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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 ... | 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}+... | 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 \geqs... | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let real numbers $x, y, z, w$ satisfy
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}... | Consider the function
$$
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_{\sub... | 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:
$$
\begin{array}{l}
1,1, \cdots, 1,... | 1018 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For a natural number $n$, let the sum of its digits be denoted as $a_{n}$, for example,
$$
\begin{array}{l}
a_{2009}=2+0+0+9=11, \\
a_{2010}=2+0+1+0=3 .
\end{array}
$$
Then $a_{1}+a_{2}+\cdots+a_{2010}=$ ( ).
(A) 28062
(B) 28065
(C) 28067
(D) 28068 | 6. D.
Consider all natural numbers from 1 to 2010 as four-digit numbers (if $n$ is less than four digits, add 0s at the beginning to make it four digits, which does not change the value of $a_{n}$).
Notice that, the number of times 1 appears in the thousands, hundreds, tens, and units place are $10^{3}$, $2 \times 10... | 28068 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 5 The number of positive integer values of $n$ that satisfy $\left|\sqrt{\frac{n}{n+2009}}-1\right|>\frac{1}{1005}$ is $\qquad$
(2009, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | Given that $n$ is a positive integer, we have
$$
\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... | 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 | 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)=... | 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... | Three, given that points $P(1, a)$ and $Q(2, 10a)$ are on the graph of the quadratic function $y = x^2 + bx - c$, we have
$$
1 + b - c = a, \quad 4 + 2b - c = 10a.
$$
Solving these equations, we get $b = 9a - 3$ and $c = 8a - 2$.
(1) From $c < b < 8a$, we know
$$
8a - 2 < 9a - 3 < 8a \Rightarrow 1 < a < 3.
$$
Since $a... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Let $p$ be a prime number greater than 2, and $k$ be a positive integer. If the graph of the function $y=x^{2}+p x+(k+1) p-4$ intersects the $x$-axis at two points, at least one of which has an integer coordinate, find the value of $k$.
---
The function is given by:
\[ y = x^2 + px + (k+1)p - 4 \]
... | From the problem, we know that the equation
$$
x^{2}+p x+(k+1) p-4=0
$$
has at least one integer root among its two roots $x_{1}, x_{2}$.
By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=-p, x_{1} x_{2}=(k+1) p-4 . \\
\text { Hence }\left(x_{1}+2\right)\left(x_{2}+2\right) \\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $a, b$ be positive integers, and satisfy $2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{15}{b}}\right)$ is an integer. Then the number of such ordered pairs $(a, b)$ is $\qquad$ pairs.
$(2009$, National Junior High School Mathematics League) | First, guess that $\sqrt{\frac{15}{a}}$ and $\sqrt{\frac{15}{b}}$ are both rational numbers.
The following is an attempt to prove this.
Let $\frac{15}{a}=A, \frac{15}{b}=B$. According to the problem, we can assume
$\sqrt{A}+\sqrt{B}=C\left(A, B, C \in \mathbf{Q}_{+}\right)$.
Thus, $\sqrt{A}=C-\sqrt{B} \Rightarrow A=(C-... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. The number of integer solutions to the inequality $\log _{6}(1+\sqrt{x})>\log _{25} x$ is $\qquad$ . | 10. 24 .
Let $\log _{25} x=t$. Then $x=25^{t}$.
Thus the original inequality $\Leftrightarrow \log _{6}\left(1+5^{t}\right)>t$
$$
\Leftrightarrow 1+5^{t}>6^{t} \Leftrightarrow\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t}>1 \text {. }
$$
Define $f(t)=\left(\frac{1}{6}\right)^{t}+\left(\frac{5}{6}\right)^{t... | 24 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Given real numbers $x, y, z$ satisfy
$$
\sqrt[5]{x-y}+\sqrt[5]{y-z}=3 \text{, and } x-z=33 \text{. }
$$
Find the value of the algebraic expression $x-2 y+z$. | Let $\sqrt[5]{x-y}=x_{1}, \sqrt[5]{y-z}=x_{2}$. Then $x_{1}+x_{2}=3, x_{1}^{5}+x_{2}^{5}=33$.
Let $x_{1} x_{2}=\lambda$. Then $x_{1} 、 x_{2}$ are the two roots of $x^{2}-3 x+\lambda=0$.
Let $x_{1}^{n}+x_{2}^{n}=a_{n}$ ( $n$ is a positive integer $)$. Then
$$
a_{n}-3 a_{n-1}+\lambda a_{n-2}=0 \text {, }
$$
i.e.,
$$
a_{... | 31 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
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 $... | 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... | 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 numbe... | 3.6028 .
Let $n$ be a natural number.
When $n \leqslant x < n+\frac{1}{3}$,
$2 n \leqslant 2 x < 2 n+\frac{2}{3}$,
$3 n \leqslant 3 x < 3 n+1$.
Thus, $M=n+2 n+3 n=6 n$.
When $n+\frac{1}{3} \leqslant x < n+\frac{1}{2}$,
$2 n+\frac{2}{3} \leqslant 2 x < 2 n+1$,
$3 n+1 \leqslant 3 x < 3 n+\frac{3}{2}$.
Thus, $M=n+2 n+(3 ... | 6028 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three. (25 points) Given a positive integer $M$ has $k$ positive divisors, among which, only two divisors are prime, and the sum of the reciprocals of these $k$ positive divisors is $\frac{342}{287}$. Find all values of $M$.
---
The text has been translated while preserving the original formatting and line breaks. | $$
\begin{array}{l}
\text{Because } 287=7 \times 41, \text{ so, 7 and 41 are exactly the two prime factors of } M. \\
\text{Let } M=7^{m} \times 41^{n}\left(m, n \in \mathbf{N}_{+}\right), d_{1}, d_{2}, \cdots, d_{k} \text{ be all the positive divisors of } M \text{ arranged in ascending order. Thus, } d_{1}=1, d_{k}=M... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Using the vertices of a regular dodecagon as the vertices of triangles, the total number of acute and obtuse triangles that can be formed is $\qquad$.
untranslated part: $\qquad$ (This part is typically left as a blank space for the answer to be filled in.) | -1.160 .
The circumference of a circle with 12 points can form $\mathrm{C}_{12}^{3}=220$ triangles; among the chords of a circle equally divided into 12 parts, there are 6 diameters, and each diameter corresponds to 10 right-angled triangles, thus yielding 60 right-angled triangles. Therefore, acute and obtuse triangle... | 160 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. $F$ is the right focus of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{12}=1$, and $P$ is a moving point on the ellipse. For the fixed point $A(-2, \sqrt{3}),|P A|+$ $2|P F|$ the minimum value is $\qquad$ . | 4. 10 .
It is known that the eccentricity of the ellipse is $\frac{1}{2}$, and the equation of the right directrix $l$ is $x=8$. The distance from point $A$ to $l$ is 10.
Let the projection of point $P$ onto $l$ be $H$. Then
$\frac{|P F|}{|P H|}=\frac{1}{2}$.
Thus, $|P A|+2|P F|=|P A|+|P H| \geqslant 10$.
The minimum ... | 10 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $S=\sqrt{1+\frac{1}{1^{2}}+\frac{1}{2^{2}}}+\sqrt{1+\frac{1}{2^{2}}+\frac{1}{3^{2}}}+$ $\cdots+\sqrt{1+\frac{1}{1999^{2}}+\frac{1}{2000^{2}}}$. Find the greatest integer not exceeding $S$.
(2000, Taiyuan Junior High School Mathematics Competition) | Prompt: Following Example 2, the largest integer not exceeding $S$ is
$$
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 {. }
$$
For any real number $x$.
For any $k$ in the set $\{1,2, \cdots, 200\}$, when $x \rightarrow k-0$, $f(x) \rightarrow-\infty$, and when $x \rightarrow k+0$, $f(x) \rightarrow+\infty$. Also, when $x \rightarrow+\infty$, $... | 2010 | Inequalities | proof | Yes | Yes | cn_contest | false |
Three, (50 points) A positive integer is called "simple" if it does not have any square factors greater than 1. Determine how many numbers in $1,2, \cdots, 2010$ are simple.
| Three, notice that $2010 < 45^2$.
Thus, if a number $n$ in the set $M = \{1, 2, \cdots, 2010\}$ is not simple, it must contain one or more square factors from the set of primes
$$
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}\righ... | 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 $\alpha \beta=1$. Therefor... | Let $\beta=\frac{\sqrt{5}-1}{2}$. Then
$$
\begin{array}{l}
\alpha+\beta=\sqrt{5}, \alpha \beta=1 . \\
\text { Therefore, } \alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta=3, \\
\alpha^{4}+\beta^{4}=\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2}=7, \\
\alpha^{8}+\beta^{8}=\left(\alpha^{4}+\beta^{4}\rig... | 2206 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Calculate $[\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}]$ (2008 appears 2008 times).
(2008, International Youth Math Invitational Competition)
【Analysis】Although there are 2008 square root operations, as long as you patiently estimate from the inside out, the pattern will naturally become apparent. | Let $a_{n}=\sqrt{2008+\sqrt{2008+\cdots+\sqrt{2008}}}$ (with $n$ 2008s, $n=1,2, \cdots$ ).
Given $44<\sqrt{2008}<45$, we have
$$
\begin{array}{l}
45^{2}<2008+44<2008+\sqrt{2008} \\
<2008+45<46^{2} .
\end{array}
$$
Thus, $45<\sqrt{2008+\sqrt{2008}}<46$, which means $45<a_{2}<46$.
Therefore, $45^{2}<2008+45<2008+a_{2}$
... | 45 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 11 Given $0<a<1$, and
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$
Then $[10 a$ ] equals $\qquad$
$(2009$, Beijing Mathematical Competition (Grade 8)) | Notice that
$0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2$.
Then $\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right]$ equals
0 or 1.
By the problem statement, 18 of these are equal to 1. Therefore,
$\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\lef... | 6 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 12 Given a positive integer $n$ less than 2006, and $\left[\frac{n}{3}\right]+\left[\frac{n}{6}\right]=\frac{n}{2}$. Then the number of such $n$ is $\qquad$.
(2006, National Junior High School Mathematics Competition) | From the properties of the Gaussian function, we know that $\left[\frac{n}{3}\right] \leqslant \frac{n}{3}$, with equality holding if and only if $\frac{n}{3}$ is an integer; $\left[\frac{n}{6}\right] \leqslant \frac{n}{6}$, with equality holding if and only if $\frac{n}{6}$ is an integer.
Therefore, $\left[\frac{n}{3}... | 334 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
2. The algebraic expression $\left[(\sqrt{6}+\sqrt{5})^{6}\right]=$ $\qquad$
(2004, "Five Sheep Cup" Mathematics Competition (Grade 9)) | Prompt: Example 1. Answer: 10581. | 10581 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The sum of all real numbers $x$ that satisfy $25\{x\}+[x]=125$ is $\qquad$
(2007, International Invitational Competition for Young Mathematicians in Cities) | Hint: $25\{x\}+[x]=125$ can be transformed into $25 x-24[x]=125$.
Example 6. Answer: 2837. | 2837 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Let
$$
\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}
$$
Then $[30 S]=(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 0
(2002, "F... | When $n \geqslant 10$, and $n$ is an integer,
$$
\begin{array}{l}
{\left[\frac{[n(n+1)-1]^{2}}{n(n+1)}\right]=\left[n(n+1)-2+\frac{1}{n(n+1)}\right]} \\
=n(n+1)-2=(n-1)(n+2) .
\end{array}
$$
Thus, $S=\frac{1}{9 \times 12}+\frac{1}{10 \times 13}+\cdots+\frac{1}{48 \times 51}$.
Notice that $\frac{1}{k(k+3)}=\frac{1}{3}\... | 2 | Algebra | MCQ | Yes | Yes | cn_contest | false |
Example 4 Algebraic Expression
$[\sqrt[3]{1 \times 2 \times 3}]+[\sqrt[3]{2 \times 3 \times 4}]+\cdots+$ $[\sqrt[3]{2000 \times 2001 \times 2002}]=(\quad)$.
(A) 2000000
(B) 2001000
(C) 2002000
(D) 2003001
(2000, "Five Sheep Cup" Mathematics Competition (Grade 9))
[Analysis] Notice that each term in the sum can be expre... | Since $k^{3}<k(k+1)(k+2)<(k+1)^{3}$,
thus, $k<\sqrt[3]{k(k+1)(k+2)}<k+1$.
Therefore, $[\sqrt[3]{k(k+1)(k+2)}]=k$.
Hence the original expression $=1+2+\cdots+2000=2001000$. | 2001000 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Let $n$ ($n<100$) be a positive integer, and there exists a positive integer $k$, such that $1 \leqslant k \leqslant n-1$, satisfying
$$
\frac{4 k+1}{2 n}+\frac{1-2 k^{2}}{n^{2}}=\frac{1}{2} \text {. }
$$
How many values of $n$ satisfy the condition? Prove your conclusion. | 1. From equation (1), we get $(2 k-n)^{2}=n+2$.
Solving, we get $k=\frac{1}{2}(n \pm \sqrt{n+2})$.
Since $k$ is an integer, therefore, $n=m^{2}-2(m \in \mathbf{N})$.
Thus, $k=\frac{1}{2}\left(m^{2} \pm m-2\right)$.
Also, $1 \leqslant k \leqslant n-1$, so
$$
1 \leqslant \frac{m^{2} \pm m-2}{2} \leqslant m^{2}-3 \text {... | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Calculate $\left[\frac{23 \times 1}{101}\right]+\left[\frac{23 \times 2}{101}\right]+\cdots+$ $\left[\frac{23 \times 100}{101}\right]$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Since 23 and 101 are coprime, for $k=1$, $2, \cdots, 100$, $\frac{23 k}{101}$ is not an integer. Therefore,
$$
\begin{array}{l}
\frac{23 k}{101}-1<\left[\frac{23 k}{101}\right]<\frac{23 k}{101}, \\
\frac{23(101-k)}{101}-1<\left[\frac{23(101-k)}{101}\right] \\
<\frac{23(101-k)}{101}. \\
\text{ Hence }\left(\frac{23 k}{1... | 1100 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Given $a^{2}(b+c)=b^{2}(a+c)=2010$, and $a \neq b$. Then $c^{2}(a+b)=$ | 1. 2010 .
Notice
$$
\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_... | 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}
... | 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.
$$
\begin{array}{l}
\text { Simplifying, we get } 3(a+3 d)+4(b+2 c)=66 . \\
\text { Therefore, }(a+3 d, b+2 c) \\
=(4 i+2,15-3 i)(i=0,1, \cdots, 5) . \\
\text { W... | 91 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
10. In a competition with 20 singers, 9 judges respectively assign them ranks from 1 to 20. It is known that for each singer, the difference between any two ranks does not exceed 3. If the sum of the ranks each singer receives is arranged in an increasing sequence: $C_{1} \leqslant C_{2} \leqslant \cdots \leqslant C_{2... | 10. 24.
If 9 judges all give a singer the first place, then $C_{1}=9$.
If two singers both get the first place, then one of them gets no less than 5 first places, while the other 4 ranks are no higher than fourth place, so $C_{1} \leqslant 5 \times 1+4 \times 4=21$.
If three singers all get the first place, then the... | 24 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. Find the smallest odd number $a$ greater than 5 that satisfies the following conditions: there exist positive integers $m_{1}, n_{1}, m_{2}, n_{2}$, such that
$$
a=m_{1}^{2}+n_{1}^{2}, a^{2}=m_{2}^{2}+n_{2}^{2} \text {, }
$$
and $m_{1}-n_{1}=m_{2}-n_{2}$. | 8. From $261=15^{2}+6^{2}, 261^{2}=189^{2}+180^{2}$,
$15-6=189-180$,
we know that 261 has the property described in the problem.
Next, we prove that 261 is the smallest odd number greater than 5 with the property described in the problem, i.e., there is no such odd number between 5 and 261.
If not, then there exists a... | 261 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four. (25 points) In a $7 \times 7$ grid, there are 64 intersection points (referred to as "nodes") where chess pieces can be placed, with at most 1 piece per point, totaling $k$ pieces. If no matter how the pieces are placed, there always exist 4 pieces such that the nodes they occupy form the four vertices of a recta... | First, as shown in Figure 2, when $k=24$, it is possible that no 4 nodes with chess pieces form the four vertices of a rectangle, so $k \geqslant 25$.
Second, we will prove that when $k=25$, there must exist 4 nodes with chess pieces that form the four vertices of a rectangle.
If not, consider the row (or column) wit... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Five. (25 points) Let positive real numbers $a$, $b$, $c$ satisfy
$$
(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}
$$ | 3 | Inequalities | proof | Yes | Yes | cn_contest | false |
1. $\left(\frac{1+\sqrt{5}}{2}\right)^{6}+\left(\frac{1-\sqrt{5}}{2}\right)^{6}=$ | Let $x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=\frac{1-\sqrt{5}}{2}$. Then $x_{1}+x_{2}=1, x_{1} x_{2}=-1$.
Thus, $x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=3$.
Therefore, $x_{1}^{6}+x_{2}^{6}=\left(x_{1}^{2}\right)^{3}+\left(x_{2}^{2}\right)^{3}$
$=\left(x_{1}^{2}+x_{2}^{2}\right)\left[\left(x_{1}^{2}\rig... | 18 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) If the sum, difference, product, and quotient of two unequal natural numbers add up to a perfect square, then such two numbers are called a "wise pair" (for example, $(8,2)$ is a wise pair, since $\left.(8+2)+(8-2)+8 \times 2+\frac{8}{2}=36=6^{2}\right)$.
If both of these natural numbers do not exce... | Three, let $(a, b)$ be a wise array, and without loss of generality, assume $a > b$.
By definition, we have
$$
(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}.
$$
Since $2 a, a b, m^{2}$ are all natural numbers, $\frac{a}{b}$ must also be a natural... | 53 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. Let the function $f(n)$ be defined on the set of positive integers, for any positive integer $n$, we have $f(f(n))=4 n+3$, and for any non-negative integer $k$, we have
$$
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} ... | 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}... | 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=... | 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 $\frac{5+\sqrt{21}}{2}+\frac{5-\sqrt{21}}{2}=5$,
$$
\frac{5+... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Given seven points on a circle, connect each pair of points. Find the minimum number of intersection points of these lines inside the circle. | We prove that these lines intersect at most three sets of three lines at a point inside the circle, and it is obvious that there cannot be four lines intersecting at a point.
First, select seven vertices of a regular octagon,
it is clear that there are three sets of three lines intersecting at a point (as shown in Figu... | 29 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 As shown in Figure $1, a / / b$, line $a$ has ten points $A_{1}, A_{2}$, $\cdots, A_{10}$, and line
$b$ has nine points $B_{1}, B_{2}, \cdots, B_{9}$
nine points. Connecting each point on
$a$ with each point on
$b$ can result in many line segments, given that no three line segments intersect at one point. How... | Solution: Take two points each on $a$ and $b$, the four points determine a unique intersection point. Taking two points from $a$ has $10 \times 9 \div 2=45$ methods, taking two points from $b$ has $9 \times 8 \div 2=36$ methods. In total, there are $45 \times 36=1620$ methods. | 1620 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. Given that $a$, $b$, and $c$ are all positive integers, and $a b c=$ 2010. Then the minimum value of $a+b+c$ is ( ).
(A) 84
(B) 82
(C) 78
(D) 76 | $$
\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}
$$
we know that the minimum value of $a+b+c$ is 78. | 78 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example: How many types of isosceles triangles with integer side lengths and a perimeter of 100 are there?
(8th "Hua Luogeng Cup" Junior Mathematics Invitational Final) | Let $a$, $b$, and $c$ be the lengths of the three sides of a triangle.
(1) $a=ba .\end{array}\right.$,
Solving, we get $25<a<33 \frac{1}{3}$.
Therefore, there are 8 possibilities.
(2) $a<b=c$.
In this case, $a=100-2 c$.
Thus, $\left\{\begin{array}{l}100-2 c<c \\ 1+2 c \leqslant 100\end{array}\right.$.
Solving, we get ... | 24 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. When $n$ is a positive integer, it is defined that
$$
n!=n \times(n-1) \times \cdots \times 2 \times 1 \text {, }
$$
which is called the factorial of $n$ (for example, $10!=10 \times 9 \times \cdots \times 2 \times 1$ $=3628800$). Therefore, in 2010!, the total number of zeros at the end is $\qquad$ | 4.501.
The number of trailing zeros depends on the number of factors of 10. Since $10=2 \times 5$, and it is clear that in 2010! the number of factors of 2 is greater than the number of factors of 5, we only need to find the number of prime factors of 5 in 2010!, which gives us the number of trailing zeros.
Among the... | 501 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 On a plane, there are seven points, and some line segments can be connected between them, so that any three points among the seven must have at least one line segment connecting a pair of them. How many line segments are needed at least? Prove your conclusion.
$(2002$, Shanghai Junior High School Mathematics ... | (1) If point $A$ is not an endpoint of any line segment, then every two of the other six points are connected by a line segment, totaling $\frac{1}{2} \times 6 \times 5=15$ line segments;
(2) If point $A$ is the endpoint of only one line segment, then every two of the five points not connected to $A$ are connected by a... | 9 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Six. (16 points) As shown in Figure 3, there is a fixed point $P$ inside $\angle M A N$. It is known that $\tan \angle M A N=3, P$ is at a distance $P D=12$ from the line $A N$, and $A D=30$. A line through $P$ intersects $A N$ and $A M$ at points $B$ and $C$ respectively. Find the minimum area of $\triangle A B C$.
-... | Six, Solution 1
As shown in Figure 7, let $D B=$
$x$, draw $C E \perp A B$,
$P F / / C A$.
Let $A E=t$. Then
$$
\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}$,... | 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$. Ther... | 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 ... | 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 ther... | 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 fin... | 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 {. }
$$
Solving for $x$ yields $x=101$.
Therefore, this track is $1 \mathrm{~m}$ lon... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. Among the seven points consisting of the center and the six vertices of a regular hexagon, if any $n$ points are taken, and among them, there must be three points that form the vertices of an equilateral triangle, then the minimum value of $n$ is $\qquad$ | -1.5 .
Consider the regular hexagon $A B C D E F$ as shown in Figure 1, with its center at $O$.
When $n=4$, take $A, C, D, F$, among which no three points can form the three vertices of an equilateral triangle.
When $n=5$, consider the following two cases:
(1) $O$ is among the five points. Consider the three pairs of ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Initially 285 Test: How many positive integers $T$ are there such that between $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$ (not including $\frac{2010}{T}$ and $\frac{2010+T}{2 T}$) there are exactly 5 different positive integers? | Solution: Obviously, when $T \geqslant 2010$,
$$
\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}
$$
From this, we know
$$
\... | 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 $2 a=7 b+k c \Rightarrow 7 b=2 a-k c$,
$$
\begin{array}{l}
a|(3 b-c) \Leftrightarrow a|(3 \times 7 b-7 c) \\
\Leftrightarrow a \mid[3(2 a-k c)-7 c] \\
\Leftrightarrow a \mid[6 a-(3 k+7) c] \\
... | 2009 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. As shown in Figure 1, in the cube $A C_{1}$ with edge length 1, points $P$ and $Q$ are moving points on edges $A D$ and $A_{1} B_{1}$, respectively. If the skew lines $B D_{1}$ and $P Q$ are perpendicular to each other, then $A P + A_{1} Q =$ $\qquad$ | 5. 1 .
As shown in Figure 5, establish a spatial rectangular coordinate system, and set
$$
\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}... | 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] ... | 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}
$$
Contradiction.... | 2004 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let $n$ be a positive integer, and $f(n)$ denote the number of integers satisfying the following conditions:
(i) Each digit $a_{i} \in\{1,2,3,4\}$, and
$a_{i} \neq a_{i+1}(i=1,2, \cdots)$;
(ii) When $n \geqslant 3$, $a_{i}-a_{i+1}$ and $a_{i+1}-a_{i+2}$ $(i=1,2, \cdots)$ have opposite signs.
(1) Find the valu... | Let $g(n)$ denote the number of wave numbers for which $a_{n}>a_{n-1}$ when $n \geqslant 2$. Then, by symmetry, we have
$$
\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}
$$
Let $m(i)$ represent the number of $(n-1)$-dig... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. At a certain charity fundraising dinner, each person ate half a plate of rice, one-third of a plate of vegetables, and one-quarter of a plate of meat. The dinner provided a total of 65 plates of food. How many people attended this fundraising dinner? | 2. 60 .
Each person ate $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ of food, and $65 \div \frac{13}{12}=60$.
Therefore, 60 people attended this charity dinner. | 60 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The number of triples of positive integers $(x, y, z)$ that satisfy $x y z=3^{2010}$ and $x \leqslant y \leqslant z<x+y$ is $\qquad$.
| 3. 336 .
Let $x=3^{a}, y=3^{b}, z=3^{c}$. Then
$$
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 {, }
$$
which contradicts $z<x+y$. Hence, $c=b$.
Thus, $a+2 b=2010$.
Therefore, $670 \leqslant b \leqslant 1005$.
Hence, the ... | 336 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. The smallest four-digit number that has exactly 14 divisors (including 1 and itself), and one of its prime factors has a units digit of 3 is $\qquad$ . | 5.1458.
Let this four-digit number be $n$.
Since $14=14 \times 1=7 \times 2$, we have $n=p^{13}$ or $p^{6} q$ (where $p$ and $q$ are different prime numbers).
If $n=p^{13}$, by the given condition, the unit digit of $p$ is 3, so, $p \geqslant 3$.
Thus, $n \geqslant 3^{13}=1594323$, which is a contradiction.
Therefore,... | 1458 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $f(x)$ represent a fourth-degree polynomial in $x$. If
$$
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 .
\e... | 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 $0<q^{2}-q+1<p^{2}$.
T... | 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
$$
\angle PQC = \angle... | 20 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
11. A farmer has 100 pigs and 100 chickens. He has four adjacent square yards, forming a $2 \times 2$ grid. The farmer wants to distribute the livestock among the yards according to the following requirements: the first row has 120 heads, the second row has 300 feet; the first column has 100 heads, the second column ha... | 11.341.
As shown in Figure 8, label the four courtyards with letters $A_{i} (i=1,2,3,4)$. Let $A_{i}$ courtyard have $x_{i}$ pigs and $y_{i}$ chickens.
From the problem, we get
every 8
$$
\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... | 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 segme... | 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 t... | 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... | 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 -3... | 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. ... | 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 t... | 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 as... | 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.
$$
\begin{array}{llllllll}
\mathbf{K} & & & & & & & \\
\mathbf{O} & \mathbf{O} & & & & & & \\
\mathbf{R} & \mathbf{R} & \mathbf{R} & & & & & ... | 10. First list the number table as shown in Figure 17, where each cell corresponds to a letter, and the number in the cell represents the number of different paths from the letter "K" to that letter. The cell at the very top is filled with 1. Starting from the second row, the number in each cell is the sum of the numbe... | 3 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $k$ be an integer greater than 1, and the sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
\begin{array}{l}
a_{0}=0, a_{1}=1, \\
a_{n+1}=k a_{n}+a_{n-1}(n=1,2, \cdots) .
\end{array}
$$
Find all $k$ that satisfy the following condition: there exist non-negative integers $l, m (l \neq m)$, and positive i... | When $k=2$, $a_{0}=0, a_{1}=1, a_{2}=2$, then from $a_{0}+2 a_{2}=a_{2}+2 a_{1}=4$, we know that taking $l=0, m=2, p=$ $2, q=1$ is sufficient.
For $k \geqslant 3$, by the recurrence relation, $\left\{a_{n}\right\}$ is a strictly increasing sequence of natural numbers and $k \mid\left(a_{n+1}-a_{n-1}\right)$.
$$
\begin... | 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 .
$$
\begin{array}{l}
\text { Original expression }=\frac{\log _{2}\left(\frac{2}{3} \cdot \frac{3}{4} \cdot \frac{4}{5} \cdot \frac{5}{6} \cdot \frac{6}{7} \cdot \frac{7}{8}\right)}{\frac{\lg 2}{\lg 3} \cdot \frac{\lg 3}{\lg 4} \cdot \frac{\lg 4}{\lg 5} \cdot \frac{\lg 5}{\lg 6} \cdot \frac{\lg 6}{\lg 7} \cdot \... | -6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
6. In a convex quadrilateral $ABCD$, $\angle BAD + \angle ADC = 240^{\circ}$, $E$ and $F$ are the midpoints of sides $AD$ and $BC$, respectively, and $EF = \sqrt{7}$. If two squares $A_1$ and $A_2$ are drawn with sides $AB$ and $CD$, respectively, and a rectangle $A_3$ is drawn with length $AB$ and width $CD$, find the... | 6.28 .
As shown in Figure 6, extend $BA$ and $CD$ to intersect at point $P$.
Given $\angle BAD + \angle ADC = 240^{\circ}$, we have
$$
\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 \... | 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
$$
\begin{array}{l}
\text { Original expression }=\sum_{i=1}^{2010} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\frac{4}{\prod_{j=1}^{2010} x_{j}} \\
=\sum_{i=1}^{2009} \frac{x_{i}}{\prod_{j=1}^{i-1} x_{j}}+\left(\frac{x_{2010}}{\prod_{j... | 4 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. Given that the circumcenter, incenter, and orthocenter of a non-isosceles acute $\triangle ABC$ are $O, I, H$ respectively, and $\angle A=60^{\circ}$. If the altitudes of $\triangle ABC$ are $AD, BE, CF$, then the ratio of the circumradius of $\triangle OIH$ to the circumradius of $\triangle DEF$ is $\qquad$ . | 4. 2 .
From $\angle B O C=\angle B I C=\angle B H C=120^{\circ}$, we know that $O, I, H, B, C$ are concyclic.
Let the circumradii of $\triangle A B C$ and $\triangle O I H$ be $R$ and $r$, respectively.
By the Law of Sines, we have
$2 R \sin A=B C=2 r \sin \angle B O C$.
Thus, $r=R$.
Let the circumradius of $\triangl... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given that $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ is a permutation of $1,2,3,4,5$, and satisfies $\left|a_{i}-a_{i+1}\right| \neq 1(i=1,2,3,4)$. Then the number of permutations $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ that meet the condition is $\qquad$. | 8. 14 .
The permutations that satisfy the conditions are:
$$
\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}
$$ | 14 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. (15 points) In a regular pentagon $A B C D E$, the diagonal $B E$ intersects diagonals $A D$ and $A C$ at points $F$ and $G$, respectively. The diagonal $B D$ intersects diagonals $C A$ and $C E$ at points $H$ and $I$, respectively. The diagonal $C E$ intersects diagonal $A D$ at point $J$. Let the set of isosceles... | 11. (1) Since all triangles formed by 10 points $A, B, C, D, E, F, G, H, I, J$ and line segments in Figure 2 are isosceles triangles,
$$
|M|=\mathrm{C}_{5}^{3}+4 \mathrm{C}_{5}^{4}+5 \mathrm{C}_{5}^{5}=35 \text{. }
$$
(2) By the pigeonhole principle, we know that among $A, B, C, D, E$, there must be three points of the... | 6 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Divide a wire of length $11 \mathrm{~cm}$ into several segments of integer centimeters, such that any three segments can form the sides of a triangle. Then the number of different ways to divide the wire is $\qquad$ (ways with the same number of segments and corresponding equal lengths are considered the same way). | 4.8.
Let the shortest two segments be $a \mathrm{~cm}, b \mathrm{~cm}$, and take any other segment $c \mathrm{~cm} (a \leqslant b \leqslant c)$. Then $b \leqslant c < a + b$.
(1) If $a = 1$, then $1 \leqslant b \leqslant c < b + 1$. Thus, $b = c$. Therefore, $b \in (11 - 1), 2b < 11 - 1$. Hence, $b$ can be $1, 2, 5$, ... | 8 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
One, (20 points) A competitive event involves several teams from two cities, A and B. It is known that city B has 8 more teams than city A, and any two teams play exactly one match. The event rules state: the winner gets 1 point, the loser gets 0 points, and there are no ties. In the end, the total score of all teams f... | Let city A have $x$ teams participating in the competition. Then city B has $x+8$ teams. When the teams from cities A and B compete against each other, city A scores a total of $y$ points, and city B scores a total of $x(x+8)-y$ points.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{2} x(x-1)+y+4 \\
=\frac{1}{2... | 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 in... | 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 cit... | 【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
$$
f(1)=1, f(2)=1 \tex... | 26703 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
11. Let $a_{n}$ be the coefficient of $x$ in the binomial expansion of $(3-\sqrt{x})^{n}(n=2,3, \cdots)$. Then
$$
\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 {. }
$$
Thus, $\sum_{n=2}^{18} \frac{3^{n}}{a_{n}}=18 \sum_{n=2}^{18} \frac{1}{n(n-1)}=17$. | 17 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. In space, there are five points, no four of which are coplanar. If several line segments are drawn such that no tetrahedron exists in the graph, then the maximum number of triangles in the graph is $\qquad$. | 14.4.
First, construct graph 6. It is easy to see that it meets the conditions and has exactly four triangles.
Now assume there exists some configuration where the number of triangles is no less than five.
If only two line segments are not connected, then these two line segments must have no common endpoints (as sho... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
18. (17 points) Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of the integers 1, 2, $\cdots, n$, and satisfy the following conditions:
(1) $a_{1}=1$;
(2) $\left|a_{i}-a_{i+1}\right| \leqslant 2(i=1,2, \cdots, n-1)$.
Let the number of such permutations be $f(n)$. Find the remainder when $f(2010)$ is divided by 3. | 18. Verifiable
$$
f(1)=1, f(2)=1, f(3)=2 \text {. }
$$
Let $n \geqslant 4$. Then $a_{2}=2$ or 3.
For $a_{2}=2$, the number of permutations is $f(n-1)$. This is because by removing the first term and reducing all subsequent terms by 1, a one-to-one correspondence can be established.
For $a_{2}=3$, if $a_{3}=2$, then $... | 1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
For each positive integer $n$, let $f(n)$ denote the last digit of $1+2+\cdots+n$ (for example, $f(1)=1$, $f(2)=3$, $f(3)=6$). Calculate the value of $f(1)+f(2)+\cdots+f(2011)$. | One, because the last digit of the sum of any 20 consecutive positive integers is 0, so,
$$
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... | 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 ... | 5 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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