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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 all $n(0 \leqslant n<n+m \leqslant 2012)$, $m \mid\left(a_{n+m}-a_{n}\right)$.
|
3. From (2), we know that for any $i, j (0 \leqslant i < j \leqslant 2012)$ and $m=3^{\alpha} \times 11^{\beta} \times 61^{\gamma} (\alpha, \beta, \gamma \in \{0,1\})$, we have
$$
a_{i} \equiv a_{j}(\bmod m) \Leftrightarrow i \equiv j(\bmod m).
$$
Let
$$
\begin{array}{l}
A_{i}=\{x \in \mathrm{N} \mid 0 \leqslant x \leqslant 2012, x \equiv i(\bmod 3)\}(i=0,1,2), \\
A_{i j}=\left\{x \in A_{i} \mid x \equiv j(\bmod 11)\right\}(j=0,1, \cdots, 10), \\
A_{i j k}=\left\{x \in A_{i j} \mid x \equiv k(\bmod 61)\right\}(k=0,1, \cdots, 60).
\end{array}
$$
From (1), we know that for $i=0,1,2$, all $a_{n} (n \in A_{i})$ are congruent modulo 3 (the remainder is denoted as $i^{\prime}$), and there is a one-to-one mapping
$$
f_{1}: A_{i} \rightarrow A_{i^{\prime}}.
$$
Similarly, there are one-to-one mappings
$$
f_{2}: A_{i j} \rightarrow A_{i^{\prime} j^{\prime}}, f_{3}: A_{i j k} \rightarrow A_{i^{\prime} j^{\prime} k^{\prime}}.
$$
Therefore, there are $3! \times 11! \times 61!$ permutations that satisfy the conditions.
Below is a method to construct all such permutations.
Let the bijection $f: (i, j, k) \rightarrow (i^{\prime}, j^{\prime}, k^{\prime})$, where
$$
0 \leqslant i, i^{\prime} \leqslant 2, 0 \leqslant j, j^{\prime} \leqslant 10, 0 \leqslant k, k^{\prime} \leqslant 60.
$$
If $n$ satisfies
$$
\left\{\begin{array}{l}
n \equiv i(\bmod 3), \\
n \equiv j(\bmod 11), \\
n \equiv k(\bmod 61),
\end{array}\right.
$$
then take $a_{n}$ to be the number that satisfies
$$
\left\{\begin{array}{l}
a_{n} \equiv i^{\prime}(\bmod 3), \\
a_{n} \equiv j^{\prime}(\bmod 11), \\
a_{n} \equiv k^{\prime}(\bmod 61).
\end{array}\right.
$$
By the Chinese Remainder Theorem, the solution is unique.
Clearly, $\left(a_{0}, a_{1}, \cdots, a_{2012}\right)$ satisfies conditions (1) and (2).
Thus, $N=3! \times 11! \times 61!$.
Therefore, $p=67$.
|
67
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 $4-\sqrt{11}$, we get
$$
\begin{array}{l}
5 x_{n}>(4-\sqrt{11}) x_{n+1}>5 x_{n}-(4-\sqrt{11}) \\
\Rightarrow 4 x_{n+1}-5 x_{n}<\sqrt{11} x_{n+1} \\
\quad<4 x_{n+1}-5 x_{n}+(4-\sqrt{11}) \\
\Rightarrow\left[\sqrt{11} x_{n+1}\right]=4 x_{n+1}-5 x_{n} \\
\Rightarrow x_{n+2}=4 x_{n+1}+\left[\sqrt{11} x_{n+1}\right] \\
\quad=4 x_{n+1}+\left(4 x_{n+1}-5 x_{n}\right) \\
\quad=8 x_{n+1}-5 x_{n} .
\end{array}
$$
Clearly, $x_{n+2} \equiv x_{n}(\bmod 2)$,
$$
x_{n+2} \equiv 3 x_{n+1}(\bmod 5) \text {. }
$$
By $x_{1}=1, x_{2}=7$, it is easy to prove using the second principle of mathematical induction that for any positive integer $n$,
$$
x_{n} \equiv 1(\bmod 2) \text {. }
$$
By $x_{2}=7$, it is easy to prove using the first principle of mathematical induction that for any positive integer $n \geq 2$,
$$
x_{n}=2 \times 3^{n-2}(\bmod 5) \text {. }
$$
Since $x_{2012} \equiv 1(\bmod 2)$,
$$
x_{2012} \equiv 2 \times 3^{2010} \equiv 2 \times 3^{2} \equiv 3(\bmod 5) \text {, }
$$
Therefore, the last digit of $x_{2012}$ is 3.
|
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), and finally return to the starting city. She finds that no matter which starting city and intermediate cities she chooses, she cannot complete her journey using only one airline. Find the maximum value of $n$.
(Liang Yingde, problem contributor)
|
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 the subgraph formed by edges of the same color has no cycles.
By a well-known conclusion, the number of edges in a simple graph without cycles is less than the number of vertices. Therefore, the number of edges of each color does not exceed $n-1$. This indicates that the total number of edges does not exceed $2(n-1)$.
On the other hand, the number of edges in a complete graph with $n$ vertices is $\frac{n(n-1)}{2}$.
Therefore, $\frac{n(n-1)}{2} \leqslant 2(n-1) \Rightarrow n \leqslant 4$.
When $n=4$, let the four cities be $A$, $B$, $C$, and $D$. If the routes $A B$, $B C$, and $C D$ are operated by the first airline, and the routes $A C$, $A D$, and $B D$ are operated by the second airline, it is easy to see that the routes operated by each airline form a chain and do not contain any cycles, i.e., this scenario satisfies the conditions of the problem.
In summary, the maximum value of $n$ is 4.
Solution 2 From the same reasoning as in Solution 1, we know that the subgraph formed by edges of the same color has no cycles.
By a well-known conclusion, a complete graph with six vertices, when colored with two colors, must contain a monochromatic triangle, i.e., a monochromatic cycle.
Therefore, $n \leqslant 5$.
We now prove: $n \neq 5$.
If not, let the five vertices be $A$, $B$, $C$, $D$, and $E$. Consider the four edges extending from vertex $A$. If at least three of these edges are the same color (e.g., $A B$, $A C$, $A D$ are the same color), then among $B C$, $B D$, and $C D$, there will be an edge of the same color as the aforementioned three edges, leading to a monochromatic triangle, which is a contradiction. If $B C$, $B D$, and $C D$ are all different colors from $A B$, $A C$, and $A D$, they will form a monochromatic triangle, which is a contradiction. If each vertex has exactly two edges of the same color, it is easy to see that the graph contains a Hamiltonian cycle, which is a contradiction.
Therefore, $n \leqslant 4$.
The rest follows the same reasoning as in Solution 1, showing that $n$ can be 4.
Hence, the maximum value of $n$ is 4.
|
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
$\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1}$
$\Rightarrow p\left|k \mathrm{C}_{4 p}^{k} \Rightarrow p\right| \mathrm{C}_{t_{p}}^{k}$
$\Rightarrow p \mid \mathrm{C}_{2012}^{k}$
If $p \mid k$, then there are only the following five cases:
$\mathrm{C}_{4 p}^{0}=\mathrm{C}_{s_{p}}^{4_{p}}=1$,
$\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}=\frac{4 p \cdot(4 p-1) \cdots \cdot(3 p+1)}{p \cdot(p-1) \cdots \cdot 1}$
$\equiv 4(\bmod p)$,
$\mathrm{C}_{4 p}^{2}=\frac{4 p \cdot(4 p-1) \cdots \cdot(3 p+1) \cdot 3 p \cdot(3 p-1) \cdots \cdot(2 p+1)}{2 p \cdot(2 p-1) \cdots \cdot(p+1) \cdot p \cdot(p-1) \cdots \cdot 1}$
$\equiv 6(\bmod p)$.
None of these five cases are multiples of $p$.
Second, consider the parity of the combination number.
First, examine the power of 2 in $n$ factorial.
Assume the binary representation of $n$ is
$n=\left(a, a_{r-1} \cdots a_{1} a_{0}\right)_{2}$
$\qquad$
The power of 2 in $n$ factorial is
$\left[\frac{n}{2}\right]+\left[\frac{n}{4}\right]+\cdots+\left[\frac{n}{2^{m}}\right]+\cdots$
$=\left(a_{r} a_{t-1} \cdots a_{2} a_{1}\right)_{2}+\left(a a_{1} a_{t-1} \cdots a_{3} a_{2}\right)_{2}+\cdots+a_{t}$
$\begin{aligned}= & a_{r}\left(2^{\prime}-1\right)+a_{r-1}\left(2^{2-1}-1\right) \\ & a_{1}\left(2^{2}-1\right)+a_{0}\left(2^{0}-1\right)\end{aligned}$
$=n-s(n)$,
where $s(n)=a_{r}+a_{t-1}+\cdots+a_{0}$ represents the sum of binary digits, which is the number of 1s.
If the combination number
$\mathrm{C}_{2012}^{k}=\frac{2012!}{k!\cdot(2012-k)!}$
$=\frac{(k+m)!}{k!\cdot m!}(m=2012-k)$
is odd, i.e., $\square$
the power of 2 in the numerator and denominator is the same
$\Rightarrow k+m-s(k+m)=k-s(k)+m-s(m)$
$\Rightarrow s(k+m)=s(k)+s(m)$,
which means the binary addition of $k+m=2012$ does not carry (each carry means changing a $(10)_{2}$ in a certain position to a $(1)_{2}$ in the previous position, reducing the digit sum by 1).
Notice that $2012=(11111011100)_{2}$, consisting of eight 1s and three 0s.
If $k+m=(11111011100)_{2}$ does not carry, then in the 0 positions, $k$ and $m$ must both be 0 (0=0+0), with only one choice; in the 1 positions, $k$ and $m$ are one 0 and one 1 (1=1+0 and 1=0+1), with two choices. Therefore, there are $2^{8}=256$ cases where the binary addition of $k+m=2012$ does not carry, meaning there are 256 combination numbers that are odd. The remaining $2013-256=1757$ are even.
Third, consider the combination number
$\mathrm{C}_{2012}^{k}=\frac{2012!}{k!\cdot(2012-k)!}=\frac{(k+m)!}{k!\cdot m!}$
is even but not a multiple of 4.
Then the power of 2 in the numerator is exactly one more than in the denominator
$\Rightarrow k+m-s(k+m)=k-s(k)+m-s(m)+1$
$\Rightarrow s(k+m)=s(k)+s(m)-1$,
which means the binary addition of $k+m=2012$ has exactly one carry, and this carry is from a lower position to an adjacent higher position. Since the lower position does not carry to the next higher position, the binary addition at the lower position must be $(1)_{2}+(1)_{2}=(10)_{2}$ (otherwise, it cannot carry to the next position), meaning $k$ and $m$ are both 1 at the lower position. The higher position, receiving the carry from the lower position, does not carry further, which can only be $0+0=0$, becoming 1 after receiving the carry, meaning $k$ and $m$ are both 0 at the higher position.
The binary addition at the two positions is
$(01)_{2}+(01)_{2}=(10)_{2}$.
From the binary representation of the sum
$k+m=2012=(11111011100)$
the carry can only be from the fifth and sixth positions or the ninth and tenth positions from the front. In any case, the two positions of the carry are fixed as $(01)_{2}+(01)_{2}=(10)_{2}$. The remaining positions do not carry, 0 positions can only be $0+0=0$, 1 positions can be $1+0=1$ or $0+1=1$, with two choices. There are $2^{7}=128$ cases for each position, totaling 256 cases. That is, there are 256 combination numbers that are even but not multiples of 4.
Thus, the combination numbers that are multiples of 4 are
$2013-256-256=1501$ (cases).
Finally, consider the cases that are not multiples of $p=503$.
$\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$ is not a multiple of 4.
$\mathrm{C}_{s_{p}}^{s}=\mathrm{C}_{4 p}^{3 p}$ has a power of 2 of
$s(p)+s(3 p)-s(4 p)=s(3 p) \geqslant 2$.
$\mathrm{C}_{s p}^{2}$ has a power of 2 of
$s(2 p)+s(2 p)-s(4 p)$
$=s(p)=s(2012)=8$.
These three combination numbers are multiples of 4 but not multiples of $p=503$.
Therefore, the combination numbers that are multiples of 4 and also multiples of $p=503$ are $1501-3=1498$.
In summary, the combination numbers that are multiples of 2012 are
|
1498
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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, there are exactly 11 pieces of clay with different weights on the table. Find the minimum value of $m$.
|
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 each group. After the first operation, there are 37 blocks weighing 1 and 37 blocks weighing 2 on the table.
In the second operation, select 36 blocks weighing 1 and 36 blocks weighing 2, and divide them into 9 groups, with 8 blocks in each group. Specifically, the $i$-th group ($1 \leqslant i \leqslant 9$) contains $i-1$ blocks weighing 2 and $9-i$ blocks weighing 1. After the second operation, there are 11 blocks of clay on the table, with weights
$$
1, 2, 9-i+2(i-1)=7+i \quad (1 \leqslant i \leqslant 9).
$$
This meets the requirement.
|
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
$$
f(x)=(x-1)(x-2)(x-3)(a x+b) \text {, }
$$
where $a, b \in \mathbf{R}$.
Therefore, $\left\{\begin{array}{l}f(4)=6(4 a+b)=6, \\ f(5)=24(5 a+b)=72 .\end{array}\right.$
Solving these, we get $a=2, b=-7$.
Thus, $f(x)=(x-1)(x-2)(x-3)(2 x-7)$.
Hence, $f(2010)$
$$
\begin{array}{l}
=(2010-1)(2010-2)(2010-3)(4020-7) \\
\equiv 2(\bmod 10) .
\end{array}
$$
Therefore, the last digit of $f(2010)$ is 2.
|
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 Mathematics Competition)
|
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-coefficient polynomial $g(x)$. Therefore,
$$
g(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right) q(x) \text {, }
$$
where $q(x)$ is an integer-coefficient polynomial.
Then $g(0)=f(0)-2010$
$$
=11-2010=-1999 \text {. }
$$
Note that 1999 is a prime number, and can be at most the product of three different integers, $-1999=(-1) \times 1 \times 1999$, so $n \leqslant 3$.
When $n=3$,
$$
\begin{array}{l}
g(x)=(x-1)(x+1)(x+1999) . \\
\text { Hence } f(x)=(x-1)(x+1)(x+1999)+2010 .
\end{array}
$$
Therefore, the maximum value of $n$ is 3.
|
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 $A^{\prime}$, $B^{\prime}$, $C^{\prime}$, $S^{\prime}$, $M^{\prime}$, $N^{\prime}$ represent the orthogonal projections of points $A$, $B$, $C$, $S$, $M$, $N$ onto a plane perpendicular to the edge $B C$, as shown in Figure 3. Points $B^{\prime}$ and $C^{\prime}$ coincide, and
$$
\begin{array}{l}
A^{\prime} M^{\prime}=M^{\prime} N^{\prime}=N^{\prime} S^{\prime}=a, A^{\prime} B^{\prime}=h_{1}, \\
M^{\prime} B^{\prime}=h_{2}, N^{\prime} B^{\prime}=h_{3}, S^{\prime} B^{\prime}=h_{4}.
\end{array}
$$
Considering that $M^{\prime} B^{\prime}$ and $N^{\prime} B^{\prime}$ are the medians of $\triangle A^{\prime} B^{\prime} N^{\prime}$ and $\triangle M^{\prime} B^{\prime} S^{\prime}$, respectively, we have
$$
\begin{array}{l}
h_{1}^{2}+h_{3}^{2}-2 h_{2}^{2}=2 a^{2}=h_{2}^{2}+h_{4}^{2}-2 h_{3}^{2} \\
\Leftrightarrow 3 h_{3}^{2}=3 h_{2}^{2}+h_{4}^{2}-h_{1}^{2} \\
\Rightarrow h_{3}=\sqrt{h_{2}^{2}+\frac{h_{4}^{2}-h_{1}^{2}}{3}} \\
\Rightarrow S_{3}=\sqrt{S_{2}^{2}+\frac{S_{4}^{2}-S_{1}^{2}}{3}}=4.
\end{array}
$$
|
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 ratio $k=\frac{B C}{A D}=\frac{1}{3}$.
It is easy to see that $\triangle A K C \backsim \triangle A O K$.
Thus, $A K^{2}=A O \cdot A C$
$$
=A O \cdot \frac{A O+O C}{A O} \cdot A O=\frac{4}{3} A O^{2} \text {, }
$$
Similarly, $D L^{2}=\frac{4}{3} O D^{2}$.
Therefore, $A K^{2}+D L^{2}=\frac{4}{3} A O^{2}+\frac{4}{3} O D^{2}$
$$
=\frac{4}{3} A D^{2}=\frac{4}{3} \times 3^{2}=12 \text {. }
$$
|
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
$$
1007 \leqslant m \leqslant 2012-400=1612 \text {. }
$$
Therefore, the number of positive integer pairs $(m, n)$ that satisfy the condition is
$$
1612-1007+1=606 \text { (pairs). }
$$
|
606
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. Divide the natural numbers from 1 to 30 into two groups, such that the product of all numbers in the first group $A$ is divisible by the product of all numbers in the second group $B$. Then the minimum value of $\frac{A}{B}$ is ( ).
(A) 1077205
(B) 1077207
(C) 1077209
(D) 1077211
|
5. A.
$$
\begin{array}{l}
\text { Given } A B=30 \times 29 \times \cdots \times 1 . \\
=2^{26} \times 3^{14} \times 5^{7} \times 7^{4} \times 11^{2} \times 13^{2} \times \\
17 \times 19 \times 23 \times 29 .
\end{array}
$$
Let $C=2^{13} \times 3^{7} \times 5^{4} \times 7^{2} \times 11 \times 13 \times 17 \times 19 \times 29$.
Since $B \mid A$, we know $C \mid A$.
Thus $\left(\frac{A}{B}\right)_{\min }=5 \times 17 \times 19 \times 23 \times 29=1077205$.
When $A=C=30 \times 29 \times \cdots \times 19 \times 17 \times 12 \times 2$,
$\frac{A}{B}$ reaches its minimum value.
|
1077205
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Given that $P(x)$ is a polynomial with integer coefficients, satisfying $P(17)=10, P(24)=17$. If the equation $P(n)=n+3$ has two distinct integer solutions $n_{1}, n_{2}$, find the value of $n_{1} n_{2}$. ${ }^{[7]}$
(2005, American Invitational Mathematics Examination)
|
【Analysis】From the conditions of the problem, we cannot determine the zeros of the integer-coefficient polynomial $P(x)$. Let's construct another polynomial function $T(x)$ such that 17 and 24 are zeros of $T(x)$, and solve the problem using the zero-product property.
Solution Let $S(x)=P(x)-x-3$. Then
$S(17)=-10, S(24)=-10$.
Let $T(x)=S(x)+10=P(x)-x+7$. Hence, 17 and 24 are zeros of the integer-coefficient polynomial $T(x)$.
Assume $T(x)=(x-17)(x-24) Q(x)$, where $Q(x)$ is an integer-coefficient polynomial.
Let $n$ be an integer root that satisfies the equation $P(n)=n+3$. Then
$$
\begin{array}{l}
S(n)=0, \\
T(n)=(n-17)(n-24) Q(n)=10 .
\end{array}
$$
Therefore, $n-17$ and $n-24$ are factors of 10.
Thus, $\left\{\begin{array}{l}n-17=2, \\ n-24=-5\end{array}\right.$ or $\left\{\begin{array}{l}n-17=5, \\ n-24=-2 .\end{array}\right.$
Solving these, we get $n_{1}=19, n_{2}=22$.
Therefore, $n_{1} n_{2}=19 \times 22=418$.
|
418
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let positive integers $k_{1} \geqslant k_{2} \geqslant \cdots \geqslant k_{n}\left(n \in \mathbf{N}_{+}\right)$, and $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{n}}=2012$.
Then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is $\qquad$
|
2. 49 .
Notice that,
$$
2012=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}+2^{4}+2^{3}+2^{2} \text {. }
$$
Also, $2^{k+1}=2^{k}+2^{k}$, and $k+1 \leqslant 2 k(k \geqslant 1$ when $)$, then the minimum value of $k_{1}+k_{2}+\cdots+k_{n}$ is
$$
10+9+8+7+6+4+3+2=49 \text {. }
$$
|
49
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Group all positive integers that are coprime with 2012 in ascending order, with the $n$-th group containing $2n-1$ numbers:
$$
\{1\},\{3,5,7\},\{9,11,13,15,17\}, \cdots \text {. }
$$
Then 2013 is in the $\qquad$ group.
|
4.32.
Notice that, $2012=2^{2} \times 503$, where $2$ and $503$ are both prime numbers.
Among the positive integers not greater than 2012, there are 1006 multiples of 2, 4 multiples of 503, and 2 multiples of $2 \times 503$. Therefore, the numbers that are not coprime with 2012 are
$$
1006+4-2=1008 \text { (numbers). }
$$
Thus, among the positive integers not greater than 2012, the numbers that are coprime with 2012 are
$$
2012-1008=1004 \text { (numbers). }
$$
Hence, 2013 is in the 1005th position.
Also, $31^{2}<1005<32^{2}$, that is
$$
\begin{array}{l}
1+3+\cdots+(2 \times 31-1)<1005 \\
<1+3+\cdots+(2 \times 32-1) .
\end{array}
$$
Thus, 2013 is in the 32nd group.
|
32
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 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 $n$.
|
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 five-digit numbers are $10016, 10049$, and the four-digit numbers are $1001, 1004, 1009$.
(2) The six-digit number with the leftmost three digits 400 is 400169, the five-digit numbers are 40016, 40049, and the four-digit numbers are $4001, 4004, 4009$.
(3) The six-digit number with the leftmost three digits 900 is 900169, the five-digit numbers are $90016, 90049$, and the four-digit numbers are $9001, 9004, 9009$.
(4) When the leftmost three digits are $144, 196, 225, 484, 625, 784$, the corresponding four-digit numbers are $1441, 1961, 2256, 4841, 6256, 7841$.
(5) The rest are all three-digit numbers. After sorting the above numbers in descending order, we find that $b_{16}=4009, b_{20}=1961$.
Thus, $b_{16}-b_{20}=2048=2^{11}$, which means $n=11$.
|
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
$$
\begin{array}{l}
n=1+2+4+\cdots+2^{2012} \\
=2^{2013}-1 \equiv 191(\bmod 1000) .
\end{array}
$$
|
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 to the sides of the triangles are
$$
\begin{array}{l}
(1,1,2),(1,2,3),(1,3,4),(1,4,4), \\
(2,2,4),(2,3,4),(3,3,3)
\end{array}
$$
assuming the circumference of the circumcircle of the regular nonagon is 9$)$.
If there are at most six distinct non-congruent red triangles, by Turán's theorem, there must be at least four pairs of congruent red triangles; if there are all seven distinct non-congruent red triangles, then particularly, there must be those corresponding to the minor arc lengths $(1,1,2)$ and $(3,3,3)$. Thus, essentially, it must be one of the two scenarios shown in Figure 5, but neither of them contains all seven distinct non-congruent red triangles, leading to a contradiction.
Therefore, there must be at least four pairs of such triangles.
Any of the scenarios in Figure 5 can serve as an example when there are four pairs.
|
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, equation (1) $\Leftrightarrow\left(1+\frac{1}{t_{k}^{2}}\right)\left(t_{k}+1\right)^{2} \geqslant 8$ $\Leftrightarrow t_{k}^{2}+2 t_{k}+2+\frac{2}{t_{k}}+\frac{1}{t_{k}^{2}} \geqslant 8$.
The last inequality is easily proven by the AM-GM inequality.
Returning to the original problem.
By the lemma, we have $\frac{1}{a_{k}^{2}} \geqslant \frac{8}{S_{k}^{2}}-\frac{1}{S_{k-1}^{2}}$.
Thus, $\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant \sum_{k=1}^{n} \frac{7}{S_{k}^{2}}+\frac{1}{S_{n}^{2}}-1$.
Hence, $1+\sum_{k=1}^{n} \frac{1}{a_{k}^{2}} \geqslant 7 \sum_{k=1}^{n} \frac{1}{S_{k}^{2}}$.
Therefore, when $\lambda=7$, the inequality holds.
Next, we show that $\lambda=7$ is the best possible.
Let $a_{k}=2^{k-1}$, and let $n \rightarrow+\infty$, it is easy to see that $\lambda \leqslant 7$.
In conclusion, $\lambda_{\text {max }}=7$.
|
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 $n=3$, $22=2^{1} \times 11^{1}$, $23=23^{1}$, $24=2^{3} \times 3^{1}$, then $22, 23, 24$ form a consecutive 3 odd group). The maximum possible value of $n$ in a consecutive $n$ odd group is $\qquad$ [1]
|
【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 both numbers that are congruent to 4 modulo 8.
Therefore, the power of 2 in their prime factorizations is even.
Thus, a connected $n$-singular group cannot contain two multiples of 4 simultaneously. However, in any 8 consecutive positive integers, there must be a number that is congruent to 0 modulo 8 and a number that is congruent to 4 modulo 8. Therefore, in any 8 consecutive positive integers, there must be two multiples of 4.
Hence, $n \leqslant 7$.
Furthermore, $29,30, \cdots, 35$ is a connected 7-singular group, so $n_{\max }=7$.
|
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 $a$ and $b$ must be even.
Since $a$ and $b$ are both prime numbers, by symmetry, we can assume $a=2$ first.
Then $2^{b} b^{2}=(b+5)(2 b+3)$.
Clearly, $b \neq 2$, so $b \geqslant 3$.
If $b>3$, then $b \geqslant 5$. Thus,
$$
2^{b} b^{2} \geqslant 2^{5} b^{2}=4 b \times 8 b>(b+5)(2 b+3) \text {. }
$$
This contradicts equation (2).
When $b=3$, equation (2) holds.
Therefore, $a=2, b=3$.
By symmetry, $a=3, b=2$ also satisfies the condition.
In summary, the number of prime pairs is 2.
|
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 $7200=2^{5} \times 3^{2} \times 5^{2}$, and both $a$ and $b$ are odd, $9 a+8 b$ is odd, meaning that the factor $2^{5}$ in the denominator remains after simplification.
Because $(b, 3)=1$, we have $(9 a+8 b, 3)=1$, meaning that the factor $3^{2}$ in the denominator remains after simplification.
Thus, when $9 a+8 b$ is a multiple of 25, the denominator of the simplified fraction $\frac{9 a+8 b}{7200}$ reaches its minimum value of 288.
A specific example can be constructed as follows.
$$
\begin{array}{l}
9 a+8 b \equiv 0(\bmod 25) \\
\Leftrightarrow 8 b \equiv 16 a(\bmod 25) \\
\Leftrightarrow b \equiv 2 a(\bmod 25) .
\end{array}
$$
Taking $a=1, b=77$ works.
|
288
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 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 $x^{2}+7=\sum_{i=1}^{4}\left(f_{i}(x)\right)^{2}$, and the degree of the polynomial $f_{i}(x)$ is at most 1.
Let $f_{i}(x)=a_{i} x+b_{i}$. Then
$$
\sum_{i=1}^{4} a_{i}^{2}=1, \sum_{i=1}^{4} a_{i} b_{i}=0, \sum_{i=1}^{4} b_{i}^{2}=7 .
$$
Let $p_{i}=a_{i}+b_{i}$, $q_{i}=a_{i}-b_{i}$. Then
$$
\begin{array}{l}
\sum_{i=1}^{4} p_{i}^{2}=\sum_{i=1}^{4}\left(a_{i}^{2}+2 a_{i} b_{i}+b_{i}^{2}\right)=8, \\
\sum_{i=1}^{4} q_{i}^{2}=\sum_{i=1}^{4}\left(a_{i}^{2}-2 a_{i} b_{i}+b_{i}^{2}\right)=8, \\
\sum_{i=1}^{4} p_{i} q_{i}=\sum_{i=1}^{4}\left(a_{i}^{2}-b_{i}^{2}\right)=-6 .
\end{array}
$$
Let the least common multiple of the denominators of the rational numbers $p_{i}$, $q_{i}$ ($i=1,2,3,4$) be $m$. Then there exist integers $x_{i}$, $y_{i}$ (after converting $p_{i}$, $q_{i}$ to a common denominator $m$, the numerators of $p_{i}$, $q_{i}$), such that
$$
\left\{\begin{array}{l}
\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}, \\
\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}, \\
\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2} .
\end{array}\right.
$$
Assume $m$ is the smallest positive integer satisfying the above system of indeterminate equations.
$$
\text { From } \sum_{i=1}^{4} x_{i}^{2}=8 m^{2} \equiv 0(\bmod 8) \text {, we know } x_{1} \text {, } x_{2} \text {, } x_{3} \text {, }
$$
$x_{4}$ are all even.
$$
\text { From } \sum_{i=1}^{4} y_{i}^{2}=8 m^{2} \equiv 0(\bmod 8) \text {, we know } y_{1} \text {, } y_{2} \text {, } y_{3} \text {, }
$$
$y_{4}$ are all even.
From $\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2} \equiv 0(\bmod 4)$, we know $m$ is also even.
Thus, $\frac{x_{i}}{2}$, $\frac{y_{i}}{2}$, $\frac{m}{2}$ also satisfy the system of indeterminate equations.
This contradicts the minimality of $m$.
|
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 $x=\frac{p}{q}\left((p, q)=1, p, q \in \mathbf{Z}_{+}\right)$.
Then $\left(\frac{p}{q}\right)^{5}=\frac{a^{2}+7}{4}$ can only have $q=1$, i.e., $x$ is a positive:
integer.
Obviously, $x \geqslant 2$.
When $x=2$,
$$
\begin{array}{l}
4 x^{5}-7=121=11^{2}, \\
4 x^{13}-7=32761=181^{2}
\end{array}
$$
satisfies the conditions.
When $x$ is an odd number,
$$
a^{2}=4 x^{5}-7 \equiv 5(\bmod 8) \text {, }
$$
which is not valid.
Assume $x$ is a positive even number below.
When $x \geqslant 4$,
$$
\begin{array}{l}
(a b)^{2}=\left(4 x^{5}-7\right)\left(4 x^{13}-7\right) \\
=16 x^{18}-28 x^{13}-28 x^{7}+49 .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
16 x^{18}-28 x^{13}-28 x^{7}+49 \\
\left(4 x^{9}-\frac{7}{2} x^{4}-1\right)^{2},
\end{array}
$$
thus $x \geqslant 4$ and being even is not valid.
In summary, only $x=2$ satisfies the conditions of the problem.
|
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 only need to find $(a+b)_{\text {min }}$.
Also, $a+b \geqslant 2 \sqrt{a b}=2 \sqrt{k+17}$
$\geqslant 2 \sqrt{18}>2 \sqrt{16}=8$.
Thus, $a+b>8$, i.e., $a+b \geqslant 9$.
If $a+b=9$, then $a=2, b=7$.
This contradicts $a b=k+17>17$.
Therefore, $a+b \geqslant 10$.
When $k=4$, $a=3, b=7$, the equality holds.
Thus, $(a+b+c+d)_{\min }=22$.
|
22
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, on a plane there are $n(n \geqslant 4)$ lines. For lines $a$ and $b$, among the remaining $n-2$ lines, if at least two lines intersect with both lines $a$ and $b$, then lines $a$ and $b$ are called a "congruent line pair"; otherwise, they are called a "separated line pair". If the number of congruent line pairs among the $n$ lines is 2012 more than the number of separated line pairs, find the minimum possible value of $n$ (the order of the lines in a pair does not matter).
|
(1) Among these $n$ lines, if there exist four lines that are pairwise non-parallel, then any two lines are coincident line pairs. However, $\mathrm{C}_{n}^{2}=2012$ has no integer solution, so there does not exist an $n$ that satisfies the condition.
(2) If the $n$ lines have only three different directions, let the number of lines in these three directions be $a$, $b$, and $c$ respectively. Assume $a \geqslant b \geqslant c$.
(i) When $a \geqslant 2, b=c=1$,
$$
\mathrm{C}_{a}^{2}+1-2 a=2012,
$$
there is no $n$ that satisfies the condition.
(ii) When $a, b \geqslant 2, c=1$,
$$
\mathrm{C}_{a}^{2}+\mathrm{C}_{b}^{2}-a b+b+a=2012,
$$
which simplifies to $(a-b)^{2}+a+b=4024$.
Let $a-b=k$. Then $a+b=4024-k^{2}$.
Thus, $a=\frac{4024+k-k^{2}}{2}, b=\frac{4024-k-k^{2}}{2}$.
Since $b \geqslant 2$, it is easy to see that $k \leqslant 62$.
Therefore, the minimum value of $a+b$ is 180.
Thus, the minimum value of $n$ is 181.
(iii) When $a, b, c \geqslant 2$,
$$
\mathrm{C}_{a}^{2}+\mathrm{C}_{b}^{2}+\mathrm{C}_{c}^{2}+a b+b c+c a=2012,
$$
which simplifies to $(a+b+c)(a+b+c-1)=4024$,
and there is no integer solution.
(3) If the $n$ lines have only two different directions, let the number of lines in these two directions be $a$ and $b$ respectively. Clearly, $a, b \geqslant 2$.
Then $\mathrm{C}_{a}^{2}+\mathrm{C}_{b}^{2}-a b=2012$, which simplifies to
$$
(a-b)^{2}-(a+b)=4024.
$$
Let $a-b=k$. Then $a+b=k^{2}-4024$.
Thus, $a=\frac{k+k^{2}-4024}{2}, b=\frac{k^{2}-k-4024}{2}$.
Since $b \geqslant 2$, it is easy to see that $k \geqslant 64$.
Therefore, the minimum value of $a+b$ is 72.
Thus, the minimum value of $n$ is 72.
In summary, the minimum value of $n$ is 72.
|
72
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 $f(0)=f(\pi)=f\left(\frac{\pi}{3}\right)=1$, i.e.,
$$
a=1, b+d=1, c=-1
$$
In this case, let $t=\cos x(-1 \leqslant t \leqslant 1)$. Then
$$
\begin{array}{l}
f(x)-1 \\
= \cos x+b \cos 2 x-\cos 3 x+d \cos 4 x-1 \\
= t+(1-d)\left(2 t^{2}-1\right)-\left(4 t^{3}-3 t\right)+ \\
d\left(8 t^{4}-8 t^{2}+1\right)-1 \\
= 2(t-1)(t+1)(2 t-1)[2 d t-(1-d)] \\
\leqslant 0
\end{array}
$$
for any real number $t \in[-1,1]$.
Thus, $d>0$, and $\frac{2 d}{2}=\frac{1-d}{1}$, i.e., $d=\frac{1}{2}$.
Therefore, the maximum value of $a+b-c+d$ is 3, and at this time
$$
(a, b, c, d)=\left(1, \frac{1}{2},-1, \frac{1}{2}\right) \text {. }
$$
|
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 (1) we know $p \mid(a+k)$ or $p \mid(a-k)$.
Clearly, $a-k \Rightarrow p \nmid(a-k) \Rightarrow p \mid(a+k)$. Let $a+k=np\left(n \in \mathbf{N}_{+}\right)$.
From $a-kp \Rightarrow n \geqslant 2 \Rightarrow k=np-a \text {. }$
$$
Substituting into equation (1) we get
$$
\begin{array}{l}
n(2a-np)=a \\
\Rightarrow a(2n-1)=n^{2}p .
\end{array}
$$
It is easy to prove that $n^{2}$ and $2n-1$ are coprime.
Thus, $n^{2} \mid a$. Let $a=mn^{2}\left(m \in \mathbf{N}_{+}\right)$.
From equation (2) we get $m(2n-1)=p$.
Since $p$ is a prime number, either $m$ or $2n-1$ must be 1.
Also, $2n-1 \geqslant 2 \times 2-1>1$, so $m=1$.
Therefore, $a=n^{2}, p=2n-1$.
Hence, $a$ meets the condition $\Leftrightarrow a=n^{2}$ and $2n-1$ is a prime number.
Given $a \geqslant 2012$, then $a \geqslant 45^{2}$.
Also, $2 \times 45-1=89$ is a prime number, thus,
$$
a_{\min }=2025
$$
|
2025
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. If a non-negative integer $m$ and the sum of its digits are both multiples of 6, then $m$ is called a "Lucky Six Number". Find the number of Lucky Six Numbers among the non-negative integers less than 2012.
|
5. Solution 1 It is easy to know that a non-negative integer is a hexagonal number if and only if its last digit is even and the sum of its digits is a multiple of 6.
For convenience, let
$$
M=\{0,1, \cdots, 2011\}
$$
write each number in $M$ as a four-digit number $\overline{a b c d}$ (when it is less than four digits, add several "0"s in front of the highest digit to make it exactly four digits), and use $f(k)$ to denote the number of hexagonal numbers in $M$ whose last digit is $k$, where $k \in\{0,2,4,6,8\}$.
For $n \in \mathbf{N}$, the number of pairs $(x, y)$ that satisfy $x+y=n$, and
$$
x, y \in\{0,1, \cdots, 9\}
$$
is denoted as $p_{n}$.
Obviously, $p_{n}=\left\{\begin{array}{ll}n+1, & n=0,1, \cdots, 9 ; \\ 19-n, & n=10,11, \cdots, 18 ; \\ 0, & n \geqslant 19 .\end{array}\right.$
First, consider all hexagonal numbers $\overline{a b c k}$ less than 2000.
If $k=0$, then when $a=0$, $b+c=0,6,12,18$; when $a=1$, $b+c=5,11,17$.
$$
\begin{array}{l}
\text { Hence } f(0)=\left(p_{0}+p_{6}+p_{12}+p_{18}\right)+\left(p_{5}+p_{11}+p_{17}\right) \\
=16+16=32 .
\end{array}
$$
If $k=2$, then when $a=0$, $b+c=4,10,16$; when $a=1$, $b+c=3,9,15$.
$$
\begin{array}{l}
\text { Hence } f(2)=\left(p_{4}+p_{10}+p_{16}\right)+\left(p_{3}+p_{9}+p_{15}\right) \\
=17+18=35 .
\end{array}
$$
If $k=4$, then when $a=0$, $b+c=2,8,14$; when $a=1$, $b+c=1,7,13$.
Hence $f(4)=\left(p_{2}+p_{8}+p_{14}\right)+\left(p_{1}+p_{7}+p_{13}\right)$
$$
=17+16=33 \text {. }
$$
When $k=6,8$, similar to the cases of $k=0,2$, we have
$$
f(6)=f(0)=32, f(8)=f(2)=35 \text {. }
$$
Therefore, the number of hexagonal numbers less than 2000 is
$$
f(0)+f(2)+f(4)+f(6)+f(8)=167 \text { (numbers). }
$$
Notice that, from 2000 to 2011, there is exactly one hexagonal number 2004. Thus, the total number of hexagonal numbers is
$$
167+1=168 \text {. }
$$
Solution 2 For a non-negative integer $n$, let $S(n)$ be the sum of its digits.
First, pair all multiples of 6 (a total of 334) among the non-negative integers less than 2000 into the following 167 pairs:
$$
\begin{array}{l}
(0,1998),(6,1992),(12,1986), \\
\cdots,(996,1002) . \\
\text { For each pair of numbers }(x, y), \text { let } \\
x=\overline{a_{1} a_{2} a_{3} a_{4}, y=b_{1} b_{2} b_{3} b_{4}}
\end{array}
$$
(Agree that when $x$ or $y$ is less than four digits, add several "0"s in front of the highest digit to make it exactly four digits). Then
$$
\begin{array}{l}
1000\left(a_{1}+b_{1}\right)+100\left(a_{2}+b_{2}\right)+ \\
10\left(a_{3}+b_{3}\right)+\left(a_{4}+b_{4}\right) \\
=x+y=1998 .
\end{array}
$$
Since $x, y$ are even, $a_{4}, b_{4} \leqslant 8$.
Therefore, $a_{4}+b_{4} \leqslant 16<18$, so $a_{4}+b_{4}=8$.
Also, since $a_{3}+b_{3} \leqslant 18<19$, we have $a_{3}+b_{3}=9$.
Similarly, $a_{2}+b_{2}=9$.
Finally, we must have $a_{1}+b_{1}=1$.
Thus, $S(x)+S(y)$
$$
\begin{array}{l}
=\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\left(a_{3}+b_{3}\right)+\left(a_{4}+b_{4}\right) \\
=1+9+9+8=27 .
\end{array}
$$
Therefore, one and only one of $S(x)$ and $S(y)$ is a multiple of 6 (because $x, y$ are both divisible by 3, so $S(x)$ and $S(y)$ are both divisible by 3). Hence, one and only one of $x, y$ is a hexagonal number.
So, the number of hexagonal numbers less than 2000 is 167.
Also, from 2000 to 2011, there is exactly one hexagonal number 2004, so the total number of hexagonal numbers is $167+1=168$.
|
168
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Find the smallest positive integer $n$ such that
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}} \\
<\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$
|
6. From the known, we must have $n \geqslant 2$ 013. At this time,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}4023, \\
\sqrt[3]{\frac{n-2013}{2011}} \geqslant \sqrt[3]{\frac{n-2011}{2013}} \\
\Leftrightarrow 2013(n-2013) \geqslant 2011(n-2011) \\
\Leftrightarrow n \geqslant 4024 .
\end{array}
$$
From equations (1) and (2), when $n \geqslant 4024$,
$$
\begin{array}{l}
\sqrt{\frac{n-2011}{2012}}-\sqrt{\frac{n-2012}{2011}}\sqrt[3]{\frac{n-2013}{2011}}-\sqrt[3]{\frac{n-2011}{2013}} .
\end{array}
$$
In summary, the smallest positive integer $n$ that satisfies the condition is
4024.
|
4024
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 $x$ to point $y$ in 2012 steps (allowing intermediate visits to points $x, y$) is even.
|
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 \geqslant n=2^{m}-1$ and any $x, y \in P_{n}$, the number of ways to jump from point $x$ to point $y$ in $k$ steps is even.
When $m=1$, the number of ways to jump is necessarily 0, so the conclusion holds.
Assume that the conclusion holds for $m=l$.
For $k \geqslant n=2^{l+1}-1$, we divide the paths from point $x$ to point $y$ in $k$ steps into three categories, and we only need to prove that the number of paths in each case is even.
(1) The path never passes through point $2^{\prime}$.
In this case, points $x$ and $y$ are on the same side of point $2^{l}$, and by the induction hypothesis, the number of paths is even.
(2) The path passes through point $2^{l}$ exactly once.
Let the $i$-th step ($i \in\{0,1, \cdots, k\}$) be the step that reaches point $2^{l}$, where $i=0$ indicates that point $x$ is point $2^{l}$, and $i=k$ indicates that point $y$ is point $2^{l}$. We need to prove that for any $i$, the number of paths is even.
Let the path be
$x, a_{1}, \cdots, a_{i-1}, 2^{l}, a_{i+1}, \cdots, a_{k-1}, y$.
We divide this path into two sub-paths: from point $x$ to $a_{i-1}$, which is $i-1$ steps; and from point $a_{i+1}$ to $y$, which is $k-i-1$ steps (for $i=0$ or $k$, there is only one sub-path, which is $k-1$ steps).
Since $k \geqslant n=2^{l+1}-1$, if $i-1<2^{l}-1$ and $k-i-1<2^{l}-1$, then
$i-1 \leqslant 2^{l}-2$ and $k-i-1 \leqslant 2^{l}-2$.
Adding these, we get $k \leqslant 2^{l+1}-2$, which is a contradiction.
Therefore, either $i-1 \geqslant 2^{l}-1$ or $k-i-1 \geqslant 2^{l}-1$ must hold.
By the induction hypothesis, the number of paths for at least one of the sub-paths is even.
By the multiplication principle, the number of paths that reach point $2^{l}$ at the $i$-th step is even.
(3) The path passes through point $2^{l}$ at least twice.
In this case, we pair the paths between the first and second visits to point $2^{l}$ by reflecting them along point $2^{l}$, so the number of such paths is necessarily even.
By mathematical induction, the proposition is proved.
In summary, the maximum value of $m$ is 10.
|
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 $k \neq 8, k \neq 9$.
Also, $2010=2^{2}+3^{2}+7^{2}+11^{2}+13^{2}+17^{2}+37^{2}$, hence $k_{\max }=7$.
|
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
$$
\begin{array}{l}
(16 a+5 a+4 a)\left[\frac{(4 x-4)^{2}}{16 a}+\frac{(-\sqrt{5} y+5)^{2}}{5 a}+\frac{(2 z+2)^{2}}{4 a}\right] \\
\geqslant(4 x-4-\sqrt{5} y+5+2 z+2)^{2} \\
\Rightarrow 25 a \geqslant(k+3)^{2} \\
\Rightarrow|k+3| \leqslant 5 \sqrt{a} .
\end{array}
$$
Since $-8 \leqslant k \leqslant 2$, we have
$$
-5 \leqslant k+3 \leqslant 5 \Rightarrow|k+3| \leqslant 5 \text {. }
$$
Therefore, $5 \sqrt{a}=5 \Rightarrow a=1$.
|
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 without loss of generality that $\tan \gamma \geqslant 1$. Then
$$
0<\tan \alpha \cdot \tan \beta \leqslant 1 \text {. }
$$
Since $\tan \alpha \cdot \tan \beta=\frac{1}{\tan \gamma}$, we have
$$
\begin{array}{l}
x+y+z=\sqrt{2}(\cos \alpha+\cos \beta+\cos \gamma) \\
=\sqrt{2}\left(\frac{1}{\sqrt{1+\tan ^{2} \alpha}}+\frac{1}{\sqrt{1+\tan ^{2} \beta}}+\frac{1}{\sqrt{1+\tan ^{2} \gamma}}\right) \\
=\sqrt{2}\left(\frac{\sqrt{1+\tan ^{2} \alpha}+\sqrt{1+\tan ^{2} \beta}}{\sqrt{1+\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \alpha \cdot \tan ^{2} \beta}}+\frac{1}{\sqrt{1+\tan ^{2} \gamma}}\right) .
\end{array}
$$
Since $\tan ^{2} \alpha+\tan ^{2} \beta \geqslant 2 \tan \alpha \cdot \tan \beta$,
$$
1+\tan ^{2} \gamma \geqslant \frac{1}{2}(1+\tan \gamma)^{2} \text {, }
$$
thus $x+y+z$
$$
\begin{array}{l}
\leqslant \sqrt{2}\left[\sqrt{1+\frac{2}{1+\tan \alpha \cdot \tan \beta}+\frac{1-\tan ^{2} \alpha \cdot \tan ^{2} \beta}{(1+\tan \alpha \cdot \tan \beta)^{2}}}+\frac{\sqrt{2}}{1+\tan \gamma}\right] \\
=\sqrt{2}\left(\sqrt{\frac{4}{1+\tan \alpha \cdot \tan \beta}}+\frac{\sqrt{2}}{1+\tan \gamma}\right) \\
=2\left(\sqrt{\frac{2}{1+\tan \alpha \cdot \tan \beta}}-\frac{1}{1+\tan \alpha \cdot \tan \beta}+1\right) \\
=-2\left(\frac{1}{\sqrt{1+\tan \alpha \cdot \tan \beta}}-\frac{\sqrt{2}}{2}\right)^{2}+3 \leqslant 3 .
\end{array}
$$
Equality holds if and only if $\alpha=\beta=\gamma=45^{\circ}$, i.e., $x=y=z=1$, in which case $(x+y+z)_{\text {max }}=3$.
|
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 (1) and (3), we get
$$
\frac{1}{a} \geqslant 3 \sqrt{3}.
$$
And $3\left(v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}\right) \leqslant\left(v_{1}+v_{2}+v_{3}\right)^{2}$
$$
\Rightarrow 3 \cdot \frac{b}{a} \leqslant \frac{1}{a^{2}} \Rightarrow 3 a b \leqslant 1.
$$
Thus, $P=\frac{5 a^{2}-6 a b+3}{a^{3}(b-a)} \geqslant \frac{5 a^{2}+1}{a^{3}(b-a)}$.
Also, $\left(v_{2}+v_{3}-v_{1}\right)\left(v_{3}+v_{1}-v_{2}\right)\left(v_{1}+v_{2}-v_{3}\right)$
$\leqslant v_{1} v_{2} v_{3}$
$\Rightarrow\left(\frac{1}{a}-2 v_{1}\right)\left(\frac{1}{a}-2 v_{2}\right)\left(\frac{1}{a}-2 v_{3}\right) \leqslant \frac{1}{a}$
$\Rightarrow 9 a^{2}-4 a b+1 \geqslant 0$
$\Rightarrow 5 a^{2}+1 \geqslant 4 a(b-a)$.
Then $P \geqslant \frac{4 a(b-a)}{a^{3}(b-a)}=\frac{4}{a^{2}} \geqslant 108$.
Therefore, when $a=\frac{\sqrt{3}}{9}, b=\sqrt{3}$, $P$ achieves its minimum value of 108.
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 As shown in Figure 3, in rectangle $A B C D$, $A B=20$, $B C=10$. If points $M$ and $N$ are taken on $A C$ and $A B$ respectively, such that the value of $B M+M N$ is minimized, find this minimum value.
|
Solve as shown in Figure 4, construct the symmetric point $B'$ of point $B$ with respect to $AC$, and connect $AB'$. Construct $B'N \perp AB$, intersecting $AC$ at point $M$, and connect $BM$. Then $BM=B'M$.
Therefore, $BM+MN=B'M+MN$.
By the shortest distance of a perpendicular segment, we know that $B'M+MN$ is minimized, so $BM+MN$ also achieves its minimum value.
It is easy to see that $\angle BAC=\angle PAC, \angle BAC=\angle ACP$.
Thus, $\angle PAC=\angle ACP$. Therefore, $PA=PC$.
Let $PA=PC=x$. Then $PD=20-x$.
In the right triangle $\triangle ADP$, by $AD^2+PD^2=AP^2$, we get $10^2+(20-x)^2=x^2$.
Solving for $x$ yields $x=12.5$.
Thus, $PD=7.5$.
It is easy to see that $\triangle ADP \sim \triangle B'AN$.
By $\frac{B'N}{AD}=\frac{AN}{PD}=\frac{AB'}{AP}=\frac{AB}{AP}$, we get $B'N=16$, which is the minimum value of $BM+MN$. At this time, $AN=12$.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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 minimum points the team needs are $\qquad$
|
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 other three teams each with 2 points. Therefore, scoring 10 points does not guarantee a place in the top four.
If a team scores 11 points but is in fifth place, then the total points of the top five teams is at least 55. Thus, the total points of the bottom three teams is at most 1. However, the total points from the three matches among these three teams should be 6, leading to a contradiction.
Therefore, scoring 11 points definitely ensures a place in the top four.
|
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 $11 \times 13 \times 17 \times 19$ is 9. Similarly, the unit digit of $21 \times 23 \times 27 \times 29, \cdots, 91 \times 93 \times 97 \times 99$ is also 9. Therefore, the unit digit of the product of all these numbers is 9.
Thus, to make the unit digit of the product of all remaining numbers on the blackboard 1, at least one more number should be erased.
Furthermore, by erasing all even numbers between $10 \sim 99$ and numbers with a unit digit of 5, as well as one number with a unit digit of 9, such as 19, the unit digit of the product of the remaining numbers can be 1.
Since there are 45 even numbers between $10 \sim 99$ and 9 numbers with a unit digit of 5, the minimum value of $n$ is
$$
45+9+1=55 \text {. }
$$
(Li Changyong provided)
|
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 $B\left(0, x_{0}+\frac{2}{x_{0}}\right)$, then $\overrightarrow{P A}=\left(\frac{1}{x_{0}},-\frac{1}{x_{0}}\right), \overrightarrow{P B}=\left(-x_{0}, 0\right)$. Therefore, $\overrightarrow{P A} \cdot \overrightarrow{P B}=\frac{1}{x_{0}}\left(-x_{0}\right)=-1$.
Solution 2 As shown in Figure 3, let $P\left(x_{0}, x_{0}+\frac{2}{x_{0}}\right)\left(x_{0}>0\right)$.
Figure 3
Then the distances from point $P$ to the line $x-y=0$ and the $y$-axis are respectively
$$
|P A|=\frac{\left|x_{0}-\left(x_{0}+\frac{2}{x_{0}}\right)\right|}{\sqrt{2}}=\frac{\sqrt{2}}{x_{0}},|P B|=x_{0} \text {. }
$$
Since $O, A, P, B$ are concyclic, we have
$$
\angle A P B=\pi-\angle A O B=\frac{3 \pi}{4} \text {. }
$$
Therefore, $\overrightarrow{P A} \cdot \overrightarrow{P B}=|\overrightarrow{P A}||\overrightarrow{P B}| \cos \frac{3 \pi}{4}=-1$.
|
-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
$$
\begin{array}{l}
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \\
\Rightarrow a^{2}-b^{2}=\frac{3}{5} c^{2} . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}} \\
=\frac{c^{2}+a^{2}-b^{2}}{c^{2}+b^{2}-a^{2}}=4 .
\end{array}
$$
Solution 2 Draw $C D \perp A B$ from point $C$, with the foot of the perpendicular at $D$. Then
$a \cos B=D B$,
$b \cos A=A D$.
From the given, we have
$D B-A D=\frac{3}{5} c$.
Figure 4
Also, $D B+D A=c$,
Solving these equations, we get
$$
\begin{array}{l}
A D=\frac{1}{5} c, D B=\frac{4}{5} c . \\
\text { Therefore, } \frac{\tan A}{\tan B}=\frac{\frac{C D}{A D}}{\frac{C D}{D B}}=\frac{D B}{A D}=4 .
\end{array}
$$
Solution 3 From the projection theorem, we have $a \cos B+b \cos A=c$.
Also, $a \cos B-b \cos A=\frac{3}{5} c$. Solving these equations, we get $a \cos B=\frac{4}{5} c, b \cos A=\frac{1}{5} c$.
Therefore, $\frac{\tan A}{\tan B}=\frac{\sin A \cdot \cos B}{\sin B \cdot \cos A}=\frac{a \cos B}{b \cos A}=4$.
|
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 $\frac{|M N|}{|A B|}$ is $\qquad$.
|
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, $\frac{|A F|+|B F|}{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}}$, which means
$$
\begin{array}{l}
\frac{|A F|+|B F|}{|A B|}=\frac{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}{\sin \frac{\pi}{3}} \\
=2 \cos \left(\theta-\frac{\pi}{3}\right) .
\end{array}
$$
As shown in Figure 5, by the definition of a parabola and the midline theorem of a trapezoid, we get
$$
\begin{array}{l}
|M N| \\
=\frac{|A F|+|B F|}{2} .
\end{array}
$$
Thus, $\frac{|M N|}{|A B|}$
$$
=\cos \left(\theta-\frac{\pi}{3}\right) \text {. }
$$
Therefore, when $\theta=\frac{\pi}{3}$, $\frac{|M N|}{|A B|}$ reaches its maximum value of 1.
Solution 2 Similarly to Solution 1, we obtain equation (1).
In $\triangle A F B$, by the Law of Cosines, we have
$$
\begin{array}{l}
|A B|^{2}=|A F|^{2}+|B F|^{2}-2|A F||B F| \cos \frac{\pi}{3} \\
=(|A F|+|B F|)^{2}-3|A F||B F| \\
\geqslant(|A F|+|B F|)^{2}-3\left(\frac{|A F|+|B F|}{2}\right)^{2} \\
=\left(\frac{|A F|+|B F|}{2}\right)^{2}=|M N|^{2} .
\end{array}
$$
Equality holds if and only if $|A F|=|B F|$.
Therefore, the maximum value of $\frac{|M N|}{|A B|}$ is 1.
|
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 . $\qquad$
|
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 $A B$, and $C M \perp A B$.
It is easy to see that $\angle P M H$ and $\angle Q M H$ are the plane angles of the dihedral angles formed by the lateral faces and the base of the regular tetrahedrons $P-A B C$ and $Q-A B C$, respectively.
$$
\begin{array}{l}
\text { Then } \angle P M H=45^{\circ} \Rightarrow P H=M H=\frac{1}{2} A H \text {. } \\
\text { By } \angle P A Q=90^{\circ}, A H \perp P Q \\
\Rightarrow A H^{2}=P H \cdot Q H \\
\Rightarrow A H^{2}=\frac{1}{2} A H \cdot Q H \\
\Rightarrow Q H=2 A H=4 M H \text {. } \\
\text { Therefore } \tan \angle Q M H=\frac{Q H}{M H}=4 .
\end{array}
$$
|
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 $n$ that satisfy $\frac{1}{4}<\sin \frac{\pi}{n}<\frac{1}{3}$ are $10, 11, 12$, and their sum is 33.
|
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 $\square$
$$
\begin{aligned}
N- & (1002+1003+\cdots+2005) \\
= & \left(\sqrt{1002^{2}+2}-1002\right)+ \\
& \left(\sqrt{1003^{2}+2}-1003\right)+\cdots+ \\
& \left(\sqrt{2005^{2}+2}-2005\right) \\
< & \frac{1}{1002}+\frac{1}{1003}+\cdots+\frac{1}{1999}+\frac{1}{2000}+\cdots+\frac{1}{2005} \\
< & \frac{998}{1002}+\frac{6}{2000}<\frac{998}{1002}+\frac{4}{1002}=1 .
\end{aligned}
$$
Therefore, the integer part of $N$ is
$$
1002+1003+\cdots+2005=1509514 .
$$
|
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 $a_{5}=1, a_{8}=2$, then $a_{6}=$ 3; then take $a_{1}=4$, then $a_{2}=5, a_{3}=8$; finally take $a_{10}=$ 7, then $a_{9}=9, a_{7}=12$.
$$
\text { Therefore, }\left(a_{4}\right)_{\min }=8+12=20 \text {. }
$$
|
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 points are changed to 2012 points, how will the situation change?
|
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,
$$
f(2 n+1)=2 n+1, f(2 n)=2 n-1 \text {. }
$$
Let the $2 n+1$ points be $A_{0}, A_{1}, \cdots, A_{2 n}$.
Since each point must connect to $2 n$ lines, and these lines must be of different colors, at least $2 n$ colors are needed.
Next, we prove that $2 n$ colors are insufficient. If there are only $2 n$ colors in total, then each point must be an endpoint of lines of each color exactly once. Suppose there are $k$ red lines, which have a total of $2 k$ distinct endpoints, then $2 k=2 n+1$, which is a contradiction.
Therefore, $f(2 n+1)>2 n$, i.e.,
$$
f(2 n+1) \geqslant 2 n+1 \text {. }
$$
Next, we use a number-theoretic construction method to show that the minimum value $2 n+1$ can be achieved.
Let $S_{0}, S_{1}, \cdots, S_{2 n}$ represent these $2 n+1$ colors, and let $\bar{x}$ denote the smallest non-negative remainder of the integer $x$ modulo $2 n+1$, i.e., $\bar{x} \in\{0,1, \cdots, 2 n\}$. For any two points $A_{i}$ and $A_{j}$ $(i \neq j)$, color the line $A_{i} A_{j}$ with the color $S_{\overline{i+j}}$. Thus, for any point $A_{k}$, any two lines extending from $A_{k}$ are of different colors. In fact, if $A_{k} A_{i}$ and $A_{k} A_{j}$ are the same color, then
$$
\begin{array}{l}
\overline{k+i}=\overline{k+j} \\
\Rightarrow k+i \equiv k+j(\bmod (2 n+1)) \\
\Rightarrow i \equiv j(\bmod (2 n+1)) .
\end{array}
$$
Since $i, j \in\{0,1, \cdots, 2 n\}$, it follows that $i=j$, which is a contradiction.
Thus, this coloring scheme meets the conditions.
Therefore, $f(2 n+1)=2 n+1$.
For $2 n+2$ points $A_{0}, A_{1}, \cdots, A_{2 n+1}$, since each point must connect to the other $2 n+1$ points with $2 n+1$ lines, and these lines must be of different colors, at least $2 n+1$ colors are needed. We need to prove that $2 n+1$ colors are sufficient.
**Construction**: Still use $S_{0}, S_{1}, \cdots, S_{2 n}$ to represent these $2 n+1$ colors, and temporarily ignore point $A_{2 n+1}$. First, color the lines between the first $2 n+1$ points $A_{0}, A_{1}, \cdots, A_{2 n}$ according to the coloring scheme for $2 n+1$ points.
Notice that for any two points $A_{i}$ and $A_{j}$ $(i \neq j)$, the line $A_{i} A_{j}$ is colored such that $i \neq j$, meaning the color codes $\overline{i+j}$ $(j \in\{0,1, \cdots, 2 n\} \backslash\{i\})$ of the $2 n$ lines extending from point $A_{i}$ form a complete residue system modulo $2 n+1$ except for one residue $\overline{i+i}=\overline{2 i}$.
Now, color the line $A_{2 n+1} A_{i}$ $(i=0,1, \cdots, 2 n)$ from point $A_{2 n+1}$ to any of the first $2 n+1$ points with the color $S_{\overline{2 i}}$. When $i \neq j$, since $2 i$ and $2 j$ are not congruent modulo $2 n+1$, the $2 n+1$ lines extending from point $A_{2 n+1}$ are all of different colors. The $2 n+1$ lines extending from other points are also all of different colors. Hence, this coloring scheme meets the conditions.
Therefore, $f(2 n+2)=2 n+1$.
Thus, $f(2 n)=2 n-1$.
Therefore, for 2011 points or 2012 points, at least 2011 colors are needed.
|
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 $[x]$ denotes the greatest integer not exceeding the real number $x$.
And $707 \sqrt{2}<1000<708 \sqrt{2}$
$$
\Rightarrow n \leqslant 707 \text {. }
$$
When $n=707$, $m=999,1000,1001$, but 1001 should be discarded.
Therefore, the number of pairs $(m, n)$ that satisfy the condition is
$$
\begin{array}{l}
\sum_{n=1}^{706}([\sqrt{2}(n+1)]-[\sqrt{2} n]+1)+2 \\
=\sum_{n=1}^{706}([\sqrt{2}(n+1)]-[\sqrt{2} n])+708 \\
=708+[707 \sqrt{2}]-[\sqrt{2}] \\
=708+999-1=1706 .
\end{array}
$$
(Chen Qian, Chen Hongfei, Yuyan Middle School, Xishui County, Huanggang City, Hubei Province, 438200)
|
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 $[-1,1]$.
$$
\begin{array}{l}
\text { Given } f(x)+f\left(2-x^{2}\right) \leqslant 0 \\
\Rightarrow f(x) \leqslant f\left(x^{2}-2\right) \\
\Rightarrow-1 \leqslant x \leqslant x^{2}-2 \leqslant 1 \\
\Rightarrow x=-1 .
\end{array}
$$
|
-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 satisfy $S(S(n))=n$ is
|
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-digit number, all satisfy the requirement, totaling 9.
(2) If $n=\overline{x y}, S(n)=\overline{y x}, S(S(n))=\overline{x y}$, totaling 81 (excluding those with $y=0$).
(3) If $n=\overline{x y z}, S(n)=\overline{y z x}, S(S(n))=\overline{z x y}$, thus, $x=y=z$, totaling 9.
(4) If $n=\overline{x y z w}, S(S(n))=\overline{z w x y}$, thus, $w=y, z=x$, totaling 9.
Therefore, the number of $n$ that satisfy the requirement is 108.
|
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
$$
\begin{array}{l}
1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7 \\
7,4,1,5,6,1,7,8,5,3,8,1,9,0,9,9 \\
8,7,5,2,7,9,6,5,1,6,7,3,0,3,3,6 \\
9,5,4,9,3,2,5,7,2,9,1,0,1,1, \cdots .
\end{array}
$$
It is not hard to see that the last digits form a periodic sequence with a period of 60.
Thus, the last digit of $a_{2013}$ is 8.
$$
\text { Also, }-1<\frac{\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^{2013}<0 \text {, so } \frac{\sqrt{5}}{5}\left(\frac{\sqrt{5}+1}{2}\right)^{2013}
$$
has an integer part of $a_{2013}-1$.
Therefore, the digits before the decimal point are 7.
(Jin Lei, Xi'an Jiaotong University Affiliated High School,
710054)
|
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 $x_{i}$ is $k_{i}$, and let $m=\max \left\{k_{1}, k_{2}, \cdots, k_{n}\right\}$. Find the minimum value of $m$.
|
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 repetitions (because some pairs of sets may have more than one common element, and some elements may belong to multiple sets).
According to the condition, the 11 sets $A_{1}, A_{2}, \cdots, A_{11}$ in the family $\Omega$ have non-empty intersections, forming $\mathrm{C}_{11}^{2}=55$ unique pairs of sets.
Therefore, $\sum_{i=1}^{n} \mathrm{C}_{k_{i}}^{2} \geqslant 55$, which means
$$
\sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \geqslant 110 \text {. }
$$
Given $\max k_{i}=m$, from the above inequality we have
$$
\begin{array}{l}
110 \leqslant \sum_{i=1}^{n} k_{i}\left(k_{i}-1\right) \\
\leqslant \sum_{i=1}^{n} k_{i}(m-1)=55(m-1) .
\end{array}
$$
Thus, $m \geqslant 3$.
If $m=\max k_{i}=3$, then for any $i$, $k_{i} \leqslant 3$, which would imply that all $k_{i}=3$.
In fact, if some $k_{i} \leqslant 2$, meaning the number of sets containing element $x_{i}$ is no more than two, then at least nine sets in the family $\Omega$ do not contain $x_{i}$. Let these sets be $A_{1}, A_{2}, \cdots, A_{9}$, and let $A_{10}$ contain $x_{i}$, denoted as $A_{10}=\left\{x_{i}, a, b, c, d\right\}$. Since $A_{10} \cap A_{i} \neq \varnothing$, then
$$
A_{10} \cap A_{i} \subset\{a, b, c, d\}(i=1,2, \cdots, 9) .
$$
Therefore, among the four elements $a, b, c, d$, there must be one element (say $a$) that belongs to at least three sets in the family $\Omega$, i.e., it belongs to at least four sets among $A_{1}, A_{2}, \cdots, A_{9}, A_{10}$, which contradicts $\max k_{i}=3$.
Thus, all $k_{i}=3$.
However, this leads to $55=\sum_{i=1}^{n} k_{i}=3 n$, which is a contradiction.
Therefore, $m>3$, i.e., $m \geqslant 4$.
Furthermore, $m=4$ is achievable.
For this, a geometric construction method is used.
The "five-element set" is associated with the pentagram in Figure 3, which has 10 vertices and the midpoints of each side, totaling 15 points, labeled as 1, 2, ..., 15.
Figure 3
Construct 11 five-element sets of $A=\{1,2, \cdots, 15\}$ as follows: the 5 points on the circumference of the pentagram, the 5 points on each of the 5 sides, and the 5 points on the "cross" shapes like $(1,4,8,6,11)$, each forming a set, resulting in 11 sets.
Therefore, the minimum value of $m$ is 4.
|
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,
$$
\begin{array}{l}
n(n+1)(n+2)(n+3)+1 \\
=\left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1 \\
=\left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 n\right)+1 \\
=\left(n^{2}+3 n+1\right)^{2} .
\end{array}
$$
Therefore, $k=3$.
|
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) $\frac{1}{a^{2 k+1}}+\frac{1}{b^{2 k+1}}+\frac{1}{c^{2 k+1}}=\frac{1}{a^{2 k+1}+b^{2 k+1}+c^{2 k+1}}$;
(3) $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k}=\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}$;
(4) $\frac{1}{a^{2 k}}+\frac{1}{b^{2 k}}+\frac{1}{c^{2 k}}=\frac{1}{a^{2 k}+b^{2 k}+c^{2 k}}$.
(2010, National High School Mathematics League Xinjiang Uygur Autonomous Region Preliminary)
|
【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 $f(a, b, c)$. For polynomial factorization problems with multiple variables, the method of selecting the main variable is usually adopted, arranging the polynomial in descending order of the main variable $a$, which generally allows for a smooth factorization.
Solution From the given condition, we have
$$
\begin{array}{l}
(b+c) a^{2}+(b+c)^{2} a+b c(b+c)=0 \\
\Rightarrow(b+c)(c+a)(a+b)=0 \\
\Rightarrow b+c=0 \text { or } c+a=0 \text { or } a+b=0
\end{array}
$$
$\Rightarrow b+c=0$ or $c+a=0$ or $a+b=0$
$\Rightarrow a, b, c$ have at least two numbers that are opposites.
Assume $a+b=0$ without loss of generality.
$$
\text { Then }\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2 k+1}=\left(\frac{1}{c}\right)^{2 k+1}=\frac{1}{c^{2 k+1}} \text {. }
$$
Therefore, equation (1) holds.
Similarly, equation (2) also holds, while equations (3) and (4) do not hold.
Thus, the number of equations that hold is 2.
|
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 $x_{k}$ belongs to exactly one A-type group and one B-type group, i.e., there are no common terms between A-type groups.
If two groups $A_{i}$ and $B_{j}$ have two common terms $x_{i}$ and $x_{t}$, then these two numbers can be combined into one term $x_{r} + x_{t}$, which would reduce the value of $n$. Therefore, we can assume that each pair of $A_{i}$ and $B_{j}$ has at most one common term.
Construct a graph model: Let the points $u_{1}, u_{2}, \cdots, u_{41}$ represent the groups $A_{1}, A_{2}, \cdots, A_{41}$, and the points $v_{1}, v_{2}, \cdots, v_{49}$ represent the groups $B_{1}, B_{2}, \cdots, B_{49}$. If the groups $A_{i}$ and $B_{j}$ have a common term, then connect a line between the corresponding points $u_{i}$ and $v_{j}$. Thus, we obtain a bipartite graph $G$ with exactly $n$ edges and 90 vertices.
Next, we prove that the graph $G$ is connected.
If the largest connected component of the graph $G$ is $G^{\prime}$, and the number of vertices in the component $G^{\prime}$ is less than 90, let there be $a$ A-type vertices $u_{k_{1}}, u_{k_{2}}, \cdots, u_{k_{4}}$ and $b$ B-type vertices $v_{x_{1}}, v_{x_{2}}, \cdots, v_{x_{1}}$ in the component $G^{\prime}$, where $a+b<90$. Then, in the corresponding A-type groups $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{n}}$ and B-type groups $B_{s_{1}}, B_{s_{2}}, \cdots, B_{s_{b}}$, each number $x_{i}$ in the A-type group $A_{k_{i}}$ must appear in some B-type group $B_{x_{j}}$, and each number $x_{j}$ in the B-type group $B_{x_{j}}$ must appear in some A-type group $A_{r_{i}}$ (otherwise, there would be an edge connecting to a vertex outside the component, leading to a contradiction).
Therefore, the sum of the numbers in the $a$ A-type groups $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{1}}$ should equal the sum of the numbers in the $b$ B-type groups $B_{x_{1}}, B_{s_{2}}, \cdots, B_{s_{b}}$, i.e., $49a = 41b$, from which it follows that $41 \mid a$ and $49 \mid b$.
Thus, $a + b \geq 41 + 49 = 90$, which is a contradiction.
Therefore, the graph $G$ is connected.
Thus, the graph $G$ has at least $90 - 1 = 89$ edges, i.e., $n \geq 89$.
On the other hand, we can construct a sequence of 89 terms $x_{1}, x_{2}, \cdots, x_{89}$ that satisfies the conditions of the problem.
For example, take
\[
\begin{array}{l}
x_{1} = x_{2} = \cdots = x_{41} = 41, \\
x_{42} = x_{43} = \cdots = x_{75} = 8, \\
x_{76} = x_{77} = x_{78} = x_{79} = 7, \\
x_{80} = x_{81} = x_{82} = x_{83} = 1, \\
x_{84} = x_{85} = 6, \\
x_{86} = x_{87} = 2, x_{88} = 5, x_{89} = 3,
\end{array}
\]
i.e., the sequence has 41 terms with a value of 41; 34 terms with a value of 8; and the remaining seven 8s can be split into seven pairs: four pairs of $\{7,1\}$, two pairs of $\{6,2\}$, and one pair of $\{5,3\}$, yielding 14 more terms.
Thus, each A-type group can consist of one 41, one 8, or one 41 and a pair of terms that sum to 8, resulting in 41 A-type groups, each with a sum of 49.
To obtain 49 B-type groups with a sum of 41, we can form one group for each of $x_{1}, x_{2}, \cdots, x_{41}$, and the remaining numbers can be combined into eight B-type groups: four groups of $\{8,8,8,8,8,1\}$, two groups of $\{8,8,8,8,7,2\}$, one group of $\{8,8,8,8,6,3\}$, and one group of $\{8,8,7,7,6,5\}$.
Thus, the minimum value of $n$ is 89.
|
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 pros in the training team?
---
The above text is the English translation of the provided Chinese text, maintaining the original format and line breaks.
|
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 greatest integer not exceeding the real number $x$) of their friend pairs. Therefore, the experts collectively dominate at least $k+(m-1) \frac{k+1}{2}$ friend pairs.
Since any friend pair can contribute to an expert only once, it will not be counted repeatedly. Thus, the above number does not exceed the total number of friend pairs in the training team, i.e.,
$$
k+(m-1) \frac{k+1}{2} \leqslant 15 k \Rightarrow m \leqslant \frac{28 k}{k+1}+1 \text {. }
$$
On the other hand, the number of team members who perform worse than the worst expert is no more than $30-m$. The worst expert also wins at least $\frac{k+1}{2}$ people, so
$$
\frac{k+1}{2} \leqslant 30-m \Rightarrow k \leqslant 59-2 m \text {. }
$$
Substituting equation (2) into equation (1) gives
$$
\begin{array}{l}
m \leqslant 28 \times \frac{59-2 m}{60-2 m}+1 \\
\Rightarrow m^{2}-59 m+856 \geqslant 0 \\
\Rightarrow m(59-m) \leqslant 856 .
\end{array}
$$
The maximum value of the positive integer $m$ that satisfies the above inequality and the condition $m \leqslant 30$ is 25, meaning the number of experts does not exceed 25.
On the other hand, it can be shown that the scenario with 25 experts is possible.
Using $1,2, \cdots, 30$ to represent the rankings of the team members (with better performance having a higher ranking).
When $m=25$, from equation (2) we get $k=9$, meaning each person has nine friends.
Now, construct a $6 \times 5$ table and define the following friendship relationships.
If two students are friends, it is only if one of the following conditions is met:
(1) They are in the same row of the first row;
(2) They are in the same column, but one of them is in the bottom row;
(3) They are in adjacent rows but different columns.
This way, each person has exactly nine friends, and the top 25 are all experts.
|
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 points in Figure 11 are not labeled).
Remove some points so that the edges are also removed. At least 12 points need to be removed, for example, removing the points in the set
$$
M=\{3,5,7,9,10,12,14,16,21,24,30,36\}
$$
At this point, the graph $G$ has $50-12=38$ points left, and these 38 numbers form the set $A$, which does not contain any of the above pairs, i.e., no pair in $A$ satisfies the condition.
Therefore, the smallest positive integer $k \geqslant 39$.
Next, we prove that $k=39$ satisfies the condition.
First, take 12 edges from the graph $G$ that have no common vertices (12 pairs of numbers):
$$
\begin{array}{l}
(9,18),(36,45),(5,20),(15,30), \\
(10,40),(8,24),(16,48),(4,12), \\
(3,6),(7,42),(21,28),(14,35),
\end{array}
$$
Each edge has exactly one vertex in the set $M$.
These 12 edges have 24 vertices in total, and the 24 numbers form a 24-element subset $B$ of the set $S$; the remaining 26 numbers in the set $S$ form the subset $C$, i.e.,
$$
S=B \cup C, B \cap C=\varnothing \text {. }
$$
Now, take any 39-element subset $T$ from the set $S$, then $T$ can have at most 26 numbers from the set $C$, which means it must have at least 13 numbers from the set $B$, and among these, there must be two numbers from the same pair of the above 12 pairs, say $(a, b)$, then $(a+b) \mid a b$.
Therefore, the smallest $k$ that satisfies the condition is 39.
|
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 new polynomial through subtraction and factoring, we can find their relationship, which is that they are equal, thus solving the evaluation of the unknown polynomial.
Solution Note that,
$$
\begin{array}{l}
a^{2}(b+c)-b^{2}(a+c) \\
=(a-b)(a b+b c+c a)=0 .
\end{array}
$$
Since $a \neq b$, then $a b+b c+c a=0$.
Therefore, $c^{2}(a+b)-b^{2}(a+c)$
$$
=(c-b)(a b+b c+c a)=0 \text {. }
$$
Thus, $c^{2}(a+b)=b^{2}(a+c)=2010$.
|
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 $\left(x_{1}-x_{2}\right)^{2}=\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}$
$=\left(-\frac{b}{a}\right)^{2}-\frac{4 c}{a}=\frac{b^{2}-4 a c}{a^{2}}$.
Assume $a>0$. Then
$A B=\left|x_{1}-x_{2}\right|=\frac{\sqrt{b^{2}-4 a c}}{a}$.
Thus $\frac{\left|4 a c-b^{2}\right|}{4 a}=\frac{\sqrt{b^{2}-4 a c}}{2 a} \Rightarrow \sqrt{b^{2}-4 a c}=2$.
Therefore, $\Delta=b^{2}-4 a c=4$.
|
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
$$
(x-2)(y-2)(z-2)
$$
$$
\begin{aligned}
= & -6(x-2)(y-2)(z-2)(x+1)^{2}(y+1)^{2}(z+1)^{2} \\
\Rightarrow & (x-2)(y-2)(z-2) . \\
& {\left[1+6(x+1)^{2}(y+1)^{2}(z+1)^{2}\right]=0 } \\
\Rightarrow & (x-2)(y-2)(z-2)=0 .
\end{aligned}
$$
Without loss of generality, let $x-2=0$.
Substituting into the given equations, we find $y=z=2$.
Thus, $2008(x-1)^{2}+2009(y-1)^{2}+2010(z-2)^{2} = 4017$.
|
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 $a_{n} \geqslant 0$ is $n=7$.
|
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
$$
A\left(\frac{5}{\cos \theta}, 0\right), B\left(0, \frac{3}{\sin \theta}\right) \text{. }
$$
Therefore, \( A B^{2} = \frac{5^{2}}{\cos ^{2} \theta} + \frac{3^{2}}{\sin ^{2} \theta} \).
By the Cauchy-Schwarz inequality, we have
$$
\begin{aligned}
& \frac{5^{2}}{\cos ^{2} \theta} + \frac{3^{2}}{\sin ^{2} \theta} \\
& = \left( \frac{5^{2}}{\cos ^{2} \theta} + \frac{3^{2}}{\sin ^{2} \theta} \right) \left( \cos ^{2} \theta + \sin ^{2} \theta \right) \\
\geqslant & (5 + 3)^{2} = 8^{2} .
\end{aligned}
$$
Thus, \( |A B| \geqslant 8 \).
The equality holds when \( \tan ^{2} \theta = \frac{3}{5} \).
|
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 $S_{\triangle F A^{\prime} B^{\prime}}=3$. Therefore,
$$
\begin{array}{l}
=2 \times \frac{1}{3} \times 1 \times 3=2 \text {. } \\
\end{array}
$$
|
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, $a=b=14$.
Considering the sums of two numbers from $1,2, \cdots, 7$ equal to 14, there are only the following four scenarios:
$$
\{1,6,7\},\{2,5,7\},\{3,4,7\},\{3,5,6\} \text {. }
$$
In each scenario, these numbers can only be placed in the even positions, and the remaining four numbers, whose sum is also 14, should be placed in the odd positions. Therefore, a total of $4 \times 6 \times 24=576$ such seven-digit numbers are obtained.
|
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 the trial root method (synthetic division). After rearranging the equation into $f(x)=0$, it is found that all coefficients are positive. If there is a rational root, it must be a negative root. The leading coefficient is 1, and the constant term is 3. If there is a rational root, it can only be -1 or -3.
Solution: The original equation can be transformed into
$$
x^{4}+3 x^{3}+6 x^{2}+7 x+3=0.
$$
It is easy to see that $x=-1$ is a solution of the equation.
$$
\begin{array}{l}
\text { Therefore, } x^{4}+3 x^{3}+6 x^{2}+7 x+3 \\
=(x+1)\left(x^{3}+2 x^{2}+4 x+3\right) \\
=(x+1)^{2}\left(x^{2}+x+3\right)=0.
\end{array}
$$
Notice that, $x^{2}+x+3=\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}>0$.
Therefore, the equation has a unique real root -1.
|
-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, $f(x, y)=x \sqrt{1-y}+y \sqrt{1-x}$
$$
\begin{array}{l}
\leqslant \sqrt{x(1-y)}+\sqrt{y(1-x)} \\
=\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta \\
=\sin (\alpha+\beta) \leqslant 1,
\end{array}
$$
The equality holds if and only if $\alpha+\beta=\frac{\pi}{2}$, and
$$
x=\sqrt{x}, y=\sqrt{y} \text {. }
$$
At this point, $\{x, y\}=\{0,1\}$.
|
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 empty?
|
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 parts must be different (otherwise, the same elements would be placed in the $x$ region); $a, b, c$ can be filled with or without numbers, and different regions will not contain the same elements (otherwise, they would be placed in $u, v, w$).
Let $\bar{u}$ represent the number of elements filled in $u$, and similarly for others.
By symmetry, we can list the cases where $\bar{u} \leqslant \bar{v} \leqslant \bar{w}$, which results in four scenarios:
(1) $(\bar{u}, \bar{v}, \bar{w})=(1,1,1)$;
(2) $(\bar{u}, \bar{v}, \bar{w})=(1,1,2)$;
(3) $(\bar{u}, \bar{v}, \bar{w})=(1,2,2)$;
(4) $(\bar{u}, \bar{v}, \bar{w})=(1,1,3)$.
For scenario (1), take one number from $M$ and place it in each of $u, v, w$, which gives $5 \times 4 \times 3=60$ ways. The remaining two numbers can be placed in $a, b, c$ arbitrarily, which gives $3^2$ ways. Therefore, scenario (1) has $60 \times 9=540$ ways.
For scenario (2), for $u, v, w$, the grid containing two numbers has three cases. For any of these cases, take two numbers from $M$ and place them in one grid, and place one number in each of the other two grids, which gives $\mathrm{C}_{5}^{2} \mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}=60$ ways. The remaining one number can be placed in one of $a, b, c$, which gives 3 ways. Therefore, scenario (2) has $3 \times 60 \times 3=540$ ways.
For scenario (3), for $u, v, w$, the grid containing one number has three cases. For any of these cases, take one number from $M$ and place it in one grid, take two numbers and place them in one grid, and place the remaining two numbers in another grid, which gives $\mathrm{C}_{5}^{1} \mathrm{C}_{4}^{2}=30$ ways. Therefore, scenario (3) has $3 \times 30=90$ ways.
For scenario (4), for $u, v, w$, the grid containing three numbers has three cases. For any of these cases, take three numbers from $M$ and place them in one grid; place one number in each of the other two grids, which gives $\mathrm{C}_{5}^{3} \mathrm{C}_{2}^{1}=20$ ways. Therefore, scenario (4) has $3 \times 20=60$ ways.
In summary, there are
$$
540+540+90+60=1230 \text{ (ways). }
$$
|
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 sphere of the tetrahedron $ABCD$ is
|
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 $\triangle AEH \cong \triangle AFH \Rightarrow \angle HAE = 30^{\circ}$.
Thus, $AH = \frac{AE}{\cos 30^{\circ}} = 2\sqrt{2}$,
$$
DH = \sqrt{AD^2 - AH^2} = 2,
$$
Therefore, $DH$ is the diameter of a sphere with radius 1.
Thus, the center $O$ of the circumscribed sphere of tetrahedron $ABCD$ lies on the extension of $DH$.
Let the radius of the circumscribed sphere be $r$. Then
$$
r^2 = (r-2)^2 + (2\sqrt{2})^2 \Rightarrow r = 3.
$$
|
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 Competition, Tianjin Preliminary Contest)
|
【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 two variables can be handled through factorization.
By the Pythagorean theorem, we have $c^{2}=a^{2}+b^{2}$.
Substituting $c=\frac{1}{3} a b-(a+b)$ into the above equation, we get
$$
\begin{array}{l}
a^{2}+b^{2} \\
=\frac{1}{9}(a b)^{2}-\frac{2}{3} a b(a+b)+a^{2}+2 a b+b^{2} .
\end{array}
$$
Simplifying, we get
$$
\begin{array}{l}
a b-6(a+b)+18=0 \\
\Rightarrow(a-6)(b-6)=18=1 \times 18=2 \times 9=3 \times 6 .
\end{array}
$$
Since $a$ and $b$ are both positive integers, we can assume without loss of generality that $a<b$.
Solving, we get
$$
(a, b, c)=(7,24,25),(8,15,17),(9,12,15) \text {. }
$$
Thus, there are three right-angled triangles that satisfy the conditions.
|
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.
$$
\begin{array}{l}
\text { Hence }\left|x_{1}-1\right|+\left|x_{2}-1\right|=2-\left(x_{1}+x_{2}\right) \\
=2(m+4) \geqslant 6 .
\end{array}
$$
|
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 line breaks and formatting.
|
$\begin{array}{l}\text { I. Draw } D G \perp A B \text { at point } G . \\ \text { Let } A C=B C=x, \\ B E=E D=C F=y\left(0<y<\frac{x}{2}\right) . \\ \text { Then } G D=\sqrt{2} y \Rightarrow A G=\sqrt{2}(x-y)=\sqrt{2} C E \text {. } \\ \text { Therefore, } \mathrm{Rt} \triangle E C F \backsim \mathrm{Rt} \triangle A G D \\ \Rightarrow \angle C E F=\angle G A D \\ \Rightarrow \angle C E F+\angle C A D \\ =\angle G A D+\angle C A D=45^{\circ} \\ \Rightarrow \tan (\angle C E F+\angle C A D)=1 .\end{array}$
|
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
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=(10-x)^{2}+(10-y)^{2}+(10-z)^{2} \\
\Rightarrow x+y+z=15 \\
\Rightarrow(x, y, z) \\
=(1,5,9),(1,6,8),(1,7,7), \\
(2,4,9),(2,5,8),(2,6,7), \\
(3,3,9),(3,4,8),(3,5,7), \\
(3,6,6),(4,4,7),(4,5,6), \\
(5,5,5) .
\end{array}
$$
Considering other possible orders of $x, y, z$, we find there are
$$
8 \times 3 \times 2 \times 1+4 \times 3+1=61
$$
ways to place the balls.
|
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 $A C \perp A B$, and the eccentricity of the ellipse is $\frac{\sqrt{u}-\sqrt{v}}{w}\left(u, v, w \in \mathbf{N}_{+}\right.$, and $\sqrt{u} 、 \sqrt{v}$ are in simplest radical form), then $u^{3} v+6 w=$ $\qquad$ .
|
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. Therefore,
$$
\text { point } A\left(-\frac{c}{2}, \frac{c}{2}\right) \text {. }
$$
Substituting its coordinates into the ellipse equation, we get
$$
\begin{array}{l}
c^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)=4 \\
\Rightarrow c^{2}\left(2 a^{2}-c^{2}\right)=4 a^{2}\left(a^{2}-c^{2}\right) \\
\Rightarrow e^{2}\left(2-e^{2}\right)=4\left(1-e^{2}\right) \\
\Rightarrow e^{4}-6 e^{2}+4=0 \\
\Rightarrow e^{2}=3 \pm \sqrt{5} .
\end{array}
$$
Since $e<1$, we have
$$
e=\sqrt{3-\sqrt{5}}=\sqrt{\frac{6-2 \sqrt{5}}{2}}=\frac{\sqrt{10}-\sqrt{2}}{2} \text {. }
$$
Therefore, $u^{3} v+6 w=10^{3} \times 2+6 \times 2=2012$.
|
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 $\frac{m}{n}=\overline{0.2012 C}$.
$$
\begin{array}{l}
\text { Hence } \frac{2012}{10000} \leqslant \frac{m}{n}<\frac{2013}{10000} \\
\Rightarrow \frac{10000}{2013}<\frac{n}{m} \leqslant \frac{10000}{2012} .
\end{array}
$$
Let $n=5 m-r$. From
$$
\begin{array}{l}
\text { Equation (1) } \Rightarrow \frac{60}{2.012} \leqslant \frac{r}{m}<\frac{65}{2013} \\
\Rightarrow \frac{2013}{13}<\frac{5 m}{r}<\frac{503}{3} \\
\Rightarrow \frac{2000}{13} r<5 m-r \leqslant \frac{500}{3} r .
\end{array}
$$
Take $r=1, m=31$, then $n=5 m-r=154$. This is the minimum value.
Thus $\left[\frac{m}{\sqrt{n}}\right]=\left[\sqrt{\frac{961}{154}}\right]=2$.
|
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 taken out, the game ends. It is known that the probability of the game ending exactly after 9 rolls is $\frac{a b^{u}}{c^{u} d^{u}}(a, b, c, d$ are different prime numbers, $u, v \in \mathbf{N}_{+}$). Find $u v+\overline{b c d}$.
|
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 $10-k$ dice are thrown, is $\frac{\mathrm{C}_{10-k}^{1} \times 1 \times 5^{9-k}}{6^{10-k}}$.
$$
\begin{array}{l}
\text { Then } \frac{a b^{u}}{c^{v} d^{u}}=\prod_{k=1}^{8} \frac{\mathrm{C}_{10-k}^{1} \times 5^{9-k}}{6^{10-k}}=\prod_{k=2}^{9} \frac{k \times 5^{k-1}}{6^{k}} \\
=\frac{9!\times 5^{36}}{6^{4}}=\frac{56 \times 5^{37}}{6^{40}}=\frac{7 \times 5^{37}}{3^{40} \times 2^{37}} \\
\Rightarrow a=7, b=5, c=3, d=2, u=37, v=40 . \\
\text { Hence } u v+\overline{b c d}=37 \times 40+532=2012 .
\end{array}
$$
|
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
$$
\begin{array}{l}
k+1 \geqslant d-k+2 \\
\Leftrightarrow k \geqslant \frac{d+1}{2} \Leftrightarrow d-k \leqslant \frac{d-1}{2} .
\end{array}
$$
First, consider the numbers in $[1,2048)$ that satisfy condition (2)
There are
$$
\begin{array}{l}
\sum_{d=1}^{10} \sum_{k=\left\{\frac{d+1}{2}\right]}^{d} \mathrm{C}_{d}^{k}=\sum_{d=1}^{10}\left[\frac{d-1}{2}\right] \\
=\sum_{a=0}^{5}\left(\sum_{i=0}^{u-1} \mathrm{C}_{2 a-1}^{i}+\sum_{i=0}^{a-1} \mathrm{C}_{2 a}^{i}\right) \\
=\sum_{a=1}^{5}\left[2^{2 a-2}+\frac{1}{2}\left(2^{2 a}-\mathrm{C}_{2 a}^{a}\right)\right] \\
=\frac{3}{4} \sum_{a=1}^{5} 4^{n}-\frac{1}{2} \sum_{a=1}^{a} \mathrm{C}_{2 a}^{a} \\
=\left(4^{5}-1\right)-\frac{1}{2} \cdot\left(\mathrm{C}_{2}^{1}+\mathrm{C}_{4}^{2}+\mathrm{C}_{6}^{3}+\mathrm{C}_{8}^{4}+\mathrm{C}_{10}^{5}\right) \\
=1023-\frac{1}{2}(2+6+20+70+252) \\
=848 \text { (numbers), }
\end{array}
$$
where $\lceil x\rceil$ represents the smallest integer not less than the real number $x$, and $[x]$ represents the largest integer not greater than the real number $x$.
Since there are 35 integers in $[2013,2048)$, and only
$$
(11111100000)_{2}=2016
$$
does not satisfy condition (2), we have
$$
N=848-34=814 \text {. }
$$
Therefore, the sum of the digits of $N$ is 13.
|
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:
$$
\begin{array}{l}
\text { (1) } a_{0}=0, a_{n}=2 n, \text { and } \\
1 \leqslant a_{k+1}-a_{k} \leqslant 3(k=0,1, \cdots, n-1) ;
\end{array}
$$
(2) There do not exist $i, j(0 \leqslant i<j \leqslant n)$ such that
$$
a_{j}-a_{i}=n \text {. }
$$
Find the value of $3 f(16)-2 f(15)+f(10)$.
|
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 point labeled 0 is black, and the black points divide the circle into $n$ arcs, each of length 1, 2, or 3;
(2) There are no two black points that are diametrically opposite, i.e., black points and white points are paired, forming diametrically opposite points.
Clearly, there cannot be three consecutive black points. Otherwise, let $A, B, C$ be three consecutive black points. Then their diametrically opposite points $A', B', C'$ would be three consecutive white points, but the arc containing these three white points would be longer than 3, which is a contradiction.
Thus, the coloring method that satisfies (1) and (2) is to color the point labeled 0 black, and color the points $1 \sim n-1$ black or white such that no three consecutive points are the same color, and then color the points $n \sim 2n-1$ accordingly (point $i$ is black $\Leftrightarrow$ point $n+i$ is white).
First, color the points of a circular arc of length $k$ such that the endpoints are black and no three consecutive points are the same color.
Let the number of such coloring methods be $g(k)$. It is easy to see that:
$$
\begin{array}{l}
g(1)=g(2)=1, \\
g(3)=2^{2}-1=3, \\
g(4)=2^{3}-4=4 .
\end{array}
$$
For $k \geqslant 5$, consider the last segment of the circular arc with black endpoints.
If its length is 3, then the number of corresponding coloring methods is $g(k-3)$;
If its length is 2, then the number of corresponding coloring methods is $g(k-2)$;
If its length is 1, then the adjacent arc length is 2 or 3, and the number of coloring methods is $g(k-3)+g(k-4)$.
Thus, $g(k)=g(k-2)+2 g(k-3)-g(k-4)$.
Next, find the number of coloring methods $f(n)$ that satisfy (1) and (2).
If point $n-1$ is black, then the number of coloring methods is $g(n-1)$.
If point $n-1$ is white and point $n$ is white, then points $n-2$ and $n+1$ are black, and point 1 is white. If point 2 is black, then the number of coloring methods is $g(n-4)$; if point 2 is white, then point 3 is black, and the number of coloring methods is $g(n-5)$. Therefore,
$$
f(n)=g(n-1)+g(n-4)+g(n-5) .
$$
Calculating each term, we get
$$
\begin{array}{l}
g(5)=6, g(6)=11, g(7)=17, \\
g(8)=27, g(9)=45, g(10)=72, \\
g(11)=116, g(12)=189, g(13)=305, \\
g(14)=493, g(15)=799 .
\end{array}
$$
From equation (1), we get
$$
\begin{array}{l}
f(16)=g(15)+g(12)+g(11) \\
=799+189+116=1104, \\
f(15)=g(14)+g(11)+g(10) \\
=493+116+72=681, \\
f(10)=g(9)+g(6)+g(5) \\
=45+11+6=62 .
\end{array}
$$
Thus, $3 f(16)-2 f(15)+f(10)=2012$.
|
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, $\Delta=\left( \pm \sqrt{\frac{9-t}{2}}\right)^{2}-4 \times \frac{3-t}{2} \geqslant 0$.
Solving this, we get $t \geqslant 1$.
Also, $\frac{9-t}{2} \geqslant 0$, which implies $t \leqslant 9$.
Thus, $1 \leqslant t \leqslant 9$.
Hence, $m=9, n=1$
$$
\Rightarrow m+n=10 \text {. }
$$
|
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 uniform speed. Initially, Barrier 1 and the ping-pong ball start from point $A$ and move along the line $l$ towards point $B$ at uniform speeds of $4 \mathrm{~cm} / \mathrm{s}$ and $5 \mathrm{~cm} / \mathrm{s}$, respectively, and assume the speed of Barrier 1 remains constant. When the ping-pong ball first hits Barrier 2, it reverses direction, and its speed decreases by $20 \%$, while Barrier 2 quickly moves $10 \mathrm{~cm}$ along the direction of $A B$, so the distance from Barrier 2 to point $A$ is now $10010 \mathrm{~cm}$. When the ping-pong ball first contacts Barrier 1 (referred to as the first contact between the ping-pong ball and Barrier 1) and reverses direction, its speed increases by $25 \%$. When the ping-pong ball hits Barrier 2 for the second time, it continues to move in the same manner, and Barrier 2 quickly moves $9 \mathrm{~cm}$ along the direction of $A B$. Each subsequent time the ping-pong ball hits Barrier 2, it continues to move in the same manner, and Barrier 2 moves $1 \mathrm{~cm}$ less than the previous time, until the number of times the ping-pong ball hits Barrier 2 exceeds 10, at which point Barrier 2 no longer moves. When the distance between Barrier 1 and Barrier 2 first equals $1 \mathrm{~cm}$, how many times has the ping-pong ball contacted Barrier 1?
|
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
$$
\begin{array}{l}
B_{0}=B, A_{n} B_{n-1}=x_{n}, A_{n} B_{n}=y_{n} . \\
\text { Then } \frac{10000}{5}+\frac{x_{1}}{5(1-20 \%)}=\frac{10000-x_{1}}{4} \\
\Rightarrow x_{1}=\frac{10000(5-4)(1-20 \%)}{4+5(1-20 \%)}=1000 .
\end{array}
$$
Thus, $y_{1}=1010$.
Similarly, $x_{2}=101, y_{2}=110$;
$$
\begin{array}{l}
x_{3}=11, y_{3}=19 ; \\
x_{4}=1.9, y_{4}=8.9 .
\end{array}
$$
When the ping-pong ball hits the 2nd board for the 5th time, the distance between the 1st board and the 2nd board is
$8.9 \times\left(1-\frac{4}{5}\right)=1.78>1$.
Thus, $x_{5}=0.89, y_{5}=6.89$.
When the ping-pong ball hits the 2nd board for the 6th time, the distance between the 1st board and the 2nd board is
6. $89 \times\left(1-\frac{4}{5}\right)=1.378>1$.
Thus, $x_{6}=0.689, y_{6}=5.689$.
When the ping-pong ball hits the 2nd board for the 7th time, the distance between the 1st board and the 2nd board is
$5.689 \times\left(1-\frac{4}{5}\right)=1.1378>1$.
Thus, $x_{1}=0.5689, y_{7}=4.5689$.
When the ping-pong ball hits the 2nd board for the 8th time, the distance between the 1st board and the 2nd board is
4. $5689 \times\left(1-\frac{4}{5}\right)=0.91378<1$.
Therefore, when the distance between the 1st board and the 2nd board is first equal to $1 \mathrm{~cm}$, the 1st board has contacted the ping-pong ball 7 times.
|
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.
$$
\text { (2) } 3 a-4 b+5=0 \text {. }
$$
Then $u=9 a^{2}+72 b+2=16 b^{2}+32 b+27$
$$
=16(b+1)^{2}+11 \text {. }
$$
Thus, when $b=-1$, the minimum value of $u$ is 11. In summary, the minimum value of $u$ is -34.
|
-34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Given
$$
\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{w^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{w^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}+\frac{z^{2}}{6^{2}-5^{2}}+\frac{w^{2}}{6^{2}-7^{2}}=1, \\
\frac{x^{2}}{8^{2}-1^{2}}+\frac{y^{2}}{8^{2}-3^{2}}+\frac{z^{2}}{8^{2}-5^{2}}+\frac{w^{2}}{8^{2}-7^{2}}=1 .
\end{array}\right.
$$
Find the value of $x^{2}+y^{2}+z^{2}+w^{2}$.
|
Solve: Consider the given system of equations as a fractional equation in terms of $t$
$$
\begin{array}{c}
\frac{x^{2}}{t-1^{2}}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1 \\
\Rightarrow\left(t-1^{2}\right)\left(t-3^{2}\right)\left(t-5^{2}\right)\left(t-7^{2}\right)- \\
x^{2}\left(t-3^{2}\right)\left(t-5^{2}\right)\left(t-7^{2}\right)- \\
y^{2}\left(t-1^{2}\right)\left(t-5^{2}\right)\left(t-7^{2}\right)- \\
z^{2}\left(t-1^{2}\right)\left(t-3^{2}\right)\left(t-7^{2}\right)- \\
w^{2}\left(t-1^{2}\right)\left(t-3^{2}\right)\left(t-5^{2}\right)=0 \\
\Rightarrow t^{4}-\left(x^{2}+y^{2}+z^{2}+w^{2}+1^{2}+3^{2}+\right. \\
\left.5^{2}+7^{2}\right) t^{3}+\cdots=0
\end{array}
$$
Obviously, the four roots of this equation are $2^{2}, 4^{2}, 6^{2}, 8^{2}$.
Thus, $2^{2}+4^{2}+6^{2}+8^{2}$
$$
=x^{2}+y^{2}+z^{2}+w^{2}+1^{2}+3^{2}+5^{2}+7^{2} \text {. }
$$
Therefore, $x^{2}+y^{2}+z^{2}+w^{2}=36$.
|
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 let me know.
|
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 $5-2 \sqrt{6}$.
$$
\begin{array}{l}
\text { Let } a=5+2 \sqrt{6}, b=5-2 \sqrt{6}, \\
u_{n}=a^{n}+b^{n}(n=1,2, \cdots) .
\end{array}
$$
It is easy to see that $a+b=10, a b=1$.
Since $a^{2}=10 a-1, b^{2}=10 b-1$, for all $n \in \mathbf{N}_{+}$, we have
$$
a^{n+2}=10 a^{n+1}-a^{n}, b^{n+2}=10 b^{n+1}-b^{n} \text {. }
$$
Adding the above two equations gives $u_{n+2}=10 u_{n+1}-u_{n}$, where,
$$
u_{1}=a+b=10, u_{2}=a^{2}+b^{2}=98 \text {. }
$$
Therefore, $u_{n}$ is a positive integer, and
$$
u_{n+4} \equiv-u_{n+2} \equiv u_{n}(\bmod 10)(n=1,2, \cdots) \text {. }
$$
Since $1006=4 \times 251+2$, we get
$$
u_{1006} \equiv u_{2} \equiv 8(\bmod 10) \text {. }
$$
Given $0<b<1$, we know
$$
\left[(\sqrt{2}+\sqrt{3})^{2012}\right]=u_{1000}-1=7(\bmod 10),
$$
which means the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$ is 7.
|
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 $x=-1$, $y_{\text {min }}=6$.
|
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 terms of $h$ and $r$, and find the value of $R$ when $r=$ $100, h=40$.
|
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, $h=(R-r)-(R-r) \cos \theta=\frac{2 r^{2}}{R-r}$
$$
\Rightarrow R=r+\frac{2 r^{2}}{h} \text {. }
$$
When $r=100, h=40$,
$$
R=100+\frac{2 \times 100^{2}}{40}=600 .
$$
|
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
$$
x_{1}+x_{2}=\left(x_{1}-1\right)+\left(x_{2}+1\right) \text {, }
$$
and
$$
\begin{array}{l}
\left(x_{1}-1\right)^{2}+\left(x_{2}+1\right)^{2} \\
=x_{1}^{2}+x_{2}^{2}+2\left(x_{2}-x_{1}\right)+2 \\
>x_{1}^{2}+x_{2}^{2} .
\end{array}
$$
Thus, when $x_{1}>1$, $x_{1}$ can be gradually adjusted to 1, at which point the value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ will increase.
Similarly, $x_{2}, x_{3}, \cdots, x_{39}$ can be gradually adjusted to 1, at which point the value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ will increase.
Therefore, when $x_{1}, x_{2}, \cdots, x_{39}$ are all 1 and $x_{40}=19$, $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ reaches its maximum value, i.e.,
$$
A=\underbrace{1^{2}+1^{2}+\cdots+1^{2}}_{39 \uparrow}+19^{2}=400 \text {. }
$$
If there exist two numbers $x_{i}$ and $x_{j}$ such that
$$
\begin{array}{l}
\quad x_{j}-x_{i} \geqslant 2(1 \leqslant i<j \leqslant 40), \\
\quad \text { then }\left(x_{i}+1\right)^{2}+\left(x_{j}-1\right)^{2} \\
\quad=x_{i}^{2}+x_{j}^{2}-2\left(x_{j}-x_{i}-1\right) \\
<x_{i}^{2}+x_{j}^{2} .
\end{array}
$$
This indicates that, in $x_{1}, x_{2}, \cdots, x_{40}$, if the difference between any two numbers is greater than 1, then by increasing the smaller number by 1 and decreasing the larger number by 1, the value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ will decrease.
Therefore, when $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ reaches its minimum value, the difference between any two numbers in $x_{1}, x_{2}, \cdots, x_{40}$ does not exceed 1.
Hence, when $x_{1}=x_{2}=\cdots=x_{22}=1$ and $x_{23}=x_{24}=\cdots=x_{40}=2$, $x_{1}^{2}+x_{2}^{2}+\cdots+x_{40}^{2}$ reaches its minimum value, i.e.,
$$
B=\frac{1^{2}+1^{2}+\cdots+1^{2}}{22 \uparrow}+\frac{2^{2}+2^{2}+\cdots+2^{2}}{18 \uparrow}=94 .
$$
Thus, $A+B=494$.
|
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
$$
\left\{\begin{array}{l}
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O A} \leqslant 1, \\
0 \leqslant \overrightarrow{O P} \cdot \overrightarrow{O B} \leqslant 2
\end{array}\right.
$$
Then the area of the figure formed by the point $Q(x+y, y)$ is
|
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
$$
\left\{\begin{array}{l}
0 \leqslant u-v \leqslant 1, \\
0 \leqslant u \leqslant 2 .
\end{array}\right.
$$
In the $u O v$ plane, draw
the plane region formed by
the point $Q(u, v)$ (the shaded part in Figure 2), its area equals 2.
|
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 $2^{4}-1=15$.
|
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 the answer.
|
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 the way back, he marks the 1st tree, the 9th tree, the 17th tree, …, each time skipping 7 trees without marking. When he gets home, the number of trees that have not been marked is
|
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 $(1+1)^{5}=32>18$, $M-2$ cannot have more than four prime factors.
If $M-2$ has only two prime factors, let
$$
\begin{array}{l}
M-2=2^{\alpha_{1}} \times 5^{\alpha_{2}} \times 67^{\alpha_{3}} . \\
\text { Then }\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right)\left(\alpha_{3}+1\right) < 18 . \\
\text { Hence }(1+2)(1+5)(1+67)(1+p) \\
=\frac{816}{335} \times 2 \times 5 \times 67 \times p .
\end{array}
$$
Solving for $p$ gives $p=3$.
Thus, $M-2=2 \times 5 \times 67 \times 3 \Rightarrow M=2010$.
|
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
$$
\begin{array}{l}
\frac{2 a^{2}+b^{2}}{c^{2}} \\
=\frac{1}{3}(2+1)\left[2\left(\frac{a}{c}\right)^{2}+\left(\frac{b}{c}\right)^{2}\right] \\
\geqslant \frac{1}{3}\left(\frac{2 a}{c}+\frac{b}{c}\right)^{2} \\
=\frac{1}{9}\left[\left(\frac{2 a}{c}+\frac{b}{c}\right)\left(\frac{2 c}{a}+\frac{c}{b}\right)\right]^{2} \\
=\frac{1}{9}\left(4+\frac{2 a}{b}+\frac{2 b}{a}+1\right)^{2} \\
\geqslant \frac{1}{9}(4+2 \times 2+1)^{2}=9,
\end{array}
$$
Equality holds if and only if $a=b=\sqrt{3} c$. Therefore, the minimum value of $\frac{2 a^{2}+b^{2}}{c^{2}}$ is 9.
|
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
$$
A=\left[\begin{array}{cccccc}
a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & a_{6} \\
f\left(a_{1}\right) & f\left(a_{2}\right) & f\left(a_{3}\right) & f\left(a_{4}\right) & f\left(a_{5}\right) & f\left(a_{6}\right)
\end{array}\right] .
$$
If the number tables $M$ and $N$ differ in at least one position, then $M$ and $N$ are considered two different number tables. The number of different number tables that satisfy the conditions is $\qquad$ (answer with a number).
|
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 mappings $f$ is equivalent to the number of ways to partition the set $\{1,2, \cdots, 6\}$ into three disjoint binary subsets. There are five elements that can be paired with 1, then three elements can be paired with the smallest remaining element, and the last two elements can be paired together, giving a total of $5 \times 3 \times 1=15$ ways. Since there are $6!$ permutations of $1,2, \cdots, 6$, the total number of valid tables is
$$
15 \times 6!=10800 \text {. }
$$
|
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
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=4, \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=5
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
x_{1}+x_{2}=4-x_{3}, \\
x_{1} x_{2}=5-x_{3}\left(x_{1}+x_{2}\right)=5-x_{3}\left(4-x_{3}\right) .
\end{array}\right.
\end{array}
$$
Thus, the quadratic equation with roots $x_{1} 、 x_{2}$ is
$$
\begin{array}{l}
x^{2}-\left(4-x_{3}\right) x-5+x_{3}\left(4-x_{3}\right)=0 \\
\Rightarrow \Delta=\left(4-x_{3}\right)^{2}-4\left[5-x_{3}\left(4-x_{3}\right)\right] \geqslant 0 \\
\Rightarrow 3 x_{3}^{2}-8 x_{3}+4 \leqslant 0 \\
\Rightarrow \frac{2}{3} \leqslant x_{3} \leqslant 2 .
\end{array}
$$
When $x_{3}=2$, $x_{1}=x_{2}=1, a=-2$.
Therefore, the maximum value of $\max \left\{x_{1}, x_{2}, x_{3}\right\}$ is 2.
|
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 equation (2) gives $x_{0} \ln a=1$, i.e.,
$$
\ln a^{x_{0}}=1 \text{. }
$$
Substituting equation (1) into equation (3) gives
$$
\ln x_{0}=1 \Rightarrow x_{0}=e \text{. }
$$
Thus, $\ln a=\frac{1}{\mathrm{e}} \Rightarrow \ln \ln a=-1$.
|
-1
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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