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Initial 281 Given that there exist $k(k \in \mathbf{N}, k \geqslant 2)$ consecutive positive integers, the mean of their squares is a perfect square. Try to find the minimum value of $k$.
|
Let the $k$ positive integers be $n+1, n+2, \cdots, n+k (n \in \mathbf{N})$. Then the mean of their squares is
$$
\begin{array}{l}
f(n, k)=\frac{1}{k} \sum_{i=1}^{k}(n+i)^{2} \\
=n^{2}+n(k+1)+\frac{(k+1)(2 k+1)}{6} .
\end{array}
$$
Since $f(n, k) \in \mathbf{Z}$, then $\frac{(k+1)(2 k+1)}{6} \in \mathbf{Z}$.
Thus, $2 \times k, 3 \times k$.
When $2<k \leqslant 13$, we have
$$
\left(n+\frac{k+1}{2}\right)^{2}<f(n, k)<\left(n+\frac{k+1}{2}+1\right)^{2} \text {. }
$$
In fact,
$$
\begin{array}{l}
\text { Equation (1) } \Leftrightarrow n^{2}+n(k+1)+\frac{k^{2}+2 k+1}{4} \\
<n^{2}+n(k+1)+\frac{(k+1)(2 k+1)}{6} \\
<n^{2}+n(k+3)+\frac{k^{2}+6 k+9}{4} \\
\Leftrightarrow \frac{k^{2}+2 k+1}{4}<\frac{2 k^{2}+3 k+1}{6}<\frac{k^{2}+6 k+9}{4}+2 n \\
\Leftrightarrow 3\left(k^{2}+2 k+1\right)<2\left(2 k^{2}+3 k+1\right) \\
<3\left(k^{2}+6 k+9\right)+24 n \\
\Leftrightarrow 1<k^{2}<12 k+25+24 n .
\end{array}
$$
The above inequality is clearly true.
At this point, $\sqrt{f(n, k)} \notin \mathbf{Z}$.
When $14<k<26$, we have
$$
\left(n+\frac{k+1}{2}\right)^{2}<f(n, k)<\left(n+\frac{k+1}{2}+2\right)^{2} \text {. }
$$
In fact,
$$
\begin{array}{l}
\text { Equation (2) } \Leftrightarrow \frac{k^{2}+2 k+1}{4}<\frac{2 k^{2}+3 k+1}{6} \\
<\frac{k^{2}+10 k+25}{4}+4 n \\
\Leftrightarrow 3\left(k^{2}+2 k+1\right)<2\left(2 k^{2}+3 k+1\right) \\
<3\left(k^{2}+10 k+25\right)+48 n \\
\Leftrightarrow 1<k^{2}<24 k+73+48 n .
\end{array}
$$
The above inequality is clearly true.
At this point, $f(n, k)=\left(n+\frac{k+1}{2}+1\right)^{2}$, i.e., $k^{2}=24 n+12 k+25$.
Then $(k-6)^{2}=24 n+61=8(3 n+7)+5$.
But the remainder of a perfect square when divided by 8 is 0 or 1, which is a contradiction.
When $k=29$,
$$
f(n, 29)=n^{2}+30 n+295=(n+15)^{2}+70 \text {, }
$$
its remainder when divided by 8 is 6 or 7, which is a contradiction.
When $k=31$,
$$
\begin{array}{c}
f(n, 31)=n^{2}+32 n+336 \\
=(n+18)^{2}-4 n+12 .
\end{array}
$$
Taking $n=3$, we get $f(3,31)=21^{2}$.
Therefore, $k_{\text {min }}=31$.
(Zhai Xiaojun, Donghai Senior High School, Jiangsu Province, 222300)
|
31
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If $m^{2}=m+1, n^{2}=n+1$, and $m \neq n$, then $m^{5}+n^{5}=$ $\qquad$
(Fourth Jiangsu Province Junior High School Mathematics Competition)
|
Given that $m$ and $n$ are the roots of the equation $x^{2}=x+1$, i.e., $x^{2}-x-1=0$.
Let $a_{k}=m^{k}+n^{k}$. Then
$$
a_{k}=a_{k-1}+a_{k-2}(k \geqslant 3) \text {. }
$$
By $a_{1}=m+n=1$,
$$
a_{2}=(m+n)^{2}-2 m n=3 \text {, }
$$
we know $a_{3}=a_{2}+a_{1}=4, a_{4}=a_{3}+a_{2}=7$.
Therefore, $m^{5}+n^{5}=a_{5}=a_{4}+a_{3}=11$.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $m$ and $n$ are rational numbers, and the equation
$$
x^{2}+m x+n=0
$$
has a root $\sqrt{5}-2$. Then $m+n=$ $\qquad$ .
(2001, National Junior High School Mathematics Competition, Tianjin Preliminary Round)
|
$$
\begin{array}{l}
(\sqrt{5}-2)^{2}+m(\sqrt{5}-2)+n=0 \\
\Rightarrow(n-2 m+9)=(4-m) \sqrt{5} \\
\Rightarrow\left\{\begin{array} { l }
{ n - 2 m + 9 = 0 , } \\
{ 4 - m = 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
m=4, \\
n=-1
\end{array}\right.\right. \\
\Rightarrow m+n=3 \text {. } \\
\end{array}
$$
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $x$ and $y$ be positive integers such that
$$
\sqrt{x-116}+\sqrt{x+100}=y \text {. }
$$
Find the maximum value of $y$.
|
Prompt: By analogy with Example 6, we can prove that $\sqrt{x-116}$ and $\sqrt{x+100}$ are both natural numbers.
Let $\sqrt{x-116}=a, \sqrt{x+100}=b$. Then $b^{2}-a^{2}=216 \Rightarrow (b+a)(b-a)=216$.
Also, $b+a \equiv (b-a) \pmod{2}$, and since 216 is even, both $b-a$ and $b+a$ are even and positive.
Thus, $b-a \geqslant 2$.
Then $y=b+a \leqslant 108$.
When $\left\{\begin{array}{l}b+a=108, \\ b-a=2\end{array}\right.$, i.e., $\left\{\begin{array}{l}b=55, \\ a=53,\end{array}\right.$, i.e., $x=53^{2}+116$, $y$ reaches its maximum value.
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Let real numbers $x, y, z, w$ satisfy
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}\right.
$$
Find the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
|
Consider the function
$$
f(x)=\prod_{i=1}^{4}\left[x-(2 i-1)^{2}\right]-\prod_{i=1}^{4}\left[x-(2 i)^{2}\right] \text {. }
$$
Then $f\left(k^{2}\right)=\prod_{i=1}^{4}\left[k^{2}-(2 i-1)^{2}\right](k=2,4,6,8)$.
By the Lagrange interpolation formula, we have
$$
f(x)=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \neq i<i<4 \\ 1 \leq j<4}} \frac{x-(2 i-1)^{2}}{(2 i-1)^{2}-(2 j-1)^{2}} .
$$
Comparing the coefficients of $x^{3}$ in equations (1) and (3), we get
$$
\begin{array}{l}
\sum_{i=1}^{4}\left[(2 i)^{2}-(2 i-1)^{2}\right] \\
=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \in i \\
1 \leqslant i<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} .
\end{array}
$$
In equation (3), let $x=k^{2}(k=2,4,6,8)$, we get
$$
\begin{array}{l}
f\left(k^{2}\right) \\
= \sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{j \in i<4 \\
1 \neq j \leqslant 4}} \frac{k^{2}-(2 j-1)^{2}}{(2 i-1)^{2}-(2 j-1)^{2}} \\
= \sum_{i=1}^{4} \frac{f\left((2 i-1)^{2}\right)}{k^{2}-(2 i-1)^{2}} \prod_{\substack{j+i<i \\
1<j \leqslant 4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} . \\
\prod_{t=1}^{4}\left[k^{2}-(2 t-1)^{2}\right] .
\end{array}
$$
From equations (2) and (4), we have
$$
\begin{array}{l}
\sum_{i=1}^{4} \frac{f\left((2 i-1)^{2}\right)}{k^{2}-(2 i-1)^{2}} \prod_{\substack{j \neq i<i \\
1 \leqslant \leqslant<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} \\
\quad=1(k=2,4,6,8) .
\end{array}
$$
By the uniqueness of the solution of the system of equations, we have
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}+w^{2} \\
=\sum_{i=1}^{4} f\left((2 i-1)^{2}\right) \prod_{\substack{1 \neq i<\\
1<j<4}}\left[(2 i-1)^{2}-(2 j-1)^{2}\right]^{-1} \\
=\sum_{i=1}^{4}\left[(2 i)^{2}-(2 i-1)^{2}\right]=36 .
\end{array}
$$
**Summary**: The construction of the function is not unique. For example, let
$$
\begin{aligned}
f(m)= & \left(\frac{x^{2}}{m-1^{2}}+\frac{y^{2}}{m-3^{2}}+\frac{z^{2}}{m-5^{2}}+\frac{w^{2}}{m-7^{2}}\right) . \\
& \prod_{i=1}^{4}\left[m-(2 i-1)^{2}\right] .
\end{aligned}
$$
Then by the Lagrange interpolation formula, we have
$$
f(m)=\sum_{i=1}^{4} f\left((2 i)^{2}\right) \prod_{\substack{j \neq i \\ 1 \leqslant j \leqslant 4}} \frac{m-(2 j)^{2}}{(2 i)^{2}-(2 j)^{2}} \text {. }
$$
Comparing the coefficients of $m^{3}$, we get the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Find the smallest positive integer $n$, such that for any sequence of $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $\sum_{i=1}^{n} a_{i}=2007$, there must be a sum of some consecutive terms equal to 30.
(Fourth China Southeast Mathematical Olympiad)
|
First, construct an integer sequence $a_{1}, a_{2}, \cdots, a_{1017}$ with 1017 terms, such that no consecutive terms sum to 30. For this, take
$$
a_{1}=a_{2}=\cdots=a_{29}=1, a_{30}=31,
$$
and $a_{30 m+i}=a_{i}(i=1,2, \cdots, 30, m \in \mathbf{N})$.
Thus, $\left\{a_{k}\right\}$ is:
$$
\begin{array}{l}
1,1, \cdots, 1,31 ; 1,1, \cdots, 1,31 ; \cdots ; \\
1,1, \cdots, 1,31 ; 1,1, \cdots, 1
\end{array}
$$
(There are 34 segments, the first 33 segments each have 30 terms, and the last segment has 27 terms, totaling 1017 terms).
Second, when the number of terms is less than 1017, it is sufficient to merge some consecutive numbers in certain segments into larger numbers.
For any positive integer sequence $a_{1}, a_{2}, \cdots, a_{1018}$ with 1018 terms that satisfies the condition $\sum_{i=1}^{1018} a_{i}=2007$, we will prove that there must be consecutive terms whose sum equals 30.
In fact, let $S_{k}=\sum_{i=1}^{k} a_{i}$. Then
$$
1 \leqslant S_{1}<S_{2}<\cdots<S_{1018}=2007 \text {. }
$$
Now consider the grouping of elements in the set $\{1,2, \cdots, 2007\}$: $\square$
$$
\begin{array}{c}
(60 k+i, 60 k+30+i), \\
(k=0,1, \cdots, 32 ; i=1,2, \cdots, 30), \\
1981,1982, \cdots, 2007 .
\end{array}
$$
There are $33 \times 30=990$ pairs and 27 numbers not in pairs. From these, any 1018 numbers taken as the values of $S_{k}$ must include two numbers from the same pair, say $\left(S_{k}, S_{k+m}\right)$. Then $S_{k+m}-S_{k}=30$, meaning that in the sequence,
$$
a_{k+1}+a_{k+2}+\cdots+a_{k+m}=30 .
$$
Therefore, the minimum value of $n$ is 1018.
|
1018
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For a natural number $n$, let the sum of its digits be denoted as $a_{n}$, for example,
$$
\begin{array}{l}
a_{2009}=2+0+0+9=11, \\
a_{2010}=2+0+1+0=3 .
\end{array}
$$
Then $a_{1}+a_{2}+\cdots+a_{2010}=$ ( ).
(A) 28062
(B) 28065
(C) 28067
(D) 28068
|
6. D.
Consider all natural numbers from 1 to 2010 as four-digit numbers (if $n$ is less than four digits, add 0s at the beginning to make it four digits, which does not change the value of $a_{n}$).
Notice that, the number of times 1 appears in the thousands, hundreds, tens, and units place are $10^{3}$, $2 \times 10^{2}$, $2 \times 10^{2}+1$, and $2 \times 10^{2}+1$ respectively. Therefore, the total number of times 1 appears is
$$
10^{3}+2 \times 10^{2} \times 3+2=1602 \text {. }
$$
The number of times 2 appears in the thousands, hundreds, tens, and units place are $11$, $2 \times 10^{2}$, $2 \times 10^{2}$, and $2 \times 10^{2}+1$ respectively. Therefore, the total number of times 2 appears is
$$
11+2 \times 10^{2} \times 3+1=612 \text {. }
$$
Similarly, the total number of times $k(k=3,4, \cdots, 9)$ appears is $2 \times 10^{2} \times 3+1=601$.
$$
\begin{array}{l}
\text { Hence } a_{1}+a_{2}+\cdots+a_{2010} \\
=1602 \times 1+612 \times 2+601(3+4+\cdots+9) \\
=28068 \text {. }
\end{array}
$$
|
28068
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 The number of positive integer values of $n$ that satisfy $\left|\sqrt{\frac{n}{n+2009}}-1\right|>\frac{1}{1005}$ is $\qquad$
(2009, I Love Mathematics Junior High School Summer Camp Mathematics Competition)
|
Given that $n$ is a positive integer, we have
$$
\begin{array}{l}
0\frac{1}{1005} \\
\quad \Rightarrow \sqrt{\frac{n}{n+2009}}<\frac{1004}{1005} .
\end{array}
$$
Let $1004=a$. Then
$$
\begin{array}{l}
\frac{n}{n+2 a+1}<\frac{a^{2}}{(a+1)^{2}} \\
\Rightarrow\left[(a+1)^{2}-a^{2}\right] n<a^{2}(2 a+1) \\
\Rightarrow(2 a+1) n<a^{2}(2 a+1) \\
\Rightarrow n<a^{2} \Rightarrow n<1004^{2} .
\end{array}
$$
Since $n$ is a positive integer, therefore, $n \leqslant 1004^{2}-1$.
Thus, the number of positive integers $n$ that satisfy the condition is $1004^{2}-1$, which is 1008015.
|
1008015
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given real numbers $x, y$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x^{3}+y^{3}=19, \\
x+y=1 .
\end{array}\right.
$$
then $x^{2}+y^{2}=$ $\qquad$
|
II, 1.13.
From $x^{3}+y^{3}=19$, we get $(x+y)\left[(x+y)^{2}-3 x y\right]=19$. Substituting $x+y=1$ yields $x y=-6$. Therefore, $x^{2}+y^{2}=(x+y)^{2}-2 x y=13$.
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Arrange several balls of two colors, red and black, in a row, requiring that both colors of balls must appear, and any two balls separated by 5 or 10 balls must be of the same color. Arrange according to this requirement, the maximum number of balls that can be placed is.
|
4. 15.
Label the positions of these balls in order as $1,2, \cdots$.
According to the problem, when $|i-j|=6$ or 11, the $i$-th ball and the $j$-th ball are the same color, denoted as $i \sim j$.
Thus, $6 \sim 12 \sim 1 \sim 7 \sim 13 \sim 2 \sim 8 \sim 14 \sim 3 \sim 9$ $\sim 15 \sim 4 \sim 10,5 \sim 11$.
Therefore, 15 balls can be placed according to the requirement.
If the number of balls is more than 15, then $10 \sim 16 \sim 5$. Thus, the colors of the balls numbered 1 to 16 are all the same. Further, it can be known that all the balls have the same color, which does not meet the requirement.
Therefore, according to this requirement, the maximum number of balls that can be placed is 15.
|
15
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Let positive integers $a, b, c (a \geqslant b \geqslant c)$ be the lengths of the sides of a triangle, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-a c-b c=13 \text {. }
$$
Find the number of triangles that meet the conditions and have a perimeter not exceeding 30.
|
Given the known equation:
$$
(a-b)^{2}+(b-c)^{2}+(a-c)^{2}=26 \text {. }
$$
Let $a-b=m, b-c=n$.
Then $a-c=m+n(m, n$ are natural numbers $)$.
Thus, equation (1) becomes
$$
m^{2}+n^{2}+m n=13 \text {. }
$$
Therefore, the pairs $(m, n)$ that satisfy equation (2) are:
$$
(m, n)=(3,1),(1,3) \text {. }
$$
(1) When $(m, n)=(3,1)$,
$$
b=c+1, a=b+3=c+4 \text {. }
$$
Since $a, b, c$ are the lengths of the sides of a triangle,
$$
b+c>a \Rightarrow(c+1)+c>c+4 \Rightarrow c>3 \text {. }
$$
Also, the perimeter of the triangle does not exceed 30, i.e.,
$$
\begin{array}{l}
a+b+c=(c+4)+(c+1)+c \leqslant 30 \\
\Rightarrow c \leqslant \frac{25}{3} .
\end{array}
$$
Thus, $3<c \leqslant \frac{25}{3}$.
Therefore, $c$ can take the values $4, 5, 6, 7, 8$, corresponding to 5 triangles that meet the conditions.
(2) When $(m, n)=(1,3)$, similarly,
$1<c \leqslant \frac{23}{3}$.
Thus, $c$ can take the values $2, 3, 4, 5, 6, 7$, corresponding to 6 triangles that meet the conditions.
In summary, the number of triangles that meet the conditions and have a perimeter not exceeding 30 is 11.
|
11
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given the quadratic function $y=x^{2}+b x-c$ whose graph passes through two points $P(1, a)$ and $Q(2,10 a)$.
(1) If $a$, $b$, and $c$ are all integers, and $c<b<8 a$, find the values of $a$, $b$, and $c$;
(2) Let the graph of the quadratic function $y=x^{2}+b x-c$ intersect the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. If the roots of the equation $x^{2}+b x-c=0$ are both integers, find the area of $\triangle A B C$.
|
Three, given that points $P(1, a)$ and $Q(2, 10a)$ are on the graph of the quadratic function $y = x^2 + bx - c$, we have
$$
1 + b - c = a, \quad 4 + 2b - c = 10a.
$$
Solving these equations, we get $b = 9a - 3$ and $c = 8a - 2$.
(1) From $c < b < 8a$, we know
$$
8a - 2 < 9a - 3 < 8a \Rightarrow 1 < a < 3.
$$
Since $a$ is an integer, we have
$$
a = 2, \quad b = 9a - 3 = 15, \quad c = 8a - 2 = 14.
$$
(2) Let $m$ and $n$ be the two integer roots of the equation, with $m \leq n$. By the relationship between roots and coefficients, we have
$$
m + n = -b = 3 - 9a, \quad mn = -c = 2 - 8a.
$$
Eliminating $a$, we get $9mn - 8(m + n) = -6$.
Multiplying both sides by 9 and factoring, we get
$$
(9m - 8)(9n - 8) = 10.
$$
Thus, $(9m - 8, 9n - 8)$
$$
= (1, 10), (2, 5), (-10, -1), (-5, -2).
$$
Solving these, we get $(m, n)$
$$
= (1, 2), \left(\frac{10}{9}, \frac{13}{9}\right), \left(-\frac{2}{9}, \frac{7}{9}\right), \left(\frac{1}{3}, \frac{2}{3}\right).
$$
Since $m$ and $n$ are integers, we have $(m, n) = (1, 2)$.
Therefore, $b = -(m + n) = -3$,
$$
c = -mn = -2.
$$
Thus, the quadratic function is
$$
y = x^2 - 3x + 2.
$$
It is easy to find that $A(1, 0)$, $B(2, 0)$, and $C(0, 2)$.
So, $S_{\triangle ABC} = \frac{1}{2} \times (2 - 1) \times 2 = 1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Let $p$ be a prime number greater than 2, and $k$ be a positive integer. If the graph of the function $y=x^{2}+p x+(k+1) p-4$ intersects the $x$-axis at two points, at least one of which has an integer coordinate, find the value of $k$.
---
The function is given by:
\[ y = x^2 + px + (k+1)p - 4 \]
To find the points where the graph intersects the $x$-axis, we set $y = 0$:
\[ x^2 + px + (k+1)p - 4 = 0 \]
This is a quadratic equation in the form:
\[ x^2 + px + (k+1)p - 4 = 0 \]
The roots of this quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = p \), and \( c = (k+1)p - 4 \).
Substituting these values into the quadratic formula, we get:
\[ x = \frac{-p \pm \sqrt{p^2 - 4 \cdot 1 \cdot ((k+1)p - 4)}}{2 \cdot 1} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4((k+1)p - 4)}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4(k+1)p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4kp - 4p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4kp - 4p + 16}}{2} \]
\[ x = \frac{-p \pm \sqrt{p^2 - 4p(k + 1) + 16}}{2} \]
For the quadratic equation to have at least one integer root, the discriminant must be a perfect square. Let's denote the discriminant by \( D \):
\[ D = p^2 - 4p(k + 1) + 16 \]
We need \( D \) to be a perfect square. Let \( D = m^2 \) for some integer \( m \):
\[ p^2 - 4p(k + 1) + 16 = m^2 \]
Rearranging the equation, we get:
\[ p^2 - 4p(k + 1) + 16 - m^2 = 0 \]
This is a quadratic equation in \( p \):
\[ p^2 - 4p(k + 1) + (16 - m^2) = 0 \]
For \( p \) to be a prime number greater than 2, the discriminant of this quadratic equation must be a perfect square. The discriminant of this quadratic equation is:
\[ \Delta = (4(k + 1))^2 - 4 \cdot 1 \cdot (16 - m^2) \]
\[ \Delta = 16(k + 1)^2 - 4(16 - m^2) \]
\[ \Delta = 16(k + 1)^2 - 64 + 4m^2 \]
\[ \Delta = 16(k + 1)^2 + 4m^2 - 64 \]
For \( \Delta \) to be a perfect square, we need:
\[ 16(k + 1)^2 + 4m^2 - 64 = n^2 \]
for some integer \( n \).
Simplifying, we get:
\[ 4(4(k + 1)^2 + m^2 - 16) = n^2 \]
\[ 4(k + 1)^2 + m^2 - 16 = \left(\frac{n}{2}\right)^2 \]
Let \( \frac{n}{2} = t \), then:
\[ 4(k + 1)^2 + m^2 - 16 = t^2 \]
We need to find integer solutions for \( k \) and \( m \) such that the above equation holds. Testing small values of \( k \):
For \( k = 1 \):
\[ 4(1 + 1)^2 + m^2 - 16 = t^2 \]
\[ 4 \cdot 4 + m^2 - 16 = t^2 \]
\[ 16 + m^2 - 16 = t^2 \]
\[ m^2 = t^2 \]
This is true for \( m = t \). Therefore, \( k = 1 \) is a solution.
Thus, the value of \( k \) is:
\[ \boxed{1} \]
|
From the problem, we know that the equation
$$
x^{2}+p x+(k+1) p-4=0
$$
has at least one integer root among its two roots $x_{1}, x_{2}$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-p, x_{1} x_{2}=(k+1) p-4 . \\
\text { Hence }\left(x_{1}+2\right)\left(x_{2}+2\right) \\
=x_{1} x_{2}+2\left(x_{1}+x_{2}\right)+4 \\
=(k-1) p .
\end{array}
$$
(1) If $k=1$, then the equation becomes
$$
x^{2}+p x+2(p-2)=0 \text {, }
$$
which has two integer roots -2 and $2-p$.
(2) If $k>1$, then $k-1>0$.
Since $x_{1}+x_{2}=-p$ is an integer, and at least one of $x_{1}, x_{2}$ is an integer, it follows that $x_{1}, x_{2}$ are both integers.
Also, since $p$ is a prime number, by equation (1), we know that $p \mid (x_{1}+2)$ or $p \mid (x_{2}+2)$.
Without loss of generality, assume $p \mid (x_{1}+2)$. Then we can set $x_{1}+2=m p$ (where $m$ is a non-zero integer).
Thus, from equation (1), we get $x_{2}+2=\frac{k-1}{m}$.
Therefore, $\left(x_{1}+2\right)+\left(x_{2}+2\right)=m p+\frac{k-1}{m}$, which means
$x_{1}+x_{2}+4=m p+\frac{k-1}{m}$.
Since $x_{1}+x_{2}=-p$, we have
$-p+4=m p+\frac{k-1}{m}$
$\Rightarrow (m+1) p+\frac{k-1}{m}=4$.
If $m$ is a positive integer, then
$(m+1) p \geqslant (1+1) \times 3=6, \frac{k-1}{m}>0$.
Thus, $(m+1) p+\frac{k-1}{m}>6$, which contradicts equation (2).
If $m$ is a negative integer, then
$(m+1) p \leqslant 0, \frac{k-1}{m}<0$.
Thus, $(m+1) p+\frac{k-1}{m}<0$, which also contradicts equation (2).
Therefore, when $k>1$, the equation
$$
x^{2}+p x+(k+1) p-4=0
$$
cannot have integer roots.
In summary, $k=1$.
(Provided by Xu Shenglin)
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 Let $a, b$ be positive integers, and satisfy $2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{15}{b}}\right)$ is an integer. Then the number of such ordered pairs $(a, b)$ is $\qquad$ pairs.
$(2009$, National Junior High School Mathematics League)
|
First, guess that $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
The following is an attempt to prove this.
Let $\frac{15}{a}=A, \frac{15}{b}=B$. According to the problem, we can assume
$\sqrt{A}+\sqrt{B}=C\left(A, B, C \in \mathbf{Q}_{+}\right)$.
Thus, $\sqrt{A}=C-\sqrt{B} \Rightarrow A=(C-\sqrt{B})^{2}$
$\Rightarrow A=C^{2}-2 C \sqrt{B}+B \Rightarrow \sqrt{B}=\frac{C^{2}+B-A}{2 C}$
$\Rightarrow \sqrt{B}$ is a rational number.
Similarly, $\sqrt{A}$ is a rational number.
Therefore, $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
Further conjecture: When $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are expressed in their simplest fractional form, the numerator is 1.
The following is an attempt to prove this.
Let $\sqrt{\frac{15}{a}}=\frac{q}{p}\left(p, q \in \mathbf{Z}_{+}\right.$, and $p$ and $q$ are coprime $)$.
Then $\left.\frac{15}{a}=\frac{q^{2}}{p^{2}} \Rightarrow a q^{2}=15 p^{2} \Rightarrow q^{2} \right\rvert\, 15 p^{2}$.
Since $p$ and $q$ are coprime, $p^{2}$ and $q^{2}$ are coprime. Hence
$q^{2} \left\lvert\, 15 \Rightarrow q^{2}=1 \Rightarrow q=1 \Rightarrow \sqrt{\frac{15}{a}}=\frac{1}{p}\right.$.
Similarly, when $\sqrt{\frac{15}{b}}$ is expressed in its simplest fractional form, the numerator is 1.
From the previous analysis, we can assume
$\sqrt{\frac{15}{a}}=\frac{1}{x}, \sqrt{\frac{15}{b}}=\frac{1}{y}\left(x, y \in \mathbf{Z}_{+}\right)$. Then $a=15 x^{2}, b=15 y^{2}$.
According to the problem, we can assume $\frac{1}{x}+\frac{1}{y}=\frac{k}{2}\left(k \in \mathbf{Z}_{+}\right)$.
Since $x \geqslant 1, y \geqslant 1$, we have
$$
\frac{k}{2}=\frac{1}{x}+\frac{1}{y} \leqslant 1+1=2 \Rightarrow k \leqslant 4 \text {. }
$$
Therefore, $k$ can only be $1,2,3,4$.
(1) When $k=1$,
$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$
$\Rightarrow(x-2)(y-2)=4$
$\Rightarrow(x, y)=(4,4)$ or $(3,6)$ or $(6,3)$;
(2) When $k=2$, $(x, y)=(2,2)$;
(3) When $k=3$, $(x, y)=(2,1)$ or $(1,2)$;
(4) When $k=4$, $(x, y)=(1,1)$.
Therefore, there are 7 such ordered pairs $(a, b)$: $(240,240),(135,540),(540,135)$, $(60,60),(60,15),(15,60),(15,15)$.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. The number of integer solutions to the inequality $\log _{6}(1+\sqrt{x})>\log _{25} x$ is $\qquad$ .
|
10. 24 .
Let $\log _{25} x=t$. Then $x=25^{t}$.
Thus the original inequality $\Leftrightarrow \log _{6}\left(1+5^{t}\right)>t$
$$
\Leftrightarrow 1+5^{t}>6^{t} \Leftrightarrow\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}>1 \text {. }
$$
Define $f(t)=\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}$. Then $f(t)$ is a decreasing function, $f(t)>1=f(1)$.
Hence $t=\log _{25} x<1$.
Therefore, $0<x<25$, a total of 24.
|
24
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given real numbers $x, y, z$ satisfy
$$
\sqrt[5]{x-y}+\sqrt[5]{y-z}=3 \text{, and } x-z=33 \text{. }
$$
Find the value of the algebraic expression $x-2 y+z$.
|
Let $\sqrt[5]{x-y}=x_{1}, \sqrt[5]{y-z}=x_{2}$. Then $x_{1}+x_{2}=3, x_{1}^{5}+x_{2}^{5}=33$.
Let $x_{1} x_{2}=\lambda$. Then $x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+\lambda=0$.
Let $x_{1}^{n}+x_{2}^{n}=a_{n}$ ( $n$ is a positive integer $)$. Then
$$
a_{n}-3 a_{n-1}+\lambda a_{n-2}=0 \text {, }
$$
i.e.,
$$
a_{n}=3 a_{n-1}-\lambda a_{n-2}(n \geqslant 3) .
$$
From $x_{1}+x_{2}=3, x_{1} x_{2}=\lambda$
$$
\begin{array}{l}
\Rightarrow a_{1}=3, a_{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=9-2 \lambda \\
\Rightarrow a_{3}=3(9-2 \lambda)-3 \lambda=27-9 \lambda \\
\Rightarrow a_{4}=3(27-9 \lambda)-\lambda(9-2 \lambda) \\
\quad=81-36 \lambda+2 \lambda^{2} \\
\Rightarrow a_{5}=3\left(81-36 \lambda+2 \lambda^{2}\right)-\lambda(27-9 \lambda)=33 \\
\Rightarrow 81-36 \lambda+2 \lambda^{2}-\lambda(9-3 \lambda)=11 \\
\Rightarrow \lambda^{2}-9 \lambda+14=0
\end{array}
$$
$$
\Rightarrow \lambda=2 \text { or } 7 \text {. }
$$
From $x^{2}-3 x+\lambda=0$ having two real roots, we get
$$
\Delta=(-3)^{2}-4 \lambda \geqslant 0 \Rightarrow \lambda \leqslant \frac{9}{4}
$$
$\Rightarrow \lambda=2 \Rightarrow x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+2=0$
$$
\Rightarrow\left\{\begin{array} { l }
{ x _ { 1 } = 1 , } \\
{ x _ { 2 } = 2 }
\end{array} \text { or } \left\{\begin{array}{l}
x_{1}=2, \\
x_{2}=1 .
\end{array}\right.\right.
$$
Since $x-2 y+z=x_{1}^{5}-x_{2}^{5}$, we have
$$
1^{5}-2^{5}=-31 \text { or } 2^{5}-1^{5}=31
$$
This is the solution.
|
31
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If real numbers $a, b$ satisfy
$$
\left(a-\sqrt{a^{2}+2010}\right)\left(b+\sqrt{b^{2}+2010}\right)+2010=0 \text {, }
$$
then $a \sqrt{b^{2}+2011}-b \sqrt{a^{2}+2011}=$ $\qquad$
|
From the given, we have
$$
\begin{array}{l}
a-\sqrt{a^{2}+2010}=-\frac{2010}{b+\sqrt{b^{2}+2010}} \\
=b-\sqrt{b^{2}+2010} .
\end{array}
$$
Similarly,
$$
b+\sqrt{b^{2}+2010}=a+\sqrt{a^{2}+2010} \text {. }
$$
Subtracting (2) from (1) and simplifying, we get $a=b$.
Therefore, the answer is 0.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given that $P$ is a point inside the circle $\odot O$ with radius 15, among all the chords passing through point $P$, 24 chords have integer lengths. Then $O P=$ $\qquad$ .
|
2. 12 .
Among the chords of circle $\odot O$ passing through point $P$, the diameter is the longest chord, and there is only one; the shortest chord is the one perpendicular to $OP$, and there is also only one. Therefore, there is only one chord each with lengths of $30$ and $18$, and two chords each with lengths of $29, 28, \cdots, 19$, making a total of 24 chords.
By the perpendicular diameter theorem and the intersecting chords theorem, we have
$$
(15+O P)(15-O P)=\frac{18}{2} \times \frac{18}{2} .
$$
Thus, $O P=12$.
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 9 Prove: $\frac{(2+\sqrt{3})^{2010}+(2-\sqrt{3})^{2010}}{2}$ is an integer, and find its remainder when divided by 4.
|
Notice
$$
\begin{array}{l}
(2+\sqrt{3})+(2-\sqrt{3})=4, \\
(2+\sqrt{3})(2-\sqrt{3})=1 .
\end{array}
$$
Therefore, $2+\sqrt{3}$ and $2-\sqrt{3}$ are the two roots of $x^{2}-4 x+1=0$.
Let $a_{n}=\frac{1}{2}(2+\sqrt{3})^{n}+\frac{1}{2}(2-\sqrt{3})^{n}$. Then
$$
a_{n}-4 a_{n-1}+a_{n-2}=0
$$
That is, $a_{n}=4 a_{n-1}-a_{n-2}(n \geqslant 3)$.
Since $a_{1}=2, a_{2}=7$ are both integers, and combining with equation (1), we know that $a_{n}$ is an integer for all positive integers $n$.
Thus, the original expression $=a_{2010}$ is an integer.
From equation (1), we know that the sequence of remainders when $\left\{a_{n}\right\}$ is divided by 4 is $2,3,2,1 ; 2,3,2,1 ; \cdots \cdots$ with a period of 4.
Therefore, its remainder when divided by 4 is 3.
|
3
|
Number Theory
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
3. $[x]$ represents the greatest integer not exceeding the real number $x$ (for example, $[\pi]=3,[-\pi]=-4,[-4]=-4$). Let $M=[x]+[2x]+[3x]$. Positive integers that cannot be expressed in the form of $M$ are called "invisible numbers". If the invisible numbers are arranged in ascending order, the 2009th invisible number is $\qquad$
|
3.6028 .
Let $n$ be a natural number.
When $n \leqslant x < n+\frac{1}{3}$,
$2 n \leqslant 2 x < 2 n+\frac{2}{3}$,
$3 n \leqslant 3 x < 3 n+1$.
Thus, $M=n+2 n+3 n=6 n$.
When $n+\frac{1}{3} \leqslant x < n+\frac{1}{2}$,
$2 n+\frac{2}{3} \leqslant 2 x < 2 n+1$,
$3 n+1 \leqslant 3 x < 3 n+\frac{3}{2}$.
Thus, $M=n+2 n+(3 n+1)=6 n+1$.
When $n+\frac{1}{2} \leqslant x < n+\frac{2}{3}$,
$2 n+1 \leqslant 2 x < 2 n+\frac{4}{3}$,
$3 n+\frac{3}{2} \leqslant 3 x < 3 n+2$.
Thus, $M=n+(2 n+1)+(3 n+1)=6 n+2$.
When $n+\frac{2}{3} \leqslant x < n+1$,
$2 n+\frac{4}{3} \leqslant 2 x < 2 n+2$,
$3 n+2 \leqslant 3 x < 3 n+3$.
Thus, $M=n+(2 n+1)+(3 n+2)=6 n+3$.
Therefore, the hidden numbers are $6 n+4$ or $6 n+5 (n \in \mathbf{N})$.
The 2009th hidden number is
$$
6 \times \frac{2009-1}{2}+4=6028
$$
|
6028
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given a positive integer $M$ has $k$ positive divisors, among which, only two divisors are prime, and the sum of the reciprocals of these $k$ positive divisors is $\frac{342}{287}$. Find all values of $M$.
---
The text has been translated while preserving the original formatting and line breaks.
|
$$
\begin{array}{l}
\text{Because } 287=7 \times 41, \text{ so, 7 and 41 are exactly the two prime factors of } M. \\
\text{Let } M=7^{m} \times 41^{n}\left(m, n \in \mathbf{N}_{+}\right), d_{1}, d_{2}, \cdots, d_{k} \text{ be all the positive divisors of } M \text{ arranged in ascending order. Thus, } d_{1}=1, d_{k}=M. \text{ Then,} \\
\frac{342}{287}=\frac{1}{d_{1}}+\frac{1}{d_{2}}+\frac{1}{d_{3}}+\cdots+\frac{1}{d_{k}} \\
=\frac{d_{k}}{d_{k}}+\frac{d_{k-1}}{d_{k}}+\frac{d_{k-2}}{d_{k}}+\cdots+\frac{d_{1}}{d_{k}} \\
=\frac{d_{1}+d_{2}+d_{3}+\cdots+d_{k}}{d_{k}} \\
=\frac{\left(1+7+7^{2}+\cdots+7^{m}\right)\left(1+41+41^{2}+\cdots+41^{n}\right)}{7^{m} \times 41^{n}} . \\
\text{Hence, } \frac{7^{m+1}-1}{6} \cdot \frac{41^{n+1}-1}{40}=342 \times 7^{m-1} \times 41^{n-1} . \\
\text{By } \left(41^{n-1}, \frac{41^{n+1}-1}{40}\right)=1, \text{ we know that } \\
41^{n-1} \mid \frac{7^{m+1}-1}{6} . \\
\text{Let } 7^{m+1}-1=41^{n-1} \times 6 t\left(t \in \mathbf{N}_{+}\right) .
\end{array}
$$
Obviously, $(t, 7)=1$. Then,
$$
\frac{41^{n+1}-1}{40}=\frac{342}{t} \times 7^{m-1} \text{. }
$$
Therefore, $t \mid 342 \Rightarrow t \leqslant 342$.
Substituting equation (1) into equation (2) and rearranging, we get
$$
\begin{array}{l}
7^{m-1}\left(41^{2} \times 7^{2}-6 \times 13680\right)=41^{2}+6 t \\
\Rightarrow 289 \times 7^{m-1}=1681+6 t \\
\Rightarrow 1681<289 \times 7^{m-1} \leqslant 1681+6 \times 342 \\
\Rightarrow 5<\frac{1681}{289}<7^{m-1} \leqslant \frac{3733}{289}<13 \\
\Rightarrow m=2 .
\end{array}
$$
Substituting into equation (3), we get $t=57$.
Substituting into equation (1), we get $n=1$.
Thus, $M=7^{m} \times 41^{n}=2009$.
(Xie Wenxiao, Huanggang High School, Hubei, 438000)
$$
|
2009
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Using the vertices of a regular dodecagon as the vertices of triangles, the total number of acute and obtuse triangles that can be formed is $\qquad$.
untranslated part: $\qquad$ (This part is typically left as a blank space for the answer to be filled in.)
|
-1.160 .
The circumference of a circle with 12 points can form $\mathrm{C}_{12}^{3}=220$ triangles; among the chords of a circle equally divided into 12 parts, there are 6 diameters, and each diameter corresponds to 10 right-angled triangles, thus yielding 60 right-angled triangles. Therefore, acute and obtuse triangles amount to $220-60=160$ (triangles).
|
160
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. $F$ is the right focus of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and $P$ is a moving point on the ellipse. For the fixed point $A(-2, \sqrt{3}),|P A|+$ $2|P F|$ the minimum value is $\qquad$ .
|
4. 10 .
It is known that the eccentricity of the ellipse is $\frac{1}{2}$, and the equation of the right directrix $l$ is $x=8$. The distance from point $A$ to $l$ is 10.
Let the projection of point $P$ onto $l$ be $H$. Then
$\frac{|P F|}{|P H|}=\frac{1}{2}$.
Thus, $|P A|+2|P F|=|P A|+|P H| \geqslant 10$.
The minimum value is achieved when point $P$ lies on the segment $A H$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+$ $\cdots+\sqrt{1+\frac{1}{1999^{2}}+\frac{1}{2000^{2}}}$. Find the greatest integer not exceeding $S$.
(2000, Taiyuan Junior High School Mathematics Competition)
|
Prompt: Following Example 2, the largest integer not exceeding $S$ is
$$
1999 .
$$
|
1999
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (40 points) Prove: The set $E$ of real numbers $x$ that satisfy the inequality
$$
\frac{1}{x-1}+\frac{2}{x-2}+\cdots+\frac{200}{x-200}>10
$$
can be expressed as the union of some mutually disjoint open intervals. Find the total length of these intervals.
|
Consider the function
$$
f(x)=\frac{1}{x-1}+\frac{2}{x-2}+\cdots+\frac{200}{x-200}-10 \text {. }
$$
For any real number $x$.
For any $k$ in the set $\{1,2, \cdots, 200\}$, when $x \rightarrow k-0$, $f(x) \rightarrow-\infty$, and when $x \rightarrow k+0$, $f(x) \rightarrow+\infty$. Also, when $x \rightarrow+\infty$, $x \rightarrow-10$. Therefore, the equation $f(x)=0$ has one solution in each of the intervals
$$
(1,2),(2,3), \cdots,(199,200),(200,+\infty).
$$
Let these 200 solutions be denoted as $x_{1}, x_{2}, \cdots, x_{200}$.
Construct the polynomial
$$
P(x)=(x-1)(x-2) \cdots(x-200) f(x) \text {. }
$$
Since $P(x)$ is a 200th-degree polynomial, the equation $P(x)=0$ has at most 200 distinct roots. Clearly, each $x_{i}$ that satisfies $f(x)=0$ is also a root of $P(x)=0$.
Therefore, $x_{1}, x_{2}, \cdots, x_{200}$ are all the roots of $P(x)=0$.
This indicates that each $x_{k}$ is the unique root in its respective interval $(k, k+1)$ $(k=1,2, \cdots, 199)$ and $(200,+\infty)$.
Thus, the solution set of the inequality $f(x)>0$ is
$$
E=\left(1, x_{1}\right) \cup\left(2, x_{2}\right) \cup \cdots \cup\left(200, x_{200}\right) \text {. }
$$
Hence, the total sum of the lengths of all intervals is
$$
\begin{aligned}
S & =\left(x_{1}-1\right)+\left(x_{2}-2\right)+\cdots+\left(x_{200}-200\right) \\
& =\left(x_{1}+x_{2}+\cdots+x_{200}\right)-(1+2+\cdots+200) \\
& =\sum_{i=1}^{200} x_{i}-10 \times 2010 .
\end{aligned}
$$
Notice that
$$
P(x)=\left[\prod_{i=1}^{200}(x-i)\right]\left(\sum_{i=1}^{200} \frac{i}{x-i}-10\right),
$$
If we expand $P(x)$, the coefficient of the highest degree term is -10.
Thus, the coefficient of $x^{199}$ in $P(x)$ is
$$
(10+1) \sum_{k=1}^{200} k=11 \times 20100 \text {. }
$$
Therefore, $\sum_{i=1}^{200} x_{i}=11 \times 2010$.
From equation (1), we get
$$
S=\sum_{i=1}^{200} x_{i}-10 \times 2010=2010 .
$$
|
2010
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (50 points) A positive integer is called "simple" if it does not have any square factors greater than 1. Determine how many numbers in $1,2, \cdots, 2010$ are simple.
|
Three, notice that $2010 < 45^2$.
Thus, if a number $n$ in the set $M = \{1, 2, \cdots, 2010\}$ is not simple, it must contain one or more square factors from the set of primes
$$
N = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43\}.
$$
Let
$$
\left[\frac{2010}{p^2}\right] = s(p), \left[\frac{2010}{p^2 q^2}\right] = s(p, q),
$$
and so on.
Then $s(2) = 502, s(3) = 223, s(5) = 80$,
$$
\begin{array}{l}
s(7) = 41, s(11) = 16, s(13) = 11, \\
s(17) = 6, s(19) = 5, s(23) = 3, \\
s(29) = s(31) = 2, \\
s(37) = s(41) = s(43) = 1.
\end{array}
$$
Using $\left[\frac{\left[\frac{a}{n}\right]}{m}\right] = \left[\frac{a}{mn}\right]$, we get
$$
\begin{array}{c}
s(2,3) = 55, s(2,5) = 20, s(2,7) = 10, \\
s(2,11) = 4, s(2,13) = 2, s(2,17) = 1, \\
s(2,19) = 1, s(2,23) = 0, s(3,5) = 8, \\
s(3,7) = 4, s(3,11) = s(3,13) = 1, \\
s(3,17) = 0, s(5,7) = 1, s(5,11) = 0; \\
s(2,3,5) = 2, s(2,3,7) = 1, \\
s(2,3,11) = s(3,5,7) = 0.
\end{array}
$$
Thus, the desired result is
$$
\begin{array}{l}
2010 - \sum_{p \in N} s(p) + \sum_{p, q \in N} s(p, q) - \sum_{p, q, r \in N} s(p, q, r) \\
= 2010 - (502 + 223 + 80 + 41 + 16 + \\
11 + 6 + 5 + 3 + 2 + 2 + 1 + 1 + 1) + \\
(55 + 20 + 10 + 4 + 2 + 1 + 1 + 8 + 4 + \\
1 + 1 + 1) - (2 + 1) \\
= 2010 - 894 + 108 - 3 = 1221.
\end{array}
$$
|
1221
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $\alpha=\frac{\sqrt{5}+1}{2}$. Then $\left[\alpha^{16}\right]=$ $\qquad$ .
(2008, "Five Sheep Cup" Mathematics Competition (Junior High School))
[Analysis] Note that $\alpha=\frac{\sqrt{5}+1}{2}$ and $\beta=\frac{\sqrt{5}-1}{2}$ $(0<\beta<1)$ can make $\alpha+\beta=\sqrt{5}$ and $\alpha \beta=1$. Therefore, the value of $\alpha^{16}+\beta^{16}$ can be calculated to make a judgment.
|
Let $\beta=\frac{\sqrt{5}-1}{2}$. Then
$$
\begin{array}{l}
\alpha+\beta=\sqrt{5}, \alpha \beta=1 . \\
\text { Therefore, } \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3, \\
\alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2}=7, \\
\alpha^{8}+\beta^{8}=\left(\alpha^{4}+\beta^{4}\right)^{2}-2(\alpha \beta)^{4}=47, \\
\alpha^{16}+\beta^{16}=\left(\alpha^{8}+\beta^{8}\right)^{2}-2(\alpha \beta)^{8}=2207 .
\end{array}
$$
Since $0<\beta<1$, we know $0<\beta^{16}<1$.
Therefore, $\left[\alpha^{16}\right]=2206$.
|
2206
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Calculate $[\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}]$ (2008 appears 2008 times).
(2008, International Youth Math Invitational Competition)
【Analysis】Although there are 2008 square root operations, as long as you patiently estimate from the inside out, the pattern will naturally become apparent.
|
Let $a_{n}=\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}$ (with $n$ 2008s, $n=1,2, \cdots$ ).
Given $44<\sqrt{2008}<45$, we have
$$
\begin{array}{l}
45^{2}<2008+44<2008+\sqrt{2008} \\
<2008+45<46^{2} .
\end{array}
$$
Thus, $45<\sqrt{2008+\sqrt{2008}}<46$, which means $45<a_{2}<46$.
Therefore, $45^{2}<2008+45<2008+a_{2}$
$$
<2008+46<46^{2} \text {. }
$$
Hence, $45<\sqrt{2008+a_{2}}<46$, which means $45<a_{3}<46$.
Similarly, $45<a_{4}<46, \cdots \cdots 45<a_{2008}<46$.
Therefore, $\left[a_{2008}\right]=45$.
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 11 Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a$ ] equals $\qquad$
$(2009$, Beijing Mathematical Competition (Grade 8))
|
Notice that
$0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$.
Then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]$ equals
0 or 1.
By the problem statement, 18 of these are equal to 1. Therefore,
$\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\left[a+\frac{11}{30}\right]=0$,
$\left[a+\frac{12}{30}\right]=\left[a+\frac{13}{30}\right]=\cdots=\left[a+\frac{29}{30}\right]=1$.
Thus, $0<a+\frac{11}{30}<1,1 \leqslant a+\frac{12}{30}<2$.
Therefore, $18 \leqslant 30 a<19 \Rightarrow 6 \leqslant 10 a<\frac{19}{3}$.
Hence, $[10 a]=6$.
|
6
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 12 Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$. Then the number of such $n$ is $\qquad$.
(2006, National Junior High School Mathematics Competition)
|
From the properties of the Gaussian function, we know that $\left[\frac{n}{3}\right] \leqslant \frac{n}{3}$, with equality holding if and only if $\frac{n}{3}$ is an integer; $\left[\frac{n}{6}\right] \leqslant \frac{n}{6}$, with equality holding if and only if $\frac{n}{6}$ is an integer.
Therefore, $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right] \leqslant \frac{n}{3}+\frac{n}{6}=\frac{n}{2}$.
Equality holds in the above inequality if and only if $\frac{n}{6}$ is an integer.
Thus, $\frac{n}{6}$ is an integer.
Since the positive integer $n$ is less than 2006, the number of $n$ that satisfy the condition is $\left[\frac{2005}{6}\right]=334$.
|
334
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. The algebraic expression $\left[(\sqrt{6}+\sqrt{5})^{6}\right]=$ $\qquad$
(2004, "Five Sheep Cup" Mathematics Competition (Grade 9))
|
Prompt: Example 1. Answer: 10581.
|
10581
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The sum of all real numbers $x$ that satisfy $25\{x\}+[x]=125$ is $\qquad$
(2007, International Invitational Competition for Young Mathematicians in Cities)
|
Hint: $25\{x\}+[x]=125$ can be transformed into $25 x-24[x]=125$.
Example 6. Answer: 2837.
|
2837
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let
$$
\begin{aligned}
S & =\frac{1}{\left[\frac{(10 \times 11-1)^{2}}{10 \times 11}\right]}+\frac{1}{\left[\frac{(11 \times 12-1)^{2}}{11 \times 12}\right]}+\cdots+ \\
& \frac{1}{\left[\frac{(49 \times 50-1)^{2}}{49 \times 50}\right]}
\end{aligned}
$$
Then $[30 S]=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 0
(2002, "Five Sheep Cup" Mathematics Competition (Grade 8))
|
When $n \geqslant 10$, and $n$ is an integer,
$$
\begin{array}{l}
{\left[\frac{[n(n+1)-1]^{2}}{n(n+1)}\right]=\left[n(n+1)-2+\frac{1}{n(n+1)}\right]} \\
=n(n+1)-2=(n-1)(n+2) .
\end{array}
$$
Thus, $S=\frac{1}{9 \times 12}+\frac{1}{10 \times 13}+\cdots+\frac{1}{48 \times 51}$.
Notice that $\frac{1}{k(k+3)}=\frac{1}{3}\left(\frac{1}{k}-\frac{1}{k+3}\right)$.
$$
\begin{array}{l}
\text { Then } 3 S=\left(\frac{1}{9}-\frac{1}{12}\right)+\left(\frac{1}{10}-\frac{1}{13}\right)+\cdots+\left(\frac{1}{48}-\frac{1}{51}\right) \\
=\frac{1}{9}+\frac{1}{10}+\frac{1}{11}-\frac{1}{49}-\frac{1}{50}-\frac{1}{51} . \\
\text { Therefore, } \frac{3}{10}-\frac{3}{40}<3 S<\frac{3}{9}-\frac{3}{60}, \\
2<\frac{9}{4}<30 S<\frac{17}{6}<3 .
\end{array}
$$
Therefore, $[30 S]=2$.
|
2
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Algebraic Expression
$[\sqrt[3]{1 \times 2 \times 3}]+[\sqrt[3]{2 \times 3 \times 4}]+\cdots+$ $[\sqrt[3]{2000 \times 2001 \times 2002}]=(\quad)$.
(A) 2000000
(B) 2001000
(C) 2002000
(D) 2003001
(2000, "Five Sheep Cup" Mathematics Competition (Grade 9))
[Analysis] Notice that each term in the sum can be expressed as $[\sqrt[3]{k(k+1)(k+2)}]$ (where $k$ is a positive integer). We need to estimate $\sqrt[3]{k(k+1)(k+2)}$.
|
Since $k^{3}<k(k+1)(k+2)<(k+1)^{3}$,
thus, $k<\sqrt[3]{k(k+1)(k+2)}<k+1$.
Therefore, $[\sqrt[3]{k(k+1)(k+2)}]=k$.
Hence the original expression $=1+2+\cdots+2000=2001000$.
|
2001000
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $n$ ($n<100$) be a positive integer, and there exists a positive integer $k$, such that $1 \leqslant k \leqslant n-1$, satisfying
$$
\frac{4 k+1}{2 n}+\frac{1-2 k^{2}}{n^{2}}=\frac{1}{2} \text {. }
$$
How many values of $n$ satisfy the condition? Prove your conclusion.
|
1. From equation (1), we get $(2 k-n)^{2}=n+2$.
Solving, we get $k=\frac{1}{2}(n \pm \sqrt{n+2})$.
Since $k$ is an integer, therefore, $n=m^{2}-2(m \in \mathbf{N})$.
Thus, $k=\frac{1}{2}\left(m^{2} \pm m-2\right)$.
Also, $1 \leqslant k \leqslant n-1$, so
$$
1 \leqslant \frac{m^{2} \pm m-2}{2} \leqslant m^{2}-3 \text {. }
$$
Solving, we get $m \geqslant 3$.
But $n<100$, hence the $n$ that satisfies the problem are $7,14,23,34,47,62,79,98$.
A total of 8.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Calculate $\left[\frac{23 \times 1}{101}\right]+\left[\frac{23 \times 2}{101}\right]+\cdots+$ $\left[\frac{23 \times 100}{101}\right]$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Since 23 and 101 are coprime, for $k=1$, $2, \cdots, 100$, $\frac{23 k}{101}$ is not an integer. Therefore,
$$
\begin{array}{l}
\frac{23 k}{101}-1<\left[\frac{23 k}{101}\right]<\frac{23 k}{101}, \\
\frac{23(101-k)}{101}-1<\left[\frac{23(101-k)}{101}\right] \\
<\frac{23(101-k)}{101}. \\
\text{ Hence }\left(\frac{23 k}{101}-1\right)+\left[\frac{23(101-k)}{101}-1\right] \\
<\left[\frac{23 k}{101}\right]+\left[\frac{23(101-k)}{101}\right] \\
<\frac{23 k}{101}+\frac{23(101-k)}{101}, \\
21<\left[\frac{23 k}{101}\right]+\left[\frac{23(101-k)}{101}\right]<23. \\
\end{array}
$$
Therefore, $\left[\frac{23 k}{101}\right]+\left[\frac{23(101-k)}{101}\right]=22$.
Thus, we can pair $\left[\frac{23 \times 1}{101}\right],\left[\frac{23 \times 2}{101}\right], \cdots$, $\left[\frac{23 \times 100}{101}\right]$ from the beginning to the end, forming 50 pairs, with each pair summing to 22.
$$
\begin{array}{l}
\text{ Hence }\left[\frac{23 \times 1}{101}\right]+\left[\frac{23 \times 2}{101}\right]+\cdots+\left[\frac{23 \times 100}{101}\right] \\
=22 \times 50=1100. \\
\end{array}
$$
|
1100
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $a^{2}(b+c)=b^{2}(a+c)=2010$, and $a \neq b$. Then $c^{2}(a+b)=$
|
1. 2010 .
Notice
$$
\begin{array}{l}
a^{2}(b+c)-b^{2}(a+c) \\
=(a-b)(a c+b c+a b)=0 .
\end{array}
$$
Since $a \neq b$, it follows that $a c+b c+a b=0$.
$$
\begin{array}{l}
\text { Then } c^{2}(a+b)-b^{2}(a+c) \\
=(c-b)(a c+b c+a b)=0 .
\end{array}
$$
Therefore, $c^{2}(a+b)=b^{2}(a+c)=2010$.
|
2010
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If real numbers $x, y$ satisfy $|x|+|y| \leqslant 1$, then the maximum value of $x^{2}-$ $xy+y^{2}$ is $\qquad$
|
2. 1 .
Notice that
$$
\begin{array}{l}
x^{2}-x y+y^{2}=\frac{1}{4}(x+y)^{2}+\frac{3}{4}(x-y)^{2} . \\
\text { Also }|x \pm y| \leqslant|x|+|y| \leqslant 1 \text {, then } \\
x^{2}-x y+y^{2} \leqslant \frac{1}{4}+\frac{3}{4}=1 .
\end{array}
$$
When $x$ and $y$ take 0 and 1 respectively, the equality holds.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In trapezoid $A B C D$, $A D / / B C, E F$ is the midline, the area ratio of quadrilateral $A E F D$ to quadrilateral $E B C F$ is $\frac{\sqrt{3}+1}{3-\sqrt{3}}$, and the area of $\triangle A B D$ is $\sqrt{3}$. Then the area of trapezoid $A B C D$ is
|
6. 2 .
Let $A D=a, B C=b$, and the height be $2 h$. The area of trapezoid $A B C D$ is $S$. Then
$$
\begin{array}{l}
E F=\frac{a+b}{2}, \frac{a}{b}=\frac{S_{\triangle A B D}}{S_{\triangle B C D}}=\frac{\sqrt{3}}{S-\sqrt{3}} . \\
\text { Also, } \frac{\sqrt{3}+1}{3-\sqrt{3}}=\frac{S_{\text {quadrilateral } A E F D}}{S_{\text {quadrilateral } E B C F}}=\frac{S+2 \sqrt{3}}{3 S-2 \sqrt{3}} .
\end{array}
$$
Applying the ratio theorem, we get $S=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Wang Qiang has four colors of small cylindrical rods, Table 1 lists the lengths of the rods of different colors.
Table 1
\begin{tabular}{|c|c|c|c|c|}
\hline Color & Green & Yellow-Red & Purple & Red \\
\hline Length & $3 \mathrm{~cm}$ & $4 \mathrm{~cm}$ & $8 \mathrm{~cm}$ & $9 \mathrm{~cm}$ \\
\hline
\end{tabular}
Now, several small rods are to be taken and joined together to form a long rod of length $2010 \mathrm{~cm}$, and each of the four colors of small rods must be used at least 81 times. Then the number of different ways to do this is.
|
9. 91.
From the problem, we have
$$
\begin{array}{l}
3(81+a)+4(81+b)+8(81+c)+9(81+d) \\
=2010,
\end{array}
$$
where $a, b, c, d$ are non-negative integers.
$$
\begin{array}{l}
\text { Simplifying, we get } 3(a+3 d)+4(b+2 c)=66 . \\
\text { Therefore, }(a+3 d, b+2 c) \\
=(4 i+2,15-3 i)(i=0,1, \cdots, 5) . \\
\text { When } a+3 d=4 i+2 \text {, }(a, d) \text { has }\left[\frac{4 i+2}{3}\right]+1
\end{array}
$$
possible values;
$$
\text { When } b+2 c=15-3 i \text {, }(b, c) \text { has }\left[\frac{15-3 i}{2}\right]+1
$$
possible values.
Therefore, the total number of different combinations is
$$
1 \times 8+3 \times 7+4 \times 5+5 \times 4+7 \times 2+8 \times 1=91 \text {. }
$$
|
91
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. In a competition with 20 singers, 9 judges respectively assign them ranks from 1 to 20. It is known that for each singer, the difference between any two ranks does not exceed 3. If the sum of the ranks each singer receives is arranged in an increasing sequence: $C_{1} \leqslant C_{2} \leqslant \cdots \leqslant C_{20}$, then the maximum value of $C_{1}$ is . $\qquad$
|
10. 24.
If 9 judges all give a singer the first place, then $C_{1}=9$.
If two singers both get the first place, then one of them gets no less than 5 first places, while the other 4 ranks are no higher than fourth place, so $C_{1} \leqslant 5 \times 1+4 \times 4=21$.
If three singers all get the first place, then their other ranks are no higher than fourth place, and their total ranks do not exceed $1 \times 9+3 \times 9+4 \times 9=72$, so $C_{1} \leqslant 24$.
If four singers all get the first place, then their total ranks do not exceed $1 \times 9+2 \times 9+3 \times 9+4 \times 9=90$, so $C_{1} \leqslant 24$.
The situation where five or more singers all get the first place is impossible, so $C_{1} \leqslant 24$.
Here is an example where $C_{1}=24$:
The judges give the three singers with rank sums $C_{1} 、 C_{2} 、 C_{3}$ the ranks $1,1,1,3,3,3,4,4,4$;
The judges give the three singers with rank sums $C_{4} 、 C_{5} 、 C_{6}$ the ranks $2,2,2,5,5,5,6,6,6$;
While the judges give the rest of the singers ranks between 7 and 20.
At this point, $C_{1}=24$.
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Find the smallest odd number $a$ greater than 5 that satisfies the following conditions: there exist positive integers $m_{1}, n_{1}, m_{2}, n_{2}$, such that
$$
a=m_{1}^{2}+n_{1}^{2}, a^{2}=m_{2}^{2}+n_{2}^{2} \text {, }
$$
and $m_{1}-n_{1}=m_{2}-n_{2}$.
|
8. From $261=15^{2}+6^{2}, 261^{2}=189^{2}+180^{2}$,
$15-6=189-180$,
we know that 261 has the property described in the problem.
Next, we prove that 261 is the smallest odd number greater than 5 with the property described in the problem, i.e., there is no such odd number between 5 and 261.
If not, then there exists an odd number $a$ between 5 and 261 and positive integers $m_{1}, n_{1}, m_{2}, n_{2}$, such that
$a=m_{1}^{2}+n_{1}^{2}, a^{2}=m_{2}^{2}+n_{2}^{2}, m_{1}-n_{1}=m_{2}-n_{2}$.
Let $m_{1}-n_{1}=l$, by symmetry, we can assume $l \geqslant 0$.
Since $a$ is odd, $m_{1}$ and $n_{1}$ have different parities.
Therefore, $l$ is an odd number.
Also, $m_{1}500>261$, a contradiction.
If $l=5$, then $a^{2} \equiv 2 n_{2}^{2}(\bmod 5)$.
Since 2 is not a quadratic residue modulo 5, $51 a$.
Also, $0 \equiv a \equiv 2 n_{1}^{2}(\bmod 5)$, so $5 \mid n_{1}$.
If $n_{1} \geqslant 10$, then $m_{1} \geqslant 15$.
Thus, $a \geqslant 15^{2}+10^{2}=325>261$, a contradiction;
If $n_{1}=5$, then
$m_{1}=10, a=10^{2}+5^{2}=125$.
But from $125^{2}=\left(n_{2}+5\right)^{2}+n_{2}^{2}$, we get
$$
\left(2 n_{2}+5\right)^{2}=5^{2} \times 1249,
$$
a contradiction.
If $l=7$, then by $a<261<16^{2}+9^{2}$, we know $n_{1} \leqslant 8$.
By enumerating $n_{1}$, we find that $a$ can only be
$65,85,109,137,169,205,245$
one of these.
From $a^{2}=\left(n_{2}+7\right)^{2}+n_{2}^{2}$, we get
$\left(2 n_{2}+7\right)^{2}=2 a^{2}-49$.
Therefore, $2 a^{2}-49$ must be a perfect square.
But by verifying each of $a=65,85,109,137,169,205,245$, we find that none of these values of $a$ satisfy the condition.
In conclusion, the smallest odd number greater than 5 with the property described in the problem is 261.
(Provided by Zhu Huawei)
|
261
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (25 points) In a $7 \times 7$ grid, there are 64 intersection points (referred to as "nodes") where chess pieces can be placed, with at most 1 piece per point, totaling $k$ pieces. If no matter how the pieces are placed, there always exist 4 pieces such that the nodes they occupy form the four vertices of a rectangle (with sides parallel to the grid lines of the chessboard). Find the minimum value of $k$.
(Zhang Limin, problem contributor)
|
First, as shown in Figure 2, when $k=24$, it is possible that no 4 nodes with chess pieces form the four vertices of a rectangle, so $k \geqslant 25$.
Second, we will prove that when $k=25$, there must exist 4 nodes with chess pieces that form the four vertices of a rectangle.
If not, consider the row (or column) with the most chess pieces among all rows and columns. Since swapping any two rows (or two columns) does not affect the result, we can assume without loss of generality that the first row has the most chess pieces.
(1) When the first row has 8 chess pieces, because
$$
25-8=17,17=8 \times 2+1 \text {, }
$$
there must be another row with 2 chess pieces. Thus, the 4 chess pieces in these two rows form a rectangle, which is a contradiction.
(2) When the first row has 7 chess pieces, if no rectangle exists, then in the remaining 7 rows of the 7 columns with chess pieces, there can be at most 7 chess pieces, totaling at most 21 chess pieces, which is a contradiction.
(3) When the first row has 6 chess pieces, if no rectangle exists, then in the 6 columns with these 6 chess pieces, there can be at most $7+6=13$ chess pieces. In the remaining 2 columns, there are 12 chess pieces $(12=7+5)$, and there must be two rows with chess pieces in these two columns, forming a rectangle with the 4 chess pieces, which is a contradiction.
(4) When the first row has 5 chess pieces, if no rectangle exists, these 5 chess pieces in the 5 columns can have at most $5+7=12$ chess pieces. In the remaining 7 rows and 3 columns, there are 13 chess pieces $(13=3 \times 4+1)$. Thus, there must be one column with 5 chess pieces. The 5 chess pieces in this column can have at most 7 chess pieces in the other 3 columns, leaving 6 chess pieces in a $2 \times 3$ grid, which must form a rectangle with 4 points, which is a contradiction.
(5) When the first row has 4 chess pieces, if no rectangle exists, these 4 chess pieces in the 4 columns can have at most $4+7=11$ chess pieces. The remaining 14 chess pieces are placed in 7 rows and 4 columns. Since $14=4 \times 3+2$, there must be one column with 4 chess pieces. Thus, the 4 chess pieces in this column can have at most $4+3=7$ chess pieces in the other 4 columns, leaving 7 chess pieces in 3 rows and 3 columns, which still form a rectangle with 4 points, which is a contradiction.
In summary, the minimum value of $k$ is 25.
|
25
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Five. (25 points) Let positive real numbers $a$, $b$, $c$ satisfy
$$
(a+2 b)(b+2 c)=9 \text {. }
$$
Prove: $\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant 3$.
(Zhang Lei)
|
Because $a, b, c > 0$, so,
$$
\begin{array}{l}
\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \\
\geqslant \frac{a+b}{2}+2 \cdot \frac{b+c}{2} \\
=\frac{1}{2}[(a+2 b)+(b+2 c)] \\
\geqslant \sqrt{(a+2 b)(b+2 c)}=3 .
\end{array}
$$
|
3
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
1. $\left(\frac{1+\sqrt{5}}{2}\right)^{6}+\left(\frac{1-\sqrt{5}}{2}\right)^{6}=$
|
Let $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$. Then $x_{1}+x_{2}=1, x_{1} x_{2}=-1$.
Thus, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=3$.
Therefore, $x_{1}^{6}+x_{2}^{6}=\left(x_{1}^{2}\right)^{3}+\left(x_{2}^{2}\right)^{3}$
$=\left(x_{1}^{2}+x_{2}^{2}\right)\left[\left(x_{1}^{2}\right)^{2}-x_{1}^{2} x_{2}^{2}+\left(x_{2}^{2}\right)^{2}\right]$
$=3\left[\left(x_{1}^{2}+x_{2}^{2}\right)^{2}-3 x_{1}^{2} x_{2}^{2}\right]$
$=3\left(3^{2}-3\right)=18$.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three, (25 points) If the sum, difference, product, and quotient of two unequal natural numbers add up to a perfect square, then such two numbers are called a "wise pair" (for example, $(8,2)$ is a wise pair, since $\left.(8+2)+(8-2)+8 \times 2+\frac{8}{2}=36=6^{2}\right)$.
If both of these natural numbers do not exceed 100, how many such wise pairs are there?
|
Three, let $(a, b)$ be a wise array, and without loss of generality, assume $a > b$.
By definition, we have
$$
(a+b)+(a-b)+a b+\frac{a}{b}=m^{2}
$$
$(a, b, m$ are natural numbers), which simplifies to
$$
2 a+a b+\frac{a}{b}=m^{2}.
$$
Since $2 a, a b, m^{2}$ are all natural numbers, $\frac{a}{b}$ must also be a natural number.
Without loss of generality, let $\frac{a}{b}=k$ (where $k$ is a natural number and $k \neq 1$). Then $a=k b$.
Substituting into equation (1) and simplifying, we get $m^{2}=(b+1)^{2} k$.
Thus, $(b+1)^{2} \mid m^{2} \Rightarrow (b+1) \mid m$.
Therefore, $k=\left(\frac{m}{b+1}\right)^{2} \geqslant 4$.
Also, $b < a \leqslant 100$, so
$1 \leqslant b \leqslant 25$, and $a=k b, 1 < k \leqslant \frac{100}{b}$.
Thus, when $b=1$, we have $k=2^{2}, 3^{2}, \cdots, 10^{2}$, so there are 9 pairs $(a, b)$ that satisfy the condition.
When $b=2$, we have $k=2^{2}, 3^{2}, \cdots, 7^{2}$, so there are 6 pairs $(a, b)$ that satisfy the condition.
Similarly, when $b=3,4$, there are 4 pairs $(a, b)$ each; when $b=5,6$, there are 3 pairs $(a, b)$ each; when $b=7,8, \cdots, 11$, there are 2 pairs $(a, b)$ each; when $b=12,13, \cdots, 25$, there is 1 pair $(a, b)$ each.
Therefore, there are a total of 53 wise arrays.
(Li Yaowen, No. 18 Middle School of Benzhuang City, Shandong Province, 277200)
|
53
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the function $f(n)$ be defined on the set of positive integers, for any positive integer $n$, we have $f(f(n))=4 n+3$, and for any non-negative integer $k$, we have
$$
f\left(2^{k}\right)=2^{k+1}+1 \text {. }
$$
Then $f(2303)=$
|
5.4607.
Notice
$$
\begin{array}{l}
2303=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4} \times 2^{3} \text {. } \\
\text { And } f(4 n+3)=f(f(f(n)))=4 f(n)+3 \text {, then } \\
f(2303) \\
=3+4 f\left(3+4 \times 3+4^{2} \times 3+4^{3} \times 2^{3}\right) \\
=\cdots \\
=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4} f\left(2^{3}\right) \\
=3+4 \times 3+4^{2} \times 3+4^{3} \times 3+4^{4}\left(2^{4}+1\right) \\
=2303+4^{4}\left(2^{4}+1-2^{3}\right)=4607 .
\end{array}
$$
|
4607
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. As shown in Figure 1, let $G$ and $H$ be the centroid and orthocenter of $\triangle ABC$ respectively, $F$ be the midpoint of segment $GH$, and the circumradius of $\triangle ABC$ be $R=1$. Then
$$
\begin{array}{l}
|\overrightarrow{A F}|^{2}+|\overrightarrow{B F}|^{2}+|\overrightarrow{C F}|^{2} \\
= \\
.
\end{array}
$$
|
6.3.
Let the circumcenter $O$ of $\triangle ABC$ be the origin of a Cartesian coordinate system. Then,
$$
\begin{array}{l}
\overrightarrow{O H}=\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}, \\
\overrightarrow{O G}=\frac{1}{3}(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C})
\end{array}
$$
$$
\text { Then } \overrightarrow{O F}=\frac{1}{2}(\overrightarrow{O G}+\overrightarrow{O H})=\frac{2}{3}(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \text {. }
$$
Therefore, $|\overrightarrow{A F}|^{2}+|\overrightarrow{B F}|^{2}+|\overrightarrow{C F}|^{2}$
$$
\begin{aligned}
= & (\overrightarrow{O A}-\overrightarrow{O F}) \cdot(\overrightarrow{O A}-\overrightarrow{O F})+ \\
& (\overrightarrow{O B}-\overrightarrow{O F}) \cdot(\overrightarrow{O B}-\overrightarrow{O F})+ \\
& (\overrightarrow{O C}-\overrightarrow{O F}) \cdot(\overrightarrow{O C}-\overrightarrow{O F}) \\
= & |\overrightarrow{O A}|^{2}+|\overrightarrow{O B}|^{2}+|\overrightarrow{O C}|^{2}- \\
& 2(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \cdot \overrightarrow{O F}+3 \overrightarrow{O F} \cdot \overrightarrow{O F} \\
= & |\overrightarrow{O A}|^{2}+|\overrightarrow{O B}|^{2}+|\overrightarrow{O C}|^{2}=3 .
\end{aligned}
$$
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sequence $\left\{a_{n}\right\}$ is defined as follows: $a_{1}=1$, and for $n \geqslant 2$, $a_{n}=\left\{\begin{array}{ll}a_{\frac{n}{2}}+1, & n \text { is even; } \\ \frac{1}{a_{n-1}}, & n \text { is odd. }\end{array}\right.$ If $a_{n}=\frac{20}{11}$, then the positive integer $n=$ $\qquad$
|
8. 198 .
From the problem, we know that when $n$ is even, $a_{n}>1$; when $n(n>1)$ is odd, $a_{n}=\frac{1}{a_{n-1}}1$, so, $n$ is even. Thus, $a_{\frac{n}{2}}=\frac{20}{11}-1=\frac{9}{11}1, \frac{n}{2}-1$ is even;
$a_{\frac{n-2}{4}}=\frac{11}{9}-1=\frac{2}{9}1, \frac{n-6}{4}$ is even;
$a_{\frac{n-6}{8}}=\frac{9}{2}-1=\frac{7}{2}>1, \frac{n-6}{8}$ is even;
$a_{\frac{n-6}{16}}=\frac{7}{2}-1=\frac{5}{2}>1, \frac{n-6}{16}$ is even;
$a_{\frac{n-6}{32}}=\frac{5}{2}-1=\frac{3}{2}>1, \frac{n-6}{32}$ is even;
$a_{\frac{n-6}{4}}=\frac{3}{2}-1=\frac{1}{2}1, \frac{n-70}{64}$ is even;
$a_{\frac{n-70}{128}}=2-1=1$.
Therefore, $\frac{n-70}{128}=1 \Rightarrow n=198$.
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
283 Find the unit digit of $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
Solve the problem, which is to find the remainder when $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]$ is divided by 10.
Let $a_{n}=\left(\frac{5+\sqrt{21}}{2}\right)^{n}+\left(\frac{5-\sqrt{21}}{2}\right)^{n}\left(n \in \mathbf{N}_{+}\right)$.
Since $\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5$,
$$
\frac{5+\sqrt{21}}{2} \cdot \frac{5-\sqrt{21}}{2}=1 \text {, }
$$
Therefore, $\frac{5+\sqrt{21}}{2} ， \frac{5-\sqrt{21}}{2}$ are the roots of the equation $x^{2}-5 x+1=0$.
Thus, $a_{n}-5 a_{n-1}+a_{n-2}=0$, which means
$$
a_{n}=5 a_{n-1}-a_{n-2}(n \geqslant 3) \text {. }
$$
Also, $a_{1}=5, a_{2}=23$ are both integers, and from equation (1), we know that $a_{n}$ is an integer for all positive integers $n$.
Hence, $a_{2010}$ is an integer.
Since $4<\sqrt{21}<5$, then
$0<5-\sqrt{21}<1$.
Therefore, $0<\frac{5-\sqrt{21}}{2}<1$.
Thus, $0<\left(\frac{5-\sqrt{21}}{2}\right)^{2010}<1$.
Consequently, $a_{2010}$ is the smallest integer greater than $\left(\frac{5+\sqrt{21}}{2}\right)^{2010}$.
Therefore, $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]=a_{2010}-1$.
From equation (1), we have
$$
\begin{array}{l}
a_{n} \equiv 5 a_{n-1}-a_{n-2} \equiv-a_{n-2}(\bmod 5) \\
\Rightarrow a_{n} \equiv-a_{n-2} \equiv-\left(-a_{n-4}\right) \equiv a_{n-4}(\bmod 5) \\
\Rightarrow a_{2010} \equiv a_{2} \equiv 23 \equiv 3(\bmod 5) .
\end{array}
$$
From equation (1), we also have
$$
\begin{aligned}
a_{n} & \equiv a_{n-1}+a_{n-2}(\bmod 2) \\
\Rightarrow & a_{n} \equiv a_{n-1}+a_{n-2} \\
& \equiv\left(a_{n-2}+a_{n-3}\right)+a_{n-2} \\
& \equiv 2 a_{n-2}+a_{n-3} \equiv a_{n-3}(\bmod 2) \\
\Rightarrow & a_{2010} \equiv a_{3} \equiv a_{2}+a_{1} \equiv 0(\bmod 2) .
\end{aligned}
$$
Thus, $a_{2010} \equiv 3(\bmod 5)$, and
$$
a_{2010} \equiv 0(\bmod 2) \text {. }
$$
Therefore, $a_{2010} \equiv 8(\bmod 10)$.
Hence, $\left[\left(\frac{5+\sqrt{21}}{2}\right)^{2010}\right]=a_{2010}-1$
$$
\equiv 7(\bmod 10) \text {. }
$$
Thus, 7 is the answer.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given seven points on a circle, connect each pair of points. Find the minimum number of intersection points of these lines inside the circle.
|
We prove that these lines intersect at most three sets of three lines at a point inside the circle, and it is obvious that there cannot be four lines intersecting at a point.
First, select seven vertices of a regular octagon,
it is clear that there are three sets of three lines intersecting at a point (as shown in Figure 3).
Second, there cannot be more than three sets of three lines intersecting at a point. We call a point other than the intersection of three lines a "remaining point".
If there are two adjacent remaining points, assume without loss of generality that these seven points are $A_{1}, A_{2}, \cdots, A_{7}$, and $A_{1}$ and $A_{2}$ are both remaining points. Then, when $A_{1}$ is a remaining point, the intersection of three lines is the intersection of $A_{2} A_{5}$, $A_{3} A_{6}$, and $A_{4} A_{7}$. When $A_{2}$ is a remaining point, the intersection of three lines is the intersection of $A_{1} A_{5}$, $A_{3} A_{6}$, and $A_{4} A_{7}$. Both intersections of three lines are the intersection of $A_{3} A_{6}$ and $A_{4} A_{7}$, hence they coincide, which is a contradiction.
Therefore, remaining points cannot be adjacent.
Thus, among the seven points, there can be at most three remaining points, meaning there can be at most three sets of three lines intersecting at a point.
In summary, the minimum number of intersection points inside the circle is $\mathrm{C}_{7}^{4}-3(3-1)=29$.
(Zhang Lei, Northeast Yucai School, 110179)
|
29
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 As shown in Figure $1, a / / b$, line $a$ has ten points $A_{1}, A_{2}$, $\cdots, A_{10}$, and line
$b$ has nine points $B_{1}, B_{2}, \cdots, B_{9}$
nine points. Connecting each point on
$a$ with each point on
$b$ can result in many line segments, given that no three line segments intersect at one point. How many intersection points do these line segments have in total?
(1996, Seoul International Mathematics Invitational Tournament China Training Team Question)
|
Solution: Take two points each on $a$ and $b$, the four points determine a unique intersection point. Taking two points from $a$ has $10 \times 9 \div 2=45$ methods, taking two points from $b$ has $9 \times 8 \div 2=36$ methods. In total, there are $45 \times 36=1620$ methods.
|
1620
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that $a$, $b$, and $c$ are all positive integers, and $a b c=$ 2010. Then the minimum value of $a+b+c$ is ( ).
(A) 84
(B) 82
(C) 78
(D) 76
|
$$
\begin{array}{l}
\text { I.1.C. } \\
\text { From } 2010=2 \times 3 \times 5 \times 67=6 \times 5 \times 67 \\
=1 \times 30 \times 67=\cdots,
\end{array}
$$
we know that the minimum value of $a+b+c$ is 78.
|
78
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example: How many types of isosceles triangles with integer side lengths and a perimeter of 100 are there?
(8th "Hua Luogeng Cup" Junior Mathematics Invitational Final)
|
Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle.
(1) $a=ba .\end{array}\right.$,
Solving, we get $25<a<33 \frac{1}{3}$.
Therefore, there are 8 possibilities.
(2) $a<b=c$.
In this case, $a=100-2 c$.
Thus, $\left\{\begin{array}{l}100-2 c<c \\ 1+2 c \leqslant 100\end{array}\right.$.
Solving, we get $33 \frac{1}{3}<c \leqslant 49 \frac{1}{2}$.
Therefore, there are 16 possibilities.
In total, there are 24 possibilities.
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. When $n$ is a positive integer, it is defined that
$$
n!=n \times(n-1) \times \cdots \times 2 \times 1 \text {, }
$$
which is called the factorial of $n$ (for example, $10!=10 \times 9 \times \cdots \times 2 \times 1$ $=3628800$). Therefore, in 2010!, the total number of zeros at the end is $\qquad$
|
4.501.
The number of trailing zeros depends on the number of factors of 10. Since $10=2 \times 5$, and it is clear that in 2010! the number of factors of 2 is greater than the number of factors of 5, we only need to find the number of prime factors of 5 in 2010!, which gives us the number of trailing zeros.
Among the integers from 1 to 2010, there are $\left[\frac{2010}{5}\right]$ multiples of 5, $\left[\frac{2010}{5^{2}}\right]$ multiples of $5^{2}$, $\qquad$
Since $5^{5}>2010$, the number of factors of 5 in 2010! is
$$
\begin{array}{l}
{\left[\frac{2010}{5}\right]+\left[\frac{2010}{5^{2}}\right]+\left[\frac{2010}{5^{3}}\right]+\left[\frac{2010}{5^{4}}\right]} \\
=402+80+16+3=501 .
\end{array}
$$
|
501
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 On a plane, there are seven points, and some line segments can be connected between them, so that any three points among the seven must have at least one line segment connecting a pair of them. How many line segments are needed at least? Prove your conclusion.
$(2002$, Shanghai Junior High School Mathematics Competition)
|
(1) If point $A$ is not an endpoint of any line segment, then every two of the other six points are connected by a line segment, totaling $\frac{1}{2} \times 6 \times 5=15$ line segments;
(2) If point $A$ is the endpoint of only one line segment, then every two of the five points not connected to $A$ are connected by a line segment, at least $\frac{1}{2} \times 5 \times 4=10$ line segments, making a total of 11 line segments;
(3) If point $A$ is the endpoint of only two line segments $A B$ and $A C$, then every two of the four points not connected to $A$ are connected by a line segment, at least $\frac{1}{2} \times 4 \times 3=6$ line segments. One more line segment must be drawn from point $B$, so at least $2+6+1=9$ line segments are needed.
(4) If each point is the endpoint of at least three line segments, then at least $\frac{1}{2} \times 7 \times 3=\frac{21}{2}$ line segments are needed. Since the number of line segments must be an integer, at least 11 line segments are needed.
Figure 2 shows that nine line segments are sufficient.
|
9
|
Combinatorics
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
Six. (16 points) As shown in Figure 3, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3, P$ is at a distance $P D=12$ from the line $A N$, and $A D=30$. A line through $P$ intersects $A N$ and $A M$ at points $B$ and $C$ respectively. Find the minimum area of $\triangle A B C$.
---
The translation is provided as requested, maintaining the original formatting and structure.
|
Six, Solution 1
As shown in Figure 7, let $D B=$
$x$, draw $C E \perp A B$,
$P F / / C A$.
Let $A E=t$. Then
$$
\begin{array}{l}
\tan \angle M A N=3 \\
\Rightarrow C E=3 t .
\end{array}
$$
Similarly, by $P D=12 \Rightarrow F D=4$.
By $\triangle A B C \backsim \triangle F B P$, we get $\frac{A B}{C E}=\frac{F B}{P D}$, i.e.,
$$
\frac{30+x}{3 t}=\frac{4+x}{12} \text {. }
$$
Thus, $t=\frac{4(30+x)}{4+x}$.
Therefore, $S_{\triangle A B C}=\frac{1}{2} A B \cdot C E$
$=\frac{1}{2}(30+x) \times \frac{12(30+x)}{4+x}$
$=6 \times \frac{(30+x)^{2}}{4+x}$.
Let $m=\frac{(30+x)^{2}}{4+x}$.
Transforming into a quadratic equation in $x$ gives
$$
x^{2}+(60-m) x+900-4 m=0 \text {. }
$$
Since $x$ is a real number, we have
$$
\Delta=(60-m)^{2}-4(900-4 m) \geqslant 0 \text {, }
$$
i.e., $m(m-104) \geqslant 0$.
Thus, $m \leqslant 0$ or $m \geqslant 104$.
Therefore, when $m=104$, we have
$\left(S_{\triangle B B C}\right)_{\text {min }}=6 \mathrm{~m}=624$.
By $\frac{(30+x)^{2}}{4+x}=104$, solving for $x$ gives $x=22$.
Solution 2 can prove that when the line through point $P$ satisfies $P B=P C$, the area of $\triangle A B C$ is minimized.
In fact, let $B_{1} C_{1}$ be any line through point $P$, forming $\triangle A B_{1} C_{1}$.
As shown in Figure 8, draw $C F$
$/ / B_{1} B$. Then
$\triangle P C F \cong \triangle P B B_{1}$.
Thus, $S_{\triangle B B C}$
$$
\begin{array}{l}
=S_{\triangle P C F}+ \\
S_{\text {quadrilateral } 1 B_{1}, P C} \\
<S_{\triangle B_{1} C_{1}} \text {. } \\
\end{array}
$$
According to the above geometric conclusion, the calculation is as follows.
Since $P D=12$, we have $C E=24$.
Since $\tan \angle M A N=3$, we have $A E=8$.
By $E D=B D$, and $E D=A D-A E=22$, we get $B D=22$.
Thus, $A B=A D+B D=52$.
$\left(S_{\triangle A B C}\right)_{\text {min }}=\frac{1}{2} A B \cdot C E=624$.
|
624
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of lattice points (points with integer coordinates) inside the region (excluding the boundary) bounded by the right branch of the hyperbola $x^{2}-y^{2}=1$ and the line $x=100$ is $\qquad$
|
3.9800 .
By symmetry, we only need to consider the situation above the $x$-axis first.
Let the line $y=k(k=1,2, \cdots, 99)$ intersect the right branch of the hyperbola at point $A_{k}$, and intersect the line $x=100$ at point $B_{k}$. Then the number of integer points inside the segment $A_{k} B_{k}$ is $99-k$. Therefore, the number of integer points in the region above the $x$-axis is
$$
\sum_{k=1}^{9}(99-k)=\sum_{k=0}^{98} k=49 \times 99 \text {. }
$$
There are 98 integer points on the $x$-axis, so the total number of integer points is
$$
2 \times 49 \times 99+98=9800 .
$$
|
9800
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let positive integers $a, b$ satisfy $1 \leqslant a < b \leqslant 100$. If there exists a positive integer $k$, such that $a b \mid\left(a^{k}+b^{k}\right)$, then the pair $(a, b)$ is called a "good pair". Find the number of all good pairs.
（Xiong Bin's problem）
|
4. Let $(a, b)=d, a=s d, b=t d,(s, t)=1(t>s)$.
Thus, $s t d^{2} \mid d^{k}\left(s^{k}+t^{k}\right)$.
Since $\left(s t, s^{k}+t^{k}\right)=1$, it follows that $s t \mid d^{k}$.
Therefore, all prime factors of $st$ can divide $d$.
If $s$ or $t$ has a prime factor $p$ not less than 11, then $p \mid d$. Thus, $p^{2} \mid a$ or $p^{2} \mid b$.
Since $p^{2}>100$, this is a contradiction.
Therefore, the prime factors of $st$ can only be $2,3,5,7$.
If $st$ has at least three of the prime factors $2,3,5,7$, then
$d \geqslant 2 \times 3 \times 5=30$.
Thus, $\max \{a, b\} \geqslant 5 d>100$, which is a contradiction.
If the set of prime factors of $st$ is $\{3,7\}$, then
$\max \{a, b\} \geqslant 7 \times 3 \times 7>100$,
which is a contradiction;
Similarly, the set of prime factors of $st$ cannot be $\{5,7\}$.
If the set of prime factors of $st$ is $\{3,5\}$, $d$ can only be 15, at this time, $s=3, t=5$, so, $(a, b)=(45,75)$. There is 1 good pair.
If the set of prime factors of $st$ is $\{2,7\}$, $d$ can only be 14, at this time, $(s, t)=(2,7),(4,7)$, so, $(a, b)=$ $(28,98),(56,98)$. There are 2 good pairs.
If the set of prime factors of $st$ is $\{2,5\}$, $d$ can only be 10,20.
When $d=10$,
$(s, t)=(2,5),(1,10),(4,5),(5,8)$;
When $d=20$, $(s, t)=(2,5),(4,5)$.
There are 6 good pairs.
If the set of prime factors of $st$ is $\{2,3\}$, $d$ can only be 6, 12, 18, 24, 30.
When $d=6$,
\[
\begin{aligned}
(s, t)= & (1,6),(1,12),(2,3),(2,9), \\
& (3,4),(3,8),(3,16),(4,9), \\
& (8,9),(9,16) ;
\end{aligned}
\]
When $d=12$,
$(s, t)=(1,6),(2,3),(3,4),(3,8)$;
When $d=18$, $(s, t)=(2,3),(3,4)$;
When $d=24$, $(s, t)=(2,3),(3,4)$;
When $d=30$, $(s, t)=(2,3)$.
There are 19 good pairs.
If the set of prime factors of $st$ is $\{7\}$, then $(s, t)=$ $(1,7)$, $d$ can only be 7,14. There are 2 good pairs.
If the set of prime factors of $st$ is $\{5\}$, then $(s, t)=$ $(1,5)$, $d$ can only be $5,10,15,20$. There are 4 good pairs.
If the set of prime factors of $st$ is $\{3\}$, then
When $(s, t)=(1,3)$, $d$ can only be $3,6, \cdots, 33$;
When $(s, t)=(1,9)$, $d$ can only be $3,6,9$;
When $(s, t)=(1,27)$, $d$ can only be 3.
There are 15 good pairs.
If the set of prime factors of $st$ is $\{2\}$, then
When $(s, t)=(1,2)$, $d$ can only be $2,4, \cdots, 50$;
When $(s, t)=(1,4)$, $d$ can only be $2,4, \cdots, 24$;
When $(s, t)=(1,8)$, $d$ can only be $2,4, \cdots, 12$;
When $(s, t)=(1,16)$, $d$ can only be $2,4,6$;
When $(s, t)=(1,32)$, $d$ can only be 2.
There are 47 good pairs.
Therefore, the total number of good pairs is
$1+2+6+19+2+4+15+47=96$ (pairs).
|
96
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given 8 points $A_{1}, A_{2}, \cdots, A_{8}$ on a circle. Find the smallest positive integer $n$, such that among any $n$ triangles with these 8 points as vertices, there must be two triangles that share a common side.
(Tao Pingsheng, problem contributor)
|
8. First, consider the maximum number of triangles with no common edges.
Connecting every pair of eight points yields $\mathrm{C}_{8}^{2}=28$ chords.
If each chord belongs to only one triangle, then these chords can form at most $r \leqslant\left[\frac{28}{3}\right]=9$ triangles with no common edges. However, if there are nine such triangles, they would have a total of 27 vertices, meaning that one of the eight points (let's say $A_{8}$) must belong to at least four triangles. The opposite sides of the four triangles sharing vertex $A_{8}$ are all lines connecting two points among $A_{1}, A_{2}, \cdots, A_{7}$. Therefore, one point (let's say $A_{k}$) must appear twice, which means the corresponding two triangles will share a common edge $A_{8} A_{k}$, leading to a contradiction. Therefore, $r \leqslant 8$.
On the other hand, it is possible to construct eight triangles such that any two of them have no common edges.
Notice that such eight triangles produce a total of 24 vertices, and each point belongs to at most three triangles. Hence, each point belongs to exactly three triangles, meaning each point has a degree of 6.
For simplicity, as shown in Figure 3, take the eight equally spaced points on the circumference as the eight vertices of graph $G$, construct an 8-vertex complete graph, and then remove four diameters. Thus, we get 24 edges, and each point belongs to six edges. Among the triangles formed by these edges, select eight isosceles triangles:
$$
(1,2,3),(3,4,5),(5,6,7),(7,8,1)
$$
and $(1,4,6),(3,6,8),(5,8,2),(7,2,4)$, which have no common edges (each set of four triangles can be obtained by rotating one of the triangles around the center appropriately).
Therefore, $r=8$.
Thus, the required minimum value $n=r+1=9$.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Robots A and B start from the starting point at the same time, moving uniformly along a hundred-meter track, and the automatic recorder shows: when A is $1 \mathrm{~m}$ away from the finish line, B is $2 \mathrm{~m}$ away from the finish line; when A reaches the finish line, B is $1.01 \mathrm{~m}$ away from the finish line. Then the actual length of this track is more than $100 \mathrm{~m}$ by $\qquad$
|
2.1.
Let the actual length of the track be $x \mathrm{~m}$, and the speeds of robots 甲 and 乙 be $v_{\text {甲 }}$ and $v_{\text {乙}}$, respectively. Thus,
$$
\frac{v_{\text {甲 }}}{v_{\text {乙 }}}=\frac{x-1}{x-2}=\frac{x}{x-1.01} \text {. }
$$
Solving for $x$ yields $x=101$.
Therefore, this track is $1 \mathrm{~m}$ longer than $100 \mathrm{~m}$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Among the seven points consisting of the center and the six vertices of a regular hexagon, if any $n$ points are taken, and among them, there must be three points that form the vertices of an equilateral triangle, then the minimum value of $n$ is $\qquad$
|
-1.5 .
Consider the regular hexagon $A B C D E F$ as shown in Figure 1, with its center at $O$.
When $n=4$, take $A, C, D, F$, among which no three points can form the three vertices of an equilateral triangle.
When $n=5$, consider the following two cases:
(1) $O$ is among the five points. Consider the three pairs of points $(A, B)$, $(C, D)$, $(E, F)$. According to the pigeonhole principle, at least one pair of points must be taken, which, together with the center $O$, can form the three vertices of an equilateral triangle.
(2) $O$ is not among the five points. Consider the two groups of points $(A, C, E)$ and $(B, D, F)$. According to the pigeonhole principle, at least one group of points must be taken, which can form the three vertices of an equilateral triangle.
In summary, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Initially 285 Test: How many positive integers $T$ are there such that between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$ (not including $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$) there are exactly 5 different positive integers?
|
Solution: Obviously, when $T \geqslant 2010$,
$$
\frac{2010}{T} \leqslant 1, \frac{2010+T}{2 T} \leqslant 1,
$$
which does not meet the requirement.
Therefore, $T < 2010$ and
$$
\begin{array}{l}
\frac{2010}{T}>\frac{2010+T}{2 T} \\
\Rightarrow \frac{2010}{T}>\frac{2010+T}{2 T} .
\end{array}
$$
From this, we know
$$
\begin{array}{l}
45,
$$
meaning that there are at most four positive integers, 6, 7, 8, 9, between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$, which is a contradiction.
In summary, $T$ is a positive integer such that $168 \leqslant T \leqslant 200$, totaling 33.
(Yu Chenjie, Class 6, Grade 2, Shangbei Junior High School, Shanghai, 200070)
|
33
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { Three, (25 points) Let three distinct prime numbers } a, b, c \text { satisfy } \\
a \text { divides }(3 b-c), b \text { divides }(a-c), c \text { divides }(2 a-7 b), \\
20<c<80 .
\end{array}
$$
Find all values of $a^{b} c$.
|
When $a>b, a>c$, we have:
$$
-7 c<-7 b<2 a-7 b<2 a<2 c \text {. }
$$
Let $2 a-7 b=k c$,
where, $k=-6,-5,-4,-3,-2,-1,0,1$.
Then $2 a=7 b+k c \Rightarrow 7 b=2 a-k c$,
$$
\begin{array}{l}
a|(3 b-c) \Leftrightarrow a|(3 \times 7 b-7 c) \\
\Leftrightarrow a \mid[3(2 a-k c)-7 c] \\
\Leftrightarrow a \mid[6 a-(3 k+7) c] \\
\Leftrightarrow a|(3 k+7) c \Leftrightarrow a|(3 k+7), \\
b|(a-c) \Leftrightarrow b|(2 a-2 c) \Leftrightarrow b \mid(7 b+k c-2 c) \\
\Leftrightarrow b|(k-2) c \Leftrightarrow b|(k-2) .
\end{array}
$$
When $k=-6$, from equations (2) and (3) we get
$$
a|(-11), b|(-8) \text {. }
$$
Thus, $a=11, b=2$.
From equation (1) we get $c=-\frac{4}{3}$ (discard).
When $k=-5$, from equations (2) and (3) we get
$a|(-8), b|(-7)$.
Thus, $a=2, b=7$.
From equation (1) we get $c=9$ (discard).
Similarly, when $k=-4,-3,-2,1$, there are no values that meet the conditions.
When $k=-1$, $a=2, b=3, c=17$, which does not satisfy $20<c<80$.
When $k=0$, $2 a=7 b$, then $217 b, 7 \mid 2 a$.
Thus, $b=2, a=7$.
From $a \mid(3 b-c)$, we get $3 b-c=a t, c=6-7 t$. Since $20<c<80$, we have $20<6-7 t<80,-10 \frac{4}{7}<t<-2$.
Only when $t=-5$, $c$ is a prime number, $c=41$. Therefore, $a^{b} c=7^{2} \times 41=2009$.
(Xie Wenxiao, Huangfeng Middle School, Hubei Province, 438000)
|
2009
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. As shown in Figure 1, in the cube $A C_{1}$ with edge length 1, points $P$ and $Q$ are moving points on edges $A D$ and $A_{1} B_{1}$, respectively. If the skew lines $B D_{1}$ and $P Q$ are perpendicular to each other, then $A P + A_{1} Q =$ $\qquad$
|
5. 1 .
As shown in Figure 5, establish a spatial rectangular coordinate system, and set
$$
\begin{array}{l}
B(1,0,0), \\
D_{1}(0,1,1), \\
P(0, a, 0), \\
Q(b, 0,1) . \\
\text { Then } \overrightarrow{B D_{1}}=(-1,1,1), \\
\overrightarrow{P Q}=(b,-a, 1) .
\end{array}
$$
$$
\text { Therefore, } 0=\overrightarrow{B D_{1}} \cdot \overrightarrow{P Q}=-b-a+1 \text {. }
$$
Thus, $a+b=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For a positive integer $x$, let $S(x)$ denote the sum of the digits of $x$. Then the maximum value of $S(x)-9[\lg x]$ is $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
8. 9 .
Let $x=\overline{a_{n} a_{n-1} \cdots a_{0}}\left(a_{0} \neq 0\right)$. Then
$$
\begin{array}{l}
10^{n} \leqslant x<10^{n+1} \\
\Rightarrow n \leqslant \lg x<n+1 \Rightarrow[\lg x]=n . \\
\text { Hence } S(x)=a_{0}+a_{1}+\cdots+a_{n} \leqslant 9(n+1) \\
=9([\lg x]+1) .
\end{array}
$$
Therefore, $S(x)-9[\lg x] \leqslant 9$.
$$
\text { Also, } S(9)-9[\lg 9]=9 \text {, then, } S(x)-9[\lg x]
$$
has a maximum value of 9.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given 2011 positive integers, the product of which is equal to their sum. How many 1s are there at least among these 2011 numbers?
|
Let the 2011 positive integers be
$a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2011}$.
From $a_{1} a_{2} \cdots a_{2011}=a_{1}+a_{2}+\cdots+a_{2011}$
$\leqslant 2011 a_{2011}$,
we know $a_{1} a_{2} \cdots a_{2010} \leqslant 2011a b \Rightarrow(a-1)(b-1)a b \Rightarrow(a-1)(b-1)<1,
\end{array}
$$
Contradiction.
If $A=(2,3,3, a, b)$ or $(3,3,3, a, b)$, the same contradiction is derived.
Therefore, $a_{2004}=1$.
(5) When $a_{2005}=a_{2006}=\cdots=a_{2010}=2, a_{2011}$ $=32$,
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=2004+2 \times 6+32 \\
=2048=a_{1} a_{2} \cdots a_{2011} .
\end{array}
$$
Combining (1)～(5), we know there are at least 2004 ones.
(Liu Donghua, Nankai High School, Tianjin, 300100)
|
2004
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $n$ be a positive integer, and $f(n)$ denote the number of integers satisfying the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and
$a_{i} \neq a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ have opposite signs.
(1) Find the value of $f(10)$;
(2) Determine the remainder when $f(2008)$ is divided by 13.
[3]
|
Let $g(n)$ denote the number of wave numbers for which $a_{n}>a_{n-1}$ when $n \geqslant 2$. Then, by symmetry, we have
$$
\begin{array}{l}
g(n)=\frac{1}{2} f(n) . \\
\text { Hence } a_{n-1}=1, a_{n}=2,3,4 ; \\
a_{n-1}=2, a_{n}=3,4 ; \\
a_{n-1}=3, a_{n}=4 .
\end{array}
$$
Let $m(i)$ represent the number of $(n-1)$-digit wave numbers where $a_{n-2}>a_{n-1}$ and the last digit is $i$. Then
$$
\begin{array}{l}
m(1)+m(2)+m(3)+m(4)=g(n-1), \\
m(4)=0 .
\end{array}
$$
When $a_{n-1}=1$, $a_{n-2}=2,3,4$. Then
$$
m(1)=g(n-2) \text {. }
$$
When $a_{n-1}=3$, $a_{n-2}=4, a_{n-3}=1,2,3$. Then $m(3)=g(n-3)$.
Thus, $g(n)=3 m(1)+2 m(2)+m(3)$
$$
=2 g(n-1)+g(n-2)-g(n-3) \text {. }
$$
It is easy to see that $g(2)=6, g(3)=14, g(4)=31$, $\cdots \cdots g(10)=4004$.
Therefore, $f(10)=2 g(10)=8008$.
Also note that the sequence of remainders when $g(n)$ is divided by 13 is:
$$
6,1,5,5,1,2,0,1,0,1,1,3,6,1,5,5, \cdots
$$
The smallest positive period is 12.
Thus, $g(2008) \equiv g(4) \equiv 5(\bmod 13)$.
Therefore, $f(2008) \equiv 10(\bmod 13)$.
|
10
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. At a certain charity fundraising dinner, each person ate half a plate of rice, one-third of a plate of vegetables, and one-quarter of a plate of meat. The dinner provided a total of 65 plates of food. How many people attended this fundraising dinner?
|
2. 60 .
Each person ate $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ of food, and $65 \div \frac{13}{12}=60$.
Therefore, 60 people attended this charity dinner.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. The number of triples of positive integers $(x, y, z)$ that satisfy $x y z=3^{2010}$ and $x \leqslant y \leqslant z<x+y$ is $\qquad$.
|
3. 336 .
Let $x=3^{a}, y=3^{b}, z=3^{c}$. Then
$$
0 \leqslant a \leqslant b \leqslant c, a+b+c=2010 \text {. }
$$
If $c \geqslant b+1$, then
$$
x+y=3^{a}+3^{b}<3^{b+1} \leqslant 3^{c}=z \text {, }
$$
which contradicts $z<x+y$. Hence, $c=b$.
Thus, $a+2 b=2010$.
Therefore, $670 \leqslant b \leqslant 1005$.
Hence, the number of such positive integer triples, which is the number of $b$, is $1005-670+1=336$.
|
336
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The smallest four-digit number that has exactly 14 divisors (including 1 and itself), and one of its prime factors has a units digit of 3 is $\qquad$ .
|
5.1458.
Let this four-digit number be $n$.
Since $14=14 \times 1=7 \times 2$, we have $n=p^{13}$ or $p^{6} q$ (where $p$ and $q$ are different prime numbers).
If $n=p^{13}$, by the given condition, the unit digit of $p$ is 3, so, $p \geqslant 3$.
Thus, $n \geqslant 3^{13}=1594323$, which is a contradiction.
Therefore, $n=p^{6} q$.
If $p \geqslant 5$, then the number of digits in $n$ will exceed 4, which is a contradiction.
If $p=3$, then $p^{6}=729$, and when $q=2$, $n=$ $p^{6} q=1458$, which satisfies the condition.
If $p=2$, then $p^{6}=64$, in this case, $q$ must be chosen from 3, 13, 23, 43, ... these prime numbers with a unit digit of 3.
Multiplying 64 by these numbers, the smallest four-digit number obtained is
$$
p^{6} q=64 \times 23=1472>1458 .
$$
Therefore, the smallest four-digit number that satisfies the condition is 1458.
|
1458
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $f(x)$ represent a fourth-degree polynomial in $x$. If
$$
f(1)=f(2)=f(3)=0, f(4)=6, f(5)=72 \text {, }
$$
then the last digit of $f(2010)$ is $\qquad$
|
6. 2 .
Since $f(1)=f(2)=f(3)=0$, the quartic polynomial can be set as
$$
\begin{array}{l}
f(x)=(x-1)(x-2)(x-3)(a x+b) . \\
\text { By } f(4)=6, f(5)=72, \text { we get } \\
\left\{\begin{array} { l }
{ 6 ( 4 a + b ) = 6 , } \\
{ 2 4 ( 5 a + b ) = 7 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=2, \\
b=-7 .
\end{array}\right.\right.
\end{array}
$$
Then $f(2010)$
$$
\begin{array}{l}
=(2010-1)(2010-2)(2010-3)(2010 \times 2-7) \\
=2009 \times 2008 \times 2007 \times 4013 .
\end{array}
$$
Therefore, the last digit of $f(2010)$ is 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $p, q$ be prime numbers, and satisfy $p^{3}+q^{3}+1=p^{2} q^{2}$. Then the maximum value of $p+q$ is
|
8.5.
Assume $q \leqslant p$. Notice that
$$
\begin{array}{l}
p^{3}+q^{3}+1=p^{2} q^{2} \\
\Rightarrow q^{3}+1=p^{2} q^{2}-p^{3} \\
\Rightarrow(q+1)\left(q^{2}-q+1\right)=p^{2}\left(q^{2}-p\right) .
\end{array}
$$
Therefore, $p^{2} \mid(q+1)\left(q^{2}-q+1\right)$.
Since $q \leqslant p$, we have $0<q^{2}-q+1<p^{2}$.
This implies $p \mid(q+1)$.
Also, since $q \leqslant p$, we have $q+1=p$.
Given that $p$ and $q$ are both prime numbers, the only solution is $p=3$ and $q=2$. Thus, the maximum value of $p+q$ is 5.
|
5
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Given $n$ positive integers (not necessarily distinct), their sum is 100, and the sum of any seven of them is less than 15. Then the minimum value of $n$ is $\qquad$ .
|
9. 50.
Let these $n$ numbers be $a_{1}, a_{2}, \cdots, a_{n}$. Then
$$
a_{1}+a_{2}+\cdots+a_{49} \leqslant 14 \times 7=98 \text {. }
$$
Thus $n \geqslant 50$, and when these 50 numbers are all 2, the condition is satisfied.
|
50
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given point $P$ inside $\triangle A B C$, satisfying $\angle A B P$ $=20^{\circ}, \angle P B C=10^{\circ}, \angle A C P=20^{\circ}$ and $\angle P C B$ $=30^{\circ}$. Then $\angle C A P=$
|
10.20.
As shown in Figure 7, construct a regular $\triangle QBC$ on side $BC$ on the same side as point $A$.
Notice that $BA$ is the angle bisector of the regular $\triangle QBC$, so, $\angle AQC = \angle ACQ = 10^{\circ}$.
Also, $CP$ is the angle bisector of the regular $\triangle QBC$, hence
$$
\angle PQC = \angle PBC = 10^{\circ}.
$$
Since points $A$ and $P$ are on the same side of $QC$, point $A$ lies on segment $PQ$.
Thus, $\angle CAP = \angle AOC + \angle ACQ = 20^{\circ}$.
|
20
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. A farmer has 100 pigs and 100 chickens. He has four adjacent square yards, forming a $2 \times 2$ grid. The farmer wants to distribute the livestock among the yards according to the following requirements: the first row has 120 heads, the second row has 300 feet; the first column has 100 heads, the second column has 320 feet. Then there are $\qquad$ different ways to distribute them.
|
11.341.
As shown in Figure 8, label the four courtyards with letters $A_{i} (i=1,2,3,4)$. Let $A_{i}$ courtyard have $x_{i}$ pigs and $y_{i}$ chickens.
From the problem, we get
every 8
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=100, \\
y_{1}+y_{2}+y_{3}+y_{4}=100, \\
x_{1}+y_{1}+x_{2}+y_{2}=120, \\
4\left(x_{3}+x_{4}\right)+2\left(y_{3}+y_{4}\right)=300, \\
x_{1}+y_{1}+x_{3}+y_{3}=100, \\
4\left(x_{2}+x_{4}\right)+2\left(y_{2}+y_{4}\right)=320 .
\end{array}\right.
$$
Solving, we get $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=(x, 30-x, 40-x, 30+x)$,
$$
\left(y_{1}, y_{2}, y_{3}, y_{4}\right)=(y, 90-y, 60-y, y-50) \text {, }
$$
where $x$ has 31 possible values from $0$ to $30$, and $y$ has 11 possible values from $50$ to $60$.
Therefore, there are $11 \times 31=341$ ways to distribute the livestock to the courtyards as required.
|
341
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Line segment $A B$ divides a square into two polygons (points $A$ and $B$ are on the sides of the square), each polygon has an inscribed circle, one of which has a radius of 6, while the other has a radius greater than 6. What is the difference between the side length of the square and twice the length of line segment $A B$?
|
If line segment $A B$ divides the square into two triangles, then $A B$ can only be the diagonal of the square. But in this case, the radii of the two incircles are equal, which contradicts the condition.
If one of the polygons is a quadrilateral, then the sum of the lengths of $A B$ and its opposite side is greater than the sum of the lengths of the other two sides, which contradicts the fact that the quadrilateral has an incircle.
Therefore, it can only be that $A B$ divides the square into a triangle and a pentagon, and the incircle of the pentagon is also the incircle of the original square (as shown in Figure 9).
As shown in Figure 10, let the incircle of $\triangle A B C$ touch $B C$, $C A$, and $A B$ at points $D$, $E$, and $F$, respectively, and the incircle of the original square, which is also the excircle of $\triangle A B C$, touch $C A$, $B C$, and $A B$ at points $P$, $Q$, and $R$, respectively. Then the side length of the original square is $C P + C Q$.
$$
\begin{aligned}
\text { Hence } & C P + C Q - 2 A B \\
= & (A P + A E + 6) + (B Q + B D + 6) - \\
& (A F + B F + A R + B R) \\
= & 12 + (A P - A R) + (B Q - B R) + \\
& (A E - A F) + (B D - B F) \\
= & 12 .
\end{aligned}
$$
|
12
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. In a candy store, candies are sold in three types of packages: small packs contain 6 candies, medium packs contain 9 candies, and large packs contain 20 candies. If you can only buy whole packs of candies, what is the maximum number of candies you cannot purchase?
|
2. Write non-negative integers in the following six-row number table:
\begin{tabular}{ccccccccc}
$\mathbf{0}$ & 6 & 12 & 18 & 24 & 30 & 36 & 42 & 48 \\
1 & 7 & 13 & 19 & 25 & 31 & 37 & 43 & 49 \\
2 & 8 & 14 & 20 & 26 & 32 & 38 & 44 & 50 \\
3 & 9 & 15 & 21 & 27 & 33 & 39 & 45 & 51 \\
4 & 10 & 16 & 22 & 28 & 34 & 40 & 46 & 52 \\
5 & 11 & 17 & 23 & 29 & 35 & 41 & 47 & 53
\end{tabular}
Each row of numbers increases by 6 from left to right.
If a number in a row can be obtained (i.e., you can buy this number of candies), then all the numbers in this row that come after it can also be obtained. Therefore, it is only necessary to find the first number in each row that can be obtained (which can be expressed in the form $9k + 20l$ where $k, l \in \mathbf{N}$), which are marked in bold in the table.
Clearly, the largest of these six numbers is 49.
Thus, the number 43, which is in the same row and immediately before it, is the largest number that cannot be obtained.
|
43
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. (40 points) Let integers $m, n$ satisfy $m \geqslant n$ and $m^{3}+n^{3}+1=4 m n$. Find the maximum value of $m-n$.
保留了原文的换行和格式，如上所示。
|
3. Let $s=m+n, p=mn$. Then $m^{3}+n^{3}=(m+n)^{3}-3m^{2}n-3mn^{2}$ $=s^{3}-3ps$.
From $s^{3}-3ps+1=4p$, we get $p=\frac{s^{3}+1}{3s+4}$.
Since $m, n$ are integers, 27p must also be an integer, and
$27p=9s^{2}-12s+16-\frac{37}{3s+4}$.
Therefore, $3s+4$ must divide 37.
Thus, $3s+4 = \pm 1$ or $\pm 37$.
If $3s+4=-1$ or -37, then $s$ is not an integer;
If $3s+4=37$, then $s=11$ and $p=36$, but there are no integers $m, n$;
If $3s+4=1$, then $s=-1$ and $p=0$.
Thus, $m=0, n=-1$.
Therefore, the maximum value of $m-n$ is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. (40 points) 64 points on a plane form an $8 \times 8$ grid. The distance between any two adjacent points in the same row or column is 1. How many rectangles with an area of 12 can be formed using four of these 64 points as vertices?
|
4. Connecting points in the same row and column can form a grid.
First, consider rectangles with sides parallel to the grid lines. For a rectangle of shape $2 \times 6$, there are two orientation choices (horizontal and vertical). One pair of opposite sides has $8-6=2$ choices, and the other pair has $8-2=6$ choices. Therefore, there are $2 \times 2 \times 6=24$ such rectangles.
Similarly, for a rectangle of shape $3 \times 4$, there are $2 \times 4 \times 5=40$ such rectangles.
Next, consider rectangles with sides parallel to the diagonals of the grid, with shapes $\sqrt{2} \times 6 \sqrt{2}$ and $2 \sqrt{2} \times 3 \sqrt{2}$.
These rectangles also have two orientation choices. First, consider the direction where the longest side extends from the bottom left to the top right. For the first shape, there is only 1 such rectangle (as shown in Figure 12); for the second shape, there are 9 such rectangles, one of which is marked in Figure 12. The top-left vertices of these 9 rectangles are marked with black dots in Figure 12. Therefore, there are $2 \times (1+9)=20$ such rectangles.
Since $12=2^{2} \times 3$, and 3 cannot be expressed as the sum of two perfect squares, there are no other shapes of rectangles.
Therefore, the total number of rectangles that meet the criteria is $24+40+20=84$.
|
84
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. (40 points) Find the largest positive integer $n$, such that there exists a unique positive integer $k$ satisfying $\frac{8}{15}<\frac{n}{n+k}<\frac{7}{13}$.
|
5. The inequality can be written as $1+\frac{7}{8}>1+\frac{k}{n}>1+\frac{6}{7}$. Or $\frac{98}{112}>\frac{k}{n}>\frac{96}{112}$.
If $n=112$, then the unique value of $k$ is 97.
Assuming $n>112$, then $\frac{98 n}{112 n}>\frac{112 k}{112 n}>\frac{96 n}{112 n}$.
Between $96 n$ and $98 n$, there are at least two numbers that are multiples of 112, in which case, the value of $k$ is not unique. Therefore, the maximum value of $n$ is 112.
|
112
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. (40 points) In each small square of a $9 \times 9$ grid, fill in a number such that each row and each column contains at most four different numbers. What is the maximum number of different numbers that can be in this grid?
|
6. Suppose this square table contains 29 different numbers. According to the pigeonhole principle, there must be a row with four different numbers, let's assume it is the first row. The remaining 25 numbers are in rows $2 \sim 9$. Similarly, according to the pigeonhole principle, there must be a row with four, let's assume it is the second row.
Next, observe the numbers in each column of the square table. Each column already has two different numbers at the top, so each column below the second row can have at most two different numbers. This way, the maximum number of different numbers is $8+9 \times 2=26$, which is a contradiction. Therefore, the number of different numbers is less than 29.
Figure 13 constructs a $9 \times 9$ square table containing 28 different numbers, with each row and each column having exactly four different numbers. Another filling method is shown in Figure 14.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (40 points) Write "KOREAIMC" in eight lines as shown in Figure 5, where the first line has one K, the second line has two O's, and so on, with the last line having eight C's.
$$
\begin{array}{llllllll}
\mathbf{K} & & & & & & & \\
\mathbf{O} & \mathbf{O} & & & & & & \\
\mathbf{R} & \mathbf{R} & \mathbf{R} & & & & & \\
\mathrm{E} & \mathbf{E} & \mathrm{E} & \mathrm{E} & & & & \\
\mathrm{A} & \mathrm{A} & \mathbf{A} & \mathrm{A} & \mathrm{A} & & & \\
\mathrm{I} & \mathrm{I} & \mathrm{I} & \mathbf{I} & \mathrm{I} & \mathrm{I} & & \\
\mathrm{M} & \mathrm{M} & \mathbf{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \\
\mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C} & \mathrm{C}
\end{array}
$$
Figure 5
Starting from the K at the top, move downward row by row to spell out "KOREAIMC". Each move can only be to the letter directly below or to a letter adjacent to the one directly below. The bold letters in Figure 5 represent one such path. If a letter is removed from Figure 5, such that the number of different remaining paths is only 516, identify the position of the removed letter.
|
10. First list the number table as shown in Figure 17, where each cell corresponds to a letter, and the number in the cell represents the number of different paths from the letter "K" to that letter. The cell at the very top is filled with 1. Starting from the second row, the number in each cell is the sum of the numbers in the cells from the row above that share a common vertex with it.
Then list
the number table as shown in Figure 18,
where each cell
also corresponds
to a letter, and the
number in the cell
indicates the number of different paths from that letter to the last row, filled in a similar manner to the table in Figure 17.
Thus, for each letter, the number table in Figure 17 gives the number of paths \( m \) from the letter "K" to that letter, while the number table in Figure 18 gives the number of paths \( n \) from that letter to the last row. If the letter is removed, the number of paths will be reduced by \( m \times n \). If the letter is not removed, the total number of different paths is 750. Since there are now only 516 paths, the total reduction is
$$
750 - 516 = 234 = 2 \times 3^{2} \times 13
$$
paths.
It is easy to see that only the numbers 13 and 26 in the fifth row of the number table in Figure 18 are multiples of 13.
At this point, \( 13 \times 9 = 117, 26 \times 9 = 234 \).
Therefore, the letter removed is the third letter "A" in the fifth row of Figure 5.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $k$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=0, a_{1}=1, \\
a_{n+1}=k a_{n}+a_{n-1}(n=1,2, \cdots) .
\end{array}
$$
Find all $k$ that satisfy the following condition: there exist non-negative integers $l, m (l \neq m)$, and positive integers $p, q$, such that
$a_{l}+k a_{0}=a_{m}+k a_{q^{*}} \quad$ (Xiong Bin)
|
When $k=2$, $a_{0}=0, a_{1}=1, a_{2}=2$, then from $a_{0}+2 a_{2}=a_{2}+2 a_{1}=4$, we know that taking $l=0, m=2, p=$ $2, q=1$ is sufficient.
For $k \geqslant 3$, by the recurrence relation, $\left\{a_{n}\right\}$ is a strictly increasing sequence of natural numbers and $k \mid\left(a_{n+1}-a_{n-1}\right)$.
$$
\begin{array}{l}
\text { then } a_{2 n} \equiv a_{0}=0(\bmod k), \\
a_{2 n+1} \equiv a_{1}=1(\bmod k)(n=0,1, \cdots) .
\end{array}
$$
If there exist $l, m \in \mathbf{N}, p, q \in \mathbf{N}_{+}, l \neq m$, satisfying $a_{l}+k a_{p}=a_{m}+k a_{q}$, without loss of generality, assume $l0$, then
$$
\begin{array}{l}
a_{l}+k a_{p} \leqslant k a_{p}+a_{p-1}=a_{p+1} \\
\leqslant a_{m}\frac{a_{l}+k a_{p}-a_{m}}{k}=a_{q} \text {. }
\end{array}
$$
From $k a_{q}+a_{m}=k a_{p}+a_{l} \geqslant k a_{p}$,
we know $a_{q} \geqslant a_{p}-\frac{a_{m}}{k} \geqslant a_{p}-\frac{a_{p}}{k}=\frac{k-1}{k} a_{p}$.
Noting that $a_{p} \geqslant k a_{p-1}$. Thus,
$a_{p}>a_{q} \geqslant \frac{k-1}{k} a_{p} \geqslant(k-1) a_{p-1} \geqslant a_{p-1}$.
From equation (4), we know $a_{q}=a_{p-1}$.
Therefore, the equalities in equations (1), (2), and (3) must all hold.
From equations (2) and (3), we get
$m=p, p=2$.
Thus, $a_{q}=a_{p-1}=a_{1}=1$.
From equation (1), we know $l=0$.
Therefore, from $a_{l}+k a_{p}=a_{m}+k a_{q}$, we get $k^{2}=k+k$, i.e., $k=2$, which is a contradiction.
Thus, $k \geqslant 3$ does not satisfy the problem's conditions.
Hence, $k$ can only be 2.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the function $f(x)=x^{2}-1$ with domain $D$, and the range is $\{-1,0,1,3\}$. Determine the maximum number of such sets $D$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
2, 1.27.
Since $f(0)=-1, f( \pm 1)=0$,
$$
f( \pm \sqrt{2})=1, f( \pm 2)=3 \text {, }
$$
Therefore, $0 \in D$; at least one from each of the sets $\{-1,1\}$, $\{-\sqrt{2}, \sqrt{2}\}$, and $\{-2,2\}$ belongs to $D$.
Thus, there are $3 \times 3 \times 3=27$ such $D$'s.
|
27
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Find the value of $\frac{\log _{2} \frac{2}{3}+\log _{2} \frac{3}{4}+\log _{2} \frac{4}{5}+\log _{2} \frac{5}{6}+\log _{2} \frac{6}{7}+\log _{2} \frac{7}{8}}{\log _{3} 3 \cdot \log _{4} 3 \cdot \log _{5} 4 \cdot \log _{8} 5 \cdot \log _{7} 6 \cdot \log _{8} 7}$.
|
2. -6 .
$$
\begin{array}{l}
\text { Original expression }=\frac{\log _{2}\left(\frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}\right)}{\frac{\lg 2}{\lg 3} \cdot \frac{\lg 3}{\lg 4} \cdot \frac{\lg 4}{\lg 5} \cdot \frac{\lg 5}{\lg 6} \cdot \frac{\lg 6}{\lg 7} \cdot \frac{\lg 7}{\lg 8}} \\
=\frac{\log _{2} \frac{2}{8}}{\frac{\lg 2}{\lg 8}}=\frac{\log _{2} 2^{-2}}{\frac{\lg 2}{3 \lg 2}}=\frac{-2}{\frac{1}{3}}=-6 .
\end{array}
$$
|
-6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In a convex quadrilateral $ABCD$, $\angle BAD + \angle ADC = 240^{\circ}$, $E$ and $F$ are the midpoints of sides $AD$ and $BC$, respectively, and $EF = \sqrt{7}$. If two squares $A_1$ and $A_2$ are drawn with sides $AB$ and $CD$, respectively, and a rectangle $A_3$ is drawn with length $AB$ and width $CD$, find the sum of the areas of the three figures $A_1$, $A_2$, and $A_3$.
|
6.28 .
As shown in Figure 6, extend $BA$ and $CD$ to intersect at point $P$.
Given $\angle BAD + \angle ADC = 240^{\circ}$, we have
$$
\angle BPC = 60^{\circ}.
$$
Connect $BD$, and take the midpoint $G$ of $BD$, then connect $EG$ and $FG$.
By the Midline Theorem of a triangle, we know
$$
EG \perp \frac{1}{2} AB, FG \perp \frac{1}{2} CD, \angle EGF = 120^{\circ}.
$$
In $\triangle EGF$, by the Law of Cosines, we get
$$
\begin{array}{l}
7 = EF^2 = EG^2 + FG^2 + EG \cdot FG \\
= \left(\frac{AB}{2}\right)^2 + \left(\frac{CD}{2}\right)^2 + \left(\frac{AB}{2}\right)\left(\frac{CD}{2}\right),
\end{array}
$$
which simplifies to $AB^2 + CD^2 + AB \times CD = 28$.
Therefore, the sum of the areas of the three figures $A_1, A_2$, and $A_3$ is 28.
|
28
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $x_{1}, x_{2}, \cdots, x_{2010}$ are all positive real numbers. Then
$$
x_{1}+\frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{1} x_{2}}+\cdots+\frac{x_{2010}}{x_{1} x_{2} \cdots x_{200}}+\frac{4}{x_{1} x_{2} \cdots x_{2010}}
$$
the minimum value is $\qquad$
|
3.4.
Starting from the last two terms, repeatedly applying the AM-GM inequality, we get
$$
\begin{array}{l}
\text { Original expression }=\sum_{i=1}^{2010} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\frac{4}{\prod_{j=1}^{2010} x_{j}} \\
=\sum_{i=1}^{2009} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\left(\frac{x_{2010}}{\prod_{j=1}^{2009} x_{j}}+\frac{4}{x_{2010} \prod_{j=1}^{2009} x_{j}}\right) \\
\geqslant \sum_{i=1}^{2009} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\frac{4}{\prod_{j=1}^{2009} x_{j}} \geqslant \cdots \geqslant x_{1}+\frac{4}{x_{1}} \geqslant 4,
\end{array}
$$
where the equality holds if and only if
$$
x_{2010}=x_{2009}=\cdots=x_{1}=2 \text {. }
$$
Therefore, the minimum value sought is 4.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that the circumcenter, incenter, and orthocenter of a non-isosceles acute $\triangle ABC$ are $O, I, H$ respectively, and $\angle A=60^{\circ}$. If the altitudes of $\triangle ABC$ are $AD, BE, CF$, then the ratio of the circumradius of $\triangle OIH$ to the circumradius of $\triangle DEF$ is $\qquad$ .
|
4. 2 .
From $\angle B O C=\angle B I C=\angle B H C=120^{\circ}$, we know that $O, I, H, B, C$ are concyclic.
Let the circumradii of $\triangle A B C$ and $\triangle O I H$ be $R$ and $r$, respectively.
By the Law of Sines, we have
$2 R \sin A=B C=2 r \sin \angle B O C$.
Thus, $r=R$.
Let the circumradius of $\triangle D E F$ be $r^{\prime}$.
Noting that $A, E, H, F$ are concyclic, and $A H$ is the diameter of this circle. By the Law of Sines, we get
$$
E F=A H \sin A=2 R \cos A \cdot \sin A=\frac{\sqrt{3}}{2} R .
$$
Since $A, F, D, C$ and $A, E, D, B$ are concyclic, respectively, we have
$\angle B D F=\angle C D E=\angle B A C=60^{\circ}$.
Thus, $\angle F D E=60^{\circ}$.
By the Law of Sines, we get
$$
E F=2 r^{\prime} \sin \angle F D E=\sqrt{3} r^{\prime} \text {. }
$$
Therefore, $r^{\prime}=\frac{R}{2}$.
Hence, $\frac{r}{r^{\prime}}=2$.
[Note] The circumcircle of $\triangle D E F$ is actually the nine-point circle of $\triangle A B C$, so $r^{\prime}=\frac{R}{2}$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ is a permutation of $1,2,3,4,5$, and satisfies $\left|a_{i}-a_{i+1}\right| \neq 1(i=1,2,3,4)$. Then the number of permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that meet the condition is $\qquad$.
|
8. 14 .
The permutations that satisfy the conditions are:
$$
\begin{array}{l}
1,3,5,2,4 ; 1,4,2,5,3 ; 2,4,1,3,5 ; \\
2,4,1,5,3 ; 2,5,3,1,4 ; 3,1,4,2,5 ; \\
3,1,5,2,4 ; 3,5,1,4,2 ; 3,5,2,4,1 ; \\
4,1,3,5,2 ; 4,2,5,1,3 ; 4,2,5,3,1 ; \\
5,2,4,1,3 ; 5,3,1,4,2 .
\end{array}
$$
|
14
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. (15 points) In a regular pentagon $A B C D E$, the diagonal $B E$ intersects diagonals $A D$ and $A C$ at points $F$ and $G$, respectively. The diagonal $B D$ intersects diagonals $C A$ and $C E$ at points $H$ and $I$, respectively. The diagonal $C E$ intersects diagonal $A D$ at point $J$. Let the set of isosceles triangles formed by the 10 points $A, B, C, D, E, F, G, H, I, J$ and the line segments in Figure 2 be denoted as $M$.
(1) Determine the number of elements in $M$;
(2) If each of the 10 points is arbitrarily colored either red or blue, does there necessarily exist an isosceles triangle in $M$ whose three vertices are the same color?
(3) If $n$ of the 10 points are arbitrarily colored red such that there necessarily exists an isosceles triangle in $M$ whose three vertices are all red, find the minimum value of $n$.
|
11. (1) Since all triangles formed by 10 points $A, B, C, D, E, F, G, H, I, J$ and line segments in Figure 2 are isosceles triangles,
$$
|M|=\mathrm{C}_{5}^{3}+4 \mathrm{C}_{5}^{4}+5 \mathrm{C}_{5}^{5}=35 \text{. }
$$
(2) By the pigeonhole principle, we know that among $A, B, C, D, E$, there must be three points of the same color, and the triangle formed by these three points belongs to $M$. Therefore, there must exist an isosceles triangle in $M$ whose three vertices are of the same color.
(3) If $n=5$, then coloring $F, G, H, I, J$ red, there will be no triangle in $M$ whose vertices are all red.
If $n=6$, when there are at least three red points among $A, B, C, D, E$, there must exist a triangle in $M$ whose vertices are all red; when there are fewer than three red points among $A, B, C, D, E$, there are at least four red points among $F, G, H, I, J$.
If there are exactly four red points among $F, G, H, I, J$, assume $F, G, H, J$ are red points, then there are at least two red points among $A, B, C, D, E$. If one of $A, B, E$ is a red point, assume $A$ is a red point, then $\triangle AFG$ is a triangle in $M$ whose vertices are all red; otherwise, $C, D$ are both red, thus, $\triangle CDH$ is a triangle in $M$ whose vertices are all red.
If $F, G, H, I, J$ are all red, then there is at least one red point among $A, B, C, D, E$. Assume $A$ is a red point, then $\triangle AFG$ is a triangle in $M$ whose vertices are all red. Therefore, the minimum value of $n$ is 6.
(Li Jianquan provided)
|
6
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Divide a wire of length $11 \mathrm{~cm}$ into several segments of integer centimeters, such that any three segments can form the sides of a triangle. Then the number of different ways to divide the wire is $\qquad$ (ways with the same number of segments and corresponding equal lengths are considered the same way).
|
4.8.
Let the shortest two segments be $a \mathrm{~cm}, b \mathrm{~cm}$, and take any other segment $c \mathrm{~cm} (a \leqslant b \leqslant c)$. Then $b \leqslant c < a + b$.
(1) If $a = 1$, then $1 \leqslant b \leqslant c < b + 1$. Thus, $b = c$. Therefore, $b \in (11 - 1), 2b < 11 - 1$. Hence, $b$ can be $1, 2, 5$, giving three ways to divide.
(2) If $a = 2$, then $2 \leqslant b \leqslant c < b + 2$. Thus, $c$ can be $b$ or $b + 1$. Let the segments of length $b \mathrm{~cm}$ and $(b + 1) \mathrm{~cm}$ (excluding the segment of length $a$) be $x$ and $y$ segments respectively, where $x \geqslant 0$, $y \geqslant 0$, and $x + y \geqslant 2$.
By the problem, $2 + bx + (b + 1)y = 11$.
Then $y = \frac{9 + x}{b + 1} - x$.
When $x = 0$, $y = \frac{9}{b + 1}$, so $b + 1$ can only be 3, giving one way to divide $(2, 3, 3, 3)$;
When $x = 1$, $y = \frac{10}{b + 1} - 1$, so $b + 1$ can only be 5, giving one way to divide $(2, 4, 5)$;
When $x = 2$, it does not meet the problem's conditions;
When $x = 3$, $y = \frac{12}{b + 1} - 3$, so $b + 1 = 3, 4$, giving two ways to divide $(2, 2, 2, 2, 3), (2, 3, 3, 3)$;
When $x \geqslant 4$, $bx \leqslant 9 \Rightarrow b \leqslant \frac{9}{x} \leqslant \frac{9}{4} < 3$, thus $b = 2$, then $2 + 2x + 3y = 11$, solving gives
$$
(x, y) = (3, 1), (0, 3),
$$
giving two ways to divide $(2, 2, 2, 2, 3), (2, 3, 3, 3)$.
(3) If $a \geqslant 3$, then $a + b + c \geqslant 9, 2a + b + c \geqslant 12$. Therefore, the wire can only be divided into 3 segments. Thus, $(a, b, c) = (3, 3, 5), (3, 4, 4)$, giving two ways to divide.
In summary, there are 8 ways to divide in total.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) A competitive event involves several teams from two cities, A and B. It is known that city B has 8 more teams than city A, and any two teams play exactly one match. The event rules state: the winner gets 1 point, the loser gets 0 points, and there are no ties. In the end, the total score of all teams from city B is 4 points more than the total score of all teams from city A. Find the minimum score of the best team from city A.
|
Let city A have $x$ teams participating in the competition. Then city B has $x+8$ teams. When the teams from cities A and B compete against each other, city A scores a total of $y$ points, and city B scores a total of $x(x+8)-y$ points.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{2} x(x-1)+y+4 \\
=\frac{1}{2}[(x+8)-1](x+8)+[x(x+8)-y] \\
\Rightarrow y=\frac{1}{2} x^{2}+8 x+12 .
\end{array}
$$
Since $y$ is an integer, $x$ must be even.
$$
\begin{array}{l}
\text { From } x(x+8)-y \geqslant 0 \\
\Rightarrow x(x+8)-\left(\frac{1}{2} x^{2}+8 x+12\right) \geqslant 0 \\
\Rightarrow x \geqslant \sqrt{24}>4 .
\end{array}
$$
Since $x$ is even, the minimum value of $x$ is 6. Clearly, $y \geqslant x+8$.
Thus, the highest score of the best team in city A is
$$
(x-1)+(x+8)=2 x+7 \text { (points). }
$$
Therefore, when $x=6$, the minimum value of $2 x+7$ is 19. Hence, the minimum value of the highest score of the best team in city A is 19.
|
19
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: With $2009^{12}$ as one of the legs, and all three sides being integers, the number of different right-angled triangles (congruent triangles are considered the same) is $\qquad$.
(2009, International Mathematics Tournament of the Cities for Young Mathematicians)
|
Let $a$, $2009^{12}$, and $c$ be the sides of a right triangle, with $c$ being the hypotenuse. Then,
$$
(c+a)(c-a)=2009^{24}=41^{24} \times 7^{48}.
$$
Since $c+a > c-a$ and they have the same parity, both must be odd.
Also, $2009^{24}=41^{24} \times 7^{48}$ has $25 \times 49$ different factors, which can be paired into $\frac{25 \times 49-1}{2}=612$ pairs, each pair forming a possible solution.
On the other hand, $2009^{12}$ clearly does not meet the requirements.
Therefore, there are 612 pairs of different integer-sided right triangles that satisfy the conditions.
|
612
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 There is a railway network between six cities, such that there is a direct railway between any two cities. On Sundays, some railways will be closed for maintenance. The railway department stipulates: after closing several sections of the railway, it must still be possible to travel by rail between any two cities (not necessarily directly). How many different railway maintenance methods meet the railway department's requirements? ${ }^{[1]}$
(2007, British Mathematical Olympiad)
|
【Analysis】The problem is equivalent to finding the number of all possible ways to connect a graph composed of six points and several edges, such that any two points are connected.
Let $f(n)$ denote the number of all possible ways to connect $n$ points such that any two points are connected. Then
$$
f(1)=1, f(2)=1 \text {. }
$$
For the case of three points, either all pairs of points are connected, or any two points are connected by selecting any two out of the three edges, so $f(3)=4$.
Since $n$ vertices can have $\mathrm{C}_{n}^{2}$ edges, there are a total of $2^{\mathrm{C}_{n}^{2}}$ different connection methods. At this point, any two vertices either have an edge or do not have an edge. Select one vertex $v$ from the $n$ vertices, and consider the set of $i(1 \leqslant i \leqslant n)$ vertices that include $v$. You can choose $i-1$ vertices from the remaining $n-1$ vertices to connect with $v$ to form this set of $i$ vertices, so there are $\mathrm{C}_{n-1}^{i-1}$ ways to select.
By the definition of $f(i)$, for the selected $i-1$ vertices and vertex $v$, there are $f(i)$ ways to ensure that any two points are connected. For the remaining $n-i$ points, there are $2^{\mathrm{C}_{n-i}^{2}}$ different connection methods.
Therefore, $2^{\mathrm{C}_{n}^{2}}=\sum_{i=1}^{n} \mathrm{C}_{n-1}^{i-1} f(i) 2^{\mathrm{C}_{n-i}^{2}}$.
Given $f(1)=1, f(2)=1, f(3)=4$, we get
$$
f(4)=38, f(5)=728, f(6)=26704 \text {. }
$$
Since among the 26704 ways, 26703 ways involve closing some sections of the railway, and 1 way involves not closing any section, the answer is 26703.
|
26703
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $a_{n}$ be the coefficient of $x$ in the binomial expansion of $(3-\sqrt{x})^{n}(n=2,3, \cdots)$. Then
$$
\sum_{n=2}^{18} \frac{3^{n}}{a_{n}}=
$$
$\qquad$
|
11. 17.
Since $a_{n}=3^{n-2} C_{n}^{2}$, therefore,
$$
\frac{3^{n}}{a_{n}}=3^{2} \times \frac{2}{n(n-1)}=\frac{18}{n(n-1)} \text {. }
$$
Thus, $\sum_{n=2}^{18} \frac{3^{n}}{a_{n}}=18 \sum_{n=2}^{18} \frac{1}{n(n-1)}=17$.
|
17
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. In space, there are five points, no four of which are coplanar. If several line segments are drawn such that no tetrahedron exists in the graph, then the maximum number of triangles in the graph is $\qquad$.
|
14.4.
First, construct graph 6. It is easy to see that it meets the conditions and has exactly four triangles.
Now assume there exists some configuration where the number of triangles is no less than five.
If only two line segments are not connected, then these two line segments must have no common endpoints (as shown in graph 6), otherwise, a tetrahedron would exist. But there are only four triangles, which is a contradiction.
If at least three line segments are not connected, when one of these line segments serves as a side of three triangles, as shown in graph 7, there are only three triangles; when each line segment serves as a side of at most two triangles, then there are at most $\left[\frac{\left(\mathrm{C}_{5}^{2}-3\right) \times 2}{3}\right]=4$ triangles.
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
18. (17 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of the integers 1, 2, $\cdots, n$, and satisfy the following conditions:
(1) $a_{1}=1$;
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2(i=1,2, \cdots, n-1)$.
Let the number of such permutations be $f(n)$. Find the remainder when $f(2010)$ is divided by 3.
|
18. Verifiable
$$
f(1)=1, f(2)=1, f(3)=2 \text {. }
$$
Let $n \geqslant 4$. Then $a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$. This is because by removing the first term and reducing all subsequent terms by 1, a one-to-one correspondence can be established.
For $a_{2}=3$, if $a_{3}=2$, then $a_{4}=4$, so the number of permutations is $f(n-3)$; if $a_{3} \neq 2$, then 2 must be placed after 4, leading to all odd numbers being in ascending order followed by all even numbers in descending order.
Therefore, $f(n)=f(n-1)+f(n-3)+1$.
Let $r(n)$ be the remainder when $f(n)$ is divided by 3. Then
$$
r(1)=r(2)=1, r(3)=2 \text {. }
$$
For $n \geqslant 4$,
$$
r(n) \equiv[r(n-1)+r(n-3)+1](\bmod 3) \text {. }
$$
Thus, $\{r(n)\}$ forms a sequence with a period of 8:
$$
1,1,2,1,0,0,2,0, \cdots \text {. }
$$
Since $2010 \equiv 2(\bmod 8)$, we have
$$
r(2010)=1,
$$
which means the remainder when $f(2010)$ is divided by 3 is 1.
|
1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For each positive integer $n$, let $f(n)$ denote the last digit of $1+2+\cdots+n$ (for example, $f(1)=1$, $f(2)=3$, $f(3)=6$). Calculate the value of $f(1)+f(2)+\cdots+f(2011)$.
|
One, because the last digit of the sum of any 20 consecutive positive integers is 0, so,
$$
f(n+20)=f(n) .
$$
Thus, it is only necessary to calculate the $f(n)$ corresponding to 1 to 20, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$f(n)$ & 1 & 3 & 6 & 0 & 5 & 1 & 8 & 6 & 5 & 5 \\
\hline$n$ & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline$f(n)$ & 6 & 8 & 1 & 5 & 0 & 6 & 3 & 1 & 0 & 0 \\
\hline
\end{tabular}
Also, $f(1)+f(2)+\cdots+f(20)=70$,
$2011=20 \times 100+11$,
then $f(1)+f(2)+\cdots+f(2011)$
$=70 \times 100+f(1)+f(2)+\cdots+f(11)$
$=7000+46=7046$.
|
7046
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there are always three numbers, the sum of any two of which is irrational.
|
Three, take four irrational numbers $\{\sqrt{2}, \sqrt{3},-\sqrt{2},-\sqrt{3}\}$, obviously they do not satisfy the condition, hence $n \geqslant 5$.
Consider five irrational numbers $a, b, c, d, e$, viewed as five points.
If the sum of two numbers is a rational number, then connect the corresponding two points with a red line. Otherwise, connect them with a blue line.
(1) There is no red triangle. Otherwise, assume without loss of generality that $a+b$, $b+c$, and $c+a$ are all rational numbers. But $(a+b)+(c+a)-(b+c)=2a$, which contradicts the fact that $a$ is irrational.
(2) There must be a monochromatic triangle. Otherwise, there must be a red circle in the graph, i.e., each point is connected to two red lines and two blue lines (if there is a point $A$ connected to at least three lines of the same color $A B, A C, A D$, then $\triangle B C D$ is a monochromatic triangle of the other color, a contradiction). Assume the five numbers on the red circle are $a, b, c, d, e$ in sequence. Then $a+b, b+c, c+d, d+e, e+a$ are rational numbers.
$$
\text { By }(a+b)-(b+c)+(c+d)-(d+e)
$$
$+(e+a)=2a$ leads to a contradiction. Thus, the monochromatic triangle must be a blue triangle.
In summary, the minimum value of $n$ is 5.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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