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3. Let $M=\{1,2, \cdots, 2009\}$. If $n \in M$, such that $S_{n}=\frac{1}{n}\left(1^{3}+2^{3}+\cdots+n^{3}\right)$ is a perfect square, then the number of such $n$ is $\qquad$.
|
3. 44 .
Notice that
$$
S_{n}=\frac{1}{n}\left[\frac{n(n+1)}{2}\right]^{2}=\frac{n(n+1)^{2}}{4} \text {. }
$$
Since $n$ and $n+1$ are coprime, if $S_{n}$ is a perfect square, then $n$ must be a perfect square. When $n$ is an even perfect square, $\frac{n}{4}$ is a perfect square integer; when $n$ is an odd perfect square, $\frac{(n+1)^{2}}{4}$ is a perfect square integer. Therefore, $n$ can take all the perfect squares in $M$, and there are 44 such numbers.
|
44
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Arrange 4 identical red balls and 4 identical blue balls in a row, with the positions numbered $1,2, \cdots$, 8 from left to right. If balls of the same color are indistinguishable, the number of arrangements where the sum of the positions of the 4 red balls is less than the sum of the positions of the 4 blue balls is $\qquad$.
|
8.31.
The sum of 8 numbers is 36. Dividing the numbers $1,2, \cdots, 8$ into two groups, there are $\frac{C_{8}^{4}}{2}=35$ ways to do so. When the sums of the two groups are not equal, the smaller sum of 4 numbers is considered the sequence of red balls, and the larger sum of 4 numbers is considered the sequence of blue balls, corresponding to a valid arrangement. Therefore, we only need to exclude the cases where the sums of the two groups are equal to 18.
Let $a_{i}(i=1,2,3,4)$ represent the 4 selected numbers, and $a_{i} \in\{1,2, \cdots, 8\}$. Without loss of generality, assume
$a_{1}18 .
$$
Similarly, $a_{2} \leqslant 4$.
Thus, we can write down 8 solutions to the equation:
$$
\begin{array}{l}
(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) .
\end{array}
$$
Therefore, the number of valid arrangements is $\frac{\mathrm{C}_{8}^{4}}{2}-\frac{8}{2}=31$.
|
31
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Let $a_{1}, a_{2}, \cdots, a_{20}$ be 20 distinct positive integers, and the set $\left\{a_{i}+a_{j} \| \leqslant i, j \leqslant 20\right\}$ contains 201 distinct elements. Find the minimum possible number of distinct elements in the set
$$
\left.|| a_{i}-a_{j}|| 1 \leqslant i, j \leqslant 20\right\}.
$$
|
The minimum number of elements in the given set is
100.
Example: Let
$$
\begin{array}{l}
a_{i}=10^{11}+10^{i}, \\
a_{10+i}=10^{11}-10^{i}(i=1,2, \cdots, 10) .
\end{array}
$$
Then $\left\{a_{i}+a_{j} \mid 1 \leqslant i \leqslant j \leqslant 20\right\}$ contains
$$
(20+19+\cdots+1)-10+1=201
$$
different elements.
$$
\begin{array}{l}
\text { And }\left\{\left|a_{i}-a_{j}\right| 11 \leqslant i, j \leqslant 20\right\} \\
=\left\{2 \times 10^{i} \mid i=1,2, \cdots, 10\right\} \cup \\
\left\{110^{i} \pm 10^{j} 111 \leqslant i, j \leqslant 20\right\}
$$
Consider the pairs $(b, c)$ (a total of 190 pairs) in $S$ such that $b > c$. By the assumption, there are at most 99 different differences, so there must be at least 91 pairs $(b, c)$ such that there exist $b^{\prime}, c^{\prime} \in S$ satisfying $b^{\prime}<b, c^{\prime}<c$, and $b-c=b^{\prime}-c^{\prime}$. For these 91 pairs $(b, c)$, they and their corresponding $b^{\prime}, c^{\prime}$ form a four-element subset $\left\{b, c, b^{\prime}, c^{\prime}\right\}$ of $S$, which can yield a good subset $\{x, y, z, w\}$ of $S$, and at most two pairs $(b, c)$ form the same subset $\{x, y, z, w\}$ (only $(b, c)=(w, z)$ or $(w, y))$. Therefore, $S$ has at least 46 good subsets.
On the other hand, the number of good subsets $\{x, y, z, w\}$ of $S$ is equal to $\sum \frac{1}{2} s_{i}\left(s_{i}-1\right)$, where $s_{i}$ is the number of pairs $(b, c)$ in $S$ such that $b+c=i\left(i \in \mathbf{N}_{+}\right)$ and $b \leqslant c$.
Note that for each $i$, each element $s$ in $S$ appears in at most one of the pairs $(b, c)$ (in fact, when $s \leqslant i-s$, $s$ appears in the pair $(s, i-s)$, and in other cases, it appears in $(i-s, s)$), so $s_{i} \leqslant 10$. Therefore, when $s_{i} \neq 0$, $1 \leqslant s_{i} \leqslant 10$.
Thus, $\frac{1}{2} s_{i}\left(s_{i}-1\right) \leqslant 5 s_{i}-5$.
Since the set $\left.\left|a_{i}+a_{j}\right| 1 \leqslant i, j \leqslant 20\right\}$ contains 201 different elements, there are 201 positive integers $i$ such that $s_{i} \geqslant 1$. Let $T$ be the set of such $i$. Using the fact that $S$ has $\mathrm{C}_{20}^{2}$ pairs $(b, c)$ such that $b<c$, and 20 pairs $(b, c)$ such that $b=c$, we have $\sum_{i \in T} s_{i}=\mathrm{C}_{20}^{2}+20=210$. Then
$$
\begin{array}{l}
\sum_{i \in T} \frac{1}{2} s_{i}\left(s_{i}-1\right) \leqslant \sum_{i \in T}\left(5 s_{i}-5\right) \\
=5(210-201)=45 .
\end{array}
$$
This contradicts the previous conclusion that $S$ has at least 46 good subsets.
Therefore, the given set must have at least 100 different elements.
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $a_{1}=1, a_{2}=2$, for $n \geqslant 2$ we have
$$
a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} .
$$
If for all positive integers $n \geqslant m$, we have $a_{n}>2+$ $\frac{2008}{2009}$, then the smallest positive integer $m$ is $\qquad$ .
|
8.4019 .
According to the problem, the recursive relationship becomes
$$
\begin{array}{l}
a_{n+1}-a_{n}=\frac{n-1}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
=\frac{n-1}{n+1}\left(a_{n}-a_{n-1}\right)(a \geqslant 2) .
\end{array}
$$
Therefore, when $n \geqslant 3$,
$$
\begin{array}{l}
a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdot \frac{2}{4} \cdot \frac{1}{3}\left(a_{2}-a_{1}\right) \\
=\frac{2}{n(n-1)}=\frac{2}{n-1}-\frac{2}{n} .
\end{array}
$$
Thus, $a_{n}-a_{2}=1-\frac{2}{n}=\frac{n-2}{n}$.
Hence, $a_{n}=a_{2}+\frac{n-2}{n}=\frac{3 n-2}{n}(n \geqslant 3)$. Let $\frac{3 n-2}{n}>2+\frac{2008}{2009}$, solving gives $n>4018$.
Therefore, the minimum value of $m$ is 4019.
|
4019
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (15 points) If all the coefficients of a polynomial are natural numbers, it is called a "natural polynomial". How many different natural polynomials $P(x)$ are there such that $P(2)=20$?
|
10. For a positive integer $n$, let $A(n)$ denote the number of distinct natural polynomials $P(x)$ that satisfy $P(2) = n$.
It is easy to prove: for any positive integer $m$, we have
$$
\begin{array}{l}
A(2 m+1)=A(2 m) \\
=A(2 m-1)+A(m) .
\end{array}
$$
In fact, for any natural polynomial $P(x)$ that satisfies $P(2)=2 m+1$, since $P(2)$ is odd, the constant term of $P(x)$ must be odd.
Let $Q(x)=P(x)-1$. Then $Q(x)$ is a natural polynomial, and $Q(2)=P(2)-1=2 m$.
Conversely, for any natural polynomial $Q(x)$ that satisfies $Q(2)=2 m$, let $P(x)=Q(x)+1$. Then $P(x)$ is a natural polynomial, and
$$
P(2)=Q(2)+1=2 m+1 \text {. }
$$
Therefore, $A(2 m+1)=A(2 m)$.
For any natural polynomial $P(x)$ that satisfies $P(2)=2 m$, if $P(0) \neq 0$, let $Q(x)=P(x)-1$. Then $Q(x)$ is a natural polynomial, and
$$
Q(2)=P(2)-1=2 m-1 \text {. }
$$
Such polynomials $P(x)$ have $A(2 m-1)$.
If $P(0)=0$, let $P(x)=x Q(x)$, then $Q(x)$ is a natural polynomial, and $2 m=P(2)=2 Q(2)$. Therefore, $Q(2)=m$, and such polynomials $P(x)$ have $A(m)$.
Thus, $A(2 m)=A(2 m-1)+A(m)$.
Therefore, equation (1) holds.
Clearly, $A(2)=2$ (i.e., $P(x)=2, x$), so,
$$
\begin{array}{l}
A(3)=A(2)=2, \\
A(5)=A(4)=2 A(2)=4, \\
A(7)=A(6)=A(3)+A(4)=2+4=6, \\
A(9)=A(8)=A(4)+A(6)=4+6=10, \\
A(11)=A(10)=A(5)+A(8)=4+10=14, \\
A(13)=A(12)=A(6)+A(10)=6+14=20, \\
A(15)=A(14)=A(7)+A(12)=6+20=26, \\
A(17)=A(16)=A(8)+A(14)=10+26=36, \\
A(19)=A(18)=A(9)+A(16)=10+36=46, \\
A(20)=A(10)+A(18)=14+46=60 .
\end{array}
$$
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If the three-digit number $\overline{a b c}$ satisfies $1 \leqslant a \leqslant b \leqslant c \leqslant 9$, then $\overline{a b c}$ is called an "uphill number". Then, the number of uphill numbers is $\qquad$
|
5. 165 .
For a fixed $b$, the number $c$ can take $10-b$ values, namely $b, b+1, \cdots, 9$; and for a fixed number $a$, the number $b$ can take $a, a+1, \cdots, 9$. Therefore, the number of three-digit increasing numbers is
$$
\begin{array}{l}
\sum_{a=1}^{9} \sum_{b=a}^{9}(10-b)=\sum_{a=1}^{9} \sum_{k=1}^{10-a} k \\
=\sum_{a=1}^{9} \frac{(10-a)(11-a)}{2} \\
=\frac{1}{2} \sum_{a=1}^{9}\left(a^{2}-21 a+110\right) \\
=\frac{1}{2}\left(\frac{9 \times 10 \times 19}{6}-21 \times \frac{9 \times 10}{2}+990\right)=165 .
\end{array}
$$
|
165
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Sure, here is the translation:
---
II. (50 points) Let $f(x)$ be a function defined on the set of natural numbers $\mathbf{N}$ and taking values in $\mathbf{N}$, satisfying: for any two distinct natural numbers $a$ and $b$,
$$
f(a) + f(b) - f(a + b) = 2009.
$$
(1) Find $f(0)$;
(2) Let $a_{1}, a_{2}, \cdots, a_{100}$ be 100 distinct natural numbers, find
$$
\begin{array}{l}
f\left(a_{1}\right) + f\left(a_{2}\right) + \cdots + f\left(a_{100}\right) - \\
f\left(a_{1} + a_{2} + \cdots + a_{100}\right);
\end{array}
$$
(3) Does there exist a function $f(x)$ that satisfies the given conditions and such that $f(2009) = 2009^2$? Prove your conclusion.
---
|
(1) Take $a=0, b=1$, then $a \neq b$.
By the condition, we have
$$
f(0)+f(1)-f(0+1)=2.009 \text {. }
$$
So, $f(0)=2009$.
(2) Proof: For any natural number $n$ greater than 1, and $n$ distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have
$$
\begin{array}{l}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{n}\right)-f\left(a_{1}+a_{2}+\cdots+a_{n}\right) \\
=2009 n-2009 .
\end{array}
$$
We use induction on $n$.
When $n=2$, the conclusion is obviously true.
Assume the conclusion holds for $n=k(k \geqslant 2)$.
Then, for $n=k+1$, let the $k+1$ distinct natural numbers be $a_{1}, a_{2}, \cdots, a_{k+1}$, where $a_{1}<a_{2}<\cdots<a_{k+1}$. By the induction hypothesis, we have
$$
\begin{array}{c}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{k+1}\right)-f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \\
=2009 k-2009+f\left(a_{2}+a_{3}+\cdots+a_{k+1}\right)+ \\
f\left(a_{1}\right)-f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) .
\end{array}
$$
Since $0 \leqslant a_{1}<a_{2}<\cdots<a_{k+1}$, we have
$$
a_{1}<a_{2}+a_{3}+\cdots+a_{k+1} \text {. }
$$
Thus, $a_{1}$ and $a_{2}+a_{3}+\cdots+a_{k+1}$ are distinct.
$$
\begin{array}{l}
\text { Then } f\left(a_{2}+a_{3}+\cdots+a_{k+1}\right)+f\left(a_{1}\right) \\
=f\left(a_{1}+a_{2}+a_{3}+\cdots+a_{k+1}\right)+2009 \text {. } \\
\text { Hence } f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{k+1}\right)- \\
f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \\
=2009 k-2009+f\left(a_{2}+a_{3}+\cdots+a_{k+1}\right)+ \\
f\left(a_{1}\right)-f\left(a_{1}+a_{2}+\cdots+a_{k+1}\right) \\
=2009 k-2009+2009 \\
=2009(k+1)-2009 \text {. } \\
\end{array}
$$
Therefore, the conclusion holds for $n=k+1$.
So, the conclusion is true.
Taking $n=100$, we get
$$
\begin{array}{l}
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{10}\right)-f\left(a_{1}+a_{2}+\cdots+a_{10}\right) \\
=200900-2009=198891 . \\
\text { (3) Let } f(x)=2008 x+2009 .
\end{array}
$$
We now prove that the function $f(x)$ defined this way satisfies the condition.
Indeed, for any two distinct natural numbers $a$ and $b$, we have
$$
\begin{array}{l}
f(a)+f(b)-f(a+b) \\
=(2008 a+2009)+(2008 b+2009)- \\
(2008 a+2008 b+2009) \\
= 2008 a+2008 b-(a+b)+2009 \\
= 2009, \\
\text { Hence } f(2009)=2008 \times 2009+2009 \\
= 2009^{2} .
\end{array}
$$
|
198891
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The afternoon class schedule needs to arrange 5 classes: Physics, Chemistry, Biology, and two self-study periods. If the first class cannot be Biology, and the last class cannot be Physics, then, the number of different ways to arrange the schedule is $\qquad$ kinds.
|
2.39.
According to the principle of inclusion-exclusion, the number of different ways to schedule classes is $\frac{5!}{2}-2 \times \frac{4!}{2}+\frac{3!}{2}=39$.
|
39
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6.50 The sum of 50 positive numbers is 231, and the sum of their squares is 2009. Then, the maximum value of the largest number among these 50 numbers is $\qquad$ .
|
6. 35 .
Let the maximum number be $x$, and the other 49 numbers be $a_{1}, a_{2}$, $\cdots, a_{49}$, and $x \geqslant a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{49}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{49} a_{i}=231-x, \\
2009=x^{2}+\sum_{i=1}^{49} a_{i}^{2} \geqslant x^{2}+\frac{1}{49}\left(\sum_{i=1}^{49} a_{i}\right)^{2} \\
=x^{2}+\frac{1}{49}(231-x)^{2},
\end{array}
$$
i.e., $50 x^{2}-462 x+231^{2} \leqslant 49 \times 2009$.
Solving this, we get $x \leqslant 35$.
Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{49}=4$, in which case $x=35$.
|
35
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $n \in \mathbf{N}$, and $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$ is a positive integer, then $n=$ $\qquad$
|
6.5.
Notice that $\sqrt{n^{2}+24}-\sqrt{n^{2}-9}$
$$
=\frac{33}{\sqrt{n^{2}+24}+\sqrt{n^{2}-9}},
$$
thus $\sqrt{n^{2}+24}+\sqrt{n^{2}-9}$ could be $1,3,11,33$. Therefore, the solution is $n=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. For a real number $x$, $[x]$ denotes the greatest integer not exceeding the real number $x$. For some integer $k$, there are exactly 2008 positive integers $n_{1}, n_{2}, \cdots, n_{2008}$, satisfying
$$
k=\left[\sqrt[3]{n_{1}}\right]=\left[\sqrt[3]{n_{2}}\right]=\cdots=\left[\sqrt[3]{n_{2008}}\right],
$$
and $k \mid n_{i}(i=1,2, \cdots, 2008)$. Then $k=$ $\qquad$
|
7.668.
If $\sqrt[3]{n}-1n \text {. }$
Also, if $k \mid n$, then $n$ can take $k^{3}, k^{3}+k, \cdots, k^{3}+$ $3 k^{2}+3 k$, a total of $3 k+4$.
Therefore, $3 k+4=2008 \Rightarrow k=668$.
|
668
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a, b \in \mathbf{N}_{+}$, when $a^{2}+b^{2}$ is divided by $a+b$, the quotient is $q$, and the remainder is $r$. Then the number of pairs $(a, b)$ that satisfy $q^{2}+r=2009$ is $\qquad$ pairs.
|
-1.0 .
From $2\left(a^{2}+b^{2}\right) \geqslant(a+b)^{2}$, we get $\frac{a^{2}+b^{2}}{a+b} \geqslant \frac{a+b}{2}$.
Then $q+1>q+\frac{r}{a+b}=\frac{a^{2}+b^{2}}{a+b}$ $\geqslant \frac{a+b}{2} \geqslant \frac{r+1}{2}$.
Thus, $r<2 q+1$.
From $q^{2}+r=2009<(q+1)^{2}$, we get $q \geqslant 44$. From $q^{2}+r=2009 \geqslant q^{2}$, we get $q \leqslant 44$. Therefore, $q=44 \Rightarrow r=2009-44^{2}=73$.
Thus, $a^{2}+b^{2}=44(a+b)+73$, which means $(a-22)^{2}+(b-22)^{2}=1041$.
Therefore, the equation has no positive integer solutions.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $a, b, c, d \in \mathbf{Z}$,
$$
f(x)=a x^{3}+b x^{2}+c x+d \text {. }
$$
Then among $A(1,1)$, $B(2009,1)$, $C(-6020,2)$, $D(2,-6020)$, at most $\qquad$ of them can be on the curve $y=f(x)$.
|
4.3.
If $A$ and $C$ are both on $y=f(x)$, then $f(1)=1, f(-6020)=2$.
It is easy to see that $\left(x_{1}-x_{2}\right) \mid \left(f\left(x_{1}\right)-f\left(x_{2}\right)\right)$. Therefore, $[1-(-6020)] \mid (f(1)-f(-6020))=-1$, which is a contradiction.
Thus, $A$ and $C$ cannot both be on $y=f(x)$.
Similarly, $B$ and $C$ cannot both be on $y=f(x)$.
Therefore, at most two points $(A, B)$ out of $A, B, C$ can be on $y=f(x)$.
In this case, $(x-1)(x-2009) \mid (f(x)-1)$.
Let $f(x)=a(x-1)(x-2009)(x-t)+1$.
Substituting $D(2,-6020)$ into equation (1) yields
$a(2-t)=3$.
Thus, when $a=1, t=-1$, $A, B, D$ are on the $y=f(x)$ defined by equation (1).
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $M=\{1,2, \cdots, 17\}$. If there are four distinct numbers $a, b, c, d \in M$, such that $a+b \equiv c+d(\bmod 17)$, then $\{a, b\}$ and $\{c, d\}$ are called a “balanced pair” of set $M$. The number of balanced pairs in set $M$ is $\qquad$.
|
8.476.
Divide the circumference into 17 equal parts, and label the points in clockwise order as $A_{1}, A_{2}, \cdots, A_{17}$. Then
$$
\begin{array}{l}
m+n \equiv k+l(\bmod 17) \\
\Leftrightarrow A_{m} A_{n} / / A_{k} A_{l} .
\end{array}
$$
Note that none of the chords connecting any pair of the 17 equally divided points are diameters, so there are 17 directions for all the chords (each parallel to the tangent at $A_{i}(i=1,2, \cdots, 17)$).
There are 8 chords parallel to the tangent at $A_{i}$, forming $C_{8}^{2}=28$ pairs of parallel chords. Considering all 17 directions, we get $28 \times 17=476$ pairs of parallel chords. Therefore, there are 476 balanced pairs in $M$.
|
476
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) Find the number of positive integers $t$ not exceeding 2009, such that for all natural numbers $n$, $\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}$ is coprime with 2009.
---
The translation maintains the original text's line breaks and formatting.
|
$$
\begin{array}{l}
\text { Three, let } \\
\begin{array}{l}
x= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1} t^{k}=\frac{1}{\sqrt{t}} \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}(\sqrt{t})^{2 k+1}, \\
\begin{array}{l}
y= \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k} t^{k}=\sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k}(\sqrt{t})^{2 k} . \\
\text { Then } \sqrt{t} x+y=\sum_{k=0}^{2 n+1} \mathrm{C}_{2 n+1}^{k}(\sqrt{t})^{k} \\
=(\sqrt{t}+1)^{2 n+1}, \\
\sqrt{t} x-y=-\sum_{k=1}^{2 n+1} \mathrm{C}_{2 n+1}^{k}(-\sqrt{t})^{k} \\
=(\sqrt{t}-1)^{2 n+1} . \\
\text { (1) }+(2)
\end{array} \\
x=\frac{1}{2 \sqrt{t}}\left[(\sqrt{t}+1)^{2 n+1}+(\sqrt{t}-1)^{2 n+1}\right]=x_{n} .
\end{array}
\end{array}
$$
Therefore, $x_{0}=1, x_{1}=3+t$,
$$
\begin{aligned}
x_{n+2}= & {\left[(\sqrt{t}+1)^{2}+(\sqrt{t}-1)^{2}\right] x_{n+1}-} \\
& (\sqrt{t}+1)^{2}(\sqrt{t}-1)^{2} x_{n},
\end{aligned}
$$
which simplifies to $x_{n+2}=2(t+1) x_{n+1}-(t-1)^{2} x_{n}$.
Multiplying (1) and (2) gives
$$
t x^{2}-y^{2}=(t-1)^{2 n+1} \text {. }
$$
Given $2009=7^{2} \times 41$, we have
$$
(x, 2009)=1 \Leftrightarrow 7 \nmid x, 41 \nmid x \text {. }
$$
(1) $\times$ (2) yields
$$
t x^{2}-y^{2}=(t-1)^{2 n+1} \text {. }
$$
(1) $7 \nmid x$. We will discuss this under modulo 7.
If $t \equiv 0$, then $x_{n} \equiv \mathrm{C}_{2 n+1}^{1}=2 n+1$, and $x_{3} \equiv 0$. This is a contradiction.
If $t \equiv 1$, then $x_{n} \equiv \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}=2^{2 n} \not\equiv 0$, which satisfies the condition.
If $t \equiv 4$, then $x_{1} \equiv 0$, which is a contradiction.
From equation (3), if $t \equiv 2,3,5,6$, then $\{x_{n}\}$ modulo 7 is a periodic sequence, and for all positive integers $n, x_{n} \neq 0$, which satisfies the condition.
Thus, the required $t \equiv 1,2,3,5,6 \pmod{7}$.
(2) $41 \nmid x$. We will discuss this under modulo 41.
If $t \equiv 0$, then $x_{n} \equiv \mathrm{C}_{2 n+1}^{1}=2 n+1$, and $x_{21} \equiv 0$, which is a contradiction.
If $t \equiv 1$, then $x_{n} \equiv \sum_{k=0}^{n} \mathrm{C}_{2 n+1}^{2 k+1}=2^{2 n} \not\equiv 0$, which satisfies the condition.
If $t \equiv -1$, then $x_{0}=1, x_{1} \equiv 2, x_{n+2} \equiv -4 x_{n}$, which satisfies the condition.
If $t \equiv 10, -20$, then $x_{2}=t^{2}+10 t+5 \equiv t^{2}+10 t-200=(t-10)(t+20) \equiv 0$, which is a contradiction.
$$
\begin{array}{l}
\text { If } t \equiv \pm 3, \pm 17, \pm 19, 6, 11, -7, -15, \\
x_{11}=\sum_{k=0}^{10} \mathrm{C}_{2 k}^{2 k+1} t^{k} \\
\equiv t^{10}+5 t^{9}-t^{8}-20 t^{7}+7 t^{6}-7 t^{5}+t^{4}+ \\
4 t^{3}+13 t^{2}+18 t-20 \\
\equiv\left(t^{2}-3^{2}\right)\left(t^{2}-17^{2}\right)\left(t^{2}-19^{2}\right)(t-6) . \\
(t-11)(t+7)(t+15) \\
\equiv 0,
\end{array}
$$
which is a contradiction.
When $t-1=s \equiv \pm 3, \pm 6, \pm 7, \pm 11, \pm 12, \pm 13, \pm 14, \pm 15, \pm 17, \pm 19$, considering the remainders of $y^{2}$ and $s^{2 n+1}$ modulo 41, we have
$$
y^{2}+(t-1)^{2 n+1}=y^{2}+s^{2 n+1} \not\equiv 0 .
$$
From equation (1), $x \neq 0$.
Thus, $t \equiv 4,7,8,12,13,14,15,16,18,20, -1, -2, -4, -5, -6, -8, -9, -10, -11, -12, -13, -14, -16, -18$ satisfy the condition.
From equation (3), if $t \equiv 2,5,9, -4, -8, -9$, then $\{x_{n}\}$ modulo 41 is a periodic sequence, and for all positive integers $n, x_{n} \neq 0$, which satisfies the condition.
Thus, the required
$$
t \equiv 1,2,4,5,7,8,9,12,13,14,15,16,18,
$$
$20, -1, -2, -4, -5, -6, -8, -9, -10, -11, -12, -13, -14, -16, -18 \pmod{41}$.
From (1), (2), and the Chinese Remainder Theorem, the number of $t$ that satisfy the condition is $\frac{5 \times 28}{7 \times 41} \times 2009=980$.
$$
|
980
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The positive integer $n$ ensures that in every $n$-element subset of the set $\{1,2, \cdots, 2008\}$, there are 2 elements (which can be the same), whose sum is a positive integer power of 2. Then the minimum value of $n$ is $\qquad$
|
8.1003.
First, if $n=1002$, take
$$
\begin{array}{l}
A=\{5,6,7\}, B=\{17,18, \cdots, 24\}, \\
C=\{33,34, \cdots, 39\}, \\
D=\{1025,1026, \cdots, 2008\} .
\end{array}
$$
Then $S=A \cup B \cup C \cup D$.
Thus, $|S|=3+8+7+984=1002$.
It is easy to verify that the sum of any two (possibly the same) elements in $S$ is not a power of 2.
Second, prove: In every 1003-element subset of the set $\{1,2, \cdots, 2008\}$, there are 2 elements (possibly the same) whose sum is a power of 2.
In fact, divide the set $\{1,2, \cdots, 2008\}$ into the union of 1003 pairwise disjoint subsets as follows:
$$
\begin{array}{l}
\{1,7\},\{2,6\},\{3,5\},\{8,24\},\{9,23\}, \\
\{10,22\}, \cdots,\{15,17\},\{25,39\}, \cdots, \\
\{31,33\},\{40,2008\},\{41,2007\}, \\
\{42,2006\}, \cdots,\{2003,1025\}, \\
\{4,16,32,1024\} .
\end{array}
$$
Each of the above sets has 2 elements (possibly the same) whose sum is a power of 2. Therefore, every 1003-element subset of $\{1,2, \cdots, 2008\}$ either contains one element from $\{4,16,32,1024\}$ or contains one of the first 1002 sets. Thus, there are 2 elements (possibly the same) whose sum is a power of 2.
In summary, the minimum value of $n$ is 1003.
|
1003
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (16 points) On the Cartesian plane, a point whose both coordinates are rational numbers is called a rational point. Find the smallest positive integer $k$ such that: for every circle that contains $k$ rational points on its circumference, the circle must contain infinitely many rational points on its circumference.
|
12. First, prove: If a circle's circumference contains 3 rational points, then the circumference must contain infinitely many rational points.
Let $\odot C_{0}$ in the plane have 3 rational points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3)$ on its circumference, with the center $C_{0}\left(x_{0}, y_{0}\right)$.
Since the perpendicular bisectors of segments $P_{1} P_{2}$ and $P_{1} P_{3}$ pass through the center $C_{0}$, we have
$$
\left\{\begin{array}{l}
\left(y_{2}-y_{1}\right)\left(y_{0}-\frac{y_{1}+y_{2}}{2}\right)+\left(x_{2}-x_{1}\right)\left(x_{0}-\frac{x_{1}+x_{2}}{2}\right)=0, \\
\left(y_{3}-y_{1}\right)\left(y_{0}-\frac{y_{1}+y_{3}}{2}\right)+\left(x_{3}-x_{1}\right)\left(x_{0}-\frac{x_{1}+x_{3}}{2}\right)=0 .
\end{array}\right.
$$
Since $x_{i} 、 y_{i}(i=1,2,3)$ are all rational numbers, the solution $\left(x_{0}, y_{0}\right)$ of the above system of linear equations in $x_{0} 、 y_{0}$ is also a rational number, i.e., $C_{0}$ is a rational point.
Suppose the coordinates of the rational point $P_{n}\left(x_{n}, y_{n}\right)$ are
$$
\begin{array}{l}
\qquad\left\{\begin{array}{l}
x_{n}=x_{0}+a_{n}\left(x_{3}-x_{0}\right)-b_{n}\left(y_{3}-y_{0}\right), \\
y_{n}=y_{0}+b_{n}\left(x_{3}-x_{0}\right)+a_{n}\left(y_{3}-y_{0}\right),
\end{array}\right. \\
\text { where, } a_{n}=\frac{n^{2}-1}{n^{2}+1}, b_{n}=\frac{2 n}{n^{2}+1}(n=4,5, \cdots) .
\end{array}
$$
Then $\left|P_{n} C_{0}\right|^{2}$
$$
\begin{aligned}
= & {\left[a_{n}\left(x_{3}-x_{0}\right)-b_{n}\left(y_{3}-y_{0}\right)\right]^{2}+} \\
& {\left[b_{n}\left(x_{3}-x_{0}\right)+a_{n}\left(y_{3}-y_{0}\right)\right]^{2} } \\
= & \left(a_{n}^{2}+b_{n}^{2}\right)\left[\left(x_{3}-x_{0}\right)^{2}+\left(y_{3}-y_{0}\right)^{2}\right] \\
= & \left(x_{3}-x_{0}\right)^{2}+\left(y_{3}-y_{0}\right)^{2}=\left|P_{3} C_{0}\right|^{2} .
\end{aligned}
$$
Hence, the points $P_{n}(n=4,5, \cdots)$ are all on the circumference of $\odot C_{0}$, i.e., the circumference of $\odot C_{0}$ contains infinitely many rational points.
Second, construct an example of a circle whose circumference contains only two rational points.
$$
C:(x-\sqrt{2})^{2}+(y-\sqrt{2})^{2}=6 .
$$
It is easy to verify that $P_{1}(-1,1)$ and $P_{2}(1,-1)$ are on the circumference of $C$. If the circumference of $C$ contains another rational point $P_{3}\left(x_{3}, y_{3}\right)$ different from $P_{1}$ and $P_{2}$, then
$$
\left(x_{3}-\sqrt{2}\right)^{2}+\left(y_{3}-\sqrt{2}\right)^{2}=6 \text {, }
$$
i.e., $x_{3}^{2}+y_{3}^{2}-2=2 \sqrt{2}\left(x_{3}+y_{3}\right)$.
Since the left side is a rational number and $\sqrt{2}$ is an irrational number, we have $x_{3}+y_{3}=0$. Thus, $x_{3}^{2}=1$. Hence, $x_{3}= \pm 1, y_{3}=$ $\mp 1$. This contradicts the assumption that $P_{3}$ is different from $P_{1}$ and $P_{2}$.
In conclusion, the minimum value of $k$ is 3.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $a, b, c \geqslant 0$, and $a b+b c+c a=\frac{1}{3}$. Prove:
$$
S=\frac{1}{a^{2}-b c+1}+\frac{1}{b^{2}-a c+1}+\frac{1}{c^{2}-a b+1} \leqslant 3 .
$$
|
Explanation: When $a=b=c=\frac{1}{3}$, $S=3$.
Also, when $a=0, b=\frac{\sqrt{3}}{3}, c=\frac{\sqrt{3}}{3}$, $S=3$ as well.
Therefore, the maximum value 3 is both conventional and unconventional, which is quite rare. Moreover, $S$ has no other extremum (the lower bound of $S$ is $\frac{5}{2}$, $S>\frac{5}{2}$ but cannot take the equality).
Below, we prove this using the graphical method.
Let $a+b+c=M$, it is easy to see that $M \geqslant 1$.
Substituting $a b+b c+a c=\frac{1}{3}$, we get
$$
\begin{aligned}
S & =\frac{1}{a^{2}+a b+a c-\frac{1}{3}+1}+ \\
& \frac{1}{b^{2}+b c+b a-\frac{1}{3}+1}+ \\
& \frac{1}{c^{2}+c a+c b-\frac{1}{3}+1} \\
& =\frac{1}{M a+\frac{2}{3}}+\frac{1}{M b+\frac{2}{3}}+\frac{1}{M c+\frac{2}{3}} .
\end{aligned}
$$
As shown in Figure 3, draw
$$
y=\frac{1}{M x+\frac{2}{3}}(x>
$$
0 ) (a part of a hyperbola).
By the properties of convex functions and decreasing nature, for any $k$ and $\varepsilon>0$ we have
$$
f(k)+f(0)>f(\varepsilon)+f(k-\varepsilon) .
$$
The left side of equation (1) is twice the length of the midline of a trapezoid with bases $f(k)$ and $f(0)$, and the right side is twice the length of the midline of a trapezoid with bases $f(\varepsilon)$ and $f(k-\varepsilon)$. Since the two midlines coincide and the left side is longer, equation (1) holds.
This indicates that when $S$ takes its maximum value, one of the variables must be 0 (let $a=0$). Then $b c=\frac{1}{3}$. Substituting into the sum, we get
$$
S=\frac{3}{2}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1} \text {. }
$$
Let $y=\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}$
$$
\begin{array}{l}
=\frac{1}{b^{2}+1}+\frac{1}{\left(\frac{1}{3 b}\right)^{2}+1} \\
=\frac{9 b^{4}+18 b^{2}+1}{9 b^{4}+10 b^{2}+1} .
\end{array}
$$
Then $(9 y-9) b^{4}+(10 y-18) b^{2}+y-1=0$.
Since $b^{2} \in \mathbf{R}$, we have
$$
\Delta=(10 y-18)^{2}-4 \times 9(y-1)^{2} \geqslant 0 \text {. }
$$
Solving this, we get $y \leqslant \frac{3}{2}(y \geqslant 3$ is discarded $)$.
Thus, $S=\frac{3}{2}+y \leqslant 3$.
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given that $a, b, c$ are positive integers, and the quadratic equation $a x^{2}+b x+c=0$ has two real roots whose absolute values are both less than $\frac{1}{3}$. Find the minimum value of $a+b+c$.
(2005, National High School Mathematics League, Fujian Province Preliminary Contest)
|
Let $x_{1}$ and $x_{2}$ be the roots of the equation $a x^{2} + b x + c = 0$. By Vieta's formulas, we have
$$
x_{1} + x_{2} = -\frac{b}{a}, \quad x_{1} x_{2} = \frac{c}{a}.
$$
Thus, $x_{1}9$.
Therefore, the roots $x_{1}$ and $x_{2}$ of $a x^{2} + b x + c = 0$ are in the interval $\left(-\frac{1}{3}, 0\right)$. Hence, we can use the knowledge of the distribution of real roots of a quadratic equation to solve the problem.
Let $f(x) = a x^{2} + b x + c$. Then
$$
\left\{\begin{array}{l}
f(0) = c > 0, \\
f\left(-\frac{1}{3}\right) = \frac{1}{9} a - \frac{1}{3} b + c > 0, \\
-\frac{1}{3} < -\frac{b}{2a} < 0, \\
\Delta = b^{2} - 4ac > 0.
\end{array}\right.
$$
From the above, we have $-\frac{1}{3} < -\frac{b}{2a} < 0$, which implies $0 < \frac{b}{a} < \frac{2}{3}$, or $0 < b < \frac{2}{3} a$. Also, from $f\left(-\frac{1}{3}\right) > 0$, we get $\frac{1}{9} a - \frac{1}{3} b + c > 0$, which simplifies to $a - 3b + 9c > 0$, or $a + 9c > 3b$.
Since $a$, $b$, and $c$ are positive integers, $a + 9c$ and $3b$ are positive integers, so $a + 9c \geq 3b + 1$.
From the above, we have $a + 9c \geq 3b + 1 \geq 6 \sqrt{ac} + 1$.
Rearranging, we get
$$
(\sqrt{a} - 3 \sqrt{c})^{2} \geq 1.
$$
Combining this with $a > 9c$, we get $\sqrt{a} - 3 \sqrt{c} \geq 1$.
Thus, $\sqrt{a} \geq 1 + 3 \sqrt{c} \geq 1 + 3 \times 1 = 4$, which implies $a \geq 16$.
When $a = 16$, we have $c = 1$. In this case, $b \geq 2 \sqrt{ac} = 8$.
Upon verification, $16 x^{2} + 8 x + 1 = 0$ satisfies the given conditions. Therefore, the minimum value of $a + b + c$ is 25.
Note 1: The most common method for solving discrete extremum problems is to first estimate the range of the object of study, determine the possible extremum, and then construct to show that the extremum can be achieved.
Note 2: In [1], from $\Delta = b^{2} - 4ac \geq 0$, we get
$$
b^{2} \geq 4ac = 4 \times \frac{a}{c} \times c^{2} > 36.
$$
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
(1) Find the minimum value of the smallest side length of a peculiar triangle;
(2) Prove that there are infinitely many isosceles peculiar triangles;
(3) How many non-isosceles peculiar triangles are there?
|
(1) Let $a, b, c (a \leqslant b \leqslant c)$ be the side lengths of a peculiar triangle. Then, by Heron's formula, we have
$$
\begin{aligned}
16 \Delta^{2}= & (a+b+c)(a+b-c) . \\
& (a-b+c)(-a+b+c) .
\end{aligned}
$$
Since $(a, b, c)=1$, at least one of $a, b, c$ must be odd. If there is an odd number of odd numbers among $a, b, c$, then $a+b+c, a+b-c, a-b+c, -a+b+c$ are all odd, which contradicts equation (1).
Therefore, exactly two of $a, b, c$ are odd.
If $a=1$, by $c1$, it leads to a contradiction.
When $c=b+1$, $b, c$ are one odd and one even.
Thus, $a, b, c$ have exactly one odd number, which is a contradiction.
If $a=4$, then $b, c$ are both odd.
By $cb-\Delta_{1}$, we have $b+\Delta_{1}=4, b-\Delta_{1}=1$, hence $2b=5$, which is a contradiction.
When $c=b+2$,
$$
p=\frac{1}{2}(a+b+c)=b+3, \\
\Delta^{2}=(b+3)(b-1) \times 3 \times 1,
$$
$$
3 \Delta_{1}^{2}=(b+3)(b-1).
$$
If $3 \mid (b+3)$, then $3 \mid b$, which contradicts the peculiar triangle. If $3 \mid (b-1)$, then $3 \mid (b+2)=c$, which also contradicts the peculiar triangle.
In summary, $a \geqslant 5$.
Moreover, $(5,5,8)$ is a peculiar triangle, so the minimum value of the smallest side length of a peculiar triangle is 5.
(2) If $m, n \in \mathbf{N}_{+}, m>n, 3 \mid m \dot{n}, (m, n)=1, m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, m^{2}+n^{2}, 2 m^{2}-2 n^{2}\right)$ is a peculiar triangle.
In fact, $\Delta=2 m n\left(m^{2}-n^{2}\right)$ is an integer.
Furthermore, since $m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, 2\right)=1$.
Thus, $\left(m^{2}+n^{2}, 2 m^{2}-2 n^{2}\right)$
$=\left(m^{2}+n^{2}, m^{2}-n^{2}\right)=\left(m^{2}+n^{2}, m^{2}\right)$
$=\left(m^{2}, n^{2}\right)=1$.
Finally, because $3 \mid m n$, and $(m, n)=1$, one of $m, n$ is a multiple of 3, so $m^{2}+n^{2}, 2 m^{2}-2 n^{2}$ are not multiples of 3.
In particular, taking $m=6 k+1, n=6$, then
$\left(36 k^{2}+12 k+37,36 k^{2}+12 k+37,72 k^{2}+24 k-70\right)$
is a peculiar triangle.
Similarly, if $m, n \in \mathbf{N}_{+}, m>n, 3 \mid \left(m^{2}-n^{2}\right)$, $(m, n)=1, m, n$ are one odd and one even, then $\left(m^{2}+n^{2}, m^{2}+n^{2}, 4 m n\right)$ is a peculiar triangle.
In particular, taking $m=6 k+1, n=2$, then
$$
\left(36 k^{2}+12 k+5,36 k^{2}+12 k+5,48 k+8\right)
$$
is a peculiar triangle.
(3) There are infinitely many non-isosceles peculiar triangles.
Take $t \equiv 5(\bmod 30)$, and let
$$
\begin{array}{l}
a=5 t^{2}, b=\frac{1}{4}\left(25 t^{4}-6 t^{2}+1\right), \\
c=\frac{1}{4}\left(25 t^{4}+6 t^{2}+1\right) .
\end{array}
$$
Since $t$ is odd, $a, b, c$ are integers, and clearly $a<b<c$.
Also, since $t$ is not a multiple of 3, $a, b, c$ are not multiples of 3.
Finally, since $5 \mid t$, $b, c$ are not multiples of 5, and thus, by $\left(t^{2}, 25 t^{4} \pm 6 t^{2}+1\right)=1$, we have $(a, b, c)=1$.
By calculation, $\Delta=\frac{1}{2} t^{2}\left(25 t^{4}-1\right)$ is an integer.
Therefore, $(a, b, c)$ is a non-isosceles peculiar triangle.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In trapezoid $A B C D$, it is known that $A D / / B C(B C>$ $A D), \angle D=90^{\circ}, B C=C D=12, \angle A B E=45^{\circ}$. If $A E=10$, then the length of $C E$ is $\qquad$
(2004, "Xinli Cup" National Junior High School Mathematics Competition)
|
(提示: Extend $D A$ to point $F$, complete the trapezoid into a square $C B F D$, then rotate $\triangle A B F$ $90^{\circ}$ around point $B$ to the position of $\triangle B C G$. It is easy to see that $\triangle A B E \cong \triangle G B E$. Therefore, $A E$ $=E G=E C+A F=10$. Let $E C=x$, then $A F=10-$ $x, D E=12-x, A D=2+x$. By the Pythagorean theorem, $(12-x)^{2}+(2+x)^{2}=10^{2}$. Solving this, we get $x=4$ or 6. )
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In trapezoid $A B C D$, it is known that $A D / / B C, A D \perp$ $C D, B C=C D=2 A D, E$ is a point on side $C D$, $\angle A B E=45^{\circ}$. Then $\tan \angle A E B=$ $\qquad$
(2007, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
(提示: Extend $D A$ to point $F$, such that $A F=A D$, to get the square $B C D F$. Then extend $A F$ to point $G$, such that $F G=C E$. Then $\triangle B F G \cong \triangle B C E$. It is also easy to prove that $\triangle A B G \cong \triangle A B E$. Let $C E=x, B C=2 a$. Then $F G=x, A F=A D=a, A E=A G=x+a, D E=2 a-x$. From $(x+a)^{2}=(2 a-x)^{2}+a^{2}$, solving gives $x=\frac{2}{3} a$. Therefore, $\tan \angle A E B=\tan \angle B G F=\frac{B F}{F G}=3$ is the answer. )
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 As shown in Figure $1, P$ is a moving point on the parabola $y^{2}=2 x$, points $B$ and $C$ are on the $y$-axis, and the circle $(x-1)^{2}+y^{2}=1$ is inscribed in $\triangle P B C$. Find the minimum value of the area of $\triangle P B C$.
(2008, National High School Mathematics Competition)
|
Explanation: By the symmetry of the parabola, we can assume
$$
P\left(2 t^{2}, 2 t\right)(t>0) \text {. }
$$
Since the equation of the circle is $x^{2}+y^{2}-2 x=0$, the equation of the chord of contact $M N$ is
$$
2 t^{2} x+2 t y-\left(x+2 t^{2}\right)=0,
$$
which simplifies to $\left(2 t^{2}-1\right) x+2 t y-2 t^{2}=0$.
Thus, the double tangents $P B$ and $P C$ are the quadratic curve passing through all common points of the double line
$$
\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0
$$
and the circle, with the equation
$$
(x-1)^{2}+y^{2}-1+\lambda\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0 .
$$
Substituting the coordinates of point $P$ into the above equation gives $\lambda=-\frac{1}{4 t^{4}}$. Therefore, the equation of the double tangents $P B$ and $P C$ can be written as
$$
(x-1)^{2}+y^{2}-1-\frac{1}{4 t^{4}}\left[\left(2 t^{2}-1\right) x+2 t y-2 t^{2}\right]^{2}=0 \text {. }
$$
Setting $x=0$ in the above equation, we get
$$
y=\frac{1}{t} y-1 \text { or } y=1-\frac{1}{t} y \text {. }
$$
Since when $t=1$, there is only one tangent $P B$ intersecting the axis, and when $01$.
Thus, $y_{B}=\frac{t}{1+t}, y_{C}=\frac{t}{1-t}$,
$$
\begin{array}{l}
B C=\frac{t}{1+t}-\frac{t}{1-t}=\frac{2 t^{2}}{t^{2}-1} . \\
\text { Hence } S_{\triangle P B C}=\frac{1}{2} B C\left|x_{P}\right|=\frac{1}{2} \cdot \frac{2 t^{2}}{t^{2}-1} \cdot 2 t^{2} \\
=\frac{2 t^{4}}{t^{2}-1}=2\left[2+\left(t^{2}-1\right)+\frac{1}{t^{2}-1}\right] \geqslant 8 .
\end{array}
$$
The equality holds if and only if $t=\sqrt{2}$. Therefore, the minimum value sought is 8.
|
8
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Quadratic Function
$$
f(x)=a x^{2}+b x+c(a, b \in \mathbf{R} \text {, and } a \neq 0)
$$
satisfies the conditions:
(1) For $x \in \mathbf{R}$, $f(x-4)=f(2-x)$, and
$$
f(x) \geqslant x \text {; }
$$
(2) For $x \in(0,2)$, $f(x) \leqslant\left(\frac{x+1}{2}\right)^{2}$;
(3) The minimum value of $f(x)$ on $\mathbf{R}$ is 0.
Find the largest $m(m>1)$, such that there exists $t \in \mathbf{R}$, for any $x \in[1, m]$, we have $f(x+t) \leqslant x$.
(2002, National High School Mathematics Competition)
|
Explanation: First, derive the analytical expression of $f(x)$ from the given conditions, then classify and discuss $m$ and $t$ to determine the value of $m$.
Since $f(x-4)=f(2-x)$, the graph of the function is symmetric about the line $x=-1$. Therefore,
$$
-\frac{b}{2 a}=-1 \Rightarrow b=2 a \text {. }
$$
From condition (3), when $x=-1$, $y=0$, i.e., $a-b+c=0$.
From conditions (1) and (2), we get $f(1) \geqslant 1, f(1) \leqslant 1$.
Thus, $f(1)=1$, i.e., $a+b+c=1$.
From the above, we can solve for $a=\frac{1}{4}, b=\frac{1}{2}, c=\frac{1}{4}$.
Therefore, $f(x)=\frac{1}{4} x^{2}+\frac{1}{2} x+\frac{1}{4}$.
Since the graph of the parabola $f(x)$ opens upwards, and the graph of $f(x+t)$ is obtained by translating the graph of $y=f(x)$ by $t$ units, to ensure that on $[1, m]$, the graph of $y=f(x+t)$ is below the graph of $y=x$, and $m$ is maximized, $1$ and $m$ should be the two roots of the equation $\frac{1}{4}(x+t+1)^{2}=x$.
Since $1$ is a root of the equation $\frac{1}{4}(x+t+1)^{2}=x$, we can solve for $t=0$ or $-4$.
Substituting $t=0$ into the original equation gives $x_{1}=x_{2}=1$ (which contradicts $m>1$);
Substituting $t=-4$ into the original equation gives $x_{1}=1, x_{2}=9$.
Therefore, $m=9$.
In summary, the maximum value of $m$ is 9.
|
9
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 A $6 \times 6$ chessboard has 36 squares. Placing chess pieces in the squares, if there are 4 pieces of the same color in a straight line (horizontal, vertical, or at a $45^{\circ}$ diagonal), it is called a "four-in-a-row". Player A places white pieces, and Player B places black pieces. If A goes first, what is the minimum number of pieces A needs to place to ensure that B cannot subsequently form a four-in-a-row ${ }^{[1]}$?
|
Solution: At least 10 chess pieces need to be placed.
This is because, to prevent the four-in-a-row formation by the black pieces placed subsequently by player B, player A must place at least one white piece in each $1 \times 4$ rectangle, meaning that at least 8 white pieces must be placed in the $2 \times 4$ rectangles on the four sides as shown in Figure 1.
If at least 2 white pieces are placed in the 4 central squares, then the number of white pieces is no less than 10.
If no white pieces are placed in the 4 central squares, then at least 2 white pieces must be placed in each of the two middle columns, at least 2 white pieces in each of the two middle rows, and at least 2 white pieces on each of the two main diagonals, and these white pieces are distinct, so at least $2 \times 6 = 12 (>10)$ white pieces must be placed.
If exactly 1 white piece is placed in the 4 central squares, by symmetry, assume it is placed at position $C$ in Figure 2, with no white pieces at $A$, $B$, or $D$. Then at least 1 white piece must be placed at $E$ and $F$, at least 1 white piece at $G$ and $H$, at least 2 white pieces on the diagonal of $A$ and $D$, and at least 1 white piece on the diagonal $M Q$, and these white pieces are distinct (at least 5 in total). Thus, by the pigeonhole principle, at least one of the two shaded areas in Figure 2 must contain 3 white pieces (assume the upper left $2 \times 4$ square contains 3 pieces), and by Figure 1, the part outside $A$, $B$, $C$, and $D$ in Figure 2 must contain at least $3 + 3 \times 2 = 9$ pieces, plus $C$, making a total of at least 10 white pieces.
Figure 3 shows one possible placement (shaded areas indicate where white pieces are placed).
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 4, in a $7 \times 8$ rectangular chessboard, a chess piece is placed at the center of each small square. If two chess pieces are in small squares that share an edge or a vertex, then these two chess pieces are said to be "connected". Now, some of the 56 chess pieces are removed so that no 5 remaining chess pieces are in a straight line (horizontally, vertically, or diagonally). What is the minimum number of chess pieces that need to be removed to meet this requirement? Explain your reasoning $^{\mid 2\rfloor}$.
---
The translation preserves the original text's formatting and structure.
|
Analysis: The difficulty ratio of this problem has significantly increased from 1, mainly due to its "apparent" asymmetry (the number of rows and columns are inconsistent). Therefore, the key to solving the problem is to find the symmetry hidden behind this asymmetry.
Inspired by Example 1, we will also start from the local perspective.
Solution: In each $1 \times 5$ rectangle, at least one chess piece must be taken. Following Example 1, as shown in Figure 5, the area outside the shaded part must have at least 10 chess pieces taken.
If at least one chess piece is taken from the shaded part in Figure 5, then the total number of chess pieces taken is no less than 11.
If no chess pieces are taken from the shaded part in Figure 5, then at least 2 chess pieces must be taken from each of the middle two columns, and at least 2 chess pieces must be taken from each of the middle three rows. Therefore, at least 10 chess pieces have been taken.
Considering the four squares (1), (2), (3), and (4) in Figure 6.
Looking diagonally, since the shaded area in Figure 6 indicates that no chess pieces have been taken, to avoid a line of five, the four squares (1), (2), (3), and (4) must all have chess pieces taken. Therefore, in Figure 6, 2 chess pieces are taken from the 3rd column and 2 chess pieces are taken from the 6th column; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 7th, and 8th columns, totaling 4 chess pieces; at least 1 chess piece is taken from the white area of each of the 1st, 2nd, 6th, and 7th rows, totaling 4 chess pieces. Therefore, 12 chess pieces are taken.
In summary, at least 11 chess pieces must be taken.
Figure 7 provides one possible method.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A bicycle tire, if installed on the front wheel, will wear out after the bicycle has traveled $5000 \mathrm{~km}$; if installed on the rear wheel, it will wear out after the bicycle has traveled $3000 \mathrm{~km}$. After traveling a certain distance, the front and rear tires can be swapped. If the front and rear tires are swapped, to make a pair of new tires on a bicycle wear out simultaneously, then the bicycle will be able to travel $\qquad$ $\mathrm{km}$.
|
II. 6.3750.
Let the total wear of each new tire when it is scrapped be $k$. Then the wear per $1 \mathrm{~km}$ for a tire installed on the front wheel is $\frac{k}{5000}$, and the wear per $1 \mathrm{~km}$ for a tire installed on the rear wheel is $\frac{k}{3000}$. Suppose a pair of new tires travels $x \mathrm{~km}$ before the positions are swapped, and $y \mathrm{~km}$ after the positions are swapped. By setting up equations based on the total wear of a tire, we have
$$
\left\{\begin{array}{l}
\frac{k x}{5000}+\frac{k y}{3000}=k, \\
\frac{k y}{5000}+\frac{k x}{3000}=k .
\end{array}\right.
$$
Adding the two equations gives
$$
\frac{k(x+y)}{5000}+\frac{k(x+y)}{3000}=2 k \text {. }
$$
Thus, $x+y=\frac{2}{\frac{1}{5000}+\frac{1}{3000}}=3750$.
|
3750
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five distinct integers satisfying the condition
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=9
$$
If $b$ is an integer root of the equation
$$
\left(x-a_{1}\right)\left(x-a_{2}\right)\left(x-a_{3}\right)\left(x-a_{4}\right)\left(x-a_{5}\right)=2009
$$
then the value of $b$ is $\qquad$.
|
8. 10 .
Notice
$$
\left(b-a_{1}\right)\left(b-a_{2}\right)\left(b-a_{3}\right)\left(b-a_{4}\right)\left(b-a_{5}\right)=2009,
$$
and $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are five different integers, so all $b-a_{1}, b-a_{2}, b-a_{3}, b-a_{4}, b-a_{5}$ are also five different integers.
$$
\begin{array}{l}
\text { Also, } 2009=1 \times(-1) \times 7 \times(-7) \times 41, \text { then } \\
b-a_{1}+b-a_{2}+b-a_{3}+b-a_{4}+b-a_{5}=41 .
\end{array}
$$
From $a_{4}+a_{2}+a_{3}+a_{4}+a_{5}=9$, we get $b=10$.
|
10
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. B. Let the positive integer $a$ satisfy $192 \mid\left(a^{3}+191\right)$, and $a<2$ 009. Find the sum of all possible positive integers $a$ that meet the condition.
|
12. B. From the problem, we have $192 \mid\left(a^{3}-1\right)$.
$$
\begin{array}{l}
\text { Also, } 192=3 \times 2^{6}, \text { and } \\
a^{3}-1=(a-1)[a(a+1)+1] \\
=(a-1) a(a+1)+(a-1) .
\end{array}
$$
Since $a(a+1)+1$ is odd, we have
$$
2^{6}\left|\left(a^{3}-1\right) \Leftrightarrow 2^{6}\right|(a-1) \text {. }
$$
Also, $3 \mid(a-1) a(a+1)$, so
$$
3\left|\left(a^{3}-1\right) \Leftrightarrow 3\right|(a-1) \text {. }
$$
Therefore, $192 \mid(a-1)$. Thus, $a=192 k+1$.
Given $0<a<2009$, then $k=0,1, \cdots, 10$.
Hence, the sum of all possible positive integers $a$ that satisfy the condition is $11+192(1+2+\cdots+10)=10571$.
|
10571
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. A. $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following conditions:
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$
and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$.
|
14. A. Let in $a_{1}, a_{2}, \cdots, a_{n}$, after removing $a_{i}(i=1$, $2, \cdots, n)$, the arithmetic mean of the remaining $n-1$ numbers is a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$
Therefore, for any $i, j(1 \leqslant i<j \leqslant n)$, we have
$$
b_{i}-b_{j}=\frac{a_{j}-a_{i}}{n-1} \text {. }
$$
Thus, $(n-1) \mid\left(a_{j}-a_{i}\right)$.
Since $b_{1}-b_{n}=\frac{a_{n}-a_{1}}{n-1}=\frac{2008}{n-1}$ is a positive integer, then $(n-1) \mid 2^{3} \times 251$.
Also, $a_{n}-1$
$$
\begin{array}{l}
=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right) \\
\geqslant(n-1)+(n-1)+\cdots+(n-1) \\
=(n-1)^{2},
\end{array}
$$
then $(n-1)^{2} \leqslant 2008$.
Thus, $n \leqslant 45$.
Combining $(n-1) \mid 2^{3} \times 251$, we know $n \leqslant 9$.
On the other hand, let
$$
\begin{array}{l}
a_{1}=8 \times 0+1, a_{2}=8 \times 1+1, \\
a_{3}=8 \times 2+1, \cdots \cdots \\
a_{8}=8 \times 7+1, a_{9}=8 \times 251+1 .
\end{array}
$$
Then these 9 numbers satisfy the problem's requirements.
In summary, the maximum value of $n$ is 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $x \in \mathbf{R}$. Then the function
$$
f(x)=\sqrt{x^{2}+1}+\sqrt{(x-12)^{2}+16}
$$
has a minimum value of $\qquad$ .
|
8. 13 .
As shown in Figure 1, take $A$ as the origin of the number line, $A B=$ 12, and then construct the perpendicular lines $A C$ and $B D$ such that
$$
\begin{array}{l}
A C=1, \\
B D=4 .
\end{array}
$$
On the number line, take point $P$ such that $A P=x$. Then
$$
f(x)=|C P|+|D P| \text {. }
$$
When points $C$, $P$, and $D$ are collinear, the value of $f$ is minimized. At this time,
$$
f_{\min }=|C D|=|A E|=\sqrt{12^{2}+5^{2}}=13 .
$$
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1$, and for each $n \in \mathbf{N}_{+}, a_{n} 、 a_{n+1}$ are the two roots of the equation $x^{2}+3 n x+b_{n}=0$. Then $\sum_{k=1}^{20} b_{k}=$ $\qquad$
|
11. 6385.
For each $n \in \mathbf{N}_{+}$, we have
$$
\begin{array}{l}
a_{n}+a_{n+1}=-3 n, \\
a_{n} a_{n+1}=b_{n} .
\end{array}
$$
Rewrite equation (1) as
$$
a_{n+1}+\frac{3(n+1)}{2}-\frac{3}{4}=-\left(a_{n}+\frac{3 n}{2}-\frac{3}{4}\right) \text {. }
$$
Therefore, $\left\{a_{n}+\frac{3 n}{2}-\frac{3}{4}\right\}$ is a geometric sequence with a common ratio of -1.
Hence, $a_{n}+\frac{3 n}{2}-\frac{3}{4}=(-1)^{n-1} \frac{7}{4}$, which means
$$
\begin{array}{l}
a_{n}=-\frac{3(2 n-1)}{4}+(-1)^{n-1} \frac{7}{4}, \\
a_{n+1}=-\frac{3(2 n+1)}{4}+(-1)^{n} \frac{7}{4} .
\end{array}
$$
Then $b_{n}=a_{n} a_{n+1}=\frac{9}{4} n^{2}-\frac{29}{8}+(-1)^{n} \frac{21}{8}$.
Therefore, $\sum_{k=1}^{20} b_{k}=6385$.
|
6385
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. From the set $M=\{1,2, \cdots, 2008\}$ consisting of the first 2008 positive integers, take a $k$-element subset $A$, such that the sum of any two numbers in $A$ cannot be divisible by their difference. Then the maximum value of $k$ is $\qquad$
|
12.670.
First, take 670 yuan set $A=\{1,4,7, \cdots, 2008\}$, where the sum of any two numbers in $A$ cannot be divisible by 3, but their difference is a multiple of 3.
Then, divide the numbers in $M$ from smallest to largest into 670 segments, each containing three numbers:
$$
1,2,3 ; 4,5,6 ; \cdots ; 2005,2006,2007 ; 2008 \text {. }
$$
If 671 numbers are taken from $A$, there must be two numbers $x$ and $y$ that come from the same segment, then $|x-y|=1$ or 2.
Note that $x-y$ and $x+y$ have the same parity, thus, $(x-y) \mid (x+y)$.
Therefore, the maximum value of $k$ is 670.
|
670
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. For the $2n$-element set $M=\{1,2, \cdots, 2n\}$, if the $n$-element sets
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}
$$
satisfy $A \cup B=M, A \cap B=\varnothing$, and $\sum_{k=1}^{n} a_{k}=\sum_{k=1}^{n} b_{k}$, then $A \cup B$ is called an "equal-sum partition" of set $M$ ($A \cup B$ and $B \cup A$ are considered the same partition). Determine the number of equal-sum partitions of the set $M=\{1,2, \cdots, 12\}$.
|
15. Solution 1: Without loss of generality, let $12 \in A$. Since when set $A$ is determined, set $B$ is uniquely determined, we only need to consider the number of sets $A$.
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{6}\right\}, a_{6}$ be the largest number.
By $1+2+\cdots+12=78$, we know
$a_{1}+a_{2}+\cdots+a_{6}=39, a_{6}=12$.
Thus, $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=27$.
Hence, $A_{1}=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ contains an odd number of
odd numbers.
(1) If $A_{1}$ contains five odd numbers, since the sum of the six odd numbers in $M$ is 36, and $27=36-9$, we have
$A_{1}=\{1,3,5,7,11\}$
At this point, we get the unique $A=\{1,3,5,7,11,12\}$.
(2) If $A$ contains three odd numbers and two even numbers, let $p$
represent the sum of the two even numbers $x_{1}, x_{2}$ in $A_{1}$, and $q$ represent the sum of the three odd numbers $y_{1}, y_{2}, y_{3}$ in $A_{1}$, then
$p \geqslant 6, q \geqslant 9$.
Thus, $q \leqslant 21, p \leqslant 18$.
We get 24 cases for $A_{4}$.
(1) When $p=6, q=21$,
$\left(x_{1}, x_{2}\right)=(2,4)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,9,11),(3,7,11),(5,7,9)$,
we get 3 cases for $A_{1}$;
(2) When $p=8, q=19$,
$\left(x_{1}, x_{2}\right)=(2,6)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,7,11),(3,5,11),(3,7,9)$,
we get 3 cases for $A_{1}$;
(3) When $p=10, q=17$,
$\left(x_{1}, x_{2}\right)=(2,8),(4,6)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,5,11),(1,7,9),(3,5,9)$,
we get 6 cases for $A_{1}$;
(4) When $p=12, q=15$,
$\left(x_{1}, x_{2}\right)=(2,10),(4,8)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,3,11),(1,5,9),(3,5,7)$,
we get 6 cases for $A_{1}$;
(5) When $p=14, q=13$,
$\left(x_{1}, x_{2}\right)=(4,10),(6,8)$,
$\left(y_{1}, y_{2}, y_{3}\right)=(1,3,9),(1,5,7)$,
we get 4 cases for $A_{1}$;
(6) When $p=16, q=11$,
$\left(x_{1}, x_{2}\right)=(6,10),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,7)$,
we get 1 case for $A_{1}$;
(7) When $p=18, q=9$,
$\left(x_{1}, x_{2}\right)=(8,10),\left(y_{1}, y_{2}, y_{3}\right)=(1,3,5)$,
we get 1 case for $A_{1}$.
(3) If $A_{1}^{\prime}$ contains one odd number and four even numbers, since the sum of the five even numbers in $M$ except 12 is $2+4+6+8+10=30$, removing one even number and adding one odd number to make the sum of the five numbers in $A_{1}$ equal to 27, we get 4 cases for $A_{1}$
$(7,2,4,6,8),(5,2,4,6,10)$,
$(3,2,4,8,10),(1,2,6,8,10)$.
In summary, there are $1+24+4=29$ cases for set $A$,
i.e., $M$ has 29 equal-sum partitions.
Solution 2: Element exchange method.
Clearly, $\sum_{i=1}^{6} a_{i}=\sum_{i=1}^{6} \cdot b_{i}$, always assume $12 \in A$.
(1) First, note one extreme partition:
$A_{0}=\{1,2,3,10,11,12\}$,
$B_{0}=\{4,5,6,7,8,9\}$.
Clearly, if the set $\{1,2,3\}$ is entirely in $A$, then the set $\{10,11,12\}$ must also be entirely in $A$.
Next, consider the case where at least one of the numbers 10, 11 is not in $A$. For this, consider element exchanges of the same number and equal sum between $A_{0}$ and $B_{0}$.
(2) $(10,1) \leftrightarrow(5,6),(4,7)$;
$(10,2) \leftrightarrow(5,7),(4,8)$;
$(10,3) \leftrightarrow(6,7),(5,8),(4,9)$;
$(10,2,3) \leftrightarrow(4,5,6)$.
We get 8 exchanges.
(3) $(11,1) \leftrightarrow(5,7),(4,8)$;
$(11,2) \leftrightarrow(6,7),(5,8),(4,9)$;
$(11,3) \leftrightarrow(6,8),(5,9)$;
$(11,1,3) \leftrightarrow(4,5,6)$;
$(11,2,3) \leftrightarrow(4,5,7)$.
We get 9 exchanges.
(4) $(10,11,1) \leftrightarrow(6,7,9),(5,8,9)$;
$(10,11,2) \leftrightarrow(6,8,9)$;
$(10,11,3) \leftrightarrow(7,8,9)$;
$(10,11,1,2) \leftrightarrow(4,5,7,8),(4,5,6,9)$;
$(10,11,1,3) \leftrightarrow(4,6,7,8),(4,5,7,9)$;
$(10,11,2,3) \leftrightarrow(5,6,7,8),(4,6,7,9)$,
$(4,5,8,9)$.
We get 11 exchanges.
Each exchange results in a new partition.
Therefore, we have a total of $1+8+9+11=29$ equal-sum partitions.
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let $x_{1}, x_{2}, \cdots, x_{6}$ be positive integers, and they satisfy the relation
$$
\begin{aligned}
x_{6} & =2288, \\
x_{n+3} & =x_{n+2}\left(x_{n+1}+2 x_{n}\right)(n=1,2,3) .
\end{aligned}
$$
Then $x_{1}+x_{2}+x_{3}=$
|
3. 8 .
Substituting $n=1,2,3$ into the relation
$$
\begin{array}{l}
x_{n+3}=x_{n+2}\left(x_{n+1}+2 x_{n}\right) . \\
\text { we get } x_{4}=x_{3}\left(x_{2}+2 x_{1}\right), \\
x_{5}=x_{4}\left(x_{3}+2 x_{2}\right)=x_{3}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right), \\
x_{6}=x_{3}^{2}\left(x_{2}+2 x_{1}\right)\left(x_{3}+2 x_{2}\right)\left(x_{2}+2 x_{1}+2\right) . \\
\text { Also, } x_{6}=2288=2^{4} \times 11 \times 13, \text { and } x_{1} 、 x_{2} 、 x_{3}
\end{array}
$$
are all positive integers, so
$$
x_{3}+2 x_{2} \geqslant 3, x_{2}+2 x_{1} \geqslant 3 \text {. }
$$
Since $x_{2}+2 x_{1}+2$ is 2 more than $x_{2}+2 x_{1}$, and $2^{4} \times 11 \times 13$ only has 13 and 11 with a difference of 2 and both are factors not less than 3, thus,
$$
x_{2}+2 x_{1}+2=13 \Rightarrow x_{2}+2 x_{1}=11 \text {. }
$$
Therefore, it can only be
$$
x_{3}^{2}\left(x_{3}+2 x_{2}\right)=2^{4}, x_{3}=1 \text { or } 2 \text {. }
$$
When $x_{3}=1$, $1+2 x_{2}=2^{4}, x_{2}$ is not an integer;
When $x_{3}=2$, we get $x_{2}=1, x_{1}=5$ which meets the conditions.
Thus, $x_{1}+x_{2}+x_{3}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given that $f(x)$ is a monotonically increasing function on $\mathbf{R}$, and for any $x \in \mathbf{R}$, $f(f(f(x)))=x$. Then $f(2009)=$ $\qquad$
|
5.2 009.
If $f(2009)2009$, then we have
$$
f(f(f(2009)))>2009,
$$
contradiction.
|
2009
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. $A, B$ two people are playing a table tennis match, with the rule that the first to win 3 games more than the opponent wins. After 13 games, $A$ finally wins with a record of 8 wins and 5 losses. Then the number of all different possible outcomes of these 13 games is $\qquad$
|
6. 243.
As shown in Figure 3, use the grid point $(m, n)$ to represent the moment when $m+n$ games have been played (with $A$ winning $m$ games and $B$ winning $n$ games). The problem then transforms into finding the number of grid paths from $(0,0)$ to $(8,5)$, where the points $(x, y)$ passed through satisfy $|x-y|<3$. Clearly, the grid paths lie within the polygon formed by the vertices $(0,0)$, $(2,0)$, $(6,4)$, $(8,5)$, $(3,5)$, and $(0,2)$ (including the boundaries). Using the labeling method, the number of grid paths that meet the conditions is 243.
|
243
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given real numbers $a, b, c$ satisfy $a+b+c=2, abc=4$.
(1) Find the minimum value of the maximum of $a, b, c$;
(2) Find the minimum value of $|a|+|b|+|c|$.
(2003, National Junior High School Mathematics Competition)
|
Solution: (1) Without loss of generality, let $a$ be the maximum of $a, b, c$, i.e., $a \geqslant b, a \geqslant c$.
From the problem, we have $a>0$, and $b+c=2-a, bc=\frac{4}{a}$.
Thus, $b, c$ are the two real roots of the quadratic equation
$$
x^{2}-(2-a) x+\frac{4}{a}=0
$$
Therefore,
$$
\begin{array}{l}
\Delta=[-(2-a)]^{2}-4 \times \frac{4}{a} \geqslant 0 \\
\Rightarrow a^{3}-4 a^{2}+4 a-16 \geqslant 0 \\
\Rightarrow\left(a^{2}+4\right)(a-4) \geqslant 0 .
\end{array}
$$
Solving this, we get $a \geqslant 4$.
When $a=4, b=c=-1$, the minimum value of the maximum among $a, b, c$ is 4.
(2) From $abc>0$, we know that $a, b, c$ are all greater than 0 or one is positive and two are negative.
(i) If $a, b, c$ are all greater than 0, then from (1) we know that the minimum value of the maximum among $a, b, c$ is 4, which contradicts $a+b+c=2$.
(ii) If $a, b, c$ are one positive and two negative, let $a>0$, $b<0$, $c<0$. Then
$$
\begin{array}{l}
|a|+|b|+|c|=a-b-c \\
=a-(2-a)=2 a-2 .
\end{array}
$$
From (1), we know $a \geqslant 4$, hence $2 a-2 \geqslant 6$.
When $a=4, b=c=-1$, the conditions of the problem are satisfied and the equality holds.
Therefore, the minimum value of $|a|+|b|+|c|$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For any real numbers $x, y$, the inequality
$$
|x-1|+|x-3|+|x-5| \geqslant k(2-|y-9|)
$$
always holds. Then the maximum value of the real number $k$ is $\qquad$
|
3. 2 .
From the geometric meaning of absolute value, we know that the minimum value of $|x-1|+|x-3|+|x-5|$ is 4 (at this time $x=3$), and the maximum value of $2-|y-9|$ is 2 (at this time $y=9$). According to the condition, we get $4 \geqslant k \cdot 2$, so, $k \leqslant 2$. Therefore, the maximum value of $k$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. $p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+p_{5}^{2}=p_{6}^{2}$ has $\qquad$ groups of positive prime solutions $\left(p_{1}, p_{2}, p_{3}, p_{4}, p_{5}, p_{6}\right)$.
|
7.5.
Obviously, $p_{6} \neq 2$.
Since $p^{2} \equiv 1$ or $4(\bmod 8)$ (where $p$ is a prime), and $p^{2} \equiv 4(\bmod 8)$ if and only if $p=2$, it follows that among $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, and $p_{5}$, there must be 4 twos (let's assume they are $p_{1}$, $p_{2}$, $p_{3}$, and $p_{4}$), and the other one is an odd prime, and $p_{6}$ is also an odd prime.
Thus, $16=p_{6}^{2}-p_{5}^{2}=\left(p_{6}+p_{5}\right)\left(p_{6}-p_{5}\right)$.
Clearly, the only solution is $p_{5}=3, p_{6}=5$.
Therefore, the number of all prime solutions is $\mathrm{C}_{5}^{4}=5$.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A classroom has desks arranged in 6 rows and 7 columns, with 40 students. Two positions in the last row are left empty, and the rest of the students are seated based on their height and vision. There are 24 students who are tall, 18 students who have good vision, and 6 students who have both conditions. It is known that if a student is short and has poor vision, they must sit in the first three rows; if a student is tall and has good vision, they must sit in the last three rows. Let the method of seating arrangement be $A$, then the number of times 2 appears in the prime factorization of $A$ is $\qquad$
|
8. 35 .
The number of students who are short and have poor eyesight is 4, the number of students who are tall and have good eyesight is 6, and the remaining students number 30 people.
The number of ways to leave two seats empty in the last row is $C_{7}^{2}$, where the power of 2 is 0;
The number of ways to arrange the students who are short and have poor eyesight is $A_{21}^{4}$, where the power of 2 is 3;
The number of ways to arrange the students who are tall and have good eyesight is $A_{19}^{6}$, where the power of 2 is 6;
The number of ways to arrange the other students is $\mathrm{A}_{30}^{30}$, where the power of 2 is $15+7+3+1=26$.
Therefore, the total power of 2 is 35.
|
35
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) On a plane, there are 32 points, with no three points being collinear. Prove: Among these 32 points, there are at least 2135 sets of four points that form the vertices of a convex quadrilateral.
|
Three, let $n=32$.
Since no three points among the $n$ points are collinear, $\mathrm{C}_{n}^{3}$ triangles can be formed by any three points among the $n$ points, among which the one with the largest area is denoted as $\triangle A B C$.
Draw lines parallel to the opposite sides through points $A, B, C$, intersecting to form $\triangle A^{\prime} B^{\prime} C^{\prime}$.
Since $\triangle A B C$ has the largest area, the remaining $n-3$ points are inside $\triangle A^{\prime} B^{\prime} C^{\prime}$.
A line through any two of these $n-3$ points, say $D, E$, intersects at most two sides of $\triangle A B C$. Without loss of generality, assume line $D E$ does not intersect side $B C$. Then $B, C, D, E$ are the four vertices of a convex quadrilateral.
Therefore, any two points among the $n-3$ points and two points among $A, B, C$ can form a convex quadrilateral.
Thus, there are at least $\mathrm{C}_{n-3}^{2}$ groups of four points $(D, E ; F, G)$ that form the vertices of a convex quadrilateral, where $F, G$ are two points among $A, B, C$, and line $D E$ does not intersect side $F G$.
Considering point $A$ and the set of the above $n-3$ points, there are at least $\mathrm{C}_{(n-2)-3}^{2}$ groups of four points that satisfy the condition.
Continuing the discussion, we know that the number of groups of four points among the $n$ points that satisfy the condition is at least
$$
\begin{array}{l}
f(n)=\sum_{k=0}^{\left.\frac{n-5}{2}\right]} \mathrm{C}_{n-3-2 k}^{2} \\
=\left\{\begin{array}{l}
\frac{(n-3)(n-1)(2 n-7)}{24}, n \text { is odd; } \\
\frac{(n-4)(n-2)(2 n-3)}{24}, n \text { is even. }
\end{array}\right.
\end{array}
$$
Substituting $n=32$ yields $f(32)=2135$.
|
2135
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
2. If
$$
\begin{array}{l}
a+b-2 \sqrt{a-1}-4 \sqrt{b-2} \\
=3 \sqrt{c-3}-\frac{1}{2} c-5,
\end{array}
$$
then, the value of $a+b+c$ is $\qquad$
|
(Tip: Refer to Example 3. Answer: 20. )
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $a$, $b$, $c$ are all real numbers, and $a+b+c=0$, $abc=2$, then the minimum value that $|a|+|b|+|c|$ can reach is $\qquad$ .
|
(Tip: Refer to Example 3. Answer: 4. )
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Find the maximum value of the function $y=\sqrt{2 x^{2}+3 x+1}+\sqrt{7-2 x^{2}-3 x}$.
|
( Hint: Refer to Example 8. Answer: The maximum value is 4 . )
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Each point in the plane is colored with one of $n$ colors, and the following conditions are satisfied:
(1) There are infinitely many points of each color, and they do not all lie on the same line;
(2) There is at least one line on which all points are exactly two colors.
Find the minimum value of $n$ such that there exist four points of different colors that are concyclic.
(3rd Northern Mathematical Olympiad Invitational)
|
Explanation: If $n=4$, then a colored plane can be constructed such that: there is exactly one point on a circle with three colors, and the rest of the points are of another color, satisfying the problem's conditions and ensuring no four points of different colors are concyclic. Therefore, $n \geqslant 5$.
When $n=5$, by condition (2), there exists a line $l$ on which exactly two colors of points lie. Without loss of generality, assume that line $l$ has only points of colors 1 and 2. By condition (1), there exist points $A$, $B$, and $C$ of colors 3, 4, and 5, respectively, that are not collinear. Let the circle passing through points $A$, $B$, and $C$ be $\odot O$.
(i) If $\odot O$ intersects line $l$, then there exist four points of different colors that are concyclic.
(ii) If $\odot O$ is separated from line $l$ and $\odot O$ has points of colors 1 and 2, then there exist four points of different colors that are concyclic.
(iii) If $\odot O$ is separated from line $l$ and $\odot O$ has no points of colors 1 and 2, as shown in Figure 11, draw a perpendicular from $O$ to line $l$ intersecting $l$ at point $D$. Assume the color of $D$ is 1, and the perpendicular intersects $\odot O$ at points $E$ and $S$. Assume the color of $E$ is 3. Consider a point $F$ on line $l$ with color 2, and let $FS$ intersect $\odot O$ at point $G$. Since $EG \perp GF$, points $D$, $E$, $G$, and $F$ are concyclic.
If $G$ is not a point of color 3, then there exist four points of different colors that are concyclic;
If $G$ is a point of color 3, and $B$, $C$ must have one point different from $S$ (assume it is $B$), $SB$ intersects line $l$ at point $H$. Since $EB \perp BH$, points $B$, $E$, $D$, and $H$ are concyclic.
If $H$ is a point of color 2, then $B$, $H$, $D$, and $E$ are four points of different colors and are concyclic;
If $H$ is a point of color 1, since
$$
SB \cdot SH = SE \cdot SD = SG \cdot SF,
$$
then $B$, $H$, $F$, and $G$ are four points of different colors and are concyclic.
In summary, when $n=5$, there exist four points of different colors that are concyclic.
Therefore, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Suppose the lengths of the three sides of a triangle are integers $l$, $m$, and $n$, and $l > m > n$. It is known that
$$
\left\{\frac{3^{l}}{10^{4}}\right\}=\left\{\frac{3^{m}}{10^{4}}\right\}=\left\{\frac{3^{n}}{10^{4}}\right\},
$$
where $\{x\}=x-[x]$, and $[x]$ represents the greatest integer not exceeding the real number $x$. Find the minimum perimeter of such a triangle.
(2003, National High School Mathematics Competition)
|
Solution: Note that, when $a \equiv b(\bmod m)(0 \leqslant b < m)$, we get $500+2 n>1000+n$.
So, $n>500$.
Thus, $n \geqslant 501, m \geqslant 1001, l \geqslant 1501$.
Therefore, $m+n+l \geqslant 3003$, which means the minimum perimeter of the triangle is 3003.
|
3003
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the minimum value of $y=\sqrt{x^{2}+1}+\sqrt{(4-x)^{2}+4}$.
|
Solution: As shown in Figure 2, construct Rt $\triangle P A C$ and Rt $\triangle P B D$ such that $A C=1, B D=2, C P=x, P D=4-x$. Thus, the original problem is transformed into: finding a point $P$ on line $l$ such that the value of $P A + P B$ is minimized.
Since $y = P A + P B \geqslant A B$, the minimum value of $y = P A + P B$ is $A B$.
Draw $B E \parallel l$, intersecting the extension of $A C$ at point $E$. Then,
$$
\begin{array}{l}
A E = A C + C E \\
= A C + B D = 3, \\
B E = C D \\
= x + 4 - x = 4 .
\end{array}
$$
Therefore, $A B = 5$. Hence, the minimum value of $y$ is 5.
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Consider rectangles in the plane with sides parallel to the coordinate axes (both length and width are greater than 0), and call such a rectangle a "box". If two boxes have a common point (including common points on the interior or boundary of the boxes), then the two boxes are said to "intersect". Find the largest positive integer $n$, such that there exist $n$ boxes $B_{1}, B_{2}, \cdots, B_{n}$, satisfying $B_{i}$ intersects $B_{j}$ if and only if $i \neq j \pm 1(\bmod n)$.
|
1. The maximum value that satisfies the condition is 6.
An example is shown in Figure 1.
Below is the proof: 6 is the maximum value.
Assume the boxes $B_{1}$, $B_{2}, \cdots, B_{n}$ satisfy the condition. Let the closed intervals corresponding to the projections of $B_{k}$ on the $x$-axis and $y$-axis be $I_{k}$ and $J_{k}$ respectively $(1 \leqslant k \leqslant n)$.
If $B_{i}$ and $B_{j}$ intersect, and the common point is $(x, y)$, then $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$. Therefore, $I_{i} \cap I_{j}$ and $J_{i} \cap J_{j}$ are non-empty. Conversely, if there exist real numbers $x$ and $y$ such that $x \in I_{i} \cap I_{j}$ and $y \in J_{i} \cap J_{j}$, then $(x, y)$ is a common point of $B_{i}$ and $B_{j}$. Thus, $B_{i}$ and $B_{j}$ do not intersect if and only if at least one of the intervals $I_{i}, I_{j}$ and $J_{i}, J_{j}$ do not intersect, i.e., either $I_{i} \cap I_{j}=\varnothing$ or $J_{i} \cap J_{j}=\varnothing$.
For convenience, if the indices of the boxes differ by 1 modulo $n$, then the two boxes or intervals are called "adjacent." Other cases are called non-adjacent.
For each $k=1,2, \cdots, n$, the adjacent boxes $B_{k}$ and $B_{k+1}$ do not intersect, so $\left(I_{k}, I_{k+1}\right)$ or $\left(J_{k}, J_{k+1}\right)(1 \leqslant k \leqslant n)$ is a pair of non-intersecting intervals. Therefore, among $\left(I_{1}, I_{2}\right), \cdots,\left(I_{n-1}, I_{n}\right),\left(I_{n}, I_{1}\right)$; $\left(J_{1}, J_{2}\right), \cdots,\left(J_{n-1}, J_{n}\right),\left(J_{n}, J_{1}\right)$, there are at least $n$ pairs of non-intersecting intervals.
Since any two non-adjacent boxes intersect, the corresponding intervals on both coordinate axes also intersect.
Proposition: Let $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$ be $n$ closed intervals on the number line, and any two non-adjacent intervals intersect. Then the number of pairs of non-intersecting intervals $\left(\Delta_{k}, \Delta_{k+1}\right)(k=1,2, \cdots, n)$ is at most 3.
Proof of the proposition: Let $\Delta_{k}=\left[a_{k}, b_{k}\right](1 \leqslant k \leqslant n)$, $\alpha=\max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be the rightmost left endpoint of $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$, and $\beta=\min \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ be the leftmost right endpoint of $\Delta_{1}, \Delta_{2}, \cdots, \Delta_{n}$. Without loss of generality, assume $\alpha=a_{2}$.
If $\alpha \leqslant \beta$, then for all $i(i=1,2, \cdots, n)$, we have $a_{i} \leqslant \alpha \leqslant \beta \leqslant b_{i}$. Each $\Delta_{i}$ contains $\alpha$, so there are no non-intersecting interval pairs $\left(\Delta_{i}, \Delta_{i+1}\right)$.
If $\beta<\alpha$, then there exists $i \in\{1,2, \cdots, n\}$ such that $\beta=b_{i}$, and $a_{i}<b_{i}=\beta<\alpha=a_{2}<b_{2}$. Thus, $\Delta_{2}$ and $\Delta_{i}$ are non-intersecting. Since $\Delta_{2}$ intersects with all other intervals except $\Delta_{1}$ and $\Delta_{3}$, $\Delta_{2}$ and $\Delta_{i}$ are non-intersecting if and only if $i=1$ or $i=3$. By symmetry, assume $i=3$, then $\beta=b_{3}$. Since each of $\Delta_{4}, \Delta_{5}, \cdots, \Delta_{n}$ intersects with $\Delta_{2}$, for $i=4,5, \cdots, n$, we have $a_{i} \leqslant \alpha \leqslant b_{i}$. Therefore, $\alpha \in \Delta_{4} \cap \Delta_{5} \cap \cdots \cap \Delta_{n}$, i.e., $\Delta_{4} \cap \Delta_{5} \cap \cdots \cap \Delta_{n} \neq \varnothing$. Similarly, $\Delta_{5}, \cdots, \Delta_{n}, \Delta_{1}$ all intersect with $\Delta_{3}$. Thus, $\beta \in \Delta_{5} \cap \cdots \cap \Delta_{n} \cap \Delta_{1}$, i.e., $\Delta_{5} \cap \cdots \cap \Delta_{n} \cap \Delta_{1} \neq \varnothing$. Therefore, the remaining interval pairs $(\Delta_{1}, \Delta_{2})$, $(\Delta_{2}, \Delta_{3})$, $(\Delta_{3}, \Delta_{4})$ are the possible non-intersecting interval pairs. Thus, there are at most 3.
By the proposition, the maximum value of the positive integer $n$ that satisfies the condition is 6.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Consider the set $S$ of all integer points (points with integer coordinates) on the coordinate plane. For each positive integer $k$, if there exists a point $C \in S$ such that the area of $\triangle A B C$ is $k$, then two distinct points $A, B \in S$ are called “$k$-friends”; if any two points in $T$ are $k$-friends, then the set $T \subset S$ is called a “$k$-clique”. Find the smallest positive integer $k$ such that there exists a $k$-clique with more than 200 elements.
|
3. First, consider the point $B \in S$ that is a $k$-friend of the point $(0,0)$. Let $B(u, v)$. Then $B$ is a $k$-friend of $(0,0)$ if and only if there exists a point $C(x, y) \in S$ such that $\frac{1}{2}|u y - v x| = k$ (here we use the area formula of $\triangle A B C$, where $A$ is the origin).
There exist integers $x, y$ such that $|u y - v x| = 2k$ if and only if the greatest common divisor (gcd) of $u, v$ is a divisor of $2k$. Therefore, the point $B(u, v) \in S$ is a $k$-friend of $(0,0)$ if and only if the gcd of $u, v$ divides $2k$.
If two points are $k$-friends, translating them to other points in $S$ will still make them $k$-friends. Thus, two points $A, B \in S, A(s, t), B(u, v)$ are $k$-friends if and only if the point $(u-s, v-t)$ is a $k$-friend of $(0,0)$, i.e., the gcd of $u-s, v-t$ divides $2k$.
Let $n$ be a positive integer that does not divide $2k$. Then the number of elements in a $k$-clique is at most $n^2$.
In fact, all points $(x, y) \in S$ are divided into $n^2$ classes modulo $n$. If a set $T (T \subset S)$ contains more than $n^2$ elements, then there exist two points $A, B \in T$ that belong to the same class. Let $A(s, t), B(u, v)$. Then
$n \mid (u-s), n \mid (v-t)$.
Thus, $n \mid d$, where $d$ is the gcd of $u-s, v-t$. Since $n$ does not divide $2k$, $d$ does not divide $2k$, and $A, B$ are not $k$-friends, so $T$ is not a $k$-clique.
Let $M(k)$ be the smallest positive integer that does not divide $2k$. Denote $M(k) = m$.
Consider the points $(x, y) \in S$ satisfying $0 \leqslant x, y \leqslant 200$.
By the definition of $M(k)$, $2k$ can be divided by $1, 2, \cdots, M(k)-1$, but not by $M(k)$. If $(M(k))^2 > 200$, then $M(k) \geqslant 15$.
(1) $M(k) = 15$, which clearly contradicts the fact that $2k$ can be divided by 3 and 5, but not by 15.
(2) $M(k) = 16$, then $2k$ can be divided by $1, 2, \cdots, 15$, and also by their least common multiple (LCM) $L$, but not by 16. Since $L$ is not a multiple of 16, $k = \frac{L}{2}$ is the smallest $k$ such that $M(k) = 16$.
(3) $M(k) \geqslant 17$, then $2k$ can be divided by the LCM of $1, 2, \cdots, 16$, which is $2L$. From $2k \geqslant 2L$, we get $k \geqslant L > \frac{L}{2}$.
In summary, the smallest $k$ that satisfies the condition is
$$
\frac{L}{2} = 180180
$$
|
180180
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$ is a root of the equation $x^{2}-5 x+1=0$. Then the unit digit of $a^{4}+a^{-4}$ is $\qquad$ .
|
ニ、1.7.
Obviously, $a^{-1}$ is another root of the equation. Then $a+a^{-1}=5$.
Thus, $a^{2}+a^{-2}=\left(a+a^{-1}\right)^{2}-2=23$, $a^{4}+a^{-4}=\left(a^{2}+a^{-2}\right)^{2}-2=527$.
|
7
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a convex quadrilateral $A B C D$, the diagonals $A C$ and $B D$ intersect at point $O$. If $S_{\triangle O A D}=4, S_{\triangle O B C}=9$, then the minimum value of the area of the convex quadrilateral $A B C D$ is
|
2. 25 .
Let $S_{\triangle A O B}=x, S_{\triangle C O D}=y$. Then
$$
\frac{4}{x}=\frac{O D}{O B}=\frac{y}{9} \Rightarrow x y=36 \text {. }
$$
Thus, $x+y \geqslant 2 \sqrt{x y}=12$.
Equality holds if and only if $x=y=6$.
Therefore, $S_{\text {quadrilateral } A B C D}=4+9+x+y \geqslant 25$.
Equality holds if and only if $x=y=6$.
|
25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. In the expansion of $\left(1+x+x^{2}+\cdots+x^{100}\right)^{3}$, after combining like terms, the coefficient of $x^{150}$ is $\qquad$ (answer with a number).
|
11.7651 .
By the polynomial multiplication rule, the problem can be transformed into finding the number of natural number solutions of the equation
$$
s+t+r=150
$$
that do not exceed 100. Clearly, the number of natural number solutions of equation (1) is $\mathrm{C}_{152}^{2}$.
Next, we find the number of natural number solutions of equation (1) that exceed 100.
Since their sum is 150, there can only be one number that exceeds 100. Without loss of generality, let $s>100$. Transform equation (1) into
$$
(s-101)+t+r=49 \text {. }
$$
Let $s^{\prime}=s-101$. Then the number of natural number solutions of the equation $s^{\prime}+t+r=49$ is $\mathrm{C}_{51}^{2}$.
Therefore, the coefficient of $x^{150}$ is $\mathrm{C}_{152}^{2}-\mathrm{C}_{3}^{1} \mathrm{C}_{51}^{2}=7651$.
|
7651
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Find the maximum value of the function $y=\sqrt{5-2 x}+\sqrt{3+2 x}$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This part is not translated as it contains instructions for the translation task itself. Here is the requested translation above.
|
Solution: From the given condition, we know $y>0$.
Notice that the variance of the two numbers $\sqrt{5-2 x}$ and $\sqrt{3+2 x}$ is
$$
\begin{array}{l}
S^{2}=\frac{1}{2}\left[(\sqrt{5-2 x})^{2}+(\sqrt{3+2 x})^{2}-2\left(\frac{y}{2}\right)^{2}\right] \\
\geqslant 0 .
\end{array}
$$
Solving this, we get $16-y^{2} \geqslant 0$, which means $y^{2} \leqslant 16$.
Therefore, $y_{\text {max }}=4$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. In a $4 \times 4$ chessboard, 8 small squares are to be colored black such that each row and each column has exactly two black squares. Then there are $\qquad$ different ways to do this (answer with a number).
|
13. 90 .
The first row can be colored with 2 black cells in $\mathrm{C}_{4}^{2}$ ways. After the first row is colored, there are the following three scenarios:
(1) The black cells in the second row are in the same columns as those in the first row, in which case, the remaining rows have only one way to be colored;
(2) The black cells in the second row are in different columns from those in the first row, in which case, the third row has $\mathrm{C}_{4}^{2}$ ways to be colored, and the coloring of the fourth row is then determined;
(3) The second row has exactly one black cell in the same column as one in the first row, and there are 4 such coloring methods. After the first and second rows are colored, the third row must have 1 black cell in a column different from the ones above, in which case, the third row has 2 ways to be colored, and the coloring of the fourth row is then determined.
Therefore, the total number of different coloring methods is
$$
6 \times(1+6+4 \times 2)=90 \text {. }
$$
|
90
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Given that $a, b$ are positive numbers, and the parabolas $y=x^{2}+a x+2 b$ and $y=x^{2}+2 b x+a$ both have common points with the $x$-axis. Then the minimum value of $a^{2}+b^{2}$ is $\qquad$
$(2000$, National Junior High School Mathematics League)
|
Solution: From the given conditions, we have
$$
a^{2}-8 b \geqslant 0,4 b^{2}-4 a \geqslant 0 \text {. }
$$
Thus, $a^{4} \geqslant 64 b^{2} \geqslant 64 a$, which means $a \geqslant 4$.
Furthermore, $b^{2} \geqslant a \geqslant 4$. Therefore, $a^{2}+b^{2} \geqslant 20$.
Moreover, when $a=4, b=2$, the parabolas $y=x^{2}+a x+2 b$ and $y=x^{2}+2 b x+a$ both intersect the $x$-axis, so the minimum value of $a^{2}+b^{2}$ is 20.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let the polynomial be
$$
\begin{aligned}
P(x)= & x^{15}-2008 x^{14}+2008 x^{13}-2008 x^{12}+ \\
& 2008 x^{11}-\cdots+2008 x^{3}-2008 x^{2}+2008 x .
\end{aligned}
$$
Then \( P(2007) = \)
|
8. 2007 .
$$
\begin{array}{c}
P(x)=(x-2007)\left(x^{14}-x^{13}+x^{12}- \\
x^{11}+\cdots+x^{2}-x\right)+x .
\end{array}
$$
Therefore, $P(2007)=2007$.
|
2007
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. For a positive integer $n \leqslant 500$, it has the property: when an element $m$ is randomly selected from the set $\{1,2, \cdots, 500\}$, the probability that $m \mid n$ is $\frac{1}{100}$. Then the maximum value of $n$ is $\qquad$
|
12. 81 .
From the problem, we know that $n$ has exactly 5 positive divisors. Let the prime factorization of $n$ be $n=p_{1}^{\alpha_{1}} p_{1}^{\alpha_{2}} \cdots p_{k}^{\alpha_{k}}$.
Then the number of positive divisors of $n$ is
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{k}+1\right)=5 \text {. }
$$
Therefore, $n$ must be of the form $p^{4}$ (where $p$ is a prime number).
Since $3^{4}=81,5^{4}=625>500$, the maximum value of $n$ is 81.
|
81
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. As shown in Figure 1, given that the circumradius $R$ of the acute triangle $\triangle ABC$ is $1$, $\angle BAC = 60^{\circ}$, and the orthocenter and circumcenter of $\triangle ABC$ are $H$ and $O$ respectively. The line segment $OH$ intersects the extension of $BC$ at point $P$. Find:
(1) The area of the concave quadrilateral $ABHC$;
(2) The value of $\mathrm{PO} \cdot \mathrm{OH}$.
|
14. (1) As shown in Figure 2, connect $A H$, and draw $O D \perp B C$ at point D.
Since $O$ is the
circumcenter of
$\triangle A B C$ and $\angle B A C$ $=60^{\circ}$, we have
$$
\begin{array}{l}
\angle B O C \\
=2 \angle B A C \\
=120^{\circ}, \\
O D=O C \cos 60^{\circ}=\frac{1}{2} .
\end{array}
$$
By the property of the Euler line, $A H=2 O D=1$.
By the Law of Sines, $B C=2 R \sin A=\sqrt{3}$, so the area of the concave quadrilateral $A B H C$ is $\frac{1}{2} A H \cdot B C=\frac{\sqrt{3}}{2}$.
(2) Since $H$ is the orthocenter of $\triangle A B C$,
$$
\angle B H C=180^{\circ}-\angle B A C=120^{\circ}=\angle B O C \text {. }
$$
Therefore, points $B, C, H, O$ are concyclic.
Thus, $P O \cdot P H=P B \cdot P C$.
Since $O$ is the circumcenter of $\triangle A B C$, and by the power of a point theorem, $P B \cdot P C=P O^{2}-R^{2}$. Hence,
$$
P O \cdot P H=P B \cdot P C=P O^{2}-R^{2} .
$$
Therefore, $P O^{2}-P O \cdot P H=1$, which means $P O \cdot O H=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
|
15. When $a=b=c=1$, we can get $k \geqslant 2$.
Below is the proof: the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$
holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \cdot \frac{1}{b}}=3 .
$$
Similarly, $b c+\frac{1}{b}+\frac{1}{c} \geqslant 3, c a+\frac{1}{c}+\frac{1}{a} \geqslant 3$.
Adding the above three inequalities, we get
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9 .
$$
In conclusion, the minimum value of $k$ is 2.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given
$$
x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}=1 \text {. }
$$
then the value of $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}$ is
|
$$
\text { II, 1. }-1 \text {. }
$$
Notice that
$$
\begin{array}{l}
x \sqrt{x^{2}+3 x+18}+x \sqrt{x^{2}-6 x+18} \\
=\frac{x^{2}\left(x^{2}+3 x+18\right)-x^{2}\left(x^{2}-6 x+18\right)}{x \sqrt{x^{2}+3 x+18}-x \sqrt{x^{2}-6 x+18}} \\
=\frac{x^{2} \cdot 9 x}{1}=9 x^{3} .
\end{array}
$$
Subtracting the above equation from the given equation, we get
$$
2 x \sqrt{x^{2}-6 x+18}=9 x^{3}-1 \text {. }
$$
Therefore, $2 x \sqrt{x^{2}-6 x+18}-9 x^{3}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Does there exist a positive integer such that its cube plus 101 is exactly a perfect square? Prove your conclusion.
|
3. If there exists a positive integer $x$ that satisfies the conditions of the problem, then $x^{3}+101=y^{2}\left(y \in \mathbf{N}_{+}\right)$.
If $x$ is even, then $y$ is odd.
Let $x=2 n, y=2 m+1(n, m \in \mathbf{N})$. Then $2 n^{3}+25=m^{2}+m$,
the left side of this equation is odd, while the right side is even, which is a contradiction.
Therefore, $x$ is odd, and $y$ is even. We can let $x=2 n+1$, $y=2 m(n, m \in \mathbf{N})$. Then
$$
\begin{array}{l}
(2 n+1)^{3}+101=(2 m)^{2} \\
\Rightarrow 8 n^{3}+12 n^{2}+6 n+102=4 m^{2} \\
\Rightarrow 4 n^{3}+6 n^{2}+3 n+51=2 m^{2} .
\end{array}
$$
Since $3 n+51$ is even and $n$ is odd, let $n=2 n_{1}-1$.
Then $x=4 n_{1}-1\left(n_{1} \in \mathbf{N}_{+}\right)$.
If $x$ divided by 3 leaves a remainder of 0, then $y^{2}$ divided by 3 leaves a remainder of 2, which is a contradiction.
If $x$ divided by 3 leaves a remainder of 1, then $y^{2}$ divided by 3 leaves a remainder of 0, and $y$ divided by 3 leaves a remainder of 0.
Let $x=3 n+1, y=3 m(n, m \in \mathbf{N})$. Then $3^{3} n^{3}+3^{3} n^{2}+3^{2} n+102=9 m^{2}$.
All terms except 102 are multiples of 9, which is a contradiction.
Therefore, $x$ divided by 3 leaves a remainder of 2. We can let
$$
x=3 n-1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
From equations (1) and (2), we have
$$
x=12 n-1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$
Since the last digit of $y^{2}$ can only be $0, 1, 4, 9, 6, 5$, and cannot be $2, 3, 7, 8$, the last digit of $x^{3}$ cannot be $1, 2, 6, 7$, which means the last digit of $x$ cannot be $1, 3, 6, 8$.
From equation (3), the smallest possible values of $x$ are
$11, 23, 35, 47, 59, 71, 83, 95, \cdots$.
From conclusion (4), the smallest possible values of $x$ cannot be
$11, 23, 71, 83$.
Therefore, the smallest possible values of $x$ are
$35, 47, 59, 95, \cdots$.
Upon inspection, when $x=35, 47, 59$, they do not meet the problem's requirements.
If $x=95$, then
$95^{3}+101=857375+101=857476=926^{2}$,
which is a perfect square.
Therefore, there exists a positive integer 95 that satisfies the conditions of the problem.
|
95
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given
$$
\begin{array}{l}
\frac{1}{1 \times \sqrt{2}+2 \sqrt{1}}+\frac{1}{2 \sqrt{3}+3 \sqrt{2}}+\cdots+ \\
\frac{1}{n \sqrt{n+1}+(n+1) \sqrt{n}}
\end{array}
$$
is greater than $\frac{19}{20}$ and less than $\frac{20}{21}$. Then the difference between the maximum and minimum values of the positive integer $n$ is
|
4.39.
Notice that
$$
\begin{array}{l}
\frac{1}{k \sqrt{k+1}+(k+1) \sqrt{k}}=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{(k+1)^{2} k-k^{2}(k+1)} \\
=\frac{(k+1) \sqrt{k}-k \sqrt{k+1}}{k^{2}+k}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}} .
\end{array}
$$
Then the original expression $=1-\frac{1}{\sqrt{n+1}}$.
Given $\frac{19}{20}<1-\frac{1}{\sqrt{n+1}}<\frac{20}{21}$, we have
$$
20<\sqrt{n+1}<21 \Rightarrow 20^{2}-1<n<21^{2}-1 \text {. }
$$
Therefore, the maximum value of the positive integer $n$ is $21^{2}-2$ and the minimum value is $20^{2}$. The difference between them is $21^{2}-20^{2}-2=41-2=39$.
|
39
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The number of positive integer values of $n$ that satisfy $\left|\sqrt{\frac{n}{n+2009}}-1\right|>\frac{1}{1005}$ is $\qquad$
|
1.1008015.
By
$$
\begin{array}{l}
1-\sqrt{\frac{n}{n+2009}}>\frac{1}{1005} \Rightarrow \sqrt{\frac{n}{n+2009}}<\frac{1004}{1005} \\
\Rightarrow \frac{n}{n+2009}<\frac{1004^2}{1005^2} \Rightarrow n<\frac{1004^2}{1005^2-1004^2} \cdot 2009=1008014.25
\end{array}
$$
Thus, the number of positive integer values of $n$ for which $1-\sqrt{\frac{n}{n+2009}}>\frac{1}{1005}$ holds is 1008015.
|
1008015
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $n$ be a positive integer. If in the $2 n+1$ consecutive natural numbers including 2009, the sum of the squares of the first $n+1$ numbers is equal to the sum of the squares of the last $n$ numbers, then the value of $n$ is $\qquad$.
|
10. 31 .
Let $m, m+1, \cdots, m+2 n$ be $2 n+1$ consecutive natural numbers that satisfy the above conditions. Then
$$
\begin{array}{l}
m^{2}+(m+1)^{2}+\cdots+(m+n)^{2} \\
=(m+n+1)^{2}+(m+n+2)^{2}+\cdots+(m+2 n)^{2} . \\
\text { Let } S_{k}=1^{2}+2^{2}+\cdots+k^{2} \text {. Then } \\
S_{m+n}-S_{m-1}=S_{m+2 n}-S_{m+n} .
\end{array}
$$
Substituting and simplifying, we get $2 m n^{2}+2 n^{3}-m^{2}+n^{2}=0$, which is $m=2 n^{2}+n$.
Also, $m \leqslant 2009 \leqslant m+2 n$, then
$$
2 n^{2}+n \leqslant 2009 \leqslant 2 n^{2}+3 n \text {. }
$$
Therefore, $n=31$.
(Xingguo Xia provided)
|
31
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given twelve red points on a circle. Find the minimum value of $n$, such that there exist $n$ triangles with red points as vertices, so that every chord with red points as endpoints is a side of one of these triangles.
（Supplied by Tao Pingsheng）
|
4. Let the set of red points be $A=\left\{A_{1}, A_{2}, \cdots, A_{12}\right\}$.
There are 11 chords passing through point $A_{1}$, and any triangle containing vertex $A_{1}$ contains exactly two chords passing through point $A_{1}$. Therefore, these 11 chords passing through point $A_{1}$ must be distributed among at least six triangles containing vertex $A_{1}$.
Similarly, the chords passing through point $A_{i} (i=2,3, \cdots, 12)$ must also be distributed among six triangles containing vertex $A_{i}$. This requires $12 \times 6=72$ triangles, and since each triangle has three vertices, they are counted three times. Therefore, at least $\frac{72}{3}=24$ triangles are needed.
Next, we prove that the lower bound of 24 can be achieved.
Without loss of generality, consider a circle with a circumference of 12, and its 12 equally divided points are red points. The number of chords with red endpoints is $\mathrm{C}_{12}^{2}=66$. If a chord subtends an arc of length $k$, we call the chord's scale $k$. Thus, there are only six scales for chords with red endpoints, where chords with scales of $1,2, \cdots, 5$ each have 12 chords, and chords with a scale of 6 have 6 chords.
This results in 24 triangles, and they satisfy the conditions of the problem.
Therefore, the minimum value of $n$ is 24.
If the scales of the chords forming the three sides of a triangle are $a, b, c (a \leqslant b \leqslant c)$, then either $a+b=c$ or $a+b+c=12$ must be satisfied.
Thus, there are only 12 possible combinations of scales for the sides of red point triangles:
$$
\begin{array}{l}
(a, b, c) \\
=(1,1,2),(2,2,4),(3,3,6),(2,5,5), \\
\quad(1,2,3),(1,3,4),(1,4,5),(1,5,6), \\
\quad(2,3,5),(2,4,6),(3,4,5),(4,4,4) .
\end{array}
$$
Here is one possible combination:
Take six each of the $(1,2,3)$, $(1,5,6)$, and $(2,3,5)$ types, and four of the $(4,4,4)$ type. In this case, exactly 66 chords are obtained, and among them, there are 12 chords each with scales of $1,2, \cdots, 5$, and 6 chords with a scale of 6.
Now, construct as follows: First, make six triangles each of the $(1,2,3)$, $(1,5,6)$, and $(2,3,5)$ types, and three triangles of the $(4,4,4)$ type, then supplement with three triangles of the $(2,4,6)$ type.
$(1,2,3)$ type six: Their vertex labels are
$\{2,3,5\},\{4,5,7\},\{6,7,9\}$,
$\{8,9,11\},\{10,11,1\},\{12,1,3\}$;
$(1,5,6)$ type six: Their vertex labels are
$\{1,2,7\},\{3,4,9\},\{5,6,11\}$,
$\{7,8,1\},\{9,10,3\},\{11,12,5\}$;
$(2,3,5)$ type six: Their vertex labels are
$\{2,4,11\},\{4,6,1\},\{6,8,3\}$,
$\{8,10,5\},\{10,12,7\},\{12,2,9\}$;
$(4,4,4)$ type three: Their vertex labels are
$\{1,5,9\},\{2,6,10\},\{3,7,11\}$;
$(2,4,6)$ type three: Their vertex labels are
$\{4,6,12\},\{8,10,4\},\{12,2,8\}$.
(The remaining triangles in each case can be obtained by rotating one of the triangles around the center of the circle as appropriate.)
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the set of all permutations $X=(x_{1}, x_{2}, \cdots, x_{9})$ of $1,2, \cdots, 9$ be $A$. For any $X \in A$, let
\[
\begin{array}{l}
f(X)=x_{1}+2 x_{2}+\cdots+9 x_{9}, \\
M=\{f(X) \mid X \in A\} .
\end{array}
\]
Find $|M|$ (where $|M|$ denotes the number of elements in the set $M$).
(Xiong Bin)
|
5. Generally prove: when $n \geqslant 4$, for all permutations $X_{n}=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ of the first $n$ positive integers $1,2, \cdots, n$, if
$$
\begin{array}{l}
f\left(X_{n}\right)=x_{1}+2 x_{2}+\cdots+n x_{n}, \\
M_{n}=\{f(X)|X \in A|,
\end{array}
$$
$$
\text { then }\left|M_{n}\right|=\frac{n^{3}-n+6}{6} \text {. }
$$
Below, we prove using mathematical induction:
$$
\begin{aligned}
M_{n}= & \left\{\frac{n(n+1)(n+2)}{6}, \frac{n(n+1)(n+2)}{6}+1,\right. \\
& \left.\cdots, \frac{n(n+1)(2 n+1)}{6}\right\} .
\end{aligned}
$$
When $n=4$, by the rearrangement inequality, the smallest element in set $M$ is $f(\{4,3,2,1\})=20$, and the largest element is $f(\{1,2,3,4\})=30$.
$$
\begin{array}{l}
\text { Also, } f(\{3,4,2,1\})=21, f(\{3,4,1,2\})=22, \\
f(\{4,2,1,3\})=23, f(\{2,3,4,1\})=24, \\
f(\{2,4,1,3\})=25, f(\{1,4,3,2\})=26, \\
f(\{1,4,2,3\})=27, f(\{2,1,4,3\})=28, \\
f(\{1,2,4,3\})=29,
\end{array}
$$
then $\left|M_{4}\right|=\{20,21, \cdots, 30\}$ has $11=\frac{4^{3}-4+6}{6}$ elements. $\square$
Therefore, the proposition holds for $n=4$.
Assume the proposition holds for $n-1(n \geqslant 5)$. Consider the case for $n$.
For any permutation $X_{n-1}=\left(x_{1}, x_{2}, \cdots, x_{n-1}\right)$ of $1,2, \cdots, n-1$, always take $x_{n}=n$, to get a permutation $x_{1}, x_{2}, \cdots, x_{n-1}, n$ of $1,2, \cdots, n$. Then
$$
\sum_{k=1}^{n} k x_{k}=n^{2}+\sum_{k=1}^{n-1} k x_{k} \text {. }
$$
By the induction hypothesis, at this point, $\sum_{k=1}^{n} k x_{k}$ takes all integers in the interval
$$
\begin{array}{l}
{\left[n^{2}+\frac{(n-1) n(n+1)}{6}, n^{2}+\frac{(n-1) n(2 n-1)}{6}\right]} \\
=\left[\frac{n\left(n^{2}+5\right)}{6}, \frac{n(n+1)(2 n+1)}{6}\right]
\end{array}
$$
Let $x_{n}=1$. Then
$$
\begin{array}{l}
\sum_{k=1}^{n} k x_{k}=n+\sum_{k=1}^{n-1} k x_{k} \\
=n+\sum_{k=1}^{n-1} k\left(x_{k}-1\right)+\frac{n(n-1)}{2} \\
=\frac{n(n+1)}{2}+\sum_{k=1}^{n-1} k\left(x_{k}-1\right) .
\end{array}
$$
By the induction hypothesis, $\sum_{k=1}^{n} k x_{k}$ takes all integers in the interval
$$
\begin{array}{l}
{\left[\frac{n(n+1)}{2}+\frac{(n-1) n(n+1)}{6},\right.} \\
\left.\frac{n(n+1)}{2}+\frac{n(n-1)(2 n-1)}{6}\right] \\
=\left[\frac{n(n+1)(n+2)}{6}, \frac{2 n\left(n^{2}+2\right)}{6}\right]
\end{array}
$$
Since $\frac{2 n\left(n^{2}+2\right)}{6} \geqslant \frac{n\left(n^{2}+5\right)}{6}$, $\sum_{k=1}^{n} k x_{k}$
takes all integers in the interval
$$
\left[\frac{n(n+1)(n+2)}{6}, \frac{n(n+1)(2 n+1)}{6}\right]
$$
Thus, the proposition holds for $n$.
By mathematical induction, the proposition is true.
$$
\begin{array}{l}
\text { Also, } \frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)(n+2)}{6} \\
=\frac{n^{3}-n}{6},
\end{array}
$$
then the number of elements in set $M_{n}$ is $\frac{n^{3}-n+6}{6}$.
In particular, when $n=9$, $|M|=\left|M_{9}\right|=121$.
|
121
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. In an $8 \times 8$ grid, what is the minimum number of small squares that need to be removed so that it is impossible to cut out a complete "T-shaped" pentomino from the remaining grid?
（Sun Wenxian, problem contributor）
Will the above translation meet your needs?
|
8. At least 14 small squares must be removed from Figure 6.
Figure 6
Figure 7
As shown in Figure 7, an $8 \times 8$ chessboard is divided into five regions. The central region must have at least two small squares removed to prevent a T-shaped pentomino from being placed. The two crossed positions are not equivalent; one is in a corner and the other is in an internal position. Removing just one of them is insufficient to prevent the T-shaped pentomino from being placed.
The following proof shows that for the four congruent regions on the boundary, each region must have at least 3 small squares removed to prevent a T-shaped pentomino from being placed.
As shown in Figure 8, using the upper right region as an example, the T-shaped part below must have 1 small square removed, and the part above must have 1 small square removed at the crossed position.
The T-shaped part below must have 1
Figure 8 small square removed, with five possible cases. If the small square at the crossed position in Figure 9 is removed, another T-shaped pentomino can still be placed; the same applies to the other cases (as shown in Figure 10). Therefore, at least 3 small squares must be removed.
Figure 9
Figure 10
In summary, for the T-shaped pentomino, at least $3 \times 4 + 2 = 14$ small squares must be removed.
(Provided by Tao Pingsheng)
|
14
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (25 points) A captain and three sailors together received 2,009 coins of the same denomination. The four agreed to distribute the coins according to the following rules: Sailor 1, Sailor 2, and Sailor 3 each write down a positive integer, denoted as \( b_{1}, b_{2}, b_{3} \), satisfying \( b_{1} \geqslant b_{2} \geqslant b_{3} \), and \( b_{1} + b_{2} + b_{3} = 2009 \); the captain, without knowing the numbers written by the sailors, divides the 2009 coins into 3 piles, with the quantities of each pile being \( a_{1}, a_{2}, a_{3} \), and \( a_{1} \geqslant a_{2} \geqslant a_{3} \). For sailor \( k \) (where \( k = 1, 2, 3 \)), if \( b_{k} < a_{k} \), they can take \( b_{k} \) coins from the \( k \)-th pile; otherwise, they cannot take any. All remaining coins go to the captain. If the captain can always ensure that he gets \( n \) coins regardless of how the three sailors write their numbers, determine the maximum value of \( n \) and prove your conclusion.
（Zhang Limin, Contributed）
|
Four, the maximum value is 673.
First, the captain can ensure getting no less than 673 gold coins.
In fact, when the captain divides the gold coins into 3 piles with numbers $671, 670, 668$,
(1) If $b_{1} \geqslant 671$, then the captain can get no less than $671+2=673$ coins;
(2) If $b_{1} < 671$, then the captain can get no less than $670+3=673$ coins.
Second, the captain cannot ensure getting more than 673 gold coins.
In fact,
(1) If $a_{1} \leqslant 671$, then $a_{2} \leqslant 671, a_{3} \geqslant 667$. When $b_{1}=a_{1}+2, b_{2}=a_{2}-1, b_{3}=a_{3}-1$, the captain can get at most $671+2=673$ coins;
(2) If $a_{1}>671$, then because
$$
a_{3} \leqslant \frac{2009-a_{1}}{2} \leqslant \frac{1337}{2}=668.5 \text {, }
$$
thus $a_{3} \leqslant 668$.
(i) When $a_{2}-a_{3} \geqslant 3$, if $b_{1}=a_{1}-1, b_{2}=$ $a_{2}-1, b_{3}=a_{3}+2$, then the captain can get at most $a_{3}+2 \leqslant$ $668+2=670$ coins;
(ii) When $a_{2}-a_{3} \leqslant 2$, if $a_{3}=1$, then
$$
a_{2} \leqslant 3, a_{1}-a_{2} \geqslant 2002 \text {, }
$$
when $b_{1}=a_{1}-2, b_{2}=a_{2}+1, b_{3}=a_{3}+1$, the captain can get at most $2+3+1=6$ coins;
$$
\begin{array}{l}
\text { If } a_{3}>1 \text {, then } \\
2 a_{2}=\left(a_{2}+a_{3}\right)+\left(a_{2}-a_{3}\right) \\
\leqslant 2009-672+2=1339 \\
\Rightarrow a_{2} \leqslant 669 .
\end{array}
$$
If $a_{3}>1$, then
Therefore, $a_{1}-a_{2} \geqslant 672-669=3$, when $b_{1}=a_{1}$ $-1, b_{2}=a_{2}+2, b_{3}=a_{3}-1$, the captain can get at most $a_{2}+2 \leqslant 671$ coins.
|
673
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the three vertices $A$, $B$, and $C$ of the right triangle $\triangle ABC$ are all on the parabola $y=x^{2}$, and the hypotenuse $AB$ is parallel to the $x$-axis. Then the height $h$ from the hypotenuse is $\qquad$.
|
3. 1 .
Let point $A\left(a, a^{2}\right)$ and $C\left(c, c^{2}\right)(|c|<a)$. Since $|AB|=|AC|$, we have $a^{2}-c^{2}=1$.
Therefore, the altitude from $C$ to the hypotenuse $AB$ is $a^{2}-c^{2}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 2, in $\triangle A B C$, it is known that $A B=9, B C$ $=8, C A=7, A D$ is the angle bisector, and a circle is drawn with $A D$ as a chord,
touching $B C$ and intersecting $A B$ and $A C$ at points $M$ and $N$, respectively. Then $M N=$
$\qquad$
|
4. 6 .
As shown in Figure 4, connect $D M$.
$$
\begin{array}{l}
\text { By } \angle B D M \\
=\angle B A D \\
=\angle C A D \\
=\angle D M N,
\end{array}
$$
we get $M N \parallel B C$.
Thus, $\triangle A M N \sim \triangle A B C$.
It is easy to know $B D=\frac{9}{9+7} \times 8=\frac{9}{2}, B M \cdot B A=B D^{2}$
$\Rightarrow B M \cdot 9=\left(\frac{9}{2}\right)^{2} \Rightarrow B M=\frac{9}{4}$
$\Rightarrow A M=9-\frac{9}{4}=\frac{3}{4} \times 9$.
Also, $\frac{M N}{B C}=\frac{A M}{A B}=\frac{\frac{3}{4} \times 9}{9}=\frac{3}{4}$, therefore,
$$
M N=\frac{3}{4} \times 8=6
$$
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the function $f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x$ has a range of $[n, M]$ on the interval $[-k, k](k>0)$, then $M+n$ $=$ $\qquad$
|
2. 2 .
Notice that
$$
f(x)=\frac{2^{x+1}}{2^{x}+1}+\sin x=1+\frac{2^{x}-1}{2^{x}+1}+\sin x \text {. }
$$
Let $g(x)=f(x)-1$. Then
$$
g(-x)=\frac{2^{-x}-1}{2^{-x}+1}+\sin (-x)=\frac{1-2^{x}}{1+2^{x}}-\sin x=-g(x)
$$
is an odd function.
Let the maximum value of $g(x)$ be $g\left(x_{0}\right)$. By the given information, we have
$$
g\left(x_{0}\right)=M-1 \text {, }
$$
and $g\left(x_{0}\right) \geqslant g(x)(x \in[-k, k])$.
Thus, $-g\left(x_{0}\right) \leqslant-g(x)=g(-x)(x \in[-k, k])$, which means $-g\left(x_{0}\right)$ is the minimum value of $g(x)$. Therefore,
$$
-g\left(x_{0}\right)=n-1 \text {. }
$$
Adding the maximum and minimum values of $g(x)$, we get
$$
0=(M-1)+(n-1) \Rightarrow M+n=2 \text {. }
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The 29th Summer Olympic Games were held in Beijing on August 8, 2008, forming a memorable number 20080808. The number of different positive divisors of 20080808 divided by 8 is $\qquad$
|
3. 8 .
From $20080808=2^{3} \times 11 \times 17 \times 31 \times 433$, we know that the number of different positive divisors of 20080808 is
$$
N=(3+1)(1+1)^{4}=8^{2} \text { (divisors). }
$$
Dividing by 8 gives 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 As shown in Figure 3, let $F$ be the focus of the parabola $y^{2}=4 x$, and $A, B$ be two points on the parabola different from the origin $O$, satisfying $\overrightarrow{F A} \cdot \overrightarrow{F B}=0$. Extend $A F, B F$ to intersect the parabola at points $C, D$ respectively. Find the minimum value of the area of quadrilateral $A B C D$.
|
As shown in Figure 3, with $F$ as the pole and $F x$ as the polar axis, the equation of the parabola is
$$
\rho=\frac{2}{1-\cos \theta} \text {. }
$$
Let point $A\left(\rho_{1}, \theta\right)(0<\theta<\pi)$. Then
$$
\begin{array}{l}
B\left(\rho_{2}, \theta+\frac{\pi}{2}\right) 、 C\left(\rho_{3}, \theta+\pi\right) 、 \\
D\left(\rho_{4}, \theta+\frac{3 \pi}{2}\right) .
\end{array}
$$
Thus, $|A C|=\rho_{1}+\rho_{3}$
$$
=\frac{2}{1-\cos \theta}+\frac{2}{1-\cos (\theta+\pi)}=\frac{4}{\sin ^{2} \theta} \text {. }
$$
Similarly, $|B D|=\frac{4}{\cos ^{2} \theta}$.
Therefore, $S_{\text {quadrilateral } A B C D}=\frac{1}{2}|A C| \cdot|B D|$
$$
=\frac{1}{2} \cdot \frac{4}{\sin ^{2} \theta} \cdot \frac{4}{\cos ^{2} \theta}=\frac{32}{\sin ^{2} 2 \theta} \text {. }
$$
When $\sin 2 \theta=1$, i.e., $\theta=\frac{\pi}{4}$, the minimum value of $S_{\text {quadrilateral } A B C D}$ is 32.
|
32
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. A line $l$ passing through the right focus $F$ of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersects the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$, and there are exactly 3 such lines, find $\lambda$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
(The polar equation of the hyperbola is
$$
\rho=\frac{2}{1-\sqrt{3} \cos \theta} \text {. }
$$
Let $A B$ be a chord passing through the right focus and intersecting only the right branch.
Then $|A B|=\rho_{1}+\rho_{2}$
$$
\begin{array}{l}
=\frac{2}{1-\sqrt{3} \cos \theta}+\frac{2}{1-\sqrt{3} \cos (\theta+\pi)} \\
=\frac{4}{1-3 \cos ^{2} \theta} \geqslant 4
\end{array}
$$
(when $\theta=\frac{\pi}{2}$, the equality holds), that is, among the chords passing through the right focus of the hyperbola and intersecting two points on the right branch, the minimum length is 4 if and only if the chord is perpendicular to the $x$-axis. Since there are exactly 3 lines that satisfy the condition, we have
(1) There is only one line that intersects both the left and right branches of the hyperbola, which must be the real axis of the hyperbola by symmetry, and there are only two lines that intersect the right branch, which can be verified not to meet the condition;
(2) There are only two lines that intersect both the left and right branches of the hyperbola, and only one line that intersects the right branch, which must be perpendicular to the $x$-axis by symmetry. In this case, $|A B|=\lambda=4$. When $\lambda=4$, it can be proven that there are two lines that intersect both the left and right branches of the hyperbola and have a length of 4, so $\lambda=4$. )
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 or 8. How many different ways are there to select such a group of eight points? Explain your reasoning.
(2001, China Mathematical Olympiad)
|
Solution: Number the points in sequence as $1, 2, \cdots, 24$, and list a $3 \times 8$ number table (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline 1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
\hline 9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
\hline 17 & 20 & 23 & 2 & 5 & 8 & 11 & 14 \\
\hline
\end{tabular}
In this table, the arc length between adjacent points in the same row is 3 (the first and last numbers are considered adjacent), and the arc length between any two points in the same column is 8. To satisfy the requirements, we must select one number from each column, and the numbers selected from adjacent columns must be in different rows (the first and eighth columns are considered adjacent).
Generalizing, consider a $3 \times n$ number table (see Table 2).
We need to select $n$ numbers such that one number is chosen from each column, and the numbers selected from adjacent columns are in different rows (the first and $n$-th columns are considered adjacent).
Table 2
\begin{tabular}{|c|c|c|c|c|}
\hline$a_{1}$ & $a_{2}$ & $a_{3}$ & $\cdots$ & $a_{n}$ \\
\hline$b_{1}$ & $b_{2}$ & $b_{3}$ & $\cdots$ & $b_{n}$ \\
\hline$c_{1}$ & $c_{2}$ & $c_{3}$ & $\cdots$ & $c_{n}$ \\
\hline
\end{tabular}
The numbers in the first, second, and third rows $\left\{a_{k}\right\}, \left\{b_{k}\right\}, \left\{c_{k}\right\}$ are colored red, yellow, and blue, respectively.
Make a disk and divide it into $n$ sectors, numbered clockwise as $1, 2, \cdots, n$.
Define a mapping: if the number selected from the $i$-th column in the table is of a certain color, then the $i$-th sector of the disk is colored that color. It is easy to see that this correspondence is one-to-one. Thus, the problem is transformed into:
Find the number of ways to color the $n$ sectors of the disk with three colors such that adjacent sectors have different colors, denoted as $x_{n}$.
It is easy to see that $x_{1}=3, x_{2}=x_{3}=6$.
When $n \geqslant 3$, $x_{n}+x_{n-1}=3 \times 2^{n-1}$, i.e.,
$$
\begin{array}{l}
(-1)^{n} x_{n}-(-1)^{n-1} x_{n-1}=-3(-2)^{n-1} . \\
\text { Let } y_{n}=(-1)^{n} x_{n}, \text { then } \\
y_{n}-y_{n-1}=-3(-2)^{n-1}, \\
y_{n-1}-y_{n-2}=-3(-2)^{n-2}, \\
\cdots \cdots . \\
y_{3}-y_{2}=-3(-2)^{2}, \\
y_{2}=-3(-2)^{1} .
\end{array}
$$
Adding these equations, we get $y_{n}=(-2)^{n}+2(n \geqslant 2)$.
Therefore, $x_{n}=2^{n}+2(-1)^{n}(n \geqslant 2)$.
When $n=8$, $x_{8}=2^{8}+2=258$, which is the number of different ways to select the eight points.
|
258
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 A deck of tri-color cards, totaling 32 cards, includes 10 cards of each color: red, yellow, and blue, numbered $1,2, \cdots, 10$; there are also a big joker and a small joker, each numbered 0. Several cards are drawn from this deck, and their scores are calculated according to the following rule: each card numbered $k$ is worth $2^{k}$ points.
If the total score of these cards is 2004, then these cards are called a "good card group." Find the number of good card groups.
$(2004$, Girls' Mathematical Olympiad)
|
Solution: This problem is referred to as the "Two Kings Problem." If an additional joker (called the "Middle King") with a value of 0 is added, and each card is still assigned a value as described, it is called the "Three Kings Problem."
Since $2004 < 2^{11}$, adding some cards with values $2^{11}, 2^{12}, \cdots$ does not affect the essence of the problem.
First, consider the Three Kings Problem.
For $n \in \{1, 2, \cdots, 2004\}$, take some numbers from
the Red Set $A = \{2^{0}, 2^{1}, \cdots, 2^{10}, \cdots\}$,
the Yellow Set $B = \{2^{0}, 2^{1}, \cdots, 2^{10}, \cdots\}$,
the Blue Set $C = \{2^{0}, 2^{1}, \cdots, 2^{10}, \cdots\}$,
such that their sum is $n$. Let the number of ways to do this be denoted as $u_{n}$. Then
$$
\begin{array}{l}
u_{1} = 3 \left(1 = 1_{A} = 1_{B} = 1_{C}\right), \\
u_{2} = 6 \left(2 = 2_{A} = 2_{B} = 2_{C} = 1_{A} + 1_{B} = 1_{B} + 1_{C} = 1_{C} + 1_{A}\right).
\end{array}
$$
Similarly, $u_{3} = 10, u_{4} = 15, u_{5} = 21, \cdots \cdots$
Define $u_{0} = 1$. Generally, we have
$$
u_{2 n} = u_{n} + 3 u_{n-1}, \quad u_{2 n+1} = 3 u_{n} + u_{n-1}.
$$
This is because
(1) For a group of cards with a total score of $2 n$.
(i) If the group contains no jokers, dividing each value by 2 results in a situation equivalent to a total score of $n$ and allowing jokers. There are $u_{n}$ such groups.
(ii) If the group contains jokers, there must be two jokers. These two jokers can be paired in three ways. Removing the jokers, the situation is equivalent to a total score of $2 n - 2$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $n - 1$ and allowing jokers. There are $3 u_{n-1}$ such groups.
Therefore, $u_{2 n} = u_{n} + 3 u_{n-1}$.
(2) For a group of cards with a total score of $2 n + 1$, it must contain an odd number of jokers.
(i) If the group contains one joker, it has three choices.
Removing the joker, the situation is equivalent to a total score of $2 n$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $n$ and allowing jokers. There are $3 u_{n}$ such groups.
(ii) If the group contains three jokers, removing the jokers, the situation is equivalent to a total score of $2 n - 2$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $n - 1$ and allowing jokers. There are $u_{n-1}$ such groups.
Therefore, $u_{2 n+1} = 3 u_{n} + u_{n-1}$.
Using induction, from (1) we get
$$
u_{n} = \frac{(n+1)(n+2)}{2}.
$$
In fact, (2) has been verified for $n < 4$.
Assume that (2) holds for $n < 2 k$. Then
$$
\begin{array}{l}
u_{2 k} = u_{k} + 3 u_{k-1} \\
= \frac{(k+1)(k+2)}{2} + 3 \cdot \frac{k(k+1)}{2} \\
= \frac{(2 k+1)(2 k+2)}{2}, \\
u_{2 k+1} = 3 u_{k} + u_{k-1} \\
= 3 \cdot \frac{(k+1)(k+2)}{2} + \frac{k(k+1)}{2} \\
= \frac{(2 k+2)(2 k+3)}{2}.
\end{array}
$$
Thus, (2) also holds for $n < 2 k + 2$.
Therefore, (2) holds for every $n$.
Now consider the Two Kings Problem.
For $n \in \{1, 2, \cdots, 2004\}$, let $a_{n}$ be the number of card groups with a total score of $n$. When $2 k \leqslant 2004$, for any card group with a total score of $2 k$:
(1) If the group contains no jokers, it is equivalent to a situation in the Three Kings Problem with a total score of $2 k$ and no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $k$ and allowing jokers. There are $u_{k}$ such groups.
(2) If the group contains jokers, there must be two jokers (both the big and small jokers are in the group). Removing the jokers, the situation is equivalent to a total score of $2 k - 2$ with no jokers. Dividing each value by 2 results in a situation equivalent to a total score of $k - 1$ and allowing jokers. According to the above discussion, there are $u_{k-1}$ such groups.
Therefore, $a_{2 k} = u_{k} + u_{k-1}$
$= \frac{(k+1)(k+2)}{2} + \frac{k(k+1)}{2}$
$= (k+1)^{2} \quad (k=1, 2, \cdots)$.
Thus, $a_{2004} = 1003^{2} = 1006009$, which is the number of good card groups.
|
1006009
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Inputting a positive integer $n$ into a machine will produce a positive integer $\frac{n(n+1)}{2}$. If 5 is input into the machine, and then the number produced is input into the machine again, the final number produced by the machine is $\qquad$ .
|
2. 120 .
The first result is $\frac{5(5+1)}{2}=15$;
The second result is $\frac{15(15+1)}{2}=120$.
|
120
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. $A, B, C$ three people pick watermelons. The sum of the number of watermelons picked by $A$ and $B$ is 6 less than that picked by $C$; the sum of the number of watermelons picked by $B$ and $C$ is 16 more than that picked by $A$; the sum of the number of watermelons picked by $C$ and $A$ is 8 more than that picked by $B$. Then the product of the number of watermelons picked by each person is $\qquad$
|
3. 60.
Let the number of watermelons collected by $A$, $B$, and $C$ be $a$, $b$, and $c$ respectively. Then
\[
\begin{array}{l}
a+b-c=-6, \\
b+c-a=16, \\
c+a-b=8 .
\end{array}
\]
By adding (1) + (2) + (3), we get $a+b+c=18$.
Thus, we have $c=12, b=5, a=1 \Rightarrow a b c=60$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. There are 19 children with ages ranging from $1 \sim 19$ years old standing in a circle. Record the difference in ages between each pair of adjacent children. The maximum possible sum of these 19 differences is . $\qquad$
|
5. 180 .
19 differences are derived from 19 equations, where each integer from $1 \sim 19$ appears twice. To maximize the sum of the differences, the larger numbers 11 and 19 should all be used as minuends, and the smaller numbers $1 \sim 9$ should all be used as subtrahends. The number 10 should be used once as a subtrahend and once as a minuend. Therefore, the maximum value is
$$
2 \sum_{i=11}^{19} i-2 \sum_{i=1}^{9} i+10-10=180 .
$$
In fact, the larger and smaller numbers can be alternately arranged in a circle, and the number 10 can be inserted at any position.
|
180
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In a certain exam, the passing score is 65 points. The overall average score of the class is 66 points, and the average score of all students who passed is 71 points, while the average score of all students who failed is 56 points. To reduce the number of failing students, the teacher adds 5 points to each student's score. After the score adjustment, the average score of all students who passed becomes 75 points, and the average score of all students who failed becomes 59 points. It is known that the number of students in the class is between 15 and 30. Therefore, the class has $\qquad$ students.
|
7. 24 .
Let the class have $n$ students. If before the bonus points, $x$ students passed the exam, then
$$
71 x+56(n-x)=66 n \Rightarrow x=\frac{2 n}{3},
$$
which means $n$ is a multiple of 3.
Let $y$ be the number of students who passed after the bonus points. Then
$$
75 y+59(n-y)=71 n \Rightarrow y=\frac{3 n}{4},
$$
which means $n$ is a multiple of 4.
In summary, $n$ is a multiple of 12.
Since $15<n<30$, therefore, $n=24$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 2, given that $A, B, C, D$ are four points on a plane that are not concyclic,
$$
\begin{array}{l}
\triangle A B D, \triangle A D C \text {, } \\
\triangle B C D, \triangle A B C
\end{array}
$$
have circumcenters
$$
\text { as } E, F, G, H \text {, }
$$
respectively. The line segments $E G, F H$
intersect at point $I$. If
$$
A I=4, B I=3 \text {, then } C I=
$$
|
8.4.
Since points $E$ and $G$ lie on the perpendicular bisector of $BD$, $EG$ is the perpendicular bisector of segment $BD$. Similarly, $FH$ is the perpendicular bisector of segment $AC$. Since $I$ is the intersection of $EG$ and $FH$, it follows that $CI = AI = 4$.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. There is a six-digit number, the sum of its digits is divisible by 26. When this six-digit number is increased by 1, the sum of the digits of the resulting number is also divisible by 26. The smallest six-digit number that satisfies the above conditions is $\qquad$
|
10. 898999.
Obviously, this six-digit number will have a carry when 1 is added.
To make the sum of the digits a multiple of 26, at least three positions need to carry over. Therefore, the last three digits of this six-digit number are all 9, and the sum of the digits of the first three is 25.
Thus, the smallest number that meets the condition is 898999.
|
898999
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. On a circle, there is 1 red point and 2009 blue points. Xiao Dan calculates the number of convex polygons where all vertices are blue points, while Xiao Dong calculates the number of convex polygons where one vertex is the red point. The difference between the two numbers they get is $\qquad$
|
11. 2017036.
For every convex $n$-sided polygon $(n \in \mathbf{N}_{+})$ calculated by Xiao Dan, there is a corresponding convex $(n+1)$-sided polygon calculated by Xiao Dong. Xiao Dong's convex $(n+1)$-sided polygon has one more red vertex than Xiao Dan's convex $n$-sided polygon. This correspondence does not hold when the polygon calculated by Xiao Dong is a triangle. Therefore, the difference in the number of polygons calculated by the two is the number of triangles calculated by Xiao Dong:
$$
\frac{2008 \times 2009}{2}=2017036 .
$$
|
2017036
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In an international chess tournament with 10 participants, each player must play exactly one game against every other player. After several games, it is found that among any three players, at least two have not yet played against each other. How many games have been played at most by this point?
|
2. Suppose there are 5 men and 5 women among the 10 players, and all matches so far have been between men and women, thus satisfying the condition, i.e., 25 matches have been played.
Next, we prove: 25 is indeed the maximum value.
Let $k$ be the number of matches played by the player who has played the most matches, denoted as $A$. The $k$ players who have played against $A$ have not played against each other, meaning each of them has played at most 10 $k$ matches. The remaining $10-k$ players have each played at most $k$ matches. Therefore, the total number of matches is at most
$$
\begin{array}{c}
\frac{k(10-k)+(10-k) k}{2}=-k^{2}+10 k \\
=25-(k-5)^{2} \leqslant 25 .
\end{array}
$$
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Find the smallest positive integer that can be expressed as the sum of the squares of four positive integers and can divide some integer of the form $2^{n}+15\left(n \in \mathbf{N}_{+}\right)$.
|
3. The smallest five positive integers that can be expressed as the sum of four positive integer squares are
$$
\begin{array}{l}
4=1+1+1+1, \\
7=4+1+1+1, \\
10=4+4+1+1, \\
12=9+1+1+1, \\
13=4+4+4+1 .
\end{array}
$$
Obviously, since $2^{n}+15$ is odd, the smallest positive integer cannot be 4, 10, or 12.
Also, $2^{n} \equiv 1,2,4 \neq-1(\bmod 7)$, so $2^{n}+15$ is not a multiple of 7.
And $2^{7}+15=143=11 \times 13$, hence the smallest positive integer that meets the condition is 13.
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. (40 points) There is a stack of cards numbered $1 \sim 15$. After arranging them in a certain order, the following two operations are performed: Place the top card on the table, then place the second card at the bottom of the stack. These two operations are repeated until all 15 cards are sequentially placed on the table. If the cards on the table are numbered from top to bottom as $1 \sim 15$ in order, what is the fourth card from the bottom of the original stack?
|
1. Assuming the cards are arranged in descending order from top to bottom as shown from left to right in Table 1. The first card originally becomes the bottom card, so it must be 15. The third card must be 14. And so on.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & $\alpha$ & 14 & $\beta$ & 13 & & 12 & & 11 & & 10 & & 9 & & 8 \\
\hline
\end{tabular}
Since the top 8 cards have been removed, the card marked “ $\alpha$ ” becomes the card after 8 and should be placed at the bottom. Therefore, the card marked “ $\beta$ ” is 7. And so on, we get Table 2.
Table 2
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & $\alpha$ & 14 & 7 & 13 & & 12 & 6 & 11 & & 10 & 5 & 9 & & 8 \\
\hline
\end{tabular}
Similarly, we can determine the original arrangement of the cards (see Table 3).
Table 3
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline 15 & 4 & 14 & 7 & 13 & 2 & 12 & 6 & 11 & 3 & 10 & 5 & 9 & 1 & 8 \\
\hline
\end{tabular}
Therefore, the fourth-to-last card is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (40 points) In each game of bridge, four players play together. It is known that the number of games played is exactly equal to the number of players, and any two players have played at least one game together. Try to find the maximum number of players.
|
9. There are $n$ players, and each game can produce 6 pairs of players.
From $\mathrm{C}_{n}^{2} \leqslant 6 n$ we get
$$
n \leqslant 13 \text {. }
$$
Next, we prove that $n=13$ is feasible.
As shown in Figure 13, number the 13
players from $0 \sim 12$.
The players in each game are as follows:
$$
\begin{array}{l}
(0,2,3,7), \\
(1,3,4,8), \\
(2,4,5,9), \\
(3,5,6,10),(4,6,7,11),(5,7,8,12), \\
(6,8,9,0),(7,9,10,1),(8,10,11,2), \\
(9,11,12,3),(10,12,0,4),(11,0,1,5), \\
(12,1,2,6) .
\end{array}
$$
The conditions are satisfied.
|
13
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In a $10 \times 10$ grid, there is a shape composed of $4 n$ $1 \times 1$ small squares, which can be covered by $n$ "田" shaped figures, or by $n$ "世" or "円" shaped figures (which can be rotated). Find the minimum value of the positive integer $n$.
(Zhu Huaiwei, problem contributor)
|
7. Let the given shapes be denoted as type $A$ and type $B$.
First, we prove that $n$ is even.
Color the $10 \times 10$ grid as shown in Figure 6.
Regardless of which 4 squares type $A$ covers, the number of black squares must be even, while for type $B$ it is odd. If $n$ is odd, the number of black squares covered by $n$ type $A$ shapes must be even; while the number of black squares covered by $n$ type $B$ shapes must be odd, leading to a contradiction. Therefore, $n$ must be even.
If $n=2$, the shapes formed by two type $A$ shapes can only be the two cases shown in Figure 7. However, neither of these can be formed by two type $B$ shapes.
Thus, $n \geq 4$.
Figure 8 shows the tiling when $n=4$.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.1. A simplest fraction is equal to the sum of two simplest fractions with denominators 600 and 700, respectively. Find the smallest possible value of the denominator of such a simplest fraction.
|
9. 1. $2^{3} \times 3 \times 7=168$.
Let the two simplest fractions be $\frac{a}{600}$ and $\frac{b}{700}$. Then $(a, 6)=(b, 7)=1$.
Thus, the sum $\frac{7 a+6 b}{4200}$ has a numerator that is coprime with 6 and 7.
Since $4200=2^{3} \times 3 \times 7 \times 5^{2}$, after canceling out the common factors, the denominator is no less than $2^{3} \times 3 \times 7=168$.
On the other hand, such a denominator can be obtained:
$$
\frac{1}{600}+\frac{3}{700}=\frac{1}{168} \text {. }
$$
|
168
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.4. In a regular 100-gon, each vertex is covered by a cloth. It is known that exactly one of the cloths hides a coin. The following action is called an operation: choose any four cloths to check if there is a coin underneath, and after each operation, the cloths are returned to their original positions, while the coin is secretly moved to the cloth under an adjacent vertex. Find the minimum number of operations required to definitely find the coin.
|
9.4.33 times.
Assume that a regular 100-gon is placed on a rotatable horizontal circular table, and the vector from the center to one vertex points due north. The initial positions of the vertices of the polygon are defined from due north in a counterclockwise direction as $0, 1, \cdots, 99$. Each operation and coin transfer will cause the table to rotate automatically by $\frac{2 \pi}{100}$ in a clockwise direction. This means the coin's position either remains unchanged or moves two positions clockwise. Therefore, the parity of the coin's position remains unchanged after each automatic rotation of the table.
If a position cannot have a coin at a certain moment, it is called "empty." Otherwise, it is called "non-empty."
Clearly, at the initial moment, the number of empty positions is 0.
Thus, each operation can increase the number of empty positions by at most 4.
Consecutive odd (even) empty positions form an odd (even) empty interval.
If an odd (even) empty interval is not a proper subset of another odd (even) empty interval, it is called "maximal."
Due to the transfer of the coin, after the table rotates,
the counterclockwise endpoint of each maximal odd (even) empty interval will become non-empty. This indicates that during the entire process of finding the coin, the transfer of the coin will not increase the number of non-empty positions only in the following situation: when all odd (even) positions are empty, and all even (odd) positions are non-empty (this situation can occur at most once). In all other cases, the number of empty positions before this operation is at most 3 more than the number of empty positions before the previous operation. Thus, after 32 operations, the number of empty positions is at most $31 \times 3 + 4 = 97$, which does not guarantee finding the coin.
Next, we show that the coin can be found in at most 33 operations.
First, check positions $0, 2, 4, 6$.
If the coin is found, the problem is solved;
If not, then after the table rotates, positions $0, 2, 4$ are empty (assuming the coin is not found in the first 32 operations).
Generally, if after the table rotates, it is known that $0, 2, \cdots, 2s$ are empty positions, then the next operation checks $2s+2, 2s+4, 2s+6, 2s+8$.
After the coin transfer and table rotation, $0, 2, \cdots, 2s+6$ are empty positions. Thus, after 16 operations and table rotations, 48 empty positions are obtained, i.e., $0, 2, \cdots, 94$.
The 17th operation checks $96, 98, 1, 3$.
After the table rotates, 51 empty positions are obtained, i.e., $0, 1, 2, 4, 6, \cdots, 98$. Next, check $3, 5, 7, 9; 9, 11, 13, 15; \cdots$.
After the 32nd operation and table rotation, 96 empty positions are obtained, i.e., $0, 1, 2, \cdots, 91, 92, 94, 96, 98$.
The 33rd operation checks the last four positions $93, 95, 97, 99$, and the coin will certainly be found.
|
33
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9.7. A diagonal of eight squares on a chessboard is called a "fence". A rook starts from a square outside the fence on the chessboard and moves, satisfying the following conditions:
(1) It stays on any square of the chessboard at most once;
(2) It never stays on a square of the fence.
Find the maximum number of times the rook can cross the fence.
|
9.7.47 times.
Let the square at the $i$-th row and $j$-th column be denoted as $(i, j)$. Suppose the eight squares occupied by the fence are $(i, i) (i=1,2, \cdots, 8)$. The non-fence squares are divided into four categories $A, B, C, D$:
$$
\begin{aligned}
A= & \{(i, j) \mid 2 \leqslant j+1 \leqslant i \leqslant 4\} \cup \\
& \{(i, j) \mid 5 \leqslant j \leqslant i-1 \leqslant 7\}, \\
B= & \{(i, j) \mid 2 \leqslant i+1 \leqslant j \leqslant 4\} \cup \\
& \{(i, j) \mid 5 \leqslant i \leqslant j-1 \leqslant 7\}, \\
C= & \{(i, j) \mid 5 \leqslant i \leqslant 8,1 \leqslant j \leqslant 4\}, \\
D= & \{(i, j) \mid 1 \leqslant i \leqslant 4,5 \leqslant j \leqslant 8\} .
\end{aligned}
$$
Since the car crosses the fence at least once in either an $A$ or $B$ square before and after each crossing, each square in $A$ or $B$ can contribute to at most two crossings. Since there are 24 squares in $A$ and $B$, the maximum number of crossings is 48.
We call the crossing between $A$ and $D$ squares an $AD$ type, and the crossing between $B$ and $C$ squares a $BC$ type.
If exactly 48 crossings are completed, then each $A$ or $B$ square corresponds to two crossings. This means that each stop before and after a crossing is exactly once in a $C$ or $D$ square, and each stop in an $A(B)$ square is followed by an $AD(BC)$ type crossing. Since it is impossible to reach a $D(C)$ square from a $C(D)$ square without crossing, the car can only complete one type of crossing, hence at most 24 crossings. This is a contradiction.
Therefore, the car can complete at most 47 crossings.
An example is as follows:
$$
\begin{array}{l}
(8,2) \xrightarrow{1}(1,2) \xrightarrow{2}(7,2) \xrightarrow{3}(7,8) \xrightarrow{4} \\
(7,3) \xrightarrow{5}(1,3) \xrightarrow{6}(8,3) \xrightarrow{7}(2,3) \xrightarrow{8} \\
(6,3) \xrightarrow{9}(6,8) \xrightarrow{10}(6,2) \xrightarrow{11}(6,7) \xrightarrow{12} \\
(6,4) \xrightarrow{13}(1,4) \xrightarrow{14}(8,4) \xrightarrow{15}(2,4) \xrightarrow{16}
\end{array}
$$
$$
\begin{array}{l}
(7,4) \xrightarrow{17}(3,4) \xrightarrow{18}(5,4) \xrightarrow{19}(5,8) \xrightarrow{20} \\
(5,3) \xrightarrow{21}(5,7) \xrightarrow{22}(5,2) \xrightarrow{23}(5,6) \xrightarrow{24} \\
(8,6) \xrightarrow{25}(1,6) \xrightarrow{26}(7,6) \xrightarrow{27}(2,6) \xrightarrow{28} \\
(2,1) \xrightarrow{29}(2,7) \xrightarrow{30}(8,7) \xrightarrow{31}(3,7) \xrightarrow{32} \\
(3,1) \xrightarrow{33}(3,6) \xrightarrow{34}(3,2) \xrightarrow{35}(3,5) \xrightarrow{36} \\
(8,5) \xrightarrow{37}(2,5) \xrightarrow{38}(7,5) \xrightarrow{39}(1,5) \xrightarrow{40} \\
(6,5) \xrightarrow{41}(4,5) \xrightarrow{42}(4,1) \xrightarrow{43}(4,6) \xrightarrow{44} \\
(4,2) \xrightarrow{45}(4,7) \xrightarrow{46}(4,3) \xrightarrow{47}(4,8) .
\end{array}
$$
|
47
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 1, given $A(-3,0)$, $B(0,-4)$, and $P$ is any point on the hyperbola $y=\frac{12}{x} (x > 0)$. A perpendicular line from point $P$ to the x-axis meets at point $C$, and a perpendicular line from point $P$ to the y-axis meets at point $D$. Then the minimum value of the area $S$ of quadrilateral $A B C D$ is ( ).
(A) 22
(B) 23
(C) 24
(D) 26
|
2. C.
Let $P\left(x, \frac{12}{x}\right)$. Then $C(x, 0)$ and $D\left(0, \frac{12}{x}\right)$.
It is easy to see that $C A=x+3, D B=\frac{12}{x}+4$, so
$$
S=\frac{1}{2} C A \cdot D B=\frac{1}{2}(x+3)\left(\frac{12}{x}+4\right) \text {. }
$$
Simplifying, we get $S=2\left(x+\frac{9}{x}\right)+12$.
Since $x>0, \frac{9}{x}>0$, we have
$$
x+\frac{9}{x}=\left(\sqrt{x}+\frac{3}{\sqrt{x}}\right)^{2}+6 \geqslant 6 \text {. }
$$
Equality holds only when $x=\frac{9}{x}$, i.e., $x=3$. Therefore, $S \geqslant 2 \times 6+12=24$.
|
24
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
2. As shown in Figure 2, the first polygon is "expanded" from an equilateral triangle, with the number of sides denoted as $a_{3}$, the second polygon is "expanded" from a square, with the number of sides denoted as $a_{4}, \cdots \cdots$ and so on. The polygon "expanded" from a regular $n(n \geqslant 3)$-sided polygon has the number of sides denoted as $a_{n}$. When $\frac{1}{a_{3}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{n}}$ equals $\frac{2007}{6030}$, the value of $n$ is $\qquad$ .
(1)
(2)
(3)
(4)
Figure 2
|
2. 2009 .
From the extended definition, we know
$$
\begin{array}{l}
a_{3}=12=3 \times 4, \\
a_{4}=20=4 \times 5, \\
\cdots \cdots \\
a_{n}=n(n+1) . \\
\text { Also, } \frac{1}{a_{n}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \\
\Rightarrow \frac{1}{a_{3}}+\frac{1}{a_{4}}+\cdots+\frac{1}{a_{n}} \\
=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right) \\
=\frac{1}{3}-\frac{1}{n+1}=\frac{2007}{6030} \\
\Rightarrow \frac{1}{n+1}=\frac{3}{6030}=\frac{1}{2010} \Rightarrow n=2009 .
\end{array}
$$
|
2009
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $41^{x}=2009,7^{y}=2009$. Then the value of $\frac{1}{x}+\frac{2}{y}$ is . $\qquad$
|
4. 1 .
From $41^{x}=2009$, we get $41^{x y}=2009^{y}$.
$$
\begin{array}{l}
\text { Also } 7^{y}=2009 \Rightarrow 7^{2 y}=2009^{2} \\
\Rightarrow 49^{y}=2009^{2} \Rightarrow 49^{x y}=2009^{2 x} .
\end{array}
$$
$$
\text { Also } 41 \times 49=2009 \Rightarrow 41^{x y} \times 49^{x y}=2009^{x y} \text {. }
$$
Therefore $2009^{y} \times 2009^{2 x}=2009^{x y}$
$$
\begin{array}{c}
\Rightarrow 2009^{2 x+y}=2009^{x y} \Rightarrow 2 x+y=x y \\
\Rightarrow \frac{2}{y}+\frac{1}{x}=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Among the natural numbers from 1 to 144, the number of ways to pick three numbers that form an increasing geometric progression with an integer common ratio is $\qquad$ .
|
3. 78.
Let the three numbers $a$, $a q$, and $a q^{2}$ form an increasing geometric sequence. Then
$$
1 \leqslant a < a q < a q^{2} \leqslant 144 \text{.}
$$
From this, we have $2 \leqslant q \leqslant 12$.
When $q$ is fixed, the number of integers $a$ such that the three numbers $a$, $a q$, and $a q^{2}$ are integers is denoted as $N(q)$.
$$
\begin{array}{l}
\text{By } a q^{2} \leqslant 144 \text{, we have } N(q) = \left[\frac{144}{q^{2}}\right] \text{.} \\
\text{Also, } N(2) = 36, N(3) = 16, N(4) = 9, \\
N(5) = 5, N(6) = 4, N(7) = N(8) = 2, \\
N(9) = N(10) = N(11) = N(12) = 1 \text{.}
\end{array}
$$
Therefore, the total number of ways is
$$
N(2) + N(3) + \cdots + N(12) = 78 \text{ (ways).}
$$
|
78
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (14 points) Let positive real numbers $x, y, z$ satisfy $xyz=1$. Try to find the maximum value of
$$
f(x, y, z)=(1-yz+z)(1-xz+x)(1-xy+y)
$$
and the values of $x, y, z$ at that time.
|
$$
\begin{array}{l}
\left\{\begin{array} { l }
{ 1 - y z + z < 0 , } \\
{ 1 - x z + x < 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+x z<1, \\
y+x y<1
\end{array}\right.\right. \\
\Rightarrow(x+x z)(y+x y)<1 \\
\Leftrightarrow x+x y+x^{2} y<0,
\end{array}
$$
Contradiction. Therefore, among $1-y z+z$, $1-x z+x$, and $1-x y+y$, at most one can be negative.
Assume all three expressions are positive. Then,
$$
\begin{array}{l}
(1-y z+z)(1-x z+x)(1-x y+y) \\
=(x-x y z+x z)(y-x y z+x y)(z-x y z+y z) \\
=(x-1+x z)(y-1+x y)(z-1+y z),
\end{array}
$$
Thus,
$$
\begin{array}{l}
f^{2}(x, y, z) \\
= {[(1-y z+z)(1-x z+x)(1-x y+y)]^{2} } \\
= {[(1-y z+z)(1-x z+x)(1-x y+y)] . } \\
{[(x-1+x z)(y-1+x y)(z-1+y z)] } \\
= {\left[z^{2}-(1-y z)^{2}\right]\left[x^{2}-(1-x z)^{2}\right]\left[y^{2}-(1-x y)^{2}\right] } \\
\leqslant(x y z)^{2}=1 .
\end{array}
$$
When $x=y=z=1$, the equality holds.
Therefore, $f(x, y, z)_{\text {max }}=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The number of integer points within the square (including the four sides) formed by the four lines $y=x+10, y=-x+10, y=$ $x-10, y=-x-10$ in the Cartesian coordinate system is $\qquad$ .
|
(Tip: By plotting, it is known that the number of integer points in each quadrant is 45. The number of integer points on the coordinate axes is 41, so, the number of integer points is $(4 \times 45+41=) 221$.
|
221
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question Arrange all positive integers $m$ whose digits are no greater than 3 in ascending order to form a sequence $\left\{a_{n}\right\}$. Then $a_{2007}=$ $\qquad$
$(2007$, National High School Mathematics League Jiangxi Province Preliminary Contest)
|
Another solution: Considering that $a_{n}$ is not continuous in decimal, and the digits of $a_{n}$ are all no greater than 3 (i.e., $0, 1, 2, 3$), then the form of $a_{n}$ is a sequence of consecutive integers in quaternary, thus simplifying the problem.
Let the set $\left\{\left(a_{n}\right)_{4}\right\}=\{$ consecutive positive integers in quaternary $\}$ $\left((x)_{k}\right.$ denotes $x$ as a $k$-ary number, for example, $(11)_{4}=$ (5) 10 ). Therefore,
$$
\begin{array}{l}
\left(a_{n}\right)_{4}=\left(a_{n-1}\right)_{4}+(1)_{10} \\
=\left(a_{n-2}\right)_{4}+(2)_{10}=\cdots=(n)_{10} .
\end{array}
$$
When $n=2007$, we have
$$
\left(a_{2007}\right)_{4}=(2007)_{10} \text {. }
$$
Therefore, $a_{2007}=133113$.
Table 1 illustrates this.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $a_{1}$ & $a_{2}$ & $a_{3}$ & $c_{4}$ & $a_{5}$ & $a_{6}$ & $a_{7}$ & $a_{8}$ & $a_{9}$ & $a_{10}$ & $1 \ldots$ & $a_{2007}$ \\
\hline Subscript & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & $\cdots$ & 2007 \\
\hline Quaternary & 1 & 2 & 3 & 10 & 11 & 12 & 13 & 20 & 21 & $n$ & $\cdots$ & 133113 \\
\hline Value of $a_{n}$ & 1 & 2 & 3 & 10 & 11 & 12 & 13 & 20 & 21 & 22 & $\cdots$ & 133113 \\
\hline
\end{tabular}
|
133113
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For example, the five-digit numbers formed by the digits $1, 2, 3, 4, 5$ with repetition, arranged in ascending order. Ask:
(1) What are the positions of 22435 and 43512?
(2) What is the 200th number? ${ }^{[1]}$
|
Analysis: Let the $n$-th number after sorting be denoted as $a_{n}$. Since the numbers do not contain 0, we cannot directly use the quinary system. Therefore, subtract 11111 from all $a_{n}$, and denote the result as $b_{n}$. Thus, we can draw the following conclusions:
(i) Each number in $b_{n}$ consists of some of the digits $0, 1, 2, 3, 4$;
(ii) The $i$-th number in $\left\{a_{n}\right\}$, denoted as $a_{i}$, when subtracted by 11111, becomes $b_{i}$, which is the $i$-th number in $\left\{b_{n}\right\}$, meaning the order of the sequence remains unchanged after the transformation;
(iii) $b_{n}$ is a non-negative integer with one, two, three, four, or five digits.
Thus, $\left\{b_{n}\right\}$ meets the requirement of being a sequence of consecutive quinary integers, and the method in the problem can be used to solve it.
Solution: (1) Note that
$$
\begin{array}{l}
22435-11111=11324, \\
43512-11111=32401 .
\end{array}
$$
Therefore, the order numbers of 22435 and 43512 are
$$
\begin{array}{l}
(11324)_{5}+1=840, \\
(32401)_{5}+1=2227 .
\end{array}
$$
(2) Since $200=(1244)_{5}+1$, the 200th
number is $1244+11111=12355$.
|
12355
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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