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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
|
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
|
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 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
|
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
|
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
|
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
|
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
|
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
|
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
|
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 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
|
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
|
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
|
$$
\left\{\begin{array}{l}
x-999 \geqslant 1000, \\
x+1 \leqslant a
\end{array}\right.
$$
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
|
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 $a_{1}, a_{2}, \cdots$ is a geometric sequence with the first term $a_{1}=a\left(a \in \mathbf{Z}_{+}\right)$ and common ratio $r\left(r \in \mathbf{Z}_{+}\right)$. Suppose
$$
\log _{4} a_{2}+\log _{4} a_{3}+\cdots+\log _{4} a_{12}=2013 .
$$
Then the number of ordered pairs $(a, r)$ that satisfy the condition is
|
6. 62.
From the given, we have $a_{n}=a r r^{n-1}$, substituting into the given equation yields
$$
a^{11} r^{66}=2^{4026} \Rightarrow a r^{6}=2^{366} \text{. }
$$
Let $a=2^{x}, r=2^{y}(x, y \in \mathrm{N})$. Then $x+6y=366$.
Thus, the number of pairs satisfying the condition is 62.
_ 171
2. $\frac{1}{3}$.
By the universal formula for two angles.
The probability of returning to the starting point after $n$ steps is
$$
\begin{array}{l}
\left.\frac{5}{3}=\frac{\cos A \cdot \mathrm{c}}{\sin A}-p_{n}\right) \\
\Rightarrow p_{n}=\frac{2}{3}\left(-\frac{1}{2}\right)^{n}+\frac{1}{3} .
\end{array}
$$
Substituting $n=10$ into the above formula yields.
|
62
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than 1. Then
$\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=$ $\qquad$ [2]
(2012, Zhejiang Province High School Mathematics Competition)
|
When the common ratio $q=1$,
$$
\begin{array}{l}
a_{n}=a_{1}, \\
\lg a_{1} \cdot \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}}=2011 .
\end{array}
$$
When the common ratio $q \neq 1$,
$$
\begin{array}{l}
\lg a_{1} \cdot \lg a_{2012}^{2011} \frac{1}{\lg a_{i} \cdot \lg a_{i+1}} \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q} \sum_{i=1}^{2011}\left(\frac{1}{\lg a_{i}}-\frac{1}{\lg a_{i+1}}\right) \\
=\frac{\lg a_{1} \cdot \lg a_{2012}}{\lg q}\left(\frac{1}{\lg a_{1}}-\frac{1}{\lg a_{2012}}\right) \\
=2011 .
\end{array}
$$
|
2011
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given positive integers $a_{1}, a_{2}, \cdots, a_{10}$ satisfy
$$
\frac{a_{j}}{a_{i}}>\frac{2}{3}(1 \leqslant i \leqslant j \leqslant 10) \text {. }
$$
Then the minimum possible value of $a_{10}$ is $\qquad$ .
|
2. 92.
From $a_{1} \geqslant 1, a_{2}>\frac{3}{2} a_{1} \geqslant \frac{3}{2}\left(a_{2} \in \mathbf{N}_{+}\right)$, we get $a_{2} \geqslant 2$.
Similarly, $a_{3}>\frac{3}{2} a_{2} \geqslant 3, a_{3} \geqslant 4$;
$$
\begin{array}{l}
a_{4}>\frac{3}{2} a_{3} \geqslant 6, a_{4} \geqslant 7 ; \\
a_{5}>\frac{3}{2} a_{4} \geqslant \frac{21}{2}, a_{5} \geqslant 11 ;
\end{array}
$$
$$
\begin{array}{l}
a_{6}>\frac{3}{2} a_{5} \geqslant \frac{33}{2}, a_{6} \geqslant 17 ; \\
a_{7}>\frac{3}{2} a_{6} \geqslant \frac{51}{2}, a_{7} \geqslant 26 ; \\
a_{8}>\frac{3}{2} a_{7} \geqslant 39, a_{8} \geqslant 40 ; \\
a_{9}>\frac{3}{2} a_{8} \geqslant 60, a_{9} \geqslant 61 ; \\
a_{10}>\frac{3}{2} a_{9} \geqslant \frac{183}{2}, a_{10} \geqslant 92 .
\end{array}
$$
Therefore, the smallest possible value of $a_{10}$ is 92.
|
92
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=1, a_{2}=2, \\
a_{n+2}=\frac{2(n+1)}{n+2} a_{n+1}-\frac{n}{n+2} a_{n}(n=1,2, \cdots) .
\end{array}
$$
If $a_{m}>2+\frac{2011}{2012}$, then the smallest positive integer $m$ is . $\qquad$
|
8.4025.
$$
\begin{array}{l}
\text { Given } a_{n+1}=\frac{2 n}{n+1} a_{n}-\frac{n-1}{n+1} a_{n-1} \\
\begin{array}{l}
\Rightarrow a_{n}-a_{n-1}=\frac{n-2}{n}\left(a_{n-1}-a_{n-2}\right) \\
=\frac{n-2}{n} \cdot \frac{n-3}{n-1}\left(a_{n-2}-a_{n-3}\right)=\cdots \\
= \frac{n-2}{n} \cdot \frac{n-3}{n-1} \cdots \cdots \frac{1}{3}\left(a_{2}-a_{1}\right) \\
= \frac{1 \times 2}{n(n-1)}=2\left(\frac{1}{n-1}-\frac{1}{n}\right) \\
\Rightarrow a_{n}=a_{1}+\sum_{k=1}^{n-1}\left(a_{k+1}-a_{k}\right) \\
=1+2\left(1-\frac{1}{n}\right)=3-\frac{2}{n} .
\end{array}
\end{array}
$$
Thus, $a_{m}>2+\frac{2011}{2012}$
$$
\begin{array}{l}
\Leftrightarrow 3-\frac{2}{m}>3-\frac{1}{2012} \\
\Leftrightarrow m>4024 .
\end{array}
$$
In summary, the smallest value of $m$ that satisfies the condition is 4025.
|
4025
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. (16 points) Given an integer $n(n \geqslant 3)$, let $f(n)$ be the minimum number of elements in a subset $A$ of the set $\left\{1,2, \cdots, 2^{n}-1\right\}$ that satisfies the following two conditions:
(i) $1 \in A, 2^{n}-1 \in A$;
(ii) Each element in subset $A$ (except 1) is the sum of two (possibly the same) elements in $A$.
(1) Find the value of $f(3)$;
(2) Prove: $f(100) \leqslant 108$.
|
12. (1) Let set $A \subseteq\left\{1,2, \cdots, 2^{3}-1\right\}$, and $A$ satisfies (i) and (ii). Then $1 \in A, 7 \in A$.
Since $\{1, m, 7\}(m=2,3, \cdots, 6)$ does not satisfy (ii), hence $|A|>3$.
Also, $\{1,2,3,7\},\{1,2,4,7\},\{1,2,5,7\}$,
$\{1,2,6,7\},\{1,3,4,7\},\{1,3,5,7\}$,
$\{1,3,6,7\},\{1,4,5,7\},\{1,4,6,7\}$,
$\{1,5,6,7\}$
do not satisfy (ii), hence $|A|>4$.
However, the set $\{1,2,4,6,7\}$ satisfies (i) and (ii), so $f(3)=5$.
(2) First, prove:
$$
f(n+1) \leqslant f(n)+2(n=3,4, \cdots) \text {. }
$$
In fact, if $A \subseteq\left\{1,2, \cdots, 2^{n}-1\right\}$ satisfies (i) and (ii), and the number of elements in set $A$ is $f(n)$. Let
$$
B=A \cup\left\{2^{n+1}-2,2^{n+1}-1\right\} \text {. }
$$
Since $2^{n+1}-2>2^{n}-1$, we have
$|B|=f(n)+2$.
Also, $2^{n+1}-2=2\left(2^{n}-1\right)$,
$$
2^{n+1}-1=1+\left(2^{n+1}-2\right) \text {, }
$$
then $B \subseteq\left\{1,2, \cdots, 2^{n+1}-1\right\}$, and set $B$ satisfies
(i) and (ii).
Thus, $f(n+1) \leqslant|B|=f(n)+2$.
Next, prove:
$$
f(2 n) \leqslant f(n)+n+1(n=3,4, \cdots) \text {. }
$$
In fact, let $A \subseteq\left\{1,2, \cdots, 2^{n}-1\right\}$ satisfy
(i) and (ii), and the number of elements in set $A$ is $f(n)$. Let
$$
\begin{aligned}
B= & A \cup\left\{2\left(2^{n}-1\right), 2^{2}\left(2^{n}-1\right), \cdots,\right. \\
& \left.2^{n}\left(2^{n}-1\right), 2^{2 n}-1\right\} .
\end{aligned}
$$
Since $2\left(2^{n}-1\right)<2^{2}\left(2^{n}-1\right)<\cdots$
$$
<2^{n}\left(2^{n}-1\right)<2^{2 n}-1 \text {, }
$$
then $B \subseteq\left\{1,2, \cdots, 2^{2 n}-1\right\}$, and
$$
\begin{array}{l}
|B|=f(n)+n+1 . \\
\text { and } 2^{k+1}\left(2^{n}-1\right) \\
=2^{k}\left(2^{n}-1\right)+2^{k}\left(2^{n}-1\right)(k=0,1, \cdots, n-1), \\
2^{2 n}-1=2^{n}\left(2^{n}-1\right)+\left(2^{n}-1\right),
\end{array}
$$
then $B$ satisfies (i) and (ii).
Thus, $f(2 n) \leqslant|B|=f(n)+n+1$.
From equations (1) and (2), we get
$$
f(2 n+1) \leqslant f(n)+n+3 \text {. }
$$
Repeatedly using equations (2) and (3), we get
$$
\begin{array}{l}
f(100) \leqslant f(50)+51 \\
\leqslant f(25)+26+51 \\
\leqslant f(12)+15+77 \\
\leqslant f(6)+7+92 \\
\leqslant f(3)+4+99=108 .
\end{array}
$$
(Xiong Hua, Gu Hongda, Liu Hongkuang, Li Dayuan, Ye Shengyang)
|
108
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Arrange the positive integers whose sum of digits is 5 in ascending order to form a sequence. Then 2012 is the $\qquad$th term of this sequence.
|
7.38.
To represent 5 as the sum of no more than four positive integers, there are six methods, that is
$$
\begin{array}{l}
5=1+4=2+3=1+1+3 \\
=1+2+2=1+1+1+2 .
\end{array}
$$
When filling them into a $1 \times 4$ grid, positions that are not filled are supplemented with 0. Then $\{5\}$ has 3 ways of filling; $\{1,4\}$ has 9 ways of filling; $\{2,3\}$ has 7 ways of filling; $\{1,1,3\}$ has 9 ways of filling; $\{1,2,2\}$ has 7 ways of filling; $\{1,1,1,2\}$ has 3 ways of filling.
There are a total of 38 ways of filling, with 2012 being the largest.
Therefore, 2012 is the 38th term.
|
38
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given that $18^{2}=324, 24^{2}=576$, they are formed by the permutation of two consecutive digits $2,3,4$ and $5,6,7$ respectively; and $66^{2}=4356$ is formed by the permutation of four consecutive digits $3, 4, 5, 6$. Then the next such square number is $\qquad$
|
8.5476.
For any square number, its last digit can only be $0, 1, 4, 5, 6, 9$, and
$$
\begin{array}{l}
(10 a)^{2}=100 a^{2}, \\
(10 a+5)^{2}=100 a^{2}+100 a+25, \\
(10 a \pm 4)^{2}=100 a^{2} \pm 80 a+16 ; \\
(10 a \pm 1)^{2}=100 a^{2} \pm 20 a+1, \\
(10 a \pm 3)^{2}=100 a^{2} \pm 60 a+9 ; \\
(10 a \pm 2)^{2}=100 a^{2} \pm 40 a+4 .
\end{array}
$$
If following the above rules, it can only be the case of
$$
3456,4536,5436,5364,3564 \text {. }
$$
But none of the above are square numbers.
The next four consecutive digits are $4,5,6,7$.
If arranged into a square number, the last digit can only be 4 or 6, from smallest to largest:
$$
4756,5476,5764,7564 .
$$
After calculation, only $5476=74^{2}$ is a square number.
|
5476
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $4^{a}-3 a^{b}=16, \log _{2} a=\frac{a+1}{b}$. Then $a^{b}=$ $\qquad$ .
|
- 1. 16.
From $\log _{2} a=\frac{a+1}{b} \Rightarrow a^{b}=2^{a+1}$.
Substituting into $4^{a}-3 a^{b}=16$, we get
$2^{2 a}-6 \times 2^{a}-16=0$.
Solving, we get $2^{a}=8$ or -2 (discard).
Therefore, $a^{b}=2^{a+1}=16$.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (16 points) Let the side lengths opposite to two interior angles of $\triangle A B C$ be $a, b, c$ respectively, and $a+b+c=16$. Find
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
the value.
|
$$
\begin{array}{l}
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 4 R^{2}\left(\sin ^{2} B \cdot \cos ^{2} \frac{C}{2}+\sin ^{2} C \cdot \cos ^{2} \frac{B}{2}+\right. \\
\left.2 \sin B \cdot \sin C \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\right. \\
\left.2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \\
= 16 R^{2} \cos \frac{B}{2} \cdot \cos \frac{C}{2}\left(\frac{1-\cos B}{2}+\right. \\
\left.\frac{1-\cos C}{2}+\frac{\cos A+\cos B+\cos C-1}{2}\right) \\
= 16 R^{2}\left(\cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\right)^{2} \\
= R^{2}(\sin A+\sin B+\sin C)^{2} \\
=\left(\frac{a+b+c}{2}\right)^{2}=64 .
\end{array}
$$
Solution 2: As shown in Figure 2, extend line segment $BC$ to points $E$ and $F$ such that $CE = AC = b$ and $BF = AB = c$. Connect $AE$ and $AF$. Let the midpoints of $AE$ and $AF$ be $M$ and $N$, respectively. Connect $BN$, $CM$, and $NM$.
$$
\begin{array}{l}
\text { Then } AM = b \cos \frac{C}{2}, AN = c \cos \frac{B}{2}, \\
\angle MAN = \angle A + \frac{1}{2}(\angle B + \angle C) = \frac{\pi}{2} + \frac{\angle A}{2}. \\
\text { Therefore, } b^{2} \cos ^{2} \frac{C}{2} + c^{2} \cos ^{2} \frac{B}{2} + 2 b \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= AM^{2} + AN^{2} - 2 AM \cdot AN \cos \angle MAN \\
= MN^{2} = \left(\frac{a+b+c}{2}\right)^{2} = 64 .
\end{array}
$$
|
64
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. Let $P$ be any point on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ other than the endpoints of the major axis, $F_{1}$ and $F_{2}$ be the left and right foci respectively, and $O$ be the center. Then
$\left|P F_{1}\right|\left|P F_{2}\right|+|O P|^{2}=$ $\qquad$ .
|
16. 25.
According to the definition of an ellipse and the cosine rule, the solution can be found.
|
25
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 (An Ancient Chinese Mathematical Problem) Emperor Taizong of Tang ordered the counting of soldiers: if 1,001 soldiers make up one battalion, then one person remains; if 1,002 soldiers make up one battalion, then four people remain. This time, the counting of soldiers has at least $\qquad$ people.
|
Let the first troop count be 1001 people per battalion, totaling $x$ battalions, then the total number of soldiers is $1001 x + 1$ people; let the second troop count be 1002 people per battalion, totaling $y$ battalions, then the total number of soldiers is $1002 y + 4$ people.
From the equality of the total number of soldiers, we get the equation
$$
1001 x + 1 = 1002 y + 4 \text{. }
$$
Rearranging gives
$$
1001(x - y) - 3 = y \text{. }
$$
Since $x$ and $y$ are both positive integers, the smallest value for $x - y$ is 1, at which point, $y = 998$.
Thus, the minimum number of soldiers in this troop count is
$$
998 \times 1002 + 4 = 1000000 \text{ (people). }
$$
|
1000000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Mother's Day is coming, and Xiao Hong, Xiao Li, and Xiao Meng went to the flower shop to buy flowers for their mothers. Xiao Hong bought 3 roses, 7 carnations, and 1 lily, and paid 14 yuan; Xiao Li bought 4 roses, 10 carnations, and 1 lily, and paid 16 yuan; Xiao Ying bought 2 stems of each of the three types of flowers. Then she should pay $\qquad$ yuan. $[1]$
|
Let the unit prices of roses, carnations, and lilies be $x$ yuan, $y$ yuan, and $\sqrt{z}$ yuan, respectively. Then,
$$
\left\{\begin{array}{l}
3 x+7 y+z=14, \\
4 x+10 y+z=16 .
\end{array}\right.
$$
Eliminating $z$ gives
$$
x=2-3 y \text{. }
$$
Substituting equation (2) into equation (1) gives
$$
z=8+2 y \text{. }
$$
From equations (2) and (3), we get
$$
x+y+z=10 \text{, }
$$
which means $2(x+y+z)=20$.
Therefore, Xiaoying should pay 20 yuan.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 In the tetrahedron $A-B C D$, it is known that
$$
\begin{array}{l}
\angle A C B=\angle C B D \\
\angle A C D=\angle A D C=\angle B C D=\angle B D C=\theta,
\end{array}
$$
and $\cos \theta=\frac{\sqrt{10}}{10}$.
If the length of edge $A B$ is $6 \sqrt{2}$, then the volume of this pyramid is
$\qquad$
|
Solve As shown in Figure 2, from the problem, we know $\triangle A C D \cong \triangle B C D$,
and $\square$
$$
\begin{array}{l}
A C=A D \\
=B C=B D .
\end{array}
$$
Then $\triangle A C D$
$\cong \triangle B C D$
$\cong \triangle C A B$
$\cong \triangle D A B$.
Therefore, $\angle A C B$
$$
\begin{array}{l}
\because \angle C B D=\angle C A D=\angle A D B, \\
A B=C D .
\end{array}
$$
Take the midpoints $E$ and $F$ of $C D$ and $A B$ respectively, and connect $A E$, $B E$, and $E F$.
Then $A E \perp C D, B E \perp C D$.
Thus, $C D \perp$ plane $A B E$.
Therefore, plane $A B E$ is the perpendicular plane of $C D$.
It is calculated that $A E=B E=9 \sqrt{2}, E F=12$. $=\frac{1}{3} \times \frac{1}{2} \times 6 \sqrt{2} \times 12 \times 6 \sqrt{2}=144$.
|
144
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Two boxes of candy have a total of 176 pieces. 16 pieces are taken from the second box and placed into the first box, at which point, the number of pieces of candy in the first box is 31 more than $m($ an integer $m>1)$ times the number of pieces of candy in the second box. Then, the first box originally had at least $\qquad$ pieces of candy.
|
Let the first box originally contain $x$ candies, and the second box originally contain $y$ candies.
According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
x+y=176, \\
x+16=m(y-16)+31 .
\end{array}\right.
$$
Rearranging, we get
$$
x+16=m(176-16-x)+31 \text {, }
$$
which simplifies to $x=\frac{160 m+15}{m+1}=160-\frac{145}{m+1}$.
Since $x$ and $m$ are both positive integers, we have:
$$
\begin{array}{l}
160-\frac{29 \times 5}{m+1} \\
\geqslant 160-29=131(m>1) .
\end{array}
$$
Therefore, the first box originally contained at least 131 candies.
|
131
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given three points $A, B, C$ in a plane satisfying
$$
|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=5,|\overrightarrow{C A}|=6 \text {. }
$$
Then the value of $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$ is ( ).
|
5. C.
From the cosine theorem, we know
$$
\begin{array}{l}
\overrightarrow{A B} \cdot \overrightarrow{B C}=-\overrightarrow{B A} \cdot \overrightarrow{B C} \\
=-\frac{|\overrightarrow{B A}|^{2}+|\overrightarrow{B C}|^{2}-|\overrightarrow{A C}|^{2}}{2} .
\end{array}
$$
Similarly,
$$
\begin{array}{l}
\overrightarrow{B C} \cdot \overrightarrow{C A}=-\frac{|\overrightarrow{C A}|^{2}+|\overrightarrow{C B}|^{2}-|\overrightarrow{A B}|^{2}}{2}, \\
\overrightarrow{C A} \cdot \overrightarrow{A B}=-\frac{|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}-|\overrightarrow{B C}|^{2}}{2} .
\end{array}
$$
Therefore, $\overrightarrow{A B} \cdot \overrightarrow{B C}+\overrightarrow{B C} \cdot \overrightarrow{C A}+\overrightarrow{C A} \cdot \overrightarrow{A B}$
$$
=-\frac{|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}+|\overrightarrow{B C}|^{2}}{2}=-35 \text {. }
$$
|
-35
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Every day at 5 PM when school is over, Xiao Ming's father always drives from home to pick him up on time and take him back. One day, the school dismissed an hour early, and Xiao Ming walked home by himself. On the way, he met his father who was coming to pick him up, and as a result, they arrived home 20 minutes earlier than usual. Then Xiao Ming walked for $\qquad$ minutes before meeting his father.
|
二、1.50 minutes.
As shown in Figure 6, Xiao Ming
starts walking home from point $A$
and meets the car coming to pick him up
at point $C$. As a result, the car returns from $C$ to $B$ 20 minutes earlier than usual. This indicates that the car takes 20 minutes to travel from $C$ to $A$ and back to $C$.
Therefore, the car takes 10 minutes to travel from $C$ to $A$.
Thus, the car meets Xiao Ming at $4:50$, meaning Xiao Ming has been walking for 50 minutes when he meets his father.
|
50
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let real numbers $x, y, z, w$ satisfy $x \geqslant y \geqslant z \geqslant w \geqslant 0$, and $5 x+4 y+3 z+6 w=100$. Denote the maximum value of $x+y+z+w$ as $a$, and the minimum value as $b$. Then $a+b=$ $\qquad$
|
2. 45 .
From $x \geqslant y \geqslant z \geqslant w \geqslant 0$, we know
$$
\begin{array}{l}
100=5 x+4 y+3 z+6 w \geqslant 4(x+y+z+w) \\
\Rightarrow x+y+z+w \leqslant 25 .
\end{array}
$$
When $x=y=z=\frac{25}{3}, w=0$, the equality holds.
$$
\begin{array}{l}
\text { Also } 100=5 x+4 y+3 z+6 w \leqslant 5(x+y+z+w) \\
\Rightarrow x+y+z+w \geqslant 20 .
\end{array}
$$
When $x=20, y=z=w=0$, the equality holds. Therefore, $a=25, b=20$.
Thus, $a+b=45$.
|
45
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. After rotating any positive integer by $180^{\circ}$, some interesting phenomena can be observed, such as 808 still being 808 after a $180^{\circ}$ rotation, 169 becoming 691 after a $180^{\circ}$ rotation, and 37 not being a number after a $180^{\circ}$ rotation. Then, among all five-digit numbers, the number of five-digit numbers that remain the same after a $180^{\circ}$ rotation is.
|
2. 60.
Among the ten digits from $0$ to $9$, $(0,0)$, $(1,1)$, $(8,8)$, and $(6,9)$ can be placed in the symmetric positions at the beginning and end of a five-digit number. When rotated $180^{\circ}$, the resulting number is the same as the original number, while other digits cannot appear in the five-digit number. Such five-digit numbers include: 18 without $(6,9)$; exactly one pair of $(6,9)$ has 30; two pairs of $(6,9)$ have 12. Therefore, the total number of such five-digit numbers is 60.
|
60
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that the sum of the first $n(n>1)$ terms of an arithmetic sequence is 2013, the common difference is 2, and the first term is an integer. Then the sum of all possible values of $n$ is . $\qquad$
|
3. 2975 .
Let the first term of the sequence be $a_{1}$, and the common difference $d=2$. Then
$$
\begin{array}{l}
S_{n}=n a_{1}+n(n-1) \\
=n\left(a_{1}+n-1\right)=2013 .
\end{array}
$$
Also, $2013=3 \times 11 \times 61$, and $n$ is a divisor of 2013, so the sum of all possible values of $n$ is
$$
(1+3) \times(1+11) \times(1+61)-1=2975 .
$$
|
2975
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=p, a_{2}=p+1, \\
a_{n+2}-2 a_{n+1}+a_{n}=n-20,
\end{array}
$$
where $p$ is a given real number, and $n$ is a positive integer. Try to find the value of $n$ that minimizes $a_{n}$.
|
10. Let $b_{n}=a_{n+1}-a_{n}$.
From the problem, we have $b_{n+1}-b_{n}=n-20$, and $b_{1}=1$. Then, $b_{n}-b_{1}=\sum_{i=1}^{n-1}\left(b_{i+1}-b_{i}\right)=\sum_{i=1}^{n-1}(i-20)$.
Thus, $b_{n}=\frac{(n-1)(n-40)}{2}+1$.
Also, $a_{3}=a_{2}+b_{2}=p-17<a_{1}<a_{2}$, so when the value of $a_{n}$ is the smallest, we have $n \geqslant 3$, and
$$
\begin{array}{l}
\left\{\begin{array}{l}
a_{n} \leqslant a_{n+1}, \\
a_{n} \leqslant a_{n-1},
\end{array}\right. \\
\left\{\begin{array}{l}
b_{n}=\frac{(n-1)(n-40)}{2}+1 \geqslant 0, \\
b_{n-1}=\frac{(n-2)(n-41)}{2}+1 \leqslant 0 .
\end{array}\right.
\end{array}
$$
Therefore, when $n=40$, the value of $a_{n}$ is the smallest.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given $x, y \in \mathbf{N}_{+}$. Find the minimum value of $\sqrt{512^{x}-7^{2}-1}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
|
Let $z=512^{x}-7^{2 y-1}$.
Obviously, $z \geqslant 0$, and
$z \equiv 1(\bmod 7), z \equiv 1(\bmod 8)$.
(1) $x=2 x_{1}-1$ is an odd number.
Then $z \equiv(-1)^{x}-1 \equiv-2 \equiv 1(\bmod 3)$.
Thus, $z \equiv 1(\bmod 3 \times 7 \times 8)$.
Let $z=1$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=1$. Then $7^{2 y-1}=512^{2 x_{1}-1}-1$.
Hence $51117^{2 y-1} \Rightarrow 7317^{2 y-2}$, which is a contradiction.
Therefore, $z \geqslant 3 \times 7 \times 8+1=169$.
Let $z=169$, i.e., $512^{2 x_{1}-1}-7^{2 y-1}=169$.
Solving this, we get $(x, y)=(1,2)$.
Thus, the smallest non-negative value of $z$ is 169.
(2) $x=2 x_{1}$ is an even number.
Then $z \equiv(-1)^{x}-1 \equiv 0(\bmod 3)$.
Thus, $z \equiv 57(\bmod 3 \times 7 \times 8)$.
Let $z=57$, i.e., $512^{2 x_{1}}-7^{2 y-1}=57$.
Let $512^{x_{1}}=s, 7^{y-1}=t$. Then $s^{2}-7 t^{2}=57$.
The fundamental solution of the equation $s^{2}-7 t^{2}=1$ is $8+3 \sqrt{7}$.
Let $s_{1}+t_{1} \sqrt{7}$ be the fundamental solution of $s^{2}-7 t^{2}=57$. Then all solutions of the equation are
$$
S_{n+1}+t_{n+1} \sqrt{7}=\left(s_{n}+t_{n} \sqrt{7}\right)(8+3 \sqrt{7}),
$$
i.e.,
$$
s_{n+1}=8 s_{n}+21 t_{n}, t_{n+1}=3 s_{n}+8 t_{n}.
$$
Solving the first equation for $t_{n}=\frac{s_{n+1}-8 s_{n}}{21}$ and substituting into the second equation, we get
$$
\begin{array}{l}
s_{n+2}=16 s_{n+1}-s_{n} \\
\Rightarrow s_{n+2}=-s_{n}(\bmod 16).
\end{array}
$$
The fundamental solution $s_{1}+t_{1} \sqrt{7}$ should satisfy
$$
\begin{array}{l}
0 \leqslant s_{1} \leqslant \sqrt{\frac{1}{2}(8+1) \times 57}, \\
0 \leqslant t_{1} \leqslant \frac{3 \sqrt{27}}{\sqrt{2(8+1)}},
\end{array}
$$
i.e., $0 \leqslant s_{1} \leqslant 16,0 \leqslant t_{1} \leqslant 5$.
Trying each one, we find only
$$
\left(s_{1}, t_{1}\right)=(8,1),(13,4).
$$
When $\left(s_{1}, t_{1}\right)=(8,1)$, by conclusion (1) we get
$$
s_{2}=85 \equiv 5(\bmod 16);
$$
When $\left(s_{1}, t_{1}\right)=(13,4)$, by conclusion (1) we get $s_{2}=188 \equiv 12(\bmod 16)$.
Thus, $s_{n} \neq 0(\bmod 16)$, which is a contradiction.
Therefore, $512^{2 x_{1}}-7^{2 y-1}=57$ has no positive integer solutions.
Thus, $512^{2 x_{1}}-7^{2 y-1} \geqslant 168+57$.
In summary, the minimum value of $\sqrt{512^{x}-7^{2 y-1}}$ is 13.
|
13
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Given
$$
A=\left\{z \mid z^{18}=1\right\} \text { and } B=\left\{\omega \mid \omega^{48}=1\right\}
$$
are sets of complex roots of unity,
$$
C=\{z w \mid z \in A, w \in B\}
$$
is also a set of complex roots of unity. How many distinct elements are there in the set $C$? ${ }^{[3]}$
|
Notice that, $z=\cos \frac{2 k \pi}{18}+\mathrm{i} \sin \frac{2 k \pi}{18}(k \in \mathbf{Z})$ (18 distinct elements),
$\omega=\cos \frac{2 t \pi}{48}+\mathrm{i} \sin \frac{2 t \pi}{48}(t \in \mathbf{Z})$ (48 distinct elements),
$$
\begin{array}{l}
z \omega=\cos \frac{2 \pi(8 k+3 t)}{144}+\mathrm{i} \sin \frac{2 \pi(8 k+3 t)}{144} . \\
\text { Let } P=\{m \mid m=8 k+3 t, t \in \mathbf{Z}\} .
\end{array}
$$
By Bézout's theorem, $P=\mathbf{Z}$.
Therefore, the set $C$ contains 144 distinct elements.
|
144
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given that the upper base, height, and lower base of a trapezoid are three consecutive positive integers, and these three numbers make the value of the polynomial $x^{3}-30 x^{2}+a x$ (where $a$ is a constant) also three consecutive positive integers in the same order. Then the area of this trapezoid is $\qquad$
|
7. 100 .
Let the upper base, height, and lower base be $n-1$, $n$, and $n+1$ respectively, and the other three consecutive integers be $m-1$, $m$, and $m+1$. Then,
$$
\begin{array}{l}
(n-1)^{3}-30(n-1)^{2}+a(n-1)=m-1, \\
n^{3}-30 n^{2}+a n=m, \\
(n+1)^{3}-30(n+1)^{2}+a(n+1)=m+1 . \\
\text { (1) }+(3)-2 \times(2) \text { gives } \\
6 n-60=0 \Rightarrow n=10 .
\end{array}
$$
Therefore, the area of the trapezoid is
$$
\frac{1}{2}(9+11) \times 10=100 \text {. }
$$
|
100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Arrange all positive integers that leave a remainder of 2 and 3 when divided by 4 in ascending order. Let $S_{n}$ denote the sum of the first $n$ terms of this sequence. Then $\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right]$ $=$ $\qquad$ ([ $x]$ denotes the greatest integer not exceeding the real number $x$).
|
8. 2025078.
Given that this sequence is
$$
2,3,6,7, \cdots, 4 n-2,4 n-1, \cdots \text {. }
$$
From $4 n-2+4 n-1=8 n-3$, we know
$$
\begin{array}{l}
S_{2 n}=5+13+\cdots+(8 n-3) \\
=4 n(n+1)-3 n=4 n^{2}+n . \\
\text { Also, } 4 n^{2}<S_{2 n}<(2 n+1)^{2} \text {, then } \\
2 n<\sqrt{S_{2 n}}<2 n+1 .
\end{array}
$$
Therefore, $\left[S_{2 n}\right]=2 n$.
Also, $S_{2 n-1}=S_{2 n}-(4 n-1)=4 n^{2}-3 n+1$, then $(2 n-1)^{2}<S_{2 n-1}<4 n^{2}$.
Therefore, $\left[S_{2 n-1}\right]=2 n-1$.
$$
\begin{array}{l}
\text { Hence }\left[\sqrt{S_{1}}\right]+\left[\sqrt{S_{2}}\right]+\cdots+\left[\sqrt{S_{2012}}\right] \\
=1+2+\cdots+2012=2025078 .
\end{array}
$$
|
2025078
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 A scientist stored the design blueprint of his time machine in a computer, setting the password to open the file as a permutation of $\{1,2, \cdots, 64\}$. He also designed a program that, when eight positive integers between $1 \sim 64$ are input each time, the computer will indicate the order (from left to right) of these eight numbers in the password. Please design an operation scheme such that the password can be determined with at most 45 inputs. [6]
|
Let the password be denoted as $a_{1} a_{2} \cdots a_{n^{2}} (n=8)$.
First, write the numbers $1, 2, \cdots, n^{2}$ arbitrarily into an $n \times n$ grid (one number per cell). After the first $n$ operations, input each row of numbers once, and rearrange the numbers in each row from left to right according to the computer's prompt.
Color the numbers in the 1st and $n$-th columns red and yellow, respectively, and color the numbers in the other columns blue.
In the $(n+1)$-th and $(n+2)$-th operations, input the numbers in the 1st and $n$-th columns, respectively, to determine the order of the red and yellow numbers in the password, and identify $a_{1}$ and $a_{n^{2}}$. Remove these two numbers from the 1st and $n$-th columns, and replace them with adjacent blue numbers.
In the $(n+3)$-th operation, input the $\frac{n}{2}-1$ leftmost (rightmost) red (yellow) numbers in the password along with the blue numbers in the 1st and $n$-th columns to determine $a_{2}$ and $a_{n^{2}-1}$. Remove these two numbers from the 1st and $n$-th columns and replace them with adjacent blue numbers.
Similarly, determine $a_{3}, a_{4}, \cdots, a_{n^{2}-2}$ using the following two methods:
Operation $A$: When the number of blue numbers in the 1st and $n$-th columns is less than $\frac{n}{2}$, input these blue numbers along with several red (yellow) numbers from the left (right) of the password to form $n$ numbers (take $\frac{n}{2}$ numbers from each of the 1st and $n$-th columns). This will determine the leftmost and rightmost remaining numbers in the password. Remove these two numbers and replace them with adjacent blue numbers.
Operation $B$: When the number of blue numbers in the 1st (or $n$-th) column is $\frac{n}{2}$, input all the numbers in the 1st (or $n$-th) column (re-color them as red (or yellow)) to determine the leftmost (or rightmost) remaining number in the password. Remove this number and replace it with an adjacent blue number.
The $(n+3)$-th operation can be considered as Operation $A$; the $(n+1)$-th and $(n+2)$-th operations can be considered as Operation $B$.
The feasibility of the operations is obvious.
Assume that Operations $A$ and $B$ are performed $x$ and $y$ times, respectively, to determine all $\{a_{i}\}$.
Since each Operation $A$ determines two numbers, each Operation $B$ determines one number, and the last operation determines $n$ or $n-1$ numbers, we have:
$$
2(x-1) + y + n - 1 \leqslant n^{2}.
$$
Consider the number of blue numbers in the 1st and $n$-th columns, which can increase by at most 2 in each Operation $A$ and decrease by $\frac{n}{2}-1$ in each Operation $B$.
If the last operation is Operation $B$, then
$$
2 + 2x \geqslant \left(\frac{n}{2} - 1\right)(y - 2) + 1;
$$
If the last operation is Operation $A$, then
$$
2 + 2(x - 1) \geqslant \left(\frac{n}{2} - 1\right)(y - 2).
$$
From equations (1) and (2), we have
$$
\begin{array}{l}
2x + y - 1 \geqslant \frac{n}{2}(y - 2) \\
\Rightarrow n^{2} - n + 2 \geqslant \frac{n}{2}(y - 2) \\
\Rightarrow \frac{y}{2} \leqslant n + \frac{2}{n}.
\end{array}
$$
Thus, the total number of operations is
$$
\begin{array}{l}
n + x + y = n + \frac{(2x + y) + y}{2} \\
\leqslant n + \frac{n^{2} - n + 3}{2} + n + \frac{2}{n} \\
= \frac{(n + 1)(n + 2) + 1}{2} + \frac{2}{n} = 45 \frac{3}{4} \\
\Rightarrow n + x + y \leqslant 45.
\end{array}
$$
|
45
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the set
$$
A=\{x \mid 5 x-a \leqslant 0, a \in \mathbf{N}\} \text {. }
$$
If $5 \in A \cap \mathbf{Z}$, then the minimum value of $a$ is
|
$$
-, 1.25
$$
From $A \left\lvert\,=\left(-\infty, \frac{a}{5}\right]\right.$, and $5 \in A \cap \mathbf{Z}$, we know
$$
\frac{a}{5} \geqslant 5 \Rightarrow a \geqslant 25 \text {. }
$$
Therefore, the minimum value of $a$ is 25.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
the graph always passes through a fixed point $A$. If point $A$ lies on the line
$$
\frac{x}{m}+\frac{y}{n}+1=0(m, n>0)
$$
then the minimum value of $3 m+n$ is
|
5. 16 .
Note that the function
$$
y=a^{x+3}-2(a>0, a \neq 1)
$$
always passes through the fixed point $(-3,-1)$.
So point $A(-3,-1)$.
Then $-\frac{3}{m}-\frac{1}{n}+1=0 \Rightarrow 1=\frac{3}{m}+\frac{1}{n}$.
Thus, $3 m+n=(3 m+n)\left(\frac{3}{m}+\frac{1}{n}\right)$
$$
=10+\frac{3 n}{m}+\frac{3 m}{n} \geqslant 16 \text {. }
$$
When and only when $m=n=4$, the equality holds. Therefore, the minimum value of $3 m+n$ is 16.
|
16
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
13. (15 points) As shown in Figure 2, in the Cartesian coordinate system, the equation of circle $\odot M$ is
$$
x^{2}+y^{2}+D x+E y+F=0,
$$
and the quadrilateral $A B C D$ inscribed in $\odot M$ has diagonals $A C$ and $B D$ that are perpendicular to each other, with $A C$ and $B D$ lying on the $x$-axis and $y$-axis, respectively.
(1) Prove: $F<0$;
(2) If the area of quadrilateral $A B C D$ is 8, the length of diagonal $A C$ is 2, and $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, find the value of $D^{2}+E^{2}-4 F$.
|
13. (1) Let $A(a, 0), C(c, 0)$.
From the problem, points $A$ and $C$ are on the negative and positive halves of the $x$-axis, respectively. Thus, $a c<0$.
When $y=0$, the equation becomes
$$
x^{2}+D x+F=0 \text{, }
$$
where the two roots of the equation are the $x$-coordinates of points $A$ and $C$.
Therefore, $x_{1} x_{c}=a c=F<0$.
(2) From the problem, we know
$S_{\text {quadrilateral } A B C D}=\frac{|\overrightarrow{A C}||\overrightarrow{B D}|}{2}=8$.
Also, $|\overrightarrow{A C}|=2$, so $|\overrightarrow{B D}|=8$.
Since $\overrightarrow{A B} \cdot \overrightarrow{A D}=0$, we have $\angle A=90^{\circ}$.
Thus, $|\overrightarrow{B D}|=2 r=8 \Rightarrow r=4$.
From the circle represented by equation (1), we have
$$
\frac{D^{2}}{4}+\frac{E^{2}}{4}-F=r^{2} \text{. }
$$
Therefore, $D^{2}+E^{2}-4 F=4 r^{2}=64$.
|
64
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
I. Fill in the Blanks (8 points each, total 64 points)
1. Among the positive integers less than 20, choose three different numbers such that their sum is divisible by 3. The number of different ways to choose these numbers is $\qquad$.
|
$-, 1.327$
$$
C_{6}^{3}+C_{6}^{3}+C_{7}^{3}+6 \times 6 \times 7=327
$$
|
327
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given a geometric sequence $\left\{a_{n}\right\}$ with all terms being positive. If $2 a_{4}+a_{3}-2 a_{2}-a_{1}=8$, then the minimum value of $2 a_{8}+a_{7}$ is $-2 x-2$ $\qquad$
|
6. 54.
Let $\left\{a_{n}\right\}$ be a geometric sequence with common ratio $q(q>0)$. Then, according to the problem,
$$
\begin{array}{l}
2 a_{2} q^{2}+a_{1} q^{2}-\left(2 a_{2}+a_{1}\right)=8 \\
\Rightarrow\left(2 a_{2}+a_{1}\right)\left(q^{2}-1\right)=8 \\
\Rightarrow 2 a_{2}+a_{1}=\frac{8}{q^{2}-1},
\end{array}
$$
and $q>1$.
Let $t=q^{2}-1$. Then
$$
\begin{array}{l}
2 a_{8}+a_{7}=2 a_{2} q^{6}+a_{1} q^{6} \\
=\left(2 a_{2}+a_{1}\right) q^{6} \\
=\frac{8 q^{6}}{q^{2}-1}=\frac{8(t+1)^{3}}{t} \\
=8\left(t^{2}+3 t+3+\frac{1}{t}\right) .
\end{array}
$$
Let $f(t)=8\left(t^{2}+3 t+3+\frac{1}{t}\right)$.
Then $f^{\prime}(t)=8\left(2 t+3-\frac{1}{t^{2}}\right)$
$$
=8 \cdot \frac{(t+1)^{2}(2 t-1)}{t^{2}} \text {. }
$$
Thus, $f(t)$ is a decreasing function on $\left(0, \frac{1}{2}\right)$ and an increasing function on $\left(\frac{1}{2},+\infty\right)$.
Therefore, $f(t)_{\min }=f\left(\frac{1}{2}\right)=54$.
|
54
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$. Then
$$
\begin{array}{l}
{\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 2012\right]} \\
=
\end{array}
$$
|
8. 18084.
When $2^{k} \leqslant x<2^{k+1}$, then $\left[\log _{2} x\right]=k$.
Given $1024=2^{10}<2012<2^{11}=2048$, we know
$$
\begin{array}{l}
{\left[\log _{2} 1024\right]+\left[\log _{2} 1025\right]+\cdots+\left[\log _{2} 2012\right]} \\
=10 \times(2012-1023)=9890 .
\end{array}
$$
$$
\begin{array}{l}
\text { Hence }\left[\log _{2} 1\right]+\left[\log _{2} 2\right]+\cdots+\left[\log _{2} 1023\right] \\
=1 \times 2+2 \times 2^{2}+\cdots+9 \times 2^{9} .
\end{array}
$$
Let $S=1 \times 2+2 \times 2^{2}+\cdots+9 \times 2^{9}$.
Then $2 S=1 \times 2^{2}+2 \times 2^{3}+\cdots+9 \times 2^{10}$.
Subtracting the above two equations, we get
$$
\begin{array}{l}
S=-2-2^{2}-\cdots-2^{9}+9 \times 2^{10} \\
=-2^{10}+2+9 \times 2^{10}=8194
\end{array}
$$
Therefore, the original expression $=8194+9890=18084$.
|
18084
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $[x]$ denote the integer part of the real number $x$. Then $[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]=(\quad)$.
(A) 146
(B) 161
(C) 210
(D) 365
|
4. C.
$$
\begin{array}{l}
{[\sqrt{1}]+[\sqrt{2}]+\cdots+[\sqrt{49}]} \\
=1 \times 3+2 \times 5+3 \times 7+4 \times 9+5 \times 11+6 \times 13+7 \\
=210
\end{array}
$$
|
210
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 From the 205 positive integers $1,2, \cdots, 205$, what is the maximum number of integers that can be selected such that for any three selected numbers $a, b, c (a<b<c)$, we have
$$
a b \neq c ?^{[1]}
$$
(2005, (Casio Cup) National Junior High School Mathematics Competition)
|
Estimate first.
Since $14 \times 15=210>205$, then $14,15, \cdots$, 205 satisfy that for any three numbers $a 、 b 、 c(a<b<c)$, we have $a b \neq c$.
Because 1 multiplied by any number equals the number itself, $1,14,15, \cdots, 205$ satisfy the condition.
Therefore, there are $205-14+1+1=193$ numbers in total.
If we select one more number from $2 \sim 13$, what would happen?
Construct the following 12 sets:
$(2,25,2 \times 25),(3,24,3 \times 24), \cdots$,
$(13,14,13 \times 14)$.
The 36 numbers in the above sets are all distinct, and the smallest number is 2, while the largest number is $13 \times 14=182<205$.
Therefore, all three numbers in each set cannot be selected.
In summary, the number of numbers that satisfy the given condition does not exceed $205-12=193$.
|
193
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2: 9 judges score 12 athletes participating in a bodybuilding competition. Each judge gives 1 point to the athlete they consider to be in 1st place, 2 points to the athlete in 2nd place, ..., and 12 points to the athlete in 12th place. The final scoring shows: the difference between the highest and lowest scores of each athlete's nine scores is no more than 3. Let the total scores of each athlete be $c_{1}, c_{2}, \cdots, c_{12}$, and $c_{1} \leqslant c_{2} \leqslant \cdots \leqslant c_{12}$. Find the maximum value of $c_{1}$.
(2004, Jiangsu Province Junior High School Mathematics Competition (Grade 9))
|
Explanation: It is impossible for 9 judges to give 1 point to five or more athletes, because among five or more athletes, at least one athlete must be rated no less than 5 by a judge. However, according to the problem, each of these five athletes is rated no more than 4 by each judge, which is a contradiction.
Therefore, 9 judges can give 1 point to at most four athletes.
We will discuss the following scenarios.
(1) If all judges give 1 point to a single athlete, then \( c_{1} = 9 \).
(2) If the nine 1 points given by the 9 judges are concentrated on two athletes, then one of these athletes must be rated 1 by at least five judges. According to the problem, the remaining judges give this athlete a score no greater than 4, thus,
\[
c_{1} \leqslant 5 \times 1 + 4 \times 4 = 21.
\]
(3) If the nine 1 points are concentrated on three athletes, then the total score of these three athletes is no more than
\[
9 \times 1 + 9 \times 3 + 9 \times 4 = 72.
\]
Thus, \( 3 c_{1} \leqslant c_{1} + c_{2} + c_{3} \leqslant 72 \).
Hence, \( c_{1} \leqslant 24 \).
(4) If the nine 1 points are distributed among four athletes, then the total score of these four athletes is
\[
9 \times 1 + 9 \times 2 + 9 \times 3 + 9 \times 4 = 90.
\]
Thus, \( 4 c_{1} \leqslant 90 \). Hence, \( c_{1} < 23 \).
In summary, \( c_{1} \leqslant 24 \).
The scenario where \( c_{1} = 24 \) is achievable. Let \( A_{i} (i = 1, 2, \cdots, 12) \) represent the athletes, and \( B_{j} (j = 1, 2, \cdots, 9) \) represent the judges. The values in Table 1 are the scores given by judge \( B_{j} \) to athlete \( A_{i} \).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & \( A_{1} \) & \( A_{2} \) & \( A_{3} \) & \( A_{4} \) & \( A_{5} \) & \( A_{6} \) & \( A_{7} \) & \( A_{8} \) & \( A_{9} \) & \( A_{10} \) & \( A_{11} \) & \( A_{12} \) \\
\hline \( B_{1} \) & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline \( B_{2} \) & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline \( B_{3} \) & 1 & 4 & 3 & 2 & 5 & 6 & 7 & 9 & 10 & 8 & 11 & 12 \\
\hline \( B_{4} \) & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline \( B_{5} \) & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline \( B_{6} \) & 4 & 3 & 1 & 5 & 2 & 7 & 9 & 6 & 8 & 11 & 10 & 12 \\
\hline \( B_{7} \) & 3 & 1 & 4 & 2 & 5 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline \( B_{8} \) & 3 & 1 & 4 & 5 & 2 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline \( B_{9} \) & 3 & 1 & 4 & 2 & 5 & 9 & 6 & 7 & 11 & 10 & 8 & 12 \\
\hline Total & 24 & 24 & 24 & 30 & 33 & 66 & 66 & 66 & 87 & 87 & 87 & 108 \\
\hline
\end{tabular}
|
24
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
II. (40 points) Given the sequence of real numbers $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=\frac{1}{3}, a_{n+1}=2 a_{n}-\left[a_{n}\right],
$$
where $[x]$ denotes the greatest integer less than or equal to the real number $x$.
$$
\text { Find } \sum_{i=1}^{2012} a_{i} \text {. }
$$
|
$$
\text { II. } a_{1}=\frac{1}{3}, a_{2}=\frac{2}{3}, a_{3}=\frac{4}{3}, a_{4}=\frac{5}{3} .
$$
By mathematical induction, it is easy to prove
$$
\left\{\begin{array}{l}
a_{2 k+1}=\frac{3 k+1}{3}, \\
a_{2 k+2}=\frac{3 k+2}{3} .
\end{array}\right.
$$
Then $a_{2 k+1}+a_{2 k+2}=2 k+1$.
Therefore, $\sum_{i=1}^{2012} a_{i}=1012036$.
|
1012036
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Choose $k$ numbers from $1,2, \cdots, 2004$, such that among the chosen $k$ numbers, there are definitely three numbers that can form the side lengths of a triangle (the three numbers must be distinct). What is the minimum value of $k$ that satisfies this condition?
|
When selecting three numbers from 1 to 2004 to form the sides of a triangle, there are too many possibilities. Instead, let's approach it from the opposite direction and list all sets of three numbers that cannot form the sides of a triangle.
First, 1, 2, 3 cannot form the sides of a triangle. Adding 5, the set \(1, 2, 3, 5\) also cannot form the sides of a triangle.
Continuing this way, the added number is exactly the sum of the last two numbers, resulting in:
\[
\begin{array}{l}
1, 2, 3, 5, 8, 13, 21 ; 34, 55, 89, 144, 233, \\
377, 610, 987, 1597
\end{array}
\]
There are 16 numbers in total. In these 16 numbers, any three chosen cannot form the sides of a triangle. If we add any other number from the remaining ones, then among these 17 numbers, we can find three numbers that can form the sides of a triangle.
Therefore, the smallest value of \(k\) that satisfies the condition is 17.
|
17
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A. Given positive integers $a$, $b$, $c$ satisfy
$$
\begin{array}{l}
a+b^{2}-2 c-2=0, \\
3 a^{2}-8 b+c=0 .
\end{array}
$$
Then the maximum value of $a b c$ is $\qquad$
|
8. A. 2013.
The two equations are simplified and rearranged to get
$$
(b-8)^{2}+6 a^{2}+a=66 \text {. }
$$
Given that $a$ is a positive integer and $6 a^{2}+a \leqslant 66$, we have $1 \leqslant a \leqslant 3$. If $a=1$, then $(b-8)^{2}=59$, which has no positive integer solutions; if $a=2$, then $(b-8)^{2}=40$, which has no positive integer solutions; if $a=3$, then $(b-8)^{2}=9$. Solving this, we get $b=11$ or 5. (1) If $b=11$, then $c=61$, thus, $a b c=3 \times 11 \times 61=2013$;
(2) If $b=5$, then $c=13$, thus, $a b c=3 \times 5 \times 13=195$.
In summary, the maximum value of $a b c$ is 2013.
|
2013
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
B. If $k$ numbers are chosen from 2, $, 8, \cdots, 101$ these 34 numbers, where the sum of at least two of them is 43, then the minimum value of $k$ is: $\qquad$
|
B. 28.
Divide the 14 numbers less than 43 into the following seven groups: $(2,41),(5,38),(8,35),(11,32)$, $(14,29),(17,26),(20,23)$.
The sum of the two numbers in each group is 43. After selecting one number from each group and then taking all numbers greater than 43, a total of 27 numbers are selected. The sum of any two of these 27 numbers is not equal to 43.
On the other hand, if 28 numbers are selected, then in the above groups, there must be one group where both numbers are selected, and their sum is 43.
In summary, the minimum value of $k$ is 28.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. A. Xiaoming volunteered to sell pens at a stationery store one day. Pencils were sold at 4 yuan each, and ballpoint pens at 7 yuan each. At the beginning, it was known that he had a total of 350 pencils and ballpoint pens. Although he did not sell them all that day, his sales revenue was 2013 yuan. Then he sold at least $\qquad$ ballpoint pens.
|
10. A. 207.
Let $x$ and $y$ represent the number of pencils and ballpoint pens sold, respectively. Then
\[
\begin{array}{l}
\left\{\begin{array}{l}
4 x+7 y=2013 ; \\
x+y=204
\end{array}\right.
\end{array}
\]
Thus, $y_{\text {min }}=207$, at which point, $x=141$.
|
207
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
12. A. As shown in Figure 5, given that $A B$ is the diameter of $\odot O$, $C$ is a point on the circumference, and $D$ is a point on the line segment $O B$ (not at the endpoints), satisfying $C D \perp A B$ and $D E \perp C O$ at point $E$. If $C E = 10$, and the lengths of $A D$ and $D B$ are both positive integers, find the length of line segment $A D$.
|
12. A. Connect $A C$ and $B C$, then $\angle A C B=90^{\circ}$.
From $\mathrm{Rt} \triangle C D E \backsim \mathrm{Rt} \triangle C O D$, we know $C E \cdot C O=C D^{2}$.
From $\mathrm{Rt} \triangle A C D \backsim \mathrm{Rt} \triangle C B D$, we know $C D^{2}=A D \cdot B D$.
Therefore, $C E \cdot C O=A D \cdot B D$.
Let $A D=a, D B=b\left(a, b \in \mathbf{N}_{+}\right)$. Then $C O=\frac{a+b}{2}$.
Given $C E=10$, substituting into equation (1) yields
$$
10 \times \frac{a+b}{2}=a b \Rightarrow(a-5)(b-5)=25 \text {. }
$$
Considering $a>b$, we can only have $a-5>b-5>0$, leading to $a-5=25, b-5=1$.
Thus, $A D=a=30$.
|
30
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and when $x \geqslant 0$,
$f(x)=2^{x}+2 x+b$ ( $b$ is a constant).
Then $f(-10)=$ $\qquad$ .
|
2. -1043 .
From the given condition, we easily know that
$$
f(0)=2^{0}+2 \times 0+b=0 \text {. }
$$
Solving for $b$ yields $b=-1$.
By the property of odd functions $f(-x)=-f(x)$, we have
$$
\begin{array}{l}
f(-10)=-f(10)=-2^{10}-2 \times 10+1 \\
=-1043 .
\end{array}
$$
|
-1043
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. As shown in Figure 3, square $ABCD$ is divided into 8 triangles of equal area. If $AG=\sqrt{50}$, then the area $S$ of square $ABCD$ is $ \qquad $.
|
4. 128 .
As shown in Figure 5, draw $K L / / D C$ through point $F$, take the midpoint $N$ of $A B$, and connect $G N$ with $A H$ intersecting at point $P$.
Let the side length of the square $A B C D$ be $a$.
Given $S_{\triangle D C I}=S_{\triangle M B H}=\frac{1}{8} S$, we know
$C I=B H=\frac{1}{4} B C=\frac{a}{4}$.
Since $S_{\triangle M D F}=2 S_{\triangle D C I}$, we have
$A D \cdot K F=2 C D \cdot C I$.
Therefore, $K F=2 C I=\frac{1}{2} a$.
Thus, $F$ is the midpoint of $D I$.
Given $S_{\triangle C A H}=S_{\triangle C H F}=S_{\triangle C F A}$, we know $G$ is the centroid of $\triangle F A H$.
Then $E$ and $P$ are the midpoints of $A F$ and $A H$ respectively, and
$$
G P=\frac{1}{2} F G \text {. }
$$
Also, $F P$ is the midline of trapezoid $A H I D$, so,
$F P=\frac{H I+A D}{2}=\frac{\frac{1}{2} a+a}{2}=\frac{3 a}{4}$
$\Rightarrow G P=\frac{1}{3} F P=\frac{a}{4}$
$\Rightarrow G N=G P+P N=\frac{a}{4}+\frac{a}{8}=\frac{3 a}{8}$.
Since $A N=\frac{a}{2}$, by the Pythagorean theorem we get
$$
\begin{array}{l}
A G^{2}=\left(\frac{a^{2}}{2}\right)+\left(\frac{3 a}{8}\right)^{2}=\frac{25 a^{2}}{64}=50 \\
\Rightarrow a^{2}=128 \Rightarrow S=128 .
\end{array}
$$
|
128
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given three distinct integers $x, y, z$ whose sum lies between 40 and 44. If $x, y, z$ form an arithmetic sequence with a common difference of $d$, and $x+y, y+z, z+x$ form a geometric sequence with a common ratio of $q$, then $d q=$ $\qquad$
|
5. 42 .
$$
\begin{array}{l}
\text { Given } x=y-d, z=y+d \\
\Rightarrow x+y=2 y-d, y+z=2 y+d \\
\Rightarrow z+x=2 y . \\
\text { Also, }(x+y)(z+x)=(y+z)^{2} \\
\Rightarrow 2 y(2 y-d)=(2 y+d)^{2} \\
\Rightarrow d(d+6 y)=0 .
\end{array}
$$
Since $d \neq 0$, we have $d=-6 y$.
$$
\begin{array}{l}
\text { Also, } 40<x+y+z=3 y<44 \\
\Rightarrow \frac{40}{3}<y<\frac{44}{3} \Rightarrow y=14 \\
\Rightarrow d=-6 y=-84 \\
\Rightarrow q=\frac{y+z}{x+y}=\frac{2 y+d}{2 y-d}=-\frac{1}{2} \\
\Rightarrow d p=42 .
\end{array}
$$
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. There are three sets of cards in red, yellow, and blue, each set containing five cards, marked with the letters $A, B, C, D, E$. If five cards are drawn from these 15 cards, with the requirement that the letters are all different and all three colors are included, then the number of different ways to draw the cards is $\qquad$ kinds.
|
4. 150.
Divide into two categories: $3, 1,1$ and $2,2,1$.
Calculate respectively:
$$
\frac{C_{3}^{1} C_{5}^{3} C_{2}^{1} C_{2}^{1} C_{1}^{1}}{A_{2}^{2}}+\frac{C_{3}^{1} C_{5}^{2} C_{2}^{1} C_{3}^{2} C_{1}^{1}}{A_{2}^{2}}=150
$$
|
150
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. (15 points) For a positive integer $n$, let $f(n)$ be the sum of the digits in the decimal representation of the number $3 n^{2}+n+1$.
(1) Find the minimum value of $f(n)$;
(2) When $n=2 \times 10^{k}-1\left(k \in \mathbf{N}_{+}\right)$, find $f(n)$;
(3) Does there exist a positive integer $n$ such that
$$
f(n)=2012 ?
$$
|
14. (1) Since $3 n^{2}+n+1$ is an odd number greater than 3, we know that $f(n) \neq 1$.
Assume $f(n)=2$. Then $3 n^{2}+n+1$ can only be a number with the first and last digits being 1 and all other digits being 0, i.e.,
$$
3 n^{2}+n+1=10^{k}+1 \text {. }
$$
Clearly, when $k=1$, $n$ does not exist.
Therefore, $k$ is an integer greater than 1.
Thus, $n(3 n+1)=2^{k} \times 5^{k}$.
Clearly, $(n, 3 n+1)=1$.
Hence, $\left\{\begin{array}{l}n=2^{k}, \\ 3 n+1=5^{k} \text {. }\end{array}\right.$
Also, $3 n+1 \leqslant 4 n=4 \times 2^{k}<5^{k}$, which is a contradiction.
Therefore, $f(n) \neq 2$.
When $n=8$,
$$
3 n^{2}+n+1=201 \text {. }
$$
Thus, $f(8)=3$.
In summary, the minimum value of $f(n)$ is 3.
(2) When $n=2 \times 10^{k}-1$, we have
$$
\begin{array}{l}
3 n^{2}+n+1=12 \times 10^{2 k}-10 \times 10^{k}+3 \\
=11 \underbrace{99 \cdots 9}_{k-1 \uparrow} \underbrace{00 \cdots 03}_{k \uparrow} .
\end{array}
$$
Thus, $f(n)=1+1+9(k-1)+3$
$$
=9 k-4\left(k \in \mathbf{N}_{+}\right) \text {. }
$$
(3) There exists a positive integer $n$ such that
$$
f(n)=2012 \text {. }
$$
In fact, let $n=2 \times 10^{k}-1$. Then $f(n)=9 k-4$.
Let $9 k-4=2012$. Solving for $k$ gives $k=224$.
Therefore, taking $n=2 \times 10^{224}-1$, we get
$$
f(n)=2012 \text {. }
$$
(Zhang Shengchun provided)
|
2012
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Let $A=\{x \mid x \geqslant 10, x \in \mathbf{N}\}, B \subseteq A$, and the elements in $B$ satisfy:
(i) The digits of any element are all different;
(ii) The sum of any two digits of any element is not equal to 9.
(1) Find the number of two-digit and three-digit numbers in $B$;
(2) Does there exist a five-digit or six-digit number in $B$;
(3) If the elements in $B$ are arranged in ascending order, find the 1081st element.
|
(1) For two-digit numbers, the digit in the tens place can be $1,2, \cdots, 9$; the digit in the units place, since it cannot be the same as the digit in the tens place and the sum of the two digits cannot be 9, has 8 possible choices for each digit in the tens place.
Therefore, the total number of two-digit numbers that meet the criteria is $9 \times 8=72$.
For three-digit numbers, first consider the digit in the hundreds place, which can be $1,2, \cdots, 9$. Next, consider the digit in the tens place, which cannot be the same as the digit in the hundreds place and the sum of the two digits cannot be 9, so there are 8 possible choices. Finally, consider the digit in the units place, which cannot be the same as the digits in the hundreds and tens places, and the sum with the digits in the hundreds and tens places cannot be 9, so there are 6 possible choices.
Therefore, the number of three-digit numbers that meet the criteria is
$$
9 \times 8 \times 6=432 \text{ (numbers). }
$$
(2) There exists a five-digit number that meets the criteria (e.g., 12340).
There does not exist a six-digit number that meets the criteria. The reason is as follows.
Following the method in (1), there are 9 possible choices for the hundred-thousands place, 8 for the ten-thousands place, 6 for the thousands place, 4 for the hundreds place, 2 for the tens place, and 0 for the units place, which is a contradiction.
(3) From (1), we know that there are 72 two-digit numbers and 432 three-digit numbers that meet the criteria.
There are also
$$
9 \times 8 \times 6 \times 4=1728
$$
numbers that meet the criteria among four-digit numbers, and among these, there are $8 \times 6 \times 4=192$ numbers with 1 in the thousands place, and 192 numbers each with 2 or 3 in the thousands place.
Since $1081-(72+432+192 \times 3)=1$, the number that meets the criteria is the smallest number with 4 in the thousands place, i.e., 4012 is the 1081st element of $B$.
|
4012
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 For a four-digit number, at most two of its digits are different. How many such four-digit numbers are there?
保留源文本的换行和格式，所以翻译结果如下：
Example 2 For a four-digit number, at most two of its digits are different.
Ask: How many such four-digit numbers are there?
|
Solution: Clearly, there are exactly 9 four-digit numbers where all four digits are the same.
Below, we consider four-digit numbers with exactly two different digits in three steps.
(1) First, consider the thousands place, which has 9 possible choices: $1,2, \cdots, 9$.
(2) Next, consider the hundreds, tens, and units places. Since there are exactly two different digits, choose 1 digit from $0,1, \cdots, 9$ in addition to the thousands place digit.
(3) After determining the two digits in the first two steps, further determine the digits in the hundreds, tens, and units places. Each of these positions has two choices, but one scenario must be excluded, where the digits in the hundreds, tens, and units places are all the same as the thousands place digit, so there are $2 \times 2 \times 2-1$ ways to choose.
In summary, there are a total of four-digit numbers
$$
9+9 \times 9(2 \times 2 \times 2-1)=576 \text { (numbers). }
$$
|
576
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Given $A \cup B \cup C=\{1,2, \cdots, 6\}$, and $A \cap B=\{1,2\},\{1,2,3,4\} \subseteq B \cup C$.
Then the number of $(A, B, C)$ that satisfy the conditions is $\qquad$ groups (different orders of $A, B, C$ are considered different groups).
|
As shown in Figure 1, for $1$ and $2$, they can belong to regions I and II, which gives $2^{2}$ possibilities; for $3$ and $4$, they can belong to $B \cup C$ except for regions I and II, i.e., regions III, IV, VI, and VII, which gives $4^{2}$ possibilities; for $5$ and $6$, they can belong to $A \cup B$ except for regions I and II, i.e., regions III, V, V, VI, and VI, which gives $5^{2}$ possibilities. Therefore, there are a total of $2^{2} \times 4^{2} \times 5^{2}=1600$ combinations.
|
1600
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 If 6 college graduates apply to three employers, and each employer hires at least one of them, then the number of different hiring scenarios is $\qquad$ kinds.
|
The number of ways to hire 3 people is $\mathrm{A}_{6}^{3}=120$; the number of ways to hire 4 people is $\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{1} \mathrm{C}_{3}^{1}}{2!} \times \mathrm{A}_{3}^{3}=540$; the number of ways to hire 5 people is
$$
\frac{C_{6}^{2} C_{4}^{2} C_{2}^{1}}{2!} \times A_{3}^{3}+\frac{C_{6}^{3} C_{3}^{1} C_{2}^{1}}{2!} \times A_{3}^{3}=900 \text { (ways); }
$$
The number of ways to hire 6 people is
$$
\begin{array}{l}
\frac{\mathrm{C}_{6}^{2} \mathrm{C}_{4}^{2} \mathrm{C}_{2}^{2}}{3!} \times \mathrm{A}_{3}^{3}+\frac{\mathrm{C}_{6}^{4} \mathrm{C}_{2}^{1} \mathrm{C}_{1}^{1}}{2!} \times \mathrm{A}_{3}^{3}+\mathrm{C}_{6}^{3} \mathrm{C}_{3}^{2} \mathrm{C}_{1}^{1} \mathrm{~A}_{3}^{3} \\
=540 \text { (ways). }
\end{array}
$$
In summary, there are $120+540+900+540=2100$ ways.
|
2100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 In a $6 \times 6$ grid, three identical red cars and three identical black cars are parked, with one car in each row and each column, and each car occupies one cell. The number of ways to park the cars is ( ).
(A) 720
(B) 20
(C) 518400
(D) 14400
|
Assume first that the three red cars are distinct and the three black cars are also distinct. The first car can obviously be placed in any of the 36 squares, giving 36 ways. The second car, which cannot be in the same row or column as the first car, has 25 ways to be placed.
Similarly, the third, fourth, fifth, and sixth cars have $16$, $9$, $4$, and $1$ ways to be placed, respectively.
Noting that the three red cars are identical and the three black cars are also identical, we have
$$
\frac{36 \times 25 \times 16 \times 9 \times 4 \times 1}{3! \times 3!} = (5!)^2 = 14400
$$
different ways. Therefore, the answer is (D).
|
14400
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7: There are 4 red cards, 3 blue cards, 2 yellow cards, and 1 white card. Cards of the same color are indistinguishable. Questions:
(1) How many ways are there to arrange these 10 cards in a row from left to right?
(2) How many ways are there to arrange the cards so that the first 3 cards from the left are of the same color?
|
(1) $\frac{10!}{4!\times 3!\times 2!\times 1!}=12600$ ways.
(2) For the left 3 cards being red, there are
$$
\frac{7!}{1!\times 3!\times 2!\times 1!}=420 \text { (ways); }
$$
For the left 3 cards being blue, there are
$$
\frac{7!}{4!\times 2!\times 1!}=105 \text { (ways). }
$$
Thus, there are $420+105=525$ ways in total.
|
525
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing order from the set $1, 2, \cdots, 14$, such that the following conditions are satisfied:
$$
a_{2}-a_{1} \geqslant 3 \text { and } a_{3}-a_{2} \geqslant 3 \text {. }
$$
Then the number of different ways to choose such numbers is $\qquad$ kinds.
|
Let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}$,
$$
a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4} \text {. }
$$
Then $x_{1}+x_{2}+x_{3}+x_{4}=14$.
Thus, the problem is transformed into finding the number of integer solutions to the equation under the conditions
$$
x_{1} \geqslant 1, x_{2} \geqslant 3, x_{3} \geqslant 3, x_{4} \geqslant 0
$$
The number of different solutions is $\mathrm{C}_{10}^{3}=120$.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $A$ and $B$ be two sets, and $(A, B)$ is called a "pair". When $A \neq B$, $(A, B)$ and $(B, A)$ are considered different pairs. Then the number of different pairs satisfying the condition $A \cup B=\{1,2,3,4\}$ is $\qquad$
|
Prompt: Following Example 3, we know there are $3^{4}=81$ pairs.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
---
Prompt: Following Example 3, we know there are $3^{4}=81$ pairs.
|
81
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5.2011 is a four-digit number whose sum of digits is 4. Then the total number of four-digit numbers whose sum of digits is 4 is $\qquad$.
|
In fact, the number of four-digit numbers $\overline{x_{1} x_{2} x_{3} x_{4}}$ is the number of integer solutions to the indeterminate equation
$$
x_{1}+x_{2}+x_{3}+x_{4}=4
$$
satisfying the conditions $x_{1} \geqslant 1, x_{2}, x_{3}, x_{4} \geqslant 0$. It is easy to see that there are $\mathrm{C}_{6}^{3}=20$ such solutions.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are 8 English letters $K, Z, A, I, G, A, K$, and $\mathrm{U}$, each written on 8 cards. Ask:
(1) How many ways are there to arrange these cards in a row?
(2) How many ways are there to arrange 7 of these cards in a row?
|
(1) $\frac{8!}{2!\times 2!}=10080$ ways.
(2) If the letter taken away is $K$ or $A$, then there are
$$
2 \times \frac{7!}{2!}=5040 \text { (ways); }
$$
If the letter taken away is $Z$, $I$, $G$, or $U$, then there are
$$
4 \times \frac{7!}{2!\times 2!}=5040 \text { (ways). }
$$
Therefore, there are $5040+5040=10080$ arrangements in total.
|
10080
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $n$ be a positive integer, and call $n$ a "good number" if the number of prime numbers not exceeding $n$ equals the number of composite numbers not exceeding $n$. Then the sum of all good numbers is ( ).
(A) 33
(B) 34
(C) 2013
(D) 2014
|
6. B.
Since 1 is neither a prime number nor a composite number, a good number must be an odd number.
Let the number of prime numbers not exceeding $n$ be $a_{n}$, and the number of composite numbers be $b_{n}$. When $n \leqslant 15$, only consider the case where $n$ is odd (as shown in Table 1).
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline$n$ & 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\
\hline$a_{n}$ & 0 & 2 & 3 & 4 & 4 & 5 & 6 & 6 \\
\hline$b_{n}$ & 0 & 0 & 1 & 2 & 4 & 5 & 6 & 8 \\
\hline
\end{tabular}
Obviously, 1, 9, 11, and 13 are good numbers.
Since $b_{15}-a_{15}=2$, when $n \geqslant 16$, based on $n=15$, every two numbers added must include one even number, which is a composite number. This means that the number of composite numbers added will not be less than the number of prime numbers added, so there must be $b_{n}-a_{n} \geqslant 2$.
Therefore, when $n \geqslant 16$, $n$ cannot be a good number.
Thus, the sum of all good numbers is
$$
1+9+11+13=34 \text {. }
$$
|
34
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
Three. (25 points) Given that $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
If positive integers $a$, $b$, and $m$ satisfy the equation
$$
(a t+m)(b t+m)=31 m
$$
find the value of $a b$.
|
Since $t$ is a root of the quadratic equation
$$
x^{2}+x-1=0
$$
$t$ is an irrational number, and $t^{2}=1-t$.
From the problem, we have
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=31 m \\
\Rightarrow a b(1-t)+m(a+b) t+m^{2}=31 m \\
\Rightarrow[m(a+b)-a b] t+\left(a b+m^{2}-31 m\right)=0 .
\end{array}
$$
Since $a, b, m$ are positive integers and $t$ is an irrational number, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
m(a+b)-a b=0, \\
a b+m^{2}-31 m=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a+b=31-m \\
a b=31 m-m^{2}
\end{array}\right.
\end{array}
$$
Therefore, $a, b$ are the two integer roots of the quadratic equation in $x$
$$
x^{2}+(m-31) x+31 m-m^{2}=0
$$
The discriminant of equation (1) is
$$
\begin{array}{l}
\Delta=(m-31)^{2}-4\left(31 m-m^{2}\right) \\
=(31-m)(31-5 m) \geqslant 0 .
\end{array}
$$
Since $a, b$ are positive integers, we have
$$
a+b=31-m>0 \text {. }
$$
Thus, $0<m \leqslant \frac{31}{5}$.
Since the discriminant $\Delta$ is a perfect square, upon verification, only $m=6$ meets the requirement.
Substituting $m=6$ yields
$$
a b=31 m-m^{2}=150 .
$$
|
150
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) Given $t=\sqrt{2}-1$. If positive integers $a$, $b$, and $m$ satisfy
$$
(a t+m)(b t+m)=17 m
$$
find the value of $a b$.
|
Given that $t=\sqrt{2}-1$, we have
$$
t^{2}=3-2 \sqrt{2} \text {. }
$$
From the problem, we know
$$
\begin{array}{l}
a b t^{2}+m(a+b) t+m^{2}=17 m \\
\Rightarrow a b(3-2 \sqrt{2})+m(a+b)(\sqrt{2}-1)+m^{2}=17 m \\
\Rightarrow \sqrt{2}[m(a+b)-2 a b]+ \\
\quad\left[3 a b-m(a+b)+m^{2}-17 m\right]=0 .
\end{array}
$$
Since $a, b, m$ are positive integers, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
m(a+b)-2 a b=0, \\
3 a b-m(a+b)+m^{2}-17 m=0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a+b=2(17-m), \\
a b=17 m-m^{2} .
\end{array}\right.
\end{array}
$$
Therefore, $a, b$ are the two integer roots of the quadratic equation in $x$:
$$
x^{2}+2(m-17) x+17 m-m^{2}=0
$$
The discriminant of equation (1) is
$$
\begin{array}{l}
\Delta=4(m-17)^{2}-4\left(17 m-m^{2}\right) \\
=4(17-m)(17-2 m) \geqslant 0 .
\end{array}
$$
Since $a, b, m$ are positive integers, we have
$$
a+b=2(17-m)>0 \text {. }
$$
Thus, $0<m \leqslant \frac{17}{2}$.
Since the discriminant $\Delta$ is a perfect square, upon verification, only $m=8$ meets the requirement.
Substituting $m=8$ into the equation, we get
$$
a b=17 m-m^{2}=72 \text {. }
$$
|
72
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the function $f(x)=\frac{3+x}{1+x}$. Let
$$
\begin{array}{l}
f(1)+f(2)+f(4)+\cdots+f(1024)=m, \\
f\left(\frac{1}{2}\right)+f\left(\frac{1}{4}\right)+\cdots+f\left(\frac{1}{1024}\right)=n .
\end{array}
$$
Then $m+n=$ . $\qquad$
|
Ni, 7.42.
From $f(x)=1+\frac{2}{1+x}$, we know $f\left(\frac{1}{x}\right)=1+\frac{2 x}{1+x}$.
Therefore, $f(x)+f\left(\frac{1}{x}\right)=4$.
Also, $f(1)=2$, so, $m+n=4 \times 10+2=42$.
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $n$ be a positive integer less than 100, and satisfies $\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n$ is an integer. Then the sum of all positive integers $n$ that meet the condition is $\qquad$
|
11. 635 .
Notice that,
$$
\frac{1}{3}\left(n^{2}-1\right)+\frac{1}{5} n=\frac{5 n^{2}+3 n-5}{15}
$$
is an integer, so, $15 \mid \left(5 n^{2}+3 n-5\right)$.
Thus, $5 \mid n$, and $3 \mid \left(n^{2}-1\right)$.
Therefore, $n=15 k+5$ or $15 k+10$.
Hence, the sum of all positive integers $n$ that satisfy the condition is
$$
\begin{array}{c}
\sum_{k=0}^{6}(15 k+5)+\sum_{k=0}^{5}(15 k+10) \\
=95+\sum_{k=0}^{5}(30 k+15) \\
=95+\frac{(15+165) \times 6}{2}=635 .
\end{array}
$$
|
635
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 8 Suppose $N$ consecutive positive integers satisfy the following conditions: the sum of the digits of the 1st number is divisible by 1, the sum of the digits of the 2nd number is divisible by 2, $\cdots$. The sum of the digits of the $N$th number is divisible by $N$. Find the maximum possible value of $N$.
|
Let the $N$ numbers be $a_{1}, a_{2}, \cdots, a_{N}$.
If $N \geqslant 22$, then among $a_{2}, a_{3}, \cdots, a_{21}$, these 20 numbers, at least two numbers have a units digit of 9, and among these two numbers, at least one has a tens digit that is not 9, let this number be $a_{i}$.
Thus, $i \leqslant 21$.
If $i$ is even, then $i \leqslant 20$.
Since $i \mid S\left(a_{i}\right)$, $S\left(a_{i}\right)$ is also even.
Since $S\left(a_{i+2}\right)=S\left(a_{i+1}\right)+1=S\left(a_{i}\right)-7$ is odd, we get $(i+2) \times S\left(a_{i+2}\right)$.
Therefore, $N \leqslant i+1 \leqslant 21$.
If $i$ is odd, then from $(i-1) \mid S\left(a_{i-1}\right)$, we get $S\left(a_{i-1}\right)$ is even, hence $S\left(a_{i+1}\right)$ is odd.
So, $(i+1) \times S\left(a_{i+1}\right)$.
Thus, $N \leqslant i \leqslant 21$.
In summary, $N \leqslant 21$.
Below is an example for 21.
$$
\begin{array}{l}
\text { By }[2,3, \cdots, 11]=9 \times 3080 \text {, we know that we can let } \\
a_{1}=38 \underset{3078 \uparrow}{99 \ldots 989} \text {, } \\
a_{i}=a_{1}+(i-1)(i=2,3, \cdots, 21) . \\
\end{array}
$$
Thus $2 \mid S\left(a_{2}\right)=9 \times 3080+2$,
$$
k \mid S\left(a_{k}\right)=9 \times 3081+k(k=3,4, \cdots, 11) \text {. }
$$
For $k=12,13, \cdots, 21$, $S\left(a_{k}\right)=k$.
Thus $k \mid S\left(a_{k}\right)$, but $22 \times S\left(a_{22}\right)=13$.
|
21
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Let $f(x)=x+\frac{1}{x}(x>0)$. If for any positive number $a$, there exist $m+1$ real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in the interval $\left[1, a+\frac{2013}{a}\right]$, such that the inequality
$$
f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right)<f\left(a_{m+1}\right)
$$
holds, find the maximum value of $m$.
|
II. 9. Let $a=\sqrt{2013}$, then there exist $m+1$ real numbers that meet the requirements in the interval $[1,2 \sqrt{2013}]$.
Notice that $[1,2 \sqrt{2013}] \subseteq\left[1, a+\frac{2013}{a}\right]$.
Therefore, we only need to consider the existence of real numbers $a_{1}, a_{2}, \cdots, a_{m+1}$ in $[1,2 \sqrt{2013}]$.
It is easy to see that $f(x)$ is an increasing function in the interval $[1,2 \sqrt{2013}]$.
Thus, $f(1) \leqslant f\left(a_{i}\right)(i=1,2, \cdots, m)$, $f\left(a_{m+1}\right) \leqslant f(2 \sqrt{2013})$.
Adding the first $m$ inequalities, we get
$$
\begin{array}{l}
m f(1) \leqslant f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{m}\right) \\
<f\left(a_{m+1}\right) \leqslant f(2 \sqrt{2013}),
\end{array}
$$
which implies $m<\sqrt{2013}+\frac{1}{4 \sqrt{2013}}<45$.
Therefore, $m \leqslant 44$.
When $m=44$, take $a_{1}=a_{2}=\cdots=a_{44}=1$, $a_{45}=2 \sqrt{2013}$, then the inequality in the problem holds.
Hence, the maximum value of $m$ is 44.
|
44
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) In an annual super football round-robin tournament, 2013 teams each play one match against every other team. Each match awards 3 points to the winner, 0 points to the loser, and 1 point to each team in the event of a draw. After the tournament, Jia told Yi the total points of his team, and Yi immediately knew the number of wins and losses of Jia's team in the entire tournament. What is the total score of Jia's team in this tournament?
|
10. Consider the case with $n$ teams.
Let the team where Jia is located win $x$ games, draw $y$ games, and lose $z$ games. Then the total score of the team is
$$
S=3 x+y \quad (x+y+z=n-1, x, y, z \geqslant 0).
$$
Consider the region $\Omega:\left\{\begin{array}{l}x+y \leqslant n-1, \\ x \geqslant 0, \\ y \geqslant 0 .\end{array}\right.$
When $S$ changes, the equation $S=3 x+y$ represents a set of parallel lines with a slope of -3. We need to determine the value of $S$ such that the line $l$ intersects the region $\Omega$ at exactly one integer point, and we call such a line $l$ a "good line."
Consider the intersection point $A(0, S)$ of the line $l$ with the $y$-axis.
When $S=0,1,2$, $l$ is a good line.
When $3 \leqslant S \leqslant n-1$, $l$ passes through at least two integer points in the region $\Omega$, so $l$ is not a good line.
When $S \geqslant n$, consider the intersection point $B\left(\frac{S}{3}, 0\right)$ of the line $l$ with the $x$-axis.
When $S=3 n-9$, $l$ passes through $B_{1}(n-3,0)$, and additionally, $l$ passes through the integer point $(n-4,3)$ in the region $\Omega$, so $l$ is not a good line.
When $S \leqslant 3 n-9$, the intersection point of $l$ with the $x$-axis lies on the segment
$O B_{1}$, and $l$ passes through at least two integer points in the region $\Omega$, so $l$ is not a good line.
When $S=3 n-4$, as shown in Figure 2, $l$ does not pass through any integer points in the region $\Omega$, so $l$ is not a good line.
When $3 n-8 \leqslant S \leqslant 3 n-5$ or $S=3 n-3$, as shown in Figure 2, $l$ passes through exactly one integer point in the region $\Omega$, so $l$ is a good line.
In summary, $S=0,1,2,3 n-8,3 n-7,3 n-6,3 n-5,3 n-3$.
Taking $n=2013$, we get
$$
S=0,1,2,6031,6032,6033,6034,6036 .
$$
|
6034
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A set of 4027 points in the plane is called a "Colombian point set", where no three points are collinear, and 2013 points are red, 2014 points are blue. Draw a set of lines in the plane, which can divide the plane into several regions. If a set of lines for a Colombian point set satisfies the following two conditions, it is called a "good line set":
(1) These lines do not pass through any point in the Colombian point set;
(2) No region contains points of both colors.
Find the minimum value of $k$, such that for any Colombian point set, there exists a good line set consisting of $k$ lines.
|
2. $k=2013$.
Solution 1 First, give an example to show that $k \geqslant 2013$.
Mark 2013 red points and 2013 blue points alternately on a circle, and color another point in the plane blue. This circle is divided into 4026 arcs, each with endpoints of different colors. If the requirement of the problem is to be met, each arc must intersect with some drawn line. Since each line can intersect the circle at most twice, at least $\frac{4026}{2}=2013$ lines are needed.
Next, prove that 2013 lines can meet the requirement.
Notice that for any two points of the same color $A$ and $B$, two lines can be used to separate them from other points. The method is: draw two lines parallel to $AB$ on either side of $AB$. As long as they are sufficiently close to $AB$, the strip between them will contain only the two colored points $A$ and $B$.
Let $P$ be the convex hull of all colored points, and consider the following two cases.
(1) Assume $P$ has a red vertex, denoted as $A$. Then a line can be drawn to separate point $A$ from all other colored points. This way, the remaining 2012 red points can be paired into 1006 pairs, and each pair can be separated from all other colored points using two parallel lines. Therefore, a total of 2013 lines can meet the requirement.
(2) Assume all vertices of $P$ are blue. Consider two adjacent vertices on $P$, denoted as $A$ and $B$. Then a line can be drawn to separate these two points from all other colored points. This way, the remaining 2012 blue points can be paired into 1006 pairs, and each pair can be separated from all other colored points using two lines. Therefore, a total of 2013 lines can meet the requirement.
[Note] The convex hull can be ignored, and only a line passing through two colored points $A$ and $B$ can be considered, such that all other colored points are on one side of this line. If $A$ and $B$ include a red point, then proceed as in (1); if $A$ and $B$ are both blue, then proceed as in (2).
If $m$ is odd, then there exists an $N$ such that for any $m \leqslant n \leqslant N$, $f(m, n)=m$.
For any $n>N$, $f(m, n)=m+1$.
Solution 2 Another proof that 2013 lines can meet the requirement.
First, give a more general conclusion: If there are $n$ marked points in the plane with no three points collinear, and these points are arbitrarily colored red or blue, then $\left[\frac{n}{2}\right]$ lines can meet the requirement of the problem, where $[x]$ denotes the greatest integer not exceeding the real number $x$.
Use induction on $n$.
When $n \leqslant 2$, the conclusion is obvious.
Assume $n \geqslant 3$.
Consider a line passing through two colored points $A$ and $B$ such that all other colored points are on one side of this line. For example, an edge of the convex hull of all colored points is such a line.
Temporarily remove points $A$ and $B$ from consideration. By the induction hypothesis, the remaining points can be separated using $\left[\frac{n}{2}\right]-1$ lines. Now, re-include points $A$ and $B$, and consider three cases.
(1) If points $A$ and $B$ are the same color, then a line parallel to $l$ can be drawn to separate points $A$ and $B$ from other colored points. Clearly, the $\left[\frac{n}{2}\right]$ lines thus obtained can meet the requirement.
(2) If points $A$ and $B$ are different colors, but are separated by some already drawn line, then the line parallel to $l$ also meets the requirement.
(3) If points $A$ and $B$ are different colors and are in the same region after drawing the $\left[\frac{n}{2}\right]-1$ lines. By the induction hypothesis, at least one color will have no other colored points in that region. Without loss of generality, assume the only blue point in that region is $A$. Then, a line can be drawn to separate point $A$ from all other colored points.
Thus, the induction step is completed.
[Note] Generalize the problem by replacing 2013 and 2014 with any positive integers $m$ and $n$, assuming $m \leqslant n$. Denote the solution to the corresponding problem as $f(m, n)$.
Following the idea of Solution 1, we get $m \leqslant f(m, n) \leqslant m+1$.
If $m$ is even, then $f(m, n)=m$.
|
2013
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $a_{1}, a_{2}, \cdots, a_{100}$ are 100 distinct positive integers. For any positive integer $i \in\{1,2, \cdots, 100\}, d_{i}$ represents the greatest common divisor of the 99 numbers $a_{j}(j \neq i)$, and $b_{i}=a_{i}+$ $d_{i}$. Question: How many different positive integers are there at least in $b_{1}, b_{2}, \cdots, b_{100}$?
|
3. Contains at least 99 different positive integers.
If we let $a_{100}=1, a_{i}=2 i(1 \leqslant i \leqslant 99)$; then $b_{1}=b_{100}=3$.
This indicates that there are at most 99 different positive integers in $b_{i}$.
Next, we prove: there are at least 99 different positive integers in $b_{i}$.
Without loss of generality, assume $a_{1}a_{j} \geqslant a_{i}+d_{i}=b_{i} .
\end{array}
$$
This indicates that $b_{i}(i \neq k)$ are all different.
|
99
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $A$ be a set of ten real-coefficient quadratic polynomials. It is known that there exist $k$ consecutive positive integers $n+1$, $n+2, \cdots, n+k$, and $f_{i}(x) \in A(1 \leqslant i \leqslant k)$, such that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence. Find the maximum possible value of $k$.
|
6. Given that $f_{1}(n+1), f_{2}(n+2), \cdots, f_{k}(n+k)$ form an arithmetic sequence, we know there exist real numbers $a$ and $b$ such that
$$
f_{i}(n+i)=a i+b .
$$
Notice that for any quadratic polynomial $f$, the equation
$$
f(n+x)=a x+b
$$
has at most two real roots. Therefore, each polynomial in $A$ appears at most twice in $f_{1}, f_{2}, \cdots, f_{k}$.
Thus, $k \leqslant 20$.
Below is an example for $k=20$.
$$
\text { Let } P_{i}(x)=[x-(2 i-1)](x-2 i)+x \text {, }
$$
where $i=1,2, \cdots, 10$.
Then $f_{2 i-1}=f_{2 i}=P_{i}$.
Hence, $f_{i}(i)=i(1 \leqslant i \leqslant 20)$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
17. Alex has 75 red cards and 75 blue cards. It is known that Alex can exchange 2 red cards for 1 silver card and 1 blue card at one stall, and can exchange 3 blue cards for 1 silver card and 1 red card at another stall. If he continues to exchange according to the above methods until he can no longer exchange any cards, how many silver cards will Alex have in the end?
(A) 62
(B) 82
( C) 83
(.D) 102
(E) 103
|
17. E.
If Alex has $a$ red cards, $b$ blue cards, and $c$ silver cards, denoted as $(a, b, c)$, then
$$
\begin{array}{l}
(75,75,0) \rightarrow(1,112,37) \\
\rightarrow(38,1,74) \rightarrow(0,20,93) \\
\rightarrow(6,2,99) \rightarrow(0,5,102) \\
\rightarrow(1,2,103) .
\end{array}
$$
Therefore, Alex has 103 silver cards.
|
103
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
23. In $\triangle A B C$, it is known that $A B=13, B C=14$, $C A=15$, points $D, E, F$ are on sides $B C, C A, D E$ respectively, satisfying $A D \perp B C, D E \perp A C, A F \perp B F$, the length of segment $D F$ is a reduced fraction $\frac{m}{n}\left(m, n \in \mathbf{N}_{+},(m, n)=1\right)$. Then $m+n=(\quad)$.
(A) 18
(B) 21
(C) 24
(D) 27
(E) 30
|
23. B.
As shown in Figure 5.
Let $p$ be the semi-perimeter of $\triangle ABC$. Then
$$
p=\frac{a+b+c}{2}=\frac{13+14+15}{2}=21.
$$
Thus, $S_{\triangle ABC}=\sqrt{p(p-a)(p-b)(p-c)}$
$$
\begin{array}{l}
=\sqrt{21 \times 8 \times 7 \times 6}=84 \\
\Rightarrow AD=\frac{2 S_{\triangle ABC}}{BC}=\frac{2 \times 84}{14}=12.
\end{array}
$$
Since $AD \perp BC$, we have
$$
BD=\sqrt{AB^2-AD^2}=\sqrt{13^2-12^2}=5.
$$
Given $BC=14$, then $DC=9$.
Since $DE \perp AC$, we have
$$
\begin{array}{l}
DE=\frac{AD \cdot DC}{AC}=\frac{12 \times 9}{15}=\frac{36}{5}, \\
AE=\frac{AD^2}{AC}=\frac{12^2}{15}=\frac{48}{5}.
\end{array}
$$
Since $\angle AFB=\angle ADB=90^\circ$, points $A, B, D, F$ are concyclic.
Therefore, $\angle AFE=\angle ABC$.
Thus, $\cot \angle AFE=\cot \angle ABC=\frac{5}{12}$.
In $\triangle AEF$,
$$
\begin{array}{l}
EF=AE \cot \angle AFE=\frac{48}{5} \times \frac{5}{12}=4 \\
\Rightarrow DF=DE-EF=\frac{36}{5}-4=\frac{16}{5} \\
\Rightarrow m=16, n=5 \\
\Rightarrow m+n=21.
\end{array}
$$
|
21
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given $x, y, z \in \mathbf{R}_{+}$,
$$
\begin{array}{l}
S=\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}, \\
T=\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1} .
\end{array}
$$
Then the minimum value of $S^{2}-T^{2}$ is
|
4. 36 .
$$
\begin{array}{l}
S^{2}-T^{2}=(S+T)(S-T) \\
=(\sqrt{x+2}+\sqrt{x+1}+\sqrt{y+5}+ \\
\quad \sqrt{y+1}+\sqrt{z+10}+\sqrt{z+1}) \cdot \\
\quad\left(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{4}{\sqrt{y+5}+\sqrt{y+1}}+\frac{9}{\sqrt{z+10}+\sqrt{z+1}}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$
The equality holds if and only if $x=\frac{7}{9}, y=\frac{55}{9}, z=15$.
|
36
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
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