problem stringlengths 2 5.64k | solution stringlengths 2 13.5k | answer stringlengths 1 43 | problem_type stringclasses 8
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3. (50 points) Find the smallest prime $p$ that satisfies $(p, N)=1$, where $N$ is the number of all $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ that meet the following conditions:
(1) $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ is a permutation of $0,1, \cdots, 2012$;
(2) For any positive divisor $m$ of 2013 and a... | 3. From (2), we know that for any $i, j (0 \leqslant i < j \leqslant 2012)$ and $m=3^{\alpha} \times 11^{\beta} \times 61^{\gamma} (\alpha, \beta, \gamma \in \{0,1\})$, we have
$$
a_{i} \equiv a_{j}(\bmod m) \Leftrightarrow i \equiv j(\bmod m).
$$
Let
$$
\begin{array}{l}
A_{i}=\{x \in \mathrm{N} \mid 0 \leqslant x \le... | 67 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. (50 points) Let the sequence $\left\{x_{n}\right\}$ satisfy
$$
x_{1}=1, x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \text {. }
$$
Find the units digit of $x_{2012}$. | 7. Clearly, $x_{2}=7$, and for any positive integer $n, x_{n}$ is a positive integer.
By the property of the floor function, we have
$$
\begin{array}{l}
4 x_{n}+\sqrt{11} x_{n}>x_{n+1}=4 x_{n}+\left[\sqrt{11} x_{n}\right] \\
>4 x_{n}+\sqrt{11} x_{n}-1 .
\end{array}
$$
Multiplying both sides of the above inequality by ... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. A country has $n(n \geqslant 3)$ cities and two airlines. There is exactly one two-way flight between every pair of cities, and this two-way flight is operated exclusively by one of the airlines. A female mathematician wants to start from a city, pass through at least two other cities (each city is visited only once... | 6. Solution 1 Consider each city as a vertex, each flight route as an edge, and each airline as a color. Then, the country's flight network can be seen as a complete graph with $n$ vertices whose edges are colored with two colors.
From the condition, we know that any cycle contains edges of both colors, meaning that t... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? | 8 . Note Quality.
First, consider whether the combination number $\mathrm{C}_{2012}^{k}$ is a multiple of $p=503$.
If $p \nmid k$, then
$\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1}$
$\Rightarrow p\left|k \mathrm{C}_{4 p}^{k} \Rightarrow p\right| \math... | 1498 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. At the beginning, there are 111 pieces of clay of equal weight on the table. Perform the following operations on the clay: First, divide a part or all of the clay into several groups, with the same number of pieces in each group, then knead the clay in each group into one piece. It is known that after $m$ operations... | 1. 2 .
Obviously, one operation can result in at most two different weights of clay blocks.
Below, we show that two operations can achieve the goal.
Assume without loss of generality that each block of clay initially weighs 1.
In the first operation, select 74 blocks and divide them into 37 groups, with two blocks in ... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Let $f(x)$ represent a quartic polynomial in $x$. If $f(1)=f(2)=f(3)=0, f(4)=6$, $f(5)=72$, then the last digit of $f(2010)$ is $\qquad$. ${ }^{3}$
(2010, International Cities Mathematics Invitational for Youth) | 【Analysis】It is given in the problem that $1,2,3$ are three zeros of the function $f(x)$, so we consider starting from the zero point form of the function.
Solution From the problem, we know
$$
f(1)=f(2)=f(3)=0 \text {. }
$$
Let the one-variable quartic polynomial function be
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {, }
... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Let $f(x)$ be a polynomial with integer coefficients, $f(0)=$
11. There exist $n$ distinct integers $x_{1}, x_{2}, \cdots, x_{n}$, such that $f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010$.
Then the maximum value of $n$ is $\qquad$ (6)
$(2010$, Xin Zhi Cup Shanghai High School Mathem... | Let $g(x)=f(x)-2010$.
$$
\begin{array}{l}
\text { By } f\left(x_{1}\right)=f\left(x_{2}\right)=\cdots=f\left(x_{n}\right)=2010, \text { we have } \\
g\left(x_{1}\right)=g\left(x_{2}\right)=\cdots=g\left(x_{n}\right)=0,
\end{array}
$$
That is, $x_{1}, x_{2}, \cdots, x_{n}$ are $n$ distinct integer roots of the integer-... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
8. On the edge $AS$ of the tetrahedron $S-ABC$, mark points $M, N$ such that $AM=MN=NS$. If the areas of $\triangle ABC$, $\triangle MBC$, and $\triangle SBC$ are $1$, $2$, and $\sqrt{37}$, respectively, find the area of $\triangle NBC$. | 8. Let the areas of $\triangle A B C$, $\triangle M B C$, $\triangle N B C$, $\triangle S B C$ be $S_{1}$, $S_{2}$, $S_{3}$, $S_{4}$, and let $h_{1}$, $h_{2}$, $h_{3}$, $h_{4}$ be the heights from these triangles to the common base $B C$, as shown in Figure 2.
Points $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $S^{\prim... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
10. Given trapezoid $A B C D$ with bases $A D=3, B C=1$, the diagonals intersect at point $O$, two circles intersect base $B C$ at points $K, L$, these two circles are tangent at point $O$, and are tangent to line $A D$ at points $A, D$ respectively. Find $A K^{2}+D L^{2}$. | 10. As shown in Figure 4, draw the common tangent of the two circles through point $O$, intersecting the lower base $AD$ at point $P$. By the properties of tangents, we have
$$
A P=O P=D P \text {. }
$$
This indicates that $\triangle A O D$ is a right triangle and is similar to $\triangle C D B$, with the similarity r... | 12 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. Given positive integers $m, n$ can be written as
$$
a_{0}+a_{1} \times 7+a_{2} \times 7^{2}+a_{3} \times 7^{3}
$$
where $a_{i} (i=0,1,2,3)$ are positive integers from 1 to 7, and
$$
m+n=2012(m>n) .
$$
Then the number of pairs $(m, n)$ that satisfy the condition is $(\quad)$.
(A) 606
(B) 608
(C) 610
(D) 612 | 4. A.
There are $7^{4}=2401$ positive integers of the given form, the largest of which is
$$
7 \times 7^{3}+7 \times 7^{2}+7 \times 7+7=2800,
$$
and the smallest is
$$
1 \times 7^{3}+1 \times 7^{2}+1 \times 7+1=400 .
$$
Since $m+n=2012(m>n)$, we have
$$
1007 \leqslant m \leqslant 2012-400=1612 \text {. }
$$
Therefo... | 606 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
5. Divide the natural numbers from 1 to 30 into two groups, such that the product of all numbers in the first group $A$ is divisible by the product of all numbers in the second group $B$. Then the minimum value of $\frac{A}{B}$ is ( ).
(A) 1077205
(B) 1077207
(C) 1077209
(D) 1077211 | 5. A.
$$
\begin{array}{l}
\text { Given } A B=30 \times 29 \times \cdots \times 1 . \\
=2^{26} \times 3^{14} \times 5^{7} \times 7^{4} \times 11^{2} \times 13^{2} \times \\
17 \times 19 \times 23 \times 29 .
\end{array}
$$
Let $C=2^{13} \times 3^{7} \times 5^{4} \times 7^{2} \times 11 \times 13 \times 17 \times 19 \ti... | 1077205 | Number Theory | MCQ | Yes | Yes | cn_contest | false |
Example 8 Given that $P(x)$ is a polynomial with integer coefficients, satisfying $P(17)=10, P(24)=17$. If the equation $P(n)=n+3$ has two distinct integer solutions $n_{1}, n_{2}$, find the value of $n_{1} n_{2}$. ${ }^{[7]}$
(2005, American Invitational Mathematics Examination) | 【Analysis】From the conditions of the problem, we cannot determine the zeros of the integer-coefficient polynomial $P(x)$. Let's construct another polynomial function $T(x)$ such that 17 and 24 are zeros of $T(x)$, and solve the problem using the zero-product property.
Solution Let $S(x)=P(x)-x-3$. Then
$S(17)=-10, S(24... | 418 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let positive integers $k_{1} \geqslant k_{2} \geqslant \cdots \geqslant k_{n}\left(n \in \mathbf{N}_{+}\right)$, and $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=2012$.
Then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is $\qquad$ | 2. 49 .
Notice that,
$$
2012=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{4}+2^{3}+2^{2} \text {. }
$$
Also, $2^{k+1}=2^{k}+2^{k}$, and $k+1 \leqslant 2 k(k \geqslant 1$ when $)$, then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is
$$
10+9+8+7+6+4+3+2=49 \text {. }
$$ | 49 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Group all positive integers that are coprime with 2012 in ascending order, with the $n$-th group containing $2n-1$ numbers:
$$
\{1\},\{3,5,7\},\{9,11,13,15,17\}, \cdots \text {. }
$$
Then 2013 is in the $\qquad$ group. | 4.32.
Notice that, $2012=2^{2} \times 503$, where $2$ and $503$ are both prime numbers.
Among the positive integers not greater than 2012, there are 1006 multiples of 2, 4 multiples of 503, and 2 multiples of $2 \times 503$. Therefore, the numbers that are not coprime with 2012 are
$$
1006+4-2=1008 \text { (numbers).... | 32 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
II. (25 points) Arrange all positive integers that satisfy the following conditions in descending order, denoted as $M$, and the $k$-th number as $b_{k}$: each number's any three consecutive digits form a non-zero perfect square. If $b_{16}-b_{20}=2^{n}$, find $n$.
Arrange all positive integers that satisfy the follow... | II. Notice that, among three-digit numbers, the perfect squares are $100, 121, 144, 169, 196, 225, 256, 289$, $324, 361, 400, 441, 484, 529, 576, 625$, $676, 729, 784, 841, 900, 961$.
For each number in $M$, consider its leftmost three digits.
(1) The six-digit number with the leftmost three digits 100 is 100169, the f... | 11 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Let $n=\sum_{a_{1}=0}^{2} \sum_{a_{2}=0}^{a_{1}} \cdots \sum_{a_{2} 012=0}^{a_{2} 011}\left(\prod_{i=1}^{2012} a_{i}\right)$. Then the remainder when $n$ is divided by 1000 is . $\qquad$ | 6. 191 .
It is evident that from $a_{1}$ to $a_{2012}$ forms a non-increasing sequence, and the maximum element does not exceed 2. Therefore, their product is a power of 2 or 0. Since each power of 2 can only be represented in one way (the sequence being non-increasing), we have
$$
\begin{array}{l}
n=1+2+4+\cdots+2^{2... | 191 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
8. If five vertices of a regular nonagon are colored red, then there are at least $\qquad$ pairs of congruent triangles (each pair of triangles has different vertex sets) whose vertices are all red. | 8. 4 .
A triangle with both vertices colored red is called a "red triangle". Thus, there are $\mathrm{C}_{5}^{3}=10$ red triangles. For a regular nonagon, the triangles formed by any three vertices are of only seven distinct types (the lengths of the minor arcs of the circumcircle of the regular nonagon corresponding ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
II. (40 points) Find the largest positive real number $\lambda$ such that for all positive integers $n$ and positive real numbers $a_{i} (i=1,2, \cdots, n)$, we have
$$
1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \lambda\left[\sum_{k=1}^{n} \frac{1}{\left(1+\sum_{s=1}^{k} a_{s}\right)^{2}}\right] .
$$ | II. Define $S_{k}=\sum_{i=1}^{k} a_{i}+1$, and supplement the definition $S_{0}=1$. First, prove a lemma.
Lemma For any positive integer $k \geqslant 1$, we have $\frac{1}{a_{k}^{2}}+\frac{1}{S_{k-1}^{2}} \geqslant \frac{8}{S_{k}^{2}}$.
Proof Let $S_{k-1}=a_{k} t_{k}$. Then $S_{k}=a_{k}\left(t_{k}+1\right)$.
Thus, equa... | 7 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 1: For $n$ consecutive positive integers, if each number is written in its standard prime factorization form, and each prime factor is raised to an odd power, such a sequence of $n$ consecutive positive integers is called a "consecutive $n$ odd group" (for example, when $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{... | 【Analysis】Notice that, in a connected $n$-singular group, if there exists a multiple of 4, then by the definition of a connected $n$-singular group, it must be a multiple of 8.
Let this number be $2^{k} A\left(k, A \in \mathbf{N}_{+}, k \geqslant 3, A\right.$ is an odd number). Then $2^{k} A+4$ and $2^{k} A-4$ are bot... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 The number of prime pairs \((a, b)\) that satisfy the equation
$$
a^{b} b^{a}=(2 a+b+1)(2 b+a+1)
$$
is \qquad (2]
(2011, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | 【Analysis】If $a$ and $b$ are both odd, then
the left side of equation (1) $\equiv 1 \times 1 \equiv 1(\bmod 2)$,
the right side of equation (1) $\equiv(2 \times 1+1+1)(2 \times 1+1+1)$ $\equiv 0(\bmod 2)$.
Clearly, the left side is not congruent to the right side $(\bmod 2)$, a contradiction.
Therefore, at least one of... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Let $X$ be the set of irreducible proper fractions with a denominator of 800, and $Y$ be the set of irreducible proper fractions with a denominator of 900, and let $A=\{x+y \mid x \in X, y \in Y\}$. Find the smallest denominator of the irreducible fractions in $A$. | 【Analysis】This problem is adapted from the 35th Russian Mathematical Olympiad question ${ }^{[1]}$.
Let $x=\frac{a}{800} \in X, y=\frac{b}{900} \in Y$, where,
$$
\begin{array}{l}
1 \leqslant a \leqslant 799, (a, 800)=1, \\
1 \leqslant b \leqslant 899, (b, 900)=1 .
\end{array}
$$
Then $x+y=\frac{9 a+8 b}{7200}$.
Since ... | 288 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Find the smallest positive integer $n$, such that there exist rational-coefficient polynomials $f_{1}, f_{2}, \cdots, f_{n}$, satisfying
$$
x^{2}+7=f_{1}^{2}(x)+f_{2}^{2}(x)+\cdots+f_{n}^{2}(x) .
$$
(51st IMO Shortlist) | 【Analysis】For the case $n=5$,
$$
x^{2}+7=x^{2}+2^{2}+1+1+1 \text {, }
$$
it meets the requirement.
Now we prove that $n \leqslant 4$ does not meet the requirement.
Assume there exist four rational coefficient polynomials $f_{1}(x)$, $f_{2}(x)$, $f_{3}(x)$, $f_{4}(x)$ (which may include the zero polynomial), such that ... | 5 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 9 Find all real numbers $x$ such that $4 x^{5}-7$ and $4 x^{13}-7$ are both perfect squares. ${ }^{[6]}$
(2008, German Mathematical Olympiad) | 【Analysis】Let
$$
4 x^{5}-7=a^{2}, 4 x^{13}-7=b^{2}(a, b \in \mathbf{N}) \text {. }
$$
Then $x^{5}=\frac{a^{2}+7}{4}>1$ is a positive rational number, and $x^{13}=\frac{b^{2}+7}{4}$ is a positive rational number.
Therefore, $x=\frac{\left(x^{5}\right)^{8}}{\left(x^{13}\right)^{3}}$ is a positive rational number.
Let $x... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given that $a, b, c, d$ are all prime numbers (allowing $a, b, c, d$ to be the same), and $abcd$ is the sum of 35 consecutive positive integers. Then the minimum value of $a+b+c+d$ is $\qquad$. ${ }^{[3]}$
(2011, Xin Zhi Cup Shanghai Junior High School Mathematics Competition) | 【Analysis】According to the problem, we set
$$
\begin{array}{l}
a b c d=k+(k+1)+\cdots+(k+34)\left(k \in \mathbf{N}_{+}\right) \\
\Rightarrow \frac{(2 k+34) \times 35}{2}=a b c d \\
\Rightarrow(k+17) \times 5 \times 7=a b c d .
\end{array}
$$
By symmetry, without loss of generality, let $c=5, d=7, a \leqslant b$.
We on... | 22 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Four, on a plane there are $n(n \geqslant 4)$ lines. For lines $a$ and $b$, among the remaining $n-2$ lines, if at least two lines intersect with both lines $a$ and $b$, then lines $a$ and $b$ are called a "congruent line pair"; otherwise, they are called a "separated line pair". If the number of congruent line pairs a... | (1) Among these $n$ lines, if there exist four lines that are pairwise non-parallel, then any two lines are coincident line pairs. However, $\mathrm{C}_{n}^{2}=2012$ has no integer solution, so there does not exist an $n$ that satisfies the condition.
(2) If the $n$ lines have only three different directions, let the n... | 72 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $a, b, c, d$ satisfy: for any real number $x$,
$a \cos x + b \cos 2x + c \cos 3x + d \cos 4x \leq 1$.
Find the maximum value of $a + b - c + d$ and the values of the real numbers $a, b, c, d$ at that time.
(Supplied by Li Shenghong) | 4. Let $f(x)=a \cos x+b \cos 2 x+$ $c \cos 3 x+d \cos 4 x$.
From $f(0)=a+b+c+d$,
$f(\pi)=-a+b-c+d$,
$f\left(\frac{\pi}{3}\right)=\frac{a}{2}-\frac{b}{2}-c-\frac{d}{2}$,
then $a+b-c+d$
$$
=f(0)+\frac{2}{3} f(\pi)+\frac{4}{3} f\left(\frac{\pi}{3}\right) \leqslant 3 .
$$
Equality holds if and only if $f(0)=f(\pi)=f\left... | 3 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 Given positive integers $a, b$ satisfy that $a-b$ is a prime number, and $ab$ is a perfect square. When $a \geqslant 2012$, find the minimum value of $a$.
保留源文本的换行和格式，直接输出翻译结果。 | Given the problem, let's set
$a-b=p(p$ is a prime number $), ab=k^{2}\left(k \in \mathbf{N}_{+}\right)$.
Then $a(a-p)=k^{2} \Rightarrow a^{2}-k^{2}=ap$
$$
\Rightarrow(a+k)(a-k)=ap \text {. }
$$
Since $a+k, a, p$ are all positive integers, we have
$$
a-k>0 \text {. }
$$
Given that $p$ is a prime number, from equation ... | 2025 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
5. If a non-negative integer $m$ and the sum of its digits are both multiples of 6, then $m$ is called a "Lucky Six Number". Find the number of Lucky Six Numbers among the non-negative integers less than 2012. | 5. Solution 1 It is easy to know that a non-negative integer is a hexagonal number if and only if its last digit is even and the sum of its digits is a multiple of 6.
For convenience, let
$$
M=\{0,1, \cdots, 2011\}
$$
write each number in $M$ as a four-digit number $\overline{a b c d}$ (when it is less than four digit... | 168 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
6. Find the smallest positive integer $n$ such that
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}} \\
<\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$ | 6. From the known, we must have $n \geqslant 2$ 013. At this time,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}4023, \\
\sqrt[3]{\frac{n-2013}{2011}} \geqslant \sqrt[3]{\frac{n-2011}{2013}} \\
\Leftrightarrow 2013(n-2013) \geqslant 2011(n-2011) \\
\Leftrightarrow n \geqslant 4024 .
\end{array}
$$
From equations (1) ... | 4024 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
8. Let $m$ be a positive integer, $n=2^{m}-1$, and the set of $n$ points on the number line be $P_{n}=\{1,2, \cdots, n\}$.
A grasshopper jumps on these points, each step moving from one point to an adjacent point. Find the maximum value of $m$ such that for any $x, y \in P_{n}$, the number of ways to jump from point $... | 8. When $m \geqslant 11$, $n=2^{m}-1>2013$.
Since there is only one way to jump from point 1 to point 2013 in 2012 steps, this is a contradiction, so $m \leqslant 10$.
We will now prove that $m=10$ satisfies the condition.
We use mathematical induction on $m$ to prove a stronger proposition: $\square$
For any $k \geq... | 10 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
1. If $a, b$ are both integers, the equation
$$
a x^{2}+b x-2008=0
$$
has two distinct roots that are prime numbers, then $3 a+b=$ $\qquad$ (2008, Taiyuan Junior High School Mathematics Competition) | Let the two prime roots of the equation be \(x_{1} 、 x_{2}\left(x_{1}<x_{2}\right)\). From the problem, we have
\[
x_{1} x_{2}=\frac{-2008}{a} \Rightarrow a x_{1} x_{2}=-2008 \text{. }
\]
It is easy to see that, \(2008=2^{3} \times 251\) (251 is a prime number).
Thus, \(x_{1}=2, x_{2}=251\).
Therefore, \(3 a+b=1000\). | 1000 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, find the maximum value of $k$; if not, please briefly explain the reason. | 提示: As the sum of the squares of the smallest 10 distinct prime numbers is
$$
\begin{array}{l}
4+9+25+49+121+169+289+361+529+841 \\
=2397>2010,
\end{array}
$$
thus, $k \leqslant 9$.
By analyzing the parity and the fact that the square of an odd number is congruent to 1 modulo 8, it is easy to prove that $k \neq 8, k \... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
3. Given that $a$ is a constant, and real numbers $x, y, z$ satisfy
$$
(x-1)^{2}+(y-\sqrt{5})^{2}+(z+1)^{2}=a
$$
when, $-8 \leqslant 4 x-\sqrt{5} y+2 z \leqslant 2$. Then $a=$ $\qquad$ | 3. 1.
Let $4 x-\sqrt{5} y+2 z=k$. Then $-8 \leqslant k \leqslant 2$.
From the given equation, we have
$$
\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}=1 \text {. }
$$
Using the Cauchy-Schwarz inequality, we get
$$
\begin{array}{l}
(16 a+5 a+4 a)\left[\frac{(4 x-4)^{2}}{16 a}+\frac{(... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Given real numbers $x, y, z \in (0, \sqrt{2})$, and satisfying
$$
\left(2-x^{2}\right)\left(2-y^{2}\right)\left(2-z^{2}\right)=x^{2} y^{2} z^{2} \text{. }
$$
Then the maximum value of $x+y+z$ is | 4.3.
$$
\begin{array}{l}
\text { Let } x=\sqrt{2} \cos \alpha, y=\sqrt{2} \cos \beta, \\
z=\sqrt{2} \cos \gamma\left(\alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)\right) \text {. }
\end{array}
$$
Then the given equation transforms to
$$
\tan \alpha \cdot \tan \beta \cdot \tan \gamma=1 \text {. }
$$
Assume wi... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $p$ and $5 p^{2}-2$ both be prime numbers: Find the value of $p$.
(2012, National Junior High School Mathematics Competition, Tianjin Preliminary Round) | It is easy to prove that when $3 \times p$, $3 \mid \left(5 p^{2}-2\right)$.
Since $5 p^{2}-2>3$, thus $5 p^{2}-2$ is not a prime number, which contradicts the given condition.
Therefore, $3 \mid p$. Then $p=3$. | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
7. Given the real-coefficient equation $a x^{3}-x^{2}+b x-1=0$ has three positive real roots. Then
$$
P=\frac{5 a^{2}-6 a b+3}{a^{3}(b-a)}
$$
the minimum value of $P$ is | 7. 108.
Let the three positive real roots of $a x^{3}-x^{2}+b x-1=0$ be $v_{1}, v_{2}, v_{3}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
v_{1}+v_{2}+v_{3}=\frac{1}{a}, \\
v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}=\frac{b}{a}, \\
v_{1} v_{2} v_{3}=\frac{1}{a} .
\end{array}
$$
From (1) and (2), we get $a>0, b>0$.
From... | 108 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 As shown in Figure 3, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value. | Solve as shown in Figure 4, construct the symmetric point $B'$ of point $B$ with respect to $AC$, and connect $AB'$. Construct $B'N \perp AB$, intersecting $AC$ at point $M$, and connect $BM$. Then $BM=B'M$.
Therefore, $BM+MN=B'M+MN$.
By the shortest distance of a perpendicular segment, we know that $B'M+MN$ is minimiz... | 16 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-2 x-m=0$, and $2 x_{1}+x_{2}=0$. Then the value of $m$ is $\qquad$ . | By the relationship between roots and coefficients, we know $x_{1}+x_{2}=2$.
Also, $2 x_{1}+x_{2}=0$, then
$$
\begin{array}{l}
x_{1}+2=0 \\
\Rightarrow x_{1}=-2 \\
\Rightarrow(-2)^{2}-2 \times(-2)-m=0 \\
\Rightarrow m=8 .
\end{array}
$$ | 8 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a, b$ be positive real numbers, $m$ be a positive integer, and satisfy
$$
\left\{\begin{array}{l}
a+b \leqslant 14, \\
a b \geqslant 48+m .
\end{array}\right.
$$
Then the value of $m$ is | 3.1.
Notice,
$$
\begin{array}{l}
14 \geqslant a+b \geqslant 2 \sqrt{a b} \geqslant 2 \sqrt{48+m} \\
\geqslant 2 \sqrt{48+1}=14 .
\end{array}
$$
Therefore, all equalities hold. Hence \( m=1 \). | 1 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
4. In a ball game competition, there are eight teams participating, and each pair of teams has to play a match. A team gets 2 points for a win, 1 point for a draw, and 0 points for a loss. If a team wants to ensure it enters the top four (i.e., its points must exceed those of at least four other teams), then the minimu... | 4. 11.
Since there are eight teams, there will be $\frac{8 \times 7}{2}=28$ matches, totaling $28 \times 2=56$ points.
If the top five teams draw with each other and each win against the bottom three teams, and the bottom three teams draw with each other, then there will be five teams each with 10 points, and the oth... | 11 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
3. (25 points) Write the 90 positive integers $10, 11, \cdots, 99$ on the blackboard, and erase $n$ of them so that the product of all the remaining numbers on the blackboard has a units digit of 1. Find the minimum value of $n$.
| 3. If the unit digit of the product of all remaining numbers on the blackboard is 1, then all even numbers between $10 \sim 99$ must be erased, and numbers with a unit digit of 5 must also be erased.
Thus, the unit digit of the remaining numbers must be one of $1, 3, 7, 9$.
Notice that the unit digit of $11 \times 13... | 55 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let $P$ be any point on the graph of the function $y=x+\frac{2}{x}(x>0)$, and draw perpendiculars from $P$ to the line $y=x$ and the $y$-axis, with the feet of the perpendiculars being $A$ and $B$ respectively. Then $\overrightarrow{P A} \cdot \overrightarrow{P B}=$ $\qquad$ | - 1. -1 .
Solution 1 Let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)$.
Then $l_{P A}: y-\left(x_{0}+\frac{2}{x_{0}}\right)=-\left(x-x_{0}\right)$, which is $y=-x+2 x_{0}+\frac{2}{x_{0}}$.
Solving the above equation with $y=x$ yields point $A\left(x_{0}+\frac{1}{x_{0}}, x_{0}+\frac{1}{x_{0}}\right)$.
Also, point $B\lef... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $\triangle A B C$ have interior angles $\angle A, \angle B, \angle C$ with opposite sides $a, b, c$ respectively, and satisfy
$$
\begin{array}{l}
a \cos B-b \cos A=\frac{3}{5} c . \\
\text { Then } \frac{\tan A}{\tan B}=
\end{array}
$$ | 2.4.
Solution 1 From the given and the cosine rule, we have
$$
\begin{array}{l}
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \\
\Rightarrow a^{2}-b^{2}=\frac{3}{5} c^{2} . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\fra... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
4. For the parabola $y^{2}=2 p x(p>0)$, the focus is $F$, and the directrix is $l$. Points $A$ and $B$ are two moving points on the parabola, and they satisfy $\angle A F B=\frac{\pi}{3}$. Let $M$ be the midpoint of segment $A B$, and let $N$ be the projection of $M$ on $l$. Then the maximum value of $\frac{|M N|}{|A B... | 4. 1 .
Solution 1 Let $\angle A B F=\theta\left(0<\theta<\frac{2 \pi}{3}\right)$. Then by the Law of Sines, we have
$$
\frac{|A F|}{\sin \theta}=\frac{|B F|}{\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}} .
$$
Thus, $\frac{|A F|+|B F|}{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}=... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Let two regular tetrahedra $P-ABC$ and $Q-ABC$ be inscribed in the same sphere. If the dihedral angle between a lateral face and the base of the regular tetrahedron $P-ABC$ is $45^{\circ}$, then the tangent value of the dihedral angle between a lateral face and the base of the regular tetrahedron $Q-ABC$ is . $\qqua... | 5.4.
As shown in Figure 6, connect $P Q$. Then $P Q \perp$ plane $A B C$, with the foot of the perpendicular $H$ being the center of the equilateral $\triangle A B C$, and $P Q$ passing through the center of the sphere $O$.
Connect $C H$ and extend it to intersect $A B$ at point $M$. Then $M$ is the midpoint of side ... | 4 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
7. The sum of all positive integers $n$ that satisfy $\frac{1}{4}<\sin \frac{\pi}{n}<\frac{1}{3}$ is . $\qquad$ | 7. 33 .
By the convexity of the sine function, we know that when $x \in\left(0, \frac{\pi}{6}\right)$, $\frac{3}{\pi} x < \sin x < x$. For example, $\frac{3}{\pi} \times \frac{\pi}{12}=\frac{1}{4}$, $\sin \frac{\pi}{10} < \frac{3}{\pi} \times \frac{\pi}{9}=\frac{1}{3}$.
Therefore, the positive integer values of $n$ t... | 33 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the function $f(x)=\sqrt{x^{2}+2}(x>0)$. Then the integer part of $N=f(1002)+f(1003)+\cdots+f(2005)$ is ( ).
(A) 1506500
(B) 1509514
(C) 4010
(D) 3013 | 6. B.
Notice,
$$
\begin{array}{l}
f(x)-x=\sqrt{x^{2}+2}-x>0, \\
f(x)-x=\sqrt{x^{2}+2}-x=\frac{2}{\sqrt{x^{2}+2}+x} \\
1002+1003+\cdots+2005 \text {, }
$$
and $\square$
$$
\begin{aligned}
N- & (1002+1003+\cdots+2005) \\
= & \left(\sqrt{1002^{2}+2}-1002\right)+ \\
& \left(\sqrt{1003^{2}+2}-1003\right)+\cdots+ \\
& \lef... | 1509514 | Algebra | MCQ | Yes | Yes | cn_contest | false |
1. Given 10 pairwise distinct positive integers $a_{1}$, $a_{2}, \cdots, a_{10}$ that satisfy the conditions
$$
\begin{array}{l}
a_{2}=a_{1}+a_{5}, a_{3}=a_{2}+a_{6}, \\
a_{4}=a_{3}+a_{7}, a_{6}=a_{5}+a_{8}, \\
a_{7}=a_{6}+a_{9}, a_{9}=a_{8}+a_{10} .
\end{array}
$$
then the minimum possible value of $a_{4}$ is | Ni, 1.20.
It is easy to get
$$
\begin{array}{l}
a_{4}=a_{3}+a_{7} \\
=a_{1}+a_{5}+a_{6}+a_{6}+a_{8}+a_{10} \\
=\left(a_{1}+a_{10}\right)+3\left(a_{5}+a_{8}\right) .
\end{array}
$$
To make $a_{4}$ the smallest, then $a_{5}$ and $a_{8}$ should be as small as possible. And they are all different, so let's take $a_{5}=1, ... | 20 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 There are 2011 points in space and no three points are collinear. Now, connect each pair of points with a line of one color, such that for any point, any two lines originating from that point are of different colors. How many different colors of lines are needed at least? Prove your conclusion. If the 2011 po... | To generalize, replace 2011 with $n(n \geqslant 2)$, and denote the minimum number of colors for the line segments as $f(n)$. It is easy to see that
$$
\begin{array}{l}
f(2)=1, f(3)=3, f(4)=3, \\
f(5)=5, \cdots \cdots
\end{array}
$$
Below, we prove that in the general case,
$$
f(2 n+1)=2 n+1, f(2 n)=2 n-1 \text {. }
$... | 2011 | Combinatorics | proof | Yes | Yes | cn_contest | false |
Three. (25 points) Let the pairs of positive integers $(m, n)$, where both are no more than 1000, satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \text {. }
$$
Find the number of all such pairs $(m, n)$. | Three, 1706.
$$
\begin{array}{l}
\text { Given } \frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \\
\Rightarrow \sqrt{2} n-1<m<\sqrt{2}(n+1) .
\end{array}
$$
For each $n$, the number of integers in the above range is
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1,
\end{array}
$$
where $[... | 1706 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. The function $f(x)=\ln \left(x+\sqrt{x^{2}+1}\right)+\arcsin x$. Then the solution set of $f(x)+f\left(2-x^{2}\right) \leqslant 0$ is $\qquad$ | $-、 1 .\{-1\}$.
It is known that the function $f(x)$ is a monotonically increasing odd function defined on $[-1,1]$.
$$
\begin{array}{l}
\text { Given } f(x)+f\left(2-x^{2}\right) \leqslant 0 \\
\Rightarrow f(x) \leqslant f\left(x^{2}-2\right) \\
\Rightarrow-1 \leqslant x \leqslant x^{2}-2 \leqslant 1 \\
\Rightarro... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
For $n \in \mathbf{N}_{+}$, define
$$
S(n)=\left[\frac{n}{10^{[18 n]}}\right]+10\left(n-10^{[i \mid n]}\left[\frac{n}{10^{\left[1 / B^{n}\right]}}\right]\right),
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then, among $1,2, \cdots, 2012$, the number of positive integers $n$ that sati... | 6. 108.
Let $t=10^{[\lg n]}$. Then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right) \text {. }
$$
Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the first digit of $n$.
We will discuss the cases below.
(1) If $n$ is a one-dig... | 108 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
For the expression $\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}$, when written as a decimal, find the digit before the decimal point. | Let $a_{n}=\frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{n}-\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{n}$.
Then $a_{1}=a_{2}=1, a_{n}=a_{n-1}+a_{n-2}(n \geqslant 3)$.
The last digits are
$$
\begin{array}{l}
1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7 \\
7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9 \\
8,7,5,2,7,9,6,5,1,6,7,3,0... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 The family of sets $\Omega$ consists of 11 five-element sets $A_{1}, A_{2}$, $\cdots, A_{11}$, where the intersection of any two sets is not empty. Let $A=\bigcup_{i=1}^{11} A_{i}=\left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$, for any $x_{i} \in A$, the number of sets in $\Omega$ that contain the element $x_{i}... | It is known that $\sum_{i=1}^{n} k_{i}=55$.
Notice that, the $k_{i}$ sets containing $x_{i}$ form
$$
\mathrm{C}_{k_{i}}^{2}=\frac{k_{i}\left(k_{i}-1\right)}{2}
$$
pairs of sets. Since the intersection of any two sets is not empty, the sum $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2}$ includes all pairs of sets, with some re... | 4 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 When $n$ is any real number and $k$ is a certain specific integer, the equation
$$
n(n+1)(n+2)(n+3)+1=\left(n^{2}+k n+1\right)^{2}
$$
holds. Then $k=$ $\qquad$ . [1]
(2010, Taiyuan Junior High School Mathematics Competition) | 【Analysis】Since the left side of the given equation is a polynomial and the right side is in the form of a product, we only need to factorize the left side. The method of factorization is to use the overall idea and the complete square formula to handle it.
Solution Note that,
$$
\begin{array}{l}
n(n+1)(n+2)(n+3)+1 \\
... | 3 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Given $a, b, c \in \mathbf{R}$, and
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c} \text {. }
$$
Then there exists an integer $k$, such that the following equations hold for:
(1) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}$;
(2... | 【Analysis】The condition equation in this problem is a fractional equation, which is relatively complex. The key to solving this problem is to simplify the relationship between the letters $a, b, c$. First, eliminate the denominator and rearrange the condition equation into the form $f(a, b, c)=0$, then factorize $f(a, ... | 2 | Algebra | proof | Yes | Yes | cn_contest | false |
For example, $5 n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ have a sum of 2009. If these $n$ numbers can be divided into 41 groups with equal sums and also into 49 groups with equal sums, find the minimum value of $n$.
| Let the 41 groups be $A_{1}, A_{2}, \cdots, A_{41}$, where the sum of the numbers in each group is 49, and we call such groups "A-type groups"; and the 49 groups be $B_{1}, B_{2}, \cdots, B_{49}$, where the sum of the numbers in each group is 41, and we call such groups "B-type groups".
Clearly, each term $x_{k}$ belo... | 89 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 6 In the Mathematical Olympiad training team, there are 30 members, each of whom has the same number of friends in the team. It is known that in a test, everyone's scores are different. If a member scores higher than the majority of their friends, they are called a "pro". Question: What is the maximum number of... | Let each team member have $k$ friends, and this exam has produced $m$ experts, the best-performing member of the team, is the best in their $k$ "friend pairs," and is naturally an expert. Each of the other experts is at least the best in $\left[\frac{k}{2}\right]+1 \geqslant \frac{k+1}{2}$ (where [x] denotes the greate... | 25 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
4. Let the set $S=\{1,2, \cdots, 50\}$. Find the smallest positive integer $k$, such that in any $k$-element subset of $S$, there exist two distinct numbers $a$ and $b$, satisfying $(a+b) \mid a b$. | First, by enumeration, we obtain 23 pairs $(a, b)$, each of which satisfies $(a+b) \mid a b$.
Construct a 50-order graph $G$ (with the set $S$ of numbers $1,2, \cdots, 50$ as vertices, and if two numbers $a, b$ belong to the above pairs, then let $a, b$ be adjacent). Thus, the graph $G$ has exactly 23 edges (isolated ... | 39 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 3 Given
$$
a^{2}(b+c)=b^{2}(a+c)=2010 \text {, and } a \neq b \text {. }
$$
Then $c^{2}(a+b)=$ $\qquad$ [2]
$(2010$, I Love Mathematics Junior High School Summer Camp Mathematics Competition) | 【Analysis】The given condition equation has the same structure as the algebraic expression to be evaluated. According to the known condition equation, it is impossible to determine the values of $a$, $b$, and $c$. We can only conjecture that there is an intrinsic relationship between $a$, $b$, and $c$. By constructing a... | 2010 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. The graph of the quadratic function $y=a x^{2}+b x+c$ intersects the $x$-axis at two points $A$ and $B$, with the vertex at $C$. If $\triangle A C B$ is a right triangle, then the value of the discriminant is $\qquad$. | 3. 4 .
As shown in Figure 4.
From the problem, we know $\Delta=b^{2}-4 a c>0$.
Let $A\left(x_{1}, 0\right), B\left(x_{2}, 0\right), C\left(-\frac{b}{2 a}, \frac{4 a c-b^{2}}{4 a}\right)$.
By Vieta's formulas, we have
$x_{1}+x_{2}=-\frac{b}{a}, x_{1} x_{2}=\frac{c}{a}$.
Then $\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_... | 4 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let real numbers $x, y, z$ simultaneously satisfy
$$
\left\{\begin{array}{l}
x^{3}+y=3 x+4, \\
2 y^{3}+z=6 y+6, \\
3 z^{3}+x=9 z+8 .
\end{array}\right.
$$
Try to find the value of $2008(x-1)^{2}+2009(y-1)^{2}+$ $2010(z-2)^{2}$. ${ }^{[3]}$
(1st Youth Mathematical Week (Zonghu Cup) Mathematical Competition) | Solve: From the given, we have
$$
\left\{\begin{array}{l}
y-2=-x^{3}+3 x+2=-(x-2)(x+1)^{2}, \\
z-2=-2 y^{3}+6 y+4=-2(y-2)(y+1)^{2}, \\
x-2=-3 z^{3}+9 z+6=-3(z-2)(z+1)^{2} .
\end{array}\right.
$$
Multiplying the above three equations, we get
$$
(x-2)(y-2)(z-2)
$$
$$
\begin{aligned}
= & -6(x-2)(y-2)(z-2)(x+1)^{2}(y+1)^{... | 4017 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
1. $a_{1}, a_{2}, a_{3}, \cdots$ is an arithmetic sequence, where $a_{1}>0, s_{n}$ represents the sum of the first $n$ terms. If $S_{3}=S_{11}$, in $S_{1}, S_{2}, S_{3}, \cdots$ the largest number is $S_{k}$, then $k=$ $\qquad$ . | -1.7 .
Let the common difference be $d$. Then
$$
\begin{array}{l}
a_{n}=a_{1}+(n-1) d . \\
\text { By } S_{3}=S_{11} \Rightarrow d=-\frac{2}{13} a_{1}<0 . \\
\text { Hence } a_{n}=a_{1}+(n-1)\left(-\frac{2}{13} a_{1}\right) \\
=\frac{a_{1}}{13}(15-2 n),
\end{array}
$$
and the largest positive integer $n$ for which $a_... | 7 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Consider a tangent line to the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$, which intersects the two symmetry axes of the ellipse at points $A$ and $B$. Then the minimum length of segment $AB$ is $\qquad$ . | 2. 8 .
Let the point of tangency be \( P(5 \cos \theta, 3 \sin \theta) \). Then the equation of the tangent line to the ellipse at point \( P \) is
$$
\frac{\cos \theta}{5} x + \frac{\sin \theta}{3} y = 1,
$$
which intersects the \( x \)-axis and \( y \)-axis at
$$
A\left(\frac{5}{\cos \theta}, 0\right), B\left(0, \f... | 8 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. In rectangle $A B C D$, it is known that $A B=2, B C=3$, $E$ and $F$ are the midpoints of $A B$ and $C D$ respectively. Rotate $\triangle F A B$ $90^{\circ}$ around $E F$ to $\triangle F A^{\prime} B^{\prime}$. Then the volume of the tetrahedron $A^{\prime} B^{\prime} C D$ is $\qquad$ . | 3. 2 .
It is known that $E F=B C=3$, and the plane of $\triangle F A^{\prime} B^{\prime}$ divides the tetrahedron $A^{\prime} B^{\prime} C D$ into two tetrahedrons of equal volume, $C F A^{\prime} B^{\prime}$ and $D F A^{\prime} B^{\prime}$. Their heights are $C F=D F=1$, and $S_{\triangle F A^{\prime} B^{\prime}}=3$.... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Use the digits $1,2, \cdots, 7$ to form a seven-digit number such that it is a multiple of 11. The number of seven-digit numbers that can be formed is $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 5. 576 .
Let $n$ be a seven-digit number satisfying the condition, and let $a$ and $b$ be the sums of the digits in the odd and even positions, respectively. Then $a+b=28$, and $a-b$ is a multiple of 11.
Since $a+b$ and $a-b$ have the same parity, they must both be even. Clearly, $|a-b| \neq 22$, so only $a-b=0$.
Thus... | 576 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$
All real roots. ${ }^{[4]}$
(2011, International Invitational Competition for Young Mathematicians in Cities) | 【Analysis】The most basic method to solve higher-degree equations is to convert them into lower-degree equations for solving, that is, to handle them as linear or quadratic equations. Factorization is the most powerful tool to achieve such a transformation. The preferred method for factoring higher-degree polynomials is... | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
7. For any $x, y \in [0,1]$, the function
$$
f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}
$$
has a maximum value of $\qquad$ . | 7. 1.
Since $x, y \in [0,1]$, then $x \leqslant \sqrt{x}, y \leqslant \sqrt{y}$.
Let $x=\sin ^{2} \alpha, y=\sin ^{2} \beta\left(\alpha, \beta \in\left[0, \frac{\pi}{2}\right]\right)$.
Thus, $f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}$
$$
\begin{array}{l}
\leqslant \sqrt{x(1-y)}+\sqrt{y(1-x)} \\
=\sin \alpha \cdot \cos \beta+\... | 1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Four, (20 points) Question: In how many different ways can the elements of the set $M=\{1,2,3,4,5\}$ be assigned to three (ordered) sets $A$, $B$, and $C$, such that each element is contained in at least one of the sets, the intersection of these three sets is empty, and the intersection of any two of these sets is not... | As shown in Figure 2, consider the seven parts divided by the Venn diagram, represented by $x, u, v, w, a, b, c$ respectively.
Now, fill the elements of $M$ into these parts. According to the problem, $x$ cannot be filled with any number, while $u, v, w$ must be filled with numbers, and the numbers filled in these par... | 1230 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
12. In the tetrahedron $ABCD$, it is known that $AD=2\sqrt{3}$, $\angle BAC=60^{\circ}$, $\angle BAD=\angle CAD=45^{\circ}$. The radius of the sphere that passes through $D$ and is tangent to the plane $ABC$ and internally tangent to the circumscribed sphere of the tetrahedron is 1, then the radius of the circumscribed... | 12.3.
As shown in Figure 3, draw a perpendicular from point $D$ to plane $ABC$, with the foot of the perpendicular being $H$. Draw $DE \perp AB$ and $DF \perp AC$, with the feet of the perpendiculars being $E$ and $F$ respectively.
Then $HE \perp AB$, $HF \perp AC$, and
$AE = AF = AD \cos 45^{\circ} = \sqrt{6}$.
From ... | 3 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 8 Let the lengths of the two legs of a right triangle be $a$ and $b$, and the length of the hypotenuse be $c$. If $a$, $b$, and $c$ are all integers, and $c=\frac{1}{3} a b-(a+b)$, find the number of right triangles that satisfy the condition. ${ }^{(6)}$
(2010, National Junior High School Mathematics Competiti... | 【Analysis】In a right-angled triangle, the three sides satisfy the Pythagorean theorem. Given the condition $c=\frac{1}{3} a b-(a+b)$, one unknown can be eliminated to obtain a quadratic equation in two variables, which is generally difficult to solve. However, the problem of integer solutions to a quadratic equation in... | 3 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
1. Let positive numbers $x, y, z$ satisfy
$$
\frac{1}{x^{3}}=\frac{8}{y^{3}}=\frac{27}{z^{3}}=\frac{k}{(x+y+z)^{3}} \text {. }
$$
Then $k=$ $\qquad$ | $$
\begin{array}{l}
\sqrt[3]{k}=\frac{x+y+z}{x}=\frac{2(x+y+z)}{y} \\
=\frac{3(x+y+z)}{z}=\frac{6(x+y+z)}{x+y+z}=6 .
\end{array}
$$
Therefore, $k=216$. | 216 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. If the equation with respect to $x$
$$
x^{2}+2(m+3) x+m^{2}+3=0
$$
has two real roots $x_{1}$ and $x_{2}$, then the minimum value of $\left|x_{1}-1\right|+\left|x_{2}-1\right|$ is $\qquad$. | 2.6 .
According to the problem, we have
$$
\begin{array}{l}
\Delta=[2(m+3)]^{2}-4\left(m^{2}+3\right) \geqslant 0 \\
\Rightarrow m \geqslant-1 .
\end{array}
$$
Then $x_{1}+x_{2}=-2(m+3)<0$.
When $x=1$, the left side of the equation is greater than 0, thus, $x_{1}$ and $x_{2}$ are on the same side of 1.
$$
\begin{arra... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
4. Let $x_{n}$ denote the unit digit of the number $n^{4}$. Then
$$
x_{1}+x_{2}+\cdots+x_{2012}=
$$
$\qquad$ | 4.6640 .
Notice that, the unit digit of $(10+n)^{4}$ is the same as that of $n^{4}$, and the unit digits of $1^{4}, 2^{4}, \cdots, 10^{4}$ are $1,6,1,6,5,6,1,6,1,0$ respectively.
Thus, $x_{1}+x_{2}+\cdots+x_{10}=33$.
Therefore, $x_{1}+x_{2}+\cdots+x_{2012}$
$$
=201 \times 33+(1+6)=6640 \text {. }
$$ | 6640 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
One. (20 points) As shown in Figure 2, in the isosceles right triangle $\triangle ABC$, $\angle C=90^{\circ}$, points $D$ and $E$ are on side $BC$, and point $F$ is on the extension of $AC$, such that $BE=ED=CF$. Find the tangent value of $\angle CEF + \angle CAD$.
---
The translation preserves the original text's li... | $\begin{array}{l}\text { I. Draw } D G \perp A B \text { at point } G . \\ \text { Let } A C=B C=x, \\ B E=E D=C F=y\left(0<y<\frac{x}{2}\right) . \\ \text { Then } G D=\sqrt{2} y \Rightarrow A G=\sqrt{2}(x-y)=\sqrt{2} C E \text {. } \\ \text { Therefore, } \mathrm{Rt} \triangle E C F \backsim \mathrm{Rt} \triangle A G... | 1 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
5. Suppose there are 10 red, 10 yellow, and 10 blue small balls. Now, all of them are to be placed into two bags, A and B, such that each bag contains balls of two colors, and the sum of the squares of the number of balls of two colors in bags A and B are equal. There are $\qquad$ ways to do this. | 5.61.
Let the number of red, yellow, and blue balls in bag A be $x, y, z (1 \leqslant x, y, z \leqslant 9)$. Then the number of balls of corresponding colors in bag B are $10-x, 10-y, 10-z$.
First, assume $x \leqslant y \leqslant z$.
From the problem, we know
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=(10-x)^{2}+(10-y)^{2... | 61 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
7. Given two lines with a slope of 1, $l_{1}$ and $l_{2}$, passing through the two foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $l_{1}$ intersects the ellipse at points $A$ and $B$, $l_{2}$ intersects the ellipse at points $C$ and $D$. If quadrilateral $\square A B C D$ satisfies $A C \pe... | 7.2012.
It is known that $\square A B C D$ is symmetric about the origin $O$. As shown in Figure 1, let $\angle A \dot{F}_{1} F_{2}=\alpha$. Then
. $\tan \alpha=1$
. $\Rightarrow \alpha=45^{\circ}$.
Since $A C \perp A B$, we know
$A C \perp A F_{1}$.
Thus, $\triangle A F_{1} O$ is an
isosceles right triangle. Therefo... | 2012 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
8. Given $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$ has a segment of digits $\overline{2012}$ in its decimal part, where $n$ is the smallest number satisfying the condition. Then $\left[\frac{m}{\sqrt{n}}\right]=$ $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$). | 8. 2 .
Let $\frac{m}{n}=\overline{A . B 2012 C}$, where $A, B, C$ are digit strings, and the lengths of $A, B$ are $k, l (k, l \in \mathbf{N})$. Then
$$
\frac{10^{6}(m-n A)-n B}{n}=\overline{0.2012 C} \triangleq \frac{a}{b},
$$
where $(a, b)=1, 1 \leqslant b \leqslant n$.
Assume $\frac{m}{n}=\overline{0.2012 C}$.
$$
... | 2 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
10. (20 points) Rolling Dice (a uniform cube, with six faces marked with $1,2,3,4,5,6$) Game rules are as follows: First roll 9 dice, take out the dice showing 1 and set them aside; on the second roll, take out the dice showing 1 from the remaining dice; $\cdots \cdots \cdots$, until no dice show 1 or all dice are take... | 10. According to the game rules, if the game ends exactly after 9 rounds, then in the first eight rounds, each time exactly 1 die shows a 1, and the ninth round ends the game regardless of whether it shows a 1 or not. Among these, the probability that exactly 1 die shows a 1 in the $k(k=1,2, \cdots, 8)$-th round, where... | 2012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Three, (50 points) Let the number of all positive integers satisfying the following conditions be $N$:
(1) less than or equal to 2,012;
(2) the number of 1s in their binary representation is at least 2 more than the number of 0s.
Find the sum of the digits of $N$.
| Three, from $2012=(11111011100)_{2}$, we know that the numbers satisfying the conditions have at most 11 digits in binary representation.
The first digit must be 1, so the number of $d+1$-digit numbers with exactly $k+1$ digits being 1 is $\mathrm{C}_{d}^{k}$, and condition (2) is equivalent to
$$
\begin{array}{l}
k+1... | 13 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Four. (50 points) Let $n \in \mathbf{N}_{+}, f(n)$ be the number of all integer sequences $\left\{a_{k} \mid k=0,1, \cdots, n\right\}$ that satisfy the following conditions:
$$
\begin{array}{l}
\text { (1) } a_{0}=0, a_{n}=2 n, \text { and } \\
1 \leqslant a_{k+1}-a_{k} \leqslant 3(k=0,1, \cdots, n-1) ;
\end{array}
$$
... | Divide a circle of length $2 \cdot n$ into $2n$ equal parts, and label the points sequentially as $0,1, \cdots, 2n$. Then color the points labeled $a_{i} (i=0,1, \cdots, n-1)$ black, and the other $n$ points white. The sequence given in the problem corresponds one-to-one with the following coloring method:
(1) The poin... | 2012 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given that $a$ and $b$ are real numbers, and $a^{2} + ab + b^{2} = 3$. If the maximum value of $a^{2} - ab + b^{2}$ is $m$, and the minimum value is $n$, find the value of $m + n$. ${ }^{\text {[2] }}$ | Let $a^{2}-a b+b^{2}=t$.
Combining this with the given equation, we get
$$
a b=\frac{3-t}{2}, a+b= \pm \sqrt{\frac{9-t}{2}} .
$$
Thus, $a$ and $b$ are the two real roots of the quadratic equation in $x$:
$$
x^{2} \pm \sqrt{\frac{9-t}{2}} x+\frac{3-t}{2}=0
$$
Therefore, $\Delta=\left( \pm \sqrt{\frac{9-t}{2}}\right)^{... | 10 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Initially 340 Given two points $A$ and $B$ on a straight line $l$, the distance between them is $10000 \mathrm{~cm}$. At points $A$ and $B$, there are two movable barriers, designated as Barrier 1 and Barrier 2, respectively. Assume there is a ping-pong ball between $A$ and $B$, moving along the straight line $l$ at a ... | Let's assume that when the ping-pong ball contacts the 2nd board for the $n$-th time, the position of the 2nd board after it moves quickly is $B_{n}$; when the ping-pong ball contacts the 1st board for the $n$-th time, its position is $A_{n}$, and we set
$$
\begin{array}{l}
B_{0}=B, A_{n} B_{n-1}=x_{n}, A_{n} B_{n}=y_{... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Example 4 Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$
Find the minimum value of $u=9 a^{2}+72 b+2$. | Notice,
$$
\begin{array}{l}
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \\
\Rightarrow(a-2 b)(3 a-4 b+5)=0 \\
\Rightarrow a-2 b=0 \text { or } 3 a-4 b+5=0 .
\end{array}
$$
(1) $a-2 b=0$.
Then $u=9 a^{2}+72 b+2=36 b^{2}+72 b+2$ $=36(b+1)^{2}-34$.
Thus, when $b=-1$, the minimum value of $u$ is -34.
$$
\text { (2) } 3 a-4 b+5=0 \t... | -34 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 1 Given
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac... | Solve: Consider the given system of equations as a fractional equation in terms of $t$
$$
\begin{array}{c}
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1 \\
\Rightarrow\left(t-1^{2}\right)\left(t-3^{2}\right)\left(t-5^{2}\right)\left(t-7^{2}\right)- \\
x^{2}\left(t-3^{2}\right... | 36 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
Example 2 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
untranslated text remains unchanged:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
This line is not translated as it seems to be an instruction and not part of the text to be translated. If you need this line translated as well, please l... | Notice,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1000} \text {. }
$$
Consider the integer-coefficient quadratic equation $x^{2}-10 x+1=0$ with $5+2 \sqrt{6}$ as one of its roots, the other root being $5-2 \sqrt{6}$.
$$
\begin{array}{l}
\text { Let } a=5+2 \sqrt{6}, b=5-2 \sqrt{6}, \\
u_{n}=a^{n}+b^{n}(n=1,2, \cdo... | 7 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Example 5 Let $x$ be a real number. Then
$$
|x-1|+|x+1|+|x+5|
$$
the minimum value is $\qquad$ (s) | Let $y=|x-1|+|x+1|+|x+5|$. When $x<-5$,
$$
y=1-x-x-1-x-5=-3 x-5 \text{; }
$$
When $-5 \leqslant x<-1$,
$$
y=1-x-x-1+x+5=-x+5 \text{; }
$$
When $-1 \leqslant x<1$,
$$
y=1-x+x+1+x+5=x+7 \text{; }
$$
When $x \geqslant 1$,
$$
y=x-1+x+1+x+5=3 x+5 \text{. }
$$
By the properties of linear functions, we know that when $x=-... | 6 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
14. (16 points) As shown in Figure 1, in a certain engineering project to measure the radius $R$ of an arc, two identical small balls are placed on the arc so that each contact point is tangent to the arc. The height difference between the balls is $h$, and the radius of the small balls is $r$. Try to express $R$ in te... | 14. Let the angle between the line connecting the centers of the two smaller circles and the line connecting the center of the larger circle be $\theta$. Then
$$
\cos \theta=\frac{(R-r)^{2}+(R-r)^{2}-(2 r)^{2}}{2(R-r)^{2}} .
$$
Also, $h=(R-r)-(R-r) \cos \theta=\frac{2 r^{2}}{R-r}$
$$
\Rightarrow R=r+\frac{2 r^{2}}{h} ... | 600 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
Example 7 Given that $x_{1}, x_{2}, \cdots, x_{40}$ are all positive integers, and $x_{1}+x_{2}+\cdots+x_{40}=58$. If the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ is $A$, and the minimum value is $B$, then $A+B=$ $\qquad$ | Solution: Since there are only a finite number of ways to write 58 as the sum of 40 positive integers, the maximum and minimum values of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ exist.
Assume without loss of generality that $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{40}$.
If $x_{1}>1$, then
$$
x_{1}+x_{2}=\lef... | 494 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
2. Let the vectors $\overrightarrow{O A}=(1,0), \overrightarrow{O B}=(1,1)$, and $O$ be the origin. A moving point $P(x, y)$ satisfies
$$
\left\{\begin{array}{l}
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O A} \leqslant 1, \\
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O B} \leqslant 2
\end{array... | 2. 2 .
From the problem, the point $P(x, y)$ satisfies
$$
\begin{array}{l}
\left\{\begin{array}{l}
0 \leqslant x \leqslant 1, \\
0 \leqslant x+y \leqslant 2 .
\end{array}\right. \\
\text { Let }\left\{\begin{array}{l}
x+y=u, \\
y=v .
\end{array}\right.
\end{array}
$$
Then the point $Q(u, v)$ satisfies
$$
\left\{\begi... | 2 | Geometry | math-word-problem | Yes | Yes | cn_contest | false |
3. Let there be a non-empty set $A \subseteq\{1,2, \cdots, 7\}$, and when $a \in A$, it must also be that $8-a \in A$. Then the number of such sets $A$ is $\qquad$ . | 3. 15 .
Find the single element or binary element set that is congruent to $8-a$ in $A$:
$$
\begin{array}{l}
A_{1}=\{4\}, A_{2}=\{1,7\}, \\
A_{3}=\{2,6\}, A_{4}=\{3,5\} .
\end{array}
$$
The problem is equivalent to finding the number of non-empty subsets of $\left\{A_{1}, A_{2}, A_{3}, A_{4}\right\}$.
Thus, there are... | 15 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
5. Given the first seven digits of an 11-digit mobile phone number are 1390931. If the remaining four digits can only be 1, 3, 5 and each must appear at least once, then there are such mobile phone numbers.
untranslated: 个.
Note: The word "个" at the end of the sentence is not translated as it is a placeholder for t... | 5. 36 . | 36 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
2. Let $(a, b)$ be real numbers. Then the minimum value of $a^{2}+a b+b^{2}-a-2 b$ is . $\qquad$ | Hint: Use the method of completing the square. The original expression can be transformed into
$$
\left(a+\frac{1}{2} b-\frac{1}{2}\right)^{2}+\frac{3}{4}(b-1)^{2}-1 \geqslant-1 \text {. }
$$
When $a+\frac{1}{2} b-\frac{1}{2}=0, b-1=0$, i.e., $a=0$, $b=1$, the original expression can achieve the minimum value of -1. | -1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
$$
\left\{\begin{array}{l}
x-999 \geqslant 1000, \\
x+1 \leqslant a
\end{array}\right.
$$
has a finite number of real solutions. Then the value of $a$ is $\qquad$ . | Ni, 1.2000.
The solution set of the inequality is $1999 \leqslant x \leqslant a-1$.
From the problem, we know $a-1=1999 \Rightarrow a=2000$. | 2000 | Inequalities | math-word-problem | Yes | Yes | cn_contest | false |
2. On the way from Xiaoming's house to the swimming pool, there are 200 trees. When going to the swimming pool and returning, Xiaoming ties red ribbons on some trees: 1: to make marks. When going to the swimming pool, he marks the 1st tree, the 6th tree, the 11th tree, …, each time skipping 4 trees without marking. On ... | 2. 140 .
According to the problem, the trees marked are the $5 x+1(x=0,1, \cdots, 39)$ and $8 y(y=1,2, \cdots, 25)$ ones, among which exactly 5 trees are marked twice.
Therefore, the number of trees that are not marked is
$$
200-(40+25-5)=140 \text {. }
$$ | 140 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
Three, (25 points) Given that the positive integer $M$ when divided by the positive integer $N$ leaves a remainder of 2, and the sum of the reciprocals of all values of $N$ is $\frac{627}{670}$, with the number of all values of $N$ being less than 16. Find all possible values of $M$.
| Obviously, $N>2, N \mid (M-2)$.
Since $670=2 \times 5 \times 67$, $M-2$ has prime factors 2, 5, and 67. Therefore, the sum of the reciprocals of all positive divisors of $M-2$ is
$$
\frac{1}{1}+\frac{1}{2}+\frac{627}{670}=\frac{816}{335} .
$$
The number of positive divisors of $M-2$ is less than 18.
Since $(1+1)^{5}=32... | 2010 | Number Theory | math-word-problem | Yes | Yes | cn_contest | false |
4. Let positive real numbers $a, b, c$ satisfy $\frac{2}{a}+\frac{1}{b}=\frac{\sqrt{3}}{c}$. Then the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is $\qquad$ . | 4. 9 .
From the given, we have $\frac{2 c}{a}+\frac{c}{b}=\sqrt{3}$.
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
$$
\begin{array}{l}
\frac{2 a^{2}+b^{2}}{c^{2}} \\
=\frac{1}{3}(2+1)\left[2\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left(\frac{2 a}{c}+... | 9 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
5. Let $a_{1}, a_{2}, \cdots, a_{6}$ be any permutation of $1,2, \cdots, 6$, and $f$ be a one-to-one mapping from $\{1,2, \cdots, 6\}$ to $\{1,2, \cdots, 6\}$, satisfying
$$
f(i) \neq i, f(f(i))=i(i=1,2, \cdots, 6) .
$$
Consider the number table
$$
A=\left[\begin{array}{cccccc}
a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & ... | 5. 10800 .
For a permutation $a_{1}, a_{2}, \cdots, a_{6}$, consider the one-to-one mapping satisfying
$$
f(i) \neq i, f(f(i))=i(i=1,2, \cdots, 6)
$$
For each such mapping $f$, the elements of set $A$ can be paired as $\{i, j\}$, such that
$$
f(i)=j, f(j)=i \text {. }
$$
Thus, for each permutation, the number of map... | 10800 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
6. Given the equation in $x$
$$
x^{3}-4 x^{2}+5 x+a=0(a \in \mathbf{R})
$$
has three real roots $x_{1}, x_{2}, x_{3}$. Then the maximum value of $\max \left\{x_{1}, x_{2}, x_{3}\right\}$ is $\qquad$ . | 6.2.
Assume $x_{3}=\max \left\{x_{1}, x_{2}, x_{3}\right\}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=4, \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=5
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{1}+x_{2}=4-x_{3}, \\
x_{1} x_{2}=5-x_{3}\left(x_{1}+x_{2}\right... | 2 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
3. Let $a>1$. Then when the graphs of $y=a^{x}$ and $y=\log _{a} x$ are tangent, $\ln \ln a=$ $\qquad$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 3. -1.
Since the two functions are inverse functions of each other and are symmetric about the line $y=x$, the point of tangency lies on $y=x$.
Let the point of tangency be $\left(x_{0}, y_{0}\right)$. Then
$$
\begin{array}{l}
x_{0}=a^{x_{0}}, \\
a^{x_{0}} \ln a=1 .
\end{array}
$$
Substituting equation (1) into equat... | -1 | Combinatorics | math-word-problem | Yes | Yes | cn_contest | false |
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