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stringlengths 1
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|---|---|---|---|---|---|---|---|---|
9. The function
$$
f(x)=\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x}
$$
has a maximum value of $\qquad$
|
9.11.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(\sqrt{2 x-7}+\sqrt{12-x}+\sqrt{44-x})^{2} \\
\leqslant(3+2+6)\left(\frac{2 x-7}{3}+\frac{12-x}{2}+\frac{44-x}{6}\right) \\
=11^{2},
\end{array}
$$
The equality holds if and only if $\frac{9}{2 x-7}=\frac{4}{12-x}=\frac{36}{44-x}$, which is when $x=8$.
Therefore, the maximum value of $f(x)$ is 11.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One. (20 points) Given the function
$$
f(x)=2 \cos x(\cos x+\sqrt{3} \sin x)-1(x \in \mathbf{R}) \text {. }
$$
(1) Find the intervals where the function $f(x)$ is monotonically increasing;
(2) Let points $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right), \cdots$, $P_{n}\left(x_{n}, y_{n}\right), \cdots$ all lie on the graph of the function $y=f(x)$, and satisfy
$$
x_{1}=\frac{\pi}{6}, x_{n+1}-x_{n}=\frac{\pi}{2}\left(n \in \mathbf{Z}_{+}\right) .
$$
Find the value of $y_{1}+y_{2}+\cdots+y_{2018}$.
|
(1) From the problem, we have
$$
f(x)=\cos 2 x+\sqrt{3} \sin 2 x=2 \sin \left(2 x+\frac{\pi}{6}\right) \text {. }
$$
Therefore, the monotonic increasing interval of $f(x)$ is
$$
\left[-\frac{\pi}{3}+k \pi, \frac{\pi}{6}+k \pi\right](k \in \mathbf{Z}) \text {. }
$$
(2) Let $t_{n}=2 x_{n}+\frac{\pi}{6}, t_{1}=2 x_{1}+\frac{\pi}{6}=\frac{\pi}{2}$.
Then $t_{n+1}-t_{n}=\pi \Rightarrow t_{n}=\left(n-\frac{1}{2}\right) \pi$.
Thus, $y_{n}=2 \sin \frac{2 n-1}{2} \pi$
$$
=\left\{\begin{array}{ll}
2, & n=2 k-1 ; \\
-2, & n=2 k
\end{array}(k \in \mathbf{Z}) .\right.
$$
Therefore, $y_{1}+y_{2}+\cdots+y_{2018}=0$.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Choose any two numbers from $2, 4, 6, 7, 8, 11, 12, 13$ to form a fraction. Then, there are $\qquad$ irreducible fractions among these fractions.
|
4. 36 .
Among 7, 11, 13, choose one number and among $2, 4, 6, 8, 12$, choose one number to form a reduced fraction, there are $2 \mathrm{C}_{3}^{1} \mathrm{C}_{5}^{1}$ $=30$ kinds; among $7, 11, 13$, choose two numbers to form a reduced fraction, there are $\mathrm{A}_{3}^{2}=6$ kinds.
There are a total of 36 different reduced fractions.
|
36
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given $f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x$. Then the set
$$
M=\left\{n \mid f\left(n^{2}-8 n-2018\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$
the number of subsets of $M$ is $\qquad$.
|
2. 1.
For any $0< x_2 < x_1$, we have
$$
\begin{array}{l}
f\left(x_{1}\right)-f\left(x_{2}\right)=\lg \frac{x_{1}}{x_{2}}-\frac{\lg \frac{x_{1}}{x_{2}}}{\lg 9}>0 .
\end{array}
$$
Thus, $f(x)$ is a decreasing function on the interval $(0,+\infty)$. Note that, $f(9)=0$.
Therefore, when $x>9$, $f(x)f(9)=0$.
Hence, $f(x)=0$ has and only has one root $x=9$.
From $f\left(n^{2}-8 n-2018\right) \geqslant 0$
$$
\Rightarrow 0<n^{2}-8 n-2018 \leqslant 9
$$
$\Rightarrow n$ has no integer solutions.
Thus, $M=\varnothing$, and the number of subsets of $M$ is 1.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. As shown in Figure 1, let $P\left(x_{p}, y_{p}\right)$ be a point on the graph of the inverse proportion function $y=\frac{2}{x}$ in the first quadrant of the Cartesian coordinate system $x O y$. Draw lines parallel to the $x$-axis and $y$-axis through point $P$, intersecting the graph of $y=\frac{10}{x}$ in the first quadrant at points $A$ and $B$, respectively. Then the area of $\triangle A O B$ is ( ).
(A) 26
(B) 24
(C) 22
(D) 20
|
3. B.
Connect $O P$.
Then $S_{\triangle A O B}=S_{\triangle A O P}+S_{\triangle P O B}+S_{\triangle A P B}$
$$
\begin{aligned}
= & \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right) y_{p}+\frac{1}{2}\left(\frac{10}{x_{p}}-y_{p}\right) x_{p}+ \\
& \frac{1}{2}\left(\frac{10}{y_{p}}-x_{p}\right)\left(\frac{10}{x_{p}}-y_{p}\right) \\
= & \frac{1}{2} \cdot \frac{100}{x_{p} y_{p}}-\frac{1}{2} x_{p} y_{p} \\
= & 24 .
\end{aligned}
$$
|
24
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
5. Arrange natural numbers whose digits sum to 11 in ascending order to form a sequence. The $m$-th number is 2018. Then $m$ is ( ).
(A) 134
(B) 143
(C) 341
(D) 413
|
5. A.
Among single-digit numbers, there are no numbers whose digit sum is 11.
Among two-digit numbers, there are 8 numbers: $29, 38, 47, 56, 65, 74, 83, 92$.
For three-digit numbers $\overline{x y z}$, when $x=1$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of 9 numbers; when $x=2$, $y$ can take 10 numbers: $0, 1, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 0$, a total of 10 numbers; similarly, when $x=3, 4, \cdots, 9$, the numbers whose digit sum is 11 are $12-x$ each. Therefore, in three-digit numbers $\overline{x y z}$, the numbers whose digit sum is 11 total $9+10+9+\cdots+3=61$.
For four-digit numbers $\overline{1 x y z}$, the numbers whose digit sum is 11 are equivalent to finding three-digit numbers $\overline{x y z}$ whose digit sum is 10. When $x=0$, $y$ can take 9 numbers: $1, 2, \cdots, 9$, and the corresponding $z$ takes $9, 8, \cdots, 1$, a total of 9 numbers. Similarly, when $x=1, 2, \cdots, 9$, the three-digit numbers $\overline{x y z}$ whose digit sum is 10 have $11-x$ each. Therefore, in four-digit numbers $\overline{1 x y z}$, the numbers whose digit sum is 11 total $9+10+9+\cdots+2=63$.
In $\overline{2 x y z}$, the numbers whose digit sum is 11, from smallest to largest, are $2009, 2018$, a total of 2 numbers.
Therefore, 2018 is the
$$
8+61+63+2=134
$$
number in the sequence.
|
134
|
Number Theory
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given that $x_{1}, x_{2}, \cdots, x_{n}$ where $x_{i}(i=1,2, \cdots, n)$ can only take one of the values $-2, 0, 1$, and satisfy
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=-17, \\
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=37 .
\end{array}
$$
Then $\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)^{2}$ is $\qquad$
|
4. 5041.
Let $x_{1}, x_{2}, \cdots, x_{n}$ have $p$ values of $x_{i}$ equal to 1, $q$ values of $x_{i}$ equal to -2, and the rest of the $x_{i}$ equal to 0. We can obtain
$$
\left\{\begin{array} { l }
{ p - 2 q = - 1 7 } \\
{ p + 4 q = 3 7 }
\end{array} \Rightarrow \left\{\begin{array}{l}
p=1, \\
q=9 .
\end{array}\right.\right.
$$
Thus, among $x_{i}(i=1,2, \cdots, n)$, one takes the value 1, nine take the value -2, and the rest take the value 0.
$$
\begin{array}{l}
\text { Therefore, }\left(x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}\right)^{2} \\
=\left(1 \times 1^{3}+9 \times(-2)^{3}\right)^{2} \\
=5041 .
\end{array}
$$
|
5041
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Among the $n$ positive integers from 1 to $n$, those with the most positive divisors are called the "prosperous numbers" among these $n$ positive integers. For example, among the positive integers from 1 to 20, the numbers with the most positive divisors are $12, 18, 20$, so $12, 18, 20$ are all prosperous numbers among the positive integers from 1 to 20. Then, the least common multiple of all the prosperous numbers among the positive integers from 1 to 100 is $\qquad$
|
5. 10080.
First, in the prime factorization of the first 100 positive integers, the maximum number of different prime factors is three. This is because the product of the smallest four primes is $2 \times 3 \times 5 \times 7=210$, which exceeds 100.
Second, to maximize the number of divisors, the prime factors should be as small as possible. The number with the maximum number of positive divisors can be determined through trial and classification enumeration, and the number with the maximum number of positive divisors is 12:
For numbers with only one prime factor, the one with the most positive divisors is $2^{6}=64$, which has seven positive divisors, less than 12;
For numbers with two different prime factors, those with 12 positive divisors are $2^{3} \times 3^{2}$ and $2^{5} \times 3$;
For numbers with three different prime factors, those with 12 positive divisors are
$$
2^{2} \times 3 \times 5, 2 \times 3^{2} \times 5, 2^{2} \times 3 \times 7 \text{. }
$$
It can be seen that among the first 100 positive integers, the numbers with 12 positive divisors are only five, and their least common multiple is
$$
2^{5} \times 3^{2} \times 5 \times 7=10080 \text{. }
$$
|
10080
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Divide the set of positive even numbers $\{2,4, \cdots\}$ into groups in ascending order, with the $n$-th group containing $3 n-2$ numbers:
$$
\{2\},\{4,6,8,10\},\{12,14, \cdots, 24\}, \cdots \text {. }
$$
Then 2018 is in the group.
|
- 1. 27 .
Let 2018 be in the $n$-th group. Since 2018 is the 1009th positive even number and according to the problem, we have
$$
\begin{array}{l}
\sum_{i=1}^{n-1}(3 i-2)<1009 \leqslant \sum_{i=1}^{n}(3 i-2) \\
\Rightarrow \frac{3(n-1)^{2}-(n-1)}{2}<1009 \leqslant \frac{3 n^{2}-n}{2} \\
\Rightarrow n=27 .
\end{array}
$$
Therefore, 2018 is located in the 27th group.
|
27
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}$ satisfies $2 S_{n}-n a_{n}=n\left(n \in \mathbf{Z}_{+}\right)$, and $a_{2}=3$.
(1) Find the general term formula of the sequence $\left\{a_{n}\right\}$;
(2) Let $b_{n}=\frac{1}{a_{n} \sqrt{a_{n+1}}+a_{n+1} \sqrt{a_{n}}}$, and $T_{n}$ be the sum of the first $n$ terms of the sequence $\left\{b_{n}\right\}$. Find the smallest positive integer $n$ such that $T_{n}>\frac{9}{20}$.
|
(1) From $2 S_{n}-n a_{n}=n$, we get
$$
2 S_{n+1}-(n+1) a_{n+1}=n+1 \text {. }
$$
Subtracting the above two equations yields
$$
2 a_{n+1}-(n+1) a_{n+1}+n a_{n}=1 \text {. }
$$
Thus, $n a_{n}-(n-1) a_{n+1}=1$,
$$
(n+1) a_{n+1}-n a_{n+2}=1 \text {. }
$$
Subtracting (2) from (1) and rearranging gives
$$
a_{n}+a_{n+2}=2 a_{n+1} \text {. }
$$
Hence, $\left\{a_{n}\right\}$ is an arithmetic sequence.
From $2 S_{1}-a_{1}=1$ and $a_{2}=3$, we get $a_{1}=1$.
Thus, the common difference of the sequence $\left\{a_{n}\right\}$ is $d=2$.
Therefore, $a_{n}=1+2(n-1)=2 n-1$.
(2) From (1), we know
$$
\begin{array}{l}
b_{n}=\frac{1}{(2 n-1) \sqrt{2 n+1}+(2 n+1) \sqrt{2 n-1}} \\
=\frac{1}{2} \times \frac{\sqrt{2 n+1}-\sqrt{2 n-1}}{\sqrt{2 n-1} \sqrt{2 n+1}} \\
=\frac{1}{2}\left(\frac{1}{\sqrt{2 n-1}}-\frac{1}{\sqrt{2 n+1}}\right) .
\end{array}
$$
Thus, $T_{n}=\frac{1}{2} \sum_{k=1}^{n}\left(\frac{1}{\sqrt{2 k-1}}-\frac{1}{\sqrt{2 k+1}}\right)$
$$
\begin{array}{l}
=\frac{1}{2}\left(1-\frac{1}{\sqrt{2 n+1}}\right)>\frac{9}{20} \\
\Rightarrow n \geqslant \frac{99}{2} .
\end{array}
$$
Therefore, the smallest positive integer $n$ that satisfies the condition is $n=50$.
|
50
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. Let $M$ be a set composed of a finite number of positive integers
$$
\begin{array}{l}
\text { such that, } M=\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}, \\
A_{i} \neq \varnothing, B_{i} \neq \varnothing(i=1,2, \cdots, 20),
\end{array}
$$
and satisfies:
(1) For any $1 \leqslant i<j \leqslant 20$,
$$
A_{i} \cap A_{j}=\varnothing, B_{i} \cap B_{j}=\varnothing \text {; }
$$
(2) For any $1 \leqslant i \leqslant 20,1 \leqslant j \leqslant 20$, if $A_{i} \cap B_{j}=\varnothing$, then $\left|A_{i} \cup B_{j}\right| \geqslant 18$.
Find the minimum number of elements in the set $M$ ( $|X|$ denotes the number of elements in the set $X$).
|
15. Let $\min _{1 \leqslant i \leqslant 20}\left\{\left|A_{i}\right|,\left|B_{i}\right|\right\}=t$.
Assume $\left|A_{1}\right|=t$,
$$
\begin{array}{l}
A_{1} \cap B_{i} \neq \varnothing(i=1,2, \cdots, k) ; \\
A_{1} \cap B_{j}=\varnothing(j=k+1, k+2, \cdots, 20) .
\end{array}
$$
Let $a_{i} \in A_{1} \cap B_{i}(i=1,2, \cdots, k)$.
By property (1), we know that $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, $\left|A_{1}\right| \geqslant k$, i.e., $t \geqslant k$.
By property (2), we know that when $j=k+1, k+2, \cdots, 20$, $\left|A_{1}\right|+\left|B_{j}\right|=\left|A_{1} \cup B_{j}\right| \geqslant 18$, i.e., $\left|B_{j}\right| \geqslant 18-t$.
$$
\begin{array}{l}
\text { Then }|M|=\left|\bigcup_{i=1}^{20} B_{i}\right|=\sum_{j=1}^{20}\left|B_{j}\right| \\
\geqslant k t+(20-k)(18-t) \\
=360+2 k t-18 k-20 t \\
=180+2(k-10)(t-9) .
\end{array}
$$
If $t \leqslant 9$, then
$k \leqslant t \leqslant 9$,
$$
|M|=180+2(k-10)(t-9) \geqslant 180 \text {; }
$$
If $t \geqslant 10$, then $|M| \geqslant 20 t \geqslant 200$.
Therefore, $|M| \geqslant 180$ always holds.
Take $A_{i}=B_{i}=\{9(i-1)+j \mid j=1,2, \cdots, 9\}$, where $i=1,2, \cdots, 20$.
Then $M=\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}=\{1,2, \cdots, 180\}$ satisfies the requirement.
In this case, $|M|=180$.
In summary, the minimum number of elements in the set $M$ is
180.
|
180
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. In $\triangle A B C$, $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively, satisfying $a^{2}+b^{2}=4-\cos ^{2} C, a b=2$. Then $S_{\triangle A B C}=$ $\qquad$
|
$-1.1$
From the problem, we have $(a-b)^{2}+\cos ^{2} C=0$.
Solving, we get $a=b=\sqrt{2}, \cos C=0$.
Therefore, $S_{\triangle A B C}=1$.
|
1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given $x, y>0$. If
$$
f(x, y)=\left(x^{2}+y^{2}+2\right)\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \text {, }
$$
then the minimum value of $f(x, y)$ is
|
3.4.
By completing the square, we get
$$
x^{2}+y^{2}+2 \geqslant(x+y)+(x y+1) \text {. }
$$
Then $f(x, y)$
$$
\begin{array}{l}
\geqslant((x+y)+(x y+1))\left(\frac{1}{x+y}+\frac{1}{x y+1}\right) \\
\geqslant(1+1)^{2}=4 .
\end{array}
$$
Equality holds if and only if $x=y=1$.
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 1 In an $8 \times 8$ chessboard, how many ways are there to select 56 squares such that: all the black squares are selected, and each row and each column has exactly seven squares selected? ? ${ }^{[1]}$
(2014, Irish Mathematical Olympiad)
|
The problem is equivalent to selecting eight white squares on the chessboard, with exactly one square selected from each row and each column.
The white squares on the chessboard are formed by the intersections of rows $1, 3, 5, 7$ and columns $1, 3, 5, 7$, resulting in a $4 \times 4$ submatrix, as well as the intersections of rows $2, 4, 6, 8$ and columns $2, 4, 6, 8$, resulting in another $4 \times 4$ submatrix. Since there are $4!$ ways to choose four white squares from each submatrix such that they are in different rows and columns, the total number of ways to make the selection is $(4!)^{2}=576$.
|
576
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
For the four-digit number $\overline{a b c d}(1 \leqslant a \leqslant 9,0 \leqslant b 、 c$ 、 $d \leqslant 9)$ : if $a>b, bd$, then $\overline{a b c d}$ is called a $P$ class number; if $ac, c<d$, 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 $\qquad$ [3] (2015, National High School Mathematics Joint Competition)
|
Let the set of all numbers of type $P$ and type $Q$ be denoted as $A$ and $B$, respectively. Further, let the set of all numbers of type $P$ that end in zero be denoted as $A_{0}$, and the set of all numbers of type $P$ that do not end in 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 $bd \geqslant 1$.
Thus, $\overline{d c b a} \in B$.
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, $N(P) - N(Q) = |A| - |B|$
$$
= \left|A_{0}\right| + \left|A_{1}\right| - |B| = \left|A_{0}\right| \text{.}
$$
For any four-digit number $\overline{a b c 0} \in A_{0}$, $b$ can take values 0, 1, ..., 9. For each $b$, since $b < a \leqslant 9$ and $b < c \leqslant 9$, $a$ and $c$ each have $9 - b$ possible values. Therefore,
$$
\left|A_{0}\right| = \sum_{b=0}^{9} (9 - b)^{2} = \sum_{k=1}^{9} k^{2} = \frac{9 \times 10 \times 19}{6} = 285 \text{.}
$$
Thus, $N(P) - N(Q) = 285$.
|
285
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 4 Let $f(x)=\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{2013!}\right]$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$). For an integer $n$, if the equation $f(x)=n$ has a real solution, then $n$ is called a "good number". Find the number of good numbers in the set $\{1,3, \cdots, 2013\}$. ${ }^{[4]}$
(2013, China Southeast Mathematical Olympiad)
|
First, point out two obvious conclusions:
(1) If $m \in \mathbf{Z}_{+}, x \in \mathbf{R}$, then $\left[\frac{x}{m}\right]=\left[\frac{[x]}{m}\right]$;
(2) For any integer $l$ and positive even number $m$, we have
$$
\left[\frac{2 l+1}{m}\right]=\left[\frac{2 l}{m}\right] \text {. }
$$
In (1), let $m=k!(k=1,2, \cdots, 2013)$ and
sum to get
$$
f(x)=\sum_{k=1}^{2013}\left[\frac{x}{k!}\right]=\sum_{k=1}^{2013}\left[\frac{[x]}{k!}\right]=f([x]) .
$$
This shows that the equation $f(x)=n$ has a real solution if and only if it has an integer solution. Therefore, we only need to consider the case where $x$ is an integer.
$$
\begin{array}{l}
\text { By } f(x+1)-f(x) \\
=[x+1]-[x]+\sum_{k=2}^{2013}\left(\left[\frac{x+1}{k!}\right]-\left[\frac{x}{k!}\right]\right) \\
\quad \geqslant 1,
\end{array}
$$
we know that $f(x)(x \in \mathbf{Z})$ is strictly increasing.
Next, find integers $a$ and $b$ such that
$$
\begin{array}{l}
f(a-1)2013,
\end{array}
$$
so $b=1173$.
Thus, the good numbers in $\{1,3, \cdots, 2013\}$ are the odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$.
In (1), let $x=2 l(l=0,1, \cdots, 586)$.
By conclusion (2), we have
$$
\left[\frac{2 l+1}{k!}\right]=\left[\frac{2 l}{k!}\right](2 \leqslant k \leqslant 2013) \text {. }
$$
Then $f(2 l+1)-f(2 l)$
$$
=1+\sum_{k=2}^{2013}\left(\left[\frac{2 l+1}{k!}\right]-\left[\frac{2 l}{k!}\right]\right)=1 \text {. }
$$
This shows that exactly one of $f(2 l)$ and $f(2 l+1)$ is odd. Therefore, there are $\frac{1174}{2}=587$ odd numbers in $\{f(0), f(1), \cdots, f(1173)\}$, i.e., there are 587 good numbers in the set $\{1,3, \cdots, 2013\}$.
|
587
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Equation
$$
x^{2}-31 x+220=2^{x}\left(31-2 x-2^{x}\right)
$$
The sum of the squares of all real roots is $\qquad$ .
|
2. 25 .
Let $y=x+2^{x}$. Then the original equation is equivalent to
$$
\begin{array}{l}
y^{2}-31 y+220=0 \\
\Rightarrow y_{1}=11, y_{2}=20 \\
\Rightarrow x_{1}+2^{x_{1}}=11 \text { and } x_{2}+2^{x_{2}}=20 .
\end{array}
$$
Since $f(x)=x+2^{x}$ is a monotonically increasing function, each equation has at most one real root, which are $3$ and $4$, respectively. Therefore, the sum of the squares of all real roots of the original equation is 25.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In the plane, there are 200 points, no three of which are collinear, and each point is labeled with one of the numbers $1, 2, 3$. All pairs of points labeled with different numbers are connected by line segments, and each line segment is labeled with a number 1, 2, or 3, which is different from the numbers at its endpoints. As a result, each of the numbers 1, 2, or 3 written on the plane appears exactly $n$ times. Then the value of $n$ is $\qquad$
|
5.199.
Let the points labeled with $1, 2, 3$ be $a, b, c$ respectively.
Thus, $a+b+c=200$, and the number of line segments labeled with $1, 2, 3$ are $bc, ca, ab$ respectively.
Then $n=a+bc=b+ca=c+ab$.
Therefore, $(a+bc)-(b+ca)=(a-b)(1-c)=0$.
Similarly, $(b-c)(1-a)=(c-a)(1-b)=0$.
If at least two of $a, b, c$ are not 1, then $a=b=c$, which contradicts $3+200$, and it is also impossible that $a=b=c=1$.
Thus, two of $a, b, c$ are 1, and the third one is 198. In this case, the equation holds, and $n=199$.
|
199
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. The sequence $\left\{a_{n}\right\}$ has nine terms, $a_{1}=a_{9}=1$, and for each $i \in\{1,2, \cdots, 8\}$, we have $\frac{a_{i+1}}{a_{i}} \in\left\{2,1,-\frac{1}{2}\right\}$. The number of such sequences is $\qquad$
(2013, National High School Mathematics League Competition)
|
Let $b_{i}=\frac{a_{i+1}}{a_{i}}(1 \leqslant i \leqslant 8)$, mapping each sequence $\left\{a_{n}\right\}$ that meets the conditions to a unique eight-term sequence $\left\{b_{n}\right\}$, where $\prod_{i=1}^{8} b_{i}=\frac{a_{9}}{a_{1}}=1$, and $b_{i} \in\left\{2,1,-\frac{1}{2}\right\}(1 \leqslant i \leqslant 8)$.
From the sign, we know that $b_{i}(1 \leqslant i \leqslant 8)$ contains an even number of $-\frac{1}{2}$. Suppose there are $2 k$ terms of $-\frac{1}{2}$, then there are $2 k$ terms of 2, and the remaining $8-4 k$ terms are 1, where $k$ can be $0, 1, 2$.
Therefore, the number of ways to choose $\left\{b_{n}\right\}$ is
$$
\sum_{k=0}^{2} \mathrm{C}_{8}^{2 k} \mathrm{C}_{8-2 k}^{2 k}=491,
$$
The number of ways to choose $\left\{a_{n}\right\}$ is also 491.
|
491
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let $n$ be a three-digit positive integer without the digit 0. If the digits of $n$ in the units, tens, and hundreds places are permuted arbitrarily, the resulting three-digit number is never a multiple of 4. Find the number of such $n$.
(54th Ukrainian Mathematical Olympiad)
|
Hint: Classify by the number of even digits (i.e., $2,4,6,8$) appearing in the three-digit code of $n$. It is known from the discussion that the number of $n$ satisfying the condition is $125+150+0+8=283$.
|
283
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $x \in\left(0, \frac{\pi}{2}\right)$. Then the minimum value of the function $y=\frac{1}{\sin ^{2} x}+\frac{12 \sqrt{3}}{\cos x}$ is $\qquad$ .
|
7. 28 .
Notice that,
$$
\begin{array}{l}
y=16\left(\frac{1}{16 \sin ^{2} x}+\sin ^{2} x\right)+16\left(\frac{3 \sqrt{3}}{4 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16\left(\frac{3 \sqrt{3}}{8 \cos x}+\frac{3 \sqrt{3}}{8 \cos x}+\cos ^{2} x\right)-16 \\
\geqslant 16 \times \frac{1}{2}+16 \times \frac{9}{4}-16=28 .
\end{array}
$$
When $x=\frac{\pi}{6}$, $y$ achieves its minimum value of 28.
|
28
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 2 Find the smallest positive integer $k$ such that for any $k$-element subset $A$ of the set $S=\{1,2, \cdots, 2012\}$, there exist three distinct elements $a, b, c$ in $S$ such that $a+b, b+c, c+a$ are all in the subset $A$.
(2012, China Mathematical Olympiad)
|
【Analysis】The condition of the problem is equivalent to being able to find three numbers $x, y, z$ in the subset $A$ such that
$$
\frac{-x+y+z}{2}, \frac{x-y+z}{2}, \frac{x+y-z}{2}
$$
are three distinct positive integers. For this, $x, y, z$ must be distinct, satisfy the triangle inequality, and their sum must be even.
A simple estimation shows that taking all the odd numbers can yield a 1006-element subset, where the sum of any three elements is odd, which is a larger counterexample to the required subset $A$. Further, it is found that adding the element 2 can result in a 1007-element counterexample. Therefore, we conjecture that $k=1008$.
Next, we prove this conjecture.
Here, we use a simple principle in combinatorial extremal problems: "The smallest $k$ for which all subsets $A$ of at least $k$ elements satisfy condition $P$" minus 1 is exactly "the largest $n$ for which there exists an $n$-element subset that does not satisfy condition $P$." Let condition $P$ be: the sum of three different numbers is even and satisfies the triangle inequality.
First, we use the pigeonhole principle to prove: any 1008-element subset $A$ of $S$ always contains three elements that satisfy condition $P$.
Since $|A|>\frac{|S|}{2}$ ensures that some pigeonhole $T$ contains more than $\frac{|T|}{2}$ elements of $A$, thus, the pigeonhole $T$ should satisfy the condition: any more than $\frac{|T|}{2}$ elements in $T$ contain three elements that satisfy condition $P$. Then $\frac{|T|}{2} \geqslant 3$, i.e., $|T| \geqslant 6$.
Next, we prove: from the set $\{3,4, \cdots, 8\}$, any four numbers contain three numbers that satisfy condition $P$.
The simplest and most direct proof is by enumeration.
List all three-element subsets of $\{3,4, \cdots, 8\}$ that satisfy condition $P$, to prove that all four-element subsets contain some three-element subset that satisfies condition $P$, it is only necessary to prove that the complements of all three-element subsets that satisfy condition $P$ contain all possible two-element subsets. Thus, find the complements of all three-element subsets that satisfy condition $P$, and list all two-element subsets of the complements, and then, check that all two-element subsets of the six-element set. Using vertices to represent the six elements and edges to represent the two-element subsets, it is easy to verify this by drawing a graph.
By translation, it can be proven that for any positive integer $n$,
$$
\{2 n+3,2 n+4, \cdots, 2 n+8\}
$$
is also a pigeonhole that satisfies the condition: the translation step is even, ensuring that the parity of the numbers at corresponding positions is the same as in the original pigeonhole, and the condition that the sum is even remains unchanged; all numbers increase, and the corresponding triangle inequality condition weakens.
Divide $S=\{1,2, \cdots, 2012\}$ into $\{1,2\}$, $\{6 n+3,6 n+4, \cdots, 6 n+8\}(n=0,1, \cdots, 334)$.
When $A \subset S$ and $A$ contains at least 1008 elements, $A$ contains at least four numbers in some pigeonhole $\{6 n+3,6 n+4, \cdots, 6 n+8\}$, thus, there are three numbers that satisfy condition $P$, and the required $a, b, c$ can be solved.
Therefore, the smallest $k=1008$.
|
1008
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
$$
\begin{array}{l}
A=\{1,2, \cdots, 99\}, \\
B=\{2 x \mid x \in A\}, \\
C=\{x \mid 2 x \in A\} .
\end{array}
$$
Then the number of elements in $B \cap C$ is $\qquad$
|
1. 24 .
From the conditions, we have
$$
\begin{array}{l}
B \cap C \\
=\{2,4, \cdots, 198\} \cap\left\{\frac{1}{2}, 1, \frac{3}{2}, 2, \cdots, \frac{99}{2}\right\} \\
=\{2,4, \cdots, 48\} .
\end{array}
$$
Therefore, the number of elements in $B \cap C$ is 24 .
|
24
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let the integer sequence $a_{1}, a_{2}, \cdots, a_{10}$ satisfy:
$$
a_{10}=3 a_{1}, a_{2}+a_{8}=2 a_{5} \text {, }
$$
and $a_{i+1} \in\left\{1+a_{i}, 2+a_{i}\right\}(i=1,2, \cdots, 9)$. Then the number of such sequences is $\qquad$
|
8. 80 .
$$
\begin{array}{l}
\text { Let } b_{i}=a_{i+1}-a_{i} \in\{1,2\}(i=1,2, \cdots, 9) \text {. } \\
\text { Then } 2 a_{1}=a_{10}-a_{1}=b_{1}+b_{2}+\cdots+b_{9}, \\
b_{2}+b_{3}+b_{4}=a_{5}-a_{2}=a_{8}-a_{5} \\
=b_{5}+b_{6}+b_{7} .
\end{array}
$$
Let $t$ represent the number of terms with value 2 among $b_{2}, b_{3}, b_{4}$. From equation (2), we know that $t$ is also the number of terms with value 2 among $b_{5}, b_{6}, b_{7}$, where $t \in\{0,1,2,3\}$.
Thus, the number of ways to choose $b_{2}, b_{3}, \cdots, b_{7}$ is
$$
\left(\mathrm{C}_{3}^{0}\right)^{2}+\left(\mathrm{C}_{3}^{1}\right)^{2}+\left(\mathrm{C}_{3}^{2}\right)^{2}+\left(\mathrm{C}_{3}^{3}\right)^{2}=20 \text {. }
$$
After fixing $b_{2}, b_{3}, \cdots, b_{7}$, we can arbitrarily assign values to $b_{8}, b_{9}$, which gives $2^{2}=4$ ways.
From equation (1), we know that $b_{1} \in\{1,2\}$ should be chosen such that $b_{1}+b_{2}+\cdots+b_{9}$ is even, and the choice of $b_{1}$ is unique, which also determines the value of the integer $a_{1}$.
Therefore, the sequence $b_{1}, b_{2}, \cdots, b_{9}$ uniquely corresponds to a sequence $a_{1}, a_{2}, \cdots, a_{10}$ that satisfies the conditions.
In summary, the number of sequences that satisfy the conditions is
$$
20 \times 4=80 \text {. }
$$
|
80
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let non-zero real numbers $a, b$ satisfy $a^{2}+b^{2}=25$. If the function $y=\frac{a x+b}{x^{2}+1}$ has a maximum value $y_{1}$ and a minimum value $y_{2}$, then $y_{1}-y_{2}=$ $\qquad$.
|
3.5.
From
$$
\begin{aligned}
y & =\frac{a x+b}{x^{2}+1} \Rightarrow y x^{2}-a x+y-b=0 \\
& \Rightarrow \Delta=a^{2}-4 y(y-b) \geqslant 0 .
\end{aligned}
$$
Thus, $y_{2}$ and $y_{1}$ are the two roots of $a^{2}-4 y(y-b)=0$, at this point,
$$
\Delta_{1}=16 b^{2}+16 a^{2}=400>0 \text {. }
$$
Therefore, $y_{2}$ and $y_{1}$ always exist.
So $y_{1}-y_{2}=\sqrt{\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}}$
$$
=\sqrt{b^{2}-\left(b^{2}-25\right)}=5 \text {. }
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Let the side length of rhombus $A_{1} A_{2} A_{3} A_{4}$ be $1, \angle A_{1} A_{2} A_{3}=$ $\frac{\pi}{6}, P$ be a point in the plane of rhombus $A_{1} A_{2} A_{3} A_{4}$. Then the minimum value of $\sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}}$ is $\qquad$
|
6. -1 .
Let the center of the rhombus be $O$. Then
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{P A_{i}} \cdot \overrightarrow{P A_{j}} \\
= & \mathrm{C}_{4}^{2}|\overrightarrow{P Q}|^{2}+\overrightarrow{P O} \cdot 3 \sum_{1 \leqslant i \leqslant 4} \overrightarrow{O A_{i}}+\sum_{1 \leqslant i<j \leqslant 4} \overrightarrow{O A_{i}} \cdot \overrightarrow{O A_{j}} \\
= & 6|\overrightarrow{P Q}|^{2}-1 \geqslant-1 .
\end{aligned}
$$
When point $P$ is at the center of the rhombus, the equality holds.
|
-1
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $P(x)=x^{5}-x^{2}+1$ have five roots $r_{1}$, $r_{2}, \cdots, r_{5}$, and $Q(x)=x^{2}+1$. Then
$$
\begin{array}{l}
Q\left(r_{1}\right) Q\left(r_{2}\right) Q\left(r_{3}\right) Q\left(r_{4}\right) Q\left(r_{5}\right) \\
=
\end{array}
$$
|
7.5.
Given that $P(x)=\prod_{j=1}^{5}\left(x-r_{j}\right)$.
$$
\begin{array}{l}
\text { Then } \prod_{j=1}^{5} Q\left(r_{j}\right)=\left(\prod_{j=1}^{5}\left(r_{j}+\mathrm{i}\right)\right)\left(\prod_{j=1}^{5}\left(r_{j}-\mathrm{i}\right)\right) \\
=P(\mathrm{i}) P(-\mathrm{i}) \\
=\left(\mathrm{i}^{5}-\mathrm{i}^{2}+1\right)\left((-\mathrm{i})^{5}-(-\mathrm{i})^{2}+1\right) \\
=(\mathrm{i}+2)(-\mathrm{i}+2)=5 .
\end{array}
$$
|
5
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Let set $S \subset\{1,2, \cdots, 200\}, S$ such that the difference between any two elements is not 4, 5, or 9. Find the maximum value of $|S|$.
untranslated portion:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
Note: The last sentence is a note to the translator and should not be included in the translated text. Here is the final translation:
4. Let set $S \subset\{1,2, \cdots, 200\}, S$ such that the difference between any two elements is not 4, 5, or 9. Find the maximum value of $|S|$.
|
Hint Consider the maximum number of numbers from $S$ that can be contained in any sequence of 13 consecutive numbers. Answer: 64.
|
64
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Given a large regular tetrahedron with an edge length of 6, a smaller regular tetrahedron is placed inside it. If the smaller tetrahedron can rotate freely within the larger one, the maximum edge length of the smaller tetrahedron is . $\qquad$
|
2. 2 .
Given that a smaller regular tetrahedron can rotate freely inside a larger regular tetrahedron, the maximum edge length of the smaller tetrahedron occurs when it is inscribed in the insphere of the larger tetrahedron.
Let the circumradius of the larger tetrahedron be $R$, and the circumradius of the smaller tetrahedron (which is the inradius of the larger tetrahedron) be $r$.
It is easy to see that, $r=\frac{1}{3} R$.
Therefore, the maximum edge length of the smaller tetrahedron is $\frac{1}{3} \times 6=2$.
|
2
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. In the expansion of $(\sqrt{3}+i)^{10}$, the sum of all odd terms is $\qquad$ .
|
3. 512 .
It is known that in the expansion of $(\sqrt{3}+\mathrm{i})^{10}$, the sum of all odd terms is the real part of the complex number.
$$
\begin{array}{l}
\text { Therefore, }(\sqrt{3}+i)^{10}=\left((-2 i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)\right)^{10} \\
=(-2 i)^{10}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{10} \\
=(-1024)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right) \\
=512-512 \sqrt{3} \mathrm{i}
\end{array}
$$
|
512
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. If $(2 \dot{x}+4)^{2 n}=\sum_{i=0}^{2 n} a_{i} x^{i}\left(n \in \mathbf{Z}_{+}\right)$, then the remainder of $\sum_{i=1}^{n} a_{2 i}$ when divided by 3 is $\qquad$
|
6.1.
Let $x=0$, we get $a_{0}=4^{2 n}$.
By setting $x=1$ and $x=-1$ respectively, and adding the two resulting equations, we get
$$
\begin{array}{l}
\sum_{i=0}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right) . \\
\text { Therefore, } \sum_{i=1}^{n} a_{2 i}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)-4^{2 n} \\
\equiv(-1)^{2 n-1}-1^{2 n} \equiv 1(\bmod 3) .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Divide a circle into a group of $n$ equal parts and color each point either red or blue. Starting from any point, record the colors of $k(k \leqslant n)$ consecutive points in a counterclockwise direction, which is called a “$k$-order color sequence” of the circle. Two $k$-order color sequences are considered different if and only if the colors at corresponding positions are different in at least one place. If any two 3-order color sequences are different, then the maximum value of $n$ is . $\qquad$
|
8. 8 .
In a 3rd-order color sequence, since each point has two color choices, there are $2 \times 2 \times 2=8$ kinds of 3rd-order color sequences.
Given that $n$ points can form $n$ 3rd-order color sequences, we know $n \leqslant 8$.
Thus, $n=8$ can be achieved.
For example, determining the colors of eight points in a counterclockwise direction as “red, red, red, blue, blue, blue, red, blue” meets the conditions.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 5 Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ modulo 3 is divisible by $x^{2}+1$.
$(2015$, Harvard-MIT Mathematics Tournament)
|
【Analysis】More explicitly, the polynomial $(x+1)^{n}-1$ needs to satisfy the equivalent condition:
There exist integer-coefficient polynomials $P$ and $Q$ such that
$$
(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x) \text {. }
$$
Assume without loss of generality that
$$
R(x)=(x+1)^{n}-1-P(x)\left(x^{2}+1\right) \text {. }
$$
Then $R(x)=3 Q(x)$, where $P(x)$ is the quotient and $R(x)$ is the remainder with coefficients all being multiples of 3.
Now consider $R(\mathrm{i})$.
Since the coefficients of the odd and even powers of $R(x)$ are all divisible by 3, the real and imaginary parts of $R(\mathrm{i})$ are also divisible by 3.
By calculation, we get
$$
\begin{array}{l}
(1+i)^{2}=2 i,(1+i)^{4}=-4, \\
(1+i)^{6}=-8 i,(1+i)^{8}=16 .
\end{array}
$$
Thus, $(1+\mathrm{i})^{8}-1=15$ has a real part and an imaginary part both divisible by 3.
Since the even powers of $1+\mathrm{i}$ are either purely real or purely imaginary and their coefficients are not divisible by 3, multiplying by $1+\mathrm{i}$ still results in an imaginary part that is not divisible by 3.
Testing $n=8$, we have
$$
\begin{aligned}
& (x+1)^{8}-1 \\
= & x^{8}+8 x^{7}+28 x^{6}+56 x^{5}+70 x^{4}+ \\
& 56 x^{3}+28 x^{2}+8 x \\
\equiv & x^{8}-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x \\
\equiv & \left(x^{2}+1\right)\left(x^{6}-x^{5}+x^{2}-x\right)(\bmod 3) .
\end{aligned}
$$
Therefore, the smallest positive integer $n$ that satisfies the condition is $n=8$.
|
8
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
15. As shown in Figure 2, two equal circles with a radius of 5 are externally tangent to each other, and both are internally tangent to a larger circle with a radius of 13, with the points of tangency being $A$ and $B$. Let $AB = \frac{m}{n}\left(m, n \in \mathbf{Z}_{+},(m, n)\right. = 1)$. Then the value of $m+n$ is ( ).
(A) 21
(B) 29
(C) 58
(D) 69
(E) 93
|
15. D.
As shown in Figure 5, let the centers of the three circles be $X$, $Y$, and $Z$.
Then points $A$ and $B$ are on the extensions of $XY$ and $XZ$, respectively, satisfying
$$
\begin{array}{l}
XY = XZ \\
= 13 - 5 = 8, \\
YZ = 5 + 5 = 10, \\
XA = XB = 13.
\end{array}
$$
Since $YZ \parallel AB \Rightarrow \triangle XYZ \sim \triangle XAB$
$$
\Rightarrow \frac{XA}{AB} = \frac{XY}{YZ} \Rightarrow AB = \frac{XA \cdot YZ}{XY} = \frac{65}{4} \text{.}
$$
Thus, $m = 65$, $n = 4$, and $m + n = 69$.
|
69
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $x, y, z$ be complex numbers, and
$$
\begin{array}{l}
x^{2}+y^{2}+z^{2}=x y+y z+z x, \\
|x+y+z|=21,|x-y|=2 \sqrt{3},|x|=3 \sqrt{3} .
\end{array}
$$
Then $|y|^{2}+|z|^{2}=$ . $\qquad$
|
7. 132.
It is easy to see that the figures corresponding to $x$, $y$, and $z$ on the complex plane form an equilateral triangle, and note that
\[
\begin{array}{l}
|x-y|^{2}+|y-z|^{2}+|z-x|^{2}+|x+y+z|^{2} \\
=3\left(|x|^{2}+|y|^{2}+|z|^{2}\right),
\end{array}
\]
Combining the conditions and $|x-y|=|y-z|=|z-x|$, we can substitute and solve.
|
132
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
In $1,2, \cdots, 100$ these 100 positive integers, remove 50 so that in the remaining positive integers, any two different $a, b$ have $a \nmid b$. Find the maximum possible value of the sum of all removed positive integers.
|
Let the remaining numbers be $a_{i}=2^{3} t_{i}$, where $i \in \{1,2, \cdots, 50\}$, $s_{i}$ is a natural number, and $t_{i}$ is an odd number.
Since for any $1 \leqslant i \neq j \leqslant 50$, we have $a_{i} \nmid a_{j}$, it follows that $t_{i} \neq t_{j}$, meaning $t_{1}, t_{2}, \cdots, t_{50}$ are 50 distinct odd numbers.
The numbers $1,2, \cdots, 100$ can be expressed in the form $2^{s} t$, where $t$ has only 50 possibilities, namely $1,3, \cdots, 99$. Therefore,
$$
\left\{t_{1}, t_{2}, \cdots, t_{50}\right\}=\{1,3, \cdots, 99\}.
$$
Without loss of generality, let $t_{i}=2 i-1$.
For convenience, we can change the indices of the remaining numbers.
Let $a_{i}=2^{s} i$, where $i \in \{1,3, \cdots, 99\}$ and $s_{i}$ is a natural number.
Set $S_{0}=\left\{a_{99}, a_{97}, \cdots, a_{35}\right\}$,
$S_{1}=\left\{a_{33}, a_{31}, \cdots, a_{13}\right\}$,
$S_{2}=\left\{a_{11}, \cdots, a_{5}\right\}$, $S_{3}=\left\{a_{3}\right\}$, $S_{4}=\left\{a_{1}\right\}$.
First, we prove that for any $a_{i} \in S_{k}$, we have
$a_{i} \geqslant 2^{k} i$.
We use mathematical induction on $k$.
When $k=0$, $a_{i}=2^{s} t \geqslant 2^{0} i$ holds.
Assume that for $k=m (0 \leqslant m \leqslant 3)$, the conclusion holds.
Then for $k=m+1$, if $a_{i}=2^{s} i \in S_{m+1}$, we have $a_{3 i} \in S_{m}$.
Since $a_{i} \nmid a_{3 i}$, by the induction hypothesis,
$a_{3 i}=2^{s_{3 i}} 3 i (s_{3 i} \geqslant m)$.
Thus, $s_{i} \geqslant s_{3 i}+1 \geqslant m+1$, which means $a_{i} \geqslant 2^{m+1} i$, and the conclusion holds.
By mathematical induction, equation (1) is proven.
Next, we prove that equation (1) holds with equality when the conditions of the problem are satisfied.
If $a_{i}, a_{j} \in S_{k}$, then $a_{i}=2^{k} i, a_{j}=2^{k} j$.
Assume $a_{i} \mid a_{j}$. Then $i \mid j$.
Since $i, j$ are both odd, it follows that $j \geqslant 3 i$.
By $a_{i}, a_{j} \in S_{k}$, we know $3 i > j$, which is a contradiction. Hence, $a_{i} \nmid a_{j}$.
Similarly, $a_{j} \nmid a_{i}$.
If $a_{i} \in S_{k}, a_{j} \in S_{l}$, without loss of generality, let $l > k$, then $a_{i}=2^{k} i, a_{j}=2^{l} j$.
Since $l > k$, it follows that $a_{i} \nmid a_{j}$.
Assume $a_{j} \mid a_{i}$. Then $j \mid i$.
By $l > k$, we know $j > i$, which is a contradiction. Hence, $a_{j} \nmid a_{i}$.
In summary, the sum of the remaining numbers is minimized when $a_{i}=2^{k} i$.
Thus, the minimum value is
$2211 + 506 + 128 + 24 + 16 = 2885$.
Therefore, the maximum possible value of the sum of all removed positive integers is $5050 - 2885 = 2165$.
|
2165
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) Color each cell of a $5 \times 5$ grid with one of five colors, such that the number of cells of each color is the same. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
Color each cell of a $5 \times 5$ grid with one of five colors, such that the number of cells of each color is the same. If two adjacent cells have different colors, their common edge is called a "separating edge." Find the minimum number of separating edges.
|
As shown in Figure 4, a $5 \times 5$ grid is divided into five parts, each colored with one of 5 colors, and at this point, there are 16 dividing edges.
Below is the proof that the number of dividing edges is at least 16.
First, let the 5 colors be denoted as $1, 2, \cdots, 5$, and let the number of rows and columns occupied by the cells of the $i$-th color be $a_{i}$ and $b_{i}$, respectively, and the length of the boundary be $c_{i}$.
For $i \in \{1, 2, \cdots, 5\}$, the $i$-th color is exactly applied to 5 cells, and it occupies $a_{i}$ rows and $b_{i}$ columns, so
$$
a_{i} b_{i} \geqslant 5.
$$
Since for each row of the $a_{i}$ rows occupied by the $i$-th color, the left and right boundaries of the $i$-th color cells in this row must be the boundaries of all the $i$-th color cells, and for each column of the $b_{i}$ columns occupied by the $i$-th color, the upper and lower boundaries of the $i$-th color cells in this column must also be the boundaries of all the $i$-th color cells, we have
$$
c_{i} \geqslant 2\left(a_{i}+b_{i}\right).
$$
Let the number of dividing edges be $f$. Since each dividing edge must be the boundary between two colors, and the boundary of the $i$-th color cells, excluding the boundary of the $5 \times 5$ grid, must be a dividing edge, we have
$$
f=\frac{1}{2}\left(\sum_{i=1}^{5} c_{i}-5 \times 4\right).
$$
From $a_{i} b_{i} \geqslant 5$, we know $a_{i}+b_{i} \geqslant 2 \sqrt{a_{i} b_{i}} \geqslant 2 \sqrt{5}$.
Since $a_{i}, b_{i} \in \mathbf{Z}$, we have
$$
a_{i}+b_{i} \geqslant 5 \Rightarrow c_{i} \geqslant 10,
$$
and equality holds if and only if the $i$-th color cells are arranged as shown in Figure 5, denoted as Figure A.
$$
\begin{array}{l}
\text { Hence } f \geqslant \frac{1}{2}(5 \times 10-5 \times 4) \\
=15,
\end{array}
$$
and the equality condition is that there are 5 Figure A's.
Notice that 5 $2 \times 2$ grids cannot be arranged without overlapping in a $5 \times 5$ grid.
Thus, the equality in (1) does not hold.
Therefore, $f \geqslant 16$.
|
16
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let positive real numbers $x, y$ satisfy
$$
x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y}=\frac{27}{4} \text {. }
$$
Then the minimum value of $P=\frac{15}{x}-\frac{3}{4 y}$ is
|
7.6.
By the AM-GM inequality for three terms, we have
$$
\begin{array}{l}
x^{2}+\frac{1}{x}=\left(x^{2}+\frac{8}{x}+\frac{8}{x}\right)-\frac{15}{x} \geqslant 12-\frac{15}{x} \\
y^{2}+\frac{1}{y}=\left(y^{2}+\frac{1}{8 y}+\frac{1}{8 y}\right)+\frac{3}{4 y} \geqslant \frac{3}{4}+\frac{3}{4 y} .
\end{array}
$$
Adding the two inequalities, we get
$$
\begin{array}{l}
\frac{27}{4}=x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y} \geqslant \frac{51}{4}+\left(\frac{3}{4 y}-\frac{15}{x}\right) \\
\Rightarrow P=\frac{15}{x}-\frac{3}{4 y} \geqslant 6 .
\end{array}
$$
Equality holds if and only if $x=2, y=\frac{1}{2}$.
Therefore, the minimum value of $P$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Let the general term formula of the sequence $\left\{a_{n}\right\}$ be $a_{n}=n^{3}-n$ $\left(n \in \mathbf{Z}_{+}\right)$, and the terms in this sequence whose unit digit is 0, arranged in ascending order, form the sequence $\left\{b_{n}\right\}$. Then the remainder when $b_{2} 018$ is divided by 7 is $\qquad$ .
|
8. 4 .
Since $a_{n}=n^{3}-n=n(n-1)(n+1)$, therefore, $a_{n}$ has a units digit of 0 if and only if the units digit of $n$ is $1, 4, 5, 6, 9, 0$. Hence, in any consecutive 10 terms of the sequence $\left\{a_{n}\right\}$, there are 6 terms whose units digit is 0.
Since $2018=336 \times 6+2,336 \times 10=3360$, the remainder 2 corresponds to a term whose units digit is 4. Therefore,
$$
\begin{array}{l}
b_{2018}=a_{3364}=3364^{3}-3364 \\
\equiv 4^{3}-4 \equiv 4(\bmod 7) .
\end{array}
$$
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. If the function
$$
f(x)=x^{2}-2 a x-2 a|x-a|+1
$$
has exactly three zeros, then the value of the real number $a$ is $\qquad$.
|
2. 1.
Let $t=|x-a|(t \geqslant 0)$.
Then the original problem is equivalent to the equation $t^{2}-2 a t+1-a^{2}=0$ having two roots $t_{1}=0, t_{2}>0$.
Upon verification, $a=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given the sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms as $S_{n}$, and
$$
a_{1}=3, S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right) \text {. }
$$
If $\left\{a_{n}\right\}$ contains three terms $a_{1} 、 a_{p} 、 a_{q}(p 、 q \in$ $\left.\mathbf{Z}_{+}, 1<p<q\right)$ that form an arithmetic sequence, then $q-p=$
|
4. 1 .
Given $S_{n}=2 a_{n}+\frac{3}{2}\left((-1)^{n}-1\right)$
$$
\Rightarrow S_{n-1}=2 a_{n-1}+\frac{3}{2}\left((-1)^{n-1}-1\right)(n \geqslant 2) \text {. }
$$
Subtracting the two equations yields $a_{n}=2 a_{n-1}-3(-1)^{n}(n \geqslant 2)$.
Let $b_{n}=\frac{a_{n}}{(-1)^{n}}$. Then
$$
\begin{array}{l}
b_{n}=-2 b_{n-1}-3 \\
\Rightarrow b_{n}+1=-2\left(b_{n-1}+1\right) \\
\quad=(-2)^{n-1}\left(b_{1}+1\right)=(-2)^{n} \\
\Rightarrow b_{n}=(-2)^{n}-1 \\
\Rightarrow a_{n}=2^{n}-(-1)^{n} .
\end{array}
$$
Assume there exists a positive integer $c$, such that $a_{1} 、 a_{n} 、 a_{n+c}$ form an arithmetic sequence. Then
$$
\begin{array}{l}
2\left(2^{n}-(-1)^{n}\right)=2^{n+c}-(-1)^{n+c}+3 \\
\Rightarrow 2^{n+1}\left(2^{c-1}-1\right)+(-1)^{n}\left(2-(-1)^{c}\right)+3=0 .
\end{array}
$$
Clearly, the above equation holds if and only if $c=1$, and $n$ is a positive odd number.
Thus, $p=n, q=n+1 \Rightarrow q-p=1$.
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. In the Cartesian coordinate system, color the set of points
$$
\left\{(m, n) \mid m, n \in \mathbf{Z}_{+}, 1 \leqslant m, n \leqslant 6\right\}
$$
red or blue. Then the number of different coloring schemes where each unit square has exactly two red vertices is $\qquad$ kinds.
|
8. 126 .
Dye the first row (points with a y-coordinate of 6), there are $2^{6}$ ways to do this, which can be divided into two cases.
(1) No two same-colored points are adjacent (i.e., red and blue alternate), there are 2 ways, and the second row can only be dyed in 2 ways, each row has only 2 ways, totaling $2^{6}$ ways;
(2) At least two adjacent same-colored points exist, there are $2^{6}-2$ ways to dye, in this case, when dyeing the second row, it can be found that the satisfactory dyeing method is unique, totaling $2^{6}-2$ ways.
In summary, the number of dyeing methods that meet the conditions is $2^{6}+2^{6}-2=126$
|
126
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. A meeting was attended by 24 representatives, and between any two representatives, they either shook hands once or did not shake hands at all. After the meeting, it was found that there were a total of 216 handshakes, and for any two representatives $P$ and $Q$ who shook hands, among the remaining 22 representatives, no more than 10 representatives shook hands with exactly one of $P$ or $Q$. A "friend circle" refers to a set of three representatives who all shook hands with each other. Find the minimum possible number of friend circles among the 24 representatives.
(Chang An)
|
7. Let the 24 representatives be $v_{1}, v_{2}, \cdots, v_{24}$, and for $i=1,2, \cdots, 24$, let $d_{i}$ denote the number of people who have shaken hands with $v_{i}$.
Define the set $E=\left\{\left\{v_{i}, v_{j}\right\} \mid v_{i}\right.$ has shaken hands with $v_{j} \}$.
For any $e=\left\{v_{i}, v_{j}\right\} \in E$, let $t(e)$ be the number of friend circles that include the set $e$.
By the given condition,
$$
\begin{array}{l}
\left(d_{i}-1\right)+\left(d_{j}-1\right) \leqslant 2 t(e)+10 \\
\Rightarrow \sum_{e=\left|v_{i}, v_{j}\right| \in E}\left(d_{i}+d_{j}\right) \leqslant \sum_{e \in E}(2 t(e)+12) .
\end{array}
$$
In $\sum_{e=\left|v_{i}, v_{j}\right| \in E}\left(d_{i}+d_{j}\right)$, the number of handshakes $d_{i}$ of representative $v_{i}$ is counted $d_{i}$ times, so the left side of equation (1) equals $\sum_{i=1}^{24} d_{i}^{2}$. Let the number of friend circles be $T$, noting that each friend circle is counted 3 times in $\sum_{e \in E} t(e)$, and the total number of edges is 216, then the right side of equation (1) equals $6 T+12 \times 216$.
$$
\begin{array}{l}
\text { Hence } \sum_{i=1}^{24} d_{i}^{2} \leqslant 6 T+12 \times 216 . \\
\text { By } \sum_{i=1}^{24} d_{i}^{2} \geqslant \frac{1}{24}\left(\sum_{i=1}^{24} d_{i}\right)^{2}=\frac{(2 \times 216)^{2}}{24} \\
\Rightarrow T \geqslant \frac{1}{6}\left(\frac{(2 \times 216)^{2}}{24}-12 \times 216\right)=864 .
\end{array}
$$
Divide the 24 representatives into four groups, each with 6 people, and let any two representatives from different groups shake hands once, while any two representatives from the same group do not shake hands.
The total number of handshakes is calculated as
$$
\frac{1}{2} \times 24 \times 18=216 \text {. }
$$
For any two representatives $P$ and $Q$ who have shaken hands, among the remaining 22 representatives, exactly 10 representatives have shaken hands with either $P$ or $Q$, which meets the condition. In this case, the number of friend circles is
$$
C_{4}^{3} \times 6^{3}=864 \text {. }
$$
In summary, the minimum possible number of friend circles among the 24 representatives is 864.
|
864
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Let integers $x, y$ satisfy $x^{2}+y^{2}4$. Then the maximum value of $x^{2}-2 x y-3 y$ is
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly.
|
3. 3 .
Using a TI calculator, we can obtain the region satisfying
$$
\left\{\begin{array}{l}
x^{2}+y^{2}4
\end{array},\right.
$$
and since we need to find the maximum value of $x^{2}-2 x y-3 y$, the integer point is in the third quadrant.
Substituting $(-2,-3)$ and $(-3,-2)$ into $x^{2}-2 x y-3 y$ and comparing the results, we find the maximum value is 3.
|
3
|
Algebra
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
11. Arrange 10 flowers in a row using red, yellow, and blue flowers (assuming there are plenty of each color), and yellow flowers cannot be adjacent. How many different arrangements are there (the 10 flowers can be of one color or two colors)?
|
11. Let $x_{n}$ be the number of different arrangements of $n$ flowers that meet the requirements. Then
$$
x_{1}=3, x_{2}=3^{2}-1=8 \text {. }
$$
When $n \geqslant 3$, let the arrangement of $n$ flowers be $a_{1}, a_{2}, \cdots, a_{n}$. If $a_{1}$ is a red or blue flower, then $a_{2}, a_{3}, \cdots, a_{n}$ is an arrangement of $n-1$ flowers that meets the conditions; if $a_{1}$ is a yellow flower, then $a_{2}$ must be a red or blue flower, in which case $a_{3}, a_{4}, \cdots, a_{n}$ is an arrangement of $n-2$ flowers that meets the conditions.
$$
\begin{array}{l}
\text { Thus, } x_{n}=2 x_{n-1}+2 x_{n-2} . \\
\text { Hence } x_{3}=2 x_{2}+2 x_{1}=22, \\
x_{4}=2 x_{3}+2 x_{2}=60, \\
x_{5}=2 x_{4}+2 x_{3}=164, \\
x_{6}=2 x_{5}+2 x_{4}=448, \\
x_{7}=2 x_{6}+2 x_{5}=1224, \\
x_{8}=2 x_{7}+2 x_{6}=3344, \\
x_{9}=2 x_{8}+2 x_{7}=9136, \\
x_{10}=2 x_{9}+2 x_{8}=24960 .
\end{array}
$$
【Note】The general term of $\left\{x_{n}\right\}$ in this problem is
$$
x_{n}=\frac{9+5 \sqrt{3}}{6}(1+\sqrt{3})^{n-1}+\frac{9-5 \sqrt{3}}{6}(1-\sqrt{3})^{n-1} .
$$
|
24960
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given real numbers $x, y$ satisfy $x^{2}+y^{2}=20$. Then the maximum value of $x y+8 x+y$ is $\qquad$ .
|
4. 42 .
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x y+8 x+y)^{2} \\
\leqslant\left(x^{2}+8^{2}+y^{2}\right)\left(y^{2}+x^{2}+1^{2}\right) \\
=84 \times 21=42^{2} .
\end{array}
$$
Therefore, the maximum value sought is 42.
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. Given the inequality $\left|a x^{2}+b x+a\right| \leqslant x$ holds for $x \in$ $[1,2]$. Then the maximum value of $3 a+b$ is $\qquad$
|
6. 3 .
From the problem, we know that $\left|a\left(x+\frac{1}{x}\right)+b\right| \leqslant 1$.
Given $x \in[1,2]$, we have $t=x+\frac{1}{x} \in\left[2, \frac{5}{2}\right]$.
Thus, $|2 a+b| \leqslant 1$, and $\left|\frac{5}{2} a+b\right| \leqslant 1$.
Therefore, $3 a+b=2\left(\frac{5}{2} a+b\right)-(2 a+b) \leqslant 3$.
When $a=4, b=-9$, the equality holds.
|
3
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. In the sequence
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]
$$
there are $\qquad$ distinct integers ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
|
7.1515.
Let the $k$-th term of the known sequence be $\left[\frac{k^{2}}{2019}\right]$. Then, when $(k+1)^{2}-k^{2} \leqslant 2019$, i.e., $k \leqslant 1009$,
$$
\begin{array}{l}
\frac{(k+1)^{2}}{2019}=\frac{k^{2}}{2019}+\frac{2 k+1}{2019} \leqslant \frac{k^{2}}{2019}+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right] \leqslant 1+\left[\frac{k^{2}}{2019}\right] \\
\Rightarrow 0 \leqslant\left[\frac{(k+1)^{2}}{2019}\right]-\left[\frac{k^{2}}{2019}\right] \leqslant 1 .
\end{array}
$$
Since $\left[\frac{1009^{2}}{2019}\right]=504$, therefore, in
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{1009^{2}}{2019}\right]
$$
the integers $0,1, \cdots, 504$ appear without omission, and these 505 integers are all distinct.
$$
\begin{array}{l}
\text { Also, when } k \geqslant 1010, \\
2 k+1=(k+1)^{2}-k^{2}>2019 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right]=\left[\frac{k^{2}+2 k+1}{2019}\right] \\
\quad \geqslant\left[\frac{k^{2}+2019}{2019}\right]=\left[\frac{k^{2}}{2019}\right]+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right]-\left[\frac{k^{2}}{2019}\right] \geqslant 1 .
\end{array}
$$
Thus, $\left[\frac{1010^{2}}{2019}\right],\left[\frac{1011^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]$ are
distinct integers, and all are greater than 504.
Therefore, the number of distinct integers in the original sequence is
$$
505+1010=1515 .
$$
|
1515
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. On each face of a cube, randomly fill in one of the numbers 1, 2, $\cdots$, 6 (the numbers on different faces are distinct). Then number the eight vertices such that the number assigned to each vertex is the product of the numbers on the three adjacent faces. The maximum value of the sum of the numbers assigned to the eight vertices is $\qquad$
|
8. 343 .
Let the numbers on the six faces be $a, b, c, d, e, f$, and $(a, b), (c, d), (e, f)$ be the numbers on the opposite faces. Thus, the sum of the numbers at the eight vertices is
$$
\begin{array}{l}
(a+b)(c+d)(e+f) \\
\leqslant\left(\frac{(a+b)+(c+d)+(e+f)}{3}\right)^{3} \\
=7^{3}=343 .
\end{array}
$$
When $a=1, b=6, c=2, d=5, e=3, f=4$, the equality holds.
Therefore, the maximum value sought is 343 .
|
343
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four. (50 points) A planet has 1000 cities $c_{1}$, $c_{2}, \cdots, c_{1000}$, and three airlines $X$, $Y$, and $Z$ provide flights between these cities. For any $1 \leqslant i<j \leqslant 1000$, exactly one airline operates a one-way flight from city $c_{i}$ to city $c_{j}$. Find the largest positive integer $n$, such that a traveler can always choose one of the airlines and complete a journey starting from a city, passing through $n-1$ cities (excluding the departure and arrival cities), and finally arriving at the $n$-th city.
|
Four, the maximum value of $n$ is 9.
For each city $c_{i}(i=1,2, \cdots, 1000)$, define a triplet of non-negative integers $\left(x_{i}, y_{i}, z_{i}\right)$ according to the following rules:
If there are no flights from company $X$ arriving at city $c_{i}$, set $x_{i}=0$; otherwise, there exists a largest positive integer $x_{i}$ such that there is a sequence of non-negative integers $k_{0}<k_{1}<\cdots<k_{x_{i}-1}<i$, satisfying that starting from city $c_{k_{0}}$, passing through cities $c_{k_{1}}, c_{k_{2}}$, $\cdots, c_{k_{x-1}}$ in sequence to reach $c_{i}$, and all routes belong to airline $X$.
|
9
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. Given four positive integers
$a, b, c, d$ satisfy:
$$
a^{2}=c(d+20), b^{2}=c(d-18) \text {. }
$$
Then the value of $d$ is $\qquad$
|
4. 180 .
Let $(a, b)=t, a=t a_{1}, b=t b_{1}$.
Then $\frac{d+20}{d-18}=\frac{c(d+20)}{c(d-18)}=\frac{a^{2}}{b^{2}}=\frac{a_{1}^{2}}{b_{1}^{2}}$ (simplest fraction).
Let $d+20=k a_{1}^{2}, d-18=k b_{1}^{2}$.
Eliminating $d$ yields
$$
\begin{array}{l}
k\left(a_{1}+b_{1}\right)\left(a_{1}-b_{1}\right)=2 \times 19 \\
\Rightarrow k=2, a_{1}+b_{1}=19, a_{1}-b_{1}=1 \\
\Rightarrow a_{1}=10, b_{1}=9 \\
\Rightarrow d=k a_{1}^{2}-20=180 .
\end{array}
$$
At this point, there exist $a=20, b=18, c=2$ that satisfy the conditions.
Therefore, $d=180$.
|
180
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Given that $a, b, c, d$ are positive integers, and $\log _{a} b=\frac{3}{2}, \log _{c} d=\frac{5}{4}, a-c=9$.
Then $a+b+c+d=$ $\qquad$
|
3. 198 .
Given $a=x^{2}, b=x^{3}, c=y^{4}, d=y^{5}$.
$$
\begin{array}{l}
\text { Given } a-c=x^{2}-y^{4}=9 \\
\Rightarrow\left(x+y^{2}\right)\left(x-y^{2}\right)=9 \\
\Rightarrow x+y^{2}=9, x-y^{2}=1 \\
\Rightarrow x=5, y^{2}=4 \\
\Rightarrow a=25, b=125, c=16, d=32 \\
\Rightarrow a+b+c+d=198 .
\end{array}
$$
|
198
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. In $\triangle A B C$, the sides opposite to $\angle A, \angle B, \angle C$ are $a, b, c$ respectively, $\angle A B C=120^{\circ}$, the angle bisector of $\angle A B C$ intersects $A C$ at point $D$, and $B D=1$. Then the minimum value of $4 a+c$ is $\qquad$
|
4.9.
From the problem, we know that $S_{\triangle A B C}=S_{\triangle A B D}+S_{\triangle B C D}$.
By the angle bisector property and the formula for the area of a triangle, we get
$$
\begin{array}{l}
\frac{1}{2} a c \sin 120^{\circ} \\
=\frac{1}{2} a \times 1 \times \sin 60^{\circ}+\frac{1}{2} c \times 1 \times \sin 60^{\circ} \\
\Rightarrow \frac{1}{a}+\frac{1}{c}=1 .
\end{array}
$$
Therefore, $4 a+c=(4 a+c)\left(\frac{1}{a}+\frac{1}{c}\right)$
$$
=5+\frac{c}{a}+\frac{4 a}{c} \geqslant 5+2 \sqrt{\frac{c}{a} \cdot \frac{4 a}{c}}=9 \text {. }
$$
Equality holds if and only if $c=2 a=3$. Therefore, the minimum value of $4 a+c$ is 9.
|
9
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
10. (20 points) In the sequence $\left\{a_{n}\right\}$, let $S_{n}=\sum_{i=1}^{n} a_{i}$ $\left(n \in \mathbf{Z}_{+}\right)$, with the convention: $S_{0}=0$. It is known that
$$
a_{k}=\left\{\begin{array}{ll}
k, & S_{k-1}<k ; \\
-k, & S_{k-1} \geqslant k
\end{array}\left(1 \leqslant k \leqslant n, k 、 n \in \mathbf{Z}_{+}\right)\right. \text {. }
$$
Find the largest positive integer $n$ not exceeding 2019 such that
$$
S_{n}=0 .
$$
|
10. Let the indices $n$ that satisfy $S_{n}=0$ be arranged in ascending order, denoted as the sequence $\left\{b_{n}\right\}$, then $b_{1}=0$.
To find the recurrence relation that $\left\{b_{n}\right\}$ should satisfy.
In fact, without loss of generality, assume $S_{b_{k}}=0$.
Thus, by Table 1, it is easy to prove by mathematical induction:
$$
\left\{\begin{array}{l}
S_{b_{k}+2 i-1}=b_{k}+2-i, \\
S_{b_{k}+2 i}=2 b_{k}+i+2
\end{array}\left(i=1,2, \cdots, b_{k}+2\right)\right. \text {. }
$$
Table 1
\begin{tabular}{|c|c|c|}
\hline$n$ & $S_{n}$ & $a_{n}$ \\
\hline$b_{k}$ & 0 & $/$ \\
\hline$b_{k}+1$ & $b_{k}+1$ & $b_{k}+1$ \\
\hline$b_{k}+2$ & $2 b_{k}+3$ & $b_{k}+2$ \\
\hline$b_{k}+3$ & $b_{k}$ & $-\left(b_{k}+3\right)$ \\
\hline$b_{k}+4$ & $2 b_{k}+4$ & $b_{k}+4$ \\
\hline$b_{k}+5$ & $b_{k}-1$ & $-\left(b_{k}+5\right)$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ \\
\hline
\end{tabular}
Let $b_{k}+2-i=0$, we have
$$
b_{k}+2 i-1=3 b_{k}+3 \text {, }
$$
which means the sequence $\left\{b_{n}\right\}$ satisfies:
$$
\begin{array}{l}
b_{k+1}=3 b_{k}+3, b_{1}=0 \\
\Rightarrow b_{n}=\frac{3^{n}-3}{2} .
\end{array}
$$
Thus, $b_{7}=1092$ is the desired result.
|
1092
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
One, (40 points) Find the smallest integer $c$, such that there exists a sequence of positive integers $\left\{a_{n}\right\}(n \geqslant 1)$ satisfying:
$$
a_{1}+a_{2}+\cdots+a_{n+1}<c a_{n}
$$
for all $n \geqslant 1$.
|
Given the problem, we have
$$
c>\frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}}.
$$
For any \( n \geqslant 1 \), we have
$$
\begin{array}{l}
nc > \frac{a_{1}+a_{2}}{a_{1}} + \frac{a_{1}+a_{2}+a_{3}}{a_{2}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n+1}}{a_{n}} \\
= n + \frac{a_{2}}{a_{1}} + \left(\frac{a_{1}}{a_{2}} + \frac{a_{3}}{a_{2}}\right) + \cdots + \\
\left(\frac{a_{1}}{a_{n}} + \frac{a_{2}}{a_{n}} + \cdots + \frac{a_{n+1}}{a_{n}}\right) \\
= n + \left(\frac{a_{2}}{a_{1}} + \frac{a_{1}}{a_{2}}\right) + \left(\frac{a_{3}}{a_{2}} + \frac{a_{2}}{a_{3}}\right) + \cdots + \\
\left(\frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}}\right) + \frac{a_{n+1}}{a_{n}} + \frac{a_{1}}{a_{3}} + \\
\frac{a_{1}+a_{2}}{a_{4}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n-2}}{a_{n}} \\
\geqslant n + 2(n-1) + \frac{a_{n+1}}{a_{n}} + \frac{a_{1}}{a_{3}} + \\
\frac{a_{1}+a_{2}}{a_{4}} + \cdots + \frac{a_{1}+a_{2}+\cdots+a_{n-2}}{a_{n}}.
\end{array}
$$
If there exists \( n \geqslant 1 \) such that \( \frac{a_{n+1}}{a_{n}} \geqslant 2 \), then
$$
\begin{array}{l}
nc > n + 2(n-1) + 2 = 3n \\
\Rightarrow c > 3 \Rightarrow c \geqslant 4.
\end{array}
$$
If for any \( n \geqslant 1 \), we have \( \frac{a_{n+1}}{a_{n}} < 2 \), then
$$
\begin{array}{l}
nc > n + 2(n-1) + \frac{1}{2^2} + \frac{1}{2^2} + \cdots + \frac{1}{2^2}.
\end{array}
$$
Taking \( n-2 > 2^2 \), we get
$$
nc > n + 2(n-1) + 2 = 3n \\
\Rightarrow c \geqslant 4.
$$
Thus, the smallest integer \( c \) is 4.
Below is a construction: take \( a_{n} = 2^{n-1} \),
$$
\begin{array}{l}
a_{1} + a_{2} + \cdots + a_{n+1} = 1 + 2 + \cdots + 2^n \\
= 2^{n+1} - 1 \\
4a_{n} = 4 \times 2^{n-1} = 2^{n+1},
\end{array}
$$
Clearly, \( a_{1} + a_{2} + \cdots + a_{n+1} < 4a_{n} \) holds.
|
4
|
Inequalities
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
4. If three angles $\alpha, \beta, \gamma$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $\qquad$
|
4. -3 .
According to the problem, $\alpha=\beta-\frac{\pi}{3}, \gamma=\beta+\frac{\pi}{3}$. Therefore, $\tan \alpha=\frac{\tan \beta-\sqrt{3}}{1+\sqrt{3} \tan \beta}, \tan \gamma=\frac{\tan \beta+\sqrt{3}}{1-\sqrt{3} \tan \beta}$.
Then, $\tan \alpha \cdot \tan \beta=\frac{\tan ^{2} \beta-\sqrt{3} \tan \beta}{1+\sqrt{3} \tan \beta}$, $\tan \beta \cdot \tan \gamma=\frac{\tan ^{2} \beta+\sqrt{3} \tan \beta}{1-\sqrt{3} \tan \beta}$, $\tan \gamma \cdot \tan \alpha=\frac{\tan ^{2} \beta-3}{1-3 \tan ^{2} \beta}$.
Thus, $\tan \alpha \cdot \tan \beta+\tan \beta \cdot \tan \gamma+\tan \gamma \cdot \tan \alpha$ $=\frac{9 \tan ^{2} \beta-3}{1-3 \tan ^{2} \beta}=-3$.
|
-3
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Let $x, y, z \in \mathbf{R}_{+}$, satisfying $x+y+z=x y z$. Then the function
$$
\begin{array}{l}
f(x, y, z) \\
=x^{2}(y z-1)+y^{2}(z x-1)+z^{2}(x y-1)
\end{array}
$$
has the minimum value of $\qquad$
|
5. 18 .
According to the conditions, we have
$$
y+z=x(y z-1) \Rightarrow y z-1=\frac{y+z}{x} \text{. }
$$
Similarly, $z x-1=\frac{z+x}{y}, x y-1=\frac{x+y}{z}$.
From $x y z=x+y+z \geqslant 3 \sqrt[3]{x y z} \Rightarrow x y z \geqslant 3 \sqrt{3}$, thus
$$
\begin{array}{l}
f(x, y, z)=2(x y+y z+z x) \\
\geqslant 2 \times 3 \sqrt[3]{(x y z)^{2}} \geqslant 18,
\end{array}
$$
When $x=y=z=\sqrt{3}$, the equality holds.
|
18
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The sequence of positive integers $\left\{a_{n}\right\}: a_{n}=3 n+2$ and $\left\{b_{n}\right\}$ $b_{n}=5 n+3(n \in \mathbf{N})$ have a common number of terms in $M=\{1,2, \cdots, 2018\}$ which is $\qquad$
|
6. 135.
It is known that 2018 is the largest common term of the two sequences within $M$. Excluding this common term, subtract 2018 from the remaining terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$, respectively, to get
$$
\left\{\overline{a_{n}}\right\}=\{3,6,9, \cdots, 2016\},
$$
which are all multiples of 3 within $M$; and
$$
\left\{\overline{b_{n}}\right\}=\{5,10,15, \cdots, 2015\},
$$
which are all multiples of 5 within $M$.
Clearly, the common terms of $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ correspond one-to-one to the common terms of $\left\{\overline{a_{n}}\right\}$ and $\left\{\overline{b_{n}}\right\}$, and these common terms are all multiples of 15 within $M$, which are $\left[\frac{2018}{15}\right]=134$, where $[x]$ denotes the greatest integer not exceeding the real number $x$. Therefore, the number of common terms sought is $134+1=135$.
|
135
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. For a positive integer $n$, let the sum of its digits be denoted as $s(n)$, and the product of its digits as $p(n)$. If $s(n) +$ $p(n) = n$ holds, then $n$ is called a "coincidence number". Therefore, the sum of all coincidence numbers is
|
8.531.
Let \( n = \overline{a_{1} a_{2} \cdots a_{k}} \left(a_{1} \neq 0\right) \).
From \( n - s(n) = p(n) \), we get
\[
\begin{array}{l}
a_{1}\left(10^{k-1}-1\right) + a_{2}\left(10^{k-2}-1\right) + \cdots + a_{k-1}(10-1) \\
= a_{1} a_{2} \cdots a_{k},
\end{array}
\]
which simplifies to \( a_{1}\left(10^{k-1}-1-a_{2} a_{3} \cdots a_{k}\right) + m = 0 \),
where \( m = a_{2}\left(10^{k-2}-1\right) + \cdots + a_{k-1}(10-1) \geqslant 0 \).
If \( k \geqslant 3 \), since
\[
10^{k-1}-1-a_{2} a_{3} \cdots a_{k} \geqslant 10^{k-1}-1-9^{k-1} > 0,
\]
this contradicts equation (1). Therefore, \( k \leqslant 2 \).
When \( k = 1 \), \( n = s(n) \), which does not meet the condition, so \( k = 2 \).
Thus, we can set \( n = \overline{a_{1} a_{2}} = 10 a_{1} + a_{2} \).
Combining \( a_{1} + a_{2} + a_{1} a_{2} = 10 a_{1} + a_{2} \), we get
\[
9 a_{1} = a_{1} a_{2} \Rightarrow a_{2} = 9 \left(a_{1} \in \{1, 2, \cdots, 9\}\right),
\]
so the complete set of coincidental numbers is \( 19, 29, \cdots, 99 \), and their sum is 531.
|
531
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given the function $f(x)=x+\frac{2}{x}$ on the interval $[1,4]$, the maximum value is $M$, and the minimum value is $m$. Then the value of $M-m$ is $\qquad$
|
ニ.7.4.
Since $f(x)$ is monotonically decreasing on the interval $[1,3]$ and monotonically increasing on the interval $[3,4]$, the minimum value of $f(x)$ is $f(3)=6$.
Also, $f(1)=10, f(4)=\frac{25}{4}$, so the maximum value of $f(x)$ is $f(1)=10$.
Therefore, $M-m=10-6=4$.
|
4
|
Calculus
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Let the line $y=k x+b$ intersect the curve $y=x^{3}-x$ at three distinct points $A, B, C$, and $|A B|=|B C|=2$. Then the value of $k$ is
|
11. 1.
Given that the curve is symmetric about the point $(0,0)$, and
$$
|A B|=|B C|=2 \text {, }
$$
we know that the line $y=k x+b$ must pass through the origin.
Thus, $b=0$.
Let $A(x, y)$. Then
$$
\begin{array}{l}
y=k x, y=x^{3}-x, \sqrt{x^{2}+y^{2}}=2 \\
\Rightarrow x=\sqrt{k+1}, y=k \sqrt{k+1} . \\
\text { Substituting into } \sqrt{x^{2}+y^{2}}=2 \text { gives } \\
(k+1)+k^{2}(k+1)=4 \\
\Rightarrow(k-1)\left(k^{2}+2 k+3\right)=0 \\
\Rightarrow k=1 .
\end{array}
$$
|
1
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given $f(x)=\frac{10}{x+1}-\frac{\sqrt{x}}{3}$. Then the number of elements in the set $M=\left\{n \in \mathbf{Z} \mid f\left(n^{2}-1\right) \geqslant 0\right\}$ is $\qquad$.
|
,- 1.6 .
From the problem, we know that $f(x)$ is monotonically decreasing on the interval $[0,+\infty)$, and $f(9)=0$.
Then $f\left(n^{2}-1\right) \geqslant f(9) \Rightarrow 1 \leqslant n^{2} \leqslant 10$.
Thus, the number of elements in set $M$ is 6.
|
6
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given $a_{k}$ as the number of integer terms in $\log _{2} k, \log _{3} k, \cdots, \log _{2018} k$. Then $\sum_{k=1}^{2018} a_{k}=$ $\qquad$
|
8.4102 .
Let $b_{m}$ be the number of integer terms in $\log _{m} 1, \log _{m} 2, \cdots, \log _{m} 2018$.
Then, $\sum_{k=1}^{2018} a_{k}=\sum_{m=2}^{2018} b_{m}$.
Notice that, $\log _{m} t$ is an integer if and only if $t$ is a power of $m$.
\[
\begin{array}{l}
\text { Then } b_{2}=11, b_{3}=7, b_{4}=6, b_{5}=b_{6}=5, \\
b_{7}=b_{8}=\cdots=b_{12}=4, \\
b_{13}=b_{14}=\cdots=b_{44}=3, \\
b_{45}=b_{46}=\cdots=b_{2018}=2 . \\
\text { Hence } \sum_{m=2}^{2018} b_{m} \\
=11+7+6+5 \times 2+4 \times 6+3 \times 32+2 \times 1974 \\
=4102 .
\end{array}
\]
Therefore, $\sum_{k=1}^{2018} a_{k}=4102$.
|
4102
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
9. (16 points) Let
$$
\begin{array}{l}
P(z)=z^{4}-(6 \mathrm{i}+6) z^{3}+24 \mathrm{iz}^{2}- \\
(18 \mathrm{i}-18) z-13 .
\end{array}
$$
Find the area of the convex quadrilateral formed by the four points in the complex plane corresponding to the four roots of $P(z)=0$.
|
II. 9. Notice that, $P(1)=0$.
Then $P(z)=(z-1)\left(z^{3}-(6 \mathrm{i}+6) z^{2}+24 \mathrm{iz}^{2}+\right.$ $(18 \mathrm{i}-5) z+13)$.
Let $Q(z)=z^{3}-(6 \mathrm{i}+5) z^{2}+(18 \mathrm{i}-5) z+13$.
Then $Q(\mathrm{i})=0$.
Hence $Q(z)=(z-\mathrm{i})\left(z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}\right)$.
Using the quadratic formula for $z^{2}-(5 \mathrm{i}+5) z+13 \mathrm{i}=0$ yields $z=3+2 \mathrm{i}$ or $2+3 \mathrm{i}$.
Thus, the four roots of $P(z)=0$ are $1, \mathrm{i}, 3+2 \mathrm{i}, 2+3 \mathrm{i}$, which form a rectangle with a length of $2 \sqrt{2}$ and a width of $\sqrt{2}$, giving an area of $\frac{1}{2} \sqrt{2} \times 2 \sqrt{2}=2$.
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given that the three sides of a triangle are consecutive natural numbers. If the largest angle is twice the smallest angle, then the perimeter of the triangle is $\qquad$ .
|
-,1.15.
Assuming the three sides of a triangle are $n-1$, $n$, and $n+1$, with the largest angle being $2 \theta$ and the smallest angle being $\theta$.
Then, by the Law of Sines, we have
$$
\frac{n-1}{\sin \theta}=\frac{n+1}{\sin 2 \theta} \Rightarrow \cos \theta=\frac{n+1}{2(n-1)} \text {. }
$$
By the Law of Cosines, we get
$$
\begin{array}{l}
\cos \theta=\frac{(n+1)^{2}+n^{2}-(n-1)^{2}}{2 n(n+1)} \\
=\frac{n+4}{2(n+1)} .
\end{array}
$$
Thus, $\cos \theta=\frac{n+1}{2(n-1)}=\frac{n+4}{2(n+1)}$
$$
\Rightarrow n=5 \text {. }
$$
Therefore, the perimeter of the triangle is 15.
|
15
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. The integer sequence $\left\{a_{i, j}\right\}(i, j \in \mathbf{N})$, where,
$$
\begin{array}{l}
a_{1, n}=n^{n}\left(n \in \mathbf{Z}_{+}\right), \\
a_{i, j}=a_{i-1, j}+a_{i-1, j+1}(i, j \geqslant 1) .
\end{array}
$$
Then the unit digit of the value taken by $a_{128,1}$ is
|
8. 4 .
By the recursive relation, we have
$$
\begin{array}{l}
a_{1,1}=1, a_{2, n}=n^{n}+(n+1)^{n+1}, \\
a_{3, n}=n^{n}+2(n+1)^{n+1}+(n+2)^{n+2} .
\end{array}
$$
Accordingly, by induction, we get
$$
\begin{array}{l}
a_{n, m}=\sum_{k=0}^{m-1} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} \\
=\sum_{k \geqslant 0} \mathrm{C}_{m-1}^{k}(n+k)^{n+k} .
\end{array}
$$
Verification: When $k=0$, $\mathrm{C}_{m}^{0}=\mathrm{C}_{m-1}^{0}$ holds, and with the identity $\mathrm{C}_{m}^{k}=\mathrm{C}_{m-1}^{k}+\mathrm{C}_{m-1}^{k-1}(k \geqslant 1)$, we can prove by induction.
Let $m=128$, by Lucas' Theorem, we have
$$
\begin{array}{l}
\mathrm{C}_{m-1}^{k}=\mathrm{C}_{127}^{k} \equiv 1(\bmod 2)(1 \leqslant k \leqslant 127), \\
\mathrm{C}_{127}^{k} \equiv 0(\bmod 5)(3 \leqslant k \leqslant 124)
\end{array}
$$
Then $a_{128,1}=\sum_{k=0}^{127} \mathrm{C}_{127}^{k}(k+1)^{k+1}$
$$
\begin{array}{l}
\equiv \sum_{k=0}^{127}(k+1)^{k+1} \equiv 0(\bmod 2), \\
a_{128,1}=\sum_{k \in\{0,1,2\}}^{127} \\
\equiv 4(\bmod 5) .
\end{array}
$$
Therefore, $a_{128,1} \equiv 4(\bmod 10)$.
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. There are five cities in a line connected by semi-circular roads, as shown in Figure 1. Each segment of the journey is from one city to another along a semi-circle. If the journey can be repeated, the total number of possible ways to start from city 5 and return to city 5 after four segments is $\qquad$.
保留源文本的换行和格式，所以翻译结果如下：
1. There are five cities in a line connected by semi-circular roads, as shown in Figure 1. Each segment of the journey is from one city to another along a semi-circle. If the journey can be repeated, the total number of possible ways to start from city 5 and return to city 5 after four segments is $\qquad$.
|
,- 1.80 .
After four segments, there are five possible ways to start from city 5 and return to city 5:
$$
\begin{array}{l}
5 \rightarrow 1 \rightarrow 5 \rightarrow 1 \rightarrow 5, \\
5 \rightarrow 1 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 5 \rightarrow 2 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 5 \rightarrow 1 \rightarrow 5, \\
5 \rightarrow 2 \rightarrow 4 \rightarrow 2 \rightarrow 5 .
\end{array}
$$
For each of the four segments, one can choose the upper semicircle or the lower semicircle, each with two options. Therefore, the total number of possible ways is $5 \times 16=80$.
|
80
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Let positive integers $m, n$ satisfy
$$
m(n-m)=-11 n+8 \text {. }
$$
Then the sum of all possible values of $m-n$ is $\qquad$
|
2. 18 .
From the problem, we have
$$
n=\frac{m^{2}+8}{m+11}=m-11+\frac{129}{m+11} \in \mathbf{Z}_{+} \text {. }
$$
Then $(m+11) \mid 129$
$$
\Rightarrow m+11=1,3,43,129 \text {. }
$$
Also, $m \in \mathbf{Z}_{+}$, checking we find that when $m=32,118$, the corresponding $n$ is a positive integer.
$$
\text { Hence }(m, n)=(32,24) \text { or }(118,108) \text {. }
$$
Therefore, the sum of the possible values is
$$
(32-24)+(118-108)=18 \text {. }
$$
|
18
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
3. Person A tosses a fair coin twice, and Person B tosses the same coin three times. If the probability that they end up with the same number of heads is written as a simplified fraction, the sum of the numerator and the denominator is $\qquad$ . (Romania)
|
3.21.
Let the outcomes of a coin landing heads up and tails up be denoted as $\mathrm{H}$ and $\mathrm{T}$, respectively.
Jia has four equally probable outcomes: HH, HT, TH, TT;
Yi has eight equally probable outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
The outcomes that match $\mathrm{HH}$ are $\mathrm{HHT}$, $\mathrm{HTH}$, $\mathrm{THH}$; the outcomes that match $\mathrm{HT}$ or $\mathrm{TH}$ are HTT, THT, TTH,
and the outcome that matches TT is TTT.
Thus, the required probability is
$$
\frac{1}{4} \times \frac{3}{8} + \frac{1}{2} \times \frac{3}{8} + \frac{1}{4} \times \frac{1}{8} = \frac{5}{16},
$$
The sum of the numerator and the denominator is $5 + 16 = 21$.
|
21
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
6. The product $1!\cdot 2!\cdot 3!\cdot \cdots \cdot 99!\cdot 100!$ ends with $\qquad$ consecutive 0s.
(Hong Kong, China, provided)
|
6. 1124.
Since in the product $1!\cdot 2!\cdot 3!\cdots \cdots 99!\cdot 100!$, there are a large number of factor 2s, the number of consecutive 0s at the end is determined by the number of factor 5s. The number of factor 5s in each factorial $x$! is shown in Table 1.
Thus, the total number of factor 5s is
$$
\begin{array}{l}
0+5(1+2+3+4+6+7+\cdots+10+ \\
12+13+\cdots+16+18+19+\cdots+22)+24 \\
=5\left(\frac{22 \times 23}{2}-5-11-17\right)+24 \\
=1124 .
\end{array}
$$
Therefore, the required value is 1124.
|
1124
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Example 3 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[3]}$
(2012, Girls' Mathematical Olympiad)
|
Notice that, $2012=4 \times 503$, where $p=503$ is a prime number, and
$$
\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1} .
$$
If $p \nmid k$, then $p \mid \mathrm{C}_{2012}^{k}$;
If $p \mid k$, then $k \in\{0, p, 2 p, 3 p, 4 p\}$.
Notice that, $\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$,
$$
\begin{array}{l}
\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}=4 \mathrm{C}_{4 p-1}^{p-1} \\
= 4 \times \frac{(4 p-1)(4 p-2) \cdots(3 p+1)}{(p-1)(p-2) \cdots 1} \\
\equiv 4(\bmod p), \\
\mathrm{C}_{4 p}^{2 p}=2 \mathrm{C}_{4 p-1}^{2 p-1} \\
= 2 \times \frac{(4 p-1)(4 p-2) \cdots(3 p+1)}{(2 p-1)(2 p-2) \cdots(p+1)} \times 3 \times \\
\frac{(3 p-1)(3 p-2) \cdots(2 p+1)}{(p-1)(p-2) \cdots 1} \\
\equiv 6(\bmod p),
\end{array}
$$
neither of which is a multiple of $p$.
If the binary representation of $n$ is
$$
n=\left(a_{r} a_{r-1} \cdots a_{0}\right)_{2}=\sum_{t=0}^{r} a_{t} 2^{t},
$$
then the power of 2 in $n!$ is
$$
\begin{array}{l}
\sum_{s=1}^{\infty}\left[\frac{n}{2^{s}}\right]=\sum_{s=1}^{\infty} \sum_{t=s}^{r} a_{t} 2^{t-s}=\sum_{t=1}^{r} \sum_{s=1}^{t} a_{t} 2^{t-s} \\
=\sum_{t=1}^{r} a_{t}\left(2^{t}-1\right)=n-\sum_{t=0}^{r} a_{t}=n-S(n),
\end{array}
$$
where $[x]$ denotes the greatest integer not exceeding the real number $x$, and $S(n)$ denotes the sum of the binary digits of $n$ (i.e., the number of 1s).
$$
\begin{array}{l}
\text { Hence } 4 \nmid \mathrm{C}_{2012}^{k} \Leftrightarrow 4 \nmid \frac{2012!}{k!\cdot(2012-k)!} \\
\Leftrightarrow 4 \nmid \frac{(k+m)!}{k!\cdot m!}(m=2012-k) \\
\Leftrightarrow k+m-S(k+m) \\
\quad=k-S(k)+m-S(m)+\varepsilon(\varepsilon \in\{0,1\}) \\
\Leftrightarrow S(k+m)=S(k)+S(m)-\varepsilon(\varepsilon \in\{0,1\}) . \\
\Leftrightarrow k+m=2012=(11111011100)_{2} \text { at most }
\end{array}
$$
has one carry.
(1) If $k+m=2012$ has no carry, then the three 0s correspond to 0s in the digits of $k$ and $m$, and the eight 1s correspond to one 0 and one 1 in the digits of $k$ and $m$. Hence, there are $2^{8}$ cases, in which $\mathrm{C}_{2012}^{k}$ is odd.
(2) If $k+m=2012$ has exactly one carry, it must occur at one of the two 10s (i.e., from the 1st and 5th positions to the 2nd and 6th positions), i.e., $01+01=10$, and the other digits follow the conclusion of (1). Hence, there are $2 \times 2^{7}=2^{8}$ cases, in which $\mathrm{C}_{2012}^{k}$ is even but not a multiple of 4.
Thus, among $\mathrm{C}_{2012}^{0}, \mathrm{C}_{2012}^{1}, \cdots, \mathrm{C}_{2012}^{2012}$, there are
$$
2013-2^{8}-2^{8}=1501
$$
multiples of 4.
Additionally, $\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$ is not a multiple of 4,
$\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}$ has a power of 2 of
$$
\begin{array}{l}
S(p)+S(3 p)-S(4 p) \\
=S(3 p)=S(1509)>2,
\end{array}
$$
$\mathrm{C}_{4 p}^{2 p}$ has a power of 2 of
$$
\begin{array}{l}
S(2 p)+S(2 p)-S(4 p)=S(p) \\
=S(503)=S(2012)=8,
\end{array}
$$
among which, there are three multiples of 4.
Therefore, among $\mathrm{C}_{2012}^{0}, \mathrm{C}_{2012}^{1}, \cdots, \mathrm{C}_{2012}^{2012}$, there are
$$
1501-3=1498
$$
multiples of 2012.
|
1498
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Let $P(x)=x^{4}+a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are real coefficients. Assume that
$$
P(1)=7, P(2)=52, P(3)=97 \text {, }
$$
then $\frac{P(9)+P(-5)}{4}=$ $\qquad$ . (Vietnam)
|
7. 1202.
Notice that, $52-7=97-52=45$,
$$
\begin{array}{l}
7=45 \times 1-38,52=45 \times 2-38, \\
97=45 \times 3-38 .
\end{array}
$$
Let $Q(x)=P(x)-45 x+38$. Then $Q(x)$ is a fourth-degree polynomial with a leading coefficient of 1, and
$$
Q(1)=Q(2)=Q(3)=0 \text {. }
$$
Thus, for some $r$,
$$
\begin{array}{l}
Q(x)=(x-1)(x-2)(x-3)(x-r) . \\
\text { Therefore, } \frac{1}{4}(P(9)+P(-5)) \\
=\frac{1}{4}(Q(9)+Q(-5))+26 \\
=\frac{1}{4}(8 \times 7 \times 6(9-r)+6 \times 7 \times 8(5+r))+26 \\
=\frac{1}{4}(6 \times 7 \times 8 \times 14)+26=1202 .
\end{array}
$$
|
1202
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
11. Given that $a$ and $b$ are real numbers, satisfying:
$$
\sqrt[3]{a}-\sqrt[3]{b}=12, \quad a b=\left(\frac{a+b+8}{6}\right)^{3} \text {. }
$$
Then $a-b=$ $\qquad$ (Proposed by Thailand)
|
11. 468 .
Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}$. Then $x-y=12, 6xy=x^{3}+y^{3}+8$.
Thus, $x^{3}+y^{3}+2^{3}-3 \times 2xy=0$
$$
\Rightarrow(x+y+2)\left(x^{2}+y^{2}+2^{2}-xy-2x-2y\right)=0 \text {. }
$$
Therefore, $x+y+2=0$, or
$$
x^{2}-xy+y^{2}-2x-2y+4=0 \text {. }
$$
Equation (1) can be rewritten as
$$
\frac{1}{2}(x-y)^{2}+\frac{1}{2}(x-2)^{2}+\frac{1}{2}(y-2)^{2}=0 \text {. }
$$
Thus, $x=y=2$, but this does not satisfy $x-y=12$.
Hence, $x+y+2=0$.
Therefore, $x=5, y=-7$.
Thus, $a=125, b=-343, a-b=468$.
|
468
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
1. Given the equation $\left|x^{2}-2 a x+b\right|=8$ has exactly three real roots, and they are the side lengths of a right triangle. Find the value of $a+b$.
(Bulgaria)
|
II. 1. Note that, for the equation
$$
x^{2}-2 a x+b-8=0
$$
the discriminant $\Delta_{1}=4\left(a^{2}-b+8\right)$; for the equation
$$
x^{2}-2 a x+b+8=0
$$
the discriminant $\Delta_{2}=4\left(a^{2}-b-8\right)$.
According to the problem, the original equation has exactly three real roots, so one of the discriminants is 0, and the other is greater than 0.
Since $\Delta_{1}>\Delta_{2}$, then $a^{2}-b-8=0$.
Thus, the repeated root of equation (2) is $a$, and the two roots of equation (1) are $a+4$ and $a-4$.
Therefore, $a^{2}+(a-4)^{2}=(a+4)^{2}$
$$
\begin{array}{l}
\Rightarrow a=16 \Rightarrow b=a^{2}-8=248 \\
\Rightarrow a+b=264 .
\end{array}
$$
|
264
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Color the six vertices of a regular pentagonal pyramid using at most six different colors, such that the two endpoints of any edge are colored differently. If one coloring method can be obtained from another by rotation, they are considered the same method. How many different coloring methods are there?
(Romania)
|
2. Since the problem does not specify the number of colors to choose from, we will calculate the coloring methods based on the actual number of colors used.
If fewer than four colors are used, the bottom face can have at most two colors. Since 5 is an odd number, the colors of the bottom face vertices cannot be alternated. In this case, there are no valid coloring methods.
If six colors are actually used, without considering rotational equivalence, there are $6!=720$ methods. After removing rotations, there are 720 $\div 5=144$ methods.
If five colors are actually used, the coloring of the bottom face vertices must be in the form $A B A C D$. By rotation, the color $B$ can be moved to a fixed vertex, so the number of coloring methods is the permutation of 5 colors, which is $5!=120$.
If four colors are actually used, the coloring of the bottom face vertices must be in the form $A B A B C$. By rotation, the color $C$ can be moved to a fixed vertex, so the number of coloring methods is the permutation of 4 colors, which is $4!=24$.
In summary, the total number of coloring methods is
$144+120+24=288$.
|
288
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. For what positive integer $k$ does $\frac{20^{k}+18^{k}}{k!}$ achieve its maximum value?
(Hong Kong, China, Contributed)
|
5.19.
Let $A_{k}=\frac{20^{k}+18^{k}}{k!}$. Then
$A_{k+1}=\frac{20^{k+1}+18^{k+1}}{(k+1)!}$,
$\frac{A_{k+1}}{A_{k}}=\frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}$.
Notice that,
$$
\begin{array}{l}
\frac{A_{k+1}}{A_{k}}>1 \Leftrightarrow \frac{20^{k+1}+18^{k+1}}{(k+1)\left(20^{k}+18^{k}\right)}>1 \\
\Leftrightarrow 20^{k+1}+18^{k+1}>(k+1)\left(20^{k}+18^{k}\right) \\
\Leftrightarrow 20^{k}(19-k)+18^{k}(17-k)>0 .
\end{array}
$$
Upon inspection, when $k=1,2, \cdots, 18$, equation (1) holds; when $k=19,20, \cdots$, equation (1) does not hold.
Therefore, when $k=19$, $A_{k}$ reaches its maximum value.
|
19
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. In the square in Figure 1, 10000 points are randomly thrown. Then the estimated number of points that fall into the shaded area (the equation of curve $C$ is $x^{2}-y=0$) is ( ).
(A) 5000
(B) 6667
(C) 7500
(D) 7854
|
5. B.
According to the problem, the area of the blank region is
$$
\int_{0}^{1} x^{2} \mathrm{~d} x=\left.\frac{1}{3} x^{3}\right|_{0} ^{1}=\frac{1}{3} \text {. }
$$
Therefore, the area of the shaded part is $\frac{2}{3}$, and the ratio of the area of the shaded part to the area of the blank part is $2: 1$.
Thus, the number of points falling into the shaded part is
$$
10000 \times \frac{2}{3} \approx 6667 .
$$
|
6667
|
Geometry
|
MCQ
|
Yes
|
Yes
|
cn_contest
| false
|
16. Nine consecutive positive integers are arranged in an increasing sequence $a_{1}, a_{2}, \cdots, a_{9}$. If $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}$ is a perfect square, and $a_{2}+a_{4}+a_{6}+a_{8}$ is a perfect cube, then the minimum value of $a_{1}+a_{2}+\cdots+a_{9}$ is $\qquad$.
|
16. 18000 .
Let these nine numbers be $a-4, a-3, \cdots, a+4$. Then $5 a=m^{2}, 4 a=n^{3}, S=9 a$. Thus, $a=\frac{m^{2}}{5}=\frac{n^{3}}{4} \Rightarrow 4 m^{2}=5 n^{3}$.
Clearly, $m$ and $n$ are both multiples of 10.
Let $m=10 m_{1}, n=10 n_{1}$, then $4 m_{1}^{2}=50 n_{1}^{3}$.
Take $m_{1}=10, n_{1}=2$. At this point,
$$
\begin{array}{l}
n=20, m=100 \\
\Rightarrow a=2000, S=9 a=18000 .
\end{array}
$$
|
18000
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
2. Laura has 2010 lamps and 2010 switches in front of her, with different switches controlling different lamps. She wants to find the correspondence between the switches and the lamps. For this, Charlie operates the switches. Each time Charlie presses some switches, and the number of lamps that light up is the same as the number of switches pressed (Note: The switches return to their original state after each operation). The questions are:
(1) What is the maximum number of different operations Charlie can perform so that Laura can accurately determine the correspondence between the lamps and the switches?
(2) If Laura operates the switches herself, what is the minimum number of operations she needs to find the correspondence between the lamps and the switches?
|
(1) Charlie selects a pair of switches $(A, B)$, each operation involves pressing them simultaneously or leaving them untouched, while other switches can be chosen arbitrarily, resulting in $2^{2009}$ different operations, but Laura cannot determine which lights are controlled by switches $A$ and $B$.
If Charlie performs $2^{2009}+1$ different operations, consider a particular switch $A$. Each operation either presses $A$ or leaves it untouched, dividing the operations into two categories, one of which must contain at least $2^{2008}+1$ operations (let's assume it's the category where $A$ is pressed). Consider the set of lights turned on by these operations, where one light is always on, indicating it is controlled by $A$.
(2) Since $2010<2^{11}$, we can use 11-bit binary numbers to number all the switches (from 00000000001 to 11111011010). Use 11 operations to number the lights (also using 11-bit binary numbers): the $k$-th operation ($k=1,2, \cdots, 11$) involves pressing the switches whose $k$-th bit is 1. Write 1 in the $k$-th bit of the number for all lights that are on, and 0 for those that are off. Thus, switches with the same number control the same light, establishing a one-to-one correspondence. Therefore, the switch-light correspondence can be determined in at most 11 operations.
|
11
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
Four, (50 points) Mr. Wanda often forgets the numbers he should remember, such as his friends' phone numbers, the password of the safe, etc. For this reason, the manufacturer specially designed a password lock for his office safe, with keys labeled $0 \sim 9$. It is known that the password of the safe is a three-digit number, but as long as two of the digits in the entered three-digit number are correct, the safe will open. Mr. Wanda has forgotten the password he set again. How many times does he need to try at least to ensure that the safe will definitely open?
|
Four, 50 times.
First, introduce several notations. Let
$E=\{1,3,5,7,9\}, F=\{0,2,4,6,8\}$
$\Phi=E \cup F=\{0,1, \cdots, 9\}$,
$\Omega=\{x y z \mid x, y, z \in \Phi\}$ (the set of all three-digit codes),
$\Omega_{11}=\{x a b \in \Omega \mid x \in \Phi, a, b \in E\}$,
$\Omega_{12}=\{a y b \in \Omega \mid y \in \Phi, a, b \in E\}$,
$\Omega_{13}=\{a b z \in \Omega \mid z \in \Phi, a, b \in E\}$,
$\Omega_{1}=\Omega_{11} \cup \Omega_{12} \cup \Omega_{13}$ (the set of all three-digit codes with at least two odd numbers),
$\Omega_{2}=\{c d w, c w d, w c d \in \Omega \mid w \in \Phi, c, d \in F\}$ (the set of all three-digit codes with at least two even numbers).
Then $\Omega=\Omega_{1} \cup \Omega_{2}$.
(1) Existence.
Prove that a subset $X$ of $\Omega$ with 50 elements can be constructed such that any three-digit code in $\Omega$ can be opened by a three-digit code in $X$.
Select 25 numbers $x_{a b}, y_{a b}, z_{a b} \in \Phi$, $a, b \in E$, and construct three subsets of $\Omega$ each with 25 elements:
$X_{11}=\left\{x_{a b} a b \in \Omega \mid a, b \in E\right\}$,
$X_{12}=\left\{a y_{a b} b \in \Omega \mid a, b \in E\right\}$,
$X_{13}=\left\{a b z_{a b} \in \Omega \mid a, b \in E\right\}$.
Then for $i=1,2,3$, any three-digit code in the set $\Omega_{1 i}$ can be opened by a three-digit code in the set $X_{1 i}$. Thus, any three-digit code in the set $\Omega_{1}$ can be opened by a three-digit code in the set
$X_{1}=X_{11} \cup X_{12} \cup X_{13}$.
Next, take special $X_{11}, X_{12}, X_{13}$ such that $X_{11} = X_{12} = X_{13}$, thus $X_{1} = X_{11} = X_{12} = X_{13}$ is a set containing 25 elements, for example:
$X_{1}=X_{11}=X_{12}=X_{13}$
$=\left\{\begin{array}{lllll}111 & 133 & 155 & 177 & 199 \\ 319 & 331 & 353 & 375 & 397 \\ 517 & 539 & 551 & 573 & 595 \\ 715 & 737 & 759 & 771 & 793 \\ 913 & 935 & 957 & 979 & 991\end{array}\right\}$
And any three-digit code in $\Omega_{1}$ can be opened by one of these 25 three-digit codes in $X_{1}$.
Similarly, a set $X_{2}$ of 25 three-digit codes composed entirely of even numbers can be constructed such that any three-digit code in $\Omega_{2}$ can be opened by a three-digit code in $X_{2}$.
Let $X=X_{1} \cup X_{2}$. By the construction of $X_{1}$ and $X_{2}$, we know $|X| = 50$, and any three-digit code in $\Omega = \Omega_{1} \cup \Omega_{2}$ can be opened by a three-digit code in $X$.
(2) 50 is the minimum.
We will prove by contradiction that if a subset $X$ of $\Omega$ satisfies that any three-digit code in $\Omega$ can be opened by a three-digit code in $X$, then $|X| \geqslant 50$.
Assume $w=|X| \leqslant 49$, and let
$X=\left\{\left(a_{1} b_{1} c_{1}\right),\left(a_{2} b_{2} c_{2}\right), \cdots,\left(a_{w} b_{w} c_{w}\right)\right\}$.
Consider the three sequences
$a_{1}, a_{2}, \cdots, a_{w} ; b_{1}, b_{2}, \cdots, b_{w} ; c_{1}, c_{2}, \cdots, c_{w}$. Then in these three sequences, at least two sequences must contain all the digits $0,1, \cdots, 9$. Otherwise, suppose the digit $a (a \in \Phi)$ does not appear in $a_{1}, a_{2}, \cdots, a_{w}$, and the digit $b (b \in \Phi)$ does not appear in $b_{1}, b_{2}, \cdots, b_{w}$. Then, the three-digit code $a b 0 \in \Omega$ cannot be opened by any three-digit code in $X$, which is a contradiction.
Without loss of generality, assume the set of different digits appearing in $a_{1}, a_{2}, \cdots, a_{w}$ is $A$, and let $x \in A$ appear the fewest times, which is $m_{1}$ times.
Similarly, the set of different digits appearing in $b_{1}, b_{2}, \cdots, b_{w}$ is $B$, and let $y \in B$ appear the fewest times, which is $m_{2}$ times; the set of different digits appearing in $c_{1}, c_{2}, \cdots, c_{w}$ is $C$, and let $z \in C$ appear the fewest times, which is $m_{3}$ times. Since at least two sequences contain all the digits $0,1, \cdots, 9$, then
$m \leqslant\left[\frac{49}{10}\right]=4$.
Consider the subset $Y$ of $X$ consisting of elements with the last digit $z$, say
$Y=\left\{\left(a_{1} b_{1} z\right),\left(a_{2} b_{2} z\right), \cdots,\left(a_{m} b_{m} z\right)\right\}$
Let $U=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$,
$V=\left\{b_{1}, b_{2}, \cdots, b_{m}\right\}$
and $U$ contains $s$ different digits, $V$ contains $t$ different digits, then
$1 \leqslant s \leqslant m, 1 \leqslant t \leqslant m$.
Note: $0,1, \cdots, 9$ have $10-s$ digits not appearing in $U$, and $10-t$ digits not appearing in $V$.
Construct the set
$Z=\{a b z \in \Omega \mid a \notin U, b \notin V\}$.
Therefore, $|Z|=(10-s)(10-t)$. For each $a b z \in Z$, there exists at least one three-digit code $\alpha \beta \gamma \in X$ that can open $a b z$.
By the construction of $Z$, we know $\alpha=a, \beta=b, \gamma \neq z$. Let
$Z^{\prime}=\{a b c \in X \mid a \notin U, b \notin V, c \neq z\}$.
Then $\left|Z^{\prime}\right| \geqslant(10-s)(10-t)$.
Construct the subset of $X$
$W=\{(a b c) \mid(a b c) \in X, a \in U\}$
Since $U$ contains $s$ different digits, and each digit appears at least $m$ times in $X$, then $|W| \geqslant m s$.
By the construction of sets $Z^{\prime}$ and $W$, it is clear that $W \cap Z^{\prime}=\varnothing$.
Then $49 \geqslant w=|X| \geqslant m s+(10-s)(10-t)$
Thus $49 \geqslant 100-10(s+t)+s t+m s$
$=2(5-s)(5-t)+50+s(m-t) \geqslant 50$,
which is a contradiction.
|
50
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
7. Given a regular 2019-gon, then, the maximum number of diagonals such that any two of them are either perpendicular or do not intersect except at endpoints.
|
7.2016 .
On one hand, no two diagonals of the regular 2019-gon $\Gamma$ are perpendicular.
Assume there exist two diagonals of the regular 2019-gon $\Gamma$ that are perpendicular, say $AB \perp CD$, and $A, B, C, D$ are four vertices of $\Gamma$. Let $E$ be a point on the perpendicular bisector of $AB$, and let $E'$ be a point on the circumcircle of $\Gamma$ such that $EE'$ is its diameter.
Since $EC = E'D$, and $C, D, E$ are vertices of the regular 2019-gon $\Gamma$, it follows that $E'$ is also a vertex of $\Gamma$.
However, there cannot be two vertices of $\Gamma$ that are symmetric with respect to the center, so the assumption is false.
On the other hand, these non-intersecting diagonals can divide the regular 2019-gon $\Gamma$ into at most 2017 triangles, thus, at most 2016 diagonals can satisfy the condition. Furthermore, by selecting all the diagonals from a single vertex of $\Gamma$, the condition is met.
In summary, at most 2016 diagonals can satisfy the condition.
|
2016
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
8. Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ with the general terms $a_{n}=2^{n}, b_{n}=5 n-2$. Then the sum of all elements in the set
$$
\left\{a_{1}, a_{2}, \cdots, a_{2019}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{2019}\right\}
$$
is $\qquad$
|
8.2184 .
Let the elements of $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ be arranged in ascending order to form the sequence $\left\{c_{n}\right\}$.
Then the problem is reduced to finding the number of solutions to the equation
$$
2^{n}=5 m-2\left(n, m \in \mathbf{Z}_{+}, n, m \leqslant 2019\right)
$$
Clearly, $m$ must be even.
Thus, we need to find the number of solutions to the equation
$$
2^{n}=10 p-2\left(n, p \in \mathbf{Z}_{+}, n \leqslant 2019, p \leqslant 1009\right)
$$
which is equivalent to finding the number of solutions to the equation
$$
2^{n-1}=5 p-1\left(n, p \in \mathbf{Z}_{+}, n \leqslant 2019, p \leqslant 1009\right)
$$
Clearly, $n=3, p=1$ satisfies the requirement.
If $n$ is even, then $4^{\frac{n}{2}}=5 m-2$, but
$$
(5-1)^{\frac{n}{2}} \equiv(-1)^{\frac{n}{2}}(\bmod 5),
$$
which is a contradiction.
Therefore, $n$ must be odd, so we need to find the number of solutions to the equation
$$
4^{q}=5 p-1\left(q, p \in \mathbf{Z}_{+}, q \leqslant 1009, p \leqslant 1009\right)
$$
By $4^{q}=(5-1)^{q} \equiv(-1)^{q}(\bmod 5)$, we know that $q$ must be odd.
Thus, $\left\{c_{n}\right\}$ is a geometric sequence with the first term 8 and common ratio 16.
$$
\text { Therefore, }\left\{a_{1}, a_{2}, \cdots, a_{2019}\right\} \cap\left\{b_{1}, b_{2}, \cdots, b_{2019}\right\}
$$
has 3 elements, which are $2^{3}, 2^{7}, 2^{11}$.
Thus, the sum of the required elements is 2184.
|
2184
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
cn_contest
| false
|
5. Periodical cicadas are insects with very long larval periods and brief adult lives. For each species of periodical cicada with larval period of 17 years, there is a similar species with a larval period of 13 years. If both the 17 -year and 13 -year species emerged in a particular location in 1900, when will they next both emerge in that location?
|
5. 2121
|
2121
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
14. Find the least common multiple of each of the following pairs of integers
a) 8,12
d) 111,303
b) 14,15
e) 256,5040
c) 28,35
f) 343,999 .
|
14. a) 24 b) 210 c) 140 d) 11211 e) 80640 f) 342657
|
140
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
10. How many ways can change be made for one dollar using
a) dimes and quarters
b) nickels. dimes, and quarters
c) pennies, nickels, dimes, and quarters?
|
10. a) 3
b) 29
c) 242
|
242
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
1. Find the values of the following sums
a) $\sum_{j=1}^{10} 2$
c) $\sum_{j=1}^{10} j^{2}$
b) $\sum_{j=1}^{10} j$
d) $\sum_{j=1}^{10} 2^{j}$.
|
1. a) 20 b) 55 c) 385 d) 2046
|
2046
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
6. Find the smallest positive integer $n$ with $\tau(n)$ equal to
a) 1
d) 6
b) 2
e) 14
c) 3
f) 100 .
|
6. a) 1
b) 2
c) 4
d) 12
e) 192
f) 45360
|
4
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
2. Find the values of the following products
a) $\prod_{j=1}^{5} 2$
c) $\prod_{j=1}^{5} j^{2}$
b) $\prod_{j=1}^{5} j$
d) $\prod_{j=1}^{5} 2^{j}$.
|
2. a) 32 b) 120 c) 14400 d) 32768
|
14400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
3. What is the ciphertext that is produced when the RSA cipher with key $(e, n)=(3,2669)$ is used to encipher the message BEST WISHES?
|
3. 12151224147100230116
|
12151224147100230116
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
3. Find $n$ ! for $n$ equal to each of the first ten positive integers.
|
3. $1,2,6,24,120,720,5040,40320,362880,3628800$
|
3628800
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
12. Let $n=2^{t} \cdot p_{1}^{t} p_{2}^{t} \cdots p_{m}^{t}$ be the prime-power factorization of $n$. Let $a$ be an integer relatively prime to $n$. Let $r_{1}, r_{2}, \ldots, r_{m}$ be primitive roots of $p_{1}^{t_{1}}, p_{2}^{t_{2}}, \ldots, p_{m}^{t}$, respectively, and let $\gamma_{1}=\operatorname{ind}_{r_{1}} a\left(\bmod p_{1}^{t_{1}}\right), \quad \gamma_{2}=\operatorname{ind}_{r_{2}} a\left(\bmod p_{2}^{t_{2}}\right)$, $\ldots, \gamma_{m}=\operatorname{ind}_{r_{-}} a\left(\bmod p_{m}^{t}\right)$. If $t_{0} \leqslant 2$, let $r_{0}$ be a primitive root of $2^{t}$, and let $\gamma_{0}=\operatorname{ind}_{r_{0}} a\left(\bmod 2^{t}\right)$. If $t_{0} \geqslant 3$, let $(\alpha, \beta)$ be the index system of $a$ modulo $2^{k}$, so that $a \equiv(-1)^{\alpha} 5^{\beta}\left(\bmod 2^{k}\right)$. Define the index system of $a$ modulo $n$ to be $\left(\gamma_{0}, \gamma_{1}, \gamma_{2}, \ldots, \gamma_{m}\right)$ if $t_{0} \leqslant 2$ and $\left(\alpha, \beta, \gamma_{1}, \gamma_{2}, \ldots, \gamma_{m}\right)$ if $t_{0} \geqslant 3$.
a) Show that if $n$ is a positive integer, then every integer has a unique index system modulo $n$.
b) Find the index systems of 17 and $41(\bmod 120)$ (in your computations, use 2 as a primitive root of the prime factor 5 of 120 ).
c) Develop rules for the index systems modulo $n$ of products and powers analogous to those for indices.
d) Use an index system modulo 60 to find the solutions of $11 x^{7} \equiv 43(\bmod 60)$
|
12. b) $(0,0,1,1),(0,0,1,4)$
d) $x \equiv 17(\bmod 60)$
|
17
|
Number Theory
|
proof
|
Yes
|
Yes
|
number_theory
| false
|
1. Find $\lambda(n)$, the minimal universal exponent of $n$, for the following values of $n$
a) 100
e) $2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7$
b) 144
f) $2^{5} \cdot 3^{2} \cdot 5^{2} \cdot 7^{3} \cdot 11^{2} \cdot 13 \cdot 17 \cdot 19$
c) 222
g) 10 !
d) 884
h) 20 !.
|
a) 20
b) 12
c) 36
d) 48
e) 180
f) 388080
g) 8640
h) 125411328000
|
48
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
1. Find the sequence of two-digit pseudo-random numbers generated using the middle-square method, taking 69 as the seed.
|
1. $69,76,77,92,46,11,12,14,19,36,29,84,5,25,62,84,5,25,62, \ldots$
|
62
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
2. Find the first ten terms of the sequence of pseudo-random numbers generated by the linear congruential method with $x_{0}=6$ and $x_{n+1} \equiv 5 x_{n}+2(\bmod 19)$. What is the period length of this generator?
|
2. $6.13,10,14,15,1,7,18,16,6,13, \ldots$ period length is 9
|
9
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 1 Find the number of prime numbers not exceeding 100.
|
We take $N=100$, the prime numbers not exceeding $\sqrt{100}=10$ are $2,3,5,7$, so by formula (20) we get
$$\begin{aligned}
\pi(100)= & 4-1+100-\left\{\left[\frac{100}{2}\right]+\left[\frac{100}{3}\right]+\left[\frac{100}{5}\right]+\left[\frac{100}{7}\right]\right\} \\
& +\left\{\left[\frac{100}{2 \cdot 3}\right]+\left[\frac{100}{2 \cdot 5}\right]+\left[\frac{100}{2 \cdot 7}\right]+\left[\frac{100}{3 \cdot 5}\right]+\left[\frac{100}{3 \cdot 7}\right]+\left[\frac{100}{5 \cdot 7}\right]\right\} \\
& -\left\{\left[\frac{100}{2 \cdot 3 \cdot 5}\right]+\left[\frac{100}{2 \cdot 3 \cdot 7}\right]+\left[\frac{100}{2 \cdot 5 \cdot 7}\right]+\left[\frac{100}{3 \cdot 5 \cdot 7}\right]\right\} \\
& +\left[\frac{100}{2 \cdot 3 \cdot 5 \cdot 7}\right] \\
= & 4-1+100-117+45-6+0=25
\end{aligned}$$
This is consistent with the result in Section 2 of Chapter 1.
|
25
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 3 (i) Find the number of integers between 1 and 500 that are not divisible by any of 5, 6, 8;
(ii) Find the number of integers between 1 and 500 that are divisible by 5, or 6, or 8.
|
Consider the sequence $A$ composed of integers from 1 to 500. Property $P_{1}$: divisible by 5, property $P_{2}$: divisible by 6, property $P_{3}$: divisible by 8. Thus, (i) is to find the number of $B^{(0)}$, and (ii) is to find the number of $A^{(0)}-B^{(0)}$.
The number of $A(1)$ is: 100; the number of $A(2)$ is: 83; the number of $A(3)$ is: 62. $A(1,2)$ consists of integers divisible by 30, with a count of 16; $A(1,3)$ consists of integers divisible by 40, with a count of 12; $A(2,3)$ consists of integers divisible by 24, with a count of 20. $A(1,2,3)$ consists of integers divisible by 120, with a count of 4. Therefore,
$$\begin{array}{c}
B^{(0)}=500-(100+83+62)+(16+20+12)-(4)=299 \\
A^{(0)}-B^{(0)}=500-299=201
\end{array}$$
|
299
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 4 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, and property $P_{3}$ is that the element is divisible by 2. Try to verify whether Theorem 7 holds.
|
$A(1)$ is $3,3,6; A(2)$ is $1,4,7; A(3)$ is $2,4,6. A(1,2)$ has no elements; $A(1,3)$ is $6; A(2,3)$ is $4. A(1,2,3)$ has no elements.$
$$B^{(0)}=B(0) \text { is } 5,5,11; B(1) \text { is } 3,3; B(2) \text { is } 1,7; B(3) \text { is } 2 \text {. }$$
$B(1,2)$ has no elements; $B(1,3)$ is $6; B(2,3)$ is $4. B(1,2,3)$ has no elements.$
$$\begin{aligned}
3 & =B^{(0)}=B(0)=A^{(0)}-A^{(1)}+A^{(2)}-A^{(3)} \\
& =10-(3+3+3)+(0+1+1)-0=3
\end{aligned}$$
This is consistent with the conclusion of Theorem 7. Similarly, the conclusion of Theorem 9 can be verified (left to the reader).
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
Example 5 Let sequence $A$ be $1,3,2,5,4,3,5,7,6,11$. Property $P_{1}$ is that the element is divisible by 3, Property $P_{2}$ is that the element leaves a remainder of 1 when divided by 3, Property $P_{3}$ is that the element is divisible by 2, Property $P_{4}$ is that the element is divisible by 3. Try to verify whether Theorem 7 holds.
|
Solve $A(1)$ is $3,3,6 ; A(2)$ is $1,4,7 ; A(3)$ is $2,4,6 ; A(4)$ is $3,3,6. A(1,2)$ has no elements; $A(1,3)$ is $6 ; A(1,4)$ is $3,3,6 ; A(2,3)$ is $4 ; A(2,4)$ has no elements; $A(3,4)$ is $6. A(1,2,3)$ has no elements; $A(1,2,4)$ has no elements; $A(1,3,4)$ is $6 ; A(2,3,4)$ has no elements; $A(1,2,3,4)$ has no elements. \square$
$B^{(0)}=B(0)$ is $5,5,11 ; B(1)$ has no elements (because property $P_{1}$ is the same as property $P_{4}$); $B(2)$ is $1,7 ; B(3)$ is $2 ; B(4)$ has no elements. $B(1,2)$ has no elements; $B(1,3)$ has no elements; $B(1,4)$ is $3,3 ; B(2,3)$ is $4 ; B(2,4)$ has no elements; $B(3,4)$ has no elements. $B(1,2,3)$ has no elements; $B(1,2,4)$ has no elements; $B(1,3,4)$ is $6 ; B(2,3,4)$ has no elements. $B(1,2,3,4)$ has no elements.
$$\begin{aligned}
3= & B^{(0)}=B(0)=A^{(0)}-A^{(1)}+A^{(2)}-A^{(3)}+A^{(4)} \\
= & 10-(3+3+3+3)+(0+1+3+1+0+1) \\
& -(0+0+1+0)+0=3
\end{aligned}$$
This is consistent with the conclusion of Theorem 7. Similarly, the conclusion of Lemma 8 can be verified (left to the reader).
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
4. Let $n$ be any given positive integer. Find the value of $\mu(n) \mu(n+1) \mu(n+2) \mu(n+3)$.
|
4. 0 , because $n, n+1, n+2, n+3$ must include one number that is divisible by 4.
|
0
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
5. Find the smallest positive integer $n$ that satisfies $\tau(n)=6$.
|
5. $n=p_{1}^{a_{1}} \cdots p_{s}^{a_{s}}, p_{1}<\cdots<p_{3} . \tau(n)=\left(\alpha_{1}+1\right) \cdots\left(\alpha_{s}+1\right)=6$. It must be that $\alpha_{1}=1$, $\alpha_{2}=2$; or $\alpha_{1}=2, \alpha_{2}=1$. Therefore, the smallest $n=2^{2} \cdot 3^{1}=12$.
|
12
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
number_theory
| false
|
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