrm{C}_{n}^{2}$.
Combining this with $x_{1}+x_{2}+\cdots+x_{n}=2 m$, we get
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=r n(n-1)+2 m \text {. }
$$
As in Example 6, we can derive the inequality
$$
(2 m)^{2} \leqslant n[m(n-1)+2 m] \text {. }
$$
Thus, the theorem is proved.
|
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
$$
m \leqslant \frac{2012}{4}(1+\sqrt{4 \times 2 \times 2011+1}) \text {, }
$$
i.e., $m \leqslant 64304$.
Therefore, the number of different distances is at least $\frac{\mathrm{C}_{2012}^{2}}{64304}$.
Thus, the number of different distances is at least 32.
[Note] This approach originates from P$\cdot$Erdös. In fact, using other methods, 32 can be improved to 45.
|
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 \( Y = \{A_{1}, A_{2}, \cdots, A_{15}\} \), then \( X + Y \) forms a bipartite graph, where \( x \in X \) is adjacent to \( A_{j} \in Y \) if and only if \( x \) is an element of \( A_{j} \).
Since the intersection of any three subsets is empty, each element in \( X \) appears in at most two subsets, i.e., the degree of each vertex in \( X \) is less than or equal to 2.
Without loss of generality, assume each vertex in \( X \) has a degree of 2 (if a vertex has a degree \( d < 2 \), we can arbitrarily add \( 2 - d \) neighbors).
Thus, the number of edges in the graph is \( 2 \times 56 = 112 \), and the graph has exactly 56 (vertices in \( X \)) angles.
For any seven vertices in \( Y \), their number of neighbors is at least 41, which means the number of angles related to these seven vertices is at least 41. Thus, for the remaining eight vertices, the number of angles formed by any two of them is at most \( 56 - 41 = 15 \).
Next, fix 14 vertices in \( Y \), among which there are \( \binom{14}{8} \) "eight-point groups," and each eight-point group has at most 15 angles, while each angle appears in \( \binom{12}{6} \) eight-point groups. Therefore, the number of angles formed by these 14 vertices is less than or equal to \( 15 \times \frac{\binom{14}{8}}{\binom{12}{6}} < 49 \).
This indicates that the number of angles related to the remaining vertex in \( Y \) is greater than 7, i.e., the degree of this vertex is at least 8.
Thus, the degree of any vertex in \( Y \) is at least 8, and the number of edges in the graph is at least \( 8 \times 15 = 120 \). This contradicts the fact that the number of edges is 112.
In conclusion, \( n = 41 \) satisfies the given requirements.
Next, we show that when \( n = 40 \), there exist 15 subsets \( A_{1}, A_{2}, \cdots, A_{15} \), such that the union of any seven subsets has at least 40 elements, but the intersection of any three subsets is empty.
In fact, consider such a bipartite graph \( X + Y \), where \( Y = \{A_{1}, A_{2}, \cdots, A_{15}\} \), and for each \( 1 \leq i \leq 7 \) and \( 8 \leq j \leq 15 \), there is exactly one vertex in \( X \) adjacent to both \( A_{i} \) and \( A_{j} \) (i.e., it forms an angle with them).
It is easy to verify that any eight vertices in \( Y \) form at most \( x(8 - x) \leq 16 \) angles, thus, the number of angles related to the remaining seven vertices is at least 40, i.e., their number of neighbors is at least 40.
In conclusion, the smallest positive integer that satisfies the requirements is 41.
|
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 Mathematics League Shandong Province Preliminary Contest)
|
Let $\alpha, \beta$ be the integer roots of the equation
$$
x^{2}+m x-36=0
$$
Assume $|\alpha| \geqslant|\beta|$. Then
$$
\begin{array}{l}
\alpha \beta=-36 \\
\Rightarrow(|\alpha|,|\beta|) \\
=(1,36)(2,18),(3,12),(4,9),(6,6) \\
\Rightarrow m= \pm 35, \pm 16, \pm 9, \pm 5,0 \\
\Rightarrow M=\{0\} \cup\{-5,5\} \cup\{-9,9\} \cup \\
\{-16,16\} \cup\{-35,35\} \text {. } \\
\end{array}
$$
By the problem statement, $A \neq \varnothing$, and the above five subsets of $M$ are either subsets of $A$ or disjoint from $A$.
Thus, there are $2^{5}-1=31$ different non-empty sets $A$.
|
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 napkins, find the maximum value of $k$.
|
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 bottom to top be
$1,2, \cdots, 2011$.
Each napkin is represented by the coordinates of its lower left corner, and the napkins are divided into four categories:
The coordinates of the first category are
$(52 i+36,52 j+1)(0 \leqslant j \leqslant i \leqslant m-2)$;
The coordinates of the second category are
$(52 i+1,52 j+36)(0 \leqslant i \leqslant j \leqslant m-2)$;
The coordinates of the third category are
$(52 i+36,52 i+36)(0 \leqslant i \leqslant m-2)$;
The coordinates of the fourth category are $(1,1)$.
In Figure 2, different shading indicates different categories of napkins.
Except for the unit squares with at least two different types of shading, the number written in all other unit squares is 1. It is easy to calculate that the number of unit squares with at least two different types of shading is
$\left(52^{2}-35^{2}\right) m-17^{2}=57392$.
Below is the proof: the number of unit squares with the same number written in them does not exceed 3986729.
For any configuration of napkins and any positive integer $M$, assume there are $g$ unit squares with numbers different from $M$. We need to prove that $g \geqslant 57392$.
A line is defined as either a row or a column.
Consider any line $l$, and let the numbers written in the adjacent unit squares be $a_{1}, a_{2}, \cdots, a_{52 m-17}$.
For $i=1,2, \cdots, 52$, let $s_{i}=\sum_{t=i(\bmod 52)} a_{t}$, where $s_{1}, s_{2}, \cdots, s_{35}$ each have $m$ terms, and $s_{36}, s_{37}, \cdots, s_{52}$ each have $m-1$ terms. Each napkin intersecting line $l$ contributes exactly 1 to each $s_{i}$.
Thus, the number of napkins intersecting $l$, denoted as $s$, satisfies
$s_{1}=s_{2}=\cdots=s_{52}=s$.
If $s>(m-1) M$, then line $l$ is called "rich"; otherwise, it is called "poor".
Assume $l$ is rich. Then each of $s_{36}, s_{37}, \cdots, s_{52}$ has at least one term greater than $M$. Consider all the unit squares corresponding to these terms, and call them "rich bad squares" for this line. Thus, each rich line has at least 17 unit squares that are rich bad squares for this line.
On the other hand, if $l$ is poor, then
$s \leqslant(m-1) M<m M$.
Thus, each of $s_{1}, s_{2}, \cdots, s_{35}$ has at least one term less than $M$.
Consider all the unit squares corresponding to these terms, and call them "poor bad squares" for this line. Thus, each poor line has at least 35 unit squares that are poor bad squares for this line.
Call the labels "small" if they are congruent to $1,2, \cdots, 35$ modulo 52, and "large" otherwise, i.e., congruent to $36,37, \cdots, 52$ modulo 52.
Considering the column numbers from left to right and the row numbers from bottom to top are $1,2, \cdots, 52 m-17$, a line is called "large" or "small" depending on whether its label is large or small.
By definition, the rich bad squares of a row belong to large columns, and the poor bad squares of a row belong to small columns, and vice versa.
On each line, place a strawberry in each bad square for that line, and additionally, place an extra strawberry in each rich bad square for each "small rich" line. A unit square can independently receive strawberries from its row and column.
Note that a unit square with one strawberry has a number different from $M$. If this unit square receives an extra strawberry, then the number in this unit square is greater than $M$. Thus, it is either in a small row or a large column, and vice versa. Assume it is in a small row, then it is not bad for this column. Thus, in this case, it has no more than two strawberries. If the extra rule does not apply to this unit square, then it also has no more than two strawberries. Thus, the total number of strawberries $N$ is at most $2 g$.
Below, we estimate $N$ using a different method.
For the $2 \times 35 \mathrm{~m}$ small lines, if it is rich, there are at least 34 strawberries; if it is poor, there are at least 35 strawberries. Therefore, in either case, there are at least 34 strawberries.
Similarly, for the $2 \times 17(m-1)$ large lines, there are at least $\min \{17,35\}=17$ strawberries.
Summing over all lines, we get
$$
\begin{array}{l}
2 g \geqslant N \geqslant 2[35 m \times 34+17(m-1) \times 17] \\
=2(1479 m-289)=2 \times 57392, \\
g \geqslant 57392 .
\end{array}
$$
Thus, $g \geqslant 57392$.
|
3986729
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (50 points) Find the smallest prime $p$ that satisfies $(p, N)=1$, where $N$ is the number of all $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ that meet the following conditions:
(1) $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ is a permutation of $0,1, \cdots, 2012$;
(2) For any positive divisor $m$ of 2013 and all $n(0 \leqslant n<n+m \leqslant 2012)$, $m \mid\left(a_{n+m}-a_{n}\right)$.
|
3. From (2), we know that for any $i, j (0 \leqslant i < j \leqslant 2012)$ and $m=3^{\alpha} \times 11^{\beta} \times 61^{\gamma} (\alpha, \beta, \gamma \in \{0,1\})$, we have
$$
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 \leqslant 2012, x \equiv i(\bmod 3)\}(i=0,1,2), \\
A_{i j}=\left\{x \in A_{i} \mid x \equiv j(\bmod 11)\right\}(j=0,1, \cdots, 10), \\
A_{i j k}=\left\{x \in A_{i j} \mid x \equiv k(\bmod 61)\right\}(k=0,1, \cdots, 60).
\end{array}
$$
From (1), we know that for $i=0,1,2$, all $a_{n} (n \in A_{i})$ are congruent modulo 3 (the remainder is denoted as $i^{\prime}$), and there is a one-to-one mapping
$$
f_{1}: A_{i} \rightarrow A_{i^{\prime}}.
$$
Similarly, there are one-to-one mappings
$$
f_{2}: A_{i j} \rightarrow A_{i^{\prime} j^{\prime}}, f_{3}: A_{i j k} \rightarrow A_{i^{\prime} j^{\prime} k^{\prime}}.
$$
Therefore, there are $3! \times 11! \times 61!$ permutations that satisfy the conditions.
Below is a method to construct all such permutations.
Let the bijection $f: (i, j, k) \rightarrow (i^{\prime}, j^{\prime}, k^{\prime})$, where
$$
0 \leqslant i, i^{\prime} \leqslant 2, 0 \leqslant j, j^{\prime} \leqslant 10, 0 \leqslant k, k^{\prime} \leqslant 60.
$$
If $n$ satisfies
$$
\left\{\begin{array}{l}
n \equiv i(\bmod 3), \\
n \equiv j(\bmod 11), \\
n \equiv k(\bmod 61),
\end{array}\right.
$$
then take $a_{n}$ to be the number that satisfies
$$
\left\{\begin{array}{l}
a_{n} \equiv i^{\prime}(\bmod 3), \\
a_{n} \equiv j^{\prime}(\bmod 11), \\
a_{n} \equiv k^{\prime}(\bmod 61).
\end{array}\right.
$$
By the Chinese Remainder Theorem, the solution is unique.
Clearly, $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ satisfies conditions (1) and (2).
Thus, $N=3! \times 11! \times 61!$.
Therefore, $p=67$.
|
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| \mathrm{C}_{t_{p}}^{k}$
$\Rightarrow p \mid \mathrm{C}_{2012}^{k}$
If $p \mid k$, then there are only the following five cases:
$\mathrm{C}_{4 p}^{0}=\mathrm{C}_{s_{p}}^{4_{p}}=1$,
$\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}=\frac{4 p \cdot(4 p-1) \cdots \cdot(3 p+1)}{p \cdot(p-1) \cdots \cdot 1}$
$\equiv 4(\bmod p)$,
$\mathrm{C}_{4 p}^{2}=\frac{4 p \cdot(4 p-1) \cdots \cdot(3 p+1) \cdot 3 p \cdot(3 p-1) \cdots \cdot(2 p+1)}{2 p \cdot(2 p-1) \cdots \cdot(p+1) \cdot p \cdot(p-1) \cdots \cdot 1}$
$\equiv 6(\bmod p)$.
None of these five cases are multiples of $p$.
Second, consider the parity of the combination number.
First, examine the power of 2 in $n$ factorial.
Assume the binary representation of $n$ is
$n=\left(a, a_{r-1} \cdots a_{1} a_{0}\right)_{2}$
$\qquad$
The power of 2 in $n$ factorial is
$\left[\frac{n}{2}\right]+\left[\frac{n}{4}\right]+\cdots+\left[\frac{n}{2^{m}}\right]+\cdots$
$=\left(a_{r} a_{t-1} \cdots a_{2} a_{1}\right)_{2}+\left(a a_{1} a_{t-1} \cdots a_{3} a_{2}\right)_{2}+\cdots+a_{t}$
$\begin{aligned}= & a_{r}\left(2^{\prime}-1\right)+a_{r-1}\left(2^{2-1}-1\right) \\ & a_{1}\left(2^{2}-1\right)+a_{0}\left(2^{0}-1\right)\end{aligned}$
$=n-s(n)$,
where $s(n)=a_{r}+a_{t-1}+\cdots+a_{0}$ represents the sum of binary digits, which is the number of 1s.
If the combination number
$\mathrm{C}_{2012}^{k}=\frac{2012!}{k!\cdot(2012-k)!}$
$=\frac{(k+m)!}{k!\cdot m!}(m=2012-k)$
is odd, i.e., $\square$
the power of 2 in the numerator and denominator is the same
$\Rightarrow k+m-s(k+m)=k-s(k)+m-s(m)$
$\Rightarrow s(k+m)=s(k)+s(m)$,
which means the binary addition of $k+m=2012$ does not carry (each carry means changing a $(10)_{2}$ in a certain position to a $(1)_{2}$ in the previous position, reducing the digit sum by 1).
Notice that $2012=(11111011100)_{2}$, consisting of eight 1s and three 0s.
If $k+m=(11111011100)_{2}$ does not carry, then in the 0 positions, $k$ and $m$ must both be 0 (0=0+0), with only one choice; in the 1 positions, $k$ and $m$ are one 0 and one 1 (1=1+0 and 1=0+1), with two choices. Therefore, there are $2^{8}=256$ cases where the binary addition of $k+m=2012$ does not carry, meaning there are 256 combination numbers that are odd. The remaining $2013-256=1757$ are even.
Third, consider the combination number
$\mathrm{C}_{2012}^{k}=\frac{2012!}{k!\cdot(2012-k)!}=\frac{(k+m)!}{k!\cdot m!}$
is even but not a multiple of 4.
Then the power of 2 in the numerator is exactly one more than in the denominator
$\Rightarrow k+m-s(k+m)=k-s(k)+m-s(m)+1$
$\Rightarrow s(k+m)=s(k)+s(m)-1$,
which means the binary addition of $k+m=2012$ has exactly one carry, and this carry is from a lower position to an adjacent higher position. Since the lower position does not carry to the next higher position, the binary addition at the lower position must be $(1)_{2}+(1)_{2}=(10)_{2}$ (otherwise, it cannot carry to the next position), meaning $k$ and $m$ are both 1 at the lower position. The higher position, receiving the carry from the lower position, does not carry further, which can only be $0+0=0$, becoming 1 after receiving the carry, meaning $k$ and $m$ are both 0 at the higher position.
The binary addition at the two positions is
$(01)_{2}+(01)_{2}=(10)_{2}$.
From the binary representation of the sum
$k+m=2012=(11111011100)$
the carry can only be from the fifth and sixth positions or the ninth and tenth positions from the front. In any case, the two positions of the carry are fixed as $(01)_{2}+(01)_{2}=(10)_{2}$. The remaining positions do not carry, 0 positions can only be $0+0=0$, 1 positions can be $1+0=1$ or $0+1=1$, with two choices. There are $2^{7}=128$ cases for each position, totaling 256 cases. That is, there are 256 combination numbers that are even but not multiples of 4.
Thus, the combination numbers that are multiples of 4 are
$2013-256-256=1501$ (cases).
Finally, consider the cases that are not multiples of $p=503$.
$\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$ is not a multiple of 4.
$\mathrm{C}_{s_{p}}^{s}=\mathrm{C}_{4 p}^{3 p}$ has a power of 2 of
$s(p)+s(3 p)-s(4 p)=s(3 p) \geqslant 2$.
$\mathrm{C}_{s p}^{2}$ has a power of 2 of
$s(2 p)+s(2 p)-s(4 p)$
$=s(p)=s(2012)=8$.
These three combination numbers are multiples of 4 but not multiples of $p=503$.
Therefore, the combination numbers that are multiples of 4 and also multiples of $p=503$ are $1501-3=1498$.
In summary, the combination numbers that are multiples of 2012 are
|
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 {. }
$$
Therefore, the number of positive integer pairs $(m, n)$ that satisfy the condition is
$$
1612-1007+1=606 \text { (pairs). }
$$
|
606
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. Divide the natural numbers from 1 to 30 into two groups, such that the product of all numbers in the first group $A$ is divisible by the product of all numbers in the second group $B$. Then the minimum value of $\frac{A}{B}$ is ( ).
(A) 1077205
(B) 1077207
(C) 1077209
(D) 1077211
|
5. A.
$$
\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}
$$
Let $C=2^{13} \times 3^{7} \times 5^{4} \times 7^{2} \times 11 \times 13 \times 17 \times 19 \times 29$.
Since $B \mid A$, we know $C \mid A$.
Thus $\left(\frac{A}{B}\right)_{\min }=5 \times 17 \times 19 \times 23 \times 29=1077205$.
When $A=C=30 \times 29 \times \cdots \times 19 \times 17 \times 12 \times 2$,
$\frac{A}{B}$ reaches its minimum value.
|
1077205
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given that $P(x)$ is a polynomial with integer coefficients, satisfying $P(17)=10, P(24)=17$. If the equation $P(n)=n+3$ has two distinct integer solutions $n_{1}, n_{2}$, find the value of $n_{1} n_{2}$. ${ }^{[7]}$
(2005, American Invitational Mathematics Examination)
|
【Analysis】From the conditions of the problem, we cannot determine the zeros of the integer-coefficient polynomial $P(x)$. Let's construct another polynomial function $T(x)$ such that 17 and 24 are zeros of $T(x)$, and solve the problem using the zero-product property.
Solution Let $S(x)=P(x)-x-3$. Then
$S(17)=-10, S(24)=-10$.
Let $T(x)=S(x)+10=P(x)-x+7$. Hence, 17 and 24 are zeros of the integer-coefficient polynomial $T(x)$.
Assume $T(x)=(x-17)(x-24) Q(x)$, where $Q(x)$ is an integer-coefficient polynomial.
Let $n$ be an integer root that satisfies the equation $P(n)=n+3$. Then
$$
\begin{array}{l}
S(n)=0, \\
T(n)=(n-17)(n-24) Q(n)=10 .
\end{array}
$$
Therefore, $n-17$ and $n-24$ are factors of 10.
Thus, $\left\{\begin{array}{l}n-17=2, \\ n-24=-5\end{array}\right.$ or $\left\{\begin{array}{l}n-17=5, \\ n-24=-2 .\end{array}\right.$
Solving these, we get $n_{1}=19, n_{2}=22$.
Therefore, $n_{1} n_{2}=19 \times 22=418$.
|
418
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let positive integers $k_{1} \geqslant k_{2} \geqslant \cdots \geqslant k_{n}\left(n \in \mathbf{N}_{+}\right)$, and $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=2012$.
Then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is $\qquad$
|
2. 49 .
Notice that,
$$
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). }
$$
Thus, among the positive integers not greater than 2012, the numbers that are coprime with 2012 are
$$
2012-1008=1004 \text { (numbers). }
$$
Hence, 2013 is in the 1005th position.
Also, $31^{2}<1005<32^{2}$, that is
$$
\begin{array}{l}
1+3+\cdots+(2 \times 31-1)<1005 \\
<1+3+\cdots+(2 \times 32-1) .
\end{array}
$$
Thus, 2013 is in the 32nd group.
|
32
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $n=\sum_{a_{1}=0}^{2} \sum_{a_{2}=0}^{a_{1}} \cdots \sum_{a_{2} 012=0}^{a_{2} 011}\left(\prod_{i=1}^{2012} a_{i}\right)$. Then the remainder when $n$ is divided by 1000 is . $\qquad$
|
6. 191 .
It is evident that from $a_{1}$ to $a_{2012}$ forms a non-increasing sequence, and the maximum element does not exceed 2. Therefore, their product is a power of 2 or 0. Since each power of 2 can only be represented in one way (the sequence being non-increasing), we have
$$
\begin{array}{l}
n=1+2+4+\cdots+2^{2012} \\
=2^{2013}-1 \equiv 191(\bmod 1000) .
\end{array}
$$
|
191
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $X$ be the set of irreducible proper fractions with a denominator of 800, and $Y$ be the set of irreducible proper fractions with a denominator of 900, and let $A=\{x+y \mid x \in X, y \in Y\}$. Find the smallest denominator of the irreducible fractions in $A$.
|
【Analysis】This problem is adapted from the 35th Russian Mathematical Olympiad question ${ }^{[1]}$.
Let $x=\frac{a}{800} \in X, y=\frac{b}{900} \in Y$, where,
$$
\begin{array}{l}
1 \leqslant a \leqslant 799, (a, 800)=1, \\
1 \leqslant b \leqslant 899, (b, 900)=1 .
\end{array}
$$
Then $x+y=\frac{9 a+8 b}{7200}$.
Since $7200=2^{5} \times 3^{2} \times 5^{2}$, and both $a$ and $b$ are odd, $9 a+8 b$ is odd, meaning that the factor $2^{5}$ in the denominator remains after simplification.
Because $(b, 3)=1$, we have $(9 a+8 b, 3)=1$, meaning that the factor $3^{2}$ in the denominator remains after simplification.
Thus, when $9 a+8 b$ is a multiple of 25, the denominator of the simplified fraction $\frac{9 a+8 b}{7200}$ reaches its minimum value of 288.
A specific example can be constructed as follows.
$$
\begin{array}{l}
9 a+8 b \equiv 0(\bmod 25) \\
\Leftrightarrow 8 b \equiv 16 a(\bmod 25) \\
\Leftrightarrow b \equiv 2 a(\bmod 25) .
\end{array}
$$
Taking $a=1, b=77$ works.
|
288
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. ${ }^{[3]}$
(2011, Xin Zhi Cup Shanghai Junior High School Mathematics Competition)
|
【Analysis】According to the problem, we set
$$
\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}
$$
By symmetry, without loss of generality, let $c=5, d=7, a \leqslant b$.
We only need to find $(a+b)_{\text {min }}$.
Also, $a+b \geqslant 2 \sqrt{a b}=2 \sqrt{k+17}$
$\geqslant 2 \sqrt{18}>2 \sqrt{16}=8$.
Thus, $a+b>8$, i.e., $a+b \geqslant 9$.
If $a+b=9$, then $a=2, b=7$.
This contradicts $a b=k+17>17$.
Therefore, $a+b \geqslant 10$.
When $k=4$, $a=3, b=7$, the equality holds.
Thus, $(a+b+c+d)_{\min }=22$.
|
22
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, on a plane there are $n(n \geqslant 4)$ lines. For lines $a$ and $b$, among the remaining $n-2$ lines, if at least two lines intersect with both lines $a$ and $b$, then lines $a$ and $b$ are called a "congruent line pair"; otherwise, they are called a "separated line pair". If the number of congruent line pairs among the $n$ lines is 2012 more than the number of separated line pairs, find the minimum possible value of $n$ (the order of the lines in a pair does not matter).
|
(1) Among these $n$ lines, if there exist four lines that are pairwise non-parallel, then any two lines are coincident line pairs. However, $\mathrm{C}_{n}^{2}=2012$ has no integer solution, so there does not exist an $n$ that satisfies the condition.
(2) If the $n$ lines have only three different directions, let the number of lines in these three directions be $a$, $b$, and $c$ respectively. Assume $a \geqslant b \geqslant c$.
(i) When $a \geqslant 2, b=c=1$,
$$
\mathrm{C}_{a}^{2}+1-2 a=2012,
$$
there is no $n$ that satisfies the condition.
(ii) When $a, b \geqslant 2, c=1$,
$$
\mathrm{C}_{a}^{2}+\mathrm{C}_{b}^{2}-a b+b+a=2012,
$$
which simplifies to $(a-b)^{2}+a+b=4024$.
Let $a-b=k$. Then $a+b=4024-k^{2}$.
Thus, $a=\frac{4024+k-k^{2}}{2}, b=\frac{4024-k-k^{2}}{2}$.
Since $b \geqslant 2$, it is easy to see that $k \leqslant 62$.
Therefore, the minimum value of $a+b$ is 180.
Thus, the minimum value of $n$ is 181.
(iii) When $a, b, c \geqslant 2$,
$$
\mathrm{C}_{a}^{2}+\mathrm{C}_{b}^{2}+\mathrm{C}_{c}^{2}+a b+b c+c a=2012,
$$
which simplifies to $(a+b+c)(a+b+c-1)=4024$,
and there is no integer solution.
(3) If the $n$ lines have only two different directions, let the number of lines in these two directions be $a$ and $b$ respectively. Clearly, $a, b \geqslant 2$.
Then $\mathrm{C}_{a}^{2}+\mathrm{C}_{b}^{2}-a b=2012$, which simplifies to
$$
(a-b)^{2}-(a+b)=4024.
$$
Let $a-b=k$. Then $a+b=k^{2}-4024$.
Thus, $a=\frac{k+k^{2}-4024}{2}, b=\frac{k^{2}-k-4024}{2}$.
Since $b \geqslant 2$, it is easy to see that $k \geqslant 64$.
Therefore, the minimum value of $a+b$ is 72.
Thus, the minimum value of $n$ is 72.
In summary, the minimum value of $n$ is 72.
|
72
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Given positive integers $a, b$ satisfy that $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$.
保留源文本的换行和格式，直接输出翻译结果。
|
Given the problem, let's set
$a-b=p(p$ is a prime number $), ab=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $a(a-p)=k^{2} \Rightarrow a^{2}-k^{2}=ap$
$$
\Rightarrow(a+k)(a-k)=ap \text {. }
$$
Since $a+k, a, p$ are all positive integers, we have
$$
a-k>0 \text {. }
$$
Given that $p$ is a prime number, from equation (1) we know $p \mid(a+k)$ or $p \mid(a-k)$.
Clearly, $a-k \Rightarrow p \nmid(a-k) \Rightarrow p \mid(a+k)$. Let $a+k=np\left(n \in \mathbf{N}_{+}\right)$.
From $a-kp \Rightarrow n \geqslant 2 \Rightarrow k=np-a \text {. }$
$$
Substituting into equation (1) we get
$$
\begin{array}{l}
n(2a-np)=a \\
\Rightarrow a(2n-1)=n^{2}p .
\end{array}
$$
It is easy to prove that $n^{2}$ and $2n-1$ are coprime.
Thus, $n^{2} \mid a$. Let $a=mn^{2}\left(m \in \mathbf{N}_{+}\right)$.
From equation (2) we get $m(2n-1)=p$.
Since $p$ is a prime number, either $m$ or $2n-1$ must be 1.
Also, $2n-1 \geqslant 2 \times 2-1>1$, so $m=1$.
Therefore, $a=n^{2}, p=2n-1$.
Hence, $a$ meets the condition $\Leftrightarrow a=n^{2}$ and $2n-1$ is a prime number.
Given $a \geqslant 2012$, then $a \geqslant 45^{2}$.
Also, $2 \times 45-1=89$ is a prime number, thus,
$$
a_{\min }=2025
$$
|
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 digits, add several "0"s in front of the highest digit to make it exactly four digits), and use $f(k)$ to denote the number of hexagonal numbers in $M$ whose last digit is $k$, where $k \in\{0,2,4,6,8\}$.
For $n \in \mathbf{N}$, the number of pairs $(x, y)$ that satisfy $x+y=n$, and
$$
x, y \in\{0,1, \cdots, 9\}
$$
is denoted as $p_{n}$.
Obviously, $p_{n}=\left\{\begin{array}{ll}n+1, & n=0,1, \cdots, 9 ; \\ 19-n, & n=10,11, \cdots, 18 ; \\ 0, & n \geqslant 19 .\end{array}\right.$
First, consider all hexagonal numbers $\overline{a b c k}$ less than 2000.
If $k=0$, then when $a=0$, $b+c=0,6,12,18$; when $a=1$, $b+c=5,11,17$.
$$
\begin{array}{l}
\text { Hence } f(0)=\left(p_{0}+p_{6}+p_{12}+p_{18}\right)+\left(p_{5}+p_{11}+p_{17}\right) \\
=16+16=32 .
\end{array}
$$
If $k=2$, then when $a=0$, $b+c=4,10,16$; when $a=1$, $b+c=3,9,15$.
$$
\begin{array}{l}
\text { Hence } f(2)=\left(p_{4}+p_{10}+p_{16}\right)+\left(p_{3}+p_{9}+p_{15}\right) \\
=17+18=35 .
\end{array}
$$
If $k=4$, then when $a=0$, $b+c=2,8,14$; when $a=1$, $b+c=1,7,13$.
Hence $f(4)=\left(p_{2}+p_{8}+p_{14}\right)+\left(p_{1}+p_{7}+p_{13}\right)$
$$
=17+16=33 \text {. }
$$
When $k=6,8$, similar to the cases of $k=0,2$, we have
$$
f(6)=f(0)=32, f(8)=f(2)=35 \text {. }
$$
Therefore, the number of hexagonal numbers less than 2000 is
$$
f(0)+f(2)+f(4)+f(6)+f(8)=167 \text { (numbers). }
$$
Notice that, from 2000 to 2011, there is exactly one hexagonal number 2004. Thus, the total number of hexagonal numbers is
$$
167+1=168 \text {. }
$$
Solution 2 For a non-negative integer $n$, let $S(n)$ be the sum of its digits.
First, pair all multiples of 6 (a total of 334) among the non-negative integers less than 2000 into the following 167 pairs:
$$
\begin{array}{l}
(0,1998),(6,1992),(12,1986), \\
\cdots,(996,1002) . \\
\text { For each pair of numbers }(x, y), \text { let } \\
x=\overline{a_{1} a_{2} a_{3} a_{4}, y=b_{1} b_{2} b_{3} b_{4}}
\end{array}
$$
(Agree that when $x$ or $y$ is less than four digits, add several "0"s in front of the highest digit to make it exactly four digits). Then
$$
\begin{array}{l}
1000\left(a_{1}+b_{1}\right)+100\left(a_{2}+b_{2}\right)+ \\
10\left(a_{3}+b_{3}\right)+\left(a_{4}+b_{4}\right) \\
=x+y=1998 .
\end{array}
$$
Since $x, y$ are even, $a_{4}, b_{4} \leqslant 8$.
Therefore, $a_{4}+b_{4} \leqslant 16<18$, so $a_{4}+b_{4}=8$.
Also, since $a_{3}+b_{3} \leqslant 18<19$, we have $a_{3}+b_{3}=9$.
Similarly, $a_{2}+b_{2}=9$.
Finally, we must have $a_{1}+b_{1}=1$.
Thus, $S(x)+S(y)$
$$
\begin{array}{l}
=\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\left(a_{3}+b_{3}\right)+\left(a_{4}+b_{4}\right) \\
=1+9+9+8=27 .
\end{array}
$$
Therefore, one and only one of $S(x)$ and $S(y)$ is a multiple of 6 (because $x, y$ are both divisible by 3, so $S(x)$ and $S(y)$ are both divisible by 3). Hence, one and only one of $x, y$ is a hexagonal number.
So, the number of hexagonal numbers less than 2000 is 167.
Also, from 2000 to 2011, there is exactly one hexagonal number 2004, so the total number of hexagonal numbers is $167+1=168$.
|
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) and (2), when $n \geqslant 4024$,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}}\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$
In summary, the smallest positive integer $n$ that satisfies the condition is
4024.
|
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 (1) and (3), we get
$$
\frac{1}{a} \geqslant 3 \sqrt{3}.
$$
And $3\left(v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}\right) \leqslant\left(v_{1}+v_{2}+v_{3}\right)^{2}$
$$
\Rightarrow 3 \cdot \frac{b}{a} \leqslant \frac{1}{a^{2}} \Rightarrow 3 a b \leqslant 1.
$$
Thus, $P=\frac{5 a^{2}-6 a b+3}{a^{3}(b-a)} \geqslant \frac{5 a^{2}+1}{a^{3}(b-a)}$.
Also, $\left(v_{2}+v_{3}-v_{1}\right)\left(v_{3}+v_{1}-v_{2}\right)\left(v_{1}+v_{2}-v_{3}\right)$
$\leqslant v_{1} v_{2} v_{3}$
$\Rightarrow\left(\frac{1}{a}-2 v_{1}\right)\left(\frac{1}{a}-2 v_{2}\right)\left(\frac{1}{a}-2 v_{3}\right) \leqslant \frac{1}{a}$
$\Rightarrow 9 a^{2}-4 a b+1 \geqslant 0$
$\Rightarrow 5 a^{2}+1 \geqslant 4 a(b-a)$.
Then $P \geqslant \frac{4 a(b-a)}{a^{3}(b-a)}=\frac{4}{a^{2}} \geqslant 108$.
Therefore, when $a=\frac{\sqrt{3}}{9}, b=\sqrt{3}$, $P$ achieves its minimum value of 108.
|
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 minimized, so $BM+MN$ also achieves its minimum value.
It is easy to see that $\angle BAC=\angle PAC, \angle BAC=\angle ACP$.
Thus, $\angle PAC=\angle ACP$. Therefore, $PA=PC$.
Let $PA=PC=x$. Then $PD=20-x$.
In the right triangle $\triangle ADP$, by $AD^2+PD^2=AP^2$, we get $10^2+(20-x)^2=x^2$.
Solving for $x$ yields $x=12.5$.
Thus, $PD=7.5$.
It is easy to see that $\triangle ADP \sim \triangle B'AN$.
By $\frac{B'N}{AD}=\frac{AN}{PD}=\frac{AB'}{AP}=\frac{AB}{AP}$, we get $B'N=16$, which is the minimum value of $BM+MN$. At this time, $AN=12$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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