problem
stringlengths 2
5.64k
| solution
stringlengths 2
13.5k
| answer
stringlengths 1
43
| problem_type
stringclasses 8
values | question_type
stringclasses 4
values | problem_is_valid
stringclasses 1
value | solution_is_valid
stringclasses 1
value | source
stringclasses 6
values | synthetic
bool 1
class |
|---|---|---|---|---|---|---|---|---|
1. Divide the sequence of positive integers $1,2, \cdots$ from left to right into segments such that the first segment has $1 \times 2$ numbers, the second segment has $2 \times 3$ numbers, $\cdots$, the $n$-th segment has $n \times(n+1)$ numbers, $\cdots$. Then 2014 is in the $\qquad$ segment.
|
$-, 1.18$.
$$
\begin{array}{l}
\text { Let } S_{n}=1 \times 2+2 \times 3+\cdots+n(n+1) \\
=\left(1^{2}+2^{2}+\cdots+n^{2}\right)+(1+2+\cdots+n) \\
=\frac{n(n+1)(n+2)}{3} .
\end{array}
$$
If 2014 is in the $(n+1)$-th segment, since there are $S_{n}$ numbers before this segment, then $S_{n}<2014 \leqslant S_{n+1}$.
Since $S_{17}=1938, S_{18}=2280$, and
$$
S_{17}<2014 \leqslant S_{18},
$$
Therefore, 2014 is in the 18th segment.
|
18
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C: x^{2}-\frac{y^{2}}{24}=1$, respectively, and let $P$ be a point on the hyperbola $C$ in the first quadrant. If $\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{4}{3}$, then the radius of the incircle of $\triangle P F_{1} F_{2}$ is . $\qquad$
|
4. 2 .
Let $\left|P F_{1}\right|=4 t$. Then $\left|P F_{2}\right|=3 t$.
Thus $4 t-3 t=\left|P F_{1}\right|-\left|P F_{2}\right|=2$
$$
\Rightarrow t=2,\left|P F_{1}\right|=8,\left|P F_{2}\right|=6 \text {. }
$$
Combining with $\left|F_{1} F_{2}\right|=10$, we know that $\triangle P F_{1} F_{2}$ is a right triangle, $P F_{1} \perp P F_{2}$.
Therefore, the inradius of $\triangle P F_{1} F_{2}$ is
$$
r=\frac{6+8-10}{2}=2 \text {. }
$$
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If the fraction $\frac{p}{q}\left(p, q \in \mathbf{Z}_{+}\right)$ is converted to a decimal as
$$
\frac{p}{q}=0.198 \cdots,
$$
then when $q$ takes the minimum value, $p+q=$ . $\qquad$
|
6. 121 .
Given $\frac{p}{q}=0.198 \cdots5 p$.
Let $q=5 p+m\left(m \in \mathbf{Z}_{+}\right)$. Then
$$
\begin{array}{l}
\frac{p}{5 p+m}=0.198 \cdots \\
\Rightarrow 0.198(5 p+m)<p<0.199(5 p+m) \\
\Rightarrow 19.8 m<p<39.8 m .
\end{array}
$$
When $m=1$, $20 \leqslant p \leqslant 39$, taking $p=20, m=1$, $q$ is minimized to 101.
Upon verification, $\frac{20}{101}=0.19801980 \cdots$ meets the requirement. Therefore, when $q$ is minimized, $p+q=121$.
|
121
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given points $A(1,-1), B(4,0), C(2,2)$, the plane region $D$ consists of all points $P(x, y)$ that satisfy
$$
\overrightarrow{A P}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}(1<\lambda \leqslant a, 1<\mu \leqslant b)
$$
If the area of region $D$ is 8, then the minimum value of $a+b$ is . $\qquad$
|
8. 4 .
As shown in Figure 3, extend $A B$ to point $N$, and extend $A C$ to point $M$, such that
$$
|A N|=a|A B|,|A M|=b|A C| .
$$
Construct $\square A B E C$ and $\square A N G M$. Then quadrilateral $E H G F$ is a parallelogram.
From the conditions, the region $D$ composed of points $P(x, y)$ is the shaded area in the figure, i.e., $\square E H G F$ (excluding the boundaries $E H$ and $E F$).
Notice that,
$$
\overrightarrow{A B}=(3,1), \overrightarrow{A C}=(1,3), \overrightarrow{B C}=(-2,2) \text {. }
$$
Thus, $|A B|=|A C|=\sqrt{10},|B C|=2 \sqrt{2}$.
Then $\cos \angle C A B=\frac{10+10-8}{2 \times \sqrt{10} \times \sqrt{10}}=\frac{3}{5}$
$$
\Rightarrow \sin \angle C A B=\frac{4}{5} \text {. }
$$
Therefore, $S_{\text {DEHGF }}$
$$
\begin{aligned}
= & (a-1) \sqrt{10}(b-1) \sqrt{10} \times \frac{4}{5}=8 \\
\Rightarrow & (a-1)(b-1)=1 \\
\Rightarrow & a+b=a+\left(\frac{1}{a-1}+1\right) \\
= & (a-1)+\frac{1}{a-1}+2 .
\end{aligned}
$$
Given $a>1, b>1$, we know that when and only when $a-1=1$, i.e., $a=b=2$, $a+b$ reaches its minimum value of 4.
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. $A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{0114}}{9}\right]$ when divided by 63 leaves a remainder of $\qquad$ ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
9. 56 .
Notice that, for any positive integer $k$, $\frac{8^{2 k-1}}{9}$ and $\frac{8^{2 k}}{9}$ are not integers, and
$$
\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}=8^{2 k-1} \text {. }
$$
Therefore, for any positive integer $k$, we have
$$
\begin{array}{l}
{\left[\frac{8^{2 k-1}}{9}\right]+\left[\frac{8^{2 k}}{9}\right]=\frac{8^{2 k-1}}{9}+\frac{8^{2 k}}{9}-1} \\
=8^{2 k-1}-1 \equiv 7(\bmod 63) . \\
\text { Hence } A=\left[\frac{8}{9}\right]+\left[\frac{8^{2}}{9}\right]+\cdots+\left[\frac{8^{2014}}{9}\right] \\
\equiv 1007 \times 7 \equiv 56(\bmod 63) .
\end{array}
$$
|
56
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For any positive integer $n$, the function $f(n)$ is the sum of the digits (i.e., the digital sum) of $n^{2}+3 n+1$ in decimal notation. Question: Does there exist an integer $n$ such that
$$
f(n)=2013 \text { or } 2014 \text { or } 2015 \text { ? }
$$
|
When $3 \mid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 1(\bmod 3) ;
$$
When $3 \nmid n$,
$$
f(n) \equiv n^{2}+3 n+1 \equiv 2(\bmod 3) \text {. }
$$
Thus, $3 \nmid f(n)$.
Since 312013, there does not exist an integer $n$ such that
$$
f(n)=2013 \text {. }
$$
If there exists an integer $n$ such that $f(n)=2014$, then by $2014 \equiv 1(\bmod 3) \Rightarrow 3 \mid n$. Hence $n^{2}+3 n+1 \equiv 1 \equiv f(n)(\bmod 9)$. But $f(n) \equiv 7(\bmod 9)$, which is a contradiction.
Therefore, there does not exist an integer $n$ such that $f(n)=2014$.
When $k>2, n=10^{k}-3$,
$$
\begin{array}{l}
n^{2}+3 n+1=\left(10^{k}-3\right)^{2}+3\left(10^{k}-3\right)+1 \\
=\underset{k-1 \uparrow}{99 \cdots 9} 7 \underset{k-1 \uparrow}{00 \cdots 01} .
\end{array}
$$
Since $2015=9 \times 223+8$, when $k=224$, i.e., $n=10^{224}-3$, $f(n)=2015$.
(Song Hongjun, Zhejiang Fuyang High School, 311400)
|
2015
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. A certain meeting has 30 participants, each of whom knows at most five others; among any five people, at least two are not acquaintances. Find the largest positive integer $k$, such that in any group of 30 people satisfying the above conditions, there always exists a group of $k$ people, none of whom are acquaintances.
(Cold Gangsong, Contributed)
|
5, hence we have
$30-|X| \leqslant 5|X|$.
Therefore, $|X| \geqslant 5$.
If $|X|=5$, then by equation (1) the equality is achieved, which means the 25 edges are distributed among the five vertices in set $X$, i.e., the neighborhood of each vertex in set $X$ is a set of five points in set $V \backslash X$.
Since $|V \backslash X|=25$, the neighborhoods of any two vertices in set $X$ do not intersect.
Let $X=\{a, b, c, d, e\}$.
Now consider the neighborhood of $a$, denoted as
$$
Y_{a}=\left\{y_{1}, y_{2}, y_{3}, y_{4}, y_{5}\right\}.
$$
By (ii), we know that there are two points in set $Y_{a}$ that are not connected, let them be $y_{1}$ and $y_{2}$.
Since the neighborhoods of any two vertices in set $X$ do not intersect, $y_{1}$ and $y_{2}$ cannot be neighbors of any of $b, c, d, e$.
Thus, $\left\{y_{1}, y_{2}, b, c, d, e\right\}$ is an independent set of graph $G$ with more than 5 elements, which is a contradiction.
This proves that $|X| \geqslant 6$.
(2) Next, we prove: there exists a graph $G$ that meets the conditions, and the maximum independent set has no more than 6 elements.
Partition the set $V$ into point sets $V_{1}, V_{2}, V_{3}$, such that
$\left|V_{i}\right|=10(i=1,2,3)$.
Let $V_{1}=\left\{A_{1}, A_{2}, \cdots, A_{5}, B_{1}, B_{2}, \cdots, B_{5}\right\}$, connect the points in set $V_{1}$ as shown in Figure 3, i.e.,
(i) $A_{i} A_{i+1}(i=1,2, \cdots, 5)$ are connected;
(ii) $B_{i} B_{i+1}(i=1,2, \cdots, 5)$ are connected;
(iii) $A_{i} B_{i}, A_{i} B_{i+1}, A_{i} B_{i-1}(i=1,2, \cdots, 5)$ are connected, where $A_{6}=A_{1}, B_{6}=B_{1}, B_{0}=B_{5}$.
The connection method for point sets $V_{2}$ and $V_{3}$ is the same as for set $V_{1}$, and for any $1 \leqslant i<j \leqslant 3$, there are no edges between point sets $V_{i}$ and $V_{j}$. Then, the degree of any vertex in graph $G$ is 5, and in graph $G$, among any five points, there are always two points that are not connected.
Now, take any maximum independent set $X$ of graph $G$.
We only need to prove: $\left|V_{1} \cap X\right| \leqslant 2$.
In fact, since $A_{i} 、 A_{i+1}(i=1,2, \cdots, 5)$ are adjacent, at most two of $A_{1}, A_{2}, \cdots, A_{5}$ belong to set $X$.
Similarly, at most two of $B_{1}, B_{2}, \cdots, B_{5}$ belong to set $X$.
If exactly two of $A_{1}, A_{2}, \cdots, A_{5}$ belong to set $X$, let them be $\left\{A_{1}, A_{3}\right\}$.
Notice that, the union of the neighborhoods of $A_{1}$ and $A_{3}$ is exactly $\left\{B_{1}, B_{2}, \cdots, B_{5}\right\}$. Hence, $B_{1}, B_{2}, \cdots, B_{5}$ do not belong to set $X$.
Similarly, if exactly two of $B_{1}, B_{2}, \cdots, B_{5}$ belong to set $X$, then $A_{1}, A_{2}, \cdots, A_{5}$ do not belong to set $X$. This proves that $\left|V_{1} \cap X\right| \leqslant 2$.
Similarly, $\left|V_{2} \cap X\right| \leqslant 2,\left|V_{3} \cap X\right| \leqslant 2$.
Thus, $|X|=|V \cap X|$
$$
=\left|V_{1} \cap X\right|+\left|V_{2} \cap X\right|+\left|V_{3} \cap X\right| \leqslant 6.
$$
Therefore, graph $G$ meets the requirements.
Combining (1) and (2), we know that the value of $k$ is 6.
|
6
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A. The minimum value of the algebraic expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$ is
|
$=、 6$. A. 13.
This problem can be transformed into finding the minimum value of the sum of distances from a point $(x, 0)$ on the $x$-axis to the points $(0,2)$ and $(12,3)$ in a Cartesian coordinate system.
Since the symmetric point of $(0,2)$ with respect to the $x$-axis is $(0,-2)$, the length of the line segment connecting $(0,-2)$ and $(12,3)$ is the minimum value of the expression $\sqrt{x^{2}+4}+\sqrt{(12-x)^{2}+9}$.
In fact, when $x=\frac{24}{5}$, the expression reaches its minimum value
$$
\sqrt{(12-0)^{2}+(3+2)^{2}}=13 \text {. }
$$
|
13
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let real numbers $x_{1}, x_{2}, \cdots, x_{2014}$ satisfy
$$
\left|x_{1}\right|=99,\left|x_{n}\right|=\left|x_{n-1}+1\right| \text {, }
$$
where, $n=2,3, \cdots, 2014$. Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$.
|
11. From the given, we have
$$
x_{n}^{2}=x_{n-1}^{2}+2 x_{n}+1(n=2,3, \cdots, 2014) \text {. }
$$
Adding the above 2013 equations, we get
$$
\begin{array}{l}
x_{2014}^{2}=x_{1}^{2}+2\left(x_{1}+x_{2}+\cdots+x_{2013}\right)+2013 \\
\Rightarrow 2\left(x_{1}+x_{2}+\cdots+x_{2014}\right) \\
\quad=x_{2014}^{2}+2 x_{2014}-2013-x_{1}^{2} \\
\quad=\left(x_{2014}+1\right)^{2}-2014-99^{2} .
\end{array}
$$
Since $x_{1}$ is odd, from the given conditions, we know that $x_{2}$ is even, $x_{3}$ is odd, $\cdots, x_{2014}$ is even.
Thus, $\left(x_{2014}+1\right)^{2} \geqslant 1$.
Therefore, $x_{1}+x_{2}+\cdots+x_{2014}$
$$
\geqslant \frac{1}{2}\left(1-2014-99^{2}\right)=-5907 \text {. }
$$
When $x_{1}=-99, x_{2}=-98, \cdots, x_{99}=-1, x_{100}$
$$
=0, x_{101}=-1, x_{102}=0, x_{103}=-1, \cdots, x_{2013}=
$$
$-1, x_{2014}=0$, the equality holds.
Hence, the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$ is -5907.
|
-5907
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Consider each permutation of $1,2, \cdots, 8$ as an eight-digit number. Then the number of eight-digit numbers that are multiples of 11 is $\qquad$
|
8. 4608 .
For each such eight-digit number $\overline{a_{1} a_{2} \cdots a_{8}}$, let
$$
A=\left\{a_{1}, a_{3}, a_{5}, a_{7}\right\}, B=\left\{a_{2}, a_{4}, a_{6}, a_{8}\right\} .
$$
Let $S(A)$ and $S(B)$ denote their digit sums, and assume $S(A) \geqslant S(B)$.
Then $S(A)+S(B)=36$.
Thus, $S(A)$ and $S(B)$ have the same parity, and $S(A)+S(B)$ and $S(A)-S(B)$ have the same parity.
Therefore, $S(A)-S(B)$ is an even number and a multiple of 11. If $S(A)-S(B)=22$, then $S(B)=7$, which is impossible (since the sum of the smallest four numbers is at least 10).
Thus, $S(A)-S(B)=0$, i.e.,
$S(A)=S(B)=18$.
Hence, $S(A)=S(B)$ has only four cases:
1278, 3456; 1368, 2457;
1467, 2358; 2367, 1458.
In each group, the four numbers can be fully permuted, resulting in $4! \times 4!$ eight-digit numbers.
Considering the two scenarios for the four numbers in odd or even positions, the total number of eight-digit numbers that satisfy the condition is
$$
4 \times 2 \times 4! \times 4! = 4608 \text{ (numbers). }
$$
|
4608
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given natural numbers $a, b, c$ whose sum is $S$, satisfying $a+b=1014, c-b=497, a>b$. Then the maximum value of $S$ is ( ).
(A) 2014
(B) 2015
(C) 2016
(D) 2017
|
-1. D.
From the given, we have $a \geqslant b+1$.
Then $1014=a+b \geqslant 2 b+1$
$$
\begin{array}{l}
\Rightarrow b \leqslant 506.5 \Rightarrow b \leqslant 506 . \\
\text { Also } S=(a+b)+(c-b)+b \\
=1014+497+b=1511+b \\
\leqslant 1511+506=2017,
\end{array}
$$
Therefore, the maximum value of $S$ is 2017.
|
2017
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given a set of data consisting of seven positive integers, the only mode is 6, and the median is 4. Then the minimum value of the sum of these seven positive integers is $\qquad$
|
2, 1.26.
Arrange these seven positive integers in ascending order, it is clear that the fourth number is 4. If 6 appears twice, then the other four numbers are $1, 2, 3, 5$, their sum is smaller, being 27; if 6 appears three times, the other three numbers are $1, 1, 2$, their sum is 26.
Therefore, the minimum sum of the seven positive integers is 26.
|
26
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{y^{2}+4}-2\right) \geqslant y>0 \text {. }
$$
Then the minimum value of $x+y$ is $\qquad$.
|
2. 2 .
Notice,
$$
\begin{array}{l}
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right) \geqslant 1 \\
\Rightarrow 2 x+\sqrt{4 x^{2}+1} \\
\geqslant \frac{1}{\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}}=\sqrt{1+\frac{4}{y^{2}}}+\frac{2}{y} .
\end{array}
$$
When $x=\frac{1}{y}$,
$$
\left(2 x+\sqrt{4 x^{2}+1}\right)\left(\sqrt{1+\frac{4}{y^{2}}}-\frac{2}{y}\right)=1 .
$$
And $f(x)=2 x+\sqrt{4 x^{2}+1}$ is monotonically increasing, so $x \geqslant \frac{1}{y}$.
Thus, $x+y \geqslant y+\frac{1}{y} \geqslant 2$.
When $x=y=1$, the equality holds.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Given the sequence $\left\{a_{n}\right\}$ satisfies
$a_{1}=1, a_{2 n}=\left\{\begin{array}{ll}a_{n}, & n \text { is even; } \\ 2 a_{n}, & n \text { is odd, }\end{array}\right.$
$a_{2 n+1}=\left\{\begin{array}{ll}2 a_{n}+1, & n \text { is even; } \\ a_{n}, & n \text { is odd. }\end{array}\right.$
Find the number of positive integers $n$ in the range $[1,2014]$ that satisfy $a_{n}=a_{2014}$.
|
Solve: From the given, we have
$$
\begin{array}{l}
a_{2014}=2 a_{1007}=2 a_{503}=2 a_{251}=2 a_{125} \\
=2\left(2 a_{62}+1\right)=2\left(4 a_{31}+1\right) \\
=2\left(4 a_{15}+1\right)=2\left(4 a_{7}+1\right) \\
=2\left(4 a_{3}+1\right)=2\left(4 a_{1}+1\right)=10 .
\end{array}
$$
In fact, $a_{n}$ can be obtained through the following process.
Write $n$ in binary, then replace consecutive "1"s with a single "1" and consecutive "0"s with a single "0". Denote this sequence as $\left\{b_{n}\right\}$.
For example, $2014=(11111011110)_{2}$,
$b_{2014}=(1010)_{2}=10$.
We will prove by mathematical induction that:
$$
a_{n}=b_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Clearly, $a_{1}=1=b_{1}, a_{2}=2=b_{2}, a_{3}=1=b_{3}$. Assume that $a_{t}=b_{t}$ holds for any positive integer $t<n$.
(1) When $n=2 k$, if $k$ is even, then $a_{2 k}=$ $a_{k}$, and the last two digits of $n=2 k$ in binary are 00, so $b_{2 k}$ $=b_{k}$, thus, $a_{2 k}=b_{2 k}$; if $k$ is odd, then $a_{2 k}=2 a_{k}$, and the last two digits of $n=2 k$ in binary are 10, so $b_{2 k}=$ $2 b_{k}$, thus, $a_{2 k}=b_{2 k}$.
(2) When $n=2 k+1$, if $k$ is even, then $a_{2 k+1}=2 a_{k}+1$, and the last two digits of $n=2 k$ in binary are 01, so $b_{2 k+1}=2 b_{k}+1$, thus, $a_{2 k+1}=b_{2 k+1}$; if $k$ is odd, then $a_{2 k+1}=a_{k}$, and the last two digits of $n=2 k$ in binary are 11, so $b_{2 k+1}=b_{k}$, thus, $a_{2 k+1}=$ $b_{2 k+1}$.
Therefore, the original problem is equivalent to
$a_{n}=b_{n}=b_{2014}=(1010)_{2}$
$=10$ the number of positive integers $n$
$\Leftrightarrow$ the number of $n$ such that when written in binary, it becomes $(1010)_{2}$ according to the rules.
Since $n<2048=2^{11}$, i.e., $n$ in binary has at most 11 digits, and numbers larger than
$$
2014=(11111011110)_{2}
$$
do not meet the requirements.
This can be understood as selecting 4 positions out of 11 to fill (1010), with 0s before the first 1 and the remaining spaces filled with the digit before the space.
The $n$ that do not meet the requirements can be understood as selecting 3 positions out of the last 5 to fill (010), with 1s before the first 0 and the remaining spaces filled with the digit before the space.
Therefore, there are $\mathrm{C}_{11}^{4}-\mathrm{C}_{5}^{3}=320$ positive integers $n$ that meet the requirements.
|
320
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$\begin{array}{l}\text { 2. If } \frac{y}{x}+\frac{x}{z}=a, \frac{z}{y}+\frac{y}{x}=b, \frac{x}{z}+\frac{z}{y}=c, \\ \text { then }(b+c-a)(c+a-b)(a+b-c)=\end{array}$
|
2.8.
Notice that,
$$
\begin{array}{l}
b+c-a=\left(\frac{z}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}\right)-\left(\frac{y}{x}+\frac{x}{z}\right) \\
=\frac{2 z}{y} .
\end{array}
$$
Similarly, $c+a-b=\frac{2 x}{z}, a+b-c=\frac{2 y}{x}$.
Therefore, $(b+c-a)(c+a-b)(a+b-c)$ $=\frac{2 z}{y} \cdot \frac{2 x}{z} \cdot \frac{2 y}{x}=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In the Cartesian coordinate plane $x O y$, the area of the region determined by the system of inequalities
$$
\left\{\begin{array}{l}
|x| \leqslant 2, \\
|y| \leqslant 2, \\
|| x|-| y|| \leqslant 1
\end{array}\right.
$$
is $\qquad$
|
6. 12 .
Obviously, the region is symmetric with respect to both the $x$-axis and the $y$-axis. Therefore, we can assume $x \geqslant 0, y \geqslant 0$.
From $\left\{\begin{array}{l}0 \leqslant x \leqslant 2, \\ 0 \leqslant y \leqslant 2, \\ -1 \leqslant x-y \leqslant 1\end{array}\right.$, draw the part of the region in the first quadrant (including the coordinate axes), and then symmetrically draw the entire region, as shown in Figure 4.
The area of the region sought is the area of a square with side length 4 minus the area of eight isosceles right triangles with leg length 1, i.e., $4^{2}-8 \times \frac{1}{2}=12$.
|
12
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that Figure 1 is the graph of an even function $f(x)$, and Figure 2 is the graph of an odd function $g(x)$.
Let the number of real roots of the equations $f(f(x))=0, f(g(x))=0$,
$$
g(g(x))=0, g(f(x))=0
$$
be $a, b, c, d$ respectively. Then
$$
a+b+c+d=
$$
$\qquad$
|
3. 30 .
From the graph, we know that the range of the function $y=f(x)$ is $[-1,1]$, and the range of $y=g(x)$ is $[-2,2]$.
Notice that, the roots of the equation $f(x)=0$ are $0, x_{1}, x_{2}$, with $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$;
The roots of the equation $g(x)=0$ are $0, x_{3}, x_{4}$, with $\left|x_{3}\right|=\left|x_{4}\right| \in(0,1)$.
Therefore, the roots of the equation $f(f(x))=0$ are the roots of the equations $f(x)=0$ or $f(x)=x_{1}$ or $f(x)=x_{2}$.
Since the range of the function $y=f(x)$ is $[-1,1]$, and $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$, the equations $f(x)=x_{1}$ and $f(x)=x_{2}$ have no roots, while the equation $f(x)=0$ has three roots. Thus, the number of roots of the equation $f(f(x))=0$ is 3.
The roots of the equation $f(g(x))=0$ are the roots of the equations $g(x)=0$ or $g(x)=x_{1}$ or $g(x)=x_{2}$.
Since the range of the function $y=g(x)$ is $[-2,2]$, and $\left|x_{1}\right|=\left|x_{2}\right| \in(1,2)$, the equations $g(x)=x_{1}$ and $g(x)=x_{2}$ each have three roots, and the equation $g(x)=0$ also has three roots. Thus, the number of roots of the equation $f(g(x))=0$ is 9.
Similarly, the equations $g(g(x))=0$ and $g(f(x))=0$ each have 9 roots.
Therefore, $a+b+c+d=3+9+9+9=30$.
|
30
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let the set $P=\{1,2, \cdots, 2014\}, A \cong P$. If any two numbers in set $A$ have a difference that is not a multiple of 99, and the sum of any two numbers is also not a multiple of 99, then the set $A$ can contain at most $\qquad$ elements.
|
5.50.
Let the set
$$
B_{i}=\{99 \times 1+i, 99 \times 2+i, \cdots, 99 \times 20+i\} \text {, }
$$
where, $i=0,1, \cdots, 34$;
$$
B_{j}=\{99 \times 1+j, 99 \times 2+j, \cdots, 99 \times 19+j\},
$$
where, $j=35,36, \cdots, 98$.
Take any $a, b \in A$.
Since the difference between any two numbers in set $A$ is not a multiple of 99, $a$ and $b$ do not belong to the same set $B_{i}$ or $B_{j}$.
Furthermore, since the sum of any two numbers in set $A$ is also not a multiple of 99, $a$ and $b$ are not a pair in the set
$$
M=\left\{\left(B_{1}, B_{98}\right),\left(B_{2}, B_{97}\right), \cdots,\left(B_{49}, B_{50}\right)\right\}
$$
Therefore, the set $A$ can have at most 50 elements, such as taking one element from each of the sets $B_{0}, B_{1}, \cdots, B_{49}$.
|
50
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given
$$
f(x)=x^{2}-53 x+196+\left|x^{2}-53 x+196\right| \text {. }
$$
Then $f(20)+f(14)=$ $\qquad$ .
|
$$
\begin{array}{l}
\text { II.7.0. } \\
\text { Let } g(x)=x^{2}-53 x+196 \\
=(x-4)(x-49)
\end{array}
$$
Then when $x<4$ or $x>49$, $g(x)>0$;
when $4 \leqslant x \leqslant 49$, $g(x) \leqslant 0$.
Thus, when $x=4,5, \cdots, 49$,
$$
f(x)=g(x)+|g(x)|=g(x)-g(x)=0 \text {. }
$$
Therefore, $f(20)+f(14)=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $0<x<\frac{\pi}{2}, \sin x-\cos x=\frac{\pi}{4}$. If $\tan x+\frac{1}{\tan x}$ can be expressed in the form $\frac{a}{b-\pi^{c}}$ ($a$, $b$, $c$ are positive integers), then $a+b+c=$ $\qquad$ .
|
8. 50 .
Squaring both sides of $\sin x-\cos x=\frac{\pi}{4}$ and rearranging yields $\sin x \cdot \cos x=\frac{16-\pi^{2}}{32}$.
Therefore, $\tan x+\frac{1}{\tan x}=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$
$$
=\frac{1}{\sin x \cdot \cos x}=\frac{32}{16-\pi^{2}} \text {. }
$$
Thus, $a=32, b=16, c=2$.
Hence, $a+b+c=50$.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
\text { 4. If } x_{1}>x_{2}>x_{3}>x_{4}>0, \text { and the inequality } \\
\log _{\frac{x_{1}}{2}} 2014+\log _{\frac{x_{2}}{x_{3}}} 2014+\log _{x_{4} \frac{1}{4}} 2014 \\
\geqslant k \log _{\frac{x_{1}}{}} 2014
\end{array}
$$
always holds, then the maximum value of the real number $k$ is . $\qquad$
|
4.9.
From the given, we have
$$
\begin{array}{l}
\frac{\ln 2014}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln 2014}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln 2014}{\ln \frac{x_{3}}{x_{4}}} \\
\geqslant k \frac{\ln 2014}{\ln \frac{x_{1}}{x_{4}}}.
\end{array}
$$
Since $x_{1}>x_{2}>x_{3}>x_{4}>0$, we have
$$
\begin{array}{l}
\ln \frac{x_{1}}{x_{2}}>0, \ln \frac{x_{2}}{x_{3}}>0, \ln \frac{x_{3}}{x_{4}}>0, \\
\ln \frac{x_{1}}{x_{4}}=\ln \frac{x_{1}}{x_{2}}+\ln \frac{x_{2}}{x_{3}}+\ln \frac{x_{3}}{x_{4}}>0. \\
\text{Therefore, } k \leqslant \frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{3}}{x_{4}}}. \\
\text{And } \frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{1}}{x_{2}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{2}}{x_{3}}}+\frac{\ln \frac{x_{1}}{x_{4}}}{\ln \frac{x_{3}}{x_{4}}} \\
=\left(\ln \frac{x_{1}}{x_{2}}+\ln \frac{x_{2}}{x_{3}}+\ln \frac{x_{3}}{x_{4}}\right) \cdot \\
\left(\frac{1}{\ln \frac{x_{1}}{x_{2}}}+\frac{1}{\ln \frac{x_{2}}{x_{3}}}+\frac{1}{\ln \frac{x_{3}}{x_{4}}}\right) \\
\geqslant 9,
\end{array}
$$
with equality holding if and only if $\ln \frac{x_{1}}{x_{2}}=\ln \frac{x_{2}}{x_{3}}=\ln \frac{x_{3}}{x_{4}}$, i.e., $\frac{x_{1}}{x_{2}}=\frac{x_{2}}{x_{3}}=\frac{x_{3}}{x_{4}}$.
Therefore, the maximum value of the real number $k$ is 9.
|
9
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let the positive integer $n$ satisfy $31 \mid\left(5^{n}+n\right)$. Then the minimum value of $n$ is $\qquad$ .
|
9. 30 .
Given $5^{3} \equiv 1(\bmod 31)$, when $n=3 k\left(k \in \mathbf{Z}_{+}\right)$, $5^{n}+n \equiv 1+n \equiv 0(\bmod 31)$, at this time, the minimum value of $n$ is $n_{\text {min }}=30$;
When $n=3 k+1\left(k \in \mathbf{Z}_{+}\right)$,
$$
5^{n}+n \equiv 5+n \equiv 0(\bmod 31),
$$
at this time, the minimum value of $n$ is $n_{\text {min }}=88$;
When $n=3 k+2\left(k \in \mathbf{Z}_{+}\right)$,
$$
5^{n}+n \equiv 25+n \equiv 0(\bmod 31),
$$
at this time, the minimum value of $n$ is $n_{\text {min }}=68$.
Therefore, the minimum value of $n$ that satisfies the condition is $n_{\min }=30$.
|
30
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given
$$
S_{n}=|n-1|+2|n-2|+\cdots+10|n-10| \text {, }
$$
where, $n \in \mathbf{Z}_{+}$. Then the minimum value of $S_{n}$ is $\qquad$
|
10. 112 .
From the problem, we know
$$
\begin{array}{l}
S_{n+1}-S_{n} \\
=|n|+2|n-1|+\cdots+10|n-9|- \\
{[|n-1|+2|n-2|+\cdots+10|n-10|] } \\
=|n|+|n-1|+\cdots+|n-9|-10|n-10| .
\end{array}
$$
When $n \geqslant 10$, $S_{n+1}-S_{n}>0$, thus, $S_{n}$ is monotonically increasing; $\square$
When $n=0$, $S_{1}-S_{0}>0$;
When $n=6$, $S_{7}-S_{6}<0$.
Therefore, $S_{n}$ is monotonically decreasing in the interval $[1,7]$ and monotonically increasing in the interval $[7,+\infty)$.
Hence, the minimum value of $S_{n}$ is $S_{7}=112$.
|
112
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. If $15 \mid \overline{\text { xaxax }}$, then the sum of all five-digit numbers $\overline{x a x a x}$ that satisfy the requirement is ( ).
(A) 50505
(B) 59595
(C) 110100
(D) 220200
|
- 1. D.
Notice that, $15| \overline{\text { xaxax }} \Leftrightarrow\left\{\begin{array}{l}5 \mid \overline{\text { xaxax }}, \\ 3 \mid \overline{\text { xaxax }} .\end{array}\right.$
From $51 \overline{\text { xaxax }} \Rightarrow x=5$;
From $31 \overline{x a x a x} \Rightarrow 31(x+a+x+a+x)$
$$
\Rightarrow 3|2 a \Rightarrow 3| a \Rightarrow a=0,3,6,9 \text {. }
$$
Thus, the five-digit numbers that meet the requirements are
$$
50505,53535,56565,59595 \text {, }
$$
their sum is 220200.
|
220200
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a_{1}, a_{2}, \cdots, a_{2015}$ be a sequence of numbers taking values from $-1, 0, 1$, satisfying
$$
\sum_{i=1}^{2015} a_{i}=5 \text {, and } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040,
$$
where $\sum_{i=1}^{n} a_{i}$ denotes the sum of $a_{1}, a_{2}, \cdots, a_{n}$.
Then the number of 1's in this sequence is $\qquad$
|
$=, 1.510$.
Let the number of -1's be $x$, and the number of 0's be $y$.
Then, from $\sum_{i=1}^{2015} a_{i}=5$, we know the number of 1's is $x+5$. Combining this with
$$
\begin{array}{c}
\text { the equation } \sum_{i=1}^{2015}\left(a_{i}+1\right)^{2}=3040, \text { we get } \\
\left\{\begin{array}{l}
x+(x+5)+y=2015, \\
4(x+5)+y=3040 .
\end{array}\right.
\end{array}
$$
Solving these equations, we get $x=505$.
Therefore, the number of 1's is 510.
|
510
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. If the integer $n$
satisfies
$$
(n+1)^{2} \mid\left(n^{2015}+1\right) \text {, }
$$
then the minimum value of $n$ is $\qquad$
|
3. -2016 .
From $n^{2015}+1=(n+1) \sum_{i=0}^{2014}(-1)^{i} n^{2014-i}$, we know
$$
\begin{array}{l}
(n+1)^{2} \mid\left(n^{2015}+1\right) \\
\left.\Leftrightarrow(n+1)\right|_{i=0} ^{2014}(-1)^{i} n^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2014}(-1)^{i}(-1)^{2014-i} \\
\Leftrightarrow(n+1) \mid \sum_{i=0}^{2014}(-1)^{2014} \\
\Leftrightarrow(n+1) \mid 2015 .
\end{array}
$$
Therefore, the minimum value of $n$ is -2016.
|
-2016
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If $x, y, z$ are real numbers, satisfying
$$
x+\frac{1}{y}=2 y+\frac{2}{z}=3 z+\frac{3}{x}=k \text{, and } x y z=3 \text{, }
$$
then $k=$
|
4. 4 .
Multiplying the three expressions yields
$$
\begin{array}{l}
k^{3}=6\left[x y z+\left(x+\frac{1}{y}\right)+\left(y+\frac{1}{z}\right)+\left(z+\frac{1}{x}\right)+\frac{1}{x y z}\right] \\
=6\left(3+k+\frac{k}{2}+\frac{k}{3}+\frac{1}{3}\right) \\
\Rightarrow k^{3}-11 k-20=0 \\
\Rightarrow(k-4)\left(k^{2}+4 k+5\right)=0 \\
\Rightarrow k=4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (20 points) If $x, y \in [0,1]$, try to find the maximum value of
$$
x \sqrt{1-y} + y \sqrt{1-x}
$$
|
$$
\begin{array}{l}
\text { I. Let } s=\sqrt{1-x}, t=\sqrt{1-y}, \\
z=x \sqrt{1-y}+y \sqrt{1-x} .
\end{array}
$$
Then $z=\left(1-s^{2}\right) t+\left(1-t^{2}\right) s=(s+t)(1-s t)$.
Also, $(1-s)(1-t)=1-(s+t)+s t \geqslant 0$, which means $s+t \leqslant 1+s t$,
thus $z \leqslant(1+s t)(1-s t)=1-s^{2} t^{2} \leqslant 1$.
The maximum value of 1 is achieved when $x=1, y=0$ or $x=0, y=1$.
|
1
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. To color the eight vertices of the cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with four different colors, such that the two endpoints of the same edge have different colors, there are a total of coloring methods.
|
8. 2652 .
First, color the four points $A, B, C, D$ above, with 84 coloring methods. Then consider the four points below, using the principle of inclusion-exclusion, the total number of methods is
$$
\begin{array}{l}
84\left\{84-C_{4}^{1}[3 \times(3+2 \times 2)+\right. \\
(3+2 \times 2)]+2\left(\frac{36}{84} \times 9+\frac{48}{84} \times 4\right)- \\
\left.C_{4}^{3}\left(\frac{36}{84} \times 3+\frac{48}{84} \times 2\right)+C_{4}^{4}\right\} \\
=2652 .
\end{array}
$$
|
2652
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[5]}$
$(2012$, Girls' Mathematical Olympiad)
|
【Analysis】Perform prime factorization $2012=2^{2} \times 503$.
First consider 503 I $\mathrm{C}_{2012}^{k}$.
By Corollary 3, we know that when and only when $k$ and $2012-k$ do not produce a carry when added in base 503, $\left(503, \mathrm{C}_{2012}^{k}\right)=1$.
Writing 2012 in base 503 gives $2012=(40)_{503}$.
Let $k=\left(a_{1} a_{2}\right)_{503}$.
No carry must have $a_{2}=0, a_{1} \in\{0,1,2,3,4\}$.
Converting to decimal, we know that when and only when
$k \in\{0,503,1006,1509,2012\}$
then $503 \nmid \mathrm{C}_{2012}^{k}$.
Next, consider the opposite.
Determine $k$ such that $v_{2}\left(\mathrm{C}_{2012}^{k}\right) \leqslant 1$.
There are two cases.
(1) $v_{2}\left(\mathrm{C}_{2012}^{k}\right)=0$.
By Corollary 3, we know that $k$ and $2012-k$ do not produce a carry when added in binary, and
$2012=(11111011100)_{2}$,
From this, it is easy to see that there are $2^{8}=256$ such $k$.
(2) $v_{2}\left(\mathrm{C}_{2012}^{k}\right)=1$.
By Corollary 2, we know that $k$ and $2012-k$ produce exactly one carry when added in binary.
First, if this carry occurs at a position where 2012's binary representation is 1, then the position to the right of this 1 must also produce a carry during the addition, resulting in at least two carries, which is a contradiction.
Thus, this single carry must occur at a position where 2012's binary representation is 0.
The following discussion is in binary.
(i) If the carry occurs at the first position from the right, note that $(11111011100)_{2}$ has a 0 at the second position from the right, so a carry must also occur at the second position, which is a contradiction.
(ii) If the carry occurs at the sixth position from the right, note that $(11111011100)_{2}$ has a 1 at the seventh position from the right, and combining this with the fact that only one carry occurs, we deduce that the seventh position of $k$ must be 0.
Thus, there are $2^{7}=128$ such $k$.
(iii) If the carry occurs at the second position from the right, similarly, there are $2^{7}=128$ such $k$.
Combining (1) and (2), there are 512 $k$ such that $v_{2}\left(\mathrm{C}_{2012}^{k}\right) \leqslant 1$.
It is easy to verify that $\mathrm{C}_{2012}^{0}=\mathrm{C}_{2012}^{2012} \equiv 1(\bmod 2)$,
$\mathrm{C}_{2012}^{503}=\mathrm{C}_{2012}^{1599} \equiv 0(\bmod 4)$,
$\mathrm{C}_{2012}^{1006} \equiv 0(\bmod 4)$.
Thus, the number of non-negative integers $k$ in the set that satisfy $2012 । \mathrm{C}_{2012}^{k}$ is
$2013-5-512+2=1498$
That is, there are 1498 elements $k$ in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012.
|
1498
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $a_{1}, a_{2}, \cdots, a_{2014}$ be a permutation of the positive integers $1,2, \cdots$, 2014. Denote
$$
S_{k}=a_{1}+a_{2}+\cdots+a_{k}(k=1,2, \cdots, 2014) \text {. }
$$
Then the maximum number of odd numbers in $S_{1}, S_{2}, \cdots, S_{2014}$ is $\qquad$
|
6.1511.
If $a_{i}(2 \leqslant i \leqslant 2014)$ is odd, then $S_{i}$ and $S_{i-1}$ have different parities. From $a_{2}$ to $a_{2014}$, there are at least $1007-1=1006$ odd numbers. Therefore, $S_{1}, S_{2}, \cdots, S_{2014}$ must change parity at least 1006 times.
Thus, $S_{1}, S_{2}, \cdots, S_{2014}$ must have at least 503 even numbers, meaning there are at most $2014-503=1511$ odd numbers.
Take $a_{1}=1, a_{2}=3, a_{3}=5, \cdots, a_{1007}=2013$, $a_{1008}=2, a_{1009}=4, \cdots, a_{2014}=2014$.
Then $S_{1}, S_{3}, \cdots, S_{1007}$ and $S_{1008} \sim S_{2014}$ are all odd, totaling $\frac{1008}{2}+1007=1511$ odd numbers.
In conclusion, the maximum number of odd numbers in $S_{1} \sim S_{2014}$ is 1511.
|
1511
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Print 90000 five-digit numbers
$$
10000,10001, \cdots, 99999
$$
on cards, with one five-digit number on each card. Some cards (such as 19806, which reads 90861 when flipped) have numbers that can be read in two different ways, causing confusion. The number of cards that will not cause confusion is $\qquad$ cards.
|
8. 88060 .
Among the ten digits $0 \sim 9$, the digits that can still represent numbers when inverted are $0, 1, 6, 8, 9$.
Since the first digit cannot be 0 and the last digit cannot be 0, the number of such five-digit numbers that can be read when inverted is $4 \times 5 \times 5 \times 5 \times 4=2000$.
Among these 2000 five-digit numbers, we need to exclude the five-digit numbers that do not cause confusion when read backwards (such as 10801, 60809).
Now, divide the digits of the five-digit number into three groups.
【First Group】The first and last digits form a group with four possibilities: $(1,1),(8,8),(6,9),(9,6)$;
【Second Group】The second and fourth digits form a group with five possibilities:
$(0,0),(1,1),(8,8),(6,9),(9,6)$;
【Third Group】The third digit can be $0, 1, 8$ three possibilities.
Thus, the number of five-digit numbers that read the same forwards and backwards is $4 \times 5 \times 3 = 60$.
Therefore, the number of five-digit numbers that do not cause confusion is $90000-(2000-60)=88060$ (numbers).
|
88060
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$
|
$=, 7 .-1$. Therefore $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$.
|
-1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let $A$ be a set composed of any 100 distinct positive integers, and let
$$
B=\left\{\left.\frac{a}{b} \right\rvert\, a 、 b \in A \text { and } a \neq b\right\},
$$
$f(A)$ denotes the number of elements in set $B$. Then the sum of the maximum and minimum values of $f(A)$ is $\qquad$ .
|
11. 10098.
From the problem, when the elements in set $B$ are pairwise coprime, $f(A)$ reaches its maximum value $\mathrm{A}_{100}^{2}=9900$; when the elements in set $B$ form a geometric sequence with a common ratio not equal to 1, $f(A)$ reaches its minimum value of $99 \times 2=198$.
Therefore, the sum of the maximum and minimum values is 10098.
|
10098
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{4028} x^{4028}$ is the expansion of $\left(x^{2}+x+2\right)^{2014}$, then
$$
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{4026}-a_{4007}-a_{4028}
$$
is $\qquad$
|
6.2 .
Let $x=\omega\left(\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)$.
Then $x^{2}+x+2=1$.
Therefore, $a_{0}+a_{1} \omega+a_{2} \omega^{2}+a_{3}+a_{4} \omega+\cdots+$ $a_{4026}+a_{4027} \omega+a_{4028} \omega^{2}=1$.
Taking the conjugate of the above equation, we get
$$
\begin{array}{l}
a_{0}+a_{1} \omega^{2}+a_{2} \omega+a_{3}+a_{4} \omega^{2}+\cdots+ \\
a_{4026}+a_{4027} \omega^{2}+a_{4028} \omega=1 .
\end{array}
$$
Adding the above two equations, we get
$$
\begin{array}{l}
2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+ \\
2 a_{405}-a_{4027}-a_{4028}=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let
$$
\begin{array}{l}
A=\{1,2, \cdots, 2014\}, \\
B_{i}=\left\{x_{i}, y_{i}\right\}(i=1,2, \cdots, t)
\end{array}
$$
be $t$ pairwise disjoint binary subsets of $A$, and satisfy the conditions
$$
\begin{array}{l}
x_{i}+y_{i} \leqslant 2014(i=1,2, \cdots, t), \\
x_{i}+y_{i} \neq x_{j}+y_{j}(1 \leqslant i<j \leqslant t) .
\end{array}
$$
Then the maximum value of $t$ is
|
8. 805 .
On the one hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\left\{x_{i}, y_{i}\right\} \cap\left\{x_{j}, y_{j}\right\}=\varnothing \text {. }
$$
Thus, $x_{1}, x_{2}, \cdots, x_{t}, y_{1}, y_{2}, \cdots, y_{t}$ are $2 t$ distinct integers. Therefore,
$$
\sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \geqslant \sum_{i=1}^{2 t} i=t(2 t+1) .
$$
On the other hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\begin{array}{l}
x_{i}+y_{i} \neq x_{j}+y_{j}, \text { and } x_{i}+y_{i} \leqslant 2014, \\
\text { then } \sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \\
\leqslant \sum_{i=1}^{t}(2015-i)=\frac{t(4029-t)}{2} .
\end{array}
$$
From equations (1) and (2), we get $t \leqslant 805$.
Below are two examples showing that $t=805$ meets the requirements.
$$
\begin{array}{l}
\text { (1) }\{k, 1208+k\}(k=1,2, \cdots, 402), \\
\{k, 403+k\}(k=403,404, \cdots, 805) ; \\
\text { (2) }\{k, 1611-2 k\}(k=1,2, \cdots, 402), \\
\{k, 2016-k\}(k=403,404, \cdots, 805) .
\end{array}
$$
|
805
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. A line $l$ is drawn through the right focus of the hyperbola $x^{2}-\frac{y^{2}}{2}=1$ intersecting the hyperbola at points $A$ and $B$. If a real number $\lambda$ makes $|A B|=\lambda$ such that there are exactly three lines $l$, then $\lambda=$ $\qquad$.
|
2.4.
Since there are an odd number of lines that satisfy the condition, by symmetry, the line perpendicular to the $x$-axis satisfies the condition, at this time,
$$
x=\sqrt{3}, y= \pm 2, \lambda=|A B|=4 \text {. }
$$
|
4
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the set $A=\{1,2,3\}, f$ and $g$ are functions from set $A$ to $A$. Then the number of function pairs $(f, g)$ whose image sets intersect at the empty set is $\qquad$ .
|
4. 42 .
When the image set of function $f$ is 1 element, if the image set of function $f$ is $\{1\}$, at this time the image set of function $g$ is a subset of $\{2,3\}$, there are $2^{3}=8$ kinds, so there are $3 \times 8=24$ pairs of functions $(f, g)$ that meet the requirements.
When the image set of function $f$ is 2 elements, if the image set of function $f$ is $\{1,2\}$, at this time $f(1) 、 f(2) 、 f(3)$ have two numbers as 1 and one number as 2, there are $2 \times 3=6$ kinds, so there are $3 \times 6=18$ pairs of functions $(f, g)$ that meet the requirements.
Therefore, the number of function pairs $(f, g)$ is 42 .
|
42
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Given $f(x)=\left(x^{2}+3 x+2\right)^{\cos \pi x}$. Then the sum of all $n$ that satisfy the equation
$$
\left|\sum_{k=1}^{n} \log _{10} f(k)\right|=1
$$
is
|
5. 21 .
It is known that for integer $x$, we have
$$
f(x)=[(x+1)(x+2)]^{(-1)^{x}} \text {. }
$$
Thus, when $n$ is odd,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2-\log _{10}(n+2) \text {; }
$$
When $n$ is even,
$$
\sum_{k=1}^{n} \log _{10} f(k)=-\log _{10} 2+\log _{10}(n+2) \text {. }
$$
Therefore, the $n$ that satisfies the condition is 3 or 18, and their sum is 21.
|
21
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) Given real numbers $a_{0}, a_{1}, \cdots, a_{2015}$, $b_{0}, b_{1}, \cdots, b_{2011}$ satisfy
$$
\begin{array}{l}
a_{n}=\frac{1}{65} \sqrt{2 n+2}+a_{n-1}, \\
b_{n}=\frac{1}{1009} \sqrt{2 n+2}-b_{n-1},
\end{array}
$$
where $n=1,2, \cdots, 2015$.
If $a_{0}=b_{2015}$, and $b_{0}=a_{2015}$, find the value of the following expression
$$
\sum_{k=1}^{2015}\left(a_{k} b_{k-1}-a_{k-1} b_{k}\right) .
$$
|
10. Notice that, for any $k=1,2 \cdots, 2015$, we have
$$
\begin{array}{l}
a_{k}-a_{k-1}=\frac{1}{65} \sqrt{2 k+2}, \\
b_{k}+b_{k-1}=\frac{1}{1009} \sqrt{2 k+2} .
\end{array}
$$
Multiplying the two equations, we get
$$
\begin{array}{l}
a_{k} b_{k}-a_{k-1} b_{k-1}+a_{k} b_{k-1}-a_{k-1} b_{k} \\
=\frac{2 k+2}{65 \times 1009} .
\end{array}
$$
Summing up, we have
$$
\begin{array}{l}
a_{2015} b_{2015}-a_{0} b_{0}+\sum_{k=1}^{2015}\left(a_{k} b_{k-1}-a_{k-1} b_{k}\right) \\
=\frac{2015(4+4032)}{2 \times 65 \times 1009}=62 .
\end{array}
$$
Therefore, the required value is 62.
|
62
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. In an All-Star basketball game, 27 players participate, each wearing a jersey with their favorite number, which is a non-negative integer. After the game, they line up in a 3-row, 9-column formation for fans to take photos. An eccentric fan only takes photos where the players in the frame form a rectangle of $a$ rows and $b$ columns $(1 \leqslant a \leqslant 3, 1 \leqslant b \leqslant 9)$ (with the rows and columns aligned with the original formation), and the sum of the jersey numbers of the players in the frame (for a single player, it is just their jersey number) is a multiple of 10. As a result, this fan only takes photos of $s$ players. Find the minimum possible value of $s$.
（Based on a problem from the 2011 National High School Mathematics Competition）
|
Prompt: $s_{\text {min }}=2$.
Assume at most one player is photographed.
By symmetry, without loss of generality, assume no players in the first row are photographed.
For $i=1,2, \cdots, 9$, let the players in the $i$-th column of the 1st, 2nd, and 3rd rows have jersey numbers $a_{i}, b_{i}, c_{i}$, respectively, and denote
$$
S_{k}=\sum_{i=1}^{k} a_{i}, T_{k}=\sum_{i=1}^{k}\left(b_{i}+c_{i}\right),
$$
where $k=0,1, \cdots, 9$.
Considering the 1st row, the 2nd to 3rd rows, and the 1st to 3rd rows, respectively, it can be proven that: $S_{0}, S_{1}, \cdots, S_{9}, T_{0}, T_{1}, \cdots, T_{9}$, and $S_{0}+T_{0}, S_{1}+T_{1}, \cdots, S_{9}+T_{9}$ all form a complete residue system modulo 10, which will lead to a contradiction.
On the other hand, suppose the players choose the numbers as shown in Table 1, it can be verified that no player with a non-zero number can be photographed. In this case, the fan exactly photographs 2 players.
Table 1
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline 1 & 1 & 1 & 2 & 1 & 1 & 1 & 1 & 0 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 1 & 1 & 1 & 0 & 1 & 1 & 1 & 1 & 2 \\
\hline
\end{tabular}
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Given
\[
\begin{aligned}
f(x, y)= & x^{3}+y^{3}+x^{2} y+x y^{2}- \\
& 3\left(x^{2}+y^{2}+x y\right)+3(x+y),
\end{aligned}
\]
and \( x, y \geqslant \frac{1}{2} \). Find the minimum value of \( f(x, y) \). ${ }^{[4]}$
(2011, Hebei Province High School Mathematics Competition)
|
【Analysis】This problem involves finding the extremum of a bivariate function, with a very complex expression, making it difficult to find a breakthrough. Observing the symmetry in the expression, we can attempt the following transformation.
First, when $x \neq y$, multiply both sides of the function by $x-y$, yielding
$$
\begin{array}{l}
(x-y) f(x, y) \\
=\left(x^{4}-3 x^{3}+3 x^{2}\right)-\left(y^{4}-3 y^{3}+3 y^{2}\right) .
\end{array}
$$
At this point, it is easy to construct the function
$$
g(x)=x^{4}-3 x^{3}+3 x^{2} \text {. }
$$
Equation (1) can be rewritten as
$$
f(x, y)=\frac{g(x)-g(y)}{x-y} \text {. }
$$
Thus, $f(x, y)$ can be viewed as the slope of the line connecting two points on the graph of $g(x)$.
Furthermore, when $x=y$,
$$
f(x, y)=4 x^{3}-9 x^{2}+6 x \text {. }
$$
Therefore, we only need to find the minimum value of the derivative function $h(x)=4 x^{3}-9 x^{2}+6 x$ of $g(x)$ for $x \geqslant \frac{1}{2}$.
It is easy to find that when $x \geqslant \frac{1}{2}$, the minimum value of $h(x)=4 x^{3}-9 x^{2}+6 x$ is 1.
Thus, the minimum value of $f(x, y)$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Given the sequence $\left\{a_{n}\right\}$:
$$
a_{1}=2, a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}(n \geqslant 1) \text {. }
$$
Determine the periodicity of the sequence $\left\{a_{n}\right\}$.
|
Solving, from $a_{1}=2$, we get
$$
a_{2}=\frac{5 a_{1}-13}{3 a_{1}-7}=3, a_{3}=\frac{5 a_{2}-13}{3 a_{2}-7}=1,
$$
which means $a_{1}, a_{2}, a_{3}$ are all distinct.
From the given conditions, we have
$$
\begin{array}{l}
a_{n+1}=\frac{5 a_{n}-13}{3 a_{n}-7}, \\
a_{n+2}=\frac{5 a_{n+1}-13}{3 a_{n+1}-7}=\frac{7 a_{n}-13}{3 a_{n}-5}, \\
a_{n+3}=\frac{5 a_{n+2}-13}{3 a_{n+2}-7}=a_{n},
\end{array}
$$
which means $a_{n+3}=a_{n}(n=1,2, \cdots)$.
In summary, the sequence $\left\{a_{n}\right\}$ is a purely periodic sequence, with the smallest positive period being 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a b c \neq 0, a+b+c=a^{2}+b^{2}+c^{2}=2 . \\
\text { Then } \frac{(1-a)^{2}}{b c}+\frac{(1-b)^{2}}{c a}+\frac{(1-c)^{2}}{a b}=
\end{array}
$$
|
7.3.
From the given, we have $a b+b c+c a=1$.
Then $b c=1-a b-a c=1-a(b+c)$ $=1-a(2-a)=(1-a)^{2}$.
Thus, the desired value is 3.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
14. A positive integer that can be expressed as the difference of squares of two positive integers is called a "wise number". For example, $9=5^{2}-4^{2}, 9$ is a wise number.
(1) Try to determine which numbers among the positive integers are wise numbers, and explain the reason;
(2) In the sequence of wise numbers arranged in ascending order, find the 2015th wise number.
|
14. (1) It is easy to know that the positive integer 1 cannot be expressed as the difference of squares of two positive integers, i.e., 1 is not a wise number.
For odd numbers greater than 1, we have
$$
2 k+1=(k+1)^{2}-k^{2}(k=1,2, \cdots),
$$
which means that all odd numbers greater than 1 are wise numbers.
When $k=2,3, \cdots$, we have
$4 k=(k+1)^{2}-(k-1)^{2}$.
When $k=1$, if $4=a^{2}-b^{2}$, then
$$
4=(a+b)(a-b) \text {, }
$$
we know that there do not exist positive integers $a, b$ that satisfy this equation, i.e., 4 is not a wise number.
Therefore, numbers greater than 4 and divisible by 4 are all wise numbers.
For positive integers of the form $4 k+2(k=0$, $1, \cdots)$, assume $4 k+2$ is a wise number. Then
$$
4 k+2=x^{2}-y^{2}=(x+y)(x-y) \text {, }
$$
where $x, y$ are positive integers.
When $x, y$ have the same parity, $4 \mid(x+y)(x-y)$, but $4 \nmid(4 k+2)$, which is a contradiction;
When $x, y$ have different parities, $(x+y)(x-y)$ is odd, but $4 k+2$ is even, which is a contradiction.
Therefore, numbers of the form $4 k+2(k=1,2, \cdots)$ are not wise numbers.
Hence, only numbers of the form $2 k+1$ and $4(k+1)(k=1,2, \cdots)$ are wise numbers.
(2) Starting from 1, the first four consecutive positive integers form a group, and by (1), only 3 is a wise number in the first group, and in each subsequent group of four numbers, three are wise numbers.
Notice that, $2016=3 \times 672$, and $4 \times 672=$ 2688, and since the first group has only 1 wise number, 2688 is the 2014th wise number.
Therefore, 2689 is the 2015th wise number.
|
2689
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. The basketball league has eight teams. Each season, each team plays two games (home and away) against each of the other teams in the league, and each team also plays four games against opponents outside the league. Therefore, in one season, the eight teams in the league play a total of ( ) games.
(A) 60
(B) 88
(C) 96
(D) 144
(E) 160
|
16. B.
There are $\mathrm{C}_{8}^{2}=28$ pairs that can be formed from the eight teams in the league.
Since each pair of teams plays both a home and an away match, the eight teams in the league play a total of $28 \times 2=56$ matches.
Since each team plays four matches against opponents outside the league, the eight teams play a total of $4 \times 8=32$ matches against teams outside the league.
Therefore, in one season, these eight teams play a total of $56+32=88$ matches.
|
88
|
Combinatorics
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
14. Let the angle between vectors $\boldsymbol{a}$ and $\boldsymbol{b}$ be $\frac{\pi}{3}$, the angle between vectors $\boldsymbol{c}-\boldsymbol{a}$ and $\boldsymbol{c}-\boldsymbol{b}$ be $\frac{2 \pi}{3}$, $|\boldsymbol{a}-\boldsymbol{b}|=5$, and $|\boldsymbol{c}-\boldsymbol{a}|=2 \sqrt{3}$. Then the maximum value of $\boldsymbol{a} \cdot \boldsymbol{c}$ is
|
14. 24 .
Let $\overrightarrow{O A}=a, \overrightarrow{O B}=b, \overrightarrow{O C}=c$. Then
$$
|\overrightarrow{A C}|=|c-a|=2 \sqrt{3},|\overrightarrow{A B}|=|a-b|=5 \text {. }
$$
Also, $\angle A O B=\frac{\pi}{3}, \angle A C B=\frac{2 \pi}{3}$, at this time, $O, A, C, B$ are concyclic.
By the Law of Sines, we get
$$
\sin \angle A B C=\frac{3}{5} \Rightarrow \cos \angle A B C=\frac{4}{5} \text {. }
$$
In $\triangle A C O$, since $\angle A O C=\angle A B C$, by the Law of Cosines, we have
$$
\begin{array}{l}
A C^{2}=|\boldsymbol{a}|^{2}+|\boldsymbol{c}|^{2}-2|\boldsymbol{a}||\boldsymbol{c}| \cos \angle A O C \\
\Rightarrow 12 \geqslant 2|\boldsymbol{a}||\boldsymbol{c}|-\frac{8}{5}|\boldsymbol{a}||\boldsymbol{c}| \\
\Rightarrow|\boldsymbol{a}||\boldsymbol{c}| \leqslant 30 \\
\Rightarrow \boldsymbol{a} \cdot \boldsymbol{c}=|\boldsymbol{a}||\boldsymbol{c}| \cos \angle A O C \leqslant 24 . \\
\text { When } \angle A C O=\frac{\pi}{4}+\frac{1}{2} \arctan \frac{4}{3} \text {, the above inequalities hold with equality. }
\end{array}
$$
When $\angle A C O=\frac{\pi}{4}+\frac{1}{2} \arctan \frac{4}{3}$, the above inequalities hold with equality.
Therefore, the maximum value of $a \cdot c$ is 24.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Allocate 24 volunteer slots to 3 schools. Then the number of allocation methods where each school gets at least one slot and the number of slots for each school is different is $\qquad$ kinds. ${ }^{[2]}$
|
Let the quotas allocated to schools A, B, and C be $x$, $y$, and $z$ respectively.
First, without considering that $x, y, z$ are pairwise distinct.
From $x+y+z=24$, we get a total number of combinations $\mathrm{C}_{23}^{2}$.
Next, we separate out the number of positive integer solutions $(x, y, z)$ that are not pairwise distinct.
From $2a+b=24$, we know $b=2c$.
Therefore, $a+c=12$, which has 11 solutions.
Among these, one solution satisfies $a=b=8$, corresponding to one triplet $(8,8,8)$;
Additionally, 10 solutions $(a, a, b) (a \neq b)$ correspond to $10 \mathrm{C}_{3}^{1}$ positive integer triplets $(x, y, z)$.
Thus, the number of pairwise distinct positive integer triplets $(x, y, z)$ is
$$
\mathrm{C}_{23}^{2}-1-10 \mathrm{C}_{3}^{1}=222 .
$$
|
222
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The sum of the ages of three people, A, B, and C, represented by $x, y, z$ is 120, and $x, y, z \in (20,60)$. Then the number of ordered triples $(x, y, z)$ is $\qquad$
|
8. 1141 .
Notice a basic conclusion:
The number of positive integer solutions $(a, b, c)$ to the indeterminate equation $a+b+c=n$ is $\mathrm{C}_{n-1}^{2}$.
Thus, the number of solutions to the indeterminate equation
$$
(x-20)+(y-20)+(z-20)=60
$$
satisfying $x, y, z>20$ is $\mathrm{C}_{59}^{2}$.
Among these $\mathrm{C}_{59}^{2}$ solutions, the number of solutions where $x \geqslant 60$ and do not meet the requirements is
$$
(x-59)+(y-20)+(z-20)=21
$$
satisfying $x \geqslant 60, y, z>20$, the number of such solutions is $\mathrm{C}_{20}^{2}$.
Similarly, the number of solutions where $y \geqslant 60$ is $\mathrm{C}_{20}^{2}$, and the number of solutions where $z \geqslant 60$ is $\mathrm{C}_{20}^{2}$.
Since no two individuals can simultaneously reach 60 years old, the number of ordered triples $(x, y, z)$ is $\mathrm{C}_{59}^{2}-3 \mathrm{C}_{20}^{2}=1141$.
|
1141
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. The sequence $\left\{a_{n}\right\}$ has 9 terms, where $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. Find the number of such sequences.
|
Prompt: Categorized Count. Answer: 491.
|
491
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
435 Given a convex polygon $F$, consider all the figures that are positively homothetic to the convex polygon $F$ and smaller than $F$. Let $n(F)$ be the minimum number of such figures (allowing translation but not rotation) needed to cover the convex polygon $F$. Find the value of $n(F)$.
|
(1) On the one hand, let the convex polygon $F$ be the parallelogram $ABCD$. It is easy to see that the convex polygon $F$ can be covered by four smaller parallelograms that are similar to the convex polygon $F$ (as shown in Figure 3).
On the other hand, for any smaller parallelogram $F_1$ that is similar to the convex polygon $F$, if $F_1$ contains point $A$, then $F_1$ cannot contain any other vertices of the convex polygon $F$ (as shown in Figure 4). This indicates that the convex polygon $F$ cannot be covered by fewer than four smaller and similar figures.
Therefore, $n(F)=4$.
(2) Let the polygon $F$ not be a parallelogram. Then $n(F) \geqslant 3$.
In fact, if the convex polygon $F$ can be covered by two smaller and similar figures, let's assume these two figures are of equal size; otherwise, the smaller one can be enlarged. Establish a Cartesian coordinate system such that the corresponding vertices of these two figures have the same y-coordinates. Consider the vertices of the convex polygon $F$ with the maximum and minimum y-coordinates, which cannot be covered by these two figures simultaneously. Hence, $n(F) \geqslant 3$.
Next, we prove that three figures are sufficient.
First, we prove a lemma.
Lemma There exists a triangle $T$ that contains the convex polygon $F$, and each side of $T$ contains a side of the convex polygon $F$.
Proof If the convex polygon $F$ is a triangle, then let $T=F$.
If the convex polygon $F$ is not a triangle, then there exist two sides of the convex polygon $F$ that are neither parallel nor intersecting. Extend these two sides until they intersect (as shown in Figure 5), resulting in a polygon $F_1$ that contains the convex polygon $F$, and the number of sides of $F_1$ is less than that of the convex polygon $F$, with each side of $F_1$ containing a side of the convex polygon $F$.
Repeating this process, the final figure $F'$ will inevitably be a parallelogram or a triangle (only these two figures do not have sides that are neither parallel nor intersecting), and $F'$ contains the convex polygon $F$, with each side of $F'$ containing a side of the convex polygon $F$.
If $F'$ is a triangle, then let $T=F'$.
If $F'$ is a parallelogram $ABCD$, since the convex polygon $F$ is not a parallelogram, $F'$ must have at least one vertex that is not a vertex of the convex polygon $F$, let's say point $A$. Let point $P$ be the point on line segment $AD$ closest to point $A$. Then $P$ is a vertex of the convex polygon $F$, and one of its sides lies on line segment $AD$. Let the other side of the convex polygon $F$ from point $P$ intersect line segment $AB$ at point $Q$. Then the triangle $T$ formed by lines $PQ$, $BC$, and $CD$ contains the convex polygon $F$, and each side of $T$ contains a side of the convex polygon $F$ (as shown in Figure 6).
Back to the original problem.
Using the lemma, assume the convex polygon $F$ is contained in $\triangle ABC$, and the sides $A_1A_2$, $B_1B_2$, and $C_1C_2$ of the convex polygon $F$ lie on segments $BC$, $AC$, and $AB$ respectively (as shown in Figure 7).
Choose any point $O$ inside the convex polygon $F$ and internal points $X$, $Y$, and $Z$ on sides $A_1A_2$, $B_1B_2$, and $C_1C_2$ respectively. Then segments $OX$, $OY$, and $OZ$ divide the convex polygon $F$ into three polygons $F_1$, $F_2$, and $F_3$. Assume polygon $F_1$ is inside quadrilateral $AZOY$.
By the selection of points $O$, $Y$, and $Z$, if $0 < k < 1$ and $k$ is infinitely close to 1, then the homothety $\varphi_1$ with center $A$ and ratio $k$ satisfies $\varphi_1$ (convex polygon $F \cup$ quadrilateral $AZOY$) contains quadrilateral $AZOY$.
Note that the part of quadrilateral $AZOY$ outside the convex polygon $F$ becomes a subset of itself after the homothety, so $\varphi_1(F)$ covers $F_1$.
Similarly, there exist two homotheties $\varphi_2$ and $\varphi_3$ with ratios less than 1 such that $\varphi_2(F)$ covers $F_2$ and $\varphi_3(F)$ covers $F_3$.
Thus, the convex polygon $F$ is covered by the union of $\varphi_1(F)$, $\varphi_2(F)$, and $\varphi_3(F)$.
Therefore, $n(F)=3$.
|
3
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, such that $a_{2}-a_{1} \geqslant 3$, and $a_{3}-a_{2} \geqslant 3$, find the number of different ways to choose them.
|
From the given information, we have
$$
\begin{aligned}
a_{1} & \leqslant a_{2}-3 \leqslant a_{3}-6 \\
& \Rightarrow 1 \leqslant a_{1}<a_{2}-2<a_{3}-4 \leqslant 10 .
\end{aligned}
$$
Substitute $\left(a_{1}, a_{2}-2, a_{3}-4\right)=(x, y, z)$.
Then the number of tuples $\left(a_{1}, a_{2}, a_{3}\right)$ equals the number of ways to choose three different elements from $\{1,2, \cdots, 10\}$, which is $\mathrm{C}_{10}^{3}=120$.
|
120
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 Add a “+” or “-” in front of each number in $1,2, \cdots, 1989$. Find the minimum non-negative algebraic sum, and write down the equation.
|
First, prove that the algebraic sum is odd.
Consider the simplest case: all filled with " + ", at this time,
$$
1+2+\cdots+1989=995 \times 1989
$$
is odd.
For the general case, it is only necessary to adjust some " + " to " - ".
Since $a+b$ and $a-b$ have the same parity, each adjustment does not change the parity of the algebraic sum, i.e., the total sum remains odd.
$$
\begin{array}{l}
\text { and } 1+(2-3-4+5)+(6-7-8+9)+ \\
\cdots+(1986-1987-1988+1989)=1 .
\end{array}
$$
Therefore, this minimum value is 1.
|
1
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 Let $a, b, c \in \mathbf{R}_{+}$. Prove:
$$
\sum_{c y c} \frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}} \leqslant 8 \text {. }
$$
|
To prove that since the inequality to be proved is homogeneous with respect to $a, b, c$, we may assume without loss of generality that $a+b+c=3$.
The original inequality is then $\sum_{\text {cyc }} \frac{(a+3)^{2}}{2 a^{2}+(3-a)^{2}} \leqslant 8$.
Consider the function $f(x)=\frac{(x+3)^{2}}{2 x^{2}+(3-x)^{2}}(0<x<3)$.
Notice that,
$$
\begin{array}{l}
f(x)=\frac{x^{2}+6 x+9}{3\left(x^{2}-2 x+3\right)}=\frac{1}{3}\left[1+\frac{8 x+6}{(x-1)^{2}+2}\right] \\
\leqslant \frac{1}{3}\left(1+\frac{8 x+6}{2}\right)=\frac{4 x+4}{3} . \\
\text { Therefore, } \sum_{\text {cyc }} f(a) \leqslant \sum_{\text {cyc }} \frac{4 a+4}{3} \\
=\frac{4(a+b+c)+12}{3}=8 .
\end{array}
$$
|
8
|
Inequalities
|
proof
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
\left(2 x^{2}+8 x+11\right)\left(y^{2}-10 y+29\right)\left(3 z^{2}-18 z+32\right) \\
\leqslant 60 .
\end{array}
$$
Then $x+y-z=$ . $\qquad$
|
8. 0 .
Original expression
$$
\begin{array}{l}
=\left[2(x+2)^{2}+3\right]\left[(y-5)^{2}+4\right]\left[3(z-3)^{2}+5\right] \\
\leqslant 60 \\
\Rightarrow x=-2, y=5, z=3 \\
\Rightarrow x+y-z=0 .
\end{array}
$$
|
0
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let $S_{n}=1+2+\cdots+n$. Then among $S_{1}, S_{2}$, $\cdots, S_{2015}$, there are. $\qquad$ that are multiples of 2015.
|
10.8 .
Obviously, $S_{n}=\frac{1}{2} n(n+1)$.
For any positive divisor $d$ of 2015, it is easy to see that in the range $1 \sim 2015$, there is exactly one $n$ that satisfies $n$ being a multiple of $d$ and $n+1$ being a multiple of $\frac{2015}{d}$. Therefore, each divisor of 2015 will generate one $n$ such that $S_{n}$ is a multiple of 2015.
Since $2015=5 \times 13 \times 31$, 2015 has a total of eight divisors.
Thus, the desired result is 8.
|
8
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 6 For any non-empty subset $X$ of the set $M=\{1,2, \cdots, 1000\}$, let $\alpha_{X}$ denote the sum of the maximum and minimum numbers in $X$. Find the arithmetic mean of all such $\alpha_{X}$.
|
【Analysis】According to the problem, the required average is
$$
f=\frac{\sum_{\varnothing \neq X \subseteq M} \alpha_{X}}{2^{1000}-1} \text {. }
$$
The key to solving this is to calculate the sum in the numerator
$$
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \text {. }
$$
To compute such an "unordered sum", one must first choose a "sequence". For this, one can consider the contribution of each element to the total sum, then accumulate, often referred to as the "contribution method"; or consider pairing two subsets so that their "sum" is easy to calculate.
Solution 1 (Contribution Method) Take any $i \in M$, then $i$ contributes to the sum (1) in two ways: $i$ contributes as the maximum number the number of times the subset $\{i\} \cup P(P \subseteq\{1,2, \cdots, i-1\})$ appears, which is $2^{i-1}$; $i$ contributes as the minimum number the number of times the subset $\{i\} \cup P(P \subseteq\{i+1, i+2, \cdots, 1000\})$ appears, which is $2^{1000-i}$.
Thus, the total contribution of $i$ to the numerator is $i\left(2^{i-1}+2^{1000-i}\right)$, and the cumulative "ordered" sum for (1) is
$$
N=\sum_{i=1}^{1000} i\left(2^{i-1}+2^{1000-i}\right) \text {. }
$$
Using the method of summing "arithmetic-geometric series" (or Abel's summation formula) we get
$$
N=1001\left(2^{1000}-1\right) \Rightarrow f=1001 \text {. }
$$
Solution 2 (Pairing Method) Take any $\varnothing \neq X \subseteq M$, define $X^{\prime}=\{1001-x \mid x \in X\}$.
If $X=X^{\prime}$, then
$\min X=\min X^{\prime}=1001-\max X$
$\Rightarrow \alpha_{X}=\max X+\min X=1001$.
If $X \neq X^{\prime}$, then
$\min X^{\prime}=1001-\max X$ and $\max X^{\prime}=1001-\min X$
$\Rightarrow \alpha_{X^{\prime}}=1001 \times 2-\alpha_{X}$
$\Rightarrow \alpha_{X}+\alpha_{X^{\prime}}=1001 \times 2$.
Let $k$ be the number of non-empty subsets among the $2^{1000}-1$ non-empty subsets that satisfy $X^{\prime}=X$. Then
$$
\begin{array}{l}
N=\sum_{\varnothing \neq X \subseteq M} \alpha_{X} \\
=1001 k+\frac{2^{1000}-1-k}{2} \times 2002 \\
=1001\left(2^{1000}-1\right) . \\
\text { Hence } f=\frac{N}{2^{1000}-1}=1001 .
\end{array}
$$
|
1001
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 7 Divide a circle with a circumference of 24 into 24 equal segments, and select eight points from the 24 points, such that the arc length between any two points is not equal to 3 and 8. Find the number of different ways to select such a group of eight points.
|
【Analysis】First analyze the essential structure of the data: Label 24 points in a clockwise direction as $1,2, \cdots, 24$. Then arrange them into the following $3 \times 8$ number table:
$$
\left(\begin{array}{cccccccc}
1 & 4 & 7 & 10 & 13 & 16 & 19 & 22 \\
9 & 12 & 15 & 18 & 21 & 24 & 3 & 6 \\
17 & 20 & 23 & 2 & 5 & 8 & 11 & 14
\end{array}\right) .
$$
In the number table, the arc length between two points represented by adjacent numbers in the same column is 8, and the arc length between two points represented by adjacent numbers in the same row is 3 (the first column and the eighth column are also adjacent, as are the first row and the third row). Therefore, any two adjacent numbers in the table cannot be taken simultaneously, meaning exactly one number is taken from each column, and the numbers taken from adjacent columns cannot be in the same row.
Generalize and construct a recursive relationship.
From a $3 \times n(n \geqslant 3)$ number table, take one number from each column such that the numbers taken from adjacent columns are not in the same row (the first column and the $n$-th column are considered adjacent), and let the number of ways to do this be $a_{n}$.
Then $a_{3}=6$.
When $n \geqslant 4$, we have
$$
a_{n}+a_{n-1}=3 \times 2^{n-1} \Rightarrow a_{n}=2^{n}+2(-1)^{n} \text {. }
$$
Recursive calculation yields $a_{8}=2^{8}+2(-1)^{8}=258$.
|
258
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Fill two $a$s and two $b$s into the 16 cells shown in Figure 3, with at most one letter per cell. If the same letters must not be in the same row or column, find the number of different ways to fill the cells.
|
Prompt: Categorized Count. Answer: 3960.
|
3960
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a positive number $x$ satisfies
$$
x^{10}+x^{5}+\frac{1}{x^{5}}+\frac{1}{x^{10}}=15250 \text {. }
$$
then the value of $x+\frac{1}{x}$ is
|
2.3.
Let $a=x^{5}+\frac{1}{x^{5}}$. Then $x^{10}+\frac{1}{x^{10}}=a^{2}-2$.
Thus, the original equation becomes
$$
\begin{array}{l}
a^{2}+a-15252=0 \\
\Rightarrow(a-123)(a+124)=0
\end{array}
$$
$\Rightarrow a=123$ or $a=-124$ (discard).
Therefore, $x^{5}+\frac{1}{x^{5}}=123$.
Let $x+\frac{1}{x}=b>0$. Then
$$
\begin{array}{l}
x^{2}+\frac{1}{x^{2}}=b^{2}-2, \\
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)\left(x^{2}-1+\frac{1}{x^{2}}\right) \\
=b\left(b^{2}-3\right)=b^{3}-3 b .
\end{array}
$$
Notice that,
$$
\begin{array}{l}
\left(x^{2}+\frac{1}{x^{2}}\right)\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x^{5}+\frac{1}{x^{5}}\right)+\left(x+\frac{1}{x}\right) . \\
\text { Hence }\left(b^{2}-2\right)\left(b^{3}-3 b\right)=123+b \\
\Rightarrow b^{5}-5 b^{3}+5 b-123=0 .
\end{array}
$$
Since $123=3 \times 41$, when $b=3$, the left and right sides of equation (1) are equal. Therefore, $b=3$ is a solution to equation (1).
Dividing the left side of equation (1) by $b-3$, we get the quotient
$$
b^{4}+3 b^{3}+16 b+41 \text {. }
$$
Since this quotient is always greater than 0, we know $b=3$, i.e., $x+\frac{1}{x}=3$.
|
3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 If a positive integer has eight positive divisors, and the sum of these eight positive divisors is 3240, then this positive integer is called a "good number". For example, 2006 is a good number, because the sum of its positive divisors $1, 2, 17, 34, 59, 118, 1003, 2006$ is 3240. Find the smallest good number.
(28th Brazilian Mathematical Olympiad)
|
Let $n=\prod_{i=1}^{k} p_{i}^{\alpha_{i}}$, where $\alpha_{i} \in \mathbf{Z}_{+}, p_{i} (i=1,2, \cdots, k)$ are prime numbers, and $p_{1}<p_{2}<\cdots<p_{k}$.
From $\tau(n)=\prod_{i=1}^{k}\left(1+\alpha_{i}\right)=8=2^{3}$, we know $k=1,2,3$.
(1) If $k=1$, then
$\alpha_{1}=7, n=p^{7}$ ($p$ is a prime number), $\sum_{i=0}^{7} p^{i}=3240$. By $\sum_{i=0}^{7} 2^{i}=255<3240<\sum_{i=0}^{7} 3^{i}=3280$, we know such a prime number $p$ does not exist.
(2) If $k=2$, then let $n=p^{3} q$ ($p, q$ are prime numbers).
Thus $\left(1+p+p^{2}+p^{3}\right)(1+q)$
$$
=3240=2^{3} \times 3^{4} \times 5.
$$
Since $1+q \geqslant 3$, we have
$$
1+p+p^{2}+p^{3} \leqslant 1080 \Rightarrow p \leqslant 7.
$$
(i) If $p=2$, then $1+q=216 \Rightarrow q=215$;
(ii) If $p=3$, then $1+q=81 \Rightarrow q=80$;
(iii) If $p=5$, then $1+q=\frac{270}{13} \notin \mathbf{Z}$;
(iv) If $p=7$, then $1+q=\frac{41}{5} \notin \mathbf{Z}$.
All these cases contradict the fact that $q$ is a prime number.
(3) If $k=3$, then let
$n=p q r (p, q, r$ are prime numbers, $p<q<r)$.
Thus $(1+p)(1+q)(1+r)=3240$.
(i) If $p=2$, then
$$
(1+q)(1+r)=1080=2^{3} \times 3^{3} \times 5 \text{ (it is easy to see that }
$$
$1+q, 1+r$ are both even numbers
$$
\begin{array}{l}
\Rightarrow(1+q, 1+r) \\
=(4,270),(6,180),(10,108),(12,90), \\
(18,60),(20,54),(30,36) \\
\Rightarrow(q, r)=(3,269),(5,179),(11,89), \\
(17,59),(19,53) \\
\Rightarrow n=2 \times 3 \times 269=1614 \text{ or } \\
n= 2 \times 5 \times 179=1790 \text{ or } \\
n= 2 \times 11 \times 89=1958 \text{ or } \\
n=2 \times 17 \times 59=2006 \text{ or } \\
n=2 \times 19 \times 53=2014.
\end{array}
$$
(ii) If $p \geqslant 3$, then $1+p, 1+q, 1+r$ are all even numbers.
$$
\text{Let } 1+p=2 p_{1}, 1+q=2 q_{1}, 1+r=2 r_{1}.
$$
Then $p_{1} q_{1} r_{1}=405=3^{4} \times 5 (2 \leqslant p_{1}<q_{1}<r_{1})$.
Thus $\left(p_{1}, q_{1}, r_{1}\right)=(3,5,27),(3,9,15)$
$$
\Rightarrow(p, q, r)=(5,9,53),(5,17,29).
$$
Since $q$ is a prime number, we get
$$
\begin{array}{l}
(p, q, r)=(5,17,29) \\
\Rightarrow n=5 \times 17 \times 29=2465.
\end{array}
$$
In summary, $n_{\min }=2 \times 3 \times 269=1614$.
|
1614
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Question 3 Let positive real numbers $a, b, c$ satisfy $ab + bc + ca = 48$. Try to find the minimum value of $f = \left(a^{2} + 5\right)\left(b^{2} + 5\right)\left(c^{2} + 5\right)$.
|
Given $A=48, k=5$, hence
$$
f \geqslant 5(48-5)^{2}=9245 \text {. }
$$
When $a=5,\{b, c\}=\left\{\frac{43+\sqrt{97}}{12}, \frac{43-\sqrt{97}}{12}\right\}$,
the equality holds.
Therefore, $f_{\min }=9245$.
|
9245
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. As shown in Figure 1, in the right $\triangle A B C$, it is known that $\angle A C B$ $=90^{\circ}, A C=21, B C=$ 28, and a square $A B D E$ is constructed outward with $A B$ as one side. The angle bisector of $\angle A C B$ intersects $D E$ at point $F$. Then the length of line segment $D F$ is $\qquad$
|
8. 15 .
As shown in Figure 6, let $CF$ intersect $AB$ and $AD$ at points $G$ and $O$ respectively.
Then $\angle BCO = 45^{\circ}$
$= \angle OAB$.
Thus, $O, A, C, B$
are concyclic.
Hence $\angle ABO$
$$
= \angle OCA = 45^{\circ}.
$$
Therefore, $OA = OB$.
Thus, $O$ is the center of the square $ABDE$.
By symmetry, $DF = AG$.
Hence $\frac{AG}{GB} = \frac{AC}{BC} = \frac{21}{28} = \frac{3}{4}$
$\Rightarrow AG = \frac{3}{7} AB = \frac{3}{7} \times 35 = 15$.
Therefore, $DF = 15$.
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. Let $d$ be a positive divisor of 2015. Then the maximum value of the unit digit of $d^{\frac{2011}{d}}$ is $\qquad$ .
|
9.7.
Notice that, $2015=5 \times 13 \times 31$.
Therefore, 2015 has eight positive divisors.
Let $G(n)$ denote the unit digit of $n$, then
$$
\begin{array}{l}
G\left(1^{2015}\right)=G\left(31^{165}\right)=1, \\
G\left(2015^{1}\right)=G\left(5^{403}\right)=G\left(65^{31}\right) \\
=G\left(155^{13}\right)=5 .
\end{array}
$$
From $G\left(3^{4}\right)=1$, we know
$$
\begin{aligned}
G\left(13^{155}\right) & =G\left(3^{155}\right)=G\left(3^{4 \times 38+3}\right) \\
=G\left(3^{3}\right) & =7, \\
G\left(403^{5}\right) & =G\left(3^{5}\right)=3 .
\end{aligned}
$$
Thus, the maximum value of the unit digit of $d^{\frac{20115}{d}}$ is 7.
|
7
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Given that $a$ and $b$ are real numbers, the system of inequalities about $x$
$$
\left\{\begin{array}{l}
20 x+a>0, \\
15 x-b \leqslant 0
\end{array}\right.
$$
has only the integer solutions $2, 3, 4$. Then the maximum value of $ab$ is . $\qquad$
|
10. -1200 .
From the conditions, we know that $-\frac{a}{20}<x \leqslant \frac{b}{15}$.
The integer solutions of the inequality system are only $2, 3, 4$, so $1 \leqslant-\frac{a}{20}<2,4 \leqslant \frac{b}{15}<5$, which means $-40<a \leqslant-20,60 \leqslant b<75$.
Therefore, $-3000<a b \leqslant-1200$.
Thus, when $a=-20, b=60$, $a b$ has the maximum value -1200.
|
-1200
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
16. (25 points) Given $A \subseteq\{1,2, \cdots, 2014\}$, let real numbers $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} 、 x_{1} 、 x_{2} 、 x_{3}$ satisfy
(i) $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3} \in\{-1,0,1\}$ and not all are 0;
(ii) $x_{1}, x_{2} 、 x_{3} \in A$;
(iii) If $x_{i}=x_{j}$, then $\lambda_{i} \lambda_{j} \neq-1(1 \leqslant i 、 j \leqslant 3)$.
If all numbers of the form $x_{1} x_{2} x_{3}$ and $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}$ are not multiples of 2014, then the set $A$ is called a "good set". Find the maximum number of elements in a good set $A$.
|
16. (1) Construct a good set $A$ with 503 elements.
Let $A=\{1,3,5, \cdots, 1005\}$.
If $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ are all non-zero, then
$$
\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3} \equiv x_{1}+x_{2}+x_{3} \equiv 1(\bmod 2) \text {. }
$$
Thus, $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}$ is odd and cannot be a multiple of 2014.
If $\lambda_{1} 、 \lambda_{2} 、 \lambda_{3}$ include 0, assume $\lambda_{3}=0$, then by condition (i), at least one of $\lambda_{1} 、 \lambda_{2}$ is non-zero.
By condition (iii), $\lambda_{1} x_{1}+\lambda_{2} x_{2} \neq 0$.
Notice that,
$$
\begin{array}{l}
\left|\lambda_{1} x_{1}+\lambda_{2} x_{2}\right| \leqslant\left|\lambda_{1} x_{1}\right|+\left|\lambda_{2} x_{2}\right| \\
\leqslant\left|x_{1}\right|+\left|x_{2}\right| \leqslant 2 \times 10051007$, then changing $\lambda_{i}$ to $-\lambda_{i}$ and $x_{i}$ to $2014-x_{i}$ results in a number that is congruent to $\lambda_{1} x_{1}+\lambda_{2} x_{2}+\lambda_{3} x_{3}$ modulo 2014.
We now discuss different cases for $r$.
1) If $d \leqslant r<2 d$, then at most $d-1$ numbers in $d+1, d+2, \cdots, d+r$ belong to the set $S$.
$$
\begin{array}{l}
\text { Hence }|S| \leqslant 1+(d-1)+d q=d q+d \\
\leqslant d q+\frac{r}{2}+\frac{d}{2}=503 .
\end{array}
$$
2) If $0 \leqslant r \leqslant d-1$, then
$$
\begin{array}{l}
|S| \leqslant 1+r+d q \leqslant d q+\frac{r}{2}+\frac{d}{2}+\frac{1}{2}=503.5 \\
\Rightarrow|S| \leqslant 503 .
\end{array}
$$
Therefore, any good set $S$ must satisfy $|S| \leqslant 503$. From (1) and (2), we know that the maximum number of elements in a good set $A$ is 503.
|
503
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given sets
$$
A=\{1,3,5,7,9\}, B=\{2,4,6,8,10\} \text {. }
$$
If set $C=\{x \mid x=a+b, a \in A, b \in B\}$, then the number of elements in set $C$ is $\qquad$ .
|
,- 1.9 .
Since $a$ is odd and $b$ is even, all elements in set $C$ are odd.
Also, the minimum value of $a+b$ is 3, and the maximum value is 19, and all odd numbers between 3 and 19 can be obtained, thus, set $C$ contains 9 elements.
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given the function
$$
f(x)=\left\{\begin{array}{ll}
0, & x<0, \\
1, & x \geqslant 0 .
\end{array}\right.
$$
Then $f(f(x))=$ $\qquad$
|
2. 1 .
Since for any $x \in \mathbf{R}$, we have $f(x) \geqslant 0$, therefore, $f(f(x))=1$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given
$$
\sin \alpha+\sqrt{3} \sin \beta=1, \cos \alpha+\sqrt{3} \cos \beta=\sqrt{3} \text {. }
$$
Then the value of $\cos (\alpha-\beta)$ is $\qquad$ .
|
3. 0 .
Squaring and adding the two known equations, we get
$$
\begin{array}{l}
4+2 \sqrt{3}(\sin \alpha \cdot \sin \beta+\cos \alpha \cdot \cos \beta)=4 \\
\Rightarrow \cos (\alpha-\beta)=0
\end{array}
$$
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. If positive integers $m, n$ satisfy $\frac{(m+n)!}{n!}=5040$, then the value of $m!n$ is $\qquad$ .
|
9. 144 .
$$
\begin{array}{l}
\text { Given } \frac{(m+n)!}{n!} \\
=(m+n)(m+n-1) \cdots(n+1), \\
5040=10 \times 9 \times 8 \times 7,
\end{array}
$$
we know $\left\{\begin{array}{l}m+n=10, \\ n+1=7\end{array} \Rightarrow\left\{\begin{array}{l}m=4, \\ n=6 .\end{array}\right.\right.$
Therefore, $m!n=144$.
|
144
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. Let the monotonic increasing sequence $\left\{a_{n}\right\}$ consist of positive integers, and $a_{7}=120, a_{n+2}=a_{n}+a_{n+1}\left(n \in \mathbf{Z}_{+}\right)$. Then $a_{8}=$ . $\qquad$
|
10. 194 .
From $a_{n+2}=a_{n}+a_{n+1}$, we get $a_{7}=5 a_{1}+8 a_{2}=120, a_{8}=8 a_{1}+13 a_{2}$. Since $(5,8)=1$, and $a_{1}, a_{2}$ are both positive integers, it follows that $8\left|a_{1}, 5\right| a_{2}$.
Let $a_{1}=8 k, a_{2}=5 m\left(k, m \in \mathbf{Z}_{+}\right)$.
Then $k+m=3$.
Also, $a_{1}<a_{2}$, so, $k=1, m=2$.
Thus, $a_{1}=8, a_{2}=10$.
Therefore, $a_{8}=8 a_{1}+13 a_{2}=194$.
|
194
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $n$ be the smallest positive integer satisfying the following conditions:
(1) $n$ is a multiple of 75;
(2) $n$ has exactly 75 positive divisors (including 1 and itself).
Find $\frac{n}{75}$.
(Eighth American Mathematical Invitational)
|
```
Given: $n=75 k=3 \times 5^{2} k$.
\[
\begin{array}{l}
\text { By } 75=3 \times 5 \times 5 \\
=(2+1)(4+1)(4+1),
\end{array}
\]
we know the number of prime factors is at most three.
By discussing the number of prime factors, we can solve to get
\[
\left(\frac{n}{75}\right)_{\min }=432 \text {. }
\]
```
|
432
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Choose three different angles from $1^{\circ}, 2^{\circ}, \cdots, 179^{\circ}$ to form the three interior angles of a triangle. There are $\qquad$ different ways to do this.
|
3.2611.
Notice that, the equation $x+y+z=180$ has $\mathrm{C}_{179}^{2}=$ 15931 sets of positive integer solutions, among which, the solution where $x=y=z$ is 1 set, and the solutions where exactly two of $x, y, z$ are equal are $88 \times 3=$ 264 sets.
Therefore, the number of selection methods is $\frac{15931-1-264}{\mathrm{~A}_{3}^{3}}=2611$ kinds.
|
2611
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $f_{0}(x)=|x|-2015$,
$$
f_{n}(x)=\left|f_{n-1}(x)\right|-1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$
Then the number of zeros of the function $y=f_{2015}(x)$ is
$\qquad$
|
8.4031 .
From the graph, it is easy to see that the function $y=f_{1}(x)$ has 4 zeros, the function $y=f_{2}(x)$ has 6 zeros, $\cdots \cdots$ and so on, the function $y=f_{2014}(x)$ has 4030 zeros. However, the intersection point of the function $y=f_{2014}(x)$ with the y-axis is $(0,1)$, therefore, the function $y=f_{2015}(x)$ has a total of 4031 zeros.
|
4031
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. For a sequence of real numbers $x_{1}, x_{2}, \cdots, x_{n}$, define its "value" as $\max _{1 \leqslant i \leqslant n}\left\{\left|x_{1}+x_{2}+\cdots+x_{i}\right|\right\}$. Given $n$ real numbers, David and George want to arrange these $n$ numbers into a sequence with low value. On one hand, diligent David examines all possible ways to find the minimum possible value $D$. On the other hand, greedy George chooses $x_{1}$ such that $\left|x_{1}\right|$ is as small as possible. From the remaining numbers, he chooses $x_{2}$ such that $\left|x_{1}+x_{2}\right|$ is as small as possible, $\cdots \cdots$ At the $i$-th step, he chooses $x_{i}$ from the remaining numbers such that $\left|x_{1}+x_{2}+\cdots+x_{i}\right|$ is as small as possible. At each step, if there are multiple numbers that give the same minimum absolute sum, George arbitrarily chooses one. Finally, the value of the sequence he gets is $G$. Find the smallest constant $c$, such that for every positive integer $n$, every array of $n$ real numbers, and every sequence George can obtain, we have
$$
G \leqslant c D \text {. }
$$
|
3. $c=2$.
If the initial numbers given are $1, -1, 2, -2$, then David arranges these four numbers as $1, -2, 2, -1$, and George can get the sequence $1, -1, 2, -2$.
Thus, $D=1, G=2$.
Therefore, $c \geqslant 2$.
We now prove: $G \leqslant 2 D$.
Let the $n$ real numbers be $x_{1}, x_{2}, \cdots, x_{n}$, and assume David and George have already obtained their arrangements. Suppose David and George get the sequences $d_{1}, d_{2}, \cdots, d_{n}$ and $g_{1}, g_{2}, \cdots, g_{n}$, respectively.
$$
\begin{array}{c}
\text { Let } M=\max _{1<i \leqslant n}\left\{\left|x_{i}\right|\right\}, \\
S=\left|x_{1}+x_{2}+\cdots+x_{n}\right|, \\
N=\max \{M, S\} .
\end{array}
$$
Then $D \geqslant S$,
$D \geqslant \frac{M}{2}$,
$G \leqslant N=\max \{M, S\}$.
From equations (1), (2), and (3), we get
$$
G \leqslant \max \{M, S\} \leqslant \max \{M, 2 S\} \leqslant 2 D \text {. }
$$
In fact, by the definition of value, equation (1) holds.
For equation (2), consider an index $i$ such that $\left|d_{i}\right|=M$.
Thus, $M=\left|d_{i}\right|$
$=\left|\left(d_{1}+d_{2}+\cdots+d_{i}\right)-\left(d_{1}+d_{2}+\cdots+d_{i-1}\right)\right|$
$\leqslant\left|d_{1}+d_{2}+\cdots+d_{i}\right|+\left|d_{1}+d_{2}+\cdots+d_{i-1}\right|$
$\leqslant 2$.
Therefore, equation (2) holds.
Finally, we prove that equation (3) holds.
Let $h_{i}=g_{1}+g_{2}+\cdots+g_{i}$.
We use mathematical induction on $i$ to prove $\left|h_{i}\right| \leqslant N$.
When $i=1$, $\left|h_{1}\right|=\left|g_{1}\right| \leqslant M \leqslant N$, so the conclusion holds.
Notice that, $\left|h_{n}\right|=S \leqslant N$, so the conclusion also holds.
Assume $\left|h_{i-1}\right| \leqslant N$, and consider two cases.
(1) Assume that $g_{i}, g_{i+1}, \cdots, g_{n}$ do not have two terms with opposite signs. Without loss of generality, assume they are all non-negative.
Then $h_{i-1} \leqslant h_{i} \leqslant \cdots \leqslant h_{n}$.
Thus, $\left|h_{i}\right| \leqslant \max \left\{\left|h_{i-1}\right|,\left|h_{n}\right|\right\} \leqslant N$.
(2) In $g_{i}, g_{i+1}, \cdots, g_{n}$, there are positive and negative numbers, so there exists an index $j \geqslant i$ such that $h_{i-1} g_{j} \leqslant 0$.
By the definition of the sequence George gets,
$$
\begin{array}{l}
\left|h_{i}\right|=\left|h_{i-1}+g_{i}\right| \leqslant\left|h_{i-1}+g_{j}\right| \\
\leqslant \max \left\{\left|h_{i-1}\right|,\left|g_{j}\right|\right\} \leqslant N .
\end{array}
$$
Thus, the conclusion holds.
This shows that equation (3) also holds.
|
2
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. There are infinitely many cards, each with a real number written on it. For each real number $x$, there is exactly one card with the number $x$ written on it. Two players each select a set of 100 cards, denoted as $A$ and $B$, such that the sets are disjoint. Formulate a rule to determine which of the two players wins, satisfying the following conditions:
(1) The winner depends only on the relative order of these 200 cards: if these 200 cards are placed face down in increasing order, and the audience is informed which card belongs to which player, but not what number is written on each card, the audience can still determine who will win;
(2) If the elements of the two sets are written in increasing order as
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{100}\right\}, B=\left\{b_{1}, b_{2}, \cdots, b_{100}\right\},
$$
where, for all $i \in\{1,2, \cdots, 100\}$, $a_{i}>b_{i}$, then $A$ defeats $B$;
(3) If three players each select a set of 100 cards, denoted as $A$, $B$, and $C$, and $A$ defeats $B$, and $B$ defeats $C$, then $A$ defeats $C$.
Question: How many such rules are there?
[Note] Two different rules mean that there exist two sets $A$ and $B$ such that in one rule, $A$ defeats $B$, and in the other rule, $B$ defeats $A$.
|
6. There are 100 ways.
To prove the more general case, where each person selects $n$ cards, there are $n$ ways that satisfy the conditions. Let $A>B$ or $B>A$ if and only if $a_{k}>b_{k}$.
Such rules satisfy all three conditions, and different $k$ correspond to different rules, hence there are at least $n$ different rules.
Next, we prove that there are no other ways to define such rules.
Assume a rule satisfies the conditions, let $k \in\{1,2$, $\cdots, n\}$ be the smallest positive integer satisfying the following property:
$$
\begin{array}{l}
A_{k}=\{1,2, \cdots, k, n+k+1, n+k+2, \cdots, 2 n\} \\
B_{k-1}$.
Thus, $U>W$ (trivial when $k=1$).
The elements in the set $V \cup W$ arranged in increasing order are the same as the elements in the set $A_{k} \cup B_{k}$ arranged in increasing order.
By the choice of $k$, $A_{k}<B_{k}$.
Thus, $V<W$.
Therefore, $X<V<W<U<Y$.
By condition (3), $X<Y$.
Hence, the conclusion holds.
|
100
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If the set $S=\{1,2,3, \cdots, 16\}$ is arbitrarily divided into $n$ subsets, then there must exist a subset in which there are elements $a, b, c$ (which can be the same), satisfying $a+b=c$. Find the maximum value of $n$.
【Note】If the subsets $A_{1}, A_{2}, \cdots, A_{n}$ of set $S$ satisfy the following conditions:
(1) $A_{i} \neq \varnothing(i=1,2, \cdots, n)$;
(2) $A_{i} \cap A_{j}=\varnothing$;
(3) $\bigcup_{i=1}^{n} A_{i}=S$,
then $A_{1}, A_{2}, \cdots, A_{n}$ is called a partition of set $S$.
（Proposed by the Problem Committee）
|
4. First, when $n=3$, assume there exists a partition of the set that does not satisfy the condition. Thus, there must be a subset with at least six elements, without loss of generality, let
$$
A=\left\{x_{1}, x_{2}, \cdots, x_{6}\right\}\left(x_{1}<x_{2}<\cdots<x_{6}\right) \text {. }
$$
Then $x_{6}-x_{1}, x_{6}-x_{2}, \cdots, x_{6}-x_{5} \notin A$. Among these, there must be three elements belonging to another subset, let
$$
x_{6}-x_{i}, x_{6}-x_{j}, x_{6}-x_{k} \in B(1 \leqslant i<j<k \leqslant 5) \text {. }
$$
Then $x_{j}-x_{i}, x_{k}-x_{j}, x_{k}-x_{i} \notin A \cup B$, but they cannot all belong to the third subset, leading to a contradiction.
Therefore, $n=3$ satisfies the condition.
Second, when $n=4$, the partition
$$
\{1\},\{2,3\},\{4,5,6,7,16\},\{8,9, \cdots, 15\}
$$
does not satisfy the condition.
In summary, $n_{\max }=3$.
[Note] $\{1,4,7,10,13,16\},\{2,3,8,9,14,15\}$, $\{5,6\},\{11,12\}$, etc., are also valid.
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. A shape obtained by removing one unit square from a $2 \times 2$ grid is called an "L-shape". In an $8 \times 8$ grid, place $k$ L-shapes, each of which can be rotated, but each L-shape must cover exactly three unit squares of the grid, and any two L-shapes must not overlap. Additionally, no other L-shape can be placed besides these $k$ L-shapes. Find the minimum value of $k$.
(Zhang Yuezhu, Contributed)
|
6. On one hand, using six straight lines to divide the large square grid into 16 smaller $2 \times 2$ grids, each $2 \times 2$ grid must have at least two cells covered by an L-shape. Therefore, the number of cells covered is no less than 32. Thus, the number of L-shapes is no less than $\left[\frac{32}{3}\right]+1$ $=11$, meaning no fewer than 11 L-shapes.
On the other hand, placing 11 L-shapes as shown in Figure 4 can meet the requirements of the problem.
In conclusion, $k_{\min }=11$.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given odd prime numbers $x, y, z$ satisfying
$$
x \mid \left(y^{5}+1\right), y \mid \left(z^{5}+1\right), z \mid \left(x^{5}+1\right) \text {. }
$$
Find the minimum value of the product $x y z$. (Cheng Chuanping, problem contributor)
|
7. Let's assume $x$ is the minimum of $x, y, z$.
(1) If $x=3$, then
$$
3^{5}+1=244=2^{2} \times 61 \Rightarrow z=61.
$$
Since $3 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{3}$.
Clearly, $5 \nmid (61^{5}+1)$.
After calculation, we find $11 \mid (61^{5}+1)$.
Thus, $y_{\text{min}}=11$.
Therefore, $(x y z)_{\min}=3 \times 11 \times 61=2013$.
This indicates that the smallest set of three consecutive odd primes greater than 2013 is $11 \times 13 \times 17=2431$.
Hence, we only need to consider the cases where $x=5$ or $x=7$.
(2) If $x=5$, note that,
$$
5^{5}+1=3126=2 \times 3 \times 521.
$$
Since $5 \times 521 > 2013$, this does not meet the requirement.
(3) If $x=7$, we estimate the value of $7^{5}+1$.
Since $7 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{7}$.
If $y=13$, then $13 \mid (z^{5}+1)$.
Thus, $z \equiv -1 \pmod{13}$.
Hence, $z_{\text{min}}=103$, and $(x y z)_{\min} > 2013$.
In summary, the minimum value of $x y z$ is 2013.
|
2013
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. A coin collector has 100 coins that look the same. The collector knows that 30 of them are genuine, and 70 are fake, and that all genuine coins weigh the same, while all fake coins weigh different and are heavier than the genuine ones. The collector has a balance scale that can be used to compare the weight of two groups of coins of the same number. To ensure finding at least one genuine coin, how many weighings are needed at minimum?
|
8. First, it is clear that 70 weighings can certainly find at least one genuine coin.
In fact, each time, one coin is placed on each side of the balance. If they weigh the same, both are genuine; if they do not, the heavier one must be a counterfeit. Thus, each weighing either finds two genuine coins or one counterfeit.
After 70 weighings, either at least two genuine coins have been found, or all 70 counterfeits have been identified. This indicates that all genuine coins can be found.
Next, we prove that when the weight of the genuine coin is \(2^{100}\), and the weight of the counterfeit coins is
\[
m_{i}=2^{100}+2^{i} \quad (1 \leqslant i \leqslant 70),
\]
69 weighings cannot guarantee finding one genuine coin.
It is easy to see that during the weighing (let's say \(k\) coins on each side), the side with the heaviest counterfeit among the \(2k\) coins will be heavier.
Assume the collector follows his plan and is informed of the results of his weighings and the weights assigned to some counterfeit coins.
The first time he weighs, designate any one of the heavier side as weight \(m_{70}\).
Generally, when he weighs, if neither side has any counterfeit coins with already assigned weights, assign the heaviest unassigned weight to any one of the counterfeit coins on the heavier side; if there are counterfeit coins with previously assigned weights on the balance, do not assign any weight to any counterfeit coin this round.
After 69 rounds, the weight \(m_{1}\) has not been assigned, and the counterfeit coin with weight \(m_{1}\) is indistinguishable from 30 genuine coins.
In summary, finding one genuine coin requires at least 70 weighings.
(Li Weiguo provided)
|
70
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The function $f(x)$ defined on $\mathbf{R}$ satisfies $f\left(\frac{2 a+b}{3}\right)=\frac{2 f(a)+f(b)}{3}(a, b \in \mathbf{R})$,
and $f(1)=1, f(4)=7$.
Then $f(2015)=$
|
$$
-, 1.4029 .
$$
From the function $f(x)$ being concave or convex in reverse on $\mathbf{R}$, we know its graph is a straight line, $f(x)=a x+b$.
Given $f(1)=1$ and $f(4)=7$, we have
$$
\begin{array}{l}
a=2, b=-1 \Rightarrow f(x)=2 x-1 \\
\Rightarrow f(2015)=4029 .
\end{array}
$$
|
4029
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. For any point $A(x, y)$ in the plane region $D$:
$$
\left\{\begin{array}{l}
x+y \leqslant 1, \\
2 x-y \geqslant-1, \\
x-2 y \leqslant 1
\end{array}\right.
$$
and a fixed point $B(a, b)$, both satisfy $\overrightarrow{O A} \cdot \overrightarrow{O B} \leqslant 1$. Then the maximum value of $a+b$ is $\qquad$
|
2. 2 .
From the problem, we know that for any $A(x, y) \in D$, we have $a x+b y \leqslant 1$.
By taking
$$
(x, y)=(1,0),(0,1),(-1,-1) \text {, }
$$
we get that the fixed point $B(a, b)$ satisfies the necessary conditions
$$
\left\{\begin{array}{l}
a \leqslant 1, \\
b \leqslant 1, \\
a+b \geqslant-1
\end{array} \quad \Rightarrow a+b \leqslant 2\right. \text {. }
$$
For the fixed point $B(1,1)$, for any $A(x, y) \in D$, we have
$$
\overrightarrow{O A} \cdot \overrightarrow{O B}=(1,1) \cdot(x, y)=x+y \leqslant 1 \text {. }
$$
When $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right)$, the equality holds.
Thus, the point $B(1,1)$ satisfies the conditions of the problem, and in this case, $a+b=2$.
Therefore, $(a+b)_{\max }=2$.
|
2
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Define the function on $\mathbf{R}$
$$
f(x)=\left\{\begin{array}{ll}
\log _{2}(1-x), & x \leqslant 0 ; \\
f(x-1)-f(x-2), & x>0 .
\end{array}\right.
$$
Then $f(2014)=$
|
3. 1.
First, consider the method for finding the function values on the interval $(0,+\infty)$.
If $x>3$, then
$$
\begin{array}{l}
f(x)=f(x-1)-f(x-2) \\
=f(x-1)-(f(x-1)+f(x-3)) \\
=-f(x-3) .
\end{array}
$$
Thus, for $x>6$, we have
$$
f(x)=-f(x-3)=f(x-6) \text {. }
$$
Therefore, $f(x)$ is a periodic function with a period of 6 on the interval $(0,+\infty)$.
$$
\begin{array}{l}
\text { Hence } f(2014)=f(4)=f(3)-f(2) \\
=-f(0)+f(-1)=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The set $X \backslash Y=\{a \mid a \in X, a \notin Y\}$ is called the difference set of set $X$ and set $Y$. Define the symmetric difference of sets $A$ and $B$ as
$$
A \Delta B=(A \backslash B) \cup(B \backslash A) \text {. }
$$
If two non-empty finite sets $S$ and $T$ satisfy $|S \Delta T|=1$, then the minimum value of $k=|S|+|T|$ is $\qquad$
|
6. 3 .
If $|S \Delta T|=1$, without loss of generality, assume
$$
\left\{\begin{array}{l}
|S \backslash T|=1, \\
|T \backslash S|=0,
\end{array}\right.
$$
i.e., $T$ is a proper subset of $S$, thus,
$$
\begin{array}{l}
|S|>|T| \geqslant 1 . \\
\text { Hence } k \geqslant|T|+1+|T| \geqslant 3 .
\end{array}
$$
$$
\text { Construct sets } S=\{1,2\}, T=\{2\} \text {. }
$$
Clearly, the condition $|S \Delta T|=1$ is satisfied, and
$$
\begin{array}{l}
k=|S|+|T|=3 . \\
\text { Therefore, } k_{\min }=3 .
\end{array}
$$
|
3
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. In an $m$ row by 10 column grid, fill each cell with either 0 or 1, such that each column contains exactly three 1s. Let the sum of the numbers in the $i$-th row ($i=1,2, \cdots, m$) be denoted as $x_{i}$, and for any two columns, there always exists a row where the cells intersecting with these two columns are both filled with 1. Let $x=\max _{1 \leqslant i \leqslant m}\left\{x_{i}\right\}$. Then $x_{\text {min }}=$ $\qquad$
|
8. 5 .
Construct an application model: "10 people go to a bookstore to buy $m$ kinds of books, each person buys 3 books, and any two of them buy at least one book in common. Find the sales volume of the book that is purchased by the most people."
Let the $i(i=1,2, \cdots, m)$-th kind of book be purchased by $x_{i}$ people, $x=\max _{1 \leqslant i \leqslant m} x_{i}$. Then
$$
\sum_{i=1}^{m} x_{i}=30 \text {, and } x_{i} \leqslant x(i=1,2, \cdots, m) \text {. }
$$
Considering the total number of "same book pairs" between any two people, we get the inequality
$$
\begin{array}{l}
\sum_{i=1}^{m} \mathrm{C}_{x_{i}}^{2} \geqslant \mathrm{C}_{10}^{2} \\
\Rightarrow \sum_{i=1}^{m} \frac{x_{i}\left(x_{i}-1\right)}{2} \geqslant 45 .
\end{array}
$$
Thus, from conclusion (1) and equation (2), we have
$$
\begin{array}{l}
45 \leqslant \frac{1}{2}(x-1) \sum_{i=1}^{m} x_{i}=15(x-1) \\
\Rightarrow x \geqslant 4 .
\end{array}
$$
If $x=4$, then all $x_{i}=4(i=1,2, \cdots, m)$, which contradicts $\sum_{i=1}^{m} x_{i}=30$. Therefore, $x \geqslant 5$.
Below, we construct a scenario where $x=5$, and the book purchased by the most people is bought by exactly 5 people.
Let the books be $B_{i}$, and the book purchasing situation for the 10 people is as follows:
$$
\begin{array}{l}
B_{1} B_{2} B_{3}, B_{3} B_{4} B_{5}, B_{1} B_{5} B_{6}, B_{1} B_{3} B_{5}, \\
B_{1} B_{2} B_{4}, B_{1} B_{4} B_{6}, B_{2} B_{4} B_{5}, B_{2} B_{5} B_{6}, \\
B_{2} B_{3} B_{6}, B_{3} B_{4} B_{6} .
\end{array}
$$
Therefore, the book purchased by the most people is bought by at least 5 people.
|
5
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a, b$ be the roots of the quadratic equation $x^{2}-x-1=0$. Then the value of $3 a^{2}+4 b+\frac{2}{a^{2}}$ is
|
From the given, we have $a b=-1, a+b=1$,
$$
\begin{array}{l}
a^{2}=a+1, b^{2}=b+1 . \\
\text { Therefore, } 3 a^{3}+4 b+\frac{2}{a^{2}}=3 a^{3}+4 b+2 b^{2} \\
=3 a^{2}+3 a+6 b+2=6(a+b)+5=11 .
\end{array}
$$
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Fill the numbers $1,2, \cdots, 36$ in a $6 \times 6$ grid, with each cell containing one number, such that the numbers in each row are in increasing order from left to right. Then the minimum value of the sum of the six numbers in the third column is $\qquad$
|
4. 63.
Let the six numbers filled in the third column, arranged in ascending order, be $A, B, C, D, E, F$.
Since the row where $A$ is located needs to fill in two numbers smaller than $A$ before it, then $A \geqslant 3$; since the row where $B$ is located needs to fill in two numbers smaller than $B$, and $A$ and the two numbers before $A$ are all smaller than $B$, then $B \geqslant 6$.
Similarly, $C \geqslant 9, D \geqslant 12, E \geqslant 15, F \geqslant 18$.
Therefore, the sum of the six numbers filled in the third column is
$$
\begin{array}{l}
A+B+C+D+E+F \\
\geqslant 3+6+9+12+15+18=63 .
\end{array}
$$
Table 1 is one way to fill the numbers so that the sum of the six numbers in the third column reaches the minimum value (the filling method for the last three columns is not unique).
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline 1 & 2 & 3 & 19 & 20 & 21 \\
\hline 4 & 5 & 6 & 25 & 27 & 29 \\
\hline 7 & 8 & 9 & 22 & 23 & 24 \\
\hline 10 & 11 & 12 & 26 & 28 & 30 \\
\hline 13 & 14 & 15 & 31 & 34 & 35 \\
\hline 16 & 17 & 18 & 32 & 33 & 36 \\
\hline
\end{tabular}
|
63
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Let $a$ and $b$ be distinct real numbers. If the quadratic function $f(x)=x^{2}+a x+b$ satisfies $f(a)=f(b)$, then the value of $f(2)$ is $\qquad$ .
|
$$
-, 1.4
$$
From the given conditions and the symmetry of the quadratic function graph, we get
$$
\begin{array}{l}
\frac{a+b}{2}=-\frac{a}{2} \Rightarrow 2 a+b=0 \\
\Rightarrow f(2)=4+2 a+b=4 .
\end{array}
$$
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the real number $\alpha$ satisfies $\cos \alpha=\tan \alpha$, then $\frac{1}{\sin \alpha}+\cos ^{4} \alpha=$ $\qquad$
|
2. 2 .
From the condition, we know that $\cos ^{2} \alpha=\sin \alpha$.
$$
\begin{array}{l}
\text { Then } \frac{1}{\sin \alpha}+\cos ^{4} \alpha=\frac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\sin \alpha}+\sin ^{2} \alpha \\
=(1+\sin \alpha)+\left(1-\cos ^{2} \alpha\right) \\
=2+\sin \alpha-\cos ^{2} \alpha=2 .
\end{array}
$$
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In the plane rectangular coordinate system $x O y$, the set of points
$$
\begin{aligned}
K= & \{(x, y) \mid(|x|+|3 y|-6) . \\
& (|3 x|+|y|-6) \leqslant 0\}
\end{aligned}
$$
corresponds to a plane region whose area is $\qquad$
|
6. 24 .
Let $K_{1}=\{(x, y)|| x|+| 3 y |-6 \leqslant 0\}$.
First, consider the part of the point set $K_{1}$ in the first quadrant, at this time, $x+3 y \leqslant 6$. Therefore, these points correspond to $\triangle O C D$ and its interior in Figure 2.
By symmetry, the region corresponding to the point set $K_{1}$ is the rhombus $A B C D$ and its interior centered at the origin $O$ in Figure 2.
Similarly, let
$$
K_{2}=\{(x, y)|| 3 x|+| y |-6 \leqslant 0\} \text {. }
$$
Then the region corresponding to the point set $K_{2}$ is the rhombus $E F G H$ and its interior centered at $O$ in Figure 2.
By the definition of the point set $K$, the plane region corresponding to $K$ is the part covered by exactly one of the point sets $K_{1}$ and $K_{2}$.
Therefore, what is required in this problem is the area $S$ of the shaded region in Figure 2.
By $l_{C D}: x+3 y=6, l_{G H}: 3 x+y=6$, we know that the intersection point of the two lines is $P\left(\frac{3}{2}, \frac{3}{2}\right)$.
By symmetry,
$$
S=8 S_{\triangle C P G}=8 \times \frac{1}{2} \times 4 \times \frac{3}{2}=24 .
$$
|
24
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For a four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c 、 d \leqslant$ $9)$, if $a>b, b<c, c>d$, then $\overline{a b c d}$ is called a $P$ class number; if $a<c, c<d$, then $\overline{a b c d}$ is called a $Q$ class number. Let $N(P)$ and $N(Q)$ represent the number of $P$ class numbers and $Q$ class numbers, respectively. Then the value of $N(P)-N(Q)$ is $\qquad$
|
8. 285.
Let the sets of all P-type numbers and Q-type numbers be denoted as $A$ and $B$, respectively. Let the set of all P-type numbers whose unit digit is zero be denoted as $A_{0}$, and the set of all P-type numbers whose unit digit is not zero be denoted as $A_{1}$.
For any four-digit number $\overline{a b c d} \in A_{1}$, map it to the four-digit number $\overline{d c b a}$.
Note that, $a > b$, and $bd \geqslant 1$.
Then, $\overline{d c b a} \in B$.
Conversely, each $\overline{d c b a} \in B$ uniquely corresponds to an element $\overline{a b c d} \in A_{1}$.
Therefore, a one-to-one correspondence is established between $A_{1}$ and $B$.
Hence, $N(P) - N(Q) = |A| - |B|$
$= |A_{0}| + |A_{1}| - |B| = |A_{0}|$.
Next, we calculate $|A_{0}|$.
For any four-digit number $\overline{a b c 0} \in A_{0}$, $b$ can take values $0, 1, \cdots, 9$. For each $b$, given $b < a \leqslant 9$ and $b < c \leqslant 9$, $a$ and $c$ each have $9 - b$ possible values.
Thus, $|A_{0}| = \sum_{b=0}^{9} (9 - b)^{2} = \sum_{k=1}^{9} k^{2} = 285$.
Therefore, $N(P) - N(Q) = 285$.
|
285
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ be an arithmetic sequence, and
$$
\sum_{i=1}^{n}\left|a_{i}+j\right|=2028(j=0,1,2,3) \text {. }
$$
Then the maximum value of the number of terms $n$ is
|
6. 52.
Since the equation $|x|=|x+1|=|x+2|$ has no solution, we have $n \geqslant 2$ and the common difference is not 0.
Assume the general term of the sequence is $a-k d(1 \leqslant k \leqslant n, d>0)$.
Construct the function $f(x)=\sum_{k=1}^{n}|x-k d|$.
The given condition is equivalent to $f(x)=2028$ having at least four distinct roots $a, a+1, a+2, a+3$, which means the graph of $y=f(x)$ intersects the line $l: y=2028$ at least four times.
Notice that the graph of $y=f(x)$ is a downward convex broken line with $n+1$ segments symmetric about the line $y=\frac{(n+1) d}{2}$. It intersects the line $l$ at least four times if and only if there is a horizontal segment of the broken line on the line $l$, which happens if and only if $n=2 m$ and $a, a+1, a+2, a+3 \in[m d,(m+1) d]$, $f(m d)=2028$, i.e., $d \geqslant 3$ and $m^{2} d=2028$.
This gives $m^{2} \leqslant \frac{2028}{3}=676 \Rightarrow m \leqslant 26$.
Clearly, when $m=26$, taking $d=3, a=78$ satisfies the conditions of the problem.
Therefore, the maximum value of $n$ is 52.
|
52
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Then the sum of the squares of all natural numbers $m$ that satisfy
$$
[(2 m+1)\{\sqrt{2 m+1}\}]=m
$$
is $\qquad$ .
|
8.0.
Let $2 m+1=n$. Then $2[n\{\sqrt{n}\}]=n-1$. Note that,
$$
\begin{array}{l}
2 n\{\sqrt{n}\}-21$ when, by $n=2[n\{\sqrt{n}\}]+1$ being odd, we know $n \geqslant 3$.
Thus, $\left|4 n-(2[\sqrt{n}]+1)^{2}\right| \leqslant \frac{4}{\sqrt{n}}+\frac{1}{n}<3$.
Therefore, $4 n-(2[\sqrt{n}]+1)^{2}= \pm 1$.
This leads to
$$
2 n-1=2[\sqrt{n}]([\sqrt{n}]+1)
$$
or $n=[\sqrt{n}]([\sqrt{n}]+1)$,
which is a contradiction.
In summary, the only positive integer $n$ that satisfies the condition is 1.
Thus, $m=0$. Hence, $m^{2}=0$.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example: $10 f(x)$ is a continuous function defined on the interval $[0,2015]$, and $f(0)=f(2015)$. Find the minimum number of real number pairs $(x, y)$ that satisfy the following conditions:
(1) $f(x)=f(y)$;
(2) $x-y \in \mathbf{Z}_{+}$.
$(2015$, Peking University Mathematics Summer Camp)
|
【Analysis】Consider a continuous function $f(x)$ on the interval $[0, n]$ that satisfies $f(0)=f(n)$. Let the minimum number of real number pairs $(x, y)$ that satisfy (1) and (2) be $F(n)$.
Obviously, $F(1)=1, F(2)=2$.
We will prove: $F(n)=n\left(n \in \mathbf{Z}_{+}\right)$.
Assume that for $n \leqslant k-1$, equation (1) holds.
We need to prove that $F(k)=k$.
First, we prove that $F(k) \geqslant k$.
In fact, take any continuous function $f(x)$ on the interval $[0, k]$ that satisfies $f(0)=f(k)$.
Without loss of generality, assume $f(0)=f(k)=0$.
If there exists $t \in\{1,2, \cdots, k-1\}$ such that $f(t)=0$, then consider the two intervals $[0, t]$ and $[t, k]$.
By the induction hypothesis, it is easy to see that
$$
\begin{array}{l}
F(k) \geqslant F(t)+F(k-t)+1 \\
=t+(k-t)+1=k+1 .
\end{array}
$$
Now assume that $f(1), f(2), \cdots, f(k-1)$ are all non-zero. Without loss of generality, assume $f(1)>0$ (if $f(1)<0$, the proof is similar).
Then
$f(k-1)-f(0)=f(k-1)>0$,
$f(k)-f(1)=-f(1)<0$.
Hence, there exists $t_0 \in (1, k-1)$ such that $f(t_0)=0$.
Since $f(k-1)-f(t_0-1)>0$, there exists $j \in (k-1, k)$ such that $f(j)=f(j+k-t_0)$.
By the induction hypothesis, the number of real number pairs $(x, y)$ that satisfy (1) and (2) in the interval $\left[j, j+k-t_0\right]$ is at least $k-t_0$.
In the interval $\left[0, t_0\right]$,
$f(0)=0, f(1)>0, f(2)>0$,
$\cdots, f(t_0-1)>0, f(t_0)<0, f(t_0)-f(t_0-m)<0$.
Then there exist $x, y \in \left[0, t_0\right]$ such that
$f(x)=f(y)$, and $x-y=m$.
Additionally, $(x, y)=(2015,0)$ is also a pair that satisfies (1) and (2).
Therefore, $F(k) \geqslant (k-t_0) + (t_0-1) + 1 = k$.
In summary, $F(k) \geqslant k$.
Next, we provide an example to show that $F(k)=k$.
In fact, any function $f(x)$ that first increases and then decreases satisfies $F(k)=k$.
This is because, for any positive integer $m \leqslant k$, the number of pairs $(x, y)$ that satisfy $f(x)=f(y)$ and $x-y=m$ is exactly one.
Thus, $F(k)=k$, i.e., for any $n \in \mathbf{Z}_{+}$, we have
$F(n)=n$.
Therefore, $F(2015)=2015$.
|
2015
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 In $\triangle ABC$ with a fixed perimeter, it is known that $AB=6$, and when vertex $C$ is at a fixed point $P$, $\cos C$ has a minimum value of $\frac{7}{25}$.
(1) Establish an appropriate coordinate system and find the equation of the locus of vertex $C$;
(2) Draw a line through point $A$ intersecting the curve from (1) at points $M$ and $N$, and find the minimum value of $|\overrightarrow{B M}||\overrightarrow{B N}|$.
|
(1) Establish a Cartesian coordinate system with the line $AB$ as the $x$-axis and the perpendicular bisector of segment $AB$ as the $y$-axis.
Let $|CA| + |CB| = 2a (a > 3)$ be a constant. Then the locus of point $C$ is an ellipse with foci at $A$ and $B$.
Thus, the focal distance $2c = |AB| = 6$.
Notice,
$$
\begin{array}{l}
\cos C = \frac{|CA|^2 + |CB|^2 - |AB|^2}{2|CA||CB|} \\
= \frac{(|CA| + |CB|)^2 - 2|CA||CB| - |AB|^2}{2|CA||CB|} \\
= \frac{2a^2 - 18}{|CA||CB|} - 1.
\end{array}
$$
Also, $|CA||CB| \leq \left(\frac{|CA| + |CB|}{2}\right)^2 = a^2$, so $\cos C \geq 1 - \frac{18}{a^2}$.
From the problem, $1 - \frac{18}{a^2} = \frac{7}{25} \Rightarrow a^2 = 25$.
At this point, $|PA| = |PB|$, and $P(0, \pm 4)$.
Therefore, the equation of the locus of point $C$ is
$$
\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0).
$$
(2) Establish a polar coordinate system with $A$ as the pole and $Ax$ as the polar axis. Then the polar equation of the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1 (y \neq 0)$ is
$$
\rho = \frac{16}{5 - 3 \cos \theta} (\theta \neq 0 \text{ and } \theta \neq \pi).
$$
Let the inclination angle of line $MN$ be $\theta$. Then
$$
|AM| = \frac{16}{5 - 3 \cos \theta}, |BM| = \frac{16}{5 + 3 \cos \theta}.
$$
By the definition of the ellipse,
$$
\begin{array}{l}
|\overrightarrow{BM}||\overrightarrow{BN}| = (10 - |AM|)(10 - |BM|) \\
= 100 - 10\left(\frac{16}{5 - 3 \cos \theta} + \frac{16}{5 + 3 \cos \theta}\right) + \\
\frac{16}{5 - 3 \cos \theta} \cdot \frac{16}{5 + 3 \cos \theta} \\
= 100 - \frac{16 \times 84}{25 - 9 \cos^2 \theta}.
\end{array}
$$
If there are no restrictions, when $\cos^2 \theta = 1$, $|\overrightarrow{BM}||\overrightarrow{BN}|$ achieves its minimum value of 16.
However, $0 < \theta < \pi$, so such $M$ and $N$ do not exist, meaning the set of minimum values of $|\overrightarrow{BM}||\overrightarrow{BN}|$ is empty.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Color each cell of a $12 \times 12$ grid either black or white, such that any $3 \times 4$ or $4 \times 3$ subgrid contains at least one black cell. Find the minimum number of black cells.
（Liang Yingde provided the problem）
|
3. The minimum number of black cells required is $n=12$.
First, prove that $n \geqslant 12$.
Since a $12 \times 12$ grid can be divided into $\frac{12 \times 12}{3 \times 4}=12$ non-overlapping $3 \times 4$ sub-grids (excluding the boundaries), according to the problem statement, there must be at least 12 black cells.
To prove that $n=12$, it is sufficient to construct an example that meets the conditions. See Figure 3.
|
12
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. How many different right-angled triangles with integer side lengths have an area that is 999 times their perimeter (considering congruent triangles as the same)? (Provided by Lin Chang)
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
|
5. Let the three sides of a right-angled triangle be \(a, b, c\) (with \(c\) being the hypotenuse).
From the Pythagorean triple formula, we have
\[
a = k \cdot 2uv, \quad b = k(u^2 - v^2), \quad c = k(u^2 + v^2),
\]
where the greatest common divisor of the three sides \(k\) is a positive integer, \(u\) and \(v\) are coprime, \(u > v\), and \(u\) and \(v\) are of opposite parity.
\[
\begin{array}{l}
\text{Then } \frac{1}{2} ab = 999(a + b + c) \\
\Leftrightarrow k^2 uv(u^2 - v^2) = 999 \times 2u(u + v) \\
\Leftrightarrow k v(u - v) = 1998 = 2 \times 3^3 \times 37.
\end{array}
\]
Notice that \(u - v\) is odd, so the factor 2 can only be assigned to \(k\) or \(v\), giving two ways; \(v\) and \(u - v\) are coprime, and the odd prime factor \(p^\alpha\) can be assigned to \(k\), \(v\), or \(u - v\) in the following ways: \((\alpha, 0, 0)\) or \((i, \alpha - i, 0)\), \((i, 0, \alpha - i) (1 \leqslant i \leqslant \alpha)\), giving \(2\alpha + 1\) ways.
By the multiplication principle, the number of ways to assign the prime factors is
\[
2(2 \times 3 + 1)(2 \times 1 + 1) = 42.
\]
Thus, there are 42 such triangles.
[Note] Generally, if the multiple is \(m = 2^\alpha p_1^{\beta_1} p_2^{\beta_2} \cdots p_n^{\beta_n}\) \((p_1, p_2, \cdots, p_n\) are distinct odd primes), then the number of such triangles is
\[
(\alpha + 2)(2\beta_1 + 1)(2\beta_2 + 1) \cdots (2\beta_n + 1).
\]
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let the 20 vertices of a regular 20-sided polygon inscribed in the unit circle in the complex plane correspond to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$. Then the number of distinct points corresponding to $z_{1}^{2015}, z_{2}^{2015}, \cdots, z_{20}^{2015}$ is $\qquad$
|
2. 4 .
Assume the vertices corresponding to the complex numbers $z_{1}, z_{2}, \cdots, z_{20}$ are arranged in a counterclockwise direction.
Let $z_{1}=\mathrm{e}^{\mathrm{i} \theta}$. Then
$$
\begin{array}{l}
z_{k+1}=\mathrm{e}^{\mathrm{i}\left(\theta+\frac{k}{10}\right)}(k=0,1, \cdots, 19), \\
z_{k+1}^{2015}=\mathrm{e}^{\mathrm{i}\left(2015 \theta+\frac{20115 k}{10}\right)}=\mathrm{e}^{\mathrm{i}\left(2015 \theta-\frac{k}{2}\right)} .
\end{array}
$$
Thus, $\left\{z_{k+1}^{2015}\right\}$ is a periodic sequence with a period of 4. Therefore, the number of different points corresponding to $z_{1}^{2015}, z_{2}^{2015}, \cdots, z_{20}^{2015}$ is 4.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. The function $f(x)$ defined on $\mathbf{R}$, for any real number $x$, satisfies
$$
\begin{array}{l}
f(x+3) \leqslant f(x)+3, \\
f(x+2) \geqslant f(x)+2,
\end{array}
$$
and $f(1)=2$. Let $a_{n}=f(n)\left(n \in \mathbf{Z}_{+}\right)$, then
$$
f(2015)=
$$
$\qquad$
|
5.2016.
Notice,
$$
\begin{array}{l}
f(x)+3 \geqslant f(x+3) \\
=f(x+1+2) \geqslant f(x+1)+2 \\
\Rightarrow f(x)+1 \geqslant f(x+1) . \\
\text { Also } f(x)+4 \leqslant f(x+2)+2 \leqslant f(x+4) \\
=f(x+1+3) \leqslant f(x+1)+3 \\
\Rightarrow f(x+1) \geqslant f(x)+1 .
\end{array}
$$
Therefore, $f(x+1)=f(x)+1$.
Thus, $f(2015)=2016$.
|
2016
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \text {. }
$$
Then the sum of the first 12 terms $S_{12}=$ $\qquad$
|
7. 104832 .
Notice that,
$$
\begin{array}{l}
a_{n}=n^{4}+6 n^{3}+11 n^{2}+6 n \\
=n(n+1)(n+2)(n+3) . \\
\text { Let } f(n)=\frac{1}{5} n(n+1)(n+2)(n+3)(n+4) .
\end{array}
$$
Then $a_{n}=f(n)-f(n-1), S_{n}=f(n)$.
Therefore, $S_{12}=f(12)=104832$.
|
104832
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.